Questions about the results about $\Delta u + e^u=0$, $3 \le n \le 9$: no finite Morse index solution, $n \ge 10$: stable radial symmetric solution

Question

I just read the celebrated paper Farina and Dancer, which talks about the following PDE in $\mathbb{R}^n$　 $$\Delta u + e^u=0.$$ They proved that when $3 \le n \le 9$, there is no finite Morse index solution, and they remark that when $n \ge 10$, there is a stable radial solution with finite Morse index, they refer to Joseph and Lundgren, but this paper talked about the equation on radial symmetric bounded domain so they directly transferred the equation into ODE. I wonder how Joseph and Lundgren implies the result they mentioned? Actually, every paper I found just remark the result and refer to other papers, if you know where I can find a proof with more details, I would be very glad!

Answer 1

It is sketched how to do so in Section 2 of E.N. Dancer - Finite Morse index solutions of exponential problems. As said there, the proof is essentially identical to the one of Theorem 2 in E.N. Dancer - Stable solutions on $\mathbb{R}^n$ and the primary branch of some non-self-adjoint convex problems. I write the argument here in a little more details. There is only one detail that makes the proof not just a few lines, I emphasise it below.

Let $n\ge 10$ and $\lambda < \lambda_*$, where $\lambda_*=\lambda_*(n)>0$ is the critical $\lambda$ as in Joseph-Lundgren. For every $\lambda<\lambda_*$ there exists a smooth nonnegative solution $u_\lambda$ of $$ \cases{-\Delta u = \lambda e^u \hspace{5pt} in \hspace{3pt} B_1(0) \,, \\ u=0 \hspace{5pt} on \hspace{3pt} \partial B_1(0) \,.}$$ Moreover $0 < M_\lambda:=\max_{B_1(0)} u_\lambda \to + \infty$ as $\lambda \nearrow \lambda_*$. Let $x_\lambda \in B_1(0) $ be the point at which the maximum $M_\lambda$ is attained. Then the sequence $u_{\lambda,\rho}(x):=u_\lambda(\rho (x-x_\lambda))- M_\lambda$ has $u_{\lambda, \rho}(x_\lambda)=0$, $u_{\lambda, \rho} \le 0$ and solves $$ -\Delta u_{\lambda, \rho} = \rho^2 \lambda e^{M_\lambda} e^{u_{\lambda, \rho}} $$ in $\Omega_{\lambda, \rho}:=\frac{1}{\rho}(B_1(0)-x_\lambda)$. Now, as $\lambda \nearrow \lambda_*$ since $M_\lambda \to +\infty$ we see that choosing $\rho= (\lambda e^{M_\lambda})^{-1/2} \to 0$ gives a solution to $-\Delta u = e^u$ in $\Omega_{\lambda}$, where $ \Omega_{\lambda} $ is $\Omega_{\lambda, \rho}$ with our specific choice of $\rho$. Now we come to the only point that requires some care.

If $\Omega_{\lambda} \nearrow \mathbb{R}^n$ by elliptic regularity on compact subsets you get a sub sequential limit that converge to the desired solution in $\mathbb{R}^n$. Moreover, since stability is a property of the solution in compact subsets (as it is tested against functions with compact support), it passes to the limit and the limit solution is also stable.

So, really the only thing to rule out is that $\Omega_\lambda$ converge to a half space instead of $\mathbb{R}^n$. This happens if $x_\lambda$ approaches $\partial B_1(0)$ faster (or at the same rate) than the rate $\rho$ at which we are blowing up. Indeed, you cannot converge to the half-space problem and the details are in the proof of Theorem 2 in the second paper by Dancer I cited above.

Answer 2

Set $$u_{\lambda,\rho}=u_\lambda(\rho(x+\frac{x_\lambda}{\rho}))-M_\lambda,$$ so $\rho(x+\frac{x_\lambda}{\rho}) \in B_1(0)$, $x \in \frac{1}{\rho}(B_1(0)-x_\lambda)=\Omega_\lambda$. Then we get $$-\Delta u_{\lambda, \rho}=\rho^2 \lambda e^{M_\lambda} e^{u_{ \lambda, \rho}}.$$ $\rho=\left(\lambda e^{M \lambda}\right)^{-1 / 2} \rightarrow 0$ gives a solution to $-\Delta u=e^u$ in $\Omega_\lambda$. And most importantly, $u_{\lambda, \rho}(0)=0$, $u_{\lambda, \rho}(0) \le 0$ in $\Omega_\lambda$. Denote $\widetilde{\Omega}$ as $\Omega_\lambda$ when $\lambda \rightarrow \lambda^{*}$. We assume that $d(0, \partial \widetilde{\Omega})$ is not bounded. Then, given a fixed ball $B_R(0)$, consider $-\Delta w=e^{u_{\lambda, \rho}}$ on $B_R(0)$ with $w=0$ on the boundary, since $e^{u_{\lambda, \rho}}$ is uniformly bounded for any $\rho$, $w$ is also uniformly bounded, we then add a constant (independent on $\rho$) to make $w >0$.

So on $B_R(0)$, ($w-u_{\lambda, \rho}$) is a positive solution of $\Delta f = 0$, then notice that $（w-u_{\lambda, \rho})(0)$ is uniformly bounded for any $\rho$, then by harnack inequality, $w-u_{\lambda, \rho}$ is uniformly bounded in $B_{R/2}(0)$, so $u_{\lambda, \rho}$ is uniformly bounded. Then take the subsequence of {$\rho \rightarrow 0$}, denote it as {$\rho_R$}, then $u_{\lambda, \rho_R} \rightarrow u_{R}$ on $B_{R/2}(0)$. Then on a bigger ball $B_{R_1}(0)$, take the subsequence of {$\rho_R$}, denote it as {$\rho_{R_1}$}, then $u_{\lambda, \rho_{R_1}} \rightarrow u_{R_1}$ on $B_{R_1/2}(0)$, and $u_{R_1}$ on $B_{R/2}$ is $u_R$, then go on with the exhaustion to infinity to get a solution of $$ -\Delta u=e^u $$ in whole $\mathbb{R}^N$.

When $d(0, \partial \widetilde{\Omega})$ is bounded, denote the point which is the most close to $0$ on $\partial\widetilde{\Omega}$ as $Z$. Choose a neighborhood of $Z$ on $\partial\widetilde{\Omega}$, denote it as $\partial N$, then move $\partial N$ along the normal vector at $Z$ to cover $0$, and denote this area as $N$. Rescale $u_\rho=-u_{\lambda, \rho} / M_{\lambda}$, then in $N_\rho$, $$M_\lambda \Delta u_\rho=1 / e^{M_\lambda u_\rho},$$ here $N_\rho$ is chosed by the same method as $N$, when $\rho \rightarrow 0$ (take subsequence), $u$ satisfies $\Delta u = 0$ in $N$, and $0 \le u \le 1$ in $N$, and $u=1$ on $\partial N$, $u=0$ inside $N$, which is against the minimum principal. So $d(0, \partial\widetilde{\Omega})$ can not be bounded.