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:

  1. (V1)

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

  2. (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 \(\liminf _{|x|\rightarrow \infty }[2V(x)+\nabla V(x)\cdot x]>0\);

  3. (F1)

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

    $$\begin{aligned} |f(t)|\le {\mathcal {C}}\left( 1+|t|^{p-1}\right) , \ \ \ \ \forall \ t\in {\mathbb {R}}; \end{aligned}$$
  4. (F2)

    \(f(t)=o(t)\) as \(t\rightarrow 0\);

  5. (F3)

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

  6. (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:

  1. (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.

  1. (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,

  1. (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 \({\mathcal {K}}*g \in L^1_{\mathrm{loc}}({\mathbb {R}}^3)\);

  2. (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. (1)

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

  2. (2)

    \(\phi _{u}\ge 0\);

  3. (3)

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

  4. (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. (5)

    \(\phi _u\in {\mathcal {D}}\);

  6. (6)

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

  7. (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:

  1. (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);

  2. (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

  1. (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)
  2. (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)
  3. (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

  1. (i)

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

  2. (ii)

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

Proof

  1. (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)
  2. (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

  1. (i)

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

  2. (ii)

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

  3. (iii)

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

  4. (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

  1. (i)

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

  2. (ii)

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

  3. (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

  1. (i)

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

  2. (ii)

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

  3. (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

  1. (i)

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

  2. (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

    $$\begin{aligned} \Gamma =\left\{ \gamma \in {\mathcal {C}}([0, 1], H^1({\mathbb {R}}^3)): \gamma (0)=0, \gamma (1)=T^2(u_1^{\infty })_{T}\right\} ; \end{aligned}$$
  3. (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 \(\lambda _0\in (1/2, 1]\);

  4. (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. \(\square \)