Existence and orbital stability of standing waves to a nonlinear Schrödinger equation with inverse square potential on the half-line

Abstract

In our work, we establish the existence of standing waves to a nonlinear Schrödinger equation with inverse-square potential on the half-line. We apply a profile decomposition argument to overcome the difficulty arising from the non-compactness of the setting. We obtain convergent minimizing sequences by comparing the problem to the problem at “infinity” (i.e., the equation without inverse square potential). Finally, we establish orbital stability/instability of the standing wave solution for mass subcritical and supercritical nonlinearities respectively.

1 Introduction

We study the existence and orbital stability of standing waves for the following nonlinear Schrödinger equation with inverse square potential on the half line

$$\begin{aligned} {\left\{ \begin{array}{ll}\displaystyle i u_t + u'' + c \frac{u}{x^2}+|u|^{p-1}u=0, \\ u(0)= u_0 \in H^1_0({\mathbb {R}}^+), \end{array}\right. } \end{aligned}$$

(1.1)

where $u: {\mathbb {R}}\times {\mathbb {R}}^+ \rightarrow {\mathbb {C}}$, $u_0: {\mathbb {R}}^+ \rightarrow {\mathbb {C}}$, $1<p<\infty$, and $0<c<1/4$.

There has been considerable interest recently in the study of the Schrödinger equation with inverse-square potential in three and higher dimensions. Classification of the so-called minimal mass blow-up solutions, global well-posedness, and stability of standing wave solutions were studied in [1, 6, 8, 22]. In the papers by Bensouilah et al. [1], and by Trachanas and Zographopoulos [22] the authors establish orbital stability of ground state solutions in the Hardy subcritical $(c<(N-2)^2/4)$ and Hardy critical $(c=(N-2)^2/4)$ case respectively for dimensions higher that three. In both cases, orbital stability is proved by showing the precompactness of minimizing sequences of the energy functional on an $L^2$ constraint. Local well-posedness was established for the two-dimensional space by Suzuki in [21], and in three and higher dimensions by Okazawa et al. in [18]. The presence of the inverse square potential in one-dimensional space has also attracted attention. In [13] H. Kovarik and F. Truc established dispersive estimates for $\partial _x^2+c/x^2$.

The dynamics of the equation is closely related to Hardy’s inequality (see [7])

$$\begin{aligned} c\int _0^\infty \frac{|u|^2}{x^2}dx\leqslant \int _0^\infty |u'|^2dx \text { for all } u \in C^\infty _0(0,\infty ), \end{aligned}$$

(1.2)

where $c\leqslant 1/4$. We introduce the Hardy functional

$$\begin{aligned} H(u)=\int _0^\infty \left( |u'|^2-\frac{c}{x^2}|u|^2\right) dx, \end{aligned}$$

which is closely related to our problem. We will mainly focus on the case $0<c<1/4$, when the natural energy space associated to (1.1) is $H^1_0({\mathbb {R}}^+)$, and the semi-norm $\left\| u'\right\| ^2_{L^2}$ is equivalent to H(u).

Let us consider the operator

$$\begin{aligned} H_c=-\frac{\partial ^2}{\partial x^2} - \frac{c}{x^2} \end{aligned}$$

acting on $C^\infty _0({\mathbb {R}}^+)$. Owing to the Hardy inequality, if $c<1/4$ the quadratic form $\langle H_c\varphi ,\varphi \rangle$ is positive definite on $C^\infty _0({\mathbb {R}}^+)$. It is natural to take the Friedrichs extension of $H_c$, thereby defining a self-adjoint operator in $L^2({\mathbb {R}}^+)$, which generates an isometry group in $H^1_0({\mathbb {R}}^+)$.

Local well-posedness for parameters $1<p<\infty$ and $0<c<\frac{1}{4}$ follows by standard arguments (see e.g. in [3] Chapter 4). In particular, the following holds.

Theorem 1.1

Let $1<p<\infty$ and $c<1/4$. For any initial value $u_0 \in H^1_0({\mathbb {R}}^+)$, there exist $T_{\mathrm {min}},T_{\mathrm {max}} \in (0,\infty ]$ and a unique maximal solution $u\in C((-T_{\mathrm {min}},T_{\mathrm {max}}),H^1_0({\mathbb {R}}^+))$ of (1.1), which satisfies for all $t\in (-T_{\mathrm {min}},T_{\mathrm {max}})$ the conservation laws

$$\begin{aligned} \left\| u(t)\right\| _{L^2}=\left\| u_0\right\| _{L^2}, \quad E(u(t))=E(u_0), \end{aligned}$$

(1.3)

where the energy is defined as

$$\begin{aligned} E(u)=\frac{1}{2}\left\| u'\right\| ^2_{L^2}-\frac{c}{2}\left\| \frac{u}{x}\right\| ^2_{L^2}-\frac{1}{p+1}\left\| u\right\| ^{p+1}_{L^{p+1}}, \textit{ for } u\in H^1_0({\mathbb {R}}^+). \end{aligned}$$

(1.4)

Moreover, the so-called blow-up alternative holds: if ${T_\mathrm {max}}<\infty$ then $\lim _{t\rightarrow T_{\mathrm {max}}}\left\| u'(t)\right\| _{L^2}=\infty$, (or $T_{\mathrm {min}}<\infty$ then $\lim _{t\rightarrow -T_{\mathrm {min}}}\left\| u'(t)\right\| _{L^2}=\infty$).

In this work we address the existence of standing wave solutions and their orbital stability/instability. By introducing the ansatz $u(t,x)=e^{i\omega t}\varphi (x)$, the standing wave equation to (1.1) reads as

$$\begin{aligned} \varphi ''+\frac{c}{x^2}\varphi -\omega \varphi +|\varphi |^{p-1}\varphi =0. \end{aligned}$$

(1.5)

First we will prove regularity of standing waves and the Pohozaev identities. To establish the existence of standing waves we carry out a minimization procedure on the Nehari manifold for the so-called action functional

$$\begin{aligned} S(v)=\frac{1}{2}\left\| v'\right\| ^2_{L^2}-\frac{c}{2}\left\| \frac{v}{x}\right\| ^2_{L^2}+\frac{\omega }{2}\left\| v\right\| ^2_{L^2}-\frac{1}{p+1}\left\| v\right\| ^{p+1}_{L^{p+1}}\quad v\in H^1_0({\mathbb {R}}^+). \end{aligned}$$

Owing to the non-compactness of the problem, we have to use a profile decomposition lemma, in the spirit of the article by Jeanjean and Tanaka [11]. To establish strong convergence of the minimizing sequence on the Nehari manifold we compare the minimization problem with the problem “at infinity”, i.e. when $c=0$. Hence, we obtain that the set of bound states is not empty:

$$\begin{aligned} {\mathcal {A}}=\{ u\in H^1_0({\mathbb {R}}^+)\setminus \{0\} : u''+c u/x^2 -\omega u + |u|^{p-1}u=0\}\ne \varnothing . \end{aligned}$$

We are in particular interested in the orbital stability/instability of ground states, i.e., solutions which minimize the action functional. We denote the set of ground sate solutions by

$$\begin{aligned} {\mathcal {G}}=\{u\in {\mathcal {A}} : S(u)\leqslant S(v) \text { for all } v \in {\mathcal {A}} \}. \end{aligned}$$

We use Lions’ concentration-compactness principle to obtain a variational characterization of ground states on an $L^2$-constraint, thereby establishing the orbital stability of the set of ground states for nonlinearities with power $1<p<5$. Finally, for $p\geqslant 5$ we establish strong instability by a convexity argument.

2 Existence of bound states

We start by investigating the standing wave equation,

$$\begin{aligned} {\left\{ \begin{array}{ll} \varphi '' + \frac{c}{x^2}\varphi - \omega \varphi + |\varphi |^{p-1}\varphi =0,\\ \varphi \in H^1_0({\mathbb {R}}^+) \setminus \{0\}. \end{array}\right. } \end{aligned}$$

(2.1)

First, we prove the regularity of solutions to (2.1) by a bootstrap argument.

Proposition 2.1

Let $\omega >0$ and $c < 1/4$. Assume $\varphi \in H^1_0({\mathbb {R}}^+)$ is a solution of (2.1) in $H^{-1}({\mathbb {R}}^+)$. Then the following statements are true

• (1) $\varphi \in W^{2,r}_0((\epsilon ,\infty ))$ for all $r\in [2, +\infty )$ and $\epsilon >0$, in particular $\varphi \in C^1((\epsilon ,\infty ))$;

• (2) The solution is exponentially bounded, that is $\mathrm {e}^{\sqrt{\omega } x}(|\varphi |+|\varphi '|)\in L^{\infty }({\mathbb {R}}^+)$;

Proof

(1) For $\varphi \in H^1_0({\mathbb {R}}^+)$ we have $\varphi \in L^q({\mathbb {R}}^+)$ for all $q\in [2,\infty ]$. We get easily that $|\varphi |^{p-1}\varphi \in L^q({\mathbb {R}}^+)$ for all $q\in [2,\infty )$. By (2.1) we have for any $\epsilon >0$ that $\varphi \in W^{2,q}_0((\epsilon ,\infty ))$ for all $q\in [2,\infty )$. By Sobolev’s embedding we get $\varphi \in C^{1,\delta }((\epsilon ,\infty ))$ for all $\delta \in (0,1)$, hence $|\varphi (x)|\rightarrow 0$, and $|\varphi '(x)|\rightarrow 0$ as $x\rightarrow \infty$.

(2) Let $\omega >0$. Changing $\varphi (x)$ to $\varphi (x)=\omega ^{1/(p-1)}\varphi (\sqrt{\omega }x)$ we may assume that $\omega =1$ in (2.1). Let $\varepsilon >0$ and $\theta _\varepsilon (x)=e^{\frac{x}{1+\varepsilon x}}$, for $x\geqslant 0$. It is easy to see that $\theta _\varepsilon$ is bounded, Lipschitz continuous, and $|\theta '_\varepsilon (x)|\leqslant \theta _\varepsilon (x)$ for all $x\in {\mathbb {R}}^+$. Additionally, $\theta _\varepsilon (x)\rightarrow e^x$ uniformly on bounded sets of ${\mathbb {R}}^+$. Taking the scalar product of the equation (2.1) with $\theta _\varepsilon \varphi \in H^1_0({\mathbb {R}}^+)$, we get

$$\begin{aligned} \mathop {{\mathrm{Re}}}\nolimits \int _{{\mathbb {R}}^+} \varphi '\cdot (\theta _\varepsilon {\bar{\varphi }})'dx - c\int _{{\mathbb {R}}^+} \theta _\varepsilon \frac{|\varphi |^2}{x^2} dx + \int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi |^2 dx = \int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi |^{p+1}dx. \end{aligned}$$

Using the inequality $\mathop {{\mathrm{Re}}}\nolimits (\varphi '(\theta _\varepsilon {\bar{\varphi }})')\geqslant \theta _\varepsilon |\varphi '|^2-\theta _\varepsilon |\varphi ||\varphi '|$ and

$$\begin{aligned} \int _{{\mathbb {R}}^+} \theta _\varepsilon |\varphi ||\varphi '|dx \leqslant \frac{1}{2}\int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi |^2dx +\frac{1}{2}\int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi '|^2dx, \end{aligned}$$

we obtain

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi '|^2dx + \frac{1}{2}\int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi |^2dx - c \int _{{\mathbb {R}}^+} \theta _\varepsilon \frac{|\varphi |^2}{x^2}dx \leqslant \int _{{\mathbb {R}}^+} \theta _\varepsilon |\varphi |^{p+1} dx. \end{aligned}$$

Let $R>0$ such that if $x>R$, then $\frac{c}{x^2}\leqslant \frac{1}{8}$ and $|\varphi (x)|^{p-1}\leqslant \frac{1}{8}$. Then we get

$$\begin{aligned}&c\int _{{\mathbb {R}}^+}\theta _\varepsilon \frac{|\varphi |^2}{x^2} dx + \int _{{\mathbb {R}}^+} \theta _\varepsilon |\varphi |^{p+1} \\&\quad \leqslant e^R \left( \int _0^R c\frac{|\varphi |^2}{x^2} dx +\int _0^R|\varphi |^{p+1}dx \right) + \frac{1}{4}\int _{{\mathbb {R}}^+} \theta _\varepsilon |\varphi |^2dx. \end{aligned}$$

From the last two inequalities it follows that

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi '|^2dx + \frac{1}{4}\int _{{\mathbb {R}}^+}\theta _\varepsilon |\varphi |^2dx \leqslant e^R \left( \int _0^R c\frac{|\varphi |^2}{x^2} dx +\int _0^R|\varphi |^{p+1}dx \right) . \end{aligned}$$

By taking $\varepsilon \downarrow 0$ we get

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^+} e^{x} |\varphi '|^2dx + \frac{1}{4}\int _{{\mathbb {R}}^+}e^{x}|\varphi |^2dx <\infty . \end{aligned}$$

Since both $\varphi$ and $\varphi '$ are Lipschitz continuous we deduce that $|\varphi (x)|e^{x}$ and $|\varphi '(x)|e^{x}$ are bounded. $\square$

We now prove that there exists a solution to (2.1). We define the action functional associated to (2.1) as follows

$$\begin{aligned} S(u)= \frac{1}{2} H(u) + \frac{\omega }{2}\left\| u\right\| ^2_{L^2} - \frac{1}{p+1}\left\| u\right\| ^{p+1}_{L^{p+1}}, \end{aligned}$$

for $c<1/4$ and $u\in H^1_0({\mathbb {R}}^+)$. Clearly, we have

$$\begin{aligned} S'(u)= - u''-\frac{c}{x^2}u + \omega u - |u|^{p-1}u. \end{aligned}$$

Therefore, to prove the existence of a solution to (2.1) amounts to show that S has a nontrivial critical point. A simple calculation yields the following identities.

