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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?

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  • $\begingroup$ 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. $\endgroup$ – MaoWao Jan 24 at 10:28
  • $\begingroup$ @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? $\endgroup$ – Mundron Schmidt Jan 25 at 21:58
  • $\begingroup$ Yes, that is correct. $\endgroup$ – MaoWao Jan 31 at 10:11

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