Ground state solutions of the non-autonomous Schrödinger–Bopp–Podolsky system

Abstract

In this paper, we consider the following non-autonomous Schrödinger–Bopp–Podolsky system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u + V(x) u + q^2\phi u = f(u)\\ -\Delta \phi + a^2 \Delta ^2 \phi = 4\pi u^2 \end{array}\right. } \hbox { in }{\mathbb {R}}^3. \end{aligned}$$

By using some original analytic techniques and new estimates of the ground state energy, we prove that this system admits a ground state solution under mild assumptions on V and f. In the final part of this paper, we give a min-max characterization of the ground state energy.

1 Introduction

Consider the following Schrödinger–Bopp–Podolsky system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u + V(x) u + q^2\phi u = f(u)\\ -\Delta \phi + a^2 \Delta ^2 \phi = 4\pi u^2 \end{array}\right. } \hbox { in }{\mathbb {R}}^3, \end{aligned}$$

(1.1)

where $u,\phi :{\mathbb {R}}^3 \rightarrow {\mathbb {R}}$, $\omega , a>0$, $q\ne 0$.

This nonlinear system appears when we couple a Schrödinger field $\psi =\psi (t,x)$ with its electromagnetic field in the Bopp–Podolsky electromagnetic theory, and, in particular, in the electrostatic case for standing waves $\psi (t,x)=e^{i\omega t} u(x)$.

System (1.1) has a strong physical meaning especially in the Bopp–Podolsky theory, developed independently by Bopp [3] and Podolsky [24]. The Bopp–Podolsky theory is a second order gauge theory for the electromagnetic field. As the Mie theory [22] and its generalizations given by Born and Infeld [4,5,6,7], it was introduced to solve the “infinity problem”, which appears in the classical Maxwell theory. In fact, by the well-known Gauss law (or Poisson equation), the electrostatic potential $\phi$ for a given charge distribution whose density is $\rho$ satisfies the equation

$$\begin{aligned} -\Delta \phi =\rho \qquad \hbox {in }{\mathbb {R}}^3. \end{aligned}$$

(1.2)

If $\rho =4\pi \delta _{x_0}$, with $x_0\in {\mathbb {R}}^3$, the fundamental solution of (1.2) is ${\mathcal {G}}(x-x_0)$, where

$$\begin{aligned} {\mathcal {G}}(x)=\frac{1}{|x|}, \end{aligned}$$

and the electrostatic energy is

$$\begin{aligned} {\mathcal {E}}_{\mathrm{M}}({\mathcal {G}}) =\frac{1}{2}\int _{{\mathbb {R}}^3} |\nabla {\mathcal {G}}|^2 =+\infty . \end{aligned}$$

Thus, Eq. (1.2) is replaced by

$$\begin{aligned} -{\text {div}}\left( \frac{\nabla \phi }{\sqrt{1-|\nabla \phi |^2}}\right) =\rho \qquad \hbox {in }{\mathbb {R}}^3 \end{aligned}$$

in the Born-Infeld theory and by

$$\begin{aligned} -\Delta \phi + a^2 \Delta ^2 \phi = \rho \qquad \hbox {in }{\mathbb {R}}^3 \end{aligned}$$

in the Bopp–Podolsky theory. In both cases, if $\rho =4\pi \delta _{x_0}$, we are able to write explicitly the solutions of the respective equations and to see that their energy is finite. In particular, when we consider the differential operator $-\Delta + a^2\Delta ^2$, we have that ${\mathcal {K}}(x-x_0)$, with

$$\begin{aligned} {\mathcal {K}}(x):=\frac{1-e^{-|x|/a}}{|x|}, \end{aligned}$$

is the fundamental solution of the equation

$$\begin{aligned} -\Delta \phi + a^2\Delta ^2\phi = 4\pi \delta _{x_0}. \end{aligned}$$

Then ${{\mathcal {K}}}$ has no singularity in $x_0$ since it satisfies

$$\begin{aligned} \lim _{x\rightarrow x_0} {\mathcal {K}}(x-x_0)=\frac{1}{a}, \end{aligned}$$

and its energy is

$$\begin{aligned} {\mathcal {E}}_{\mathrm{BP}}({\mathcal {K}}) =\frac{1}{2}\int _{{\mathbb {R}}^3} |\nabla {\mathcal {K}}|^2 +\frac{a^2}{2} \int _{{\mathbb {R}}^3} |\Delta {\mathcal {K}}|^2 <+\infty . \end{aligned}$$

Moreover, the Bopp–Podolsky theory may be interpreted as an effective theory for short distances (see [20]), while for large distances it is experimentally indistinguishable from the Maxwell theory. Thus, the Bopp–Podolsky parameter $a>0$, which has dimension of the inverse of mass, can be interpreted as a cut-off distance or can be linked to an effective radius for the electron. For more physical details we refer the reader to the recent papers [1, 2, 9, 10, 16, 17] and to references therein.

The differential operator $-\Delta +\Delta ^2$ appears in various different interesting mathematical and physical situations; see [19] and the references therein.

Before stating our results, few preliminaries are in order. We introduce here the space ${\mathcal {D}}$ as the completion of ${\mathcal {C}}^{\infty }_{c}({\mathbb {R}}^{3})$ with respect to the norm $\sqrt{\Vert \nabla \phi \Vert _{2}^2+a^2\Vert \Delta \phi \Vert _2^2}$; see Sect. 2 for more properties on this space.

For fixed $a>0$ and $q\ne 0$, we say that a pair $( u, \phi )\in H^{1}({\mathbb {R}}^{3})\times {\mathcal {D}}$ is a solution of problem (1.1) if

$$\begin{aligned} \displaystyle \int _{{\mathbb {R}}^3}\left[ \nabla u \nabla v +V(x)u v\right] \mathrm {d}x+q^2\int _{{\mathbb {R}}^3} \phi u v\mathrm {d}x&=\int _{{\mathbb {R}}^3}f(u) v\mathrm {d}x, \ \ \ \ \forall \ v\in H^{1}({\mathbb {R}}^{3}), \\ \displaystyle \int _{{\mathbb {R}}^3} \nabla \phi \nabla \xi \mathrm {d}x +a^{2}\int _{{\mathbb {R}}^3} \Delta \phi \Delta \xi \mathrm {d}x&= 4\pi \int _{{\mathbb {R}}^3} \phi u^{2}\mathrm {d}x,\ \ \ \ \forall \ \xi \in {\mathcal {D}}. \end{aligned}$$

We say that a solution $( u,\phi )$ is nontrivial whenever $u \not \equiv 0$; a solution is called a ground state solution if its energy is minimal among all nontrivial solutions. As described in Sect. 2, to solve problem (1.1) is equivalent to solving

$$\begin{aligned} -\Delta u +V(x)u+q^{2}\left( \frac{1-e^{-|x|/a}}{|x|}*u^2\right) u = f(u) \quad \text{ in } {\mathbb {R}}^{3}, \end{aligned}$$

(1.3)

whose solutions correspond to critical points of the energy functional defined in $H^1({\mathbb {R}}^3)$ by

$$\begin{aligned} {\mathcal {I}}(u)=\frac{1}{2}\int _{{\mathbb {R}}^3}\left[ |\nabla u|^2+V(x)u^2\right] \mathrm {d}x +\frac{q^2}{4} \int _{{\mathbb {R}}^3}\left( \frac{1-e^{-|x|/a}}{|x|}*u^2\right) u^2\mathrm {d}x-\int _{{\mathbb {R}}^3}F(u)\mathrm {d}x, \end{aligned}$$

(1.4)

where $F(u)=\int _0^uf(t)\mathrm {d}t$. A solution is called a ground state solution if its energy is minimal among all nontrivial solutions.

In this paper, we also consider the following “limit” system with a general nonlinearity f

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u + V_{\infty }u + q^2\phi u = f(u)\\ -\Delta \phi + a^2 \Delta ^2 \phi = 4\pi u^2 \end{array}\right. } \hbox { in }{\mathbb {R}}^3. \end{aligned}$$

(1.5)

To the best of our knowledge, there is no result on the existence of ground state solutions for systems (1.1) and (1.5). Inspired by [11, 12, 14, 25], we will seek a ground state solution of Nehari–Poho$\breve{\mathrm{z}}$aev type for systems (1.1) and (1.5).

To state our results, we introduce the following assumptions:

• (V1) $V\in {\mathcal {C}}({\mathbb {R}}^3, [0, \infty ))$ and $V_{\infty }:=\lim _{|y|\rightarrow \infty }V(y)= \sup _{x\in {\mathbb {R}}^3}V(x)>0$;

• (V2) $V\in {\mathcal {C}}^1({\mathbb {R}}^3, {\mathbb {R}})$, $\nabla V(x)\cdot x\in L^{\infty }({\mathbb {R}}^3)$, $2V(x)+\nabla V(x)\cdot x \ge 0$ and $\liminf _{|x|\rightarrow \infty }[2V(x)+\nabla V(x)\cdot x]>0$;

• (F1) $f\in {\mathcal {C}}({\mathbb {R}}, {\mathbb {R}})$, and there exist constants ${\mathcal {C}}>0$ and $p\in (2,6)$ such that

  $$\begin{aligned} |f(t)|\le {\mathcal {C}}\left( 1+|t|^{p-1}\right) , \ \ \ \ \forall \ t\in {\mathbb {R}}; \end{aligned}$$

• (F2) $f(t)=o(t)$ as $t\rightarrow 0$;

• (F3) $F(t)\ge 0$ for all $t\in {\mathbb {R}}$ and $\lim _{|t|\rightarrow \infty }\frac{F(t)}{|t|^3}=\infty$;

• (F4) the function $\frac{2f(t)t-3F(t)}{t^3}$ is nondecreasing on $(-\infty ,0)$ and $(0,+\infty )$.

Our first result is as follows.

Theorem 1.1

Assume that (V1), (V2) and (F1)–(F4) hold. Then problem (1.1) admits a ground state solution.

Remark 1.2

There are many functions which satisfy (V1) and (V2). An example is given by $V(x) = 1- \frac{\sin ^2 |x|}{1+ |x|}$.

For the constant potential case, we replace the monotonicity condition (F4) with the super-quadratic condition which is easier to verify:

• (F5) $f(t)t\ge 3F(t)$ for all $t\in {\mathbb {R}}$, and there exist $\kappa >3/2$ and $r_0, {\mathcal {C}}_0>0$ such that

  $$\begin{aligned} \left| \frac{f(t)}{t}\right| ^{\kappa }\le {\mathcal {C}}_0[f(t)t-3F(t)], \ \ \ \ \forall \ |t|\ge r_0. \end{aligned}$$

Our second result is as follows.

Theorem 1.3

Assume that (F1)–(F3) and (F5) hold. Then problem (1.5) admits a ground state solution.

Finally, we give the min-max property of the ground state energy of ${\mathcal {I}}$. To this end, we introduce the following monotonicity condition.

• (V3) $V\in {\mathcal {C}}^1({\mathbb {R}}^3)$, and the function $t\mapsto t^2[V(tx)-\nabla V(tx)\cdot (tx)]$ is increasing on $(0, +\infty )$ for every $x\in {\mathbb {R}}^3$.

We define the Nehari–Poho$\breve{\mathrm{z}}$aev manifold as follows:

$$\begin{aligned} {\mathcal {M}}= \{u\in H^1({\mathbb {R}}^3){\setminus }\{0\} : {\mathcal {J}}(u):=2{\mathcal {I}}'(u)[u]-{\mathcal {P}}(u)=0\}, \end{aligned}$$

(1.6)

where ${\mathcal {P}}(u)$ is the Poho$\breve{\mathrm{z}}$aev functional of (1.3) defined by

$$\begin{aligned} {\mathcal {P}}(u):= & {} \frac{1}{2}\Vert \nabla u\Vert _2^2+\frac{1}{2}\int _{{{\mathbb {R}}}^3}\left[ 3V(x)+\nabla V(x)\cdot x\right] u^2\mathrm {d}x -3\int _{{{\mathbb {R}}}^3}F(u)\mathrm {d}x\nonumber \\&+\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3} \left[ 5\frac{1-e^{-\frac{|x-y|}{a}}}{|x-y|/a} + e^{-\frac{|x-y|}{a}}\right] u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned}$$

(1.7)

If $u\in H^1({\mathbb {R}}^3)$ is a critical point of ${\mathcal {I}}$, then u satisfies ${\mathcal {P}}(u)=0$; see [18, A.14] for more details. Then every nontrivial solution of (1.1) is contained in ${\mathcal {M}}$. In this direction, we have the following theorem.

Theorem 1.4

Assume that (V1), (V3), (F1)–(F4) hold. Then problem (1.1) admits a ground state solution ${\bar{u}}\in H^1({\mathbb {R}}^3)$ such that

$$\begin{aligned} {\mathcal {I}}({\bar{u}})=\inf _{{\mathcal {M}}}{\mathcal {I}}=\inf _{u\in H^1({\mathbb {R}}^3){\setminus }\{0\}}\max _{t> 0}{\mathcal {I}}(t^2u_t)>0, \end{aligned}$$

where $u_t(x):=u(tx)$.

Remark 1.5

We observe that the function $V(x)=1-\frac{1}{\left( 1+|x|\right) ^\alpha }$ with $\alpha >0$ satisfies hypotheses (V1) and (V3).

For the limiting problem related to (1.3), that is, (1.3) with $V(x)\equiv V_{\infty }$, we further weaken (F4) to the following condition:

(F4$'$):

there exists a constant $\theta \in [0,1)$ such that the function $\frac{4f(t)t-6F(t)-\theta V_{\infty } t}{2t^3}$ is nondecreasing on $(-\infty ,0)$ and $(0,+\infty )$.

To state the following result, we define the energy functional in $H^1({\mathbb {R}}^3)$ by

$$\begin{aligned} {\mathcal {I}}^{\infty }(u)=\frac{1}{2}\int _{{\mathbb {R}}^3}\left[ |\nabla u|^2+V_{\infty }u^2\right] \mathrm {d}x +\frac{q^2}{4} \int _{{\mathbb {R}}^3}\left( \frac{1-e^{-|x|/a}}{|x|}*u^2\right) u^2\mathrm {d}x-\int _{{\mathbb {R}}^3}F(u)\mathrm {d}x, \end{aligned}$$

(1.8)

and the Nehari–Poho$\breve{\mathrm{z}}$aev manifold by

$$\begin{aligned} {\mathcal {M}}^{\infty }:= \{u\in H^1({\mathbb {R}}^3){\setminus }\{0\} : {\mathcal {J}}^{\infty }(u):=2({\mathcal {I}}^{\infty })'(u)[u]-{\mathcal {P}}^{\infty }(u)=0\}, \end{aligned}$$

(1.9)

where ${\mathcal {P}}^{\infty }(u)$ is the Poho$\breve{\mathrm{z}}$aev functional defined by

$$\begin{aligned} {\mathcal {P}}^\infty (u):= & {} \frac{1}{2}\Vert \nabla u\Vert _2^2+\frac{3}{2}\int _{{{\mathbb {R}}}^3}V_\infty u^2\mathrm {d}x -3\int _{{{\mathbb {R}}}^3}F(u)\mathrm {d}x\nonumber \\&+\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3} \left[ 5\frac{1-e^{-\frac{|x-y|}{a}}}{|x-y|/a} + e^{-\frac{|x-y|}{a}}\right] u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned}$$

We have the following corollary.