Lemma 2.2

Assume $p>1$, $\omega >0$ and $c< 1/4$. Let $\varphi \in H^1_0({\mathbb {R}}^+)$ be a solution of (2.1) in $H^{-1}({\mathbb {R}}^+)$. Then the following identities are true:

$$\begin{aligned} \left\| \varphi '\right\| ^2_{L^2} - c\left\| \frac{\varphi }{x}\right\| ^2_{L^2} + \omega \left\| \varphi \right\| ^2_{L^2}-\left\| \varphi \right\| ^{p+1}_{L^{p+1}}&=0, \end{aligned}$$

(2.2)

$$\begin{aligned} \left\| \varphi '\right\| ^2_{L^2} - c\left\| \frac{\varphi }{x}\right\| ^2_{L^2} - \frac{p-1}{2(p+1)}\left\| \varphi \right\| ^{p+1}_{L^{p+1}}&= 0. \end{aligned}$$

(2.3)

Proof

We obtain the first equality by multiplying (2.1) by ${\bar{\varphi }}$ and integrating over ${\mathbb {R}}^+$.

To prove the second equality, let us put $\varphi _{\lambda }(x)=\lambda ^{1/2}\varphi (\lambda x)$ for $\lambda >0$. We have that

$$\begin{aligned} S(\varphi _\lambda )=\frac{\lambda ^2}{2}\left\| \varphi '\right\| ^2_{L^2}-\frac{\lambda ^2 c}{2}\left\| \frac{\varphi }{x}\right\| ^2_{L^2}+\frac{\omega }{2}\left\| \varphi \right\| ^2_{L^2}-\frac{\lambda ^{(p-1)/2}}{p+1}\left\| \varphi \right\| ^{p+1}_{L^{p+1}}, \end{aligned}$$

from which we get

$$\begin{aligned} \frac{\partial }{\partial \lambda } S(\varphi _\lambda )\Big |_{\lambda =1}=\left\| \varphi '\right\| ^2_{L^2} - c \left\| \frac{\varphi }{x} \right\| _{L^2}^2-\frac{p-1}{2(p+1)}\left\| \varphi \right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

We also have that

$$\begin{aligned} \frac{\partial }{\partial \lambda }S(\varphi _\lambda )\Big |_{\lambda =1} = \left\langle S'(\varphi ), \frac{\partial \varphi _\lambda }{\partial \lambda }\Big |_{\lambda =1} \right\rangle . \end{aligned}$$

Now $\frac{\partial \varphi _\lambda }{\partial \lambda }\Big |_{\lambda =1}=\frac{1}{2} \varphi + x \varphi '$ is in $H^1({\mathbb {R}}^+)$, since $\varphi$ and $\varphi '$ are exponentially decaying at infinity by Proposition 2.1. We obtain that the right hand-side is well-defined. Since $\varphi$ is a critical point of S, we obtain $S'(\varphi )=0$, which concludes the proof. $\square$

Remark 2.3

Since (2.2) and (2.3) hold for solutions of (2.1), it follows for $\omega \ne 0$ that

$$\begin{aligned} \omega \left\| \varphi \right\| ^2_{L^2}=\frac{p+3}{2(p+1)}\left\| \varphi \right\| ^{p+1}_{L^{p+1}}>0. \end{aligned}$$

Hence, non-trivial solution of (2.1) exists only if $\omega >0$.

Let us define for all $u\in H^1_0({\mathbb {R}}^+)$ the following functional:

$$\begin{aligned} J(u)= (S'(u),u)_{H^{-1},H^1_0} =H(u)+ \omega \left\| u\right\| ^2_{L^2}-\left\| u\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

It follows from Lemma 2.2, that ${\mathcal {N}}= \{ u\in H^1_0({\mathbb {R}}^+)\setminus \{0\} : J(u)=0 \}$ contains all nontrivial critical points of S. We aim to show that the infimum of the following minimization problem is attained

$$\begin{aligned} m=\inf \{S(u) : u \in {\mathcal {N}}\}=\frac{p-1}{2(p+1)}\inf \{\left\| u\right\| ^{p+1}_{L^{p+1}}:u \in {\mathcal {N}} \}. \end{aligned}$$

(2.4)

First we prove the following lemma.

Lemma 2.4

${\mathcal {N}}$ is nonempty, and $m>0$.

Proof

Let $u\in H^1_0({\mathbb {R}}^+)\setminus \{0\}$. Take

$$\begin{aligned} t(u)=\left( \frac{H(u)+\omega \left\| u\right\| _{L^2}^2}{\left\| u\right\| ^{p+1}_{L^{p+1}}}\right) ^{1/(p-1)}. \end{aligned}$$

By simple calculation, we get that $J(t(u)u)=0$, hence $t(u)u\in {\mathcal {N}}$. We see that

$$\begin{aligned} m=\inf _{u\in {\mathcal {N}}} S(u)=\inf _{u\in {\mathcal {N}}}\left( S(u)-\frac{1}{p+1}J(u) \right) =\frac{p-1}{2(p+1)} \inf _{u\in {\mathcal {N}}} (H(u)+\omega \left\| u\right\| ^2_{L^2}). \end{aligned}$$

It follows from Sobolev’s and Hardy’s inequalities, that there exists $C>0$ such that

$$\begin{aligned} H(u) + \omega \left\| u\right\| ^2_{L^2} =\left\| u\right\| ^{p+1}_{L^{p+1}} \leqslant C (H(u) + \omega \left\| u\right\| ^2_{L^2})^{(p+1)/2}, \end{aligned}$$

for all $u \in {\mathcal {N}}$. Hence,

$$\begin{aligned} \left( \frac{1}{C}\right) ^{2/(p-1)}\leqslant H(u) + \omega \left\| u\right\| ^2_{L^2} \text { for all } u\in H^1_0({\mathbb {R}}^+), \end{aligned}$$

which implies that

$$\begin{aligned} m\geqslant \frac{p-1}{2(p+1)}\left( \frac{1}{C}\right) ^{2/(p-1)}>0. \end{aligned}$$

$\square$

Lemma 2.5

Let $c<1/4$, and $p>1$. Then if $u\in H^1_0({\mathbb {R}}^+)$ is a minimizer of (2.4), then |u| is also a minimizer. In particular, we can search for the minimizers of (2.4) among the non-negative, real-valued functions of $H^1_0({\mathbb {R}}^+)$.

Proof

Let $u\in H^1_0({\mathbb {R}}^+)$ be a solution of the minimization problem (2.4). It is well-known that if $u\in H^1_0({\mathbb {R}}^+)$ then $|u|\in H^1_0({\mathbb {R}}^+)$ and $\left\| |u|'\right\| _{L^2}\leqslant \left\| u'\right\| _{L^2}$. Moreover, $\left\| |u|\right\| _{L^{p+1}}=\left\| u\right\| _{L^{p+1}}$. Therefore, $J(|u|)\leqslant J(u)$. Hence there exists a $\lambda \in (0,1]$ such that $J(\lambda |u|)=J(u)=0$. Then

$$\begin{aligned} m\leqslant S(\lambda |u|)=\frac{p-1}{2(p+1)} \left\| \lambda u\right\| ^{p+1}_{L^{p+1}}\leqslant \frac{p-1}{2(p+1)} \left\| u\right\| ^{p+1}_{L^{p+1}}=m. \end{aligned}$$

Hence $\lambda =1$, $J(|u|)=0$, and $S(|u|)=m$. $\square$

Let $m\in {\mathbb {R}}$. We say that $\{u_n\}_{n\in {\mathbb {N}}}$ is a Palais-Smale sequence for S at level m, if

$$\begin{aligned} S(u_n)\rightarrow m, \quad S'(u_n)\rightarrow 0 \textit{ in } H^{-1}({\mathbb {R}}^+), \end{aligned}$$

as $n\rightarrow \infty$.

Lemma 2.6

Let $c<1/4$, and $p>1$. There exists a bounded Palais-Smale sequence $\{u_n\}_{n\in {\mathbb {N}}}\subset {\mathcal {N}}$ for S at the level m. Namely, there is a sequence $\{u_n\}_{n\in {\mathbb {N}}}\subset {\mathcal {N}}$ bounded in $H^1({\mathbb {R}}^+)$ such that, as $n\rightarrow \infty$,

$$\begin{aligned} S(u_n)\rightarrow m, \quad S'(u_n)\rightarrow 0 \textit{ in } H^{-1}({\mathbb {R}}^+). \end{aligned}$$

Proof

Since ${\mathcal {N}}$ is a closed manifold in $H^1_0({\mathbb {R}}^+)$, it is a complete metric space. Hence, Ekeland’s variational principle (see pp. 51–53 in [20]) directly yields the existence of a Palais-Smale sequence at level m in ${\mathcal {N}}$.

We now show that if $\{u_n\}_{n\in {\mathbb {N}}} \subset {\mathcal {N}}$ and $\left\| u_n\right\| ^2_{H^1} \rightarrow \infty$, then $S(u_n)\rightarrow \infty$. Indeed, since $u_n\in {\mathcal {N}}$ from Hardy’s inequality we get that

$$\begin{aligned} S(u_n)= & {} \frac{p-1}{2(p+1)}(H(u_n)+\omega \left\| u_n\right\| ^2_{L^2}) \\\geqslant & {} \frac{p-1}{2(p+1)}(\min \{1,(1-4c)\}\left\| u'_n\right\| ^2_{L^2}+\omega \left\| u_n\right\| ^2_{L^2}). \end{aligned}$$

Therefore, any Palais-Smale sequence $\{u_n\}_{n\in {\mathbb {N}}}$ is bounded in $H^1_0({\mathbb {R}}^+)$. $\square$

Before proceeding to our next lemma, let us recall some classical results, see e.g. [3], concerning the case $c=0$. It is well-known that the set of solutions of

$$\begin{aligned} q''-\omega q+|q|^{p-1}q=0, \quad \omega >0, \quad q \in H^1({\mathbb {R}}) \end{aligned}$$

(2.5)

is given by $\{e^{i\theta }q(\cdot + y): y\in {\mathbb {R}}, \theta \in {\mathbb {R}}\}$, where q is a symmetric, positive solution of (2.5), explicitly given by

$$\begin{aligned} q(x)=\left( \frac{(p+1)\omega }{2}\text {sech}^2\left( \frac{(p-1)\sqrt{\omega }}{2}x \right) \right) ^{1/(p-1)}. \end{aligned}$$

(2.6)

Moreover, up to translation and phase invariance, it is the unique solution of the minimization problem

$$\begin{aligned} m^\infty&=\inf \{S^\infty (u): u\in H^1({\mathbb {R}})\setminus \{0\}, J^\infty (u)=0\}\\&=\frac{p-1}{2(p+1)}\inf \{\left\| u\right\| ^{p+1}_{L^{p+1}({\mathbb {R}})}: u\in H^1({\mathbb {R}})\setminus \{0\}, J^\infty (u)=0\}, \end{aligned}$$

where the functionals $S^\infty$ and $J^\infty$ are defined by

$$\begin{aligned} S^\infty (u)&=\frac{1}{2}\left\| u'\right\| ^2_{L^2({\mathbb {R}})}+\frac{\omega }{2}\left\| u\right\| ^2_{L^2({\mathbb {R}})}-\frac{1}{p+1}\left\| u\right\| ^{p+1}_{L^{p+1}({\mathbb {R}})},\\ J^\infty (u)&=\left\| u'\right\| ^2_{L^2({\mathbb {R}})}+\omega \left\| u\right\| ^2_{L^2({\mathbb {R}})}-\left\| u\right\| ^{p+1}_{L^{p+1}({\mathbb {R}})}. \end{aligned}$$

Lemma 2.7

Let $0<c<1/4$, and $p>1$. Then $m<m^\infty$.

Proof

It is not hard to see that $m\leqslant m^\infty$, we only need to prove that $m\ne m^\infty$. Let us first note that if $u\in H^1_0({\mathbb {R}}^+)\setminus \{0\}$ and $J(u)<0$, then $m<{\tilde{S}}(u)$, where

$$\begin{aligned} {\tilde{S}}(u)=\frac{p-1}{2(p+1)}\left( H(u)+\omega \left\| u\right\| ^2_{L^2} \right) . \end{aligned}$$

Indeed, if $J(u)<0$, then let us define

$$\begin{aligned} t(u)=\left( \frac{H(u)+\omega \left\| u\right\| ^2_{L^2}}{\left\| u\right\| ^{p+1}_{L^{p+1}}}\right) ^{1/(p-1)}. \end{aligned}$$

Hence $t(u) \in (0,1)$, $t(u) u\in {\mathcal {N}}$, and

$$\begin{aligned} m\leqslant {\tilde{S}}(t(u) u)=t^2(u){\tilde{S}}(u)< {\tilde{S}}(u). \end{aligned}$$

Now let us define $\psi _A(x)=q(x+A)-q(x-A)$ for $x\geqslant 0$. For large enough A we obtain the following estimates (see Lemma 5.1 in the Appendix):

$$\begin{aligned} \int _0^\infty |\psi _A'|^2dx&= \int _{-\infty }^\infty |q'|^2dx+O\left( \left( 2A+\frac{1}{\sqrt{\omega }}\right) e^{-2\sqrt{\omega } A}\right) ,\\ \int _0^\infty |\psi _A|^2dx&= \int _{-\infty }^\infty |q|^2dx +O\left( \left( 2A+\frac{1}{\sqrt{\omega }}\right) e^{-2\sqrt{\omega } A}\right) ,\\ \int _0^\infty \frac{|\psi _A|^2}{x^2}dx&\leqslant \frac{4}{A^2}\int _{-\infty }^\infty |q|^2dx +O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) ,\\ \int _0^\infty |\psi _A|^{p+1}dx&= \int _{-\infty }^\infty |q|^{p+1}dx + O\left( e^{-2\sqrt{\omega } A}\right) . \end{aligned}$$

Since $0<c<1/4$, we obtain for $A>0$ large enough

$$\begin{aligned} J(\psi _A)&\leqslant \left\| q'\right\| ^2_{L^2({\mathbb {R}})}+\omega \left\| q\right\| ^2_{L^2({\mathbb {R}})}-\left\| q\right\| ^{p+1}_{L^{p+1}({\mathbb {R}})}-\frac{4c}{A^2}\left\| q\right\| ^2_{L^2({\mathbb {R}})}+O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) \\&=-\frac{4c}{A^2}\left\| q\right\| ^2_{L^2({\mathbb {R}})}+O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) <0, \end{aligned}$$

and

$$\begin{aligned} {\tilde{S}}(\psi _A)&\leqslant \frac{p-1}{2(p+1)}\left( \left\| q'\right\| ^2_{L^2({\mathbb {R}})}+\omega \left\| q\right\| ^2_{L^2({\mathbb {R}})}-\frac{4c}{A^2}\left\| q\right\| ^2_{L^2({\mathbb {R}})}\right) +O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) \\&=m^\infty -\frac{p-1}{2(p+1)}\frac{4c}{A^2}\left\| q\right\| ^2_{L^2({\mathbb {R}})}+O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) < m^\infty . \end{aligned}$$

Since $J(\psi _A)<0$, we get

$$\begin{aligned} m<{\tilde{S}}(\psi _A)< m^\infty , \end{aligned}$$

which concludes the proof. $\square$

We need the following lemma, which describes the behavior of bounded Palais-Smale sequences. We note that $H^1_0({\mathbb {R}}^+)$ functions can be extended to functions in $H^1({\mathbb {R}})$ by setting $u\equiv 0$ on ${\mathbb {R}}^-$. The proof of the following statement is presented in the appendix.

