WikiWaves - User contributions [en]

Category:Multipole Methods for Linear Water Waves

2010-02-19T00:19:13Z

Sean Curry: /* With the Free Surface BC */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

<math>\psi_n\,</math> can be expanded in a power series:

<center><math>\psi_n = \sum_{m=0}^\infty A_{mn}r^m e^{im\theta},\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the substiuting the identity <center><math>e^{\omega z} e^{-i\omega x} = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m e^{im\theta}
</math></center>
into the expression for <math>\psi_n\,</math>. The identity is derived by expanding <math> e^{\omega z}e^{-i\omega x}e^{\omega f} = e^{\omega(z+f-ix)} \,</math> as a power series and using <math>z+f = -r\cos\theta \,\, and \,\, x=r\sin\theta</math>.

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_n</math> for n>0.
This accounts for the second linearly independent solution for n>0.

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n\,</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center><center><math>
+\frac{K}{n}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships: (need a picture)

<center><math>r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2},</math></center>
<center><math>x = r \sin(\theta) = r_1 \sin(\theta_1),\,</math></center>
<center><math>-z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1).</math></center>

Resulting in the expression:
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{K}{n}\frac{e^{in\theta}}{r^n} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} - \frac{K}{n}\frac{e^{-in\theta_1}}{r_1^n}</math></center>
ie.
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \left( \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} \right) + \frac{K}{n} \left( \frac{e^{in\theta}}{r^n} - \frac{e^{-in\theta_1}}{r_1^n} \right)</math></center>

Once again to obtain a complete set we also use the compex conjugates of these potentials.

These idntities have been used:

<math> -\frac{e^{-in\theta_1}}{r_1^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega </math>
and
<math> \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} = \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \beta_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-17T01:21:40Z

Sean Curry: /* With the Free Surface BC */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_{m=0}^\infty A_{mn}r^m e^{im\theta},\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} e^{i\omega x} = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m e^{im\theta}
</math></center>

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n\,</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center><center><math>
+\frac{K}{n}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships: (need a picture)

<center><math>r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2},</math></center>
<center><math>x = r \sin(\theta) = r_1 \sin(\theta_1),\,</math></center>
<center><math>-z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1).</math></center>

Resulting in the expression:
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{K}{n}\frac{e^{in\theta}}{r^n} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} - \frac{K}{n}\frac{e^{-in\theta_1}}{r_1^n}</math></center>
ie.
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \left( \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} \right) + \frac{K}{n} \left( \frac{e^{in\theta}}{r^n} - \frac{e^{-in\theta_1}}{r_1^n} \right)</math></center>

Once again to obtain a complete set we also use the compex conjugates of these potentials.

These idntities have been used:

<math> -\frac{e^{-in\theta_1}}{r_1^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega </math>
and
<math> \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} = \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \beta_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-17T00:50:02Z

Sean Curry: /* Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth) */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_{m=0}^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n\,</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center><center><math>
+\frac{K}{n}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships: (need a picture)

<center><math>r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2},</math></center>
<center><math>x = r \sin(\theta) = r_1 \sin(\theta_1),\,</math></center>
<center><math>-z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1).</math></center>

Resulting in the expression:
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{K}{n}\frac{e^{in\theta}}{r^n} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} - \frac{K}{n}\frac{e^{-in\theta_1}}{r_1^n}</math></center>
ie.
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \left( \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} \right) + \frac{K}{n} \left( \frac{e^{in\theta}}{r^n} - \frac{e^{-in\theta_1}}{r_1^n} \right)</math></center>

Once again to obtain a complete set we also use the compex conjugates of these potentials.

These idntities have been used:

<math> -\frac{e^{-in\theta_1}}{r_1^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega </math>
and
<math> \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} = \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \beta_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-17T00:27:30Z

Sean Curry: /* With the Free Surface BC */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_{m=0}^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n\,</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center><center><math>
+\frac{K}{n}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships: (need a picture)

<center><math>r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2},</math></center>
<center><math>x = r \sin(\theta) = r_1 \sin(\theta_1),\,</math></center>
<center><math>-z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1).</math></center>

Resulting in the expression:
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{K}{n}\frac{e^{in\theta}}{r^n} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} - \frac{K}{n}\frac{e^{-in\theta_1}}{r_1^n}</math></center>
ie.
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \left( \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} \right) + \frac{K}{n} \left( \frac{e^{in\theta}}{r^n} - \frac{e^{-in\theta_1}}{r_1^n} \right)</math></center>

Once again to obtain a complete set we also use the compex conjugates of these potentials.

These idntities have been used:

<math> -\frac{e^{-in\theta_1}}{r_1^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega </math>
and
<math> \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} = \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T21:41:52Z

Sean Curry: /* Wave-Free Potentials */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_m=0^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n\,</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center><center><math>
+\frac{K}{n}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships: (need a picture)

<center><math>r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2},</math></center>
<center><math>x = r \sin(\theta) = r_1 \sin(\theta_1),\,</math></center>
<center><math>-z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1).</math></center>

Resulting in the expression:
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{K}{n}\frac{e^{in\theta}}{r^n} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} - \frac{K}{n}\frac{e^{-in\theta_1}}{r_1^n}</math></center>
ie.
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \left( \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} \right) + \frac{K}{n} \left( \frac{e^{in\theta}}{r^n} - \frac{e^{-in\theta_1}}{r_1^n} \right)</math></center>

Once again to obtain a complete set we also use the compex conjugates of these potentials.

These idntities have been used:

<math> -\frac{e^{-in\theta_1}}{r_1^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega </math>
and
<math> \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} = \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T21:38:52Z

Sean Curry: /* Wave-Free Potentials */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_m=0^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega + \frac{K}{n}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships: (need a picture)

<center><math>r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2},</math></center>
<center><math>x = r \sin(\theta) = r_1 \sin(\theta_1),\,</math></center>
<center><math>-z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1).</math></center>

Resulting in the expression:
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{K}{n}\frac{e^{in\theta}}{r^n} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} - \frac{K}{n}\frac{e^{-in\theta_1}}{r_1^n}</math></center>
ie.
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \left( \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} \right) + \frac{K}{n} \left( \frac{e^{in\theta}}{r^n} - \frac{e^{-in\theta_1}}{r_1^n} \right)</math></center>

Once again to obtain a complete set we also use the compex conjugates of these potentials.

These idntities have been used:

<math> -\frac{e^{-in\theta_1}}{r_1^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega </math>
and
<math> \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} = \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T21:30:45Z

Sean Curry: /* Wave-Free Potentials */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_m=0^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega + \frac{K}{n}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships: (need a picture)

<center><math>r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2},</math></center>
<center><math>x = r \sin(\theta) = r_1 \sin(\theta_1),\,</math></center>
<center><math>-z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1).</math></center>

Resulting in the expression:
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{K}{n}\frac{e^{in\theta}}{r^n} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} - \frac{K}{n}\frac{e^{-in\theta_1}}{r_1^n}</math></center>
ie.
<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \left( \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{e^{-i(n+1)\theta_1}}{r_1^{(n+1)}} \right) + \frac{K}{n} \left( \frac{e^{in\theta}}{r^n} - \frac{e^{-in\theta_1}}{r_1^n} \right)</math></center>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T21:18:05Z

Sean Curry: /* With the Free Surface BC */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_m=0^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Wave-Free Potentials ==

The combination <math>\phi_{n+1} + Kn^{-1}\phi_n</math>, n=1,2,3,... corresponds to a wave free singularity ie. no waves are radiated to infinity so the potential dies off in the far field.

<center><math>\phi_{n+1} + Kn^{-1}\phi_n = \frac{e^{i(n+1)\theta}}{r^{(n+1)}} + \frac{(-1)^{(n+1)}}{n!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n} e^{\omega (z-f)} e^{-i\omega x}d\omega + Kn^{-1}\left( \frac{e^{in\theta}}{r^n} + \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega \right)</math></center>

This can be simplified by using the coordinate relationships:

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T20:51:12Z

Sean Curry: /* With the Free Surface BC */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} </math></center>
<center><math>\Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) </math></center>

Hence

<center><math>A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

Working to get <math>\hat{f}_n(\omega)</math>:

We can write:
<center><math>f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega</math></center>
where
<center><math>g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
</math></center>
We can rewrite <math>f_n(x)\,</math> as an inverse fourier transform by using the euler identity and the odd and even properties of sine and cosine. Allowing us to obtain <math>\hat{f}_n(\omega)</math>:
<center><math>f_n(x) = \int_0^\infty g(\omega)\cos(\omega x) d\omega -i \int_0^\infty g(\omega)\sin(\omega x) d\omega</math></center>
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)\cos(\omega x) d\omega -\frac{i}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)\sin(\omega x) d\omega</math></center>
Since <math>g(|\omega|)\,</math> is even, its integral against sine over all x is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace -i*sine for the second).
<center><math>\Rightarrow f_n(x) = \frac{1}{2}\int_{-\infty}^\infty g(|\omega|)e^{-i\omega x} d\omega + \frac{1}{2} \int_{-\infty}^\infty \frac{\omega}{|\omega|}g(|\omega|)e^{-i\omega x} d\omega</math></center>
Finally, combining the integrals, <math>\hat{f}_n(\omega)</math> emerges:
<center><math>\Rightarrow f_n(x) = \int_{-\infty}^\infty \frac{1}{2}\left( g(|\omega|) + \frac{\omega}{|\omega|}g(|\omega|)\right) e^{-i\omega x} d\omega</math></center>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_m=0^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T20:13:26Z

Sean Curry: /* With the Free Surface BC */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For n>0
<center><math>\phi_n = \frac{e^{in\theta}}{r^n} + \psi_n</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots</math></center>

For n=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For n>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_n(x) = (\partial_z - K)\psi_n|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{in\theta}}{r^n}\right)|_{z=0}
</math>

<math> \Rightarrow f_n(x) = \frac{(-1)^n}{(n-1)!} \int_0^\infty(K+\omega)\omega^{n-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{in\theta}}{r^n} = \frac{(-1)^n}{(n-1)!} \int_0^\infty\omega^{n-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_n\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_n = \int_{-\infty}^\infty \psi_n e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_n = \omega^2\hat{\psi}_n \Rightarrow \hat{\psi}_n = A_n(\omega)e^{\omega z} + B_n(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} = (\omega - K)A(\omega) = \hat{f}_n(\omega)</math></center>

Hence

<center><math>A(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_n = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_n e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_n(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For n>0 the form of <math>\hat{f}_n(\omega)</math> can be obtained from the expansion of <math>f_n(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_n\,</math> easily simplifies to:

<center><math>\psi_n = \frac{(-1)^n}{(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{n-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for n>0. Note: the integral above goes under the singularity at <math>\omega=K</math>

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

<math>\psi_m\,</math> can be expanded in a power series:

<center><math>\psi_m = \sum_m=0^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center>

where
<center><math>A_{mn}= \frac{(-1)^{m+n}}{m!(n-1)!} \int_0^\infty \frac{\omega + K}{\omega - K} \omega^{m+n-1} e^{-2\omega f} d\omega
</math></center>
Note: the integral above goes under the singularity at <math>\omega=K</math>

Which comes from the identity (to be derived)
<center><math>e^{\omega z} \cos(\omega x) = e^{-\omega f}\sum_{m=0}^\infty \frac{(-1)^m}{m!} \omega^m r^m\cos(m\theta)
</math></center>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T19:25:31Z

Sean Curry: /* With the Free Surface BC */

{{incomplete pages}}

== Multipole Expansions ==

Multipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated separately.

