WikiWaves - User contributions [en]

Standard Notation

2012-07-26T19:59:12Z

Adi Kurniawan: /* Greek letters */

This is a list of standard notation with definitions. If you find notation which does not appear here or non-standard notation please
feel free to highlight this, or better still try and fix it. The material on these webpages was taken from a variety of sources and we
know the notation is currently not always consistent between pages.

== Latin Letters ==

* <math>A</math> is the wave amplitude
* <math> c \,(=\omega / k) </math> or sometime <math>c_p</math> is the wave phase velocity
* <math> c_g = \frac{\mathrm{d} \omega}{\mathrm{d} k} </math> is the wave group velocity
* <math>d</math> is a water depth parameter
* <math>D</math> is the modulus of rigidity for a plate
* <math>e^{i\omega t}</math> is the time dependence in frequency domain
* <math>E</math> is the Young's modulus
* <math> \mathcal{E}(t) </math> is the energy density
* <math>g</math> is the acceleration due to gravity
* <math>h</math> is the water depth (with the bottom at <math>z=-h</math>)
* <math>\mathbf{i}</math> is the unit vector in the <math>x</math> direction
* <math>\mathrm{Im}</math> is the imaginary part of a complex argument
* <math>\mathbf{j}</math> is the unit vector in the <math>y</math> direction
* <math>\mathbf{k}</math> is the unit vector in the <math>z</math> direction
* <math> k </math> is the wave number
* <math>k_n</math> are the roots of the dispersion eqution
* <math>\mathcal{L}</math> is the linear operator at the body surface
* <math>\mathcal{M}</math> is the momentum
* <math>\mathbf{n}</math> is the outward normal
* <math>\frac{\partial\phi}{\partial n}</math> is <math>\nabla\phi\cdot\mathbf{n}</math>
* <math>P</math> is the pressure (<math>P_1</math>, <math>P_2</math> etc are the first, second order pressures)
* <math>\mathcal{P}(t)</math> the energy flux is the rate of change of energy density <math> \mathcal{E}(t) </math>
* <math>\mathbf{r}</math> vector in the horizontal directions only <math>(x,y)</math>
* <math>R</math> is the radius of a cylinder
* <math>\mathrm{Re}</math> is the real part of a complex argument
* <math>S_F</math> is the free surface
* <math>t</math> is the time
* <math> T \,(= 2\pi / \omega)</math> is the wave period
* <math>U</math> is the forward speed
* <math>U_n</math> is the normal derivative of the moving surface of a volume
* <math> V_n = \mathbf{n} \cdot \nabla \Phi </math> is the flow in the normal direction for potential <math>\Phi</math>
* <math>\mathbf{v}</math> is the flow velocity vector at <math>\mathbf{x}</math>
* <math>\mathbf{x}</math> is the fixed Eulerian vector <math>(x,y,z)</math>
* <math>x</math> and <math>y</math> are in the horizontal plane with <math>z</math> pointing vertically upward and the free surface is at <math>z=0</math>
* <math>\bar{x}</math> is the <math>x</math> coordinate in a moving frame.
* <math>X_n(x)</math> is an eigenfunction arising from separation of variables in the <math>x</math> direction.
* <math>Z(z)</math> is an eigenfunction arising from separation of variables in the <math>z</math> direction.

== Greek letters ==

* <math>\alpha</math> is free surface constant <math>\alpha = \omega^2/g</math>
* <math>\mathcal{E}</math> is the energy
* <math>\zeta</math> is the displacement of the surface
* <math>\xi</math> any other displacement, most usually a body in the fluid
* <math>\eta</math> any other displacement, most usually a body in the fluid
* <math> \lambda \,(= 2\pi/k) </math> is the wave length
* <math>\rho</math> is the fluid density (sometimes also string density).
* <math>\rho_i</math> is the plate density
* <math>\phi\,</math> is the velocity potential in the frequency domain
* <math>\phi^{\mathrm{I}}\,</math> is the incident potential
* <math>\phi^{\mathrm{D}}\,</math> is the diffracted potential
* <math>\phi^{\mathrm{S}}\,</math> is the scattered potential (<math>\phi^{\mathrm{S}}
= \phi^{\mathrm{I}}+\phi^{\mathrm{D}}\,</math>)
* <math>\phi_{m}^{\mathrm{R}}\,</math> is the radiated potential (for the <math>m</math> mode
* <math>\Phi\,</math> is the velocity potential in the time domain
* <math>\bar{\Phi}\,</math> is the velocity potential in the time domain for a moving coordinate system
* <math>\omega</math> is the wave/angular frequency
* <math>\Omega\,</math> is the fluid region
* <math>\partial \Omega</math> is the boundary of fluid region, <math>\partial\Omega_F</math> is the free surface, <math>\partial\Omega_B</math> is the body surface.

== Other notation, style etc. ==

* We prefer <math>\partial_x\phi</math> etc. for all derivatives or <math>\frac{\partial\phi}{\partial x}</math>. Try to avoid <math>\phi_x\,</math> or <math>\phi^{\prime}</math>
* We prefer <math>\mathrm{d}x\,\!</math> etc. for differentials. Avoid <math>dx\,\!</math>
* <math>\mathrm{Re}\,\!</math> and <math>\mathrm{Im}\,\!</math> for the real and imaginary parts.
* We use two equals signs for the first heading (rather than a single) following [http://www.wikipedia.org wikipedia] style, then three etc.

[[Category:Administration]]

Solution of Wave-Body Flows, Green's Theorem

2012-07-26T19:55:16Z

Adi Kurniawan: /* Boundary-value problem */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Solution of Wave-Body Flows, Green's Theorem
| next chapter = [[Rankine Intergral Equations For Ship Flows]]
| previous chapter = [[Seakeeping In Random Waves]]
}}

{{incomplete pages}}

==Introduction==

Two types of wave body interaction problems are encountered frequently in applications and solved by the methods described in this section: zero-speed linear wave body interactions in the frequency domain in 2D and 3D, forward-speed seakeeping problems in the frequency or time domain in three dimensions (linear and nonlinear).

A consensus has been reached over the past two decades that the most efficient and robust solution methods are based on Green's Theorem using either a wave-source potential or the Rankine source as the Green function. The numerical solution of the resulting integral equations in practice is in almost all cases carried out by panel methods.

==Frequency-domain radiation-diffraction. U = 0==

===Boundary-value problem===

Green's Theorem generates a boundary integral equation for the complex potential <math>\phi\,</math> over the body boundary <math>S_B\,</math> for the proper choice of the Green function:

<center><math> \iint_S \left( \phi_1 \frac{\partial\phi_2}{\partial n} - \phi_2 \frac{\partial\phi_1}{\partial n} \right) \mathrm{d}S = 0 \, </math></center>

For any <math>\phi_1, \ \phi_2\,</math> that solve the Laplace equation in a closed volume <math>V\,</math>.

Define the volume <math>V\,</math> and <math>S\,</math> as follows:

The fluid volume <math>V\,</math> is enclosed by the union of several surfaces

<center><math> S \equiv S_B + S_F + S_\infty + S_H + S_E </math></center>

<math> S_B \, </math>: mean position of body surface 
<math> S_F \, </math>: mean position of the free surface 
<math> S_\infty \, </math>: Bounding cylindrical surface with radius <math> R = \left( x^2 +y^2 \right)^{1/2} \, </math>. Will be allowed to expand after the statement of Green's Theorem 
<math> S_H \, </math>: Seafloor (assumed flat) of a surface which will be allowed to approach <math> Z=-\infty\,</math> 
<math> S_E \, </math>: Spherical surface with radius <math> V = \epsilon \ , </math> centered at point <math> \vec\xi \, </math> in the fluid domain 
<math> \vec{n}\,</math>: Unit normal vector on <math>S\,</math>, at point <math>\vec{x}\,</math> on <math>S\,</math>

Define two velocity potentials <math>\phi_i(\vec{x})\,</math>:

<math> \phi_1(\vec{x}) = \phi(\vec{x}) \equiv \,</math> Unknown complex radiation or diffraction potential 
<math> \phi_2(\vec{x}) = G(\vec{x};\vec{\xi}) \equiv \,</math> Green function value at point <math>\vec{x}\,</math> due to a singularity centered at point <math>\vec{\xi}\,</math>.

Two types of Green functions will be used:

Rankine source: <math> \nabla_x^2 G = 0 \, </math>

<center><math> G(\vec{x};\vec{\xi}) = - \frac{1}{4\pi} \left|\vec{x}-\vec{\xi}\right|^{-1} = - \frac{1}{4\pi r} \, </math></center>
<center><math> = - \frac{1}{4\pi} \left\{ (x-\xi)^2 + (Y-n)^2 + (Z-\zeta)^2 \right\}^{-1/2} \, </math></center>

Note that the flux of fluid emitted from <math>\vec{\xi}\,</math> is equal to <math>1\,</math>.

This Rankine source and its gradient with respect to <math>\vec\xi\,</math> (dipoles) is the Green function that will be used in the ship seakeeping problem.

===Havelock's wave source potential===

...Also known as the <math>U=0\,</math> wave Green function in the frequency domain.

Satisfies the free surface condition and near <math>\vec\xi=0\,</math> behaves like a Rankine source:

The following choice for <math>G(\vec{X};\vec{\xi})\,</math> satisfies the Laplace equation and the free-surface condition:

<center><math> F(\vec{X},\vec{\xi}) = - \frac{1}{4\pi} \left( \frac{1}{r} + \frac{1}{r_1} \right) - \frac{K}{2\pi} \int_0^\infty \frac{\mathrm{d}u}{u-K} e^{u(Z+\zeta)} J_0(uR) </math></center>

where:
<center><math> K = \frac{\omega^2}{g} \, </math></center>
<center><math> R^2 = (X-\xi)^2 + (Y-n)^2 \, </math></center>
<center><math> J_0(uR) = \mbox{Bessel Function of order zero} \,</math></center>
<center><math> \mbox{Contour indented above pole} n = K \, </math></center>

Verify that with respect to the argument <math>\vec{X}\,</math>, the velocity potential <math> \phi_2(\vec{X}) \equiv G(\vec{X};\vec{\xi})\,</math> satisfies the free surface condition:

<center><math> \frac{\partial\phi_2}{\partial Z} - K \phi_2 = 0, \quad Z=0 </math></center>

<center><math> \phi_2 \sim - \frac{1}{4\pi} r^{-1}, \quad \vec{X} \to \vec{\xi} </math></center>

As <math> KR\to\infty\,</math>:

<center><math> G \sim - \frac{i}{2} K e^{K(Z+\zeta)} H_0^{(2)}(KR) </math></center>

where <math> H_0^{(2)} (KR)\,</math> is the Hankel function of the second kind and order zero.

At <math> KR\to\infty\,</math>:

<center><math> H_0^{(2)}(KR) \sim \sqrt{\frac{2}{\pi K R}} e^{-i \left( KR-\frac{\pi}{4} \right)} + O \left( \frac{1}{R} \right) </math></center>

Therefore the real velocity potential

<center><math> \mathbb{G} = \mathrm{Re} \left\{ G e^{i\omega t} \right\} </math></center>

Represents outgoing ring waves of the form <math>\propto e^{i(\omega T -KR)}\,</math> hence satisfying the radiation condition.

A similar far-field radiation condition is satisfied by the velocity potential <math> \phi_1(\vec{X})\equiv\phi(\vec{X})\,</math>

<center><math> \phi_1 \sim \frac{\mathbb{A}(\theta)}{(KR)^{1/2}} e^{KZ-iKR} + O\left(\frac{1}{R}\right) </math></center>

It follows that on <math>S_\infty\,</math>:

<center><math> \frac{\partial\phi_1}{\partial n} = \frac{\partial\phi_1}{\partial R} = - i K \phi_1 \, </math></center>
<center><math> \frac{\partial\phi_2}{\partial n} = \frac{\partial\phi_2}{\partial R} = - i K \phi_2 \, </math></center>

Therefore:

<center><math> \phi_1 \frac{\partial\phi_2}{\partial n} - \phi_2 \frac{\partial\phi_1}{\partial n} = - i K \left( \phi_1 \phi_2 - \phi_2 \phi_1 \right) = 0 </math></center>

with errors that decay like <math>R^{-3/2}\,</math>, hence faster than <math>R\,</math>, which is the rate at which the surface <math> S_\infty\,</math> grows as <math>R\to\infty\,</math>.

On <math>S_F(Z=0)\,</math>:

<center><math> \frac{\partial\phi_1}{\partial n} = \frac{\partial\phi_1}{\partial Z}, \quad \frac{\partial\phi_2}{\partial n} = \frac{\partial\phi_2}{\partial Z} </math></center>
<center><math> \phi_1 \frac{\partial\phi_2}{\partial n} - \phi_2 \frac{\partial\phi_1}{\partial n} = \phi_1 \frac{\partial\phi_2}{\partial Z} - \phi_2\frac{\partial\phi_1}{\partial Z} = Y \left( \phi_1\phi_2 - \phi_2\phi_1\right) = 0 </math></center>

It follows that upon application of Green's Theorem on the unknown potential <math> \phi_1 \equiv \phi\,</math> and the wave Green function <math> \phi_2 \equiv G\,</math> only the integrals over <math> S_B\,</math> and <math>S_\epsilon\,</math> survive.

Over <math> S_H\,</math>, either <math> \frac{\partial\phi_1}{\partial n} = \frac{\partial \phi_2}{\partial n} = 0 \,</math> by virtue of the boundary condition if the water depth is finite or <math>\frac{\partial\phi_1}{\partial Z} \to 0, \ \frac{\partial\phi_2}{\partial Z}\to 0 \,</math> as <math> Z\to - \infty\,</math> by virtue of the vanishing of the respective flow velocities at large depths.

There remains to interpret and evaluate the integral over <math> S_\epsilon\,</math> and <math>S_B\,</math>. Start with <math> S_\epsilon\,</math>:

<center><math> I_\epsilon = \iint_{S_\epsilon} \left( \phi_1 \frac{\partial\phi_2}{\partial n} - \phi_2 \frac{\partial\phi_1}{\partial n} \right) \mathrm{d}S </math></center>

or:

<center><math> I_\epsilon = \iint_{S_\epsilon} \left( \phi \frac{\partial G}{\partial n} - G \frac{\partial\phi}{\partial n} \right) \mathrm{d}S_X </math></center>

Note that the integral over <math> S_\epsilon\,</math> is over the <math> \vec{X}\,</math> variable with <math> \vec{\xi}\,</math> being the fixed point where the source is centered.

Near <math>\vec{\xi}\,</math>:

<center><math> G \sim - \frac{1}{4\pi r}, \quad \frac{\partial G}{\partial n} = - \frac{\partial G}{\partial r} \sim \frac{1}{4\pi r^2} </math></center>
<center><math> \phi \to \phi(\vec{\xi}) = \phi(\vec{X}) \ \mbox{as} \ \epsilon \to 0, \vec{X} \to \vec{\xi} </math></center>

In the limit as <math> r\to 0 \,</math> the integrand over the sphere <math>S_\epsilon\,</math> becomes spherically symmetric and with vanishing errors

<center><math> I_\epsilon \to 4 \pi r^2 \left[ \phi(\vec{\xi}) \frac{1}{4 \pi r^2} + G \frac{\partial\phi}{\partial r} \right] = \phi(\vec{\xi}) </math></center>

In summary:

<center><math> \phi(\vec{\xi}) + \iint_{S_B} \left[ \phi(\vec{X}) \frac{\partial G(\vec{X};\vec{\xi})}{\partial n_X} - G(\vec{X}; \vec{\xi}) \frac{\partial\phi}{\partial n_X} \right] \mathrm{d} S_n = 0 </math></center>

on <math> S_B: \quad \frac{\partial\phi}{\partial n_X} = V(X) = \,</math> known from the boundary condition of the radiation and diffraction problems.

It follows that a relationship is obtained between the value of <math> \phi(\vec{\xi})\,</math> at some point in the fluid domain and its values <math> \phi_(\vec{X})\,</math> and normal derivatives over the body boundary:

<center><math> \phi(\vec{\xi}) + \iint_{S_B} \phi(\vec{X}) \frac{\partial G ( \vec{X}; \vec{\xi} )} {\partial n_X} \mathrm{d} S_X = \iint_{S_B} G (\vec{X};\vec{\xi}) V(\vec{X}) \mathrm{d} S_X </math></center>

Stated differently, knowledge of <math> \phi\,</math> and <math> \frac{\partial\phi}{\partial n}\,</math> over the body boundary allows the determination of <math> \phi\,</math> and upon differentiation of <math>\nabla\phi\,</math> in the fluid domain.

In order to determine <math> \phi(\vec{X})\,</math> on the body boundary <math> S_B\,</math>, simply allow <math>\vec{\xi}\to S_B\,</math> in which case the sphere <math> S_\epsilon\,</math> becomes a <math>\frac{1}{2}\,</math> sphere as <math> \epsilon\to 0 \, </math>:

Note that <math> \vec{\xi}\,</math> is a fixed point where the point source is centered and <math> \vec{X}\,</math> is a dummy integration variable moving over the body boundary <math>S_B\,</math>.

The reduction of Green's Theorem derived above survives almost identically with a factor of <math>\frac{1}{2}\,</math> now multiplying the <math>I_\epsilon\,</math> integral since only <math> \frac{1}{2}\,</math> of the <math>S_\epsilon\,</math> surface lies in the fluid domain in the limit as <math>\epsilon\to 0 \, </math> and for a body surface which is smooth. It follows that:

<center><math> \frac{1}{2} \phi(\vec{\xi}) + \iint_{S_B}\phi(\vec{X}) \frac{\partial G (\vec{X}; \vec{\xi})}{\partial n_X} \mathrm{d} S_X = \iint_{S_B} G(\vec{X}; \vec{\xi}) V(\vec{X}) \mathrm{d} S_X </math></center>

where now both <math>\vec{X}\,</math> and <math>\vec{\xi}\,</math> lie no the body surface. This becomes an integral equation for <math>\phi(\vec{X})\,</math> over a surface <math> S_B\,</math> of boundary extent. Its solution is carried out with panel methods described below.

The interpretation of the derivative under the integral sign is as follows:

<center><math> \frac{\partial G}{\partial n_X} \equiv \vec{n}_X \cdot \nabla_X G(\vec{X};\vec{\xi}) \equiv \left( n_1 \frac{\partial}{\partial X} + n_2 \frac{\partial}{\partial Y} + n_3 \frac{\partial}{\partial Z} \right) G (\vec{X}; \vec{\xi}) </math></center>

where derivatives are taken with respect to the first argument for a point source centered at point <math> \vec{\xi}\,</math>.

==Infinite domain potential flow solutions==

In the absence of the free surface, the derivation of the Green integral equation remains almost unchanged using <math>G\,</math>:

<center><math> \phi_2(\vec{X}) = - \frac{1}{4\pi} \left| \vec{X} - \vec{\xi} \right|^{-1} \equiv G(\vec{X}; \vec{\xi}) </math></center>

The Rankine source as the Green function and using the property that as <math> R\to \infty\,</math>

For closed boundaries <math>S_B\,</math> with no shed wakes responsible for lifting effects the resulting integral equation for <math>\phi(\vec{X})\,</math> over the body boundary becomes:

<center><math> \frac{1}{2} \phi(\vec{\xi}) + \iint_{S_B} \phi(\vec{X}) \frac{\partial G(\vec{X};\vec{\xi})}{\partial n_X} \mathrm{d} S_X = \iint_{S_B} G(\vec{X};\vec{\xi}) V(\vec{X}) \mathrm{d} S_X </math></center>

with

<center><math> V(\vec{X}) = \frac{\partial\phi}{\partial n}, \quad \mbox{on} \ S_B \, </math></center>

===Example===
Uniform flow past <math> S_B\,</math>:
<center><math> \Phi = U X + \phi, \ \frac{\partial\Phi}{\partial n} = 0, \ \mbox{on} \ S_B \, </math></center>
<center><math> \Longrightarrow \frac{\partial\phi}{\partial n} = -\frac{\partial}{\partial n} \left(U X\right) = - n_1 U \equiv V(\vec{X}) </math></center>

So the RHS of the Green equation becomes:

<center><math> RHS = \iint_{S_B} G(\vec{X};\vec{\xi}) \left(-Un_1\right) \mathrm{d} S_X \, </math></center>.

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

Memory Effect Function

2010-11-11T19:26:32Z

Adi Kurniawan: /* Integro-differential equation method */

{{incomplete pages}}

==Frequency-domain equation of motion==

The radiation force coefficients also play a significant role in determining the motion of a floating structure possibly subject to mooring and applied forces. To obtain the correct equation of motion in the frequency domain it is important to adopt the approach similar to that taken by [[McIver 2005]], i.e. keep the initial condition terms in the Fourier transform operations. Therefore, if the Fourier transform of the velocity is <math>v_{\mu}(\omega)</math> then the Fourier transform of the acceleration is given by

<center><math>

\int_{0}^{\infty}\ddot{X}_{\mu}(t)e^{i\omega t}\mathrm{d}t=-i\omega v_{\mu}(\omega) - V_{\mu}(0)

</math></center>

the time-derivative of the potential obeys a similar relation

<center><math>

\int_{0}^{\infty}\frac{\partial\Phi}{\partial t}e^{i\omega t}\mathrm{d}t=-i\omega \phi_{\mu}(\omega) - \Phi(\mathbf{x},0).

</math></center>

Therefore, the ini\partial conditions of the structure <math>(X_{\mu}(0),V_{\mu}(0))</math> and of the potential <math>\Phi(\mathbf{x},0)</math> will be present in the Fourier transform of the equation motion which will be referred to as the frequency-domain equation of motion. If the frequency domain potential is decomposed in the same manner as before then the Fourier transform of the equation of motion for the structure is
<center><math>
M\left[-V_{\mu}(0)-i\omega v_{\mu}(\omega)\right]
=-\rho\iint_{\Gamma}(-\Phi(\mathbf{x},0))n_{\mu}\mathrm{d}S+F^{S}_{\mu}(\omega)+
i\omega \sum_{\nu} (f_{\mu\nu}(\omega)+i\gamma_{\mu\nu}/\omega)v_{\nu}(\omega)-
\sum_{\nu} c_{\mu\nu}\frac{v(\omega)+X(0)}{-i\omega}+f^{A}_{\mu}(\omega)
</math></center>
for <math>\mu=1,\ldots,6</math>, where <math>f^{A}_{\mu}(\omega)</math> is the Fourier transform of the applied force <math>F_{\mu}(t)</math> in equation~(\ref{linearisedmotion}). Although it is assumed that <math>\Phi(\mathbf{x},t)=0</math> for <math>t<0</math>, for a non-zero initial velocity <math>\lim_{t^{+}\rightarrow 0}\Phi(\mathbf{x},t)\neq 0</math> because of the impulsive pressure on the free-surface resulting from the initial motion of the pressure. Therefore, the first term on the right-hand side of the above equation is not identically zero, rather is given by
<center><math>
\rho\iint_{\Gamma}(-\Phi(\mathbf{x},0))n_{\mu}\mathrm{d}S=a(\infty)V(0)
</math></center>
[[Mei 1983]]
where <math>a(\infty)=\lim_{\omega\rightarrow\infty}a(\omega)</math> is the infinite frequency added mass.

The frequency-domain equation is usually re-expressed in the following form
<center><math>
\sum_{\nu} \{c_{\mu\nu}-\omega^{2}\left[M_{\mu\mu}\delta_{\mu\nu}+a_{\mu\nu}(\omega)+i(b_{\mu\nu}+\gamma_{\mu\nu})/\omega\right]\}v_{\nu}(\omega)
= -i\omega\left[ F_{\mu}^{S}(\omega)+f_{\mu}(\omega)+(M_{\mu\mu}+a_{\mu\mu}(\infty))V_{\mu}(0)\right]-\sum_{\nu}c_{\mu\nu}X_{\nu}(0)
</math></center>
for <math>\mu=1,\ldots,6</math>. If it is assumed that the structure can only move in one mode of motion and that there is no incident wave or applied force then the complex velocity will be given by
<center><math>
v_{\mu}(\omega)=\frac{-i\omega(M_{\mu\mu}+a_{\mu\mu}(\infty))V_{\mu}(0)-c_{\mu\mu}X_{\mu}(0)}{c_{\mu\mu}-\omega^{2}\left[M_{\mu\mu}+a_{\mu\mu}(\omega)+i(b_{\mu\mu}+\gamma_{\mu\mu})/\omega\right]}.
</math></center>

As described by [[McIver 2006]], the zero of the denominator gives the location of the pole referred to as the motion resonance. The complex force coefficient <math>f_{\mu\nu}(\omega)</math> will also possess a pole, however this resonance is annulled by the frequency-dependence of the complex velocity (see [[McIver 2006]]) and so the motion, in the absence of incident waves, is dominated by the motion resonance term. Solving for <math>v(\omega)</math> then the time-domain velocity can be recovered using the inverse transform
<center><math>
V(t)=\frac{1}{\pi} Re \int^{\infty}_{0}v(\omega)e^{-i\omega t} \mathrm{d}t.
</math></center>

==Integro-differential equation method==

The widespread availability of numerical and analytical methods for determining hydrodynamic coefficients in the frequency domain facilitated the development of a solution method for the motion of a floating body in the time domain. The principles of this method are described by \citeasnoun{ccmei}~in~\S7.11 based on work by \citeasnoun{cummins} and \citeasnoun{wehausen1971} among others. The method requires the consideration of the fluid response to a displacement given impulsively to a structure denoted by <math>\delta(t-\tau)V_{\alpha}(\tau)</math> where <math>\dot{X_{\alpha}}=V_{\alpha}</math>. The resultant impulse response functions for the velocity potential and the hydrodynamic force are written as the sum of an infinite frequency component and a time-dependent component using the Cummins' decomposition. To obtain the fluid response to a general continuous velocity function <math>V_{\alpha}(t)</math> an integral of the impulse response over the range <math>-\infty<\tau<\infty</math> must be calculated. The hydrodynamic force on the structure due to the forced motion of the structure is then shown to be

<center><math>(11)

\mathcal{F}_{\beta}^{R}(t)=-m_{\beta\alpha}(\infty)\ddot{X_{\alpha}}-\int^{t}_{-\infty}L_{\beta\alpha}(t-\tau)\ddot{X}_{\alpha}(\tau)\,\mathrm{d}\tau

</math></center>

where <math>m_{\beta\alpha}(\infty)</math> is an infinite frequency added mass coefficient and <math>L_{\beta\alpha}(t-\tau)</math> is referred to as the impulse response function.

By considering the velocity of the body to be time-harmonic, i.e. <math>V_{\alpha}(t)=Re\{ v_{\alpha} e^{-i\omega t} \}</math>, the impulse response function <math>L_{\beta\alpha}</math> can be related to the added mass and damping coefficients by comparing the force expression~(11) to (5). The comparison yields the following relations

<center><math>
\begin{align}

a_{\alpha\beta}-m_{\alpha\beta}(\infty)&=\int^{\infty}_{0}L_{\alpha\beta}(\tau)\cos\omega\tau\, \mathrm{d}\tau \\

b_{\alpha\beta}&=\omega\int^{\infty}_{0}L_{\alpha\beta}(\tau)\sin\omega\tau\, \mathrm{d}\tau

\end{align}
</math></center>

and so if the added mass or damping coefficients are known for all frequencies <math>0\leq\omega\leq\infty</math>, the impulse response or memory function <math>L_{\alpha\beta}(t)</math> can be found by inverse cosine or sine transform. Therefore, it is be possible to determine the radiation force due to a non-harmonic forcing by using frequency-domain information. The exciting force can also be determined in a similar manner; to obtain it the incident wave-form and the radiation impulse response function are required.

By expressing the total hydrodynamic force in terms of the radiation force~(11) and the exciting force, the equation of motion~(\ref{linearisedmotion}) for transient motions of a floating structure becomes

<center><math>

\left[M_{\alpha\beta}+m_{\alpha\beta}(\infty)\right]\ddot{X}_{\beta} + \int^{t}_{-\infty}L_{\beta\alpha}(t-\tau)\ddot{X}_{\alpha}(\tau)\,\mathrm{d}\tau + c_{\alpha\beta}X_{\beta}=\mathcal{F}^{D}_{\alpha}+F_{\mu}(t),

</math></center>
for <math>\alpha=1,\ldots,6</math>, where repeated indices implies summation. This set of integro-differential equations is to be solved given the position and velocity of the body. It is assumed that the impulse response function and infinite frequency added mass are known from Fourier transforms of the frequency domain hydrodynamic coefficients.

Thus, it is possible, in principle, to calculate the transient response of a floating structure from the frequency-domain responses. Numerically, an accurate calculation of transient response requires that the hydrodynamic coefficients be computed at a large number of discrete values of the frequency <math>\omega</math> on the interval <math>[0,\infty)</math>. This time-domain solution method has been in use since the work by Cummins and Wehausen. As numerical methods and computing power have improved, more complex problems can be solved to a higher degree of accuracy. A recent application of the method can be found in the investigations into latching control by~\citeasnoun{babarit2006}. The principal drawback of the application of this method is that the motion of the fluid cannot be determined from the computational results. Therefore, if the fluid response is required, for example in the investigation of trapped modes, further computations are necessary.

Linear Wave-Body Interaction

2010-11-06T12:11:41Z

Adi Kurniawan: /* Haskind relations of exciting forces */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Linear Wave-Body Interaction
| next chapter = [[Long Wavelength Approximations]]
| previous chapter = [[Ship Kelvin Wake]]
}}

{{incomplete pages}}

[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]

We consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] in the
[[Frequency Domain Problem|Frequency Domain]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem).

== Introduction ==

We derive here the equations of motion for a body in [[Linear Plane Progressive Regular Waves]] in the frequency domain in
two dimensions. We begin with the equations in the time domian. The simplest problems is [[Waves reflecting off a vertical wall]]

== Equations for a Floating Body in the Time Domain ==

We begin with the equations for a floating two-dimensional body in the time domain.

{{equations of motion time domain without body condition}}

{{two dimensional floating body time domain}}

More details can be found in [[:Category:Time-Dependent Linear Water Waves|Time-Dependent Linear Water Waves]]

== Equations for a Floating Body in the Frequency Domain ==

The dynamic condition is the equation of motion for the structure in the [[Frequency Domain Problem|frequency domain]]
can be found from the time domain equations and we introduce the following notation
<center><math>
\xi_{\nu} = \zeta_{\nu}e^{-\mathrm{i}\omega t}\,
</math></center>
This give us
{{standard linear wave scattering equations without body condition}}
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=\mathrm{i}\omega\rho\iint_{\partial\Omega}\phi n_{\mu}\, \mathrm{d}S
- \sum_{\nu} C_{\mu\nu}\zeta_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>
The equations of motion for <math> \zeta_\nu\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions. We use the standard numbering of the modes of motion.

== Equations for a Fixed Body in Frequency Domain ==

The equations for a fixed body are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for a rigid body}}
plus the radiation conditions.

We decompose the potential as
<math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} \,,
</math>
where <math>\phi^{\mathrm{I}}</math> is the incident potential and <math>\phi^{\mathrm{D}}</math>
is the diffracted potential. The boundary condition for the diffracted potential is
<center><math>
\begin{align}
\Delta\phi^{\mathrm{D}}&=0, &-h<z<0,\,\,\mathbf{x} \in \Omega \\
\partial_n\phi^{\mathrm{D}} &= 0, &z=-h, \\
\partial_n \phi^{\mathrm{D}} &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial\Omega_{F},
\end{align}
</math></center>
plus
<center><math>
\partial_n \phi^{\mathrm{D}} = - \partial_n \phi^{\mathrm{I}},\,\, \mathbf{x} \in \partial\Omega_{B},
</math></center>

Code to calculate the solution (using a slighly modified method) can be found in
[[Boundary Element Method for a Fixed Body in Finite Depth]]

== Equations for the Radiation Potential in Frequency Domain ==

We decompose the body motion into the rigid body modes of motion. Associated with
each of these modes is a potential which must be solved for.
The equations for the radiation potential in the frequency domain are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for the radiation modes}}

{{sommerfeld radiation condition two dimensions for radiation}}

Code to calculate the radiation potential can be found in
[[Boundary Element Method for the Radiation Potential in Finite Depth]]

We denote the solution for each of the radiation potentials by
<math>\phi_\nu^{\mathrm{R}}</math> and the total potential is written as
<center><math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_\nu \zeta_\nu \phi_\nu^{\mathrm{R}}
</math>
</center>

== Final System of Equations ==

We substitute the expansion for the potential into the equations in the frequency domain and we obtain
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
\left(\phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_{\nu} \zeta_\nu \phi_{\nu}^{\mathrm{R}}\right) \mathbf{n}_{\mu}\, dS
- \sum_{\nu} C_{\mu\nu}\xi_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>

{{added mass damping and force matrices definition}}

Then the equations can be expressed as follows.
<center><math> \left[-\omega^2 \left(\mathbf{M} + \mathbf{A} \right) +
\mathrm{i}\omega \mathbf{B} + \mathbf{C} \right] \vec{\zeta} = \mathbf{f} </math></center>
where <math>\mathbf{M}</math> is the mass matrix, <math>\mathbf{A}</math> is the added mass matrix,
<math>\mathbf{B}</math> is the damping matrix, <math>\mathbf{C}</math> is the hydrostatic matrix,
<math>\vec{\zeta}</math> is the vector of body displacements and <math>\mathbf{f}</math> is the force.

The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the two dimensional problem for the sake of clarity.

== Symmetric body ==

For a body which is [[:Category:Symmetry in Two Dimensions|Symmetric in Two Dimensions]]
the Heave is decoupled from Surge and Roll.
In other words the Surge and Roll motions do not influence Heave and vice versa.

== Matlab Code ==

*A program to solve for pitch and heave and only for two geometries can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/rigid_body_motion.m rigid_body_motion.m]

* a program to calculate the solution for a specific geometry (with plot as output as shown) can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/wave_bem_example_floating.m wave_bem_example_floating.m]

[[Image:Wave_bem_example_floating_RT2.jpg|300px|right|thumb|The reflection (solid line) and transmission (dashed line)
for a dock for heave and pitch (red), heave only (blue) and pitch only (black)]]

=== Additional code ===

This program requires
* A program to calculate the geometery [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/circlebody_twod.m circlebody_twod.m]
* {{fixed body bem code}}
* {{floating body radiation code}}
* {{free surface dispersion equation code}}
* {{boundary element code}}

== Symmetry-reciprocity relations ==

It will be shown that
<center><math> A_{ij}(\omega) = A_{ji}(\omega) \,</math></center>
<center><math> B_{ij}(\omega) = B_{ji}(\omega) \,</math></center>
Along the same lines it will be shown that the exciting force <math>\mathbf{X}_j\,</math> can be expressed in terms of <math> \psi_j\,</math> circumventing the solution for the diffraction potential.
The core result needed for the proof of the above properties is [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
<center><math> \iint_S \left( \psi_1 \frac{\partial\psi_2}{\partial n} - \psi_2 \frac{\partial\psi_1}{\partial n} \right) \mathrm{d}S = 0 \,</math></center>
where <math>\nabla^2 \psi_i=0</math>.

{{energy_region_plates}}

[[Image:Symmetry_boundary.jpg|thumb|right|600px|Boundary]]

In the surface wave-body problem define the closed surfaces as shown in figure on the right.
Let <math> \phi_j\,</math> be rediation or diffraction potentials. Over the boundaries <math>S^\pm\,</math> we have:

<center><math> S^+: \quad \phi_j \ \sim \ \frac{igA_j^+}{\omega} e^{Kz-iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial \phi_j}{\partial x} \ \sim \ -iK\phi_j \,</math></center>

<center><math> S^-: \quad \phi_j \ \sim \ \frac{igA_j^-}{\omega} e^{Kz+iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = - \frac{\partial \phi_j}{\partial x} \ \sim \ - iK\phi_j \,</math></center>

On <math> S_F: \qquad \frac{\partial\phi_j}{\partial z} = K\phi_j, \qquad \frac{\partial \Phi_j}{\partial n} = \frac{\partial \phi_j}{\partial z} </math>

On <math> S_\infty: \qquad \left| \phi_j \right|, \quad \left| \nabla \phi_j \right| \to 0 </math>.

Applying Green's identity to any pair of the radiation potentials <math> \psi_i, \psi_j \,</math>:

<center><math> \iint_{S_B} \left[ \psi_i \frac{\partial\psi_j}{\partial n} - \psi_j \frac{\partial\psi_i}{\partial n} \right] \mathrm{d}S = - \iint_{S_F} \left[ \psi_i \frac{\partial\psi_j}{\partial z} - \psi_j \frac{\partial\psi_i}{\partial z} \right] \mathrm{d}S </math></center>

<center><math> - \iint_{S_+} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S
+ \iint_{S_-} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S = 0 </math></center>

It follows that:

<center><math> \iint_{S_B} \psi_i \frac{\partial\psi_j}{\partial n} \mathrm{d}S = \iint_{S_B} \psi_j \frac{\partial\psi_i}{\partial n} \mathrm{d}S </math></center>
or
<center><math> A_{ij}(\omega) = A_{ji}(\omega), \qquad B_{ij}(\omega) = B_{ji}(\omega). \,</math></center>

== Haskind relations of exciting forces ==

<center><math>
\begin{align}
\mathbf{X}_i(\omega) &= - i\omega\rho\iint_{S_B} (\phi_I + \phi_7) n_i \mathrm{d}S \\
&= - \rho \iint_{S_B} (\phi_I + \phi_7) \frac{\partial \phi_i}{\partial n} \mathrm{d}S
\end{align}
</math></center>

where the radiation velocity potential <math> \phi_i \,</math> is known to satisfy:

<center><math> \frac{\partial\phi_i}{\partial n} = i\omega n_i, \quad \mbox{on} \ S_B </math></center>

and

<center><math> \frac{\partial\phi_7}{\partial n} = \frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \ S_B </math></center>

Both <math> \phi_i\,</math> and <math> \phi_7\,</math> satisfy the condition of outgoing waves at infinity. By virtue of [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]

<center><math> \iint_{S_B} \phi_7 \frac{\partial\phi_i}{\partial n} \mathrm{d}S = \iint_{S_B} \phi_i \frac{\partial\phi_7}{\partial n} \mathrm{d}S = -\iint_{S_B} \phi_i \frac{\partial\phi_I}{\partial n} \mathrm{d}S </math></center>

The Haskind expression for the exciting force follows:

<center><math> \mathbf{X}_i(\omega) = \rho \iint_{S_B} \left[ \phi_I \frac{\partial\phi_i}{\partial n} - \phi_i \frac{\partial\phi_I}{\partial n} \right] \mathrm{d}S </math></center>

The symmetry of the <math> A_{ij}(\omega), B_{ij}(\omega) \,</math> matrices applies in 2D and 3D. The application of Green's Theorem in 3D is very similar using the far-field representation for the potential <math> \phi_j\,</math>

<center><math> \phi_j \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O\left(\frac{1}{R^{3/2}}\right) </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial\phi_j}{\partial R} \sim - i K \phi_j + O\left(\frac{1}{R^{3/2}}\right) </math></center>

where <math> R \,</math> is a radius from the body out to infinity and the <math> R^{-\frac{1}{2}} \,</math> decay arises from the energy conservation principle. Details of the 3D proof may be found in [[Mei 1983]] and [[Wehausen and Laitone 1960]]

The use of the Haskind relations for the exciting forces does not require the solution of the diffraction problem. This is convenient and often more accurate.

The Haskind relations take other forms which will not be presented here but are detailed in [[Wehausen and Laitone 1960]]. The ones that are used in practice relate the exciting forces to the damping coefficients.

These take the form:

2D: <math> B_{ii} = \frac{\left| \mathbf{X}_i \right|^2}{2\rho g V_g}, \quad V_g = \frac{g}{2\omega}, </math> Deep water

3D: <math> B_{33} = \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2 \,</math> --- Heave

(Axisymmetric bodies) <math> B_{22} = \frac{K}{8\rho g V_g} \left| \mathbf{X}_2 \right|^2 \,</math> --- Sway

So knowledge of <math> \mathbf{X}_i(\omega)\,</math> allows the direct evaluation of the diagonal damping coefficients. These expressions are useful in deriving theoretical results in wave-body interactions to be discussed later.

The two-dimensional theory of wave-body interactions in the frequency domain extends to three dimencions very directly with little difficulty.

The statement of the 6 d.o.f. seakeeping problem is:

<center><math> \sum_{j=1}^6 \left[ - \omega^2 \left( M_{ij} + A_{ij} \right) + i \omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_j, \quad i=1,\cdots,6 </math></center>

where

<center><math> M_{ij}: \mbox{Body inertia matrix including moments of inertia for rotational modes. For details refer to MH} \, </math></center>

<center><math> A_{ij}(\omega): \mbox{Added mass matrix} \,</math></center>

<center><math> B_{ij}(\omega): \mbox{Damping matrix} \, </math></center>

<center><math> C_{ij}: \mbox{Hydrostatic and static inertia restoring matrix. For details refer to MH} \, </math></center>

<center><math> \mathbf{X}_i(\omega): \mbox{Wave exciting forces and moments} </math></center>

At zero speed the definitions of the added-mass, damping matrices and exciting forces are identical to those in two dimensions.

The boundary value problems satisfied by the radiation potentials <math>\phi_j, \ j=1,\cdots,6 \,</math> and the diffraction potential <math> \phi_7 \,</math> are as follows:

Free-surface condition:

<center><math> -\omega^2 \phi_j + g \frac{\partial\phi_j}{\partial Z} = 0, \quad z=0 \quad j=1,\cdots,7 </math></center>

Body-boundary conditions:

<center><math>
\begin{align}
\frac{\partial\phi_7}{\partial n} &= -\frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \quad S_B \\
\phi_I &= \frac{i g A}{\omega} e^{Kz-iKx\cos\beta-iKy\sin\beta+i\omega t} \\
\frac{\partial\phi_j}{\partial n} &= i\omega n_j, \quad j=1,\cdots,6 \, \\
n_j &= \begin{cases}
n_j, & j=1,2,3 \\
\left( \vec{x} \times \vec{n} \right), & j=4,5,6
\end{cases}
\end{align}
</math></center>

<center><math> j=1: \ \mbox{Surge} \qquad j=2: \ \mbox{Sway} \qquad j=3: \ \mbox{Heave} </math></center>
<center><math> j=4: \ \mbox{Roll} \qquad j=5: \ \mbox{Pitch} \qquad j=6: \ \mbox{Yaw} </math></center>

At large distances from the body the velocity potentials satisfy the radiation condition:

<center><math> \phi_j (R,\theta) \sim \frac{A_j(\theta)}{\sqrt{R}} e^{Kz-iKR} + O \left( \frac{1}{R^{3/2}} \right) </math></center>

with

<center><math> K = \frac{\omega^2}{g}. \,</math></center>

This radiation condition is essential for the formulation and solution of the boundary value problems for <math>\phi_j\,</math> using panel methods which are the standard solution technique at zero and forward speed.

Qualitative behaviour of the forces, coefficients and motions of floating bodies

<center><math> - \omega^2 \phi + g \phi_Z =0 \quad \begin{cases}
\phi_Z=0,\quad \omega=0 \\
\phi=0, \quad \omega \to \infty
\end{cases} </math></center>

<center><math> B_{33}(\omega), \sim \omega, \mbox{at low} \ \omega \,</math></center>

The 2D Heave added mass is singular at low frequencies. It is finite in 3D

The 2D Heave damping coefficient is decaying to zero linearly in 2D and superlinearly in 3D. A two-dimensional section is a better wavemaker than a three-dimensional one

A 2D section oscillating in Sway is less effective a wavemaker at low frequencies than the same section oscillating in Heave

The zero-frequency limit of the Sway added mass is finite and similar to the infinite frequency limit of the Heave added mass.

In long waves the Heave exciting force tends to the Heave restoring coefficient times the ambient wave amplitude the free surface behaves like a flat surface moving up and down.

In long waves the Sway exciting force tends to zero. Proof will follow

In short waves all forces tend to zero.

Pitch exciting moment (same applies to Roll) tends to zero. Long waves have a small slope which is proportional to <math> KA</math>, where <math> K\,</math> is the wave number and <math> A\,</math> is the wave amplitude.

Prove that to leading order for <math>KA\to 0 \,</math>:

<center><math> \left| X_S(\omega) \right| \sim KA C_{55}\,</math></center>

where <math>C_{55}\,</math> is the Pitch (<math> C_{44} \,</math> for Roll) hydrostatic restoring coefficient. [NB: very long waves look like a flat surface inclined at <math> KA\,</math> ].

== Body motions in regular waves ==

Heave:

<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega)}{-\omega^2(A_{33} + M) + i\omega B_{33} +C_{33} } </math></center>

Resonance:

<center><math> \omega^2 = \frac{C_{33}}{M+A_{33}} = \frac{\rho g A \omega}{M + A_{33} (\omega)} </math></center>

In principle the above equation is nonlinear for <math>\omega\,</math>. Will be approximated below

At resonance:
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega^*)}{i\omega^* B_{33}(\omega^*)} \,</math></center>

Invoking the relation between the damping coefficient and the exicting force in 3D:

<center><math> \frac{\left| \Pi_3 \right|}{A} = \frac{\left| \mathbf{X}_3(\omega) \right|}{\omega \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2}, \quad V_g=\frac{g}{2\omega} </math></center>

<center><math> =\frac{2\rho g}{\omega^3 \left|\mathbf{X}_3(\omega)\right|}, \quad \mbox{at resonance} </math></center>

This counter-intuitive result shows that for a body undergoing a pure Heave oscillation, the modulus of the Heave response at resonance is inversely proportional to the modulus of the Heave exciting force.

Viscous effects not discussed here may affect Heave response at resonance

The behavior of the Sway response can be found in an analagous manner,

-----

This article is based on the MIT open course notes and the original articles can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here] and
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C5323823-0180-45EA-B165-15856948A0A2/0/lecture10.pdf here]

Linear Wave-Body Interaction

2010-11-06T12:06:11Z

Adi Kurniawan: /* Haskind relations of exciting forces */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Linear Wave-Body Interaction
| next chapter = [[Long Wavelength Approximations]]
| previous chapter = [[Ship Kelvin Wake]]
}}

{{incomplete pages}}

[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]

We consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] in the
[[Frequency Domain Problem|Frequency Domain]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem).

== Introduction ==

We derive here the equations of motion for a body in [[Linear Plane Progressive Regular Waves]] in the frequency domain in
two dimensions. We begin with the equations in the time domian. The simplest problems is [[Waves reflecting off a vertical wall]]

== Equations for a Floating Body in the Time Domain ==

We begin with the equations for a floating two-dimensional body in the time domain.

{{equations of motion time domain without body condition}}

{{two dimensional floating body time domain}}

More details can be found in [[:Category:Time-Dependent Linear Water Waves|Time-Dependent Linear Water Waves]]

== Equations for a Floating Body in the Frequency Domain ==

The dynamic condition is the equation of motion for the structure in the [[Frequency Domain Problem|frequency domain]]
can be found from the time domain equations and we introduce the following notation
<center><math>
\xi_{\nu} = \zeta_{\nu}e^{-\mathrm{i}\omega t}\,
</math></center>
This give us
{{standard linear wave scattering equations without body condition}}
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=\mathrm{i}\omega\rho\iint_{\partial\Omega}\phi n_{\mu}\, \mathrm{d}S
- \sum_{\nu} C_{\mu\nu}\zeta_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>
The equations of motion for <math> \zeta_\nu\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions. We use the standard numbering of the modes of motion.

== Equations for a Fixed Body in Frequency Domain ==

The equations for a fixed body are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for a rigid body}}
plus the radiation conditions.

We decompose the potential as
<math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} \,,
</math>
where <math>\phi^{\mathrm{I}}</math> is the incident potential and <math>\phi^{\mathrm{D}}</math>
is the diffracted potential. The boundary condition for the diffracted potential is
<center><math>
\begin{align}
\Delta\phi^{\mathrm{D}}&=0, &-h<z<0,\,\,\mathbf{x} \in \Omega \\
\partial_n\phi^{\mathrm{D}} &= 0, &z=-h, \\
\partial_n \phi^{\mathrm{D}} &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial\Omega_{F},
\end{align}
</math></center>
plus
<center><math>
\partial_n \phi^{\mathrm{D}} = - \partial_n \phi^{\mathrm{I}},\,\, \mathbf{x} \in \partial\Omega_{B},
</math></center>

Code to calculate the solution (using a slighly modified method) can be found in
[[Boundary Element Method for a Fixed Body in Finite Depth]]

== Equations for the Radiation Potential in Frequency Domain ==

We decompose the body motion into the rigid body modes of motion. Associated with
each of these modes is a potential which must be solved for.
The equations for the radiation potential in the frequency domain are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for the radiation modes}}

{{sommerfeld radiation condition two dimensions for radiation}}

Code to calculate the radiation potential can be found in
[[Boundary Element Method for the Radiation Potential in Finite Depth]]

We denote the solution for each of the radiation potentials by
<math>\phi_\nu^{\mathrm{R}}</math> and the total potential is written as
<center><math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_\nu \zeta_\nu \phi_\nu^{\mathrm{R}}
</math>
</center>

== Final System of Equations ==

We substitute the expansion for the potential into the equations in the frequency domain and we obtain
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
\left(\phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_{\nu} \zeta_\nu \phi_{\nu}^{\mathrm{R}}\right) \mathbf{n}_{\mu}\, dS
- \sum_{\nu} C_{\mu\nu}\xi_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>

{{added mass damping and force matrices definition}}

Then the equations can be expressed as follows.
<center><math> \left[-\omega^2 \left(\mathbf{M} + \mathbf{A} \right) +
\mathrm{i}\omega \mathbf{B} + \mathbf{C} \right] \vec{\zeta} = \mathbf{f} </math></center>
where <math>\mathbf{M}</math> is the mass matrix, <math>\mathbf{A}</math> is the added mass matrix,
<math>\mathbf{B}</math> is the damping matrix, <math>\mathbf{C}</math> is the hydrostatic matrix,
<math>\vec{\zeta}</math> is the vector of body displacements and <math>\mathbf{f}</math> is the force.

The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the two dimensional problem for the sake of clarity.

== Symmetric body ==

For a body which is [[:Category:Symmetry in Two Dimensions|Symmetric in Two Dimensions]]
the Heave is decoupled from Surge and Roll.
In other words the Surge and Roll motions do not influence Heave and vice versa.

== Matlab Code ==

*A program to solve for pitch and heave and only for two geometries can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/rigid_body_motion.m rigid_body_motion.m]

* a program to calculate the solution for a specific geometry (with plot as output as shown) can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/wave_bem_example_floating.m wave_bem_example_floating.m]

[[Image:Wave_bem_example_floating_RT2.jpg|300px|right|thumb|The reflection (solid line) and transmission (dashed line)
for a dock for heave and pitch (red), heave only (blue) and pitch only (black)]]

=== Additional code ===

This program requires
* A program to calculate the geometery [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/circlebody_twod.m circlebody_twod.m]
* {{fixed body bem code}}
* {{floating body radiation code}}
* {{free surface dispersion equation code}}
* {{boundary element code}}

== Symmetry-reciprocity relations ==

It will be shown that
<center><math> A_{ij}(\omega) = A_{ji}(\omega) \,</math></center>
<center><math> B_{ij}(\omega) = B_{ji}(\omega) \,</math></center>
Along the same lines it will be shown that the exciting force <math>\mathbf{X}_j\,</math> can be expressed in terms of <math> \psi_j\,</math> circumventing the solution for the diffraction potential.
The core result needed for the proof of the above properties is [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
<center><math> \iint_S \left( \psi_1 \frac{\partial\psi_2}{\partial n} - \psi_2 \frac{\partial\psi_1}{\partial n} \right) \mathrm{d}S = 0 \,</math></center>
where <math>\nabla^2 \psi_i=0</math>.

{{energy_region_plates}}

[[Image:Symmetry_boundary.jpg|thumb|right|600px|Boundary]]

In the surface wave-body problem define the closed surfaces as shown in figure on the right.
Let <math> \phi_j\,</math> be rediation or diffraction potentials. Over the boundaries <math>S^\pm\,</math> we have:

<center><math> S^+: \quad \phi_j \ \sim \ \frac{igA_j^+}{\omega} e^{Kz-iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial \phi_j}{\partial x} \ \sim \ -iK\phi_j \,</math></center>

<center><math> S^-: \quad \phi_j \ \sim \ \frac{igA_j^-}{\omega} e^{Kz+iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = - \frac{\partial \phi_j}{\partial x} \ \sim \ - iK\phi_j \,</math></center>

On <math> S_F: \qquad \frac{\partial\phi_j}{\partial z} = K\phi_j, \qquad \frac{\partial \Phi_j}{\partial n} = \frac{\partial \phi_j}{\partial z} </math>

On <math> S_\infty: \qquad \left| \phi_j \right|, \quad \left| \nabla \phi_j \right| \to 0 </math>.

Applying Green's identity to any pair of the radiation potentials <math> \psi_i, \psi_j \,</math>:

<center><math> \iint_{S_B} \left[ \psi_i \frac{\partial\psi_j}{\partial n} - \psi_j \frac{\partial\psi_i}{\partial n} \right] \mathrm{d}S = - \iint_{S_F} \left[ \psi_i \frac{\partial\psi_j}{\partial z} - \psi_j \frac{\partial\psi_i}{\partial z} \right] \mathrm{d}S </math></center>

<center><math> - \iint_{S_+} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S
+ \iint_{S_-} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S = 0 </math></center>

It follows that:

<center><math> \iint_{S_B} \psi_i \frac{\partial\psi_j}{\partial n} \mathrm{d}S = \iint_{S_B} \psi_j \frac{\partial\psi_i}{\partial n} \mathrm{d}S </math></center>
or
<center><math> A_{ij}(\omega) = A_{ji}(\omega), \qquad B_{ij}(\omega) = B_{ji}(\omega). \,</math></center>

== Haskind relations of exciting forces ==

<center><math>
\begin{align}
\mathbf{X}_i(\omega) &= - i\omega\rho\iint_{S_B} (\phi_I + \phi_7) n_i \mathrm{d}S \\
&= - \rho \iint_{S_B} (\phi_I + \phi_7) \frac{\partial \phi_i}{\partial n} \mathrm{d}S
\end{align}
</math></center>

where the radiation velocity potential <math> \phi_i \,</math> is known to satisfy:

<center><math> \frac{\partial\phi_i}{\partial n} = i\omega n_i, \quad \mbox{on} \ S_B </math></center>

and

<center><math> \frac{\partial\phi_7}{\partial n} = \frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \ S_B </math></center>

Both <math> \phi_i\,</math> and <math> \phi_7\,</math> satisfy the condition of outgoing waves at infinity. By virtue of [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]

<center><math> \iint_{S_B} \phi_7 \frac{\partial\phi_i}{\partial n} \mathrm{d}S = \iint_{S_B} \phi_i \frac{\partial\phi_7}{\partial n} \mathrm{d}S = -\iint_{S_B} \phi_i \frac{\partial\phi_I}{\partial n} \mathrm{d}S </math></center>

The Haskind expression for the exciting force follows:

<center><math> \mathbf{X}_i(\omega) = \rho \iint_{S_B} \left[ \phi_I \frac{\partial\phi_i}{\partial n} - \phi_i \frac{\partial\phi_I}{\partial n} \right] \mathrm{d}S </math></center>

The symmetry of the <math> A_{ij}(\omega), B_{ij}(\omega) \,</math> matrices applies in 2D and 3D. The application of Green's Theorem in 3D is very similar using the far-field representation for the potential <math> \phi_j\,</math>

<center><math> \phi_j \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O\left(\frac{1}{R^{3/2}}\right) </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial\phi_j}{\partial R} \sim - i K \phi_j + O\left(\frac{1}{R^{3/2}}\right) </math></center>

where <math> R \,</math> is a radius from the body out to infinity and the <math> R^{-\frac{1}{2}} \,</math> decay arises from the energy conservation principle. Details of the 3D proof may be found in [[Mei 1983]] and [[Wehausen and Laitone 1960]]

The use of the Haskind relations for the exciting forces does not require the solution of the diffraction problem. This is convenient and often more accurate.

The Haskind relations take other forms which will not be presented here but are detailed in [[Wehausen and Laitone 1960]]. The ones that are used in practice relate the exciting forces to the damping coefficients.

These take the form:

2D: <math> B_{ii} = \frac{\left| \mathbf{X}_i \right|^2}{2\rho g V_g}, \quad V_g = \frac{g}{2\omega}, </math> Deep water

3D: <math> B_{33} = \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2 \,</math> --- Heave

(Axisymmetric bodies) <math> B_{22} = \frac{K}{8\rho g V_g} \left| \mathbf{X}_2 \right|^2 \,</math> --- Sway

So knowledge of <math> \mathbf{X}_i(\omega)\,</math> allows the direct evaluation of the diagonal damping coefficients. These expressions are useful in deriving theoretical results in wave-body interactions to be discussed later.

The two-dimensional theory of wave-body interactions in the frequency domain extends to three dimencions very directly with little difficulty.

The statement of the 6 d.o.f. seakeeping problem is:

<center><math> \sum_{j=1}^6 \left[ - \omega^2 \left( M_{ij} + A_{ij} \right) + i \omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_j, \quad i=1,\cdots,6 </math></center>

where

<center><math> M_{ij}: \mbox{Body inertia matrix including moments of inertia for rotational modes. For details refer to MH} \, </math></center>

<center><math> A_{ij}(\omega): \mbox{Added mass matrix} \,</math></center>

<center><math> B_{ij}(\omega): \mbox{Damping matrix} \, </math></center>

<center><math> C_{ij}: \mbox{Hydrostatic and static inertia restoring matrix. For details refer to MH} \, </math></center>

<center><math> \mathbf{X}_i(\omega): \mbox{Wave exciting forces and moments} </math></center>

At zero speed the definitions of the added-mass, damping matrices and exciting forces are identical to those in two dimensions.

