Difference between revisions of "Category:Multipole Methods for Linear Water Waves"
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We write | We write | ||
− | <math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math> | + | <center><math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math></center> |
− | <math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math> | + | <center><math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math></center> |
Where | Where | ||
− | <math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math> | + | <center><math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math></center> |
We want | We want | ||
− | <math>\partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots</math> | + | <center><math>\partial_z \phi |_{z=f} = K\phi|_{z=f} \quad \forall m = 0,1,2, \ldots</math></center> |
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For m=0 | 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> | + | <center><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></center> | ||
Using | Using | ||
− | <math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x))\frac{d\mu}{\mu}</math> | + | <center><math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu|z+f|}\cos(\mu x))\frac{d\mu}{\mu}</math></center> |
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For m>0 | 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) | + | <center><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></center> | ||
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So our psi functions satisfy | So our psi functions satisfy | ||
− | <math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots \quad \ldots (1)</math> | + | <center><math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots \quad \ldots (1)</math></center> |
− | <math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math> | + | <center><math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math></center> |
− | <math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = f_m(x) \quad \ldots (3)</math> | + | <center><math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = f_m(x) \quad \ldots (3)</math></center> |
Taking a Fourier transform in x | Taking a Fourier transform in x | ||
− | <math>\hat{\psi}_m = \int_{-\infty}^\infty \psi_m e^{i\omega x}dx</math> | + | <center><math>\hat{\psi}_m = \int_{-\infty}^\infty \psi_m e^{i\omega x}dx</math></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><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> | ||
− | <math>but \quad (2) \Rightarrow B_m(\omega) = 0 \quad \forall m</math> | + | <center><math>but \quad (2) \Rightarrow B_m(\omega) = 0 \quad \forall m</math></center> |
− | <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> | + | <center><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></center> |
Hence | Hence | ||
− | <math>A(\omega) = \frac{\hat{f}_m(\omega)}{\omega - K} e^{-\omega f} </math> | + | <center><math>A(\omega) = \frac{\hat{f}_m(\omega)}{\omega - K} e^{-\omega f} </math></center> |
and | 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> | + | <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-f)} e^{-i\omega x}d\omega</math></center> |
== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)== | == Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)== |
Revision as of 01:02, 9 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
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:
for m=0 we can write:
and for [math]\displaystyle{ m\neq 0: }[/math]
Hence the general solution can be expressed as:
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
Where
We want
For m=0
Using
For m>0
So our psi functions satisfy
Taking a Fourier transform in x
Hence
and
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]
Hydrodynamic Forces
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