Lemma 2.8

Let $\{u_n\}_{n\in {\mathbb {N}}}\subset H^1_0({\mathbb {R}}^+)$ be a bounded Palais-Smale sequence for S at level m. Then there exists a subsequence still denoted by $\{u_n\}_{n\in {\mathbb {N}}}$, a $u_0 \in H^1_0({\mathbb {R}}^+)$ solution of

$$\begin{aligned} \varphi ''+\frac{c}{x^2}\varphi -\omega \varphi +|\varphi |^{p-1}\varphi =0, \end{aligned}$$

an integer $k\geqslant 0$, $\{x_n^i\}_{i=1}^k\subset {\mathbb {R}}^+$, and nontrivial solutions $q_i$ of (2.5) satisfying

$$\begin{aligned}&u_n \rightharpoonup u_0 \quad \textit{weakly in} \quad H^1_0({\mathbb {R}}^+), \\&S(u_n)\rightarrow S(u_0)+\sum _{i=1}^kS^\infty (q_i), \\&u_n-(u_0+\sum _{i=1}^kq_i(x-x_n^i))\rightarrow 0 \quad \textit{strongly in}\quad H^1({\mathbb {R}}),\\&|x_n^i|\rightarrow \infty ,\quad |x_n^i-x_n^j|\rightarrow \infty \quad for \quad 1\leqslant i\ne j\leqslant k, \end{aligned}$$

where in case $k=0$, the above holds without $q_i$ and $x_n^i$.

We only need to show that the critical point of S provided by Lemma 2.8 is non-trivial.

Theorem 2.9

Let $0<c<1/4$. Then there exists $u\in {\mathcal {N}}\setminus \{0\}$, $u\geqslant 0$ a.e., such that $S(u)=m$.

Proof

We only have to prove that the $\{u_n\}_{n\in {\mathbb {N}}}$ bounded Palais-Smale sequence obtained in Lemma 2.6 admits a strongly convergent subsequence. Assume that it is not the case. Using Lemma 2.8 we see that $k\geqslant 1$ and $u_n$ is weakly convergent to $u_0$ in $H^1_0({\mathbb {R}}^+)$ up to a subsequence. Then

$$\begin{aligned} m=\lim _{n\rightarrow \infty } S(u_n)\geqslant S(u_0)+S^\infty (q)=S(u_0)+m^\infty . \end{aligned}$$

Now, $S(u_0)\geqslant 0$ since $J(u_0)=0$. Thus $m\geqslant m^\infty$, which contradicts Lemma 2.7. Hence $k=0$ and $u_n\rightarrow u_0$ in $H^1_0({\mathbb {R}}^+)$. $\square$

Lemma 2.10

Let $p>1$ and $\omega >0$. There exists a $\mu >0$ such that

$$\begin{aligned} \int _0^\infty |u|^2dx= \mu , \text { for every } u \in {\mathcal {G}}. \end{aligned}$$

The mass of ground state solutions is $\mu =\frac{m}{\omega }\frac{p+3}{p-1}$. Moreover, we have

$$\begin{aligned} \left\| u\right\| ^{p+1}_{L^{p+1}}=\frac{2(p+1)}{p-1}m, \text { and } H(u)=m \text { for every } u\in {\mathcal {G}}. \end{aligned}$$

Proof

Since $u \in {\mathcal {G}}$ is a solution of (2.1), it satisfies (2.2) and (2.3). By subtracting the two identities we get

$$\begin{aligned} \omega \left\| u\right\| _{L^2}^2=\frac{p+3}{2(p+1)}\left\| u\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

(2.7)

Additionally, since u is a ground state solution, it also solves the minimization problem (2.4). From (2.4) and (2.3) we get

$$\begin{aligned} \omega \left\| u\right\| ^2_{L^2}+\frac{p-5}{2(p+1)}\left\| u\right\| _{L^{p+1}}^{p+1}=2m. \end{aligned}$$

(2.8)

From (2.7) and (2.8) it follows

$$\begin{aligned} \left\| u\right\| ^2_{L^2}=\frac{m}{\omega }\frac{p+3}{p-1}>0. \end{aligned}$$

Thus, let $\mu =\frac{m}{\omega }\frac{p+3}{p-1}$. Now it follows from (2.4) and (2.3) that

$$\begin{aligned} \left\| u\right\| ^{p+1}_{L^{p+1}}=\frac{2(p+1)}{p-1}m, \text { and } H(u)=m \text { for every } u\in {\mathcal {G}}. \end{aligned}$$

which concludes the proof. $\square$

3 Stability

In this section we consider nonlinearities with $1<p<5$. Our aim is to prove orbital stability of the standing waves. To do so, we investigate the minimization problem:

$$\begin{aligned} I=\inf \{E(u) : u \in \Gamma \}, \end{aligned}$$

(3.1)

where

$$\begin{aligned} \Gamma = \{ u\in H^1_0({\mathbb {R}}^+) : \left\| u\right\| _{L^2}^2=\mu \}. \end{aligned}$$

and the energy E is defined by (1.4). We will rely on a of Lions’ concentration-compactness principle [15] and the arguments by Cazenave and Lions [4], see also in [3]. The main problem is to obtain compactness of minimizing sequences owing to the absence of translation invariance. We define the problem at infinity by

$$\begin{aligned} I^\infty =\inf \{ E^\infty (u) : u\in H^1({\mathbb {R}}) \text { and } \left\| u\right\| ^2_{L^2}=\mu \}, \end{aligned}$$

(3.2)

where

$$\begin{aligned} E^\infty (u)=\frac{1}{2}\int _{{\mathbb {R}}}|u'|^2dx-\frac{1}{p+1}\int _{\mathbb {R}}|u|^{p+1}dx. \end{aligned}$$

We recall some well-known facts about the minimization problem (3.2) (see [3, Chapter 8.]). For every $\mu >0$, there exists a unique, positive, symmetric function $q=q(\mu ) \in H^1({\mathbb {R}})$, such that

$$\begin{aligned} \left\| q\right\| _{L^2}=\mu , \quad E^\infty (q)=I^\infty , \end{aligned}$$

and q solves the nonlinear equation

$$\begin{aligned} q''-\lambda q+|q|^{p-1}q=0, \end{aligned}$$

where $\lambda =\lambda (\mu )$. Moreover, there exists $M>0$ such that

$$\begin{aligned} e^{\sqrt{\lambda } |x|}|q(x)|\leqslant M \textit{ and } e^{\sqrt{\lambda } |x|}|q'(x)|\leqslant M. \end{aligned}$$

We proceed by proving the following lemma:

Lemma 3.1

If $0<c<1/4$, then the following inequality holds:

$$\begin{aligned} I<I^\infty . \end{aligned}$$

Proof

For $A>0$, let C(A) be a normalizing factor specified later. Let us define

$$\begin{aligned} \Psi _A(x)=C(A)(q(x+A)-q(x-A)) \text { for } x\geqslant 0. \end{aligned}$$

Since q is even, we obtain $\Psi _A \in H^1_0({\mathbb {R}}^+)$ and

$$\begin{aligned}&\int _0^\infty |\Psi _A(x)|^2dx= C^2(A)\left( \int _{-\infty }^\infty |q|^2dx -\int _{-\infty }^\infty q(x+A)q(x-A)dx\right) . \end{aligned}$$

We estimate the second integral by (see Lemma 5.1)

$$\begin{aligned} \int _{-\infty }^\infty q(x+A)q(x-A)dx = O\left( \left( 2A+\frac{1}{\sqrt{\lambda }}\right) e^{-2\sqrt{\lambda } A}\right) . \end{aligned}$$

We define

$$\begin{aligned} C(A)=\left( \frac{\mu }{\mu -\int _{-\infty }^\infty q(x+A)q(x-A)dx}\right) ^{1/2}. \end{aligned}$$

C(A) is a continuous function of A, $C(A)\geqslant 1$, and $C(A)\rightarrow 1$ exponentially fast as $A\rightarrow \infty$. Thus, $\left\| \Psi _A\right\| _{L^2}=\mu$ for all $A>0$. By Lemma 5.1 in the Appendix, we obtain for $A>0$ large enough that

$$\begin{aligned} \int _0^\infty |\Psi _A'|^2dx&= C^2(A)\int _{-\infty }^\infty |q'|dx+O\left( \left( 2A+\frac{1}{\sqrt{\lambda }}\right) e^{-2\sqrt{\lambda } A}\right) ,\\ \int _0^\infty \frac{|\Psi _A|^2}{x^2}dx&\leqslant \frac{4C^2(A)}{A^2}\int _0^\infty |\Psi _A|^2dx +O\left( \frac{1}{A^2}e^{-\sqrt{\lambda } A}\right) ,\\ \int _0^\infty |\Psi _A|^{p+1}dx&= C^{p+1}(A)\int _{-\infty }^\infty |q|^{p+1}dx + O(e^{-2\sqrt{\lambda } A}). \end{aligned}$$

Hence for A large enough we get

$$\begin{aligned} E(\Psi _A)&=\frac{1}{2}\int _0^\infty |\Psi _A'|^2dx -\frac{c}{2}\int _0^\infty \frac{|\Psi _A|^2}{x^2}dx- \frac{1}{p+1}\int _0^\infty |\Psi _A|^{p+1}dx\\&\leqslant C^2(A)\left( \frac{1}{2}\int _{-\infty }^\infty |q'|^2dx- \frac{C^{p-1}(A)}{p+1}\int _{-\infty }^\infty |q|^{p+1}dx \right) \\&\quad -\frac{c}{2}\frac{4C^2(A)}{A^2}\int _0^\infty |\Psi _A|^2dx +O\left( \frac{1}{A^2}e^{-\sqrt{\lambda } A}\right) . \end{aligned}$$

Owing to the exponential decay of the last term, for large A we get

$$\begin{aligned} E(\Psi _A)\leqslant E(q) -\frac{2c}{A^2}\mu =I^\infty -\frac{2c}{A^2}\mu . \end{aligned}$$

Since $0<c<1/4$ we get that $E(\Psi _A)<I^\infty$, which concludes the proof. $\square$

We need the following version of the concentration-compactness principle. The proof follows the same way as in the classical case (see [15]).

Lemma 3.2

Let $0<c<1/4$, and $\{ u_n\}_{n\in {\mathbb {N}}} \subset H^1_0({\mathbb {R}}^+)$ be a sequence satisfying

$$\begin{aligned} \lim _{n\rightarrow \infty } \left\| u_n\right\| ^2_{L^2} = M \text { and } \lim _{n\rightarrow \infty } H(u_n) < \infty . \end{aligned}$$

Then there exists a subsequence $\{u_n\}_{n\in {\mathbb {N}}}$ such that it satisfies one of the following alternatives.

(Vanishing) $\lim _{n\rightarrow \infty }\left\| u_n\right\| _{L^p}\rightarrow 0$ for all $p\in (2,\infty )$.

(Dichotomy) There are sequences $\{v_n\}_{n\in {\mathbb {N}}}, \{w_n\}_{n\in {\mathbb {N}}}$ in $H^1_0({\mathbb {R}}^+)$ and a constant $\alpha \in (0,1)$ such that:

• (1) ${{\,\mathrm{dist}\,}}({{\,\mathrm{supp}\,}}(v_n),{{\,\mathrm{supp}\,}}(w_n))\rightarrow \infty$;

• (2) $|v_n|+|w_n|\leqslant |u_n|$;

• (3) $\sup _{n\in {\mathbb {N}}}(\left\| v_n\right\| _{H^1}+\left\| w_n\right\| _{H^1})<\infty$;

• (4) $\left\| v_n\right\| _{L^2}^2\rightarrow \alpha M$ and $\left\| w_n\right\| _{L^2}^2\rightarrow (1-\alpha )M$ as $n\rightarrow \infty$;

• (5) $\lim _{n\rightarrow \infty }\left| \int _0^\infty |u_n|^qdx-\int _0^\infty |v_n|^qdx-\int _0^\infty |w_n|^qdx\right| =0$ for all $q\in [2,\infty )$;

• (6) $\liminf _{n\rightarrow \infty }\{H(u_n)-H(v_n)-H(w_n)\}\geqslant 0$.

(Compactness) There exists a sequence $y_n \in {\mathbb {R}}^+$, such that for any $\varepsilon >0$ there is an $R>0$ with the property that

$$\begin{aligned} \int _{(y_n-R,y_n+R)\cap {\mathbb {R}}^+} |u_n|^2\geqslant M-\varepsilon . \end{aligned}$$

for all $n\in {\mathbb {N}}$.