The boundary value problems satisfied by the radiation potentials <math>\phi_j, \ j=1,\cdots,6 \,</math> and the diffraction potential <math> \phi_7 \,</math> are as follows:

Free-surface condition:

<center><math> -\omega^2 \phi_j + g \frac{\partial\phi_j}{\partial Z} = 0, \quad z=0 \quad j=1,\cdots,7 </math></center>

Body-boundary conditions:

<center><math> \frac{\partial\phi_7}{\partial n} = -\frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \quad S_B </math></center>

<center><math> \phi_I = \frac{i g A}{\omega} e^{Kz-iKx\cos\beta-iKy\sin\beta+i\omega t} </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = i\omega n_j, \quad j=1,\cdots,6 \, </math></center>

<center><math> n_j = \begin{cases}
n_j, & j=1,2,3 \\
\left( \vec{x} \times \vec{n} \right), & j=4,5,6
\end{cases} </math></center>

<center><math> j=1: \ \mbox{Surge} \qquad j=2: \ \mbox{Sway} \qquad j=3: \ \mbox{Heave} </math></center>
<center><math> j=4: \ \mbox{Roll} \qquad j=5: \ \mbox{Pitch} \qquad j=6: \ \mbox{Yaw} </math></center>

At large distances from the body the velocity potentials satisfy the radiation condition:

<center><math> \phi_j (R,\theta) \sim \frac{A_j(\theta)}{\sqrt{R}} e^{Kz-iKR} + O \left( \frac{1}{R^{3/2}} \right) </math></center>

with

<center><math> K = \frac{\omega^2}{g}. \,</math></center>

This radiation condition is essential for the formulation and solution of the boundary value problems for <math>\phi_j\,</math> using panel methods which are the standard solution technique at zero and forward speed.

Qualitative behaviour of the forces, coefficients and motions of floating bodies

<center><math> - \omega^2 \phi + g \phi_Z =0 \quad \begin{cases}
\phi_Z=0,\quad \omega=0 \\
\phi=0, \quad \omega \to \infty
\end{cases} </math></center>

<center><math> B_{33}(\omega), \sim \omega, \mbox{at low} \ \omega \,</math></center>

The 2D Heave added mass is singular at low frequencies. It is finite in 3D

The 2D Heave damping coefficient is decaying to zero linearly in 2D and superlinearly in 3D. A two-dimensional section is a better wavemaker than a three-dimensional one

A 2D section oscillating in Sway is less effective a wavemaker at low frequencies than the same section oscillating in Heave

The zero-frequency limit of the Sway added mass is finite and similar to the infinite frequency limit of the Heave added mass.

In long waves the Heave exciting force tends to the Heave restoring coefficient times the ambient wave amplitude the free surface behaves like a flat surface moving up and down.

In long waves the Sway exciting force tends to zero. Proof will follow

In short waves all forces tend to zero.

Pitch exciting moment (same applies to Roll) tends to zero. Long waves have a small slope which is proportional to <math> KA</math>, where <math> K\,</math> is the wave number and <math> A\,</math> is the wave amplitude.

Prove that to leading order for <math>KA\to 0 \,</math>:

<center><math> \left| X_S(\omega) \right| \sim KA C_{55}\,</math></center>

where <math>C_{55}\,</math> is the Pitch (<math> C_{44} \,</math> for Roll) hydrostatic restoring coefficient. [NB: very long waves look like a flat surface inclined at <math> KA\,</math> ].

== Body motions in regular waves ==

Heave:

<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega)}{-\omega^2(A_{33} + M) + i\omega B_{33} +C_{33} } </math></center>

Resonance:

<center><math> \omega^2 = \frac{C_{33}}{M+A_{33}} = \frac{\rho g A \omega}{M + A_{33} (\omega)} </math></center>

In principle the above equation is nonlinear for <math>\omega\,</math>. Will be approximated below

At resonance:
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega^*)}{i\omega^* B_{33}(\omega^*)} \,</math></center>

Invoking the relation between the damping coefficient and the exicting force in 3D:

<center><math> \frac{\left| \Pi_3 \right|}{A} = \frac{\left| \mathbf{X}_3(\omega) \right|}{\omega \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2}, \quad V_g=\frac{g}{2\omega} </math></center>

<center><math> =\frac{2\rho g}{\omega^3 \left|\mathbf{X}_3(\omega)\right|}, \quad \mbox{at resonance} </math></center>

This counter-intuitive result shows that for a body undergoing a pure Heave oscillation, the modulus of the Heave response at resonance is inversely proportional to the modulus of the Heave exciting force.

Viscous effects not discussed here may affect Heave response at resonance

The behavior of the Sway response can be found in an analagous manner,

-----

This article is based on the MIT open course notes and the original articles can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here] and
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C5323823-0180-45EA-B165-15856948A0A2/0/lecture10.pdf here]

Linear Wave-Body Interaction

2010-11-06T11:41:59Z

Adi Kurniawan: /* Equations for a Fixed Body in Frequency Domain */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Linear Wave-Body Interaction
| next chapter = [[Long Wavelength Approximations]]
| previous chapter = [[Ship Kelvin Wake]]
}}

{{incomplete pages}}

[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]

We consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] in the
[[Frequency Domain Problem|Frequency Domain]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem).

== Introduction ==

We derive here the equations of motion for a body in [[Linear Plane Progressive Regular Waves]] in the frequency domain in
two dimensions. We begin with the equations in the time domian. The simplest problems is [[Waves reflecting off a vertical wall]]

== Equations for a Floating Body in the Time Domain ==

We begin with the equations for a floating two-dimensional body in the time domain.

{{equations of motion time domain without body condition}}

{{two dimensional floating body time domain}}

More details can be found in [[:Category:Time-Dependent Linear Water Waves|Time-Dependent Linear Water Waves]]

== Equations for a Floating Body in the Frequency Domain ==

The dynamic condition is the equation of motion for the structure in the [[Frequency Domain Problem|frequency domain]]
can be found from the time domain equations and we introduce the following notation
<center><math>
\xi_{\nu} = \zeta_{\nu}e^{-\mathrm{i}\omega t}\,
</math></center>
This give us
{{standard linear wave scattering equations without body condition}}
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=\mathrm{i}\omega\rho\iint_{\partial\Omega}\phi n_{\mu}\, \mathrm{d}S
- \sum_{\nu} C_{\mu\nu}\zeta_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>
The equations of motion for <math> \zeta_\nu\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions. We use the standard numbering of the modes of motion.

== Equations for a Fixed Body in Frequency Domain ==

The equations for a fixed body are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for a rigid body}}
plus the radiation conditions.

We decompose the potential as
<math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} \,,
</math>
where <math>\phi^{\mathrm{I}}</math> is the incident potential and <math>\phi^{\mathrm{D}}</math>
is the diffracted potential. The boundary condition for the diffracted potential is
<center><math>
\begin{align}
\Delta\phi^{\mathrm{D}}&=0, &-h<z<0,\,\,\mathbf{x} \in \Omega \\
\partial_n\phi^{\mathrm{D}} &= 0, &z=-h, \\
\partial_n \phi^{\mathrm{D}} &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial\Omega_{F},
\end{align}
</math></center>
plus
<center><math>
\partial_n \phi^{\mathrm{D}} = - \partial_n \phi^{\mathrm{I}},\,\, \mathbf{x} \in \partial\Omega_{B},
</math></center>

Code to calculate the solution (using a slighly modified method) can be found in
[[Boundary Element Method for a Fixed Body in Finite Depth]]

== Equations for the Radiation Potential in Frequency Domain ==

We decompose the body motion into the rigid body modes of motion. Associated with
each of these modes is a potential which must be solved for.
The equations for the radiation potential in the frequency domain are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for the radiation modes}}

{{sommerfeld radiation condition two dimensions for radiation}}

Code to calculate the radiation potential can be found in
[[Boundary Element Method for the Radiation Potential in Finite Depth]]

We denote the solution for each of the radiation potentials by
<math>\phi_\nu^{\mathrm{R}}</math> and the total potential is written as
<center><math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_\nu \zeta_\nu \phi_\nu^{\mathrm{R}}
</math>
</center>

== Final System of Equations ==

We substitute the expansion for the potential into the equations in the frequency domain and we obtain
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
\left(\phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_{\nu} \zeta_\nu \phi_{\nu}^{\mathrm{R}}\right) \mathbf{n}_{\mu}\, dS
- \sum_{\nu} C_{\mu\nu}\xi_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>

{{added mass damping and force matrices definition}}

Then the equations can be expressed as follows.
<center><math> \left[-\omega^2 \left(\mathbf{M} + \mathbf{A} \right) +
\mathrm{i}\omega \mathbf{B} + \mathbf{C} \right] \vec{\zeta} = \mathbf{f} </math></center>
where <math>\mathbf{M}</math> is the mass matrix, <math>\mathbf{A}</math> is the added mass matrix,
<math>\mathbf{B}</math> is the damping matrix, <math>\mathbf{C}</math> is the hydrostatic matrix,
<math>\vec{\zeta}</math> is the vector of body displacements and <math>\mathbf{f}</math> is the force.

The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the two dimensional problem for the sake of clarity.

== Symmetric body ==

For a body which is [[:Category:Symmetry in Two Dimensions|Symmetric in Two Dimensions]]
the Heave is decoupled from Surge and Roll.
In other words the Surge and Roll motions do not influence Heave and vice versa.

== Matlab Code ==

*A program to solve for pitch and heave and only for two geometries can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/rigid_body_motion.m rigid_body_motion.m]

* a program to calculate the solution for a specific geometry (with plot as output as shown) can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/wave_bem_example_floating.m wave_bem_example_floating.m]

[[Image:Wave_bem_example_floating_RT2.jpg|300px|right|thumb|The reflection (solid line) and transmission (dashed line)
for a dock for heave and pitch (red), heave only (blue) and pitch only (black)]]

=== Additional code ===

This program requires
* A program to calculate the geometery [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/circlebody_twod.m circlebody_twod.m]
* {{fixed body bem code}}
* {{floating body radiation code}}
* {{free surface dispersion equation code}}
* {{boundary element code}}

== Symmetry-reciprocity relations ==

It will be shown that
<center><math> A_{ij}(\omega) = A_{ji}(\omega) \,</math></center>
<center><math> B_{ij}(\omega) = B_{ji}(\omega) \,</math></center>
Along the same lines it will be shown that the exciting force <math>\mathbf{X}_j\,</math> can be expressed in terms of <math> \psi_j\,</math> circumventing the solution for the diffraction potential.
The core result needed for the proof of the above properties is [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
<center><math> \iint_S \left( \psi_1 \frac{\partial\psi_2}{\partial n} - \psi_2 \frac{\partial\psi_1}{\partial n} \right) \mathrm{d}S = 0 \,</math></center>
where <math>\nabla^2 \psi_i=0</math>.

{{energy_region_plates}}

[[Image:Symmetry_boundary.jpg|thumb|right|600px|Boundary]]

In the surface wave-body problem define the closed surfaces as shown in figure on the right.
Let <math> \phi_j\,</math> be rediation or diffraction potentials. Over the boundaries <math>S^\pm\,</math> we have:

<center><math> S^+: \quad \phi_j \ \sim \ \frac{igA_j^+}{\omega} e^{Kz-iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial \phi_j}{\partial x} \ \sim \ -iK\phi_j \,</math></center>

<center><math> S^-: \quad \phi_j \ \sim \ \frac{igA_j^-}{\omega} e^{Kz+iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = - \frac{\partial \phi_j}{\partial x} \ \sim \ - iK\phi_j \,</math></center>

On <math> S_F: \qquad \frac{\partial\phi_j}{\partial z} = K\phi_j, \qquad \frac{\partial \Phi_j}{\partial n} = \frac{\partial \phi_j}{\partial z} </math>

On <math> S_\infty: \qquad \left| \phi_j \right|, \quad \left| \nabla \phi_j \right| \to 0 </math>.

Applying Green's identity to any pair of the radiation potentials <math> \psi_i, \psi_j \,</math>:

<center><math> \iint_{S_B} \left[ \psi_i \frac{\partial\psi_j}{\partial n} - \psi_j \frac{\partial\psi_i}{\partial n} \right] \mathrm{d}S = - \iint_{S_F} \left[ \psi_i \frac{\partial\psi_j}{\partial z} - \psi_j \frac{\partial\psi_i}{\partial z} \right] \mathrm{d}S </math></center>

<center><math> - \iint_{S_+} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S
+ \iint_{S_-} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S = 0 </math></center>

It follows that:

<center><math> \iint_{S_B} \psi_i \frac{\partial\psi_j}{\partial n} \mathrm{d}S = \iint_{S_B} \psi_j \frac{\partial\psi_i}{\partial n} \mathrm{d}S </math></center>
or
<center><math> A_{ij}(\omega) = A_{ji}(\omega), \qquad B_{ij}(\omega) = B_{ji}(\omega). \,</math></center>

== Haskind relations of exciting forces ==

<center><math> \mathbf{X}_i(\omega) = - i\omega\rho\iint_{S_B} (\phi_I + \phi_7) n_i \mathrm{d}S </math></center>

<center><math> = - \rho \iint_{S_B} (\phi_I + \phi_7) \frac{\partial \phi_i}{\partial n} \mathrm{d}S </math></center>

where the radiation velocity potential <math> \phi_i \,</math> is known to satisfy:

<center><math> \frac{\partial\phi_i}{\partial n} = i\omega n_i, \quad \mbox{on} \ S_B </math></center>

and

<center><math> \frac{\partial\phi_7}{\partial n} = \frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \ S_B </math></center>

Both <math> \phi_i\,</math> and <math> \phi_7\,</math> satisfy the condition of outgoing waves at infinity. By virtue of [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]

<center><math> \iint_{S_B} \phi_7 \frac{\partial\phi_i}{\partial n} \mathrm{d}S = \iint_{S_B} \phi_i \frac{\partial\phi_7}{\partial n} \mathrm{d}S = -\iint_{S_B} \phi_i \frac{\partial\phi_I}{\partial n} \mathrm{d}S </math></center>

The Haskind expression for the exciting force follows:

<center><math> \mathbf{X}_i(\omega) = \rho \iint_{S_B} \left[ \phi_I \frac{\partial\phi_i}{\partial n} - \phi_i \frac{\partial\phi_I}{\partial n} \right] \mathrm{d}S </math></center>

The symmetry of the <math> A_{ij}(\omega), B_{ij}(\omega) \,</math> matrices applies in 2D and 3D. The application of Green's Theorem in 3D is very similar using the far-field representation for the potential <math> \phi_j\,</math>

<center><math> \phi_j \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O\left(\frac{1}{R^{3/2}}\right) </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial\phi_j}{\partial R} \sim - i K \phi_j + O\left(\frac{1}{R^{3/2}}\right) </math></center>

where <math> R \,</math> is a radius from the body out to infinity and the <math> R^{-\frac{1}{2}} \,</math> decay arises from the energy conservation principle. Details of the 3D proof may be found in [[Mei 1983]] and [[Wehausen and Laitone 1960]]

The use of the Haskind relations for the exciting forces does not require the solution of the diffraction problem. This is convenient and often more accurate.

The Haskind relations take other forms which will not be presented here but are detailed in [[Wehausen and Laitone 1960]]. The ones that are used in practice relate the exciting forces to the damping coefficients.

These take the form:

2D: <math> B_{ii} = \frac{\left| \mathbf{X}_i \right|^2}{2\rho g V_g}, \quad V_g = \frac{g}{2\omega}, </math> Deep water

3D: <math> B_{33} = \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2 \,</math> --- Heave

(Axisymmetric bodies) <math> B_{22} = \frac{K}{8\rho g V_g} \left| \mathbf{X}_2 \right|^2 \,</math> --- Sway

So knowledge of <math> \mathbf{X}_i(\omega)\,</math> allows the direct evaluation of the diagonal damping coefficients. These expressions are useful in deriving theoretical results in wave-body interactions to be discussed later.

The two-dimensional theory of wave-body interactions in the frequency domain extends to three dimencions very directly with little difficulty.

The statement of the 6 d.o.f. seakeeping problem is:

<center><math> \sum_{j=1}^6 \left[ - \omega^2 \left( M_{ij} + A_{ij} \right) + i \omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_j, \quad i=1,\cdots,6 </math></center>

where

<center><math> M_{ij}(\omega): \mbox{Body inertia matrix including moments of inertia for rotational modes. For details refer to MH} \, </math></center>

<center><math> A_{ij}(\omega): \mbox{Added mass matrix} \,</math></center>

<center><math> B_{ij}(\omega): \mbox{Damping matrix} \, </math></center>

<center><math> C_{ij}: \mbox{Hydrostatic and static inertia restoring matrix. For details refer to MH} \, </math></center>

<center><math> \mathbf{X}_i(\omega): \mbox{Wave exciting forces and moments} </math></center>

At zero speed the definitions of the added-mass, damping matrices and exciting forces are identical to those in two dimensions.

The boundary value problems satisfied by the radiation potentials <math>\phi_j, \ j=1,\cdots,6 \,</math> and the diffraction potential <math> \phi_7 \,</math> are as follows:

Free-surface condition:

<center><math> -\omega^2 \phi_j + g \frac{\partial\phi_j}{\partial Z} = 0, \quad z=0 \quad j=1,\cdots,7 </math></center>

Body-boundary conditions:

<center><math> \frac{\partial\phi_7}{\partial n} = -\frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \quad S_B </math></center>

<center><math> \phi_I = \frac{i g A}{\omega} e^{KZ-iKX\cos\beta-iKY\sin\beta+i\omega t} </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = i\omega n_j, \quad j=1,\cdots,6 \, </math></center>

<center><math> n_j = \begin{Bmatrix}
& n_j, \qquad & j=1,2,3 \\
& \left( \vec{X} \times \vec{n} \right)_{j+3}, \quad j=4,5,6
\end{Bmatrix} </math></center>

<center><math> i=1: \ \mbox{Surge} \qquad i=2: \ \mbox{Sway} \qquad i=3: \ \mbox{Heave} </math></center>
<center><math> i=4: \ \mbox{Roll} \qquad i=5: \ \mbox{Pitch} \qquad i=6: \ \mbox{Yaw} </math></center>

At large distances from the body the velocity potentials satisfy the radiation condition:

<center><math> \phi_j (R,\theta) \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O \left( \frac{1}{R^{3/2}} \right) </math></center>

with

<center><math> K = \frac{\omega^2}{g}. \,</math></center>

This radiation condition is essential for the formulation and solution of the boundary value problems for <math>\phi_j\,</math> using panel methods which are the standard solution technique at zero and forward speed.

Qualitative behaviour of the forces, coefficients and motions of floating bodies

<center><math> - \omega^2 \phi + g \phi_Z =0 \quad \begin{cases}
\phi_Z=0,\quad \omega=0 \\
\phi=0, \quad \omega \to \infty
\end{cases} </math></center>

<center><math> B_{33}(\omega), \sim \omega, \mbox{at low} \ \omega \,</math></center>

The 2D Heave added mass is singular at low frequencies. It is finite in 3D

The 2D Heave damping coefficient is decaying to zero linearly in 2D and superlinearly in 3D. A two-dimensional section is a better wavemaker than a three-dimensional one

A 2D section oscillating in Sway is less effective a wavemaker at low frequencies than the same section oscillating in Heave

The zero-frequency limit of the Sway added mass is finite and similar to the infinite frequency limit of the Heave added mass.

In long waves the Heave exciting force tends to the Heave restoring coefficient times the ambient wave amplitude the free surface behaves like a flat surface moving up and down.

In long waves the Sway exciting force tends to zero. Proof will follow

In short waves all forces tend to zero.

Pitch exciting moment (same applies to Roll) tends to zero. Long waves have a small slope which is proportional to <math> KA</math>, where <math> K\,</math> is the wave number and <math> A\,</math> is the wave amplitude.

Prove that to leading order for <math>KA\to 0 \,</math>:

<center><math> \left| X_S(\omega) \right| \sim KA C_{55}\,</math></center>

where <math>C_{55}\,</math> is the Pitch (<math> C_{44} \,</math> for Roll) hydrostatic restoring coefficient. [NB: very long waves look like a flat surface inclined at <math> KA\,</math> ].

== Body motions in regular waves ==

Heave:

<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega)}{-\omega^2(A_{33} + M) + i\omega B_{33} +C_{33} } </math></center>

Resonance:

<center><math> \omega^2 = \frac{C_{33}}{M+A_{33}} = \frac{\rho g A \omega}{M + A_{33} (\omega)} </math></center>

In principle the above equation is nonlinear for <math>\omega\,</math>. Will be approximated below

At resonance:
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega^*)}{i\omega^* B_{33}(\omega^*)} \,</math></center>

Invoking the relation between the damping coefficient and the exicting force in 3D:

<center><math> \frac{\left| \Pi_3 \right|}{A} = \frac{\left| \mathbf{X}_3(\omega) \right|}{\omega \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2}, \quad V_g=\frac{g}{2\omega} </math></center>

<center><math> =\frac{2\rho g}{\omega^3 \left|\mathbf{X}_3(\omega)\right|}, \quad \mbox{at resonance} </math></center>

This counter-intuitive result shows that for a body undergoing a pure Heave oscillation, the modulus of the Heave response at resonance is inversely proportional to the modulus of the Heave exciting force.

Viscous effects not discussed here may affect Heave response at resonance

The behavior of the Sway response can be found in an analagous manner,

-----

This article is based on the MIT open course notes and the original articles can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here] and
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C5323823-0180-45EA-B165-15856948A0A2/0/lecture10.pdf here]

Linear Wave-Body Interaction

2010-11-06T11:39:02Z

Adi Kurniawan: /* Haskind relations of exciting forces */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Linear Wave-Body Interaction
| next chapter = [[Long Wavelength Approximations]]
| previous chapter = [[Ship Kelvin Wake]]
}}

{{incomplete pages}}

[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]

We consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] in the
[[Frequency Domain Problem|Frequency Domain]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem).

== Introduction ==

We derive here the equations of motion for a body in [[Linear Plane Progressive Regular Waves]] in the frequency domain in
two dimensions. We begin with the equations in the time domian. The simplest problems is [[Waves reflecting off a vertical wall]]

== Equations for a Floating Body in the Time Domain ==

We begin with the equations for a floating two-dimensional body in the time domain.

{{equations of motion time domain without body condition}}

{{two dimensional floating body time domain}}

More details can be found in [[:Category:Time-Dependent Linear Water Waves|Time-Dependent Linear Water Waves]]

== Equations for a Floating Body in the Frequency Domain ==

The dynamic condition is the equation of motion for the structure in the [[Frequency Domain Problem|frequency domain]]
can be found from the time domain equations and we introduce the following notation
<center><math>
\xi_{\nu} = \zeta_{\nu}e^{-\mathrm{i}\omega t}\,
</math></center>
This give us
{{standard linear wave scattering equations without body condition}}
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=\mathrm{i}\omega\rho\iint_{\partial\Omega}\phi n_{\mu}\, \mathrm{d}S
- \sum_{\nu} C_{\mu\nu}\zeta_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>
The equations of motion for <math> \zeta_\nu\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions. We use the standard numbering of the modes of motion.

== Equations for a Fixed Body in Frequency Domain ==

The equations for a fixed body are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for a rigid body}}
plus the radiation conditions.

We decompose the potential as
<center>
<math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}}
</math>
</center>
where <math>\phi^{\mathrm{I}}</math> is the incident potential and <math>\phi^{\mathrm{D}}</math>
is the diffracted potential. The boundary condition for the diffracted potential is
<center><math>
\Delta\phi^{\mathrm{D}}=0, \, -h<z<0,\,\,\,\mathbf{x} \in \Omega
</math></center>
<center><math>
\partial_n\phi^{\mathrm{D}} = 0, \, z=-h,
</math></center>
<center><math>
\partial_n \phi^{\mathrm{D}} = \alpha \phi,\,z=0,\,\,\mathbf{x} \in \partial\Omega_{F},
</math></center>
plus
<center><math>
\partial_n \phi^{\mathrm{D}} = - \partial_n \phi^{\mathrm{I}},\,\, \mathbf{x} \in \partial\Omega_{B},
</math></center>

Code to calculate the solution (using a slighly modified method) can be found in
[[Boundary Element Method for a Fixed Body in Finite Depth]]

== Equations for the Radiation Potential in Frequency Domain ==

We decompose the body motion into the rigid body modes of motion. Associated with
each of these modes is a potential which must be solved for.
The equations for the radiation potential in the frequency domain are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for the radiation modes}}

{{sommerfeld radiation condition two dimensions for radiation}}

Code to calculate the radiation potential can be found in
[[Boundary Element Method for the Radiation Potential in Finite Depth]]

We denote the solution for each of the radiation potentials by
<math>\phi_\nu^{\mathrm{R}}</math> and the total potential is written as
<center><math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_\nu \zeta_\nu \phi_\nu^{\mathrm{R}}
</math>
</center>

== Final System of Equations ==

We substitute the expansion for the potential into the equations in the frequency domain and we obtain
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
\left(\phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_{\nu} \zeta_\nu \phi_{\nu}^{\mathrm{R}}\right) \mathbf{n}_{\mu}\, dS
- \sum_{\nu} C_{\mu\nu}\xi_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>

{{added mass damping and force matrices definition}}

Then the equations can be expressed as follows.
<center><math> \left[-\omega^2 \left(\mathbf{M} + \mathbf{A} \right) +
\mathrm{i}\omega \mathbf{B} + \mathbf{C} \right] \vec{\zeta} = \mathbf{f} </math></center>
where <math>\mathbf{M}</math> is the mass matrix, <math>\mathbf{A}</math> is the added mass matrix,
<math>\mathbf{B}</math> is the damping matrix, <math>\mathbf{C}</math> is the hydrostatic matrix,
<math>\vec{\zeta}</math> is the vector of body displacements and <math>\mathbf{f}</math> is the force.

The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the two dimensional problem for the sake of clarity.

== Symmetric body ==

For a body which is [[:Category:Symmetry in Two Dimensions|Symmetric in Two Dimensions]]
the Heave is decoupled from Surge and Roll.
In other words the Surge and Roll motions do not influence Heave and vice versa.

== Matlab Code ==

*A program to solve for pitch and heave and only for two geometries can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/rigid_body_motion.m rigid_body_motion.m]

* a program to calculate the solution for a specific geometry (with plot as output as shown) can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/wave_bem_example_floating.m wave_bem_example_floating.m]

[[Image:Wave_bem_example_floating_RT2.jpg|300px|right|thumb|The reflection (solid line) and transmission (dashed line)
for a dock for heave and pitch (red), heave only (blue) and pitch only (black)]]

=== Additional code ===

This program requires
* A program to calculate the geometery [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/circlebody_twod.m circlebody_twod.m]
* {{fixed body bem code}}
* {{floating body radiation code}}
* {{free surface dispersion equation code}}
* {{boundary element code}}

== Symmetry-reciprocity relations ==

It will be shown that
<center><math> A_{ij}(\omega) = A_{ji}(\omega) \,</math></center>
<center><math> B_{ij}(\omega) = B_{ji}(\omega) \,</math></center>
Along the same lines it will be shown that the exciting force <math>\mathbf{X}_j\,</math> can be expressed in terms of <math> \psi_j\,</math> circumventing the solution for the diffraction potential.
The core result needed for the proof of the above properties is [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
<center><math> \iint_S \left( \psi_1 \frac{\partial\psi_2}{\partial n} - \psi_2 \frac{\partial\psi_1}{\partial n} \right) \mathrm{d}S = 0 \,</math></center>
where <math>\nabla^2 \psi_i=0</math>.

{{energy_region_plates}}

[[Image:Symmetry_boundary.jpg|thumb|right|600px|Boundary]]

In the surface wave-body problem define the closed surfaces as shown in figure on the right.
Let <math> \phi_j\,</math> be rediation or diffraction potentials. Over the boundaries <math>S^\pm\,</math> we have:

<center><math> S^+: \quad \phi_j \ \sim \ \frac{igA_j^+}{\omega} e^{Kz-iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial \phi_j}{\partial x} \ \sim \ -iK\phi_j \,</math></center>

<center><math> S^-: \quad \phi_j \ \sim \ \frac{igA_j^-}{\omega} e^{Kz+iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = - \frac{\partial \phi_j}{\partial x} \ \sim \ - iK\phi_j \,</math></center>

On <math> S_F: \qquad \frac{\partial\phi_j}{\partial z} = K\phi_j, \qquad \frac{\partial \Phi_j}{\partial n} = \frac{\partial \phi_j}{\partial z} </math>

On <math> S_\infty: \qquad \left| \phi_j \right|, \quad \left| \nabla \phi_j \right| \to 0 </math>.

Applying Green's identity to any pair of the radiation potentials <math> \psi_i, \psi_j \,</math>:

<center><math> \iint_{S_B} \left[ \psi_i \frac{\partial\psi_j}{\partial n} - \psi_j \frac{\partial\psi_i}{\partial n} \right] \mathrm{d}S = - \iint_{S_F} \left[ \psi_i \frac{\partial\psi_j}{\partial z} - \psi_j \frac{\partial\psi_i}{\partial z} \right] \mathrm{d}S </math></center>

<center><math> - \iint_{S_+} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S
+ \iint_{S_-} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S = 0 </math></center>

It follows that:

<center><math> \iint_{S_B} \psi_i \frac{\partial\psi_j}{\partial n} \mathrm{d}S = \iint_{S_B} \psi_j \frac{\partial\psi_i}{\partial n} \mathrm{d}S </math></center>
or
<center><math> A_{ij}(\omega) = A_{ji}(\omega), \qquad B_{ij}(\omega) = B_{ji}(\omega). \,</math></center>

== Haskind relations of exciting forces ==

<center><math> \mathbf{X}_i(\omega) = - i\omega\rho\iint_{S_B} (\phi_I + \phi_7) n_i \mathrm{d}S </math></center>

<center><math> = - \rho \iint_{S_B} (\phi_I + \phi_7) \frac{\partial \phi_i}{\partial n} \mathrm{d}S </math></center>

where the radiation velocity potential <math> \phi_i \,</math> is known to satisfy:

<center><math> \frac{\partial\phi_i}{\partial n} = i\omega n_i, \quad \mbox{on} \ S_B </math></center>

and

<center><math> \frac{\partial\phi_7}{\partial n} = \frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \ S_B </math></center>

Both <math> \phi_i\,</math> and <math> \phi_7\,</math> satisfy the condition of outgoing waves at infinity. By virtue of [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]

<center><math> \iint_{S_B} \phi_7 \frac{\partial\phi_i}{\partial n} \mathrm{d}S = \iint_{S_B} \phi_i \frac{\partial\phi_7}{\partial n} \mathrm{d}S = -\iint_{S_B} \phi_i \frac{\partial\phi_I}{\partial n} \mathrm{d}S </math></center>

The Haskind expression for the exciting force follows:

<center><math> \mathbf{X}_i(\omega) = \rho \iint_{S_B} \left[ \phi_I \frac{\partial\phi_i}{\partial n} - \phi_i \frac{\partial\phi_I}{\partial n} \right] \mathrm{d}S </math></center>

The symmetry of the <math> A_{ij}(\omega), B_{ij}(\omega) \,</math> matrices applies in 2D and 3D. The application of Green's Theorem in 3D is very similar using the far-field representation for the potential <math> \phi_j\,</math>

<center><math> \phi_j \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O\left(\frac{1}{R^{3/2}}\right) </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial\phi_j}{\partial R} \sim - i K \phi_j + O\left(\frac{1}{R^{3/2}}\right) </math></center>

where <math> R \,</math> is a radius from the body out to infinity and the <math> R^{-\frac{1}{2}} \,</math> decay arises from the energy conservation principle. Details of the 3D proof may be found in [[Mei 1983]] and [[Wehausen and Laitone 1960]]

The use of the Haskind relations for the exciting forces does not require the solution of the diffraction problem. This is convenient and often more accurate.

The Haskind relations take other forms which will not be presented here but are detailed in [[Wehausen and Laitone 1960]]. The ones that are used in practice relate the exciting forces to the damping coefficients.

These take the form:

2D: <math> B_{ii} = \frac{\left| \mathbf{X}_i \right|^2}{2\rho g V_g}, \quad V_g = \frac{g}{2\omega}, </math> Deep water

3D: <math> B_{33} = \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2 \,</math> --- Heave

(Axisymmetric bodies) <math> B_{22} = \frac{K}{8\rho g V_g} \left| \mathbf{X}_2 \right|^2 \,</math> --- Sway

So knowledge of <math> \mathbf{X}_i(\omega)\,</math> allows the direct evaluation of the diagonal damping coefficients. These expressions are useful in deriving theoretical results in wave-body interactions to be discussed later.

The two-dimensional theory of wave-body interactions in the frequency domain extends to three dimencions very directly with little difficulty.

The statement of the 6 d.o.f. seakeeping problem is:

<center><math> \sum_{j=1}^6 \left[ - \omega^2 \left( M_{ij} + A_{ij} \right) + i \omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_j, \quad i=1,\cdots,6 </math></center>

where

<center><math> M_{ij}(\omega): \mbox{Body inertia matrix including moments of inertia for rotational modes. For details refer to MH} \, </math></center>

<center><math> A_{ij}(\omega): \mbox{Added mass matrix} \,</math></center>

<center><math> B_{ij}(\omega): \mbox{Damping matrix} \, </math></center>

<center><math> C_{ij}: \mbox{Hydrostatic and static inertia restoring matrix. For details refer to MH} \, </math></center>

<center><math> \mathbf{X}_i(\omega): \mbox{Wave exciting forces and moments} </math></center>

At zero speed the definitions of the added-mass, damping matrices and exciting forces are identical to those in two dimensions.

The boundary value problems satisfied by the radiation potentials <math>\phi_j, \ j=1,\cdots,6 \,</math> and the diffraction potential <math> \phi_7 \,</math> are as follows:

Free-surface condition:

<center><math> -\omega^2 \phi_j + g \frac{\partial\phi_j}{\partial Z} = 0, \quad z=0 \quad j=1,\cdots,7 </math></center>

Body-boundary conditions:

<center><math> \frac{\partial\phi_7}{\partial n} = -\frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \quad S_B </math></center>

<center><math> \phi_I = \frac{i g A}{\omega} e^{KZ-iKX\cos\beta-iKY\sin\beta+i\omega t} </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = i\omega n_j, \quad j=1,\cdots,6 \, </math></center>

<center><math> n_j = \begin{Bmatrix}
& n_j, \qquad & j=1,2,3 \\
& \left( \vec{X} \times \vec{n} \right)_{j+3}, \quad j=4,5,6
\end{Bmatrix} </math></center>

<center><math> i=1: \ \mbox{Surge} \qquad i=2: \ \mbox{Sway} \qquad i=3: \ \mbox{Heave} </math></center>
<center><math> i=4: \ \mbox{Roll} \qquad i=5: \ \mbox{Pitch} \qquad i=6: \ \mbox{Yaw} </math></center>

At large distances from the body the velocity potentials satisfy the radiation condition:

<center><math> \phi_j (R,\theta) \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O \left( \frac{1}{R^{3/2}} \right) </math></center>

with

<center><math> K = \frac{\omega^2}{g}. \,</math></center>

This radiation condition is essential for the formulation and solution of the boundary value problems for <math>\phi_j\,</math> using panel methods which are the standard solution technique at zero and forward speed.

Qualitative behaviour of the forces, coefficients and motions of floating bodies

<center><math> - \omega^2 \phi + g \phi_Z =0 \quad \begin{cases}
\phi_Z=0,\quad \omega=0 \\
\phi=0, \quad \omega \to \infty
\end{cases} </math></center>

<center><math> B_{33}(\omega), \sim \omega, \mbox{at low} \ \omega \,</math></center>

The 2D Heave added mass is singular at low frequencies. It is finite in 3D

The 2D Heave damping coefficient is decaying to zero linearly in 2D and superlinearly in 3D. A two-dimensional section is a better wavemaker than a three-dimensional one

A 2D section oscillating in Sway is less effective a wavemaker at low frequencies than the same section oscillating in Heave

The zero-frequency limit of the Sway added mass is finite and similar to the infinite frequency limit of the Heave added mass.

In long waves the Heave exciting force tends to the Heave restoring coefficient times the ambient wave amplitude the free surface behaves like a flat surface moving up and down.

In long waves the Sway exciting force tends to zero. Proof will follow

In short waves all forces tend to zero.

Pitch exciting moment (same applies to Roll) tends to zero. Long waves have a small slope which is proportional to <math> KA</math>, where <math> K\,</math> is the wave number and <math> A\,</math> is the wave amplitude.

Prove that to leading order for <math>KA\to 0 \,</math>:

<center><math> \left| X_S(\omega) \right| \sim KA C_{55}\,</math></center>

where <math>C_{55}\,</math> is the Pitch (<math> C_{44} \,</math> for Roll) hydrostatic restoring coefficient. [NB: very long waves look like a flat surface inclined at <math> KA\,</math> ].

== Body motions in regular waves ==

Heave:

<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega)}{-\omega^2(A_{33} + M) + i\omega B_{33} +C_{33} } </math></center>

Resonance:

<center><math> \omega^2 = \frac{C_{33}}{M+A_{33}} = \frac{\rho g A \omega}{M + A_{33} (\omega)} </math></center>

In principle the above equation is nonlinear for <math>\omega\,</math>. Will be approximated below

At resonance:
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega^*)}{i\omega^* B_{33}(\omega^*)} \,</math></center>

Invoking the relation between the damping coefficient and the exicting force in 3D:

<center><math> \frac{\left| \Pi_3 \right|}{A} = \frac{\left| \mathbf{X}_3(\omega) \right|}{\omega \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2}, \quad V_g=\frac{g}{2\omega} </math></center>

<center><math> =\frac{2\rho g}{\omega^3 \left|\mathbf{X}_3(\omega)\right|}, \quad \mbox{at resonance} </math></center>

This counter-intuitive result shows that for a body undergoing a pure Heave oscillation, the modulus of the Heave response at resonance is inversely proportional to the modulus of the Heave exciting force.

Viscous effects not discussed here may affect Heave response at resonance

The behavior of the Sway response can be found in an analagous manner,

-----

This article is based on the MIT open course notes and the original articles can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here] and
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C5323823-0180-45EA-B165-15856948A0A2/0/lecture10.pdf here]

Ship Kelvin Wake

2010-11-06T11:37:03Z

Adi Kurniawan: /* Kelvin wake */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Ship Kelvin Wake
| next chapter = [[Linear Wave-Body Interaction]]
| previous chapter = [[Wavemaker Theory]]
}}

{{incomplete pages}}

= Introduction =

[[Image:Wake.avon.gorge.arp.750pix.jpg|thumb|right|450px|Wake created behind a ship]]

A ship moving over the surface of undisturbed water sets up waves emanating from the bow and stern of the ship. The waves created by the ship consist of divergent and transverse waves. The divergent wave are observed as the [http://en.wikipedia.org/wiki/Wake wake] of a ship with a series of diagonal or oblique crests moving outwardly from the point of disturbance. These wave were first studied by [http://en.wikipedia.org/wiki/Lord_Kelvin Lord Kelvin]

== Translating Coordinate System ==

We have a the standard fixed coordinate system <math>x, y, z</math> and a moving coordinate systems
which is moving in the <math>x</math> direction with speed <math>U</math>. We denote the moving coordinate
systems in the <math>x</math> direction by
<center><math>
\bar{x} = x + U t
</math></center>

Let <math> \Phi(\mathbf{x},t) \,</math> be the velocity potential describing the potential flow
generated by the ship relative to the earth frame.
The same potential expressed relative to the ship frame is <math> \bar{\Phi}(\mathbf{\bar{x}},t) \, </math>.
The relation between the two potentials is given by the identity
<center><math> \Phi(x,y,z,t) = \bar{\Phi}(\bar{x},y,z,t)
= \bar{\Phi}(x-Ut,y,z,t) \,</math></center>
where the relation between the coordinates of the two coordinate systems has been introduced.
Note the time dependence occurs in two places in <math> \bar{\Phi} \,</math> and in one place in <math> \Phi \,</math>. The governing equations are always derived relative to the earth coordinate system and time derivatives are initially taken on <math> \Phi \, </math>.
Therefore
<center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \bar{\phi} ( x-Ut,y,z,t)
= \frac{\partial\phi}{\partial t} - U \frac{\partial\phi}{\partial x} </math></center>

All time derivatives of the earth fixed velocity potential <math> \Phi \, </math> which appear in the free surface condition and the Bernoulli equation can be expressed in terms of derivatives of <math> \bar{\Phi} \, </math> using the Galilean transformation derived above.

If the flow is steady relative to the ship fixed coordinate system
<center><math> \frac{\partial\bar{\Phi}}{\partial t} = 0 \, </math></center>
but
<center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = -U \frac{\partial\bar{\Phi}}{\partial x} \,</math></center>
or, the ship wake is stationary relative to the ship but not relative to an observed on the beach.

== Kelvin wake ==

[[Image:Kelvin_wake.jpg|thumb|right|600px|Diagram of the Kelvin Wake]]

Local view of Kelvin wake consists approximately of a [[Linear Plane Progressive Regular Waves| plane progressive wave group]] propagating in direction <math> \theta\,\!</math>.
As noted above surface wave systems of general form always consist of combinations of plane progressive waves of different frequencies and directions. The same model will apply to the ship kelvin wake.
Relative to the earth frame, the local plane wave in [[Infinite Depth]] takes the form

<center><math> \Phi = \frac{\mathrm{i}gA}{\omega} e^{kz-\mathrm{i}k(x\cos\theta+y\sin\theta)+ \mathrm{i}\omega t} </math></center>

Relative to the ship frame
<center><math>bar{\Phi} = \frac{igA}{\omega} e^{kz-\mathrm{i}k(x\cos\theta+y\sin\theta)-\mathrm{i}(kU\cos\theta-\omega)t} </math></center>

But relative to the ship frame waves are stationary, so we must have:

<center><math> kU\cos\theta = \omega \,</math></center>
or
<center><math> \frac{\omega}{k} = C_p = U \cos \theta \, </math></center>
This implies the following
* The phase velocity of the waves in the kelvin wake propagating in direction <math>\theta\,</math> must be equal to <math> U\cos\theta\,</math>, otherwise they cannot be stationary relative to the ship.
* Relative to the earth system the frequency of a local system propagating in direction <math> \theta \, </math> is given by the relation <math> \omega = kU \cos \theta \, </math>

Relative to the earth system the [[Infinite Depth]] [[Dispersion Relation for a Free Surface]] states
<center><math> \omega^2 = g k \, </math></center>
so that
<center><math> \lambda(\theta) = \frac{2\pi U^2 \cos^2 \theta}{g} \, </math></center>

This is the wavelength of waves in a Kelvin wake propagating in direction <math> \theta \, </math> which
are stationary relative to the ship.

== Application of the Group velocity ==

An observer sitting on an earth fixed frame observes a local wave system propagating in direction <math> \theta\,</math> travelling at its [[Wave Energy Density and Flux|group velocity]] <math>\frac{d\omega}{dK}\,</math> by virtue of the Rayleigh device which states that we need to focus on the speed of the energy density (<math> \sim \, </math> wave amplitude) rather than the speed of wave crests. So, relative to the earth fixed inclined coordinate system <math> (X', Y') \,</math>:

<center><math> \frac{X'}{t} = V_g = \frac{d\omega}{dK} \, </math></center>

Or

<center><math> X' = \frac{d\omega}{dK} t \ \Longrightarrow \ \frac{d}{dK} (K X' - \omega t) = 0 </math></center>

<center><math> X' = X \cos \theta + Y \sin \theta = x \cos\theta + y\sin\theta + Ut \cos\theta \,</math></center>

So:

<center><math> KX' - \omega t = K ( x\cos\theta + y\sin\theta ) + (KU\cos\theta - \omega) t \,</math></center>
However <math>(KU\cos\theta - \omega) t =0</math> so that the Rayleigh condition for the velocity of the group takes the form:

<center><math> \frac{d}{dK} [ K(\theta) (x\cos\theta + y\sin\theta) ] = 0 \, </math></center>

By virtue of the dispersion relation derived above:

<center><math> K(\theta) = \frac{g}{U^2 \cos^2 \theta} \, </math></center>

It follows from the chain rule of differentiation that Rayleigh's condition is:

<center><math> \frac{d}{d\theta} \left[ \frac{g}{U^2\cos^2\theta} (x\cos\theta + y\sin\theta) ]\right] = 0 \, </math></center>

At the position of the Kelvin waves which are locally observed by an observer at the beach.

* So the "visible" waves in the wake of a ship are wave groups which must travel at the local group velocity. These conditions translate into the above equation which will be solved and discussed next. More discussion and a more mathematical derivation based on the [http://en.wikipedia.org/wiki/Principle_of_stationary_phase principle of stationary phase] can be found in [[Newman 1977]].

== Solution of the equation for angle ==

[[Image:Graphical_image.jpg|thumb|right|600px|Graphical image of the equations]]

The solution of the above equation will produce a relation between <math> \frac{y}{x} \,</math> and <math> \theta\,</math>. So local waves in a Kelvin wake can only propagate in a certain direction <math> \theta\,</math>, given <math> \frac{y}{x} \, </math>.

Simple algebra leads to:

<center><math> \frac{y}{x} = - \frac{\cos\theta\sin\theta}{1+sin^2\theta} = \frac{y}{x}(\theta)
</math></center>
which implies that
* <math>\frac{y}{x}(\theta) \,</math> is anti-symmetric about <math> \theta = 0 \, </math> and each part corresponds to the Kelvin wake in the port and starboard sides of the vessel. The physics on
either side is identical due to symmetry.

* <math> \theta = 0 \,</math>: waves propagating in the same direction as the ship. These waves can only exist at <math> Y = 0\, </math> as seen above.

* <math> \theta = \frac{\pi}{2} \, </math>: waves propagating at a <math> 90^\circ\,</math> angle relative to the ship direction of forward translation.

* <math> \theta = 35^\circ 16' \, </math>: (or 35,26°) waves propagating at an angle <math> \theta = 35^\circ 16' \, </math> relative to the ship axis. These are waves seen at the caustic of the Kelvin wake.

Let the solution of <math> \frac{y}{x}(\theta) \, </math> be of the form, when inverted:

<center><math> \mbox{Region I}: \qquad \theta = f_1 (\frac{y}{x}) \, </math></center>

<center><math> \mbox{Region II}: \qquad \theta = f_2 (\frac{y}{x}) \, </math></center>

Note that observable waves cannot exist for values of <math> \frac{y}{x}\,</math> that exceed the value shown in the figure or <math> \left. \frac{y}{x} \right|_{Max} = 2^{-3/2} \,</math>. This translates into a value for the corresponding angle equal to <math> 19^\circ 28' \, </math> (or 19,47°) which is the angle of the caustic for any speed <math> U\,</math>.

[[Image:Kelvin_wave_image.jpg|thumb|right|600px|"transverse" and "divergent" wave systems in the Kelvin wake]]

The crests of the wave system trailing a ship, the Kelvin wake, are curves of constant phase of:

<center><math> \frac{x\cos\theta+y\sin\theta}{\cos^2\theta} = \mathbf{C} \, </math></center>

In <math> \mbox{Region I} \, </math>:

<center><math> \mathbf{C} = \frac{x\cos f_1(\frac{y}{x}) + y\sin f_1 (\frac{y}{x})}{\cos^2 f_1 (\frac{y}{x})} \equiv G_1 (\frac{y}{x}) </math></center>

In <math> \mbox{Region II} \, </math>:

<center><math> \mathbf{C} = \frac{x\cos f_2(\frac{y}{x}) + y\sin f_2 (\frac{y}{x})}{\cos^2 f_2 (\frac{y}{x})} \equiv G_2 (\frac{y}{x}) </math></center>

Plotting these curves we obtain a visual graph of the "transverse" and "divergent" wave systems in the Kelvin wake.

------

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/80D65339-8DD7-4909-A936-117BE9C97F5A/0/lecture8.pdf here]

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

[[Category:Linear Water-Wave Theory]]

Wave Momentum Flux

2010-11-06T11:35:15Z

Adi Kurniawan: /* Mean horizontal momentum flux due to a Plane Progressive Regular Wave */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Wave Momentum Flux
| next chapter = [[Wavemaker Theory]]
| previous chapter = [[Wave Energy Density and Flux]]
}}

{{incomplete pages}}

== Introduction ==

The momentum is important to determine the forces.

== Momentum flux in potential flow ==

The momentum flux (the time derivative of the [http://en.wikipedia.org/wiki/Momentum momentum] <math>\mathcal{M}</math>) is given by
<center><math> \frac{\mathrm{d}\mathcal{M}(t)}{\mathrm{d}t} = \rho \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \mathbf{v} \mathrm{d}V
= \rho \iiint_{\Omega(t)} \frac{\partial\mathbf{v}}{\partial t} \mathrm{d}V + \rho \iint_{\partial\Omega(t)} \mathbf{v} U_n \mathrm{d}S,
</math></center>
where <math> U_n \, </math> is the outward normal velocity of the surface <math> \partial\Omega(t)\, </math>.
This equation follows from the [http://en.wikipedia.org/wiki/Reynolds_transport_theorem transport theorem].

We begin with [http://en.wikipedia.org/wiki/Euler_equations Euler's equation] in the absence of viscosity
<center><math>
\frac{\partial\mathbf{v}}{\partial t} + (\mathbf{v} \cdot \nabla ) \mathbf{v} =
- \nabla \left(\frac{P}{\rho} + g z\right)
</math></center>
where we have defined the direction of gravity to be in the negative <math>z</math>
direction.
We may recast the momentum flux in the form
<center><math>
\frac{\mathrm{d}\mathcal{M}(t)}{\mathrm{d}t} = - \rho \iiint_{\Omega(t)} \left( \nabla ( \frac{P}{\rho} + g z ) + ( \mathbf{v} \cdot \nabla ) \mathbf{v} \right)
\mathrm{d}V + \rho \iint_{\partial\Omega(t)} \mathbf{v} U_n \mathrm{d}S
</math></center>
So far <math> \Omega(t)\, </math> is an arbitrary closed time dependent volume bounded by the time dependent surface <math> \partial\Omega(t)\, </math>.
We have however defined the gravitational acceleration to be in the negative <math>z</math> direction.

=== Simplification ===

We use the following vector theorem
<center><math> (\mathbf{v} \cdot \nabla ) \mathbf{v} = \nabla ( \frac{1}{2} \mathbf{v} \cdot \mathbf{v} ) - \mathbf{v} \times (\nabla \times \mathbf{v} ) </math></center>
If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and we can use Gauss's vector theorem:
<center><math> \iiint_{\Omega(t)} \nabla ( \frac{1}{2} \mathbf{v} \cdot \mathbf{v} ) \mathrm{d}V = \frac{1}{2} \iint_{\partial\Omega(t)}
\left(\mathbf{v} \cdot \mathbf{v} \right) \mathbf{n} \mathrm{d}S </math></center>
In potential flow it follows that
<center><math> \iint_{\partial\Omega(t)} \frac{1}{2} ( \mathbf{v} \cdot \mathbf{v} ) \mathbf{n} \mathrm{d}S = \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi \mathrm{d}S
= \iint_{\partial\Omega(t)} V_n \mathbf{v} \mathrm{d}S </math></center>
since for <math> \nabla^2 \Phi = 0 \, </math>;
<center><math> \iiint_{\partial\Omega} \frac{1}{2} ( \nabla\Phi \cdot \nabla\Phi) \mathbf{n} \mathrm{d}S \equiv \iint_{\partial\Omega} \frac{\partial\Phi}{\partial n} \nabla\Phi \mathrm{d}S. </math></center>
In the above <math> \mathbf{n} \, </math> is the unit vector pointing out of the volume <math> \Omega(t) </math> and <math> V_n = \mathbf{n} \cdot \nabla \Phi </math>

Upon substitution in the momentum flux formula, we obtain:
<center><math> \frac{\mathrm{d}\mathcal{M}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega(t)} \left( ( \frac{P}{\rho} + gz ) \mathbf{n} + \mathbf{v} (V_n - U_n )\right) \mathrm{d}S </math></center>

This formula is of central importance in potential flow marine hydrodynamics because the rate of change of the linear momentum defined above is just <math> \pm \,</math> the force acting on the fluid volume. When its mean value can be shown to vanish, important force expressions on solid boundaries follow and will be derived in what follows.

== Hydrostatic Term ==

Consider separately the term in the momentum flux expression involving the hydrostatic pressure:
<center><math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} = - \rho \iint_{\partial\Omega} gz \mathbf{n},
</math></center>
We break the boundary up into the free surface, ends, body surface, and sea floor at infinite depth, i,.e.
<math>\partial\Omega=\partial\Omega_F + \partial\Omega^{\pm} + \partial\Omega_B + \partial\Omega_{\infty} </math>
The integral over the body surface, assuming a fully submerged body is:
<center><math> \frac{\mathrm{d}\mathcal{M}_{H,B}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega_B} gz \mathbf{n} = \rho g V \mathbf{k} </math></center>
where <math>V</math> is the volume. This follows from the vector theorem of Gauss and is the [http://en.wikipedia.org/wiki/Displacement_%28fluid%29 principle of Archimedes]. That is,
the momentum flux is equal to the buoyancy force.

We may therefore consider the second part of the integral involving wave effects independently and in the absence of the body, assumed fully submerged. In the case of a surface piercing body and in the fully nonlinear case matters are more complex. Consider the application of the momentum conservation theorem in the case of a submerged or floating body in steep waves.

[[Image:Momentum.png|600px|right|thumb|Momentum boundaries]]

Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial. Note that unlike the [[Wave Energy Density and Flux|energy conservation principle]], the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component. The integral of the hydrostatic term over the remaining surfaces leads to:
<center><math> \frac{\mathrm{d} \mathcal{M}_{H,S}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega_F+\partial\Omega^+ + \partial\Omega^- +\partial\Omega_{\infty}} gz \mathbf{n} \mathrm{d}S = - \rho g V_{\text{Fluid}} \mathbf{k} </math></center>
where <math>V_{\mbox{Fluid}} </math> is the fluid volume.
This is simply the static weight of the volume of fluid bounded by <math> \partial\Omega_F, \partial\Omega^+, \partial\Omega^- \,</math> and <math> \partial\Omega_{\infty}.</math> With no waves present, this is simply the weight of the ocean water "column" bounded by <math> \partial\Omega\, </math> which does not concern us here. This weight does not change in principle when waves are present at least when <math> \partial\Omega^+, \partial\Omega^- \,</math> are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "rational" to apply the momentum conservation theorem over the "linearized" volume <math> V_L(t) \, </math> which is perfectly possible within the framework derived above. In this case <math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} \,</math> is exactly equal to the static weight of the water column and can be ignored in the wave-body interaction problem.

On <math> S_F; P=P_a=0 \, </math> and hence all terms within the free-surface integral and over <math> S_{\infty} \, </math> (seafloor) can be neglected. It follows that:
<center><math> \frac{\mathrm{d} \mathcal{M}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega^\pm +\partial\Omega_B} \left[ \frac{P}{\rho} \mathbf{n} + \mathbf{v} (V_n -U_n) \right] \mathrm{d}S </math></center>

Note that the free surface integrals also vanish for the horizontal component since the hydrostatic force is always vertical. This momentum flux formula is of central importance in wave-body interactions and has many important applications, some of which are discussed bellow. Note that the mathematical derivations involved in its proof apply equally when the volume <math> \Omega </math> and its enclosed surface are selected to be at their linearized positions. In such a case it is essential to set <math> U_n=0\,</math> and <math> V_n \ne 0 </math>. Let the math take over and suggest the proper expression for the force. In the fully nonlinear case, <math>U_n \ne 0 \, </math> on <math> \partial\Omega_F\, </math> and <math> P=0 \,</math> on <math> \partial\Omega_F \, </math>!

On a solid boundary:
<center><math> U_n =V_n \,</math></center>
and
<center><math> \overrightarrow{F}_B = \iint_{S_B} P \vec{n} dS </math></center>
With <math> \vec{n} \,</math> pointing inside the body. We may therefore recast the momentum conservation theorem in the form:
<center><math> \overrightarrow{F}_B (t) = - \frac{\mathrm{d} \overrightarrow{M}}{\mathrm{d}t} - \rho \iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n) \right] \mathrm{d}S </math></center>
Where <math> S^\pm \, </math> are fluid boundaries at some distance from the body. If the volume of fluid surrounded by the body, free surface and the furfaces <math> S^\pm \, </math> does not grow in time, then the momentum of the enclosed fluid cannot grow either, so <math> \overrightarrow{M}(t) \, </math> is a stationary physical quantity. It is a well known result that the mean value in time of the time derivative of a stationary quantity is zero. So:
<center><math> {\overline{\frac{\mathrm{d}\overrightarrow{M}}{\mathrm{d}t}}}^t = 0 </math></center>

 Proof :

<center><math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \frac{\mathrm{d}F{\tau}}{\mathrm{d}\tau} \mathrm{d}\tau = \lim_{T\to\infty} \frac{1}{2T} [F(T) - F(-T) ] </math></center>

Since <math> F(\pm \tau) \, </math> must be bounded for a stationary signal <math> F(t) \, </math>, it follows that <math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = 0 \, </math>.

Taking mean values, it follows that:

<center><math> {\overline{\overrightarrow{F}_B (t)}}^t = - \rho \ {\overline{\iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n ) \right] \mathrm{d}S}}^t </math></center>

This is the fundamental formula underlying the definition of the mean wave drift forces acting on floating bodies. Such forces are very imprtant for stationary floating structures and can be expressed in terms of integrals of wave effects over control surfaces <math> S^\pm \, </math> which may be located at infinity.

The extension of the above formula for <math> {\overline{\overrightarrow{F}_B}}^t \, </math> in three dimensions is trivial. Simply replace <math> S^\pm \, </math> by <math> S_{\infty} \, </math>, a control surface at infinity. Common choices are a vertical cylindrical boundary or two vertical planes paraller to the axis of forward motion of a ship.