Corollary 1.6

Assume that (F1)–(F3) and (F4$'$) hold. Then problem (1.5) admits a ground state solution ${\bar{u}}\in H^1({\mathbb {R}}^3)$ such that

$$\begin{aligned} {\mathcal {I}}^{\infty }({\bar{u}})=\inf _{{\mathcal {M}}^{\infty }}{\mathcal {I}}^{\infty } =\inf _{u\in H^1({\mathbb {R}}^3){\setminus }\{0\}}\max _{t> 0}{\mathcal {I}}^{\infty }(t^2u_t)>0. \end{aligned}$$

Remark 1.7

Our more general conditions (F1)–(F4) or (F4$'$) on the function f(u) allow many other examples different to the pure power nonlinearity considered in [18]. For example, the function $f(u) = 3|u|u\ln (1+u^2) + \frac{2|u|^3u}{1+u^2}$ satisfies (F1)–(F4). The function $f(u) = a|u|^{3/2}u+b|u|^{1/2}u$ with a, $b > 0$ satisfies (F1)–(F3) and (F4$'$) with $\theta = \frac{2}{3}$ when $15\sqrt{10}a \ge 14 b^{3/2} > 0$ but it does not fulfill (F4).

To prove Theorem 1.4, that is, to obtain a ground solution for Eq. (1.1) with (V1) and (V3), we first choose a minimizing sequence $\{u_n\}$ of ${\mathcal {I}}$ on ${\mathcal {M}}$, which satisfies

$$\begin{aligned} {\mathcal {I}}(u_n)\rightarrow m:=\inf _{{\mathcal {M}}}{\mathcal {I}}, \ \ \ \ \ {\mathcal {P}}(u_n)= 0. \end{aligned}$$

(1.10)

Next, we show that the sequence $\{u_n\}$ is bounded in $H^1({\mathbb {R}}^3)$.

Due to lack of global compactness and adequate information on ${\mathcal {I}}'(u_n)$ and in order to avoid relying the radial compactness, we establish a crucial inequality related to ${\mathcal {I}}(u)$, ${\mathcal {I}}(u_t)$ and ${\mathcal {J}}(u)$ (Lemma 3.4), which plays a crucial role in our arguments, see Lemmas 3.8, 3.9, 3.13, 3.14 and 4.5 . With the help of this inequality, we then can recover the compactness for the minimizing sequence $\{u_n\}$ and show that $\{u_n\}$ converges weakly to some ${\bar{u}}\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ and ${\mathcal {I}}({\bar{u}})=\inf _{{\mathcal {M}}}{\mathcal {I}}$ by using Lions’ concentration-compactness, the “least energy squeeze approach” and some subtle analysis. Finally, we take advantage of a quantitative deformation lemma and the intermediate value theorem to show that ${\bar{u}}$ is a critical point of ${\mathcal {I}}$, as the Lagrange multiplier theorem does not work, because ${\mathcal {M}}$ is not a ${\mathcal {C}}^1$-manifold, .

To prove Theorem 1.1, we use the monotonicity technique explored by Jeanjean [21] to parameterize the nonlinearity f. In such a way, we build a parametrization of the energy functional associated to (1.1) and give some energy relations of problems (1.1) and (1.5) which play a key role in getting the critical point of (1.1), see Lemma 4.5. Moreover, in order to show that a critical point associated to the parametrization functional is indeed a solution to the original problem, we also need give a delicate estimation for the parametrization problem. Finally, we study the constant potential case by using weaker conditions.

Throughout the paper we make use of the following notations:

• Under (V1), $H^1({\mathbb {R}}^3)$ denotes the Sobolev space equipped with the inner product and norm

  $$\begin{aligned} (u, v)=\int _{{\mathbb {R}}^3}[\nabla u \nabla v+V(x)uv]\mathrm {d}x, \ \ \Vert u\Vert =(u, u)^{1/2}, \ \ \forall \ u,v\in H^1({\mathbb {R}}^3); \end{aligned}$$

• $L^s({\mathbb {R}}^3) (1\le s< \infty )$ denotes the Lebesgue space with the norm $\Vert u\Vert _s =\left( \int _{{\mathbb {R}}^3}|u|^s\mathrm {d}x\right) ^{1/s}$;

• For any $x\in {\mathbb {R}}^3$ and $r>0$, $B_r(x):=\{y\in {\mathbb {R}}^3: |y-x|<r \}$;

• $S=\inf _{u\in D^{1,2}({\mathbb {R}}^3){\setminus }\{0\}}\Vert \nabla u\Vert _2^2/\Vert u\Vert _6^2$;

• $C_1, C_2,\cdots$ denote positive constants possibly different in different places.

2 Variational setting

We start with some preliminary basic results. Let us consider the nonlinear Schrödinger Lagrangian density

$$\begin{aligned} {\mathcal {L}}_{\mathrm{Sc}} =i\hbar {\bar{\psi }} \partial _t \psi -\frac{\hbar ^2}{2m} |\nabla \psi |^2 + 2F(\psi ), \end{aligned}$$

where $\psi :{\mathbb {R}}\times {\mathbb {R}}^3\rightarrow {\mathbb {C}}$, $\hbar ,m>0$, and let $(\phi ,\mathbf{A})$ be the gauge potential of the electromagnetic field $(\mathbf{E},\mathbf{H})$, namely $\phi :{\mathbb {R}}^3\rightarrow {\mathbb {R}}$ and $\mathbf{A}:{\mathbb {R}}^3\rightarrow {\mathbb {R}}^3$ satisfy

$$\begin{aligned} {\mathbf {E}}=-\nabla \phi -\frac{1}{c}\partial _t{\mathbf {A}}, \qquad {\mathbf {H}}=\nabla \times {\mathbf {A}}. \end{aligned}$$

The coupling of the field $\psi$ with the electromagnetic field $(\mathbf{E},\mathbf{H})$ through the minimal coupling rule, namely the study of the interaction between $\psi$ and its own electromagnetic field, can be obtained by replacing in ${\mathcal {L}}_{\mathrm{Sc}}$ the derivatives $\partial _t$ and $\nabla$ respectively with the covariant ones

$$\begin{aligned} D_{t}= \partial _{t}+\frac{iq}{\hbar } \phi , \qquad {\mathbf {D}}=\nabla -\frac{iq}{\hbar c} {\mathbf {A}}, \end{aligned}$$

q being a coupling constant. This leads to consider

$$\begin{aligned} {\mathcal {L}}_{\mathrm{CSc}}&= i \hbar {\overline{\psi }} D_{t}\psi -\frac{\hbar ^{2}}{2m} |{\mathbf {D}} \psi |^{2}+2F(\psi )\\&= i\hbar \overline{\psi } \left( \partial _t +\frac{iq}{\hbar } \phi \right) \psi -\frac{\hbar ^2}{2m} \left| \left( \nabla - \frac{iq}{\hbar c}{} \mathbf{A}\right) \psi \right| ^2 + 2F(\psi ). \end{aligned}$$

Now, to get the total Lagrangian density, we have to add to ${\mathcal {L}}_{\mathrm{CSc}}$ the Lagrangian density of the electromagnetic field.

The Bopp–Podolsky Lagrangian density (see [24, Formula (3.9)]) is

$$\begin{aligned} {\mathcal {L}}_{\mathrm{BP}}&= \frac{1}{8\pi }\left\{ |{\mathbf {E}}|^2 - |{\mathbf {H}}|^2 + a^2 \left[ ({\text {div}} {\mathbf {E}})^2 - \left| \nabla \times {\mathbf {H}} - \frac{1}{c} \partial _t{\mathbf {E}}\right| ^2\right] \right\} \\&= \frac{1}{8\pi }\left\{ |\nabla \phi +\frac{1}{c}\partial _t{\mathbf {A}}|^2 - |\nabla \times {\mathbf {A}}|^2 \right. \\&\qquad \left. + a^2 \left[ \left( \Delta \phi +\frac{1}{c}{\text {div}} \partial _t {\mathbf {A}}\right) ^2 - \left| \nabla \times \nabla \times {\mathbf {A}} + \frac{1}{c} \partial _t(\nabla \phi +\frac{1}{c}\partial _t{\mathbf {A}})\right| ^2\right] \right\} . \end{aligned}$$

Thus, the total action is

$$\begin{aligned} {\mathcal {S}}(\psi ,\phi ,{\mathbf {A}})=\int _{{\mathbb {R}}^3}{\mathcal {L}} \mathrm {d}x\mathrm {d}t \end{aligned}$$

where ${\mathcal {L}}:={\mathcal {L}}_{\mathrm{CSc}} + {\mathcal {L}}_{\mathrm{BP}}$ is the total Lagrangian density.

Let ${\mathcal {D}}$ be the completion of ${\mathcal {C}}_c^\infty ({\mathbb {R}}^3)$ with respect to the norm $\Vert \cdot \Vert _{{\mathcal {D}}}$ induced by the scalar product

$$\begin{aligned} \langle \varphi ,\psi \rangle _{{\mathcal {D}}} := \int _{{\mathbb {R}}^3}\nabla \varphi \nabla \psi \mathrm {d}x + a^2 \int _{{\mathbb {R}}^3}\Delta \varphi \Delta \psi \mathrm {d}x. \end{aligned}$$

Then ${\mathcal {D}}$ is a Hilbert space continuously embedded into $D^{1,2}({\mathbb {R}}^3)$ and consequently in $L^6({\mathbb {R}}^3)$.

We notice the following auxiliary properties; see Lemmas 3.1 and 3.2 in [18].

Lemma 2.1

The space ${\mathcal {D}}$ is continuously embedded in $L^\infty ({\mathbb {R}}^3)$.

The next property gives a useful characterization of the space ${\mathcal {D}}$.

Lemma 2.2

The space ${\mathcal {C}}^{\infty }_{c}({\mathbb {R}}^{3})$ is dense in

$$\begin{aligned} {\mathcal {A}} := \left\{ \phi \in D^{1,2}({\mathbb {R}}^{3}) : \Delta \phi \in L^{2}({\mathbb {R}}^{3})\right\} \end{aligned}$$

normed by $\sqrt{\langle \phi ,\phi \rangle _{{\mathcal {D}}}}$ and, therefore, ${\mathcal {D}}={\mathcal {A}}$.

For every fixed $u\in H^1({\mathbb {R}}^3)$, the Riesz representation theorem implies that there is a unique solution $\phi _u\in {\mathcal {D}}$ of the second equation in (1.1). To write explicitly such a solution (see also [24, Formula (2.6)]), we consider

$$\begin{aligned} {\mathcal {K}}(x)=\frac{1-e^{-|x|/a}}{|x|}. \end{aligned}$$

We have the following fundamental properties.

Lemma 2.3

[18, Lemma 3.3] For all $y\in {\mathbb {R}}^3$, ${\mathcal {K}}(\cdot -y)$ solves in the sense of distributions

$$\begin{aligned} -\Delta \phi +a^2\Delta ^2 \phi = 4\pi \delta _y. \end{aligned}$$

Moreover,

• (i) if $g\in L^1_{\mathrm{loc}}({\mathbb {R}}^3)$ and, for a.e. $x\in {\mathbb {R}}^3$, the map $y\in {\mathbb {R}}^3\mapsto g(y)/|x-y|$ is summable, then ${\mathcal {K}}*g \in L^1_{\mathrm{loc}}({\mathbb {R}}^3)$;

• (ii) if $f\in L^s ({\mathbb {R}}^3)$ with $1\le s< 3/2$, then ${\mathcal {K}}*g\in L^q({\mathbb {R}}^3)$ for $q\in (3s/(3-2s),+\infty ]$.

In both cases, ${\mathcal {K}}*g$ solves

$$\begin{aligned} -\Delta \phi +a^2\Delta ^2 \phi = 4\pi g \end{aligned}$$

(2.1)

in the sense of distributions, and we have the following distributional derivatives:

$$\begin{aligned} \nabla ({\mathcal {K}}*g)= (\nabla {\mathcal {K}})*g \quad \hbox { and }\quad \Delta ({\mathcal {K}}*g)= (\Delta {\mathcal {K}})*g \quad \hbox { a.e. in } {\mathbb {R}}^3. \end{aligned}$$

Fix $u\in H^1({\mathbb {R}}^3)$, the unique solution in ${\mathcal {D}}$ of the second equation in (1.1) is

$$\begin{aligned} \phi _u:={\mathcal {K}}*u^2. \end{aligned}$$

(2.2)

Actually the following useful properties hold.

Lemma 2.4

[18, Lemma 3.4] For every $u\in H^{1}({\mathbb {R}}^{3})$ we have:

• (1) for every $y\in {\mathbb {R}}^3$, $\phi _{u( \cdot +y)} = \phi _{u}( \cdot +y)$;

• (2) $\phi _{u}\ge 0$;

• (3) for every $s\in (3,+\infty ]$, $\phi _{u}\in L^{s}({\mathbb {R}}^{3})\cap {\mathcal {C}}_{0}({\mathbb {R}}^{3})$;

• (4) for every $s\in (3/2,+\infty ]$, $\nabla \phi _{u} = \nabla {\mathcal {K}} * u^{2}\in L^{s}({\mathbb {R}}^{3})\cap {\mathcal {C}}_{0}({\mathbb {R}}^{3})$;

• (5) $\phi _u\in {\mathcal {D}}$;

• (6) $\Vert \phi _{u}\Vert _{6}\le C \Vert u\Vert ^{2}$;

• (7) $\phi _{u}$ is the unique minimizer of the functional

  $$\begin{aligned} E(\phi ) = \frac{1}{2} \Vert \nabla \phi \Vert _{2}^{2} +\frac{a^{2}}{2} \Vert \Delta \phi \Vert _{2}^{2}-\int _{{\mathbb {R}}^3}\phi u^{2}\mathrm {d}x, \quad \phi \in {\mathcal {D}}. \end{aligned}$$

Moreover, if $v_{n}\rightharpoonup v$ in $H^{1}({\mathbb {R}}^{3})$, then $\phi _{v_{n}} \rightharpoonup \phi _{v}$ in ${\mathcal {D}}$.

