# Positive solutions of Schrödinger-Kirchhoff-Poisson system without compact condition

## Abstract

### Purpose

The existence of positive solutions for a class of nonlinear Schrödinger-Kirchhoff-Poisson systems.

### Methods

Variational method.

### Results

Some results on the existence of positive solutions.

## 1 Introduction

In this paper, we are concerned with the existence of positive solutions for the following nonlinear Schrödinger-Kirchhoff-Poisson system:

\begin{aligned} (\mathcal{SK}) \textstyle\begin{cases} - (a+b { \int_{\mathbb {R}^{3}}}\vert \nabla u\vert ^{2} )\triangle u+u-qK(x)\phi u=f(x)\vert u\vert ^{p-1}u, & x\in \mathbb {R}^{3} , \\ -\triangle\phi=qK(x)u^{2}, & x\in \mathbb {R}^{3}, \end{cases}\displaystyle \end{aligned}

by using the variational methods, where $$a>0$$ and $$b\geqslant0$$ are constants, $$1< p<5$$, $$q>0$$, and $$K, f: \mathbb {R}^{3} \rightarrow \mathbb {R}$$ are two nonnegative functions.

The Kirchhoff equation

$$-\biggl(a+b { \int_{\mathbb {R}^{3}}}\vert \nabla u\vert ^{2}\biggr)\triangle u+u=f(x,u),\quad x\in \mathbb {R}^{3},$$
(1.1)

which is related to the stationary analogue of the equation

$$\rho\frac{\partial^{2}u}{\partial t^{2}}- \biggl( \frac{P_{0}}{h}+\frac {E}{2L} \int_{0}^{L} \biggl\vert \frac{\partial u}{\partial x} \biggr\vert ^{2}\,dx \biggr) \frac{\partial^{2}u}{\partial x^{2}}=0,$$
(1.2)

arises in many mathematical physics contexts. Equation (1.2) was proposed by Kirchhoff  in 1883 as an extension of the classical D’Alembert wave equation for free vibrations of elastic strings. Kirchhoff’s model considers the changes in the length of the string produced by transverse vibrations.

The Schrödinger-Poisson system

$$\textstyle\begin{cases} -\triangle u+u+qK(x)\phi u=f(x,u), & x\in \mathbb {R}^{3}, \\ -\triangle\phi=K(x)u^{2}, & x\in \mathbb {R}^{3}, \end{cases}$$
(1.3)

describes a charged wave interacting with its own electrostatic field . We also refer the readers to [3, 4] and the references therein for more mathematical and physical background of (1.3). The Schödinger-Kirchhoff-Poisson system $$(\mathcal{SK})$$ is a more generalized Kirchhoff-type system of (1.3).

In recent years, there have been enormous results on the existence and multiplicity of solutions of problem (1.3) for $$q>0$$ (see e.g. ). To the best of our knowledge, there are a few articles on the existence of solutions to problem (1.3) for $$q<0$$. Recently, in , the author proved that problem (1.3) has a positive ground state solution for $$q<0$$ and $$f(x,u)=a(x)\vert u\vert ^{p-1}u$$, $$3< p<5$$. In , the author proved that problem (1.3) has a positive solution for $$q<0$$. When $$q\equiv1$$, the problem $$(\mathcal{S}\mathcal{K})$$ can be reduced to

$$\textstyle\begin{cases} -(a+b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2})\triangle u+u-K(x)\phi u=f(x)\vert u\vert ^{p-1}u, & x\in \mathbb {R}^{3}, \\ -\triangle\phi=K(x)u^{2}, & x\in \mathbb {R}^{3}. \end{cases}$$
(1.4)

Li and Ye  proved that problem (1.4) has a a positive ground state solution for $$K(x)\equiv0$$, $$f(x)\equiv1$$, $$2< p<5$$.

Motivated by the works mentioned, we consider system $$(\mathcal{SK})$$ with the Kirchhoff term $$(\int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2})\triangle u$$, which implies that the equation in $$(\mathcal{SK})$$ is no longer a pointwise identity and is different from the classical Schrödinger-Poisson system. We must emphasize that the combined effects of the nonlocal term it contains and the negative coefficient at ϕu make problem (1.4) an interesting problem.

To obtain our main results, we first require some conditions on f and K:

($$f_{1}$$):

$$f(x)\geqslant f_{\infty} =\lim_{\vert x\vert \rightarrow+\infty}f(x)>0$$, and $$\alpha(x)=f(x)-f_{\infty} \in L^{\frac {6}{5-p}}(\mathbb {R}^{3})$$;

($$f_{2}$$):

$$(\nabla f,x)\in L^{\frac{6}{5-p}}(\mathbb {R}^{3})\setminus\{0\}$$, and $$(\nabla f,x)\geqslant0$$;

($$K_{1}$$):

$$K\in L^{2}(\mathbb {R}^{3})$$;

($$K_{2}$$):

$$(\nabla K,x)\in L^{2}(\mathbb {R}^{3})\setminus\{0\}$$;

($$K_{3}$$):

$$K(x)\geqslant K_{\infty}=\lim_{\vert x\vert \rightarrow +\infty}K(x)\geqslant0$$, and $$K-K_{\infty} \in L^{2}(\mathbb {R}^{3})$$;

($$K_{4}$$):

$$K(x)+(\nabla K,x)\geqslant0$$, $$x\in \mathbb {R}^{3}$$.

## 2 Conclusion

In this paper, we get the existence of a positive solution for all $$1< p<5$$. Our main results are as follows.

### Theorem 2.1

Let ($$f_{1}$$), ($$f_{2}$$), ($$K_{1}$$), and ($$K_{2}$$) hold. Then there exists $$q_{0}>0$$ such that, for any $$0< q< q_{0}$$, problem $$(\mathcal{S}\mathcal{K})$$ admits a positive solution $$(u,\phi_{u}) \in H^{1}(\mathbb {R}^{3})\times D^{1,2}(\mathbb {R}^{3})$$ for all $$1< p<5$$.

### Theorem 2.2

Let ($$f_{1}$$), ($$f_{2}$$), and ($$K_{2}$$)-($$K_{4}$$) hold. Then problem (1.4) admits a positive ground state solution $$(u,\phi_{u}) \in H^{1}(\mathbb {R}^{3})\times D^{1,2}(\mathbb {R}^{3})$$ for all $$1< p<5$$.

As a consequence of Theorem 2.2, we have that the problem

\begin{aligned} \textstyle\begin{cases} -(a+b { \int_{\mathbb {R}^{3}}}\vert \nabla u\vert ^{2})\triangle u+u=\vert u\vert ^{p-1}u, & x\in \mathbb {R}^{3}, \\ u\in H^{1}({\mathbb {R}}^{3}), & x\in \mathbb {R}^{3}, \end{cases}\displaystyle \end{aligned}

has a positive ground state solution for all $$1< p<5$$. This can be viewed as an extension of a recent result of Li and Ye  concerning the existence of positive ground state solutions.

The paper is organized as follows. In Section 3, we describe the notation and preliminaries. In Section 4, we give a proof of Theorem 2.1. In Section 5, we give a proof of Theorem 2.2.

## 3 Notation and preliminaries

• Let $$a>0$$ is fixed, and let $$H^{1}(\mathbb {R}^{3})$$ be the usual Sobolev space endowed with the scalar product and norm

$$(u,v)= \int_{\mathbb {R}^{3}}(a\nabla u\nabla v+uv) \qquad \mbox{and}\qquad \Vert u\Vert = \biggl( \int_{\mathbb {R}^{3}}\bigl(a\vert \nabla u\vert ^{2}+u^{2} \bigr) \biggr)^{{\frac{1}{2}}}.$$
• $$H_{r}^{1}(\mathbb {R}^{3}):= \{u:u\in H^{1}(\mathbb {R}^{3}),u(x)=u(\vert x\vert ) \}$$.

• $$D^{1,2}(\mathbb {R}^{3})$$ is the completion of $$C_{0}^{\infty}(\mathbb {R}^{3})$$ with respect to the norm

$$\Vert u\Vert _{D^{1,2}}= \biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{{\frac{1}{2}}}.$$
• For any $$u\in H^{1}(\mathbb {R}^{3})$$ and $$t>0$$, we define $${u}_{t}(x)=t^{\frac{1}{2}}u(\frac{x}{t})$$.

• For any $$z\in{ \mathbb {R}^{3}}$$ and $$\rho>0$$, $$B_{\rho}(z)$$ denotes the ball of radius ρ centered at z.

• For any $$s\in[1,+\infty)$$, $$\vert \cdot \vert _{s}$$ denotes the usual norm of the Lebesgue space $$L^{s}({\mathbb {R}^{3}})$$.

• $$S_{1}$$ is the best Sobolev constant for the embedding of $$H^{1}(\mathbb {R}^{3})$$ in $$L^{6}(\mathbb {R}^{3})$$, that is,

$$S_{1}=\operatorname{inf}_{u\in H^{1}(\mathbb {R}^{3})\backslash\{0\}}\frac {\Vert u\Vert }{\vert u\vert _{6}}.$$
• $$S_{2}$$ is the best Sobolev constant for the embedding of $$D^{1,2}(\mathbb {R}^{3})$$ in $$L^{6}(\mathbb {R}^{3})$$, that is,

$$S_{2}=\operatorname{inf}_{u\in D^{1,2}(\mathbb {R}^{3})\backslash\{0\} }\frac{\Vert u\Vert _{D^{1,2}}}{\vert u\vert _{6}}.$$
• $$H^{-1}(\mathbb {R}^{3})$$ denotes the conjugate Sobolev space of $$H^{1}({\mathbb {R}^{3}})$$.

• C and $$C_{i}$$, $$i\in \mathbb {N}$$, are various positive constants.

It is well known that problem $$(\mathcal{SK})$$ can be reduced to a nonlinear Schrödinger-Kirchhoff equation with a nonlocal term. For any $$u \in H^{1}({\mathbb {R}^{3}})$$, define the linear functional $$L_{u}$$ in $$D^{1,2}(\mathbb {R}^{3})$$ by

$$L_{u}(v)= \int_{\mathbb {R}^{3}}K(x)u^{2}v.$$

Then ($$K_{1}$$) or ($$K_{3}$$) and the Hölder and Sobolev inequalities imply

\begin{aligned} \bigl\vert L_{u}(v)\bigr\vert \leqslant C\Vert u \Vert ^{2}\cdot \Vert v\Vert _{D^{1,2}}. \end{aligned}
(3.1)

Hence, by the Lax-Milgram theorem there exists a unique $$\phi_{u}\in D^{1,2}(\mathbb {R}^{3})$$ such that

$$\int_{\mathbb {R}^{3}}\nabla\phi_{u}\nabla v = q \int_{\mathbb {R}^{3}}K(x)u^{2}v,\quad \forall v\in D^{1,2} \bigl(\mathbb {R}^{3}\bigr),$$
(3.2)

that is, $$\phi_{u}$$ is a weak solution of $$-\triangle\phi=qK(x)u^{2}$$, and

$$\phi_{u}(x)=\frac{q}{4\pi} \int_{\mathbb {R}^{3}}\frac{K(y)u^{2}(y)}{\vert x-y\vert }\,dy.$$
(3.3)

Moreover, $$\Vert \phi_{u}\Vert _{D^{1,2}(\mathbb {R}^{3})}^{2}=q\int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}$$ and $$\phi_{u}>0$$ when $$u\neq0$$. Then (3.3) inserted into the first equation of $$(\mathcal{SK})$$ gives

$$-\biggl(a+b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}\biggr)\triangle u + u - qK(x)\phi_{u}u =f(x)\vert u\vert ^{p-1}u,\quad x \in \mathbb {R}^{3}.$$
(3.4)

Problem (3.4) is variational, and its solutions are the critical points of the functional defined in $$H^{1}(\mathbb {R}^{3})$$ by

$$I_{q}(u)=\frac{1}{2}\Vert u \Vert ^{2}+ \frac{b}{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}-\frac{q}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}- \frac{1}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}.$$
(3.5)

It is clear to see that $$I_{q}$$ is well defined on $$H^{1}(\mathbb {R}^{3})$$ and is of class $$C^{1}$$, and for any $$u,v\in H^{1}(\mathbb {R}^{3})$$,

\begin{aligned} \bigl\langle I_{q}'(u),v \bigr\rangle =& \biggl(a+b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr) \int_{\mathbb {R}^{3}}\nabla u\nabla v + \int_{\mathbb {R}^{3}} uv \\ &{}-q \int_{\mathbb {R}^{3}}K(x)\phi_{u}uv- \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p-1}uv. \end{aligned}
(3.6)

Thus, if $$u\in H^{1}(\mathbb {R}^{3})$$ is a critical point of $$I_{q}$$, then the pair $$(u,\phi_{u})$$ with $$\phi_{u}$$ as in (3.3) is a solution of the problem $$(\mathcal{SK})$$.

Define the operator $$\Phi: H^{1}(\mathbb {R}^{3})\rightarrow D^{1,2}(\mathbb {R}^{3})$$ as $$\Phi[u]=\phi_{u}$$. The following lemma shows that the operator Φ possesses the property.

### Lemma 3.1

(See )

If K satisfies ($$K_{1}$$) or ($$K_{3}$$), then

1. (i)

Φ is continuous;

2. (ii)

Φ maps bounded sets into bounded sets;

3. (iii)

If $$u_{n}\rightharpoonup u$$ in $$H^{1}(\mathbb {R}^{3})$$, then $$\Phi [u_{n}]\rightharpoonup\Phi[u]$$ in $$D^{1,2}(\mathbb {R}^{3})$$.

