Difference between revisions of "Category:Multipole Methods for Linear Water Waves"
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<math>\psi_m\,</math> can be expanded in a power series: | <math>\psi_m\,</math> can be expanded in a power series: | ||
− | <center><math>\psi_m = \ | + | <center><math>\psi_m = \sum_{m=0}^\infty A_{mn}r^m \cos(m\theta),\,\,\,\textrm{for}\,\,\, m>0</math> </center> |
where | where | ||
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<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) | <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> | </math></center> | ||
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== Wave-Free Potentials == | == Wave-Free Potentials == |
Revision as of 00:27, 17 February 2010
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:
(note that the last expression can be obtained from combining the expressions:
where [math]\displaystyle{ \alpha = \omega^2/g \, }[/math])
In two-dimensions the Sommerfeld Radiation Condition is
where [math]\displaystyle{ \phi^{\mathrm{{I}}} }[/math] is the incident potential.
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]\displaystyle{ a }[/math] centered at the origin. The equation in polar coordinates is
[math]\displaystyle{ \partial_n \phi |_{r=a} = 0 }[/math]
Using separation of variables
we write
[math]\displaystyle{ \phi = R(r)\Theta(\theta).\, }[/math] Substituting into Laplace's equations gives
The equation for [math]\displaystyle{ \Theta\, }[/math] is
When m=0 the solution is
The equation for R is
This is a standard DE, to solve we substitute R = r^n giving:
This gives two independent solutions for all m except m = 0, where we only get the constant solution.
Noting that:
So along with the obvious constant function we have two independant sotutions. Thus for m=0 we can write:
and for [math]\displaystyle{ m\neq 0: }[/math]
The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:
Hence the general solution can be expressed as:
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
and for m=0
Where
We want the multipoles to satisfy the free surface condition
For n=0 the free surface condition gives our boundary condition for [math]\displaystyle{ \psi_0\, }[/math]:
where we used the expansion
For n>0 the free surface condition yields the boundary condition for [math]\displaystyle{ \psi_m\, }[/math]:
where we used the expansion (valid for z>-f)
So our psi functions satisfy the following problem:
We solve for [math]\displaystyle{ \hat{\psi}_n\, }[/math] by taking a Fourier transform in x to simplify Laplace's equation
Hence
and we obtaion [math]\displaystyle{ \psi\, }[/math] by the inverse Fourier transform:
For n>0 the form of [math]\displaystyle{ \hat{f}_n(\omega) }[/math] can be obtained from the expansion of [math]\displaystyle{ f_n(x)\, }[/math] labelled [math]\displaystyle{ (*)\, }[/math] above by rewriting it as a Fourier transform. Substituting the result into the expression above, [math]\displaystyle{ \psi_n\, }[/math] easily simplifies to:
for n>0. Note: the integral above goes under the singularity at [math]\displaystyle{ \omega=K }[/math]
Working to get [math]\displaystyle{ \hat{f}_n(\omega) }[/math]:
We can write:
where
We can rewrite [math]\displaystyle{ 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]\displaystyle{ \hat{f}_n(\omega) }[/math]:
Since [math]\displaystyle{ 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).
Finally, combining the integrals, [math]\displaystyle{ \hat{f}_n(\omega) }[/math] emerges:
[math]\displaystyle{ \psi_0 \, }[/math] remains to be derived.
In order to have a complete set to expand our flud potential we need to include [math]\displaystyle{ \bar{\phi}_m }[/math] for m>0.
This accounts for the second linearly independent solution for m>0.
[math]\displaystyle{ \psi_m\, }[/math] can be expanded in a power series:
where
Note: the integral above goes under the singularity at [math]\displaystyle{ \omega=K }[/math]
Which comes from the identity (to be derived)
Wave-Free Potentials
The combination [math]\displaystyle{ \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.
This can be simplified by using the coordinate relationships: (need a picture)
Resulting in the expression:
ie.
Once again to obtain a complete set we also use the compex conjugates of these potentials.
These idntities have been used:
[math]\displaystyle{ -\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]\displaystyle{ \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.
where
and [math]\displaystyle{ \phi^+, \,\, \phi^- }[/math] are the symmetric and antisymmetric parts of [math]\displaystyle{ \phi \, }[/math] respectively.
For a wave incident from the left we have
and
Converting [math]\displaystyle{ \phi^+\, }[/math] and [math]\displaystyle{ \phi^-\, }[/math] into the polar coordinate system:
The above power series converge for r < 2f and the coefficients are given by
Applying the body boundary condition [math]\displaystyle{ \partial_r \phi^\pm |_{r=a} = 0 }[/math] and noting that the cosines are orthogonal over [math]\displaystyle{ \theta \in (-\pi,\pi] }[/math] gives the results
From which we can see that [math]\displaystyle{ \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]\displaystyle{ \alpha_n }[/math]
Hence we can calculate [math]\displaystyle{ \phi\, }[/math] from:
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:
Giving:
Hydrodynamic Forces
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