== Applications ==

=== Mean horizontal momentum flux due to a [[Plane Progressive Regular Waves| Plane Progressive Regular Wave]] ===

[[Image:Plane_wave_momentum.png|thumb|right|600px|Plane progressive wave momentum]]

We can determine the mean horizontal momentum flux due to a [[Plane Progressive Regular Waves| Plane Progressive Regular Wave]] with surface displacement
<center><math> \zeta = A \cos (\omega t - k x) \,</math></center>
where <math>\omega</math> and <math>k</math> are related by the [[Dispersion Relation for a Free Surface]]
<math> \omega^2 = gk \tanh kh \, </math>

The momentum flux across <math> \Omega^+ \, </math> is given by

<center><math> \frac{\mathrm{d}M_x}{\mathrm{d}t} = - \int_{-h}^{\zeta} ( P + \rho u^2 ) \mathrm{d}z </math></center>
The pressure is given by [[Conservation Laws and Boundary Conditions|Bernoulli's equation]]
<center><math> P = - \rho \frac{\partial\Phi}{\partial t} - \frac{1}{2} \rho ( u^2 + v^2 ) - \rho g z </math></center>
so that the momentum flux is
<center><math>
\begin{align}
\frac{dM_x}{dt} &= \rho \int_{-h}^{\zeta} \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} ( v^2 - u^2 ) + g z \right] \mathrm{d}z \\
&= \rho \left( \int_{-h}^{0} + \int_{0}^{\zeta} \right) \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} (v^2 - u^2) + gz \right] \mathrm{d}z
\end{align}
</math></center>

Taking mean values in time and keeping terms of <math> O(A^2) </math> we obtain:

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = {\overline{\rho \zeta(t) \left. \frac{\partial\Phi}{\partial t} \right |_{h=0}}}^t + \frac{1}{2} \rho \ {\overline{\int_{-h}^{0} (v^2 -u^2 ) \mathrm{d}z}}^t + {\overline{\frac{1}{2} \rho g \zeta^2}}^t </math></center>

Invoking the [[Linear and Second-Order Wave Theory|linearized dynamic free surface condition]] we obtain

<center><math> \left. \frac{\partial\Phi}{\partial t} \right |_{z=0} = - g \zeta </math></center>

It follows that:

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = {-\frac{1}{2} \rho g \zeta^2 (t)}^t + \frac{1}{2} \rho {\overline{\int_{-h}^0 (v^2-u^2)\mathrm{d}z}}^t </math></center>

In deep water <math> \overline{v^2} = \overline{u^2} \, </math> and the second term is identically zero, so

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = - \frac{1}{2} \rho g {\overline{\zeta^2 (t)}}^t = - \frac{1}{4} \rho g A^2 </math></center>

In water of finite depth the wave particle trajectories are elliptical with the mean horizontal velocities larger than the mean vertical velocities. So:

<center><math> \overline{V_H^2} < \overline{U_H^2} </math></center>

Therefore, in finite depth the modulus of the mean momentum flux is higher than in deep water for the same A.
So the mean horizontal momentum flux due to a plane progressive wave against its direction of propagation and equal to <math> - \frac{1}{4} \rho g A^2 </math>.

=== [[Wavemaker Theory]] ===

[[Image:Wave_maker_momentum.png|600px|right|thumb|Wavemaker momentum]]
Consider a wave maker shown in the figure generating a wave of amplitude <math>A</math> at infinity.

What is the mean horizontal force on the wavemaker? From the momentum conservation theorem the mean horizontal flux of momentum to the left must flow into the wavemaker. This mean flux translates into a mean horizontal force in the same direction, as shown in the figure. Not an easy conclusion without using some basic fluid mechanics!

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/A939D3A9-1F49-46F5-BEB5-0F6646CE340E/0/lecture5.pdf here]

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

Wave Momentum Flux

2010-11-06T11:33:46Z

Adi Kurniawan: /* Hydrostatic Term */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Wave Momentum Flux
| next chapter = [[Wavemaker Theory]]
| previous chapter = [[Wave Energy Density and Flux]]
}}

{{incomplete pages}}

== Introduction ==

The momentum is important to determine the forces.

== Momentum flux in potential flow ==

The momentum flux (the time derivative of the [http://en.wikipedia.org/wiki/Momentum momentum] <math>\mathcal{M}</math>) is given by
<center><math> \frac{\mathrm{d}\mathcal{M}(t)}{\mathrm{d}t} = \rho \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \mathbf{v} \mathrm{d}V
= \rho \iiint_{\Omega(t)} \frac{\partial\mathbf{v}}{\partial t} \mathrm{d}V + \rho \iint_{\partial\Omega(t)} \mathbf{v} U_n \mathrm{d}S,
</math></center>
where <math> U_n \, </math> is the outward normal velocity of the surface <math> \partial\Omega(t)\, </math>.
This equation follows from the [http://en.wikipedia.org/wiki/Reynolds_transport_theorem transport theorem].

We begin with [http://en.wikipedia.org/wiki/Euler_equations Euler's equation] in the absence of viscosity
<center><math>
\frac{\partial\mathbf{v}}{\partial t} + (\mathbf{v} \cdot \nabla ) \mathbf{v} =
- \nabla \left(\frac{P}{\rho} + g z\right)
</math></center>
where we have defined the direction of gravity to be in the negative <math>z</math>
direction.
We may recast the momentum flux in the form
<center><math>
\frac{\mathrm{d}\mathcal{M}(t)}{\mathrm{d}t} = - \rho \iiint_{\Omega(t)} \left( \nabla ( \frac{P}{\rho} + g z ) + ( \mathbf{v} \cdot \nabla ) \mathbf{v} \right)
\mathrm{d}V + \rho \iint_{\partial\Omega(t)} \mathbf{v} U_n \mathrm{d}S
</math></center>
So far <math> \Omega(t)\, </math> is an arbitrary closed time dependent volume bounded by the time dependent surface <math> \partial\Omega(t)\, </math>.
We have however defined the gravitational acceleration to be in the negative <math>z</math> direction.

=== Simplification ===

We use the following vector theorem
<center><math> (\mathbf{v} \cdot \nabla ) \mathbf{v} = \nabla ( \frac{1}{2} \mathbf{v} \cdot \mathbf{v} ) - \mathbf{v} \times (\nabla \times \mathbf{v} ) </math></center>
If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and we can use Gauss's vector theorem:
<center><math> \iiint_{\Omega(t)} \nabla ( \frac{1}{2} \mathbf{v} \cdot \mathbf{v} ) \mathrm{d}V = \frac{1}{2} \iint_{\partial\Omega(t)}
\left(\mathbf{v} \cdot \mathbf{v} \right) \mathbf{n} \mathrm{d}S </math></center>
In potential flow it follows that
<center><math> \iint_{\partial\Omega(t)} \frac{1}{2} ( \mathbf{v} \cdot \mathbf{v} ) \mathbf{n} \mathrm{d}S = \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi \mathrm{d}S
= \iint_{\partial\Omega(t)} V_n \mathbf{v} \mathrm{d}S </math></center>
since for <math> \nabla^2 \Phi = 0 \, </math>;
<center><math> \iiint_{\partial\Omega} \frac{1}{2} ( \nabla\Phi \cdot \nabla\Phi) \mathbf{n} \mathrm{d}S \equiv \iint_{\partial\Omega} \frac{\partial\Phi}{\partial n} \nabla\Phi \mathrm{d}S. </math></center>
In the above <math> \mathbf{n} \, </math> is the unit vector pointing out of the volume <math> \Omega(t) </math> and <math> V_n = \mathbf{n} \cdot \nabla \Phi </math>

Upon substitution in the momentum flux formula, we obtain:
<center><math> \frac{\mathrm{d}\mathcal{M}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega(t)} \left( ( \frac{P}{\rho} + gz ) \mathbf{n} + \mathbf{v} (V_n - U_n )\right) \mathrm{d}S </math></center>

This formula is of central importance in potential flow marine hydrodynamics because the rate of change of the linear momentum defined above is just <math> \pm \,</math> the force acting on the fluid volume. When its mean value can be shown to vanish, important force expressions on solid boundaries follow and will be derived in what follows.

== Hydrostatic Term ==

Consider separately the term in the momentum flux expression involving the hydrostatic pressure:
<center><math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} = - \rho \iint_{\partial\Omega} gz \mathbf{n},
</math></center>
We break the boundary up into the free surface, ends, body surface, and sea floor at infinite depth, i,.e.
<math>\partial\Omega=\partial\Omega_F + \partial\Omega^{\pm} + \partial\Omega_B + \partial\Omega_{\infty} </math>
The integral over the body surface, assuming a fully submerged body is:
<center><math> \frac{\mathrm{d}\mathcal{M}_{H,B}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega_B} gz \mathbf{n} = \rho g V \mathbf{k} </math></center>
where <math>V</math> is the volume. This follows from the vector theorem of Gauss and is the [http://en.wikipedia.org/wiki/Displacement_%28fluid%29 principle of Archimedes]. That is,
the momentum flux is equal to the buoyancy force.

We may therefore consider the second part of the integral involving wave effects independently and in the absence of the body, assumed fully submerged. In the case of a surface piercing body and in the fully nonlinear case matters are more complex. Consider the application of the momentum conservation theorem in the case of a submerged or floating body in steep waves.

[[Image:Momentum.png|600px|right|thumb|Momentum boundaries]]

Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial. Note that unlike the [[Wave Energy Density and Flux|energy conservation principle]], the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component. The integral of the hydrostatic term over the remaining surfaces leads to:
<center><math> \frac{\mathrm{d} \mathcal{M}_{H,S}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega_F+\partial\Omega^+ + \partial\Omega^- +\partial\Omega_{\infty}} gz \mathbf{n} \mathrm{d}S = - \rho g V_{\text{Fluid}} \mathbf{k} </math></center>
where <math>V_{\mbox{Fluid}} </math> is the fluid volume.
This is simply the static weight of the volume of fluid bounded by <math> \partial\Omega_F, \partial\Omega^+, \partial\Omega^- \,</math> and <math> \partial\Omega_{\infty}.</math> With no waves present, this is simply the weight of the ocean water "column" bounded by <math> \partial\Omega\, </math> which does not concern us here. This weight does not change in principle when waves are present at least when <math> \partial\Omega^+, \partial\Omega^- \,</math> are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "rational" to apply the momentum conservation theorem over the "linearized" volume <math> V_L(t) \, </math> which is perfectly possible within the framework derived above. In this case <math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} \,</math> is exactly equal to the static weight of the water column and can be ignored in the wave-body interaction problem.

On <math> S_F; P=P_a=0 \, </math> and hence all terms within the free-surface integral and over <math> S_{\infty} \, </math> (seafloor) can be neglected. It follows that:
<center><math> \frac{\mathrm{d} \mathcal{M}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega^\pm +\partial\Omega_B} \left[ \frac{P}{\rho} \mathbf{n} + \mathbf{v} (V_n -U_n) \right] \mathrm{d}S </math></center>

Note that the free surface integrals also vanish for the horizontal component since the hydrostatic force is always vertical. This momentum flux formula is of central importance in wave-body interactions and has many important applications, some of which are discussed bellow. Note that the mathematical derivations involved in its proof apply equally when the volume <math> \Omega </math> and its enclosed surface are selected to be at their linearized positions. In such a case it is essential to set <math> U_n=0\,</math> and <math> V_n \ne 0 </math>. Let the math take over and suggest the proper expression for the force. In the fully nonlinear case, <math>U_n \ne 0 \, </math> on <math> \partial\Omega_F\, </math> and <math> P=0 \,</math> on <math> \partial\Omega_F \, </math>!

On a solid boundary:
<center><math> U_n =V_n \,</math></center>
and
<center><math> \overrightarrow{F}_B = \iint_{S_B} P \vec{n} dS </math></center>
With <math> \vec{n} \,</math> pointing inside the body. We may therefore recast the momentum conservation theorem in the form:
<center><math> \overrightarrow{F}_B (t) = - \frac{\mathrm{d} \overrightarrow{M}}{\mathrm{d}t} - \rho \iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n) \right] \mathrm{d}S </math></center>
Where <math> S^\pm \, </math> are fluid boundaries at some distance from the body. If the volume of fluid surrounded by the body, free surface and the furfaces <math> S^\pm \, </math> does not grow in time, then the momentum of the enclosed fluid cannot grow either, so <math> \overrightarrow{M}(t) \, </math> is a stationary physical quantity. It is a well known result that the mean value in time of the time derivative of a stationary quantity is zero. So:
<center><math> {\overline{\frac{\mathrm{d}\overrightarrow{M}}{\mathrm{d}t}}}^t = 0 </math></center>

 Proof :

<center><math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \frac{\mathrm{d}F{\tau}}{\mathrm{d}\tau} \mathrm{d}\tau = \lim_{T\to\infty} \frac{1}{2T} [F(T) - F(-T) ] </math></center>

Since <math> F(\pm \tau) \, </math> must be bounded for a stationary signal <math> F(t) \, </math>, it follows that <math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = 0 \, </math>.

Taking mean values, it follows that:

<center><math> {\overline{\overrightarrow{F}_B (t)}}^t = - \rho \ {\overline{\iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n ) \right] \mathrm{d}S}}^t </math></center>

This is the fundamental formula underlying the definition of the mean wave drift forces acting on floating bodies. Such forces are very imprtant for stationary floating structures and can be expressed in terms of integrals of wave effects over control surfaces <math> S^\pm \, </math> which may be located at infinity.

The extension of the above formula for <math> {\overline{\overrightarrow{F}_B}}^t \, </math> in three dimensions is trivial. Simply replace <math> S^\pm \, </math> by <math> S_{\infty} \, </math>, a control surface at infinity. Common choices are a vertical cylindrical boundary or two vertical planes paraller to the axis of forward motion of a ship.

== Applications ==

=== Mean horizontal momentum flux due to a [[Plane Progressive Regular Waves| Plane Progressive Regular Wave]] ===

[[Image:Plane_wave_momentum.png|thumb|right|600px|Plane progressive wave momentum]]

We can determine the mean horizontal momentum flux due to a [[Plane Progressive Regular Waves| Plane Progressive Regular Wave]] with surface displacement
<center><math> \zeta = A \cos (\omega t - k x) \,</math></center>
where <math>\omega</math> and <math>k</math> are related by the [[Dispersion Relation for a Free Surface]]
<math> \omega^2 = gk \tanh kh \, </math>

The momentum flux across <math> \Omega^+ \, </math> is given by

<center><math> \frac{\mathrm{d}M_x}{\mathrm{d}t} = - \int_{-h}^{\zeta} ( P + \rho u^2 ) \mathrm{d}z </math></center>
The pressure is given by [[Conservation Laws and Boundary Conditions|Bernoulli's equation]]
<center><math> P = - \rho \frac{\partial\Phi}{\partial t} - \frac{1}{2} \rho ( u^2 + v^2 ) - \rho g z </math></center>
so that the momentum flux is
<center><math> \frac{dM_x}{dt} = \rho \int_{-h}^{\zeta} \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} ( v^2 - u^2 ) + g z \right] \mathrm{d}z = \rho \left( \int_{-h}^{0} + \int_{0}^{\zeta} \right) \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} (v^2 - u^2) + gz \right] \mathrm{d}z </math></center>

Taking mean values in time and keeping terms of <math> O(A^2) </math> we obtain:

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = {\overline{\rho \zeta(t) \left. \frac{\partial\Phi}{\partial t} \right |_{h=0}}}^t + \frac{1}{2} \rho \ {\overline{\int_{-h}^{0} (v^2 -u^2 ) \mathrm{d}z}}^t + {\overline{\frac{1}{2} \rho g \zeta^2}}^t </math></center>

Invoking the [[Linear and Second-Order Wave Theory|linearized dynamic free surface condition]] we obtain

<center><math> \left. \frac{\partial\Phi}{\partial t} \right |_{z=0} = - g \zeta </math></center>

It follows that:

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = {-\frac{1}{2} \rho g \zeta^2 (t)}^t + \frac{1}{2} \rho {\overline{\int_{-h}^0 (v^2-u^2)\mathrm{d}z}}^t </math></center>

In deep water <math> \overline{v^2} = \overline{u^2} \, </math> and the second term is identically zero, so

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = - \frac{1}{2} \rho g {\overline{\zeta^2 (t)}}^t = - \frac{1}{4} \rho g A^2 </math></center>

In water of finite depth the wave particle trajectories are elliptical with the mean horizontal velocities larger than the mean vertical velocities. So:

<center><math> \overline{V_H^2} < \overline{U_H^2} </math></center>

Therefore, in finite depth the modulus of the mean momentum flux is higher than in deep water for the same A.
So the mean horizontal momentum flux due to a plane progressive wave against its direction of propagation and equal to <math> - \frac{1}{4} \rho g A^2 </math>.

=== [[Wavemaker Theory]] ===

[[Image:Wave_maker_momentum.png|600px|right|thumb|Wavemaker momentum]]
Consider a wave maker shown in the figure generating a wave of amplitude <math>A</math> at infinity.

What is the mean horizontal force on the wavemaker? From the momentum conservation theorem the mean horizontal flux of momentum to the left must flow into the wavemaker. This mean flux translates into a mean horizontal force in the same direction, as shown in the figure. Not an easy conclusion without using some basic fluid mechanics!

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/A939D3A9-1F49-46F5-BEB5-0F6646CE340E/0/lecture5.pdf here]

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

Wave Momentum Flux

2010-11-06T11:31:55Z

Adi Kurniawan: /* Hydrostatic Term */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Wave Momentum Flux
| next chapter = [[Wavemaker Theory]]
| previous chapter = [[Wave Energy Density and Flux]]
}}

{{incomplete pages}}

== Introduction ==

The momentum is important to determine the forces.

== Momentum flux in potential flow ==

The momentum flux (the time derivative of the [http://en.wikipedia.org/wiki/Momentum momentum] <math>\mathcal{M}</math>) is given by
<center><math> \frac{\mathrm{d}\mathcal{M}(t)}{\mathrm{d}t} = \rho \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \mathbf{v} \mathrm{d}V
= \rho \iiint_{\Omega(t)} \frac{\partial\mathbf{v}}{\partial t} \mathrm{d}V + \rho \iint_{\partial\Omega(t)} \mathbf{v} U_n \mathrm{d}S,
</math></center>
where <math> U_n \, </math> is the outward normal velocity of the surface <math> \partial\Omega(t)\, </math>.
This equation follows from the [http://en.wikipedia.org/wiki/Reynolds_transport_theorem transport theorem].

We begin with [http://en.wikipedia.org/wiki/Euler_equations Euler's equation] in the absence of viscosity
<center><math>
\frac{\partial\mathbf{v}}{\partial t} + (\mathbf{v} \cdot \nabla ) \mathbf{v} =
- \nabla \left(\frac{P}{\rho} + g z\right)
</math></center>
where we have defined the direction of gravity to be in the negative <math>z</math>
direction.
We may recast the momentum flux in the form
<center><math>
\frac{\mathrm{d}\mathcal{M}(t)}{\mathrm{d}t} = - \rho \iiint_{\Omega(t)} \left( \nabla ( \frac{P}{\rho} + g z ) + ( \mathbf{v} \cdot \nabla ) \mathbf{v} \right)
\mathrm{d}V + \rho \iint_{\partial\Omega(t)} \mathbf{v} U_n \mathrm{d}S
</math></center>
So far <math> \Omega(t)\, </math> is an arbitrary closed time dependent volume bounded by the time dependent surface <math> \partial\Omega(t)\, </math>.
We have however defined the gravitational acceleration to be in the negative <math>z</math> direction.

=== Simplification ===

We use the following vector theorem
<center><math> (\mathbf{v} \cdot \nabla ) \mathbf{v} = \nabla ( \frac{1}{2} \mathbf{v} \cdot \mathbf{v} ) - \mathbf{v} \times (\nabla \times \mathbf{v} ) </math></center>
If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and we can use Gauss's vector theorem:
<center><math> \iiint_{\Omega(t)} \nabla ( \frac{1}{2} \mathbf{v} \cdot \mathbf{v} ) \mathrm{d}V = \frac{1}{2} \iint_{\partial\Omega(t)}
\left(\mathbf{v} \cdot \mathbf{v} \right) \mathbf{n} \mathrm{d}S </math></center>
In potential flow it follows that
<center><math> \iint_{\partial\Omega(t)} \frac{1}{2} ( \mathbf{v} \cdot \mathbf{v} ) \mathbf{n} \mathrm{d}S = \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi \mathrm{d}S
= \iint_{\partial\Omega(t)} V_n \mathbf{v} \mathrm{d}S </math></center>
since for <math> \nabla^2 \Phi = 0 \, </math>;
<center><math> \iiint_{\partial\Omega} \frac{1}{2} ( \nabla\Phi \cdot \nabla\Phi) \mathbf{n} \mathrm{d}S \equiv \iint_{\partial\Omega} \frac{\partial\Phi}{\partial n} \nabla\Phi \mathrm{d}S. </math></center>
In the above <math> \mathbf{n} \, </math> is the unit vector pointing out of the volume <math> \Omega(t) </math> and <math> V_n = \mathbf{n} \cdot \nabla \Phi </math>

Upon substitution in the momentum flux formula, we obtain:
<center><math> \frac{\mathrm{d}\mathcal{M}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega(t)} \left( ( \frac{P}{\rho} + gz ) \mathbf{n} + \mathbf{v} (V_n - U_n )\right) \mathrm{d}S </math></center>

This formula is of central importance in potential flow marine hydrodynamics because the rate of change of the linear momentum defined above is just <math> \pm \,</math> the force acting on the fluid volume. When its mean value can be shown to vanish, important force expressions on solid boundaries follow and will be derived in what follows.

== Hydrostatic Term ==

Consider separately the term in the momentum flux expression involving the hydrostatic pressure:
<center><math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} = - \rho \iint_{\partial\Omega} gz \mathbf{n},
</math></center>
We break the boundary up into the free surface, ends, body surface, and sea floor at infinite depth, i,.e.
<math>\partial\Omega=\partial\Omega_F + \partial\Omega^{\pm} + \partial\Omega_B + \partial\Omega_{\infty} </math>
The integral over the body surface, assuming a fully submerged body is:
<center><math> \frac{\mathrm{d}\mathcal{M}_{H,B}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega_B} gz \mathbf{n} = \rho g V \mathbf{k} </math></center>
where <math>V</math> is the volume. This follows from the vector theorem of Gauss and is the [http://en.wikipedia.org/wiki/Displacement_%28fluid%29 principle of Archimedes]. That is,
the momentum flux is equal to the buoyancy force.

We may therefore consider the second part of the integral involving wave effects independently and in the absence of the body, assumed fully submerged. In the case of a surface piercing body and in the fully nonlinear case matters are more complex. Consider the application of the momentum conservation theorem in the case of a submerged or floating body in steep waves.

[[Image:Momentum.png|600px|right|thumb|Momentum boundaries]]

Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial. Note that unlike the [[Wave Energy Density and Flux|energy conservation principle]], the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component. The integral of the hydrostatic term over the remaining surfaces leads to:
<center><math> \frac{d \mathcal{M}_{H,S}}{dt} = - \rho \iint_{\partial\Omega_F+\partial\Omega^+ + \partial\Omega^- +\partial\Omega_{\infty}} gz \mathbf{n} \mathrm{d}S = - \rho g V_{\mbox{Fluid}} \mathbf{k} </math></center>
where <math>V_{\mbox{Fluid}} </math> is the fluid volume.
This is simply the static weight of the volume of fluid bounded by <math> \partial\Omega_F, \partial\Omega^+, \partial\Omega^- \,</math> and <math> \partial\Omega_{\infty}.</math> With no waves present, this is simply the weight of the ocean water "column" bounded by <math> \partial\Omega\, </math> which does not concern us here. This weight does not change in principle when waves are present at least when <math> \partial\Omega^+, \partial\Omega^- \,</math> are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "rational" to apply the momentum conservation theorem over the "linearized" volume <math> V_L(t) \, </math> which is perfectly possible within the framework derived above. In this case <math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} \,</math> is exactly equal to the static weight of the water column and can be ignored in the wave-body interaction problem.

On <math> S_F; P=P_a=0 \, </math> and hence all terms within the free-surface integral and over <math> S_{\infty} \, </math> (seafloor) can be neglected. It follows that:
<center><math> \frac{\mathrm{d} \mathcal{M}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega^\pm +\partial\Omega_B} \left[ \frac{P}{\rho} \mathbf{n} + \mathbf{v} (V_n -U_n) \right] \mathrm{d}S </math></center>

Note that the free surface integrals also vanish for the horizontal component since the hydrostatic force is always vertical. This momentum flux formula is of central importance in wave-body interactions and has many important applications, some of which are discussed bellow. Note that the mathematical derivations involved in its proof apply equally when the volume <math> \Omega </math> and its enclosed surface are selected to be at their linearized positions. In such a case it is essential to set <math> U_n=0\,</math> and <math> V_n \ne 0 </math>. Let the math take over and suggest the proper expression for the force. In the fully nonlinear case, <math>U_n \ne 0 \, </math> on <math> \partial\Omega_F\, </math> and <math> P=0 \,</math> on <math> \partial\Omega_F \, </math>!

On a solid boundary:
<center><math> U_n =V_n \,</math></center>
and
<center><math> \overrightarrow{F}_B = \iint_{S_B} P \vec{n} dS </math></center>
With <math> \vec{n} \,</math> pointing inside the body. We may therefore recast the momentum conservation theorem in the form:
<center><math> \overrightarrow{F}_B (t) = - \frac{\mathrm{d} \overrightarrow{M}}{\mathrm{d}t} - \rho \iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n) \right] \mathrm{d}S </math></center>
Where <math> S^\pm \, </math> are fluid boundaries at some distance from the body. If the volume of fluid surrounded by the body, free surface and the furfaces <math> S^\pm \, </math> does not grow in time, then the momentum of the enclosed fluid cannot grow either, so <math> \overrightarrow{M}(t) \, </math> is a stationary physical quantity. It is a well known result that the mean value in time of the time derivative of a stationary quantity is zero. So:
<center><math> {\overline{\frac{\mathrm{d}\overrightarrow{M}}{\mathrm{d}t}}}^t = 0 </math></center>

 Proof :

<center><math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \frac{\mathrm{d}F{\tau}}{\mathrm{d}\tau} \mathrm{d}\tau = \lim_{T\to\infty} \frac{1}{2T} [F(T) - F(-T) ] </math></center>

Since <math> F(\pm \tau) \, </math> must be bounded for a stationary signal <math> F(t) \, </math>, it follows that <math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = 0 \, </math>.

Taking mean values, it follows that:

<center><math> {\overline{\overrightarrow{F}_B (t)}}^t = - \rho \ {\overline{\iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n ) \right] \mathrm{d}S}}^t </math></center>

This is the fundamental formula underlying the definition of the mean wave drift forces acting on floating bodies. Such forces are very imprtant for stationary floating structures and can be expressed in terms of integrals of wave effects over control surfaces <math> S^\pm \, </math> which may be located at infinity.

The extension of the above formula for <math> {\overline{\overrightarrow{F}_B}}^t \, </math> in three dimensions is trivial. Simply replace <math> S^\pm \, </math> by <math> S_{\infty} \, </math>, a control surface at infinity. Common choices are a vertical cylindrical boundary or two vertical planes paraller to the axis of forward motion of a ship.

== Applications ==

=== Mean horizontal momentum flux due to a [[Plane Progressive Regular Waves| Plane Progressive Regular Wave]] ===

[[Image:Plane_wave_momentum.png|thumb|right|600px|Plane progressive wave momentum]]

We can determine the mean horizontal momentum flux due to a [[Plane Progressive Regular Waves| Plane Progressive Regular Wave]] with surface displacement
<center><math> \zeta = A \cos (\omega t - k x) \,</math></center>
where <math>\omega</math> and <math>k</math> are related by the [[Dispersion Relation for a Free Surface]]
<math> \omega^2 = gk \tanh kh \, </math>

The momentum flux across <math> \Omega^+ \, </math> is given by

<center><math> \frac{\mathrm{d}M_x}{\mathrm{d}t} = - \int_{-h}^{\zeta} ( P + \rho u^2 ) \mathrm{d}z </math></center>
The pressure is given by [[Conservation Laws and Boundary Conditions|Bernoulli's equation]]
<center><math> P = - \rho \frac{\partial\Phi}{\partial t} - \frac{1}{2} \rho ( u^2 + v^2 ) - \rho g z </math></center>
so that the momentum flux is
<center><math> \frac{dM_x}{dt} = \rho \int_{-h}^{\zeta} \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} ( v^2 - u^2 ) + g z \right] \mathrm{d}z = \rho \left( \int_{-h}^{0} + \int_{0}^{\zeta} \right) \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} (v^2 - u^2) + gz \right] \mathrm{d}z </math></center>

Taking mean values in time and keeping terms of <math> O(A^2) </math> we obtain:

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = {\overline{\rho \zeta(t) \left. \frac{\partial\Phi}{\partial t} \right |_{h=0}}}^t + \frac{1}{2} \rho \ {\overline{\int_{-h}^{0} (v^2 -u^2 ) \mathrm{d}z}}^t + {\overline{\frac{1}{2} \rho g \zeta^2}}^t </math></center>

Invoking the [[Linear and Second-Order Wave Theory|linearized dynamic free surface condition]] we obtain

<center><math> \left. \frac{\partial\Phi}{\partial t} \right |_{z=0} = - g \zeta </math></center>

It follows that:

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = {-\frac{1}{2} \rho g \zeta^2 (t)}^t + \frac{1}{2} \rho {\overline{\int_{-h}^0 (v^2-u^2)\mathrm{d}z}}^t </math></center>

In deep water <math> \overline{v^2} = \overline{u^2} \, </math> and the second term is identically zero, so

<center><math> {\overline{\frac{\mathrm{d}M_x}{\mathrm{d}t}}}^t = - \frac{1}{2} \rho g {\overline{\zeta^2 (t)}}^t = - \frac{1}{4} \rho g A^2 </math></center>

In water of finite depth the wave particle trajectories are elliptical with the mean horizontal velocities larger than the mean vertical velocities. So:

<center><math> \overline{V_H^2} < \overline{U_H^2} </math></center>

Therefore, in finite depth the modulus of the mean momentum flux is higher than in deep water for the same A.
So the mean horizontal momentum flux due to a plane progressive wave against its direction of propagation and equal to <math> - \frac{1}{4} \rho g A^2 </math>.

=== [[Wavemaker Theory]] ===

[[Image:Wave_maker_momentum.png|600px|right|thumb|Wavemaker momentum]]
Consider a wave maker shown in the figure generating a wave of amplitude <math>A</math> at infinity.

What is the mean horizontal force on the wavemaker? From the momentum conservation theorem the mean horizontal flux of momentum to the left must flow into the wavemaker. This mean flux translates into a mean horizontal force in the same direction, as shown in the figure. Not an easy conclusion without using some basic fluid mechanics!

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/A939D3A9-1F49-46F5-BEB5-0F6646CE340E/0/lecture5.pdf here]

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

Wave Energy Density and Flux

2010-11-06T11:29:18Z

Adi Kurniawan: /* Rayleigh's proof of the group velocity formula */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = [[Wave Energy Density and Flux]]
| next chapter = [[Wave Momentum Flux]]
| previous chapter = [[Linear Plane Progressive Regular Waves]]
}}

{{complete pages}}

== Introduction ==

We are interested in the transport of energy by ocean waves. It is important to realise that under the assumptions of linear theory, there is no net motion of particles, but there is a transport of energy (as would be expected). The energy consists of two parts, one kinetic due to the motion of the fluid and the other potential due to the variation in the fluid height. It is the resonance between these two energies which gives rise to the wave motion. The situation is analogous to
[http://en.wikipedia.org/wiki/Simple_harmonic_motion Simple Harmonic Motion] but more complicated.

== Energy per volume ==

[[Image:Energy_volume.png|thumb|thumb|right|600px|Energy Volume]]

We begin by defining
<math> \mathcal{E}(t) </math> as the energy in control volume <math> \Omega(t) </math> given by
<center><math> \mathcal{E}(t) = \rho \iiint_\Omega \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}V </math></center>
where <math>\rho</math> is the fluid density, <math>g</math> is the acceleration due to gravity and <math>\mathbf{v}</math> is the vector of fluid velocity.
The mean energy over a unit horizontal surface area <math> S </math> is given by
<center><math> \overline{\mathcal{E}} = \overline{\frac{\mathcal{E}(t)}{S}} = \rho \overline{ \int_{-h}^{\zeta(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}z} = \frac{1}{2} \rho \overline{ \int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - h^2 ) } </math></center>
where <math> \zeta(t) \, </math> is free surface elevation and the overbar denotes average (which will be important when we consider waves).
Note that we are considering water of constant [[Finite Depth]].
We can ignore the term <math> -\frac{1}{2} \rho g h^2 \, </math> which represents the potential energy of the ocean at rest.

The remaining perturbation component is the sum of the kinetic and potential energy components, that is
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} </math></center>
where
<center><math> \overline{\mathcal{E}_{kin}} = \frac{1}{2} \rho \overline{\int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z}, \qquad |\mathbf{v}|^2
= \nabla\Phi \cdot \nabla \Phi = \partial_x^2\Phi + \partial_z^2\Phi </math></center>
and
<center><math> \overline{\mathcal{E}_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)} </math></center>
Note that we are assuming only two dimensions <math>x</math> and <math>z</math>.

=== Energy in [[Linear Plane Progressive Regular Waves]] ===

Consider now as a special case of [[Linear Plane Progressive Regular Waves]] by the velocity potential in [[Infinite Depth]] water (for simplicity). The velocity potential throughout the fluid domain is then given by
<center><math> \Phi = \mathrm{Re} \{ \frac{igA}{\omega} e^{kz-ikx+i\omega t} \} </math></center>
The components is the <math>x</math> and <math>z</math> directions are given by
<center><math> \partial_x\Phi = \mathrm{Re} \{ \frac{igA}{\omega} (-ik) e^{kz-ikx+i\omega t} \} = A \mathrm{Re} \{ \omega e^{kz-ikx+i\omega t} \} </math></center>
and
<center><math> \partial_z\Phi = \mathrm{Re} \{ \frac{igA}{\omega} k e^{kz-ikx+i\omega t} \}
= A \mathrm{Re} \{ i \omega e^{kz-ikx+i\omega t} \} </math></center>
respectively.
We require the following Lemma which is easily proved. If
<center><math> \mathrm{Re} \{ A e^{i\omega t} \} = A(t) \,\!</math></center>
and
<center><math> \mathrm{Re} \{ B e^{i\omega t} \} = B(t) \,\!</math></center>
then it follows that
<center><math> \overline{A(t)B(t)} = \frac{1}{2} \mathrm{Re} \{ A B^* \}. </math></center>

This allows us to write the following expression
<center><math>
\begin{align}
\overline{\mathcal{E}_{kin}} &= \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta )
\left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z \\
&= \frac{1}{2} \rho \int_{-\infty}^0 \overline{ \left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z + O (A^3) \\
&= \rho \frac{\omega^2 A^2}{4k} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ k=\omega^2/g \\
\overline{\mathcal{E}_{pot}} &= \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 .
\end{align}
</math></center>
Note that it is a standard feature of linear oscillations that the average potential and kinetic energies are equal. Hence
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} = \frac{1}{2} \rho g A^2 </math></center>

== Energy Flux ==

[[Image:volume_normal.png|thumb|right|600px|Moving Volume]]

''Energy flux'' <math>\mathcal{P}(t)</math> is the rate of change of energy density <math> \mathcal{E}(t) </math>. It is the flux of energy which is critical to ocean waves. While the individual fluid particles do not move the waves carry energy. We begin by deriving the energy flux in general conditions.

<center><math> \mathcal{P}(t) \equiv \frac{\mathrm{d}\mathcal{E}(t)}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \epsilon(t) \mathrm{d}V
= \iint_{\partial\Omega(t)}
\frac{\partial \epsilon(t)}{\partial t} \mathrm{d}V + \iint_{\partial\Omega(t)} \epsilon(t) U_n \mathrm{d} S </math></center>
(the last result following from the transport theorem) where <math> U_n</math> is the normal velocity of surface <math> \partial\Omega(t) </math> outwards of the enclosed volume <math> \Omega(t)</math>.
We know that
<center><math>
\begin{align}
\frac{\partial \epsilon}{\partial t} &= \frac{\partial}{\partial t} \{ \frac{1}{2} \rho |\mathbf{v}|^2 + \rho g z \} \\
&= \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) \\
&= \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi
\end{align}
</math></center>
so that
<center><math> \mathcal{P}(t) = \rho \iiint_{\Omega(t)} \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) \mathrm{d}V
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
Invoking the scalar form of Gauss's theorem in the first term, we obtain:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \nabla \Phi \cdot \mathbf{n} \mathrm{d}S
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
where <math>\mathbf{n}</math> is the unit normal.

An alternative form for the energy flux <math> \mathcal{P}(t) \, </math> crossing the closed control surface <math> \partial\Omega(t) \, </math>
is obtained by invoking Bernoulli's equation in the second term. Recall that:
<center><math> \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gz = 0 </math></center>
at any point in the fluid domain and on the boundary.
Here we allowed <math> \ P_a </math> the atmospheric pressure to be non-zero for the sake of physical clarity. Upon substitution in
the equation above for <math> \mathcal{P}(t) </math> we obtain the alternate form:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}s
- \rho \iint_{\partial\Omega(t)} \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n \mathrm{d}s </math></center>
where <math>\frac{\partial\phi}{\partial n}</math> is <math>\nabla\phi\cdot\mathbf{n}</math>
So the energy flux across <math> \partial\Omega(t)\, </math> is given by the terms under the integral sign. They can be collected in the more compact form:
<center><math> \mathcal{P}(t) = \iint \left\{ \rho \frac{\partial\Phi}{\partial t} \left( \frac{\partial\Phi}{\partial n} - U_n \right)
- ( P - P_a) U_n \right\} \mathrm{d}s </math></center>
Note that <math> \mathcal{P}(t) \, </math> measures the energy flux into the volume <math> \Omega(t) \, </math> or the rate of growth of the energy density <math> \mathcal{E}(t)\, </math>.

=== Energy transfer for each boundary ===

Break <math> \partial\Omega(t) \, </math> into its components and derive specialized forms of <math> \mathcal{P}(t) \, </math> pertinent to each.

* <math> \partial\Omega_F </math> nonlinear position of the free surface. On this <math> \partial_n\Phi= U_n; </math> and <math> P = P_a </math>, so the fluid pressure is equal to the atmospheric pressure. Therefore over <math> \partial\Omega_F </math> <math>\mathcal{P}(t)=0</math>, as expected, i.e. there is no energy flow into the atmosphere.
*<math> \partial\Omega_B</math> is the non-moving solid boundary, <math> U_n = </math> and <math> \frac{\partial\Phi}{\partial n} = U_n=0</math> which is the no-normal flux condition.
*<math> \partial\Omega^\pm</math> which are the fluid boundaries fixed in space relative to an earth frame <math> U_n = 0 </math> and <math>, \frac{\partial\Phi}{\partial n} \ne 0</math>
*<math> \partial\Omega_U </math> the fluid boundaries moving with velocity <math>\mathbf{u}</math> relative to an earth frame.
*<math> U_n = \mathbf{u} \cdot \mathbf{n}, \quad \frac{\partial\Phi}{\partial n} \ne 0 </math>. This case will be of interest for ships moving with constant velocity <math> U </math>.

The formula derived above are very general for potential flows with a free surface and solid boundaries. We are now ready to apply them to plane progressive waves.

== Surface Wave Problem ==

We are ready now to apply the above formulas to the surface wave problem.

=== Energy flux across a vertical fluid boundary fixed in space ===

<center><math>
\begin{align}
\mathcal{P}(t) &= - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z \\
&= - \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z \\
&= - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}z + O(A^3)
\end{align}
</math></center>

Mean energy flux for a [[Linear Plane Progressive Regular Waves]] follows upon substitution of the regular wave velocity potential and taking mean values:
<center><math> \overline{\mathcal{P}} = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z = \frac{1}{2} \rho g A^2 \left( \frac{1}{2} \frac{g}{\omega} \right) </math></center>
where <math>A</math> is the wave amplitude.
or
<center><math> \overline{\mathcal{P}} = \overline{\mathcal{E}} c_g, \quad c_g = \mbox{group velocity} = \frac{1}{2} c_p </math></center>.
It follows from this that the mean energy flux of a plane progressive wave is the product of its mean energy density times
the [http://en.wikipedia.org/wiki/Group_velocity group velocity] of deep water waves.

A more formal proof that this is the velocity with which the energy flux of plane progressive waves propagates is to consider what needs to be the horizontal velocity <math>U_n \equiv U </math> of a fluid boundary so that the mean energy flux across it vanishes?

This can be found from the solution of the following equation:

<center><math> \overline{\mathcal{P}(t)} = \rho \ {\overline{\int_{-\infty}^0 \partial_t\Phi \partial_x\Phi \mathrm{d}z}} - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}} = 0 </math></center>

Where terms of <math> O(A^3) </math> have been neglected. Note that within linear theory, energy density and energy flux are quantities of <math>O(A^2)</math>. If higher-order terms are kept then we need to consider the treatment of second-order surface wave theory, at least.
Solving the above equation for <math> U </math> we obtain:
<center><math> U = \dfrac{\rho \ {\overline{\int_{-\infty}^0 \dfrac{\partial\Phi}{\partial t} \dfrac{\partial\Phi}{\partial x} \mathrm{d}z}}}
{{\overline{\int_{-\infty}^0 \left( \dfrac{P}{\rho} + \dfrac{\partial\Phi}{\partial t} \right) \mathrm{d}z}}} </math></center>

Upon substitution of the plane progressive wave velocity potential and definition of pressure from Bernoulli's equation we obtain:
<center><math> U \equiv c_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} c_p </math></center>
Note that <math> U \equiv c_g </math> by definition. If the above exercise is repeated in water of finite depth the solution for <math> U </math> after some algebra is:

<center><math> U = c_g = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) c_p </math></center>

with

<center><math> \omega^2 = gk \tanh kh\,</math></center>

It may be shown that the group velocity <math> c_g </math> is given by the relation

<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d} k} </math></center>

=== Rayleigh's proof of the group velocity formula ===

This relation follows from the very elegant "device" due to [http://en.wikipedia.org/wiki/John_Strutt%2C_3rd_Baron_Rayleigh Rayleigh] which applies to any wave form:
Consider two plane progressive waves of nearly equal frequencies and hence wavenumbers. Their joint wave elevation is given by

<center><math> \zeta(x,t) = A \cos ( \omega_1 t - k_1 x) + A \cos ( \omega_2 t - k_2 x) \, </math></center>

where the amplitude is assumed to be common and

<center><math>
\begin{align}
\omega_2 &= \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 \\
k_2 &= k_1 + \Delta k, \quad | \Delta k | \ll k_1 , k_2
\end{align}
</math></center>

Converting into complex notation:

<center><math>
\begin{align}
\zeta(x,t) &= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_2 t - i k_2 x} \} \\
&= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_1 t - i k_1 x + i \Delta\omega t - i \Delta k x} \} \\
&= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) \}
\end{align}
</math></center>

The combined wave elevation <math> \zeta \,</math> vanishes identically where <math> \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) = 0 \, </math>,
i.e.
<center><math> e^{i(\Delta\omega t - \Delta k x)} = -1 \,</math></center>
or when
<center><math> \Delta \omega t - \Delta k x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots </math></center>

Solving for <math> x(t) \, </math> we obtain:

<center><math> x(t) = \frac{1}{\Delta k} \{ (2n+1)\pi + t \Delta\omega \} </math></center>

For values of <math> x(t)\, </math> given above, <math> \zeta = 0 \, </math>. These are the nodes of the bi-chromatic wave train where at all times the elevation vanishes and hence the evergy density is zero. The wave group has the form of consecutive packets separated by nodes.
The speed of the nodes is <math> \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\Delta\omega}{\Delta k} \to \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>
and the energy trapped within two consecutive nodes cannot escape so it must travel at the group velocity: <math> c_g = \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>.
Note that Rayleigh's proof applies equally to waves in finite depth or deep water and in principle to any propagating wave form.
In finite depth it can be shown after some algebra that
<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d}k} = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) \frac{\omega}{k} </math></center>

== Summary ==

The formulae for the energy flux derived above are very general and for potential flow nonlinear surface waves that are not breaking constitute the energy conservation principle.
Energy flux (power) input into the fluid domain by any mechanism, wavemaker, wind (in a conservative manner), a ship or any floating body must be "retrieved" at some distance away. Deriving expressions of the energy flux retrieved at "infinity" is a powerful method for estimating the wave resistance of ships (more on this later), the wave damping of floating bodies, etc.
Yet, the only general way of evaluating wave forces on floating bodies (moving or not) or solid boundaries is by applying the [[Wave Momentum Flux| Wave Momentum]].

-----
This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/courses/mechanical-engineering/2-24-ocean-wave-interaction-with-ships-and-offshore-energy-systems-13-022-spring-2002/lecture-notes/lecture4.pdf here]

Wave Energy Density and Flux

2010-11-06T11:27:14Z

Adi Kurniawan: /* Energy flux across a vertical fluid boundary fixed in space */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = [[Wave Energy Density and Flux]]
| next chapter = [[Wave Momentum Flux]]
| previous chapter = [[Linear Plane Progressive Regular Waves]]
}}

{{complete pages}}

== Introduction ==

We are interested in the transport of energy by ocean waves. It is important to realise that under the assumptions of linear theory, there is no net motion of particles, but there is a transport of energy (as would be expected). The energy consists of two parts, one kinetic due to the motion of the fluid and the other potential due to the variation in the fluid height. It is the resonance between these two energies which gives rise to the wave motion. The situation is analogous to
[http://en.wikipedia.org/wiki/Simple_harmonic_motion Simple Harmonic Motion] but more complicated.

== Energy per volume ==

[[Image:Energy_volume.png|thumb|thumb|right|600px|Energy Volume]]

We begin by defining
<math> \mathcal{E}(t) </math> as the energy in control volume <math> \Omega(t) </math> given by
<center><math> \mathcal{E}(t) = \rho \iiint_\Omega \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}V </math></center>
where <math>\rho</math> is the fluid density, <math>g</math> is the acceleration due to gravity and <math>\mathbf{v}</math> is the vector of fluid velocity.
The mean energy over a unit horizontal surface area <math> S </math> is given by
<center><math> \overline{\mathcal{E}} = \overline{\frac{\mathcal{E}(t)}{S}} = \rho \overline{ \int_{-h}^{\zeta(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}z} = \frac{1}{2} \rho \overline{ \int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - h^2 ) } </math></center>
where <math> \zeta(t) \, </math> is free surface elevation and the overbar denotes average (which will be important when we consider waves).
Note that we are considering water of constant [[Finite Depth]].
We can ignore the term <math> -\frac{1}{2} \rho g h^2 \, </math> which represents the potential energy of the ocean at rest.

The remaining perturbation component is the sum of the kinetic and potential energy components, that is
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} </math></center>
where
<center><math> \overline{\mathcal{E}_{kin}} = \frac{1}{2} \rho \overline{\int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z}, \qquad |\mathbf{v}|^2
= \nabla\Phi \cdot \nabla \Phi = \partial_x^2\Phi + \partial_z^2\Phi </math></center>
and
<center><math> \overline{\mathcal{E}_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)} </math></center>
Note that we are assuming only two dimensions <math>x</math> and <math>z</math>.

=== Energy in [[Linear Plane Progressive Regular Waves]] ===

Consider now as a special case of [[Linear Plane Progressive Regular Waves]] by the velocity potential in [[Infinite Depth]] water (for simplicity). The velocity potential throughout the fluid domain is then given by
<center><math> \Phi = \mathrm{Re} \{ \frac{igA}{\omega} e^{kz-ikx+i\omega t} \} </math></center>
The components is the <math>x</math> and <math>z</math> directions are given by
<center><math> \partial_x\Phi = \mathrm{Re} \{ \frac{igA}{\omega} (-ik) e^{kz-ikx+i\omega t} \} = A \mathrm{Re} \{ \omega e^{kz-ikx+i\omega t} \} </math></center>
and
<center><math> \partial_z\Phi = \mathrm{Re} \{ \frac{igA}{\omega} k e^{kz-ikx+i\omega t} \}
= A \mathrm{Re} \{ i \omega e^{kz-ikx+i\omega t} \} </math></center>
respectively.
We require the following Lemma which is easily proved. If
<center><math> \mathrm{Re} \{ A e^{i\omega t} \} = A(t) \,\!</math></center>
and
<center><math> \mathrm{Re} \{ B e^{i\omega t} \} = B(t) \,\!</math></center>
then it follows that
<center><math> \overline{A(t)B(t)} = \frac{1}{2} \mathrm{Re} \{ A B^* \}. </math></center>

This allows us to write the following expression
<center><math>
\begin{align}
\overline{\mathcal{E}_{kin}} &= \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta )
\left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z \\
&= \frac{1}{2} \rho \int_{-\infty}^0 \overline{ \left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z + O (A^3) \\
&= \rho \frac{\omega^2 A^2}{4k} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ k=\omega^2/g \\
\overline{\mathcal{E}_{pot}} &= \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 .
\end{align}
</math></center>
Note that it is a standard feature of linear oscillations that the average potential and kinetic energies are equal. Hence
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} = \frac{1}{2} \rho g A^2 </math></center>

== Energy Flux ==

[[Image:volume_normal.png|thumb|right|600px|Moving Volume]]

''Energy flux'' <math>\mathcal{P}(t)</math> is the rate of change of energy density <math> \mathcal{E}(t) </math>. It is the flux of energy which is critical to ocean waves. While the individual fluid particles do not move the waves carry energy. We begin by deriving the energy flux in general conditions.

<center><math> \mathcal{P}(t) \equiv \frac{\mathrm{d}\mathcal{E}(t)}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \epsilon(t) \mathrm{d}V
= \iint_{\partial\Omega(t)}
\frac{\partial \epsilon(t)}{\partial t} \mathrm{d}V + \iint_{\partial\Omega(t)} \epsilon(t) U_n \mathrm{d} S </math></center>
(the last result following from the transport theorem) where <math> U_n</math> is the normal velocity of surface <math> \partial\Omega(t) </math> outwards of the enclosed volume <math> \Omega(t)</math>.
We know that
<center><math>
\begin{align}
\frac{\partial \epsilon}{\partial t} &= \frac{\partial}{\partial t} \{ \frac{1}{2} \rho |\mathbf{v}|^2 + \rho g z \} \\
&= \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) \\
&= \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi
\end{align}
</math></center>
so that
<center><math> \mathcal{P}(t) = \rho \iiint_{\Omega(t)} \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) \mathrm{d}V
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
Invoking the scalar form of Gauss's theorem in the first term, we obtain:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \nabla \Phi \cdot \mathbf{n} \mathrm{d}S
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
where <math>\mathbf{n}</math> is the unit normal.

An alternative form for the energy flux <math> \mathcal{P}(t) \, </math> crossing the closed control surface <math> \partial\Omega(t) \, </math>
is obtained by invoking Bernoulli's equation in the second term. Recall that:
<center><math> \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gz = 0 </math></center>
at any point in the fluid domain and on the boundary.
Here we allowed <math> \ P_a </math> the atmospheric pressure to be non-zero for the sake of physical clarity. Upon substitution in
the equation above for <math> \mathcal{P}(t) </math> we obtain the alternate form:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}s
- \rho \iint_{\partial\Omega(t)} \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n \mathrm{d}s </math></center>
where <math>\frac{\partial\phi}{\partial n}</math> is <math>\nabla\phi\cdot\mathbf{n}</math>
So the energy flux across <math> \partial\Omega(t)\, </math> is given by the terms under the integral sign. They can be collected in the more compact form:
<center><math> \mathcal{P}(t) = \iint \left\{ \rho \frac{\partial\Phi}{\partial t} \left( \frac{\partial\Phi}{\partial n} - U_n \right)
- ( P - P_a) U_n \right\} \mathrm{d}s </math></center>
Note that <math> \mathcal{P}(t) \, </math> measures the energy flux into the volume <math> \Omega(t) \, </math> or the rate of growth of the energy density <math> \mathcal{E}(t)\, </math>.

=== Energy transfer for each boundary ===

Break <math> \partial\Omega(t) \, </math> into its components and derive specialized forms of <math> \mathcal{P}(t) \, </math> pertinent to each.

* <math> \partial\Omega_F </math> nonlinear position of the free surface. On this <math> \partial_n\Phi= U_n; </math> and <math> P = P_a </math>, so the fluid pressure is equal to the atmospheric pressure. Therefore over <math> \partial\Omega_F </math> <math>\mathcal{P}(t)=0</math>, as expected, i.e. there is no energy flow into the atmosphere.
*<math> \partial\Omega_B</math> is the non-moving solid boundary, <math> U_n = </math> and <math> \frac{\partial\Phi}{\partial n} = U_n=0</math> which is the no-normal flux condition.
*<math> \partial\Omega^\pm</math> which are the fluid boundaries fixed in space relative to an earth frame <math> U_n = 0 </math> and <math>, \frac{\partial\Phi}{\partial n} \ne 0</math>
*<math> \partial\Omega_U </math> the fluid boundaries moving with velocity <math>\mathbf{u}</math> relative to an earth frame.
*<math> U_n = \mathbf{u} \cdot \mathbf{n}, \quad \frac{\partial\Phi}{\partial n} \ne 0 </math>. This case will be of interest for ships moving with constant velocity <math> U </math>.

The formula derived above are very general for potential flows with a free surface and solid boundaries. We are now ready to apply them to plane progressive waves.

== Surface Wave Problem ==

We are ready now to apply the above formulas to the surface wave problem.

=== Energy flux across a vertical fluid boundary fixed in space ===

<center><math>
\begin{align}
\mathcal{P}(t) &= - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z \\
&= - \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z \\
&= - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}z + O(A^3)
\end{align}
</math></center>

Mean energy flux for a [[Linear Plane Progressive Regular Waves]] follows upon substitution of the regular wave velocity potential and taking mean values:
<center><math> \overline{\mathcal{P}} = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z = \frac{1}{2} \rho g A^2 \left( \frac{1}{2} \frac{g}{\omega} \right) </math></center>
where <math>A</math> is the wave amplitude.
or
<center><math> \overline{\mathcal{P}} = \overline{\mathcal{E}} c_g, \quad c_g = \mbox{group velocity} = \frac{1}{2} c_p </math></center>.
It follows from this that the mean energy flux of a plane progressive wave is the product of its mean energy density times
the [http://en.wikipedia.org/wiki/Group_velocity group velocity] of deep water waves.

A more formal proof that this is the velocity with which the energy flux of plane progressive waves propagates is to consider what needs to be the horizontal velocity <math>U_n \equiv U </math> of a fluid boundary so that the mean energy flux across it vanishes?

This can be found from the solution of the following equation:

<center><math> \overline{\mathcal{P}(t)} = \rho \ {\overline{\int_{-\infty}^0 \partial_t\Phi \partial_x\Phi \mathrm{d}z}} - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}} = 0 </math></center>

Where terms of <math> O(A^3) </math> have been neglected. Note that within linear theory, energy density and energy flux are quantities of <math>O(A^2)</math>. If higher-order terms are kept then we need to consider the treatment of second-order surface wave theory, at least.
Solving the above equation for <math> U </math> we obtain:
<center><math> U = \dfrac{\rho \ {\overline{\int_{-\infty}^0 \dfrac{\partial\Phi}{\partial t} \dfrac{\partial\Phi}{\partial x} \mathrm{d}z}}}
{{\overline{\int_{-\infty}^0 \left( \dfrac{P}{\rho} + \dfrac{\partial\Phi}{\partial t} \right) \mathrm{d}z}}} </math></center>

Upon substitution of the plane progressive wave velocity potential and definition of pressure from Bernoulli's equation we obtain:
<center><math> U \equiv c_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} c_p </math></center>
Note that <math> U \equiv c_g </math> by definition. If the above exercise is repeated in water of finite depth the solution for <math> U </math> after some algebra is:

<center><math> U = c_g = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) c_p </math></center>

with

<center><math> \omega^2 = gk \tanh kh\,</math></center>

It may be shown that the group velocity <math> c_g </math> is given by the relation

<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d} k} </math></center>

=== Rayleigh's proof of the group velocity formula ===

This relation follows from the very elegant "device" due to [http://en.wikipedia.org/wiki/John_Strutt%2C_3rd_Baron_Rayleigh Rayleigh] which applies to any wave form:
Consider two plane progressive waves of nearly equal frequencies and hence wavenumbers. Their joint wave elevation is given by

<center><math> \zeta(x,t) = A \cos ( \omega_1 t - k_1 x) + A \cos ( \omega_2 t - k_2 x) \, </math></center>

where the amplitude is assumed to be common and:

<center><math> \omega_2 = \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 </math></center>

<center><math> k_2 = k_1 + \Delta k, \quad | \Delta k | \ll k_1 , k_2 </math></center>

Converting into complex notation:

<center><math> \zeta(x,t) = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_2 t - i k_2 x} \}
= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_1 t - i k_1 x + i \Delta\omega t - i \Delta k x} \} </math></center> 
<center><math> = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) \} </math></center>

The combined wave elevation <math> \zeta \,</math> vanishes identically where <math> \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) = 0 \, </math>,
i.e.
<center><math> e^{i(\Delta\omega t - \Delta k x)} = -1 \,</math></center>
or when
<center><math> \Delta \omega t - \Delta k x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots </math></center>

Solving for <math> x(t) \, </math> we obtain:

<center><math> x(t) = \frac{1}{\Delta k} \{ (2n+1)\pi + t \Delta\omega \} </math></center>

For values of <math> x(t)\, </math> given above, <math> \zeta = 0 \, </math>. These are the nodes of the bi-chromatic wave train where at all times the elevation vanishes and hence the evergy density is zero. The wave group has the form of consecutive packets separated by nodes.
The speed of the nodes is <math> \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\Delta\omega}{\Delta k} \to \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>
and the energy trapped within two consecutive nodes cannot escape so it must travel at the group velocity: <math> c_g = \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>.
Note that Rayleigh's proof applies equally to waves in finite depth or deep water and in principle to any propagating wave form.
In finite depth it can be shown after some algebra that
<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d}k} = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) \frac{\omega}{k} </math></center>

== Summary ==

The formulae for the energy flux derived above are very general and for potential flow nonlinear surface waves that are not breaking constitute the energy conservation principle.
Energy flux (power) input into the fluid domain by any mechanism, wavemaker, wind (in a conservative manner), a ship or any floating body must be "retrieved" at some distance away. Deriving expressions of the energy flux retrieved at "infinity" is a powerful method for estimating the wave resistance of ships (more on this later), the wave damping of floating bodies, etc.
Yet, the only general way of evaluating wave forces on floating bodies (moving or not) or solid boundaries is by applying the [[Wave Momentum Flux| Wave Momentum]].