To obtain the boundedness of (PS) sequences, we recall the indirect approach developed by Jeanjean .

### Proposition 3.1

Let ($$X, \Vert \cdot \Vert$$) be a real Banach space, and let $$J\subset \mathbb {R}^{+}$$ be an interval. Consider the following family of $$C^{1}$$ functionals on X:

$$I_{\lambda}(u)=A(u)-\lambda B(u),\quad \forall \lambda\in J,$$

with $$B(u)\geqslant0$$ and either $$A(u)\rightarrow+\infty$$ or $$B(u)\rightarrow+\infty$$ as $$\Vert u\Vert \rightarrow+\infty$$ and such that $$I_{\lambda}(0)=0$$. For every $$\lambda\in J$$, set

$$\Gamma_{\lambda}:=\bigl\{ \gamma\in C\bigl([0,1],X\bigr)\mid \gamma(0)=0,I_{\lambda}\bigl( \gamma(1)\bigr)< 0\bigr\} \neq\emptyset$$

and

$$c_{\lambda}:= \operatorname{inf}_{\gamma\in\Gamma_{\lambda}}\max _{t \in[0,1]}I_{\lambda}\bigl(\gamma(t)\bigr)> 0 .$$
(3.7)

Then, for almost every $$\lambda\in J$$, there is a bounded $$(\mathrm{PS})_{c_{\lambda}}$$ sequence $$\{u_{n}\}\subset X$$.

### Lemma 3.2

(See , Lemma 2.3)

Under the assumptions of Proposition  3.1, the map $$\lambda\rightarrow c_{\lambda}$$ is nonincreasing and left-continuous.

## 4 A mountain pass solution

In this section, let ($$f_{1}$$), ($$f_{2}$$), ($$K_{1}$$), and ($$K_{2}$$) hold, and let $$1< p<5$$. Using ($$K_{1}$$), (3.2), and the Sobolev inequality, we have

$$\Vert \phi_{u}\Vert _{{D^{1,2}}}\leqslant q \vert K\vert _{2}S_{2}^{-1}\cdot \vert u \vert _{6}^{2},\qquad \Vert \phi_{u} \Vert _{{D^{1,2}}}\leqslant q \vert K\vert _{2}S_{2}^{-1}S_{1}^{-2} \cdot \Vert u\Vert ^{2},$$
(4.1)

and

$$\int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \leqslant q \vert K\vert _{2}^{2}S_{2}^{-2} \cdot \vert u\vert _{6}^{4},\qquad \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \leqslant q \vert K\vert _{2}^{2}S_{2}^{-2}S_{1}^{-4} \cdot \Vert u\Vert ^{4}.$$
(4.2)

To overcome the difficulty of finding bounded (PS) sequences for the associated function $$I_{q}$$, we use the cut-off function $$\chi \in C^{\infty}_{0}(\mathbb {R}^{+},\mathbb {R})$$ satisfying

$$\textstyle\begin{cases} \chi(s)=1, & s\in[0,1] , \\ 0\leqslant\chi(s)\leqslant1, & s\in(1,2),\\ \chi(s)=0, & s\in[2,+\infty),\\ \vert \chi'\vert _{\infty}\leqslant2, \end{cases}$$
(4.3)

and study the following modified functional $$I_{q}^{T}:H^{1}(\mathbb {R}^{3})\rightarrow \mathbb {R}$$:

$$I_{q}^{T}(u)=\frac{1}{2}\Vert u \Vert ^{2}+ \frac{b}{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}-\frac{q}{4}\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}- \frac {1}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}.$$

Letting $$J=[ \frac{1}{2},1]$$, we consider the following family of functionals on $$X=H^{1}(\mathbb {R}^{3})$$:

\begin{aligned} I_{q,\lambda}^{T}(u) =& \frac{1}{2}\Vert u\Vert ^{2}+ \frac{b}{4}\biggl( \int _{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\ &{}-\frac{q\lambda}{4}\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int _{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} - \frac{\lambda}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}. \end{aligned}
(4.4)

Then $$I_{q,\lambda}^{T}(u)=A(u)-\lambda B(u)$$, where

$$A(u)=\frac{1}{2}\Vert u\Vert ^{2}+ \frac{b}{4} \biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}\rightarrow+\infty\quad \mbox{as } \Vert u\Vert \rightarrow+ \infty$$

and

$$B(u)=\frac{q}{4}\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}+ \frac{1}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}\geqslant0.$$

Moreover,

\begin{aligned} \bigl\langle \bigl(I_{q,\lambda}^{T} \bigr)'(u),v\bigr\rangle =& \biggl(a +b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr) \int_{\mathbb {R}^{3}}\nabla u\nabla v + \int_{\mathbb {R}^{3}} uv \\ &{}-\frac{3q\lambda}{2T^{6}}\chi'\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \int_{\mathbb {R}^{3}}\vert u\vert ^{4}uv \\ &{}-\lambda q\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u}uv-\lambda \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p-1}uv. \end{aligned}
(4.5)

The following lemma implies that $$I_{q,\lambda}^{T}$$ satisfies the conditions of Proposition 3.1.

### Lemma 4.1

1. (i)

For any $$q_{0}>0$$, there exists a constant $$\eta>0$$ such that $$c_{\lambda}\geqslant\eta>0$$ for all $$\lambda\in J$$ and $$q\in (0,q_{0}]$$;

2. (ii)

$$\Gamma_{\lambda}\neq\emptyset$$ for all $$\lambda\in J$$.

### Proof

(i) For any $$\lambda\in J$$ and $$q_{0}>0$$, from ($$f_{1}$$), ($$K_{1}$$), (4.2)-(4.3), and the Hölder and Sobolev inequalities it follows that

\begin{aligned} I_{q,\lambda}^{T}(u) =&\frac{1}{2}\Vert u \Vert ^{2}+ \frac{b}{4}\biggl( \int _{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\ &{}-\frac{q\lambda}{4}\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int _{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} - \frac{\lambda}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1} \\ \geqslant&\frac{1}{2}\Vert u\Vert ^{2}- \frac{q}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}- \frac{1}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1} \\ \geqslant&\frac{1}{2}\Vert u\Vert ^{2}-C \bigl(\Vert u \Vert ^{4}+\Vert u\Vert ^{p+1}\bigr), \quad \forall q \in(0,q_{0}]. \end{aligned}

Since $$1< p<5$$, there exists $$\rho>0$$ such that $$I_{q,\lambda}^{T}(u)> 0$$ for $$0<\Vert u\Vert \leqslant\rho$$. In particular, there exists $$\eta>0$$ such that $$I_{q,\lambda }^{T}(u)\geqslant\eta$$ for $$\Vert u\Vert =\rho$$. Now fix $$\lambda \in J$$ and $$\gamma\in\Gamma_{\lambda}$$. Since $$I_{q,\lambda}^{T}(\gamma(1))<0$$, $$\Vert \gamma(1)\Vert >\rho$$. By $$\gamma(0)=0$$ and the continuity of γ we can deduce that there exists $$t_{\gamma}\in(0,1)$$ such that $$\Vert \gamma(t_{\gamma})\Vert =\rho$$. Hence, for any $$\lambda\in J$$,

$$c_{\lambda}\geqslant \operatorname{inf}_{\gamma\in\Gamma_{\lambda}}I_{q,\lambda}^{T} \bigl(\gamma (t_{\gamma})\bigr)\geqslant\eta>0.$$

(ii) Set $$w\in H^{1}(\mathbb {R}^{3})\backslash\{0\}$$ with $$\vert w\vert _{6}=1$$. Define $$\gamma:[0,1]\rightarrow H^{1}(\mathbb {R}^{3})$$ as

\begin{aligned} \gamma(t)= \textstyle\begin{cases} 0, & t=0 , \\ \overline{w}_{t}, & t\in(0,1], \end{cases}\displaystyle \end{aligned}

where $$\overline{w}={w}_{\theta}$$ and $$\theta>2T$$. It is clear to see that γ is a continuous path from 0 to . Moreover, for all $$\lambda\in J$$, it follows from (4.3) that

\begin{aligned} I_{q,\lambda}^{T}\bigl( \gamma(1)\bigr) \leqslant&\frac{a}{2}{\theta}^{2} \int_{\mathbb {R}^{3}}\vert \nabla w\vert ^{2}+ \frac{1}{2}{\theta}^{4} \int_{\mathbb {R}^{3}}w^{2} \\ &{}+ \frac{b}{4}{\theta}^{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla w\vert ^{2} \biggr)^{2}-\frac{1}{2(p+1)}{\theta} ^{\frac{p+7}{2}} \int_{\mathbb {R}^{3}}f_{\infty} \vert w\vert ^{p+1}. \end{aligned}
(4.6)

As $$1< p<5$$, we get $$I_{q,\lambda}^{T}(\gamma(1))<0$$ for θ large enough. □

### Lemma 4.2

For any $$\lambda\in J$$, each bounded (PS) sequence of the functional $$I_{q,\lambda}^{T}$$ admits a convergent subsequence.

### Proof

Let $$\lambda\in J$$, and let $$\{u_{n}\}$$ be a bounded (PS) sequence for $$I_{q,\lambda}^{T}$$, that is,

$$\sup_{n\in \mathbb {N}}\bigl\vert I_{q,\lambda}^{T}(u_{n}) \bigr\vert < \infty,\qquad \bigl(I_{q,\lambda}^{T}\bigr)'(u_{n}) \rightarrow0 \quad \mbox{in } H^{-1}\bigl(\mathbb {R}^{3}\bigr).$$
(4.7)

Up to a subsequence, we may suppose that there exists $$u\in H^{1}(\mathbb {R}^{3})$$ such that

\begin{aligned}& \begin{aligned} &\mbox{(a) } u_{n} \rightharpoonup u \quad \mbox{in } H^{1}\bigl(\mathbb {R}^{3}\bigr), \\ &\mbox{(b) } u_{n}\rightarrow u \quad \mbox{in } L_{\mathrm{loc}}^{s} \bigl(\mathbb {R}^{3}\bigr) \mbox{ for } s\in[2,6), \mbox{and } \\ &\mbox{(c) } u_{n}(x)\rightarrow u(x)\quad \mbox{a.e. in } \mathbb {R}^{3}. \end{aligned} \end{aligned}
(4.8)

Now we prove that

$$\int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} \rightarrow \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}$$
(4.9)

and

$$\int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}w\rightarrow \int_{\mathbb {R}^{3}}K(x)\phi_{u}uw,\quad \forall w\in H^{1}\bigl(\mathbb {R}^{3}\bigr).$$
(4.10)

First, by $$u_{n}\rightharpoonup u$$ in $$H^{1}(\mathbb {R}^{3})$$ and Lemma 3.1(iii) we have that

$$\phi_{u_{n}}\rightharpoonup\phi_{u}\quad \mbox{in } D^{1,2}\bigl(\mathbb {R}^{3}\bigr).$$
(4.11)

Moreover, in view of the Sobolev embedding theorem and Lemma 3.1(iii), we deduce that

\begin{aligned}& \begin{aligned} &\mbox{(a) } u_{n} \rightharpoonup u\quad \mbox{in } L^{6}\bigl(\mathbb {R}^{3}\bigr), \\ &\mbox{(b) } u_{n}^{2}\rightharpoonup u^{2} \quad \mbox{in } L_{\mathrm{loc}}^{3}\bigl(\mathbb {R}^{3}\bigr),\quad\text{and} \\ &\mbox{(c) } \phi_{u_{n}}\rightharpoonup\phi_{u}\quad \mbox{in } L_{\mathrm{loc}}^{6}\bigl(\mathbb {R}^{3}\bigr). \end{aligned} \end{aligned}
(4.12)

Thus, given $$\varepsilon>0$$, it follows from (4.11) that

$$\biggl\vert \int_{\mathbb {R}^{3}}K(x) (\phi_{u_{n}}-\phi_{u})u^{2} \biggr\vert < \varepsilon,$$
(4.13)

and for any fixed $$w\in H^{1}(\mathbb {R}^{3})$$, we have

$$\biggl\vert \int_{\mathbb {R}^{3}}K(x)\phi_{u}(u_{n}-u)w \biggr\vert < \varepsilon.$$
(4.14)