Multipoles satisfy:

{{Standard linear wave scattering equations without body condition}}

{{sommerfeld radiation condition two dimensions}}

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

We start by developing the theory for Laplace's equation without the free surface for
a disk of radius <math>a</math> centered at the origin. The equation in polar coordinates is
<center><math>
\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad
\mathrm{for} \quad r>a
</math></center>

<center>
<math>\partial_n \phi |_{r=a} = 0
</math>
</center>

<center><math>
|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty
</math></center>

Using [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
we write
<math>\phi = R(r)\Theta(\theta).\,</math> Substituting into Laplace's equations gives

<center>
<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2
</math></center>

The equation for <math>\Theta\,</math> is

<center><math>\quad \partial_\theta^2 \Theta = -m^2\Theta </math></center>

<center><math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math></center>

When m=0 the solution is

<center><math> \Theta = A(0) + B(0)\theta \,</math></center>

<center> <math>\Theta(\theta + 2\pi) = \Theta(\theta) \Rightarrow m \in Z \,\, and \,\, B(0) = 0\,</math></center>

The equation for R is

<center><math> \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math></center>

This is a standard DE, to solve we substitute R = r^n giving:

<center><math>n(n-1)r^n + nr^n = m^2r^n \Rightarrow n^2 = m^2 \Rightarrow n= \pm m</math></center>

This gives two independent solutions for all m except m = 0, where we only get the constant solution.

Noting that:

<center><math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 </math> </center>

So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:

<center><math>R = C\left(0\right) + D(0) \ln(r)</math></center>

and for <math>m\neq 0:</math>

<center><math>R = C\left(m\right)r^m + D(m)r^{-m}</math></center>

The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:

<center><math>|\nabla\phi| \rightarrow 0 \,\, as \,\, r \rightarrow \infty \Rightarrow\partial_r R \rightarrow 0 \,\, as \,\, r \rightarrow \infty </math></center>

<center><math>\Rightarrow C(m)= 0 \,\, \forall m > 0 \,\,\, and \,\,\, D(m)=0 \,\, \forall m<0</math></center>

Hence the general solution can be expressed as:

<center><math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math></center>

We don't need to include the negative m terms because of the symmetry of the sine and cosine eigenfunctions means these solutions are not independent.

The solution can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

Note that the origin is in the free surface directly above the center of the disc for cartesian coordinates. However the origin for the polar coordinates is still at the center of the disk.

We add a solution of Laplace's equation to the eigenfunctions for the infinite domain problem and require that these new functions(multipoles) satisfy the free surface condition.

For m>0
<center><math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m</math></center>
and for m=0
<center><math>\phi_0 = \ln(r) + \psi_0 \,</math></center>

Where

<center><math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math></center>

We want the multipoles to satisfy the free surface condition

<center><math>\partial_z \phi |_{z=0} = K\phi|_{z=0} \quad \forall m = 0,1,2, \ldots</math></center>

For m=0 the free surface condition gives our boundary condition for <math>\psi_0\,</math>:

<center><math>f_0(x) = (\partial_z - K)\psi_0|_{z=0} = (K - \partial_z) \left(ln\left(\frac{r}{a}\right)\right)|_{z=0}
</math>

<math>\Rightarrow f_0(x) = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-\omega f}\cos(\omega x)d\omega
</math></center>

where we used the expansion

<center><math>\ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega}</math></center>

For m>0 the free surface condition yields the boundary condition for <math>\psi_m\,</math>:

<center><math>f_m(x) = (\partial_z - K)\psi_m|_{z=0} = \left(K - \partial_z \right) \left(\frac{e^{im\theta}}{r^m}\right)|_{z=0}
</math>

<math> \Rightarrow f_m(x) = \frac{(-1)^m}{(m-1)!} \int_0^\infty(K+\omega)\omega^{m-1}e^{-\omega f}e^{-i\omega x} d\omega \quad \quad (*)
</math></center>

where we used the expansion (valid for z>-f)

<center><math> \frac{e^{im\theta}}{r^m} = \frac{(-1)^m}{(m-1)!} \int_0^\infty\omega^{m-1}e^{-\omega(z+f) }e^{-i\omega x} d\omega
</math></center>

So our psi functions satisfy the following problem:

<center><math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots \quad \ldots (1)</math></center>

<center><math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center>

<center><math>\partial_z \psi_m |_{z=0} - K\psi_m|_{z=0} = f_m(x) \quad \ldots (3)</math></center>

We solve for <math>\hat{\psi}_m\,</math> by taking a Fourier transform in x to simplify Laplace's equation

<center><math>\hat{\psi}_m = \int_{-\infty}^\infty \psi_m e^{i\omega x}dx</math></center>

<center><math>(1) \Rightarrow \partial_z^2 \hat{\psi}_m = \omega^2\hat{\psi}_m \Rightarrow \hat{\psi}_m = A_m(\omega)e^{\omega z} + B_m(\omega)e^{-\omega z}
</math></center>

<center><math>but \quad (2) \Rightarrow B_m(\omega) = 0 \quad \forall m</math></center>

<center><math>(3) \Rightarrow \partial_z \hat{\psi}_m |_{z=0} - K\hat{\psi}_m|_{z=0} = (\omega - K)\hat{\psi}_m|_{z=0} = (\omega - K)A(\omega) = \hat{f}_m(\omega)</math></center>

Hence

<center><math>A(\omega) = \frac{\hat{f}_m(\omega)}{\omega - K} </math></center>

and we obtaion <math>\psi\,</math> by the inverse Fourier transform:

<center><math>\psi_m = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_m e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_m(\omega)}{\omega - K} e^{\omega z} e^{-i\omega x}d\omega</math></center>

For m>0 the form of <math>\hat{f}_m(\omega)</math> can be obtained from the expansion of <math>f_m(x)\,</math> labelled <math>(*)\,</math> above by rewriting it as a Fourier transform. Substituting the result into the expression above, <math>\psi_m\,</math> easily simplifies to:

<center><math>\psi_m = \frac{(-1)^m}{(m-1)!} \int_{-\infty}^\infty \frac{\omega + K}{\omega - K} \omega^{m-1} e^{\omega (z-f)} e^{-i\omega x}d\omega
</math></center>

for m>0.

<math>\psi_0 \,</math> remains to be derived.

In order to have a complete set to expand our flud potential we need to include <math>\bar{\phi}_m</math> for m>0.
This accounts for the second linearly independent solution for m>0.

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<center><math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-\,</math> </center>

where

<center><math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math></center>

and <math>\phi^+, \,\, \phi^-</math> are the symmetric and antisymmetric parts of <math>\phi \,</math> respectively.

For a wave incident from the left we have

<center><math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math></center>

<center><math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

<center><math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math></center>

and <center><math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math></center>

Converting <math>\phi^+\,</math> and <math>\phi^-\,</math> into the polar coordinate system:

<center><math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math></center>

<center><math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math></center>

The above power series converge for r < 2f and the coefficients are given by

<center><math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math></center>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<center><math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

<center><math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math></center>

From which we can see that <math>\beta_n = -i\alpha_n\,</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi\,</math> from:

<center><math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math></center>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<center><math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

<center><math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math></center>

Giving:

<center><math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math></center>

== Hydrodynamic Forces ==

[[Category:Linear Water-Wave Theory]]

Category:Multipole Methods for Linear Water Waves

2010-02-11T02:45:04Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-10T21:15:14Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-10T21:11:33Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-10T02:57:15Z

Sean Curry: /* Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth) */

Category:Multipole Methods for Linear Water Waves

2010-02-10T02:51:36Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-10T02:50:54Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-10T02:35:37Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-10T02:25:34Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-09T21:11:57Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-09T01:49:46Z

Sean Curry: /* 2D Exterior Neumann Problem */

Category:Multipole Methods for Linear Water Waves

2010-02-09T01:46:30Z

Sean Curry: /* 2D Exterior Neumann Problem */

Category:Multipole Methods for Linear Water Waves

2010-02-09T01:22:38Z

Sean Curry: /* 2D Exterior Neumann Problem */

Category:Multipole Methods for Linear Water Waves

2010-02-09T01:02:26Z

Sean Curry: /* With the Free Surface BC */

Category:Multipole Methods for Linear Water Waves

2010-02-09T00:58:44Z

Sean Curry: /* 2D Exterior Neumann Problem */

User talk:Sean Curry

2010-02-08T02:14:26Z

Sean Curry: /* With the Free Surface BC */

== Multipole Expansions ==

Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated seperately.

Multipoles satisfy:

- The feild equation

<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

- The free surface boundary condition

<math>\partial _z \phi = K\phi \quad on \quad z=0</math>

where the z-axis is oriented vertically upwards and the zero is on the free surface

- The bed boundary condition

<math>\partial_n \phi = 0 \quad on \quad z=-h(x,y)</math>

- A radiation condition ensuring the potentials represent outgoing waves in the far feild. In two dimensions:

<math> \lim _{kx \rightarrow \pm \infty} \left( \frac{\partial \phi}{\partial x} \mp ik\phi \right) = 0 </math>

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

Outside a disk of radius a centered at the origin in an infinite domain:

<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

<math>\partial_n \phi |_{r=a} = 0</math>

<math>|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>Let \quad \phi = R(r)\Theta(\theta)</math>

Seperating variables we get:

<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>

<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>

<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

we substitute R = r^n giving:

<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

<math>R = C\left(0\right) + D(0) \ln(r)</math>

and for <math>m\neq 0:</math>

<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>

<math>Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>

Hence the general solution can be expressed as:

<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

Which can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

We write

<math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math>

<math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math>

Where

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

We want

<math>\partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots</math>

For m=0

<math>\partial_z \psi_0 |_{z=f} - K\psi_0|_{z=f} = K\ln\left(\frac{r}{a}\right)|_{z=f} - \partial_z \left(ln\left(\frac{r}{a}\right)\right)|_{z=f} = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-2\mu f}\cos(\mu x)d\mu = f_0(x)</math>

Using

<math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x))\frac{d\mu}{\mu}</math>

For m>0

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = K \frac{e^{im\theta}}{r^m} |_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x) \quad \quad \quad \ldots</math> needs simplifying

So our psi functions satisfy

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots \quad \ldots (1)</math>

<math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math>

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = f_m(x) \quad \ldots (3)</math>

Taking a Fourier transform in x

<math>\hat{\psi}_m = \int_{-\infty}^\infty \psi_m e^{i\omega x}dx</math>

<math>(1) \Rightarrow \partial_z^2 \hat{\psi}_m = \omega^2\hat{\psi}_m \Rightarrow \hat{\psi}_m = A_m(\omega)e^{\omega z} + B_m(\omega)e^{-\omega z}</math>

<math>but \quad (2) \Rightarrow B_m(\omega) = 0 \quad \forall m</math>

<math>(3) \Rightarrow \partial_z \hat{\psi}_m |_{z=f} - K\hat{\psi}_m|_{z=f} = \hat{f}_m(\omega) = (\omega - K)\hat{\psi}_m|_{z=f} = (\omega - K)A(\omega)e^{\omega f}</math>

Hence

<math>A(\omega) = \frac{\hat{f}_m(\omega)}{\omega - K} e^{-\omega f} </math>

and

<math>\psi_m = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_m e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_m(\omega)}{\omega - K} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-</math>

where

<math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math>

and <math>\phi^+</math>, <math>\phi^-</math> are the symmetric and antisymmetric parts of <math>\phi</math> respectively.

For a wave incident from the left we have

<math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math>,

<math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>,

<math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math>,

and <math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>.

Converting <math>\phi^+</math> and <math>\phi^-</math> into the polar coordinate system:

<math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math>

<math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math>

The above power series converge for r < 2f and the coefficients are given by

<math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

<math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

From which we can see that <math>\beta_n = -i\alpha_n</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi</math> from:

<math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

<math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

Giving:

<math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math>

== Hydrodynamic Forces ==

User talk:Sean Curry

2010-02-07T22:44:59Z

Sean Curry: /* With the Free Surface BC */

== Multipole Expansions ==

Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated seperately.