-----
This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/courses/mechanical-engineering/2-24-ocean-wave-interaction-with-ships-and-offshore-energy-systems-13-022-spring-2002/lecture-notes/lecture4.pdf here]

Wave Energy Density and Flux

2010-11-06T11:22:55Z

Adi Kurniawan: /* Energy Flux */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = [[Wave Energy Density and Flux]]
| next chapter = [[Wave Momentum Flux]]
| previous chapter = [[Linear Plane Progressive Regular Waves]]
}}

{{complete pages}}

== Introduction ==

We are interested in the transport of energy by ocean waves. It is important to realise that under the assumptions of linear theory, there is no net motion of particles, but there is a transport of energy (as would be expected). The energy consists of two parts, one kinetic due to the motion of the fluid and the other potential due to the variation in the fluid height. It is the resonance between these two energies which gives rise to the wave motion. The situation is analogous to
[http://en.wikipedia.org/wiki/Simple_harmonic_motion Simple Harmonic Motion] but more complicated.

== Energy per volume ==

[[Image:Energy_volume.png|thumb|thumb|right|600px|Energy Volume]]

We begin by defining
<math> \mathcal{E}(t) </math> as the energy in control volume <math> \Omega(t) </math> given by
<center><math> \mathcal{E}(t) = \rho \iiint_\Omega \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}V </math></center>
where <math>\rho</math> is the fluid density, <math>g</math> is the acceleration due to gravity and <math>\mathbf{v}</math> is the vector of fluid velocity.
The mean energy over a unit horizontal surface area <math> S </math> is given by
<center><math> \overline{\mathcal{E}} = \overline{\frac{\mathcal{E}(t)}{S}} = \rho \overline{ \int_{-h}^{\zeta(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}z} = \frac{1}{2} \rho \overline{ \int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - h^2 ) } </math></center>
where <math> \zeta(t) \, </math> is free surface elevation and the overbar denotes average (which will be important when we consider waves).
Note that we are considering water of constant [[Finite Depth]].
We can ignore the term <math> -\frac{1}{2} \rho g h^2 \, </math> which represents the potential energy of the ocean at rest.

The remaining perturbation component is the sum of the kinetic and potential energy components, that is
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} </math></center>
where
<center><math> \overline{\mathcal{E}_{kin}} = \frac{1}{2} \rho \overline{\int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z}, \qquad |\mathbf{v}|^2
= \nabla\Phi \cdot \nabla \Phi = \partial_x^2\Phi + \partial_z^2\Phi </math></center>
and
<center><math> \overline{\mathcal{E}_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)} </math></center>
Note that we are assuming only two dimensions <math>x</math> and <math>z</math>.

=== Energy in [[Linear Plane Progressive Regular Waves]] ===

Consider now as a special case of [[Linear Plane Progressive Regular Waves]] by the velocity potential in [[Infinite Depth]] water (for simplicity). The velocity potential throughout the fluid domain is then given by
<center><math> \Phi = \mathrm{Re} \{ \frac{igA}{\omega} e^{kz-ikx+i\omega t} \} </math></center>
The components is the <math>x</math> and <math>z</math> directions are given by
<center><math> \partial_x\Phi = \mathrm{Re} \{ \frac{igA}{\omega} (-ik) e^{kz-ikx+i\omega t} \} = A \mathrm{Re} \{ \omega e^{kz-ikx+i\omega t} \} </math></center>
and
<center><math> \partial_z\Phi = \mathrm{Re} \{ \frac{igA}{\omega} k e^{kz-ikx+i\omega t} \}
= A \mathrm{Re} \{ i \omega e^{kz-ikx+i\omega t} \} </math></center>
respectively.
We require the following Lemma which is easily proved. If
<center><math> \mathrm{Re} \{ A e^{i\omega t} \} = A(t) \,\!</math></center>
and
<center><math> \mathrm{Re} \{ B e^{i\omega t} \} = B(t) \,\!</math></center>
then it follows that
<center><math> \overline{A(t)B(t)} = \frac{1}{2} \mathrm{Re} \{ A B^* \}. </math></center>

This allows us to write the following expression
<center><math>
\begin{align}
\overline{\mathcal{E}_{kin}} &= \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta )
\left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z \\
&= \frac{1}{2} \rho \int_{-\infty}^0 \overline{ \left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z + O (A^3) \\
&= \rho \frac{\omega^2 A^2}{4k} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ k=\omega^2/g \\
\overline{\mathcal{E}_{pot}} &= \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 .
\end{align}
</math></center>
Note that it is a standard feature of linear oscillations that the average potential and kinetic energies are equal. Hence
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} = \frac{1}{2} \rho g A^2 </math></center>

== Energy Flux ==

[[Image:volume_normal.png|thumb|right|600px|Moving Volume]]

''Energy flux'' <math>\mathcal{P}(t)</math> is the rate of change of energy density <math> \mathcal{E}(t) </math>. It is the flux of energy which is critical to ocean waves. While the individual fluid particles do not move the waves carry energy. We begin by deriving the energy flux in general conditions.

<center><math> \mathcal{P}(t) \equiv \frac{\mathrm{d}\mathcal{E}(t)}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \epsilon(t) \mathrm{d}V
= \iint_{\partial\Omega(t)}
\frac{\partial \epsilon(t)}{\partial t} \mathrm{d}V + \iint_{\partial\Omega(t)} \epsilon(t) U_n \mathrm{d} S </math></center>
(the last result following from the transport theorem) where <math> U_n</math> is the normal velocity of surface <math> \partial\Omega(t) </math> outwards of the enclosed volume <math> \Omega(t)</math>.
We know that
<center><math>
\begin{align}
\frac{\partial \epsilon}{\partial t} &= \frac{\partial}{\partial t} \{ \frac{1}{2} \rho |\mathbf{v}|^2 + \rho g z \} \\
&= \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) \\
&= \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi
\end{align}
</math></center>
so that
<center><math> \mathcal{P}(t) = \rho \iiint_{\Omega(t)} \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) \mathrm{d}V
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
Invoking the scalar form of Gauss's theorem in the first term, we obtain:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \nabla \Phi \cdot \mathbf{n} \mathrm{d}S
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
where <math>\mathbf{n}</math> is the unit normal.

An alternative form for the energy flux <math> \mathcal{P}(t) \, </math> crossing the closed control surface <math> \partial\Omega(t) \, </math>
is obtained by invoking Bernoulli's equation in the second term. Recall that:
<center><math> \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gz = 0 </math></center>
at any point in the fluid domain and on the boundary.
Here we allowed <math> \ P_a </math> the atmospheric pressure to be non-zero for the sake of physical clarity. Upon substitution in
the equation above for <math> \mathcal{P}(t) </math> we obtain the alternate form:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}s
- \rho \iint_{\partial\Omega(t)} \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n \mathrm{d}s </math></center>
where <math>\frac{\partial\phi}{\partial n}</math> is <math>\nabla\phi\cdot\mathbf{n}</math>
So the energy flux across <math> \partial\Omega(t)\, </math> is given by the terms under the integral sign. They can be collected in the more compact form:
<center><math> \mathcal{P}(t) = \iint \left\{ \rho \frac{\partial\Phi}{\partial t} \left( \frac{\partial\Phi}{\partial n} - U_n \right)
- ( P - P_a) U_n \right\} \mathrm{d}s </math></center>
Note that <math> \mathcal{P}(t) \, </math> measures the energy flux into the volume <math> \Omega(t) \, </math> or the rate of growth of the energy density <math> \mathcal{E}(t)\, </math>.

=== Energy transfer for each boundary ===

Break <math> \partial\Omega(t) \, </math> into its components and derive specialized forms of <math> \mathcal{P}(t) \, </math> pertinent to each.

* <math> \partial\Omega_F </math> nonlinear position of the free surface. On this <math> \partial_n\Phi= U_n; </math> and <math> P = P_a </math>, so the fluid pressure is equal to the atmospheric pressure. Therefore over <math> \partial\Omega_F </math> <math>\mathcal{P}(t)=0</math>, as expected, i.e. there is no energy flow into the atmosphere.
*<math> \partial\Omega_B</math> is the non-moving solid boundary, <math> U_n = </math> and <math> \frac{\partial\Phi}{\partial n} = U_n=0</math> which is the no-normal flux condition.
*<math> \partial\Omega^\pm</math> which are the fluid boundaries fixed in space relative to an earth frame <math> U_n = 0 </math> and <math>, \frac{\partial\Phi}{\partial n} \ne 0</math>
*<math> \partial\Omega_U </math> the fluid boundaries moving with velocity <math>\mathbf{u}</math> relative to an earth frame.
*<math> U_n = \mathbf{u} \cdot \mathbf{n}, \quad \frac{\partial\Phi}{\partial n} \ne 0 </math>. This case will be of interest for ships moving with constant velocity <math> U </math>.

The formula derived above are very general for potential flows with a free surface and solid boundaries. We are now ready to apply them to plane progressive waves.

== Surface Wave Problem ==

We are ready now to apply the above formulas to the surface wave problem.

=== Energy flux across a vertical fluid boundary fixed in space ===

<center><math> \mathcal{P}(t) = - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z =
- \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z
= - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}z + O(A^3) </math></center>

Mean energy flux for a [[Linear Plane Progressive Regular Waves]] follows upon substitution of the regular wave velocity potential and taking mean values:
<center><math> \overline{\mathcal{P}} = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z = \frac{1}{2} \rho g A^2 \left( \frac{1}{2} \frac{g}{\omega} \right) </math></center>
where <math>A</math> is the wave amplitude.
or
<center><math> \overline{\mathcal{P}} = \overline{\mathcal{E}} c_g, \quad c_g = \mbox{group velocity} = \frac{1}{2} c_p </math></center>.
It follows from this that the mean energy flux of a plane progressive wave is the product of its mean energy density times
the [http://en.wikipedia.org/wiki/Group_velocity group velocity] of deep water waves.

A more formal proof that this is the velocity with which the energy flux of plane progressive waves propagates is to consider what needs to be the horizontal velocity <math>U_n \equiv U </math> of a fluid boundary so that the mean energy flux across it vanishes?

This can be found from the solution of the following equation:

<center><math> \overline{\mathcal{P}(t)} = \rho \ {\overline{\int_{-\infty}^0 \partial_t\Phi \partial_x\Phi \mathrm{d}z}} - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}} = 0 </math></center>

Where terms of <math> O(A^3) </math> have been neglected. Note that within linear theory, energy density and energy flux are quantities of <math>O(A^2)</math>. If higher-order terms are kept then we need to consider the treatment of second-order surface wave theory, at least.
Solving the above equation for <math> U </math> we obtain:
<center><math> U = \frac{\rho \ {\overline{\int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z}}}
{{\overline{\int_{-\infty}^0 \left( \frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}}} </math></center>

Upon substitution of the plane progressive wave velocity potential and definition of pressure from Bernoulli's equation we obtain:
<center><math> U \equiv c_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} c_p </math></center>
Note that <math> U \equiv c_g </math> by definition. If the above exercise is repeated in water of finite depth the solution for <math> U </math> after some algebra is:

<center><math> U = c_g = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) c_p </math></center>

with

<center><math> \omega^2 = gk \tanh kh\,</math></center>

It may be shown that the group velocity <math> c_g </math> is given by the relation

<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d} k} </math></center>

=== Rayleigh's proof of the group velocity formula ===

This relation follows from the very elegant "device" due to [http://en.wikipedia.org/wiki/John_Strutt%2C_3rd_Baron_Rayleigh Rayleigh] which applies to any wave form:
Consider two plane progressive waves of nearly equal frequencies and hence wavenumbers. Their joint wave elevation is given by

<center><math> \zeta(x,t) = A \cos ( \omega_1 t - k_1 x) + A \cos ( \omega_2 t - k_2 x) \, </math></center>

where the amplitude is assumed to be common and:

<center><math> \omega_2 = \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 </math></center>

<center><math> k_2 = k_1 + \Delta k, \quad | \Delta k | \ll k_1 , k_2 </math></center>

Converting into complex notation:

<center><math> \zeta(x,t) = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_2 t - i k_2 x} \}
= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_1 t - i k_1 x + i \Delta\omega t - i \Delta k x} \} </math></center> 
<center><math> = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) \} </math></center>

The combined wave elevation <math> \zeta \,</math> vanishes identically where <math> \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) = 0 \, </math>,
i.e.
<center><math> e^{i(\Delta\omega t - \Delta k x)} = -1 \,</math></center>
or when
<center><math> \Delta \omega t - \Delta k x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots </math></center>

Solving for <math> x(t) \, </math> we obtain:

<center><math> x(t) = \frac{1}{\Delta k} \{ (2n+1)\pi + t \Delta\omega \} </math></center>

For values of <math> x(t)\, </math> given above, <math> \zeta = 0 \, </math>. These are the nodes of the bi-chromatic wave train where at all times the elevation vanishes and hence the evergy density is zero. The wave group has the form of consecutive packets separated by nodes.
The speed of the nodes is <math> \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\Delta\omega}{\Delta k} \to \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>
and the energy trapped within two consecutive nodes cannot escape so it must travel at the group velocity: <math> c_g = \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>.
Note that Rayleigh's proof applies equally to waves in finite depth or deep water and in principle to any propagating wave form.
In finite depth it can be shown after some algebra that
<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d}k} = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) \frac{\omega}{k} </math></center>

== Summary ==

The formulae for the energy flux derived above are very general and for potential flow nonlinear surface waves that are not breaking constitute the energy conservation principle.
Energy flux (power) input into the fluid domain by any mechanism, wavemaker, wind (in a conservative manner), a ship or any floating body must be "retrieved" at some distance away. Deriving expressions of the energy flux retrieved at "infinity" is a powerful method for estimating the wave resistance of ships (more on this later), the wave damping of floating bodies, etc.
Yet, the only general way of evaluating wave forces on floating bodies (moving or not) or solid boundaries is by applying the [[Wave Momentum Flux| Wave Momentum]].

-----
This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/courses/mechanical-engineering/2-24-ocean-wave-interaction-with-ships-and-offshore-energy-systems-13-022-spring-2002/lecture-notes/lecture4.pdf here]

Wave Energy Density and Flux

2010-11-06T11:19:26Z

Adi Kurniawan: /* Energy in Linear Plane Progressive Regular Waves */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = [[Wave Energy Density and Flux]]
| next chapter = [[Wave Momentum Flux]]
| previous chapter = [[Linear Plane Progressive Regular Waves]]
}}

{{complete pages}}

== Introduction ==

We are interested in the transport of energy by ocean waves. It is important to realise that under the assumptions of linear theory, there is no net motion of particles, but there is a transport of energy (as would be expected). The energy consists of two parts, one kinetic due to the motion of the fluid and the other potential due to the variation in the fluid height. It is the resonance between these two energies which gives rise to the wave motion. The situation is analogous to
[http://en.wikipedia.org/wiki/Simple_harmonic_motion Simple Harmonic Motion] but more complicated.

== Energy per volume ==

[[Image:Energy_volume.png|thumb|thumb|right|600px|Energy Volume]]

We begin by defining
<math> \mathcal{E}(t) </math> as the energy in control volume <math> \Omega(t) </math> given by
<center><math> \mathcal{E}(t) = \rho \iiint_\Omega \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}V </math></center>
where <math>\rho</math> is the fluid density, <math>g</math> is the acceleration due to gravity and <math>\mathbf{v}</math> is the vector of fluid velocity.
The mean energy over a unit horizontal surface area <math> S </math> is given by
<center><math> \overline{\mathcal{E}} = \overline{\frac{\mathcal{E}(t)}{S}} = \rho \overline{ \int_{-h}^{\zeta(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}z} = \frac{1}{2} \rho \overline{ \int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - h^2 ) } </math></center>
where <math> \zeta(t) \, </math> is free surface elevation and the overbar denotes average (which will be important when we consider waves).
Note that we are considering water of constant [[Finite Depth]].
We can ignore the term <math> -\frac{1}{2} \rho g h^2 \, </math> which represents the potential energy of the ocean at rest.

The remaining perturbation component is the sum of the kinetic and potential energy components, that is
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} </math></center>
where
<center><math> \overline{\mathcal{E}_{kin}} = \frac{1}{2} \rho \overline{\int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z}, \qquad |\mathbf{v}|^2
= \nabla\Phi \cdot \nabla \Phi = \partial_x^2\Phi + \partial_z^2\Phi </math></center>
and
<center><math> \overline{\mathcal{E}_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)} </math></center>
Note that we are assuming only two dimensions <math>x</math> and <math>z</math>.

=== Energy in [[Linear Plane Progressive Regular Waves]] ===

Consider now as a special case of [[Linear Plane Progressive Regular Waves]] by the velocity potential in [[Infinite Depth]] water (for simplicity). The velocity potential throughout the fluid domain is then given by
<center><math> \Phi = \mathrm{Re} \{ \frac{igA}{\omega} e^{kz-ikx+i\omega t} \} </math></center>
The components is the <math>x</math> and <math>z</math> directions are given by
<center><math> \partial_x\Phi = \mathrm{Re} \{ \frac{igA}{\omega} (-ik) e^{kz-ikx+i\omega t} \} = A \mathrm{Re} \{ \omega e^{kz-ikx+i\omega t} \} </math></center>
and
<center><math> \partial_z\Phi = \mathrm{Re} \{ \frac{igA}{\omega} k e^{kz-ikx+i\omega t} \}
= A \mathrm{Re} \{ i \omega e^{kz-ikx+i\omega t} \} </math></center>
respectively.
We require the following Lemma which is easily proved. If
<center><math> \mathrm{Re} \{ A e^{i\omega t} \} = A(t) \,\!</math></center>
and
<center><math> \mathrm{Re} \{ B e^{i\omega t} \} = B(t) \,\!</math></center>
then it follows that
<center><math> \overline{A(t)B(t)} = \frac{1}{2} \mathrm{Re} \{ A B^* \}. </math></center>

This allows us to write the following expression
<center><math>
\begin{align}
\overline{\mathcal{E}_{kin}} &= \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta )
\left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z \\
&= \frac{1}{2} \rho \int_{-\infty}^0 \overline{ \left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z + O (A^3) \\
&= \rho \frac{\omega^2 A^2}{4k} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ k=\omega^2/g \\
\overline{\mathcal{E}_{pot}} &= \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 .
\end{align}
</math></center>
Note that it is a standard feature of linear oscillations that the average potential and kinetic energies are equal. Hence
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} = \frac{1}{2} \rho g A^2 </math></center>

== Energy Flux ==

[[Image:volume_normal.png|thumb|right|600px|Moving Volume]]

''Energy flux'' <math>\mathcal{P}(t)</math> is the rate of change of energy density <math> \mathcal{E}(t) </math>. It is the flux of energy which is critical to ocean waves. While the individual fluid particles do not move the waves carry energy. We begin by deriving the energy flux in general conditions.

<center><math> \mathcal{P}(t) \equiv \frac{\mathrm{d}\mathcal{E}(t)}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \epsilon(t) \mathrm{d}V
= \iint_{\partial\Omega(t)}
\frac{\partial \epsilon(t)}{\partial t} \mathrm{d}V + \iint_{\partial\Omega(t)} \epsilon(t) U_n \mathrm{d} S </math></center>
(the last result following from the transport theorem) where <math> U_n</math> is the normal velocity of surface <math> \partial\Omega(t) </math> outwards of the enclosed volume <math> \Omega(t)</math>.
We know that
<center><math> \frac{\partial \epsilon}{\partial t} = \frac{\partial}{\partial t} \{ \frac{1}{2} \rho |\mathbf{v}|^2 + \rho g z \}
= \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) </math></center>
<center><math> = \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi </math></center>
so that
<center><math> \mathcal{P}(t) = \rho \iiint_{\Omega(t)} \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) \mathrm{d}V
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
Invoking the scalar form of Gauss's theorem in the first term, we obtain:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \nabla \Phi \cdot \mathbf{n} \mathrm{d}S
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
where <math>\mathbf{n}</math> is the unit normal.

An alternative form for the energy flux <math> \mathcal{P}(t) \, </math> crossing the closed control surface <math> \partial\Omega(t) \, </math>
is obtained by invoking Bernoulli's equation in the second term. Recall that:
<center><math> \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gz = 0 </math></center>
at any point in the fluid domain and on the boundary.
Here we allowed <math> \ P_a </math> the atmospheric pressure to be non-zero for the sake of physical clarity. Upon substitution in
the equation above for <math> \mathcal{P}(t) </math> we obtain the alternate form:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}s
- \rho \iint_{\partial\Omega(t)} \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n \mathrm{d}s </math></center>
where <math>\frac{\partial\phi}{\partial n}</math> is <math>\nabla\phi\cdot\mathbf{n}</math>
So the energy flux across <math> \partial\Omega(t)\, </math> is given by the terms under the integral sign. They can be collected in the more compact form:
<center><math> \mathcal{P}(t) = \iint \left\{ \rho \frac{\partial\Phi}{\partial t} \left( \frac{\partial\Phi}{\partial n} - U_n \right)
- ( P - P_a) U_n \right\} \mathrm{d}s </math></center>
Note that <math> \mathcal{P}(t) \, </math> measures the energy flux into the volume <math> \Omega(t) \, </math> or the rate of growth of the energy density <math> \mathcal{E}(t)\, </math>.

=== Energy transfer for each boundary ===

Break <math> \partial\Omega(t) \, </math> into its components and derive specialized forms of <math> \mathcal{P}(t) \, </math> pertinent to each.

* <math> \partial\Omega_F </math> nonlinear position of the free surface. On this <math> \partial_n\Phi= U_n; </math> and <math> P = P_a </math>, so the fluid pressure is equal to the atmospheric pressure. Therefore over <math> \partial\Omega_F </math> <math>\mathcal{P}(t)=0</math>, as expected, i.e. there is no energy flow into the atmosphere.
*<math> \partial\Omega_B</math> is the non-moving solid boundary, <math> U_n = </math> and <math> \frac{\partial\Phi}{\partial n} = U_n=0</math> which is the no-normal flux condition.
*<math> \partial\Omega^\pm</math> which are the fluid boundaries fixed in space relative to an earth frame <math> U_n = 0 </math> and <math>, \frac{\partial\Phi}{\partial n} \ne 0</math>
*<math> \partial\Omega_U </math> the fluid boundaries moving with velocity <math>\mathbf{u}</math> relative to an earth frame.
*<math> U_n = \mathbf{u} \cdot \mathbf{n}, \quad \frac{\partial\Phi}{\partial n} \ne 0 </math>. This case will be of interest for ships moving with constant velocity <math> U </math>.

The formula derived above are very general for potential flows with a free surface and solid boundaries. We are now ready to apply them to plane progressive waves.

== Surface Wave Problem ==

We are ready now to apply the above formulas to the surface wave problem.

=== Energy flux across a vertical fluid boundary fixed in space ===

<center><math> \mathcal{P}(t) = - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z =
- \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z
= - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}z + O(A^3) </math></center>

Mean energy flux for a [[Linear Plane Progressive Regular Waves]] follows upon substitution of the regular wave velocity potential and taking mean values:
<center><math> \overline{\mathcal{P}} = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z = \frac{1}{2} \rho g A^2 \left( \frac{1}{2} \frac{g}{\omega} \right) </math></center>
where <math>A</math> is the wave amplitude.
or
<center><math> \overline{\mathcal{P}} = \overline{\mathcal{E}} c_g, \quad c_g = \mbox{group velocity} = \frac{1}{2} c_p </math></center>.
It follows from this that the mean energy flux of a plane progressive wave is the product of its mean energy density times
the [http://en.wikipedia.org/wiki/Group_velocity group velocity] of deep water waves.

A more formal proof that this is the velocity with which the energy flux of plane progressive waves propagates is to consider what needs to be the horizontal velocity <math>U_n \equiv U </math> of a fluid boundary so that the mean energy flux across it vanishes?

This can be found from the solution of the following equation:

<center><math> \overline{\mathcal{P}(t)} = \rho \ {\overline{\int_{-\infty}^0 \partial_t\Phi \partial_x\Phi \mathrm{d}z}} - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}} = 0 </math></center>

Where terms of <math> O(A^3) </math> have been neglected. Note that within linear theory, energy density and energy flux are quantities of <math>O(A^2)</math>. If higher-order terms are kept then we need to consider the treatment of second-order surface wave theory, at least.
Solving the above equation for <math> U </math> we obtain:
<center><math> U = \frac{\rho \ {\overline{\int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z}}}
{{\overline{\int_{-\infty}^0 \left( \frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}}} </math></center>

Upon substitution of the plane progressive wave velocity potential and definition of pressure from Bernoulli's equation we obtain:
<center><math> U \equiv c_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} c_p </math></center>
Note that <math> U \equiv c_g </math> by definition. If the above exercise is repeated in water of finite depth the solution for <math> U </math> after some algebra is:

<center><math> U = c_g = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) c_p </math></center>

with

<center><math> \omega^2 = gk \tanh kh\,</math></center>

It may be shown that the group velocity <math> c_g </math> is given by the relation

<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d} k} </math></center>

=== Rayleigh's proof of the group velocity formula ===

This relation follows from the very elegant "device" due to [http://en.wikipedia.org/wiki/John_Strutt%2C_3rd_Baron_Rayleigh Rayleigh] which applies to any wave form:
Consider two plane progressive waves of nearly equal frequencies and hence wavenumbers. Their joint wave elevation is given by

<center><math> \zeta(x,t) = A \cos ( \omega_1 t - k_1 x) + A \cos ( \omega_2 t - k_2 x) \, </math></center>

where the amplitude is assumed to be common and:

<center><math> \omega_2 = \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 </math></center>

<center><math> k_2 = k_1 + \Delta k, \quad | \Delta k | \ll k_1 , k_2 </math></center>

Converting into complex notation:

<center><math> \zeta(x,t) = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_2 t - i k_2 x} \}
= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_1 t - i k_1 x + i \Delta\omega t - i \Delta k x} \} </math></center> 
<center><math> = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) \} </math></center>

The combined wave elevation <math> \zeta \,</math> vanishes identically where <math> \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) = 0 \, </math>,
i.e.
<center><math> e^{i(\Delta\omega t - \Delta k x)} = -1 \,</math></center>
or when
<center><math> \Delta \omega t - \Delta k x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots </math></center>

Solving for <math> x(t) \, </math> we obtain:

<center><math> x(t) = \frac{1}{\Delta k} \{ (2n+1)\pi + t \Delta\omega \} </math></center>

For values of <math> x(t)\, </math> given above, <math> \zeta = 0 \, </math>. These are the nodes of the bi-chromatic wave train where at all times the elevation vanishes and hence the evergy density is zero. The wave group has the form of consecutive packets separated by nodes.
The speed of the nodes is <math> \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\Delta\omega}{\Delta k} \to \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>
and the energy trapped within two consecutive nodes cannot escape so it must travel at the group velocity: <math> c_g = \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>.
Note that Rayleigh's proof applies equally to waves in finite depth or deep water and in principle to any propagating wave form.
In finite depth it can be shown after some algebra that
<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d}k} = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) \frac{\omega}{k} </math></center>

== Summary ==

The formulae for the energy flux derived above are very general and for potential flow nonlinear surface waves that are not breaking constitute the energy conservation principle.
Energy flux (power) input into the fluid domain by any mechanism, wavemaker, wind (in a conservative manner), a ship or any floating body must be "retrieved" at some distance away. Deriving expressions of the energy flux retrieved at "infinity" is a powerful method for estimating the wave resistance of ships (more on this later), the wave damping of floating bodies, etc.
Yet, the only general way of evaluating wave forces on floating bodies (moving or not) or solid boundaries is by applying the [[Wave Momentum Flux| Wave Momentum]].

-----
This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/courses/mechanical-engineering/2-24-ocean-wave-interaction-with-ships-and-offshore-energy-systems-13-022-spring-2002/lecture-notes/lecture4.pdf here]

Wave Energy Density and Flux

2010-11-06T11:19:00Z

Adi Kurniawan: /* Energy in Linear Plane Progressive Regular Waves */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = [[Wave Energy Density and Flux]]
| next chapter = [[Wave Momentum Flux]]
| previous chapter = [[Linear Plane Progressive Regular Waves]]
}}

{{complete pages}}

== Introduction ==

We are interested in the transport of energy by ocean waves. It is important to realise that under the assumptions of linear theory, there is no net motion of particles, but there is a transport of energy (as would be expected). The energy consists of two parts, one kinetic due to the motion of the fluid and the other potential due to the variation in the fluid height. It is the resonance between these two energies which gives rise to the wave motion. The situation is analogous to
[http://en.wikipedia.org/wiki/Simple_harmonic_motion Simple Harmonic Motion] but more complicated.

== Energy per volume ==

[[Image:Energy_volume.png|thumb|thumb|right|600px|Energy Volume]]

We begin by defining
<math> \mathcal{E}(t) </math> as the energy in control volume <math> \Omega(t) </math> given by
<center><math> \mathcal{E}(t) = \rho \iiint_\Omega \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}V </math></center>
where <math>\rho</math> is the fluid density, <math>g</math> is the acceleration due to gravity and <math>\mathbf{v}</math> is the vector of fluid velocity.
The mean energy over a unit horizontal surface area <math> S </math> is given by
<center><math> \overline{\mathcal{E}} = \overline{\frac{\mathcal{E}(t)}{S}} = \rho \overline{ \int_{-h}^{\zeta(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}z} = \frac{1}{2} \rho \overline{ \int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - h^2 ) } </math></center>
where <math> \zeta(t) \, </math> is free surface elevation and the overbar denotes average (which will be important when we consider waves).
Note that we are considering water of constant [[Finite Depth]].
We can ignore the term <math> -\frac{1}{2} \rho g h^2 \, </math> which represents the potential energy of the ocean at rest.

The remaining perturbation component is the sum of the kinetic and potential energy components, that is
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} </math></center>
where
<center><math> \overline{\mathcal{E}_{kin}} = \frac{1}{2} \rho \overline{\int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z}, \qquad |\mathbf{v}|^2
= \nabla\Phi \cdot \nabla \Phi = \partial_x^2\Phi + \partial_z^2\Phi </math></center>
and
<center><math> \overline{\mathcal{E}_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)} </math></center>
Note that we are assuming only two dimensions <math>x</math> and <math>z</math>.

=== Energy in [[Linear Plane Progressive Regular Waves]] ===

Consider now as a special case of [[Linear Plane Progressive Regular Waves]] by the velocity potential in [[Infinite Depth]] water (for simplicity). The velocity potential throughout the fluid domain is then given by
<center><math> \Phi = \mathrm{Re} \{ \frac{igA}{\omega} e^{kz-ikx+i\omega t} \} </math></center>
The components is the <math>x</math> and <math>z</math> directions are given by
<center><math> \partial_x\Phi = \mathrm{Re} \{ \frac{igA}{\omega} (-ik) e^{kz-ikx+i\omega t} \} = A \mathrm{Re} \{ \omega e^{kz-ikx+i\omega t} \} </math></center>
and
<center><math> \partial_z\Phi = \mathrm{Re} \{ \frac{igA}{\omega} k e^{kz-ikx+i\omega t} \}
= A \mathrm{Re} \{ i \omega e^{kz-ikx+i\omega t} \} </math></center>
respectively.
We require the following Lemma which is easily proved. If
<center><math> \mathrm{Re} \{ A e^{i\omega t} \} = A(t) \,\!</math></center>
and
<center><math> \mathrm{Re} \{ B e^{i\omega t} \} = B(t) \,\!</math></center>
then it follows that
<center><math> \overline{A(t)B(t)} = \frac{1}{2} \mathrm{Re} \{ A B^* \} </math></center>.

This allows us to write the following expression
<center><math>
\begin{align}
\overline{\mathcal{E}_{kin}} &= \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta )
\left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z \\
&= \frac{1}{2} \rho \int_{-\infty}^0 \overline{ \left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z + O (A^3) \\
&= \rho \frac{\omega^2 A^2}{4k} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ k=\omega^2/g \\
\overline{\mathcal{E}_{pot}} &= \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 .
\end{align}
</math></center>
Note that it is a standard feature of linear oscillations that the average potential and kinetic energies are equal. Hence
<center><math> \overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} = \frac{1}{2} \rho g A^2 </math></center>

== Energy Flux ==

[[Image:volume_normal.png|thumb|right|600px|Moving Volume]]

''Energy flux'' <math>\mathcal{P}(t)</math> is the rate of change of energy density <math> \mathcal{E}(t) </math>. It is the flux of energy which is critical to ocean waves. While the individual fluid particles do not move the waves carry energy. We begin by deriving the energy flux in general conditions.

<center><math> \mathcal{P}(t) \equiv \frac{\mathrm{d}\mathcal{E}(t)}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \epsilon(t) \mathrm{d}V
= \iint_{\partial\Omega(t)}
\frac{\partial \epsilon(t)}{\partial t} \mathrm{d}V + \iint_{\partial\Omega(t)} \epsilon(t) U_n \mathrm{d} S </math></center>
(the last result following from the transport theorem) where <math> U_n</math> is the normal velocity of surface <math> \partial\Omega(t) </math> outwards of the enclosed volume <math> \Omega(t)</math>.
We know that
<center><math> \frac{\partial \epsilon}{\partial t} = \frac{\partial}{\partial t} \{ \frac{1}{2} \rho |\mathbf{v}|^2 + \rho g z \}
= \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) </math></center>
<center><math> = \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi </math></center>
so that
<center><math> \mathcal{P}(t) = \rho \iiint_{\Omega(t)} \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) \mathrm{d}V
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
Invoking the scalar form of Gauss's theorem in the first term, we obtain:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \nabla \Phi \cdot \mathbf{n} \mathrm{d}S
+ \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S </math></center>
where <math>\mathbf{n}</math> is the unit normal.

An alternative form for the energy flux <math> \mathcal{P}(t) \, </math> crossing the closed control surface <math> \partial\Omega(t) \, </math>
is obtained by invoking Bernoulli's equation in the second term. Recall that:
<center><math> \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gz = 0 </math></center>
at any point in the fluid domain and on the boundary.
Here we allowed <math> \ P_a </math> the atmospheric pressure to be non-zero for the sake of physical clarity. Upon substitution in
the equation above for <math> \mathcal{P}(t) </math> we obtain the alternate form:
<center><math> \mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}s
- \rho \iint_{\partial\Omega(t)} \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n \mathrm{d}s </math></center>
where <math>\frac{\partial\phi}{\partial n}</math> is <math>\nabla\phi\cdot\mathbf{n}</math>
So the energy flux across <math> \partial\Omega(t)\, </math> is given by the terms under the integral sign. They can be collected in the more compact form:
<center><math> \mathcal{P}(t) = \iint \left\{ \rho \frac{\partial\Phi}{\partial t} \left( \frac{\partial\Phi}{\partial n} - U_n \right)
- ( P - P_a) U_n \right\} \mathrm{d}s </math></center>
Note that <math> \mathcal{P}(t) \, </math> measures the energy flux into the volume <math> \Omega(t) \, </math> or the rate of growth of the energy density <math> \mathcal{E}(t)\, </math>.

=== Energy transfer for each boundary ===

Break <math> \partial\Omega(t) \, </math> into its components and derive specialized forms of <math> \mathcal{P}(t) \, </math> pertinent to each.

* <math> \partial\Omega_F </math> nonlinear position of the free surface. On this <math> \partial_n\Phi= U_n; </math> and <math> P = P_a </math>, so the fluid pressure is equal to the atmospheric pressure. Therefore over <math> \partial\Omega_F </math> <math>\mathcal{P}(t)=0</math>, as expected, i.e. there is no energy flow into the atmosphere.
*<math> \partial\Omega_B</math> is the non-moving solid boundary, <math> U_n = </math> and <math> \frac{\partial\Phi}{\partial n} = U_n=0</math> which is the no-normal flux condition.
*<math> \partial\Omega^\pm</math> which are the fluid boundaries fixed in space relative to an earth frame <math> U_n = 0 </math> and <math>, \frac{\partial\Phi}{\partial n} \ne 0</math>
*<math> \partial\Omega_U </math> the fluid boundaries moving with velocity <math>\mathbf{u}</math> relative to an earth frame.
*<math> U_n = \mathbf{u} \cdot \mathbf{n}, \quad \frac{\partial\Phi}{\partial n} \ne 0 </math>. This case will be of interest for ships moving with constant velocity <math> U </math>.

The formula derived above are very general for potential flows with a free surface and solid boundaries. We are now ready to apply them to plane progressive waves.

== Surface Wave Problem ==

We are ready now to apply the above formulas to the surface wave problem.

=== Energy flux across a vertical fluid boundary fixed in space ===

<center><math> \mathcal{P}(t) = - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z =
- \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z
= - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}z + O(A^3) </math></center>

Mean energy flux for a [[Linear Plane Progressive Regular Waves]] follows upon substitution of the regular wave velocity potential and taking mean values:
<center><math> \overline{\mathcal{P}} = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z = \frac{1}{2} \rho g A^2 \left( \frac{1}{2} \frac{g}{\omega} \right) </math></center>
where <math>A</math> is the wave amplitude.
or
<center><math> \overline{\mathcal{P}} = \overline{\mathcal{E}} c_g, \quad c_g = \mbox{group velocity} = \frac{1}{2} c_p </math></center>.
It follows from this that the mean energy flux of a plane progressive wave is the product of its mean energy density times
the [http://en.wikipedia.org/wiki/Group_velocity group velocity] of deep water waves.

A more formal proof that this is the velocity with which the energy flux of plane progressive waves propagates is to consider what needs to be the horizontal velocity <math>U_n \equiv U </math> of a fluid boundary so that the mean energy flux across it vanishes?

This can be found from the solution of the following equation:

<center><math> \overline{\mathcal{P}(t)} = \rho \ {\overline{\int_{-\infty}^0 \partial_t\Phi \partial_x\Phi \mathrm{d}z}} - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}} = 0 </math></center>

Where terms of <math> O(A^3) </math> have been neglected. Note that within linear theory, energy density and energy flux are quantities of <math>O(A^2)</math>. If higher-order terms are kept then we need to consider the treatment of second-order surface wave theory, at least.
Solving the above equation for <math> U </math> we obtain:
<center><math> U = \frac{\rho \ {\overline{\int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z}}}
{{\overline{\int_{-\infty}^0 \left( \frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}}} </math></center>

Upon substitution of the plane progressive wave velocity potential and definition of pressure from Bernoulli's equation we obtain:
<center><math> U \equiv c_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} c_p </math></center>
Note that <math> U \equiv c_g </math> by definition. If the above exercise is repeated in water of finite depth the solution for <math> U </math> after some algebra is:

<center><math> U = c_g = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) c_p </math></center>

with

<center><math> \omega^2 = gk \tanh kh\,</math></center>

It may be shown that the group velocity <math> c_g </math> is given by the relation

<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d} k} </math></center>

=== Rayleigh's proof of the group velocity formula ===

This relation follows from the very elegant "device" due to [http://en.wikipedia.org/wiki/John_Strutt%2C_3rd_Baron_Rayleigh Rayleigh] which applies to any wave form:
Consider two plane progressive waves of nearly equal frequencies and hence wavenumbers. Their joint wave elevation is given by

<center><math> \zeta(x,t) = A \cos ( \omega_1 t - k_1 x) + A \cos ( \omega_2 t - k_2 x) \, </math></center>

where the amplitude is assumed to be common and:

<center><math> \omega_2 = \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 </math></center>

<center><math> k_2 = k_1 + \Delta k, \quad | \Delta k | \ll k_1 , k_2 </math></center>

Converting into complex notation:

<center><math> \zeta(x,t) = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_2 t - i k_2 x} \}
= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_1 t - i k_1 x + i \Delta\omega t - i \Delta k x} \} </math></center> 
<center><math> = A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) \} </math></center>

The combined wave elevation <math> \zeta \,</math> vanishes identically where <math> \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) = 0 \, </math>,
i.e.
<center><math> e^{i(\Delta\omega t - \Delta k x)} = -1 \,</math></center>
or when
<center><math> \Delta \omega t - \Delta k x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots </math></center>

Solving for <math> x(t) \, </math> we obtain:

<center><math> x(t) = \frac{1}{\Delta k} \{ (2n+1)\pi + t \Delta\omega \} </math></center>

For values of <math> x(t)\, </math> given above, <math> \zeta = 0 \, </math>. These are the nodes of the bi-chromatic wave train where at all times the elevation vanishes and hence the evergy density is zero. The wave group has the form of consecutive packets separated by nodes.
The speed of the nodes is <math> \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\Delta\omega}{\Delta k} \to \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>
and the energy trapped within two consecutive nodes cannot escape so it must travel at the group velocity: <math> c_g = \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}\omega}{\mathrm{d}k} \, </math>.
Note that Rayleigh's proof applies equally to waves in finite depth or deep water and in principle to any propagating wave form.
In finite depth it can be shown after some algebra that
<center><math> c_g = \frac{\mathrm{d}\omega}{\mathrm{d}k} = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) \frac{\omega}{k} </math></center>

== Summary ==

The formulae for the energy flux derived above are very general and for potential flow nonlinear surface waves that are not breaking constitute the energy conservation principle.
Energy flux (power) input into the fluid domain by any mechanism, wavemaker, wind (in a conservative manner), a ship or any floating body must be "retrieved" at some distance away. Deriving expressions of the energy flux retrieved at "infinity" is a powerful method for estimating the wave resistance of ships (more on this later), the wave damping of floating bodies, etc.
Yet, the only general way of evaluating wave forces on floating bodies (moving or not) or solid boundaries is by applying the [[Wave Momentum Flux| Wave Momentum]].

-----
This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/courses/mechanical-engineering/2-24-ocean-wave-interaction-with-ships-and-offshore-energy-systems-13-022-spring-2002/lecture-notes/lecture4.pdf here]

Linear and Second-Order Wave Theory

2010-11-06T11:11:18Z

Adi Kurniawan: /* Kinematic condition */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = [[Linear and Second-Order Wave Theory]]
| next chapter = [[Linear Plane Progressive Regular Waves]]
| previous chapter = [[Conservation Laws and Boundary Conditions]]
}}

{{complete pages}}

We saw in [[Conservation Laws and Boundary Conditions]] that the potential flow model for wave propagation is given Laplace's equation plus the free-surface conditions. In this section we present the linear and second order theory for these equations. The linear theory is valid for small wave heights and the second order theory is an improvement on this. However, neither of these theories work for very steep waves and of course the potential theory breaks down once the wave begins to break and completely different methods are required in this situation.

== Linearization of Free-surface Conditions ==

We use [http://en.wikipedia.org/wiki/Perturbation_theory perturbation theory] to expand the solution as follows
<center><math>
\begin{align}
\zeta &= \zeta_1 + \zeta_2 + \zeta_3 + \cdots \\
\Phi &= \Phi_1 + \Phi_2 + \Phi_3 + \cdots
\end{align}
</math></center>
where we are assuming that there exists a small parameter (the wave slope) and that with respect to this the
<math>\Phi_i</math> is proportional to <math>\epsilon^i</math> or <math>O(\epsilon^i)</math>. We then derive the boundary value problem for <math> \zeta_i,\Phi_i </math>. Rarely we need to go beyond <math> i = 3 </math> (in fact it is unlikely that the terms beyond
this will improve the accuracy.

In this section we will only derive the free-surface conditions up to second order. Remember that <math>\nabla^2 \Phi_i =0</math> for all <math>i</math>
We expand the kinematic and dynamic free surface conditions about the <math>z=0</math> plane and derive statements for the unknown pairs <math> (\Phi_1,\zeta_1 </math> and <math> (\Phi_2, \zeta_2) </math> at <math> z=0 </math>. The same technique can be used to linearize the body boundary condition at <math> U=0 </math> (zero speed) and <math> U>0 </math> (forward speed).

== Kinematic condition ==

The fully non-linear kinematic condition was derived in [[Conservation Laws and Boundary Conditions]] and we begin with this equation
<center><math> \left ( \frac{\partial \zeta}{\partial t} + \nabla \Phi \cdot \nabla \zeta \right )_{z=\zeta} = \left ( \frac{\partial \Phi}{\partial z} \right )_{z=\zeta} </math></center>
We expand this equation about <math>\zeta = 0</math>, which we can do because we have assumed that the slope is small. In fact the slope is our parameter <math>\epsilon</math>. It is obvious at this point that the theory does not apply to very steep waves. This gives us the following equation
<center><math>
\left( \frac{\partial\zeta}{\partial t} + \nabla \Phi \cdot \nabla \zeta \right)_{z=0} + \zeta \frac{\partial}{\partial z} \left( \frac{\partial\zeta}{\partial t} + \nabla \Phi \cdot \nabla \zeta \right)_{z=0} + \;\cdots = \left( \frac{\partial\Phi}{\partial z} \right)_{z=0} + \zeta \left( \frac{\partial^2 \Phi}{\partial z^2} \right)_{z=0} + \;\cdots
</math></center>
where we have only taken the first order expansion. We then substitute our expressions
<center> <math>
\begin{align}
\zeta &= \zeta_1 + \zeta_2 + \cdots \\
\Phi &= \Phi_1 + \Phi_2 + \cdots
\end{align}
</math></center>
and keep terms of <math>\ O(\varepsilon), \ O(\varepsilon^2)</math>, remembering that <math>\zeta_1\Phi_1</math> is <math>O(\varepsilon^2)</math> etc.

== Dynamic condition ==

The fully non-linear Dynamic condition was derived in [[Conservation Laws and Boundary Conditions]] and is given by
<center><math>
\zeta (x,y,t) = -\frac{1}{g} \left( \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right)_{z=\zeta}
</math></center>

<center><math>
\left . \begin{matrix}
\zeta = \dfrac{1}{g} \left( \dfrac{\partial\Phi}{\partial t} + \dfrac{1}{2} \nabla\Phi \cdot \nabla\Phi \right)_{z=0}\\
\dfrac{1}{g} \zeta \dfrac{\partial}{\partial z} \left( \dfrac{\partial\Phi}{\partial t} + \dfrac{1}{2} \nabla\Phi \cdot \nabla\Phi \right)_{z=0} + \cdots
\end{matrix} \right \}
\begin{matrix}
\zeta = \zeta_1 +\zeta_2 + \cdots \\
\Phi = \Phi_1 + \Phi_2 + \cdots
\end{matrix}
</math></center>

== Linear problem ==

The linear problem is the <math>O(\varepsilon)</math> problem derived by equating the terms which are proportional to <math>\epsilon</math>.

This can be done straight forwardly and gives the following expressions

<center><math> \frac{\partial\zeta_1}{\partial t} = \frac{\partial\Phi_1}{\partial z} , \ z=0; </math></center>
which follows from the Kinematic equation and
<center><math> \zeta_1 = -\frac{1}{g} \frac{\partial\Phi_1}{\partial t}, \ z=0; </math></center>
which follows from the Dynamic equation. These are the linear free surface conditions.

=== Derivation using Bernoulli's equation ===

The pressure from Bernoulli, <math> \omega </math> constant terms set equal to zero, at a fixed point in the fluid domain at <math> \mathbf{x}=(x,y,z) </math> is given by:
<center><math> P = - \rho \left( \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \cdots \nabla\Phi + gz \right); </math></center>
When then make the perturbation expansion for the potential and the pressure
<center><math> \Phi = \Phi_1 + \Phi_2 + \cdots </math></center>
and
<center><math> P = P_0 + P_1 + P_2 + \cdots </math></center>.

This allows us to derive
<center><math> P_0 = -\rho g z \,, </math></center>
which is called the Hydrostatic pressure and
<center><math> P_1 = - \rho \frac{\partial\Phi_1}{\partial t} </math></center>
which is the linear pressure.

=== Classical linear free surface condition ===

If we eliminate <math> \zeta_1 </math> from the kinematic and dynamic free surface conditions, we obtain the classical linear free surface condition:
<center><math> \begin{cases}
\dfrac{\partial^2\Phi_1}{\partial t^2} + g \dfrac{\partial\Phi_1}{\partial z} = 0, \qquad z=0\\
\zeta_1 = - \dfrac{1}{g} \dfrac{\partial\Phi_1}{\partial t}, \qquad z=0
\end{cases} </math></center>

With:

<center><math> P_1 = - \rho \frac{\partial\Phi_1}{\partial t}, \qquad \mbox{at some fixed point} \ \mathbf{x} </math></center>

Note that on <math> z=0, \ P_1 \ne 0 </math> in fact it can obtained from the expressions above in the form

<center><math> P_1 = -\rho g \zeta_1, \qquad z=0 </math></center>

So linear theory states that the linear perturbation pressure on the <math> z=0 \, </math> plane due to a surface wave disturbance is equal to the positive (negative) "hydrostatic" pressure induced by the positive (negative) wave elevation <math> \zeta_1 \, </math>.

== Second-order problem ==

The second order equations can also be derived straight forwardly. The kinematic condition is
<center><math> \frac{\partial\zeta_2}{\partial t} + \nabla\Phi_1 \cdot \nabla\zeta_1 = \frac{\partial\Phi_2}{\partial z} + \zeta_1 \frac{\partial^2 \Phi_1}{\partial z^2}, \quad z=0 </math></center>
and the dynamic condition
<center><math> \zeta_2 = - \frac{1}{g} \left( \frac{\partial\Phi_2}{\partial t} + \frac{1}{2} \nabla\Phi_1 \cdot \nabla\Phi_1 \right)_{z=0} - \frac{1}{g} \zeta_1 \frac{\partial^2\Phi_1}{\partial z \partial t}, \quad z=0 </math></center>

Alternatively, the known linear terms may be moved in the right-hand side as forcing functions, leading to:

=== Kinematic second-order condition ===

<center>
<math> \frac{\partial\zeta_2}{\partial t} - \frac{\partial\Phi_2}{\partial z} = \zeta_1 \frac{\partial^2 \Phi_1}{\partial z^2} - \nabla\Phi_1 \cdot \nabla\zeta_1; \quad z=0 </math>
</center>

=== Dynamic second-order condition ===

<center>
<math> \zeta_2 + \frac{1}{g} \frac{\partial\Phi_2}{\partial t} = - \frac{1}{g} \left( \frac{1}{2} \nabla\Phi_1 \cdot \nabla\Phi_1 + \zeta_1 \frac{\partial^2\Phi_1}{\partial z \partial t} \right)_{z=0} </math>
</center>
where the second order pressure is given by
<center>
<math> P_2 = -\rho \left( \frac{\partial\Phi_2}{\partial t} + \frac{1}{2} \nabla\Phi_1 \cdot \nabla\Phi_1 \right); \quad \mbox{at} \ \mathbf{x} </math>
</center>

The very attractive feature of second order surface wave theory is that it allows the prior solution of the linear problem which is often possible analytically and numerically.
The linear solution is then used as a forcing function for the solution of the second order problem. This is often possible analytically and in most cases numerically in the absence or presence of bodies.
Linear and second-order theories are also very appropriate to use for the modeling of surface waves as stochastic processes.
Both theories are very useful in practice, particularly in connection with wave-body interactions.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/0B7683D3-9B31-453E-B98F-9F71A3C36C58/0/lecture2.pdf here]

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

[[Category:Linear Water-Wave Theory]]
[[Category:Nonlinear Water-Wave Theory]]

Linear and Second-Order Wave Theory

2010-11-06T11:07:15Z

Adi Kurniawan: /* Linearization of Free-surface Conditions */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = [[Linear and Second-Order Wave Theory]]
| next chapter = [[Linear Plane Progressive Regular Waves]]
| previous chapter = [[Conservation Laws and Boundary Conditions]]
}}

{{complete pages}}

We saw in [[Conservation Laws and Boundary Conditions]] that the potential flow model for wave propagation is given Laplace's equation plus the free-surface conditions. In this section we present the linear and second order theory for these equations. The linear theory is valid for small wave heights and the second order theory is an improvement on this. However, neither of these theories work for very steep waves and of course the potential theory breaks down once the wave begins to break and completely different methods are required in this situation.

== Linearization of Free-surface Conditions ==

We use [http://en.wikipedia.org/wiki/Perturbation_theory perturbation theory] to expand the solution as follows
<center><math>
\begin{align}
\zeta &= \zeta_1 + \zeta_2 + \zeta_3 + \cdots \\
\Phi &= \Phi_1 + \Phi_2 + \Phi_3 + \cdots
\end{align}
</math></center>
where we are assuming that there exists a small parameter (the wave slope) and that with respect to this the
<math>\Phi_i</math> is proportional to <math>\epsilon^i</math> or <math>O(\epsilon^i)</math>. We then derive the boundary value problem for <math> \zeta_i,\Phi_i </math>. Rarely we need to go beyond <math> i = 3 </math> (in fact it is unlikely that the terms beyond
this will improve the accuracy.

In this section we will only derive the free-surface conditions up to second order. Remember that <math>\nabla^2 \Phi_i =0</math> for all <math>i</math>
We expand the kinematic and dynamic free surface conditions about the <math>z=0</math> plane and derive statements for the unknown pairs <math> (\Phi_1,\zeta_1 </math> and <math> (\Phi_2, \zeta_2) </math> at <math> z=0 </math>. The same technique can be used to linearize the body boundary condition at <math> U=0 </math> (zero speed) and <math> U>0 </math> (forward speed).

== Kinematic condition ==

The fully non-linear kinematic condition was derived in [[Conservation Laws and Boundary Conditions]] and we begin with this equation
<center><math> \left ( \frac{\partial \zeta}{\partial t} + \nabla \Phi \cdot \nabla \zeta \right )_{z=\zeta} = \left ( \frac{\partial \Phi}{\partial z} \right )_{z=\zeta} </math></center>
We expand this equation about <math>\zeta = 0</math>, which we can do because we have assumed that the slope is small. In fact the slope is our parameter <math>\epsilon</math>. It is obvious at this point that the theory does not apply to very steep waves. This gives us the following equation
<center><math> \left( \frac{\partial\zeta}{\partial t} + \nabla \Phi \cdot \nabla \zeta \right)_{z=0} + \zeta \frac{\partial}{\partial z} \left( \frac{\partial\zeta}{\partial t} + \nabla \Phi \cdot \nabla \zeta \right)_{z=0} + \cdots </math></center> 
<center><math> = \left( \frac{\partial\Phi}{\partial z} \right)_{z=0} + \zeta \left( \frac{\partial^2 \Phi}{\partial z^2} \right)_{z=0} + \cdots </math></center>
where we have only taken the first order expansion. We then substitute our expressions
<center> <math>
\begin{matrix}
\zeta = \zeta_1 + \zeta_2 + \cdots \\
\Phi = \Phi_1 + \Phi_2 + \cdots
\end{matrix}
</math></center>
and keep terms of <math>\ O(\varepsilon), \ O(\varepsilon^2)</math>, remembering that <math>\zeta_1\Phi_1</math> is <math>O(\varepsilon^2)</math> etc.

== Dynamic condition ==

The fully non-linear Dynamic condition was derived in [[Conservation Laws and Boundary Conditions]] and is given by
<center><math>
\zeta (x,y,t) = -\frac{1}{g} \left( \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right)_{z=\zeta}
</math></center>

<center><math>
\left . \begin{matrix}
\zeta = \dfrac{1}{g} \left( \dfrac{\partial\Phi}{\partial t} + \dfrac{1}{2} \nabla\Phi \cdot \nabla\Phi \right)_{z=0}\\
\dfrac{1}{g} \zeta \dfrac{\partial}{\partial z} \left( \dfrac{\partial\Phi}{\partial t} + \dfrac{1}{2} \nabla\Phi \cdot \nabla\Phi \right)_{z=0} + \cdots
\end{matrix} \right \}
\begin{matrix}
\zeta = \zeta_1 +\zeta_2 + \cdots \\
\Phi = \Phi_1 + \Phi_2 + \cdots
\end{matrix}
</math></center>

== Linear problem ==

The linear problem is the <math>O(\varepsilon)</math> problem derived by equating the terms which are proportional to <math>\epsilon</math>.

This can be done straight forwardly and gives the following expressions

<center><math> \frac{\partial\zeta_1}{\partial t} = \frac{\partial\Phi_1}{\partial z} , \ z=0; </math></center>
which follows from the Kinematic equation and
<center><math> \zeta_1 = -\frac{1}{g} \frac{\partial\Phi_1}{\partial t}, \ z=0; </math></center>
which follows from the Dynamic equation. These are the linear free surface conditions.

=== Derivation using Bernoulli's equation ===

The pressure from Bernoulli, <math> \omega </math> constant terms set equal to zero, at a fixed point in the fluid domain at <math> \mathbf{x}=(x,y,z) </math> is given by:
<center><math> P = - \rho \left( \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \cdots \nabla\Phi + gz \right); </math></center>
When then make the perturbation expansion for the potential and the pressure
<center><math> \Phi = \Phi_1 + \Phi_2 + \cdots </math></center>
and
<center><math> P = P_0 + P_1 + P_2 + \cdots </math></center>.

This allows us to derive
<center><math> P_0 = -\rho g z \,, </math></center>
which is called the Hydrostatic pressure and
<center><math> P_1 = - \rho \frac{\partial\Phi_1}{\partial t} </math></center>
which is the linear pressure.