Since $$K\in L^{2}(\mathbb {R}^{3})$$, there exists $$\rho={\rho}(\varepsilon)>0$$ such that

$$\vert K\vert _{2,\mathbb {R}^{3}\setminus B_{\rho}(0)}< \varepsilon,\quad \forall \rho\geqslant \overline{\rho},$$
(4.15)

and

\begin{aligned}& \bigl\vert u_{n}^{2}-u^{2}\bigr\vert _{3,B_{\rho}(0)}< \varepsilon, \end{aligned}
(4.16)
\begin{aligned}& \vert \phi_{u_{n}}-\phi_{u} \vert _{6,B_{\rho}(0)}< \varepsilon \end{aligned}
(4.17)

for large n. Hence, by (4.12), (4.13), (4.15), and (4.16) we deduce

\begin{aligned}& \biggl\vert \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2}- \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \biggr\vert \\& \quad \leqslant \biggl\vert \int_{\mathbb {R}^{3}}K(x)\phi _{u_{n}}\bigl(u_{n}^{2}-u^{2} \bigr) \biggr\vert + \biggl\vert \int_{\mathbb {R}^{3}}K(x) (\phi_{u_{n}}-\phi_{u})u^{2} \biggr\vert \\& \quad \leqslant \vert \phi_{u_{n}}\vert _{6}\cdot \vert K\vert _{2,\mathbb {R}^{3}\setminus B_{\rho}(0)}\cdot\bigl\vert u_{n}^{2}-u^{2} \bigr\vert _{3} +\vert \phi_{u_{n}}\vert _{6}\cdot \vert K\vert _{2}\cdot \bigl\vert u_{n}^{2}-u^{2} \bigr\vert _{3,B_{\rho}(0)}+\varepsilon \\& \quad \leqslant C\varepsilon \end{aligned}

for large n. Analogously, by (4.12), (4.14), (4.15), and (4.17) we infer

\begin{aligned}& \biggl\vert \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}w- \int_{\mathbb {R}^{3}}K(x)\phi_{u}uw \biggr\vert \\& \quad \leqslant \biggl\vert \int_{\mathbb {R}^{3}}K(x) (\phi_{u_{n}}-\phi _{u})u_{n}w \biggr\vert + \biggl\vert \int_{\mathbb {R}^{3}}K(x)\phi _{u}(u_{n}-u)w \biggr\vert \\& \quad \leqslant \vert K\vert _{2,\mathbb {R}^{3}\setminus B_{\rho}(0)}\cdot \vert \phi_{u_{n}}- \phi_{u}\vert _{6}\cdot \vert u_{n}\vert _{6}\cdot \vert w\vert _{6} \\& \qquad {}+\vert K\vert _{2}\vert \cdot \vert \phi_{u_{n}}- \phi_{u}\vert _{6,B_{\rho}(0)}\cdot \vert u_{n} \vert _{6}\cdot \vert w\vert _{6}+\varepsilon \\& \quad \leqslant C\varepsilon \end{aligned}

for large n. This completes the proof of (4.9) and (4.10). Similarly, we can get

\begin{aligned}& \int_{\mathbb {R}^{3}}\alpha(x)\vert u_{n}\vert ^{p-1}u_{n}^{2}= \int_{\mathbb {R}^{3}}\alpha(x)\vert u\vert ^{p-1}u^{2}+o(1), \end{aligned}
(4.18)
\begin{aligned}& \int_{\mathbb {R}^{3}}\alpha(x)\vert u_{n}\vert ^{p-1}u_{n}w= \int_{\mathbb {R}^{3}}\alpha(x)\vert u\vert ^{p-1}uw+o(1),\quad \forall w\in H^{1}\bigl(\mathbb {R}^{3}\bigr), \end{aligned}
(4.19)

and

$$\int_{\mathbb {R}^{3}}\vert u_{n}\vert ^{p+1}= \int_{\mathbb {R}^{3}}\vert u_{n}\vert ^{p-1}u_{n}u+o(1).$$
(4.20)

Thus, it follows from (4.18)-(4.20) that

\begin{aligned} \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p-1}u_{n}(u_{n}-u)=o(1). \end{aligned}
(4.21)

Moreover, by (4.7)-(4.10) and (4.3) we have

\begin{aligned} o(1) =&\bigl\langle \bigl(I_{q,\lambda}^{T} \bigr)'(u_{n})-\bigl(I_{q,\lambda }^{T} \bigr)'(u),u_{n}-u\bigr\rangle \\ =&\Vert u_{n}-u\Vert ^{2}+b \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \int_{\mathbb {R}^{3}} \nabla u_{n}\nabla(u_{n}-u) \\ &{}-b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \int_{\mathbb {R}^{3}} \nabla u\nabla(u_{n}-u) \\ &{}-\frac{3q\lambda}{2T^{6}}\chi'\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} \int_{\mathbb {R}^{3}}\vert u_{n}\vert ^{4}u_{n}(u_{n}-u) \\ &{}+\frac{3q\lambda}{2T^{6}}\chi'\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \int_{\mathbb {R}^{3}}\vert u\vert ^{4}u(u_{n}-u) \\ &{}-q\lambda\chi\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}(u_{n}-u) \\ &{}+q\lambda\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u}u(u_{n}-u) \\ &{}-q\lambda \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p-1}u_{n}(u_{n}-u) \\ &{}+q\lambda \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p-1}u(u_{n}-u). \end{aligned}
(4.22)

By (4.8)(a) and (4.3) we deduce

\begin{aligned}& \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \int_{\mathbb {R}^{3}} \nabla u \nabla(u_{n}-u)=o(1),\qquad \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p-1}u(u_{n}-u)=o(1), \end{aligned}
(4.23)
\begin{aligned}& \chi'\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{ \mathbb {R}^{3}}K(x)\phi_{u}u^{2} \int_{ \mathbb {R}^{3}}\vert u\vert ^{4}u(u_{n}-u)=o(1), \end{aligned}
(4.24)

and

$$\int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \int_{\mathbb {R}^{3}} \nabla u_{n} \nabla(u_{n}-u) = \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \int_{\mathbb {R}^{3}} \bigl\vert \nabla(u_{n}-u)\bigr\vert ^{2}+o(1).$$
(4.25)

It follows (4.3), (4.9), and (4.10) that

$$\chi'\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} \int_{\mathbb {R}^{3}}\vert u_{n}\vert ^{4}u_{n}(u_{n}-u)=o(1).$$
(4.26)

Therefore, it follows from (4.21)-(4.26) that

$$\Vert u_{n}-u\Vert ^{2}+b \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \int_{\mathbb {R}^{3}} \bigl\vert \nabla(u_{n}-u)\bigr\vert ^{2}=o(1).$$

This implies $$\Vert u_{n}-u\Vert ^{2}=o(1)$$. Thus, $$u_{n}\rightarrow u$$ in $$H^{1}(\mathbb {R}^{3})$$. □

### Lemma 4.3

For almost every $$\lambda\in J$$, there exists $$u^{\lambda}\in H^{1}(\mathbb {R}^{3})\backslash\{0\}$$ such that

$$I_{q,\lambda}^{T}\bigl(u^{\lambda}\bigr)=c_{\lambda},\qquad \bigl(I_{q,\lambda }^{T}\bigr)'\bigl(u^{\lambda} \bigr)=0.$$

### Proof

By Lemma 3.1, Lemma 3.2, and Proposition 3.1, for almost every $$\lambda\in J$$, there exists a bounded $$(\mathrm{PS})_{c_{\lambda }}$$ sequence $$\{u_{n}^{\lambda}\}\subset H^{1}(\mathbb {R}^{3})$$. Up to a subsequence, by Lemma 4.2 we can suppose that there exists $$u^{\lambda} \in H^{1}(\mathbb {R}^{3})$$ such that $$u_{n}^{\lambda} \rightarrow u^{\lambda}$$ in $$H^{1}(\mathbb {R}^{3})$$. Then we have $$I_{q,\lambda}^{T}(u^{\lambda})=c_{\lambda}$$ and $$(I_{q,\lambda}^{T})'(u^{\lambda})=0$$. □

### Lemma 4.4

(Pohoz̆aev identity)

If $$u \in H^{1}(\mathbb {R}^{3})$$ is a weak solution of

\begin{aligned}& - \biggl(a +b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)\triangle u +u-q\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr)K(x)\phi_{u}u \\& \qquad {}-\frac{3q\lambda}{2T^{6}}\chi'\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr)\vert u \vert ^{4}u \int_{\mathbb {R}^{3}}K(x)\phi _{u}u^{2} \\& \quad =\lambda f(x) \vert u\vert ^{p-1}u, \end{aligned}

then we have

\begin{aligned} 0 =&P(u) \\ :=&\frac{a}{2}{ \int_{\mathbb {R}^{3}}}\vert \nabla u\vert ^{2} + \frac{3}{2}{ \int_{\mathbb {R}^{3}}}u^{2} +\frac{b}{2}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\ &{}-\frac{5q\lambda}{4}\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int _{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \\ &{}-\frac{3q\lambda}{4}\frac{\vert u\vert _{6}^{6}}{T^{6}}\chi'\biggl( \frac {\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \\ &{}-\frac{q\lambda}{4}\chi\biggl(\frac{\vert u\vert _{6}^{6}}{T^{6}}\biggr) \int _{\mathbb {R}^{3}}(\nabla K,x)\phi_{u}u^{2} \\ &{}-\frac{3\lambda}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}- \frac{\lambda}{p+1} \int_{\mathbb {R}^{3}}(\nabla f,x)\vert u\vert ^{p+1}, \end{aligned}
(4.27)

where $$\phi_{u}$$ is defined by (3.3).

### Proof

The proof is standard, and we omit it (see ). □

### Lemma 4.5

Let $$m_{i}$$ ($$i=1,2,3$$) be positive constants, $$p>1$$, and $$g(t)=m_{1}t^{2}+m_{2}t^{4}-m_{3}t^{3+\frac{p+1}{2}}$$, $$t\geqslant 0$$. Then g has a unique positive critical point, which corresponds to its maximum.

### Proof

The proof is similar to that of Lemma 3.3 in  and is elementary. We omit the proof. □

### Lemma 4.6

Let $$\lambda_{n} \in J$$, and let $$u_{n}$$ be a critical point of $$I_{q,\lambda_{n}}^{T}$$ at level $$c_{\lambda_{n}}$$ for every $$n\in \mathbb {N}$$. Then, for $$T>0$$ sufficiently large, there exists $$q_{0}=q_{0}(T)$$ such that, for any $$0< q< q_{0}$$, up to a subsequence, $$\vert u_{n}\vert _{6}\leqslant T$$ for all $$n\in \mathbb {N}$$.

### Proof

We argue by contradiction. First of all, it follows from $$(I_{q,\lambda_{n}}^{T})'(u_{n})=0$$ and Lemma 4.4 that $$u_{n}$$ satisfies the following Pohoz̆aev identity:

\begin{aligned}& \frac{a}{2}{ \int_{\mathbb {R}^{3}}}\vert \nabla u_{n}\vert ^{2} +\frac {3}{2}{ \int_{\mathbb {R}^{3}}}u_{n}^{2} +\frac{b}{2} \biggl( \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \biggr)^{2}-\frac {5q\lambda_{n}}{4}\chi\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int _{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} \\& \quad {}-\frac{3q\lambda_{n}}{4}\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\chi '\biggl( \frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2}- \frac{q\lambda_{n}}{4}\chi \biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}(\nabla K,x)\phi_{u_{n}}u_{n}^{2} \\& \quad {}-\frac{3\lambda_{n} }{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1}- \frac{\lambda_{n} }{p+1} \int_{\mathbb {R}^{3}}(\nabla f,x)\vert u_{n} \vert ^{p+1}=0. \end{aligned}
(4.28)

Since $$I_{q,\lambda_{n}}^{T}(u_{n})=c_{\lambda_{n}}$$, we have

\begin{aligned} c_{\lambda_{n}} =&\frac{1}{2} \Vert u_{n}\Vert ^{2}+ \frac{b}{4}\biggl( \int _{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \biggr)^{2} \\ &{}-\frac{q\lambda_{n}}{4}\chi\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int _{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} -\frac{\lambda_{n}}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1}. \end{aligned}
(4.29)

Hence, by (4.28) and (4.29) we obtain

\begin{aligned}& a \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2}+ \frac{b}{4}\biggl( \int _{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \biggr)^{2}+\frac{q\lambda _{n}}{2}\chi\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} \\& \quad {}+\frac{3q\lambda_{n}}{4}\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\chi '\biggl( \frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi _{u_{n}}u_{n}^{2} +\frac{q\lambda_{n}}{4}\chi\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int _{\mathbb {R}^{3}}(\nabla K,x)\phi_{u_{n}}u_{n}^{2} \\& \quad {} +\frac{\lambda_{n} }{p+1} \int_{\mathbb {R}^{3}}(\nabla f,x)\vert u_{n} \vert ^{p+1}=3c_{\lambda_{n}}. \end{aligned}

Moreover, combining ($$f_{2}$$), ($$K_{2}$$), (4.1), (4.2), (4.3), (4.30), Lemma 3.1(ii), and the Sobolev embedding theorem, we deduce

\begin{aligned} a \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \leqslant&3c_{\lambda_{n}} -\frac{3q\lambda_{n}}{4}\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\chi '\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi _{u_{n}}u_{n}^{2} \\ &{}-\frac{q\lambda_{n}}{4}\chi\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}(\nabla K,x)\phi_{u_{n}}u_{n}^{2} -\frac{\lambda_{n} }{p+1} \int_{\mathbb {R}^{3}}(\nabla f,x)\vert u_{n}\vert ^{p+1} \\ \leqslant&3c_{\lambda_{n}}+C_{1}\frac{q^{2}}{T^{6}} \chi'\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr)\vert u_{n} \vert _{6}^{10} +C_{2}q^{2}\chi\biggl( \frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr)\vert u_{n}\vert _{6}^{4}. \end{aligned}