Multipoles satisfy:

- The feild equation

<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

- The free surface boundary condition

<math>\partial _z \phi = K\phi \quad on \quad z=0</math>

where the z-axis is oriented vertically upwards and the zero is on the free surface

- The bed boundary condition

<math>\partial_n \phi = 0 \quad on \quad z=-h(x,y)</math>

- A radiation condition ensuring the potentials represent outgoing waves in the far feild. In two dimensions:

<math> \lim _{kx \rightarrow \pm \infty} \left( \frac{\partial \phi}{\partial x} \mp ik\phi \right) = 0 </math>

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

Outside a disk of radius a centered at the origin in an infinite domain:

<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

<math>\partial_n \phi |_{r=a} = 0</math>

<math>|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>Let \quad \phi = R(r)\Theta(\theta)</math>

Seperating variables we get:

<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>

<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>

<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

we substitute R = r^n giving:

<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

<math>R = C\left(0\right) + D(0) \ln(r)</math>

and for <math>m\neq 0:</math>

<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>

<math>Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>

Hence the general solution can be expressed as:

<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

Which can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

We write

<math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math>

<math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math>

Where

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

We want

<math>\partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots</math>

For m=0

<math>\partial_z \psi_0 |_{z=f} - K\psi_0|_{z=f} = K\ln\left(\frac{r}{a}\right)|_{z=f} - \partial_z \left(ln\left(\frac{r}{a}\right)\right)|_{z=f} = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-2\mu f}\cos(\mu x)d\mu = f_0(x)</math>

Using

<math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x)\frac{d\mu}{\mu})</math>

For m>0

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = K \frac{e^{im\theta}}{r^m} |_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x) \quad \quad \quad \ldots</math> needs simplifying

So our psi functions satisfy

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots \quad \ldots (1)</math>

<math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math>

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = f_m(x) \quad \ldots (3)</math>

Taking a Fourier transform in x

<math>\hat{\psi}_m = \int_{-\infty}^\infty \psi_m e^{i\omega x}dx</math>

<math>(1) \Rightarrow \partial_z^2 \hat{\psi}_m = \omega^2\hat{\psi}_m \Rightarrow \hat{\psi}_m = A_m(\omega)e^{\omega z} + B_m(\omega)e^{-\omega z}</math>

<math>but \quad (2) \Rightarrow B_m(\omega) = 0 \quad \forall m</math>

<math>(3) \Rightarrow \partial_z \hat{\psi}_m |_{z=f} - K\hat{\psi}_m|_{z=f} = \hat{f}_m(\omega) = (\omega - K)\hat{\psi}_m|_{z=f} = (\omega - K)A(\omega)e^{\omega f}</math>

Hence

<math>A(\omega) = \frac{\hat{f}_m(\omega)}{\omega - K} e^{-\omega f} </math>

and

<math>\psi_m = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\psi}_m e^{-i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\hat{f}_m(\omega)}{\omega - K} e^{\omega (z-f)} e^{-i\omega x}d\omega</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-</math>

where

<math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math>

and <math>\phi^+</math>, <math>\phi^-</math> are the symmetric and antisymmetric parts of <math>\phi</math> respectively.

For a wave incident from the left we have

<math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math>,

<math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>,

<math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math>,

and <math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>.

Converting <math>\phi^+</math> and <math>\phi^-</math> into the polar coordinate system:

<math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math>

<math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math>

The above power series converge for r < 2f and the coefficients are given by

<math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

<math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

From which we can see that <math>\beta_n = -i\alpha_n</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi</math> from:

<math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

<math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

Giving:

<math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math>

== Hydrodynamic Forces ==

User talk:Sean Curry

2010-02-07T21:50:04Z

Sean Curry: /* With the Free Surface BC */

== Multipole Expansions ==

Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated seperately.

Multipoles satisfy:

- The feild equation

<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

- The free surface boundary condition

<math>\partial _z \phi = K\phi \quad on \quad z=0</math>

where the z-axis is oriented vertically upwards and the zero is on the free surface

- The bed boundary condition

<math>\partial_n \phi = 0 \quad on \quad z=-h(x,y)</math>

- A radiation condition ensuring the potentials represent outgoing waves in the far feild. In two dimensions:

<math> \lim _{kx \rightarrow \pm \infty} \left( \frac{\partial \phi}{\partial x} \mp ik\phi \right) = 0 </math>

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

Outside a disk of radius a centered at the origin in an infinite domain:

<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

<math>\partial_n \phi |_{r=a} = 0</math>

<math>|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>Let \quad \phi = R(r)\Theta(\theta)</math>

Seperating variables we get:

<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>

<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>

<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

we substitute R = r^n giving:

<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

<math>R = C\left(0\right) + D(0) \ln(r)</math>

and for <math>m\neq 0:</math>

<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>

<math>Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>

Hence the general solution can be expressed as:

<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

Which can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

We write

<math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math>

<math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math>

Where

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

We want

<math>\partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots</math>

For m=0

<math>\partial_z \psi_0 |_{z=f} - K\psi_0|_{z=f} = K\ln\left(\frac{r}{a}\right)|_{z=f} - \partial_z \left(ln\left(\frac{r}{a}\right)\right)|_{z=f} = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-2\mu f}\cos(\mu x)d\mu = f_0(x)</math>

Using

<math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x)\frac{d\mu}{\mu})</math>

For m>0

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = K \frac{e^{im\theta}}{r^m} |_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x) \quad \quad \quad \ldots</math> needs simplifying

So our psi functions satisfy

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

<math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty</math>

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = f_m(x) </math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-</math>

where

<math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math>

and <math>\phi^+</math>, <math>\phi^-</math> are the symmetric and antisymmetric parts of <math>\phi</math> respectively.

For a wave incident from the left we have

<math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math>,

<math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>,

<math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math>,

and <math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>.

Converting <math>\phi^+</math> and <math>\phi^-</math> into the polar coordinate system:

<math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math>

<math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math>

The above power series converge for r < 2f and the coefficients are given by

<math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

<math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

From which we can see that <math>\beta_n = -i\alpha_n</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi</math> from:

<math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

<math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

Giving:

<math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math>

== Hydrodynamic Forces ==

User talk:Sean Curry

2010-02-07T21:41:02Z

Sean Curry: /* With the Free Surface BC */

== Multipole Expansions ==

Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated seperately.

Multipoles satisfy:

- The feild equation

<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

- The free surface boundary condition

<math>\partial _z \phi = K\phi \quad on \quad z=0</math>

where the z-axis is oriented vertically upwards and the zero is on the free surface

- The bed boundary condition

<math>\partial_n \phi = 0 \quad on \quad z=-h(x,y)</math>

- A radiation condition ensuring the potentials represent outgoing waves in the far feild. In two dimensions:

<math> \lim _{kx \rightarrow \pm \infty} \left( \frac{\partial \phi}{\partial x} \mp ik\phi \right) = 0 </math>

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

Outside a disk of radius a centered at the origin in an infinite domain:

<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

<math>\partial_n \phi |_{r=a} = 0</math>

<math>|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>Let \quad \phi = R(r)\Theta(\theta)</math>

Seperating variables we get:

<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>

<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>

<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

we substitute R = r^n giving:

<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

<math>R = C\left(0\right) + D(0) \ln(r)</math>

and for <math>m\neq 0:</math>

<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>

<math>Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>

Hence the general solution can be expressed as:

<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

Which can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

We write

<math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math>

<math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math>

Where

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

We want

<math>\partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots</math>

For m=0

<math>\partial_z \psi_0 |_{z=f} - K\psi_0|_{z=f} = K\ln\left(\frac{r}{a}\right)|_{z=f} - \partial_z \left(ln\left(\frac{r}{a}\right)\right)|_{z=f} = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-2\mu f}\cos(\mu x)d\mu = f_0(x)</math>

Using

<math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x)\frac{d\mu}{\mu})</math>

For m>0

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = K \frac{e^{im\theta}}{r^m} |_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x) \quad \quad \quad \ldots</math> needs simplifying

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-</math>

where

<math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math>

and <math>\phi^+</math>, <math>\phi^-</math> are the symmetric and antisymmetric parts of <math>\phi</math> respectively.

For a wave incident from the left we have

<math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math>,

<math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>,

<math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math>,

and <math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>.

Converting <math>\phi^+</math> and <math>\phi^-</math> into the polar coordinate system:

<math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math>

<math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math>

The above power series converge for r < 2f and the coefficients are given by

<math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

<math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

From which we can see that <math>\beta_n = -i\alpha_n</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi</math> from:

<math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

<math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

Giving:

<math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math>

== Hydrodynamic Forces ==

User talk:Sean Curry

2010-02-07T21:39:37Z

Sean Curry: /* With the Free Surface BC */

== Multipole Expansions ==

Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated seperately.

Multipoles satisfy:

- The feild equation

<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

- The free surface boundary condition

<math>\partial _z \phi = K\phi \quad on \quad z=0</math>

where the z-axis is oriented vertically upwards and the zero is on the free surface

- The bed boundary condition

<math>\partial_n \phi = 0 \quad on \quad z=-h(x,y)</math>

- A radiation condition ensuring the potentials represent outgoing waves in the far feild. In two dimensions:

<math> \lim _{kx \rightarrow \pm \infty} \left( \frac{\partial \phi}{\partial x} \mp ik\phi \right) = 0 </math>

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

Outside a disk of radius a centered at the origin in an infinite domain:

<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

<math>\partial_n \phi |_{r=a} = 0</math>

<math>|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>Let \quad \phi = R(r)\Theta(\theta)</math>

Seperating variables we get:

<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>

<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>

<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

we substitute R = r^n giving:

<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

<math>R = C\left(0\right) + D(0) \ln(r)</math>

and for <math>m\neq 0:</math>

<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>

<math>Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>

Hence the general solution can be expressed as:

<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

Which can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

We write

<math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math>

<math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math>

Where

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

We want

<math>\partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots</math>

For m=0

<math>\partial_z \psi_0 |_{z=f} - K\psi_0|_{z=f} = K\ln\left(\frac{r}{a}\right)|_{z=f} - \partial_z \left(ln\left(\frac{r}{a}\right)\right)|_{z=f} = K \ln\left(\frac{\sqrt{x^2 + f^2}}{a}\right) - \int_0^\infty e^{-2\mu f}\cos(\mu x)d\mu = f_0(x)</math>

Using

<math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x)\frac{d\mu}{\mu})</math>

For m>0

<math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = K \frac{e^{im\theta}}{r^m} |_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x)</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-</math>

where

<math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math>

and <math>\phi^+</math>, <math>\phi^-</math> are the symmetric and antisymmetric parts of <math>\phi</math> respectively.

For a wave incident from the left we have

<math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math>,

<math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>,

<math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math>,

and <math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>.

Converting <math>\phi^+</math> and <math>\phi^-</math> into the polar coordinate system:

<math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math>

<math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math>

The above power series converge for r < 2f and the coefficients are given by

<math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

<math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

From which we can see that <math>\beta_n = -i\alpha_n</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi</math> from:

<math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

<math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

Giving:

<math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math>

== Hydrodynamic Forces ==

User talk:Sean Curry

2010-02-07T21:14:35Z

Sean Curry: /* 2D Exterior Neumann Problem */

== Multipole Expansions ==

Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated seperately.