=== Classical linear free surface condition ===

If we eliminate <math> \zeta_1 </math> from the kinematic and dynamic free surface conditions, we obtain the classical linear free surface condition:
<center><math> \begin{cases}
\dfrac{\partial^2\Phi_1}{\partial t^2} + g \dfrac{\partial\Phi_1}{\partial z} = 0, \qquad z=0\\
\zeta_1 = - \dfrac{1}{g} \dfrac{\partial\Phi_1}{\partial t}, \qquad z=0
\end{cases} </math></center>

With:

<center><math> P_1 = - \rho \frac{\partial\Phi_1}{\partial t}, \qquad \mbox{at some fixed point} \ \mathbf{x} </math></center>

Note that on <math> z=0, \ P_1 \ne 0 </math> in fact it can obtained from the expressions above in the form

<center><math> P_1 = -\rho g \zeta_1, \qquad z=0 </math></center>

So linear theory states that the linear perturbation pressure on the <math> z=0 \, </math> plane due to a surface wave disturbance is equal to the positive (negative) "hydrostatic" pressure induced by the positive (negative) wave elevation <math> \zeta_1 \, </math>.

== Second-order problem ==

The second order equations can also be derived straight forwardly. The kinematic condition is
<center><math> \frac{\partial\zeta_2}{\partial t} + \nabla\Phi_1 \cdot \nabla\zeta_1 = \frac{\partial\Phi_2}{\partial z} + \zeta_1 \frac{\partial^2 \Phi_1}{\partial z^2}, \quad z=0 </math></center>
and the dynamic condition
<center><math> \zeta_2 = - \frac{1}{g} \left( \frac{\partial\Phi_2}{\partial t} + \frac{1}{2} \nabla\Phi_1 \cdot \nabla\Phi_1 \right)_{z=0} - \frac{1}{g} \zeta_1 \frac{\partial^2\Phi_1}{\partial z \partial t}, \quad z=0 </math></center>

Alternatively, the known linear terms may be moved in the right-hand side as forcing functions, leading to:

=== Kinematic second-order condition ===

<center>
<math> \frac{\partial\zeta_2}{\partial t} - \frac{\partial\Phi_2}{\partial z} = \zeta_1 \frac{\partial^2 \Phi_1}{\partial z^2} - \nabla\Phi_1 \cdot \nabla\zeta_1; \quad z=0 </math>
</center>

=== Dynamic second-order condition ===

<center>
<math> \zeta_2 + \frac{1}{g} \frac{\partial\Phi_2}{\partial t} = - \frac{1}{g} \left( \frac{1}{2} \nabla\Phi_1 \cdot \nabla\Phi_1 + \zeta_1 \frac{\partial^2\Phi_1}{\partial z \partial t} \right)_{z=0} </math>
</center>
where the second order pressure is given by
<center>
<math> P_2 = -\rho \left( \frac{\partial\Phi_2}{\partial t} + \frac{1}{2} \nabla\Phi_1 \cdot \nabla\Phi_1 \right); \quad \mbox{at} \ \mathbf{x} </math>
</center>

The very attractive feature of second order surface wave theory is that it allows the prior solution of the linear problem which is often possible analytically and numerically.
The linear solution is then used as a forcing function for the solution of the second order problem. This is often possible analytically and in most cases numerically in the absence or presence of bodies.
Linear and second-order theories are also very appropriate to use for the modeling of surface waves as stochastic processes.
Both theories are very useful in practice, particularly in connection with wave-body interactions.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/0B7683D3-9B31-453E-B98F-9F71A3C36C58/0/lecture2.pdf here]

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

[[Category:Linear Water-Wave Theory]]
[[Category:Nonlinear Water-Wave Theory]]

Conservation Laws and Boundary Conditions

2010-11-06T11:06:17Z

Adi Kurniawan: /* Gauss theorem */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Conservation Laws and Boundary Conditions
| next chapter = [[Linear and Second-Order Wave Theory]]
| previous chapter =
}}

{{complete pages}}

We begin by deriving the equations of motion used to model ocean waves. The equations of motion of a general fluid are given by the celebrated [http://en.wikipedia.org/wiki/Navier_Stokes Navier Stokes equations]. However, for the large scale processes that occur in ocean waves many simplifications are possible.

== Coordinate system and velocity potential ==

[[Image:Coordinate_system.png|right|thumb|500px|Coordinate System]]

We begin by defining the coordinate system.

<center><math>
\begin{matrix}
&(x,y,z) &: &\mbox{Coordinate system} \\
&\mathbf{x} &: &\mbox{Fixed Eulerian Vector} \\
& \mathbf{v} &: &\mbox{Flow Velocity Vector at} \ \mathbf{x} \\
&\zeta &: &\mbox{Free Surface Elevation} \\
&\mathbf{g} &: &\mbox{Acceleration due to gravity}
\end{matrix} </math></center>

At the moment we have not yet defined the region of space which is occupied by the fluid. However, the free surface plays a very important role in the propagation of ocean waves.

The most important assumption we make is that the fluid is an [http://en.wikipedia.org/wiki/Viscosity ideal fluid], i.e. there are no shear stresses due to viscosity and that the flow is [http://en.wikipedia.org/wiki/Irrotational irrotational]. This means that

<center><math>\nabla \times \mathbf{v} = 0</math></center>

We now introduce the very important concept of the velocity potential. Essentially this allows us to express the solution as a function of a scalar (a function which has a single value as opposed to a vector function which has multiple values) rather than a vector throughout the fluid domain. There is an important theorem in vector calculus [http://en.wikipedia.org/wiki/Irrotational_vector_field] that if <math>\nabla \times \mathbf{v} = 0</math> then we can express the irrotational vector as the gradient of a scalar function, i.e.
<center><math>
\mathbf{v} = \nabla \Phi
</math></center>
where <math>\Phi(\mathbf{x},t)</math> is called the [http://en.wikipedia.org/wiki/Velocity_potential velocity potential].

It turns out that the potential flow model of surface wave propagation and wave-body interactions is very accurate for the kind of length and time scales which are important in the ocean. It is important to realise, however, that we have made considerable simplifications and that certain processes, most notably wave breaking, are in no way covered by this theory. In fact, the process of wave breaking is extremely complicated and is much less well understood than the potential flow model.

== Conservation of mass ==

The key equation we will solve to understand ocean waves is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation] which is derived very simply from the expression for the velocity potential. We assume that the fluid is incompressible so that conservation of mass gives us the following condition
<center><math> \nabla \cdot \mathbf{v} = 0 </math></center>
This condition in turn implies, using the definition of the velocity potential that
<center><math>
\nabla \cdot \nabla \Phi = 0 \Rightarrow \nabla^2 \Phi = 0
</math></center>
or
<center><math>
\partial_x^2 \Phi + \partial_y^2\Phi + \partial_z^2\Phi = 0,
</math></center>
which is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation].

== Conservation of linear momentum ==

We begin with [http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29 Euler's equation] in the absence of viscosity
<center> <math>
\partial_t \mathbf{v} + (\mathbf{v}\cdot \nabla)\mathbf{v}= - \frac1{\rho} \nabla P + \mathbf{g}
</math> </center>
where
<math> P(\mathbf{x}, t) </math> is the fluid Pressure at <math>(\mathbf{x}, t)</math> and
<math> \mathbf{g}= - \mathbf{k} g </math> is the acceleration due to gravity where
<math> \mathbf{k} </math> is the unit vector pointing in the positive <math>z</math>-direction (so we are now setting the <math>z</math> coordinate to point in the vertical direction). Finally <math> \rho </math> is the water density.

We then use the following vector identity
<center><math> (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac 1{2} \nabla (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v}\times ( \nabla \times \mathbf{v}) </math></center>
and since we have irrotational flow (i.e. <math> \nabla \times \mathbf{v}= 0 </math>) Euler's equation becomes
<center><math> \partial_t \mathbf{v} + \frac{1}{2} \nabla (\mathbf{v} \cdot \mathbf{v}) = - \frac 1{\rho} \nabla P - \nabla (g z) </math></center>
where we have used <math> \nabla z = \mathbf{k} </math>.
We now substitute <math> \mathbf{v}= \nabla \Phi </math> and we obtain
<center><math> \nabla (\partial_t \Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z ) = 0. </math></center>
We now observe that if
<center><math> \nabla F( \mathbf{x}, t) =0 \quad \Longrightarrow \quad F (\mathbf{x}, t) = C </math></center>
where <math> C </math> is an arbitrary constant.

==== Bernoulli's equation ====

[http://en.wikipedia.org/wiki/Bernoulli%27s_equation Bernoulli's equation] follows from the equation above.
<center><math> \partial_t \Phi + \frac 1{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z = C </math></center>
or
<center><math> \frac{P}{\rho} = - \partial_t\Phi-\frac{1}{2}\nabla\Phi \cdot \nabla \Phi - g z + C </math></center>

The value of the constant <math> C </math> is immaterial (it can be thought of as defining the reference pressure.
It is also worth noting that the
angular momentum conservation principle is contained in
<math> \nabla \times \mathbf{v} = 0. </math>
In particular, if the particles are modelled as spheres, this equation implies no angular velocity at all times.

== Derivation of nonlinear free-surface condition ==

A very important result is the boundary condition at the free surface of the fluid and air. There are two conditions which relate the free surface displacement <math>\zeta(x,y,t)</math> and the velocity potential <math>\Phi(x,y,z,t)</math> at the free surface. The dynamic condition is derived from the Bernoulli's equation and the kinematic condition is derived from the equations linking the fact that the velocity vector at the surface is given by the gradient of the potential. We will present two methods to derive these equations.

=== Method I ===

We derive the dynamic condition directly from Bernoulli's equation.
On <math> z=\zeta; \ P=P_a \equiv \mbox{Atmospheric Pressure} </math>.
This allows us to rewrite Bernoulli's equation as
<center><math> \frac{P_a}{\rho}=-\frac{\partial\Phi}{\partial t}-\frac{1}{2}\nabla\Phi\cdot\nabla\Phi-g\zeta+C \qquad \mbox{on} \ z=\zeta(x,y,t) </math></center>
We will simplify this equation by showing that we are free to set the pressure to any value.

The kinematic condition is derived as follows.
On <math>z=\zeta</math> The mathematical function
<center><math>z-\zeta(x,y,t)\equiv\tilde{f}(x,y,z,t)</math></center>

is always zero when tracing a fluid particle on the free surface. So the [http://en.wikipedia.org/wiki/Total_derivative substantial or total derivative] of <math>\tilde{f}</math> must vanish, thus

<center><math> \frac{D\tilde{f}}{Dt}=0=\left (\partial_t + \mathbf{v} \cdot \nabla \right ) \tilde{f}=0, \qquad \mbox{on} \ z=\zeta </math></center>

Expanding, we obtain
<center><math> \left (\partial_t + \mathbf{v} \cdot \nabla \right ) (z-\zeta) =0, \qquad \mbox{on} \ z=\zeta </math></center>
which in turn implies that
<center><math> \partial_t\zeta + \partial_x\Phi \partial_x\zeta + \partial_y\Phi \partial_y\zeta = \partial_z\Phi, \qquad \mbox{on} \ z=\zeta </math></center>
which is the Kinematic free-surface condition.

We have already derived the dynamic condtion from Bernoulli's equation
<center>
<math> \partial_t\Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + g \zeta = C - \frac{P_a}{\rho}, \qquad \mbox{on} \ z=\zeta </math></center>

Constants in Bernoulli's equation may be set equal to zero when we are eventually interested in integrating pressures over closed or open boundaries (floating or submerged bodies) to obtain forces & moments. This follows from a simple application of one of the two Gauss vector theorems.

==== Gauss theorem ====

[[Image:Force_coordinates2.png|right|thumb|500px|Force coordinates]]

We need to use the following theorems often called [http://en.wikipedia.org/wiki/Gauss_theorem Gauss theorem] although more properly known as the divergence theorem.
We begin with the vector version. If
<math> \mathbf{n} </math> is the unit normal vector pointing inside the volume <math> V </math> with surface <math>S</math> and
<math> f(\mathbf{x})</math> is an arbitrary sufficiently differentiable scalar function, then
<center>
<math> \iiint_{\Omega} \nabla f \mathrm{d}v = -\iint_{\partial\Omega} f_{s} \mathbf{n} \mathrm{d}s </math>
</center>

Note the three scalar identities that follow:
<center><math>
\begin{align}
\iiint_{{\Omega}} \partial_x f \mathrm{d}v &= - \iint_{\partial\Omega} f n_1 \mathrm{d}s \\
\iiint_{{\Omega}} \partial_y f \mathrm{d}v &= - \iint_{\partial\Omega} f n_2 \mathrm{d}s \\
\iiint_{{\Omega}} \partial_z f \mathrm{d}v &= - \iint_{\partial\Omega} f n_3 \mathrm{d}s .
\end{align}
</math></center>

The scalar version is as follows where
<math> \mathbf{v}</math> is an arbitrary sufficiently differentiable vector function
<center>
<math> \iiint_{\Omega} \nabla \cdot \mathbf{v} = - \iint_{\partial\Omega} \mathbf{v} \cdot \mathbf{n} \mathrm{d}s </math></center>
The scalar identity is often used to prove mass conservation principle.

==== Definition of force and moment in terms of fluid pressure ====

The force <math>\mathbf{F}</math> is given by
<center><math>
\mathbf{F} = \iint_{\partial\Omega} P\mathbf{n}\mathrm{d}s
</math></center>
where <math>P</math> is the pressure and the moment <math>\mathbf{M}</math> is given by
<center><math>
\mathbf{M} = \iint_{\partial\Omega} P(\mathbf{x}\times\mathbf{n})\mathrm{d}s
</math></center>

It follows from the Gauss theorem that if <math> P = C </math> the force and moment over a closed boundary S vanish identically. Hence without loss of generality in the context of wave body interactions we will set <math> C=0 </math>. It follows that the dynamic free surface condition takes the form

<center><math> \zeta (x,y,t) = - \frac{1}{g} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right \}, \qquad z=\zeta </math></center>

=== Method II ===

When tracing a fluid particle on the free surface the hydrodynamic pressure given by Bernoulli (after the constant <math>C</math> has been set equal to zero) must vanish as we follow the particle:

<center><math> \frac{D}{Dt} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi+gz \right \} =0, \qquad z=\zeta </math></center>
or
<center><math> \left ( \partial_t + \mathbf{v} \cdot \nabla \right ) \left ( \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi +gz \right ) =0, \qquad z=\zeta </math></center>

This condition also follows upon elimination of <math>\zeta</math> from the kinematic & dynamic conditions derived under method I.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/1814747D-3A05-45A1-BDE5-2CEF40DEA25F/0/lecture1.pdf here]

[[Category:Linear Water-Wave Theory]]

Template:Frequency domain equations for a floating plate

2010-11-06T11:03:02Z

Adi Kurniawan:

If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to
<math>\exp (-\mathrm{i}\omega t)\,</math> the equations become
<center><math>
\begin{align}
-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 \\
\rho g\zeta - \mathrm{i}\omega\rho \phi &= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, &z=0 \\
\Delta \phi &= 0, &-h<z<0 \\
\partial_z \phi &= 0, &z=-h,
\end{align}
</math></center>
where
<math>\zeta</math> is the surface displacement and <math>\phi</math> is the velocity potential in the frequency domain.

These equations can be simplified by defining <math>\alpha = \omega^2/g</math>,
<math>\beta = D/\rho g</math> and <math>\gamma = \rho_i h/\rho</math> to obtain
<center><math>
\begin{align}
\Delta \phi &= 0, &-h < z \leq 0 \\
\partial_z \phi &= 0, &z = - h \\
\beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{\alpha}\phi, &z = 0 \\
-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 .
\end{align}
</math></center>

Template:Incident plane wave 2d definition

2010-11-06T10:53:50Z

Adi Kurniawan:

<math>\phi^{\mathrm{I}}\,</math>
is a plane wave travelling in the <math>x</math> direction,
<center><math>
\phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \,
</math></center>
where <math>A </math> is the wave amplitude (in potential) <math>\mathrm{i} k </math> is
the positive imaginary solution of the [[Dispersion Relation for a Free Surface]]
(note we are assuming that the time dependence is of the form <math>\exp(-\mathrm{i}\omega t) </math>)
and
<center><math>
\phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h}
</math></center>

Template:General body boundary condition

2010-11-06T10:52:33Z

Adi Kurniawan:

<center><math>
\partial_n\phi = \mathcal{L}\phi, \quad \mathbf{x}\in\partial\Omega_B,
</math></center>
where <math>\mathcal{L}</math> is a linear
operator which relates the normal and potential on the body surface through the physics
of the body.

Template:Standard linear wave scattering equations without body condition

2010-11-06T10:40:05Z

Adi Kurniawan:

<center><math>
\begin{align}
\Delta\phi &=0, &-h<z<0,\,\,\mathbf{x} \in \Omega \\
\partial_z\phi &= 0, &z=-h, \\
\partial_z \phi &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}},
\end{align}
</math></center>

(note that the last expression can be obtained from combining the expressions:
<center><math>
\begin{align}
\partial_z \phi &= -\mathrm{i} \omega \zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \\
\mathrm{i} \omega \phi &= g\zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}},
\end{align}
</math></center>

where <math>\alpha = \omega^2/g \,</math>)

Burgers Equation

2010-11-06T10:23:40Z

Adi Kurniawan: /* Exact Solution of Burgers equations */

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u</math> and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-vu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\dfrac{\mathrm{d}u}{\mathrm{d}\xi } &=&w \\
\dfrac{\mathrm{d}w}{\mathrm{d}\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime }=c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu^{\prime
}-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u^{\prime }-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} .
We note that the wave speed
is
<center><math>
v=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik\left( u^{2}\right) -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn to get
<center><math>
\partial _{t}u=\partial _{x}\left( u^{2}\right)
</math></center>
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>

{| class="wikitable"
|-
! Phase plane for a travelling wave solution
! Numerical solution of Burgers equation
|-
| [[Image:Burgers_phase.jpg|thumb|right|500px|Phase plane for a travelling wave solution of Burgers equation]]
| [[Image:File-Burgers2.gif|thumb|right|500px| Numerical solution of Burgers equation]]

|}

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}+\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{\mathrm{d}}{\mathrm{d}x}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( s\right) \mathrm{d}s\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] \mathrm{d}y
\end{matrix}</math></center>
Which can be expressed as
<center><math>
\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] \mathrm{d}y
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) \mathrm{d}s+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \dfrac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&\dfrac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\dfrac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] \mathrm{d}y}
\end{matrix}</math></center>

Burgers Equation

2010-11-06T10:22:33Z

Adi Kurniawan: /* Exact Solution of Burgers equations */

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u</math> and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-vu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\dfrac{\mathrm{d}u}{\mathrm{d}\xi } &=&w \\
\dfrac{\mathrm{d}w}{\mathrm{d}\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime }=c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu^{\prime
}-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u^{\prime }-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} .
We note that the wave speed
is
<center><math>
v=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik\left( u^{2}\right) -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn to get
<center><math>
\partial _{t}u=\partial _{x}\left( u^{2}\right)
</math></center>
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>

{| class="wikitable"
|-
! Phase plane for a travelling wave solution
! Numerical solution of Burgers equation
|-
| [[Image:Burgers_phase.jpg|thumb|right|500px|Phase plane for a travelling wave solution of Burgers equation]]
| [[Image:File-Burgers2.gif|thumb|right|500px| Numerical solution of Burgers equation]]

|}

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}+\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{\mathrm{d}}{\mathrm{d}x}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( s\right) \mathrm{d}s\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] \mathrm{d}y
\end{matrix}</math></center>
Which can be expressed as
<center><math>
\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] \mathrm{d}y
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) \mathrm{d}s+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \frac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&\frac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\frac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] \mathrm{d}y}
\end{matrix}</math></center>

Burgers Equation

2010-11-06T10:21:18Z

Adi Kurniawan: /* Travelling Wave Solution */

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u</math> and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-vu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\dfrac{\mathrm{d}u}{\mathrm{d}\xi } &=&w \\
\dfrac{\mathrm{d}w}{\mathrm{d}\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime }=c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu^{\prime
}-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u^{\prime }-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} .
We note that the wave speed
is
<center><math>
v=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik\left( u^{2}\right) -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn to get
<center><math>
\partial _{t}u=\partial _{x}\left( u^{2}\right)
</math></center>
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>

{| class="wikitable"
|-
! Phase plane for a travelling wave solution
! Numerical solution of Burgers equation
|-
| [[Image:Burgers_phase.jpg|thumb|right|500px|Phase plane for a travelling wave solution of Burgers equation]]
| [[Image:File-Burgers2.gif|thumb|right|500px| Numerical solution of Burgers equation]]

|}

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}+\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{d}{dx}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( s\right) ds\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] dy
\end{matrix}</math></center>
Which can be expressed as
<center><math>
\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] dy
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) ds+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \frac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&\frac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\frac{f}{2\nu }\right] dy}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] dy}
\end{matrix}</math></center>

Reaction-Diffusion Systems

2010-11-06T10:19:55Z

Adi Kurniawan: /* Diffusion */

{{nonlinear waves course
| chapter title = Reaction-Diffusion Systems
| next chapter = [[Burgers Equation]]
| previous chapter = [[Example Calculations for the KdV and IST]]
}}

{{complete pages}}

We present here a brief theory of reaction diffusion waves.

== Law of Mass Action ==

The law of mass action states that equation rates are proportional to the concentration
of reacting species and the ratio in which they combined. It is discussed in detail in
[[Billingham and King 2000]]. We will present here a few simple examples.

=== Example 1: Simple Decay ===

Suppose we have of chemical <math>P</math> which decays to <math>A</math>, i.e.
<center>
<math>P \to A</math>
</center>
with rate <math>k[P]</math> where <math>[P]</math> denotes concentration. Then if we
set <math>p=[P]</math> and <math>a = [A] </math> we obtain the equations
<center>
<math> \frac{\mathrm{d}p}{\mathrm{d}t} = -kp\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}a}{\mathrm{d}t} = kp</math>
</center>
which has solution
<center>
<math>
p = p_0 e^{-kt}\,\,\,\textrm{and}\,\,\, a = a_0 + p_0(1-e^{-kt})
</math>
</center>
where <math>a_0</math> and <math>p_0</math> are the values of <math>a</math>
<math>p</math> repectively at <math>t=0</math>.

=== Example 2: Quadratic Autocatalysis ===

This example will be important when we consider reaction diffusion problems.
We consider the reaction
<center>
<math>A + B \to 2B</math>
</center>
with rate proportional to <math>k[A][B]</math>. If we define <math>a = [A]</math>
and <math>b = [B]</math> we obtain the following equations
<center>
<math> \frac{\mathrm{d}a}{\mathrm{d}t} = -kab\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}b}{\mathrm{d}t} = kab</math>
</center>
We can solve these equations by observing that
<center>
<math>
\frac{\mathrm{d}(a+b)}{\mathrm{d}t} = 0
</math>
</center>
so that <math>a + b = a_0 + b_0</math>. We can then eliminate <math>a</math> to obtain
<center>
<math>
\frac{\mathrm{d}b}{\mathrm{d}t} = k(a_0 + b_0 - b)b
</math>
</center>
which is separable with solution
<center>
<math>
b = \frac{b_0(a_0 + b_0)e^{k(a_0 + b_0)t}}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
and
<center>
<math>
a = \frac{a_0(a_0 + b_0)}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
Note that <math>a\to 0</math> and <math>b\to a_0 + b_0</math>
as <math>t\to \infty.</math>

== Diffusion ==

The equation for spatially homogeneous diffusion of a chemical with concentration
<math>c</math> is
<center>
<math>
\partial_t c = D\nabla^2 c
</math>
</center>
which is the [http://en.wikipedia.org/wiki/Heat_equation heat equation]. We will consider this in
only one spatial dimension. Consider it on the boundary <math>-\infty < x < \infty</math>. In this case
we can solve by the [http://en.wikipedia.org/wiki/Fourier_transform Fourier transform] and obtain
<center>
<math>
\partial_t \hat{c} = -D k^2 \hat{c}
</math>
</center>
where <math>\hat{c}</math> is the Fourier transform of <math>c</math>. This has solution
<center>
<math>
\hat{c} = \hat{c}_0 e^{-D k^2 t}
</math>
</center>
We can find the inverse transform using convolution and obtain
<center>
<math>
c(x,t) = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{(x-s)^2/4Dt}\mathrm{d}s
</math>
</center>

=== Solution of the dispersion equation using FFT ===

We can solve the dispersion equation using the discrete Fourier transform and
its closely related numerical implementation the FFT (Fast Fourier Transform).
We have already met the FFT [[Numerical Solution of the KdV]] but we consider it here in more detail.
We consider the concentration
on the finite domain <math>-L \leq x \leq L</math> and use a
[http://en.wikipedia.org/wiki/Fourier_series Fourier series] expansion
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(t) e^{\mathrm{i} k_n x}
</math>
</center>
where <math>k_n = \pi n /L </math>. If we substitute this into the diffusion equation we obtain
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(0)e^{-k_n^2 D t} e^{\mathrm{i} k_n x}
</math>
</center>
Note that this is not the same solution as we obtained on the infinite domain because
of the boundary conditions on the finite domain. The coefficients <math>\hat{c}_n(0)</math>
are found using the initial conditions so that
<center>
<math>
\hat{c}_n(0) = \frac{1}{2L} \int_{-L}^{L} e^{-\mathrm{i} k_n x} c_0(x) \mathrm{d}x
</math>
</center>

The key to the numerical solution of this equation is the use of the FFT. We begin by discretising the
domain into a series of <math>N</math> points <math>x_m = -L + 2Lm/N </math>. We then use this to
approximate the integral above and obtain
<center>
<math>
\hat{c}_n(0) = \frac{1}{N} \sum_{m=0}^{N-1} e^{-\mathrm{i} k_n x_m} c_0(x_m)
</math>
</center>
<center>
<math>
= \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} e^{\mathrm{i} \pi n} c_0(x_m)
</math>
</center>

We also get
<center>
<math>
c(x_m,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(0) e^{-k_n^2 D t} e^{\mathrm{i} k_n x_m}
</math>
</center>
<center>
<math>
= \sum_{n=-\infty}^{\infty} \hat{c}_n(0) e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N} e^{-\mathrm{i} \pi n}
</math>
</center>
but we know that
<center>
<math>
\hat{c}_n(0) e^{-\mathrm{i} \pi n} = \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} c_0(x_m)
</math>
</center>

=== The discrete Fourier transform ===
The
[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]
of a sequence of ''2N'' complex numbers ''c''0, ..., ''c''''2N''−1 is transformed into the sequence of ''N'' complex numbers <math>\hat{c}</math>0, ..., <math>\hat{c}</math>''N''−1 by the DFT according to the formula:
<center>
<math>\hat{c}_m = \sum_{n=0}^{N-1} c_n e^{-2\pi \mathrm{i}mn/N} \quad \quad m = 0, \dots, N-1</math>
</center>

We denote the transform by the symbol <math>\mathcal{F}</math>, as in <math>\mathbf{X} = \mathcal{F} \left \{ \mathbf{x} \right \} </math> or <math>\mathcal{F} \left ( \mathbf{x} \right )</math> or <math>\mathcal{F} \mathbf{x}</math>.

The '''inverse discrete Fourier transform (IDFT)''' is given by
<center>
<math>c_n = \frac{1}{2N} \sum_{m=0}^{N-1} \hat{c}_m e^{2\pi \mathrm{i}mn/N} \quad \quad n = 0,\dots,N-1.</math>
</center>

Therefore we can write
<center>
<math>
c(x_m,t) = \mathcal{F} \left\{ e^{-k_n^2 D t} \mathcal{F}^{-1} \left\{ c_0(x_m) \right\}
\right\}
</math>
</center>

The only difficulty is that we need to define carefully the values of
<math>k_n</math>

The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]
or '''FFT''' algorithm. A naive implementation of the discrete Fourier transform above (or its inverse)
will involve order <math>N^2</math> operations. Using FFT algorithms, this can be reduced to order <math>N \log(N)</math>. This is an incredible speed up, for example
if N = 1024, FFT algorithms are more efficient by a factor of 147. This is the reason FFT
algorithms are used so extensively.

== Reaction Diffusion Equations ==

We consider an auto catalytic reaction where the chemical species also diffuse. In this
case the equations are
<center>
<math>\partial_t a = D\partial_x^2 a - kab</math>
</center>
<center>
<math>\partial_t b = D\partial_x^2 b + kab</math>
</center>
We can non-dimensionalise these equations scaling the variables as
<center>
<math>
z = x/x^*\,\,\,\tau = t/t^*\,\,\,\alpha = a/a_0\,\,\,\beta = b/a_0
</math>
</center>
So that the equations become
<center>
<math>
\frac{1}{t^*}\partial_\tau \alpha = \frac{a_0}{(x^*)^2}D\partial_z^2 \alpha
- k a_0^2 \alpha\beta
</math>
</center>
<center>
<math>
\frac{1}{t^*}\partial_\tau \beta = \frac{a_0}{(x^*)^2}D\partial_z^2 \beta
+ k a_0^2 \alpha\beta
</math>
</center>
If we choose
<center>
<math>
x^* = \sqrt{\frac{D}{ka_0}}\,\,\,t^*=\frac{1}{ka_0}
</math>
</center>
then we obtain the system
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
-\alpha\beta
</math>
</center>
<center>
<math>
\partial_\tau \beta =\partial_z^2 \beta
+ \alpha\beta
</math>
</center>

=== Solution via split step method ===

[[Image:Reaction diffusion.gif|thumb|right|500px|Solutions for <math>\alpha(z,0) =1
\, \beta(z,0) = \exp(-10z^2)</math>]]

We can solve this equations numerically using a
[http://en.wikipedia.org/wiki/Split-step_method split step method]. We assume
that at time <math>\tau</math> we know <math>\alpha(z,\tau)</math>
and <math>\beta(z,\tau)</math>. We then solve first the following equation
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
</math>
</center>
from <math>\tau</math> to <math>\tau + \Delta\tau</math>
(which we can do exactly using the spectral methods just discussed for
the dispersion equation). We write this solution as
<math>\tilde{\alpha}(z,\tau + \Delta\tau)</math>
Then we solve
<center>
<math>
\partial_\tau \alpha = -\alpha\beta
</math>
</center>
by assuming that <math>\beta</math> is constant and subject to the boundary
condition that <math>\alpha(z,\tau) = \tilde{\alpha}(z,\tau + \Delta\tau)</math>.
This gives
<center>
<math>
\alpha(z,\tau + \Delta\tau) = e^{-\beta(z,\tau) \Delta\tau} \tilde{\alpha}(z,\tau+ \Delta\tau)\,
</math>
</center>
and we do likewise for the equation for <math>\beta</math>. Note that
while both steps are exact the result from the split step method is an
approximation with error which becomes smaller as the step size becomes
smaller.

We can easily implement this split step method in matlab and we obtain
a pair of travelling waves.

== Travelling Waves solution ==

When we solve the equations we found the solution formed travelling waves and
we now consider this phenomena in detail.

We define a new coordinate <math>y = z - v\tau</math> (so we will consider only
waves travelling to the right, although we could analyse waves travelling to
the left in a similar fashion). We seek stationary solutions in
<math>\alpha(y)</math> and <math>\beta(y)</math> which satisfy
<center>
<math>
\frac{\mathrm{d}^2 \alpha}{\mathrm{d}y^2} + v \frac{\mathrm{d} \alpha}{\mathrm{d}y} = \alpha\beta
</math>
</center>
and
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} = -\alpha\beta
</math>
</center>
If we add these equations we obtain
<center>
<math>
\frac{\mathrm{d}^2 (\alpha+\beta)}{\mathrm{d}y^2} + v \frac{\mathrm{d} (\alpha+\beta)}{\mathrm{d}y} = 0
</math>
</center>
so that <math>\alpha + \beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions
are that as <math>y\to\infty </math> <math>\alpha = 1</math> and
<math>\beta = 0</math>. We also require the solution to be bounded
as <math>y\to-\infty</math> so that <math>\alpha + \beta = 1</math>.
This means that, since <math>\alpha \geq 0</math>, we must have
<math>0\leq \beta \leq 1</math>.
We can then obtain the following equation
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} + \beta(1-\beta)= 0
</math>
</center>
which we can write as the system of first order equations.

[[Image:R_d_phase_portrait.jpg|thumb|right|500px|Phase portrait for out system showing
the equilibrium points and the heteroclinic connection]]

We define the variable <math>\gamma = \frac{\mathrm{d}\beta}{\mathrm{d}y}</math>
and we obtain
<center><math>
\begin{align}
\frac{\mathrm{d}\beta}{\mathrm{d} y} &= \gamma&\\
\frac{\mathrm{d}\gamma}{\mathrm{d} y} &= -v\gamma + \beta(\beta -1)& \\
\end{align}
</math></center>
This dynamical system has equilibrium points at <math>(0,0)</math>
and <math>(1,0)</math>. We can analyse these equilibrium points by
linearization. The Jacobian matrix is
<center>
<math>
J =\begin{pmatrix}
0 & 1 \\
-1 + 2\beta & -v
\end{pmatrix}
</math>
</center>

We can easily see that the Jacobian evaluated at our first equilibrium point is
<center>
<math>
J_{(0,0)} =\begin{pmatrix}
0 & 1 \\
-1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\mu_{\pm} = -1/2 (v \mp \sqrt{v^2-4})</math>. Therefore
this point is a nodal sink (possibly a spiral)

[[Image:R_d_wave.gif|right|Travelling wave solution for v=2 and the position on
the heteroclinic connection]]

Additionally,
<center>
<math>
J_{(1,0)} =\begin{pmatrix}
0 & 1 \\
1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\lambda_{\pm} = -1/2 (v \mp \sqrt{v^2+4})</math>.
This is a a saddle point. The unstable and stable
separatrices leave the equilibrium point at <math>(1,0)</math> in the directions
<math> \begin{pmatrix}\lambda_{\pm} \\ 1\end{pmatrix}</math>. The only path on which <math>\beta</math> is bounded
as <math>y\to-\infty</math> are the unstable separatrices. Also, only the
unstable separatrix which enters the region <math>\beta<1</math> is physically meaningful.

To find a travelling wave we need to find a heteroclinic connection
between the two equilibrium points which also has to satisfy the conditions
that <math>0\leq \beta \leq 1</math>.

We need to show that the heteroclinic connection does not cross the <math>\beta</math> axis.
Consider the region
<center>
<math>
R = \left\{(\beta,\gamma)\,|\, \beta<1,\,-k\beta<\gamma<0\right\}
</math>
</center>
On the line <math>\beta = 1,d\beta/dy<0</math> and hence all flow it into <math>R</math>.
On the line <math>\gamma = 0, d\gamma/dy < 0</math> for <math>0<\beta<1</math>. On the line
<math>\gamma = -k \beta</math> we know that <math>d\beta/dy < 0</math> so that integral paths
enter the region if and only if <math>d\gamma/d\beta < \gamma/\beta</math>. We know that
<center>
<math>
d\gamma/d\beta - \gamma/\beta = -v - \frac{\beta(1-\beta)}{\gamma} -\frac{\gamma}{\beta}
= \frac{1}{k} (k^2 - vk +1 -\beta)
</math>
</center>
when <math>\gamma = -k\beta</math>. Therefore we need to find a value of <math>k</math>
so that <math>k^2 - vk +1 < 0</math>, which is possible provided <math>v\geq 2</math>, for example
<math>k = \dfrac{v}{2}</math>.
[[Category:Simple Nonlinear Waves]]

Numerical Solution of the KdV

2010-11-06T10:12:09Z

Adi Kurniawan: Reverted edits by Adi Kurniawan (Talk) to last revision by Meylan

{{nonlinear waves course
| chapter title = Numerical Solution of the KdV
| next chapter = [[Conservation Laws for the KdV]]
| previous chapter = [[Introduction to KdV]]
}}

== Introduction ==

We present here a method to solve the KdV equation numerically. There are
many different methods to solve the KdV and we use here a spectral method
which has been found to work well. Spectral methods work by using the
Fourier transform (or some varient of it) to calculate the derivative.

==Fourier Transform==

Recall that the Fourier transform is given by
<center><math>
\mathcal{F}\left[ f(x)\right] =\hat{f}\left( k\right) =\int_{-\infty
}^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
</math></center>
and the inverse Fourier transform is given by
<center><math>
f\left( x\right) =\mathcal{F}^{-1}\left[ \hat{f}\left( k\right) \right] =
\frac{1}{2\pi }\int_{-\infty }^{\infty }\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
(note that there are other ways of writing this transform). The most
important property of the Fourier transform is that
<center><math>
\begin{matrix}
\int_{-\infty }^{\infty }\left( \partial _{x}f\left( x\right) \right)
\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)
\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\
&=&-k\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
\end{matrix}
</math></center>
where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm
\infty .</math> This means that the Fourier transform converts differentiation to
multiplication by <math>k</math>.

==Solution for the Linearized KdV==

We begin with a simple example. Suppose we want to solve the linearized KdV
equation
<center><math>
\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty <x<\infty
</math></center>
subject to solve initial conditions
<center><math>
u\left( x,0\right) =f\left( x\right)
</math></center>
We can solve this equation by taking the Fourier transform. and obtain
<center><math>
\partial _{t}\hat{u}-ik^{3}\hat{u}=0
</math></center>
So that
<center><math>
\hat{u}\left( k,t\right) =\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}k^{3}t}
</math></center>
Therefore
<center><math>
u\left( x,t\right) = \frac{1}{2\pi}
\int_{-\infty}^{\infty }\hat{f}\left( k\right)
\mathrm{e}^{\mathrm{i}k^{3}t}\mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>

==Numerical Implementation Using the FFT==

The Fast Fourier Transform (FFT) is a method to calculate the fourier
transform efficiently for discrete sets of points. These points need to be
evenly spaced in the <math>x</math> plane and are given by
<center><math>
x_{n}=x_{0}+n\Delta x,\ \ 0\leq n\leq N-1
</math></center>
(note they can start at any <math>x</math> value). For the FFT to be as efficient as
possible <math>N</math> should be a power of <math>2.</math> Corresponding to the discrete set of
points in the <math>x</math> domain is a discrete set of points in the <math>k</math> plane given
by
<center><math>
k_{n}=\left\{
\begin{matrix}
n\Delta k,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right) \Delta k,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right.
</math></center>
where <math>\Delta k=2\pi /(N\Delta x).</math> Note that this numbering seems slighly
odd and is due to aliasing. We are not that interested in the frequency
domain solution but we need to make sure that we select the correct values
of <math>k</math> for our numerical code.

==Numerical Code for the Linear KdV==

Here is the code to solve the linear KdV using MATLAB

N = 1024;

t=0.1;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

f = exp(-x.^2);

f_hat = fft(f);

u = real(ifft(f_hat.*exp(i*k.^3*t)));

==Numerical Solution of the KdV==

It turns out that a method to solve the KdV equation can be derived using
spectral methods. We begin with the KdV equation written as
<center><math>
\partial _{t}u+3\partial _{x}\left( u\right) ^{2}+\partial _{x}^{3}u=0.
</math></center>
The Fourier transform of the KdV is therefore
<center><math>
\partial _{t}\hat{u}+3ik\widehat{\left( u^{2}\right)} -ik^{3}\hat{u}=0
</math></center>
We solve this equation by a split step method. We write the equation as
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)} +ik^{3}\hat{u}
</math></center>
We can solve
<center><math>
\partial _{t}\hat{u}=ik^{3}\hat{u}
</math></center>
exactly while the equation
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)}
</math></center>
needs to be solved by time stepping. The idea of the split step method is to
solve alternatively each of these equations when stepping from <math>t</math> to <math>
t+\Delta t.</math> Therefore we solve
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
-3ik \Delta t\widehat{\left( u^{2}\right)}
\end{matrix}</math></center>
Note that we are using Euler's method to time step and that the solution
could be improved by using a better method, such as the Runge-Kutta 4
method.

The only slighly tricky thing is that we have both <math>u</math> and <math>\hat{u}</math> in the
equation, but we can simply use the Fourier transform to connect these. The
equation then becomes
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
- 3ik\Delta t\left( \mathcal{F}\left( \left( \mathcal{F}^{-1}\left[ \hat{u}_{1}\left( k,t+\Delta t\right)\right]
\right) ^{2}\right) \right)
\end{matrix}</math></center>

==Numerical Code to solve the KdV by the split step method==

Here is some code to solve the KdV using MATLAB

N = 256;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

c=16;

u = 1/2*c*(sech(sqrt(c)/2*(x+8))).^2;

delta_t = 0.4/N^2;

tmax = 0.1; nmax = round(tmax/delta_t);

U = fft(u);

for n = 1:nmax

% first we solve the linear part

U = U.*exp(1i*k.^3*delta_t);

%then we solve the non linear part

U = U - delta_t*(3i*k.*fft(real(ifft(U)).^2));

end

== Example Calculations ==
We cosider the evolution of the KdV with two solitons as
initial condition as given below.
<center><math>
u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)
+ \, 2 \mathrm{sech}^{2}\left(\left(
x +1\right) \right)</math></center>

{| class="wikitable"
|-
! Animation
! Three-dimensional plot.
|-
| [[Image:Two_soliton.gif|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.]]
| [[Image:Two_soliton.jpg|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.
Note the phase shift which occurs in the soliton interaction.]]

|}

Numerical Solution of the KdV

2010-11-06T10:11:37Z

Adi Kurniawan: /* Numerical Code for the Linear KdV */

{{nonlinear waves course
| chapter title = Numerical Solution of the KdV
| next chapter = [[Conservation Laws for the KdV]]
| previous chapter = [[Introduction to KdV]]
}}

== Introduction ==

We present here a method to solve the KdV equation numerically. There are
many different methods to solve the KdV and we use here a spectral method
which has been found to work well. Spectral methods work by using the
Fourier transform (or some varient of it) to calculate the derivative.

==Fourier Transform==

Recall that the Fourier transform is given by
<center><math>
\mathcal{F}\left[ f(x)\right] =\hat{f}\left( k\right) =\int_{-\infty
}^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
</math></center>
and the inverse Fourier transform is given by
<center><math>
f\left( x\right) =\mathcal{F}^{-1}\left[ \hat{f}\left( k\right) \right] =
\frac{1}{2\pi }\int_{-\infty }^{\infty }\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
(note that there are other ways of writing this transform). The most
important property of the Fourier transform is that
<center><math>
\begin{matrix}
\int_{-\infty }^{\infty }\left( \partial _{x}f\left( x\right) \right)
\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)
\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\
&=&-k\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
\end{matrix}
</math></center>
where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm
\infty .</math> This means that the Fourier transform converts differentiation to
multiplication by <math>k</math>.

==Solution for the Linearized KdV==

We begin with a simple example. Suppose we want to solve the linearized KdV
equation
<center><math>
\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty <x<\infty
</math></center>
subject to solve initial conditions
<center><math>
u\left( x,0\right) =f\left( x\right)
</math></center>
We can solve this equation by taking the Fourier transform. and obtain
<center><math>
\partial _{t}\hat{u}-ik^{3}\hat{u}=0
</math></center>
So that
<center><math>
\hat{u}\left( k,t\right) =\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}k^{3}t}
</math></center>
Therefore
<center><math>
u\left( x,t\right) = \frac{1}{2\pi}
\int_{-\infty}^{\infty }\hat{f}\left( k\right)
\mathrm{e}^{\mathrm{i}k^{3}t}\mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>

==Numerical Implementation Using the FFT==

The Fast Fourier Transform (FFT) is a method to calculate the fourier
transform efficiently for discrete sets of points. These points need to be
evenly spaced in the <math>x</math> plane and are given by
<center><math>
x_{n}=x_{0}+n\Delta x,\ \ 0\leq n\leq N-1
</math></center>
(note they can start at any <math>x</math> value). For the FFT to be as efficient as
possible <math>N</math> should be a power of <math>2.</math> Corresponding to the discrete set of
points in the <math>x</math> domain is a discrete set of points in the <math>k</math> plane given
by
<center><math>
k_{n}=\left\{
\begin{matrix}
n\Delta k,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right) \Delta k,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right.
</math></center>
where <math>\Delta k=2\pi /(N\Delta x).</math> Note that this numbering seems slighly
odd and is due to aliasing. We are not that interested in the frequency
domain solution but we need to make sure that we select the correct values
of <math>k</math> for our numerical code.

==Numerical Code for the Linear KdV==

Here is the code to solve the linear KdV using MATLAB

<syntaxhighlight lang="matlab">
N = 1024;

t=0.1;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

f = exp(-x.^2);

f_hat = fft(f);

u = real(ifft(f_hat.*exp(i*k.^3*t)));
</syntaxhighlight>

==Numerical Solution of the KdV==

It turns out that a method to solve the KdV equation can be derived using
spectral methods. We begin with the KdV equation written as
<center><math>
\partial _{t}u+3\partial _{x}\left( u\right) ^{2}+\partial _{x}^{3}u=0.
</math></center>
The Fourier transform of the KdV is therefore
<center><math>
\partial _{t}\hat{u}+3ik\widehat{\left( u^{2}\right)} -ik^{3}\hat{u}=0
</math></center>
We solve this equation by a split step method. We write the equation as
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)} +ik^{3}\hat{u}
</math></center>
We can solve
<center><math>
\partial _{t}\hat{u}=ik^{3}\hat{u}
</math></center>
exactly while the equation
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)}
</math></center>
needs to be solved by time stepping. The idea of the split step method is to
solve alternatively each of these equations when stepping from <math>t</math> to <math>
t+\Delta t.</math> Therefore we solve
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
-3ik \Delta t\widehat{\left( u^{2}\right)}
\end{matrix}</math></center>
Note that we are using Euler's method to time step and that the solution
could be improved by using a better method, such as the Runge-Kutta 4
method.

The only slighly tricky thing is that we have both <math>u</math> and <math>\hat{u}</math> in the
equation, but we can simply use the Fourier transform to connect these. The
equation then becomes
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
- 3ik\Delta t\left( \mathcal{F}\left( \left( \mathcal{F}^{-1}\left[ \hat{u}_{1}\left( k,t+\Delta t\right)\right]
\right) ^{2}\right) \right)
\end{matrix}</math></center>

==Numerical Code to solve the KdV by the split step method==

Here is some code to solve the KdV using MATLAB

N = 256;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

c=16;

u = 1/2*c*(sech(sqrt(c)/2*(x+8))).^2;

delta_t = 0.4/N^2;

tmax = 0.1; nmax = round(tmax/delta_t);

U = fft(u);

for n = 1:nmax

% first we solve the linear part

U = U.*exp(1i*k.^3*delta_t);

%then we solve the non linear part

U = U - delta_t*(3i*k.*fft(real(ifft(U)).^2));

end

== Example Calculations ==
We cosider the evolution of the KdV with two solitons as
initial condition as given below.
<center><math>
u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)
+ \, 2 \mathrm{sech}^{2}\left(\left(
x +1\right) \right)</math></center>

{| class="wikitable"
|-
! Animation
! Three-dimensional plot.
|-
| [[Image:Two_soliton.gif|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.]]
| [[Image:Two_soliton.jpg|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.
Note the phase shift which occurs in the soliton interaction.]]

|}

KdV Equation Derivation

2010-11-06T09:56:31Z

Adi Kurniawan: /* Derivation */

{{complete pages}}

We consider the method of derivation of KdV Equation in the concept of [http://www.wikiwaves.org/index.php/Nonlinear_Shallow_Water_Waves Nonlinear Shallow Water Waves].

==Introduction==

In the analysis of [http://www.wikiwaves.org/index.php/Nonlinear_Shallow_Water_Waves Nonlinear Shallow Water Waves] equations we see that there are two important geometrical parameters, <math>\epsilon = \frac{h}{\lambda}</math> and <math>\alpha=\frac{a}{h}</math> involved. By choosing appropriate magnitudes for <math>\epsilon</math> and <math>\alpha</math>, we can consider a theory in which dispersion and nonlinearity are in balance. The Korteweg-de Vries Equation verifies the relation between dispersion and nonlinearity properties.

==Derivation==

We begin with the equations for waves on water,

<center>
<math>
\begin{matrix}
&\Phi_{xx} + \Phi_{yy} &= 0 \quad &-\infin<x<\infin, 0 \le y \le \eta(x,t) \\
\end{matrix}
</math>
</center>

Provided that at <math>y=\eta(x,t)=h+aH(x,t)</math> we have,

<center><math>
\begin{matrix}
&\Phi_{y} &= &\eta_t + \Phi_x \eta_x \\
&\Phi_t + \frac{1}{2}({\Phi_x}^2 + {\Phi_y}^2) + g\eta &= &B(t)\\
&\Phi_y = 0 &, &y = 0
\end{matrix}
</math></center>

To make these equations dimensionless, we use the scaled variables,
<center><math>
\bar{x}=\frac{x}{\lambda}, \quad \bar{y}=\frac{y}{h}, \quad \bar{\Phi}=\frac{h\Phi}{\lambda a \sqrt{gh}}, \quad \bar{t}=\frac{t\sqrt{gh}}{\lambda}
</math></center>

where <math>\sqrt{gh}</math> is defined as linear wave speed in shallow water. Hence the dimensionless system is,

<center><math>
\begin{matrix}
&\epsilon^2 {\bar{\Phi}}_{\bar{x}\bar{x}} + {\bar{\Phi}}_{\bar{y}\bar{y}} &= &0 \\ \\
&{\bar{\Phi}}_{\bar{y}} &= &\epsilon^2(H_{\bar{t}}+\alpha {\bar{\Phi}}_{\bar{x}} H_{\bar{x}}) \\ \\
&{\bar{\Phi}}_{\bar{t}} + \frac{1}{2}\alpha ({{\bar{\Phi}}_{\bar{x}}}^2 + \epsilon^2 {{\bar{\Phi}}_{\bar{y}}}^2) + H &= &(B(t)-gh) / ag \\ \\
&{\bar{\Phi}}_{\bar{y}} = 0 &, &\bar{y} = 0
\end{matrix}
</math></center>

where <math>\epsilon = \frac{h}{\lambda}</math> and <math>\alpha=\frac{a}{h}</math> are two small parameters which are given in this problem.

In the next step we use the transform <math>\bar{\Phi} \to \bar{\Phi} + \int\limits_{0}^{\bar{t}}(\frac{B(s) - gh}{ag})\mathrm{d}s</math> and introduce further transformation to remove <math>\epsilon</math> from the equations,

<center><math>
z = \frac{\alpha^{1 / 2}}{\epsilon}(\bar{x}-\bar{t}), \quad \tau = \frac{\alpha^{3/2}}{\epsilon}\bar{t}, \quad \Psi = \frac{\alpha^{1/2}}{\epsilon}\bar{\Phi}
</math></center>

The key idea is that <math>\frac{\alpha^{1 / 2}}{\epsilon}</math> is <math>O(1)</math>.

Hence,

<center><math>
\begin{matrix}
&\alpha \Psi_{zz} + \Psi_{\bar{y}\bar{y}} = 0 & -\infin < z <\infin , 0 \le \bar{y} \le 1 + \alpha H(z,\tau) &(1) \\ \\
&\Psi_{\bar{y}} = \alpha (-H_z+\alpha H_{\tau} + \alpha \Psi_z H_z) & y=1+ \alpha H(z,\tau) &(2) \\ \\
&H - \Psi_z + \alpha \Psi_{\tau} + \frac{1}{2} ({\Psi_{\bar{y}}}^2+\alpha {\Psi_z}^2)=0 &y=1+ \alpha H(z,\tau) &(3) \\ \\
&\Psi_{\bar{y}} = 0 &\bar{y}=0 &(4)
\end{matrix}
</math></center>

The boundary condition (4) expresses <math>\Psi</math> at the flat bed, <math>\bar{y}=0</math>. The boundary condition (3) is Bernoulli equation and (2) is kinematic boundary condition. Now we use asymptotic expansions of the form,

<center><math>
\begin{matrix}
&\Psi &= &\Psi_0 + \alpha \Psi_1 + {\alpha}^2 \Psi_2 + o({\alpha}^3) &(5)\\ \\
&H &= &H_0 + \alpha H_1 + o(\alpha^2) &(6)
\end{matrix}
</math></center>

to derive an equation for each <math>H_i</math> according to the boundary conditions (2) to (4).

* Derivation of <math>H_i</math>'s:

Substituting (5) and (6), (1) must be true for all powers of <math>\alpha</math>. Therefore,

<center><math>
\begin{matrix}
&O(\alpha^0) &: &\Psi_{0, \bar{y}\bar{y}} = 0 &\rArr &\Psi_0 = B_0(z, \tau) \\ \\
&O(\alpha) &: &\Psi_{1, \bar{y}\bar{y}} = -\Psi_{0, zz} &\rArr &\Psi_1 = -\frac{1}{2}{\bar{y}}^2 B_{0, zz}+B_1(z, \tau) \\ \\
&O(\alpha^2) &: &\Psi_{2, \bar{y}\bar{y}} = -\Psi_{1, zz} &\rArr &\Psi_2 = \frac{1}{24}{\bar{y}}^4B_{0,zzzz}-\frac{1}{2}{\bar{y}}^2 B_{1,zz}+ B_2(z, \tau)
\end{matrix}
</math></center>

Now at leading order the Bernoulli and kinematic equations, (3) and (2), gives,

<center><math>
\begin{matrix}
&H_0(z,\tau) = \Psi_{0,z} = B_{0,z} &(a) \\ \\
&H_1-B_{1,z}+\frac{1}{2}B_{0,zzz}+B_{0,\tau}+\frac{1}{2}B^2_{0,z} = 0 &(b) \\ \\
&-H_0B_{0,zz}+\frac{1}{6}B_{0,zzzz}-B_{1,zz} = -H_{1,z}+H_{0,\tau}+B_{0,z}H_{0,z} &(c)
\end{matrix}
</math></center>

Differentiating (b) and eliminating <math>H_1</math> and <math>B_1</math> from (c) allow us to write,

<center><math>
-H_0B_{0,zz}-\frac{1}{3}B_{0,zzzz}-B_{0,z\tau}-B_{0,z}B_{0,zz} = H_{0,\tau}+B_{0,z}H_{0,z}
</math></center>

Finally, (a) gives <math>B_0</math> in terms of <math>H_0</math> and hence

<center><math>
2H_{0,\tau}+3H_0H_{0,z}+\frac{1}{3}H_{0,zzz}=0
</math></center>

which is named Korteweg-de Vries (KdV) equation.

==Interpretation==

KdV equation includes dispersive effects through the term <math>H_{0,zzz}</math> and nonlinear effects through the term <math>H_0H_{0,z}</math> and governs the behavior of the small amplitude waves, with <math>\alpha<<1</math>. It is reasonable to ask when and where the independent variables, <math>z</math> and <math>\tau</math>, are of <math>O(1)</math> in order to determine more precisely the region in physical space where the KdV equation is valid as an approximation of the actual flow. According to the definition of <math>z</math> and <math>\tau</math>, if <math>\alpha=O(\epsilon^2)</math>, then <math>\bar{t}>>1</math> and <math>\bar{x}=\bar{t}+O(1)</math>. This leads us to interpret any waveform that arises as a solution of the KdV equation as the large time limit of an initial value problem.

For solution of KdV equation please refer [http://www.wikiwaves.org/index.php/KdV_Equation_Solutions here.]

[[Category:Nonlinear Water-Wave Theory]]

Nonlinear Shallow Water Waves

2010-11-06T09:51:24Z

Adi Kurniawan: /* Accelerating Piston */

{{nonlinear waves course
| chapter title = Nonlinear Shallow Water Waves
| next chapter = [[Introduction to KdV]]
| previous chapter = [[Traffic Waves]]
}}

{{complete pages}}

== Introduction ==

We assume that water is incompressible,
viscous effects are negligible and that the typical wave lengths are much larger than the water depth.
This allows us to assume [[:Category:Shallow Depth|Shallow Depth]]. We assume that the problem has not variation
in either the <math>y</math> or <math>z</math> direction. The fluid is governed by two parameters,
<math>u</math>, the velocity of the water, and <math>h</math> the water depth (note that this is not the still water depth since the problem
is nonlinear).

The theory we present here is discussed in [[Stoker 1957]], [[Billingham and King 2000]] and [[Johnson 1997]].

== Equations of Motion ==

The equation for the conservation of mass can derived by considering a a region <math>[x,x+\Delta x]</math>
Conservation of mass then implies that
<center>
<math>
\partial_t \int_x^{x + \Delta x} \rho h(s,t) \mathrm{d}s = \rho h(x,t)u(x,t) - \rho h(x+\Delta x,t)u(x+\Delta x,t)
</math>
</center>
If we take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t h(x ,t) + \partial_x (h(x ,t)u(x ,t)) = 0
</math>
</center>

A second equation comes from conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x}^{x + \Delta x}
+ \int_0^{h(x)} P(x,z,t) \mathrm{d}z -
\int_0^{h(x + \Delta x)} P(x+\Delta x,z,t) \mathrm{d}z
</math>
</center>
where <math>\rho \,</math> denotes density, and the pressure <math>P \,</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). This then gives us
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= -\left. \rho u^2 h \right|_{x}^{x + \Delta x}
+ \frac{1}{2}\rho g {h(x)}^2 -
\frac{1}{2}\rho g {h(x + \Delta x)}^2
</math>
</center>
If we then take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t \left( h u \right)
+ \partial_x \left(u^2 h + \frac{1}{2} gh^2\right) =0
</math>
</center>
We can simplify this using the equation derived from conservation of mass to
to obtain
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>

The equations
<center>
<math>
\partial_t h + u \partial_x h + h \partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>
are called the nonlinear shallow water equations. They determine the horizontal water velocity and the local water depth.

We can rewrite them in terms of the local wave speed <math>c(x, t) = \sqrt{gh(x, t)}</math> as follows:
<center>
<math>2\partial_t c + 2u\partial_x c + c\partial_x u = 0
</math>
</center>
<center>
<math>\partial_t u + u\partial_x u + 2c \partial_x c = 0
</math>
</center>
These equation are almost identical to those of compressible gas dynamics. Much of our understanding
of the equations for water have been found by researchers studying compressible gas dynamics.