By the definition of $$c_{\lambda_{n}}$$ and by (4.6) we have

\begin{aligned} c_{\lambda_{n}} \leqslant&\max _{{\theta}\geqslant0 } I_{q,\lambda_{n}}^{T}({w}_{\theta}) \\ \leqslant&\max _{{{\theta}}\geqslant0 } \biggl[\frac {a}{2}{\theta} ^{2} \int_{\mathbb {R}^{3}}\vert \nabla w\vert ^{2}+ \frac {1}{2}{\theta} ^{4} \int_{\mathbb {R}^{3}} w^{2} + \frac{b}{4}{\theta} ^{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla w\vert ^{2} \biggr)^{2} \\ &{}-\frac{q\lambda_{n}}{4}\chi\biggl(\frac{{\theta}^{6}}{T^{6}}\biggr) \int_{\mathbb {R}^{3}}K(x)\phi_{w_{\theta}}w_{\theta}^{2} -\frac{\lambda_{n}}{p+1}{\theta} ^{3+\frac{p+1}{2}} \int_{\mathbb {R}^{3}}\vert w\vert ^{p+1} \biggr] \\ \leqslant&\max _{{\theta}\geqslant0 } \biggl[\frac {a}{2}{\theta} ^{2} \int_{\mathbb {R}^{3}}\vert \nabla w\vert ^{2}+ \frac {1}{2}{\theta} ^{4} \int_{\mathbb {R}^{3}}\vert w\vert ^{2} + \frac{b}{4}{ \theta} ^{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla w\vert ^{2} \biggr)^{2} \\ &{}-\frac{\lambda_{n}}{p+1}{\theta} ^{3+\frac{p+1}{2}} \int_{\mathbb {R}^{3}}\vert w\vert ^{p+1} \biggr] +\max _{{\theta}\geqslant0 } \biggl[\frac{q\lambda _{n}}{4}\chi\biggl(\frac{{\theta}^{6}}{T^{6}} \biggr) \int_{\mathbb {R}^{3}}K(x)\phi _{w_{\theta}}w_{\theta}^{2} \biggr] \\ =&A_{1}+A_{2}(T), \end{aligned}

where

\begin{aligned}& A_{1}=\max _{{\theta}\geqslant0 } \biggl[\frac {a}{2}{ \theta} ^{2} \int_{ \mathbb {R}^{3}}\vert \nabla w\vert ^{2}+ \frac {1}{2}{\theta} ^{4} \int_{ \mathbb {R}^{3}}\vert w\vert ^{2} + \frac{b}{4}{ \theta} ^{4}\biggl( \int_{ \mathbb {R}^{3}}\vert \nabla w\vert ^{2} \biggr)^{2} \\& \hphantom{A_{1}=}{}-\frac{\lambda_{n}}{p+1}{\theta} ^{3+\frac{p+1}{2}} \int_{ \mathbb {R}^{3}}\vert w\vert ^{p+1} \biggr], \\& A_{2}(T)=\max _{{\theta}\geqslant0 } \biggl[ \frac{q\lambda _{n}}{4}\chi\biggl(\frac{{\theta}^{6}}{T^{6}}\biggr) \int_{ \mathbb {R}^{3}}K(x)\phi _{w_{\theta}}w_{\theta}^{2} \biggr]. \end{aligned}

By Lemma 4.5, $$A_{1}$$ is a finite number. If $${\theta} ^{6}\geqslant {2T^{6}}$$, then $$A_{2}(T)=0$$. Otherwise, it follows from (4.2) that

\begin{aligned} A_{2}(T) \leqslant& \frac{q}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{w_{\theta }}w_{\theta}^{2} \\ \leqslant&\frac{q^{2}}{4}\vert K\vert _{2}^{2}S_{2}^{-2} \cdot \vert w_{\theta} \vert _{6}^{4} \\ =& \frac{q^{2}}{4}\vert K\vert _{2}^{2}S_{2}^{-2}{ \theta}^{4}\leqslant C_{3}q^{2}T^{4}. \end{aligned}
(4.30)

We also have

\begin{aligned}& C_{1}\frac{q^{2}}{T^{6}}\chi'\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}} \biggr)\vert u_{n}\vert _{6}^{10}\leqslant C_{4}q^{2}T^{4}, \\& C_{2}q^{2} \chi\biggl(\frac{\vert u_{n}\vert _{6}^{6}}{T^{6}}\biggr)\vert u_{n}\vert _{6}^{4} \leqslant C_{5}q^{2}T^{4}. \end{aligned}

Then we deduce

$$a \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \leqslant 3A_{1}+C_{6}q^{2}T^{4}.$$
(4.31)

By the Sobolev embedding theorem we have

$$\vert u\vert _{6}^{2}\leqslant C \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \leqslant C_{7}+C_{8}q^{2}T^{4}.$$
(4.32)

We suppose by contradiction that there exists no subsequence of $$\{ u_{n}\}$$ that is uniformly bounded by T. Then we can assume that $$\vert u_{n}\vert _{6}>T$$, $$n\in \mathbb {N}$$. Therefore by (4.32) we conclude that

$$T^{2}< \vert u_{n}\vert _{6}^{2} \leqslant C \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \leqslant C_{7}+C_{8}q^{2}T^{4},$$

which is not true for T large and q small enough. Indeed, we can find $$T_{0}>0$$ such that $$T_{0}^{2}>C_{7}+1$$ and $$q_{0}=q_{0}(T_{0})$$ such that $$C_{8}q^{2}T^{4}<1$$ for any $$0< q< q_{0}$$. □

### Proof of Theorem 2.1

Let T, $$q_{0}$$ be as in Lemma 4.6 and fix $$0 < q < q_{0}$$. According to Lemma 4.3, there exist sequences $$\{\lambda_{n}\} \subset J$$ and $$\{u_{n}\} \subset H^{1}(\mathbb {R}^{3})$$ such that

$$\lambda_{n}\rightarrow1^{-},\qquad I_{q,\lambda_{n}}^{T}(u_{n})=c_{\lambda _{n}},\qquad \bigl(I_{q,\lambda_{n}}^{T}\bigr)'(u_{n})=0,$$
(4.33)

where $$c_{\lambda_{n}}$$ is defined by (3.7). We will prove that $$\{u_{n}\}_{n}$$ is a bounded (PS) sequence for $$I_{q}=I_{q,1}^{T}$$. By Lemma 4.6 we know that

$$\vert u_{n}\vert _{6}\leqslant T.$$
(4.34)

Using the Hölder inequality, we get

\begin{aligned} \int_{\mathbb {R}^{3}} \vert u_{n}\vert ^{p+1} \leqslant\biggl( \int_{\mathbb {R}^{3}} \vert u_{n}\vert ^{6} \biggr)^{\frac{p-1}{4}}\biggl( \int_{\mathbb {R}^{3}} u_{n}^{2}\biggr)^{\frac{5-p}{4}} \leqslant T^{\frac{3(p-1)}{2}}\biggl( \int_{\mathbb {R}^{3}} u_{n}^{2}\biggr)^{\frac{5-p}{4}}. \end{aligned}
(4.35)

Since $$\langle(I_{q,\lambda}^{T})'(u_{n}),u_{n}\rangle=0$$, it follows from ($$f_{1}$$), (4.2), (4.3), (4.5), (4.34), and (4.35) that

\begin{aligned} \int_{\mathbb {R}^{3}} u_{n}^{2} =&\lambda q \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2}+ \lambda \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1} \\ &{}- \biggl(a +b \int_{\mathbb {R}^{3}}\vert \nabla u_{n} \vert ^{2} \biggr) \int _{\mathbb {R}^{3}}\vert \nabla u_{n} \vert ^{2} \\ \leqslant& q \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2}+ \int_{\mathbb {R}^{3}}\alpha(x)\vert u_{n}\vert ^{p+1}+ \int_{\mathbb {R}^{3}}f_{\infty }\vert u_{n} \vert ^{p+1} \\ \leqslant& C_{9}q^{2}T^{4}+C_{10}T^{p+1}+f_{\infty} T^{\frac {3(p-1)}{2}}\biggl( \int_{\mathbb {R}^{3}} u_{n}^{2}\biggr)^{\frac{5-p}{4}}. \end{aligned}

As $$1< p<5$$, $$\int_{\mathbb {R}^{3}} u_{n}^{2}$$ is bounded. Combining (4.31), we deduce that $$\{u_{n}\}$$ is bounded in $$H^{1}(\mathbb {R}^{3})$$.

On the other hand, by (4.34), (4.3), and (4.4) we have

$$I_{q,\lambda_{n}}^{T}(u_{n})=\frac{1}{2}\Vert u_{n}\Vert ^{2}+ \frac {b}{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \biggr)^{2}-\frac {q\lambda_{n}}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} -\frac{ \lambda_{n} }{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1}.$$

Since $$\lambda_{n}\rightarrow1^{-}$$, we can prove that $$\{u_{n}\}$$ is a (PS) sequence for $$I_{q}=I_{q,1}^{T}$$ by similar arguments as in Theorem 1.1 of . We finish as in Lemma 4.3. □

## 5 A ground state solution

To apply the global compactness lemma to solve problem (1.4), first of all, we need to consider the existence of ground state solutions of the associated ‘limit problem’ of (1.4), which is given as

$$\textstyle\begin{cases} -(a+b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2})\triangle u+u-K_{\infty}\phi_{u} u=f_{\infty} \vert u\vert ^{p-1}u, & x\in \mathbb {R}^{3}, \\ -\triangle\phi_{u} =K_{\infty}u^{2}, & x\in \mathbb {R}^{3}, \end{cases}$$
(5.1)

and the corresponding least energy of the associated limited functional

$$I_{{\infty}}(u)=\frac{1}{2}\Vert u \Vert ^{2}+ \frac{b}{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}- \frac{1}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac{1}{p+1} \int _{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1}.$$
(5.2)

Set

$$\mathcal{M}_{\infty}:= \bigl\{ u\in H^{1}\bigl( \mathbb {R}^{3}\bigr)\backslash\{0\} \vert G_{\infty}(u)=0 \bigr\} ,$$

where

\begin{aligned}& G_{\infty}(u)=\frac{1}{2}\bigl\langle I_{\infty}'(u),u \bigr\rangle + P_{\infty }(u) \\& \hphantom{G_{\infty}(u)}=a \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+2 \int_{\mathbb {R}^{3}}u^{2} + b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\& \hphantom{G_{\infty}(u)=}{}-\frac{7}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac {p+7}{2(p+1)} \int_{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1}, \\& P_{\infty}(u)=\frac{a}{2} \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+ \frac{3}{2} \int_{\mathbb {R}^{3}}u^{2} + \frac{b}{2}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\& \hphantom{P_{\infty}(u)=}{}-\frac{5}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac{3}{p+1} \int_{\mathbb {R}^{3}}f_{\infty }\vert u\vert ^{p+1}. \end{aligned}

We remark that

$$I_{\infty}\vert_{\mathcal{M}_{\infty}}(u)=\frac{a}{4} \int_{\mathbb {R}^{3}} \vert \nabla u\vert ^{2} + \frac{3}{16} \int_{\mathbb {R}^{3}}K_{\infty }\phi_{u}u^{2}+ \frac{p-1}{8(p+1)} \int_{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1}$$
(5.3)

and

\begin{aligned} I_{\infty}\vert_{\mathcal{M}_{\infty}}(u) =& \frac{p+3}{2(p+7)}a \int _{\mathbb {R}^{3}} \vert \nabla u\vert ^{2} + \frac{p-1}{2(p+7)} \int_{\mathbb {R}^{3}}u^{2} \\ &{}+\frac{p-1}{4(p+7)}b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} +\frac{7-p}{4(p+7)} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}. \end{aligned}
(5.4)

### Lemma 5.1

$$I_{\infty}$$ is not bounded from below.

### Proof

For any $$u\in H^{1}(\mathbb {R}^{3})\setminus\{0\}$$, we have

\begin{aligned} I_{\infty}(u_{t}) =&\frac{a}{2}t^{2} \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+ \frac{1}{2}t^{4} \int_{\mathbb {R}^{3}}u^{2} + \frac{b}{4}t^{4} \biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\ &{}-\frac{1}{4}t^{7} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac {1}{p+1}t ^{\frac{p+7}{2}} \int_{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1}. \end{aligned}

As $$1< p<5$$, $$I_{\infty}(u_{t})\rightarrow-\infty$$ as $$t\rightarrow +\infty$$. □

### Lemma 5.2

Let $$m_{i}$$ ($$i=1,2,3,4$$) be positive constants, $$1< p<5$$, and $$g(t)=m_{1}t^{2}+m_{2}t^{4}-m_{3}t^{7}-m_{4}t^{\frac{p+7}{2}}$$ for $$t\geqslant0$$. Then g has a unique positive critical point, which corresponds to its maximum.

### Proof

The proof is similar to Lemma 3.3 of , and we omit it. □

### Lemma 5.3

For any $$u\in H^{1}(\mathbb {R}^{3})\setminus\{0\}$$, there is a unique $${\overline{t}}>0$$ such that $$u_{\overline{t}}\in\mathcal{M}_{\infty }$$. Moreover, $$I_{\infty}(u_{\overline{t}})=\max _{t>0 } I_{\infty}(u_{t})$$.

### Proof

For any $$u\in H^{1}(\mathbb {R}^{3})\setminus\{0\}$$ and $$t\geqslant0$$, we denote

\begin{aligned} \gamma(t) \triangleq& I_{\infty}(u_{t}) \\ =&\frac{a}{2}t^{2} \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+ \frac{1}{2}t^{4} \int_{\mathbb {R}^{3}}u^{2} + \frac{b}{4}t^{4} \biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\ &{}-\frac{1}{4}t^{7} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac {1}{p+1}t ^{\frac{p+7}{2}} \int_{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1}. \end{aligned}

By Lemma 5.2, γ has a unique critical point $$\overline{t}>0$$ corresponding to its maximum. Then $$\gamma(\overline{t})=\max _{t>0 }\gamma(t)$$ and $$\gamma'(\overline{t})=0$$. It follows that $$I_{\infty}(u_{\overline{t}})=\max _{t>0 } I_{\infty}(u_{t})$$, $$G_{\infty}(u_{\overline{t}})=\overline{t}\gamma '(\overline{t})=0$$ a.e., and $$u_{\overline{t}} \in{\mathcal{M}_{\infty }}$$. □

Set

\begin{aligned}& c_{1}:= \operatorname{inf}_{\gamma\in\Gamma}\max _{t \in [0,1]}I_{\infty}\bigl(\eta(t)\bigr), \\& c_{2}:= \operatorname{inf}_{u \in H^{1}(\mathbb {R}^{3})\setminus\{0\}}\max _{t >0}I_{\infty }(u_{t}), \\& c_{3}:= \operatorname{inf}_{u \in\mathcal{M}_{\infty}}I_{\infty}(u), \\& c_{4}:= \operatorname{inf}_{u \in H^{1}_{r}(\mathbb {R}^{3})\cap\mathcal {M}_{\infty}}I_{\infty}(u), \end{aligned}

where

$$\Gamma:=\bigl\{ \eta\in C\bigl([0,1],H^{1}\bigl(\mathbb {R}^{3} \bigr)\bigr)\mid\eta(0)=0,I_{\infty}\bigl( \eta(1)\bigr)< 0\bigr\} .$$

The constant $$c_{3}$$ turns out to be a nontrivial number, as we will prove in the next lemma, which contains the statement of the main properties of $$\mathcal{M}_{{\infty}}$$.