Under hypotheses (V1), (F1) and (F2), the energy functional defined in $H^{1}({\mathbb {R}}^{3})\times {\mathcal {D}}$ by

$$\begin{aligned} {\mathcal {S}}(u,\phi )= & {} \frac{1}{2}\int _{{\mathbb {R}}^3}\left[ |\nabla u|_2^2+V(x)u^2\right] \mathrm {d}x + \frac{q^{2}}{2} \int _{{\mathbb {R}}^3} \phi u^2\mathrm {d}x\nonumber \\&-\frac{q^{2}}{16\pi } \Vert \nabla \phi \Vert _2^2 -\frac{a^2q^{2}}{16\pi } \Vert \Delta \phi \Vert _2^2-\int _{{\mathbb {R}}^3}F(u)\mathrm {d}x \end{aligned}$$

(2.3)

is continuously differentiable and its critical points correspond to the weak solutions of problem (1.1). Indeed, if $(u, \phi )\in H^{1}({\mathbb {R}}^{3})\times {\mathcal {D}}$ is a critical point of ${\mathcal {S}}$, then

$$\begin{aligned} 0= & {} \partial _{u} {\mathcal {S}}( u,\phi )[v] =\int _{{\mathbb {R}}^3}\left[ \nabla u \nabla v+V(x)uv\right] \mathrm {d}x\\&+q^{2}\int _{{\mathbb {R}}^3}\phi u v\mathrm {d}x -\int _{{\mathbb {R}}^3}f(u)v\mathrm {d}x, \ \ \ \ \forall \ v\in H^1({\mathbb {R}}^3) \end{aligned}$$

and

$$\begin{aligned} 0=\partial _{\phi } {\mathcal {S}}( u,\phi )[\xi ]=\frac{ q^{2}}{2} \int _{{\mathbb {R}}^3}{u}^{2} \xi \mathrm {d}x - \frac{q^{2}}{8\pi }\int _{{\mathbb {R}}^3}\nabla \phi \nabla \xi \mathrm {d}x - \frac{a^2 q^{2}}{8\pi }\int _{{\mathbb {R}}^3}\Delta \phi \Delta \xi \mathrm {d}x,\ \ \ \forall \ \xi \in {\mathcal {D}}. \end{aligned}$$

(2.4)

In order to avoid the difficulty generated by the strongly indefiniteness of the functional ${\mathcal {S}}$, we apply a reduction procedure. Noting that $\partial _{\phi } {\mathcal {S}}$ is a ${\mathcal {C}}^{1}$ functional, if $G_{\Phi }$ is the graph of the map $\Phi : u\in H^{1}({\mathbb {R}}^{3})\mapsto \phi _{u}\in {\mathcal {D}}$, an application of the implicit function theorem gives

$$\begin{aligned} G_{\Phi }=\left\{ (u,\phi )\in H^{1}({\mathbb {R}}^{3})\times {\mathcal {D}}: \partial _{\phi } {\mathcal {S}}(u,\phi ) = 0\right\} \quad \text { and } \quad \Phi \in {\mathcal {C}}^{1}(H^{1}({\mathbb {R}}^{3}), {\mathcal {D}} ). \end{aligned}$$

Jointly with (2.3) and (2.4), the functional ${\mathcal {I}}(u):={\mathcal {S}}( u,\phi _u)$ has the reduced form

$$\begin{aligned} {\mathcal {I}}(u)=\frac{1}{2}\int _{{\mathbb {R}}^3}\left[ |\nabla u|^2+V(x)u^2\right] \mathrm {d}x +\frac{q^2}{4} \int _{{\mathbb {R}}^3}\phi _u u^2\mathrm {d}x-\int _{{\mathbb {R}}^3}F(u)\mathrm {d}x, \end{aligned}$$

(2.5)

which is of class ${\mathcal {C}}^{1}$ on $H^{1}({\mathbb {R}}^{3})$ and, for all $u,v\in H^{1}({\mathbb {R}}^{3})$

$$\begin{aligned} {\mathcal {I}}'(u)[v]= & {} \partial _{u}{\mathcal {S}}(u,\Phi (u))[v]+ \partial _{\phi }{\mathcal {S}}(u,\Phi (u))\circ \Phi '(u) [v]\nonumber \\= & {} \partial _{u} {\mathcal {S}}(u,\Phi (u))[v] \nonumber \\= & {} \int _{{\mathbb {R}}^3}\left[ \nabla u \nabla v+V(x)uv\right] \mathrm {d}x+q^{2}\int _{{\mathbb {R}}^3}\phi _{u}uv\mathrm {d}x -\int _{{\mathbb {R}}^3}f(u)v\mathrm {d}x. \end{aligned}$$

(2.6)

Moreover, the following statements are equivalent:

• (i) the pair $(u,\phi )\in H^{1}({\mathbb {R}}^{3})\times {\mathcal {D}}$ is a critical point of ${\mathcal {S}}$, that is, $(u,\phi )$ is a solution of problem (1.1);

• (ii) u is a critical point of ${\mathcal {I}}$ and $\phi =\phi _{u}$.

Hence, if $u\in H^1({\mathbb {R}}^3)$ is a critical point of ${\mathcal {I}}$, then the pair $(u, \phi _{u})$ is a solution of (1.1). For the sake of simplicity, in many cases we just say $u\in H^1({\mathbb {R}}^3)$, instead of $(u, \phi _{u})\in H^1({\mathbb {R}}^3)\times {\mathcal {D}}$, is a solution of (1.1).

3 Proof of Theorem 1.3

In this section, we give the proof of Theorem 1.3.

By a simple calculation, we have the following two lemmas.

Lemma 3.1

Let $b>0$. Then

$$\begin{aligned} h(t):=t^3\left[ e^{-\frac{b}{t}}-e^{-b}\right] +\frac{1-t^3}{3}be^{-b}\ge 0, \ \ \ \ \forall \ t>0 \end{aligned}$$

(3.1)

and

$$\begin{aligned} 1-e^{-b}-\frac{1}{3}be^{-b}\ge 0. \end{aligned}$$

(3.2)

Lemma 3.2

• (i) Assume that (V1) and (V3) hold. Then

  $$\begin{aligned} 3\left[ V(x)-tV(t^{-1}x)\right] -(1-t^3)[V(x)-\nabla V(x)\cdot x] > 0, \ \ \ \ \forall \ t\in [0, 1)\cup (1, +\infty ). \end{aligned}$$

  (3.3)

• (ii) Assume that (F1) and (F4) hold. Then

  $$\begin{aligned} \frac{2(1-t^3)}{3}f(\tau )\tau +(t^3-2)F(\tau )+\frac{1}{t^3}F(t^2\tau )\ge 0, \ \ \ \ \forall \ t> 0, \ \tau \in {\mathbb {R}}. \end{aligned}$$

  (3.4)

• (iii) Assume that (F1) and (F4$'$) hold. Then

  $$\begin{aligned}&\frac{2(1-t^3)}{3}f(\tau )\tau +(t^3-2)F(\tau )+\frac{1}{t^3}F(t^2\tau )\nonumber \\&\quad +\frac{\theta _0}{6}(1-t)^2(2+t)V_{\infty }\tau ^2\ge 0, \ \ \ \ \forall \ t> 0, \ \tau \in {\mathbb {R}}. \end{aligned}$$

  (3.5)

Note that if $t\rightarrow 0$ in (3.4) and (3.5), then

$$\begin{aligned} f(\tau )\tau -3F(\tau )\ge 0, \ \ \ \ \forall \ \tau \in {\mathbb {R}}\end{aligned}$$

(3.6)

and

$$\begin{aligned} f(\tau )\tau -3F(\tau )+\frac{\theta V_\infty }{2}\tau ^2 \ge 0, \ \ \ \ \forall \ \tau \in {\mathbb {R}}. \end{aligned}$$

(3.7)

Lemma 3.3

Assume that (V1) and (V3) hold. Then

$$\begin{aligned} |\nabla V(x)\cdot x|\rightarrow 0\ \ \text{ as }\ |x|\rightarrow \infty . \end{aligned}$$

(3.8)

Proof

Arguing by contradiction, we assume that there exist a sequence $\{x_n\}\subset {\mathbb {R}}^3$ and $\delta >0$ such that

$$\begin{aligned} |x_n|\rightarrow \infty , \ \ \text{ and } \ \ \nabla V(x_n)\cdot x_n\ge \delta \ \text{ or }\ \nabla V(x_n)\cdot x_n\le -\delta \ \ \ \ \forall \ n\in {\mathbb {N}}. \end{aligned}$$

Now, we distinguish two cases: i) $\nabla V(x_n)\cdot x_n\ge \delta$ for all $n\in {\mathbb {N}}$ and ii) $\nabla V(x_n)\cdot x_n\le -\delta$ for all $n\in {\mathbb {N}}$.

Case i) $\nabla V(x_n)\cdot x_n\ge \delta$ for all $n\in {\mathbb {N}}$. In this case, by (3.3), one has

$$\begin{aligned} \delta\le & {} \nabla V(x_n)\cdot x_n\nonumber \\< & {} V(x_n)+\frac{3}{t^3-1}[V(x_n)-tV(t^{-1}x_n)]\nonumber \\= & {} V(x_n)+\frac{3(1-t)}{t^3-1}V(x_n)+\frac{3t}{t^3-1}[V(x_n)-V(t^{-1}x_n)]\nonumber \\= & {} \frac{(t-1)(t+2)}{t^2+t+1}V(x_n)+\frac{3t}{t^3-1}[V(x_n)-V(t^{-1}x_n)], \ \ \forall \ t>1. \end{aligned}$$

(3.9)

Since

$$\begin{aligned} \lim _{|t|\rightarrow 1}\frac{(t-1)(t+2)}{t^2+t+1}=0, \end{aligned}$$

(3.10)

there exists $t_1>1$ such that

$$\begin{aligned} \frac{(t_1-1)(t_1+2)}{t_1^2+t_1+1}V_{\infty }<\frac{\delta }{2}. \end{aligned}$$

(3.11)

Then it follows from (V1), (3.9) and (3.11) that

$$\begin{aligned} \delta \le \frac{(t_1-1)(t_1+2)}{t_1^2+t_1+1}V_{\infty }+\frac{3t_1}{t_1^3-1}[V(x_n)-V(t_1^{-1}x_n)] \le \frac{\delta }{2}+o(1), \end{aligned}$$

(3.12)

which is an obvious contradiction.

Case ii) $\nabla V(x_n)\cdot x_n\le -\delta$ for all $n\in {\mathbb {N}}$. In this case, (3.3) gives

$$\begin{aligned} -\delta\ge & {} \nabla V(x_n)\cdot x_n\nonumber \\> & {} V(x_n)+\frac{3}{1-t^3}[tV(t^{-1}x_n)-V(x_n)]\nonumber \\= & {} V(x_n)+\frac{3(t-1)}{1-t^3}V(x_n)+\frac{3t}{1-t^3}[V(t^{-1}x_n)-V(x_n)]\nonumber \\= & {} \frac{(t-1)(t+2)}{t^2+t+1}V(x_n)+\frac{3t}{1-t^3}[V(t^{-1}x_n)-V(x_n)], \ \ \forall \ 0<t<1.\nonumber \\ \end{aligned}$$

(3.13)

From (3.10), there exists $0<t_2<1$ such that

$$\begin{aligned} \frac{(t_2-1)(t_2+2)}{t_2^2+t_2+1}V_{\infty }>-\frac{\delta }{2}. \end{aligned}$$

(3.14)

Then it follows from (V1), (3.13) and (3.14) that

$$\begin{aligned} -\delta \ge \frac{(t_2-1)(t_2+2)}{t_2^2+t_2+1}V_{\infty }+\frac{3t_2}{1-t_2^3}[V(t_2^{-1}x_n)-V(x_n)] \ge -\frac{\delta }{2}+o(1),\nonumber \\ \end{aligned}$$

(3.15)

which is again an obvious contradiction. This completes the proof. $\square$

Since ${\mathcal {J}}(u)=2{\mathcal {I}}'(u)[u]-{\mathcal {P}}(u)$ for $u\in H^1({\mathbb {R}}^3)$, we have

$$\begin{aligned} {\mathcal {J}}(u)= & {} \frac{3}{2}\Vert \nabla u\Vert _2^2+\frac{1}{2}\int _{{{\mathbb {R}}}^3}[V(x)-\nabla V(x)\cdot x]u^2\mathrm {d}x -\int _{{{\mathbb {R}}}^3}[2f(u)u-3F(u)]\mathrm {d}x\nonumber \\&+\frac{3q^2}{4}\int _{{\mathbb {R}}^3}\phi _{u}u^2\mathrm {d}x -\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned}$$

(3.16)

Define the function

$$\begin{aligned} \beta (x,t):=3\left[ V(x)-tV(t^{-1}x)\right] -(1-t^3)[V(x)-\nabla V(x)\cdot x], \ \ \ \ \forall \ x\in {\mathbb {R}}^3,\ t >0. \end{aligned}$$

(3.17)

Lemma 3.4

Assume that (V1), (V3), (F1) and (F4) hold. Then

$$\begin{aligned} {\mathcal {I}}(u) \ge {\mathcal {I}}\left( t^2u_t\right) +\frac{1-t^3}{3}{\mathcal {J}}(u) +\frac{1}{6}\int _{{{\mathbb {R}}}^3}\beta (x, t)u^2\mathrm {d}x, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3), \ t> 0, \end{aligned}$$

(3.18)

where $u_t(x)=u(tx)$.

Proof

For $u\in H^1({\mathbb {R}}^3)$ and $t> 0$, one has

$$\begin{aligned} {\mathcal {I}}\left( t^2u_t\right)= & {} \frac{t^3}{2}\Vert \nabla u\Vert _2^2+\frac{t}{2}\int _{{{\mathbb {R}}}^3}V(t^{-1}x)u^2\mathrm {d}x\nonumber \\&+\frac{q^2t^3}{4}\int _{{{\mathbb {R}}}^3}\int _{{{\mathbb {R}}}^3}\frac{1-e^{-\frac{|x-y|}{ta}}}{|x-y|} u^2(x)u^2(y)\mathrm {d}x\mathrm {d}y\nonumber \\&\ \ -\frac{1}{t^3}\int _{{{\mathbb {R}}}^3}F(t^2u)\mathrm {d}x. \end{aligned}$$

(3.19)

Thus, (2.5), (3.1), (3.3), (3.4), (3.16), (3.17) and (3.19) imply that for all $u\in H^1({\mathbb {R}}^3)$ and all $t> 0$

$$\begin{aligned}&{\mathcal {I}}(u)-{\mathcal {I}}\left( t^2u_t\right) \\&\quad = \frac{1-t^3}{2}\Vert \nabla u\Vert _2^2+\frac{1}{2}\int _{{{\mathbb {R}}}^3}\left[ V(x)-tV(t^{-1}x)\right] u^2\mathrm {d}x +\int _{{{\mathbb {R}}}^3}\left[ \frac{1}{t^3}F(t^2u)-F(u)\right] \mathrm {d}x\\&\qquad +\frac{q^2}{4}\int _{{{\mathbb {R}}}^3}\int _{{{\mathbb {R}}}^3}\frac{1-e^{-\frac{|x-y|}{a}} -t^3\left( 1-e^{\frac{-|x-y|}{at}}\right) }{|x-y|}u^2(x)u^2(y)\mathrm {d}x\mathrm {d}y\\&\quad = \frac{1-t^3}{3}{\mathcal {J}}(u)+\frac{1}{6}\int _{{{\mathbb {R}}}^3} \left\{ 3\left[ V(x)-tV(t^{-1}x)\right] -(1-t^3)[V(x)-\nabla V(x)\cdot x]\right\} u^2\mathrm {d}x\\&\qquad \ \ +\int _{{{\mathbb {R}}}^3}\left[ \frac{2(1-t^3)}{3}f(u)u+(t^3-2)F(u)+\frac{1}{t^3}F(t^2u)\right] \mathrm {d}x\\&\qquad \ \ +\frac{3q^2}{4}\int _{{{\mathbb {R}}}^3}\int _{{{\mathbb {R}}}^3}\frac{t^3\left[ e^{-\frac{|x-y|}{at}}-e^{-\frac{|x-y|}{a}}\right] +(1-t^3)\frac{|x-y|}{3a}e^{-\frac{|x-y|}{a}}}{|x-y|}u^2(x)u^2(y)\mathrm {d}x\mathrm {d}y\\&\quad \ge \frac{1-t^3}{3}{\mathcal {J}}(u)+\frac{1}{6}\int _{{{\mathbb {R}}}^3}\beta (x, t)u^2\mathrm {d}x. \end{aligned}$$

This shows (3.18). $\square$

Remark that (3.18) with $t\rightarrow 0$ gives

$$\begin{aligned} {\mathcal {I}}(u) \ge \frac{1}{3}{\mathcal {J}}(u)+\frac{1}{6}\int _{{{\mathbb {R}}}^3} \left[ 2V(x)+\nabla V(x)\cdot x\right] u^2\mathrm {d}x, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3). \end{aligned}$$

(3.20)

For the limiting problem, corresponding to (2.5) and (3.16), we define the following functionals in $H^1({\mathbb {R}}^3)$:

$$\begin{aligned} {\mathcal {I}}^{\infty }(u) = \frac{1}{2}\int _{{\mathbb {R}}^3}\left( |\nabla u|^2+V_{\infty }u^2\right) \mathrm {d}x +\frac{q^2}{4}\int _{{{\mathbb {R}}}^3}\phi _uu^2\mathrm {d}x-\int _{{\mathbb {R}}^3}F(u)\mathrm {d}x \end{aligned}$$

(3.21)

and

$$\begin{aligned} {\mathcal {J}}^{\infty }(u)= & {} \frac{3}{2}\Vert \nabla u\Vert _2^2+\frac{V_{\infty }}{2}\Vert u\Vert _2^2 -\int _{{{\mathbb {R}}}^3}[2f(u)u-3F(u)]\mathrm {d}x\nonumber \\&+\frac{3q^2}{4}\int _{{\mathbb {R}}^3}\phi _{u}u^2\mathrm {d}x -\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned}$$

(3.22)

From Lemma 3.4, we deduce the following two properties.

