Nonautonomous and non periodic Schrödinger equation with indefinite linear part

Abstract

The existence of solution of the nonlinear Schrödinger equation

$$\begin{aligned} \begin{array}{lc} -\Delta u + V(x) u = f(x,u),&\end{array} \end{aligned}$$

is stablished in $\mathbb {R}^N$, where V changes sign and f is an asymptotically linear function at infinity, with V and f non periodic in x. Spectral theory, a classical linking theorem and interaction between translated solutions of the problem at infinity are employed.

Appendix

Let $\partial B_1$ be the boundary of $B_1$, where $B_1$ is the open ball of radius 1 in a finite dimensional space spanned by the functions $u_0^+(\cdot -y),\phi _1,\ldots ,\phi _k$. We want to prove that

$$\begin{aligned} \displaystyle \lim _{R\rightarrow \infty } \displaystyle \int _{\mathbb {R}^N}\left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,Ru)}{(Ru)^2} \right) u^2dx = 0, \end{aligned}$$

uniformly for $u\in \partial B_1$. Indeed, for each $R=n\in \mathbb {N}$, consider $J_n:\partial B_1 \rightarrow \mathbb {R}$ the functional given by $J_n(u) = \displaystyle \int _{\mathbb {R}^N}\left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,nu)}{(nu)^2} \right) u^2dx$. The continuity of the function F shows that $J_n$ is a continuous functional for each fixed n. Hypothesis $(f_4)$ and equivalence of the norms $\Vert \cdot \Vert$ and $\Vert \cdot \Vert _E$ show that there exists a constant $C>0$ such that

$$\begin{aligned} 0 \le J_n(u)= \displaystyle \int _{\mathbb {R}^N}\left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,nu)}{(nu)^2} \right) u^2dx \le a_0\Vert u\Vert ^2_{E} \le C \end{aligned}$$

for all $u\in \partial B_1$, where $a_0=\sup _{\mathbb {R}^N} a(x)$. Hence the continuity of the functional $J_n$ in the compact set $\partial B_1$ ensures that, for each fixed n, the functional $J_n$ assumes its maximum at $u_n\in \partial B_1$. Consider $(u_n)$ the sequence of these maxima. Since $\Vert u_n\Vert =1$ for each n and the space spanned by the functions $u_0^+(\cdot -y),\phi _1,\ldots ,\phi _k$ is finite dimensional, there exists $\overline{u}\in \partial B_1$ such that, up to a subsequence,

$$\begin{aligned} u_n\displaystyle \rightarrow \overline{u} \ \ \ \text {as} \ \ \ n\rightarrow \infty \end{aligned}$$

(5.1)

strongly in the norm $\Vert \cdot \Vert$. For all $u\in \partial B_1$ and for each n

$$\begin{aligned} 0\le J_n(u) \le J_n(u_n), \end{aligned}$$

that is,

$$\begin{aligned} 0 \le \displaystyle \int _{\mathbb {R}^N}\left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,nu)}{(nu)^2} \right) u^2dx \le \displaystyle \int _{\mathbb {R}^N}\left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(nu_n)}{(nu_n)^2} \right) u_n^2dx \end{aligned}$$

(5.2)

for all u and for each n. Taking the limit $n\rightarrow \infty$, firstly, note that

$$\begin{aligned} u_n(x) \rightarrow \overline{u}(x) \ \ \ \text {a.e in} \ \ \ \mathbb {R}^N. \end{aligned}$$

Thus, if $\overline{u}(x)\ne 0$, it follows that $|n\overline{u}(x)|\rightarrow \infty$ if $n\rightarrow \infty$. Hence hypothesis $(f_4)$ yields

$$\begin{aligned} \left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,nu_n(x))}{(nu_n(x))^2} \right) u_n(x)^2 \rightarrow 0 \end{aligned}$$

(5.3)

if $n\rightarrow \infty$. If $\overline{u}(x) = 0$, we also have (5.3). By the strongly convergence in (5.1), there exist a function $\overline{h}\in L^1(\mathbb {R}^N)$ such that, up to a subsequence,

$$\begin{aligned} 0\le \left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,nu_n(x))}{(nu_n(x))^2} \right) u_n(x)^2 \le a_0|u^2_n(x)| \le a_0\overline{h}(x) \in L^1(\mathbb {R}^N). \end{aligned}$$

(5.4)

Finally, by (5.3) and (5.4), Lebesgue Dominated Convergence Theorem ensures that

$$\begin{aligned} \displaystyle \lim _{n\rightarrow \infty }\displaystyle \int _{\mathbb {R}^N}\left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,nu_n)}{(nu_n)^2} \right) u_n^2dx = 0. \end{aligned}$$

Therefore taking $n\rightarrow \infty$ in (5.2), we have

$$\begin{aligned} \displaystyle \lim _{n\rightarrow \infty }\displaystyle \int _{\mathbb {R}^N}\left( \displaystyle \frac{a(x)}{2} - \displaystyle \frac{F(x,nu)}{(nu)^2} \right) u^2dx = 0 \end{aligned}$$

uniformly for $u\in \partial B_1$. $\square$