Stable standing waves of nonlinear Schrödinger equations with potentials and general nonlinearities

Abstract

The existence and nonexistence of the minimizer of the $L^2$-constraint minimization problem $e(\alpha ):=\inf \{ E(u)|\ u\in H^1(\mathbb {R}^N),\ \left\| u\right\| _{L^2(\mathbb {R}^N)}^2=\alpha \}$ are studied. Here,

$$\begin{aligned} E(u):=\frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2+V(x) |u|^2dx-\int _{\mathbb {R}^N}F(|u|)dx, \end{aligned}$$

$V(x)\in C(\mathbb {R}^N)$, $0\not \equiv V(x)\le 0$, $V(x)\rightarrow 0$ ($|x|\rightarrow \infty$) and $F(s) = \int _0^s f(t) dt$ is a rather general nonlinearity. We show that there exists $\alpha _0\ge 0$ such that $e(\alpha )$ is attained for $\alpha >\alpha _0$ and $e(\alpha )$ is not attained for $0<\alpha <\alpha _0$. We study differences between the cases $V(x)\not \equiv 0$ and $V(x)\equiv 0$, and obtain sufficient conditions for $\alpha _0=0$. In particular, if $N=1,2$, then $\alpha _0=0$, and hence $e(\alpha )$ is attained for all $\alpha >0$.

Appendix A: Nonexistence of minimizer

We consider the following case:

• (V4) $0\not \equiv V(x)\ge 0$ and $\lim _{|x|\rightarrow \infty }V(x)=0$.

Theorem A.1

Suppose (V4) and the following (F11):

• (F11) $f(s)\le f(|s|)$ for $s\in \mathbb {R}$, $f(s)\ge 0$ for $s\ge 0$, $|f(s)|\le C(|s|+|s|^{{p_\mathrm{c}}})$, $\lim _{s\rightarrow \infty }f(s)/s^{{p_\mathrm{c}}}=0$.

Then $e(\alpha )=e_{\infty }(\alpha )$ for $\alpha \ge 0$ and $e(\alpha )$ is not attained for $\alpha >0$.

The assumption (F11) is weaker than (F1)–(F5).

Proof

First, we show that $e(\alpha )=e_{\infty }(\alpha )$. Since $V(x)\ge 0$, we see that $e(\alpha )\ge e_{\infty }(\alpha )$. On the other hand, for any $u\in M(\alpha )$ and $n\in \mathbb {N}$, we obtain

$$\begin{aligned} e(\alpha )\le E(u(\,\cdot \,-n\mathbf{e}_1))=E_{\infty }(u)+\frac{1}{2}\int _{\mathbb {R}^N}V(x+n\mathbf{e}_1)|u|^2dx. \end{aligned}$$

Letting $n\rightarrow \infty$, we obtain $e(\alpha )\le E_{\infty }(u)$. Since u is arbitrary, we see that $e(\alpha )\le e_{\infty }(\alpha )$. Thus, $e(\alpha )=e_{\infty }(\alpha )$.

Second, we show by contradiction that $e(\alpha )$ is not attained. Suppose on the contrary that $e(\alpha )$ is attained by $u_0\in H\cap M(\alpha )$. By Remark 2.7, we may assume $u_0 \ge 0$. Since $E \in C^1( H_\mathbb {R},\mathbb {R})$ due to (F11), there exists a $\lambda \in \mathbb {R}$ such that $-\Delta u_0 + (V(x) + 2 \lambda )_+ u_0 \ge -\Delta u_0+(V(x)+2\lambda )u_0=f(u_0)\ge 0$ in $\mathbb {R}^N$. Thus, the weak Harnack inequality [14, Theorem 8.18] yields $u_0>0$ in $\mathbb {R}^N$. Using this fact and $0\not \equiv V(x)\ge 0$, we obtain

$$\begin{aligned} e(\alpha )=E(u_0)=E_{\infty }(u_0)+\frac{1}{2}\int _{\mathbb {R}^N}V(x)u_0^2dx>E_{\infty }(u_0)\ge e_{\infty }(\alpha ). \end{aligned}$$

This is a contradiction, because $e(\alpha )=e_{\infty }(\alpha )$. Therefore, $e(\alpha )$ has no minimizer. $\square$