Corollary 3.5

Assume that (V1), (V3), (F1) and (F4) hold. Then for $u\in {\mathcal {M}}$

$$\begin{aligned} {\mathcal {I}}(u) = \max _{t> 0} {\mathcal {I}}\left( t^2u_t\right) . \end{aligned}$$

Corollary 3.6

Assume that (F1) and (F4) hold. Then

$$\begin{aligned} {\mathcal {I}}^{\infty }(u) \ge {\mathcal {I}}^{\infty }\left( t^2u_t\right) +\frac{1-t^3}{3}{\mathcal {J}}^{\infty }(u) +\frac{(1-t)^2(2+t)}{6}V_{\infty }\Vert u\Vert _2^2, \ \ \forall \ u\in H^1({\mathbb {R}}^3), \ t> 0. \end{aligned}$$

(3.23)

By using (3.5) instead of (3.4), as in the proof of Lemma 3.4, we have the following lemma.

Lemma 3.7

Assume that (F1) and (F4$'$) hold. Then

$$\begin{aligned} {\mathcal {I}}^{\infty }(u)\ge & {} {\mathcal {I}}^{\infty }\left( t^2u_t\right) +\frac{1-t^3}{3}{\mathcal {J}}^{\infty }(u)\nonumber \\&+\frac{(1-\theta )(1-t)^2(2+t)}{6}V_{\infty }\Vert u\Vert _2^2, \ \ \forall \ u\in H^1({\mathbb {R}}^3), \ t> 0. \end{aligned}$$

(3.24)

Lemma 3.8

Assume that (V1), (V3) and (F1)–(F4) hold. Then for any $u\in H^1({\mathbb {R}}^3){\setminus }\{0\}$, there exists a unique $t_u>0$ such that $t_u^2u_{t_u}\in {\mathcal {M}}$.

Proof

Let $u\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ be fixed and define the function $\zeta (t):={\mathcal {I}}(t^2u_t)$ on $(0, \infty )$. Using (3.16) and (1.6), it is easily checked that

$$\begin{aligned} \zeta '(t)=0 \ \ \Leftrightarrow \ \ \frac{1}{t}{\mathcal {J}}(t^2u_t)=0 \ \ \Leftrightarrow \ \ t^2u_t\in {\mathcal {M}}. \end{aligned}$$

By (V1) and (F1)–(F3), we have $\lim _{t\rightarrow 0^+}\zeta (t)=0$, $\zeta (t)>0$ for $t>0$ small and $\zeta (t)<0$ for t large. Therefore, $\max _{t\in (0, \infty )}\zeta (t)$ is achieved at $t_0=t_u>0$, so that $\zeta '(t_0)=0$ and $t_0^2u_{t_0}\in {\mathcal {M}}$.

Next, we claim that $t_u$ is unique for any $u\in H^1({\mathbb {R}}^3){\setminus }\{0\}$. In fact, for any given $u\in H^1({\mathbb {R}}^3){\setminus }\{0\}$, let $t_1, t_2>0$ be such that $\zeta '(t_1)= \zeta '(t_2)=0$. Then ${\mathcal {J}}(t_1^2u_{t_1})={\mathcal {J}}(t_2^2u_{t_2})=0$. Jointly with (3.18), we have

$$\begin{aligned} {\mathcal {I}}(t_1^2u_{t_1})\ge & {} {\mathcal {I}}(t_2^2u_{t_2})+\frac{t_1^3-t_2^3}{3t_1^3}{\mathcal {J}}(t_1^2u_{t_1}) +\frac{t_1}{6}\int _{{{\mathbb {R}}}^3}\beta (x, t_2/t_1)u^2\mathrm {d}x\nonumber \\= & {} {\mathcal {I}}(t_2^2u_{t_2})+\frac{t_1}{6}\int _{{{\mathbb {R}}}^3}\beta (x, t_2/t_1)u^2\mathrm {d}x \end{aligned}$$

(3.25)

and

$$\begin{aligned} {\mathcal {I}}(t_2^2u_{t_2})\ge & {} {\mathcal {I}}(t_1^2u_{t_1})+\frac{t_2^3-t_1^3}{3t_2^3}{\mathcal {J}}(t_2^2u_{t_2}) +\frac{t_2}{6}\int _{{{\mathbb {R}}}^3}\beta (x, t_1/t_2)u^2\mathrm {d}x\nonumber \\= & {} {\mathcal {I}}(t_1^2u_{t_1})+\frac{t_2}{6}\int _{{{\mathbb {R}}}^3}\beta (x, t_1/t_2)u^2\mathrm {d}x. \end{aligned}$$

(3.26)

Then (3.1), (3.25) and (3.25) give $t_1=t_2$. Therefore, $t_u> 0$ is unique for any $u\in H^1({\mathbb {R}}^3){\setminus }\{0\}$. $\square$

Combining Corollary 3.5 with Lemma 3.8, w obtain the following min-max property.

Lemma 3.9

Assume that (V1), (V3) and (F1)–(F4) hold. Then

$$\begin{aligned} m=\inf _{{\mathcal {M}}}{\mathcal {I}}=\inf _{u\in H^1({\mathbb {R}}^3){\setminus }\{0\}}\max _{t> 0}{\mathcal {I}}(t^2u_t). \end{aligned}$$

Lemma 3.10

Assume that (V1), (V3) and (F1)–(F4) hold. Then

• (i) there exists $\rho >0$ such that $\Vert u\Vert \ge \rho , \ \forall \ u\in {\mathcal {M}}$;

• (ii) $m=\inf _{{\mathcal {M}}} {\mathcal {I}}>0$.

Proof

• (i). In view of [13, Lemma 2.5], if V satisfies (V1) and (V3), then there exist $\varrho _1, \varrho _2>0$ such that

  $$\begin{aligned}&2V(x)+\nabla V(x)\cdot x \ge \varrho _1, \ \ \ \ \forall \ x\in {\mathbb {R}}^3, \end{aligned}$$

  (3.27)
  $$\begin{aligned}&V(x)-\nabla V(x)\cdot x \ge \varrho _2, \ \ \ \ \forall \ x\in {\mathbb {R}}^3. \end{aligned}$$

  (3.28)

  Since ${\mathcal {J}}(u)=0$ for $u\in {\mathcal {M}}$, by (3.2), (3.16), (3.28) and the Sobolev embedding theorem, we have

  $$\begin{aligned} \frac{\min \{3,\varrho _2\}}{2}\Vert u\Vert ^2\le & {} \frac{3}{2}\Vert \nabla u\Vert _2^2+\frac{1}{2}\int _{{{\mathbb {R}}}^3}[V(x)-\nabla V(x)\cdot x]u^2\mathrm {d}x\nonumber \\&\ \ +\frac{3q^2}{4}\int _{{{\mathbb {R}}}^3}\int _{{{\mathbb {R}}}^3} \frac{1-e^{-\frac{|x-y|}{a}} -\frac{|x-y|}{3a}e^{-\frac{|x-y|}{a}}}{|x-y|}u^2(x)u^2(y)\mathrm {d}x\mathrm {d}y\nonumber \\\le & {} \int _{{{\mathbb {R}}}^3}[2f(u)u-3F(u)]\mathrm {d}x\nonumber \\\le & {} \frac{\min \{3,\varrho _2\}}{4}\Vert u\Vert ^{2}+C_1\Vert u\Vert ^{p}, \ \ \ \ \forall \ u\in {\mathcal {M}}, \end{aligned}$$

  which implies

  $$\begin{aligned} \Vert u\Vert \ge \rho :=\left( \frac{\min \{3,\varrho _2\}}{4C_1}\right) ^{1/(p-2)}, \ \ \ \ \forall \ u\in {\mathcal {M}}. \end{aligned}$$

  (3.29)

• (ii). Let $\{u_n\}\subset {\mathcal {M}}$ be such that ${\mathcal {I}}(u_n)\rightarrow m$. There are two possible cases: 1) $\inf _{n\in {\mathbb {N}}}\Vert u_n\Vert _2>0$ and 2) $\inf _{n\in {\mathbb {N}}}\Vert u_n\Vert _2=0$.

  Case 1) $\inf _{n\in {\mathbb {N}}}\Vert u_n\Vert _2:=\rho _1>0$. In this case, (3.20) and (3.27) yield

  $$\begin{aligned} m+o(1)={\mathcal {I}}(u_n)={\mathcal {I}}(u_n)-\frac{1}{3}{\mathcal {J}}(u_n)\ge \frac{\varrho _1}{6}\rho _1^2>0. \end{aligned}$$

  (3.30)

  Case 2) $\inf _{n\in {\mathbb {N}}}\Vert u_n\Vert _2=0$. By (3.29), passing to a subsequence, we have

  $$\begin{aligned} \Vert u_n\Vert _2\rightarrow 0, \ \ \ \ \Vert \nabla u_n\Vert _2\ge \frac{1}{2}\rho . \end{aligned}$$

  (3.31)

  Let $t_n=\Vert \nabla u_n\Vert _2^{-2/3}$. Then (3.31) implies that $\{t_n\}$ is bounded. Using (F1), (F2) and the Sobolev inequality, there exists $C_2>0$ such that

  $$\begin{aligned} \left| \int _{{\mathbb {R}}^3}F(u)\mathrm {d}x\right| \le C_2\Vert u\Vert _2^2+\frac{S^3}{4}\Vert u\Vert _6^{6} \le C_2\Vert u\Vert _2^2+\frac{1}{4}\Vert \nabla u\Vert _2^{6}, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3). \end{aligned}$$

  (3.32)

  Since ${\mathcal {J}}(u_n)=0$ for all $n\in {\mathbb {N}}$, then (3.18), (3.19), (3.31) and (3.32) give

  $$\begin{aligned} m+o(1)= & {} {\mathcal {I}}(u_n)\ge {\mathcal {I}}(t_n^2(u_n)_{t_n})\\= & {} \frac{t_n^3}{2}\Vert \nabla u_n\Vert _2^2+\frac{t_n}{2}\int _{{{\mathbb {R}}}^3}V(t_n^{-1}x)u_n^2\mathrm {d}x\\&+\frac{q^2t_n^3}{4}\int _{{{\mathbb {R}}}^3}\int _{{{\mathbb {R}}}^3}\frac{1-e^{-\frac{|x-y|}{at_n}}}{|x-y|} u_n^2(x)u_n^2(y)\mathrm {d}x\mathrm {d}y\\&\ \ -\frac{1}{t_n^3}\int _{{{\mathbb {R}}}^3}F(t_n^2u_n)\mathrm {d}x\\\ge & {} \frac{t_n^3}{2}\Vert \nabla u_n\Vert _2^2-C_2t_n\Vert u_n\Vert _2^2-\frac{t_n^9}{4}\Vert \nabla u_n\Vert _2^6 \nonumber \\= & {} \frac{1}{4}t_n^3\Vert \nabla u_n\Vert _2^2\left[ 2-\left( t_n^3\Vert \nabla u_n\Vert _2^2\right) ^{2}\right] +o(1)=\frac{1}{4}+o(1). \end{aligned}$$

  Cases 1) and 2) show that $m=\inf _{{\mathcal {M}}}{\mathcal {I}}>0$. This completes the proof. $\square$

Lemma 3.11

Assume that (V1), (V3) and (F1)–(F4) hold. Then $m^{\infty }:=\inf _{{\mathcal {M}}^{\infty }} {\mathcal {I}}^{\infty }\ge m$.

Proof

Arguing by contradiction, suppose that $m>m^{\infty }$. Let $\varepsilon :=m-m^{\infty }$. Then there exists $u_{\varepsilon }^{\infty }$ such that

$$\begin{aligned} u_{\varepsilon }^{\infty }\in {\mathcal {M}}^{\infty }\ \ \mathrm{and}\ \ m^{\infty }+\frac{\varepsilon }{2}> {\mathcal {I}}^{\infty }(u_{\varepsilon }^{\infty }). \end{aligned}$$

(3.33)

In view of Lemma 3.8, there exists $t_\varepsilon >0$ such that $t_\varepsilon ^2(u_{\varepsilon }^{\infty })_{t_\varepsilon }\in {\mathcal {M}}$. Thus, it follows from (V1), (2.5), (3.3), (3.21), (3.24) and (3.33) that

$$\begin{aligned} m^{\infty }+\frac{\varepsilon }{2}> {\mathcal {I}}^{\infty }(u_{\varepsilon }^{\infty }) \ge {\mathcal {I}}^{\infty }(t_\varepsilon ^2(u_{\varepsilon }^{\infty })_{t_\varepsilon }) \ge {\mathcal {I}}(t_\varepsilon ^2(u_{\varepsilon }^{\infty })_{t_\varepsilon }) \ge m. \end{aligned}$$

This contradiction shows that $m^{\infty }\ge m$. $\square$

By combining [18, Lemma B.2] and [23, 26], we obtain the following Brezis-Lieb type lemma, see [8].

