Strong Instability of Standing Waves for the Nonlinear Schrödinger Equation in Trapped Dipolar Quantum Gases

Abstract

In this paper, we study the strong instability of standing waves for the nonlinear Schrödinger equation arising in trapped dipolar quantum gases. Two cases are considered: the first when the system is free, the second when a partial/complete harmonic potential is added. In the free case, we present a new argument to prove that the ground state standing waves are strongly unstable by blow-up. In the second case, if $\partial ^2_\mu S_\omega (Q^{\mu }_\omega )|_{\mu =1}\le 0$, we deduce that the ground state standing wave $u(t,x)=e^{i\omega t}Q_\omega (x)$ is strongly unstable by blow-up, where $S_\omega$ is the action, and $Q_\omega ^{\mu }=\mu ^{3/2}Q_\omega (\mu x)$ is the $L^2$-invariant scaling.

1 Introduction

Recently, the dipolar Bose–Einstein condensate, i.e., a condensate made out of particles possessing a permanent electric or magnetic dipole moment, has attracted much attention; see, e.g., [1,2,3,4,5,6]. Within the realm of the Gross-Pitaevskii (mean field) approximation, the corresponding Bose-Einstein condensates are described by the following Schrödinger equation, see, e.g., [1, 2, 6, 7],

$$\begin{aligned} i\hbar \partial _tu=-\frac{\hbar ^2}{2m}\Delta u+W(x)u+ U_0|u|^{2}u+(V_{dip}*|u|^{2})u,~~~ x\in {\mathbb {R}}^3, ~~t>0, \end{aligned}$$

(1.1)

where $\hbar$ is the Planck constant, m is the mass of a dipolar particle, $*$ denotes the convolution, and W(x) is a partial/complete harmonic potential defined by

$$\begin{aligned} W(x)=a^2(x_1^2+\cdots +x_d^2), \end{aligned}$$

(1.2)

where $1 \le d \le 3$ and a is the trapping frequency. $U_0=4\pi \hbar ^2a_s/m$ denotes the local interaction between dipoles in the condensate with $a_s$ the s-wave scattering length (positive for repulsive interaction and negative for attractive interaction). The long-range dipolar interaction potential between two dipoles is given by

$$\begin{aligned} V_{dip}(x)=\frac{\mu _0\mu ^2_{dip}}{4\pi }\frac{1-3cos^2\theta }{|x|^3},~~~ x\in {\mathbb {R}}^3, \end{aligned}$$

(1.3)

where $\mu _0$ is the vacuum magnetic permeability, $\mu _{dip}$ is the permanent magnetic dipole moment, and $\theta$ is the angle between $x\in {\mathbb {R}}^3$ and the dipole axis ${\mathbf {n}}\in {\mathbb {R}}^3$, with $|{\mathbf {n}}|=1$. In other words, $\theta$ is defined by

$$\begin{aligned} \cos \theta =\frac{x\cdot {\mathbf {n}}}{|x|}. \end{aligned}$$

In order to simplify the mathematical analysis, we rescale (1.1) into the following dimensionless Schrödinger equation:

$$\begin{aligned} \left\{ \begin{array}{l} i\partial _tu=-\frac{1}{2}\Delta u+W(x)u+ \lambda _1|u|^{2}u+\lambda _2(K*|u|^{2})u,~~~~ (t,x)\in [0,T^*)\times {\mathbb {R}}^3, \\ u(0,x) = u_0 (x) . \end{array} \right. \end{aligned}$$

(1.4)

The dimensionless long-range dipolar interaction potential K(x) is given by

$$\begin{aligned} K(x)=\frac{1-3\cos ^2\theta }{|x|^3},~~x\in {\mathbb {R}}^3. \end{aligned}$$

To simplify notation, we shall from now on assume, without restriction of generality, that ${\mathbf {n}}=(0,0,1)$. The dipole-interaction kernel K then reads

$$\begin{aligned} K(x)=\frac{x_1^2+x_2^2-2x_3^2}{|x|^5},~~x\in {\mathbb {R}}^3. \end{aligned}$$

We focus on the case when $\lambda _1$ and $\lambda _2 \in {\mathbb {R}}$ fulfill the following conditions:

$$\begin{aligned} \lambda _1< \left\{ \begin{array}{l} \frac{4}{3}\pi \lambda _2, ~~~~if~~\lambda _2 >0 ,\\ -\frac{8}{3}\pi \lambda _2,~~~~~~if~~\lambda _2 <0 .\\ \end{array} \right. \end{aligned}$$

(1.5)

These conditions, which, following the terminology introduced in [8], define the unstable regime. Denoting the Fourier transform of u by ${\widehat{u}}=\int _{{\mathbb {R}}^3}e^{ix\cdot \xi }u(x)dx$, the Fourier transform of K is given by

$$\begin{aligned} {\widehat{K}}(\xi )=\frac{4}{3}\pi \frac{2\xi ^2_3-\xi ^2_1-\xi ^2_2}{|\xi |^2}\in \left[ -\frac{4}{3}\pi ,\frac{8}{3}\pi \right] , \end{aligned}$$

see [8]. Then, thanks to the Plancherel identity, one gets

$$\begin{aligned} \lambda _1\int _{{\mathbb {R}}^3} |u(x)|^4 dx +\lambda _2\int _{{\mathbb {R}}^3} (K *|u|^2)(x)|u(x)|^2 dx=\frac{1}{(2\pi )^3}\int _{{\mathbb {R}}^3} (\lambda _1+\lambda _2{\widehat{K}}(\xi ))|\widehat{|u|^2}(\xi )|^2d\xi . \end{aligned}$$

Here and in the following, for notational convenience, we denote

$$\begin{aligned} F(u)=-\lambda _1\int _{{\mathbb {R}}^3} |u(x)|^4 dx -\lambda _2\int _{{\mathbb {R}}^3} (K *|u|^2)(x)|u(x)|^2 dx. \end{aligned}$$

Notice that F(u) may be negative for some $u\in H^1$. In order to guarantee $F(u)>0$ for all $u\in H^1$, $\lambda _1$ and $\lambda _2 \in {\mathbb {R}}$ satisfy the following stronger condition

$$\begin{aligned} \lambda _1< \left\{ \begin{array}{l} -\frac{8}{3}\pi \lambda _2, ~~~~if~~\lambda _2 >0 ,\\ \frac{4}{3}\pi \lambda _2,~~~~~~if~~\lambda _2 <0 .\\ \end{array} \right. \end{aligned}$$

(1.6)

The Cauchy problem for (1.4) is locally well-posed in the energy space X, see [8]. Here, the energy space X for (1.4) is defined by

$$\begin{aligned} X:=\{v\in H^1,~~and~~\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx<\infty \} \end{aligned}$$

with the norm

$$\begin{aligned} \Vert v\Vert _{X}=\left( \Vert \nabla v\Vert _{L^2}^2+\Vert v\Vert _{L^2}^2+\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx\right) ^{1/2}. \end{aligned}$$

When $\omega \in (-d,\infty )$, by Heisenberg’s inequality

$$\begin{aligned} \Vert v\Vert _{L^2}^2\le 2\Vert x_jv\Vert _{L^2}\Vert \partial _{x_j}v\Vert _{L^2},~~for~1\le j\le 3, \end{aligned}$$

there exist positive constants $C_1(\omega )$ and $C_2(\omega )$ such that

$$\begin{aligned} C_1(\omega )\Vert v\Vert _{X}^2\le \Vert \nabla v\Vert _{L^2}^2+\omega \Vert v\Vert _{L^2}^2+\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx \le C_2(\omega )\Vert v\Vert _{X}^2 \end{aligned}$$

for all $v\in X$.

Because of important applications of Eq. (1.4) in physics, it has received much attention both from physics (see [9, 10]) and mathematics (see [1, 2, 8, 11,12,13,14,15,16,17,18,19]). In particular, the existence and the stability of standing waves in stable regime have been established in [2, 11, 14]. The standing wave of (1.4) is a solution of the form $e^{i\omega t}Q_\omega$, where $\omega \in {\mathbb {R}}$ is a frequency and $Q_\omega \in X\setminus \{0\}$ is a nontrivial solution to the elliptic equation

$$\begin{aligned} -\frac{1}{2} \Delta Q_\omega + \omega Q_\omega +W(x)Q_\omega +\lambda _1 |Q_\omega |^2 Q_\omega + \lambda _2 (K *|Q_\omega |^2)Q_\omega =0. \end{aligned}$$

(1.7)

The natural energy corresponding to (1.4) is

$$\begin{aligned} E(u(t))= \frac{1}{2}\int _{{\mathbb {R}}^3} |\nabla u(t,x)|^2 dx +\int _{{\mathbb {R}}^3} W(x)|u(t,x)|^2 dx-\frac{1}{2}F(u(t)). \end{aligned}$$

(1.8)

Therefore, (1.7) can be written as $S'_\omega (Q_\omega )=0$, where

$$\begin{aligned} S_\omega (Q):=E(Q)+\omega \Vert Q\Vert _{L^2}^2, \end{aligned}$$