### Lemma 5.4

1. (i)

There exists $$\sigma>0$$ such that $$\Vert u\Vert \geqslant\sigma$$ for all $$u \in\mathcal{M}_{{\infty}}$$;

2. (ii)

$$I_{{\infty}}$$ is bounded from below on $$\mathcal{M}_{{\infty}}$$ by a positive constant, i.e. $$c_{3}>0$$;

3. (iii)

u is a free critical point of $$I_{\infty}$$ if and only if u is a critical point of $$I_{\infty}$$ constrained on  $$\mathcal{M}_{\infty}$$.

### Proof

(i) Let $$u \in\mathcal{M}_{\infty}$$. By the Hölder inequality we see that

\begin{aligned} 0 =&a \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+2 \int_{\mathbb {R}^{3}}u^{2} + b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\ &{}-\frac{7}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac {p+7}{2(p+1)} \int_{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1} \\ \geqslant&\Vert u\Vert ^{2} -C \bigl(\Vert u \Vert ^{4}+\Vert u\Vert ^{p+1}\bigr). \end{aligned}

As $$1< p<5$$, there exists $$\sigma>0$$ such that

$$\Vert u\Vert \geqslant\sigma>0,\quad \forall u \in\mathcal {M}_{\infty}.$$
(5.5)

(ii) It follows from (5.4) and (5.5) that

\begin{aligned} I_{\infty}(u) =&\frac{p+3}{2(p+7)}a \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+ \frac{p-1}{2(p+7)} \int_{\mathbb {R}^{3}}u^{2} \\ &{}+ \frac{p-1}{4(p+7)}b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}+\frac{7-p}{4(p+7)} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2} \\ \geqslant&\frac{p-1}{2(p+7)}\sigma^{2}>0. \end{aligned}

Thus, $$c_{3}>0$$.

(iii) The proof consists of two steps.

Step 1. $$G'_{\infty}(u)\neq0$$ for any $$u\in\mathcal {M}_{\infty}$$, and hence $$\mathcal{M}_{\infty}$$ is a $$C^{1}$$-manifold.

We will prove this by contradiction. Suppose that $$G'_{\infty}(u) =0$$ for some $$u\in\mathcal{M}_{\infty}$$. Denote

\begin{aligned}& k\triangleq I_{\infty}(u),\qquad \alpha\triangleq a \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2},\qquad \beta \triangleq \int_{\mathbb {R}^{3}}u^{2}, \\& \mu\triangleq b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2},\qquad \nu \triangleq \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2},\qquad \delta\triangleq \int_{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1}. \end{aligned}

In a weak sense, the equation $$G'_{\infty}(u) =0$$ can be written as

\begin{aligned}& -2\biggl(a+2b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}\biggr)\triangle u +4u-7K_{\infty}\phi_{u}u \\& \quad =\frac{p+7}{2}f_{\infty} \vert u\vert ^{p-1}u, \end{aligned}
(5.6)

where $$\phi_{u}(x)= \frac{K_{\infty}}{4\pi}\int_{\mathbb {R}^{3}}\frac {u^{2}(y)}{\vert x-y\vert }\,dy$$. Combining $$G_{\infty}(u)=0$$, (5.6), and its Pohoz̆ave identity, we have

$$\textstyle\begin{cases} \frac{1}{2}\alpha+ \frac{1}{2}\beta+ \frac{1}{4}\mu- \frac{1}{4}\nu - \frac{1}{p+1}\delta=k>0,\\ \alpha+ 2\beta+ \mu- \frac{7}{4}\nu- \frac{p+7}{2(p+1)}\delta=0,\\ 2\alpha+ 4\beta+ 4\mu- 7\nu- \frac{p+7}{2}\delta=0,\\ \alpha+ 6\beta+ 2\mu- \frac{21}{4}\nu- \frac{3(p+7)}{2(p+1)}\delta=0. \end{cases}$$

Thus, we can conclude that

$$\nu+\frac{(p+3)(p-1)}{1+p}\delta=-32k< 0,$$

which is impossible since ν and δ are positive. Thus, $$G'_{\infty}(u)\neq0$$ for any $$u\in\mathcal{M}_{\infty}$$, and $$\mathcal{M}_{\infty}$$ is a $$C^{1}$$-manifold by the implicit function theorem.

Step 2. Every critical point of $$I_{\infty}\vert_{\mathcal {M}_{\infty}}$$ is the critical point of $$I_{\infty}$$ in $$H^{1}(\mathbb {R}^{3})$$.

If u is a critical point of $$I_{\infty}\vert_{\mathcal{M}_{\infty }}$$, then there exists a Lagrange multiplier $$\lambda\in \mathbb {R}$$ such that

$$I'_{\infty}(u)=\lambda G'_{\infty}(u).$$

We claim that $$\lambda=0$$. The equation $$I'_{\infty}(u)-\lambda G'_{\infty}(u)=0$$ can be written in the weak sense as

\begin{aligned}& -\biggl(a+b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}\biggr)\triangle u +u-K_{\infty}\phi_{u}u-f_{\infty} \vert u \vert ^{p-1}u \\& \quad =\lambda\biggl[-2\biggl(a+2b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}\biggr)\triangle u +4u-7K_{\infty}\phi_{u}u-\frac{p+7}{2}f_{\infty } \vert u\vert ^{p-1}u\biggr], \end{aligned}

that is, u solves the equations $$-\triangle\phi=K_{\infty}u^{2}$$ and

\begin{aligned}& - \biggl((2\lambda-1)a+(4\lambda-1)b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)\triangle u+(4\lambda-1)u \\& \qquad {}-(7\lambda-1)K_{\infty}\phi_{u}u \\& \quad =\biggl[\frac{p+7}{2} \lambda-1\biggr]f_{\infty }\vert u\vert ^{p-1}u \end{aligned}

for $$x\in \mathbb {R}^{3}$$. Using the notation in Step 1, we have that

$$\textstyle\begin{cases} \frac{1}{2}\alpha+ \frac{1}{2}\beta+ \frac{1}{4}\mu- \frac {1}{p+1}\delta= \frac{1}{4}\nu+k>0,\\ \alpha+ 2\beta+ \mu - \frac{p+7}{2(p+1)}\delta= \frac{7}{4}\nu,\\ (2\lambda-1)\alpha+ (4\lambda-1)\beta+ (4\lambda-1)\mu- [\frac {p+7}{2}\lambda-1]\delta=(7\lambda-1)\nu,\\ \frac{2\lambda-1}{2}\alpha+ \frac{3(4\lambda-1)}{2}\beta+ \frac {4\lambda-1}{2}\mu -\frac{3(p+7)\lambda-6}{2(p+1)}\delta=\frac{5(7\lambda-1)}{4}\nu. \end{cases}$$
(5.7)

The coefficient matrix of (5.7) is

$$\left({\begin{matrix} {\frac{1}{2}} & \frac{1}{2} & \frac{1}{4} & -\frac{1}{p+1}\\ {1} & 2 & 1 & -\frac{p+7}{2(p+1)}\\ {2\lambda-1} & 4\lambda-1 &4\lambda-1 & 1-\frac{p+7}{2}\lambda\\ {\frac{2\lambda-1}{2}} & \frac{3(4\lambda-1)}{2} & \frac{4\lambda-1}{2} {}& -\frac{3(p+7)\lambda-6}{2(p+1)} \end{matrix}}\right),$$

and its determinant is

$$\det{A}=\frac{\lambda(1-4\lambda)(p+3)(p-1)}{16(p+1)}.$$

As $$1< p<5$$,

$$\det{A}=0\quad \Longleftrightarrow\quad \lambda=0,\quad \lambda=\frac{1}{4}.$$

Now we prove that $$\lambda=0$$ by excluding the other two possibilities:

1. (1)

If $$\lambda=\frac{1}{4}$$, then the last equation of (5.7) is

$$\frac{1}{4}\alpha+ \frac{15}{16}\nu+ \frac{3(p-1)}{8(p+1)}\delta=0.$$

This is a contradiction since $$\alpha>0$$, $$\delta>0$$, $$\nu\geqslant0$$, and $$1< p<5$$.

2. (2)

If $$\lambda\neq0$$, $$\lambda\neq\frac{1}{4}$$, then the linear system (5.7) has a unique solution. We obtain

$$\delta=-\frac{(p+1)(32k+15\nu)}{(p+3)(p-1)}.$$

As $$1< p<5$$, $$k>0$$ and $$\nu\geqslant0$$, we have $$\delta<0$$. This is impossible since $$\delta>0$$.  □

### Lemma 5.5

$$c\triangleq c_{1} =c_{2}=c_{3}=c_{4}$$.

### Proof

(i) $$c_{1} =c_{2}=c_{3}$$. The proof is similar to the argument of Nehari manifold method .

(ii) $$c_{2}=c_{4}$$. It is clear that $$c_{2}=c_{3}\leqslant c_{4}$$. We next prove that $$c_{2}\geqslant c_{4}$$. For any $$u \in H^{1}(\mathbb {R}^{3})\setminus\{0\}$$, it follows from (5.3) that there exists a unique $${\overline{t}}>0$$ such that $$u_{\overline {t}}\in\mathcal{M}_{\infty}$$ and $$I_{\infty}(u_{\overline {t}})=\max _{t>0 } I_{\infty}(u_{t})$$. Similarly, for $$u^{*}$$, the Schwarz symmetric arrangement of u, there exists a unique $$t^{*}>0$$ such that $$u^{*}_{t^{*}} \in\mathcal{M}_{\infty}$$, that is, $$u^{*}_{t^{*}} \in\mathcal{M}_{\infty} \cap H^{1}_{r}(\mathbb {R}^{3})$$. Note that

$$\int_{\mathbb {R}^{3}}\bigl\vert \nabla u^{*}\bigr\vert ^{2}\leqslant \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2},\qquad \int_{\mathbb {R}^{3}}\bigl\vert u^{*}\bigr\vert ^{2}= \int_{\mathbb {R}^{3}}\vert u\vert ^{2},\qquad \int_{\mathbb {R}^{3}}\bigl\vert u^{*}\bigr\vert ^{p+1}= \int_{\mathbb {R}^{3}}\vert u\vert ^{p+1},$$

and (see )

$$\int_{\mathbb {R}^{3}}\phi_{u}u^{2}\leqslant \int_{\mathbb {R}^{3}}\phi_{u^{*}}\bigl(u^{*} \bigr)^{2}.$$

Then we have that

$$I_{\infty}(u_{t^{*}})\geqslant I_{\infty} \bigl(u^{*}_{t^{*}}\bigr).$$

By the preceding we deduce that

$$c_{2}=\operatorname{inf}_{u \in H^{1}(\mathbb {R}^{3})\setminus\{0\}}\max _{t >0}I_{\infty }(u_{t})\geqslant I_{\infty}(u_{t_{0}}) \geqslant I_{\infty }\bigl(u^{*}_{t^{*}}\bigr)\geqslant \operatorname{inf}_{u \in H^{1}_{r}(\mathbb {R}^{3})\cap\mathcal{M}_{\infty }}I_{\infty}(u)=c_{4}.$$

□

### Lemma 5.6

(See )

Let $$\{u_{n}\}$$ be a sequence satisfying $$u_{n}\rightharpoonup u$$ in $$H_{r}^{1}(\mathbb {R}^{3})$$. Then

$$\int_{\mathbb {R}^{3}}\phi_{u_{n}}u_{n}^{2} \rightarrow \int_{\mathbb {R}^{3}}\phi_{u}u^{2}.$$

### Theorem 5.1

Problem (5.1) admits a positive ground state solution $$(u,\phi_{u}) \in H^{1}(\mathbb {R}^{3})\times D^{1,2}(\mathbb {R}^{3})$$.