Lemma 3.12

Assume that (V1), (V2), (F1) and (F2) hold. If $u_n\rightharpoonup {\bar{u}}$ in $H^1({\mathbb {R}}^3)$, then up to a subsequence

$$\begin{aligned} {\mathcal {I}}(u_n)= & {} {\mathcal {I}}({\bar{u}})+{\mathcal {I}}(u_n-{\bar{u}})+o(1),\ \ \ \nonumber \\ {\mathcal {J}}(u_n)= & {} {\mathcal {J}}({\bar{u}})+{\mathcal {J}}(u_n-{\bar{u}})+o(1) \end{aligned}$$

(3.34)

$$\begin{aligned} {\mathcal {I}}'(u_n)= & {} {\mathcal {I}}'({\bar{u}})+{\mathcal {I}}'(u_n-{\bar{u}})+o(1), \end{aligned}$$

(3.35)

$$\begin{aligned} {\mathcal {I}}'(u_n)[u_n]= & {} {\mathcal {I}}'({\bar{u}})[{\bar{u}}]+{\mathcal {I}}'(u_n-{\bar{u}})[u_n-{\bar{u}}]+o(1). \end{aligned}$$

(3.36)

Lemma 3.13

Assume that (V1), (V3) and (F1)–(F4) hold. Then m is achieved.

Proof

Let $\{u_n\}\subset {\mathcal {M}}$ be such that ${\mathcal {I}}(u_n)\rightarrow m$. Since ${\mathcal {J}}(u_n)=0$, then (3.20) and (3.27) yield

$$\begin{aligned} m+o(1)= & {} {\mathcal {I}}(u_n)={\mathcal {I}}(u_n)-\frac{1}{3}{\mathcal {J}}(u_n)\nonumber \\\ge & {} \frac{1}{6}\int _{{{\mathbb {R}}}^3}[2V(x)+\nabla V(x)\cdot x]u_n^2\mathrm {d}x \ge \frac{\varrho _1}{6}\Vert u_n\Vert _2^2. \end{aligned}$$

(3.37)

This shows that $\{\Vert u_n\Vert _2\}$ is bounded. Now we assert that $\{\Vert \nabla u_n\Vert _2\}$ is also bounded. Arguing by contradiction, suppose that $\Vert \nabla u_n\Vert _2 \rightarrow \infty$. From (F1), (F2) and the Sobolev inequality, there exists $C_2>0$ such that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^3}F(u)\mathrm {d}x\right|\le & {} C_2\Vert u\Vert _2^2+\frac{1}{2(8m)^{2}}S^{3}\Vert u\Vert _6^{6} \le C_2\Vert u\Vert _2^2\nonumber \\&+\frac{1}{2(8m)^{2}}\Vert \nabla u\Vert _2^{6}, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3). \end{aligned}$$

(3.38)

Let $t_n=\left( 8m/\Vert \nabla u_n\Vert _2^2\right) ^{1/3}$. Since ${\mathcal {J}}(u_n)=0$, it follows from (3.18), (3.19) and (3.38) that

$$\begin{aligned} m+o(1)= & {} {\mathcal {I}}(u_n) \ge {\mathcal {I}}(t_n^2(u_n)_{t_n})\nonumber \\= & {} \frac{t_n^3}{2}\Vert \nabla u_n\Vert _2^2+\frac{t_n}{2}\int _{{{\mathbb {R}}}^3}V(t_n^{-1}x)u_n^2\mathrm {d}x\nonumber \\&+\frac{q^2t_n^3}{4}\int _{{{\mathbb {R}}}^3}\int _{{{\mathbb {R}}}^3}\frac{1-e^{-\frac{|x-y|}{at_n}}}{|x-y|} u_n^2(x)u_n^2(y)\mathrm {d}x\mathrm {d}y\nonumber \\&\ \ -\frac{1}{t_n^3}\int _{{{\mathbb {R}}}^3}F(t_n^2u_n)\mathrm {d}x\nonumber \\\ge & {} \frac{t_n^3}{2}\Vert \nabla u_n\Vert _2^2-C_2t_n\Vert u_n\Vert _2^2 -\frac{1}{4(8m)^{2}}\left( t_n^3\Vert \nabla u_n\Vert _2^2\right) ^{3}\nonumber \\= & {} \frac{1}{2}t_n^3\Vert \nabla u_n\Vert _2^2\left[ 1-\frac{1}{2} \left( \frac{t_n^3\Vert \nabla u_n\Vert _2^2}{8m}\right) ^{2}\right] +o(1)\nonumber \\= & {} 2m+o(1). \end{aligned}$$

(3.39)

This contradiction shows that $\{\Vert \nabla u_n\Vert _2\}$ is also bounded and the assertion holds. Hence $\{u_n\}$ is bounded in $H^1({\mathbb {R}}^3)$. Thus, there exists ${\bar{u}}\in H^1({\mathbb {R}}^3)$ such that, passing to a subsequence, $u_n\rightharpoonup {\bar{u}}$ in $H^1({\mathbb {R}}^3)$, $u_n\rightarrow {\bar{u}}$ in $L_{\mathrm {loc}}^s({\mathbb {R}}^3)$ for all $1\le s<6$ and $u_n\rightarrow {\bar{u}}$ a.e. in ${\mathbb {R}}^3$. There are two possible cases: i) ${\bar{u}}=0$ and ii) ${\bar{u}}\ne 0$.

Case i) ${\bar{u}}=0$, i.e. $u_n\rightharpoonup 0$ in $H^1({\mathbb {R}}^3)$, $u_n\rightarrow 0$ in $L_{\mathrm {loc}}^s({\mathbb {R}}^3)$ for all $1\le s<6$ and $u_n\rightarrow 0$ a.e. in ${\mathbb {R}}^3$. Using (V1) and (3.8), it is easily checked that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}[V_{\infty }-V(x)]u_n^2\mathrm {d}x= \lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}\nabla V(x)\cdot xu_n^2\mathrm {d}x=0. \end{aligned}$$

(3.40)

From (2.5), (3.16), (3.21), (3.22) and (3.40), we derive

$$\begin{aligned} {\mathcal {I}}^{\infty }(u_n)\rightarrow m \ \ \hbox {and}\ \ {\mathcal {J}}^{\infty }(u_n)\rightarrow 0. \end{aligned}$$

(3.41)

From [26, Lemma 1.21], we deduce that there exist $\delta >0$ and a sequence $\{y_n\}\subset {\mathbb {R}}^3$ such that $\int _{B_1(y_n)}|u_n|^2\mathrm {d}x> \delta$. Let ${\hat{u}}_n(x)=u_n(x+y_n)$. Then we have $\Vert {\hat{u}}_n\Vert =\Vert u_n\Vert$ and

$$\begin{aligned} {\mathcal {J}}^{\infty }({\hat{u}}_n)= o(1), \ \ \ \ {\mathcal {I}}^{\infty }({\hat{u}}_n)\rightarrow m, \ \ \ \ \int _{B_1(0)}|{\hat{u}}_n|^2\mathrm {d}x> \delta . \end{aligned}$$

(3.42)

Therefore, there exists ${\hat{u}}\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that, passing to a subsequence,

$$\begin{aligned} \left\{ \begin{array}{ll} {\hat{u}}_n\rightharpoonup {\hat{u}}, &{} \text{ in } \ H^1({\mathbb {R}}^3); \\ {\hat{u}}_n \rightarrow {\hat{u}}, &{} \text{ in } \ L_{\mathrm {loc}}^s({\mathbb {R}}^3), \ \forall \ s\in [1, 6);\\ {\hat{u}}_n\rightarrow {\hat{u}}, &{} \text{ a.e. } \text{ in } \ {\mathbb {R}}^3. \end{array} \right. \end{aligned}$$

(3.43)

Let $w_n={\hat{u}}_n-{\hat{u}}$. Then (3.43) and Lemma 3.12 yield

$$\begin{aligned} {\mathcal {I}}^{\infty }({\hat{u}}_n)={\mathcal {I}}^{\infty }({\hat{u}})+{\mathcal {I}}^{\infty }(w_n)+o(1),\ \ \ \ {\mathcal {J}}^{\infty }({\hat{u}}_n)={\mathcal {J}}^{\infty }({\hat{u}})+{\mathcal {J}}^{\infty }(w_n)+o(1).\nonumber \\ \end{aligned}$$

(3.44)

We define the functional $\Psi ^{\infty }: H^1({\mathbb {R}}^3)\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \begin{aligned} \Psi ^{\infty }(u)&= {\mathcal {I}}^{\infty }(u)-\frac{1}{3}{\mathcal {J}}^{\infty }(u)\\&= \frac{V_{\infty }}{3}\Vert u\Vert _2^2+\frac{2}{3}\int _{{\mathbb {R}}^3}[f(u)u-3F(u)]\mathrm {d}x\\&\quad +\frac{q^2}{12a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x)u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned} \end{aligned}$$

(3.45)

By (3.21), (3.22), (3.42), (3.44) and (3.45), we have

$$\begin{aligned} \Psi ^{\infty }(w_n)=m-\Psi ^{\infty }({\hat{u}})+o(1), \ \ \hbox {and} \ \ {\mathcal {J}}^{\infty }(w_n) = -{\mathcal {J}}^{\infty }({\hat{u}})+o(1). \end{aligned}$$

(3.46)

If there exists a subsequence $\{w_{n_i}\}$ of $\{w_n\}$ such that $w_{n_i}=0$, then

$$\begin{aligned} {\mathcal {I}}^{\infty }({\hat{u}})=m \ \ \hbox {and}\ \ {\mathcal {J}}^{\infty }({\hat{u}})=0. \end{aligned}$$

(3.47)

Thus, we assume that $w_n\ne 0$ for all $n\in {\mathbb {N}}$. We claim that ${\mathcal {J}}^{\infty }({\hat{u}})\le 0$. Otherwise, if ${\mathcal {J}}^{\infty }({\hat{u}})>0$, then (3.46) implies ${\mathcal {J}}^{\infty }(w_n) < 0$ for large n. In view of Lemma 3.8, there exists $t_n>0$ such that $t_n^2(w_n)_{t_n}\in {\mathcal {M}}^{\infty }$ for large n. From (3.21), (3.22), (3.23), (3.46) and Lemma 3.11, we obtain

$$\begin{aligned} m-\Psi ^{\infty }({\hat{u}})+o(1)= & {} \Psi ^{\infty }(w_n)={\mathcal {I}}^{\infty }(w_n)-\frac{1}{3}{\mathcal {J}}^{\infty }(w_n)\nonumber \\\ge & {} {\mathcal {I}}^{\infty }\left( t_n^2(w_n)_{t_n}\right) -\frac{t_n^3}{3}{\mathcal {J}}^{\infty }(w_n) +\frac{(1-t_n)^2(2+t_n)V_{\infty }}{6}\Vert w_n\Vert _2^2\nonumber \\\ge & {} m^{\infty }-\frac{t_n^3}{3}{\mathcal {J}}^{\infty }(w_n)+\frac{(1-t_n)^2(2+t_n)V_{\infty }}{6}\Vert w_n\Vert _2^2\nonumber \\> & {} m, \end{aligned}$$

which contradicts the fact that $\Psi ^{\infty }({\hat{u}})>0$. Hence, ${\mathcal {J}}^{\infty }({\hat{u}})\le 0$ and the claim holds. In view of Lemma 3.8, there exists $t_{\infty }>0$ such that $t_{\infty }^2{\hat{u}}_{t_{\infty }}\in {\mathcal {M}}^{\infty }$. Now (3.23), (3.41), (3.42), (3.45), Fatou’s lemma and Lemma 3.11 yield

$$\begin{aligned} m= & {} \lim _{n\rightarrow \infty } \left[ {\mathcal {I}}^{\infty }({\hat{u}}_n) -\frac{1}{3}{\mathcal {J}}^{\infty }({\hat{u}}_n)\right] \nonumber \\= & {} \lim _{n\rightarrow \infty } \Psi ^{\infty }({\hat{u}}_n)\ge \Psi ^{\infty }({\hat{u}}) ={\mathcal {I}}^{\infty }({\hat{u}})-\frac{1}{3}{\mathcal {J}}^{\infty }({\hat{u}})\nonumber \\\ge & {} {\mathcal {I}}^{\infty }\left( t_{\infty }^2{\hat{u}}_{t_{\infty }}\right) -\frac{t_{\infty }^3}{3}{\mathcal {J}}^{\infty }({\hat{u}}) +\frac{(1-t_{\infty })^2(2+t_{\infty })V_{\infty }}{6}\Vert {\hat{u}}\Vert _2^2\nonumber \\\ge & {} m^{\infty }-\frac{t_{\infty }^3}{3}{\mathcal {J}}^{\infty }({\hat{u}}) +\frac{(1-t_{\infty })^2(2+t_{\infty })V_{\infty }}{6}\Vert {\hat{u}}\Vert _2^2\ge m, \end{aligned}$$

which implies again the validity of (3.47) also in this case. In view of Lemma 3.8, there exists ${\hat{t}}>0$ such that ${\hat{t}}^2{\hat{u}}_{{\hat{t}}}\in {\mathcal {M}}$. Moreover, it follows from (V1), (2.5), (3.21), (3.47) and Corollary 3.5 that

$$\begin{aligned} m\le {\mathcal {I}}({\hat{t}}^2{\hat{u}}_{{\hat{t}}})\le {\mathcal {I}}^{\infty }({\hat{t}}^2{\hat{u}}_{{\hat{t}}}) \le {\mathcal {I}}^{\infty }({\hat{u}})=m. \end{aligned}$$

This shows that m is achieved at ${\hat{t}}^2{\hat{u}}_{{\hat{t}}}\in {\mathcal {M}}$.

Case ii) ${\bar{u}}\ne 0$. We define the functional $\Psi : H^1({\mathbb {R}}^3)\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \begin{aligned} \Psi (u)&= {\mathcal {I}}(u)-\frac{1}{3}{\mathcal {J}}(u)\\&= \frac{1}{6}\int _{{{\mathbb {R}}}^3}[2V(x)+\nabla V(x)\cdot x]u_{n}^2\mathrm {d}x +\frac{2}{3}\int _{{\mathbb {R}}^3}[f(u)u-3F(u)]\mathrm {d}x\\&\quad \ \ +\frac{q^2}{12a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x)u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned} \end{aligned}$$

(3.48)

In this case, similarly to the proof of (3.47), by using ${\mathcal {I}}, {\mathcal {J}}$ and $\Psi$ instead of ${\mathcal {I}}^{\infty }, {\mathcal {J}}^{\infty }$ and $\Psi ^{\infty }$, we deduce that ${\mathcal {I}}({\bar{u}})=m$ and ${\mathcal {J}}({\bar{u}})=0$. $\square$

Lemma 3.14

Assume that (V1), (V3) and (F1)–(F4) hold. If ${\bar{u}}\in {\mathcal {M}}$ and ${\mathcal {I}}({\bar{u}})=m$, then ${\bar{u}}$ is a critical point of ${\mathcal {I}}$.

