# Eigenvalues and eigenfunctions of the Laplace operator on entire plane

According to the answers in the the following questions: How to prove the spectrum of the Laplace operator? and What is spectrum for Laplacian in $$\mathbb{R}^n$$ , the spectrum of the Laplace operator $$\Delta :H^2(\mathbb{R}^2)\subset L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$$ is in fact $$\sigma(\Delta)=(-\infty,0].$$ However, I was not able to find a discussion on the eigenvalues of $$\Delta$$.

The set of eigenvalues $$\sigma_p(\Delta)$$ (also called point spectrum) is known to be contained in $$\sigma(\Delta)$$ and one can have $$\sigma_p(\Delta)\subsetneq \sigma(\Delta)$$. Indeed, by taking the Fourier transform $$\mathcal{F}:L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$$ of the eigenvalue problem one has $$\Delta u(x) = \lambda u(x),\;\;\forall x\in \Bbb R^2 \;\;\;\overset{\mathcal F}{\longrightarrow}\;\;\;\;-4\pi^2|\xi|^2\hat u(\xi) =\lambda \hat u(\xi), \;\;\;\forall\xi\in \Bbb R^2,$$ and this can only be satisfied by $$\hat u=u=0$$. This means that the only eigenvalue-eigenvector pair in this setting is $$(\lambda,u)=(0,0)$$ . Also, the same argument applies when $$\Delta$$ is seen as $$\Delta:W^{m,p}(\mathbb{R}^2)\subset L^p(\mathbb{R}^2)\to L^p(\mathbb{R}^2)$$ with $$p\in [1,2)$$ and $$\mathcal F:L^p(\mathbb{R}^2)\to L^{p^*}(\mathbb{R}^2)$$ with $$1/p+1/p^*=1$$.

Question 1. What happens when $$p>2$$ and the Fourier transform becomes distribution valued, so that the above elementary argument cannot be applied directly?

It seems if $$u\in C^2(\mathbb{R}^2)$$ is in fact an eigenvalue of $$\Delta$$, then it cannot be in $$L^p(\mathbb{R}^2)$$ for any $$p\in [1,2]$$. Depending on the answer to Q1, this might also hold for $$p>2$$. In any case, it seems that the $$L^p$$ framework is not suitable for this problem.

Question 2. On what space(s) could one define the domain of $$\Delta$$ to obtain non-trivial eigenvalues?

Edit. The crossed out sentence should be replaced by: "The point spectrum $$\sigma_p(\Delta)$$ is therefore empty."

The point spectrum coincides with the spectrum minus 0 if $$p>2n/(n-1)$$ and it is empty in the remaining cases ($$n$$ is the dimension). This is proved in G. Talenti: "Spectrum of the Laplace operator acting in $$L^p(R^n)$$", Indam, Symposia Mathematica vol VII, Academic Press 1971.
• Thank you very much for the reference. However, it remains to show that for $p>2n/(n-2)$ (so in our case $p>4$) and $\Delta:W^{m,p}\to L^p$ , the spectrum is still given by $\sigma(\Delta)=(\infty,0]$. I have a proof of this using the Fourier transform that works, but only for $1\leq p\leq 2$. However, in our case, we need to study the spectrum for $p>4$! Jan 29 at 10:31
• @JochenGlueck Jochen you are right, the argument works for the spectrum minus 0. Of course, there is a way of avoiding semigroups. For $l\lambda$ not a negative number (and not 0), write formally the inverse through the Fourier transform and then use Mikhlin to check the boundedness. If $\lambda<0$ one shows it is an approximate eigenvlaue (the functions $e^{iax}$ should be an approximate eigenvector). For the point spectrum, one first show that if there is an eigenfuntions, then there is a radial one, by averaging, and then ends with a Bessel equation whose asymptotic is known. Jan 29 at 12:26
• @JochenGlueck neither was I and this is actually a remarkable property! On another another, I think I made a mistake by calling the trivial pair $(\lambda, \hat u)=(0,0)$ an eingenpair. Jan 29 at 12:51
• @JochenGlueck Since you are interested I go on. If $\Delta u+k^2 u=0$ ($K \neq 0$) and $u$ is radial, then setting $u(r)=r^{(1-n)/2}w(r)$ you get a Bessel equation whose solutions at infinity oscillate lile $\sin$. Then the asympotics for $v$ at inifinity is like $r^{(1-n)/2}$ which is in $L^p$ for $p>2n(n-1)$. Jan 29 at 12:52