== Linearized Equations ==

We can linearize these equations by assuming that <math>u</math> is small and that
<math>h=h_0 + \zeta </math> where <math>h_0</math> is the average water depth
and <math>\zeta</math> is also assumed small. This gives us
<center>
<math>
\partial_t \zeta + h_0\partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + g \partial_x \zeta = 0
</math>
</center>
These linear shallow water equations which can be
derived from the linear equations for water of finite
depth and taking the limit of small depth (see [[:Category:Shallow Depth|Shallow Depth]]).

== Characteristics ==

The equations possess characteristics.
Adding and subtracting the two equations above we obtain
<center>
<math>
\frac{\partial (u \pm 2c)}{\partial t}+ (u \pm c)\frac{\partial (u \pm 2c)}{\partial x} = 0
</math>
</center>
This means that
on the <math>\;C_+</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_+}{\mathrm{d} t} = u + c = u + \sqrt{gh}
</math>
</center>
the <math>\;C_+</math> invariant
<center>
<math>
R_+ = u + 2c = u + 2\sqrt{gh}
</math>
</center>
is a constant,
and on the <math>\;C_-</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - c = u - \sqrt{gh}
</math>
</center>
the <math>\;C_-</math> invariant
<center>
<math>
R_- = u - 2c = u - 2\sqrt{gh}
</math>
</center>
is a constant.

The functions <math>R_{\pm} (u ,c) = u \pm 2c</math>, are called the Riemannian invariants.

== Simple Waves ==

The problem as formulated can be solved by advancing the solution along the characteristics, but
this will in general be quite difficult analytically. However, there is a special class of problems,
called ''Simple Waves'' in which the solution only changes on one characteristic.
They are best illustrated through some examples. Note that the characteristic can meet forming
a shock, which is called a [http://en.wikipedia.org/wiki/Tidal_bore bore] or a
[http://en.wikipedia.org/wiki/Hydraulic_jump hydraulic jump]
when it occurs on the surface of the water.

=== The dam break problem ===
Assume the water occupies the region <math>{x < 0 ; 0 < z < h_0 }</math> initially held back by a dam at <math>x = 0</math>.
At <math>t = 0</math>, the dam is removed (breaks). What is the height of the water
<math>h(x,t) \,</math> for <math>t > 0? \,</math> The initial condition is therefore
<center>
<math>h(x,0) = \begin{cases}
h_0, & x < 0 \\
0, & x > 0
\end{cases}
</math>
</center>
<center>
<math>u(x ,0) = 0. \,
</math>
</center>
On the characteristic that originates at <math>t = 0</math> for <math>x < 0</math>,
<center>
<math>
R_\pm = u \pm 2\sqrt{gh} = \pm 2\sqrt{gh_0} = \pm 2c_0
</math>
</center>
where <math>c_0 = \sqrt{gh_0}</math> is the initial (linear) wave speed.

Therefore, if a <math>C_+</math> and a <math>C_-</math> characteristic from this region intersect, then
<center>
<math>
u + 2\sqrt{gh} = 2c_0 , \;\mathrm{and}\; u - 2\sqrt{gh} = -2c_0
</math>
</center>
and hence, <math>u = 0</math> and <math>h = h_0</math>.
Moreover,
<center>
<math>
\frac{\mathrm{d} X_\pm}{\mathrm{d} t} = u \pm \sqrt{gh} = \pm c_0
</math>
</center>
so these characteristics are straight lines in the region <math>\big\{x < -c_0 t \big\}</math>
(the undisturbed region).

The <math>\;C_+</math> characteristic leave the region a<math>\big\{x < -c_0 t \big\}</math>
and enter <math>\big\{x > -c_0 t \big\}</math>. For now we will assume that these characteristics fill the domain
(and show that this is true shortly).
For <math>\big\{x > -c_0 t \big\}</math>, the <math>C_-</math> characteristics are given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - \sqrt{gh}
</math>
</center>
and on each of the <math>C_-</math> characteristics <math>R_- = u - 2\sqrt{gh}</math> is constant.
However, since this region is filled with <math>C_+</math> characteristics where <math>R_+ = u + 2\sqrt{gh} = 2c_0</math>, <math>u</math> and <math>h</math>
must be constant on each <math>C_-</math> characteristic.
This also means that the <math>C_-</math> characteristics must be straight lines.

Since the fluid occupies <math>\big\{x < 0 \big\}</math> at <math>t = 0</math>,
these <math>C_-</math> characteristics must start at the origin, with
<center>
<math>
X_-(t) = \left(u - \sqrt{gh}\right)t
</math>
</center>
which in turn implies that
<center>
<math>
u - \sqrt{gh} = \frac{x}{t}
</math>
</center>
We also have <math>R_+ = u + 2\sqrt{gh} = 2c_0</math> from the
<math>C_+</math> characteristics. We can solve these equations
at each point in <math>\big\{x > -c_0 t \big\}</math>. Solving for <math>u</math> and <math>h</math> gives
<center>
<math>
h(x, t) =
\begin{cases}
\frac{h_0}{9}\left(2 - \frac{x}{c_0 t}\right)^2, \quad -c_0 t < x< 2 c_0 t,\\
h_0,\quad x <-c_0 t,
\end{cases}
</math>
</center>
<center>
<math>
u(x, t) =
\begin{cases}
\frac{2}{3} \left (c_0 + \frac{x}{t} \right ), \quad -c_0 t < x< 2 c_0 t,\\
0,\quad x <-c_0 t.
\end{cases}
</math>
</center>
Where we have assumed that, since <math>h = 0</math> for <math>x = 2c_0 t,</math>
the <math>C_+</math> characteristic only exist in the region <math>\big\{x < 2c_0 t \big\}</math>,
We will verify this by explicitly calculating them.

It remains to determine the <math>C_+</math> characteristic, which originated in <math>\big\{x < 0 \big\}</math>,
and show they
will fill the domain <math>\big\{x < 2c_0 t \big\}</math>. For <math>\big\{x < -c_0 t \big\}</math>,
the <math>\;C_+</math> characteristics are straight lines with slope <math>c_0</math> and are given by
<center>
<math>
X_+ (t) = -x_0 + c_0 t, \quad \left(x_0 > 0,\;\; t < \frac{x_0}{2c_0}\right)
</math>
</center>
When <math>t = \frac{x_0}{2c_0},\;\;\ X_{+} (t) = -c_0 t</math> so that for
<center>
<math>
t > \frac{x_0}{2c_0}, \quad \frac{\mathrm{d} X_{+} (t)}{\mathrm{d} t} = u + \sqrt{gh}
</math>
</center>
and substituting the solution we found for <math>h</math> and <math>u</math>
<center>
<math>
\frac{\mathrm{d}X_{+} (t)}{\mathrm{d} t} = \frac{4}{3}c_0 + \frac{X_{+} (t)}{3t}
</math>
</center>
Solving this ODE subject to <math>X_+ \left(\frac{x_0}{2c_0}\right) = -\frac{x_0}{2}</math> gives
<center>
<math>
X_+ (t) = 2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\;\;
</math>
</center>
the equation for a characteristic curve.
The curves indeed fill the domain <math>\big\{x < 2c_0 t \big\}</math>
and all satisfy <math>\big\{X_+ (t) < 2c_0 t \big\}</math>. To summarize, the
<math>C^{+}</math> characteristics are given by
<center>
<math>
X_+ (t) =
\begin{cases}
2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\quad t> x_0/2 c_0\\
-x_0 + c_0 t, \quad 0\ < t < x_0/2 c_0
\end{cases}
</math>
</center>

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Characteristics_dam_break.jpg|thumb|right|500px|Characteristics for the dam
break problem, blue for <math>C_{+}</math> and red for <math>C_{-}</math>. The solid red lines
show the curves <math>x=-c_0 t</math> and <math>x=2c_0 t</math> (note we have assumed
here that <math>c_0 =1</math>]]
| [[Image:Dambreak.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the Dam Break problem]]
|}

== Shocks ==

For a unique solution two exist there must be a single <math>C_+</math> and <math>C_-</math>
characteristic through each point. When two characteristics of the same kind meet we
have a shock forming.

=== Accelerating Piston ===

We now consider the problem of water initially at rest occupying the
half space <math>x>0</math> which is initially at rest. At <math>t=0</math>
the piston at <math>x=0</math> begins to move to the right with constant
acceleration <math>a</math> so that the position of the piston is given by
<math>\frac{1}{2}at^2</math>.

We assume that the <math>C_-</math> characteristics which originate in the
water at <math>t=0</math> fill the fluid. On these characteristics
<center>
<math>
R_- = u - 2 c = -2c_0 \,
</math>
</center>
and this condition must hold throughout the fluid.
On the <math>C_+</math> characteristics we know that
<math> R_+ = u + 2 c </math> must be a constant and hence on the
<math>C_+</math> characteristics <math>u</math> and <math>c</math>
must be constant and hence the <math>C_+</math> characteristics must
be straight lines. Note that this does not mean that the
<math>C_+</math> characteristics have the same slope and there is no
requirement that the <math>C_-</math> characteristics are straight lines.

The <math>C_+</math> characteristic originate from the fluid
or from the front of the piston. We consider those which originate from
the piston. The <math>C_+</math> characteristic which originates from
the piston at <math>t=t_0</math> must satisfy
<center>
<math>
u+2c = a t_0 + 2c_{\text{plate}} \,
</math>
</center>
where <math> a t_0</math> is the velocity of the piston at time <math>t=t_0</math>
and <math>c_{\text{plate}}</math> is the speed (related to height) at the plate.
We know that <math> R_- = u - 2 c =-2c_0</math> through out the fluid, so that if
we solve this at the plate (where <math>u=at_0</math> and <math>c=c_{\text{plate}}</math>)
then we get
<center><math>
c_{\text{plate}} = at_0/2 + c_0\,
</math></center>

On the <math>C_+</math> characteristics <math>u</math> and <math>c</math>
are constant and therefore
<center>
<math>
\frac{\mathrm{d}X_+}{\mathrm{d}t} = u+c = at_0 + \left( \frac{1}{2}at_0 + c_0 \right)
</math>
</center>
Hence
<center>
<math>
X_+(t,t_0) = \left( \frac{3}{2} a t_0 + c_0 \right) t - c_0 t_0 -a t_0^2
</math>
</center>
using the condition <math>X_+(t_0,t_0) = \frac{1}{2} a t_0^2</math> (the initial value which
comes from the position of the piston at <math>t=t_0</math>).
The slope of these lines increases and eventually meet to form a shock.
We find this point of intersection by considering neighboring characteristics
and determining when they first intersect.
<center>
<math>
X_+(t,t_0 + \Delta t) = X_+(t,t_0) + \Delta t \frac{\partial X_+}{\partial t_0} (t,t_0)
</math>
</center>
It follows that neighbouring characteristics will meet when
<center>
<math>
\frac{\partial X_+}{\partial t_0} (t,t_0) = 0
</math>
</center>
which implies that
<center>
<math>
t = \frac{2c_0}{3a} + \frac{4}{3}t_0
</math>
</center>
The first time that a shock forms is the minimum value of this equation.
For this piston example, this occurs when <math>t_0 = 0</math> and the value
of <math>t</math> is <math>t = 2c_0/(3a)</math>. At this point
a shock is formed and we can no longer find a unique solution by following the
characteristics.

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Accelerating_piston.jpg|thumb|right|500px| <math>C_{+}</math> characteristics for the
accelerating piston, red undisturbed, blue from the piston and the green line shows the transition.]]
| [[Image:Accelerating_piston2.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the
accelerating piston.]]
|}

=== Piston Moving with Constant Velocity ===

This example is also known as the Moving Wall Problem, and is connected to Shallow Water Bores.

We consider the case of a piston, with positive constant velocity <math>V</math> (which is initially at <math>x=0</math>), advancing into a semi-infinite expanse of
fluid that is initially at rest with depth <math>h_0</math>.

The <math>C_+</math> characteristics which originate in the fluid
at <math>t=0</math> have slope
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0}</math>
</center>
and the <math>C_+</math> characteristic which originate at the piston at
<math>t=0</math> must satisfy
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0} + \frac{3}{2} V </math>
</center>
so that these two characteristics will intersect at <math>t=0</math>.
Therefore a shock forms immediately and we can track this by determining the
speed of the shock

=== Speed of the shock ===

We need the conservation equations in integral form to determine the speed
of the shock. Conservation of mass, written as an integral is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h \mathrm{d}x
+ \left. \rho u h \right|_{x_2}^{x_1} =0
</math>
</center>
If the shock is located at <math>s(t)</math>
which we assume is located between <math>x_1</math>
and <math>x_2</math>, then
<center>
<math>
\partial_t \int_{x_1}^{x_2} h \mathrm{d}x
= \partial_t \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) h \mathrm{d}x
= \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) \partial_t h \mathrm{d}x + h^{+} \partial_t s(t) - h^{-}\partial_t s(t),
</math>
</center>
where <math>h^+</math> is the height on the right (positive) side of
the jump and <math>h^-</math> is the height on the left (negative) side.
If we take the limit as <math>x_1\to x_2</math> we then obtain the following
identity
<center>
<math>
h^{+}\partial_t s(t) - h^{-}\partial_t s(t) -
u^{+} h^{+} + u^{-} h^{-} = 0
</math>
</center>
where <math>u^+</math> is the height on the right (positive) side of
the jump and <math>u^-</math> is the height on the left (negative) side.

We now need to consider the equation for conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x_1}^{x_2}
+ \int_0^{h(x_1)} P(x_1,z,t) \mathrm{d}z -
\int_0^{h(x_2)} P(x_2,z,t) \mathrm{d}z
</math>
</center>
where the pressure <math>P</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). We can apply a similar argument as before to obtain
<center>
<math>
h^{+}u^{+}\partial_t s(t) - h^{-}u^{-}\partial_t s(t) =
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2
</math>
</center>

=== Hydraulic Jump ===

For a hydraulic jump, <math>\dot{s}(t) = 0</math>, which means that we must solve
<center>
<math>
u^{+} h^{+} - u^{-} h^{-} = 0 \,
</math>
</center>
<center>
<math>
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2 =0
</math>
</center>
If we introduce the variables
<center>
<math>
H = \frac{h^{+}}{h^{-}}
</math>
</center>
and
<center>
<math>
\mathrm{Fr} = \frac{u^{-}}{\sqrt{gh^{-}}}
</math>
</center>
where <math>\mathrm{Fr}</math> is the ''Froude'' number
which is equivalent to the Mach number for gas dynamics.
Then we obtain
<center>
<math>
H^2 -1 = 2 \mathrm{Fr}^2 \left(1 - \frac{1}{H}\right)
</math>
</center>
This expression has the roots
<center>
<math>
H=1, \quad H=\frac{1}{2}\left(-1\pm\sqrt{1 + 8 \mathrm{Fr}^2}\right)
</math>
</center>
The only physically meaningful solution is the root which satisfies
<math>H>1</math>. This is only true providing <math>\mathrm{Fr} > 1</math>, which means
that we can only obtain a hydraulic jump if the flow is supercritical.

Below is a video of a hydraulic jump. You can clearly see the point where the flow is changing from supercritical to subcritical (look for the small turbulent region in the channel)
{{#ev:youtube|5etwhZ0d2GU}}

=== Shallow Water Bore ===
We now consider a bore, in which
the shock wave advances into still water.
We denote the fluid speed by <math>V = u^{-}</math>.
We denote the height on the wall side
by <math>h_1</math> and the height on the other side must be <math>h_0</math>, i.e.
<math>h^{+} = h_0</math>, <math>h^{-} = h_1</math>, <math>u^{+} = 0</math>.
This means that
<center>
<math>
h_{0}\partial_t s(t) - h_{1}\partial_t s(t)
+ V h_{1}= 0
</math>
</center>
and
<center>
<math>
- h_{1}V\partial_t s(t) = - \left(V\right)^2 h_{1}
+ \frac{1}{2} g \left(h_{0}\right)^2 - \frac{1}{2} g \left(h_{1}\right)^2
</math>
</center>
which can be solved to obtain the shock speed and the height of the moving fluid.

Below is a video of surfing on the [http://en.wikipedia.org/wiki/Severn Severn] bore, do not believe everything they
say. You might also want to check out the [http://en.wikipedia.org/wiki/Pororoca Pororoca]
a tidal bore on the Amazon.
{{#ev:youtube|0jtkyuuMgVQ}}

[[Category:Simple Nonlinear Waves]]
[[Category:Nonlinear Water-Wave Theory]]
[[Category:Shallow Depth]]

Traffic Waves

2010-11-06T09:49:53Z

Adi Kurniawan: /* Speed of the shock */

{{nonlinear waves course
| chapter title = Traffic Waves
| next chapter = [[Nonlinear Shallow Water Waves]]
| previous chapter = [[Method of Characteristics for Linear Equations]]
}}

{{complete pages}}

We consider here some simple equations which model traffic flow. This problem is discussed in
[[Billingham and King 2000]].

[[Category:Reference]]

== Equations ==

We consider a single lane of road, and we measure distance along the road with
the variable <math>x</math> and <math>t</math> is time.
We define the following variables
<center><math>
\begin{matrix}
&\rho(x,t) &: &\mbox{car density (cars/km)} \\
& v(\rho) &: &\mbox{car velocity (km/hour)} \\
& q(x,t) =\rho v &: &\mbox{car flow rate (cars/hour)} \\
\end{matrix}
</math></center>

If we consider a finite length of road <math>x_1\leq x \leq x_2</math> then the net flow of cars
in and out must be balanced by the change in density. This means that
<center><math>
\frac{\partial}{\partial t} \int_{x_1}^{x_2} \rho(x,t) \mathrm{d}x = -q(x_2,t) + q(x_1,t)
</math></center>
We now consider continuous densities (which is obviously an approximation) and
set <math>x_2 = x_1 + \Delta x</math> and we obtain
<center><math>
\frac{\partial}{\partial t} \rho(x_1,t) = -\frac{q(x_2,t) + q(x_1,t)}{\Delta x}
</math></center>
and if we take the limit as <math>\Delta x \to 0</math> we obtain the differential equation
<center><math>
\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} = 0
</math></center>
Note that this equation has been derived purely from the need to conserve cars (it is a conservation equation) and
is not possible to solve this equation until we have derived a connection between <math>\rho</math>
and <math>q</math>.

== Equation for <math>\rho</math> only ==

At the moment we assume that we have some expression for <math>v(\rho)</math>
If we substitute the expression for <math>q = v\rho</math> into our differential equation we obtain
<center><math>
\frac{\partial \rho}{\partial t} + \frac{\partial }{\partial x} \left(v(\rho)\rho\right) = 0
</math></center>
which gives us
<center><math>
\frac{\partial \rho}{\partial t} + \left(v^{\prime}(\rho)\rho + v(\rho)\right)
\frac{\partial \rho }{\partial x} = 0
</math></center>
or
<center><math>
\frac{\partial \rho}{\partial t} + c(\rho)\frac{\partial \rho }{\partial x} = 0
</math></center>
where <math>c(\rho) = \left(v^{\prime}(\rho)\rho + v(\rho)\right)</math>
is the '''kinematic wave speed'''. Note that this is not the speed of the cars, but
the speed at which disturbances in the density travel.

== A simple relationship between <math>\rho</math> and <math>q</math> ==

The relationship between <math>\rho</math> and <math>q</math> is an equation of state and
there is no ''exact'' equation since it depends on many unknowns. One of the
simplest relationship between <math>\rho</math> and <math>q</math> is derived from
the following assumptions

* When the density <math>\rho = 0</math> the speed is <math>v=v_0</math>
* When the density is <math>\rho = \rho_{\max} </math> the speed is <math>v=0</math>
* The speed is a linear function of <math>\rho</math> between these two values.

This also gives good fit with measured data. We will either consider the general case or use this simple
relationship. Using this we obtain
<center><math>
v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
</math></center>
The flux of cars is given by
<center><math>
q = \rho v(\rho) = v_0\frac{\rho(\rho_{\max} - \rho)}{\rho_{\max}}
</math></center>
and the wave speed is
<center><math>
c(\rho) = \left(v^{\prime}(\rho)\rho + v(\rho)\right) = -\frac{v_0}{\rho_{\max}}\rho + v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
= v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>

{| class="wikitable"
|-
! <math>v</math>
! <math>q</math>
! <math>c</math>
|-
| [[Image:Velocity.jpg|thumb|350px|<math>v(\rho)</math> versus <math>\rho</math>]]
| [[Image:Q_flux.jpg|thumb|350px|<math>q(\rho)</math> versus <math>\rho</math>]]
| [[Image:C_speed.jpg|thumb|350px|<math>\rho(\rho)</math> versus <math>\rho</math>]]
|}

== Small Amplitude Disturbances ==

We can linearise the model by assuming that the variation in density is small so
that we can write
<center><math>
\rho = \rho_0 + \tilde{\rho}
</math></center>
where we assume that <math>\tilde{\rho}</math> is small. This allows us to write the equations as
<center><math>
\frac{\partial \tilde{\rho}}{\partial t} + c(\rho_0) \frac{\partial \tilde{\rho}}{\partial x} = 0
</math></center>
where the main difference between this and the full equation is that the wave speed <math>c</math> is
a constant. This is the linearised equation. Note that this linearisation does not give a good model because
traffic density does not vary only a small amount about some mean (as is the case for accoustic waves where the
density of air is roughly constant).

Under these assumptions the solution to the equation is
<center><math>
\tilde{\rho} = f(x - c(\rho_0)t)
</math></center>
where <math>f</math> is determined by the initial condition. This represents
disturbances which travel with speed <math>c(\rho_0)</math> in the positive
<math>x</math> direction.

We now consider the '''characteristic curves''' which are curves along which the density
<math>\rho</math> is a constant. These are give by
<center><math>
x = X(t) = x_0 + c(\rho_0) t.\,
</math></center>
which are just straight lines of constant slope. We will see shortly that the full (nonlinear)
equations also possess characteristics.

== Nonlinear Initial Value Problem ==

We wish to solve
<center><math>
\frac{\partial \rho}{\partial t} + c(\rho) \frac{\partial \rho}{\partial x} = 0
</math></center>
subject to the initial conditions
<center><math>
\rho(x,0) = \rho_0(x) \,
</math></center>
It turns out that the concept of characteristic curves is very important for this problem.

If we want <math>\rho(X(t),t)</math> to be a constant then we require
<center><math>
\frac{\mathrm{d}}{\mathrm{d}t}\rho(X(t),t) = \frac{\mathrm{d} X}{\mathrm{d}t} \frac{\partial \rho}{\partial x} +
\frac{\partial \rho}{\partial t} = 0.
</math></center>
Comparing this to the governing partial differential equation we can see that we require
<center><math>
\frac{\mathrm{d} X}{\mathrm{d}t} = c(\rho)
</math></center>
This means that the characteristics are straight lines (since <math>\rho</math> is constant) with
slope given by <math> c(\rho_0(x_0))</math> so that the equation for the characteristics is
<center><math>
X(t) = x_0 + c(\rho_0(x_0))t \,
</math></center>

This does not allow us to write down a solution to the initial value problem,
all we can do is write
<center><math>
\rho(x_0 + c(\rho_0(x_0))t,t) = \rho_0(x_0)\,
</math></center>
which allows us to calculate the solution stepping forward in time, but not to determine the solution given
a value of <math>(x,t)</math> (because we have no way of knowing what <math>c(\rho_0(x_0))</math> is.

The characteristics are a family of straight lines which will all have different slopes. If two characteristics
meet, our solution method will break down because there will be two values of the density <math>\rho</math>.
This gives rise to a '''shock'''. It turns
out that this the formation of shocks is a product of the equations themselves and not with the solution method.
We will see shortly that special methods are required to treat these shocks.

=== Case when no shocks are formed ===

The characteristic curves will fill the space without meeting provided that the wave speed <math>c(\rho_0)</math>
is a monotonically increasing function of the distance <math>x</math>. If we work with our previous model we
have
<center><math>
v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
</math></center>
and
<center><math>
c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>
so that <math>c</math> is a monotonically decreasing function of density <math>\rho</math>. This means
that the wave speed <math>c(\rho_0)</math> will be
a monotonically increasing function of the distance <math>x</math> if an only if the density is a
monotonically decreasing function. In this case the solution can be calculated straightforwardly
by expansion of the initial density.

==== No shock example ====

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by <math>\rho_0 = 1/2(1- \tanh(x))</math>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Traffic example1 rho.jpg|thumb|350px| <math>\rho_0 = 1/2(1- \tanh(x))</math> ]]
| [[Image:Traffic_example1_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Traffic_example1_characteristics.jpg|thumb|350px|Characterisitics for <math>\rho_0 = 1/2(1- \tanh(x))</math>]]
| [[Image:Traffic_example11.gif|thumb|350px|<math>\rho(x,t) for </math><math>\rho_0 = 1/2(1- \tanh(x))</math>]]
|}

==== Riemann problem and the expansion fan ====

We can consider a simple problem in which there is a jump in the initial density
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < 0 \\
\rho_{R},& x > 0
\end{matrix}
\right.
</math></center>
where <math>\rho_{L} > \rho_{R}</math> so that we do not form a shock. In this case
the characteristics on each side of <math>x=0</math> have a different slope and the
question is what happens between. It is easiest to think about the following problem
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < -\epsilon \\
\frac{\rho_{R}-\rho_{L}}{2\epsilon}x + \frac{\rho_{R}+\rho_{L}}{2} & -\epsilon \leq x \leq \epsilon \\
\rho_{R},& x > \epsilon
\end{matrix}
\right.
</math></center>
We can then see that we have lines of uniformly varying slope for <math>-\epsilon<x<\epsilon</math>
with slope between <math>c(\rho_L)</math> and <math>c(\rho_R)</math>. If we then take the limit
as <math>\epsilon \to 0</math> we obtain an expansion fan emanating from <math>x=0</math>.

If we assume that
<center><math>
c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>
then we know that on the lines of the expansion fan (which all start at <math>x=0</math>) we have
<math>c(\rho) = x/t</math>. We can rearrange this and solve for <math>x</math> and obtain
<math>\rho =\frac{ 1}{2} \rho_{\max} (1-x/v_0 t)</math>

The solution is then given by
<center><math>
\rho(x,t) =
\left\{
\begin{matrix}
\rho_{L},& x < c(\rho_L) t\\
\frac{ \rho_{\max}}{2} (1-x/v_0 t),& c(\rho_L) t \leq x \leq c(\rho_R) t\\
\rho_{R},& x > c(\rho_R) t
\end{matrix}
\right.
</math></center>
This solution is known as an '''expansion fan'''.

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
0.6,& x < 0 \\
0.3,& x > 0
\end{matrix}
\right.
</math></center>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Expansion_fan_rho.jpg|thumb|350px| <math>\rho_0</math> ]]
| [[Image:Expansion_fan_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Expansion_fan_characteristics.jpg|thumb|350px|Characterisitics]]
| [[Image:Expansion_fan1.gif|thumb|350px|<math>\rho(x,t)</math>]]
|}

=== Shocks ===

So far we have only considered the case when <math>c(x_0)</math> is monotonically increasing so that
two characteristics never cross. We now consider the case when characteristics can meet.
A movie of this case is shown below.

{{#ev:youtube|Suugn-p5C1M}}

We can easily see that
the first characteristics to meet will be neighbouring characteristics. Consider two characteristics
with
<center><math>
X_1(t) = x_0 + c(x_0)t\,
</math></center>
<center><math>
X_2(t) = x_0 + \delta x + c(x_0+\delta x)t\,
</math></center>
Then these curves will meet at time <math>T</math> where
<center><math>
x_0 + c(x_0)T = x_0 + \delta x + c(x_0+\delta x)T\,
</math></center>
which implies that
<center><math>
T = -\frac{1}{c^{\prime}(x_0)}
</math></center>
Note the following
* If <math>c^{\prime}(x) > 0 </math> then no shock will form.
* The shock first forms at the minimum positive value of
<math> - \frac{1}{c^{\prime}(x)} </math> for <math> -\infty < x <\infty </math>.

==== Shock Fitting ====

If we calculate the solution using our formula
<center><math>
\rho(x_0 + c(\rho_0(x_0))t,t) = \rho(x_0)\,
</math></center>
then we find that the solution becomes multivalued in the case when a shock forms.
We then have to fit a shock. One way to do this is by imposing the condition that equal
areas are removed and added when we chose the position of the shock.
This corresponds to the condition that
the number of cars must be conserved

==== Speed of the shock ====

If we consider the case when there is a shock at <math>s(t)</math> with
<math>\rho = \rho^{-}</math> at <math>s=s^{-}</math>
and
<math>\rho = \rho^{+}</math> at <math>s=s^{+}</math>
(where <math>s^{-}</math>
is just less than s(t) and <math>s^{+}</math>
is just greater than s(t) ). If we substitute
this into the governing integral equation we obtain
<center><math>
\frac{\partial}{\partial t} \left( \int_{x_1}^{s(t)} + \int_{s(t)}^{x_2}\right)
\rho(x,t)\mathrm{d}x = q(x_1,t) - q(x_2,t)
</math></center>
and hence
<center><math>
\int_{x_1}^{x_2}
\frac{\partial \rho(x,t)}{\partial t} \mathrm{d}x + \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{-}
- \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{+} = q(x_1,t) - q(x_2,t)
</math></center>
If we now take the limit as <math>x_1\to x_2</math> we obtain
<center><math>
\frac{\mathrm{d}s}{\mathrm{d}t}\rho^{-}
- \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{+} = q(\rho^{-}) - q(\rho^{+})
</math></center>
so that
<center><math>
\frac{\mathrm{d}s}{\mathrm{d}t} = \frac{q(\rho^{-}) - q(\rho^{+})}
{\rho^{-} - \rho^{+}}
</math></center>

==== Shock example ====

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by <math>\rho_0 = 1/2(1 + \tanh(x))</math>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Traffic example2 rho.jpg|thumb|350px| <math>\rho_0 = 1/2(1+ \tanh(x))</math> ]]
| [[Image:Traffic_example2_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Traffic_example2_characteristics.jpg|thumb|350px|Characterisitics for <math>\rho_0 = 1/2(1+ \tanh(x))</math>]]
| [[Image:Traffic_example2.gif|thumb|350px|<math>\rho(x,t) for </math><math>\rho_0 = 1/2(1+ \tanh(x))</math> Dotted solution is without shock fitting.]]
|}

==== Riemann problem ====

We now consider the Riemann problem
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < 0 \\
\rho_{R},& x > 0
\end{matrix}
\right.
</math></center>
where <math>\rho_{L} < \rho_{R}</math>. In this case a shock forms immediately and
the characteristics terminate at the shock. The shock moves with constant speed given by
the equation for the motion of the shock (or can be found by the equal areas rule). We obtain
<center><math>
\rho(x,t) =
\left\{
\begin{matrix}
\rho_{L},& x < \frac{1}{2} \left(c(\rho_{L}) + c(\rho_{R}) \right) t \\
\rho_{R},& x > \frac{1}{2} \left(c(\rho_{L}) + c(\rho_{R}) \right) t
\end{matrix}
\right.
</math></center>

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
0.3,& x < 0 \\
0.6,& x > 0
\end{matrix}
\right.
</math></center>

The figures below show the initial density, the initial speed,
the characteristics, and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Shock_rho.jpg|thumb|350px| <math>\rho_0</math> ]]
| [[Image:Shock_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Shock_characteristics.jpg|thumb|350px|Characterisitics with shock shown in green.]]
| [[Image:Shock3.gif|thumb|350px|<math>\rho(x,t)</math> with the red dotted line showing the solution without shock fitting]]
|}

[[Category:Simple Nonlinear Waves]]

Connection betwen KdV and the Schrodinger Equation

2010-11-06T09:48:24Z

Adi Kurniawan: /* Reflectionless Potential */

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknows <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>mathbf{C}</math> are given by
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This gives us
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{-2k_{1}x + 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

Connection betwen KdV and the Schrodinger Equation

2010-11-06T09:47:39Z

Adi Kurniawan: /* Scattering Data */

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknows <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>mathbf{C}</math> are given by
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This gives us
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{-2k_{1}x + 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

Properties of the Linear Schrodinger Equation

2010-11-05T15:13:21Z

Adi Kurniawan: /* Example 1: \delta function potential */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example 1: <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) = u_0 \delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \partial_x^2 w +\delta(x) w + \lambda w \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma<x<\varsigma\\
a_{2}e^{-kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\varsigma\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma<x<\varsigma\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Linear Wave-Body Interaction

2010-11-05T15:10:30Z

Adi Kurniawan: /* Haskind relations of exciting forces */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Linear Wave-Body Interaction
| next chapter = [[Long Wavelength Approximations]]
| previous chapter = [[Ship Kelvin Wake]]
}}

{{incomplete pages}}

[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]

We consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] in the
[[Frequency Domain Problem|Frequency Domain]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem).

== Introduction ==

We derive here the equations of motion for a body in [[Linear Plane Progressive Regular Waves]] in the frequency domain in
two dimensions. We begin with the equations in the time domian. The simplest problems is [[Waves reflecting off a vertical wall]]

== Equations for a Floating Body in the Time Domain ==

We begin with the equations for a floating two-dimensional body in the time domain.

{{equations of motion time domain without body condition}}

{{two dimensional floating body time domain}}

More details can be found in [[:Category:Time-Dependent Linear Water Waves|Time-Dependent Linear Water Waves]]

== Equations for a Floating Body in the Frequency Domain ==

The dynamic condition is the equation of motion for the structure in the [[Frequency Domain Problem|frequency domain]]
can be found from the time domain equations and we introduce the following notation
<center><math>
\xi_{\nu} = \zeta_{\nu}e^{-\mathrm{i}\omega t}\,
</math></center>
This give us
{{standard linear wave scattering equations without body condition}}
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=\mathrm{i}\omega\rho\iint_{\partial\Omega}\phi n_{\mu}\, \mathrm{d}S
- \sum_{\nu} C_{\mu\nu}\zeta_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>
The equations of motion for <math> \zeta_\nu\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions. We use the standard numbering of the modes of motion.

== Equations for a Fixed Body in Frequency Domain ==

The equations for a fixed body are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for a rigid body}}
plus the radiation conditions.

We decompose the potential as
<center>
<math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}}
</math>
</center>
where <math>\phi^{\mathrm{I}}</math> is the incident potential and <math>\phi^{\mathrm{D}}</math>
is the diffracted potential. The boundary condition for the diffracted potential is
<center><math>
\Delta\phi^{\mathrm{D}}=0, \, -h<z<0,\,\,\,\mathbf{x} \in \Omega
</math></center>
<center><math>
\partial_n\phi^{\mathrm{D}} = 0, \, z=-h,
</math></center>
<center><math>
\partial_n \phi^{\mathrm{D}} = \alpha \phi,\,z=0,\,\,\mathbf{x} \in \partial\Omega_{F},
</math></center>
plus
<center><math>
\partial_n \phi^{\mathrm{D}} = - \partial_n \phi^{\mathrm{I}},\,\, \mathbf{x} \in \partial\Omega_{B},
</math></center>

Code to calculate the solution (using a slighly modified method) can be found in
[[Boundary Element Method for a Fixed Body in Finite Depth]]

== Equations for the Radiation Potential in Frequency Domain ==

We decompose the body motion into the rigid body modes of motion. Associated with
each of these modes is a potential which must be solved for.
The equations for the radiation potential in the frequency domain are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for the radiation modes}}

{{sommerfeld radiation condition two dimensions for radiation}}

Code to calculate the radiation potential can be found in
[[Boundary Element Method for the Radiation Potential in Finite Depth]]

We denote the solution for each of the radiation potentials by
<math>\phi_\nu^{\mathrm{R}}</math> and the total potential is written as
<center><math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_\nu \zeta_\nu \phi_\nu^{\mathrm{R}}
</math>
</center>

== Final System of Equations ==

We substitute the expansion for the potential into the equations in the frequency domain and we obtain
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
\left(\phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_{\nu} \zeta_\nu \phi_{\nu}^{\mathrm{R}}\right) \mathbf{n}_{\mu}\, dS
- \sum_{\nu} C_{\mu\nu}\xi_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>

{{added mass damping and force matrices definition}}

Then the equations can be expressed as follows.
<center><math> \left[-\omega^2 \left(\mathbf{M} + \mathbf{A} \right) +
\mathrm{i}\omega \mathbf{B} + \mathbf{C} \right] \vec{\zeta} = \mathbf{f} </math></center>
where <math>\mathbf{M}</math> is the mass matrix, <math>\mathbf{A}</math> is the added mass matrix,
<math>\mathbf{B}</math> is the damping matrix, <math>\mathbf{C}</math> is the hydrostatic matrix,
<math>\vec{\zeta}</math> is the vector of body displacements and <math>\mathbf{f}</math> is the force.

The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the two dimensional problem for the sake of clarity.

== Symmetric body ==

For a body which is [[:Category:Symmetry in Two Dimensions|Symmetric in Two Dimensions]]
the Heave is decoupled from Surge and Roll.
In other words the Surge and Roll motions do not influence Heave and vice versa.

== Matlab Code ==

*A program to solve for pitch and heave and only for two geometries can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/rigid_body_motion.m rigid_body_motion.m]

* a program to calculate the solution for a specific geometry (with plot as output as shown) can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/wave_bem_example_floating.m wave_bem_example_floating.m]

[[Image:Wave_bem_example_floating_RT2.jpg|300px|right|thumb|The reflection (solid line) and transmission (dashed line)
for a dock for heave and pitch (red), heave only (blue) and pitch only (black)]]

=== Additional code ===

This program requires
* A program to calculate the geometery [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/circlebody_twod.m circlebody_twod.m]
* {{fixed body bem code}}
* {{floating body radiation code}}
* {{free surface dispersion equation code}}
* {{boundary element code}}

== Symmetry-reciprocity relations ==

It will be shown that
<center><math> A_{ij}(\omega) = A_{ji}(\omega) \,</math></center>
<center><math> B_{ij}(\omega) = B_{ji}(\omega) \,</math></center>
Along the same lines it will be shown that the exciting force <math>\mathbf{X}_j\,</math> can be expressed in terms of <math> \psi_j\,</math> circumventing the solution for the diffraction potential.
The core result needed for the proof of the above properties is [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
<center><math> \iint_S \left( \psi_1 \frac{\partial\psi_2}{\partial n} - \psi_2 \frac{\partial\psi_1}{\partial n} \right) \mathrm{d}S = 0 \,</math></center>
where <math>\nabla^2 \psi_i=0</math>.

{{energy_region_plates}}

[[Image:Symmetry_boundary.jpg|thumb|right|600px|Boundary]]

In the surface wave-body problem define the closed surfaces as shown in figure on the right.
Let <math> \phi_j\,</math> be rediation or diffraction potentials. Over the boundaries <math>S^\pm\,</math> we have:

<center><math> S^+: \quad \phi_j \ \sim \ \frac{igA_j^+}{\omega} e^{Kz-iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial \phi_j}{\partial x} \ \sim \ -iK\phi_j \,</math></center>

<center><math> S^-: \quad \phi_j \ \sim \ \frac{igA_j^-}{\omega} e^{Kz+iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = - \frac{\partial \phi_j}{\partial x} \ \sim \ - iK\phi_j \,</math></center>

On <math> S_F: \qquad \frac{\partial\phi_j}{\partial z} = K\phi_j, \qquad \frac{\partial \Phi_j}{\partial n} = \frac{\partial \phi_j}{\partial z} </math>

On <math> S_\infty: \qquad \left| \phi_j \right|, \quad \left| \nabla \phi_j \right| \to 0 </math>.

Applying Green's identity to any pair of the radiation potentials <math> \psi_i, \psi_j \,</math>:

<center><math> \iint_{S_B} \left[ \psi_i \frac{\partial\psi_j}{\partial n} - \psi_j \frac{\partial\psi_i}{\partial n} \right] \mathrm{d}S = - \iint_{S_F} \left[ \psi_i \frac{\partial\psi_j}{\partial z} - \psi_j \frac{\partial\psi_i}{\partial z} \right] \mathrm{d}S </math></center>

<center><math> - \iint_{S_+} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S
+ \iint_{S_-} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S = 0 </math></center>

It follows that:

<center><math> \iint_{S_B} \psi_i \frac{\partial\psi_j}{\partial n} \mathrm{d}S = \iint_{S_B} \psi_j \frac{\partial\psi_i}{\partial n} \mathrm{d}S </math></center>
or
<center><math> A_{ij}(\omega) = A_{ji}(\omega), \qquad B_{ij}(\omega) = B_{ji}(\omega). \,</math></center>

== Haskind relations of exciting forces ==

<center><math> \mathbf{X}_i(\omega) = - i\omega\rho\iint_{S_B} (\phi_I + \phi_7) n_i \mathrm{d}S </math></center>

<center><math> = - \rho \iint_{S_B} (\phi_I + \phi_7) \frac{\partial \phi_i}{\partial n} \mathrm{d}S </math></center>

where the radiation velocity potential <math> \phi_i \,</math> is known to satisfy:

<center><math> \frac{\partial\phi_i}{\partial n} = i\omega n_i, \quad \mbox{on} \ S_B </math></center>

and

<center><math> \frac{\partial\phi_7}{\partial n} = \frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \ S_B </math></center>

Both <math> \phi_i\,</math> and <math> \phi_7\,</math> satisfy the condition of outgoing waves at infinity. By virtue of [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]

<center><math> \iint_{S_B} \phi_7 \frac{\partial\phi_i}{\partial n} \mathrm{d}S = \iint_{S_B} \phi_i \frac{\partial\phi_7}{\partial n} \mathrm{d}S = -\iint_{S_B} \phi_i \frac{\partial\phi_I}{\partial n} \mathrm{d}S </math></center>

The Haskind expression for the exciting force follows:

<center><math> \mathbf{X}_i(\omega) = \rho \iint_{S_B} \left[ \phi_I \frac{\partial\phi_i}{\partial n} - \phi_i \frac{\partial\phi_I}{\partial n} \right] \mathrm{d}S </math></center>

The symmetry of the <math> A_{ij}(\omega), B_{ij}(\omega) \,</math> matrices applies in 2D and 3D. The application of Green's Theorem in 3D is very similar using the far-field representation for the potential <math> \phi_j\,</math>

<center><math> \phi_j \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O\left(\frac{1}{R^{3/2}}\right) </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial\phi_j}{\partial R} \sim - i K \phi_j + O\left(\frac{1}{R^{3/2}}\right) </math></center>

where <math> R \,</math> is a radius from the body out to infinity and the <math> R^{-\frac{1}{2}} \,</math> decay arises from the energy conservation principle. Details of the 3D proof may be found in [[Mei 1983]] and [[Wehausen and Laitone 1960]]

The use of the Haskind relations for the exciting forces does not require the solution of the diffraction problem. This is convenient and often more accurate.

The Haskind relations take other forms which will not be presented here but are detailed in [[Wehausen and Laitone 1960]]. The ones that are used in practice relate the exciting forces to the damping coefficients.

These take the form:

2D: <math> B_{ii} = \frac{\left| \mathbf{X}_i \right|^2}{2\rho g V_g}, \quad V_g = \frac{g}{2\omega}, </math> Deep water

3D: <math> B_{33} = \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2 \,</math> --- Heave

(Axisymmetric bodies) <math> B_{22} = \frac{K}{8\rho g V_g} \left| \mathbf{X}_2 \right|^2 \,</math> --- Sway

So knowledge of <math> \mathbf{X}_i(\omega)\,</math> allows the direct evaluation of the diagonal damping coefficients. These expressions are useful in deriving theoretical results in wave-body interactions to be discussed later.

The two-dimensional theory of wave-body interactions in the frequency domain extends to three dimencions very directly with little difficulty.

The statement of the 6 d.o.f. seakeeping problem is:

<center><math> \sum_{j=1}^6 \left[ - \omega^2 \left( M_{ij} + A_{ij} \right) + i \omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_j, \quad i=1,\cdots,6 </math></center>

Where:

<center><math> \mbox{Body inertia matrix including moments of inertia for rotational modes. For details refer to MH} \, </math></center>

<center><math> A_{ij}(\omega): \mbox{Added mass matrix} \,</math></center>

<center><math> B_{ij}(\omega): \mbox{Damping matrix} \, </math></center>

<center><math> C_{ij}: \mbox{Hydrostatic and static inertia restoring matrix. For details refer to MH} \, </math></center>

<center><math> \mathbf{X}_i(\omega): \mbox{Wave exciting forces and moments} </math></center>

At zero speed the definitions of the added-mass, damping matrices and exciting forces are identical to those in two dimensions.

The boundary value problems satisfied by the radiation potentials <math>\phi_j, \ j=1,\cdots,6 \,</math> and the diffraction potential <math> \phi_7 \,</math> are as follows:

Free-surface condition:

<center><math> -\omega^2 \phi_j + g \frac{\partial\phi_j}{\partial Z} = 0, \quad z=0 \quad j=1,\cdots,7 </math></center>

Body-boundary conditions:

<center><math> \frac{\partial\phi_7}{\partial n} = -\frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \quad S_B </math></center>

<center><math> \phi_I = \frac{i g A}{\omega} e^{KZ-iKX\cos\beta-iKY\sin\beta+i\omega t} </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = i\omega n_j, \quad j=1,\cdots,6 \, </math></center>

<center><math> n_j = \begin{Bmatrix}
& n_j, \qquad & j=1,2,3 \\
& \left( \vec{X} \times \vec{n} \right)_{j+3}, \quad j=4,5,6
\end{Bmatrix} </math></center>

<center><math> i=1: \ \mbox{Surge} \qquad i=2: \ \mbox{Sway} \qquad i=3: \ \mbox{Heave} </math></center>
<center><math> i=4: \ \mbox{Roll} \qquad i=5: \ \mbox{Pitch} \qquad i=6: \ \mbox{Yaw} </math></center>

At large distances from the body the velocity potentials satisfy the radiation condition:

<center><math> \phi_j (R,\theta) \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O \left( \frac{1}{R^{3/2}} \right) </math></center>

with

<center><math> K = \frac{\omega^2}{g}. \,</math></center>

This radiation condition is essential for the formulation and solution of the boundary value problems for <math>\phi_j\,</math> using panel methods which are the standard solution technique at zero and forward speed.

Qualitative behaviour of the forces, coefficients and motions of floating bodies

<center><math> - \omega^2 \phi + g \phi_Z =0 \quad \begin{cases}
\phi_Z=0,\quad \omega=0 \\
\phi=0, \quad \omega \to \infty
\end{cases} </math></center>

<center><math> B_{33}(\omega), \sim \omega, \mbox{at low} \ \omega \,</math></center>

The 2D Heave added mass is singular at low frequencies. It is finite in 3D

The 2D Heave damping coefficient is decaying to zero linearly in 2D and superlinearly in 3D. A two-dimensional section is a better wavemaker than a three-dimensional one

A 2D section oscillating in Sway is less effective a wavemaker at low frequencies than the same section oscillating in Heave

The zero-frequency limit of the Sway added mass is finite and similar to the infinite frequency limit of the Heave added mass.

In long waves the Heave exciting force tends to the Heave restoring coefficient times the ambient wave amplitude the free surface behaves like a flat surface moving up and down.

In long waves the Sway exciting force tends to zero. Proof will follow

In short waves all forces tend to zero.

Pitch exciting moment (same applies to Roll) tends to zero. Long waves have a small slope which is proportional to <math> KA</math>, where <math> K\,</math> is the wave number and <math> A\,</math> is the wave amplitude.

Prove that to leading order for <math>KA\to 0 \,</math>:

<center><math> \left| X_S(\omega) \right| \sim KA C_{55}\,</math></center>

where <math>C_{55}\,</math> is the Pitch (<math> C_{44} \,</math> for Roll) hydrostatic restoring coefficient. [NB: very long waves look like a flat surface inclined at <math> KA\,</math> ].

== Body motions in regular waves ==

Heave:

<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega)}{-\omega^2(A_{33} + M) + i\omega B_{33} +C_{33} } </math></center>

Resonance:

<center><math> \omega^2 = \frac{C_{33}}{M+A_{33}} = \frac{\rho g A \omega}{M + A_{33} (\omega)} </math></center>

In principle the above equation is nonlinear for <math>\omega\,</math>. Will be approximated below

At resonance:
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega^*)}{i\omega^* B_{33}(\omega^*)} \,</math></center>

Invoking the relation between the damping coefficient and the exicting force in 3D:

<center><math> \frac{\left| \Pi_3 \right|}{A} = \frac{\left| \mathbf{X}_3(\omega) \right|}{\omega \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2}, \quad V_g=\frac{g}{2\omega} </math></center>

<center><math> =\frac{2\rho g}{\omega^3 \left|\mathbf{X}_3(\omega)\right|}, \quad \mbox{at resonance} </math></center>

This counter-intuitive result shows that for a body undergoing a pure Heave oscillation, the modulus of the Heave response at resonance is inversely proportional to the modulus of the Heave exciting force.

Viscous effects not discussed here may affect Heave response at resonance

The behavior of the Sway response can be found in an analagous manner,

-----

This article is based on the MIT open course notes and the original articles can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here] and
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C5323823-0180-45EA-B165-15856948A0A2/0/lecture10.pdf here]

Linear Wave-Body Interaction

2010-11-05T15:09:48Z

Adi Kurniawan: /* Haskind relations of exciting forces */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Linear Wave-Body Interaction
| next chapter = [[Long Wavelength Approximations]]
| previous chapter = [[Ship Kelvin Wake]]
}}

{{incomplete pages}}

[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]

We consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] in the
[[Frequency Domain Problem|Frequency Domain]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem).

== Introduction ==

We derive here the equations of motion for a body in [[Linear Plane Progressive Regular Waves]] in the frequency domain in
two dimensions. We begin with the equations in the time domian. The simplest problems is [[Waves reflecting off a vertical wall]]

== Equations for a Floating Body in the Time Domain ==

We begin with the equations for a floating two-dimensional body in the time domain.

{{equations of motion time domain without body condition}}

{{two dimensional floating body time domain}}

More details can be found in [[:Category:Time-Dependent Linear Water Waves|Time-Dependent Linear Water Waves]]

== Equations for a Floating Body in the Frequency Domain ==

The dynamic condition is the equation of motion for the structure in the [[Frequency Domain Problem|frequency domain]]
can be found from the time domain equations and we introduce the following notation
<center><math>
\xi_{\nu} = \zeta_{\nu}e^{-\mathrm{i}\omega t}\,
</math></center>
This give us
{{standard linear wave scattering equations without body condition}}
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=\mathrm{i}\omega\rho\iint_{\partial\Omega}\phi n_{\mu}\, \mathrm{d}S
- \sum_{\nu} C_{\mu\nu}\zeta_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>
The equations of motion for <math> \zeta_\nu\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions. We use the standard numbering of the modes of motion.

== Equations for a Fixed Body in Frequency Domain ==

The equations for a fixed body are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for a rigid body}}
plus the radiation conditions.

We decompose the potential as
<center>
<math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}}
</math>
</center>
where <math>\phi^{\mathrm{I}}</math> is the incident potential and <math>\phi^{\mathrm{D}}</math>
is the diffracted potential. The boundary condition for the diffracted potential is
<center><math>
\Delta\phi^{\mathrm{D}}=0, \, -h<z<0,\,\,\,\mathbf{x} \in \Omega
</math></center>
<center><math>
\partial_n\phi^{\mathrm{D}} = 0, \, z=-h,
</math></center>
<center><math>
\partial_n \phi^{\mathrm{D}} = \alpha \phi,\,z=0,\,\,\mathbf{x} \in \partial\Omega_{F},
</math></center>
plus
<center><math>
\partial_n \phi^{\mathrm{D}} = - \partial_n \phi^{\mathrm{I}},\,\, \mathbf{x} \in \partial\Omega_{B},
</math></center>

Code to calculate the solution (using a slighly modified method) can be found in
[[Boundary Element Method for a Fixed Body in Finite Depth]]

== Equations for the Radiation Potential in Frequency Domain ==

We decompose the body motion into the rigid body modes of motion. Associated with
each of these modes is a potential which must be solved for.
The equations for the radiation potential in the frequency domain are

{{standard linear wave scattering equations without body condition}}
{{frequency domain equations for the radiation modes}}

{{sommerfeld radiation condition two dimensions for radiation}}

Code to calculate the radiation potential can be found in
[[Boundary Element Method for the Radiation Potential in Finite Depth]]

We denote the solution for each of the radiation potentials by
<math>\phi_\nu^{\mathrm{R}}</math> and the total potential is written as
<center><math>
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_\nu \zeta_\nu \phi_\nu^{\mathrm{R}}
</math>
</center>

== Final System of Equations ==

We substitute the expansion for the potential into the equations in the frequency domain and we obtain
<center><math>
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
\left(\phi^{\mathrm{I}} + \phi^{\mathrm{D}} +
\sum_{\nu} \zeta_\nu \phi_{\nu}^{\mathrm{R}}\right) \mathbf{n}_{\mu}\, dS
- \sum_{\nu} C_{\mu\nu}\xi_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
</math></center>

{{added mass damping and force matrices definition}}

Then the equations can be expressed as follows.
<center><math> \left[-\omega^2 \left(\mathbf{M} + \mathbf{A} \right) +
\mathrm{i}\omega \mathbf{B} + \mathbf{C} \right] \vec{\zeta} = \mathbf{f} </math></center>
where <math>\mathbf{M}</math> is the mass matrix, <math>\mathbf{A}</math> is the added mass matrix,
<math>\mathbf{B}</math> is the damping matrix, <math>\mathbf{C}</math> is the hydrostatic matrix,
<math>\vec{\zeta}</math> is the vector of body displacements and <math>\mathbf{f}</math> is the force.

The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the two dimensional problem for the sake of clarity.

== Symmetric body ==

For a body which is [[:Category:Symmetry in Two Dimensions|Symmetric in Two Dimensions]]
the Heave is decoupled from Surge and Roll.
In other words the Surge and Roll motions do not influence Heave and vice versa.

== Matlab Code ==

*A program to solve for pitch and heave and only for two geometries can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/rigid_body_motion.m rigid_body_motion.m]

* a program to calculate the solution for a specific geometry (with plot as output as shown) can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/wave_bem_example_floating.m wave_bem_example_floating.m]

[[Image:Wave_bem_example_floating_RT2.jpg|300px|right|thumb|The reflection (solid line) and transmission (dashed line)
for a dock for heave and pitch (red), heave only (blue) and pitch only (black)]]

=== Additional code ===

This program requires
* A program to calculate the geometery [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/circlebody_twod.m circlebody_twod.m]
* {{fixed body bem code}}
* {{floating body radiation code}}
* {{free surface dispersion equation code}}
* {{boundary element code}}

== Symmetry-reciprocity relations ==

It will be shown that
<center><math> A_{ij}(\omega) = A_{ji}(\omega) \,</math></center>
<center><math> B_{ij}(\omega) = B_{ji}(\omega) \,</math></center>
Along the same lines it will be shown that the exciting force <math>\mathbf{X}_j\,</math> can be expressed in terms of <math> \psi_j\,</math> circumventing the solution for the diffraction potential.
The core result needed for the proof of the above properties is [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
<center><math> \iint_S \left( \psi_1 \frac{\partial\psi_2}{\partial n} - \psi_2 \frac{\partial\psi_1}{\partial n} \right) \mathrm{d}S = 0 \,</math></center>
where <math>\nabla^2 \psi_i=0</math>.

{{energy_region_plates}}

[[Image:Symmetry_boundary.jpg|thumb|right|600px|Boundary]]

In the surface wave-body problem define the closed surfaces as shown in figure on the right.
Let <math> \phi_j\,</math> be rediation or diffraction potentials. Over the boundaries <math>S^\pm\,</math> we have:

<center><math> S^+: \quad \phi_j \ \sim \ \frac{igA_j^+}{\omega} e^{Kz-iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial \phi_j}{\partial x} \ \sim \ -iK\phi_j \,</math></center>

<center><math> S^-: \quad \phi_j \ \sim \ \frac{igA_j^-}{\omega} e^{Kz+iKx} \,</math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = - \frac{\partial \phi_j}{\partial x} \ \sim \ - iK\phi_j \,</math></center>

On <math> S_F: \qquad \frac{\partial\phi_j}{\partial z} = K\phi_j, \qquad \frac{\partial \Phi_j}{\partial n} = \frac{\partial \phi_j}{\partial z} </math>

On <math> S_\infty: \qquad \left| \phi_j \right|, \quad \left| \nabla \phi_j \right| \to 0 </math>.