Proof

Assume that ${\mathcal {I}}'({\bar{u}})\ne 0$. Then there exist $\delta >0$ and $\rho >0$ such that

$$\begin{aligned} \Vert u-{\bar{u}}\Vert \le 3\delta \Rightarrow \Vert {\mathcal {I}}'(u)\Vert \ge \rho . \end{aligned}$$

It is easy to check that

$$\begin{aligned} \lim _{t\rightarrow 1}\left\| t^2{\bar{u}}_t-{\bar{u}}\right\| =0. \end{aligned}$$

Then there exists $\delta _1>0$ such that

$$\begin{aligned} |t-1|<\delta _1\Rightarrow \Vert t^2{\bar{u}}_t-{\bar{u}}\Vert < \delta . \end{aligned}$$

(3.49)

Using (V1), (V3) and (F1)–(F3), it is easy to prove that there exist $T_1\in (0,1)$ and $T_2\in (1,\infty )$ such that

$$\begin{aligned} {\mathcal {J}}\left( T_1^{2}{\bar{u}}_{T_1}\right) >0,\ \ \ \ \ \ {\mathcal {J}}\left( T_2^{2}{\bar{u}}_{T_2}\right) <0. \end{aligned}$$

(3.50)

In view of Lemma 3.4, we have

$$\begin{aligned} {\mathcal {I}}\left( t^2{\bar{u}}_t\right) \le {\mathcal {I}}({\bar{u}})-\frac{1}{6}\int _{{{\mathbb {R}}}^3}\beta (x, t){\bar{u}}^2\mathrm {d}x, \ \ \ \ \forall \ t>0. \end{aligned}$$

(3.51)

The rest of the proof is similar to that of [11, Lemma 2.14]. For the sake of completeness, we give the details. Let

$$\begin{aligned} \beta _0:=\min \left\{ \int _{{\mathbb {R}}^3}\beta (T_1,x){\bar{u}}^2\mathrm {d}x, \int _{{\mathbb {R}}^3}\beta (T_2,x){\bar{u}}^2\mathrm {d}x\right\} , \end{aligned}$$

and $\varepsilon :=\min \{\beta _0/24, 1, \rho \delta /8\}$. From [26, Lemma 2.3], there exists a deformation $\eta \in {\mathcal {C}}([0, 1]\times H^1({\mathbb {R}}^3), H^1({\mathbb {R}}^3))$ such that

• (i) $\eta (1, u)=u$ if ${\mathcal {I}}(u)<m-2\varepsilon$ or ${\mathcal {I}}(u)>m+2\varepsilon$;

• (ii) $\eta \left( 1, {\mathcal {I}}^{m+\varepsilon }\cap B({\bar{u}}, \delta )\right) \subset {\mathcal {I}}^{m-\varepsilon }$;

• (iii) ${\mathcal {I}}(\eta (1, u))\le {\mathcal {I}}(u), \ \forall \ u\in H^1({\mathbb {R}}^3)$;

• (iv) $\eta (1, u)$ is a homeomorphism of $H^1({\mathbb {R}}^3)$.

Note that Corollary 3.5 implies that ${\mathcal {I}}\left( t^2{\bar{u}}_t\right) \le {\mathcal {I}}({\bar{u}})=m$ for all $t> 0$. Then (3.49) and ii) give

$$\begin{aligned} {\mathcal {I}}\left( \eta \left( 1, t^2{\bar{u}}_t\right) \right) \le m-\varepsilon , \ \ \ \ \forall \ t> 0, \ \ |t-1|< \delta _1. \end{aligned}$$

(3.52)

On the other hand, (3.51) and iii) yield

$$\begin{aligned} {\mathcal {I}}\left( \eta \left( 1, t^2{\bar{u}}_t\right) \right)\le & {} {\mathcal {I}}\left( t^2{\bar{u}}_t\right) \le m-\frac{1}{6}\int _{{\mathbb {R}}^3}\beta (t,x){\bar{u}}^2\mathrm {d}x \nonumber \\\le & {} m-\frac{\delta _2}{6}, \ \ \ \ \forall \ t> 0, \ \ |t-1|\ge \delta _1, \end{aligned}$$

(3.53)

where

$$\begin{aligned} \delta _2:=\min \left\{ \int _{{\mathbb {R}}^3}\beta (1-\delta _1,x){\bar{u}}^2\mathrm {d}x, \int _{{\mathbb {R}}^3}\beta (1+\delta _1,x){\bar{u}}^2\mathrm {d}x\right\} >0. \end{aligned}$$

Combining (3.52) with (3.53), we have

$$\begin{aligned} \max _{t\in [T_1, T_2]}{\mathcal {I}}\left( \eta \left( 1, t^2{\bar{u}}_t\right) \right) <m. \end{aligned}$$

(3.54)

Define the function $\Psi _0(t):={\mathcal {J}}\left( \eta \left( 1, t^2{\bar{u}}_t\right) \right)$ for all $t> 0$. It follows from (3.51) and i) that $\eta (1, t^2{\bar{u}}_t)=t^2{\bar{u}}_t$ for $t=T_1$ and $t=T_2$, which, together with (3.50), implies

$$\begin{aligned} \Psi _0(T_1)={\mathcal {J}}\left( T_1^2{\bar{u}}_{T_1}\right) >0, \ \ \ \ \Psi _0(T_2)={\mathcal {J}}\left( T_2^2{\bar{u}}_{T_2}\right) <0. \end{aligned}$$

Since $\Psi _0(t)$ is continuous on $(0, \infty )$, then we have that $\eta \left( 1, t^2{\bar{u}}_t\right) \cap {\mathcal {M}}\ne \emptyset$ for some $t_0\in [T_1, T_2]$, contradicting the definition of m. $\square$

Proof of Theorem 1.4

In view of Lemmas 3.13 and 3.14, there exists ${\bar{u}}\in {\mathcal {M}}$ such that

$$\begin{aligned} {\mathcal {I}}({\bar{u}})=m=\inf _{u\in H^1({\mathbb {R}}^3){\setminus }\{0\}}\max _{t> 0}{\mathcal {I}}(t^2u_t), \ \ \ \ {\mathcal {I}}'({\bar{u}})=0. \end{aligned}$$

This shows that ${\bar{u}}$ is a ground state solution of (1.1) such that ${\mathcal {I}}({\bar{u}})=m=\inf _{{\mathcal {M}}}{\mathcal {I}}$. $\square$

Remark 3.15

As in the proof of Theorem  1.4, by replacing Lemma 3.4 with Lemma 3.7, we then obtain Corollary 1.6.

4 Proof of Theorem 1.1

In this section, we give the proof of Theorem 1.1. Without loss of generality, we consider that $V(x)\not \equiv V_{\infty }$.

Proposition 4.1

[21] Let X be a Banach space and let $J\subset {\mathbb {R}}^+$ be an interval, and

$$\begin{aligned} \Phi _{\lambda }(u)=A(u)-\lambda B(u), \ \ \ \ \forall \ \lambda \in J, \end{aligned}$$

be a family of ${\mathcal {C}}^1$-functionals on X such that

• (i) either $A(u)\rightarrow +\infty$ or $B(u)\rightarrow +\infty$, as $\Vert u\Vert \rightarrow \infty$;

• (ii) B maps every bounded set of X into a set of ${\mathbb {R}}$ bounded below;

• (iii) there are two points $v_1, v_2$ in X such that

  $$\begin{aligned} {\tilde{c}}_{\lambda }:=\inf _{\gamma \in {\tilde{\Gamma }}}\max _{t\in [0, 1]}\Phi _{\lambda }(\gamma (t))>\max \{\Phi _{\lambda }(v_1), \Phi _{\lambda }(v_2)\}, \end{aligned}$$

  (4.1)

where

$$\begin{aligned} {\tilde{\Gamma }}=\left\{ \gamma \in {\mathcal {C}}([0, 1], X): \gamma (0)=v_1, \gamma (1)=v_2\right\} . \end{aligned}$$

Then, for almost every $\lambda \in J$, there exists a sequence $\{u_n(\lambda )\}$ such that

• (i) $\{u_n(\lambda )\}$ is bounded in X;

• (ii) $\Phi _{\lambda }(u_n(\lambda ))\rightarrow {\tilde{c}}_{\lambda }$;

• (iii) $\Phi _{\lambda }'(u_n(\lambda ))\rightarrow 0$ in $X^*$, where $X^*$ is the dual of X.

For $\lambda \in [1/2, 1]$ we introduce two families of ${\mathcal {C}}^1$-functionals on $H^1({\mathbb {R}}^3)$ defined by

$$\begin{aligned} {\mathcal {I}}_{\lambda }(u)&:=\frac{1}{2}\int _{{{\mathbb {R}}}^3}\left( |\nabla u|^2+V(x)u^2\right) \mathrm {d}x+\frac{q^2}{4}\int _{{{\mathbb {R}}}^3}\phi _{u}(x)u^2\mathrm {d}x -\lambda \int _{{\mathbb {R}}^3}F(u)\mathrm {d}x, \end{aligned}$$

(4.2)

$$\begin{aligned} {\mathcal {I}}_{\lambda }^{\infty }(u)&:=\frac{1}{2}\int _{{{\mathbb {R}}}^3}\left( |\nabla u|^2+V_{\infty }u^2\right) \mathrm {d}x +\frac{q^2}{4}\int _{{{\mathbb {R}}}^3}\phi _{u}(x)u^2\mathrm {d}x -\lambda \int _{{\mathbb {R}}^3}F(u)\mathrm {d}x. \end{aligned}$$

(4.3)

In view of [18, A.14], we obtain the following useful identity.

Lemma 4.2

Assume that (V1), (V2) and (F1)–(F3) hold. Let u be a critical point of ${\mathcal {I}}_{\lambda }$ in $H^1({\mathbb {R}}^3)$, then the following Poho$\breve{\mathrm{z}}$aev-type identity holds

$$\begin{aligned} {\mathcal {P}}_{\lambda }(u):= & {} \frac{1}{2}\Vert \nabla u\Vert _2^2+\frac{1}{2}\int _{{{\mathbb {R}}}^3}\left[ 3V(x)+\nabla V(x)\cdot x\right] u^2\mathrm {d}x -3\lambda \int _{{{\mathbb {R}}}^3}F(u)\mathrm {d}x\nonumber \\&+\frac{5q^2}{4}\int _{{\mathbb {R}}^3}\phi _{u}u^2\mathrm {d}x +\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y =0.\nonumber \\ \end{aligned}$$

(4.4)

Let us set ${\mathcal {J}}_{\lambda }(u):=2 {\mathcal {I}}_{\lambda }'(u)[u]-{\mathcal {P}}_{\lambda }(u)$ for all $\lambda \in [1/2, 1]$. Then

$$\begin{aligned} {\mathcal {J}}_{\lambda }(u)= & {} \frac{3}{2}\Vert \nabla u\Vert _2^2+\frac{1}{2}\int _{{{\mathbb {R}}}^3} \left[ V(x)-\nabla V(x)\cdot x\right] u^2\mathrm {d}x -\lambda \int _{{{\mathbb {R}}}^3}[2f(u)u-3F(u)]\mathrm {d}x\nonumber \\&+\frac{3q^2}{4}\int _{{\mathbb {R}}^3}\phi _{u}u^2\mathrm {d}x -\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned}$$

(4.5)

Similarly, for all $\lambda \in [1/2,1]$, if u is a critical point of ${\mathcal {I}}^{\infty }_{\lambda }$, then u satisfies the following Poho$\breve{\mathrm{z}}$aev-type identity:

$$\begin{aligned} {\mathcal {P}}_{\lambda }^{\infty }(u)&:= \frac{1}{2}\Vert \nabla u\Vert _2^2+\frac{3V_{\infty }}{2}\int _{{{\mathbb {R}}}^3}\Vert u\Vert _2^2 -3\lambda \int _{{{\mathbb {R}}}^3}F(u)\mathrm {d}x\nonumber \\&\quad \ \ \ \ +\frac{5q^2}{4}\int _{{\mathbb {R}}^3}\phi _{u}u^2\mathrm {d}x +\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y. =0, \end{aligned}$$

(4.6)

We also let

$$\begin{aligned} {\mathcal {J}}_{\lambda }^{\infty }(u)&= \frac{3}{2}\Vert \nabla u\Vert _2^2+\frac{V_{\infty }}{2}\Vert u\Vert _2^2 -\lambda \int _{{{\mathbb {R}}}^3}[2f(u)u-3F(u)]\mathrm {d}x\nonumber \\&\quad +\frac{3q^2}{4a}\int _{{\mathbb {R}}^3}\phi _{u}u^2\mathrm {d}x -\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x) u^2(y)\mathrm {d}x\mathrm {d}y. \end{aligned}$$

(4.7)

Define

$$\begin{aligned} {\mathcal {M}}_{\lambda }^{\infty }:=\{u\in H^1({\mathbb {R}}^3){\setminus }\{0\}: {\mathcal {J}}_{\lambda }^{\infty }(u)=0\}, \ \ \ \ m_{\lambda }^{\infty }:=\inf _{{\mathcal {M}}_{\lambda }^{\infty }}{\mathcal {I}}_{\lambda }^{\infty }. \end{aligned}$$

(4.8)

By Lemma 3.7, we have the following lemma.

Lemma 4.3

Assume that (F1), (F3) and (F4) hold. Then

$$\begin{aligned} {\mathcal {I}}_{\lambda }^{\infty }(u)\ge & {} {\mathcal {I}}_{\lambda }^{\infty }(t^2u_t)+\frac{1-t^3}{3}{\mathcal {J}}_{\lambda }^{\infty }(u)\nonumber \\&+\frac{(1-t)^2(2+t)}{6}V_{\infty }\Vert u\Vert _2^2, \ \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3), \ t > 0. \end{aligned}$$

(4.9)

In view of Corollary 1.6, ${\mathcal {I}}_1^{\infty }={\mathcal {I}}^{\infty }$ has a minimizer $u_1^{\infty }\ne 0$ on ${\mathcal {M}}_1^{\infty }={\mathcal {M}}^{\infty }$, i.e.

$$\begin{aligned} u_1^{\infty }\in {\mathcal {M}}_1^{\infty }, \ \ \ \ ({\mathcal {I}}_1^{\infty })'(u_1^{\infty })=0 \ \ \ \ \text{ and } \ \ \ \ m_1^{\infty }={\mathcal {I}}_1^{\infty }(u_1^{\infty }). \end{aligned}$$

(4.10)

Noting that (1.5) is autonomous, $V\in {\mathcal {C}}({\mathbb {R}}^3, {\mathbb {R}})$ and $V(x)\le V_{\infty }$ but $V(x)\not \equiv V_{\infty }$, we can find ${\bar{x}}\in {\mathbb {R}}^3$ and ${\bar{r}}>0$ such that

$$\begin{aligned} V_{\infty }-V(x)>0, \ \ |u_1^{\infty }(x)|>0\ \ \ \ a.e. \ |x-{\bar{x}}|\le {\bar{r}} \end{aligned}$$

(4.11)

after suitable translations to $u_1^{\infty }$.