We are now in a position to prove the following lemma.

Lemma 3.3

Let $1<p<5$, $0<c<1/4$, and $\omega >0$. Then the infimum in (3.1) is attained. Additionally, all minimizing sequences are relatively compact, that is if $\{u_n\}_{n\in {\mathbb {N}}}$ satisfies $\left\| u_n\right\| _{L^2}^2\rightarrow \mu$ and $E(u_n)\rightarrow I$ then there exists a subsequence $\{u_n\}_{n\in {\mathbb {N}}}$ which converges to a minimizer $u\in H^1_0({\mathbb {R}}^+)$.

Proof

Step 1. We first show that $0>I>-\infty$. Let $u\in \Gamma$. For $\lambda >0$, we define $u_\lambda (x)=\lambda ^{1/2}u(\lambda x) \in \Gamma$. Clearly,

$$\begin{aligned} E(u_\lambda )=\frac{\lambda ^2}{2}\left\| u'\right\| ^2_{L^2}-\frac{c\lambda ^2}{2}\int _0^\infty \frac{|u|^2}{x^2}dx-\frac{\lambda ^{(p-1)/2}}{p+1}\left\| u\right\| ^{p+1}_{L^{p+1}} \end{aligned}$$

Since $1<p<5$, we can choose a small $\lambda >0$ such that $E(u_\lambda )<0$. Hence $I<0$.

Since $c\in (0,1/4)$, we have $H(u)\sim \left\| u'\right\| ^2_{L^2}$. We get from the Gagliardo-Nirenberg inequality that there exists $C>0$ such that for all $u\in H^1_0({\mathbb {R}}^+)$

$$\begin{aligned} \int _0^\infty |u|^{p+1}dx\leqslant C H(u)^{\frac{p-1}{4}}\left( \int _0^\infty |u|^2dx\right) ^{1+\frac{p-1}{4}}. \end{aligned}$$

Since $1<p<5$, this yields that there exists $\delta >0$ and $K>0$ such that

$$\begin{aligned} E(u)\geqslant \delta \left\| u\right\| ^2_{H^1}-K \text { for all } u \in \Gamma , \end{aligned}$$

(3.3)

from which follows that $I>-\infty$.

Every minimizing sequence is bounded in $H^1_0({\mathbb {R}}^+)$ and bounded from below in $L^{p+1}({\mathbb {R}}^+)$. Indeed, let $\{u_n\}_{n\in {\mathbb {N}}}\subset \Gamma$ be a minimizing sequence, then by (3.3) it is bounded in $H^1_0({\mathbb {R}}^+)$. Furthermore, for n large enough we have $E(u_n)<I/2$, thus

$$\begin{aligned} \left\| u_n\right\| _{L^{p+1}}^{p+1}>-\frac{p+1}{2}I. \end{aligned}$$

(3.4)

Now $I<0$, hence the result follows.

Step 2. We now verify that all minimizing sequences have a subsequence which converges to a limit u in $H^1_0({\mathbb {R}}^+)$. Let $\{u_n\}_{n\in {\mathbb {N}}}$ satisfy $\left\| u_n\right\| ^2_{L^2}\rightarrow \mu$ and $E(u_n)\rightarrow I$. Since every minimizing sequence is bounded in $H^1_0({\mathbb {R}}^+)$, $\{u_n\}_{n\in {\mathbb {N}}}$ has a weak-limit $u\in L^p({\mathbb {R}}^+)$ . We can apply the concentration-compactness principle (see Lemma 3.2) to the sequence $\{u_n\}_{n\in {\mathbb {N}}}$. We note that since the sequence is bounded from below in $L^{p+1}({\mathbb {R}}^+)$ vanishing cannot occur.

Now let us assume that dichotomy occurs. Let $\alpha \in (0,1)$, $\{v_n\}_{n\in {\mathbb {N}}}$ and $\{w_n\}_{n\in {\mathbb {N}}}$ sequences as in Lemma 3.2. It follows from (5) and (6) of Lemma 3.2 that

$$\begin{aligned} \liminf _{n\rightarrow \infty }(E(u_n)-E(v_n)-E(w_n))\geqslant 0, \end{aligned}$$

hence

$$\begin{aligned} \limsup _{n\rightarrow \infty }(E(v_n)+E(w_n))\leqslant I. \end{aligned}$$

(3.5)

Observe that for $u\in H^1_0({\mathbb {R}}^+)$, and $a>0$, we have

$$\begin{aligned} E(u)=\frac{1}{a^2}E(au)+\frac{a^{p-1}-1}{p+1}\int _0^\infty |u|^{p+1}dx. \end{aligned}$$

Let $a_n=\sqrt{\mu }/\left\| v_n\right\| _{L^2}$ and $b_k^2=\sqrt{\mu }/\left\| w_n\right\| _{L^2}$. Hence, $a_nv_n \in \Gamma$ and $b_nw_n\in \Gamma$, which implies

$$\begin{aligned} E(v_n)&\geqslant \frac{I}{a^2_n}+\frac{a_n^{p-1}-1}{p+1}\int _0^\infty |v_n|^{p+1}dx, \\ E(w_n)&\geqslant \frac{I}{b^2_n}+\frac{b_n^{p-1}-1}{p+1}\int _0^\infty |w_n|^{p+1}dx. \end{aligned}$$

Therefore

$$\begin{aligned} E(v_n)+E(w_n)\geqslant I(a_n^{-2}+b_n^{-2})+\frac{a_n^{p-1}}{p+1}\int _0^\infty |v_n|^{p+1}+\frac{b_n^{p-1}}{p+1}\int _0^\infty |w_n|^{p+1}. \end{aligned}$$

Now we observe $a_n^{-2}\rightarrow \alpha$ and $b_n^{-2}\rightarrow (1-\alpha )$ by (4) of Lemma 3.2. Since $\alpha \in (0,1)$, we get that $\theta = \min \{\alpha ^{-(p-1)/2};(1-\alpha )^{-(p-1)/2}) \}>1$. Property (5) of Lemma 3.2 and (3.4) implies

$$\begin{aligned} \liminf _{n\rightarrow \infty }(E(v_n)+E(w_n))\geqslant I+ \frac{\theta -1}{p+1}\liminf _{n\rightarrow \infty }\int _0^\infty |u_n|^{p+1}dx, \geqslant I+\frac{\theta -1}{2}>I, \end{aligned}$$

which contradicts (3.5). Hence the following holds: there exists a sequence $y_n\in {\mathbb {R}}^+$, such that for any $\varepsilon >0$ there exists $R>0$ with the property that

$$\begin{aligned} \int _{(y_n-R,y_n+R)\cap {\mathbb {R}}^+}|u_n|^2\geqslant \mu -\varepsilon . \end{aligned}$$

(3.6)

for all $n\in {\mathbb {N}}$.

We now show that $\{y_n\}_{n\in {\mathbb {N}}}$ is bounded in ${\mathbb {R}}^+$. First we show that if $y_n\rightarrow \infty$, then

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _0^\infty \frac{|u_n|^2}{x^2}dx=0. \end{aligned}$$

(3.7)

Let us assume by contradiction that

$$\begin{aligned} \int _0^\infty \frac{|u_n|^2}{x^2}dx\geqslant \delta >0, \end{aligned}$$

(3.8)

which implies together with Hardy’s inequality that

$$\begin{aligned} H(u_n)\geqslant (1/4-c)\delta . \end{aligned}$$

(3.9)

Let us take $\xi \in C^\infty ({\mathbb {R}}^+)$, such that for ${\tilde{R}}>0$ and $a>0$ we have that $\xi (r)=1$ for $0\leqslant r \leqslant {\tilde{R}}$, $\xi (r)=0$ for $r\geqslant {\tilde{R}}+a$, and $\left\| \xi '\right\| _{L^\infty }\leqslant 2/a$. We introduce $u_{n,1}=u_n\cdot \xi$ and $u_{n,2}=u_n\cdot (1-\xi )$. Clearly, $u_{n,1}\in H^1_0({\mathbb {R}}^+)$, $u_{n,2}\in H^1_0({\mathbb {R}}^+)$ and $u_n=u_{n,1}+u_{n,2}$. Moreover, the following inequalities hold

$$\begin{aligned} |u'_{n,1}|^2&\leqslant 2(4a^{-2}|u_n|^2+|u'_n|^2), \\ |u'_{n,2}|^2&\leqslant 2(4a^{-2}|u_n|^2+|u'_n|^2). \end{aligned}$$

We obtain by direct calculation that

$$\begin{aligned} E(u_n)=E(u_{n,1})+E(u_{n,2})+\rho _n \end{aligned}$$

where

$$\begin{aligned} \rho _n&=\frac{1}{2}\int _{{\tilde{R}}}^{{\tilde{R}}+a} \left[ (|u_n'|^2-|u'_{n,1}|^2-|u'_{n,2}|^2)-\frac{c}{x^2}(|u_n|^2-|u_{n,1}|^2-|u_{n,2}|^2) \right] dx \\ {}&\quad -\frac{1}{p+1}\int _{{\tilde{R}}}^{{\tilde{R}}+a}(|u_n|^{p+1}-|u_{n,1}|^{p+1}-|u_{n,2}|^{p+1})dx. \end{aligned}$$

We show that there exists ${\tilde{R}}>0$ and $a>1$, such that for n large enough $|\rho _n|\leqslant (1/4-c)\frac{\delta }{4}$. First we observe by the properties of the cut-off that

$$\begin{aligned} \left| \frac{1}{2}\int _{{\tilde{R}}}^{{\tilde{R}}+a} (|u_n'|^2-|u'_{n,1}|^2-|u'_{n,2}|^2)dx\right| \leqslant \frac{5}{2} \int _{{\tilde{R}}}^{{\tilde{R}}+a}|u_n'|^2dx+\frac{8}{a^2}\int _{{\tilde{R}}}^{{\tilde{R}}+a}|u_n|^2dx. \end{aligned}$$

We claim that there exist ${\tilde{R}}>0$ and $a>1$ such that for a subsequence $\{u_{n_k}\}$ we have

$$\begin{aligned} \int _{{\tilde{R}}}^{{\tilde{R}}+a}|u'_{n_k}|^2dx< \frac{1}{20}(1/4-c)\delta . \end{aligned}$$

(3.10)

Suppose that this claim does not hold, that is for all $R>0$, $a>1$ there exists $k\in {\mathbb {N}}$ such that for all $n\geqslant k$ the following holds

$$\begin{aligned} \int _R^{R+a}|u'_n|^2dx\geqslant \frac{1}{20}(1/4-c)\delta . \end{aligned}$$

Let $(R_1,R_1+a_1)$. There exists $k_1\in {\mathbb {N}}$, such that for all $n\geqslant k_1$ we have

$$\begin{aligned} \int _{R_1}^{R_1+a_1}|u'_n|^2dx\geqslant \frac{1}{20}(1/4-c)\delta . \end{aligned}$$

Now let $R_2>R_1+a_1$ and $a_2>1$. Then by our assumption there exists $k_2\in {\mathbb {N}}$, such that for all $n\geqslant k_2$ it holds that

$$\begin{aligned} \int _{R_2}^{R_2+a_2}|u'_n|^2dx\geqslant \frac{1}{20}(1/4-c)\delta . \end{aligned}$$

Hence, there exists a subsequence $\{v_{n_k}\}_{k\in {\mathbb {N}}}$ such that for all $j\in \{1,2\}$ it holds that

$$\begin{aligned} \int _{R_j}^{R_j+a_j}|u'_{n_k}|^2dx\geqslant \frac{1}{20}(1/4-c)\delta \end{aligned}$$

for all $k\in {\mathbb {N}}$. Therefore, we can construct for all $l\in {\mathbb {N}}$ a subsequence $\{u_{n_k}\}_{k\in {\mathbb {N}}}$, such that for all $1\leqslant j\leqslant l$ there are disjoint intervals $A_j=(R_j,R_j+a_j)$, such that

$$\begin{aligned} \int _{A_j}|u'_{n_k}|^2dx\geqslant \frac{1}{20}(1/4-c)\delta . \end{aligned}$$

Hence for all $l\in {\mathbb {N}}$ there exists a subsequence $\{ u_{n_k}\}_{k\in {\mathbb {N}}}$, such that for all $k\in {\mathbb {N}}$ we have

$$\begin{aligned} \int _0^\infty |u'_{n_k}|^2dx\geqslant \sum _{j=1}^{l}\int _{A_j}|u'_{n_k}|^2dx\geqslant \frac{l}{20}(1/4-c)\delta . \end{aligned}$$

This implies that $\int _0^\infty |u'_{n_k}|^2dx\rightarrow \infty$, which is a contradiction since $\{u_n\}_{n\in {\mathbb {N}}}$ is bounded in $H^1_0({\mathbb {R}}^+)$. Hence the assertion (3.10) is true. Now we note that

$$\begin{aligned} \int _0^R |u_n|^{p+1}dx \leqslant \left\| u_n\right\| ^{p-1}_{L^\infty }\int _0^R|u_n|^2dx. \end{aligned}$$

Since $\{u_n\}_{n\in {\mathbb {N}}}$ is bounded in $L^\infty ({\mathbb {R}}^+)$, in view of (3.6) we obtain for $R>0$ given in (3.6) that

$$\begin{aligned} \int _0^R|u_n|^2dx \rightarrow 0 \quad \text {implies} \quad \int _0^R |u_n|^{p+1}dx\rightarrow 0. \end{aligned}$$

(3.11)

