Energy conservation in the solution of the Helmholtz equation

Question

This might be a silly question, but I know very little about the theoretical properties finite elements, so here goes. Suppose you were to solve the Helmholtz equation (let's say in 2D) with a spatially varying wave speed using finite elements. For reference the equation is $$ [\nabla^2 + k^2(\vec{r})]\psi(\vec{r}) = f(\vec{r}) $$ This being a scattering problem, the source term is limited to, say, Dirichlet values on the boundary of the domain. My question is, when the domain is large (say ten or more of wavelength (I consider this large, it might not be)), how well does energy conservation hold? More precisely, what kind of guarantees can you place on conservation of energy as a function of domain size?

Assume for now linear elements (in practice, we use finite differences, is there a huge difference?) since $k(\vec{r})$ is piecewise constant, so higher orders aren't terribly helpful. I hope I have not revealed too much of my ignorance on this subject!

Edit: I am referring to energy in the "intuitive" sense. I work mostly with Maxwell's equations, which in 2D at constant frequency becomes the Helmholtz equation. I believe the mathematical definition of energy flow at a point is $\Psi\nabla \Psi$, or something to that effect (it should be the Poynting vector). Energy conservation would say that in a source-free region, the integral of the energy in the surface normal direction along the boundary of the region should be identically zero. Intuitively, I should see waves obeying the inverse square law (or whatever its analog is in 2D) from a point source exactly.

Also, I realize there is a difference between finite differences and finite elements. If you could comment on both, that would be even better.

Answer 1

Mathematically, you have the Diriclet energy: $$ E = \int (-|\nabla\psi|^2+k^2|\psi|^2-f\psi^*-f^*\psi)d^Dx $$ whose minimisation gives you the Helmholtz equation. The natural energy current would be: $$ j = \frac{1}{2i}(\psi^*\nabla \psi-\psi\nabla \psi^*) $$ which satisfies: $$ \nabla \cdot j = 0 $$ from the Helmholtz equation, so you have exact local energy conservation.

More abstractly, this arises from Noether's theorem. Your equation (or equivalently your energy functional) is invariant by a change of phase in the function: $$ \psi\to e^{i\phi }\psi $$ You can therefore apply the general recipe to cook up a conserved local current.

Note that depending on the physical meaning of $\psi$, this local conservation law may or may not correspond to physical energy. For example, if it is supposed to be a first quantisation wave function, then $j$ is rather a current of probability.

Hope this helps.