By (V1), we have $V_{\max }:=\max _{x\in {\mathbb {R}}^3}V(x)\in (0,\infty )$. Let

$$\begin{aligned} {\mathcal {I}}_{\lambda }^{*}(u)= & {} \frac{1}{2}\int _{{\mathbb {R}}^3}\left( |\nabla u|^2+V_{\max }u^2\right) \mathrm {d}x +\frac{q^2}{4}\int _{{{\mathbb {R}}}^3}\phi _{u}(x)u^2\mathrm {d}x\nonumber \\&-\lambda \int _{{\mathbb {R}}^3}F(u)\mathrm {d}x. \end{aligned}$$

(4.12)

Then it follows from (3.19) and (4.10) that there exists $T>0$ such that

$$\begin{aligned} {\mathcal {I}}^*_{1/2}\left( t^2 (u_1^{\infty })_{t}\right) <0, \ \ \ \ \forall \ t\ge T. \end{aligned}$$

(4.13)

Lemma 4.4

Assume that (V1), (V2) and (F1)–(F3) hold. Then

• (i) there exists $T>0$, independent of $\lambda$, such that ${\mathcal {I}}_{\lambda }(T^2(u_1^{\infty })_{T})<0$ for all $\lambda \in [1/2, 1]$;

• (ii) there exists a positive constant $\kappa _0$, independent of $\lambda$, such that for all $\lambda \in [1/2, 1]$,

  $$\begin{aligned} c_{\lambda }:=\inf _{\gamma \in \Gamma }\max _{t\in [0, 1]}{\mathcal {I}}_{\lambda }(\gamma (t))\ge \kappa _0 >\max \{{\mathcal {I}}_{\lambda }(0), {\mathcal {I}}_{\lambda }(T^2(u_1^{\infty })_{T})\}, \end{aligned}$$

  where

  $$\begin{aligned} \Gamma =\left\{ \gamma \in {\mathcal {C}}([0, 1], H^1({\mathbb {R}}^3)): \gamma (0)=0, \gamma (1)=T^2(u_1^{\infty })_{T}\right\} ; \end{aligned}$$

• (iii) $c_{\lambda }$ is bounded for $\lambda \in [1/2, 1]$ and $\limsup _{\lambda \rightarrow \lambda _0}c_{\lambda }\le c_{\lambda _0}$ for all $\lambda _0\in (1/2, 1]$;

• (iv) if f further satisfies (F4), then $m_{\lambda }^{\infty }$ are non-increasing on $\lambda \in [1/2, 1]$.

The proof of Lemma 4.4 is standard, so we omit it. Moreover, similarly to proof of [15, Lemma 4.5], we have the following lemma.

Lemma 4.5

Assume that (V1), (V2) and (F1)–(F4) hold. Then there exists ${\bar{\lambda }}\in [1/2, 1)$ such that $c_{\lambda }<m_{\lambda }^{\infty }$ for all $\lambda \in ({\bar{\lambda }}, 1]$.

Lemma 4.6

Assume that (V1), (V2) and (F1)–(F4) hold. Then for almost every $\lambda \in ({\bar{\lambda }},1]$, there exists $u_{\lambda }\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that

$$\begin{aligned} {\mathcal {I}}_{\lambda }'(u_{\lambda })=0, \ \ \ \ {\mathcal {I}}_{\lambda }(u_{\lambda }) = c_{\lambda }. \end{aligned}$$

(4.14)

Proof

By Proposition 4.1, for almost every $\lambda \in [1/2,1]$, there exists a bounded sequence $\{u_n(\lambda )\} \subset H^1({\mathbb {R}}^3)$, which we denote it by $\{u_n\}$ for simplicity, such that

$$\begin{aligned} {\mathcal {I}}_{\lambda }(u_n)\rightarrow c_{\lambda }>0, \ \ \ \ {\mathcal {I}}_{\lambda }'(u_n) \rightarrow 0. \end{aligned}$$

(4.15)

Similarly to the proof of [18, Lemma 4.5], using Lemma 3.12, we then deduce that there exist $u_{\lambda }\in H^1({\mathbb {R}}^3)$, an integer $l\in {\mathbb {N}}\cup \{0\}$, a sequence $\{y_n^k\}\subset {\mathbb {R}}^3$ and $w^k\in H^1({\mathbb {R}}^3)$ for $1\le k\le l$ such that $u_n\rightharpoonup u_{\lambda }$ in $H^1({\mathbb {R}}^3)$, ${\mathcal {I}}_{\lambda }'(u_{\lambda })=0$, $({\mathcal {I}}_{\lambda }^{\infty })'(w^k)=0$ and ${\mathcal {I}}_{\lambda }^{\infty }(w^k)\ge m_{\lambda }^{\infty }$ for $1\le k\le l$,

$$\begin{aligned} \left\| u_n-u_{\lambda }-\sum _{k=1}^lw^k(\cdot +y_n^k)\right\| \rightarrow 0\ \ \mathrm{and} \ \ {\mathcal {I}}_{\lambda }(u_n)\rightarrow {\mathcal {I}}_{\lambda }(u_{\lambda })+\sum _{i=1}^{l}{\mathcal {I}}_{\lambda }^{\infty }(w^i).\nonumber \\ \end{aligned}$$

(4.16)

Since ${\mathcal {I}}_{\lambda }'(u_{\lambda })=0$, then ${\mathcal {J}}_{\lambda }(u_{\lambda })=0$. It follows from (V2), (3.6), (4.2) and (4.5) that

$$\begin{aligned} {\mathcal {I}}_{\lambda }(u_{\lambda })= & {} {\mathcal {I}}_{\lambda }(u_{\lambda })-\frac{1}{3}{\mathcal {J}}_{\lambda }(u_{\lambda })\nonumber \\= & {} \frac{1}{6}\int _{{{\mathbb {R}}}^3}[2V(x)+\nabla V(x)\cdot x]u_{\lambda }^2\mathrm {d}x +\frac{q^2}{12a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u^2(x)u^2(y)\mathrm {d}x\mathrm {d}y\nonumber \\&\ \ +\frac{2\lambda }{3}\int _{{{\mathbb {R}}}^3}[f(u_{\lambda })u_{\lambda }-3F(u_{\lambda })]\mathrm {d}x \ge 0. \end{aligned}$$

(4.17)

If $l\ne 0$, then

$$\begin{aligned} c_{\lambda }=\lim _{n\rightarrow \infty }{\mathcal {I}}_{\lambda }(u_n)= {\mathcal {I}}_{\lambda }(u_{\lambda })+\sum _{i=1}^{l}{\mathcal {I}}_{\lambda }^{\infty }(w^i) \ge m_{\lambda }^{\infty }, \ \ \ \ \forall \ \lambda \in ({\bar{\lambda }}, 1], \end{aligned}$$

which contradicts Lemma 4.5. Thus, $l = 0$, and (4.16) implies that $u_n\rightarrow u_{\lambda }$ in $H^1({\mathbb {R}}^3)$ and ${\mathcal {I}}_{\lambda }(u_{\lambda })=c_{\lambda }$ for almost every $\lambda \in ({\bar{\lambda }},1]$. $\square$

Lemma 4.7

Assume that (V1), (V2) and (F1)–(F4) hold. Then there exists ${\bar{u}}\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that

$$\begin{aligned} {\mathcal {I}}'({\bar{u}})=0, \ \ \ \ 0<{\mathcal {I}}({\bar{u}})\le c_1. \end{aligned}$$

(4.18)

Proof

In view of Lemma 4.4 (ii) and (iii) and Lemma 4.6, there exist two sequences $\{\lambda _n\}\subset ({\bar{\lambda }}, 1]$ and $\{u_{\lambda _n}\}\subset H^1({\mathbb {R}}^3)$, which we denoted it by $\{u_n\}$ for brevity, such that

$$\begin{aligned} \lambda _n\rightarrow 1, \ \ \ \ c_{\lambda _n}\rightarrow c_*> 0, \ \ \ \ {\mathcal {I}}_{\lambda _n}'(u_n)=0, \ \ \ \ {\mathcal {I}}_{\lambda _n}(u_n) = c_{\lambda _n}. \end{aligned}$$

(4.19)

Now we assert that $\{u_n\}$ is bounded in $H^1({\mathbb {R}}^3)$.

By (4.2), (4.5), (4.19) and Lemma 4.4 (iii), one has

$$\begin{aligned} C_1\ge & {} c_{\lambda _n}= {\mathcal {I}}_{\lambda _n}(u_n)-\frac{1}{3}{\mathcal {J}}_{\lambda _n}(u_n)\nonumber \\= & {} \frac{1}{6}\int _{{{\mathbb {R}}}^3}[2V(x)+\nabla V(x)\cdot x]u_{n}^2\mathrm {d}x +\frac{q^2}{12a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u_n^2(x)u_n^2(y)\mathrm {d}x\mathrm {d}y\nonumber \\&\ \ +\frac{2\lambda _n}{3}\int _{{{\mathbb {R}}}^3}[f(u_{n})u_{n}-3F(u_{n})]\mathrm {d}x. \end{aligned}$$

(4.20)

By (V2), there exist constants $\varrho _0,R_0>0$ such that

$$\begin{aligned} 2V(x)+\nabla V(x)\cdot x\ge \varrho _0, \ \ \ \ \forall \ |x|\ge R_0. \end{aligned}$$

(4.21)

Then it follows from (3.6), (4.20) and (4.21) that

$$\begin{aligned} C_1 \ge \frac{\varrho _0}{6}\int _{|x|\ge R_0}u_{n}^2\mathrm {d}x +\frac{q^2e^{-\frac{2R_0}{a}}}{12a}\left( \int _{|x|<R_0}u_n^2\mathrm {d}x\right) ^2, \end{aligned}$$

(4.22)

which implies that $\{\Vert u_n\Vert _2\}$ is bounded.

Next, we prove that $\{\Vert \nabla u_n\Vert _2\}$ is also bounded. Arguing by contradiction, suppose that $\Vert \nabla u_n\Vert _2 \rightarrow \infty$. By (V1), (V2), (4.22) and Lemma 4.4 (iii), one has

$$\begin{aligned} c_{\lambda _n}+\int _{{{\mathbb {R}}}^3}[V_{\infty }-V(x)+|\nabla V(x)\cdot x|]u_n^2\mathrm {d}x\le M_0 \end{aligned}$$

(4.23)

for some constant $M_0>0$. Let $t_n=\min \left\{ 1, 2(M_0/\Vert \nabla u_n\Vert _2^2)^{1/3}\right\}$. Then $t_n \rightarrow 0$. Thus, it follows from (4.2), (4.3), (4.5), (4.7), (4.9) and (4.23) that

$$\begin{aligned} {\mathcal {I}}_{\lambda _n}^{\infty }(t_n^2(u_n)_{t_n})\le & {} {\mathcal {I}}_{\lambda _n}^{\infty }(u_n)-\frac{1-t_n^3}{3}{\mathcal {J}}_{\lambda _n}^{\infty }(u_n)\nonumber \\= & {} {\mathcal {I}}_{\lambda _n}(u_n)+\frac{1}{2}\int _{{{\mathbb {R}}}^3}[V_{\infty }-V(x)]u_n^2\mathrm {d}x\nonumber \\&\ \ -\frac{1-t_n^3}{3}\left[ {\mathcal {J}}_{\lambda _n}(u_n) +\frac{1}{2}\int _{{{\mathbb {R}}}^3}[V_{\infty }-V(x)+\nabla V(x)\cdot x ]u_n^2\mathrm {d}x\right] \nonumber \\\le & {} c_{\lambda _n}+\int _{{{\mathbb {R}}}^3}[V_{\infty }-V(x)+|\nabla V(x)\cdot x |]u_n^2\mathrm {d}x\le M_0. \end{aligned}$$

(4.24)

As in the proof of (3.39), we then deduce a contradiction by using (4.24). Hence, $\{u_n\}$ is bounded in $H^1({\mathbb {R}}^3)$, and the assertion holds.

Similarly to the proof of Lemma 4.6, there exists ${\bar{u}}\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that (4.18) holds. $\square$

Proof of Theorems 1.1

Define

$$\begin{aligned} {\mathcal {K}}:=\left\{ u\in H^1({\mathbb {R}}^3){\setminus }\{0\} : {\mathcal {I}}'(u)=0\right\} , \ \ \ \ {\hat{m}}:=\inf _{u\in {\mathcal {K}}}{\mathcal {I}}(u). \end{aligned}$$

Then Lemma 4.7 shows that ${\mathcal {K}}\ne \emptyset$ and ${\hat{m}}\le c_1$. For any $u\in {\mathcal {K}}$, (3.16), (4.5) and Lemma 4.2 imply ${\mathcal {J}}(u)={\mathcal {J}}_1(u)=2{\mathcal {I}}'(u)[u]-{\mathcal {P}}(u)=0$. By (2.5), (3.16) and (4.21), one has

$$\begin{aligned} {\mathcal {I}}(u)={\mathcal {I}}(u)-\frac{1}{3}{\mathcal {J}}(u)\ge \frac{\varrho _0}{6}\int _{|x|\ge R_0}u^2\mathrm {d}x +\frac{q^2e^{-\frac{2R_0}{a}}}{12a}\left( \int _{|x|<R_0}u^2\mathrm {d}x\right) ^2>0, \ \ \ \ \forall \ u\in {\mathcal {K}}, \end{aligned}$$

which implies ${\hat{m}}\ge 0$. Since ${\mathcal {I}}'(u)[u]=0$ for $u\in {\mathcal {K}}$, we then deduce from (F1), (F2) and the Sobolev embedding theorem that there exists $\alpha _0>0$ such that

$$\begin{aligned} \Vert u\Vert \ge \alpha _0, \ \ \ \ \forall \ u\in {\mathcal {K}}. \end{aligned}$$

(4.25)

Let $\{u_n\}\subset {\mathcal {K}}$ be such that ${\mathcal {I}}'(u_n)=0$ and ${\mathcal {I}}(u_n) \rightarrow {\hat{m}}$. In view of Lemma 4.5, we have ${\hat{m}}\le c_1<m_1^{\infty }$. Similarly to the proof of Lemma 4.6, we deduce that there exists ${\hat{u}}\in H^1({\mathbb {R}}^3)$ such that $u_n\rightarrow {\hat{u}}$ in $H^1({\mathbb {R}}^3)$, ${\mathcal {I}}'({\hat{u}})=0$ and ${\mathcal {I}}({\hat{u}}) = {\hat{m}}$. Moreover, (4.25) leads to ${\hat{u}}\ne 0$. Hence, ${\hat{u}}\in H^1({\mathbb {R}}^3)$ is a ground state solution of (1.1). $\square$

Proof of Theorems 1.3

As in the proof of Lemma 4.6, for almost every $\lambda \in [1/2,1]$, there exists a bounded sequence $\{u_n(\lambda )\} \subset H^1({\mathbb {R}}^3)$, which we denote it by $\{u_n\}$ for simplicity, and a positive constant $\kappa _0^{\infty }$, independent of $\lambda$, such that

$$\begin{aligned} {\mathcal {I}}_{\lambda }^{\infty }(u_n)\rightarrow c_{\lambda }^{\infty }\ge \kappa _0^{\infty }, \ \ \ \ ({\mathcal {I}}_{\lambda }^{\infty })'(u_n) \rightarrow 0. \end{aligned}$$

(4.26)

