Timeline for A naive question about the stable solution and Morse index of elliptic PDE
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 5, 2023 at 20:09 | comment | added | Elio Li | Sorry for the unclear question, I rephrase a bit and mention some examples here. mathoverflow.net/questions/455946/… | |
| Oct 5, 2023 at 19:03 | comment | added | Elio Li | Besides, I'm still confused about some points. Stable solution can be maximum point, minimum point or saddle point. If $Q_u(\varphi) < 0 $ strictly, will it be enough to show that $u$ is a local maximum? If so, if for any $\varphi \in C_c^1(\Omega)$, $Q_u(\varphi) < 0 $, which means infinite Morse index, and $u$ is the local maximum. | |
| Oct 5, 2023 at 18:54 | comment | added | Elio Li | 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? | |
| Oct 5, 2023 at 9:20 | comment | added | Michele Caselli | 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 $\mathbb{R}$). | |
| Oct 5, 2023 at 5:27 | comment | added | Elio Li | 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. | |
| Oct 5, 2023 at 4:52 | comment | added | Elio Li | 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? | |
| Oct 4, 2023 at 10:09 | comment | added | Michele Caselli | 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. | |
| Oct 4, 2023 at 5:13 | history | edited | Elio Li | CC BY-SA 4.0 |
added 80 characters in body
|
| Oct 4, 2023 at 4:37 | history | edited | Elio Li | CC BY-SA 4.0 |
added 291 characters in body
|
| Oct 3, 2023 at 4:24 | history | asked | Elio Li | CC BY-SA 4.0 |