(1.9)

is the action functional. We also define the following functionals

$$\begin{aligned} P_\omega (Q):=&\partial _\mu S_\omega (\mu Q)|_{\mu =1}=\Vert \nabla Q\Vert _{L^2}^2+2\omega \Vert Q\Vert _{L^2}^2 +2\int _{{\mathbb {R}}^3} W(x)|Q(x)|^2 dx-2F(Q), \end{aligned}$$

(1.10)

$$\begin{aligned} I(Q):=&\partial _\mu S_\omega (Q^{\mu })|_{\mu =1}=\Vert \nabla Q\Vert _{L^2}^2 -2\int _{{\mathbb {R}}^3} W(x)|Q(x)|^2 dx-\frac{3}{2}F(Q), \end{aligned}$$

(1.11)

where $Q^\mu (x):=\mu ^{3/2}Q(\mu x)$. We denote the set of non-trivial solutions of (1.7) by

$$\begin{aligned} {\mathcal {N}}_{\omega }:=\{Q_\omega \in X\backslash \{0\}:~~ S'_\omega (Q_\omega )=0\}. \end{aligned}$$

Definition 1.1

(Ground states). A function $Q_\omega \in {\mathcal {N}}_{\omega }$ is called a ground state for (1.7) if it is a minimizer of $S_\omega$ over the set ${\mathcal {N}}_{\omega }$. The set of ground states is denoted by ${\mathcal {G}}_{\omega }$. In particular,

$$\begin{aligned} {\mathcal {G}}_{\omega }=\{Q_\omega \in {\mathcal {N}}_{\omega },~~S_\omega (Q_\omega )\le S_\omega (v_\omega ),~\forall v_\omega \in {\mathcal {N}}_{\omega }\}. \end{aligned}$$

The strong instability of standing waves for the nonlinear Schrödinger equation was first studied by Berestycki and Cazenave in [20] (see also [21]). Later, Le Coz in [22] gave an alternative, simple proof of the classical result of Berestycki and Cazenave. The key point is to establish the finite time blow-up by using the variational characterization of ground states as minimizers of the action functional and the virial identity. More precisely, based on the variational characterization of ground states on the Pohozaev manifold ${\mathcal {N}}:=\{v\in H^1,~~I(v)=0\}$ or the Nehari manifold, ones can obtain the key estimate $I(u(t))\le 2(S_\omega (u_0)-S_\omega (Q_\omega ))$, where $Q_\omega$ is the corresponding ground state solution. Then, it follows from the virial identity and the choice of initial data $u_0$ that

$$\begin{aligned} \frac{d^2}{dt^2}\int _{{\mathbb {R}}^N} |x|^2 |u(t,x)|^2dx=8I(u(t))\le 16(S_\omega (u_0)-S_\omega (Q_\omega ))<0. \end{aligned}$$

This implies that the solution u(t) blows up in a finite time. Thus, ones can prove the strong instability of ground state standing waves, see [21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36].

In general, these arguments rely heavily on the properties of function $f(\mu )=S_\omega (Q^\mu )$, where $S_\omega (Q)$ is the corresponding action functional, $Q^\mu (x):=\mu ^{\frac{N}{2}}Q(\mu x)$. More precisely, if equation $f'(\mu )=0$ has unique solution $\mu _0>0$, which is the maximum point of $f(\mu )$ on $(0,\infty )$, ones can establish the variational characterization of ground states in the Pohozaev manifold ${\mathcal {N}}$. In particular, if $Q_\omega$ is the corresponding ground state, it easily follows from Pohozaev’s identities that $f'(1)=\partial _\mu S_\omega (Q_\omega ^\mu )|_{\mu =1}=0$ and $f''(1)=\partial _\mu ^2 S_\omega (Q_\omega ^\mu )|_{\mu =1}<0$. In particular, $S_\omega (Q_\omega ^\mu )<S_\omega (Q_\omega )$ and $I(Q_\omega ^\mu )<0$ for all $\mu >1$. Therefore, the strong instability of standing wave can be easily obtained.

For the nonlinear Schrödinger equation (1.4), when the system is free, i.e., $a=0$, it easily follows that equation $f'(\mu )=0$ has unique solution $\mu _0>0$, and $f(\mu )$ has the unique maximum point on $(0,\infty )$. In particular, if $Q_\omega$ is the ground state of (1.7) with $a=0$, it easily follows from Pohozaev’s identities that $f'(1)=\partial _\mu S_\omega (Q_\omega ^\mu )|_{\mu =1}=0$ and $f''(1)=\partial _\mu ^2 S_\omega (Q_\omega ^\mu )|_{\mu =1}<0$. Moreover, $S_\omega (Q_\omega ^\mu )<S_\omega (Q_\omega )$ and $I(Q_\omega ^\mu )<0$ for all $\mu >1$. Therefore, the strong instability of standing wave can be obtained by the classical argument, see [13, 15].

In all previous papers, the study of the strong instability relies on the variational characterization of ground states as minimizers of the action functional. When $W(x)=0$, we do not need the variational characterization of ground states. By some careful analysis for the characterization of ground states, we will obtain a new blow-up criterion. Based on this blow-up criterion, we can prove the strong instability of standing waves. Our main results are as follows:

Lemma 1.2

Let $W=0$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $u_0\in \Sigma := \{v\in H^1~and~|x|v \in L^2\}$ and satisfies

$$\begin{aligned} E(u_0)\Vert u_0\Vert _{L^2}^2<E(Q)\Vert Q\Vert _{L^2}^2, \end{aligned}$$

(1.12)

and

$$\begin{aligned} \Vert \nabla u_0\Vert _{L^2}\Vert u_0\Vert _{L^2} >\Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}, \end{aligned}$$

(1.13)

where Q is the ground state of the elliptic equation

$$\begin{aligned} -\frac{1}{2} \Delta Q + Q +\lambda _1 |Q|^2 Q + \lambda _2 (K *|Q|^2)Q =0. \end{aligned}$$

(1.14)

Then the solution u(t) of (1.4) with the initial data $u_0$ blows up in finite time.

When $W=0$, let $Q_\omega (x)=\omega ^{\frac{1}{2}}Q(\omega ^{\frac{1}{2}}x)$ in (1.7), then Q satisfies Eq. (1.14). In addition, a direct computation shows that

$$\begin{aligned} \Vert Q_\omega \Vert _{L^{2}}=&\omega ^{-\frac{1}{4}}\Vert Q\Vert _{L^{2}},~~\Vert \nabla Q_\omega \Vert _{L^{2}}=\omega ^{\frac{1}{4}}\Vert \nabla Q\Vert _{L^{2}}, \end{aligned}$$

(1.15)

$$\begin{aligned}&-\lambda _1 \Vert Q_\omega \Vert _{L^4}^4 - \lambda _2 \int _{{\mathbb {R}}^3}(K *|Q_\omega |^2)(x)|Q_\omega (x)|^2dx\nonumber \\ =&-\lambda _1 \omega ^{\frac{1}{2}}\Vert Q\Vert _{L^4}^4 - \lambda _2 \omega ^{\frac{1}{2}}\int _{{\mathbb {R}}^3}(K *|Q|^2)(x)|Q(x)|^2dx. \end{aligned}$$

(1.16)

These imply

$$\begin{aligned} E(Q_\omega )^{1/2}\Vert Q_\omega \Vert _{L^{2}}=E(Q)^{1/2}\Vert Q\Vert _{L^{2}}, \end{aligned}$$

(1.17)

and

$$\begin{aligned} \Vert \nabla Q_\omega \Vert _{L^{2}} \Vert Q_\omega \Vert _{L^{2}}=\Vert \nabla Q\Vert _{L^{2}} \Vert Q\Vert _{L^{2}}. \end{aligned}$$

(1.18)

In fact, the quantities $E(u_\mu )^{1/2}\Vert u_\mu \Vert _{L^2}$ and $\Vert \nabla u_\mu \Vert _{L^{2}} \Vert u_\mu \Vert _{L^2}$ are also scaling invariant of (1.4).

Theorem 1.3

Let $\omega >0$, $W=0$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6), $Q_\omega$ be the ground state related to (1.7). Then, the standing wave $u(t,x)=e^{i\omega t}Q_\omega (x)$ is strongly unstable in the following sense: there exists $\{u_{0,n}\}\subset H^1$ such that $u_{0,n}\rightarrow Q_\omega$ in $H^1$ as $n \rightarrow \infty$ and the corresponding solution $u_n$ of (1.4) with initial data $u_{0,n}$ blows up in finite time for any $n\ge 1$.