Multipoles satisfy:

- The feild equation

<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

- The free surface boundary condition

<math>\partial _z \phi = K\phi \quad on \quad z=0</math>

where the z-axis is oriented vertically upwards and the zero is on the free surface

- The bed boundary condition

<math>\partial_n \phi = 0 \quad on \quad z=-h(x,y)</math>

- A radiation condition ensuring the potentials represent outgoing waves in the far feild. In two dimensions:

<math> \lim _{kx \rightarrow \pm \infty} \left( \frac{\partial \phi}{\partial x} \mp ik\phi \right) = 0 </math>

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

Outside a disk of radius a centered at the origin in an infinite domain:

<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

<math>\partial_n \phi |_{r=a} = 0</math>

<math>|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>Let \quad \phi = R(r)\Theta(\theta)</math>

Seperating variables we get:

<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>

<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>

<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

we substitute R = r^n giving:

<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

<math>R = C\left(0\right) + D(0) \ln(r)</math>

and for <math>m\neq 0:</math>

<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>

<math>Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>

Hence the general solution can be expressed as:

<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

Which can also be expressed simply in terms of complex exponentials.

== With the Free Surface BC ==

When we introduce the free surface BC we need to find a new set of eigenfunctions. We can use eigenfunctions from the infinite domain problem to construct them.

We write

<math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math>

<math>\phi_0 = \ln\frac{r}{a} + \psi_0 </math>

Where

<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-</math>

where

<math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math>

and <math>\phi^+</math>, <math>\phi^-</math> are the symmetric and antisymmetric parts of <math>\phi</math> respectively.

For a wave incident from the left we have

<math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math>,

<math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>,

<math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math>,

and <math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>.

Converting <math>\phi^+</math> and <math>\phi^-</math> into the polar coordinate system:

<math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math>

<math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math>

The above power series converge for r < 2f and the coefficients are given by

<math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

<math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

From which we can see that <math>\beta_n = -i\alpha_n</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi</math> from:

<math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

<math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

Giving:

<math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math>

== Hydrodynamic Forces ==

User talk:Sean Curry

2010-02-04T03:50:15Z

Sean Curry: /* 2D Exterior Neumann Problem */

== Multipole Expansions ==

Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed within any structures that are present. Multipoles can be constructed with their singularity submerged or on the free surface and these are treated seperately.

Multipoles satisfy:

- The feild equation

<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

- The free surface boundary condition

<math>\partial _z \phi = K\phi \quad on \quad z=0</math>

where the z-axis is oriented vertically upwards and the zero is on the free surface

- The bed boundary condition

<math>\partial_n \phi = 0 \quad on \quad z=-h(x,y)</math>

- A radiation condition ensuring the potentials represent outgoing waves in the far feild. In two dimensions:

<math> \lim _{kx \rightarrow \pm \infty} \left( \frac{\partial \phi}{\partial x} \mp ik\phi \right) = 0 </math>

A linear combination of these multipoles can then be made to satisfy the body boundary conditions.

== 2D Exterior Neumann Problem ==

Outside a disk of radius a centered at the origin in an infinite domain:

<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

<math>\partial_n \phi |_{r=a} = 0</math>

<math>|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>Let \quad \phi = R(r)\Theta(\theta)</math>

Seperating variables we get:

<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>

<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>

<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

we substitute R = r^n giving:

<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

<math>R = C\left(0\right) + D(0) \ln(r)</math>

and for <math>m\neq 0:</math>

<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>

<math>Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>

Hence the general solution can be expressed as:

<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

Which can also be expressed simply in terms of complex exponentials

== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

***Picture to show problem set up***

The potential for the scattering problem splits as incident and diffracted potentials and also into symmetric and antisymmetric parts.

<math>\phi = \phi_I + \phi_D = \phi^+ + \phi^-</math>

where

<math> \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} </math>

and <math>\phi^+</math>, <math>\phi^-</math> are the symmetric and antisymmetric parts of <math>\phi</math> respectively.

For a wave incident from the left we have

<math>\phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta}</math>,

<math>\phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>,

<math>\phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+</math>,

and <math>\phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- </math>.

Converting <math>\phi^+</math> and <math>\phi^-</math> into the polar coordinate system:

<math>\phi^+ = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \cos(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\cos(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \cos(m\theta)\right)</math>

<math>\phi^- = ie^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} \sin(n\theta) + \sum_{n=1}^{\infty}a^n \alpha_n \left(\frac{\sin(n\theta)}{r^n} + \sum_{m=0}^\infty A_{mn}^+ r^m \sin(m\theta)\right)</math>

The above power series converge for r < 2f and the coefficients are given by

<math> A_{mn} = \frac{(-1)^{m+n}}{m!(n-1)!} \quad SingularIntegral_0^\infty \quad \frac{\mu + K}{\mu - K} \mu ^{m+n-1} e^{-2\mu f}d\mu </math>

Applying the body boundary condition <math>\partial_r \phi^\pm |_{r=a} = 0</math> and noting that the cosines are orthogonal over <math>\theta \in (-\pi,\pi]</math> gives the results

<math> \alpha_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \alpha_n = e^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

<math> \beta_m - \sum_{n=1}^{\infty}a^{n+m} A_{mn} \beta_n = -ie^{-Kf}\frac {(-Ka)^m}{m!}; m=1,2,3,\ldots</math>

From which we can see that <math>\beta_n = -i\alpha_n</math>.

The above sums can be truncated at the Nth term yielding NxN system of linear equations that can be solved for the coefficients <math>\alpha_n</math>

Hence we can calculate <math>\phi</math> from:

<math>\phi = \phi^+ + \phi^- = \phi_I + \sum_{n=1}^{\infty}a^n \alpha_n ( \phi^+_n + \phi^-_n )</math>

This expression also gives the values of the reflection and transmission coefficients when we substitute the far field expressions for the symmetric and antisymmetric multipoles. In the far field:

<math> \phi^+_n \sim \frac{2\pi i (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

<math > \phi^-_n \sim \mp \frac{2\pi (-K)^n}{(n-1)!} e^{K(z-f)} e^{\pm Kx} </math>

Giving:

<math>R=0, \quad T = 1 + 4 \pi i e^{-Kf} \sum_{n=1}^{\infty} \alpha_n \frac{(-Ka)^n}{(n-1)!}
</math>

== Hydrodynamic Forces ==

User talk:Sean Curry

2010-02-04T02:49:43Z

Sean Curry: /* 2D Exterior Neumann Problem */

User talk:Sean Curry

2010-02-04T02:10:23Z

Sean Curry: /* 2D Exterior Neumann Problem */

User talk:Sean Curry

2010-02-03T23:49:54Z

Sean Curry: /* Multipole Expansions */

User talk:Sean Curry

2010-01-20T09:13:25Z

Sean Curry: /* Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth) */

User talk:Sean Curry

2010-01-20T09:11:52Z

Sean Curry: /* Multipole Expansions */

User talk:Sean Curry

2010-01-18T03:06:06Z

Sean Curry: /* Multipole Expansions */

User talk:Sean Curry

2010-01-18T03:03:43Z

Sean Curry: /* Multipole Expansions */

User talk:Sean Curry

2010-01-18T03:00:51Z

Sean Curry: Created page with '== Multipole Expansions == Moltipole expansions are a technique used in linear water wave theory where the potential is represented as a sum of singularities (multipoles) placed…'

Variable Depth Shallow Water Wave Equation

2009-03-01T01:32:10Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,:) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\omega^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\omega^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\omega^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = -\frac{(b-a)}{2L}\partial_xh(x) - \omega^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u(x,\omega) = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\omega^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{-L}^{L}\,f u_k\,dx}{(\omega^{2}-\lambda_k) \int_{-L}^{L}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta(x,\omega)=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) \quad (6)</math>
,with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x \zeta |_L </math>.

We choose a basis of the solution space for any particular <math>\omega</math> to be <math>\{\zeta_1,\,\zeta_2\}</math>, where <math> \zeta_1 </math> is the solution to the BVP(a=1,b=0) and <math> \zeta_1 </math> is the solution to the BVP(a=0,b=1). The functions <math> \zeta_1 </math> and <math> \zeta_2 </math> can be calculated for any <math> \omega </math> from (6).

The aim here is to construct a matrix '''S''' such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of '''S''' must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So '''S''' is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the area of constant depth on the left hand side there is a potential of the form <math> e^{i\omega x} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{-i\omega x} </math> and <math> Te^{i\omega x} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-i\omega L}+Re^{i\omega L}, \quad \zeta |_L = Te^{i\omega L}, \quad \partial_x \zeta |_{-L} = i\omega e^{-i\omega L}-i\omega Re^{i\omega L}, \quad \partial_x \zeta |_L = i\omega Te^{i\omega L} </math>
</center>

<center>
<math> \begin{pmatrix} i\omega e^{-i\omega L}-i\omega Re^{i\omega L} \\ i\omega Te^{i\omega L} \end{pmatrix} = S\begin{pmatrix} e^{-i\omega L}+Re^{i\omega L} \\ Te^{i\omega L} \end{pmatrix} </math>
</center>

Knowing '''S''' these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem(<math>a_+=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b_+=Te^{i\omega L}</math>). Taking a linear combination of the solutions already calculated (<math> a_+\zeta_1(x,\omega) + b_+\zeta_2(x,\omega) </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a generalised eigenfunction potential for the whole real axis which we denote as <math> \zeta_+(x,\omega)</math>.

Independent generalised eigenfunctons (which we denote as <math> \zeta_-(x,\omega)</math>)can be found by considering the potential <math> e^{-i\omega x} </math> on the right hand region of constant depth. The corresponding reflected and transmitted potentials from the variable depth area are of the form <math> Re^{i\omega x} </math> and <math> Te^{-i\omega x} </math> respectively. Again knowing '''S''' we can solve for R and T and hence find the numerical boundary conditions (<math>a_-=Te^{i\omega L}</math> and <math>b_-=e^{-i\omega L}+Re^{i\omega L}</math>).

We come out with:

<center><math>
\zeta_+(x,\omega) = \left\{ \begin{matrix}
{e^{i\omega x}+Re^{-i\omega x}, \quad \mbox{ for } x<-L}
\\ {a_+\zeta_1(x,\omega) + b_+\zeta_2(x,\omega),\quad \mbox{ for } -L\leq x \leq L}
\\ {Te^{i\omega x}, \quad \mbox{ for } x>L}
\end{matrix} \right.
</math></center>

where R and T are found by solving:

<center><math>
\begin{pmatrix} (S_{11} + i\omega)e^{i\omega L} & S_{12}e^{i\omega L} \\
S_{21}e^{i\omega L} & (S_{22}-i\omega)e^{i\omega L} \end{pmatrix} \begin{pmatrix}R\\T\end{pmatrix}
= \begin{pmatrix} (i\omega - S_{11}) e^{-i\omega L} \\ -S_{21}e^{-i\omega L} \end{pmatrix}
</math></center>

and <math>a_+=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b_+=Te^{i\omega L}</math>

Note that the values of R and T for \zeta_- are different from those for \zeta_+ (Although they are related through some identities). For \zeta_- we have:

<center><math>
\zeta_-(x,\omega) = \left\{ \begin{matrix}
{Te^{-i\omega x}, \quad \mbox{ for } x<-L}
\\ {a_-\zeta_1(x,\omega) + b_-\zeta_2(x,\omega),\quad \mbox{ for } -L\leq x \leq L}
\\ {e^{-i\omega x}+Re^{i\omega x}, \quad \mbox{ for } x>L}
\end{matrix} \right.
</math></center>

where R and T are found by solving:

<center><math>
\begin{pmatrix} S_{12}e^{i\omega L} & (S_{11}+i\omega)e^{i\omega L} \\
(S_{22}-i\omega)e^{i\omega L} & S_{21}e^{i\omega L}\end{pmatrix} \begin{pmatrix}R\\T\end{pmatrix}
= \begin{pmatrix} S_{12}e^{-i\omega L} \\ -(i\omega + S_{21})e^{-i\omega L} \end{pmatrix}
</math></center>

Note a_+, b_+, a_- and b_- are found frm their corresponding R and T values.

