Non trivial Eigenvectors of the Laplacian in $H^1$

Question

Let $N$ be natural and $\Omega \subset \mathbb R^N$ be open and bounded. I want to show that $$f-\Delta f=0$$ in $D'(\Omega)$ has non trivial solutions in $H^1$. So far I've managed to show solutions to this PDE (for general open $\Omega$) are orthogonal to $H^1_0$ in $H^1$ and using Fourier transform I've found the solutions for different $\Omega \subseteq \mathbb R$. I'm still not sure how to prove existence for general bounded open $\Omega$.

Answer 1

$H_0^1(\Omega)$ is a proper, closed subspace of $H^1(\Omega)$, therefore its orthogonal complement $H_0^1(\Omega)^{\perp}$ is non-trivial. Since this is the space of solutions, there exists at least a nontrivial solution.

In fact, we can show that for any $h\in H^1(\Omega)$ there exists a uniqe solution to $$\begin{cases}\Delta u= u & \Omega \\ u=h & \partial \Omega. \end{cases} $$

Fix $h\in H^1(\Omega)\setminus H_0^1(\Omega)$. Let $$H^1_h(\Omega):=\left\{u\in H^1(\Omega):u-h\in H_0^1(\Omega)\right\}$$ Thus $0\notin H^1_h(\Omega)$, and $H_h^1(\Omega)$ is closed in $H^1(\Omega)$. Since an element $u\in H^1(\Omega)$ solves the equation if and only if $u\in H_0^1(\Omega)^{\perp}$, we just have to show that $$H^1_h(\Omega)\cap H_0^1(\Omega)^{\perp}\neq \emptyset. $$ An element of this set is the element of $H_h^1(\Omega)$ with the smallest $H^1$ norm. To show that it is the only element, notice that if $u_1,u_2\in H_h^1(\Omega)\cap H_0^1(\Omega)^{\perp}$, then $u_1-u_2\in H_0^1(\Omega)\cap H_0^1(\Omega)^{\perp}=\left\{0\right\}$.