User talk:Sean Curry

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]\displaystyle{ \nabla^2 \phi = 0 }[/math]throughout the fluid domain

- The free surface boundary condition

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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\gt a }[/math]

[math]\displaystyle{ \partial_n \phi |_{r=a} = 0 }[/math]

[math]\displaystyle{ |\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty }[/math]

[math]\displaystyle{ Let \quad \phi = R(r)\Theta(\theta) }[/math]

Seperating variables we get:

[math]\displaystyle{ \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]\displaystyle{ For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta }[/math]

[math]\displaystyle{ \Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 }[/math]

[math]\displaystyle{ \mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta }[/math]

[math]\displaystyle{ Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0 }[/math]

[math]\displaystyle{ For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R }[/math]

we substitute R = r^n giving:

[math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ R = C\left(0\right) + D(0) \ln(r) }[/math]

and for [math]\displaystyle{ m\neq 0: }[/math]

[math]\displaystyle{ R = C\left(m\right)r^m + D(m)r^{-m} }[/math]

[math]\displaystyle{ Since\quad|\nabla\phi| \rightarrow 0 \quad as \quad r \rightarrow \infty, }[/math]

[math]\displaystyle{ \partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty }[/math]

[math]\displaystyle{ \Rightarrow C(m)= 0 \quad\forall m \neq 0 }[/math]

Hence the general solution can be expressed as:

[math]\displaystyle{ \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]\displaystyle{ \phi_m = \frac{e^{im\theta}}{r^m} + \psi_m }[/math]

[math]\displaystyle{ \phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 }[/math]

Where

[math]\displaystyle{ \nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots }[/math]

We want

[math]\displaystyle{ \partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots }[/math]

For m=0

[math]\displaystyle{ \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]\displaystyle{ \ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x))\frac{d\mu}{\mu} }[/math]

For m>0

[math]\displaystyle{ \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]\displaystyle{ \nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots \quad \ldots (1) }[/math]
 
 [math]\displaystyle{ \psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2) }[/math]
 
 [math]\displaystyle{ \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]\displaystyle{ \hat{\psi}_m = \int_{-\infty}^\infty \psi_m e^{i\omega x}dx }[/math]

[math]\displaystyle{ (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]\displaystyle{ but \quad (2) \Rightarrow B_m(\omega) = 0 \quad \forall m }[/math]

[math]\displaystyle{ (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]\displaystyle{ A(\omega) = \frac{\hat{f}_m(\omega)}{\omega - K} e^{-\omega f} }[/math]

and

[math]\displaystyle{ \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]\displaystyle{ \phi = \phi_I + \phi_D = \phi^+ + \phi^- }[/math]

where

[math]\displaystyle{ \partial_r \phi_I |_{r=a} = - \partial_r \phi_D |_{r=a} }[/math]

and [math]\displaystyle{ \phi^+ }[/math], [math]\displaystyle{ \phi^- }[/math] are the symmetric and antisymmetric parts of [math]\displaystyle{ \phi }[/math] respectively.

For a wave incident from the left we have

[math]\displaystyle{ \phi_I = e^{Kz} e^{iKx} = e^{-Kf}\sum_{n=0}^\infty \frac{(-Kr)^n }{n!} e^{-in\theta} }[/math],

[math]\displaystyle{ \phi_D = \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- }[/math],

[math]\displaystyle{ \phi^+ = e^{Kz} \cos(Kx) + \sum_{n=1}^{\infty}a^n \alpha_n \phi_n^+ }[/math],

and [math]\displaystyle{ \phi^- = ie^{Kz} \sin(Kx) + \sum_{n=1}^{\infty}a^n \beta_n \phi_n^- }[/math].

Converting [math]\displaystyle{ \phi^+ }[/math] and [math]\displaystyle{ \phi^- }[/math] into the polar coordinate system:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ 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]\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

[math]\displaystyle{ \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]\displaystyle{ \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]\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:

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

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

Giving:

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