MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\bar{u} \not\equiv 0$?

share|cite|improve this question
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 King May 3 '11 at 21:18

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.

share|cite|improve this answer
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 King May 3 '11 at 21:57
I don't actually see why the convergence should be strong! – Mercy King 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 King 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 King May 5 '11 at 10:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.