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For example, for $$ -\Delta u=f(u) \quad \text { in } \Omega, $$ we call a solution is stable if $$ Q_u(\varphi):=\int_{\Omega}|\nabla \varphi|^2 d x-\int_{\Omega} f^{\prime}(u) \varphi^2 d x \geq 0, \quad \forall \varphi \in C_c^1(\Omega) . $$ Here $Q_u(\varphi)$ is the second variation of its energy functional.

And the Morse index is defined by the maximal dimension of space $K_\varphi$ in which $Q_u(\varphi) <0$.

I want to ask a naive question about the stable solution. $Q_u(\varphi) \ge0$ means local minimum, the Morse index is defined by the maximal dimension of space $K_\varphi$ in which $Q_u(\varphi) <0$, if we consider infinite Morse index, it should be possible that $Q_u(\varphi) <0, \forall \varphi \in C_c^1(\Omega)$, so a local maximum. Is there something I misunderstood, thank you!

My thoughts: For the questions that we have proved no stable solution (so no local minimum), maybe we should set $v=-u$ and study the $$ -\Delta v=-f(-v) \quad \text { in } \Omega $$ again. I think the method should be the same, so local minimum of $u$ is the local maximum of $v$, local maximum of $u$ is the local minimum of $v$, so no local maximum for $u$?

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    $\begingroup$ Can you rephrase precisely the question you would like to know the answer to? I am not sure I understand what you are asking for. In any case, $Q_u(\varphi) \ge 0$ (that is, stable solution) does not mean local minimum. Even for functions of one variable $x\mapsto -x^{100}$ has a critical point at $x=0$ which is "stable", meaning has second derivative equal to zero. $\endgroup$
    – Michele Caselli
    Commented Oct 4, 2023 at 10:09
  • $\begingroup$ Thanks for your reply! You mean that $Q_u(\varphi) \ge 0$ doesn't imply that $u$ is the local minimum of the functional corresponding to the PDE? Because I have always understood the concept this way, ($Q_u(\varphi) \ge 0$ means $u$ is local minimum, $Q_u(\varphi) \le 0$ means local maximum, $Q_u(\varphi) \ge 0$ for some $\varphi$ and $Q_u(\varphi) < 0$ for some $\varphi$ means saddle point.) So I'm confused that the stable solution only means local minimum, Morse index is defined by the maximal dimension of space in which $Q_u(\varphi)<0$. so infinite index solution could be local maximum? $\endgroup$
    – Elio Li
    Commented Oct 5, 2023 at 4:52
  • $\begingroup$ I think I may misunderstood some ideas, because there are examples of multivariable functions whose Hessian is $0$ on $p$ but $p$ is a saddle point, for example $z = x^4 − y^4$ on $p=(0,0)$. So $Q_u(\varphi) \geq 0$ doesn't mean local minimum. But I can't come up with an example. $\endgroup$
    – Elio Li
    Commented Oct 5, 2023 at 5:27
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    $\begingroup$ Stable solution just means that (at a critical point $u$) you cannot decrease the energy at second-order, this does not mean you increase it! Not only a stable solution does not have to be a local minimum, it could be a local maximum (e.g. $f(x)=-x^{100}$ on $\mathbb{R}$). $\endgroup$
    – Michele Caselli
    Commented Oct 5, 2023 at 9:20
  • $\begingroup$ Thanks for your comment. The example for the stable points of functions being local maximum or saddle points is natural, but the examples of stable solutions being local maximum or saddle point of functionals are a bit unusual for me, do you know some classical examples? $\endgroup$
    – Elio Li
    Commented Oct 5, 2023 at 18:54

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