# Help with a proof (2-D wave propagation stuff)

1. Aug 14, 2007

### Heavytortoise

I'm reading through an older paper on acoustics with one of those "it can be readily seen that" assertions. Of course, to save my life I can't verify the author's assertion. Here goes: starting with the Helmholtz equation $$\nabla^2\psi+k^2\psi=0$$, decompose $$\psi$$ with the 2-D cylindrical basis functions $$\psi_n = H^{(1)}_n(kr)\cos n\theta$$. Then with the surface integral definitions

$$\hat{Q}_{mn} = \frac{1}{4}\int n\cdot \mbox{Re}(\psi_m)\nabla\psi_ndS, \qquad Q_{mn} = \frac{1}{4}\int n\cdot \nabla(\mbox{Re}(\psi_m))\psi_ndS$$

he asserts that by applying the divergence theorem to the difference $$\hat{Q}-Q$$ we end up with $$i$$ times the identity matrix. Pretty quickly I reach the expression

$$\hat{Q}_{mn}-Q_{mn}=\frac{1}{4}\int_V(Re\psi_m)\nabla^2\psi_n- \psi_n\nabla^2(Re\psi_m)dV$$

without using any of the properties of $$\psi$$, where the point $$r=0$$ is not included in the volume of integration. But in the next step, where I substitute $$k^2\psi_n$$ for $$\nabla^2\psi_n$$, all I end up doing is 'proving' that the whole expression is equal to zero, which makes no physical sense. Can anyone see what I'm missing here?