Branch continuation inside the essential spectrum for the nonlinear Schrödinger equation

Abstract

We consider the nonlinear stationary Schrödinger equation

$$\begin{aligned} -\Delta u -\lambda u= Q(x)|u|^{p-2}u, \qquad \text {in }\mathbb {R}^N \end{aligned}$$

in the case where $N \ge 3$, p is a superlinear, subcritical exponent, Q is a bounded, nonnegative and nontrivial weight function with compact support in $\mathbb {R}^N$ and $\lambda \in \mathbb {R}$ is a parameter. Under further restrictions either on the exponent p or on the shape of Q, we establish the existence of a continuous branch $\mathcal {C}$ of nontrivial solutions to this equation which intersects $\{\lambda \} \times L^{s}(\mathbb {R}^N)$ for every $\lambda \in (-\infty , \lambda _Q)$ and $s> \frac{2N}{N-1}$. Here, $\lambda _Q>0$ is an explicit positive constant which only depends on N and $\text {diam}(\text {supp }Q)$. In particular, the set of values $\lambda$ along the branch enters the essential spectrum of the operator $-\Delta$.

Appendix

In this section, we add a result from general topology which has been used in the proof of Theorem 5.2. It is a variant of a classical Lemma by Whyburn (see [44, Theorem (9.1)]). For similar variants which have inspired the following proposition, see [26, 27].

Proposition 6.1

Let (X, d) be a locally compact metric space, $z_* \in X$, and let ${\mathcal C}_n \subset X$, $n \in \mathbb {N}$ be connected subsets satisfying the following assumptions.

• (i) There exist points $z_n \in {\mathcal C}_n$, $n \in \mathbb {N}$ such that $z_n \rightarrow z_* \in X$ as $n \rightarrow \infty$.

• (ii) The sets $\bigcup _{n \ge m} {\mathcal C}_n$, $m \in \mathbb {N}$ are not relatively compact in (X, d).

Then, the connected component ${\mathcal C}\subset X$ of X which contains $z_*$ is not relatively compact.

For the proof of this proposition, we need the following well-known result (see [44]).

Lemma 6.2

Suppose that (X, d) is a compact metric space, A and B are disjoint closed subsets of X, and suppose that no connected component of X intersects both A and B. Then, there exist two disjoint compact subsets $X_A, X_B \subset X$ such that $A \subset X_A$, $B \subset X_B$ and $X = X_A \cup X_B$.

Proof of Proposition 6.1

We suppose by contradiction that ${\mathcal C}$ is relatively compact. Since, by definition, ${\mathcal C}\subset X$ is closed, it follows that ${\mathcal C}$ is compact. Since X is locally compact, there exists a compact neighborhood $V \subset X$ of the set ${\mathcal C}$. Then, ${\mathcal C}$ and $\partial V$ are non-intersecting closed subsets contained in the compact metric space (V, d), and the maximal connectedness of ${\mathcal C}$ implies that that there does not exist a connected component of V which intersects ${\mathcal C}$ and $\partial V$. By Lemma 6.2, there exist disjoint compact subsets $X_A, X_B \subset V$ such that ${\mathcal C}\subset X_A$, $\partial V \subset X_B$ and

$$\begin{aligned} V = X_A \cup X_B. \end{aligned}$$

(6.1)

We may then choose a compact neighborhood $V_1 \subset X$ of $X_A$ such that $V_1 \cap X_B= \varnothing$, and we consider the compact set $V_2= V \cap V_1$. We have

$$\begin{aligned} \partial V_2 \subset [\partial V \cap V_1] \cup [\partial V_1 \cap V] \subset V \cap V_1, \end{aligned}$$

and thus, it follows that

$$\begin{aligned} \partial V_2 \cap X_A= \varnothing = \partial V_2 \cap X_B. \end{aligned}$$

Consequently,

$$\begin{aligned} \partial V_2 = \varnothing \end{aligned}$$

(6.2)

by (6.1), which implies, in particular, that $V_2$ is also open in X. On the other hand, since $z_* \in {\mathcal C}\subset X_A \subset V_2$, there exists $n_0 \in \mathbb {N}$ such that $z_n \in V_2$ for $n \ge n_0$, which means that ${\mathcal C}_n \cap V_2 \not = \varnothing$ for $n \ge n_0$. The connectedness of ${\mathcal C}_n$ and (6.2) then imply that

$$\begin{aligned} {\mathcal C}_n \subset V_2 \qquad \text {for } \;n \ge n_0, \end{aligned}$$

but this contradicts assumption (ii) since $V_2$ is compact. Thus, the proof is finished. $\square$