Next, we analyze the strong instability of (1.4) with a partial/complete harmonic potential. The strong instability of standing waves has been studied under various assumptions for other kinds of Schrödinger equations with a harmonic potential, see [34, 35, 37,38,39,40,41]. In fact, there exist some essential differences in the analysis of the strong instability for (1.4) between $W(x)=0$ and $W(x)\ne 0$. For example, for any $Q\in X$, we define

$$\begin{aligned} f(\mu ):=S_\omega (Q^\mu )=&\frac{\mu ^2}{2}\int _{{\mathbb {R}}^3} |\nabla Q(x)|^2 dx+\omega \int _{{\mathbb {R}}^3} |Q(x)|^2 dx \nonumber \\ {}&+\mu ^{-2}\int _{{\mathbb {R}}^3} W(x)|Q(x)|^2 dx-\frac{\mu ^3}{2}F(Q), \end{aligned}$$

(1.19)

where $Q^\mu (x):=\mu ^{\frac{3}{2}}Q(\mu x)$. It is obvious that $f(\mu )\rightarrow +\infty$, as $\mu \rightarrow 0^+$. Thus, there is no any maximum point of $f(\mu )$ on $(0,\infty )$. It is hard to establish the variational characterization of ground states in the Pohozaev manifold ${\mathcal {N}}:=\{v\in X,~~I(v)=0\}$. On the other hand, we define

$$\begin{aligned} f_1(\mu ):=&S_\omega (\mu Q)=\frac{\mu ^2}{2}\int _{{\mathbb {R}}^3} |\nabla Q(x)|^2 dx+\omega \mu ^2\int _{{\mathbb {R}}^3} |Q(x)|^2 dx\\&+\mu ^{2}\int _{{\mathbb {R}}^3} W(x)|Q(x)|^2 dx-\frac{\mu ^4}{2}F(Q). \end{aligned}$$

It is obvious that equation $f_1'(\mu )=0$ has the unique solution $\mu _0>0$, and $f_1(\mu )$ has the maximum point on $(0,\infty )$. Based on this fact, we can obtain the variational characterization of ground states in the Nehari manifold. But it is difficult to obtain the key estimate $I(u(t))\le 2(S_\omega (u_0)-S_\omega (Q_\omega ))$.

Let $Q_\omega$ be the ground state solution of (1.7), it follows from Pohozaev identities related to (1.7) that $I(Q_\omega )=0$, namely $\partial _\mu S_\omega (Q_\omega ^\mu )|_{\mu =1}=0$. In order to study the strong instability, we need to construct a sequence $\{u_{0,n}\}\subset X$ such that $u_{0,n}\rightarrow Q_\omega$ in X as $n\rightarrow \infty$, and the solution $u_n(t)$ with initial data $u_{0,n}$ blows up in finite time. Due to $\mu \partial _\mu S_\omega (Q_\omega ^\mu )=I(Q_\omega ^\mu )$ and $\frac{d^2}{dt^2}\int _{{\mathbb {R}}^N} |x|^2 |u(t,x)|^2dx=2I(u(t))$, ones often choose $u_{0,n}=Q_\omega ^{\mu _n}= \mu _n^{\frac{3}{2}}Q(\mu _n x)$ for some $\mu _n>1$ and $\mu _n\rightarrow 1$ as $n\rightarrow \infty$. In order to prove that the solution $u_n(t)$ with initial data $u_{0,n}$ blows up in finite time, we need $I(u_{0,n})=I(Q_\omega ^{\mu _n})<0$. But, for the nonlinear Schrödinger equation (1.4) with $W(x)\ne 0$, $I(Q_\omega ^{\mu _n})$ may be larger than zero. In fact, it is easy to check that $I(\mu _n^bQ_\omega (\mu _n^cx))$ with $b,c\ne 0$ may be larger than zero. Therefore, we overcome this difficulty by assuming that $\mu =1$ is the local maximum point of $f(\mu )$, i.e., $\partial _\mu ^2 S_\omega (Q_\omega ^\mu )|_{\mu =1}\le 0$, which implies that $I(Q_\omega ^{\mu })<0$ and $S_\omega (Q_\omega ^\mu )<S_\omega (Q_\omega )$ for all $\mu >1$. Under this assumption, we can prove the strong instability of standing waves for (1.4). Because $\mu =1$ is not always the maximum point of $f(\mu )$, our argument will be more complicated.

Theorem 1.4

Let $\omega >0$, $W\ne 0$, $1\le d\le 3$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $Q_\omega$ is the ground state related to (1.7) and $\partial _\mu ^2S(Q_\omega ^\mu )|_{\mu =1}\le 0$. Then, the standing wave $u(t,x)=e^{i\omega t}Q_\omega (x)$ is strongly unstable in the following sense: there exists $\{u_{0,n}\}\subset X$ such that $u_{0,n}\rightarrow Q_\omega$ in X as $n \rightarrow \infty$ and the corresponding solution $u_n$ of (1.4) with initial data $u_{0,n}$ blows up in finite time for any $n\ge 1$.

Remark

When the trapping potential $a^2|x|^2$ is small, Bellazzini and Jeanjean in [13] proved that normalized ground state standing waves are orbitally stable, see Theorem 1.7 in [13]. Here, we prove that ground state standing waves are strongly unstable by blow-up under the assumption $\partial _\mu ^2S(Q_\omega ^\mu )|_{\mu =1}\le 0$. However, it is not clear whether the normalized ground state standing waves constructed by Bellazzini and Jeanjean are ground states in the sense of Definition 1.1. Moreover, the relation between them for the classical nonlinear Schrödinger equation with a harmonic potential in the $L^2$-supercritical case is also not clear, see [37, 42].

In Lemma 5.2, we prove that there exists $\omega _*>0$ such that $\partial ^2_\mu S_\omega (Q_\omega ^{\mu })|_{\mu =1}\le 0$ for all $\omega >\omega _*$. Therefore, we have the following corollary.

Corollary 1.5

Let $W\ne 0$, $1\le d\le 3$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $Q_\omega$ is the ground state related to (1.7). Then, there exists $\omega _*>0$ such that when $\omega >\omega _*$, the standing wave $u(t,x)=e^{i\omega t}Q_\omega (x)$ is strongly unstable by blow-up.

This paper is organized as follows. Some preliminary results are given in Sect. 2. In Sect. 3, we will establish the variational characterization of the ground states. In Sect. 4, we will prove Theorems 1.2 and 1.3. In Sect. 5, we will prove Theorem 1.4.

2 Preliminaries

In this section, we recall some preliminary results that will be used later. Firstly, let us recall the local theory for the Cauchy problem (1.4) established in [8].

Lemma 2.1

[8] For $\lambda _1,\lambda _2\in {\mathbb {R}}$, $u_0\in X$, there exists $T=T(\Vert u_0\Vert _{X})$ such that (1.4) admits a unique solution $u \in C\left( [0,T],X\right)$. Let $[0,T^{*})$ be the maximal time interval on which the solution u is well-defined, if $T^{*}< \infty$, then $\Vert u(t)\Vert _{X}\rightarrow \infty$ as $t \rightarrow T^*$ Moreover, the solution u(t) enjoys conservation of mass and energy, i.e., $\Vert u(t)\Vert _{L^2} = \Vert u_0\Vert _{L^2}$ and $E(u(t))= E(u_0)$ for all $t\in [0,T^*)$, where E(u(t)) is defined by (1.8).

The following sharp Gagliardo–Nirenberg type inequality has been established in [11], which is vital in the study of blow-up criteria and instability of standing waves.

Lemma 2.2

Let $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.5). Then for all $u \in H^1$,

$$\begin{aligned} F(u)=-\lambda _1 \int _{{\mathbb {R}}^3} |u(x)|^4 dx - \lambda _2\int _{{\mathbb {R}}^3} (K *|u|^2)(x)|u(x)|^2dx \le C_{opt} \Vert \nabla u\Vert _{L^2}^3 \Vert u\Vert _{L^2} \end{aligned}$$

(2.1)

where the best constant $C_{opt}$ is given by

$$\begin{aligned} C_{opt} = \frac{4}{6^{3/2} \Vert Q\Vert _{L^2}^2}, \end{aligned}$$

(2.2)

where Q is the ground state of the elliptic Eq. (1.14). Moreover, the following Pohozaev’s identities hold true:

$$\begin{aligned} \Vert \nabla Q\Vert ^2_{L^2} =\frac{3}{2}F(Q) = 6 \Vert Q\Vert ^2_{L^2}. \end{aligned}$$

(2.3)

From (1.8) and (2.3), we can obtain the following useful results.

$$\begin{aligned} E(Q)=\frac{1}{6}\Vert \nabla Q\Vert _{L^2}^2=\frac{1}{4}F(Q)=\Vert Q\Vert _{L^2}^2. \end{aligned}$$

(2.4)

In addition, (2.1) can be rewritten as

$$\begin{aligned} F(u)^{2/3} \le C_Q\Vert u\Vert _{L^2}^{2/3}\Vert \nabla u\Vert _{L^2}^2, \end{aligned}$$

(2.5)

where

$$\begin{aligned} C_Q=(C_{opt})^{\frac{2}{3}}=\frac{F(Q)^{\frac{2}{3}}}{\Vert Q\Vert _{L^2}^{2/3} \Vert \nabla Q\Vert _{L^2}^2}. \end{aligned}$$

(2.6)

Applying (2.4), (2.6) is equivalent to

$$\begin{aligned} C_Q=\frac{4^{2/3}}{6} \frac{1}{\Vert Q\Vert _{L^2}^{2/3}E(Q)^{\frac{1}{3}}}. \end{aligned}$$

(2.7)

In order to prove the existence of blow-up solution, we need the following virial identity, which can be proved by a similar argument as that in [21, 43].