For large n we have ${\tilde{R}}+a<y_n-R$, since $y_n\rightarrow \infty$ by our assumption. Now (3.11) implies

$$\begin{aligned}&\left| \frac{8}{a^2}\int _{{\tilde{R}}}^{{\tilde{R}}+a}|u_n|^2dx\right| +\left| \int _{{\tilde{R}}}^{{\tilde{R}}+a}\frac{c}{x^2}(|u_n|^2-|u_{n,1}|^2-|u_{n,2}|^2) dx\right| \nonumber \\&\qquad +\left| \frac{1}{p+1}\int _{{\tilde{R}}}^{{\tilde{R}}+a}(|u_n|^{p+1}-|u_{n,1}|^{p+1}-|u_{n,2}|^{p+1})dx\right| \nonumber \\&\quad \leqslant \left| \frac{8}{a^2}\int _{{\tilde{R}}}^{{\tilde{R}}+a}|u_n|^2dx\right| +\frac{c}{{\tilde{R}}^2}\left| \int _{{\tilde{R}}}^{{\tilde{R}}+a}|u_n|^2(1-\xi ^2-(1-\xi )^2) dx\right| \nonumber \\&\qquad +\left| \frac{1}{p+1}\int _{{\tilde{R}}}^{{\tilde{R}}+a}|u_n|^{p+1}(1-\xi ^{p+1}-(1-\xi )^{p+1})dx\right| \nonumber \\&\quad \leqslant \frac{(1/4-c)\delta }{8}. \end{aligned}$$

(3.12)

for large n. Now (3.10) and (3.12) implies

$$\begin{aligned} |\rho _n|\leqslant \frac{(1/4-c)\delta }{4}. \end{aligned}$$

(3.13)

Let us observe that $\left\| u_{n,1}\right\| _{L^{p+1}}\rightarrow 0$ by (3.11). Hence

$$\begin{aligned} E(u_{n,1})=\frac{1}{2}H(u_{n,1})+o(1). \end{aligned}$$

Now let us notice that $\mathrm {supp} (u_{n,2}) \subset ({\tilde{R}}, \infty )$. Moreover, in view of (3.6),

$$\begin{aligned} \int _0^\infty |u_{n,2}|^2dx = \int _{y_n-R}^\infty |u_{n,2}|^2dx+o(1). \end{aligned}$$

Hence

$$\begin{aligned} \int _0^\infty \frac{|u_{n,2}|^2}{x^2}dx= \int _{y_n-R}^\infty \frac{|u_{n,2}|^2}{x^2}dx + o(1)\leqslant \frac{\mu }{|y_n-R|^2}. \end{aligned}$$

Now $y_n\rightarrow \infty$ implies that

$$\begin{aligned} E(u_{n,2})= E^\infty (u_{n,2})+o(1). \end{aligned}$$

Thus,

$$\begin{aligned} E(u_n)=\frac{1}{2}H(u_{n,1})+E^\infty (u_{n,2})+\rho _n+o(1). \end{aligned}$$

From the properties of the cut-off and (3.6), we get

$$\begin{aligned} \left\| u_{n,2}\right\| _{L^2}^2=\left\| u_n\right\| ^2_{L^2}-\left\| u_{n,1}\right\| ^2_{L^2}-2\mathop {{\mathrm{Re}}}\nolimits \int _{R'}^{R'+a} u_{n,1}{\bar{u}}_{n,2}dx \rightarrow \mu . \end{aligned}$$

Since $\frac{1}{2}H(u_{n,1})+\rho _n>0$ by (3.9) and (3.13), we obtain

$$\begin{aligned} I=\lim _{n\rightarrow \infty }E(u_n)\geqslant \lim _{n\rightarrow \infty }E^\infty (u_{n,2})\geqslant I^\infty . \end{aligned}$$

which is a contradiction, hence (3.7) follows.

Now, from (3.7) we obtain

$$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \frac{1}{2}\int _0^\infty |u'_n|^2dx-\frac{c}{2}\int _0^\infty \frac{|u_n|^2}{x^2}dx-\frac{1}{p+1}\int _0^\infty |u_n|^{p+1}dx\right) = \\&\qquad =\lim _{n\rightarrow \infty } \left( \frac{1}{2}\int _0^\infty |u'_n|^2dx-\frac{1}{p+1}\int _0^\infty |u_n|^{p+1}dx\right) . \end{aligned}$$

Hence

$$\begin{aligned} I\geqslant I^\infty , \end{aligned}$$

which is again a contradiction. Thus $\{y_n\}_{n\in {\mathbb {N}}}$ is bounded and has an accumulation point $y^*\in {\mathbb {R}}^+$. Therefore, it follows that for any $\varepsilon >0$ there is $R>0$ such that

$$\begin{aligned} \int _0^R|u_n|^2\geqslant \mu -\varepsilon . \end{aligned}$$

for all $n\in {\mathbb {N}}$. Hence $u_n\rightarrow u$ strongly in $L^{2}({\mathbb {R}}^+)$. Moreover, since $\{u_n\}$ is bounded in $H^1_0({\mathbb {R}}^+)$ it is also strongly convergent in $L^{p+1}({\mathbb {R}}^+)$. By the weak-lower semicontinuity of H (see [17]), it follows that $E(u)\leqslant \lim _{n\rightarrow \infty }E(u_n)=I$. Hence $E(u)=I$, and $E(u_n)\rightarrow E(u)$ implies that $H(u_n)\rightarrow H(u)$, which concludes that proof. $\square$

Remark 3.4

If $c<0$, the infimum is not attained on the $L^2$ constraint. Indeed, let us assume that there exists $v\in H^1_0({\mathbb {R}}^+)$, such that $\left\| v\right\| ^2_{L^2}=\mu$ and $E(v)=I$. Then taking translates of v, i.e. $v(\cdot -y)$ for $y>0$, we get $E(v(\cdot -y))<I$, which is a contradiction.

Lemma 3.5

Let $0<c<1/4$, $\omega >0$ and $1<p<5$. Let $\mu$ be defined by Lemma 2.10. Then $u\in H^1_0({\mathbb {R}}^+)$ is a ground state solution of (2.1) if and only if u solves the minimization problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u \in \Gamma , \\ S(u)=\inf \{S(v): v\in \Gamma \}. \end{array}\right. } \end{aligned}$$

(3.14)

Proof

Step 1. Let us first define

$$\begin{aligned} m_{{\mathcal {A}}}=\inf \{S(u) : u \in {\mathcal {A}}\}, \end{aligned}$$

and

$$\begin{aligned} m_{\Gamma }=\inf \{S(u) : u \in \Gamma \}. \end{aligned}$$

If $u\in {\mathcal {G}}$, then $S(u)=m_\Gamma$. By Lemma 2.10 we know that $u\in \Gamma$, hence $m_{{\mathcal {A}}}\leqslant m_\Gamma$.

Step 2. We claim that every solution of (3.14) belongs to ${\mathcal {A}}$. Indeed, let us consider a solution u to (3.14). There exists a Lagrange multiplier $\lambda _1\in {\mathbb {R}}$ such that $S'(u)=\lambda _1 u$. Hence there exists $\lambda \in {\mathbb {R}}$ such that

$$\begin{aligned} -u'' -\frac{c}{x^2}u+ \lambda \omega u= |u|^{p-1}u. \end{aligned}$$

(3.15)

Indeed, since u is a solution of (3.14), and for $\lambda >0$ let

$$\begin{aligned} u_\lambda (x)=\lambda ^{1/2}u(\lambda x). \end{aligned}$$

We have $u_\lambda \in \Gamma$. Since $u_1$ is a solution of (3.14), we get from (3.15) and Lemma 2.2 that

$$\begin{aligned} \frac{\partial }{\partial \lambda }S(u_\lambda )|_{\lambda =1}=\left\| u'\right\| ^2_{L^2}-c\left\| \frac{u}{x}\right\| ^2_{L^2}-\frac{p-1}{2(p+1)} \left\| u\right\| ^{p+1}_{L^{p+1}}=0. \end{aligned}$$

(3.16)

We can deduce directly from (3.15) and (3.16) that

$$\begin{aligned} \lambda \omega \mu =\frac{p+3}{p-1} H(u), \end{aligned}$$

which implies that $\lambda >0$. Let us define v by

$$\begin{aligned} u(x)=\lambda ^{1/(p-1)}v(\lambda ^{1/2}x). \end{aligned}$$

By (3.16), $v\in {\mathcal {A}}$, hence

$$\begin{aligned} S(v)\geqslant m_{{\mathcal {A}}}. \end{aligned}$$

We obtain simple calculation that

$$\begin{aligned} m_{\Gamma }=S(u)=\lambda ^{2/(p-1)+1/2}S(v) + (1-\lambda )\frac{\omega \mu }{2}. \end{aligned}$$

Hence,

$$\begin{aligned} m_{{\mathcal {A}}}\geqslant \lambda ^{\frac{2}{p-1}+\frac{1}{2}}m_{{\mathcal {A}}}+(1-\lambda )\frac{\omega \mu }{2}. \end{aligned}$$

Since u is a solution of (3.15), we obtain from Lemma 2.2 that $m_{{\mathcal {A}}}\geqslant 0$. By Lemma 2.2 and Lemma 2.10 we have that

$$\begin{aligned} \frac{\omega \mu }{2}=\left( \frac{2}{p-1}+\frac{1}{2} \right) m_{{\mathcal {A}}}, \end{aligned}$$

hence

$$\begin{aligned} 0\geqslant \lambda ^{\frac{2}{p-1}+\frac{1}{2}} - \lambda \left( \frac{2}{p-1}+\frac{1}{2} \right) + \left( \frac{2}{p-1}-\frac{3}{2} \right) . \end{aligned}$$

The right hand side is always strictly positive, except if $\lambda =1$. Thus, $\lambda =1$, which implies together with (3.16) that $u\in {\mathcal {A}}$.

Step 3. It follows from Step 2, that $m_\Gamma \leqslant m_{{\mathcal {A}}}$, hence $m_\Gamma =m_{{\mathcal {A}}}$. In particular, it follows that if $u\in {\mathcal {G}}$, then $u \in \Gamma$ and $S(u)=m_{{\mathcal {A}}}$, thus u satisfies (3.14). Conversely, let u be the solution of (3.14). Then by Step 2 $u\in {\mathcal {A}}$, and $S(u)=m_\Gamma =m_{{\mathcal {A}}}$, hence $u \in {\mathcal {G}}$. $\square$

Theorem 3.6

Let $0<c<1/4$, $\omega >0$, and $1<p<5$. If $\varphi$ is a ground state solution of (2.1), then the standing wave $u(t,x)=e^{i\omega t}\varphi (x)$ is an orbitally stable solution of (1.1), i.e. for all $\varepsilon >0$ there is $\delta >0$, such that if $u(0)\in H^1_0({\mathbb {R}}^+)$ satisfies $\left\| \varphi -u(0)\right\| _{H^1}<\delta$, then the corresponding maximal solution u of (1.1) satisfies

$$\begin{aligned} \sup _{t\in {\mathbb {R}}}\inf _{\theta \in {\mathbb {R}}}\left\| u(t)-e^{i\theta }\varphi \right\| _{H^1}<\varepsilon . \end{aligned}$$

Proof

Assume by contradiction that there exist a sequence $\{\varphi _n\}_{n\in {\mathbb {N}}}\subset H^1_0({\mathbb {R}}^+)$, a sequence $\{t_n\}_{n\in {\mathbb {N}}}\subset {\mathbb {R}}$, and $\varepsilon >0$, such that

$$\begin{aligned} \lim _{n\rightarrow \infty } \left\| \varphi _n-\varphi \right\| _{H^1}=0, \end{aligned}$$

and the corresponding maximal solution $u_n$ of (1.1) with initial value $\varphi _n$ satisfies

$$\begin{aligned} \inf _{\theta \in {\mathbb {R}}}\left\| u_n(t_n)-e^{i\theta }\varphi \right\| _{H^1}\geqslant \varepsilon . \end{aligned}$$

Set $v_n=u_n(t_n)$. Applying Lemma 3.5, we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\inf _{\varphi \in {\mathcal {G}}}\left\| v_n-\varphi \right\| _{H^1}\geqslant \varepsilon . \end{aligned}$$

(3.17)

By the conservation of charge and energy, we obtain

$$\begin{aligned} \left\| v_n\right\| ^2_{L^2}\rightarrow \mu , \textit{ and } E(v_n)\rightarrow I. \end{aligned}$$

Hence $\{v_n\}_{n\in {\mathbb {N}}}$ is a minimizing sequence of (3.1). It follows from Lemma 3.3, that there exists a solution u of the problem (3.1), such that $\left\| v_n-u\right\| _{H^1}\rightarrow 0$. By Lemma 3.5 we obtain that $u\in {\mathcal {G}}$, which contradicts (3.17). $\square$

4 Instability

In this section we assume that $p \geqslant 5$. Let us define for $v\in H^1_0({\mathbb {R}}^+)$ the functional

$$\begin{aligned} Q(v)= \left\| v'\right\| ^2_{L^2}-c\left\| \frac{v}{x}\right\| ^2_{L^2}-\frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

In Lemma 2.2 we have shown that if v is a solution of (2.1), then $Q(v)=0$. First, we prove the virial identities.

Proposition 4.1

Let $u_0\in H^1_0({\mathbb {R}}^+)$ be such that $x u_0 \in L^2({\mathbb {R}}^+)$ and u be the corresponding maximal solution to (1.1). Then $x u(t) \in L^2 ({\mathbb {R}}^+)$ for any $t\in (-T_{\mathrm {min}},T_{\mathrm {max}})$. Moreover, the following identities hold for all $v\in H^1_0({\mathbb {R}}^+)$:

$$\begin{aligned} \frac{\partial }{\partial t} \left\| x u(t)\right\| ^2_{L^2}&= 4\mathop {{\mathrm{Im}}}\nolimits \int _0^\infty {\bar{u}}(t) x u'(t) dx, \\ \frac{\partial ^2}{\partial t^2} \left\| x u(t)\right\| ^2_{L^2}&= 8 Q(u(t)). \end{aligned}$$

Proof

The proof follows the same line as in [6]. $\square$

Proposition 4.2

Let $p\geqslant 5$ and let $u_0\in H^1_0({\mathbb {R}}^+)$ be such that

$$\begin{aligned} xu_0\in L^2({\mathbb {R}}^+) \text { and } E(u_0)<0. \end{aligned}$$

Then the maximal solution u to (1.1) with initial condition $u_0$ blows up in finite time.

