Generalize the conception of 'stable' solution and 'stable outside a compact set' solution of elliptic PDE from $\mathbb{R}^n$ to closed manifold

Question

I want to talk about generalizing the conception of 'stable' solution and 'stable outside the compact set' solution of elliptic PDE from $\mathbb{R}^n$ to closed manifold.

I'm reading $\Delta u +e^u=0$ in $\mathbb{R}^n$, it said that, a solution $u \in C^2(\Omega )$ of $-\Delta u=\mathrm{e}^u$ is stable in $\Omega$ if: $$ \forall \psi \in C_c^1(\Omega) \quad Q_u(\psi):=\int_{\Omega}|\nabla \psi|^2-\mathrm{e}^u \psi^2 \geqslant 0 . $$ Here $Q_u$ is the second variation of the energy function corresponding to the PDE.

What I'm thinking is that for a PDE on closed manifold, can I also calculate like this? (Compute the second variation of the energy functional), because the manifold itself is compact, so we don't need the test function with compact support and the classical solution of the PDE must be bounded. If we compute like this and get that there is no stable solution, can we say that the possible solution is the saddle point of the energy functional?

If we can do like this, here come another question, when discussing the finite Morse index solution, people always studying the 'stable outside the compact set' solution, the definition is: $$ \forall \psi \in C_c^1(\Omega \backslash \mathcal{K}) \quad Q_u \ge 0$$ here $\mathcal{K} \in \Omega$ and $Q_u $ is the second variation.

Because the finite Morse index solution is stable outside a compact set, but for closd manifold it's hard to define 'outside a compact set', so for Morse index not equal to 0, we can only study its Morse index by studying its negative eigenvalue, right?

Answer 1

There are never smooth solutions to $-\Delta u = e^u$ on a closed manifold $M$, just integrate the PDE on $M$ and you get $\int_M e^u =0$ and this never happens.

For other type of PDEs that admit solutions you can for sure talk about stability on $M$ and having finite Morse index on $M$. It is not anymore true that "finite Morse index solution are stable outside a compact set" though. On $\mathbb{R}^n$ I think this is proved by contradiction since if they were unstable outside all balls $\{ B_k(0) \}_{k\in \mathbb{N}}$ then you would find infinitely many independent (disjoint support even) directions where the second variation is negative, i.e. the solution would have infinite index. Clearly this does not work on a closed $M$, since the exhaustion by balls in the contradiction argument eventually becomes equal to $M$.

But there are sort of local analogues of the same argument on manifolds. If you do the same argument with shrinking annuli $\{B_{1/2^k}(p) \setminus , B_{1/2^{k+1}}(p) \}_{k\in \mathbb{N}}$ you essentially find that index-$k$ solutions must locally stable everywhere outside $k$-points on $M$ where "all the instability" is concentrated.