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I was trying to read the paper "Über das asymptotische Verhalten der Lösungen von $\Delta u+\lambda u =0$ in unendlichen Gebieten" by Franz Rellich (MR17816, Zbl 0028.16401).

  • Since it is in German is there any English version available?
  • Also one of the key results of paper seems to be the non-existence of $L^2$ solution of the equation $\Delta f=-\lambda f$ in the set $\{x\in\mathbb{R}^n:|x|>R_0\}$?.

I am just curious to know whether this holds true for any other smooth Riemannian manifold or for eigenfunctions of some other operator other than a Laplacian? Any intuition/comment will also be very helpful.

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  • $\begingroup$ Unfortunately I do not read german, too. Having a look at the proof I think it is the usual one (maybe he was the first) using expansion in spherical harmonics and asymptotic behavior of Bessel functions. $\endgroup$ Commented Sep 12 at 12:55
  • $\begingroup$ And any comment regarding the last question? $\endgroup$
    – Emmie
    Commented Sep 12 at 14:16
  • $\begingroup$ I do not know and looks delicate. The result is false for $\lambda=0$ where one can construct solutions tending to 0 at infinty as any power. $\endgroup$ Commented Sep 12 at 14:29

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This is more a long comment than an answer, to show that $L^2$ solutions may exist in some cases. A simple example is $D^2+D+k^2$ in 1d which has solution $e^{-\alpha x}$ with $2\alpha=1+ \sqrt{1-4k^2}$ ($x >0)$. This is due to the drift term. A more involved example is for $|x| \geq 1$ the pure second order operator $$ L=\Delta+a \sum_{i,j}\frac{x_ix_j}{|x|^2}D_{ij} \quad a>-1. $$ If $u$ is radial and $Lu+k^2 u=0$ then $(1+a)D_{rr}u+\frac{n-1}{r}D_r u +k^2 u=0$ or $D_{rr}u+\frac{n-1}{(1+a)r}D_r u +\lambda^2 u=0$ with $\lambda^2=k^2/(1+a)$. This is reduced to a Bessel type equation as for the Laplacian writing $u=r^{\frac{1-n}{2(1+a)}}v$ and $v$ is asymptotic to a $\sin r$ for large $r$ since $\lambda^2>0$. This gives $$u^2 \leq \frac{C}{r^{\frac{n-1}{1+a}}}$$ so that $u \in L^2$ at infinity if $-1<a<0$ is properly chosen.

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  • $\begingroup$ True. Again sorry for asking but can one expect a similar sort of result at least for bi-laplacian which is a higher-order generalisation of laplacian. I actually couldn't read the above-mentioned paper as it's in German but in such eigenvalue problems, one way people do an estimate is by lifting it to a harmonic function by adding one extra variable. As that seems not possible even for bi-Laplacian, I don't know whether there are any bypass of lifting techniques for other operators. Again if you have any idea, any comment would be helpful. $\endgroup$
    – Emmie
    Commented Sep 12 at 17:40
  • $\begingroup$ I did not try the trick of lifting, could work. For higher order operaots I do not know but radial computations could work for the bilaplacian....with a lot of effort. $\endgroup$ Commented Sep 12 at 17:52
  • $\begingroup$ Actually, I was thinking about it yesterday, the issue is this kind of function $e^{-\lambda t}f(x)$ where $(x,t)\in \mathbb{R}^n\times(0,\infty)$ which works for eigenvalue problem for laplacian to lift it to harmonic function doesn't work for higher-order case. So, I am not sure what then could be possible candidates for lifting actually exponential is a nice candidate as it's always non-negative but I am not sure what could be possible candidate for lifting in this case? $\endgroup$
    – Emmie
    Commented Sep 13 at 8:49
  • $\begingroup$ As you mentioned that trick of lifting could work for higher order, any idea what would be a possible candidate for lifting? It seems $e^{\lambda t}f(x)$ kind of functions not going to work. $\endgroup$
    – Emmie
    Commented Sep 18 at 9:39
  • $\begingroup$ I do not know how to use lifting for the problem you asked, even for the Laplacian. The exponential factor can be very large at infinity, or one shoud restrict on a strip. Do you know how to use in the simplest case? For higher order one could use a complex lambda, for example $\lambda^4=-k^2$ $\endgroup$ Commented Sep 18 at 12:06

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