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:
[math]\displaystyle{
\begin{align}
\Delta\phi &=0, &-h\lt z\lt 0,\,\,\mathbf{x} \in \Omega \\
\partial_z\phi &= 0, &z=-h, \\
\partial_z \phi &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}},
\end{align}
}[/math]
(note that the last expression can be obtained from combining the expressions:
[math]\displaystyle{
\begin{align}
\partial_z \phi &= -\mathrm{i} \omega \zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \\
\mathrm{i} \omega \phi &= g\zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}},
\end{align}
}[/math]
where [math]\displaystyle{ \alpha = \omega^2/g \, }[/math])
In two-dimensions the Sommerfeld Radiation Condition is
[math]\displaystyle{
\left( \frac{\partial}{\partial|x|} - \mathrm{i} k \right)
(\phi-\phi^{\mathrm{{I}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
}[/math]
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.
Motivation for Multipoles
We present here the motivation for multipoles. For the case of Laplace's equation in an infinite
region surrounding a disk we may construct the solution very simply using a separation of variables
solution.
Consider Laplace's equation for
a disk of radius [math]\displaystyle{ a }[/math] centered at the origin in an infinite medium. The Laplace's equation in polar coordinates is
[math]\displaystyle{
\Delta \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\gt a
}[/math]
We also have boundary conditions at the disk which we assume are Neumann given by
[math]\displaystyle{ \partial_n \phi |_{r=a} = 0
}[/math]
and we have a decay condition at infinity
[math]\displaystyle{
|\nabla\phi| \rightarrow 0 \quad \mathrm{as} \quad r \rightarrow \infty.
}[/math]
Using separation of variables
we write
[math]\displaystyle{ \phi(r,\theta) = R(r)\Theta(\theta).\, }[/math] Substituting into Laplace's equations gives
[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]
The equation for [math]\displaystyle{ \Theta\, }[/math] is
[math]\displaystyle{ \quad \frac{d^2\Theta}{d\theta^2} = -m^2\Theta }[/math]
[math]\displaystyle{ \Rightarrow \quad \Theta_m(\theta) = A_m \cos(m\theta) + B_m \sin(m\theta) \quad \mathrm{for} \quad m \gt 0 }[/math]
When [math]\displaystyle{ m=0\, }[/math] the solution is
[math]\displaystyle{ \Theta_0(\theta) = A_0 + B_0\theta, \quad \mathrm{we \ know \ that} \quad \Theta_0(0) = \Theta_0(2\pi) \, }[/math]
[math]\displaystyle{ A_0 = A_0 + B_02\pi \quad \therefore \quad B_0 = 0 \quad \mathrm{and} \quad \Theta_0(\theta) = A_0 \, }[/math]
The equation for [math]\displaystyle{ R\, }[/math] is
[math]\displaystyle{ \quad r^2\frac{d^2R}{dr^2} + r\frac{dR}{dr} = m^2 R \quad \mathrm{for} \quad m \gt 0 }[/math]
This is a standard differential equation, to solve we substitute [math]\displaystyle{ R(r)=r^n\, }[/math] 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 [math]\displaystyle{ m\gt 0\, }[/math].
[math]\displaystyle{ \quad R_m(r) = C_m r^m + D_m r^{-m} }[/math]
Now
[math]\displaystyle{ r^2\frac{d^2R}{dr^2} + r\frac{dR}{dr} = 0 \quad \mathrm{for} \quad m = 0 }[/math]
[math]\displaystyle{ \Rightarrow \quad \frac{d}{dr} \left(r\frac{dR}{dr}\right) = 0, \quad \quad \frac{dR}{dr} = \frac{1}{r}D_0 }[/math]
So the solution for [math]\displaystyle{ m=0\, }[/math] is
[math]\displaystyle{ \quad R_0(r) = C_0 + D_0 \ln|r| }[/math]
The derivatives of the radial eigenfunctions must go to zero indicating zero fluid flow at infinity:
[math]\displaystyle{ |\nabla\phi| \rightarrow 0 \quad \mathrm{as} \quad r \rightarrow \infty \quad \quad \Rightarrow \quad \quad \frac{dR}{dr} \rightarrow 0 \quad \mathrm{as} \quad r \rightarrow \infty }[/math]
[math]\displaystyle{ \quad \frac{dR_m}{dr} = \frac{C_m m}{r}r^m - \frac{D_m m}{r}r^{-m} }[/math]
[math]\displaystyle{ C_m r^m - \frac{D_m}{r^m} \rightarrow 0 \quad \mathrm{as} \quad r \rightarrow \infty \quad \therefore \quad \mathrm{if} \quad m \gt 0 \quad \mathrm{then} \quad C_m = 0. }[/math]
[math]\displaystyle{ \quad R_m(r) = \frac{D_m}{r^m} }[/math]
Hence the general solution can be expressed as:
[math]\displaystyle{ \phi = R_0(r)\Theta_0(\theta) + \sum_{m=1}^\infty R_m(r)\Theta_m(\theta) }[/math]
[math]\displaystyle{ \phi = R_0(r)\Theta_0(\theta) + \sum_{m=1}^\infty R_m(r)\Theta_m(\theta) }[/math]
[math]\displaystyle{ \phi = [C_0 + D_0 \ln(r)]A_0 + \sum_{m=1}^\infty \frac{D_m}{r^m}\left[ A_m\cos(m\theta) + B_m\sin(m\theta) \right] }[/math]
[math]\displaystyle{ \phi = E_0 + F_0 \ln(r) + \sum_{m=1}^\infty \left[ E_m\frac{\cos(m\theta)}{r^m} + F_m\frac{\sin(m\theta)}{r^m} \right] }[/math]
The solution can also be expressed simply in terms of complex exponentials.