Lemma 2.3

Let $u_0\in \Sigma$ and u(t) be the corresponding solution of (1.4). Then $u(t)\in \Sigma$ for all $t\in [0,T^*)$, the function $y(t) :=\int _{{\mathbb {R}}^3} |x|^2|u(t,x)|^2 dx$ belongs to $C^2[0,T^*)$, and

$$\begin{aligned} y''(t)=&\,2\int _{{\mathbb {R}}^3} |\nabla u(t,x)|^2 dx-4\int _{{\mathbb {R}}^3} W(x)|u(t,x)|^2 dx\nonumber \\&+ 3\lambda _1\int _{{\mathbb {R}}^3} |u(t,x)|^4 dx -\lambda _2\int _{{\mathbb {R}}^3} \left( (\nabla K \cdot x)*|u|^2\right) (t,x)|u(t,x)|^2 dx\nonumber \\ =&\,2\int _{{\mathbb {R}}^3} |\nabla u(t,x)|^2 dx -4\int _{{\mathbb {R}}^3}W(x)|u(t,x)|^2 dx-3F(u(t))\nonumber \\ =&\,6E(u_0)-\Vert \nabla u(t)\Vert ^2_{L^2}-10\int _{{\mathbb {R}}^3} W(x)|u(t,x)|^2 dx. \end{aligned}$$

(2.8)

In this paper, we also need the so called Brezis-Lieb’s lemma, see [11, 44].

Lemma 2.4

Let $0<p<\infty$. Suppose that $f_n\rightarrow f$ almost everywhere and $\{f_n\}$ is a bounded sequence in $L^p$, then

$$\begin{aligned}&\lim _{n\rightarrow \infty }(\Vert f_n\Vert _{L^{p}}^{p}-\Vert f_n-f\Vert _{L^{p}}^{p}-\Vert f\Vert _{L^{p}}^{p})=0,\\&\quad \lim _{n\rightarrow \infty }\left( \int _{{\mathbb {R}}^3} \left( K *|f_n|^2\right) (x)|f_n(x)|^2 dx-\int _{{\mathbb {R}}^3} \left( K *|f_n-f|^2\right) (x)|(f_n-f)(x)|^2 dx\right) \\&\quad =\int _{{\mathbb {R}}^3} \left( K *|f|^2\right) (x)|f(x)|^2 dx. \end{aligned}$$

Finally, we recall the following Pohozaev identities related to (1.7), see [11].

Lemma 2.5

If $Q\in X$ and satisfies Eq. (1.7), then the following properties hold:

$$\begin{aligned} \Vert \nabla Q\Vert _{L^2}^2+2\omega \Vert Q\Vert _{L^2}^2 +2\int _{{\mathbb {R}}^3} W(x)|Q(x)|^2 dx-2F(Q)=0, \end{aligned}$$

(2.9)

and

$$\begin{aligned} \frac{1}{2}\Vert \nabla Q\Vert _{L^2}^2+2\omega \Vert Q\Vert _{L^2}^2+3\int _{{\mathbb {R}}^3} W(x)|Q(x)|^2 dx -\frac{5}{4}F(Q)=0. \end{aligned}$$

(2.10)

3 Characterizations of the Ground States

In this subsection, we will establish the following characterization of the ground state related to (1.7).

Lemma 3.1

Let $1\le d<3$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that a sequence $\{u_n\}$ is bounded in X, and satisfies

$$\begin{aligned} \liminf _{n\rightarrow \infty }F(u_n)>0. \end{aligned}$$

(3.1)

Then, there exist a sequence $\{y^n\}$ in ${\mathbb {R}}^{3-d}$ and $u\in X \backslash \{0\}$ such that

$$\begin{aligned}&u_n(x_1,x_2-y_1^n,x_3-y_2^n)\rightharpoonup u~~in~X,~~if~d=1, \\&u_n(x_1,x_2,x_3-y_1^n)\rightharpoonup u~~in~X,~~if~d=2. \end{aligned}$$

Proof

We first deduce from Plancherel’s formula that

$$\begin{aligned} \Vert u\Vert _{L^4}^4=\Vert \varphi \Vert _{L^2}^2=\frac{1}{(2\pi )^3}\Vert {\widehat{\varphi }}\Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned} F(u)=-\frac{1}{(2\pi )^3} \int _{{\mathbb {R}}^3}(\lambda _1+\lambda _2{\widehat{K}}(\xi ))|\varphi (\xi )|^2d\xi , \end{aligned}$$

where $\varphi =|u|^2$. From Lemma 2.3 in [8], we have ${\widehat{K}}\in [-\frac{4\pi }{3},\frac{8\pi }{3}]$. Therefore, under the assumption (1.6), it follows that

$$\begin{aligned}&F(u_n)\le \left( \frac{4\pi }{3}\lambda _2-\lambda _1\right) \Vert u_n\Vert _{L^4}^4,~~if~~\lambda _2>0, \\&F(u_n)\le \left( -\frac{8\pi }{3}\lambda _2-\lambda _1\right) \Vert u_n\Vert _{L^4}^4,~~if~~\lambda _2<0. \end{aligned}$$

This, together with (3.1), implies that $\liminf _{n\rightarrow \infty }\Vert u_n\Vert _{L^4}^4>0$. Thus, this lemma follows from Lemma 3.3 in [42]. $\square$

Based on this lemma, we can establish the following characterization of the ground state related to (1.7).

Proposition 3.2

Let $1\le d\le 3$, $\omega \in (-d,\infty )$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Then $Q_\omega \in {\mathcal {G}}_{\omega }$ if and only if $Q_\omega$ solves the following minimization problem:

$$\begin{aligned} d(\omega )=inf\{S_\omega (v):~v\in X\backslash \{0\},~~P_\omega (v)= 0\}. \end{aligned}$$

(3.2)

To prove this proposition, we firstly define the following functional

$$\begin{aligned} H_\omega (v)&=S_\omega (v)-\frac{1}{4}P_\omega (v)\nonumber \\&=\frac{1}{4}\Vert \nabla v\Vert _{L^2}^2+\frac{\omega }{2}\Vert v\Vert _{L^2}^2+\frac{1}{2}\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx. \end{aligned}$$

(3.3)

In order to solve the minimization problem (3.2), we define an equivalent minimization problem:

$$\begin{aligned} d_1(\omega )=inf\{H_\omega (v):~v\in X\backslash \{0\},~~P_\omega (v)\le 0\}. \end{aligned}$$

(3.4)

Lemma 3.3

Let $1\le d\le 3$, $\omega \in (-d,\infty )$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Then there exists $u\in X\backslash \{0\}$ such that $P_\omega (u)=0$ and $H_\omega (u)=d_1(\omega )$.

Proof

When $d=3$, since the embedding $X\hookrightarrow L^q$ with $q\in [2,6)$ is compact, this lemma can be easily proved. So we only prove the case $1\le d<3$. We first show that $d_1(\omega )>0$. By $P_\omega (v)\le 0$, we have

$$\begin{aligned}&\Vert \nabla v\Vert _{L^2}^2+2\omega \Vert v\Vert _{L^2}^2 +2\int _{{\mathbb {R}}^3} W(x)|v(x)|^2 dx\\&\quad \le 2F(v)\le C(\Vert \nabla v\Vert _{L^2}^2+\omega \Vert v\Vert _{L^2}^2 +\int _{{\mathbb {R}}^3} W(x)|v(x)|^2 dx)^2, \end{aligned}$$

namely,

$$\begin{aligned} 1\le C(\Vert \nabla v\Vert _{L^2}^2+\omega \Vert v\Vert _{L^2}^2 +\int _{{\mathbb {R}}^3} W(x)|v(x)|^2 dx). \end{aligned}$$

Taking the infimum over v, we have $d_1(\omega )>0$.

We now show the minimizing problem (3.4) is attained. Let $\{v_n\}$ be a minimizing sequence for (3.4), i.e., $\{v_n\}\subseteq X\backslash \{0\}$, $P_\omega (v_n)\le 0$ and $H_\omega (v_n)\rightarrow d_1(\omega )$ as $n\rightarrow \infty$. This implies that $\{v_n\}$ is bounded in X. In addition, it follows from $P_\omega (v_n)\le 0$ that

$$\begin{aligned} 4d_1(\omega )\le 4H_\omega (v_n)\le \Vert \nabla v_n\Vert _{L^2}^2+2\omega \Vert v_n\Vert _{L^2}^2 +2\int _{{\mathbb {R}}^3} W(x)|v_n(x)|^2 dx\le 2F(v_n). \end{aligned}$$

(3.5)

This implies that $\liminf _{n\rightarrow \infty }F(v_n)>0$. Applying Lemma 3.1, there exist a subsequence, still denoted by $\{v_n\}$ and $u\in X\backslash \{0\}$ such that

$$\begin{aligned} u_n:=\tau _{y^n}v_n\rightharpoonup u\ne 0~~weakly~in ~X, \end{aligned}$$

for some $\{y^n\}\subseteq {\mathbb {R}}^{3-d}$. Here we define

$$\begin{aligned} \tau _{y}v(x)=v(x_1,x_2-y_1,x_3-y_2)~~for~d=1~~and~~\tau _{y}v(x)=v(x_1,x_2,x_3-y_1)~~for~d=2. \end{aligned}$$

Moreover, we deduce from Lemma 2.4 that

$$\begin{aligned} P_\omega (u_n)-P_\omega (u_n-u)-P_\omega (u)\rightarrow 0, \end{aligned}$$

(3.6)

and

$$\begin{aligned} H_\omega (u_n)-H_\omega (u_n-u)-H_\omega (u)\rightarrow 0. \end{aligned}$$

(3.7)

Now, we claim that $P_\omega (u)\le 0$. If not, it follows from (3.6) and $P_\omega (u_n)> 0$ that $P_\omega (u_n-u)\le 0$ for sufficiently large n. Thus, by the definition of $d_1(\omega )$, it follows that

$$\begin{aligned} H_\omega (u_n-u)\ge d_1(\omega ), \end{aligned}$$

which, together with $H_\omega (u_n)\rightarrow d_1(\omega )$, implies that

$$\begin{aligned} H_\omega (u)\le 0, \end{aligned}$$

which is a contradiction. We thus obtain $P_\omega (u)\le 0$.