### Proof

By Lemmas 5.4-5.5 we only need to prove that c is attained at some $$u\in H^{1}_{r}(\mathbb {R}^{3})\cap\mathcal{M}_{\infty}$$. Letting $$\{u_{n}\}\subset H^{1}_{r}(\mathbb {R}^{3})\cap\mathcal{M}_{\infty }$$ be a minimizing sequence for $$I_{\infty}$$, it follows from (5.4) that

\begin{aligned} c+1 >&\frac{p+3}{2(p+7)}a \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} + \frac{p-1}{2(p+7)} \int_{\mathbb {R}^{3}}u_{n}^{2} \\ &{}+\frac{p-1}{4(p+7)}b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \biggr)^{2} +\frac{7-p}{4(p+7)} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u_{n}}u_{n}^{2} \end{aligned}

for n large enough. Therefore, $$\{u_{n}\}$$ is bounded in $$H_{r}^{1}(\mathbb {R}^{3})$$. Passing, if necessary, to a subsequence, we may assume that there exists a function $$\tilde{u}\in H_{r}^{1}(\mathbb {R}^{3})$$ such that

\begin{aligned} & u_{n}\rightharpoonup \tilde{u} \quad \mbox{in } H_{r}^{1}\bigl(\mathbb {R}^{3} \bigr), \\ & u_{n}\rightarrow\tilde{u}\quad \mbox{in } L^{s}\bigl( \mathbb {R}^{3}\bigr) \mbox{ for } s\in(2,6), \mbox{and } \\ & u_{n}\rightarrow\tilde{u}\quad \mbox{a.e. in } \mathbb {R}^{3}. \end{aligned}
(5.8)

Now we show that $$u_{n} \rightarrow\tilde{u}$$ in $$H^{1}(\mathbb {R}^{3})$$. Indeed, it follows from (5.8), Fatou’s lemma, and Lemma 5.6 that

\begin{aligned}& a \int_{\mathbb {R}^{3}}\vert \nabla\tilde{u}\vert ^{2}\leqslant \liminf _{n\rightarrow\infty} a \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2}, \\& \int_{\mathbb {R}^{3}}\vert \tilde{u}\vert ^{2}\leqslant \liminf _{n\rightarrow\infty} \int_{\mathbb {R}^{3}}\vert u_{n}\vert ^{2}, \\& b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla\tilde{u}\vert ^{2} \biggr)^{2}\leqslant \liminf _{n\rightarrow\infty} b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \biggr)^{2}, \end{aligned}

and

\begin{aligned}& \int_{\mathbb {R}^{3}}\vert \tilde{u}\vert ^{p+1}= \liminf _{n\rightarrow\infty} \int_{\mathbb {R}^{3}}\vert u_{n}\vert ^{p+1}, \\& \int _{\mathbb {R}^{3}}\phi_{\tilde{u}}\tilde{u}^{2}= \liminf _{n\rightarrow\infty} \int_{\mathbb {R}^{3}}\phi_{u_{n}}u_{n}^{2}. \end{aligned}

Clearly, we have $$G_{\infty}(\tilde{u})\leqslant0$$. Therefore, by Lemma 5.3 there exists a unique $$t_{0}\in(0,1]$$ such that $$\tilde {u}_{t_{0}}\in\mathcal{M}_{\infty}$$. If $$t_{0}\in(0,1)$$, then it follows from (5.3) that

$$I_{\infty}(\tilde{u}_{t_{0}})< \lim _{n\rightarrow \infty}I_{\infty}(u_{n})=c,$$

which is impossible. Thus $$t_{0}=1$$, and c is attained at $$\tilde{u} \in\mathcal{M}_{\infty}$$.

By the standard regularity arguments as in the proof of Theorem 1.4 of  we see that ũ is a positive ground state solution for problem (5.1). □

Assume that ($$f_{1}$$)-($$f_{2}$$) and ($$K_{2}$$)-($$K_{4}$$) hold. We apply Proposition 3.1 to prove Theorem 2.2. Set $$J=[\frac{1}{2},1]$$. We consider a family of functionals on $$H^{1}(\mathbb {R}^{3})$$,

$$I_{\lambda}(u)=\frac{1}{2}\Vert u \Vert ^{2}+ \frac{b}{4}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}- \frac{\lambda}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}- \frac{\lambda }{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}.$$
(5.9)

Then $$I_{\lambda}(u)=A(u)-\lambda B(u)$$, where

$$A(u)=\frac{1}{2}\Vert u\Vert ^{2}+ \frac{b}{4} \biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}\rightarrow+\infty\quad \mbox{as } n\rightarrow+\infty$$

and

$$B(u)=\frac{1}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}+ \frac{1}{p+1} \int _{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}\geqslant0.$$

### Lemma 5.7

Assume that ($$f_{1}$$)-($$f_{2}$$) and ($$K_{3}$$) hold. Then

1. (i)

there exists $$v\in H^{1}(\mathbb {R}^{3})\setminus\{0\}$$ such that $$I_{\lambda}(v)< 0$$ for all $$\lambda\in J$$;

2. (ii)

$$c_{\lambda}:=\operatorname{inf}_{\gamma\in\Gamma_{\lambda}}\max _{t\in[0,1]}I_{\lambda}(\gamma(t))>0$$ for all $$\lambda\in J$$, where

$$\Gamma_{\lambda}:= \bigl\{ \gamma\in C\bigl([0,1],H^{1}\bigl( \mathbb {R}^{3}\bigr)\bigr)\mid\gamma (0)=0,\gamma(1)=v \bigr\} .$$

### Proof

The proof is similar to those of Lemma 3.1 and Lemma 3.2, and we omit it. □

By Proposition 3.1 we see that, for any $$\lambda\in[ \frac{1}{2}, 1]$$, the associated limit problem

$$\textstyle\begin{cases} -(a+b \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2})\triangle u+u-\lambda K_{\infty} \phi u=\lambda f_{\infty} \vert u\vert ^{p-1}u, & x\in \mathbb {R}^{3}, \\ -\triangle\phi=K_{\infty}u^{2}, & x\in \mathbb {R}^{3}, \end{cases}$$
(5.10)

has a positive ground state solution $$(u_{\lambda},\phi_{u_{\lambda}}) \in H^{1}(\mathbb {R}^{3})\times D^{1,2}(\mathbb {R}^{3})$$, that is, for any $$\lambda \in[\frac{1}{2}, 1]$$,

$$m_{\lambda}^{\infty}:=\operatorname{inf}\bigl\{ I_{\lambda}^{\infty}(u),u \in\mathcal {M}_{\lambda}^{\infty}\bigr\}$$
(5.11)

is achieved at $$u_{\lambda}\in\mathcal{M}_{\lambda}^{\infty} \triangleq\{u\in H^{1}(\mathbb {R}^{3})\backslash\{0\}: G_{\lambda}^{\infty }(u)=0\}$$, where

\begin{aligned} G_{\lambda}^{\infty}(u) =&a \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+2 \int_{\mathbb {R}^{3}}u^{2}+ b\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2} \\ &{}- \frac{7\lambda}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac {p+7}{2(p+1)}\lambda \int_{\mathbb {R}^{3}}f_{\infty} \vert u\vert ^{p+1}. \end{aligned}
(5.12)

### Lemma 5.8

Assume that ($$f_{1}$$)-($$f_{2}$$) and ($$K_{2}$$)-($$K_{3}$$) hold and that $$K\not\equiv K_{\infty}$$ or $$f\not \equiv f_{\infty}$$. Then $$c_{\lambda}< m_{\lambda}^{\infty}$$ for any $$\lambda\in J$$.

### Proof

Let $$u_{\lambda}$$ be the minimizer of $$m_{\lambda }^{\infty}$$. By Lemma 5.3 we see that $$I_{\lambda}^{\infty}(u_{\lambda})= \max _{t>0 } I_{\lambda}^{\infty}(t^{\frac{1}{2}}u_{\lambda}(\frac{x}{t}))$$. Then, choosing $$v(x)=(u_{\lambda})_{t_{0}}$$ for $$t_{0}$$ large enough in Lemma 5.7(i), we have that, for any $$\lambda\in J$$, there exists $$\widehat{t}\in(0,t_{0})$$ such that

\begin{aligned} c_{\lambda} =&\operatorname{inf}_{\gamma\in\Gamma_{\lambda}}\max _{t\in[0,1]}I_{\lambda} \bigl(\gamma(t)\bigr) \\ \leqslant&\max _{0< t< t_{0} } I_{\lambda}\bigl((u_{\lambda})_{t_{0}}\bigr) \\ =&I_{\lambda} \bigl((u_{\lambda})_{\widehat{t}}\bigr). \end{aligned}

By the assumptions on K and f, since $$u_{\lambda}>0$$, we derive

$$I_{\lambda}\bigl((u_{\lambda})_{\widehat{t}}\bigr)< I_{\lambda}^{\infty } \bigl((u_{\lambda})_{\widehat{t}}\bigr).$$

Thus, we conclude that

$$c_{\lambda}< I_{\lambda}^{\infty}\bigl((u_{\lambda})_{\widehat{t}} \bigr)\leqslant I_{\lambda}^{\infty}(u_{\lambda})=m_{\lambda}^{\infty}.$$

□

To prove that the functional $$I_{\lambda}$$ satisfies $$(\mathrm{PS})_{c_{\lambda }}$$ condition for a.e. $$\lambda\in J$$, we need the following global compactness lemma, which is suitable for Kirchhoff equations.

### Lemma 5.9

Assume that ($$f_{1}$$), ($$f_{2}$$), and ($$K_{3}$$) hold. Then for $$c > 0$$, $$\lambda\in J$$, and a bounded $$(\mathrm{PS})_{c}$$ sequence $$\{ u_{n}\}\subset H^{1}(\mathbb {R}^{3})$$ of $$I_{\lambda}$$, there exist $$u\in H^{1}(\mathbb {R}^{3})$$ and $$A \in \mathbb {R}$$ such that $$J'_{\lambda}(u)=0$$, where

\begin{aligned} J_{\lambda}(u) =&\frac{a+bA^{2}}{2} \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+ \frac{1}{2} \int_{\mathbb {R}^{3}}u^{2} \\ &{}-\frac{\lambda}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}- \frac{\lambda }{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}. \end{aligned}
(5.13)

Moreover, there exists a finite (possibly, empty) set $$\{ w^{1},w^{2},\ldots, w^{l}\}\subset H^{1}(\mathbb {R}^{3})$$ of nontrivial positive solutions of

$$-\bigl(a+bA^{2}\bigr)\triangle u+u-\lambda K_{\infty}\phi_{u}u=\lambda f_{\infty } \vert u \vert ^{p-1}u$$
(5.14)

and $$\{y_{n}^{1},y_{n}^{2},\ldots,y_{n}^{l}\}\subset \mathbb {R}^{3}$$ such that

\begin{aligned}& \bigl\vert y_{n}^{k}\bigr\vert \rightarrow\infty,\qquad \bigl\vert y_{n}^{k}-y_{n}^{k'}\bigr\vert \rightarrow\infty,\quad k\neq k', n\rightarrow\infty, \\& c+\frac{bA^{4}}{4}=J_{\lambda}(u)+\sum_{k=1}^{l}J_{\lambda}^{\infty} \bigl(w^{k}\bigr), \end{aligned}

and

\begin{aligned}& \Biggl\Vert u_{n}-u-\sum_{k=1}^{l}w^{k} \bigl(\cdot-y_{n}^{k}\bigr)\Biggr\Vert \rightarrow0, \\& A^{2}=\vert \nabla u\vert _{2}^{2}+\sum _{k=1}^{l}\bigl\vert \nabla w^{k}\bigr\vert _{2}^{2}, \end{aligned}

where

$$J_{\lambda}^{\infty}(u)=\frac{a+bA^{2}}{2} \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}+ \frac{1}{2} \int_{\mathbb {R}^{3}}u^{2}- \frac{\lambda}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{u}u^{2}- \frac{\lambda }{p+1} \int_{\mathbb {R}^{3}} f_{\infty} \vert u\vert ^{p+1}.$$
(5.15)

### Proof

The proof is similar to that of Lemma 3.4 in . Since $$\{u_{n}\}$$ is bounded in $$H^{1}( \mathbb {R}^{3})$$, there exist $$u\in H^{1}( \mathbb {R}^{3})$$ and $$A\in \mathbb {R}$$ such that

$$u_{n}\rightharpoonup{u}\quad \mbox{in } H^{1} \bigl(\mathbb {R}^{3}\bigr) \quad \mbox{and}\quad \int_{ \mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \rightarrow A^{2}.$$
(5.16)

Moreover, using ($$f_{1}$$)-($$f_{2}$$) and ($$K_{2}$$)-($$K_{3}$$), we have

$$K\phi_{u_{n}}u_{n}\rightarrow K \phi_{u}u \quad \mbox{and}\quad f\vert u_{n}\vert ^{p-1}u_{n} \rightarrow f\vert u\vert ^{p-1}u \quad \mbox{in } H^{-1}\bigl( \mathbb {R}^{3}\bigr).$$
(5.17)

Therefore, $$I'_{\lambda}(u_{n})\rightarrow0$$ implies that

\begin{aligned}& \int_{ \mathbb {R}^{3}}(a\nabla u\nabla\varphi+u\varphi)+bA^{2} \int_{ \mathbb {R}^{3}}\nabla u\nabla\varphi-\lambda \int_{ \mathbb {R}^{3}}K(x)\phi _{u}u\varphi \\& \quad {}-\lambda \int_{ \mathbb {R}^{3}}f(x)\vert u\vert ^{p-1}u\varphi=0,\quad \forall \varphi\in H^{1}\bigl( \mathbb {R}^{3}\bigr), \end{aligned}

that is, $$J'_{\lambda}(u)=0$$. Since

\begin{aligned} J_{\lambda}(u_{n}) =&\frac{a+bA^{2}}{2} \int_{ \mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2}+ \frac{1}{2} \int_{ \mathbb {R}^{3}}\vert u_{n}\vert ^{2} \\ &{}-\frac{\lambda}{4} \int_{ \mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2}- \frac {\lambda}{p+1} \int_{ \mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1} \\ =&\frac{a}{2} \int_{ \mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2}+ \frac {1}{2} \int_{ \mathbb {R}^{3}}\vert u_{n}\vert ^{2}+ \frac{b}{4}\biggl( \int_{ \mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \biggr)^{2} \\ &{}- \frac{\lambda}{4} \int_{ \mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} \\ &{}-\frac{\lambda}{p+1} \int_{ \mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1}+ \frac{bA^{2}}{4} \int_{ \mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2}+o(1) \\ =&I_{\lambda}(u_{n})+\frac{bA^{4}}{4}+o(1) \end{aligned}

and

\begin{aligned} \bigl\langle J'_{\lambda}(u_{n}),\varphi\bigr\rangle =&\bigl(a+bA^{2}\bigr) \int_{ \mathbb {R}^{3}}\nabla u_{n}\nabla\varphi+ \int_{ \mathbb {R}^{3}} u_{n}\varphi \\ &{}-\lambda \int_{ \mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}\varphi-\lambda \int_{ \mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p-1}u_{n} \varphi \\ =&a \int_{ \mathbb {R}^{3}}\nabla u_{n}\nabla\varphi+ \int_{ \mathbb {R}^{3}} u_{n}\varphi +b \int_{ \mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2} \int_{ \mathbb {R}^{3}}\nabla u_{n}\nabla\varphi \\ &{}-\lambda \int_{ \mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}\varphi-\lambda \int_{ \mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p-1}u_{n} \varphi+o(1) \\ =&\bigl\langle I'_{\lambda}(u_{n}),\varphi\bigr\rangle +o(1), \end{aligned}

we conclude that

$$J_{\lambda}(u_{n})\rightarrow c+\frac{bA^{4}}{4},\qquad J'_{\lambda }(u_{n})\rightarrow0 \quad \mbox{in } H^{-1}\bigl( \mathbb {R}^{3}\bigr).$$

We next show that

$$\Biggl\Vert u_{n}-u-\sum_{k=1}^{l}w^{k} \bigl(\cdot-y_{n}^{k}\bigr)\Biggr\Vert \rightarrow0,$$

where either $$l=0$$ or $$l>0$$.