Proof

First, let us note that

$$\begin{aligned} Q(u(t))=2E(u(t))+\frac{5-p}{2(p+1)}\left\| u(t)\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

Since $p\geqslant 5$, we get by the conservation of the energy that

$$\begin{aligned} Q(u(t))\leqslant 2E(u_0)<0 \text { for all } t\in (-T_{\mathrm {min}},T_{\mathrm {max}}). \end{aligned}$$

Hence, Proposition 4.1 implies that

$$\begin{aligned} \frac{\partial ^2}{\partial t^2} \left\| xu(t)\right\| ^2_{L^2}\leqslant 16 E(u_0) \text { for all } t\in (-T_{\mathrm {min}},T_{\mathrm {max}}). \end{aligned}$$

Integrating twice, we get

$$\begin{aligned} \left\| xu(t)\right\| ^2_{L^2}\leqslant 8E(u_0)t^2 +\left( 4\mathop {{\mathrm{Im}}}\nolimits \int _0^\infty {\bar{u}}_0 x u'_0dx \right) t +\left\| xu_0\right\| ^2_{L^2} \end{aligned}$$

(4.1)

The main coefficient of the second order polynomial on the right hand side is negative. Thus, it is negative for |t| large, what contradicts with $\left\| xu(t)\right\| ^2_{L^2}\geqslant 0$ for all t. Therefore, $-T_{\mathrm {min}}>-\infty$ and $T_{\mathrm {max}}<+\infty$. $\square$

Theorem 4.3

Assume that $\omega >0$ and $p=5$. Then for any solution $\varphi \in H^1_0({\mathbb {R}}^+)$ of (2.1) the standing wave $e^{i\omega t} \varphi (x)$ is unstable by blow-up.

Proof

Since $p=5$, we have for all $v\in H^1_0({\mathbb {R}}^+)$, that $2E(v)=Q(v)$. Hence from Lemma 2.2 we get that

$$\begin{aligned} E(\varphi )=0. \end{aligned}$$

Let us define $\varphi _{n,0}=\left( 1+\frac{1}{n}\right) \varphi$. It is easy to see that $E(\varphi _{n,0})<0$. By Lemma 2.1 we know that $x \varphi _{n,0}\in L^2({\mathbb {R}}^+)$. The conclusion follows from Proposition 4.2. $\square$

Theorem 4.4

Let $p>5$. Then for any ground state solution $\varphi$ to (2.1), the corresponding standing wave $e^{i\omega t}\varphi (x)$ is orbitally unstable.

We need to prove a series of Lemmas to establish Theorem 4.4.

Lemma 4.5

Let $v\in H^1_0({\mathbb {R}}^+)\setminus \{0\}$ such that $Q(v)\leqslant 0$, and set $v_\lambda (x)=\lambda ^{1/2}v(\lambda x)$ for $\lambda >0$. Then there exists $\lambda ^* \in (0,1]$ such that the following assertions hold:

• (1) $Q(v_{\lambda ^*})=0$.

• (2) $\lambda ^*=1$ if and only if $Q(v)=0$.

• (3) $\frac{\partial }{\partial \lambda } S(v_\lambda ) =\frac{1}{\lambda } Q(v_\lambda )$.

• (4) $\frac{\partial }{\partial \lambda } S(v_\lambda ) >0$ for all $\lambda \in (0,\lambda ^*)$, and $\frac{\partial }{\partial \lambda } S(v_\lambda ) <0$ for all $\lambda \in (\lambda ^*, +\infty )$.

• (5) The function $(\lambda ^*, +\infty )\ni \lambda \mapsto S(v_\lambda )$ is concave.

Proof

We get that by the scaling properties of $\lambda \mapsto Q(v_\lambda )$ that

$$\begin{aligned} Q(v_\lambda )=\lambda ^2\left\| v'\right\| ^2_{L^2}-\lambda ^2 c \left\| \frac{v}{x}\right\| ^2_{L^2}-\lambda ^{\frac{p-1}{2}}\frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

We get from the Hardy inequality that for $c\in (0,1/4)$

$$\begin{aligned}&(1-4c)\lambda ^2\left\| v'\right\| ^2_{L^2}-\lambda ^{\frac{p-1}{2}}\frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}} \\&\quad \leqslant Q(v_\lambda )\leqslant \lambda ^2\left\| v'\right\| ^2_{L^2}-\lambda ^{\frac{p-1}{2}}\frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

Since $p>5$, there exists $\lambda \in (0,1]$ small enough, such that $Q(v_\lambda )>0$. Hence, there exists $\lambda ^*\in (0,1]$, such that $Q(v_{\lambda ^*})=0$. This proves (1). To prove (2), we first note that if $\lambda ^*=1$, then clearly $Q(v)=0$. Now assume that $Q(v)=0$. Then

$$\begin{aligned} Q(v_\lambda )&=\lambda ^2Q(v)+(\lambda ^2-\lambda ^{\frac{p-1}{2}})\frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}} \\&=(\lambda ^2-\lambda ^{\frac{p-1}{2}})\frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}}, \end{aligned}$$

which is positive for all $\lambda \in (0,1)$, since $p>5$. Hence, (2) follows. (3) follows form simple calculation:

$$\begin{aligned} \frac{\partial }{\partial \lambda } S(v_\lambda )&=\lambda \left\| v'\right\| ^2_{L^2}-\lambda c\left\| \frac{v}{x}\right\| ^2_{L^2}-\lambda ^{\frac{p-1}{2}-1}\frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}} \\&=\frac{1}{\lambda }Q(v_\lambda ). \end{aligned}$$

To show (4), we note that

$$\begin{aligned} Q(v_\lambda )=\frac{\lambda ^2}{(\lambda ^*)^2} Q(v_{\lambda ^*}) + \lambda ^2\left( (\lambda ^*)^{\frac{p-5}{2}} -\lambda ^{\frac{p-5}{2}} \right) \frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

Since $p>5$ and $Q(v_{\lambda ^*})=0$, we get that $\lambda >\lambda ^*$ implies $Q(v_\lambda )<0$, and $\lambda <\lambda ^*$ implies $Q(v_\lambda )>0$. This and (3), implies (4).

Finally, we get by simple calculation that

$$\begin{aligned} \frac{\partial ^2}{\partial \lambda ^2} S(v_\lambda )= \frac{1}{\lambda ^2} Q(v_\lambda )-\lambda ^{\frac{p-5}{2}}\left( \frac{p-1}{2}-2 \right) \frac{p-1}{2(p+1)}\left\| v\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

Since $p>5$, we obtain for $\lambda >\lambda ^*$ that $\frac{\partial ^2}{\partial \lambda ^2} S(v_\lambda )<0$ which concludes the proof of (5). $\square$

To prove orbital instability we prove a new variational characterization of the ground state. Let us define the following set

$$\begin{aligned} {\mathcal {M}}=\{v\in H^1_0({\mathbb {R}}^+)\setminus \{0\}: Q(v)=0, J(v)\leqslant 0\}, \end{aligned}$$

and the corresponding minimization problem

$$\begin{aligned} d=\inf _{W\in {\mathcal {M}}} S(W). \end{aligned}$$

Then we have the following.

Lemma 4.6

The following equality holds:

$$\begin{aligned} m=d, \end{aligned}$$

where m is defined by (2.4).

Proof

Let $v\in {\mathcal {G}}$. Since v solves (2.1), by Lemma 2.2 we have that $Q(v)=J(v)=0$, hence ${\mathcal {G}}\subset {\mathcal {M}}$, and

$$\begin{aligned} d \leqslant m. \end{aligned}$$

Let now $v\in {\mathcal {M}}$. Assume first, that $J(v)=0$. In this case $v\in {\mathcal {N}}$, and $m\leqslant S(v)$. Let us assume that $J(v)<0$. Then for $v_\lambda (x)=\lambda ^{1/2}v(\lambda x)$ we have

$$\begin{aligned} J(v_\lambda )=\lambda ^2\left\| v'\right\| ^2_{L^2}-\lambda ^2c\left\| \frac{v}{x}\right\| ^2_{L^2}+\omega \left\| v\right\| ^2_{L^2}-\lambda ^{(p-1)/2}\left\| v\right\| ^{p+1}_{L^{p+1}}, \end{aligned}$$

and $\lim _{\lambda \downarrow 0} J(v_\lambda )>\omega \left\| v\right\| ^2_{L^2}$, thus there exists $\lambda _1\in (0,1)$, such that $J(v_{\lambda _1})=0$. By Proposition 2.9

$$\begin{aligned} m\leqslant S(v_{\lambda _1}). \end{aligned}$$

From $Q(v)=0$ and Lemma 4.5 we have

$$\begin{aligned} S(v_{\lambda _1})\leqslant S(v), \end{aligned}$$

hence $m\leqslant S(v)$ for all $v\in {\mathcal {M}}$. Therefore $m\leqslant d$, which concludes the proof. $\square$

We now define the manifold

$$\begin{aligned} {\mathcal {J}}=\{u\in H^1_0({\mathbb {R}}^+)\setminus \{0\}: J(u)<0, Q(u)<0, S(u)< d\}. \end{aligned}$$

We will prove the invariance of ${\mathcal {J}}$ under the flow of (1.1).

Lemma 4.7

Let $u_0 \in {\mathcal {J}}$ and $u\in C((-T_{\mathrm {min}},T_{\mathrm {max}}), H^1_0({\mathbb {R}}^+))$ the corresponding solution to (1.1). Then $u(t)\in {\mathcal {J}}$ for all $t\in (-T_{\mathrm {min}}, T_{\mathrm {max}})$.

Proof

Let $u_0\in {\mathcal {J}}$ and $u\in C((-T_{\mathrm {min}}, T_{\mathrm {max}}), H^1_0({\mathbb {R}}^+))$ the corresponding maximal solution. Since S is conserved under the flow of (1.1) we have for all $t\in (-T_{\mathrm {min}}, T_{\mathrm {max}})$ that

$$\begin{aligned} S(u(t))=S(u_0)<d. \end{aligned}$$

We prove the assertion by contradiction. Suppose that there exists $t\in (-T_{\mathrm {min}}, T_{\mathrm {max}})$ such that

$$\begin{aligned} J(u(t))\geqslant 0. \end{aligned}$$

Then, since J and u are continuous, there exists $t_0\in (-T_{\mathrm {min}}, T_{\mathrm {max}})$ such that

$$\begin{aligned} J(u(t_0))=0, \end{aligned}$$

thus $u(t_0) \in {\mathcal {N}}$. Then by Proposition 2.9 we have that

$$\begin{aligned} S(u(t_0))\geqslant d, \end{aligned}$$

which is a contradiction, thus $J(u(t))<0$ for all $t \in (-T_{\mathrm {min}},T_{\mathrm {max}})$. Let us suppose now that for some $t\in (-T_{\mathrm {min}}, T_{\mathrm {max}})$ we have

$$\begin{aligned} Q(u(t))\geqslant 0. \end{aligned}$$

Again, by continuity, there exists $t_1 \in (-T_{\mathrm {min}}, T_{\mathrm {max}})$ such that

$$\begin{aligned} Q(u(t_1))=0. \end{aligned}$$

Hence we that $Q(u(t_1))=0$, and $J(u(t_1))<0$. Therefore, by Lemma 4.6

$$\begin{aligned} S(u(t_1))\geqslant d, \end{aligned}$$

which is a contradiction. Hence,

$$\begin{aligned} Q(u(t))<0 \end{aligned}$$

for all $t \in (-T_{\mathrm {min}},T_{\mathrm {max}})$, which concludes the proof. $\square$

Lemma 4.8

Let $u_0 \in {\mathcal {J}}$ and $u\in C((-T_{\mathrm {min}},T_{\mathrm {max}}),H^1_0({\mathbb {R}}^+))$. Then there exists $\varepsilon >0$ such that $Q(u(t))\leqslant -\varepsilon$ for all $t\in (-T_{\mathrm {min}},T_{\mathrm {max}})$.

Proof

Let $u_0 \in {\mathcal {J}}$ and let us define $v:=u(t)$ and $v_\lambda (x)=\lambda ^{1/2}v(\lambda x)$. By Lemma 4.5, there exists $\lambda _0<1$ such that $Q(v_{\lambda ^*})=0$. If $J(v_{\lambda ^*})\leqslant 0$, then by Lemma 4.7 we get $S(v_{\lambda ^*})\geqslant m.$ On the other hand, if $J(v_{\lambda ^*})>0$, there exists $\lambda _1\in (\lambda ^*,1)$, such that $J(\lambda _1)=0$ and we replace $\lambda ^*$ with $\lambda _1$. In this case, by Lemma 4.6 we get $S(v_{\lambda ^*})\geqslant m$. In conclusion, in both cases we obtain

$$\begin{aligned} S(v_{\lambda ^*})\geqslant d. \end{aligned}$$

(4.2)

By Lemma 4.5 we know that $\lambda \mapsto S(v_\lambda )$ is concave on $(\lambda ^*, +\infty )$, thus

$$\begin{aligned} S(v)-S(v_{\lambda ^*})\geqslant (1-\lambda ^*)\frac{\partial }{\partial \lambda } S(v_\lambda )\Big |_{\lambda =1}. \end{aligned}$$

(4.3)

From Lemma 4.5 we have

$$\begin{aligned} \frac{\partial }{\partial \lambda }S(v_\lambda )\Big |_{\lambda =1}=Q(v). \end{aligned}$$

(4.4)

Moreover, since $Q(v)<0$ and $\lambda ^*\in (0,1)$, we have

$$\begin{aligned} (1-\lambda ^*)Q(v) > Q(v). \end{aligned}$$

(4.5)

Combining (4.2)–(4.5), we obtain

$$\begin{aligned} S(v)-d > Q(v). \end{aligned}$$

Define $-\varepsilon = S(v)-d$. Then $\varepsilon >0$, since $v\in {\mathcal {J}}$. Owing to the conservation of the energy and mass, $\varepsilon >0$ is independent from t, which concludes the proof. $\square$

Lemma 4.9

Let us take $u_0\in {\mathcal {J}}$ such that $x u_0 \in L^2({\mathbb {R}}^+)$. Then the maximal solution $u\in C((-T_{\mathrm {min}},T_{\mathrm {max}}),H^1_0({\mathbb {R}}^+))$ corresponding to the initial value problem (1.1) blows up in finite time.