Furthermore, we deduce from the definition of $d_1(\omega )$ and the weak lower semicontinuity of norm that

$$\begin{aligned} d_1(\omega )\le H_\omega (u)\le \liminf _{n\rightarrow \infty }H_\omega (u_n)=d_1(\omega ). \end{aligned}$$

This yields that

$$\begin{aligned} H_\omega (u)=d_1(\omega ). \end{aligned}$$

Finally, we show that $P_\omega (u)=0$. Suppose that $P_\omega (u)<0$ and set

$$\begin{aligned} f(\mu ):=P_\omega (\mu u)=\mu ^2\Vert \nabla u\Vert _{L^2}^2+2\omega \mu ^2\Vert u\Vert _{L^2}^2 +2\mu ^2\int _{{\mathbb {R}}^3} W(x)|u(x)|^2 dx-2\mu ^4F(u). \end{aligned}$$

Then, $f(\mu )>0$ for sufficiently small $\mu >0$ and $f(1)=P_\omega (u)<0$. Therefore, there exists $\mu _0\in (0,1)$ such that $P_\omega (\mu _0 u)=0$. Then, it follows that

$$\begin{aligned} H_\omega (\mu _0 u)&=\frac{\mu _0^2}{4}\Vert \nabla u\Vert _{L^2}^2+\frac{\omega \mu _0^2}{2}\Vert u\Vert _{L^2}^2+\frac{\mu _0^2}{2}\int _{{\mathbb {R}}^3}W(x)|u(x)|^2dx \\ {}&<H_\omega (u)=d_1(\omega ), \end{aligned}$$

which contradicts the definition of $d_1(\omega )$. Hence, we have $P_\omega (u)=0$. $\square$

By the fact $d(\omega )=d_1(\omega )$ and this lemma, we can obtain the following Corollary.

Corollary 3.4

Let $1\le d\le 3$, $\omega \in (-d,\infty )$. Then there exists $u\in X \{0\}$, $P_\omega (u)=0$ and $S_\omega (u)=d(\omega )$.

We now denote the set of all minimizers of (3.2) by

$$\begin{aligned} {\mathcal {M}}_{\omega }=\{u\in X\backslash \{0\},~~S_\omega (u)=d(\omega ),~P_\omega (u)=0\}. \end{aligned}$$

Lemma 3.5

Let $1\le d\le 3$, $\omega \in (-d,\infty )$. Then ${\mathcal {M}}_{\omega }\subseteq {\mathcal {G}}_{\omega }$.

Proof

Let $u\in {\mathcal {M}}_\omega$. Then, there exists a Lagrange multiplier $\mu \in {\mathbb {R}}$ such that

$$\begin{aligned} S'_\omega (u)=\mu P'_\omega (u) \end{aligned}$$

Thus, we have

$$\begin{aligned} \mu \langle P'_\omega (u), u\rangle =\langle S'_\omega (u), u\rangle =P_\omega (u)=0. \end{aligned}$$

On the other hand,

$$\begin{aligned} \langle P'_\omega (u), u\rangle&=\partial _\mu P_\omega (\mu u)|_{\mu =1}\\ {}&= 2\Vert \nabla u\Vert _{L^2}^2+4\omega \Vert u\Vert _{L^2}^2 +4\int _{{\mathbb {R}}^3} W(x)|u(x)|^2 dx-8F(u)\\ {}&= -4F(u)<0. \end{aligned}$$

Thus, it follows that $\mu =0$ and $S'_\omega (u)=0$. This yields $u\in {\mathcal {N}}_\omega$. To prove $u\in {\mathcal {G}}_\omega$, it remains to show that $S_\omega (u)\le S_\omega (v)$ for all $v\in {\mathcal {N}}_\omega$. To see this, notice that

$$\begin{aligned} P_\omega (v)=\langle S'_\omega (v), v\rangle =0, \end{aligned}$$

for all $v\in {\mathcal {N}}_\omega$. By definition of $d(\omega )$, we have $S_\omega (u)\le S_\omega (v)$. Thus, $u\in {\mathcal {G}}_\omega$. $\square$

Lemma 3.6

Let $1\le d\le 3$, $\omega \in (-d,\infty )$. Then ${\mathcal {G}}_\omega \subset {\mathcal {M}}_\omega$.

Proof

Let $u\in {\mathcal {G}}_\omega$. Since ${\mathcal {M}}_\omega$ is not empty, we take $v\in {\mathcal {M}}_\omega$. By Lemma 3.5, $v\in {\mathcal {G}}_\omega$. In particular, $S_\omega (u)=S_\omega (v)$. Since $v\in {\mathcal {M}}_\omega$, we get

$$\begin{aligned} S_\omega (u)=S_\omega (v)=d(\omega ). \end{aligned}$$

It remains to show that $P_\omega (u)=0$. Since $u\in {\mathcal {N}}_\omega$, we have $S'_\omega (u)=0$, hence $P_\omega (u)=\langle S'_\omega (u), u\rangle =0$. Therefore, $u\in {\mathcal {M}}_\omega$. $\square$

Proof of Proposition 3.2

Proposition 3.2 follows immediately from Corollary 3.4, Lemmas 3.5 and 3.6.

Based on Proposition 3.2., we can also establish the following characterizations of the ground state related to (1.7).

Lemma 3.7

Let $1\le d\le 3$, $\omega \in (-d,\infty )$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $Q_\omega$ is the ground state related to (1.7). Then we have

$$\begin{aligned} F(Q_\omega )&=\inf \{F(v),~v\in X\backslash \{0\},~P_\omega (v)=0\}\nonumber \\ {}&=\inf \{F(v),~v\in X\backslash \{0\},~P_\omega (v)\le 0\}, \end{aligned}$$

(3.8)

and

$$\begin{aligned} S_\omega (Q_\omega )=\inf \{S_\omega (v),~v\in X\backslash \{0\},~F(v)=F(Q_\omega )\}. \end{aligned}$$

(3.9)

Proof

Since we have

$$\begin{aligned} S_\omega (v)=\frac{1}{2}P_\omega (v)+\frac{1}{2}F(v), \end{aligned}$$

we see that

$$\begin{aligned} d(\omega )&=\inf \{S_\omega (v),~v\in X\backslash \{0\},~P_\omega (v)=0\}\\&=\inf \left\{ \frac{1}{2}F(v),~v\in X\backslash \{0\},~P_\omega (v)= 0\right\} . \end{aligned}$$

This implies that $d(\omega )=S_\omega (Q_\omega )=\frac{1}{2}F(Q_\omega )$. We set

$$\begin{aligned} d_1(\omega ):=\inf \left\{ \frac{1}{2}F(v),~v\in X\backslash \{0\},~P_\omega (v)\le 0\right\} . \end{aligned}$$

Since it is clear that $d_1(\omega )\le d(\omega )$, we show $d(\omega )\le d_1(\omega )$. For any $v\in X\backslash \{0\}$ satisfying $P_\omega (v)<0$, there exists $\mu _0\in (0,1)$ such that $P_\omega (\mu _0v)=0$. Thus, we have

$$\begin{aligned} d(\omega )\le \frac{1}{2}F(\mu _0v)<\frac{1}{2}F(v). \end{aligned}$$

Hence, we have $d(\omega )\le d_1(\omega )$.

Finally, we put $d_2(\omega ):=\inf \{S_\omega (v),~v\in X\backslash \{0\},~F(v)=F(Q_\omega )\}.$ Since $d_2(\omega )\le S_\omega (Q_\omega )$, it suffices to prove $d_2(\omega )\ge S_\omega (Q_\omega )$. By (3.8), for any $v\in X\backslash \{0\}$ satisfying $F(v)=F(Q_\omega )$, we have $P_\omega (v)\ge 0$. Thus, we have

$$\begin{aligned} S_\omega (v)\ge \frac{1}{2}F(v)=\frac{1}{2}F(Q_\omega )=S_\omega (Q_\omega ). \end{aligned}$$

Therefore, we obtain $d_2(\omega )\ge S_\omega (Q_\omega )$. $\square$

4 Strong Instability in the Case $W=0$

In this section, we will prove the strong instability of standing waves for (1.4) without harmonic potential. Firstly, we prove Lemma 1.2.

