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.
[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.
We consider here the theory only in two dimensions.
The multipoles are singular at a point [math]\displaystyle{ \mathbf{x^{\prime}} }[/math] and they have various orders of singularity.
For [math]\displaystyle{ n=0 }[/math]
[math]\displaystyle{ \phi_0 \to \ln r \,\mathrm{as}\, \mathbf{x} \to \mathbf{x}^{\prime} }[/math]
where [math]\displaystyle{ (r,\theta) }[/math] are the polar coordinates centered at [math]\displaystyle{ \mathbf{x^{\prime}} }[/math]
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] (refer to Figure 1) centered at the origin in an infinite medium. The Laplace's equation in polar coordinates is
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
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.
Here [math]\displaystyle{ r\, }[/math] and [math]\displaystyle{ \theta\, }[/math] are polar coordinates defined by [math]\displaystyle{ z + f = -r\cos\theta\, }[/math], [math]\displaystyle{ \quad \, }[/math][math]\displaystyle{ x = r\sin\theta \, }[/math], note also 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, as shown in Figure 1.
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.
The singularity at [math]\displaystyle{ (0,-f)\, }[/math] (refer to Figure 1) is not of the form [math]\displaystyle{ \ln r\, }[/math] since this would give a potential that corresponds to a pulsating source of fluid. From this point on Multipoles with higher order with a submerged singularity are only considered for this problem
For [math]\displaystyle{ n\gt 0\, }[/math] the free surface condition yields the boundary condition for [math]\displaystyle{ \psi_n\, }[/math]:
We solve for [math]\displaystyle{ \hat{\psi}_n\, }[/math] by taking a Fourier transform in [math]\displaystyle{ x\, }[/math] to simplify Laplace's equation
For [math]\displaystyle{ n\gt 0\, }[/math] the form of [math]\displaystyle{ \hat{f}_n(|\mu|) }[/math] can be obtained from [math]\displaystyle{ f_n(x)\, }[/math][math]\displaystyle{ (3)\, }[/math] by rewriting it as a Fourier transform. Refer to Appendix A2 for confirmation that changing the limits of integration with different constant outside of integration for [math]\displaystyle{ \psi_n(x,z)\, }[/math] gives the same result. Now substituting the result into [math]\displaystyle{ (5)\, }[/math], [math]\displaystyle{ \psi_n\, }[/math] easily simplifies to:
Note: the integral is singular at [math]\displaystyle{ \mu=k\, }[/math]. Refer to Appendix A1 on steps required to numerically integrate this singular integral.
This result is arrived at by expanding the exponential part of the integral of [math]\displaystyle{ \psi_n(x,z)\, }[/math]. So we have
This identity is then substituted into the expression for [math]\displaystyle{ \psi_n\, }[/math] to complete the power series.
In order to have a complete set to expand our fluid potential we need to include [math]\displaystyle{ \bar{\phi}_n \, }[/math] for [math]\displaystyle{ n\gt 0 \, }[/math].
This accounts for the second linearly independent solution for [math]\displaystyle{ n\gt 0 \, }[/math].
Wave-Free Potentials
The combination [math]\displaystyle{ \phi_{n+1}(x,z) + kn^{-1}\phi_n(x,z)\, }[/math], [math]\displaystyle{ n=1,2,3,...\, }[/math] corresponds to a wave free singularity i.e. no waves are radiated to infinity so the potential dies off in the far field.
where [math]\displaystyle{ \psi_n(x,z) \, }[/math] is defined by [math]\displaystyle{ (6)\, }[/math].
The above relationship can be simplified by using the coordinate relationships defined below. The coordinates that are use are illustrated in Figure 2.
The above power series converge for [math]\displaystyle{ r \lt 2f \, }[/math] where [math]\displaystyle{ A^+_{mn} \, }[/math] and [math]\displaystyle{ A^-_{mn} \, }[/math] are both defined by [math]\displaystyle{ (7)\, }[/math].
Applying the body boundary condition [math]\displaystyle{ \partial_r \phi^{s,a} |_{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:
[math]\displaystyle{ \phi^s_n(x,z) \sim \frac{2\pi i (-k)^n}{(n-1)!} e^{k(z-f)} e^{\pm ikx} \quad \mathrm{as} \quad x \rightarrow \pm \infty }[/math]
The first integral is no longer singular and can be numerically evaluated using an adaptive Gauss-Kronrod quadrature. The second integral remains singular and can be transformed into a series of known functions as follows:
now if we let [math]\displaystyle{ t = -\xi + \upsilon \, }[/math] it follows that [math]\displaystyle{ dt = -d \xi \, }[/math]. The limits of integration follow suit, at [math]\displaystyle{ \xi = 0 \, }[/math][math]\displaystyle{ t = \upsilon \, }[/math] and at [math]\displaystyle{ \xi = \infty \, }[/math][math]\displaystyle{ t = -\infty \, }[/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(\mu)\, }[/math]:
[math]\displaystyle{ f_n(x) = \int_0^\infty g_n(\mu)\cos\mu x \ d\mu -i \int_0^\infty g_n(\mu)\sin\mu x \ d\mu }[/math]
[math]\displaystyle{ = \frac{1}{2}\int_{-\infty}^\infty g_n(|\mu|)\cos\mu x \ d\mu -\frac{i}{2} \int_{-\infty}^\infty \frac{\mu}{|\mu|}g_n(|\mu|)\sin\mu x \ d\mu }[/math]
Since [math]\displaystyle{ g_n(|\mu|)\, }[/math] is even, its integral against sine over all [math]\displaystyle{ x\, }[/math] is zero so we can replace the cosine with a complex exponential in the first integral (similarly we replace (-i sine) for the second).