Now we have the generalised eigenfunctions <math>\zeta_+(x,\omega)</math> and <math>\zeta_-(x,\omega)</math>, which have the orthogonality relationsips:

<center><math>
\int_{-\infty}^{\infty}\zeta_+(x,\omega)\zeta_-(x,\omega')dx= 0 \qquad (7)
</math></center>
<center><math>
\int_{-\infty}^{\infty}\zeta_+(x,\omega)\zeta_+(x,\omega')dx= 2 \pi \delta(\omega-\omega') \qquad (8)
</math></center>
<center><math>
\int_{-\infty}^{\infty}\zeta_-(x,\omega)\zeta_-(x,\omega')dx= 2 \pi \delta(\omega-\omega') \qquad (9)
</math></center>

For any particular <math>\omega </math> the general solution to the differential equation can be written as:

<center><math>
\zeta(x,t,\omega) = cos(\omega t)\left(c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega)\right) + \frac{sin(\omega t)}{\omega}\left(c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega)\right)
</math></center>

The general solution to the PDE is therefore:

<center><math>
\zeta(x,t) = \int_{0}^{\infty} \zeta(x,t,\omega) d \omega
</math></center>
<center><math>
\implies \zeta(x,t) = \int_{0}^{\infty} \left\{ cos(\omega t)\left(c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega)\right) + \frac{sin(\omega t)}{\omega}\left(c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega)\right)
\right\} d \omega \qquad (10)
</math></center>

Giving the initial conditions:

<center><math>
F(x) = \zeta(x,0) = \int_{0}^{\infty} \left\{ c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega) \right\} d \omega
</math></center>
<center><math>
G(x) = \partial_t \zeta(x,0) = \int_{0}^{\infty} \left\{ c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega) \right\} d \omega
</math></center>

Using identities (7),(8) and (9) we can show:
<center><math>
c_1(\omega) = \frac{1}{2\pi}\langle F(x), \zeta_+(x,\omega) \rangle
</math></center>
<center><math>
d_1(\omega) = \frac{1}{2\pi}\langle F(x), \zeta_-(x,\omega) \rangle
</math></center>
<center><math>
c_2(\omega) = \frac{1}{2\pi}\langle G(x), \zeta_+(x,\omega) \rangle
</math></center>
<center><math>
d_2(\omega) = \frac{1}{2\pi}\langle G(x), \zeta_-(x,\omega) \rangle
</math></center>

Which we can use in combination with (10) to solve the IVP.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-25T03:54:17Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,:) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\omega^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\omega^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\omega^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = -\frac{(b-a)}{2L}\partial_xh(x) - \omega^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u(x,\omega) = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\omega^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{-L}^{L}\,f u_k\,dx}{(\omega^{2}-\lambda_k) \int_{-L}^{L}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta(x,\omega)=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) \quad (6)</math>
,with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x \zeta |_L </math>.

We choose a basis of the solution space for any particular <math>\omega</math> to be <math>\{\zeta_1,\,\zeta_2\}</math>, where <math> \zeta_1 </math> is the solution to the BVP(a=1,b=0) and <math> \zeta_1 </math> is the solution to the BVP(a=0,b=1). The functions <math> \zeta_1 </math> and <math> \zeta_2 </math> can be calculated for any <math> \omega </math> from (6).

The aim here is to construct a matrix '''S''' such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of '''S''' must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So '''S''' is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the area of constant depth on the left hand side there is a potential of the form <math> e^{i\omega x} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{-i\omega x} </math> and <math> Te^{i\omega x} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-i\omega L}+Re^{i\omega L}, \quad \zeta |_L = Te^{i\omega L}, \quad \partial_x \zeta |_{-L} = i\omega e^{-i\omega L}-i\omega Re^{i\omega L}, \quad \partial_x \zeta |_L = i\omega Te^{i\omega L} </math>
</center>

<center>
<math> \begin{pmatrix} i\omega e^{-i\omega L}-i\omega Re^{i\omega L} \\ i\omega Te^{i\omega L} \end{pmatrix} = S\begin{pmatrix} e^{-i\omega L}+Re^{i\omega L} \\ Te^{i\omega L} \end{pmatrix} </math>
</center>

Knowing '''S''' these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem(<math>a_+=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b_+=Te^{i\omega L}</math>). Taking a linear combination of the solutions already calculated (<math> a_+\zeta_1(x,\omega) + b_+\zeta_2(x,\omega) </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a generalised eigenfunction potential for the whole real axis which we denote as <math> \zeta_+(x,\omega)</math>.

Independent generalised eigenfunctons (which we denote as <math> \zeta_-(x,\omega)</math>)can be found by considering the potential <math> e^{-i\omega x} </math> on the right hand region of constant depth. The corresponding reflected and transmitted potentials from the variable depth area are of the form <math> Re^{i\omega x} </math> and <math> Te^{-i\omega x} </math> respectively. Again knowing '''S''' we can solve for R and T and hence find the numerical boundary conditions (<math>a_-=Te^{i\omega L}</math> and <math>b_-=e^{-i\omega L}+Re^{i\omega L}</math>).

We come out with:

<center><math>
\zeta_+(x,\omega) = \left\{ \begin{matrix}
{e^{i\omega x}+Re^{-i\omega x}, \quad \mbox{ for } x<-L}
\\ {a_+\zeta_1(x,\omega) + b_+\zeta_2(x,\omega),\quad \mbox{ for } -L\leq x \leq L}
\\ {Te^{i\omega x}, \quad \mbox{ for } x>L}
\end{matrix} \right.
</math></center>

where R and T are found by solving:

<center><math>
\begin{pmatrix} (S_{11} + i\omega)e^{i\omega L} & S_{12}e^{i\omega L} \\
S_{21}e^{i\omega L} & (S_{22}-i\omega)e^{i\omega L} \end{pmatrix} \begin{pmatrix}R\\T\end{pmatrix}
= \begin{pmatrix} (i\omega - S_{11}) e^{-i\omega L} \\ -S_{21}e^{-i\omega L} \end{pmatrix}
</math></center>

and <math>a_+=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b_+=Te^{i\omega L}</math>

Note that the values of R and T for \zeta_- are different from those for \zeta_+ (Although they are related through some identities). For \zeta_- we have:

<center><math>
\zeta_-(x,\omega) = \left\{ \begin{matrix}
{Te^{-i\omega x}, \quad \mbox{ for } x<-L}
\\ {a_-\zeta_1(x,\omega) + b_-\zeta_2(x,\omega),\quad \mbox{ for } -L\leq x \leq L}
\\ {e^{-i\omega x}+Re^{i\omega x}, \quad \mbox{ for } x>L}
\end{matrix} \right.
</math></center>

where R and T are found by solving:

<center><math>
\begin{pmatrix} S_{12}e^{i\omega L} & (S_{11}+i\omega)e^{i\omega L} \\
(S_{22}-i\omega)e^{i\omega L} & S_{21}e^{i\omega L}\end{pmatrix} \begin{pmatrix}R\\T\end{pmatrix}
= \begin{pmatrix} S_{12}e^{-i\omega L} \\ -(i\omega + S_{21})e^{-i\omega L} \end{pmatrix}
</math></center>

Note a_+, b_+, a_- and b_- are found frm their corresponding R and T values.

Now we have the generalised eigenfunctions <math>\zeta_+(x,\omega)</math> and <math>\zeta_-(x,\omega)</math>, which have the orthogonality relationsips:

<center><math>
\int_{-\infty}^{\infty}\zeta_+(x,\omega)\zeta_-(x,\omega')dx= 0 \qquad (7)
</math></center>
<center><math>
\int_{-\infty}^{\infty}\zeta_+(x,\omega)\zeta_+(x,\omega')dx= 2 \pi \delta(\omega-\omega') \qquad (8)
</math></center>
<center><math>
\int_{-\infty}^{\infty}\zeta_-(x,\omega)\zeta_-(x,\omega')dx= 2 \pi \delta(\omega-\omega') \qquad (9)
</math></center>

For any particular \omega the general solution to the differential equation can be written as:

<center><math>
\zeta(x,t,\omega) = cos(\omega t)\left(c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega)\right) + \frac{sin(\omega t)}{\omega}\left(c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega)\right)
</math></center>

The general solution to the PDE is therefore:

<center><math>
\zeta(x,t) = \int_{0}^{\infty} \zeta(x,t,\omega) d \omega
</math></center>
<center><math>
\implies \zeta(x,t) = \int_{0}^{\infty} \left\{ cos(\omega t)\left(c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega)\right) + \frac{sin(\omega t)}{\omega}\left(c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega)\right)
\right\} d \omega \qquad (10)
</math></center>

Giving the initial conditions:

<center><math>
F(x) = \zeta(x,0) = \int_{0}^{\infty} \left\{ c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega) \right\} d \omega
</math></center>
<center><math>
G(x) = \partial_t \zeta(x,0) = \int_{0}^{\infty} \left\{ c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega) \right\} d \omega
</math></center>

Using identities (7),(8) and (9) we can show:
<center><math>
c_1(\omega) = \frac{1}{2\pi}\langle F(x), \zeta_+(x,\omega) \rangle
</math></center>
<center><math>
d_1(\omega) = \frac{1}{2\pi}\langle F(x), \zeta_-(x,\omega) \rangle
</math></center>
<center><math>
c_2(\omega) = \frac{1}{2\pi}\langle G(x), \zeta_+(x,\omega) \rangle
</math></center>
<center><math>
d_2(\omega) = \frac{1}{2\pi}\langle G(x), \zeta_-(x,\omega) \rangle
</math></center>

Which we can use in combination with (10) to solve the IVP.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-25T03:49:25Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,:) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\omega^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\omega^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\omega^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \omega^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u(x,\omega) = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\omega^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{-L}^{L}\,f u_k\,dx}{(\omega^{2}-\lambda_k) \int_{-L}^{L}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta(x,\omega)=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) \quad (6)</math>
,with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x \zeta |_L </math>.

We choose a basis of the solution space for any particular <math>\omega</math> to be <math>\{\zeta_1,\,\zeta_2\}</math>, where <math> \zeta_1 </math> is the solution to the BVP(a=1,b=0) and <math> \zeta_1 </math> is the solution to the BVP(a=0,b=1). The functions <math> \zeta_1 </math> and <math> \zeta_2 </math> can be calculated for any <math> \omega </math> from (6).

The aim here is to construct a matrix '''S''' such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of '''S''' must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So '''S''' is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the area of constant depth on the left hand side there is a potential of the form <math> e^{i\omega x} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{-i\omega x} </math> and <math> Te^{i\omega x} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-i\omega L}+Re^{i\omega L}, \quad \zeta |_L = Te^{i\omega L}, \quad \partial_x \zeta |_{-L} = i\omega e^{-i\omega L}-i\omega Re^{i\omega L}, \quad \partial_x \zeta |_L = i\omega Te^{i\omega L} </math>
</center>

<center>
<math> \begin{pmatrix} i\omega e^{-i\omega L}-i\omega Re^{i\omega L} \\ i\omega Te^{i\omega L} \end{pmatrix} = S\begin{pmatrix} e^{-i\omega L}+Re^{i\omega L} \\ Te^{i\omega L} \end{pmatrix} </math>
</center>

Knowing '''S''' these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem(<math>a_+=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b_+=Te^{i\omega L}</math>). Taking a linear combination of the solutions already calculated (<math> a_+\zeta_1(x,\omega) + b_+\zeta_2(x,\omega) </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a generalised eigenfunction potential for the whole real axis which we denote as <math> \zeta_+(x,\omega)</math>.

