Difference between revisions of "Template:Removing the depth dependence"
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<math> | <math> | ||
| − | \phi(x,y,z) = \frac{\ | + | \phi(x,y,z) = \frac{\cosh \big( k (z+h) \big)}{\cosh(k h)} \bar{\phi}(x,y) |
</math> | </math> | ||
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Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]] | Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]] | ||
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| − | <math>\nabla^2 \ | + | <math>\nabla^2 \bar{\phi} + k^2 \bar{\phi} = 0 </math> |
</center> | </center> | ||
| − | in the region not occupied by the scatterers. | + | in the region not occupied by the scatterers. |
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Latest revision as of 00:04, 9 June 2010
If we have a problem in which all the scatterers are of constant cross sections so that
[math]\displaystyle{ \partial\Omega = \partial\hat{\Omega} \times z\in[-h,0] }[/math]
where [math]\displaystyle{ \partial\hat{\Omega} }[/math] is a function only of [math]\displaystyle{ x,y }[/math] i.e. the boundary of the scattering bodies is uniform with respect to depth. We can remove the depth dependence separation of variables and obtain that the dependence on depth is given by
[math]\displaystyle{ \phi(x,y,z) = \frac{\cosh \big( k (z+h) \big)}{\cosh(k h)} \bar{\phi}(x,y) }[/math]
Since [math]\displaystyle{ \phi }[/math] satisfies Laplace's Equation, then [math]\displaystyle{ \Phi }[/math] satisfies Helmholtz's Equation
[math]\displaystyle{ \nabla^2 \bar{\phi} + k^2 \bar{\phi} = 0 }[/math]
in the region not occupied by the scatterers.
