# stability estimate for the Helmholtz equation using Dirichlet boundary conditions.

Given the following problem on a square domain $$\Omega$$

\begin{align} -\Delta u - \kappa^2 u &= f \quad \text{in } \Omega, \\ u &= 0 \quad \text{ on } \partial\Omega, \end{align}

where $$\kappa>0$$. The weak formulation is to find $$u \in H_0^1(\Omega)$$ such that $$b(u,v) = (f,v) \quad \forall v \in H_0^1(\Omega)$$ where $$b(u,v) = \int_\Omega (\nabla u \nabla v - \kappa^2 ) dx.$$ I'm looking to see if there is any kind of reference to the assumption that for any $$f \in L^2(\Omega)$$, the problem has a unique solution $$u \in H^1_0(\Omega)$$ and there exists a $$C_\text{Stab} > 0$$ such that

$$\|u\|_\kappa \le C_\text{Stab} \|f\|$$ where $$\|u\|_\kappa^2 = \|\nabla u \|^2 + \kappa \|\nabla u \|^2.$$ I'm not exactly certain that such an assumption is even possible with Dirichlet boundary conditions. The only references I can find are using mixed boundary conditions, or some form of scattering problem. Any help would be greatly appreciated.

If $$\kappa^2$$ is a Dirichlet eigenvalue for $$\Delta$$, there will be no uniqueness. For example if $$f=0$$, then you get the trivial solution and the eigenfunction corresponding to $$\kappa^2$$ (see also Fredholm alternative).
For other $$\kappa^2$$ you can use the resolvent, i.e., the inverse of $$(-\Delta - \kappa^2)$$ to solve the equation. The resolvent is a bounded operator and there are some generic bounds for its norm in terms of the distance to the spectrum, see this post.
• Yeah, thats what I feared is probably the case. I think the assumption works for general elliptic indefinite and non-self adjoint problems, but not so much for Helmholtz. What if the problem was set as $div (A \nabla u) - \kappa^2u =f$? Where $A$ is SPD? I'm trying to adapt some theory that was applied to the general elliptic PDE, but using a more complex scenario applied to Helmhotlz. Trying to use the same asumption seemed like a good way of ensuring a unique solution for the problem. What if $\kappa^2 = c(x)$ where $c(x) \in L^\infty(\Omega)$? Thanks for the help. Dec 20, 2023 at 23:14