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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.

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1 Answer 1

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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.

This is also discussed in Evans' book on PDEs, page 305.

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  • $\begingroup$ 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. $\endgroup$
    – Kernow1
    Dec 20, 2023 at 23:14

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