Independent generalised eigenfunctons (which we denote as <math> \zeta_-(x,\omega)</math>)can be found by considering the potential <math> e^{-i\omega x} </math> on the right hand region of constant depth. The corresponding reflected and transmitted potentials from the variable depth area are of the form <math> Re^{i\omega x} </math> and <math> Te^{-i\omega x} </math> respectively. Again knowing '''S''' we can solve for R and T and hence find the numerical boundary conditions (<math>a_-=Te^{i\omega L}</math> and <math>b_-=e^{-i\omega L}+Re^{i\omega L}</math>).

We come out with:

<center><math>
\zeta_+(x,\omega) = \left\{ \begin{matrix}
{e^{i\omega x}+Re^{-i\omega x}, \quad \mbox{ for } x<-L}
\\ {a_+\zeta_1(x,\omega) + b_+\zeta_2(x,\omega),\quad \mbox{ for } -L\leq x \leq L}
\\ {Te^{i\omega x}, \quad \mbox{ for } x>L}
\end{matrix} \right.
</math></center>

where R and T are found by solving:

<center><math>
\begin{pmatrix} (S_{11} + i\omega)e^{i\omega L} & S_{12}e^{i\omega L} \\
S_{21}e^{i\omega L} & (S_{22}-i\omega)e^{i\omega L} \end{pmatrix} \begin{pmatrix}R\\T\end{pmatrix}
= \begin{pmatrix} (i\omega - S_{11}) e^{-i\omega L} \\ -S_{21}e^{-i\omega L} \end{pmatrix}
</math></center>

and <math>a_+=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b_+=Te^{i\omega L}</math>

Note that the values of R and T for \zeta_- are different from those for \zeta_+ (Although they are related through some identities). For \zeta_- we have:

<center><math>
\zeta_-(x,\omega) = \left\{ \begin{matrix}
{Te^{-i\omega x}, \quad \mbox{ for } x<-L}
\\ {a_-\zeta_1(x,\omega) + b_-\zeta_2(x,\omega),\quad \mbox{ for } -L\leq x \leq L}
\\ {e^{-i\omega x}+Re^{i\omega x}, \quad \mbox{ for } x>L}
\end{matrix} \right.
</math></center>

where R and T are found by solving:

<center><math>
\begin{pmatrix} S_{12}e^{i\omega L} & (S_{11}+i\omega)e^{i\omega L} \\
(S_{22}-i\omega)e^{i\omega L} & S_{21}e^{i\omega L}\end{pmatrix} \begin{pmatrix}R\\T\end{pmatrix}
= \begin{pmatrix} S_{12}e^{-i\omega L} \\ -(i\omega + S_{21})e^{-i\omega L} \end{pmatrix}
</math></center>

Note a_+, b_+, a_- and b_- are found frm their corresponding R and T values.

Now we have the generalised eigenfunctions <math>\zeta_+(x,\omega)</math> and <math>\zeta_-(x,\omega)</math>, which have the orthogonality relationsips:

<center><math>
\int_{-\infty}^{\infty}\zeta_+(x,\omega)\zeta_-(x,\omega')dx= 0 \qquad (7)
</math></center>
<center><math>
\int_{-\infty}^{\infty}\zeta_+(x,\omega)\zeta_+(x,\omega')dx= 2 \pi \delta(\omega-\omega') \qquad (8)
</math></center>
<center><math>
\int_{-\infty}^{\infty}\zeta_-(x,\omega)\zeta_-(x,\omega')dx= 2 \pi \delta(\omega-\omega') \qquad (9)
</math></center>

For any particular \omega the general solution to the differential equation can be written as:

<center><math>
\zeta(x,t,\omega) = cos(\omega t)\left(c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega)\right) + \frac{sin(\omega t)}{\omega}\left(c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega)\right)
</math></center>

The general solution to the PDE is therefore:

<center><math>
\zeta(x,t) = \int_{0}^{\infty} \zeta(x,t,\omega) d \omega
</math></center>
<center><math>
\implies \zeta(x,t) = \int_{0}^{\infty} \left\{ cos(\omega t)\left(c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega)\right) + \frac{sin(\omega t)}{\omega}\left(c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega)\right)
\right\} d \omega \qquad (10)
</math></center>

Giving the initial conditions:

<center><math>
F(x) = \zeta(x,0) = \int_{0}^{\infty} \left\{ c_1(\omega)\zeta_+(x,\omega)+d_1(\omega)\zeta_-(x,\omega) \right\} d \omega
</math></center>
<center><math>
G(x) = \partial_t \zeta(x,0) = \int_{0}^{\infty} \left\{ c_2(\omega)\zeta_+(x,\omega)+d_2(\omega)\zeta_-(x,\omega) \right\} d \omega
</math></center>

Using identities (7),(8) and (9) we can show:
<center><math>
c_1(\omega) = \frac{1}{2\pi}\langle F(x), \zeta_+(x,\omega) \rangle
</math></center>
<center><math>
d_1(\omega) = \frac{1}{2\pi}\langle F(x), \zeta_-(x,\omega) \rangle
</math></center>
<center><math>
c_2(\omega) = \frac{1}{2\pi}\langle G(x), \zeta_+(x,\omega) \rangle
</math></center>
<center><math>
d_2(\omega) = \frac{1}{2\pi}\langle G(x), \zeta_-(x,\omega) \rangle
</math></center>

Which we can use in combination with (10) to solve the IVP.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-24T21:11:28Z

Sean Curry: /* Matlab code */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,:) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\omega^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\omega^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\omega^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \omega^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u(x,\omega) = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\omega^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{-L}^{L}\,f u_k\,dx}{(\omega^{2}-\lambda_k) \int_{-L}^{L}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta(x,\omega)=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) \quad (6)</math>
,with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x \zeta |_L </math>.

We choose a basis of the solution space for any particular <math>\omega</math> to be <math>\{\zeta_1,\,\zeta_2\}</math>, where <math> \zeta_1 </math> is the solution to the BVP(a=1,b=0) and <math> \zeta_1 </math> is the solution to the BVP(a=0,b=1). The functions <math> \zeta_1 </math> and <math> \zeta_2 </math> can be calculated for any <math> \omega </math> from (6).

The aim here is to construct a matrix '''S''' such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of '''S''' must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So '''S''' is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the area of constant depth on the left hand side there is a potential of the form <math> e^{i\omega x} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{-i\omega x} </math> and <math> Te^{i\omega x} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-i\omega L}+Re^{i\omega L}, \quad \zeta |_L = Te^{i\omega L}, \quad \partial_x \zeta |_{-L} = i\omega e^{-i\omega L}-i\omega Re^{i\omega L}, \quad \partial_x \zeta |_L = i\omega Te^{i\omega L} </math>
</center>

<center>
<math> \begin{pmatrix} i\omega e^{-i\omega L}-i\omega Re^{i\omega L} \\ i\omega Te^{i\omega L} \end{pmatrix} = S\begin{pmatrix} e^{-i\omega L}+Re^{i\omega L} \\ Te^{i\omega L} \end{pmatrix} </math>
</center>

Knowing '''S''' these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem(<math>a=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b=Te^{i\omega L}</math>). Taking a linear combination of the solutions already calculated (<math> a\zeta_1(x,\omega) + b\zeta_2(x,\omega) </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a generalised eigenfunction potential for the whole real axis which we denote as <math> \zeta_+(x,\omega)</math>.

Independent generalised eigenfunctons can be found by considering the potential <math> e^{-i\omega x} </math> on the right hand region of constant depth. The corresponding reflected and transmitted potentials from the variable depth area are of the form <math> Re^{i\omega x} </math> and <math> Te^{-i\omega x} </math> respectively. Again knowing '''S''' we can solve for R and T and hence find the numerical boundary conditions (<math>a=Te^{i\omega L}</math> and <math>b=e^{-i\omega L}+Re^{i\omega L}</math>).

If a waveform <math> f(x) </math> is travelling in from <math> -\infty </math> Taking the fourier transform gives
<center>
<math> \hat{f}(\omega)=\int_{-\infty}^{\infty}\,f(x)e^{2\pi i \omega} \,dx </math>
</center>

and then inverting <math> \hat{f}(\omega) \zeta(x) </math> with respect to <math> t </math> gives
<center>
<math> f(x,t) = \int_{-\infty}^{\infty}\,\hat{f}(\omega) \zeta(x) e^{-2\pi i \omega t} \,d\omega </math>
</center>

which is the time dependent solution for the wave form.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-24T02:52:27Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,1:length(x)) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\omega^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\omega^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\omega^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \omega^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u(x,\omega) = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\omega^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{-L}^{L}\,f u_k\,dx}{(\omega^{2}-\lambda_k) \int_{-L}^{L}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta(x,\omega)=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) \quad (6)</math>
,with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x \zeta |_L </math>.

We choose a basis of the solution space for any particular <math>\omega</math> to be <math>\{\zeta_1,\,\zeta_2\}</math>, where <math> \zeta_1 </math> is the solution to the BVP(a=1,b=0) and <math> \zeta_1 </math> is the solution to the BVP(a=0,b=1). The functions <math> \zeta_1 </math> and <math> \zeta_2 </math> can be calculated for any <math> \omega </math> from (6).

The aim here is to construct a matrix '''S''' such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of '''S''' must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So '''S''' is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the area of constant depth on the left hand side there is a potential of the form <math> e^{i\omega x} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{-i\omega x} </math> and <math> Te^{i\omega x} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-i\omega L}+Re^{i\omega L}, \quad \zeta |_L = Te^{i\omega L}, \quad \partial_x \zeta |_{-L} = i\omega e^{-i\omega L}-i\omega Re^{i\omega L}, \quad \partial_x \zeta |_L = i\omega Te^{i\omega L} </math>
</center>

<center>
<math> \begin{pmatrix} i\omega e^{-i\omega L}-i\omega Re^{i\omega L} \\ i\omega Te^{i\omega L} \end{pmatrix} = S\begin{pmatrix} e^{-i\omega L}+Re^{i\omega L} \\ Te^{i\omega L} \end{pmatrix} </math>
</center>

Knowing '''S''' these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem(<math>a=e^{-i\omega L}+Re^{i\omega L}</math> and <math> b=Te^{i\omega L}</math>). Taking a linear combination of the solutions already calculated (<math> a\zeta_1(x,\omega) + b\zeta_2(x,\omega) </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a generalised eigenfunction potential for the whole real axis which we denote as <math> \zeta_+(x,\omega)</math>.

Independent generalised eigenfunctons can be found by considering the potential <math> e^{-i\omega x} </math> on the right hand region of constant depth. The corresponding reflected and transmitted potentials from the variable depth area are of the form <math> Re^{i\omega x} </math> and <math> Te^{-i\omega x} </math> respectively. Again knowing '''S''' we can solve for R and T and hence find the numerical boundary conditions (<math>a=Te^{i\omega L}</math> and <math>b=e^{-i\omega L}+Re^{i\omega L}</math>).

If a waveform <math> f(x) </math> is travelling in from <math> -\infty </math> Taking the fourier transform gives
<center>
<math> \hat{f}(\omega)=\int_{-\infty}^{\infty}\,f(x)e^{2\pi i \omega} \,dx </math>
</center>

and then inverting <math> \hat{f}(\omega) \zeta(x) </math> with respect to <math> t </math> gives
<center>
<math> f(x,t) = \int_{-\infty}^{\infty}\,\hat{f}(\omega) \zeta(x) e^{-2\pi i \omega t} \,d\omega </math>
</center>

which is the time dependent solution for the wave form.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-24T02:26:58Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,1:length(x)) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\omega^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\omega^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\omega^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \omega^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad (4) </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u(x,\omega) = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\omega^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{-L}^{L}\,f u_k\,dx}{(\omega^{2}-\lambda_k) \int_{-L}^{L}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta(x,\omega)=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) </math> with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x \zeta |_L </math>.