Proof of Lemma 1.2

We firstly recall the sharp constant in Gagliardo-Nirenberg inequality (2.1)

$$\begin{aligned} C_{opt}=\frac{F(Q)}{\Vert \nabla Q\Vert _{L^2}^3 \Vert Q\Vert _{L^2}}. \end{aligned}$$

(4.1)

By (2.3), we can rewrite $C_{opt}$ as

$$\begin{aligned} C_{opt}=\frac{2}{3}\frac{1}{\Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}}. \end{aligned}$$

(4.2)

By a direct calculation, we also have

$$\begin{aligned} E(Q)\Vert Q\Vert _{L^2}^{2}=\frac{1}{6} \Vert \nabla Q\Vert ^2_{L^2}\Vert Q\Vert _{L^2}^2. \end{aligned}$$

(4.3)

Multiplying both sides of E(u(t)) by $\Vert u(t)\Vert _{L^2}^2$, we deduce from (2.1) that

$$\begin{aligned} E(u(t))\Vert u(t)\Vert _{L^2}^{2}&=\frac{1}{2}\int _{{\mathbb {R}}^3} |\nabla u|^2 dx\Vert u(t)\Vert _{L^2}^{2} - \frac{F(u(t))}{2}\Vert u(t)\Vert _{L^2}^{2} \\&\ge \frac{1}{2}(\Vert \nabla u(t)\Vert _{L^2}\Vert u(t)\Vert _{L^2})^2 - \frac{C_{opt}}{2}(\Vert \nabla u(t)\Vert _{L^2}\Vert u(t)\Vert _{L^2})^3 \\&=f(\Vert \nabla u(t)\Vert _{L^2}\Vert u(t)\Vert _{L^2}), \end{aligned}$$

where $f(x):=\frac{1}{2}x^2-\frac{C_{opt}}{2}x^{3}$. It is easy to see that f is increasing on $(0,x_0)$ and decreasing on $(x_0,\infty )$, where

$$\begin{aligned} x_0=\frac{1}{C_{opt}}=\frac{3}{2} \Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}. \end{aligned}$$

It follows from (4.2) and (4.3) that

$$\begin{aligned} f(\Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}) =E(Q)\Vert Q\Vert _{L^2}^2. \end{aligned}$$

Thus, the conservation of mass and energy together with (1.12) imply

$$\begin{aligned} f(\Vert \nabla u(t)\Vert _{L^2}\Vert u(t)\Vert _{L^2})&\le E(u(t))\Vert u(t)\Vert _{L^2}^2=E(u_0)\Vert u_0\Vert _{L^2}^2\\&<E(Q)\Vert Q\Vert _{L^2}^2=f(\Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}), \end{aligned}$$

for all $t\in [0,T^*)$.

Applying the continuity argument, it easily follows that

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2}\Vert u(t)\Vert _{L^2}<\Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}, \end{aligned}$$

(4.4)

for all $t\in [0,T^*)$. This implies that the solution u(t) of (1.4) exists globally.

Similarly, applying the continuity argument and (1.13), we can obtain

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2}\Vert u(t)\Vert _{L^2}>\Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}. \end{aligned}$$

(4.5)

for any $t\in [0,T^*)$. On the other hand, since $E(u_0)\Vert u_0\Vert _{L^2}^2<E(Q)\Vert Q\Vert _{L^2}^2$, we pick $\eta >0$ small enough so that

$$\begin{aligned} E(u_0)\Vert u_0\Vert _{L^2}^2\le (1-\eta )E(Q)\Vert Q\Vert _{L^2}^2. \end{aligned}$$

Thus, by the conservation of energy, (4.3) and (4.5), we have

$$\begin{aligned} I(u(t))\Vert u(t)\Vert _{L^2}^2 =&\, 3E(u(t))\Vert u(t)\Vert _{L^2}^2-\frac{1}{2}\Vert \nabla u(t)\Vert _{L^2}^2 \Vert u(t)\Vert _{L^2}^2\\=&\, 3E(u_0)\Vert u_0\Vert _{L^2}^2-\frac{1}{2} \left( \Vert \nabla u(t)\Vert _{L^2}\Vert u(t)\Vert _{L^2}\right) ^2\\\le&\, 3(1-\eta )E(Q)\Vert Q\Vert _{L^2}^2-\frac{1}{2} \left( \Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}\right) ^2\\ =&\, -3 \eta E(Q)\Vert Q\Vert _{L^2}^2, \end{aligned}$$

for all $t\in [0,T^*)$. This implies $y''(t)=2 I(u(t))\le - 6\eta E(Q)\cdot \frac{\Vert Q\Vert _{L^2}^2}{\Vert u_0\Vert _{L^2}}$. Thus, by a standard argument, it follows that the solution u(t) of (1.4) blows up in finite time. This completes the proof.

Proof of Theorem 1.3

Let $Q_\omega$ be the ground state related to (1.7) with $a=0$, a direct computation shows

$$\begin{aligned} S_\omega (Q^{\mu }_\omega )= \frac{\mu ^2}{2}\Vert \nabla Q_\omega \Vert _{L^2}^2+\omega \Vert Q_\omega \Vert _{L^2}^2 -\frac{\mu ^3}{2}F(Q_\omega ), \end{aligned}$$

and

$$\begin{aligned} \partial _\mu S_\omega (Q^{\mu }_\omega )= \mu \Vert \nabla Q_\omega \Vert _{L^2}^2 -\frac{3\mu ^2}{2}F(Q_\omega )=\frac{I(Q^\mu _\omega )}{\mu }. \end{aligned}$$

It is easy to see that the equation $\partial _\mu S_\omega (Q^{\mu }_\omega )=0$ has a unique non-zero solution

$$\begin{aligned} \frac{2\Vert \nabla Q_\omega \Vert _{L^2}^2}{3\mu F(Q_\omega )}=1. \end{aligned}$$

The last inequality comes from the fact that $I(Q_\omega )=0$, which follows from Pohozaev’s identities (2.3). We thus obtain

$$\begin{aligned} \left\{ \begin{array}{l} \partial _\mu S_\omega (Q_\omega ^{\mu })>0~~if~\mu \in (0,1), \\ \partial _\mu S_\omega (Q_\omega ^{\mu })<0~~if~\mu \in (1,\infty ). \end{array} \right. \end{aligned}$$

This implies that $S_\omega (Q_\omega ^{\mu })<S_\omega (Q_\omega )$ for any $\mu >0$ and $\mu \ne 1$. This, together with $\Vert Q^{\mu }_\omega \Vert _{L^2}=\Vert Q_\omega \Vert _{L^2}$, implies that for any $\mu >1$

$$\begin{aligned} E(Q^{\mu }_\omega )<E(Q_\omega ). \end{aligned}$$

(4.6)

Let $\mu _n>1$ such that $\lim _{n\rightarrow \infty }\mu _n=1$. We take the initial data

$$\begin{aligned} u_{0,n}(x)=Q^{\mu _n}_\omega (x)=\mu _n^{\frac{3}{2}}Q_\omega (\mu _nx). \end{aligned}$$

By Brezis-Lieb’s lemma, we have $u_{0,n}\rightarrow Q_\omega$ in $H^1$ as $n \rightarrow \infty$. We deduce from (4.6) that

$$\begin{aligned} E(u_{0,n})<E(Q_\omega ), \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla u_{0,n}\Vert _{L^2}=\mu _n\Vert \nabla Q_\omega \Vert _{L^2} >\Vert \nabla Q_\omega \Vert _{L^2}. \end{aligned}$$

Thus, by $\Vert u_{0,n}\Vert _{L^2}=\Vert Q_\omega \Vert _{L^2}$, (1.17) and (1.18), we have

$$\begin{aligned} E(u_{0,n})^{\frac{1}{2}}\Vert u_{0,n}\Vert _{L^2}< E(Q_\omega )^{\frac{1}{2}}\Vert Q_\omega \Vert _{L^2}=E(Q)^{\frac{1}{2}}\Vert Q\Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla u_{0,n}\Vert _{L^2}\Vert u_{0,n}\Vert _{L^2} >\Vert \nabla Q_\omega \Vert _{L^2}\Vert Q_\omega \Vert _{L^2}=\Vert \nabla Q\Vert _{L^2}\Vert Q\Vert _{L^2}. \end{aligned}$$

On the other hand, since $Q_\omega$ is exponentially decaying, $u_{0,n}\in \Sigma$. Thus, applying Lemma 1.2, the solution $u_n$ of (1.4) with initial data $u_{0,n}$ blows up in finite time. This completes the proof.

5 The Proof of Theorem 1.4

In this section, we will prove the strong instability of standing waves for (1.4) with a partial/complete harmonic potential. Firstly, we establish a key estimate under the assumption $\partial _\mu ^2S_\omega (Q_\omega ^\mu )|_{\mu =1}\le 0$.

Lemma 5.1

Let $1\le d\le 3$, $\omega \in (-d,\infty )$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $Q_\omega$ is the ground state related to (1.7) and $\partial _\mu ^2S_\omega (Q_\omega ^\mu )|_{\mu =1}\le 0$. If $v\in X$, and satisfies

$$\begin{aligned} I(v)\le 0,~~~\Vert v\Vert _{L^2}=\Vert Q_\omega \Vert _{L^2},~~~F(v)>F(Q_\omega ), \end{aligned}$$

(5.1)

then $S_\omega (Q_\omega )\le S_\omega (v)-\frac{1}{2}I(v)$.