Applying Green's identity to any pair of the radiation potentials <math> \psi_i, \psi_j \,</math>:

<center><math> \iint_{S_B} \left[ \psi_i \frac{\partial\psi_j}{\partial n} - \psi_j \frac{\partial\psi_i}{\partial n} \right] \mathrm{d}S = - \iint_{S_F} \left[ \psi_i \frac{\partial\psi_j}{\partial z} - \psi_j \frac{\partial\psi_i}{\partial z} \right] \mathrm{d}S </math></center>

<center><math> - \iint_{S_+} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S
+ \iint_{S_-} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S = 0 </math></center>

It follows that:

<center><math> \iint_{S_B} \psi_i \frac{\partial\psi_j}{\partial n} \mathrm{d}S = \iint_{S_B} \psi_j \frac{\partial\psi_i}{\partial n} \mathrm{d}S </math></center>
or
<center><math> A_{ij}(\omega) = A_{ji}(\omega), \qquad B_{ij}(\omega) = B_{ji}(\omega). \,</math></center>

== Haskind relations of exciting forces ==

<center><math> \mathbf{X}_i(\omega) = - i\omega\rho\iint_{S_B} (\phi_I + \phi_7) n_i \mathrm{d}S </math></center>

<center><math> = - \rho \iint_{S_B} (\phi_I + \phi_7) \frac{\partial \phi_i}{\partial n} \mathrm{d}S </math></center>

where the radiation velocity potential <math> \phi_i \,</math> is known to satisfy:

<center><math> \frac{\partial\phi_i}{\partial n} = i\omega n_i, \quad \mbox{on} \ S_B </math></center>

and

<center><math> \frac{\partial\phi_7}{\partial n} = \frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \ S_B </math></center>

Both <math> \phi_i\,</math> and <math> \phi_7\,</math> satisfy the condition of outgoing waves at infinity. By virtue of [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]

<center><math> \iint_{S_B} \phi_7 \frac{\partial\phi_i}{\partial n} \mathrm{d}S = \iint_{S_B} \phi_i \frac{\partial\phi_7}{\partial n} \mathrm{d}S = -\iint_{S_B} \phi_i \frac{\partial\phi_I}{\partial n} \mathrm{d}S </math></center>

The Haskind expression for the exciting force follows:

<center><math> \mathbf{X}_i(\omega) = \rho \iint_{S_B} \left[ \phi_I \frac{\partial\phi_i}{\partial n} - \phi_i \frac{\partial\phi_I}{\partial n} \right] \mathrm{d}S </math></center>

The symmetry of the <math> A_{ij}(\omega), B_{ij}(\omega) \,</math> matrices applies in 2D and 3D. The application of Green's Theorem in 3D is very similar using the far-field representation for the potential <math> \phi_j\,</math>

<center><math> \phi_j \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O\left(\frac{1}{R^{3/2}}\right) </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial\phi_j}{\partial R} \sim - i K \phi_j + O\left(\frac{1}{R^{3/2}}\right) </math></center>

where <math> R \,</math> is a radius from the body out to infinity and the <math> R^{-\frac{1}{2}} \,</math> decay arises from the energy conservation principle. Details of the 3D proof may be found in [[Mei 1983]] and [[Wehausen and Laitone 1960]]

The use of the Haskind relations for the exciting forces does not require the solution of the diffraction problem. This is convenient and often more accurate.

The Haskind relations take other forms which will not be presented here but are detailed in [[Wehausen and Laitone 1960]]. The ones that are used in practice relate the exciting forces to the damping coefficients.

These take the form:

2D: <math> B_ii = \frac{\left| \mathbf{X}_i \right|^2}{2\rho g V_g}, \quad V_g = \frac{g}{2\omega}, </math> Deep water

3D: <math> B_33 = \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2 \,</math> --- Heave

(Axisymmetric bodies) <math> B_22 = \frac{K}{8\rho g V_g} \left| \mathbf{X}_2 \right|^2 \,</math> --- Sway

So knowledge of <math> \mathbf{X}_i(\omega)\,</math> allows the direct evaluation of the diagonal damping coefficients. These expressions are useful in deriving theoretical results in wave-body interactions to be discussed later.

The two-dimensional theory of wave-body interactions in the frequency domain extends to three dimencions very directly with little difficulty.

The statement of the 6 d.o.f. seakeeping problem is:

<center><math> \sum_{j=1}^6 \left[ - \omega^2 \left( M_{ij} + A_{ij} \right) + i \omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_j, \quad i=1,\cdots,6 </math></center>

Where:

<center><math> \mbox{Body inertia matrix including moments of inertia for rotational modes. For details refer to MH} \, </math></center>

<center><math> A_{ij}(\omega): \mbox{Added mass matrix} \,</math></center>

<center><math> B_{ij}(\omega): \mbox{Damping matrix} \, </math></center>

<center><math> C_{ij}: \mbox{Hydrostatic and static inertia restoring matrix. For details refer to MH} \, </math></center>

<center><math> \mathbf{X}_i(\omega): \mbox{Wave exciting forces and moments} </math></center>

At zero speed the definitions of the added-mass, damping matrices and exciting forces are identical to those in two dimensions.

The boundary value problems satisfied by the radiation potentials <math>\phi_j, \ j=1,\cdots,6 \,</math> and the diffraction potential <math> \phi_7 \,</math> are as follows:

Free-surface condition:

<center><math> -\omega^2 \phi_j + g \frac{\partial\phi_j}{\partial Z} = 0, \quad z=0 \quad j=1,\cdots,7 </math></center>

Body-boundary conditions:

<center><math> \frac{\partial\phi_7}{\partial n} = -\frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \quad S_B </math></center>

<center><math> \phi_I = \frac{i g A}{\omega} e^{KZ-iKX\cos\beta-iKY\sin\beta+i\omega t} </math></center>

<center><math> \frac{\partial\phi_j}{\partial n} = i\omega n_j, \quad j=1,\cdots,6 \, </math></center>

<center><math> n_j = \begin{Bmatrix}
& n_j, \qquad & j=1,2,3 \\
& \left( \vec{X} \times \vec{n} \right)_{j+3}, \quad j=4,5,6
\end{Bmatrix} </math></center>

<center><math> i=1: \ \mbox{Surge} \qquad i=2: \ \mbox{Sway} \qquad i=3: \ \mbox{Heave} </math></center>
<center><math> i=4: \ \mbox{Roll} \qquad i=5: \ \mbox{Pitch} \qquad i=6: \ \mbox{Yaw} </math></center>

At large distances from the body the velocity potentials satisfy the radiation condition:

<center><math> \phi_j (R,\theta) \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O \left( \frac{1}{R^{3/2}} \right) </math></center>

with

<center><math> K = \frac{\omega^2}{g}. \,</math></center>

This radiation condition is essential for the formulation and solution of the boundary value problems for <math>\phi_j\,</math> using panel methods which are the standard solution technique at zero and forward speed.

Qualitative behaviour of the forces, coefficients and motions of floating bodies

<center><math> - \omega^2 \phi + g \phi_Z =0 \quad \begin{cases}
\phi_Z=0,\quad \omega=0 \\
\phi=0, \quad \omega \to \infty
\end{cases} </math></center>

<center><math> B_{33}(\omega), \sim \omega, \mbox{at low} \ \omega \,</math></center>

The 2D Heave added mass is singular at low frequencies. It is finite in 3D

The 2D Heave damping coefficient is decaying to zero linearly in 2D and superlinearly in 3D. A two-dimensional section is a better wavemaker than a three-dimensional one

A 2D section oscillating in Sway is less effective a wavemaker at low frequencies than the same section oscillating in Heave

The zero-frequency limit of the Sway added mass is finite and similar to the infinite frequency limit of the Heave added mass.

In long waves the Heave exciting force tends to the Heave restoring coefficient times the ambient wave amplitude the free surface behaves like a flat surface moving up and down.

In long waves the Sway exciting force tends to zero. Proof will follow

In short waves all forces tend to zero.

Pitch exciting moment (same applies to Roll) tends to zero. Long waves have a small slope which is proportional to <math> KA</math>, where <math> K\,</math> is the wave number and <math> A\,</math> is the wave amplitude.

Prove that to leading order for <math>KA\to 0 \,</math>:

<center><math> \left| X_S(\omega) \right| \sim KA C_{55}\,</math></center>

where <math>C_{55}\,</math> is the Pitch (<math> C_{44} \,</math> for Roll) hydrostatic restoring coefficient. [NB: very long waves look like a flat surface inclined at <math> KA\,</math> ].

== Body motions in regular waves ==

Heave:

<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega)}{-\omega^2(A_{33} + M) + i\omega B_{33} +C_{33} } </math></center>

Resonance:

<center><math> \omega^2 = \frac{C_{33}}{M+A_{33}} = \frac{\rho g A \omega}{M + A_{33} (\omega)} </math></center>

In principle the above equation is nonlinear for <math>\omega\,</math>. Will be approximated below

At resonance:
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega^*)}{i\omega^* B_{33}(\omega^*)} \,</math></center>

Invoking the relation between the damping coefficient and the exicting force in 3D:

<center><math> \frac{\left| \Pi_3 \right|}{A} = \frac{\left| \mathbf{X}_3(\omega) \right|}{\omega \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2}, \quad V_g=\frac{g}{2\omega} </math></center>

<center><math> =\frac{2\rho g}{\omega^3 \left|\mathbf{X}_3(\omega)\right|}, \quad \mbox{at resonance} </math></center>

This counter-intuitive result shows that for a body undergoing a pure Heave oscillation, the modulus of the Heave response at resonance is inversely proportional to the modulus of the Heave exciting force.

Viscous effects not discussed here may affect Heave response at resonance

The behavior of the Sway response can be found in an analagous manner,

-----

This article is based on the MIT open course notes and the original articles can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here] and
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C5323823-0180-45EA-B165-15856948A0A2/0/lecture10.pdf here]

Conservation Laws and Boundary Conditions

2010-09-02T14:08:34Z

Adi Kurniawan: /* Definition of force and moment in terms of fluid pressure */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Conservation Laws and Boundary Conditions
| next chapter = [[Linear and Second-Order Wave Theory]]
| previous chapter =
}}

{{complete pages}}

We begin by deriving the equations of motion used to model ocean waves. The equations of motion of a general fluid are given by the celebrated [http://en.wikipedia.org/wiki/Navier_Stokes Navier Stokes equations]. However, for the large scale processes that occur in ocean waves many simplifications are possible.

== Coordinate system and velocity potential ==

[[Image:Coordinate_system.png|right|thumb|500px|Coordinate System]]

We begin by defining the coordinate system.

<center><math>
\begin{matrix}
&(x,y,z) &: &\mbox{Coordinate system} \\
&\mathbf{x} &: &\mbox{Fixed Eulerian Vector} \\
& \mathbf{v} &: &\mbox{Flow Velocity Vector at} \ \mathbf{x} \\
&\zeta &: &\mbox{Free Surface Elevation} \\
&\mathbf{g} &: &\mbox{Acceleration due to gravity}
\end{matrix} </math></center>

At the moment we have not yet defined the region of space which is occupied by the fluid. However, the free surface plays a very important role in the propagation of ocean waves.

The most important assumption we make is that the fluid is an [http://en.wikipedia.org/wiki/Viscosity ideal fluid], i.e. there are no shear stresses due to viscosity and that the flow is [http://en.wikipedia.org/wiki/Irrotational irrotational]. This means that

<center><math>\nabla \times \mathbf{v} = 0</math></center>

We now introduce the very important concept of the velocity potential. Essentially this allows us to express the solution as a function of a scalar (a function which has a single value as opposed to a vector function which has multiple values) rather than a vector throughout the fluid domain. There is an important theorem in vector calculus [http://en.wikipedia.org/wiki/Irrotational_vector_field] that if <math>\nabla \times \mathbf{v} = 0</math> then we can express the irrotational vector as the gradient of a scalar function, i.e.
<center><math>
\mathbf{v} = \nabla \Phi
</math></center>
where <math>\Phi(\mathbf{x},t)</math> is called the [http://en.wikipedia.org/wiki/Velocity_potential velocity potential].

It turns out that the potential flow model of surface wave propagation and wave-body interactions is very accurate for the kind of length and time scales which are important in the ocean. It is important to realise, however, that we have made considerable simplifications and that certain processes, most notably wave breaking, are in no way covered by this theory. In fact, the process of wave breaking is extremely complicated and is much less well understood than the potential flow model.

== Conservation of mass ==

The key equation we will solve to understand ocean waves is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation] which is derived very simply from the expression for the velocity potential. We assume that the fluid is incompressible so that conservation of mass gives us the following condition
<center><math> \nabla \cdot \mathbf{v} = 0 </math></center>
This condition in turn implies, using the definition of the velocity potential that
<center><math>
\nabla \cdot \nabla \Phi = 0 \Rightarrow \nabla^2 \Phi = 0
</math></center>
or
<center><math>
\partial_x^2 \Phi + \partial_y^2\Phi + \partial_z^2\Phi = 0,
</math></center>
which is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation].

== Conservation of linear momentum ==

We begin with [http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29 Euler's equation] in the absence of viscosity
<center> <math>
\partial_t \mathbf{v} + (\mathbf{v}\cdot \nabla)\mathbf{v}= - \frac1{\rho} \nabla P + \mathbf{g}
</math> </center>
where
<math> P(\mathbf{x}, t) </math> is the fluid Pressure at <math>(\mathbf{x}, t)</math> and
<math> \mathbf{g}= - \mathbf{k} g </math> is the acceleration due to gravity where
<math> \mathbf{k} </math> is the unit vector pointing in the positive <math>z</math>-direction (so we are now setting the <math>z</math> coordinate to point in the vertical direction). Finally <math> \rho </math> is the water density.

We then use the following vector identity
<center><math> (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac 1{2} \nabla (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v}\times ( \nabla \times \mathbf{v}) </math></center>
and since we have irrotational flow (i.e. <math> \nabla \times \mathbf{v}= 0 </math>) Euler's equation becomes
<center><math> \partial_t \mathbf{v} + \frac{1}{2} \nabla (\mathbf{v} \cdot \mathbf{v}) = - \frac 1{\rho} \nabla P - \nabla (g z) </math></center>
where we have used <math> \nabla z = \mathbf{k} </math>.
We now substitute <math> \mathbf{v}= \nabla \Phi </math> and we obtain
<center><math> \nabla (\partial_t \Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z ) = 0. </math></center>
We now observe that if
<center><math> \nabla F( \mathbf{x}, t) =0 \quad \Longrightarrow \quad F (\mathbf{x}, t) = C </math></center>
where <math> C </math> is an arbitrary constant.

==== Bernoulli's equation ====

[http://en.wikipedia.org/wiki/Bernoulli%27s_equation Bernoulli's equation] follows from the equation above.
<center><math> \partial_t \Phi + \frac 1{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z = C </math></center>
or
<center><math> \frac{P}{\rho} = - \partial_t\Phi-\frac{1}{2}\nabla\Phi \cdot \nabla \Phi - g z + C </math></center>

The value of the constant <math> C </math> is immaterial (it can be thought of as defining the reference pressure.
It is also worth noting that the
angular momentum conservation principle is contained in
<math> \nabla \times \mathbf{v} = 0. </math>
In particular, if the particles are modelled as spheres, this equation implies no angular velocity at all times.

== Derivation of nonlinear free-surface condition ==

A very important result is the boundary condition at the free surface of the fluid and air. There are two conditions which relate the free surface displacement <math>\zeta(x,y,t)</math> and the velocity potential <math>\Phi(x,y,z,t)</math> at the free surface. The dynamic condition is derived from the Bernoulli's equation and the kinematic condition is derived from the equations linking the fact that the velocity vector at the surface is given by the gradient of the potential. We will present two methods to derive these equations.

=== Method I ===

We derive the dynamic condition directly from Bernoulli's equation.
On <math> z=\zeta; \ P=P_a \equiv \mbox{Atmospheric Pressure} </math>.
This allows us to rewrite Bernoulli's equation as
<center><math> \frac{P_a}{\rho}=-\frac{\partial\Phi}{\partial t}-\frac{1}{2}\nabla\Phi\cdot\nabla\Phi-g\zeta+C \qquad \mbox{on} \ z=\zeta(x,y,t) </math></center>
We will simplify this equation by showing that we are free to set the pressure to any value.

The kinematic condition is derived as follows.
On <math>z=\zeta</math> The mathematical function
<center><math>z-\zeta(x,y,t)\equiv\tilde{f}(x,y,z,t)</math></center>

is always zero when tracing a fluid particle on the free surface. So the [http://en.wikipedia.org/wiki/Total_derivative substantial or total derivative] of <math>\tilde{f}</math> must vanish, thus

<center><math> \frac{D\tilde{f}}{Dt}=0=\left (\partial_t + \mathbf{v} \cdot \nabla \right ) \tilde{f}=0, \qquad \mbox{on} \ z=\zeta </math></center>

Expanding, we obtain
<center><math> \left (\partial_t + \mathbf{v} \cdot \nabla \right ) (z-\zeta) =0, \qquad \mbox{on} \ z=\zeta </math></center>
which in turn implies that
<center><math> \partial_t\zeta + \partial_x\Phi \partial_x\zeta + \partial_y\Phi \partial_y\zeta = \partial_z\Phi, \qquad \mbox{on} \ z=\zeta </math></center>
which is the Kinematic free-surface condition.

We have already derived the dynamic condtion from Bernoulli's equation
<center>
<math> \partial_t\Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + g \zeta = C - \frac{P_a}{\rho}, \qquad \mbox{on} \ z=\zeta </math></center>

Constants in Bernoulli's equation may be set equal to zero when we are eventually interested in integrating pressures over closed or open boundaries (floating or submerged bodies) to obtain forces & moments. This follows from a simple application of one of the two Gauss vector theorems.

==== Gauss theorem ====

[[Image:Force_coordinates2.png|right|thumb|500px|Force coordinates]]

We need to use the following theorems often called [http://en.wikipedia.org/wiki/Gauss_theorem Gauss theorem] although more properly known as the divergence theorem.
We begin with the vector version. If
<math> \mathbf{n} </math> is the unit normal vector pointing inside the volume <math> V </math> with surface <math>S</math> and
<math> f(\mathbf{x})</math> is an arbitrary sufficiently differentiable scalar function, then
<center>
<math> \iiint_{\Omega} \nabla f \mathrm{d}v = -\iint_{\partial\Omega} f_{s} \mathbf{n} \mathrm{d}s </math>
</center>

Note the three scalar identities that follow:
<center><math> \iiint_{{\Omega}} \partial_x f \mathrm{d}v = - \iint_{\partial\Omega} f n_1 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_y f \mathrm{d}v = - \iint_{\partial\Omega} f n_2 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_z f \mathrm{d}v = - \iint_{\partial\Omega} f n_3 \mathrm{d}s </math></center>

The scalar version is as follows where
<math> \mathbf{v}</math> is an arbitrary sufficiently differentiable vector function
<center>
<math> \iiint_{\Omega} \nabla \cdot \mathbf{v} = - \iint_{\partial\Omega} \mathbf{v} \cdot \mathbf{n} \mathrm{d}s </math></center>
The scalar identity is often used to prove mass conservation principle.

==== Definition of force and moment in terms of fluid pressure ====

The force <math>\mathbf{F}</math> is given by
<center><math>
\mathbf{F} = \iint_{\partial\Omega} P\mathbf{n}\mathrm{d}s
</math></center>
where <math>P</math> is the pressure and the moment <math>\mathbf{M}</math> is given by
<center><math>
\mathbf{M} = \iint_{\partial\Omega} P(\mathbf{x}\times\mathbf{n})\mathrm{d}s
</math></center>

It follows from the Gauss theorem that if <math> P = C </math> the force and moment over a closed boundary S vanish identically. Hence without loss of generality in the context of wave body interactions we will set <math> C=0 </math>. It follows that the dynamic free surface condition takes the form

<center><math> \zeta (x,y,t) = - \frac{1}{g} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right \}, \qquad z=\zeta </math></center>

=== Method II ===

When tracing a fluid particle on the free surface the hydrodynamic pressure given by Bernoulli (after the constant <math>C</math> has been set equal to zero) must vanish as we follow the particle:

<center><math> \frac{D}{Dt} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi+gz \right \} =0, \qquad z=\zeta </math></center>
or
<center><math> \left ( \partial_t + \mathbf{v} \cdot \nabla \right ) \left ( \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi +gz \right ) =0, \qquad z=\zeta </math></center>

This condition also follows upon elimination of <math>\zeta</math> from the kinematic & dynamic conditions derived under method I.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/1814747D-3A05-45A1-BDE5-2CEF40DEA25F/0/lecture1.pdf here]

[[Category:Linear Water-Wave Theory]]

Conservation Laws and Boundary Conditions

2010-09-02T14:07:41Z

Adi Kurniawan: /* Definition of force and moment in terms of fluid pressure */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Conservation Laws and Boundary Conditions
| next chapter = [[Linear and Second-Order Wave Theory]]
| previous chapter =
}}

{{complete pages}}

We begin by deriving the equations of motion used to model ocean waves. The equations of motion of a general fluid are given by the celebrated [http://en.wikipedia.org/wiki/Navier_Stokes Navier Stokes equations]. However, for the large scale processes that occur in ocean waves many simplifications are possible.

== Coordinate system and velocity potential ==

[[Image:Coordinate_system.png|right|thumb|500px|Coordinate System]]

We begin by defining the coordinate system.

<center><math>
\begin{matrix}
&(x,y,z) &: &\mbox{Coordinate system} \\
&\mathbf{x} &: &\mbox{Fixed Eulerian Vector} \\
& \mathbf{v} &: &\mbox{Flow Velocity Vector at} \ \mathbf{x} \\
&\zeta &: &\mbox{Free Surface Elevation} \\
&\mathbf{g} &: &\mbox{Acceleration due to gravity}
\end{matrix} </math></center>

At the moment we have not yet defined the region of space which is occupied by the fluid. However, the free surface plays a very important role in the propagation of ocean waves.

The most important assumption we make is that the fluid is an [http://en.wikipedia.org/wiki/Viscosity ideal fluid], i.e. there are no shear stresses due to viscosity and that the flow is [http://en.wikipedia.org/wiki/Irrotational irrotational]. This means that

<center><math>\nabla \times \mathbf{v} = 0</math></center>

We now introduce the very important concept of the velocity potential. Essentially this allows us to express the solution as a function of a scalar (a function which has a single value as opposed to a vector function which has multiple values) rather than a vector throughout the fluid domain. There is an important theorem in vector calculus [http://en.wikipedia.org/wiki/Irrotational_vector_field] that if <math>\nabla \times \mathbf{v} = 0</math> then we can express the irrotational vector as the gradient of a scalar function, i.e.
<center><math>
\mathbf{v} = \nabla \Phi
</math></center>
where <math>\Phi(\mathbf{x},t)</math> is called the [http://en.wikipedia.org/wiki/Velocity_potential velocity potential].

It turns out that the potential flow model of surface wave propagation and wave-body interactions is very accurate for the kind of length and time scales which are important in the ocean. It is important to realise, however, that we have made considerable simplifications and that certain processes, most notably wave breaking, are in no way covered by this theory. In fact, the process of wave breaking is extremely complicated and is much less well understood than the potential flow model.

== Conservation of mass ==

The key equation we will solve to understand ocean waves is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation] which is derived very simply from the expression for the velocity potential. We assume that the fluid is incompressible so that conservation of mass gives us the following condition
<center><math> \nabla \cdot \mathbf{v} = 0 </math></center>
This condition in turn implies, using the definition of the velocity potential that
<center><math>
\nabla \cdot \nabla \Phi = 0 \Rightarrow \nabla^2 \Phi = 0
</math></center>
or
<center><math>
\partial_x^2 \Phi + \partial_y^2\Phi + \partial_z^2\Phi = 0,
</math></center>
which is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation].

== Conservation of linear momentum ==

We begin with [http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29 Euler's equation] in the absence of viscosity
<center> <math>
\partial_t \mathbf{v} + (\mathbf{v}\cdot \nabla)\mathbf{v}= - \frac1{\rho} \nabla P + \mathbf{g}
</math> </center>
where
<math> P(\mathbf{x}, t) </math> is the fluid Pressure at <math>(\mathbf{x}, t)</math> and
<math> \mathbf{g}= - \mathbf{k} g </math> is the acceleration due to gravity where
<math> \mathbf{k} </math> is the unit vector pointing in the positive <math>z</math>-direction (so we are now setting the <math>z</math> coordinate to point in the vertical direction). Finally <math> \rho </math> is the water density.

We then use the following vector identity
<center><math> (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac 1{2} \nabla (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v}\times ( \nabla \times \mathbf{v}) </math></center>
and since we have irrotational flow (i.e. <math> \nabla \times \mathbf{v}= 0 </math>) Euler's equation becomes
<center><math> \partial_t \mathbf{v} + \frac{1}{2} \nabla (\mathbf{v} \cdot \mathbf{v}) = - \frac 1{\rho} \nabla P - \nabla (g z) </math></center>
where we have used <math> \nabla z = \mathbf{k} </math>.
We now substitute <math> \mathbf{v}= \nabla \Phi </math> and we obtain
<center><math> \nabla (\partial_t \Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z ) = 0. </math></center>
We now observe that if
<center><math> \nabla F( \mathbf{x}, t) =0 \quad \Longrightarrow \quad F (\mathbf{x}, t) = C </math></center>
where <math> C </math> is an arbitrary constant.

==== Bernoulli's equation ====

[http://en.wikipedia.org/wiki/Bernoulli%27s_equation Bernoulli's equation] follows from the equation above.
<center><math> \partial_t \Phi + \frac 1{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z = C </math></center>
or
<center><math> \frac{P}{\rho} = - \partial_t\Phi-\frac{1}{2}\nabla\Phi \cdot \nabla \Phi - g z + C </math></center>

The value of the constant <math> C </math> is immaterial (it can be thought of as defining the reference pressure.
It is also worth noting that the
angular momentum conservation principle is contained in
<math> \nabla \times \mathbf{v} = 0. </math>
In particular, if the particles are modelled as spheres, this equation implies no angular velocity at all times.

== Derivation of nonlinear free-surface condition ==

A very important result is the boundary condition at the free surface of the fluid and air. There are two conditions which relate the free surface displacement <math>\zeta(x,y,t)</math> and the velocity potential <math>\Phi(x,y,z,t)</math> at the free surface. The dynamic condition is derived from the Bernoulli's equation and the kinematic condition is derived from the equations linking the fact that the velocity vector at the surface is given by the gradient of the potential. We will present two methods to derive these equations.

=== Method I ===

We derive the dynamic condition directly from Bernoulli's equation.
On <math> z=\zeta; \ P=P_a \equiv \mbox{Atmospheric Pressure} </math>.
This allows us to rewrite Bernoulli's equation as
<center><math> \frac{P_a}{\rho}=-\frac{\partial\Phi}{\partial t}-\frac{1}{2}\nabla\Phi\cdot\nabla\Phi-g\zeta+C \qquad \mbox{on} \ z=\zeta(x,y,t) </math></center>
We will simplify this equation by showing that we are free to set the pressure to any value.

The kinematic condition is derived as follows.
On <math>z=\zeta</math> The mathematical function
<center><math>z-\zeta(x,y,t)\equiv\tilde{f}(x,y,z,t)</math></center>

is always zero when tracing a fluid particle on the free surface. So the [http://en.wikipedia.org/wiki/Total_derivative substantial or total derivative] of <math>\tilde{f}</math> must vanish, thus

<center><math> \frac{D\tilde{f}}{Dt}=0=\left (\partial_t + \mathbf{v} \cdot \nabla \right ) \tilde{f}=0, \qquad \mbox{on} \ z=\zeta </math></center>

Expanding, we obtain
<center><math> \left (\partial_t + \mathbf{v} \cdot \nabla \right ) (z-\zeta) =0, \qquad \mbox{on} \ z=\zeta </math></center>
which in turn implies that
<center><math> \partial_t\zeta + \partial_x\Phi \partial_x\zeta + \partial_y\Phi \partial_y\zeta = \partial_z\Phi, \qquad \mbox{on} \ z=\zeta </math></center>
which is the Kinematic free-surface condition.

We have already derived the dynamic condtion from Bernoulli's equation
<center>
<math> \partial_t\Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + g \zeta = C - \frac{P_a}{\rho}, \qquad \mbox{on} \ z=\zeta </math></center>

Constants in Bernoulli's equation may be set equal to zero when we are eventually interested in integrating pressures over closed or open boundaries (floating or submerged bodies) to obtain forces & moments. This follows from a simple application of one of the two Gauss vector theorems.

==== Gauss theorem ====

[[Image:Force_coordinates2.png|right|thumb|500px|Force coordinates]]

We need to use the following theorems often called [http://en.wikipedia.org/wiki/Gauss_theorem Gauss theorem] although more properly known as the divergence theorem.
We begin with the vector version. If
<math> \mathbf{n} </math> is the unit normal vector pointing inside the volume <math> V </math> with surface <math>S</math> and
<math> f(\mathbf{x})</math> is an arbitrary sufficiently differentiable scalar function, then
<center>
<math> \iiint_{\Omega} \nabla f \mathrm{d}v = -\iint_{\partial\Omega} f_{s} \mathbf{n} \mathrm{d}s </math>
</center>

Note the three scalar identities that follow:
<center><math> \iiint_{{\Omega}} \partial_x f \mathrm{d}v = - \iint_{\partial\Omega} f n_1 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_y f \mathrm{d}v = - \iint_{\partial\Omega} f n_2 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_z f \mathrm{d}v = - \iint_{\partial\Omega} f n_3 \mathrm{d}s </math></center>

The scalar version is as follows where
<math> \mathbf{v}</math> is an arbitrary sufficiently differentiable vector function
<center>
<math> \iiint_{\Omega} \nabla \cdot \mathbf{v} = - \iint_{\partial\Omega} \mathbf{v} \cdot \mathbf{n} \mathrm{d}s </math></center>
The scalar identity is often used to prove mass conservation principle.

==== Definition of force and moment in terms of fluid pressure ====

The force <math>\mathbf{F}</math> is given by
<center><math>
\mathbf{F} = \iint_{\partial\Omega} P\mathbf{n}\mathrm{d}s
</math></center>
where <math>P</math> is the pressure and the moment <math>\mathbf{M}</math> is given by
<center><math>
\mathbf{M} = \iint_{\partial\Omega} P(\mathbf{x}\times\mathbf{n})\mathrm{d}s
</math></center>

It follows from the Gauss theorem that if <math> \P = C </math> the force and moment over a closed boundary S vanish identically. Hence without loss of generality in the context of wave body interactions we will set <math> C=0 </math>. It follows that the dynamic free surface condition takes the form

<center><math> \zeta (x,y,t) = - \frac{1}{g} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right \}, \qquad z=\zeta </math></center>

=== Method II ===

When tracing a fluid particle on the free surface the hydrodynamic pressure given by Bernoulli (after the constant <math>C</math> has been set equal to zero) must vanish as we follow the particle:

<center><math> \frac{D}{Dt} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi+gz \right \} =0, \qquad z=\zeta </math></center>
or
<center><math> \left ( \partial_t + \mathbf{v} \cdot \nabla \right ) \left ( \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi +gz \right ) =0, \qquad z=\zeta </math></center>

This condition also follows upon elimination of <math>\zeta</math> from the kinematic & dynamic conditions derived under method I.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/1814747D-3A05-45A1-BDE5-2CEF40DEA25F/0/lecture1.pdf here]

[[Category:Linear Water-Wave Theory]]

Conservation Laws and Boundary Conditions

2010-08-28T07:11:52Z

Adi Kurniawan: /* Method I */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Conservation Laws and Boundary Conditions
| next chapter = [[Linear and Second-Order Wave Theory]]
| previous chapter =
}}

{{complete pages}}

We begin by deriving the equations of motion used to model ocean waves. The equations of motion of a general fluid are given by the celebrated [http://en.wikipedia.org/wiki/Navier_Stokes Navier Stokes equations]. However, for the large scale processes that occur in ocean waves many simplifications are possible.

== Coordinate system and velocity potential ==

[[Image:Coordinate_system.png|right|thumb|500px|Coordinate System]]

We begin by defining the coordinate system.

<center><math>
\begin{matrix}
&(x,y,z) &: &\mbox{Coordinate system} \\
&\mathbf{x} &: &\mbox{Fixed Eulerian Vector} \\
& \mathbf{v} &: &\mbox{Flow Velocity Vector at} \ \mathbf{x} \\
&\zeta &: &\mbox{Free Surface Elevation} \\
&\mathbf{g} &: &\mbox{Acceleration due to gravity}
\end{matrix} </math></center>

At the moment we have not yet defined the region of space which is occupied by the fluid. However, the free surface plays a very important role in the propagation of ocean waves.

The most important assumption we make is that the fluid is an [http://en.wikipedia.org/wiki/Viscosity ideal fluid], i.e. there are no shear stresses due to viscosity and that the flow is [http://en.wikipedia.org/wiki/Irrotational irrotational]. This means that

<center><math>\nabla \times \mathbf{v} = 0</math></center>

We now introduce the very important concept of the velocity potential. Essentially this allows us to express the solution as a function of a scalar (a function which has a single value as opposed to a vector function which has multiple values) rather than a vector throughout the fluid domain. There is an important theorem in vector calculus [http://en.wikipedia.org/wiki/Irrotational_vector_field] that if <math>\nabla \times \mathbf{v} = 0</math> then we can express the irrotational vector as the gradient of a scalar function, i.e.
<center><math>
\mathbf{v} = \nabla \Phi
</math></center>
where <math>\Phi(\mathbf{x},t)</math> is called the [http://en.wikipedia.org/wiki/Velocity_potential velocity potential].

It turns out that the potential flow model of surface wave propagation and wave-body interactions is very accurate for the kind of length and time scales which are important in the ocean. It is important to realise, however, that we have made considerable simplifications and that certain processes, most notably wave breaking, are in no way covered by this theory. In fact, the process of wave breaking is extremely complicated and is much less well understood than the potential flow model.

== Conservation of mass ==

The key equation we will solve to understand ocean waves is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation] which is derived very simply from the expression for the velocity potential. We assume that the fluid is incompressible so that conservation of mass gives us the following condition
<center><math> \nabla \cdot \mathbf{v} = 0 </math></center>
This condition in turn implies, using the definition of the velocity potential that
<center><math>
\nabla \cdot \nabla \Phi = 0 \Rightarrow \nabla^2 \Phi = 0
</math></center>
or
<center><math>
\partial_x^2 \Phi + \partial_y^2\Phi + \partial_z^2\Phi = 0,
</math></center>
which is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation].

== Conservation of linear momentum ==

We begin with [http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29 Euler's equation] in the absence of viscosity
<center> <math>
\partial_t \mathbf{v} + (\mathbf{v}\cdot \nabla)\mathbf{v}= - \frac1{\rho} \nabla P + \mathbf{g}
</math> </center>
where
<math> P(\mathbf{x}, t) </math> is the fluid Pressure at <math>(\mathbf{x}, t)</math> and
<math> \mathbf{g}= - \mathbf{k} g </math> is the acceleration due to gravity where
<math> \mathbf{k} </math> is the unit vector pointing in the positive <math>z</math>-direction (so we are now setting the <math>z</math> coordinate to point in the vertical direction). Finally <math> \rho </math> is the water density.

We then use the following vector identity
<center><math> (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac 1{2} \nabla (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v}\times ( \nabla \times \mathbf{v}) </math></center>
and since we have irrotational flow (i.e. <math> \nabla \times \mathbf{v}= 0 </math>) Euler's equation becomes
<center><math> \partial_t \mathbf{v} + \frac{1}{2} \nabla (\mathbf{v} \cdot \mathbf{v}) = - \frac 1{\rho} \nabla P - \nabla (g z) </math></center>
where we have used <math> \nabla z = \mathbf{k} </math>.
We now substitute <math> \mathbf{v}= \nabla \Phi </math> and we obtain
<center><math> \nabla (\partial_t \Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z ) = 0. </math></center>
We now observe that if
<center><math> \nabla F( \mathbf{x}, t) =0 \quad \Longrightarrow \quad F (\mathbf{x}, t) = C </math></center>
where <math> C </math> is an arbitrary constant.

==== Bernoulli's equation ====

[http://en.wikipedia.org/wiki/Bernoulli%27s_equation Bernoulli's equation] follows from the equation above.
<center><math> \partial_t \Phi + \frac 1{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z = C </math></center>
or
<center><math> \frac{P}{\rho} = - \partial_t\Phi-\frac{1}{2}\nabla\Phi \cdot \nabla \Phi - g z + C </math></center>

The value of the constant <math> C </math> is immaterial (it can be thought of as defining the reference pressure.
It is also worth noting that the
angular momentum conservation principle is contained in
<math> \nabla \times \mathbf{v} = 0. </math>
In particular, if the particles are modelled as spheres, this equation implies no angular velocity at all times.

== Derivation of nonlinear free-surface condition ==

A very important result is the boundary condition at the free surface of the fluid and air. There are two conditions which relate the free surface displacement <math>\zeta(x,y,t)</math> and the velocity potential <math>\Phi(x,y,z,t)</math> at the free surface. The dynamic condition is derived from the Bernoulli's equation and the kinematic condition is derived from the equations linking the fact that the velocity vector at the surface is given by the gradient of the potential. We will present two methods to derive these equations.

=== Method I ===

We derive the dynamic condition directly from Bernoulli's equation.
On <math> z=\zeta; \ P=P_a \equiv \mbox{Atmospheric Pressure} </math>.
This allows us to rewrite Bernoulli's equation as
<center><math> \frac{P_a}{\rho}=-\frac{\partial\Phi}{\partial t}-\frac{1}{2}\nabla\Phi\cdot\nabla\Phi-g\zeta+C \qquad \mbox{on} \ z=\zeta(x,y,t) </math></center>
We will simplify this equation by showing that we are free to set the pressure to any value.

The kinematic condition is derived as follows.
On <math>z=\zeta</math> The mathematical function
<center><math>z-\zeta(x,y,t)\equiv\tilde{f}(x,y,z,t)</math></center>

is always zero when tracing a fluid particle on the free surface. So the [http://en.wikipedia.org/wiki/Total_derivative substantial or total derivative] of <math>\tilde{f}</math> must vanish, thus

<center><math> \frac{D\tilde{f}}{Dt}=0=\left (\partial_t + \mathbf{v} \cdot \nabla \right ) \tilde{f}=0, \qquad \mbox{on} \ z=\zeta </math></center>

Expanding, we obtain
<center><math> \left (\partial_t + \mathbf{v} \cdot \nabla \right ) (z-\zeta) =0, \qquad \mbox{on} \ z=\zeta </math></center>
which in turn implies that
<center><math> \partial_t\zeta + \partial_x\Phi \partial_x\zeta + \partial_y\Phi \partial_y\zeta = \partial_z\Phi, \qquad \mbox{on} \ z=\zeta </math></center>
which is the Kinematic free-surface condition.

We have already derived the dynamic condtion from Bernoulli's equation
<center>
<math> \partial_t\Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + g \zeta = C - \frac{P_a}{\rho}, \qquad \mbox{on} \ z=\zeta </math></center>

Constants in Bernoulli's equation may be set equal to zero when we are eventually interested in integrating pressures over closed or open boundaries (floating or submerged bodies) to obtain forces & moments. This follows from a simple application of one of the two Gauss vector theorems.

==== Gauss theorem ====

[[Image:Force_coordinates2.png|right|thumb|500px|Force coordinates]]

We need to use the following theorems often called [http://en.wikipedia.org/wiki/Gauss_theorem Gauss theorem] although more properly known as the divergence theorem.
We begin with the vector version. If
<math> \mathbf{n} </math> is the unit normal vector pointing inside the volume <math> V </math> with surface <math>S</math> and
<math> f(\mathbf{x})</math> is an arbitrary sufficiently differentiable scalar function, then
<center>
<math> \iiint_{\Omega} \nabla f \mathrm{d}v = -\iint_{\partial\Omega} f_{s} \mathbf{n} \mathrm{d}s </math>
</center>

Note the three scalar identities that follow:
<center><math> \iiint_{{\Omega}} \partial_x f \mathrm{d}v = - \iint_{\partial\Omega} f n_1 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_y f \mathrm{d}v = - \iint_{\partial\Omega} f n_2 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_z f \mathrm{d}v = - \iint_{\partial\Omega} f n_3 \mathrm{d}s </math></center>

The scalar version is as follows where
<math> \mathbf{v}</math> is an arbitrary sufficiently differentiable vector function
<center>
<math> \iiint_{\Omega} \nabla \cdot \mathbf{v} = - \iint_{\partial\Omega} \mathbf{v} \cdot \mathbf{n} \mathrm{d}s </math></center>
The scalar identity is often used to prove mass conservation principle.

==== Definition of force and moment in terms of fluid pressure ====

The force <math>\mathbf{F}</math> is given by
<center><math>
\mathbf{F} = \iint_{\partial\Omega} P\mathbf{n}\mathrm{d}s
</math></center>
where <math>P</math> is the pressure and the moment <math>\mathbf{M}</math> is given by
<center><math>
\mathbf{M} = \iint_{\partial\Omega} P(\mathbf{x}\times\mathbf{v})\mathrm{d}s
</math></center>

It follows from the Gauss theorem that if <math> \rho = C </math> the force and moment over a closed boundary S vanish identically. Hence without loss of generality in the context of wave body interactions we will set <math> C=0 </math>. It follows that the dynamic free surface condition takes the form

<center><math> \zeta (x,y,t) = - \frac{1}{g} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right \}, \qquad z=\zeta </math></center>

=== Method II ===

When tracing a fluid particle on the free surface the hydrodynamic pressure given by Bernoulli (after the constant <math>C</math> has been set equal to zero) must vanish as we follow the particle:

<center><math> \frac{D}{Dt} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi+gz \right \} =0, \qquad z=\zeta </math></center>
or
<center><math> \left ( \partial_t + \mathbf{v} \cdot \nabla \right ) \left ( \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi +gz \right ) =0, \qquad z=\zeta </math></center>

This condition also follows upon elimination of <math>\zeta</math> from the kinematic & dynamic conditions derived under method I.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/1814747D-3A05-45A1-BDE5-2CEF40DEA25F/0/lecture1.pdf here]

[[Category:Linear Water-Wave Theory]]

Conservation Laws and Boundary Conditions

2010-08-28T07:03:26Z

Adi Kurniawan: /* Method I */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Conservation Laws and Boundary Conditions
| next chapter = [[Linear and Second-Order Wave Theory]]
| previous chapter =
}}

{{complete pages}}

We begin by deriving the equations of motion used to model ocean waves. The equations of motion of a general fluid are given by the celebrated [http://en.wikipedia.org/wiki/Navier_Stokes Navier Stokes equations]. However, for the large scale processes that occur in ocean waves many simplifications are possible.

== Coordinate system and velocity potential ==

[[Image:Coordinate_system.png|right|thumb|500px|Coordinate System]]

We begin by defining the coordinate system.

<center><math>
\begin{matrix}
&(x,y,z) &: &\mbox{Coordinate system} \\
&\mathbf{x} &: &\mbox{Fixed Eulerian Vector} \\
& \mathbf{v} &: &\mbox{Flow Velocity Vector at} \ \mathbf{x} \\
&\zeta &: &\mbox{Free Surface Elevation} \\
&\mathbf{g} &: &\mbox{Acceleration due to gravity}
\end{matrix} </math></center>

At the moment we have not yet defined the region of space which is occupied by the fluid. However, the free surface plays a very important role in the propagation of ocean waves.

The most important assumption we make is that the fluid is an [http://en.wikipedia.org/wiki/Viscosity ideal fluid], i.e. there are no shear stresses due to viscosity and that the flow is [http://en.wikipedia.org/wiki/Irrotational irrotational]. This means that

<center><math>\nabla \times \mathbf{v} = 0</math></center>

We now introduce the very important concept of the velocity potential. Essentially this allows us to express the solution as a function of a scalar (a function which has a single value as opposed to a vector function which has multiple values) rather than a vector throughout the fluid domain. There is an important theorem in vector calculus [http://en.wikipedia.org/wiki/Irrotational_vector_field] that if <math>\nabla \times \mathbf{v} = 0</math> then we can express the irrotational vector as the gradient of a scalar function, i.e.
<center><math>
\mathbf{v} = \nabla \Phi
</math></center>
where <math>\Phi(\mathbf{x},t)</math> is called the [http://en.wikipedia.org/wiki/Velocity_potential velocity potential].

It turns out that the potential flow model of surface wave propagation and wave-body interactions is very accurate for the kind of length and time scales which are important in the ocean. It is important to realise, however, that we have made considerable simplifications and that certain processes, most notably wave breaking, are in no way covered by this theory. In fact, the process of wave breaking is extremely complicated and is much less well understood than the potential flow model.

== Conservation of mass ==

The key equation we will solve to understand ocean waves is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation] which is derived very simply from the expression for the velocity potential. We assume that the fluid is incompressible so that conservation of mass gives us the following condition
<center><math> \nabla \cdot \mathbf{v} = 0 </math></center>
This condition in turn implies, using the definition of the velocity potential that
<center><math>
\nabla \cdot \nabla \Phi = 0 \Rightarrow \nabla^2 \Phi = 0
</math></center>
or
<center><math>
\partial_x^2 \Phi + \partial_y^2\Phi + \partial_z^2\Phi = 0,
</math></center>
which is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation].

== Conservation of linear momentum ==

We begin with [http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29 Euler's equation] in the absence of viscosity
<center> <math>
\partial_t \mathbf{v} + (\mathbf{v}\cdot \nabla)\mathbf{v}= - \frac1{\rho} \nabla P + \mathbf{g}
</math> </center>
where
<math> P(\mathbf{x}, t) </math> is the fluid Pressure at <math>(\mathbf{x}, t)</math> and
<math> \mathbf{g}= - \mathbf{k} g </math> is the acceleration due to gravity where
<math> \mathbf{k} </math> is the unit vector pointing in the positive <math>z</math>-direction (so we are now setting the <math>z</math> coordinate to point in the vertical direction). Finally <math> \rho </math> is the water density.

We then use the following vector identity
<center><math> (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac 1{2} \nabla (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v}\times ( \nabla \times \mathbf{v}) </math></center>
and since we have irrotational flow (i.e. <math> \nabla \times \mathbf{v}= 0 </math>) Euler's equation becomes
<center><math> \partial_t \mathbf{v} + \frac{1}{2} \nabla (\mathbf{v} \cdot \mathbf{v}) = - \frac 1{\rho} \nabla P - \nabla (g z) </math></center>
where we have used <math> \nabla z = \mathbf{k} </math>.
We now substitute <math> \mathbf{v}= \nabla \Phi </math> and we obtain
<center><math> \nabla (\partial_t \Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z ) = 0. </math></center>
We now observe that if
<center><math> \nabla F( \mathbf{x}, t) =0 \quad \Longrightarrow \quad F (\mathbf{x}, t) = C </math></center>
where <math> C </math> is an arbitrary constant.

==== Bernoulli's equation ====

[http://en.wikipedia.org/wiki/Bernoulli%27s_equation Bernoulli's equation] follows from the equation above.
<center><math> \partial_t \Phi + \frac 1{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z = C </math></center>
or
<center><math> \frac{P}{\rho} = - \partial_t\Phi-\frac{1}{2}\nabla\Phi \cdot \nabla \Phi - g z + C </math></center>

The value of the constant <math> C </math> is immaterial (it can be thought of as defining the reference pressure.
It is also worth noting that the
angular momentum conservation principle is contained in
<math> \nabla \times \mathbf{v} = 0. </math>
In particular, if the particles are modelled as spheres, this equation implies no angular velocity at all times.

== Derivation of nonlinear free-surface condition ==

A very important result is the boundary condition at the free surface of the fluid and air. There are two conditions which relate the free surface displacement <math>\zeta(x,y,t)</math> and the velocity potential <math>\Phi(x,y,z,t)</math> at the free surface. The dynamic condition is derived from the Bernoulli's equation and the kinematic condition is derived from the equations linking the fact that the velocity vector at the surface is given by the gradient of the potential. We will present two methods to derive these equations.

=== Method I ===

We derive the dynamic condition directly from Bernoulli's equation.
On <math> z=\zeta; \ P=P_a \equiv \mbox{Atmospheric Pressure} </math>.
This allows us to rewrite Bernoulli's equation as
<center><math> \frac{P_a}{\rho}=-\frac{\partial\Phi}{\partial t}-\frac{1}{2}\nabla\Phi\cdot\nabla\Phi-g\zeta+C \qquad \mbox{on} \ z=\zeta(x,y,t) </math></center>
We will simplify this equation by showing that we are free to set the pressure to any value.

The kinematic condition is derived as follows.
On <math>z=\zeta</math> The mathematical function
<center><math>z-\zeta(x,y,t)\equiv\tilde{f}(x,y,z,t)</math></center>

is always zero when tracing a fluid particle on the free surface. So the [http://en.wikipedia.org/wiki/Total_derivative substantial or total derivative] of <math>\tilde{f}</math> must vanish, thus

<center><math> \frac{D\tilde{f}}{Dt}=0=\left (\partial_t + \mathbf{v} \cdot \nabla \right ) \tilde{f}=0, \qquad \mbox{on} \ z=\zeta </math></center>

Expanding, we obtain
<center><math> \left (\partial_t + \mathbf{v} \cdot \nabla \right ) (z-\zeta) =0, \qquad \mbox{on} \ z=\zeta </math></center>
which in turn implies that
<center><math> \partial_t\zeta + \partial_x\Phi \partial_x\zeta + \partial_y\Phi \partial_y\zeta = \partial_z\Phi, \, z=\zeta </math></center>
which is the Kinematic free-surface condition.

We have already derived the dynamic condtion from Bernoulli's equation
<center>
<math> \partial_t\Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + g \zeta = \mathbb{C} - \frac{P_a}{\rho}, \ z=\zeta </math></center>

Constants in Bernoulli's equation may be set equal to zero when we are eventually interested in integrating pressures over closed or open boundaries (floating or submerged bodies) to obtain forces & moments. This follows from a simple application of one of the two Gauss vector theorems.

==== Gauss theorem ====

[[Image:Force_coordinates2.png|right|thumb|500px|Force coordinates]]

We need to use the following theorems often called [http://en.wikipedia.org/wiki/Gauss_theorem Gauss theorem] although more properly known as the divergence theorem.
We begin with the vector version. If
<math> \mathbf{n} </math> is the unit normal vector pointing inside the volume <math> V </math> with surface <math>S</math> and
<math> f(\mathbf{x})</math> is an arbitrary sufficiently differentiable scalar function, then
<center>
<math> \iiint_{\Omega} \nabla f \mathrm{d}v = -\iint_{\partial\Omega} f_{s} \mathbf{n} \mathrm{d}s </math>
</center>

Note the three scalar identities that follow:
<center><math> \iiint_{{\Omega}} \partial_x f \mathrm{d}v = - \iint_{\partial\Omega} f n_1 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_y f \mathrm{d}v = - \iint_{\partial\Omega} f n_2 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_z f \mathrm{d}v = - \iint_{\partial\Omega} f n_3 \mathrm{d}s </math></center>

The scalar version is as follows where
<math> \mathbf{v}</math> is an arbitrary sufficiently differentiable vector function
<center>
<math> \iiint_{\Omega} \nabla \cdot \mathbf{v} = - \iint_{\partial\Omega} \mathbf{v} \cdot \mathbf{n} \mathrm{d}s </math></center>
The scalar identity is often used to prove mass conservation principle.

==== Definition of force and moment in terms of fluid pressure ====

The force <math>\mathbf{F}</math> is given by
<center><math>
\mathbf{F} = \iint_{\partial\Omega} P\mathbf{n}\mathrm{d}s
</math></center>
where <math>P</math> is the pressure and the moment <math>\mathbf{M}</math> is given by
<center><math>
\mathbf{M} = \iint_{\partial\Omega} P(\mathbf{x}\times\mathbf{v})\mathrm{d}s
</math></center>

It follows from the Gauss theorem that if <math> \rho = C </math> the force and moment over a closed boundary S vanish identically. Hence without loss of generality in the context of wave body interactions we will set <math> C=0 </math>. It follows that the dynamic free surface condition takes the form

<center><math> \zeta (x,y,t) = - \frac{1}{g} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right \}, \qquad z=\zeta </math></center>

=== Method II ===

When tracing a fluid particle on the free surface the hydrodynamic pressure given by Bernoulli (after the constant <math>C</math> has been set equal to zero) must vanish as we follow the particle:

<center><math> \frac{D}{Dt} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi+gz \right \} =0, \qquad z=\zeta </math></center>
or
<center><math> \left ( \partial_t + \mathbf{v} \cdot \nabla \right ) \left ( \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi +gz \right ) =0, \qquad z=\zeta </math></center>

This condition also follows upon elimination of <math>\zeta</math> from the kinematic & dynamic conditions derived under method I.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/1814747D-3A05-45A1-BDE5-2CEF40DEA25F/0/lecture1.pdf here]

[[Category:Linear Water-Wave Theory]]

Conservation Laws and Boundary Conditions

2010-08-28T07:01:39Z

Adi Kurniawan: /* Method I */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Conservation Laws and Boundary Conditions
| next chapter = [[Linear and Second-Order Wave Theory]]
| previous chapter =
}}

{{complete pages}}

We begin by deriving the equations of motion used to model ocean waves. The equations of motion of a general fluid are given by the celebrated [http://en.wikipedia.org/wiki/Navier_Stokes Navier Stokes equations]. However, for the large scale processes that occur in ocean waves many simplifications are possible.

== Coordinate system and velocity potential ==

[[Image:Coordinate_system.png|right|thumb|500px|Coordinate System]]

We begin by defining the coordinate system.

<center><math>
\begin{matrix}
&(x,y,z) &: &\mbox{Coordinate system} \\
&\mathbf{x} &: &\mbox{Fixed Eulerian Vector} \\
& \mathbf{v} &: &\mbox{Flow Velocity Vector at} \ \mathbf{x} \\
&\zeta &: &\mbox{Free Surface Elevation} \\
&\mathbf{g} &: &\mbox{Acceleration due to gravity}
\end{matrix} </math></center>

At the moment we have not yet defined the region of space which is occupied by the fluid. However, the free surface plays a very important role in the propagation of ocean waves.

The most important assumption we make is that the fluid is an [http://en.wikipedia.org/wiki/Viscosity ideal fluid], i.e. there are no shear stresses due to viscosity and that the flow is [http://en.wikipedia.org/wiki/Irrotational irrotational]. This means that

<center><math>\nabla \times \mathbf{v} = 0</math></center>

We now introduce the very important concept of the velocity potential. Essentially this allows us to express the solution as a function of a scalar (a function which has a single value as opposed to a vector function which has multiple values) rather than a vector throughout the fluid domain. There is an important theorem in vector calculus [http://en.wikipedia.org/wiki/Irrotational_vector_field] that if <math>\nabla \times \mathbf{v} = 0</math> then we can express the irrotational vector as the gradient of a scalar function, i.e.
<center><math>
\mathbf{v} = \nabla \Phi
</math></center>
where <math>\Phi(\mathbf{x},t)</math> is called the [http://en.wikipedia.org/wiki/Velocity_potential velocity potential].

It turns out that the potential flow model of surface wave propagation and wave-body interactions is very accurate for the kind of length and time scales which are important in the ocean. It is important to realise, however, that we have made considerable simplifications and that certain processes, most notably wave breaking, are in no way covered by this theory. In fact, the process of wave breaking is extremely complicated and is much less well understood than the potential flow model.

== Conservation of mass ==

The key equation we will solve to understand ocean waves is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation] which is derived very simply from the expression for the velocity potential. We assume that the fluid is incompressible so that conservation of mass gives us the following condition
<center><math> \nabla \cdot \mathbf{v} = 0 </math></center>
This condition in turn implies, using the definition of the velocity potential that
<center><math>
\nabla \cdot \nabla \Phi = 0 \Rightarrow \nabla^2 \Phi = 0
</math></center>
or
<center><math>
\partial_x^2 \Phi + \partial_y^2\Phi + \partial_z^2\Phi = 0,
</math></center>
which is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation].

== Conservation of linear momentum ==

We begin with [http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29 Euler's equation] in the absence of viscosity
<center> <math>
\partial_t \mathbf{v} + (\mathbf{v}\cdot \nabla)\mathbf{v}= - \frac1{\rho} \nabla P + \mathbf{g}
</math> </center>
where
<math> P(\mathbf{x}, t) </math> is the fluid Pressure at <math>(\mathbf{x}, t)</math> and
<math> \mathbf{g}= - \mathbf{k} g </math> is the acceleration due to gravity where
<math> \mathbf{k} </math> is the unit vector pointing in the positive <math>z</math>-direction (so we are now setting the <math>z</math> coordinate to point in the vertical direction). Finally <math> \rho </math> is the water density.

We then use the following vector identity
<center><math> (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac 1{2} \nabla (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v}\times ( \nabla \times \mathbf{v}) </math></center>
and since we have irrotational flow (i.e. <math> \nabla \times \mathbf{v}= 0 </math>) Euler's equation becomes
<center><math> \partial_t \mathbf{v} + \frac{1}{2} \nabla (\mathbf{v} \cdot \mathbf{v}) = - \frac 1{\rho} \nabla P - \nabla (g z) </math></center>
where we have used <math> \nabla z = \mathbf{k} </math>.
We now substitute <math> \mathbf{v}= \nabla \Phi </math> and we obtain
<center><math> \nabla (\partial_t \Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z ) = 0. </math></center>
We now observe that if
<center><math> \nabla F( \mathbf{x}, t) =0 \quad \Longrightarrow \quad F (\mathbf{x}, t) = C </math></center>
where <math> C </math> is an arbitrary constant.

==== Bernoulli's equation ====

[http://en.wikipedia.org/wiki/Bernoulli%27s_equation Bernoulli's equation] follows from the equation above.
<center><math> \partial_t \Phi + \frac 1{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z = C </math></center>
or
<center><math> \frac{P}{\rho} = - \partial_t\Phi-\frac{1}{2}\nabla\Phi \cdot \nabla \Phi - g z + C </math></center>

The value of the constant <math> C </math> is immaterial (it can be thought of as defining the reference pressure.
It is also worth noting that the
angular momentum conservation principle is contained in
<math> \nabla \times \mathbf{v} = 0. </math>
In particular, if the particles are modelled as spheres, this equation implies no angular velocity at all times.

== Derivation of nonlinear free-surface condition ==

A very important result is the boundary condition at the free surface of the fluid and air. There are two conditions which relate the free surface displacement <math>\zeta(x,y,t)</math> and the velocity potential <math>\Phi(x,y,z,t)</math> at the free surface. The dynamic condition is derived from the Bernoulli's equation and the kinematic condition is derived from the equations linking the fact that the velocity vector at the surface is given by the gradient of the potential. We will present two methods to derive these equations.

=== Method I ===

We derive the dynamic condition directly from Bernoulli's equation.
On <math> z=\zeta; \ P=P_a \equiv \mbox{Atmospheric Pressure} </math>.
This allows us to rewrite Bernoulli's equation as
<center><math> \frac{P_a}{\rho}=-\frac{\partial\Phi}{\partial t}-\frac{1}{2}\nabla\Phi\cdot\nabla\Phi-g\zeta+C \qquad \mbox{on} \ z=\zeta(x,y,t) </math></center>
We will simplify this equation by showing that we are free to set the pressure to any value.

