# Showing a solution of elliptic PDe is non-degenerate

Dear Mathoverflowers:

I am interested in radial positive solutions of $-\Delta u(r) = r^\alpha u(r)^p$ in the unit ball in $R^N$ with $u=0$ on the boundary.

Here $p>1$ and $\alpha >0$. (There is a positive solution provided $p<\frac{N+2+2\alpha}{N-2}$, Ni 82).

I am interested in when I can say the associated linearized operator $L:= -\Delta - p r^\alpha u(r)^{p-1}$ does not have zero as an eigenvalue.

Are there any standard methods for attempting to show this?

craig

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Since $p>1$ if you know that $u(r)\leq 1$ then $u(r)$ is a supersolution to your linearized operator $L$ (technically $u(r)$ is a supersolution to $-L$). In this case, since $L$ is a Schrodinger operator there is a classical result of Fischer-Colbrie and Schoen * that the operator $L$ has non-negative spectrum (with positivity unless $u$ is a solution to $L$). Of course I don't know if $u(r)\leq 1$ is plausible... *"The structure of complete stable minimal surfaces in 3-manifolds of nonnegative scalar curvature" Comm. Pure Appl. Math. Vol. 30, 199-211 1980. – Rbega Dec 12 '11 at 20:27
I must be missing something but I don't see where the $u(r) \le 1$ is helping anything. As mentioned below I know the first eigenvalue of $L$ is negative, so i think this will cause a problem. I will dig up reference. Thanks. – greg Dec 13 '11 at 1:52
You're right I was a little to quick with my computation you would need $u\geq p$ to get a super solution, which is of course ridiculous. – Rbega Dec 13 '11 at 2:36

$L$ is invertible if $pr^\alpha u(r)^{p-1}$ is small in a certain sense, cf. Gilbrag-Trudinger. In the general case I don't know anything except to mention the Fredholm or spectral theories. If you need a result for a particular $u$ (and $\alpha$ and $p$) perhaps there are some rigorous numerical methods. I would be interested in other answers.

Update: One potential approach would be to show first that any solution must be radially symmetric and then use ODE methods to the equation obtained.

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Thanks for response. I believe here you are refering to the result that says if the $L^\frac{N}{2}$ norm of the quantity in question is small with regards to Sobolev imbedding constant, then $L$ has the maximum principle? In any case I don't think this will work since the first eigenvalue of $L$ is negative. I will look into your other commments. Thanks again. – greg Dec 13 '11 at 1:46
The first eigenvalue of $L=-\Delta-V$ is not negative, if $V$ is small enough. There is a smallness condition involving the width of the domain etc. Intuitively, the first eigenvalue of $-\Delta$ is strictly positive, so one still has a room for subtracting something small. – timur Dec 13 '11 at 3:18
I meant in the particular case where $L = -\Delta - r^\alpha u(r)^{p-1}$ and where $u$ is a positive solution of the given PDE. If the first eigenvalue was non-negative then one would obtain $\int | \nabla \psi|^2 \ge \int p r^\alpha u^{p-1} \psi^2$ for all $\psi$ which are zero on the boundary. Putting $\psi=u$ into this inequality would show that $u=0$. – greg Dec 13 '11 at 3:40
@greg, have you looked at ODE methods? – timur Dec 13 '11 at 13:57
I have looked at ODE methods in the sense that I have played around with radial functions and taken derivatives and....but no real explicit radial methods... What I suspect is that there is no radial solution $v$ of $L(v)=0$. The reason I think this is that in this case a change of variables can remove the term $r^\alpha$ from all the equations, and then there are known results. Let me explain where this is coming from, which i do below, since maybe I don't even need what I am asking for. – greg Dec 13 '11 at 14:16

Here I explain where the original problem is coming from, since it is quite possible I don't need what I am asking for.

I am interested in obtaining positive solutions of the equation $-\Delta u(x) + t H(u) = g(t,x) u^p$ in $B$ with $u=0$ on the boundary of $B$. Here $H$ is some second order linear elliptic operator, $t \in R$. My interest is in obtaining positive solutions $u$ for a various values of $t$ close to zero (I don't care if I can only get solutions for $t$ on one side of zero. Also $g(0,x) =|x|^\alpha$. Also $p$ is in general supercritical. So what I was attempting to do what perturb off of the radial positive solution using the implicit function theorem, and this was why I needed the non-degeneracy. But I would be happy to use another abstract result that would allow me to obtain solutions by perturbing off of the radial solution.