[math]\displaystyle{ \phi(r,\theta) = E_0 + F_0 \ln(r) + \sum_{n=1}^\infty \left[ G_n\frac{e^{in\theta}}{r^n} + H_n\frac{e^{-in\theta}}{r^n} \right] }[/math]
With the Free Surface Boundary Condition
When we introduce the free surface boundary condition 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
[math]\displaystyle{ \phi_n = \frac{e^{in\theta}}{r^n} + \psi_n }[/math]
and for m=0
[math]\displaystyle{ \phi_0 = \ln(r) + \psi_0 \, }[/math]
Where
[math]\displaystyle{ \nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots }[/math]
We want the multipoles to satisfy the free surface condition
[math]\displaystyle{ \partial_z \phi_n |_{z=0} = K\phi_n|_{z=0} \quad \forall n = 0,1,2, \ldots }[/math]
For n=0 the free surface condition gives our boundary condition for [math]\displaystyle{ \psi_0\, }[/math]:
[math]\displaystyle{ 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]\displaystyle{ \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]
where we used the expansion
[math]\displaystyle{ \ln(r) = \int_0^\infty (e^{-\omega}- e^{-\omega|z+f|}\cos(\omega x))\frac{d\omega}{\omega} }[/math]
For n>0 the free surface condition yields the boundary condition for [math]\displaystyle{ \psi_m\, }[/math]:
[math]\displaystyle{ 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]\displaystyle{ \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]
where we used the expansion (valid for z>-f)
[math]\displaystyle{ \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]
So our psi functions satisfy the following problem:
[math]\displaystyle{ \nabla^2 \psi_n = 0 \quad for \quad n = 0,1,2, \ldots \quad \ldots (1) }[/math]
[math]\displaystyle{ \psi_n \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2) }[/math]
[math]\displaystyle{ \partial_z \psi_n |_{z=0} - K\psi_n|_{z=0} = f_n(x) \quad \ldots (3) }[/math]
We solve for [math]\displaystyle{ \hat{\psi}_n\, }[/math] by taking a Fourier transform in x to simplify Laplace's equation
[math]\displaystyle{ \hat{\psi}_n(\omega,z) = \int_{-\infty}^\infty \psi_n(x,z) e^{i\omega x}dx }[/math]
[math]\displaystyle{ (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]
[math]\displaystyle{ but \quad (2) \Rightarrow B_n(\omega) = 0 \quad \forall n }[/math]
[math]\displaystyle{ (3) \Rightarrow \partial_z \hat{\psi}_n |_{z=0} - K\hat{\psi}_n|_{z=0} = (\omega - K)\hat{\psi}_n|_{z=0} }[/math]
[math]\displaystyle{ \Rightarrow \hat{f}_n(\omega) = (\omega - K)A_n(\omega) }[/math]
Hence
[math]\displaystyle{ A_n(\omega) = \frac{\hat{f}_n(\omega)}{\omega - K} }[/math]
and we obtaion [math]\displaystyle{ \psi\, }[/math] by the inverse Fourier transform:
[math]\displaystyle{ \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]
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:
[math]\displaystyle{ \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]
for n>0. Note: the integral above goes under the singularity at [math]\displaystyle{ \omega=K }[/math]
[math]\displaystyle{ \psi_n\, }[/math] can be expanded in a power series:
[math]\displaystyle{ \psi_n = \sum_{m=0}^\infty A_{mn}r^m e^{im\theta},\,\,\,\textrm{for}\,\,\, m\gt 0 }[/math]
where
[math]\displaystyle{ 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]
Note: the integral above goes under the singularity at [math]\displaystyle{ \omega=K }[/math]
Which comes from the substiuting the identity
[math]\displaystyle{ 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]
into the expression for [math]\displaystyle{ \psi_n\, }[/math]. The identity is derived by expanding [math]\displaystyle{ e^{\omega z}e^{-i\omega x}e^{\omega f} = e^{\omega(z+f-ix)} \, }[/math] as a power series and using [math]\displaystyle{ z+f = -r\cos\theta \,\, and \,\, x=r\sin\theta }[/math].
Working to get [math]\displaystyle{ \hat{f}_n(\omega) }[/math]:
We can write:
[math]\displaystyle{ f_n(x) = \int_0^\infty g(\omega)e^{-i\omega x} d\omega }[/math]
where
[math]\displaystyle{ g(\omega)= \frac{(-1)^n}{(n-1)!}(K+\omega)\omega^{n-1}e^{-\omega f}
}[/math]
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]:
[math]\displaystyle{ 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]
[math]\displaystyle{ \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]
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).
[math]\displaystyle{ \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]
Finally, combining the integrals, [math]\displaystyle{ \hat{f}_n(\omega) }[/math] emerges:
[math]\displaystyle{ \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]
[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}_n }[/math] for n>0.
This accounts for the second linearly independent solution for n>0.
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.
[math]\displaystyle{ \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][math]\displaystyle{
+\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]
This can be simplified by using the coordinate relationships: (need a picture)
[math]\displaystyle{ r = \sqrt{x^2 + (z+f)^2}, \quad r_1 = \sqrt{x^2 + (z-f)^2}, }[/math]
[math]\displaystyle{ x = r \sin(\theta) = r_1 \sin(\theta_1),\, }[/math]
[math]\displaystyle{ -z-f = r \cos(\theta), \quad z-f = r_1 \cos(\theta_1). }[/math]
Resulting in the expression:
[math]\displaystyle{ \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]
ie.
[math]\displaystyle{ \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]
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.
[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^+, \,\, \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 \beta_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]
Hydrodynamic Forces