We choose a basis of the solution space to be <math>\{\zeta_1,\,\zeta_2\}</math>, where <math> \zeta_1 </math> is the solution to the BVP(a=1,b=0) and <math> \zeta_1 </math> is the solution to the BVP(a=0,b=1).

The aim here is to construct a matrix '''S''' such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of '''S''' must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So '''S''' is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the area of constant depth on the left hand side there is a potential of the form <math> e^{i\omega x} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{-i\omega x} </math> and <math> Te^{i\omega x} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-i\omega L}+Re^{i\omega L}, \quad \zeta |_L = Te^{i\omega L}, \quad \partial_x \zeta |_{-L} = i\omega e^{-i\omega L}-i\omega Re^{i\omega L}, \quad \partial_x \zeta |_L = i\omega Te^{i\omega L} </math>
</center>

<center>
<math> \begin{pmatrix} i\omega e^{-i\omega L}-i\omega Re^{i\omega L} \\ i\omega Te^{i\omega L} \end{pmatrix} = S\begin{pmatrix} e^{-i\omega L}+Re^{i\omega L} \\ Te^{i\omega L} \end{pmatrix} </math>
</center>

Knowing '''S''' these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem(<math>a=e^{-i\omega L}+Re^{i\omega L} and b=Te^{i\omega L}</math>). Taking a linear combination of the solutions already calculated (<math> a\zeta_1(x,\omega) + b\zeta_2(x,\omega) </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a generalised eigenfunction potential for the whole real axis which we denote as <math> \zeta_+(x,\omega)</math>.

Independent generalised eigenfunctons can be found by considering the potential <math> e^{-i\omega x} </math> on the right hand region of constant depth. The corresponding reflected and transmitted potentials from the variable depth area are of the form <math> Re^{i\omega x} </math> and <math> Te^{-i\omega x} </math> respectively.

If a waveform <math> f(x) </math> is travelling in from <math> -\infty </math> Taking the fourier transform gives
<center>
<math> \hat{f}(\omega)=\int_{-\infty}^{\infty}\,f(x)e^{2\pi i \omega} \,dx </math>
</center>

and then inverting <math> \hat{f}(\omega) \zeta(x) </math> with respect to <math> t </math> gives
<center>
<math> f(x,t) = \int_{-\infty}^{\infty}\,\hat{f}(\omega) \zeta(x) e^{-2\pi i \omega t} \,d\omega </math>
</center>

which is the time dependent solution for the wave form.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-17T21:27:48Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,1:length(x)) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\omega^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\omega^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\omega^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\omega^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \omega^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad (4) </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \omega^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\omega^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{0}^{1}\,f u_k\,dx}{(\omega^{2}-\lambda_k) \int_{0}^{1}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) </math> with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x \zeta |_L </math>.

The aim here is to construct a matrix <math> S </math> such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of <math> S </math> must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So <math>S</math> is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the areas of constant depth there is a potential of the form <math> e^{i\omega x} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{i\omega x} </math> and <math> Te^{i\omega x} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-i\omega L}+Re^{i\omega L}, \quad \zeta |_L = Te^{i\omega L}, \quad \partial_x \zeta |_{-L} = i\omega e^{-i\omega L}-i\omega Re^{i\omega L}, \quad \partial_x \zeta |_L = i\omega Te^{i\omega L} </math>
</center>

Knowing <math> S </math> these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem. Taking a linear combination of the solutions already calculated (<math> \zeta_1 </math> and <math> \zeta_2 </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a potential for the whole real axis.

If a waveform <math> f(x) </math> is travelling in from <math> -\infty </math> Taking the fourier transform gives
<center>
<math> \hat{f}(\omega)=\int_{-\infty}^{\infty}\,f(x)e^{2\pi i \omega} \,dx </math>
</center>

and then inverting <math> \hat{f}(\omega) \zeta(x) </math> with respect to <math> t </math> gives
<center>
<math> f(x,t) = \int_{-\infty}^{\infty}\,\hat{f}(\omega) \zeta(x) e^{-2\pi i \omega t} \,d\omega </math>
</center>

which is the time dependent solution for the wave form.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-16T23:28:12Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,1:length(x)) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\kappa^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\kappa^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\kappa^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\kappa^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \kappa^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad (4) </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \kappa^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \kappa^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\kappa^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{0}^{1}\,f u_k\,dx}{(\kappa^{2}-\lambda_k) \int_{0}^{1}\, \rho u_{k}^{2}\,dx} </math>
</center>

The coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta=\frac{(b-a)}{2L}x + \frac{b+a}{2}+\sum_{n=1}^{\infty}c_{n} \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right) </math> with <math> \zeta |_{-L}=a \quad \zeta |_{L}=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_{-L} </math> and <math> \partial_x |zeta |_L </math>.

The aim here is to construct a matrix <math> S </math> such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_{-L} \\ \zeta |_L \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_{-L} \\ \partial_x \zeta |_L \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of <math> S </math> must be <math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \\ \partial_x \zeta_1 |_L \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_2 |_L \end{pmatrix} </math>. So <math>S</math> is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_{-L} \, \partial_x \zeta_2 |_{-L} \\ \partial_x \zeta_1 |_L \, \partial_x \zeta_2 |_L \end{pmatrix} </math>
</center>

Now for the areas of constant depth there is a potential of the form <math> e^{ikx} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{ikx} </math> and <math> Te^{ikx} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_{-L} = e^{-ikL}+Re^{ikL}, \quad \zeta |_L = Te^{ikL}, \quad \partial_x \zeta |_{-L} = ike^{-ikL}-ikRe^{ikL}, \quad \partial_x \zeta |_L = ikTe^{ikL} </math>
</center>

Knowing <math> S </math> these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem. Taking a linear combination of the solutions already calculated (<math> \zeta_1 </math> and <math> \zeta_2 </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a potential for the whole real axis.

If a waveform <math> f(x) </math> is travelling in from <math> -\infty </math> Taking the fourier transform gives
<center>
<math> \hat{f}(k)=\int_{-\infty}^{\infty}\,f(x)e^{2\pi i k} \,dx </math>
</center>

and then inverting <math> \hat{f}(k) \zeta(x) </math> with respect to <math> t </math> gives
<center>
<math> f(x,t) = \int_{-\infty}^{\infty}\,\hat{f}(k) \zeta(x) e^{-2\pi i k t} \,dk </math>
</center>

which is the time dependent solution for the wave form.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-16T23:11:28Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,1:length(x)) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\kappa^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\kappa^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\kappa^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\kappa^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \kappa^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad (4) </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center><math> \sum_{k=1}^{\infty} (\partial_x(h(x)\partial_x u_k) + \kappa^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (-\lambda_k\rho u_k + \kappa^{2}\rho u_k)b_k = f(x) </math></center>
<center><math> \implies \sum_{k=1}^{\infty} (\kappa^{2}-\lambda_k) b_k \rho u_k = f(x) \quad (5) </math></center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{0}^{1}\,f u_k\,dx}{(\kappa^{2}-\lambda_k) \int_{0}^{1}\, \rho u_{k}^{2}\,dx} </math>
</center>

Also to find the coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta=(b-a)x+a+\sum_{n=1}^{\infty}c_{n} \sin(n\pi x) </math> with <math> \zeta |_0=a \quad \zeta |_1=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_0 </math> and <math> \partial_x |zeta |_1 </math>.

The aim here is to construct a matrix <math> S </math> such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_0 \\ \zeta |_1 \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_0 \\ \partial_x \zeta |_1 \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of <math> S </math> must be <math> \begin{pmatrix} \partial_x \zeta_1 |_0 \\ \partial_x \zeta_1 |_1 \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_0 \\ \partial_x \zeta_2 |_1 \end{pmatrix} </math>. So <math>S</math> is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_0 \, \partial_x \zeta_2 |_0 \\ \partial_x \zeta_1 |_1 \, \partial_x \zeta_2 |_1 \end{pmatrix} </math>
</center>

Now for the areas of constant depth there is a potential of the form <math> e^{ikx} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{ikx} </math> and <math> Te^{ikx} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_0 = 1+R, \quad \zeta |_1 = Te^{ik}, \quad \partial_x \zeta |_0 = ik(1-R), \quad \partial_x \zeta |_1 = ikTe^{ik} </math>
</center>

Knowing <math> S </math> these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem. Taking a linear combination of the solutions already calculated (<math> \zeta_1 </math> and <math> \zeta_2 </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a potential for the whole real axis.

If a waveform <math> f(x) </math> is travelling in from <math> -\infty </math> Taking the fourier transform gives
<center>
<math> \hat{f}(k)=\int_{-\infty}^{\infty}\,f(x)e^{2\pi i k} \,dx </math>
</center>

and then inverting <math> \hat{f}(k) \zeta(x) </math> with respect to <math> t </math> gives
<center>
<math> f(x,t) = \int_{-\infty}^{\infty}\,\hat{f}(k) \zeta(x) e^{-2\pi i k t} \,dk </math>
</center>

which is the time dependent solution for the wave form.

[[Category:Shallow Depth]]

Variable Depth Shallow Water Wave Equation

2009-02-16T22:48:45Z

Sean Curry: /* Solution using Separation of Variables */

== Introduction ==

We consider here the problem of waves reflected by a region of variable depth in
an otherwise uniform depth region assuming the equations of [[:Category:Shallow Depth|Shallow Depth]].

== Equations ==

We begin with the shallow depth equation
{{shallow depth one dimension}}

== Waves in a finite basin ==

We consider the problem of waves in a finite basin <math>-L<x<L</math>. At the edge of the basin the boundary conditions are
<center><math>
\left.\partial_x \zeta\right|_{x=-L} = \left.\partial_x \zeta\right|_{x=L} =0
</math></center>.

We solve the equations by expanding in the modes for the basin which satisfy
<center><math>
\partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n \rho(x) \zeta_n ,
</math></center>
normalised so that
<center><math>
\int_{-L}^L \rho\zeta_n \zeta_m = \delta_{mn}.
</math></center>
The solution is then given by
<center><math>
\zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t )
</math></center>
<center><math>
+ \sum_{n=1} ^{\infty} \left(\int_{-L}^L \rho(x)\zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}}
</math></center>
where we have assumed that <math>\lambda_0 = 0</math>.