Proof

Firstly, we assume that $\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx \ge \int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx$. In this case, we note that

$$\begin{aligned} S_\omega (v)-\frac{1}{2}I(v)=\omega \Vert v\Vert _{L^2}^2+2\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx +\frac{1}{4}F(v). \end{aligned}$$

(5.2)

Thus, we deduce from $I(Q_\omega )=0$ that

$$\begin{aligned} S_\omega (Q_\omega )=&\,\omega \Vert Q_\omega \Vert _{L^2}^2+2\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx +\frac{1}{4}F(Q_\omega )\\\le&\,\omega \Vert v\Vert _{L^2}^2+2\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx +\frac{1}{4}F(v)\\ =&\,S_\omega (v)-\frac{1}{2}I(v), \end{aligned}$$

which is the desired estimate.

Next, we consider the case $\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx < \int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx$. Let $\mu _0$ satisfy

$$\begin{aligned} F(v^{\mu _0})=\mu _0^{3}F(v)=F(Q_\omega ). \end{aligned}$$

Then, it follows from (5.1) that $0<\mu _0<1$. In addition, applying Lemma 3.7, it holds that

$$\begin{aligned} S_\omega (Q_\omega )\le S_\omega (v^{\mu _0}). \end{aligned}$$

(5.3)

In order to establish the key estimate, we define

$$\begin{aligned} g(\mu ):=S_\omega (v^\mu )-\frac{\mu ^2}{2}I(v)=\omega \Vert v\Vert _{L^2}^2+(\mu ^{2}+\mu ^{-2}) \int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx +\frac{3 \mu ^{2}-2\mu ^{3}}{4}F(v), \end{aligned}$$

where $\mu >0$. If $g(\mu _0)\le g(1)$, we deduce from $I(v)\le 0$ and (5.3) that

$$\begin{aligned} S_\omega (Q_\omega )\le S_\omega (v^{\mu _0})\le S_\omega (v^{\mu _0})-\frac{\mu _0^{2}}{2}I(v)=g(\mu _0)\le g(1)=S_\omega (v)-\frac{I(v)}{2}, \end{aligned}$$

which is the desired estimate.

Thus, we only prove $g(\mu _0)\le g(1)$, which is equivalent to

$$\begin{aligned} 4(\mu _0^{2}+\mu _0^{-2}-2)\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx\le (1-3\mu _0^{2}+2\mu _0^3)F(v). \end{aligned}$$

(5.4)

Note that $I(Q_\omega )=0$ and $\partial _\mu ^2S_\omega (Q_\omega ^\mu )|_{\mu =1}\le 0$ and

$$\begin{aligned} \partial _\mu ^2S(v^\mu )|_{\mu =1}-I(v)=8\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx -\frac{3}{2}F(v). \end{aligned}$$

(5.5)

Thus, we can obtain

$$\begin{aligned} 16\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx \le 3F(Q_\omega ). \end{aligned}$$

This, together with (5.4) implies that

$$\begin{aligned} 16(\mu _0^{2}+\mu _0^{-2}-2)\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx&=16(\mu _0-\mu _0^{-1})^2\int _{{\mathbb {R}}^3}W(x)|v(x)|^2dx\\&\le 16(\mu _0-\mu _0^{-1})^2\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx\\&\le 3(\mu _0-\mu _0^{-1})^2F(Q_\omega )\\&= 3(\mu _0-\mu _0^{-1})^2\mu _0^{3}F(v). \end{aligned}$$

Thus, in order to prove (5.4), we need only show that

$$\begin{aligned} 3(\mu _0-\mu _0^{-1})^2\mu _0^{3}\le 4(1-3\mu _0^{2}+2\mu _0^3). \end{aligned}$$

(5.6)

Here, we put $2\beta =3$ and define

$$\begin{aligned} h(\mu ):= \beta -1-\beta \mu +\mu ^{\beta }-\frac{\beta (\beta -1)}{2}\mu ^{\beta -1}(\mu -1)^2. \end{aligned}$$

for $\mu >0$. Then, (5.6) is equivalent to $h(\mu _0^2)\ge 0$. By the Taylor expansion of $\mu ^\beta$ at $\mu =1$, we have

$$\begin{aligned} h(\mu _0^2)=\frac{\beta (\beta -1)}{2}(\mu ^2_0-1)^2\{\xi ^{\beta -2}-(\mu ^2_0)^{\beta -1}\} \end{aligned}$$

for some $\xi \in (\mu _0^2,1)$. Since $\beta >1$ and $\mu _0^2<\xi <1$, we have

$$\begin{aligned} (\mu ^2_0)^{\beta -1}\le \xi ^{\beta -1}\le \xi ^{\beta -2}, \end{aligned}$$

and obtain $h(\mu ^2_0)\ge 0$. Thus, we have (5.4) and $g(\mu _0)\le g(1)$. This completes the proof. $\square$

Lemma 5.2

Let $1\le d\le 3$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $Q_\omega$ is the ground state related to (1.7). Then, there exists $\omega _*>0$ such that $\partial ^2_\mu S_\omega (Q_\omega ^{\mu })|_{\mu =1}\le 0$ for all $\omega >\omega _*$.

Proof

Let $Q_\omega \in {\mathcal {G}}_\omega$, it follows from Lemma 2.5 that $I(Q_\omega )=0$. We consequently obtain

$$\begin{aligned} \partial ^2_\mu S_\omega (Q_\omega ^{\mu })|_{\mu =1}=8\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx -\frac{3}{2} F(Q_\omega ). \end{aligned}$$

Thus, $\partial ^2_\mu S_\omega (Q_\omega ^{\mu })|_{\mu =1}\le 0$ if and only if

$$\begin{aligned} \frac{\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx }{F(Q_\omega )}<\frac{3}{16}. \end{aligned}$$

So it is sufficient to prove that

$$\begin{aligned} \lim _{\omega \rightarrow \infty }\frac{\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx }{F(Q_\omega )}=0. \end{aligned}$$

Let $Q_\omega (x)=\omega ^{\frac{1}{2}}{\tilde{Q}}_\omega (\sqrt{\omega }x)$, then ${\tilde{Q}}_\omega$ satisfies

$$\begin{aligned} -\frac{1}{2} \Delta {\tilde{Q}}_\omega +{\tilde{Q}}_\omega +\omega ^{-2}W(x){\tilde{Q}}_\omega +\lambda _1 |{\tilde{Q}}_\omega |^2 {\tilde{Q}}_\omega + \lambda _2 (K *|{\tilde{Q}}_\omega |^2){\tilde{Q}}_\omega =0 \end{aligned}$$

Since

$$\begin{aligned} \frac{\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx }{F(Q_\omega )}=\omega ^{-2} \frac{\int _{{\mathbb {R}}^3}W(x)|{\tilde{Q}}_\omega (x)|^2dx }{F({\tilde{Q}}_\omega )}, \end{aligned}$$

it is sufficient to prove that

$$\begin{aligned} \lim _{\omega \rightarrow \infty }\omega ^{-2} \frac{\int _{{\mathbb {R}}^3}W(x)|{\tilde{Q}}_\omega (x)|^2dx }{F({\tilde{Q}}_\omega )}=0. \end{aligned}$$

Let $V\in H^1\setminus \{0\}$ be a ground state solution to the elliptic problem

$$\begin{aligned} -\frac{1}{2} \Delta V+V +\lambda _1 |V|^2 V + \lambda _2 (K *|V|^2)V=0, \end{aligned}$$

then

$$\begin{aligned} S_0(V)=\inf \{S_0(v),~~v\in H^1\setminus \{0\},~~{\tilde{P}}_0(v)=0\}, \end{aligned}$$

where

$$\begin{aligned} S_0(v)=\frac{1}{2}\Vert \nabla v\Vert _{L^2}^2+\Vert v\Vert _{L^2}^2 -\frac{1}{2}F(v), \end{aligned}$$

and

$$\begin{aligned} {\tilde{P}}_0(v)=\Vert \nabla v\Vert _{L^2}^2+2\Vert v\Vert _{L^2}^2 -2F(v). \end{aligned}$$

Then, by a similar argument as that in Lemma 3.7, we have

$$\begin{aligned} F(V) =\inf \left\{ F(v),~~v\in H^1\setminus \{0\},~~{\tilde{P}}_0(v)\le 0\right\} , \end{aligned}$$

(5.7)

and

$$\begin{aligned} F({\tilde{Q}}_\omega ) =\inf \left\{ F(v),~~v\in X\setminus \{0\},~~{\tilde{P}}_\omega (v)\le 0\right\} , \end{aligned}$$

(5.8)

where

$$\begin{aligned} {\tilde{P}}_\omega (v)=\Vert \nabla v\Vert _{L^2}^2+2\Vert v\Vert _{L^2}^2+2\omega ^{-2}\int _{{\mathbb {R}}^3}W(x)|v(x)|^{2}dx -2F(v). \end{aligned}$$

In addition, we infer from ${\tilde{P}}_0(V)=0$ that for $\mu >1$

$$\begin{aligned} {\tilde{P}}_\omega (\mu V)=\mu ^2\left( 2(1-\mu ^2) F(V)+2\omega ^{-2}\int _{{\mathbb {R}}^3}W(x)|V(x)|^{2}dx\right) . \end{aligned}$$

Then, for any $\mu >1$, there exists $\omega (\mu )$ such that ${\tilde{P}}_\omega (\mu V)<0$ for all $\omega >\omega (\mu )$. This and (5.8) imply that

$$\begin{aligned} F({\tilde{Q}}_\omega )\le \mu ^{4}F(V). \end{aligned}$$

(5.9)

