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Given the Helmholtz equation

$$ \begin{cases} \Delta u + k^2 u &= 0 & \text{ in } \Omega = \{(r,\phi) \mid r_0 < r < r_1 \} \\ u(r,\phi) &= \cos(n\phi)& \text{ on } \Gamma_1 = \{(r,\phi) \mid r=r_0 \} \\ \frac{\partial u}{\partial r} &= iku & \text{ on } \Gamma_2 = \{(r,\phi) \mid r=r_0 \} \end{cases} $$

satisfying the Sommerfeld radiation condition.

Using separation of variables ($u(r,\phi) = R(r)\cdot \Phi(\phi)$) I got the solution

$$u(r,\phi) = \cos(n\phi) \frac{H_n^{(1)}(kr)}{H_n^{(1)}(kr_0)} $$

where $H_n^{(1)}$ denotes the Hankel function of the first kind with parameter $n$.

Can anyone confirm this? I'm trying to test a FD-solver for this equation but for that I need the exact solution, and I was unable to find one using maple.

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Just plug the alleged solution in to the differential equation and boundary conditions, and simplify. In Maple, you might try:

with(VectorCalculus);
SetCoordinates(polar(r,phi));
pde:= Laplacian(u(r,phi)) + k^2*u(r,phi);
alleged:= cos(n*phi)*HankelH1(n,k*r)/HankelH1(n,k*r0);
simplify(eval(pde, u(r,phi)=alleged));

$$0$$

But I don't get your condition on $\frac{\partial u}{\partial r}$.

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  • $\begingroup$ Thank you very much, I suspected there must be an error. $\endgroup$ – flawr Apr 4 '17 at 22:07

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