Step 1: Set $$u_{n}^{1}=u_{n}-u$$. By (5.15) and the Bresiz-Lieb lemma we obtain that

1. (A.1)

$$\vert \nabla u_{n}^{1}\vert _{2}^{2}=\vert \nabla u_{n}\vert _{2}^{2}-\vert \nabla u\vert _{2}^{2}+o(1)$$,

2. (B.1)

$$\vert u_{n}^{1}\vert _{2}^{2}=\vert u_{n}\vert _{2}^{2}-\vert u\vert _{2}^{2}+o(1)$$,

3. (C.1)

$$J_{\lambda}^{\infty}(u_{n}^{1})\rightarrow c+\frac {bA^{4}}{4}-J_{\lambda}(u)$$, and

4. (D.1)

$$(J_{\lambda}^{\infty})'(u_{n}^{1})\rightarrow0$$ in $$H^{-1}( \mathbb {R}^{3})$$.

Let

$$\sigma^{1}:= \limsup _{n\rightarrow+\infty} \sup _{y\in \mathbb {R}^{3}} \int_{B_{1}(y)}\bigl\vert u_{n}^{1}\bigr\vert ^{2}.$$

Similarly to (4.10), we have that

$$\int_{ \mathbb {R}^{3}}\phi_{u_{n}^{1}}u_{n}^{1} \varphi\rightarrow0.$$
(5.18)

Vanishing: If $$\sigma^{1}=0$$, then it follows from Lemma 1.21 of  that $$u_{n}^{1}\rightarrow0$$ in $$L^{s}( \mathbb {R}^{3})$$ for $$s\in(2,2^{*})$$. Thus, $$\int_{ \mathbb {R}^{3}}\vert u_{n}^{1}\vert ^{p-1}u_{n}^{1}\rightarrow0$$ in $$H^{-1}( \mathbb {R}^{3})$$. Since $$(J_{\lambda}^{\infty})'(u_{n}^{1})\rightarrow0$$, we get that $$u_{n}^{1}\rightarrow0$$ in $$H^{1}( \mathbb {R}^{3})$$, and the proof is completed.

Nonvanishing: If $$\sigma^{1}>0$$, then there exists a sequence $$\{ y_{n}^{1}\}\in \mathbb {R}^{3}$$ such that

$$\int_{B_{1}(y_{n}^{1})}\bigl\vert u_{n}^{1}\bigr\vert ^{2}>\frac{\sigma^{1}}{2}.$$

Set $$w_{n}^{1}\triangleq u_{n}^{1}(\cdot+y_{n}^{1})$$. Then $$\{ w_{n}^{1}\}$$ is bounded, and we may assume that $$w_{n}^{1}\rightharpoonup w^{1}$$ in $$H^{1}( \mathbb {R}^{3})$$. Hence $$(J_{\lambda}^{\infty})'(w^{1})=0$$. As

$$\int_{B_{1}(0)}\bigl\vert w_{n}^{1}\bigr\vert ^{2}>\frac{\sigma^{1}}{2},$$

we see that $$w^{1}\neq0$$. Moreover, $$u_{n}^{1}\rightharpoonup0$$ in $$H^{1}( \mathbb {R}^{3})$$ implies that $$\{y_{n}^{1}\}$$ is unbounded. Hence, we may assume that $$\vert y_{n}^{1}\vert \rightarrow+\infty$$.

Step 2: Setting $$u_{n}^{2}=u_{n}-u-w^{1}(\cdot-y_{n}^{1})$$, we can similarly get that

1. (A.2)

$$\vert \nabla u_{n}^{2}\vert _{2}^{2}=\vert \nabla u_{n}\vert _{2}^{2}-\vert \nabla u\vert _{2}^{2}-\vert \nabla w^{1} \vert _{2}^{2}+o(1)$$,

2. (B.2)

$$\vert u_{n}^{2}\vert _{2}^{2}=\vert u_{n}\vert _{2}^{2}-\vert u\vert _{2}^{2}-\vert w^{1} \vert _{2}^{2}+o(1)$$,

3. (C.2)

$$J_{\lambda}^{\infty}(u_{n}^{2})\rightarrow c+\frac {bA^{4}}{4}-J_{\lambda}(u)-J_{\lambda}^{\infty}(w^{1})$$,

4. (D.2)

$$(J_{\lambda}^{\infty})'(u_{n}^{2})\rightarrow0$$ in $$H^{-1}( \mathbb {R}^{3})$$.

Similarly to the arguments in Step 1, let

$$\sigma^{2}:= \limsup _{n\rightarrow+\infty} \sup _{y\in \mathbb {R}^{3}} \int_{B_{1}(y)}\bigl\vert u_{n}^{2}\bigr\vert ^{2}.$$

If vanishing occurs, then $$\Vert u_{n}^{2}\Vert \rightarrow0$$, that is, $$\Vert u_{n}-u-w^{1}(\cdot-y_{n}^{1})\Vert \rightarrow0$$. Moreover, it follows from (5.16) and from (A.2) and (C.2) that

\begin{aligned}& A^{2}=\vert \nabla u\vert _{2}^{2}+\bigl\vert \nabla w^{1}\bigr\vert _{2}^{2},\\& c+ \frac{bA^{4}}{4}=J_{\lambda}(u)+J_{\lambda}^{\infty} \bigl(w^{1}\bigr). \end{aligned}

If nonvanishing occurs, then there exist a sequence $$\{y_{n}^{2}\}\in \mathbb {R}^{3}$$ and a nontrivial $$w^{2}\in H^{1}( \mathbb {R}^{3})$$ such that $$w_{n}^{2}\triangleq u_{n}^{2}(\cdot+y_{n}^{2}) \rightharpoonup w^{2}$$ in $$H^{1}( \mathbb {R}^{3})$$. Then by (D.2) we have that $$(J_{\lambda }^{\infty})'(w^{2})=0$$. Furthermore, $$u_{n}^{2}\rightharpoonup0$$ in $$H^{1}( \mathbb {R}^{3})$$ implies that $$\vert y_{n}^{2}\vert \rightarrow\infty$$ and $$\vert y_{n}^{2}-y_{n}^{1}\vert \rightarrow\infty$$.

We next proceed by iteration. Recall that if $$w^{k}$$ is a nontrivial solution of $$I_{\lambda}^{\infty}$$, then $$I_{\lambda}^{\infty}(w^{k})>0$$. So there exists some finite $$l\in \mathbb {N}$$ such that only the vanishing case occurs in Step l. Then the lemma is proved. □

### Lemma 5.10

Assume that ($$f_{1}$$)-($$f_{2}$$) and ($$K_{2}$$)-($$K_{4}$$) hold. For $$\lambda\in J$$, let $$\{u_{n}\}\subset H^{1}(\mathbb {R}^{3})$$ be a bounded $$(\mathrm{PS})_{c_{\lambda }}$$ sequence of $$I_{\lambda}$$. Then there exists a nontrivial $$u_{\lambda} \in H^{1}(\mathbb {R}^{3})$$ such that

$$u_{n}\rightarrow u_{\lambda} \quad \textit{in } H^{1}\bigl( \mathbb {R}^{3}\bigr).$$

### Proof

By Lemma 5.9 we know that, for $$\lambda\in J$$, there exist $$u_{\lambda} \in H^{1}(\mathbb {R}^{3})$$ and $$A_{\lambda} \in \mathbb {R}$$ such that

$$u_{n} \rightharpoonup u_{\lambda} \quad \mbox{in } H^{1} \bigl(\mathbb {R}^{3}\bigr),\qquad \int_{\mathbb {R}^{3}}\vert \nabla u_{n} \vert ^{2} \rightarrow A_{\lambda}^{2},\qquad J'_{\lambda}(u_{\lambda})=0.$$

Now we prove that $$u_{n}\rightarrow u_{\lambda}$$ in $$H^{1}(\mathbb {R}^{3})$$, that is, $$\{w_{1},w_{2},\ldots,w_{l}\}=\emptyset$$. By contradiction we assume that there exist a positive integer l and $$\{y_{n}^{k}\}_{k=1}^{l}\subset \mathbb {R}^{3}$$ with $$\vert y_{n}^{k}\vert \rightarrow\infty$$ as $$n\rightarrow \infty$$ for each $$1\leqslant k \leqslant l$$, and nontrivial positive solutions $$w^{1},w^{2},\ldots, w^{l}$$ of problem (5.14) such that

\begin{aligned}& c_{\lambda}+\frac{bA_{\lambda}^{4}}{4}=J_{\lambda}(u_{\lambda})+\sum _{k=1}^{l}J_{\lambda}^{\infty} \bigl(w^{k}\bigr), \\& \Biggl\Vert u_{n}-u_{\lambda}-\sum _{k=1}^{l}w^{k}\bigl(\cdot -y_{n}^{k}\bigr)\Biggr\Vert \rightarrow0, \end{aligned}

and

$$A_{\lambda}^{2}=\vert \nabla u_{\lambda} \vert _{2}^{2}+\sum_{k=1}^{l} \bigl\vert \nabla w^{k}\bigr\vert _{2}^{2}.$$
(5.19)

In the weak sense, the equation $$J'_{\lambda}(u_{\lambda})=0$$ can be written as

\begin{aligned} -\bigl(a+bA_{\lambda}^{2}\bigr)\Delta u_{\lambda}+u_{\lambda}-\lambda K(x)\phi_{u_{\lambda}}u_{\lambda}= \lambda f(x)\vert u_{\lambda} \vert ^{p-1}u_{\lambda}, \end{aligned}
(5.20)

where $$\phi_{u_{\lambda}}= \frac{1}{4\pi}\int_{\mathbb {R}^{3}}\frac {K(y)u_{\lambda}^{2}(y)}{\vert x-y\vert }\,dy$$. Denote

$$\textstyle\begin{cases} \alpha\triangleq a\int_{\mathbb {R}^{3}}\vert \nabla u_{\lambda} \vert ^{2},\qquad \beta\triangleq\int_{\mathbb {R}^{3}}\vert u_{\lambda} \vert ^{2},\qquad \mu\triangleq bA_{\lambda}^{2}\int_{\mathbb {R}^{3}}\vert \nabla u_{\lambda} \vert ^{2},\\ \nu\triangleq\int_{\mathbb {R}^{3}}K(x)\phi_{u_{\lambda}}u_{\lambda}^{2},\qquad \delta\triangleq\int_{\mathbb {R}^{3}}f(x)\vert u_{\lambda} \vert ^{p+1},\\ \overline{\nu}\triangleq\int_{\mathbb {R}^{3}}(\nabla K,x)\phi_{u_{\lambda }}u_{\lambda}^{2},\qquad \overline{\delta}\triangleq\int_{\mathbb {R}^{3}}(\nabla f,x)\vert u_{\lambda} \vert ^{p+1}. \end{cases}$$

It is clear that α, β, μ are nonnegative. Then ($$f_{1}$$)-($$f_{2}$$) and ($$K_{3}$$) imply that δ, δ̅, ν are nonnegative too. Then the Pohožaev identity of (5.20) and $$\langle J'_{\lambda }(u_{\lambda}),u_{\lambda}\rangle=0$$ can be written as follows:

$$\textstyle\begin{cases} \frac{1}{2}\alpha+\frac{3}{2}\beta+\frac{1}{2}\mu-\frac{5\lambda }{4}\nu-\frac{3\lambda}{p+1}\delta-\frac{\lambda}{4}\overline{\nu }-\frac{\lambda}{p+1}\overline{\delta}=0,\\ \frac{1}{2}\alpha+\frac{1}{2}\beta+\frac{1}{4}\mu-\frac{\lambda }{4}\nu-\frac{\lambda}{p+1}\delta=J_{\lambda}(u_{\lambda})-\frac {1}{4}\mu. \end{cases}$$

Note that ($$K_{4}$$) implies that $$\nu+\overline{\nu}$$ is nonnegative. Then we conclude

$$3\biggl(J_{\lambda}(u_{\lambda})-\frac{1}{4}\mu\biggr)=\alpha+ \frac{1}{4}\mu+\frac {\lambda}{2}(\nu+\overline{\nu})+\frac{\lambda}{p+1} \bar{\delta}\geqslant0.$$

