# About positive eigenvectors of $-\Delta$ on $\mathbb R^N$

Im interested whether there exists positive solutions of $$(*)\begin{cases} -\Delta u=\lambda u\\ u\in H^1(\mathbb R^N) \end{cases}$$ for some $$\lambda\in\mathbb R$$. Here, we hve $$-\Delta=-\sum\partial_i^2$$.

On the one hand, it es well known that in case $$N=1$$ the equation $$-u''=\lambda u$$ has the general solution $$\begin{cases} a\sin(\sqrt{\lambda}x)+b\cos(\sqrt{\lambda}x) & \text{ if }\lambda>0\\ at+b & \text{ if }\lambda=0\\ ae^{\sqrt{|\lambda|}x}+be^{-\sqrt{|\lambda|}x} & \text{ if }\lambda<0 \end{cases}$$ But obviously non of these are square integrable. So I claimed

There is no nontrivial and positive solution of $$(*)$$ for $$N\leq 4$$.

I would appreciate if you can verify if my proof it correct or point out if I missed a point.

Proof:

Let be $$\lambda\geq 0$$ and $$u\in H^1(\mathbb R^N)$$ a positive solution of $$(*)$$. Then we get $$-\Delta u=\lambda u\geq 0$$. In this case, there are theorems which claims for $$N\geq 4$$ that $$u$$ has to be constantly zero.

So, we have to consider the case $$\lambda<0$$. A solution of $$(*)$$ is a critical point of the function $$I(u):=\frac12|\nabla u|_2^2-\frac{\lambda}2|u|_2^2$$, where $$|~\cdot~|_2$$ denotes the $$L^2(\mathbb R^N)$$-norm. But then we get $$0=\nabla I(u)[u]=|\nabla u|_2^2-\lambda|u|_2^2=|\nabla u|_2^2+|\lambda|\cdot|u|_2^2.$$ But this yields $$u\equiv 0$$.

End of the proof

Ok, how about the eigenvectors of $$-\Delta$$ for $$N\geq 2$$? The conclusion is, that the eigenvectors of $$-\Delta$$ have either sign changes or they are not in $$L^2(\mathbb R^N)$$. Are there some eigenvectors explicitly known?

• The Laplacian has no eigenfunctions in $L^2$. This can be easily verified using Fourier transform. See also this closely related question: math.stackexchange.com/questions/790401/… and probably many others on this site. – MaoWao Jan 24 at 10:28
• @MaoWao Nice, I see. So, assume $f$ is an $L^2$ eigenfunction to an eigenvalue $\lambda\in\mathbb C$, then using Fourier transformation gives $(|\xi|^2-\lambda)\hat f=0$, which implies $\hat f=0$ and then $f=0$. Is it correct? – Mundron Schmidt Jan 25 at 21:58
• Yes, that is correct. – MaoWao Jan 31 at 10:11