# Convergence of mountain pass solutions of $-\Delta u+u=u|u|^{p-2}$

Consider the following equation in $\mathbb{R}^N, N \ge 3$: $$(E) \quad -\Delta u +u=|u|^{p-2}u,$$ where $2 < p < 2^{*} =2N/(N-2)$.

Denote by $J: H^1(\mathbb{R}^N) \to \mathbb{R}$ the functional that's naturally associated to (E), and by $J_k$ its restriction to $H^1_0(B_k(0))$, $k$ positive integer. Each $J_k$ yields a (non-trivial) mountain pass solution $u_k \in H^1_0(B_k)$, and the sequence $(u_k)$ is uniformly bounded in $H^1(\mathbb{R}^N)$. We may assume that $u_k$ converges weakly to some $\bar{u} \in H^1(\mathbb{R}^N)$.

Question: Is it true that $\bar{u} \not\equiv 0$?

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Evans only shows how the MPT works on a simple example. Note that one cannot apply the MPT to (E) due to the faillure of the PS condition. So this is really a different problem! –  Mercy May 3 '11 at 21:18

## 1 Answer

If I am not grossly mistaken, with your notations, set $c_k := \inf_{\gamma \in \Gamma_k}\max_{u \in \gamma([0,1])} J_k(u)$ (where $\Gamma_k$ is the set of paths $\gamma : [0,1] \to H^1_0(B_k(0))$ such that $\gamma(0) = 0$ and $\gamma(1) = \varphi$ for some fixed $\varphi \in H^1_0(B_1)$ such that $J(\varphi) <0$ and $\Vert\varphi\Vert_{H^1_{0}}$ is large enough. Denote also by $c_\infty$ the analogous mountain pass value for $k=\infty$

Since upon extending a function by zero one can consider that $H^1_0(B_k(0)) \subset H^1_0(B_{k+1}(0))$, one has also $\Gamma_k \subset \Gamma_{k+1}$ and thus $0 < c_\infty \leq c_{k+1} \leq c_k$. From this one may conclude that indeed $u_k \to {\overline u} \not\equiv 0$ and that $J({\overline u})=c_{\infty}$, so that ${\overline u}$ is a non trivial solution to equation (E).

I am adding the lines below after having read the comments.

Sorry for having been so\dots elliptic in my answer. I should have added the following details: first we choose $\varphi \geq 0$, so that the critical points $u_{k}$ are nonnegative. Then according to Gidas-Ni-Nirenberg's result $u_{k}$ has a spherical symmetry, that is $u_{k}$ is radial. Next, since $$\Vert \nabla u_{k}\Vert^2 + \Vert u_{k}\Vert^2 = \Vert u_{k}\Vert_{p}^{p} \lesssim \Vert \nabla u_{k}\Vert^{\theta p}\Vert u_{k}\Vert^{(1-\theta)p},$$ (using Gagliardo-Nirenberg inequality, for some $0<\theta<1$) then one deduces that there is $R_{0} >0$ such that for all $k \geq1$ we have $\Vert u_{k}\Vert_{p}^{p} = \Vert \nabla u_{k}\Vert^2 + \Vert u_{k}\Vert^2 \geq R_{0}^2$. Using the fact that the imbedding $H^1_{\rm rad}({\Bbb R}^n) \subset L^{p}({\Bbb R}^n)$ is compact, one infers that $u_{k}\to {\overline u}$ strongly in $L^{p}$, and weakly in $H^1$. In particular $\Vert {\overline u}\Vert_{p}^{p}\geq R_{0}^2$, and thus ${\overline u}\not\equiv 0$. Using the equation satisfied by $u_{k}$ and the strong convergence in $L^{p}$, one checks easily that ${\overline u}$ is solution to (E), and hence $$\Vert \nabla {\overline u}\Vert^2 + \Vert {\overline u}\Vert^2 = \Vert {\overline u}\Vert_{p}^{p},$$ yielding also that the convergence in $H^1$ is strong.

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Passing to the limit only gives that $J(\bar{u}) \le c_{\infty}$ by weak lower semi-continuity. Actually, if $\bar{u} \not\equiv 0$, then $J(\bar{u}) \ge c_{\infty}$. –  Mercy May 3 '11 at 21:57
I don't actually see why the convergence should be strong! –  Mercy May 3 '11 at 22:22
I'm sorry, I should have made my statement more precise. I am actually not interested in the radial case since it's well understood. Here are some details: ---- Let $G$ be the subgroup of $O(N)$ that leaves invariant the $x_N$-axis; then $G=O(N-1)\times\mathbb{Z}_2$. Define an action of $G$ on $H^1$ by $$\sigma.u=su\circ\sigma \quad \forall \sigma=(g,s) \in G.$$ Set $$E=\{u \in H^1| \quad \sigma.u=u \quad \forall \sigma \in G\},$$ and $$$E_k=\{u \in H^1_0(B_k(0))| \quad \sigma.u=u \quad \forall \sigma \in G\}.$$ Actually we look at$J$as a functional on$E$. So,$u_k \in E_k$. – Mercy May 4 '11 at 15:56 I see your point. In this case, instead of taking$\varphi$nonnegative one has to take it invariant under$G$, that is in the space$E$, and see whether the imbedding$E \subset L^p$is compact. Another approach would be to use P.L. Lions' concentration-compactness method in order to prove that the sequence$u_k \in E$does not vanish. – Otared Kavian May 4 '11 at 16:32 I have tried those two options: [1] the imbedding$E \subset L^p$isn't compact. Indeed, if$0 \not\equiv u \in E$, then the sequence$u_k=u(+ke_N)$, where$e_N=(0,\ldots, 0,1) \in \mathbb{R}^N$, is bounded but possesses no convergent subsequence. [2] If we use Lions' concentration-compactness principle, we end up with "a translanted" non-trivial limit$\bar{u}$which may not belong to$E\$. –  Mercy May 5 '11 at 10:29