How to calculate the first eigenvalue of laplace in $ \mathbb{R}_+^2 $?

Question

For the operator $ -\Delta $, we can consider the problems of eigenvalues and eigenfunctions as follows. That is, we can consider the Dirichlet problem \begin{align} \left\{\begin{matrix} -\Delta u=\lambda u&\text{in}&\Omega,\\ u=0&\text{on}&\partial\Omega. \end{matrix}\right. \end{align} It is well known that the first eigenvalue of this problem is given by \begin{align} \lambda_1=\inf_{v\in H_0^1(\Omega),v\neq 0}\frac{\int_{\Omega}|\nabla v|^2dx}{\int_{\Omega}|v|^2dx}. \end{align} I want calculate the first eigenvalue for the operator \begin{align} -\Delta=-\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial y^2} \end{align} in the half plane $ \mathbb{R}_+^2=\{(x,y):y>0\} $. I want to use the results in the unit disk but cannot go on. Can you give me some references or hints?

Answer 1

There is no first eigenvalue in this case. If $\Omega$ were bounded then what you say is correct; however, this is not necessarily true if $\Omega$ is unbounded it is here.

Indeed, let $\lambda \in \mathbb R$. Then $$u(x,y) = \begin{cases} \sin(\sqrt \lambda y), &\text{if } \lambda >0 \\ y, &\text{if } \lambda = 0 \\ \sinh(\sqrt{-\lambda } y), &\text{if } \lambda <0 \end{cases} $$ satisfy the eigenvalue problem in the halfplane. Hence, $\lambda $ is an eigenvalue for all $ \lambda \in \mathbb R$.

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You might argue that none of the above solutions are in $H^1_0(\mathbb R^2_+)$. However, if $u$ solves the eigenvalue problem and is $ H^1_0(\mathbb R^2_+)$ then I believe it is possible to prove that $u \equiv 0$ in $\mathbb R^2_+$. I don't have time right now but I'll update this post in the future with a proof of this.