=== Calculation of <math>\zeta_n</math> ===

We can calculate the eigenfunctions <math>\zeta_n</math> by an expansion in the modes for the case of uniform depth.
We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of
<center>
<math>J[\zeta] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left(\partial_x \zeta\right)^2 - \lambda \rho(x) \zeta^2 \right\}</math>
</center>
subject to the boundary conditions that the normal derivative vanishes (where <math>\lambda</math> is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth <math>h=1</math>
<center>
<math>
\zeta = \sum_{n=0}^{N} a_n \psi_n(x)
</math>
</center>
where
<center>
<math>
\psi_n = \frac{1}{\sqrt{L}} \cos( n \pi (\frac{1}{2L}x + \frac{1}{2})),\,\,n\ne 1
</math>
</center>
<center>
<math>
\psi_0 = \frac{1}{\sqrt{2L}},\,
</math>
</center>
and substitute this expansion into the variational equation we obtain
<center><math>
J[\vec{a}] = \int_{-L}^L \frac{1}{2}\left\{ h(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right)^2 - \lambda \rho(x) \left(\sum_{n=0}^{N} a_n \psi_n(x)\right)^2 \right\}
</math></center>
<center><math>
\frac{dJ}{da_m} = \int_{-L}^L \frac{1}{2}\left\{ h(x)2\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) - \lambda \rho(x)2\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\} = 0
</math></center>
<center><math>
\int_{-L}^L \left\{ h(x)\partial_x \psi_m(x)\left( \sum_{n=0}^{N} a_n \partial_x \psi_n(x)\right) \right\} = \int_{-L}^L \left\{ \lambda \rho(x)\psi_m(x)\left(\sum_{n=0}^{N} a_n \psi_n(x)\right) \right\}
</math></center>
<center><math>
\sum_{n=0}^{N} a_n \int_{-L}^L \left\{ h(x)\partial_x \psi_n(x)\partial_x \psi_m(x)\right\} = \lambda \sum_{n=0}^{N} a_n \int_{-L}^L \left\{ \rho(x)\psi_n(x)\psi_m(x) \right\}
</math></center>
this equation can be rewritten using matrices as
<center>
<math>
\mathbf{K} \vec{a} = \lambda\mathbf{M} \vec{a}
</math>
</center>
where the elements of the matrices '''K''' and '''M''' are
<center>
<math>
\mathbf{K}_{mn} = \int_{-L}^L \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\}
</math>
</center>
and
<center><math>
\mathbf{M}_{mn} = \int_{-L}^L \left\{\rho(x)\psi_n(x)\psi_m(x)\right\}
</math></center>

=== Matlab code ===
'''Example of the method of numerical integration; using a weighting matrix to implement Simpson's Rule:'''

x = linspace(-4,4,500).';

f = exp(-10*(x.^2));

g = sin(x);

h = x(2)-x(1);

simp = 2*h/3*ones(1,length(x));

simp(2:2:length(x)-1) = 4*h/3;

simp(1) = h/3;

simp(end) = h/3;

simp = diag(simp);

%The integral of f(x)^2 from (by Simpson's Rule) -4 to 4 is:

int1 = transpose(f)*simp*f;

%The integral of f(x)*g(x) from -4 to 4 is:

int2 = transpose(f)*simp*g;

'''Multiple integrals can be executed in one expression:'''

%NUMERICALLY CALCULATING FOURIER COEFICIENTS

%Make a row vector of x values

x = linspace(-pi,pi,200);

%Define a row vector of the values of the function to be fitted by a Fourier sine series

f = 5*x.^3 - 3*x;

%Simpson's Rule weighting matrix:

h = x(2)-x(1);
simp = 2*h/3*ones(1,length(x));
simp(2:2:length(x)-1) = 4*h/3;
simp(1) = h/3;
simp(end) = h/3;
simp = diag(simp);

%Create a matrix M with the fourier functions, x increasing along the rows, n increasing down the columns

M = zeros(round(length(x)/2),length(x));

for n = 1:length(M(:,1))

M(n,1:length(x)) = sin(n*x)/sqrt(pi);

end

%Calculate the column vector containing the coefficients of the sine series

fourier_coefficients = M*simp*transpose(f);

%Calculate a row vector of the fitting function evaluated at the values of x

fourier_sine_fit = transpose(fourier_coefficients)*M;

%Plot to check

plot(x,f,x,fourier_sine_fit)

== Waves in an infinite basin ==

We assume that the density <math>\rho</math> and the depth <math>h</math> are constant and equal to one outside the region <math>-L<x<L</math>.
We can therefore write the wave as

== Solution using Separation of Variables ==

Taking a separable solution gives the eigenvalue problem
<center>
<math>\partial_x \left( h(x) \partial_x\zeta \right) = -\kappa^{2}\rho(x)\zeta \quad (1)</math>
</center>

Given boundary conditions <math>\zeta \mid_{-L} = a</math> and <math>\zeta \mid_L = b</math> we can take <math> \zeta = \zeta_p + u </math> With <math> \zeta_p = \frac{(b-a)}{2L}x + \frac{b+a}{2} </math> satisfying the boundary conditions and <math> u </math> satisfying <math> u |_{-L} = u |_L = 0</math>

Substituting this form into (1) gives
<center>
<math> \partial_x(h(x)\partial_x \zeta_p)+\partial_x(h(x)\partial_xu) = -\kappa^{2}\left(\zeta_p+u\right)</math>
</center>
Or, on rearranging
<center>
<math> \partial_x(h(x)\partial_xu)+\kappa^{2}u = -\partial_x(h(x)\partial_x \zeta_p)-\kappa^{2}\zeta_p = f(x)\quad (2)</math>
</center>
so <center><math> f(x) = -\frac{(b-a)}{2L}\partial_xh(x) - \kappa^2 \left(\frac{(b-a)}{2L}x + \frac{b+a}{2}\right) </math>
</center>
Now consider the homogenous Sturm-Liouville problem for u
<center>
<math> \partial_x(h(x)\partial_xu)+\lambda u = 0\quad u|_0=u|_1=0 \quad (3)</math>
</center>
By Sturm-Liouville theory this has an infinite set of eigenvalues <math> \lambda_k </math> with corresponding eigenfunctions <math> u_k </math>. Also since <math> u_k|_0=u_k|_1=0\quad \forall k </math> Each <math> u_k</math> can be expanded as a fourier series in terms of sine functions.
<center>
<math> u_k = \sum_{n=1}^{\infty} a_{n,k} \psi_n</math>
</center>
where<center>
<math>\psi_n = \sin\left(\frac{n\pi}{2}\left(\frac{x}{L} + 1\right)\right)</math>
</center>
Transforming (3) into the equivalent variational problem gives
<center>
<math> J[u] = \int_{-L}^{L}\,hu'^{2}-\lambda \rho u^{2} \, dx \quad (4) </math>
</center>
Substituting the fourier expansion for <math>u_k</math> into (4):
<center>
<math> J = \int_{-L}^{L}\,h\left(\sum_{n=1}^{\infty} a_{n} \psi_n'\right)^{2}-\lambda \rho \left(\sum_{n=1}^{\infty} a_{n} \psi_n\right)^{2} \, dx \quad (4) </math></center>
Since <math>u_k</math> minimises J, we require <math> \frac{\partial J}{\partial a_{n}}=0 \quad \forall n </math>
<center>
<math> \frac{\partial J}{\partial a_{m}}=\int_{-L}^{L}\left\{2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'-2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_n\right\}dx=0</math>
</center>
<center>
<math> \implies \int_{-L}^{L}2h\psi_m'\sum_{n=1}^{\infty} a_{n}\psi_n'dx= \int_{-L}^{L}2\lambda\rho \psi_m \sum_{n=1}^{\infty} a_{n} \psi_ndx</math>
</center>
<center><math> \implies \sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}h\psi_m'\psi_n'dx= \lambda\sum_{n=1}^{\infty}a_{n}\int_{-L}^{L}\rho \psi_m \psi_ndx</math></center>

By defining a vector <math> \textbf{a} = \left(a_{n}\right)</math> and matrices <math>K_{(n,m)} = \int_{-L}^{L}h(x)\psi_m'(x)\psi_n'(x)dx </math> and <math>M_{(n,m)} = \int_{-L}^{L}\rho(x)\psi_m(x)\psi_n(x)dx </math> we have the linear system <math> K\textbf{a} = \lambda M\textbf{a} </math> which returns the eigenvalues and eigenvectors of equation (3), with eigenvectors <math> \textbf{a} </math> representing coefficient vectors of the fourier expansions of eigenfunctions.

If we now construct <math> u = \sum_{k=1}^{\infty} b_k u_k </math> and substitute this into equation (2) we get
<center>
<math> \sum_{k=1}^{\infty} (\kappa^{2}-\lambda_k) b_k u_k = -(b-a)\partial_xh(x)-\kappa^{2}\left((b-a)x+a\right) \quad (5) </math>
</center>
And defining the RHS of equation (5) as <math> f(x) </math>, a known function, we can retrieve the coefficients <math> b_k </math> by integrating against <math> u_k </math>
<center>
<math> b_k = \frac{\int_{0}^{1}\,f u_k\,dx}{(\kappa^{2}-\lambda_k) \int_{0}^{1}\, u_{k}^{2}\,dx} </math>
</center>

Also to find the coefficients <math> c_n </math> of the fourier expansion of u are just <math> \sum_{k=1}^{\infty}a_{n,k}b_{k} </math> with <math> a_{n.k} </math> being the <math>n</math>th coefficient of the <math>k</math>th eigenfunction of the Sturm-Liouville problem.

So <math> \zeta=(b-a)x+a+\sum_{n=1}^{\infty}c_{n} \sin(n\pi x) </math> with <math> \zeta |_0=a \quad \zeta |_1=b </math> and, given <math> a </math> and <math> b </math> explicitly differentiating <math> \zeta </math> gives <math> \partial_x \zeta |_0 </math> and <math> \partial_x |zeta |_1 </math>.

The aim here is to construct a matrix <math> S </math> such that, given <math> a </math> and <math> b </math>
<center>
<math> S \begin{pmatrix} \zeta |_0 \\ \zeta |_1 \end{pmatrix}=\begin{pmatrix} \partial_x \zeta |_0 \\ \partial_x \zeta |_1 \end{pmatrix}</math>
</center>

Taking <math> a=1,\,b=0 </math> to give <math> \zeta_1 </math> shows that the first column of <math> S </math> must be <math> \begin{pmatrix} \partial_x \zeta_1 |_0 \\ \partial_x \zeta_1 |_1 \end{pmatrix} </math> and likewise taking <math> a=0,\,b=1 </math> to give <math> \zeta_2 </math> shows the second column must be <math> \begin{pmatrix} \partial_x \zeta_2 |_0 \\ \partial_x \zeta_2 |_1 \end{pmatrix} </math>. So <math>S</math> is given by
<center>
<math> \begin{pmatrix} \partial_x \zeta_1 |_0 \, \partial_x \zeta_2 |_0 \\ \partial_x \zeta_1 |_1 \, \partial_x \zeta_2 |_1 \end{pmatrix} </math>
</center>

Now for the areas of constant depth there is a potential of the form <math> e^{ikx} </math> which, creates reflected and transmitted potentials from the variable depth area of the form <math> Re^{ikx} </math> and <math> Te^{ikx} </math> respectively where the magnitudes of <math> R </math> and <math> T </math> are unknown. We can calculate that the boundary conditions for <math> \zeta </math> must be
<center>
<math> \zeta |_0 = 1+R, \quad \zeta |_1 = Te^{ik}, \quad \partial_x \zeta |_0 = ik(1-R), \quad \partial_x \zeta |_1 = ikTe^{ik} </math>
</center>

Knowing <math> S </math> these boundary conditions can be solved for <math> R</math> and <math>T</math>, which in turn gives actual numerical boundary conditions to the original problem. Taking a linear combination of the solutions already calculated (<math> \zeta_1 </math> and <math> \zeta_2 </math>) will provide the solution for these new boundary conditions. This solution, along with the potentials outside this region gives a potential for the whole real axis.

If a waveform <math> f(x) </math> is travelling in from <math> -\infty </math> Taking the fourier transform gives
<center>
<math> \hat{f}(k)=\int_{-\infty}^{\infty}\,f(x)e^{2\pi i k} \,dx </math>
</center>

and then inverting <math> \hat{f}(k) \zeta(x) </math> with respect to <math> t </math> gives
<center>
<math> f(x,t) = \int_{-\infty}^{\infty}\,\hat{f}(k) \zeta(x) e^{-2\pi i k t} \,dk </math>
</center>

which is the time dependent solution for the wave form.

[[Category:Shallow Depth]]