not sure if this meets the standards here in this forum. I was dealing with the following functional $I(u)=\frac{1}{p}\int_{\Omega}|\nabla u|^{p}dx-\frac{1}{q}\lambda\int_{\Omega}|u|^qdx$ for $p \geq 2$ and $p-1<q<p^*$, $p^*=\frac{Np}{N-p}$, $p<N$, $\Omega$ is a bounded domain in $\mathbb{R}^N$. I was interested to check the Morse index of $I$. Towards this, I computed the double derivative which is as follows. \begin{eqnarray}I''(u)(w,w)&=\int_{\Omega}\{|\nabla u|^{p-2}|\nabla w|^2dx+2(p-2)|\nabla u|^{p-4}(\nabla u.\nabla w)^2\}dx\nonumber\\ &-\lambda \int_{\Omega}(q-1)|u|^{q-2}|w|^2dx.\end{eqnarray} I wasn't able to figure out what will be the largest dimension of a subspace of $W_0^{1,p}(\Omega)$ such that $I"(u)$ is nonnegative?. Is it at least possible to do so?.
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1$\begingroup$ same question on MSE math.stackexchange.com/questions/3483440/… $\endgroup$– Calvin KhorCommented Dec 21, 2019 at 8:32
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1$\begingroup$ You may consult the paper ''Morse index and uniqueness for positive solutions of radial $p$-Laplace equations" by Aftalion and Pacella in Trans of AMS (2004). $\endgroup$– GabSCommented Jan 7, 2020 at 16:54
1 Answer
I guess you mean "the largest dimension of a subspace of $W_0^{1,p}(\Omega)$ such that $I''(u)$ is negative"? Otherwise, the largest dimension equals infinity, due to the spectral theory of the linearized $p$-Laplacian, see, e.g., Theorem 1.1 in [Castorina, Esposito, Sciunzi; 2011].
Moreover, the choice of $W_0^{1,p}(\Omega)$ doesn't seem to be optimal for $p \neq 2$, see again [Castorina, Esposito, Sciunzi; 2011].
Finally, in general, the Morse index depends on $u$. E.g., if $u$ is a local minimizer of $I$ (i.e., in the case $p-1<q<p$), then the Morse index equals $0$. On the other hand, if $p=2$, $q>p$, and $\Omega$ is a ball, then the Morse index of the positive radial solution is $1$.