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Feb 2 at 7:37 comment added Elio Li Thanks for your reply, I get it. But this made me confused, I know some applications where the stable solution is used to model the burning or the explosion of the fuel, you mean that this concept is almost useless in the well-posedness of elliptic PDE? And if we prove that there is no stable solution, then the only conclusion we can draw is that if there are solutions, they must be the saddle point or local maximum of the energy functional, right?
Feb 1 at 16:09 comment added Giuseppe Negro I believe you should think in finite-dimensional terms. Consider a function $F\colon \mathbb R^2\to \mathbb R$. This is your "energy functional". Its critical points are the "solutions" and its Hessian matrix is the "second variation". You can make examples of this form for all your questions. For example, it is immediate to prove by these examples that knowing the sign of the second variation at critical points typically yields no information on the global behaviour of the energy functional.
Feb 1 at 12:59 comment added Liviu Nicolaescu A local minimum of the energy functional yields a stable solution. The functional need not have to be convex to have a local minimum. The stable solutions are the one "observable in real life". The existence of a stable solution does not help very much elicidating the shape of the energy functional.
Feb 1 at 12:35 history asked Elio Li CC BY-SA 4.0