Using (F1), (F2), (4.26) and [26, Lemma 1.21], we can prove that there exists a sequence $y_n\in {\mathbb {R}}^3$ such that $\int _{B_1(y_n)}|u_n|^2\mathrm {d}x> 0$. Let ${\bar{u}}_n(x)=u_n(x+y_n)$. Then $\Vert {\bar{u}}_n\Vert =\Vert u_n\Vert$ and there exists ${\bar{u}}\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that ${\bar{u}}_n\rightharpoonup {\tilde{u}}$ in $H^1({\mathbb {R}}^3)$. Note that

$$\begin{aligned} {\mathcal {I}}_{\lambda }^{\infty }({\bar{u}}_n)\rightarrow c_{\lambda }^{\infty }\ge \kappa _0^{\infty }, \ \ \ \ ({\mathcal {I}}_{\lambda }^{\infty })'({\bar{u}}_n) \rightarrow 0. \end{aligned}$$

(4.27)

By a standard argument, for almost every $\lambda \in [1/2,1]$, there exists $u_{\lambda }\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that

$$\begin{aligned} ({\mathcal {I}}_{\lambda }^{\infty })'(u_{\lambda })=0, \ \ \ \ {\mathcal {I}}_{\lambda }^{\infty }(u_{\lambda }) = c_{\lambda }^{\infty }\ge \kappa _0^{\infty }. \end{aligned}$$

(4.28)

From (4.28), there exist two sequences $\{\lambda _n\}\subset [1/2, 1]$ and $\{u_{\lambda _n}\}\subset H^1({\mathbb {R}}^3)$, which we denote the latter by $\{u_n\}$, such that

$$\begin{aligned} \lambda _n\rightarrow 1, \ \ \ \ \kappa _0^{\infty }\le c_{\lambda _n}^{\infty }\rightarrow c^{\infty },\ \ \ \ ({\mathcal {I}}_{\lambda _n}^{\infty })'(u_n)=0, \ \ \ \ {\mathcal {I}}_{\lambda _n}^{\infty }(u_n) = c_{\lambda _n}^{\infty }. \end{aligned}$$

(4.29)

Similarly to (4.20), we have

$$\begin{aligned} C_2\ge & {} c_{\lambda _n}^{\infty }= {\mathcal {I}}_{\lambda _n}^{\infty }(u_n)-\frac{1}{3}{\mathcal {J}}_{\lambda _n}^{\infty }(u_n)\nonumber \\= & {} \frac{V_{\infty }}{3}\Vert u_{n}\Vert _2^2 +\frac{q^2}{12a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u_n^2(x)u_n^2(y)\mathrm {d}x\mathrm {d}y\nonumber \\&\ \ +\frac{2\lambda _n}{3}\int _{{{\mathbb {R}}}^3}[f(u_{n})u_{n}-3F(u_{n})]\mathrm {d}x, \end{aligned}$$

(4.30)

which implies

$$\begin{aligned} \Vert u_{n}\Vert _2^2\le C_3, \ \ \ \ \int _{{{\mathbb {R}}}^3}[f(u_{n})u_{n}-3F(u_{n})]\mathrm {d}x\le C_4, \end{aligned}$$

(4.31)

and

$$\begin{aligned} \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}u_n^2(x)u_n^2(y)\mathrm {d}x\mathrm {d}y\le C_5, \end{aligned}$$

(4.32)

Next, we claim that $\{\Vert \nabla u_n\Vert _2\}$ is also bounded. Arguing by contradiction, suppose that $\Vert \nabla u_n\Vert _2\rightarrow \infty$. Set $v_n=u_n/\Vert u_n\Vert$, then $\Vert v_n\Vert =1$, and (4.31) implies $\Vert v_n\Vert _2\rightarrow 0$. If $\delta _0:=\limsup _{n\rightarrow \infty }\sup _{y\in {\mathbb {R}}^3}\int _{B_1(y)}|v_n|^2\mathrm {d}x=0$, then by [26, Lemma 1.21], $v_n\rightarrow 0$ in $L^{s}({\mathbb {R}}^3)$ for $2<s<6$.

Since $\Vert v_n\Vert _2\rightarrow 0$, we have

$$\begin{aligned} \int _{0<|u_n|\le r_0}\frac{f(u_n)}{u_n}v_n^2\mathrm {d}x\le C_6\Vert v_n\Vert _2^2=o(1). \end{aligned}$$

(4.33)

Set $\kappa '=\kappa /(\kappa -1)$. Then (F5), (4.31) and the Hölder inequality yield

$$\begin{aligned} \int _{|u_n|>r_0}\frac{f(u_n)}{u_n}v_n^2\mathrm {d}x\le & {} \left[ \int _{|u_n|>r_0} \left| \frac{f(u_n)}{u_n}\right| ^{\kappa }\mathrm {d}x\right] ^{1/\kappa } \Vert v_n\Vert _{2\kappa '}^2\nonumber \\\le & {} C_7\left( \int _{|u_n|>r_0} [f(u_n)u_n-3F(u_n)]\mathrm {d}x\right) ^{1/\kappa } \Vert v_n\Vert _{2\kappa '}^2\nonumber \\\le & {} C_8\Vert v_n\Vert _{2\kappa '}^2 =o(1). \end{aligned}$$

(4.34)

Since $({\mathcal {I}}_{\lambda _n}^{\infty })'(u_n)[u_n]=0$ by (4.29), then (4.33) and (4.34) yield

$$\begin{aligned} 1\le & {} \frac{1}{\Vert u_n\Vert ^2}\left[ \int _{{{\mathbb {R}}}^3}\left( |\nabla u_n|^2+V_{\infty }u_n^2\right) \mathrm {d}x +q^2\int _{{{\mathbb {R}}}^3}\phi _{u_n}(x)u_n^2\mathrm {d}x\right] \nonumber \\= & {} \lambda _n\int _{{{\mathbb {R}}}^3}\frac{f(u_n)}{u_n}v_n^2\mathrm {d}x =o(1). \end{aligned}$$

This contradiction shows that $\delta _0=\limsup _{n\rightarrow \infty }\sup _{y\in {\mathbb {R}}^3} \int _{B_1(y)}|v_n|^2\mathrm {d}x>0$. Going if necessary to a subsequence, we may assume that there exists a sequence $\{y_n\}\subset {\mathbb {R}}^3$ such that $\int _{B_{1}(y_n)}|v_n|^2\mathrm {d}x> \frac{\delta _0}{2}$ for all $n\in {\mathbb {N}}$. Let $w_n(x)=v_n(x+y_n)$. Then $\Vert w_n\Vert =\Vert v_n\Vert =1$, and for all $n\in {\mathbb {N}}$

$$\begin{aligned} \int _{B_1(0)}|w_n|^2\mathrm {d}x> \frac{\delta _0}{2}. \end{aligned}$$

(4.35)

Then there exists $w\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that, passing to a subsequence, $w_n\rightharpoonup w$ in $H^1({\mathbb {R}}^3)$, $w_n\rightarrow w$ in $L^{s}_{\mathrm {loc}}({\mathbb {R}}^3)$ for all $1 \le s<6$, $w_n\rightarrow w$ a.e. in ${\mathbb {R}}^3$. Let us define ${\tilde{u}}_n(x)=u_n(x+y_n)$. Then ${\tilde{u}}_n/\Vert u_n\Vert =w_n\rightarrow w$ a.e. in ${\mathbb {R}}^3$ and $w\ne 0$. For $x\in \{y\in {\mathbb {R}}^3 : w(y)\ne 0\}$, we have $\lim _{n\rightarrow \infty }|{\tilde{u}}_n(x)|=\infty$. By (F1) and (F2), there exists $M_1>0$ such that

$$\begin{aligned} F(t)+M_1t^2\ge 0, \ \ \ \ \forall \ t\in {\mathbb {R}}. \end{aligned}$$

(4.36)

Note that (4.29) and (4.32) lead to

$$\begin{aligned} \lambda _n\rightarrow 1, \ \ \ \ \kappa _0^{\infty }\le c_{\lambda _n}^{\infty }\rightarrow c^{\infty },\ \ \ \ ({\mathcal {I}}_{\lambda _n}^{\infty })'({\tilde{u}}_n)=0, \ \ \ \ {\mathcal {I}}_{\lambda _n}^{\infty }({\tilde{u}}_n) = c_{\lambda _n}^{\infty } \end{aligned}$$

(4.37)

and

$$\begin{aligned} \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}}{\tilde{u}}_n^2(x){\tilde{u}}_n^2(y)\mathrm {d}x\mathrm {d}y\le C_5. \end{aligned}$$

(4.38)

From (F3), (4.3), (4.6), (4.37), (4.38), Lemma 4.2 and Fatou’s lemma, we derive

$$\begin{aligned} 0= & {} \lim _{n\rightarrow \infty }\frac{{\mathcal {I}}_{\lambda }^{\infty }({\tilde{u}}_n) -\frac{1}{5}{\mathcal {P}}_{\lambda }^{\infty }({\tilde{u}}_n)}{\Vert {\tilde{u}}_n(x)\Vert ^3}\nonumber \\= & {} \lim _{n\rightarrow \infty }\left\{ \frac{1}{5\Vert {\tilde{u}}_n(x)\Vert ^3}\left[ 2\Vert \nabla {\tilde{u}}_n\Vert _2^2 +V_{\infty }\Vert {\tilde{u}}_n\Vert _2^2-\frac{q^2}{4a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3} e^{-\frac{|x-y|}{a}}{\tilde{u}}^2(x) {\tilde{u}}^2(y)\mathrm {d}x\mathrm {d}y\right] \right. \nonumber \\&\ \ -\left. \frac{2\lambda _n}{5\Vert u_n\Vert ^3}\int _{{{\mathbb {R}}}^3}F({\tilde{u}}_n)\mathrm {d}x\right\} \nonumber \\\le & {} -\frac{1}{5}\liminf _{n\rightarrow \infty }\int _{{{\mathbb {R}}}^3}\frac{F({\tilde{u}}_n)+M_1{\tilde{u}}_n^2}{|{\tilde{u}}_n|^3}w_n^3\mathrm {d}x =- \infty . \end{aligned}$$

This contradiction shows that $\{u_n\}$ is bounded in $H^1({\mathbb {R}}^3)$ and the claim holds.

As in the proof of Lemma 4.6, there exists ${\bar{u}}\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that

$$\begin{aligned} ({\mathcal {I}}^{\infty })'({\bar{u}})=0, \ \ \ \ 0<{\mathcal {I}}^{\infty }({\bar{u}})\le c_1^{\infty }. \end{aligned}$$

Set

$$\begin{aligned} {\mathcal {K}}^{\infty }:=\left\{ u\in H^1({\mathbb {R}}^3){\setminus }\{0\} : ({\mathcal {I}}^{\infty })'(u)=0\right\} , \ \ \ \ {\hat{m}}^{\infty }:=\inf _{u\in {\mathcal {K}}^{\infty }}{\mathcal {I}}^{\infty }(u). \end{aligned}$$

The above argument shows that ${\mathcal {K}}^{\infty }\ne \emptyset$.

For any $u\in {\mathcal {K}}^{\infty }$, Lemma 4.2 implies ${\mathcal {J}}^{\infty }(u)=2({\mathcal {I}}^{\infty })'(u)[u]-{\mathcal {P}}^{\infty }(u)=0$. By (F5) and (3.45), we have

$$\begin{aligned} {\mathcal {I}}^{\infty }(u)={\mathcal {I}}^{\infty }(u)-\frac{1}{3}{\mathcal {J}}^{\infty }(u)\ge \frac{V_{\infty }}{3}\Vert u\Vert _2^2>0, \ \ \ \ \forall \ u\in {\mathcal {K}}^{\infty }, \end{aligned}$$

which implies ${\hat{m}}^{\infty }\ge 0$. Since $({\mathcal {I}}^{\infty })'(u)[u]=0$ for $u\in {\mathcal {K}}^{\infty }$, we easily deduce from (F1), (F2) and the Sobolev embedding theorem that there exists $\alpha _{\infty }>0$ such that

$$\begin{aligned} \Vert u\Vert \ge \alpha _{\infty }, \ \ \ \ \forall \ u\in {\mathcal {K}}^{\infty }. \end{aligned}$$

(4.39)

Let $\{u_n\}\subset {\mathcal {K}}^{\infty }$ be such that $({\mathcal {I}}^{\infty })'(u_n)=0$ and ${\mathcal {I}}^{\infty }(u_n) \rightarrow {\hat{m}}^{\infty }$. Since $({\mathcal {I}}^{\infty })'(u_n)[u_n]=0$, we can deduce from (4.39) and [26, Lemma 1.21] that $\{u_n\}$ is non-vanishing, and so up to a subsequence, there exists a sequence $\{y_n\}\subset {\mathbb {R}}^3$ such that $\int _{B_{1}(y_n)}|u_n|^2\mathrm {d}x>0$. Let ${\hat{u}}_n(x)=v_n(x+y_n)$. Then there exists ${\hat{u}}\in H^1({\mathbb {R}}^3){\setminus }\{0\}$ such that $u_n\rightharpoonup {\hat{u}}$ in $H^1({\mathbb {R}}^3)$, $({\mathcal {I}}^{\infty })'({\hat{u}})=0$ and ${\mathcal {I}}^{\infty }({\hat{u}})\ge {\hat{m}}^{\infty }$. Moreover, it follows from (F5), (3.21), (3.22) and Fatou’s lemma that

$$\begin{aligned} {\hat{m}}^{\infty }= & {} \lim _{n\rightarrow \infty }\left[ {\mathcal {I}}^{\infty }({\hat{u}}_n)-\frac{1}{3}{\mathcal {J}}^{\infty }({\hat{u}}_n)\right] \nonumber \\= & {} \lim _{n\rightarrow \infty }\left[ \frac{V_{\infty }}{3}\Vert {\hat{u}}_{n}\Vert _2^2 +\frac{q^2}{12a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}} {\hat{u}}_n^2(x){\hat{u}}_n^2(y)\mathrm {d}x\mathrm {d}y\right. \nonumber \\&\ \ +\left. \frac{2}{3}\int _{{{\mathbb {R}}}^3} [f({\hat{u}}_{n}){\hat{u}}_{n}-3F({\hat{u}}_{n})]\mathrm {d}x\right] \nonumber \\\ge & {} \frac{V_{\infty }}{3}\Vert {\hat{u}}\Vert _2^2 +\frac{q^2}{12a}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}e^{-\frac{|x-y|}{a}} {\hat{u}}^2(x){\hat{u}}^2(y)\mathrm {d}x\mathrm {d}y\nonumber \\&\ \ +\frac{2}{3}\int _{{{\mathbb {R}}}^3}[f({\hat{u}}){\hat{u}}-3F({\hat{u}})]\mathrm {d}x\nonumber \\= & {} {\mathcal {I}}^{\infty }({\hat{u}})-\frac{1}{3}{\mathcal {J}}^{\infty }({\hat{u}})={\mathcal {I}}^{\infty }({\hat{u}})\ge {\hat{m}}^{\infty }, \end{aligned}$$

which implies ${\mathcal {I}}^{\infty }({\hat{u}})= {\hat{m}}^{\infty }$. Hence, ${\hat{u}}\in H^1({\mathbb {R}}^3)$ is a ground state solution of problem (1.5). The proof is now complete. $\square$