Exercise on Laplacian Operator plus Constant

Question

$Let\;L:=\Delta + c\;where\;c:=constant\;and\;U\;an\;open\;and\;bounded\;subset\;of\;\mathbb R^3)$

$Prove\;for\;any\;solution\;u\;of\;Lu=0\;\;(where\;u\in \mathcal C^2(\bar U))\; that \; if\;u=0\;on\;\partial U\; then\; u=0\;everywhere\;in\;U\;when\;c \lt 0.\\Furthermore\;show\;when\;c \gt 0\; that\;there\;are\;solutions\;of\;L\;which\;are\;equal\;to\;0\;on\;a\;sphere\;but\;not\;at\;its\;interior$

Well,I'm trying to solve the above problem but I've been stuck!

I 've only proven that $u(y)=\frac{\sqrt c r}{sin(\sqrt c r)} \frac{1}{4 πr^2} {\int}_{\vert x-y \vert =r} u(x) \;dS_x\;$ where $u\;$ is a solution of $Lu=0\;$ when $sin(\sqrt c r) \neq 0.$ Since this reminds me enough, the mean value theorem, I thought to use some kind of "maximum principle" but I have no idea how to proceed with this operator.

I would appreciate if somebody could help me through this.. Hints of course are welcome!

Thanks in advance!!

Answer 1

If $\Delta u + cu=0$ with $c<0$, then both $\Delta u$ and $u$ have the same sign. Suppose $u$ has a point of local maximum in $\Omega$; then the second derivatives are nonpositive there, hence $\Delta u\le 0$. This shows that $u$ cannot have a strictly positive maximum. Similarly, it cannot have a strictly negative minimum.

To find a nontrivial solution on a ball when $c>0$, one can separate the variables in spherical coordinates and study the resulting ODE. Which is something that was done already, so I would look up spherical Bessel functions and pick the sumplest one: in spherical coordinates, $u(\rho) = \sin(\rho )/\rho$ satisfies $\Delta u + u=0$ (to be verified by a direct computation), and vanishes on the sphere $\rho=\pi$.