Proof

From Lemma 4.8 we know that there exists $\varepsilon >0$ such that

$$\begin{aligned} Q(u(t))< -\varepsilon \text { for } t\in (-T_{\mathrm {min}},T_{\mathrm {max}}). \end{aligned}$$

From Proposition 4.1 we know that $\frac{\partial ^2}{\partial t^2}\left\| xu(t)\right\| ^2_{L^2}=8Q(u(t))$, and by integration we get

$$\begin{aligned} \left\| xu(t)\right\| ^2_{L^2}\leqslant -4\varepsilon t^2 + C_1t+C_2. \end{aligned}$$

(4.6)

The right hand side of (4.6) is negative for large |t|, which contradicts with $\left\| x u(t)\right\| ^2_{L^2}>0$ for all t. Therefore, $T_{\mathrm {min}}>-\infty$ and $T_{\mathrm {max}}<\infty$ and by local well-posedness it follows that

$$\begin{aligned} \lim _{t\downarrow -T_{\mathrm {min}}}\left\| u(t)\right\| _{H^1}=+\infty , \text { and } \lim _{t\uparrow T_{\mathrm {max}}} \left\| u(t)\right\| _{H^1}=+\infty . \end{aligned}$$

$\square$

Proof of Theorem 4.4

Let $\varphi \in {\mathcal {G}}$. Owing to Lemma 4.9, it suffices to show that there exists a sequence $\{\varphi _\lambda \}\subset {\mathcal {J}}$, which converges to $\varphi$ in $H^1_0({\mathbb {R}}^+)$. Let us put $\varphi _\lambda (x)=\lambda ^{1/2}\varphi (\lambda x)$. By Lemma 4.5$\{\varphi _\lambda \}\subset {\mathcal {J}}$ for all $\lambda \in (0,1)$. Additionally, by Proposition 2.1, $\varphi$ decays exponentially at infinity, and so does $\varphi _\lambda$. Therefore, $x\varphi _\lambda \in L^2({\mathbb {R}}^+)$. Clearly, $\varphi _\lambda \rightarrow \varphi$ as $\lambda \rightarrow 0$, and by Lemma 4.9 the maximal solution of (1.1) corresponding to $\varphi _\lambda$, blows up in finite time for all $\lambda \in (0,1)$. Hence, the conclusion follows. $\square$

Appendix

We prove the following Lemma:

Lemma 5.1

Let $\psi _A(x)=q(x+A)-q(x-A)$, where q is (2.6). Then $\psi _A\in H^1_0({\mathbb {R}}^+)$ and for large $A>0$, we have the following approximations:

$$\begin{aligned} \int _0^\infty |\psi _A'|^2dx&= \int _{-\infty }^\infty |q'|^2dx+O\left( \left( 2A+\frac{1}{\sqrt{\omega }}\right) e^{-2\sqrt{\omega } A}\right) , \end{aligned}$$

(5.1)

$$\begin{aligned} \int _0^\infty |\psi _A|^2dx&= \int _{-\infty }^\infty |q|^2dx +O\left( \left( 2A+\frac{1}{\sqrt{\omega }}\right) e^{-2\sqrt{\omega } A}\right) , \end{aligned}$$

(5.2)

$$\begin{aligned} \int _0^\infty \frac{|\psi _A(x)|^2}{x^2}&\lesssim \frac{1}{A^2}\int _{-\infty }^\infty |q|^2dx+ O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) , \end{aligned}$$

(5.3)

$$\begin{aligned} \int _0^\infty |\psi _A(x)|^{p+1}dx&=\int _{-\infty }^\infty |q|^{p+1}dx + O(e^{-2\sqrt{\omega } A}). \end{aligned}$$

(5.4)

Proof

We will use the fact that $q(x)\leqslant Me^{-\sqrt{\omega }|x|}$ and $q'(x)\leqslant Me^{-\sqrt{\omega }|x|}$ for some $M>0$.

We get (5.1) by using the symmetry of q and $q'$:

$$\begin{aligned} \int _0^\infty |\psi _A'|^2dx=\int _{-\infty }^\infty |q'|^2dx -\int _{-\infty }^\infty q'(x+A)q'(x-A)dx. \end{aligned}$$

We estimate the second term by

$$\begin{aligned}&\left| \int _{-\infty }^\infty q'(x+A)q'(x-A)dx\right| \\&\quad \lesssim \int _{-\infty }^\infty e^{-\sqrt{\omega }|x+A|-\sqrt{\omega }|x-A|}dx=\left( \left( 2A+\frac{1}{\sqrt{\omega }}\right) e^{-2\sqrt{\omega } A} \right) , \end{aligned}$$

hence (5.1) follows. We get (5.2) the same way.

We now show (5.3). From Hardy’s inequality we get

$$\begin{aligned} \int _0^{A/2}\frac{|\psi _A|^2}{x^2}dx\leqslant 4\int _0^{A/2}|\psi '_A(x)|^2=O(e^{-\sqrt{\omega } A}). \end{aligned}$$

Moreover, we have

$$\begin{aligned} \int _{A/2}^\infty \frac{|\psi _A(x)|^2}{x^2}dx \leqslant \frac{4}{A^2}\int _{A/2}^\infty |\psi _A|^2dx=\frac{4}{A^2}\int _{-\infty }^\infty |q|^2+O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) . \end{aligned}$$

Hence

$$\begin{aligned} \int _0^\infty \frac{|\psi _A|^2}{x^2}dx&=\int _0^{A/2}\frac{|\psi _A|^2}{x^2}dx+\int ^\infty _{A/2}\frac{|\psi _A|^2}{x^2}dx \\&\leqslant \frac{4}{A^2}\int _{-\infty }^\infty |q|^2dx+O\left( \frac{1}{A^2}e^{-\sqrt{\omega } A}\right) , \end{aligned}$$

which is the estimate in (5.3).

To show (5.4), we use the fact that

$$\begin{aligned} |q(x-A)-q(x+A)|^{p+1}=q^{p+1}(x-A)-(p+1)q^p(x-A)q(x+A)+O(q^2(x+A)). \end{aligned}$$

We get

$$\begin{aligned} \int _0^\infty q^{p+1}(x-A)dx&=\int _{-\infty }^\infty q^{p+1}(x)dx-\int _{-\infty }^{-A}q^{p+1}(x)dx \\&=\int _{-\infty }^\infty q^{p+1}(x)dx+O(e^{-\sqrt{\omega }(p+1)A}), \\ \int _0^\infty q^p(x-A)q(x+A)dx&\lesssim \int _0^\infty e^{-\sqrt{\omega } p|x-A|-\sqrt{\omega }|x+A|}dx =O\left( e^{-2\sqrt{\omega } A}\right) , \\ \int _0^\infty O(q^2(x+A))dx&= O(e^{-2\sqrt{\omega } A}). \end{aligned}$$

Hence

$$\begin{aligned} \int _0^\infty |\psi _A(x)|^{p+1}dx= \int _{-\infty }^\infty |q|^{p+1}dx+O(e^{-2\sqrt{\omega } A}). \end{aligned}$$

This concludes the proof. $\square$

We now state the proof of Lemma 2.8. The proof follows the arguments of the paper [11], with some important modifications. We introduce the norm

$$\begin{aligned} \left\| u\right\| ^2=\int _0^\infty \left( |u'|^2-c\frac{|u|^2}{x^2}+\omega |u|^2\right) dx, \end{aligned}$$

which is equivalent to the standard norm on $H^1_0({\mathbb {R}}^+)$ if $0<c<1/4$.

Proof of Lemma 2.8

Step 1. There exists $u_0\in H^1_0({\mathbb {R}}^+)$, such that, up to a subsequence, $u_n$ is weakly convergent to $u_0$ in $H^1_0({\mathbb {R}}^+)$, and $S'(u_0)=0.$

Since $\{u_n\}_{n\in {\mathbb {N}}}$ is bounded in $H^1_0({\mathbb {R}}^+)$, it admits a weakly convergent subsequence in $H^1_0({\mathbb {R}}^+)$ with a weak limit $u_0\in H^1_0({\mathbb {R}}^+)$. We only need to show that $S'(u_0)=0$. Since by our assumption $S'(u_n)\rightarrow 0$, it suffices to show that for all $\varphi \in C^\infty _0({\mathbb {R}}^+)$ we have

$$\begin{aligned} S'(u_n)\varphi -S'(u_0)\varphi \rightarrow 0. \end{aligned}$$

Indeed, we have

$$\begin{aligned} S'(u_n)\varphi -S'(u_0)\varphi&= \mathop {{\mathrm{Re}}}\nolimits \int ^\infty _0 (u_n'-u_0'){\bar{\varphi }}'dx-c\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty \frac{(u_n-u_0){\bar{\varphi }}}{x^2}dx \\&\quad +\omega \mathop {{\mathrm{Re}}}\nolimits \int _0^\infty (u_n-u_0){\bar{\varphi }} dx\\&\quad -\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty (|u_n|^{p-1}u_n-|u_0|^{p-1}u_0){\bar{\varphi }} dx. \end{aligned}$$

Since $u_n\rightharpoonup u_0$ in $H^1_0({\mathbb {R}}^+)$ and strongly in $L^q_{\mathrm {loc}}({\mathbb {R}}^+)$ for all $q\geqslant 1$, our statement follows.

Let us set $v_n=u_n-u_0$.

Step 2. Assume that

$$\begin{aligned} \sup _{z\in {\mathbb {R}}^+}\int _{B_1(z)}|v_n|^2dx\rightarrow 0, \end{aligned}$$

(5.5)

where $B_1(z)$ is the unit ball centered at z. Then $u_n\rightarrow u_0$ strongly in $H^1_0({\mathbb {R}}^+)$, and Lemma 2.8holds with $k=0$.

Using the fact that $S'(u_0)=0$, we get

$$\begin{aligned} S'(u_n)v_n&=\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty u'_n{\bar{v}}'_n dx -c\mathop {{\mathrm{Re}}}\nolimits \int ^\infty _0 \frac{u_n{\bar{v}}_n}{x^2}dx+\omega \mathop {{\mathrm{Re}}}\nolimits \int _0^\infty u_n{\bar{v}}_ndx \\&\quad -\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty |u_n|^{p-1}u_n{\bar{v}}_ndx=\\&=\left\| v_n\right\| ^2+\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty (|u_0|^{p-1}u_0-|u_n|^{p-1}u_n){\bar{v}}_n dx. \end{aligned}$$

Hence,

$$\begin{aligned} \left\| v_n\right\| ^2=S'(u_n)v_n+\mathop {{\mathrm{Re}}}\nolimits \int ^\infty _0(|u_n|^{p-1}u_n-|u_0|^{p-1}u_0){\bar{v}}_ndx. \end{aligned}$$

We recall that $S'(u_n)\rightarrow 0$. Hölder’s inequality implies that

$$\begin{aligned} \left| \int _0^\infty |u_n|^{p-1}u_nv_n dx\right| \leqslant \left\| u_n\right\| _{L^{p+1}}^p\left\| v_n\right\| _{L^{p+1}}. \end{aligned}$$

Assumption (5.5) and Lemma 1.1 in [16] implies that $\left\| v_n\right\| _{L^{p+1}}\rightarrow 0$. Hence

$$\begin{aligned} \mathop {{\mathrm{Re}}}\nolimits \int _0^\infty |u_n|^{p-1}u_n{\bar{v}}_n dx \rightarrow 0. \end{aligned}$$

We obtain similarly that $\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty |u_0|^{p-1}u_0{\bar{v}}_n dx \rightarrow 0$, hence $\left\| v_n\right\| ^2\rightarrow 0$, which completes the proof of Step 2.

Step 3. Assume that there exist $\{z_n\}_{n\in {\mathbb {N}}}\subset {\mathbb {R}}^+$ and $d>0$ , such that

$$\begin{aligned} \int _{B_1(z_n)}|v_n|^2 dx \rightarrow d. \end{aligned}$$

(5.6)

Then, up to a subsequence, we have for $q\in H^1({\mathbb {R}})$ , that (i) $z_n\rightarrow \infty$ , (ii) $u_n(\cdot +z_n)\rightharpoonup q \ne 0$ in $H^1({\mathbb {R}})$ , and (iii) ${S^\infty }'(q)=0$ .

To show (i), let us assume by contradiction that $\{z_n\}_{n\in {\mathbb {N}}}$ has an accumulation point $z^*\in {\mathbb {R}}^+$. Then for a subsequence of $\{v_n\}_{n\in {\mathbb {N}}}$ we have

$$\begin{aligned} \int _{B_2(z^*)}|v_n|^2dx \geqslant d. \end{aligned}$$

Since $v_n\rightharpoonup 0$ in $H^1_0({\mathbb {R}}^+)$, we have $v_n \rightarrow 0$ in $L^2(B_2(z^*))$, which implies that

$$\begin{aligned} d \leqslant \lim _{n\rightarrow \infty }\int _{B_2(z^*)}|v_n|^2dx=0, \end{aligned}$$

which is a contradiction, hence (i) holds.

Since $u_n(\cdot + z_n)$ is bounded in $H^1({\mathbb {R}})$ the re exists $q\in H^1({\mathbb {R}})$ such that $u_n(\cdot +z_n)$ converges weakly to q in $H^1({\mathbb {R}})$. We only need to show that $q\ne 0$. Since $u_0(\cdot + z_n) \rightharpoonup 0$ in $H^1({\mathbb {R}})$, we have that $v_n(\cdot + z_n) \rightharpoonup q$ in $H^1({\mathbb {R}})$, and in $L^2_{\mathrm {loc}}({\mathbb {R}})$ in particular. Hence

$$\begin{aligned} \int _{B_1(0)}|q(x)|^2dx=\lim _{n\rightarrow \infty }\int _{B_1(0)}|v_n(x+z_n)|^2dx=\int _{B_1(z_n)}|v_n(y)|^2dy\geqslant d>0. \end{aligned}$$

This implies that $q\ne 0$.

