Now I have a nonlinear elliptic PDE, if I do scaling such as $u=\lambda v$ and then get a new PDE, then I turn to study the new equation since it has a simpler form. My question is easy to understand: if I prove that for the new PDE, all the solutions are not the local minimum of the energy functional (of the new PDE), can we say that all the solution of the PDE before scaling are also not the local minimum of the energy functional (of the old PDE)?
I have checked their second derivations, they are not the same since we have the scaling. But they are basically the same question, I think it's natural to ask whether the type of critical points will be the same as well?
Ok, after seeing some of the comments, I think the answer to your question is yes.
Let $$ I[v] = \int_\Omega L(\nabla u, u,x)\, dx $$ be some functional which we’d like to minimise over a set of admissible functions $\mathcal A$. My interpretation of your question is: if there are no stable local minimums of $I$ if and only if there are no stable local minimums of $I_\lambda$ with $I_\lambda [v]:=I[\lambda v]$ and $\lambda >0$. (Correct me if this is wrong)
For simplicity, assume that $\mathcal A$ is a vector space (often this is the case). Then $u$ is a local minimum of $I_\lambda$ if $$ \frac{d}{d\varepsilon}\bigg \vert_{\varepsilon=0} I_\lambda [u+\varepsilon \eta]=0 \text{ and } \frac{d^2}{d\varepsilon^2}\bigg \vert_{\varepsilon=0} I_\lambda [u+\varepsilon \eta]\geq 0 $$ for all $\eta \in \mathcal A$. But clearly, $$\frac{d}{d\varepsilon}\bigg \vert_{\varepsilon=0} I_\lambda [u+\varepsilon \eta]= \lambda\frac{d}{d\varepsilon}\bigg \vert_{\varepsilon=0} I[\lambda u+\varepsilon \eta] \text{ and } \frac{d^2}{d\varepsilon^2}\bigg \vert_{\varepsilon=0} I_\lambda [u+\varepsilon \eta]=\lambda^2 \frac{d^2}{d\varepsilon^2}\bigg \vert_{\varepsilon=0} I [\lambda u+\varepsilon \eta] ,$$ so $u$ is a local minimum of $I_\lambda$ if and only if $\lambda u$ is a local minimum of $I$.