The kinematic condition is derived as follows.
On <math>z=\zeta</math> The mathematical function
<center><math>z-\zeta(x,y,t)\equiv\tilde{f}(x,y,z,t)</math></center>

is always zero when tracing a fluid particle on the free surface. So the [http://en.wikipedia.org/wiki/Total_derivative substantial or total derivative] of <math>\tilde{f}</math> must vanish, thus

<center><math> \frac{D\tilde{f}}{Dt}=0=\left (\partial_t + \mathbf{v} \cdot \nabla \right ) \tilde{f}=0, \qquad \mbox{on} \ z=\zeta </math></center>

Expanding we obtain:
<center><math> \left (\partial_t + \mathbf{v} \cdot \nabla \right ) (z-\zeta) =0, \qquad \mbox{on} \ z=\zeta </math></center>
which in turn implies that
<center><math> \partial_t\zeta + \partial_x\Phi \partial_x\zeta + \partial_y\Phi \partial_y\zeta = \partial_z\Phi, \, z=\zeta </math></center>
which is the Kinematic free-surface condition.

We have already derived the dynamic condtion from Bernoulli's equation
<center>
<math> \partial_t\Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + g \zeta = \mathbb{C} - \frac{P_a}{\rho}, \ z=\zeta </math></center>

Constants in Bernoulli's equation may be set equal to zero when we are eventually interested in integrating pressures over closed or open boundaries (floating or submerged bodies) to obtain forces & moments. This follows from a simple application of one of the two Gauss vector theorems.

==== Gauss theorem ====

[[Image:Force_coordinates2.png|right|thumb|500px|Force coordinates]]

We need to use the following theorems often called [http://en.wikipedia.org/wiki/Gauss_theorem Gauss theorem] although more properly known as the divergence theorem.
We begin with the vector version. If
<math> \mathbf{n} </math> is the unit normal vector pointing inside the volume <math> V </math> with surface <math>S</math> and
<math> f(\mathbf{x})</math> is an arbitrary sufficiently differentiable scalar function, then
<center>
<math> \iiint_{\Omega} \nabla f \mathrm{d}v = -\iint_{\partial\Omega} f_{s} \mathbf{n} \mathrm{d}s </math>
</center>

Note the three scalar identities that follow:
<center><math> \iiint_{{\Omega}} \partial_x f \mathrm{d}v = - \iint_{\partial\Omega} f n_1 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_y f \mathrm{d}v = - \iint_{\partial\Omega} f n_2 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_z f \mathrm{d}v = - \iint_{\partial\Omega} f n_3 \mathrm{d}s </math></center>

The scalar version is as follows where
<math> \mathbf{v}</math> is an arbitrary sufficiently differentiable vector function
<center>
<math> \iiint_{\Omega} \nabla \cdot \mathbf{v} = - \iint_{\partial\Omega} \mathbf{v} \cdot \mathbf{n} \mathrm{d}s </math></center>
The scalar identity is often used to prove mass conservation principle.

==== Definition of force and moment in terms of fluid pressure ====

The force <math>\mathbf{F}</math> is given by
<center><math>
\mathbf{F} = \iint_{\partial\Omega} P\mathbf{n}\mathrm{d}s
</math></center>
where <math>P</math> is the pressure and the moment <math>\mathbf{M}</math> is given by
<center><math>
\mathbf{M} = \iint_{\partial\Omega} P(\mathbf{x}\times\mathbf{v})\mathrm{d}s
</math></center>

It follows from the Gauss theorem that if <math> \rho = C </math> the force and moment over a closed boundary S vanish identically. Hence without loss of generality in the context of wave body interactions we will set <math> C=0 </math>. It follows that the dynamic free surface condition takes the form

<center><math> \zeta (x,y,t) = - \frac{1}{g} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right \}, \qquad z=\zeta </math></center>

=== Method II ===

When tracing a fluid particle on the free surface the hydrodynamic pressure given by Bernoulli (after the constant <math>C</math> has been set equal to zero) must vanish as we follow the particle:

<center><math> \frac{D}{Dt} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi+gz \right \} =0, \qquad z=\zeta </math></center>
or
<center><math> \left ( \partial_t + \mathbf{v} \cdot \nabla \right ) \left ( \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi +gz \right ) =0, \qquad z=\zeta </math></center>

This condition also follows upon elimination of <math>\zeta</math> from the kinematic & dynamic conditions derived under method I.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/1814747D-3A05-45A1-BDE5-2CEF40DEA25F/0/lecture1.pdf here]

[[Category:Linear Water-Wave Theory]]

Conservation Laws and Boundary Conditions

2010-08-28T06:58:25Z

Adi Kurniawan: /* Conservation of linear momentum */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Conservation Laws and Boundary Conditions
| next chapter = [[Linear and Second-Order Wave Theory]]
| previous chapter =
}}

{{complete pages}}

We begin by deriving the equations of motion used to model ocean waves. The equations of motion of a general fluid are given by the celebrated [http://en.wikipedia.org/wiki/Navier_Stokes Navier Stokes equations]. However, for the large scale processes that occur in ocean waves many simplifications are possible.

== Coordinate system and velocity potential ==

[[Image:Coordinate_system.png|right|thumb|500px|Coordinate System]]

We begin by defining the coordinate system.

<center><math>
\begin{matrix}
&(x,y,z) &: &\mbox{Coordinate system} \\
&\mathbf{x} &: &\mbox{Fixed Eulerian Vector} \\
& \mathbf{v} &: &\mbox{Flow Velocity Vector at} \ \mathbf{x} \\
&\zeta &: &\mbox{Free Surface Elevation} \\
&\mathbf{g} &: &\mbox{Acceleration due to gravity}
\end{matrix} </math></center>

At the moment we have not yet defined the region of space which is occupied by the fluid. However, the free surface plays a very important role in the propagation of ocean waves.

The most important assumption we make is that the fluid is an [http://en.wikipedia.org/wiki/Viscosity ideal fluid], i.e. there are no shear stresses due to viscosity and that the flow is [http://en.wikipedia.org/wiki/Irrotational irrotational]. This means that

<center><math>\nabla \times \mathbf{v} = 0</math></center>

We now introduce the very important concept of the velocity potential. Essentially this allows us to express the solution as a function of a scalar (a function which has a single value as opposed to a vector function which has multiple values) rather than a vector throughout the fluid domain. There is an important theorem in vector calculus [http://en.wikipedia.org/wiki/Irrotational_vector_field] that if <math>\nabla \times \mathbf{v} = 0</math> then we can express the irrotational vector as the gradient of a scalar function, i.e.
<center><math>
\mathbf{v} = \nabla \Phi
</math></center>
where <math>\Phi(\mathbf{x},t)</math> is called the [http://en.wikipedia.org/wiki/Velocity_potential velocity potential].

It turns out that the potential flow model of surface wave propagation and wave-body interactions is very accurate for the kind of length and time scales which are important in the ocean. It is important to realise, however, that we have made considerable simplifications and that certain processes, most notably wave breaking, are in no way covered by this theory. In fact, the process of wave breaking is extremely complicated and is much less well understood than the potential flow model.

== Conservation of mass ==

The key equation we will solve to understand ocean waves is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation] which is derived very simply from the expression for the velocity potential. We assume that the fluid is incompressible so that conservation of mass gives us the following condition
<center><math> \nabla \cdot \mathbf{v} = 0 </math></center>
This condition in turn implies, using the definition of the velocity potential that
<center><math>
\nabla \cdot \nabla \Phi = 0 \Rightarrow \nabla^2 \Phi = 0
</math></center>
or
<center><math>
\partial_x^2 \Phi + \partial_y^2\Phi + \partial_z^2\Phi = 0,
</math></center>
which is [http://en.wikipedia.org/wiki/Laplaces_equation Laplace's equation].

== Conservation of linear momentum ==

We begin with [http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29 Euler's equation] in the absence of viscosity
<center> <math>
\partial_t \mathbf{v} + (\mathbf{v}\cdot \nabla)\mathbf{v}= - \frac1{\rho} \nabla P + \mathbf{g}
</math> </center>
where
<math> P(\mathbf{x}, t) </math> is the fluid Pressure at <math>(\mathbf{x}, t)</math> and
<math> \mathbf{g}= - \mathbf{k} g </math> is the acceleration due to gravity where
<math> \mathbf{k} </math> is the unit vector pointing in the positive <math>z</math>-direction (so we are now setting the <math>z</math> coordinate to point in the vertical direction). Finally <math> \rho </math> is the water density.

We then use the following vector identity
<center><math> (\mathbf{v} \cdot \nabla) \mathbf{v} = \frac 1{2} \nabla (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v}\times ( \nabla \times \mathbf{v}) </math></center>
and since we have irrotational flow (i.e. <math> \nabla \times \mathbf{v}= 0 </math>) Euler's equation becomes
<center><math> \partial_t \mathbf{v} + \frac{1}{2} \nabla (\mathbf{v} \cdot \mathbf{v}) = - \frac 1{\rho} \nabla P - \nabla (g z) </math></center>
where we have used <math> \nabla z = \mathbf{k} </math>.
We now substitute <math> \mathbf{v}= \nabla \Phi </math> and we obtain
<center><math> \nabla (\partial_t \Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z ) = 0. </math></center>
We now observe that if
<center><math> \nabla F( \mathbf{x}, t) =0 \quad \Longrightarrow \quad F (\mathbf{x}, t) = C </math></center>
where <math> C </math> is an arbitrary constant.

==== Bernoulli's equation ====

[http://en.wikipedia.org/wiki/Bernoulli%27s_equation Bernoulli's equation] follows from the equation above.
<center><math> \partial_t \Phi + \frac 1{2} \nabla \Phi \cdot \nabla \Phi + \frac {P}{\rho} + g z = C </math></center>
or
<center><math> \frac{P}{\rho} = - \partial_t\Phi-\frac{1}{2}\nabla\Phi \cdot \nabla \Phi - g z + C </math></center>

The value of the constant <math> C </math> is immaterial (it can be thought of as defining the reference pressure.
It is also worth noting that the
angular momentum conservation principle is contained in
<math> \nabla \times \mathbf{v} = 0. </math>
In particular, if the particles are modelled as spheres, this equation implies no angular velocity at all times.

== Derivation of nonlinear free-surface condition ==

A very important result is the boundary condition at the free surface of the fluid and air. There are two conditions which relate the free surface displacement <math>\zeta(x,y,t)</math> and the velocity potential <math>\Phi(x,y,z,t)</math> at the free surface. The dynamic condition is derived from the Bernoulli's equation and the kinematic condition is derived from the equations linking the fact that the velocity vector at the surface is given by the gradient of the potential. We will present two methods to derive these equations.

=== Method I ===

We derive the dynamic condition directly from Bernoulli's equation.
On <math> z=\zeta; \ P=P_a \equiv \mbox{Atmospheric Pressure} </math>.
This allows us to rewrite Bernoulli's equation as
<center><math> \frac{P_a}{\rho}=-\frac{\partial\Phi}{\partial t}-\frac{1}{2}\nabla\Phi\cdot\nabla\Phi-g\zeta+\mathbb{C} \qquad \mbox{on} \ z=\zeta(x,y,t) </math></center>
We will simplify this equation by showing that we are free to set the pressure to any value.

The kinematic condition is derived as follows.
On <math>z=\zeta</math> The mathematical function
<center><math>z-\zeta(x,y,t)\equiv\tilde{f}(x,y,z,t)</math></center>

is always zero when tracing a fluid particle on the free surface. So the [http://en.wikipedia.org/wiki/Total_derivative substantial or total derivative] of <math>\tilde{f}</math> must vanish, thus

<center><math> \frac{D\tilde{f}}{Dt}=0=\left (\partial_t + \mathbf{v} \cdot \nabla \right ) \tilde{f}=0, \qquad \mbox{on} \ z=\zeta </math></center>

Expanding we obtain:
<center><math> \left (\partial_t + \mathbf{v} \cdot \nabla \right ) (z-\zeta) =0, \qquad \mbox{on} \ z=\zeta </math></center>
which in turn implies that
<center><math> \partial_t\zeta + \partial_x\Phi \partial_x\zeta + \partial_y\Phi \partial_y\zeta = \partial_z\Phi, \, z=\zeta </math></center>
which is the Kinematic free-surface condition.

We have already derived the dynamic condtion from Bernoulli's equation
<center>
<math> \partial_t\Phi + \frac{1}{2} \nabla \Phi \cdot \nabla \Phi + g \zeta = \mathbb{C} - \frac{P_a}{\rho}, \ z=\zeta </math></center>

Constants in Bernoulli's equation may be set equal to zero when we are eventually interested in integrating pressures over closed or open boundaries (floating or submerged bodies) to obtain forces & moments. This follows from a simple application of one of the two Gauss vector theorems.

==== Gauss theorem ====

[[Image:Force_coordinates2.png|right|thumb|500px|Force coordinates]]

We need to use the following theorems often called [http://en.wikipedia.org/wiki/Gauss_theorem Gauss theorem] although more properly known as the divergence theorem.
We begin with the vector version. If
<math> \mathbf{n} </math> is the unit normal vector pointing inside the volume <math> V </math> with surface <math>S</math> and
<math> f(\mathbf{x})</math> is an arbitrary sufficiently differentiable scalar function, then
<center>
<math> \iiint_{\Omega} \nabla f \mathrm{d}v = -\iint_{\partial\Omega} f_{s} \mathbf{n} \mathrm{d}s </math>
</center>

Note the three scalar identities that follow:
<center><math> \iiint_{{\Omega}} \partial_x f \mathrm{d}v = - \iint_{\partial\Omega} f n_1 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_y f \mathrm{d}v = - \iint_{\partial\Omega} f n_2 \mathrm{d}s </math></center> 
<center><math> \iiint_{{\Omega}} \partial_z f \mathrm{d}v = - \iint_{\partial\Omega} f n_3 \mathrm{d}s </math></center>

The scalar version is as follows where
<math> \mathbf{v}</math> is an arbitrary sufficiently differentiable vector function
<center>
<math> \iiint_{\Omega} \nabla \cdot \mathbf{v} = - \iint_{\partial\Omega} \mathbf{v} \cdot \mathbf{n} \mathrm{d}s </math></center>
The scalar identity is often used to prove mass conservation principle.

==== Definition of force and moment in terms of fluid pressure ====

The force <math>\mathbf{F}</math> is given by
<center><math>
\mathbf{F} = \iint_{\partial\Omega} P\mathbf{n}\mathrm{d}s
</math></center>
where <math>P</math> is the pressure and the moment <math>\mathbf{M}</math> is given by
<center><math>
\mathbf{M} = \iint_{\partial\Omega} P(\mathbf{x}\times\mathbf{v})\mathrm{d}s
</math></center>

It follows from the Gauss theorem that if <math> \rho = C </math> the force and moment over a closed boundary S vanish identically. Hence without loss of generality in the context of wave body interactions we will set <math> C=0 </math>. It follows that the dynamic free surface condition takes the form

<center><math> \zeta (x,y,t) = - \frac{1}{g} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi \right \}, \qquad z=\zeta </math></center>

=== Method II ===

When tracing a fluid particle on the free surface the hydrodynamic pressure given by Bernoulli (after the constant <math>C</math> has been set equal to zero) must vanish as we follow the particle:

<center><math> \frac{D}{Dt} \left \{ \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi+gz \right \} =0, \qquad z=\zeta </math></center>
or
<center><math> \left ( \partial_t + \mathbf{v} \cdot \nabla \right ) \left ( \partial_t\Phi + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi +gz \right ) =0, \qquad z=\zeta </math></center>

This condition also follows upon elimination of <math>\zeta</math> from the kinematic & dynamic conditions derived under method I.

-----

This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/1814747D-3A05-45A1-BDE5-2CEF40DEA25F/0/lecture1.pdf here]

[[Category:Linear Water-Wave Theory]]

Why you should contribute

2010-08-28T06:50:51Z

Adi Kurniawan:

If you read this wiki you should contribute. Basically, we need readers to make contributions for this wiki to really work. No matter how little you know about the mathematical theory you will find typos and small errors on pages as you read them. If you start to work through the mathematics you will find errors in the equations, gaps in the derivations or points which confuse you and where you had to think hard to make progress.

If you make a contribution to fix these typos, add in extra information, or make requests you will be contributing in a very meaningful way. We know it is a slight hassle to [[Sign up instructions|sign up]] (and we only require this because of problems with spam) but it is not all that difficult and does not require you to divulge your mother's middle name or equivalent.

Finally, we notice every edit and it makes us feel great that you have gone to all this trouble.

Introduction to KdV

2010-08-20T10:34:44Z

Adi Kurniawan:

{{nonlinear waves course
| chapter title = Introduction to KdV
| next chapter = [[Numerical Solution of the KdV]]
| previous chapter = [[Nonlinear Shallow Water Waves]]
}}

{{complete pages}}

The KdV (Korteweg-De Vries) equation is one of the most important non-linear
pde's. It was originally derived to model shallow water waves with weak
nonlinearities, but it has a wide variety of applications. The derivation of
the KdV is given in [[KdV Equation Derivation]]. The KdV equation
is written as
<center><math>
\partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0.
</math></center>

More information about it can be found at [http://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation Korteweg de Vries equation]

==Travelling Wave Solution==

The KdV equation posesses travelling wave solutions. One particular
travelling wave solution is called a soltion and it was discovered
experimentally by [http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell]
in 1834. However, it was not understood
theoretically until the work of [http://en.wikipedia.org/wiki/Diederik_Korteweg Korteweg] and
[http://en.wikipedia.org/wiki/Gustav_de_Vries de Vries] in 1895.

We begin with the assumption that the wave travels with contant form, i.e.
is of the form
<center><math>
u\left( x,t\right) =f\left( x-ct\right)
</math></center>
Note that in this equation the parameter <math>c</math> is an unknown as is the
function <math>f.</math>
Only very special values of <math>c</math> and <math>f</math> will give travelling
waves.
We introduce the coordinate <math>\zeta = x - ct</math>.
If we substitute this expression into the KdV equation we obtain
<center><math>
-c\frac{\mathrm{d}f}{\mathrm{d}\zeta}+6f\frac{\mathrm{d}f}{\mathrm{d}\zeta}+\frac{\mathrm{d}^{3}f}{\mathrm{d}\zeta^{3}}=0
</math></center>
We can integrate this with respect to <math>\zeta</math> to obtain
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
where <math>A</math> is a constant of integration.

If think about this equation as Newton's second law in a potential well <math>
V(f) </math> for which the equation is
<center><math>
\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f}
</math></center>
then the potential well is given by
<center><math>
V\left( f\right) =-A_{0}-A_{1}f-c\frac{f^{2}}{2}+f^{3}
</math></center>
Therefore our equation for <math>f</math> may be thought of as the motion of a particle
in a cubic well.

The constant <math>A_0</math> has no effect on our solution so we can set it to be zero.
We can choose the constant <math>A_{1}=0</math> and then we have a
maximum at <math>f=0</math>. There is a solution which rolls from this at <math>t=-\infty </math>
and then runs up the other side and finally returns to the maximum at <math>
t=\infty .</math> This corresponds to a solitary wave solution.

We can also think about the equation as a first order system using <math>
f^{^{\prime }}=v.</math> This gives us
<center><math>\begin{matrix}
\frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2} \\
\frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v
\end{matrix}</math></center>
If we chose <math>A_{1}=0</math> then we obtain two equilibria at <math>(f,v)=\left(
0,0\right) </math> and <math>(3/c,0).</math> If we analysis these equilibria we find the
first is a saddle and the second is a nonlinear center (it is neither repelling nor
attracting). There is a
homoclinic connection which connects the equilibrium point at the origin. This holoclinic
connection represents the solitary wave. Within this homoclinic connection
lie periodic orbits which represent the cnoidal waves.

We can also integrate the equation
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}}=A_{1}
</math></center>
by multiplying by <math>f^{\prime }</math>and integrating. This gives us
<center><math>
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f + c\frac{
f^{2}}{2}-f^{3} = -V(f)
</math></center>
It is no coincidence that the right hand side is the potential energy, because this
is nothing more that the equation for conservation of energy (or the first
integral of the Lagrangian system) which does not depend on <math>\zeta</math>.

This is a separable equation and the only challenge is to integrate
<center><math>
\int \frac{1}{\sqrt{-V(f)}} \mathrm{d}f.
</math></center>

==Formula for the solitary wave==

We know that the solitary wave solution is found when <math>
A_{0}=A_{1}=0.</math> This gives us
<center><math>
\left( f^{^{\prime }}\right) ^{2}=f^{2}\left( c-2f\right)
</math></center>
This can be solved by separation of variables to give
<center><math>
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta
</math></center>
We then substitute
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( s\right)
</math></center>
and note that
<center><math>
c-2f=c\left( 1-\mathrm{sech}^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right)
</math></center>
and that
<center><math>
\frac{\mathrm{d}f}{\mathrm{d}s}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) }
</math></center>
This means that
<center><math>\begin{matrix}
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt{c}}\int \frac{\sinh \left(
f\right) }{\mathrm{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
f\right) }\mathrm{d}s \\
&=&-\frac{2}{\sqrt{c}}s
\end{matrix}</math></center>
This gives us
<center><math>
-\frac{2}{\sqrt{s}}=\zeta+a
</math></center>
Therefore
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( \zeta+a\right) \right)
</math></center>
Of course we assumed that <math>x=x-ct</math> so the formula for the solitary wave is
given by
<center><math>
f\left( x-ct\right) =\frac{1}{2}c\,\mathrm{sech}^{2}\left[ \frac{\sqrt{c}}{2}\left(
x-ct+a\right) \right]
</math></center>
Note that a solution exists for each
<math>c</math>, and that the amplitude is proportional to <math>c.</math> All of this was
discovered experimentally by
[http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell].

==Formula for the cnodal wave==

If we consider the case when the solution oscillates between two values <math>F_2<F_3</math>
(which we can assume are also roots of <math>V(f)</math> without loss of generality) then
we can integrate the equation to obtain
<center><math>
f\left( \zeta\right) =F_2+(F_3 - F_2) \mathrm{cn}^{2}\left( \gamma \zeta ;k\right)
</math></center>
where <math>cn</math> is a Jacobi Elliptic function and
<math>\gamma</math> and <math>k</math> are constants which depend on <math>c</math>.
We can write this equation as
<center><math>
f\left( x-ct\right) =a+b \mathrm{cn}^{2}\left( \gamma (x-ct);k\right)
</math></center>
where <math>a=k^2\gamma^2</math> and <math>c = 6b + 4(2k^2 -1)\gamma^2</math>.
These waves are known as [http://en.wikipedia.org/wiki/Cnoidal_wave cnoidal waves].

In the limit the two solutions agree. We also obtain a sinusoidal solution in the limit of
small amplitude.

Nonlinear Shallow Water Waves

2010-08-20T10:32:30Z

Adi Kurniawan:

{{nonlinear waves course
| chapter title = Nonlinear Shallow Water Waves
| next chapter = [[Introduction to KdV]]
| previous chapter = [[Traffic Waves]]
}}

{{complete pages}}

== Introduction ==

We assume that water is incompressible,
viscous effects are negligible and that the typical wave lengths are much larger than the water depth.
This allows us to assume [[:Category:Shallow Depth|Shallow Depth]]. We assume that the problem has not variation
in either the <math>y</math> or <math>z</math> direction. The fluid is governed by two parameters,
<math>u</math>, the velocity of the water, and <math>h</math> the water depth (note that this is not the still water depth since the problem
is nonlinear).

The theory we present here is discussed in [[Stoker 1957]], [[Billingham and King 2000]] and [[Johnson 1997]].

== Equations of Motion ==

The equation for the conservation of mass can derived by considering a a region <math>[x,x+\Delta x]</math>
Conservation of mass then implies that
<center>
<math>
\partial_t \int_x^{x + \Delta x} \rho h(s,t) \mathrm{d}s = \rho h(x,t)u(x,t) - \rho h(x+\Delta x,t)u(x+\Delta x,t)
</math>
</center>
If we take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t h(x ,t) + \partial_x (h(x ,t)u(x ,t)) = 0
</math>
</center>

A second equation comes from conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x}^{x + \Delta x}
+ \int_0^{h(x)} P(x,z,t) \mathrm{d}z -
\int_0^{h(x + \Delta x)} P(x+\Delta x,z,t) \mathrm{d}z
</math>
</center>
where <math>\rho \,</math> denotes density, and the pressure <math>P \,</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). This then gives us
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= -\left. \rho u^2 h \right|_{x}^{x + \Delta x}
+ \frac{1}{2}\rho g {h(x)}^2 -
\frac{1}{2}\rho g {h(x + \Delta x)}^2
</math>
</center>
If we then take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t \left( h u \right)
+ \partial_x \left(u^2 h + \frac{1}{2} gh^2\right) =0
</math>
</center>
We can simplify this using the equation derived from conservation of mass to
to obtain
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>

The equations
<center>
<math>
\partial_t h + u \partial_x h + h \partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>
are called the nonlinear shallow water equations. They determine the horizontal water velocity and the local water depth.

We can rewrite them in terms of the local wave speed <math>c(x, t) = \sqrt{gh(x, t)}</math> as follows:
<center>
<math>2\partial_t c + 2u\partial_x c + c\partial_x u = 0
</math>
</center>
<center>
<math>\partial_t u + u\partial_x u + 2c \partial_x c = 0
</math>
</center>
These equation are almost identical to those of compressible gas dynamics. Much of our understanding
of the equations for water have been found by researchers studying compressible gas dynamics.

== Linearized Equations ==

We can linearize these equations by assuming that <math>u</math> is small and that
<math>h=h_0 + \zeta </math> where <math>h_0</math> is the average water depth
and <math>\zeta</math> is also assumed small. This gives us
<center>
<math>
\partial_t \zeta + h_0\partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + g \partial_x \zeta = 0
</math>
</center>
These linear shallow water equations which can be
derived from the linear equations for water of finite
depth and taking the limit of small depth (see [[:Category:Shallow Depth|Shallow Depth]]).

== Characteristics ==

The equations possess characteristics.
Adding and subtracting the two equations above we obtain
<center>
<math>
\frac{\partial (u \pm 2c)}{\partial t}+ (u \pm c)\frac{\partial (u \pm 2c)}{\partial x} = 0
</math>
</center>
This means that
on the <math>\;C_+</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_+}{\mathrm{d} t} = u + c = u + \sqrt{gh}
</math>
</center>
the <math>\;C_+</math> invariant
<center>
<math>
R_+ = u + 2c = u + 2\sqrt{gh}
</math>
</center>
is a constant,
and on the <math>\;C_-</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - c = u - \sqrt{gh}
</math>
</center>
the <math>\;C_-</math> invariant
<center>
<math>
R_- = u - 2c = u - 2\sqrt{gh}
</math>
</center>
is a constant.

The functions <math>R_{\pm} (u ,c) = u \pm 2c</math>, are called the Riemannian invariants.

== Simple Waves ==

The problem as formulated can be solved by advancing the solution along the characteristics, but
this will in general be quite difficult analytically. However, there is a special class of problems,
called ''Simple Waves'' in which the solution only changes on one characteristic.
They are best illustrated through some examples. Note that the characteristic can meet forming
a shock, which is called a [http://en.wikipedia.org/wiki/Tidal_bore bore] or a
[http://en.wikipedia.org/wiki/Hydraulic_jump hydraulic jump]
when it occurs on the surface of the water.

=== The dam break problem ===
Assume the water occupies the region <math>{x < 0 ; 0 < z < h_0 }</math> initially held back by a dam at <math>x = 0</math>.
At <math>t = 0</math>, the dam is removed (breaks). What is the height of the water
<math>h(x,t) \,</math> for <math>t > 0? \,</math> The initial condition is therefore
<center>
<math>h(x,0) = \begin{cases}
h_0, & x < 0 \\
0, & x > 0
\end{cases}
</math>
</center>
<center>
<math>u(x ,0) = 0. \,
</math>
</center>
On the characteristic that originates at <math>t = 0</math> for <math>x < 0</math>,
<center>
<math>
R_\pm = u \pm 2\sqrt{gh} = \pm 2\sqrt{gh_0} = \pm 2c_0
</math>
</center>
where <math>c_0 = \sqrt{gh_0}</math> is the initial (linear) wave speed.

Therefore, if a <math>C_+</math> and a <math>C_-</math> characteristic from this region intersect, then
<center>
<math>
u + 2\sqrt{gh} = 2c_0 , \;\mathrm{and}\; u - 2\sqrt{gh} = -2c_0
</math>
</center>
and hence, <math>u = 0</math> and <math>h = h_0</math>.
Moreover,
<center>
<math>
\frac{\mathrm{d} X_\pm}{\mathrm{d} t} = u \pm \sqrt{gh} = \pm c_0
</math>
</center>
so these characteristics are straight lines in the region <math>\big\{x < -c_0 t \big\}</math>
(the undisturbed region).

The <math>\;C_+</math> characteristic leave the region a<math>\big\{x < -c_0 t \big\}</math>
and enter <math>\big\{x > -c_0 t \big\}</math>. For now we will assume that these characteristics fill the domain
(and show that this is true shortly).
For <math>\big\{x > -c_0 t \big\}</math>, the <math>C_-</math> characteristics are given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - \sqrt{gh}
</math>
</center>
and on each of the <math>C_-</math> characteristics <math>R_- = u - 2\sqrt{gh}</math> is constant.
However, since this region is filled with <math>C_+</math> characteristics where <math>R_+ = u + 2\sqrt{gh} = 2c_0</math>, <math>u</math> and <math>h</math>
must be constant on each <math>C_-</math> characteristic.
This also means that the <math>C_-</math> characteristics must be straight lines.

Since the fluid occupies <math>\big\{x < 0 \big\}</math> at <math>t = 0</math>,
these <math>C_-</math> characteristics must start at the origin, with
<center>
<math>
X_-(t) = \left(u - \sqrt{gh}\right)t
</math>
</center>
which in turn implies that
<center>
<math>
u - \sqrt{gh} = \frac{x}{t}
</math>
</center>
We also have <math>R_+ = u + 2\sqrt{gh} = 2c_0</math> from the
<math>C_+</math> characteristics. We can solve these equations
at each point in <math>\big\{x > -c_0 t \big\}</math>. Solving for <math>u</math> and <math>h</math> gives
<center>
<math>
h(x, t) =
\begin{cases}
\frac{h_0}{9}\left(2 - \frac{x}{c_0 t}\right)^2, \quad -c_0 t < x< 2 c_0 t,\\
h_0,\quad x <-c_0 t,
\end{cases}
</math>
</center>
<center>
<math>
u(x, t) =
\begin{cases}
\frac{2}{3} \left (c_0 + \frac{x}{t} \right ), \quad -c_0 t < x< 2 c_0 t,\\
0,\quad x <-c_0 t.
\end{cases}
</math>
</center>
Where we have assumed that, since <math>h = 0</math> for <math>x = 2c_0 t,</math>
the <math>C_+</math> characteristic only exist in the region <math>\big\{x < 2c_0 t \big\}</math>,
We will verify this by explicitly calculating them.

It remains to determine the <math>C_+</math> characteristic, which originated in <math>\big\{x < 0 \big\}</math>,
and show they
will fill the domain <math>\big\{x < 2c_0 t \big\}</math>. For <math>\big\{x < -c_0 t \big\}</math>,
the <math>\;C_+</math> characteristics are straight lines with slope <math>c_0</math> and are given by
<center>
<math>
X_+ (t) = -x_0 + c_0 t, \quad \left(x_0 > 0,\;\; t < \frac{x_0}{2c_0}\right)
</math>
</center>
When <math>t = \frac{x_0}{2c_0},\;\;\ X_{+} (t) = -c_0 t</math> so that for
<center>
<math>
t > \frac{x_0}{2c_0}, \quad \frac{\mathrm{d} X_{+} (t)}{\mathrm{d} t} = u + \sqrt{gh}
</math>
</center>
and substituting the solution we found for <math>h</math> and <math>u</math>
<center>
<math>
\frac{\mathrm{d}X_{+} (t)}{\mathrm{d} t} = \frac{4}{3}c_0 + \frac{X_{+} (t)}{3t}
</math>
</center>
Solving this ODE subject to <math>X_+ \left(\frac{x_0}{2c_0}\right) = -\frac{x_0}{2}</math> gives
<center>
<math>
X_+ (t) = 2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\;\;
</math>
</center>
the equation for a characteristic curve.
The curves indeed fill the domain <math>\big\{x < 2c_0 t \big\}</math>
and all satisfy <math>\big\{X_+ (t) < 2c_0 t \big\}</math>. To summarize, the
<math>C^{+}</math> characteristics are given by
<center>
<math>
X_+ (t) =
\begin{cases}
2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\quad t> x_0/2 c_0\\
-x_0 + c_0 t, \quad 0\ < t < x_0/2 c_0
\end{cases}
</math>
</center>

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Characteristics_dam_break.jpg|thumb|right|500px|Characteristics for the dam
break problem, blue for <math>C_{+}</math> and red for <math>C_{-}</math>. The solid red lines
show the curves <math>x=-c_0 t</math> and <math>x=2c_0 t</math> (note we have assumed
here that <math>c_0 =1</math>]]
| [[Image:Dambreak.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the Dam Break problem]]
|}

== Shocks ==

For a unique solution two exist there must be a single <math>C_+</math> and <math>C_-</math>
characteristic through each point. When two characteristics of the same kind meet we
have a shock forming.

=== Accelerating Piston ===

We now consider the problem of water initially at rest occupying the
half space <math>x>0</math> which is initially at rest. At <math>t=0</math>
the piston at <math>x=0</math> begins to move to the right with constant
acceleration <math>a</math> so that the position of the piston is given by
<math>\frac{1}{2}at^2</math>.

We assume that the <math>C_-</math> characteristics which originate in the
water at <math>t=0</math> fill the fluid. On these characteristics
<center>
<math>
R_- = u - 2 c = -2c_0 \,
</math>
</center>
and this condition must hold throughout the fluid.
On the <math>C_+</math> characteristics we know that
<math> R_+ = u + 2 c </math> must be a constant and hence on the
<math>C_+</math> characteristics <math>u</math> and <math>c</math>
must be constant and hence the <math>C_+</math> characteristics must
be straight lines. Note that this does not mean that the
<math>C_+</math> characteristics have the same slope and there is no
requirement that the <math>C_-</math> characteristics are straight lines.

The <math>C_+</math> characteristic originate from the fluid
or from the front of the piston. We consider those which originate from
the piston. The <math>C_+</math> characteristic which originates from
the piston at <math>t=t_0</math> must satisfy
<center>
<math>
u+2c = a t_0 + 2c_{\textrm{plate}} \,
</math>
</center>
where <math> a t_0</math> is the velocity of the piston at time <math>t=t_0</math>
and <math>c_{\textrm{plate}}</math> is the speed (related to height) at the plate.
We know that <math> R_- = u - 2 c =-2c_0</math> through out the fluid, so that if
we solve this at the plate (where <math>u=at_0</math> and <math>c=c_{\textrm{plate}}</math>)
then we get <math>c_{\textrm{plate}} = at_0/2 + c_0</math>

On the <math>C_+</math> characteristics <math>u</math> and <math>c</math>
are constant and therefore
<center>
<math>
\frac{\mathrm{d}X_+}{\mathrm{d}t} = u+c = at_0 + \left( \frac{1}{2}at_0 + c_0 \right)
</math>
</center>
Hence
<center>
<math>
X_+(t,t_0) = \left( \frac{3}{2} a t_0 + c_0 \right) t - c_0 t_0 -a t_0^2
</math>
</center>
using the condition <math>X_+(t_0) = \frac{1}{2} a t_0^2</math> (the initial value which
comes from the position of the piston at <math>t=t_0</math>).
The slope of these lines increases and eventually meet to form a shock.
We find this point of intersection by considering neighboring characteristics
and determining when they first intersect.
<center>
<math>
X_+(t,t_0 + \Delta t) = X_+(t,t_0) + \Delta t \frac{\partial X_+}{\partial t_0} (t,t_0)
</math>
</center>
It follows that neighbouring characteristics will meet when
<center>
<math>
\frac{\partial X_+}{\partial t_0} (t,t_0) = 0
</math>
</center>
which implies that
<center>
<math>
t = \frac{2c_0}{3a} + \frac{4}{3}t_0
</math>
</center>
The first time that a shock forms is the minimum value of this equation.
For this piston example, this occurs when <math>t_0 = 0</math> and the value
of <math>t</math> is <math>t = 2c_0/(3a)</math>. At this point
a shock is formed and we can no longer find a unique solution by following the
characteristics.

=== Piston Moving with Constant Velocity ===

This example is also known as the Moving Wall Problem, and is connected to Shallow Water Bores.

We consider the case of a piston, with positive constant velocity <math>V</math> (which is initially at <math>x=0</math>), advancing into a semi-infinite expanse of
fluid that is initially at rest with depth <math>h_0</math>.

The <math>C_+</math> characteristics which originate in the fluid
at <math>t=0</math> have slope
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0}</math>
</center>
and the <math>C_+</math> characteristic which originate at the piston at
<math>t=0</math> must satisfy
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0} + V </math>
</center>
so that these two characteristics will intersect at <math>t=0</math>.
Therefore a shock forms immediately and we can track this by determining the
speed of the shock

=== Speed of the shock ===

We need the conservation equations in integral form to determine the speed
of the shock. Conservation of mass, written as an integral is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h \mathrm{d}x
+ \left. \rho u h \right|_{x_2}^{x_1} =0
</math>
</center>
If the shock is located at <math>s(t)</math>
which we assume is located between <math>x_1</math>
and <math>x_2</math>, then
<center>
<math>
\partial_t \int_{x_1}^{x_2} h \mathrm{d}x
= \partial_t \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) h \mathrm{d}x
= \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) \partial_t h \mathrm{d}x + h^{+} \partial_t s(t) - h^{-}\partial_t s(t),
</math>
</center>
where <math>h^+</math> is the height on the right (positive) side of
the jump and <math>h^-</math> is the height on the left (negative) side.
If we take the limit as <math>x_1\to x_2</math> we then obtain the following
identity
<center>
<math>
h^{+}\partial_t s(t) - h^{-}\partial_t s(t) -
u^{+} h^{+} + u^{-} h^{-} = 0
</math>
</center>
where <math>u^+</math> is the height on the right (positive) side of
the jump and <math>u^-</math> is the height on the left (negative) side.

We now need to consider the equation for conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x_1}^{x_2}
+ \int_0^{h(x_1)} P(x_1,z,t) \mathrm{d}z -
\int_0^{h(x_2)} P(x_2,z,t) \mathrm{d}z
</math>
</center>
where the pressure <math>P</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). We can apply a similar argument as before to obtain
<center>
<math>
h^{+}u^{+}\partial_t s(t) - h^{-}u^{-}\partial_t s(t) =
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2
</math>
</center>

=== Hydraulic Jump ===

For a hydraulic jump, <math>\dot{s}(t) = 0</math>, which means that we must solve
<center>
<math>
u^{+} h^{+} - u^{-} h^{-} = 0 \,
</math>
</center>
<center>
<math>
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2 =0
</math>
</center>
If we introduce the variables
<center>
<math>
H = \frac{h^{+}}{h^{-}}
</math>
</center>
and
<center>
<math>
\mathrm{Fr} = \frac{u^{-}}{\sqrt{gh^{-}}}
</math>
</center>
where <math>\mathrm{Fr}</math> is the ''Froude'' number
which is equivalent to the Mach number for gas dynamics.
Then we obtain
<center>
<math>
H^2 -1 = 2 \mathrm{Fr}^2 \left(1 - \frac{1}{H}\right)
</math>
</center>
This expression has the roots
<center>
<math>
H=1, \quad H=\frac{1}{2}\left(-1\pm\sqrt{1 + 8 \mathrm{Fr}^2}\right)
</math>
</center>
The only physically meaningful solution is the root which satisfies
<math>H>1</math>. This is only true providing <math>\mathrm{Fr} > 1</math>, which means
that we can only obtain a hydraulic jump if the flow is supercritical.

Below is a video of a hydraulic jump. You can clearly see the point where the flow is changing from supercritical to subcritical (look for the small turbulent region in the channel)
{{#ev:youtube|5etwhZ0d2GU}}

=== Shallow Water Bore ===
We now consider a bore, in which
the shock wave advances into still water.
We denote the fluid speed by <math>V = u^{-}</math>.
We denote the height on the wall side
by <math>h_1</math> and the height on the other side must be <math>h_0</math>, i.e.
<math>h^{+} = h_0</math>, <math>h^{-} = h_1</math>, <math>u^{+} = 0</math>.
This means that
<center>
<math>
h_{0}\partial_t s(t) - h_{1}\partial_t s(t)
+ V h_{1}= 0
</math>
</center>
and
<center>
<math>
- h_{1}V\partial_t s(t) = - \left(V\right)^2 h_{1}
+ \frac{1}{2} g \left(h_{0}\right)^2 - \frac{1}{2} g \left(h_{1}\right)^2
</math>
</center>
which can be solved to obtain the shock speed and the height of the moving fluid.

Below is a video of surfing on the [http://en.wikipedia.org/wiki/Severn Severn] bore, do not believe everything they
say. You might also want to check out the [http://en.wikipedia.org/wiki/Pororoca Pororoca]
a tidal bore on the Amazon.
{{#ev:youtube|0jtkyuuMgVQ}}

[[Category:Simple Nonlinear Waves]]
[[Category:Nonlinear Water-Wave Theory]]
[[Category:Shallow Depth]]

Traffic Waves

2010-08-20T10:30:12Z

Adi Kurniawan:

{{nonlinear waves course
| chapter title = Traffic Waves
| next chapter = [[Nonlinear Shallow Water Waves]]
| previous chapter = [[Method of Characteristics for Linear Equations]]
}}

{{complete pages}}

We consider here some simple equations which model traffic flow. This problem is discussed in
[[Billingham and King 2000]].

[[Category:Reference]]

== Equations ==

We consider a single lane of road, and we measure distance along the road with
the variable <math>x</math> and <math>t</math> is time.
We define the following variables
<center><math>
\begin{matrix}
&\rho(x,t) &: &\mbox{car density (cars/km)} \\
& v(\rho) &: &\mbox{car velocity (km/hour)} \\
& q(x,t) =\rho v &: &\mbox{car flow rate (cars/hour)} \\
\end{matrix}
</math></center>

If we consider a finite length of road <math>x_1\leq x \leq x_2</math> then the net flow of cars
in and out must be balanced by the change in density. This means that
<center><math>
\frac{\partial}{\partial t} \int_{x_1}^{x_2} \rho(x,t) \mathrm{d}x = -q(x_2,t) + q(x_1,t)
</math></center>
We now consider continuous densities (which is obviously an approximation) and
set <math>x_2 = x_1 + \Delta x</math> and we obtain
<center><math>
\frac{\partial}{\partial t} \rho(x_1,t) = -\frac{q(x_2,t) + q(x_1,t)}{\Delta x}
</math></center>
and if we take the limit as <math>\Delta x \to 0</math> we obtain the differential equation
<center><math>
\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} = 0
</math></center>
Note that this equation has been derived purely from the need to conserve cars (it is a conservation equation) and
is not possible to solve this equation until we have derived a connection between <math>\rho</math>
and <math>q</math>.

== Equation for <math>\rho</math> only ==

At the moment we assume that we have some expression for <math>v(\rho)</math>
If we substitute the expression for <math>q = v\rho</math> into our differential equation we obtain
<center><math>
\frac{\partial \rho}{\partial t} + \frac{\partial }{\partial x} \left(v(\rho)\rho\right) = 0
</math></center>
which gives us
<center><math>
\frac{\partial \rho}{\partial t} + \left(v^{\prime}(\rho)\rho + v(\rho)\right)
\frac{\partial \rho }{\partial x} = 0
</math></center>
or
<center><math>
\frac{\partial \rho}{\partial t} + c(\rho)\frac{\partial \rho }{\partial x} = 0
</math></center>
where <math>c(\rho) = \left(v^{\prime}(\rho)\rho + v(\rho)\right)</math>
is the '''kinematic wave speed'''. Note that this is not the speed of the cars, but
the speed at which disturbances in the density travel.

== A simple relationship between <math>\rho</math> and <math>q</math> ==

The relationship between <math>\rho</math> and <math>q</math> is an equation of state and
there is no ''exact'' equation since it depends on many unknowns. One of the
simplest relationship between <math>\rho</math> and <math>q</math> is derived from
the following assumptions

* When the density <math>\rho = 0</math> the speed is <math>v=v_0</math>
* When the density is <math>\rho = \rho_{\max} </math> the speed is <math>v=0</math>
* The speed is a linear function of <math>\rho</math> between these two values.

This also gives good fit with measured data. We will either consider the general case or use this simple
relationship. Using this we obtain
<center><math>
v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
</math></center>
The flux of cars is given by
<center><math>
q = \rho v(\rho) = v_0\frac{\rho(\rho_{\max} - \rho)}{\rho_{\max}}
</math></center>
and the wave speed is
<center><math>
c(\rho) = \left(v^{\prime}(\rho)\rho + v(\rho)\right) = -\frac{v_0}{\rho_{\max}}\rho + v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
= v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>

{| class="wikitable"
|-
! <math>v</math>
! <math>q</math>
! <math>c</math>
|-
| [[Image:Velocity.jpg|thumb|350px|<math>v(\rho)</math> versus <math>\rho</math>]]
| [[Image:Q_flux.jpg|thumb|350px|<math>q(\rho)</math> versus <math>\rho</math>]]
| [[Image:C_speed.jpg|thumb|350px|<math>\rho(\rho)</math> versus <math>\rho</math>]]
|}

== Small Amplitude Disturbances ==

We can linearise the model by assuming that the variation in density is small so
that we can write
<center><math>
\rho = \rho_0 + \tilde{\rho}
</math></center>
where we assume that <math>\tilde{\rho}</math> is small. This allows us to write the equations as
<center><math>
\frac{\partial \tilde{\rho}}{\partial t} + c(\rho_0) \frac{\partial \tilde{\rho}}{\partial x} = 0
</math></center>
where the main difference between this and the full equation is that the wave speed <math>c</math> is
a constant. This is the linearised equation. Note that this linearisation does not give a good model because
traffic density does not vary only a small amount about some mean (as is the case for accoustic waves where the
density of air is roughly constant).

Under these assumptions the solution to the equation is
<center><math>
\tilde{\rho} = f(x - c(\rho_0)t)
</math></center>
where <math>f</math> is determined by the initial condition. This represents
disturbances which travel with speed <math>c(\rho_0)</math> in the positive
<math>x</math> direction.

We now consider the '''characteristic curves''' which are curves along which the density
<math>\rho</math> is a constant. These are give by
<center><math>
x = X(t) = x_0 + c(\rho_0) t.\,
</math></center>
which are just straight lines of constant slope. We will see shortly that the full (nonlinear)
equations also possess characteristics.

== Nonlinear Initial Value Problem ==

We wish to solve
<center><math>
\frac{\partial \rho}{\partial t} + c(\rho) \frac{\partial \rho}{\partial x} = 0
</math></center>
subject to the initial conditions
<center><math>
\rho(x,0) = \rho_0(x) \,
</math></center>
It turns out that the concept of characteristic curves is very important for this problem.

If we want <math>\rho(X(t),t)</math> to be a constant then we require
<center><math>
\frac{\mathrm{d}}{\mathrm{d}t}\rho(X(t),t) = \frac{\mathrm{d} X}{\mathrm{d}t} \frac{\partial \rho}{\partial x} +
\frac{\partial \rho}{\partial t} = 0.
</math></center>
Comparing this to the governing partial differential equation we can see that we require
<center><math>
\frac{\mathrm{d} X}{\mathrm{d}t} = c(\rho)
</math></center>
This means that the characteristics are straight lines (since <math>\rho</math> is constant) with
slope given by <math> c(\rho_0(x_0))</math> so that the equation for the characteristics is
<center><math>
X(t) = x_0 + c(\rho_0(x_0))t \,
</math></center>

This does not allow us to write down a solution to the initial value problem,
all we can do is write
<center><math>
\rho(x_0 + c(\rho_0(x_0))t,t) = \rho_0(x_0)\,
</math></center>
which allows us to calculate the solution stepping forward in time, but not to determine the solution given
a value of <math>(x,t)</math> (because we have no way of knowing what <math>c(\rho_0(x_0))</math> is.

The characteristics are a family of straight lines which will all have different slopes. If two characteristics
meet, our solution method will break down because there will be two values of the density <math>\rho</math>.
This gives rise to a '''shock'''. It turns
out that this the formation of shocks is a product of the equations themselves and not with the solution method.
We will see shortly that special methods are required to treat these shocks.

=== Case when no shocks are formed ===

The characteristic curves will fill the space without meeting provided that the wave speed <math>c(\rho_0)</math>
is a monotonically increasing function of the distance <math>x</math>. If we work with our previous model we
have
<center><math>
v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
</math></center>
and
<center><math>
c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>
so that <math>c</math> is a monotonically decreasing function of density <math>\rho</math>. This means
that the wave speed <math>c(\rho_0)</math> will be
a monotonically increasing function of the distance <math>x</math> if an only if the density is a
monotonically decreasing function. In this case the solution can be calculated straightforwardly
by expansion of the initial density.

==== No shock example ====

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by <math>\rho_0 = 1/2(1- \tanh(x))</math>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Traffic example1 rho.jpg|thumb|350px| <math>\rho_0 = 1/2(1- \tanh(x))</math> ]]
| [[Image:Traffic_example1_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Traffic_example1_characteristics.jpg|thumb|350px|Characterisitics for <math>\rho_0 = 1/2(1- \tanh(x))</math>]]
| [[Image:Traffic_example11.gif|thumb|350px|<math>\rho(x,t) for </math><math>\rho_0 = 1/2(1- \tanh(x))</math>]]
|}

==== Riemann problem and the expansion fan ====

We can consider a simple problem in which there is a jump in the initial density
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < 0 \\
\rho_{R},& x > 0
\end{matrix}
\right.
</math></center>
where <math>\rho_{L} > \rho_{R}</math> so that we do not form a shock. In this case
the characteristics on each side of <math>x=0</math> have a different slope and the
question is what happens between. It is easiest to think about the following problem
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < -\epsilon \\
\frac{\rho_{R}-\rho_{L}}{2\epsilon}x + \frac{\rho_{R}+\rho_{L}}{2} & -\epsilon \leq x \leq \epsilon \\
\rho_{R},& x > \epsilon
\end{matrix}
\right.
</math></center>
We can then see that we have lines of uniformly varying slope for <math>-\epsilon<x<\epsilon</math>
with slope between <math>c(\rho_L)</math> and <math>c(\rho_R)</math>. If we then take the limit
as <math>\epsilon \to 0</math> we obtain an expansion fan emanating from <math>x=0</math>.

If we assume that
<center><math>
c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>
then we know that on the lines of the expansion fan (which all start at <math>x=0</math>) we have
<math>c(\rho) = x/t</math>. We can rearrange this and solve for <math>x</math> and obtain
<math>\rho =\frac{ 1}{2} \rho_{\max} (1-x/v_0 t)</math>

The solution is then given by
<center><math>
\rho(x,t) =
\left\{
\begin{matrix}
\rho_{L},& x < c(\rho_L) t\\
\frac{ \rho_{\max}}{2} (1-x/v_0 t),& c(\rho_L) t \leq x \leq c(\rho_R) t\\
\rho_{R},& x > c(\rho_R) t
\end{matrix}
\right.
</math></center>
This solution is known as an '''expansion fan'''.

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
0.6,& x < 0 \\
0.3,& x > 0
\end{matrix}
\right.
</math></center>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Expansion_fan_rho.jpg|thumb|350px| <math>\rho_0</math> ]]
| [[Image:Expansion_fan_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Expansion_fan_characteristics.jpg|thumb|350px|Characterisitics]]
| [[Image:Expansion_fan1.gif|thumb|350px|<math>\rho(x,t)</math>]]
|}

=== Shocks ===

So far we have only considered the case when <math>c(x_0)</math> is monotonically increasing so that
two characteristics never cross. We now consider the case when characteristics can meet.
A movie of this case is shown below.

{{#ev:youtube|Suugn-p5C1M}}

We can easily see that
the first characteristics to meet will be neighbouring characteristics. Consider two characteristics
with
<center><math>
X_1(t) = x_0 + c(x_0)t\,
</math></center>
<center><math>
X_2(t) = x_0 + \delta x + c(x_0+\delta x)t\,
</math></center>
Then these curves will meet at time <math>T</math> where
<center><math>
x_0 + c(x_0)T = x_0 + \delta x + c(x_0+\delta x)T\,
</math></center>
which implies that
<center><math>
T = -\frac{1}{c^{\prime}(x_0)}
</math></center>
Note the following
* If <math>c^{\prime}(x) > 0 </math> then no shock will form.
* The shock first forms at the minimum positive value of
<math> - \frac{1}{c^{\prime}(x)} </math> for <math> -\infty < x <\infty </math>.

==== Shock Fitting ====

If we calculate the solution using our formula
<center><math>
\rho(x_0 + c(\rho_0(x_0))t,t) = \rho(x_0)\,
</math></center>
then we find that the solution becomes multivalued in the case when a shock forms.
We then have to fit a shock. One way to do this is by imposing the condition that equal
areas are removed and added when we chose the position of the shock.
This corresponds to the condition that
the number of cars must be conserved

==== Speed of the shock ====

If we consider the case when there is a shock at <math>s(t)</math> with
<math>\rho = \rho^{-}</math> at <math>s=s^{-}</math>
and
<math>\rho = \rho^{+}</math> at <math>s=s^{+}</math>
(where <math>s^{-}</math>
is just less than s(t) and <math>s^{+}</math>
is just greater than s(t) ). If we substitute
this into the governing integral equation we obtain
<center><math>
\frac{\partial}{\partial t} \left( \int_{x_1}^{s(t)} + \int_{s(t)}^{x_2}\right)
\rho(x,t)\mathrm{d}x = q(x_1,t) - q(x_2,t)
</math></center>
and hence
<center><math>
\int_{x_1}^{x_2}
\frac{\partial \rho(x,t)}{\partial t} dx + \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{-}
- \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{+} = q(x_1,t) - q(x_2,t)
</math></center>
If we now take the limit as <math>x_1\to x_2</math> we obtain
<center><math>
\frac{\mathrm{d}s}{\mathrm{d}t}\rho^{-}
- \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{+} = q(\rho^{-}) - q(\rho^{+})
</math></center>
so that
<center><math>
\frac{\mathrm{d}s}{\mathrm{d}t} = \frac{q(\rho^{-}) - q(\rho^{+})}
{\rho^{-} - \rho^{+}}
</math></center>

==== Shock example ====

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by <math>\rho_0 = 1/2(1 + \tanh(x))</math>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Traffic example2 rho.jpg|thumb|350px| <math>\rho_0 = 1/2(1+ \tanh(x))</math> ]]
| [[Image:Traffic_example2_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Traffic_example2_characteristics.jpg|thumb|350px|Characterisitics for <math>\rho_0 = 1/2(1+ \tanh(x))</math>]]
| [[Image:Traffic_example2.gif|thumb|350px|<math>\rho(x,t) for </math><math>\rho_0 = 1/2(1+ \tanh(x))</math> Dotted solution is without shock fitting.]]
|}

==== Riemann problem ====

We now consider the Riemann problem
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < 0 \\
\rho_{R},& x > 0
\end{matrix}
\right.
</math></center>
where <math>\rho_{L} < \rho_{R}</math>. In this case a shock forms immediately and
the characteristics terminate at the shock. The shock moves with constant speed given by
the equation for the motion of the shock (or can be found by the equal areas rule). We obtain
<center><math>
\rho(x,t) =
\left\{
\begin{matrix}
\rho_{L},& x < \frac{1}{2} \left(c(\rho_{L}) + c(\rho_{R}) \right) t \\
\rho_{R},& x > \frac{1}{2} \left(c(\rho_{L}) + c(\rho_{R}) \right) t
\end{matrix}
\right.
</math></center>

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
0.3,& x < 0 \\
0.6,& x > 0
\end{matrix}
\right.
</math></center>

The figures below show the initial density, the initial speed,
the characteristics, and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Shock_rho.jpg|thumb|350px| <math>\rho_0</math> ]]
| [[Image:Shock_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Shock_characteristics.jpg|thumb|350px|Characterisitics with shock shown in green.]]
| [[Image:Shock3.gif|thumb|350px|<math>\rho(x,t)</math> with the red dotted line showing the solution without shock fitting]]
|}

[[Category:789]]
[[Category:Simple Nonlinear Waves]]

Method of Characteristics for Linear Equations

2010-08-19T14:35:11Z

Adi Kurniawan:

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} U}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>
</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

User talk:Meylan

2010-08-19T13:49:59Z

Adi Kurniawan: /* Boundary Element Method for a Fixed Body in Finite Depth */

Dr. Meylan, what about putting links which would be useful for water waves community under '''Links'''?
[[User:Adi Kurniawan|Adi Kurniawan]] 15:44, 6 April 2009 (NZST)

== Boundary Element Method for a Fixed Body in Finite Depth ==

1) Sections 4.2 and 4.3 seem to be the same.

2) I tried running the Matlab code, but the plot of R and T did not look like the one shown in the figure. --[[User:Adi Kurniawan|Adi Kurniawan]] 13:49, 19 August 2010 (UTC)

User talk:Meylan

2010-08-19T13:49:21Z

Adi Kurniawan: /* Boundary Element Method for a Fixed Body in Finite Depth */ new section

Dr. Meylan, what about putting links which would be useful for water waves community under '''Links'''?
[[User:Adi Kurniawan|Adi Kurniawan]] 15:44, 6 April 2009 (NZST)

== Boundary Element Method for a Fixed Body in Finite Depth ==

1) Sections 4.2 and 4.3 seem to be the same.
2) I tried running the Matlab code, but the plot of R and T did not look like the one shown in the figure.