We finally show (iii). We define ${\tilde{u}}(\cdot )=u_n(\cdot +z_n)$. We obtain, similarly as in Step 1, that for any $\varphi \in C^\infty _0({\mathbb {R}})$,

$$\begin{aligned} {S^\infty }'({\tilde{u}}_n)\varphi -{S^\infty }'(q)\varphi \rightarrow 0. \end{aligned}$$

It remains to show that ${S^\infty }'({\tilde{u}}_n)\varphi \rightarrow 0$. For any fixed $\varphi \in C^\infty _0({\mathbb {R}})$, $\varphi (\cdot -z_n)$ is in $H^1_0({\mathbb {R}}^+)$ for sufficiently big $n\in {\mathbb {N}}$. Hence, we obtain

$$\begin{aligned} S'(u_n)\varphi (\cdot -z_n)&=\mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty u'_n(x+z_n){\bar{\varphi }}'_n(x) dx-c\mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty \frac{u_n(x+z_n){\bar{\varphi }}(x)}{(x+z_n)^2}dx\\&\quad +\omega \mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty u_n(x+z_n){\bar{\varphi }}(x)dx \\&\quad -\mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty |u_n(x+z_n)|^{p-1}u_n(x+z_n){\bar{\varphi }}(x)dx. \end{aligned}$$

Since $S'(u_n)\rightarrow 0$ and $\varphi (\cdot - z_n)$ is bounded in $H^1({\mathbb {R}})$, it follows

$$\begin{aligned} \mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty&{\tilde{u}}'_n(x){\bar{\varphi }}'_n(x) dx-c\mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty \frac{{\tilde{u}}_n(x){\bar{\varphi }}(x)}{(x+z_n)^2}dx \\&\quad +\omega \mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty {\tilde{u}}_n(x){\bar{\varphi }}(x)dx-\mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty |{\tilde{u}}_n(x)|^{p-1}{\tilde{u}}_n(x){\bar{\varphi }}(x)dx\rightarrow 0. \end{aligned}$$

Moreover, since $u_n$ is bounded in $L^\infty$, and $\varphi$ is compactly supported, we get

$$\begin{aligned}&\left| \mathop {{\mathrm{Re}}}\nolimits \int _{-z_n}^\infty \frac{{\tilde{u}}_n(x){\bar{\varphi }}(x)}{(x+z_n)^2}dx\right| \\&\quad = \left| \mathop {{\mathrm{Re}}}\nolimits \int _0^\infty \frac{u_n(x){\bar{\varphi }}(x-z_n)}{x^2}dx\right| \leqslant \frac{1}{(z_n-\inf \{{{\,\mathrm{supp}\,}}(\varphi )\})^2}\left\| u_n\varphi \right\| _{L^\infty }\rightarrow 0, \end{aligned}$$

Thus

$$\begin{aligned} {S^\infty }'({\tilde{u}}_n)\varphi&=\mathop {{\mathrm{Re}}}\nolimits \int _{-\infty }^\infty {\tilde{u}}'_n(x){\bar{\varphi }}'_n(x) dx+\omega \mathop {{\mathrm{Re}}}\nolimits \int _{-\infty }^\infty {\tilde{u}}_n(x){\bar{\varphi }}(x)dx \\&\quad -\mathop {{\mathrm{Re}}}\nolimits \int _{-\infty }^\infty |{\tilde{u}}_n(x)|^{p-1}{\tilde{u}}_n(x){\bar{\varphi }}(x)dx\rightarrow 0, \end{aligned}$$

which concludes the proof of Step 3.

Step 4. Suppose there exist $k\geqslant 1$, $\{ x_n^i\}\subset {\mathbb {R}}^+$, $q_i\in H^1({\mathbb {R}})$ for $1\leqslant i\leqslant k$ , such that

$$\begin{aligned} x_n^i\rightarrow \infty , \quad |x_n^i-x_n^{j}|\rightarrow \infty \textit{ if } i\ne j,\\ u_n(\cdot + x_n^i)\rightarrow q_i\ne 0, \textit{ for all } 1\leqslant i \leqslant k,\\ {S^\infty }'(q_i)=0. \end{aligned}$$

Then

(1) If $\sup _{z\in {\mathbb {R}}^+}\int _{B_1(z)}|u_n-u_0-\sum _{i=1}^kq_i(\cdot -x_n^i)|^2dx\rightarrow 0$ then

$$\begin{aligned} \left\| u_n-u_0-\sum _{i=1}^kq_i(\cdot -x_n^i)\right\| _{H^1}\rightarrow 0. \end{aligned}$$

(2) If there exist $\{z_n\}\subset {\mathbb {R}}^+$ and $d>0$, such that

$$\begin{aligned} \int _{B_1(z_n)}\left| u_n-u_0-\sum _{i=1}^kq_i(\cdot -x_n^i)\right| ^2dx \rightarrow d, \end{aligned}$$

then, up to a subsequence, it follows that

$$\begin{aligned} (i)\textit{ } z_n\rightarrow \infty , \textit{ and } |z_n-x_n^i|\rightarrow \infty \textit{ for all } 1\leqslant i \leqslant k,\\ (ii)\textit{ } u_n(\cdot +z_n )\rightharpoonup q_{i+1} \quad (iii)\textit{ } {S^\infty }'(q_{i+1})=0. \end{aligned}$$

Suppose assumption (1) holds. We introduce $\xi _n=u_n-u_0-\sum _{i=1}^kq^a_i(\cdot - x_n^i)$, where $q^a_i$ is a suitable cut-off of $q_i$, such that ${{\,\mathrm{supp}\,}}(q^a_i)\subset (0,\infty )$. This is possible owing to the exponential decay of $q_i$ at infinity, and $x_n^i\rightarrow \infty$ as $n\rightarrow \infty$ for all i. We get

$$\begin{aligned} S'(u_n)\xi _n&=\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty u'_n{\bar{\xi }}'_ndx-c\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty \frac{u_n{\bar{\xi }}_n}{x^2}dx+\omega \mathop {{\mathrm{Re}}}\nolimits \int _0^\infty u_n{\bar{\xi }}_ndx \\&\quad -\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty |u_n|^{p-1}u_n{\bar{\xi }}_ndx\\&=\left\| \xi _n\right\| ^2+\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty (u'_0+\sum _{i=1}^k {q^a_i}'(\cdot -x_n^i)){\bar{\xi }}'_ndx \\&\quad +\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty \left( \omega -\frac{c}{x^2}\right) \left( u_0+\sum _{i=1}^kq^a_i(\cdot -x_n^i)\right) {\bar{\xi }}_ndx \\&\quad -\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty |u_n|^{p-1}u_n{\bar{\xi }}_ndx. \end{aligned}$$

Since $S'(u_0)\xi _n=0$, we get

$$\begin{aligned} S'(u_n)\xi _n&=\left\| \xi _n\right\| ^2+\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty (|u_0|^{p-1}u_0-|u_n|^{p-1}u_n){\bar{\xi }}_ndx \\&\quad + \mathop {{\mathrm{Re}}}\nolimits \int _0^\infty \sum _{i=1}^k{q^a_i}'(\cdot -x_n^i){\bar{\xi }}'_ndx \\&\quad +\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty \left( \omega -\frac{c}{x^2} \right) \sum _{i=1}^kq^a_i(\cdot -x_n^i){\bar{\xi }}_ndx. \end{aligned}$$

Using the fact that $\left\| \xi _n\right\| _{L^{p+1}}\rightarrow 0$ by Lemma 1.1 in [16], we get that the second term of the right hand side converges to zero. Now, from the weak convergence of $\xi _n$ to zero and that $S'(u_n)\rightarrow 0$,we obtain that $\left\| \xi _n\right\| \rightarrow 0$.

Suppose now that assumption (2) holds. Then (i) and (ii) follows as in Step 3. To show (ii), let us set ${\tilde{u}}_n=u_n(\cdot + z_n)$. We note that

$$\begin{aligned} {S^\infty }'({\tilde{u}}_n)\varphi - {S^\infty }'(q)\varphi \rightarrow 0, \end{aligned}$$

for all $\varphi \in C^\infty _0({\mathbb {R}})$. Now ${S^\infty }'({\tilde{u}}_n)\rightarrow 0$ follows similarly as in Step 3, which concludes the proof.

Step 5. Conclusion By Step 1 we know that $u_n\rightharpoonup u_0$ and $S'(u_0)=0$. Hence (i) of Lemma 2.8 is verified. If the assumption of Step 2 holds, then Lemma 2.8 is true with $k=0$. Otherwise, the assumption of Step 3 holds. We have to iterate Step 4. We only need to show that assumption 1 of Step 4 occurs after a finite number of iterations. Let us notice that

$$\begin{aligned}&\left\| u_n-u_0-\sum _{i=1}^kq_i(\cdot - x_i^n)\right\| ^2_{H^1} \\&\quad =\left\| u_n\right\| ^2_{H^1}+\left\| u_0\right\| ^2_{H^1}+\sum _{i=1}^k\left\| q_i\right\| ^2_{H^1}-2\left\langle u_n,u_0+\sum _{i=1}^kq_i(\cdot -x_i^n)\right\rangle _{H^1}. \end{aligned}$$

Moreover, since $u_n\rightharpoonup u_0$ and $u_n(\cdot + x_i^n)\rightharpoonup q_i$, we get for the last term that

$$\begin{aligned} \left\langle u_n,u_0+\sum _{i=1}^kq_i(\cdot -x_i^n)\right\rangle _{H^1} \rightarrow \left\| u_0\right\| ^2_{H^1}+\sum _{i=1}^k\left\| q_i\right\| ^2_{H^1}, \end{aligned}$$

Now since $u_n$ converges weakly to $u_0$, we obtain for $k\geqslant 1$ that

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| u_n\right\| ^2_{H^1}-\left\| u_0\right\| ^2_{H^1}-\sum _{i=1}^k\left\| q_i\right\| ^2_{H^1}=\lim _{n\rightarrow \infty }\left\| u_n-u_0-\sum _{i=1}^kq_i(\cdot - x_i^n)\right\| ^2_{H^1}\geqslant 0. \end{aligned}$$

Since $q_i$ is a nontrivial critical point of $S^\infty$, it is true that $\left\| q_i\right\| _{H^1}\geqslant \epsilon >0$. Hence, after a finite number of iterations assumption 1 of Step 4 must occur.

Finally, we have to verify that

$$\begin{aligned} S(u_n)\rightarrow S(u_0)+\sum _{i=1}^kS^\infty (q_i). \end{aligned}$$

We first show that

$$\begin{aligned} S(u_n)\rightarrow S(u_0)+S^\infty (v_n). \end{aligned}$$

(5.7)

A straightforward calculation gives

$$\begin{aligned} S(u_n)&=S(u_0)+S^\infty (v_n)+\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty u'_0({\bar{u}}'_n-{\bar{u}}'_0)dx-c\mathop {{\mathrm{Re}}}\nolimits \int _0^\infty \frac{u_0({\bar{u}}_n-{\bar{u}}_0)}{x^2}dx \\&\quad +\omega \mathop {{\mathrm{Re}}}\nolimits \int _0^\infty u_0({\bar{u}}_n-{\bar{u}}_0)dx -\frac{c}{2}\int _0^\infty \frac{|u_n-u_0|^2}{x^2}dx \\&\quad +\frac{1}{p+1}\left( \left\| u_n-u_0\right\| ^{p+1}_{L^{p+1}}-\left\| u_n\right\| ^{p+1}_{L^{p+1}}+\left\| u_n\right\| ^{p+1}_{L^{p+1}}\right) \end{aligned}$$

From a lemma by Brezis and Lieb (see e.g. Lemme 4.6 [12]) we have

$$\begin{aligned} \int ^\infty _0|u_n-u_0|^{p+1}dx-\int ^\infty _0|u_n|^{p+1}dx+\int ^\infty _0|u_0|^{p+1}dx\rightarrow 0. \end{aligned}$$

Hence (5.7) follows. It only remains to show that

$$\begin{aligned} S^\infty (v_n)\rightarrow \sum _{i=1}^kS^\infty (q_i). \end{aligned}$$

We calculate

$$\begin{aligned} S(v_n)&=\frac{1}{2}\left\| v_n-\sum _{i=1}^kq_i(\cdot - x_i^n)\right\| ^2_{H^1}+\frac{1}{2}\left\| \sum _{i=1}^kq_i(\cdot - x_i^n)\right\| ^2_{H^1} \\&\quad +\left\langle v_n-\sum _{i=1}^kq_i(\cdot -x_i^n),\sum _{i=1}^kq_i(\cdot - x_i^n) \right\rangle _{H^1} -\frac{1}{p+1}\left\| \sum _{i=1}^kq_i(\cdot - x_i^n)\right\| ^{p+1}_{L^{p+1}} \\&\quad -\frac{1}{p+1}\left\| v_n\right\| ^{p+1}_{L^{p+1}}+\frac{1}{p+1}\left\| \sum _{i=1}^kq_i(\cdot -x_i^n)\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

We have shown that $v_n-\sum _{i=1}^kq_i(\cdot - x_i^n)\rightarrow 0$ strongly in $H^1$. Hence the first and third term above converges to zero as $n\rightarrow \infty$. By using Sobolev’s inequality and $\left\| A-B\right\| \geqslant |\left\| A\right\| -\left\| B\right\| |$ we have

$$\begin{aligned} \left\| \sum _{i=1}^kq_i(\cdot -x_i^n)\right\| ^{p+1}_{L^{p+1}}-\left\| v_n\right\| ^{p+1}_{L^{p+1}}\rightarrow 0, \end{aligned}$$

which concludes the proof. $\square$