On the other hand, we deduce from ${\tilde{P}}_\omega ({\tilde{Q}}_\omega )=0$ that

$$\begin{aligned} {\tilde{P}}_0(\mu {\tilde{Q}}_\omega )=\mu ^2\left( 2(1-\mu ^2) F({\tilde{Q}}_\omega ) -2\omega ^{-2}\int _{{\mathbb {R}}^3}W(x)|{\tilde{Q}}_\omega (x)|^{2}dx\right) . \end{aligned}$$

Then, for any $\mu >1$, ${\tilde{P}}_0(\mu {\tilde{Q}}_\omega )<0$. We consequently deduce from (5.7) and (5.9) that for any $\omega >\omega (\mu )$,

$$\begin{aligned} \mu ^{-4}F(V)\le F({\tilde{Q}}_\omega )\le \mu ^{4}F(V). \end{aligned}$$

Since $\mu >1$ is arbitrary, we have

$$\begin{aligned} \lim _{\omega \rightarrow \infty }F({\tilde{Q}}_\omega )=F(V). \end{aligned}$$

(5.10)

Notice that

$$\begin{aligned} {\tilde{P}}_0(\mu {\tilde{Q}}_\omega )=\mu ^2\Vert \nabla {\tilde{Q}}_\omega \Vert _{L^2}^2+2\mu ^2\Vert {\tilde{Q}}_\omega \Vert _{L^2}^2-2\mu ^4 F({\tilde{Q}}_\omega ), \end{aligned}$$

there exists $\mu (\omega )>0$ such that ${\tilde{P}}_0(\mu (\omega ) {\tilde{Q}}_\omega )=0$. This and (5.7) yield that

$$\begin{aligned} F(V)\le \mu (\omega )^4 F({\tilde{Q}}_\omega ). \end{aligned}$$

This implies that $\liminf _{\omega \rightarrow \infty }\mu (\omega )\ge 1$ and

$$\begin{aligned} \liminf _{\omega \rightarrow \infty }{\tilde{P}}_0( {\tilde{Q}}_\omega ) =\liminf _{\omega \rightarrow \infty }2(\mu (\omega )^2-1) F({\tilde{Q}}_\omega )\ge 0. \end{aligned}$$

(5.11)

On the other hand, we deduce from ${\tilde{P}}_\omega ( {\tilde{Q}}_\omega )=0$ that ${\tilde{P}}_0( {\tilde{Q}}_\omega )<0$. Combining this and (5.11), it follows that

$$\begin{aligned} \lim _{\omega \rightarrow \infty }{\tilde{P}}_0( {\tilde{Q}}_\omega )=0. \end{aligned}$$

We consequently obtain that

$$\begin{aligned} \omega ^{-2}\int _{{\mathbb {R}}^3}W(x)|{\tilde{Q}}_\omega (x)|^{2}dx=-{\tilde{P}}_0( {\tilde{Q}}_\omega )+{\tilde{P}}_\omega ( {\tilde{Q}}_\omega )=-{\tilde{P}}_0( {\tilde{Q}}_\omega )\rightarrow 0, \end{aligned}$$

as $\omega \rightarrow \infty$. Thus, we see from (5.10) that

$$\begin{aligned} \lim _{\omega \rightarrow \infty }\omega ^{-2} \frac{\int _{{\mathbb {R}}^3}W(x)|{\tilde{Q}}_\omega (x)|^2dx }{F({\tilde{Q}}_\omega )}=0. \end{aligned}$$

This completes the proof. $\square$

Next, we define a set

$$\begin{aligned} {\mathcal {B}}:=\{v\in X;~S_\omega (v)<S_\omega (Q_\omega ),~\Vert v\Vert _{L^2}=\Vert Q_\omega \Vert _{L^2},~F(v)>F(Q_\omega ),~I(v)<0\}, \end{aligned}$$

which is invariant under the flow of (1.4).

Lemma 5.3

Let $1\le d\le 3$, $\omega \in (-d,\infty )$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $Q_\omega$ is the ground state related to (1.7) and $\partial _\mu ^2S(Q_\omega ^\mu )|_{\mu =1}\le 0$. If $u_0\in {\mathcal {B}}$, then the solution u(t) to (1.4) with initial data $u_0$ belongs to ${\mathcal {B}}$.

Proof

Let $u_0\in {\mathcal {B}}$, we deduce from the conservations of mass and energy that

$$\begin{aligned} S_\omega (u(t))=S_\omega (u_0)<S_\omega (Q_\omega ), \end{aligned}$$

(5.12)

and

$$\begin{aligned} \Vert u(t)\Vert _{L^2}=\Vert u_0\Vert _{L^2}=\Vert Q_\omega \Vert _{L^2}, \end{aligned}$$

(5.13)

for any $t\in [0,T^*)$. In addition, by the continuity of the function $t\mapsto F(u(t))$ and Lemma 3.7, if there exists $t_0\in [0,T^*)$ so that $F(u(t_0))= F(Q_\omega )$, then $S_\omega (u(t_0))\ge S_\omega (Q_\omega )$ which contradicts with (5.12). Therefore, we have $F(u(t))>F(Q_\omega )$ for any $t\in [0,T^*)$.

Next, we need only show that if $I(u_0)<0$, then $I(u(t))<0$ for all $t\in [0,T^*)$. Let us prove this by contradiction. If not, there exists $t_0\in [0,T^*)$ such that $I(u(t_0))=0$. On the other hand, due to $\Vert u(t_0)\Vert _{L^2}=\Vert Q_\omega \Vert _{L^2}$, $F(u(t_0))>F(Q_\omega )$, it follows from Lemma 5.1 that

$$\begin{aligned} S_\omega (Q_\omega )\le S_\omega (u(t_0))-\frac{I(u(t_0))}{2}=S_\omega (u(t_0)), \end{aligned}$$

(5.14)

which is a contradiction with (5.12) and (5.13). This completes the proof. $\square$

Lemma 5.4

Let $1\le d\le 3$, $\omega \in (-d,\infty )$, $\lambda _1,\lambda _2\in {\mathbb {R}}$ and satisfy the assumption (1.6). Assume that $Q_\omega$ is the ground state related to (1.7). If $\partial _\mu ^2S_\omega (Q_\omega ^\mu )|_{\mu =1}\le 0$, then $Q_\omega ^\mu \in {\mathcal {B}}$ for all $\mu >1$.

Proof

By some basic calculations, it easily follows that

$$\begin{aligned} \Vert Q_\omega ^\mu \Vert _{L^2}=\Vert Q_\omega \Vert _{L^2}~~and~~ F(Q_\omega ^\mu )=\mu ^3F(Q_\omega )>F(Q_\omega ) \end{aligned}$$

for $\mu >1$. Next, let $f(\mu )$ be defined by (1.19), then,

$$\begin{aligned} f''(\mu )=\Vert \nabla Q_\omega \Vert _{L^2}^2+6\mu ^{-4}\int _{{\mathbb {R}}^3}W(x)|Q_\omega (x)|^2dx -3\mu F(Q_\omega )\le f''(1), \end{aligned}$$

for all $\mu >1$. Therefore, we deduce from $\partial _\mu ^2S_\omega (Q_\omega ^\mu )|_{\mu =1}\le 0$ that

$$\begin{aligned} f''(\mu )\le f''(1)\le 0, \end{aligned}$$

for all $\mu >1$. This, together with Pohozaev identities related to (1.7), implies that

$$\begin{aligned} \frac{I(Q_\omega ^\mu )}{\mu }= f'(\mu )< f'(1)=I(Q_\omega )=0, \end{aligned}$$

and

$$\begin{aligned} S_\omega (Q_\omega ^\mu )<S_\omega (Q_\omega ), \end{aligned}$$

for all $\mu >1$. Thus, $Q_\omega ^\mu \in {\mathcal {B}}$ for all $\mu >1$. $\square$

Proof of Theorem 1.4

Let $\mu _n>1$ such that $\lim _{n\rightarrow \infty }\mu _n=1$ and $u_{0,n}(x)=Q^{\mu _n}_\omega (x)=\mu _n^{\frac{3}{2}}Q_\omega (\mu _nx)$. Assume that $u_n(t)$ is the corresponding solution to (1.4) with initial data $u_{0,n}$. By Lemma 5.4, we have $u_{0,n}\in {\mathcal {B}}$. Since $Q_\omega$ is exponentially decaying, $u_{0,n} \in {\mathcal {B}}\cap \Sigma$. By Lemma 5.3, $u_n(t)\in {\mathcal {B}}\cap \Sigma$ for all $t\in [0,T^*)$. On the other hand, it follows from Lemmas 2.3 and 5.1 that

$$\begin{aligned} \frac{d^2}{dt^2}\int _{{\mathbb {R}}^3} |x|^2 |u_n(t,x)|^2dx=2I(u(t))\le 4(S(u_n(t))-S(Q_\omega ))=4(S(u_{0,n})-S(Q_\omega ))<0, \end{aligned}$$

for all $t\in [0,T^*)$. This implies that the solution $u_n(t)$ with initial data $u_{0,n}$ blows up in finite time. This completes the proof.