Hence, we derive

$$J_{\lambda}(u_{\lambda})\geqslant\frac{1}{4}bA_{\lambda}^{2} \int_{\mathbb {R}^{3}}\vert \nabla u_{\lambda} \vert ^{2}.$$
(5.21)

On the other hand, for each nontrivial positive solution $$w^{k}$$ ($$k=1,2,\ldots, l$$) of problem (5.14), we have the following Pohožaev identity:

\begin{aligned} \widetilde{P}_{\lambda}^{\infty}\bigl(w^{k}\bigr) \triangleq&\frac{a+bA_{\lambda }^{2}}{2} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2}+\frac {3}{2} \int_{\mathbb {R}^{3}}\bigl\vert w^{k}\bigr\vert ^{2} \\ &{}-\frac{5\lambda}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{w^{k}}\bigl\vert w^{k}\bigr\vert ^{2}-\frac{3\lambda}{p+1} \int_{\mathbb {R}^{3}}f_{\infty}\bigl\vert w^{k}\bigr\vert ^{p+1}=0. \end{aligned}

It follows from (5.19) that

\begin{aligned} 0 =&\frac{1}{2}\bigl\langle \bigl(J_{\lambda}^{\infty} \bigr)'\bigl(w^{k}\bigr),w^{k} \bigr\rangle + \widetilde{P}_{\lambda}^{\infty}\bigl(w^{k}\bigr) \\ =&\bigl(a+bA_{\lambda}^{2}\bigr) \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2}+2 \int_{\mathbb {R}^{3}}\bigl\vert w^{k}\bigr\vert ^{2} \\ &{}-\frac{7\lambda}{4} \int_{\mathbb {R}^{3}}K_{\infty}\phi_{w^{k}}\bigl\vert w^{k}\bigr\vert ^{2}-\frac{(p+7)\lambda}{2(p+1)} \int_{\mathbb {R}^{3}}f_{\infty }\bigl\vert w^{k}\bigr\vert ^{p+1} \\ \geqslant& G_{\lambda}^{\infty}\bigl(w^{k}\bigr). \end{aligned}

Then there exists $$t_{k}\in(0,1]$$ such that $$(w^{k})_{t_{k}}\in\mathcal {M}_{\lambda}^{\infty}$$, that is, $$G_{\lambda}^{\infty}((w^{k})_{t_{k}})=0$$. By direct calculation we obtain

\begin{aligned} J_{\lambda}^{\infty} \bigl(w^{k}\bigr) =&\frac{a(p+3)}{2(p+7)} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2}+\frac{p-1}{2(p+7)} \int_{\mathbb {R}^{3}}\bigl\vert w^{k}\bigr\vert ^{2}+\frac{b(p-1)}{4(p+7)}A_{\lambda }^{2} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2} \\ &{}+\frac{(7-p)\lambda}{4(p+7)} \int_{\mathbb {R}^{3}}K_{\infty}\phi _{w^{k}}\bigl\vert w^{k}\bigr\vert ^{2}+\frac{bA_{\lambda}^{2}}{4} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2} \\ \geqslant&\frac{a(p+3)}{2(p+7)}t_{k}^{2} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2}+\frac{p-1}{2(p+7)}t_{k}^{4} \int_{\mathbb {R}^{3}}\bigl\vert w^{k}\bigr\vert ^{2}+\frac{b(p-1)}{4(p+7)}t_{k}^{4}\biggl( \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2}\biggr)^{2} \\ &{}+\frac{(7-p)\lambda}{4(p+7)}t_{k}^{7} \int_{\mathbb {R}^{3}}\phi _{w^{k}}\bigl\vert w^{k} \bigr\vert ^{2}+\frac{bA_{\lambda}^{2}}{4} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2} \\ =&I_{\lambda}^{\infty}\bigl(\bigl(w^{k} \bigr)_{t_{k}}\bigr)-\frac{2}{p+7}G_{\lambda}^{\infty } \bigl(\bigl(w^{k}\bigr)_{t_{k}}\bigr) +\frac{bA_{\lambda}^{2}}{4} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2} \\ \geqslant& m_{\lambda}^{\infty}+\frac{bA_{\lambda}^{2}}{4} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2}. \end{aligned}
(5.22)

By (5.19)-(5.22) we have

\begin{aligned} c_{\lambda}+\frac{bA_{\lambda}^{4}}{4} =&J_{\lambda}(u_{\lambda})+\sum _{k=1}^{l}J_{\lambda}^{\infty} \bigl(w^{k}\bigr) \\ \geqslant&\frac{bA_{\lambda}^{2}}{4} \int_{\mathbb {R}^{3}}\vert \nabla u_{\lambda} \vert ^{2}+lm_{\lambda}^{\infty}+ \frac{bA_{\lambda }^{2}}{4}\sum_{k=1}^{l} \int_{\mathbb {R}^{3}}\bigl\vert \nabla w^{k}\bigr\vert ^{2} \\ \geqslant& m_{\lambda}^{\infty}+\frac{bA_{\lambda}^{4}}{4}. \end{aligned}

Hence we get that $$c_{\lambda}\geqslant m_{\lambda}^{\infty}$$, which contradicts to Lemma 5.8. So $$u_{n}\rightarrow u_{\lambda}$$ in $$H^{1}(\mathbb {R}^{3})$$. As a consequence, we obtain that $$I_{\lambda }'(u_{\lambda})=0$$ and $$I_{\lambda}(u_{\lambda})=c_{\lambda}$$. □

To prove that problem (1.4) has a positive ground state solution, we define

$$m:=\operatorname{inf}\bigl\{ I_{1}(u),u\in\mathcal{S}\bigr\} ,\quad \mbox{where } \mathcal{S}:=\bigl\{ u\in H^{1}\bigl(\mathbb {R}^{3}\bigr) \backslash\{0\}, I_{1}'(u)=0\bigr\} .$$

### Proof of Theorem 2.2

We complete the proof in three steps.

Step 1. $$\mathcal{S}\neq\emptyset$$.

By Lemma 5.7, Lemma 5.10, and Proposition 3.1, for a.e. $$\lambda\in J$$, there exists $$u_{\lambda}\in H^{1}(\mathbb {R}^{3})\backslash\{0\}$$ such that $$I_{\lambda }'(u_{\lambda})=0$$ and $$I_{\lambda}(u_{\lambda})=c_{\lambda}$$. Choosing a sequence $$\{\lambda_{n}\}\subset J$$ satisfying $$\lambda _{n}\rightarrow1^{-}$$, we have a sequence $$\{u_{{\lambda}_{n}}\}\subset H^{1}(\mathbb {R}^{3})\backslash\{0\}$$ such that $$I_{{\lambda}_{n}}'(u_{{\lambda}_{n}})=0$$ and $$I_{{\lambda }_{n}}(u_{{\lambda}_{n}})=c_{{\lambda}_{n}}$$. For simplicity, denoting $$\{u_{n}\}$$ instead of $$\{u_{{\lambda}_{n}}\}$$, we next show that $$\{u_{n}\}$$ is bounded in $$H^{1}(\mathbb {R}^{3})$$.

Denote

$$\textstyle\begin{cases} \alpha_{n}\triangleq a\int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2},\qquad \beta_{n} \triangleq\int_{\mathbb {R}^{3}}\vert u_{n}\vert ^{2},\qquad \mu_{n} \triangleq b(\int_{\mathbb {R}^{3}}\vert \nabla u_{n}\vert ^{2})^{2},\\ \nu_{n} \triangleq\int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2},\qquad \delta _{n}\triangleq\int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1},\\ \overline{\nu}_{n}\triangleq\int_{\mathbb {R}^{3}}(\nabla K,x)\phi _{u_{n}}u_{n}^{2},\qquad \overline{\delta}_{n}\triangleq\int_{\mathbb {R}^{3}}(\nabla f,x)\vert u_{n}\vert ^{p+1}. \end{cases}$$

Then we have

$$\textstyle\begin{cases} \frac{1}{2}\alpha_{n} +\frac{3}{2}\beta_{n}+\frac{1}{2}\mu_{n}-\frac {5\lambda_{n}}{4}\nu_{n} -\frac{3\lambda_{n}}{p+1}\delta_{n}-\frac {\lambda_{n}}{4}\overline{\nu}_{n}-\frac{\lambda_{n}}{p+1}\overline {\delta}_{n}=0,\\ \frac{1}{2}\alpha_{n} +\frac{1}{2}\beta_{n}+\frac{1}{4}\mu_{n}-\frac {\lambda_{n}}{4}\nu_{n} -\frac{\lambda_{n}}{p+1}\delta_{n}=c_{{\lambda _{n}}},\\ \alpha_{n}+\beta_{n}+\mu_{n}-\lambda_{n}\nu_{n}-\lambda_{n}\delta_{n}=0. \end{cases}$$

Hence, we have

$$\alpha_{n} +\frac{\lambda_{n}}{4}\nu_{n} + \frac{(p-1)\lambda _{n}}{2(p+1)}\delta_{n}+\frac{\lambda_{n}}{2}(\nu_{n}+ \overline{\nu }_{n})+\frac{\lambda_{n}}{p+1}\overline{\delta}_{n}=4c_{\lambda _{n}} \leqslant4c_{\frac{1}{2}}.$$
(5.23)

Note that $$\alpha_{n}$$, $$\beta_{n}$$, $$\mu_{n}$$ are nonnegative. Conditions ($$f_{1}$$)-($$f_{2}$$), ($$K_{3}$$), and ($$K_{4}$$) imply that $$\delta_{n}$$, $$\overline {\delta}_{n}$$, $$\nu_{n}$$, $$\nu_{n}+\overline{\nu}_{n}$$ are nonnegative too. Then we conclude that $$\delta_{n}$$ and $$\nu_{n}$$ are bounded. Moreover, by the third equation we have that $$\alpha_{n} +\beta_{n}$$ is bounded, that is, $$\{u_{n}\}$$ is bounded in $$H^{1}(\mathbb {R}^{3})$$. Therefore, we get

\begin{aligned}& \lim _{n\rightarrow\infty}I_{1}(u_{n}) \\& \quad = \lim _{n\rightarrow\infty} \biggl(I_{{\lambda}_{n}}(u_{n}) +({\lambda}_{n}-1) \biggl(\frac{1}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}^{2} +\frac{1}{p+1} \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p+1} \biggr) \biggr) \\& \quad =\lim _{n\rightarrow\infty}c_{{\lambda}_{n}}=c_{1} \end{aligned}

and

\begin{aligned}& \lim _{n\rightarrow\infty}\bigl\langle I'_{1}(u_{n}), \varphi\bigr\rangle \\& \quad = \lim _{n\rightarrow\infty}\bigl\langle I'_{{\lambda }_{n}}(u_{n}), \varphi\bigr\rangle +({\lambda}_{n}-1) \biggl( \int_{\mathbb {R}^{3}}K(x)\phi_{u_{n}}u_{n}\varphi + \int_{\mathbb {R}^{3}}f(x)\vert u_{n}\vert ^{p-1}u_{n} \varphi\biggr) \\& \quad =0, \end{aligned}

that is, $$\{u_{n}\}$$ is a bounded $$(\mathrm{PS})_{c_{1}}$$ sequence for $$I_{1}$$. By Lemma 5.10 there exists $$u_{0}\in H^{1}(\mathbb {R}^{3})\backslash\{0\}$$ such that $$I_{1}'(u_{0})=0$$ and $$I_{1}(u_{0})=c_{1}$$.

Step 2. $$0< m<\infty$$.

It is clear that $$m\leqslant c_{1}<\infty$$. Next, we prove $$m\geqslant0$$. For all $$u\in\mathcal{S}$$, we have $$\langle I_{1}'(u),u\rangle=0$$. Then by a standard argument we get that $$\Vert u\Vert \geqslant\delta$$ for some $$\delta>0$$ (see Lemma 5.4(i)). On the other hand, by the Pohoz̆aev identity of problem (1.4),

\begin{aligned} 0 =&P_{1}(u) \\ \triangleq&\frac{a}{2}{ \int_{\mathbb {R}^{3}}}\vert \nabla u\vert ^{2} + \frac{3}{2}{ \int_{\mathbb {R}^{3}}}u^{2} +\frac{b}{2}\biggl( \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} \biggr)^{2}-\frac {5}{4} \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2} \\ &{}-\frac{1}{4} \int_{\mathbb {R}^{3}}(\nabla K,x)\phi_{u}u^{2}- \frac{3}{p+1} \int _{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1}- \frac{1}{p+1} \int_{\mathbb {R}^{3}}(\nabla f,x)\vert u\vert ^{p+1}, \end{aligned}

where $$\phi_{u}(x)= \frac{1}{4\pi}\int_{\mathbb {R}^{3}}\frac {K(y)u^{2}(y)}{\vert x-y\vert }\,dy$$, we have

\begin{aligned} I_{1}(u) =&I_{1}(u)-\frac{1}{8} \bigl[\bigl\langle I'_{1}(u),u \bigr\rangle +2{P}_{1}(u) \bigr] \\ =&\frac{a}{4} \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2} + \frac {1}{8} \int_{\mathbb {R}^{3}}K(x)\phi_{u}u^{2}+ \frac{p-1}{8(p+1)} \int_{\mathbb {R}^{3}}f(x)\vert u\vert ^{p+1} \\ &{}+\frac{1}{16} \int_{\mathbb {R}^{3}} \bigl(K(x)+(\nabla K,x) \bigr)\phi_{u}u^{2}+ \frac{1}{4(p+1)} \int_{\mathbb {R}^{3}}(\nabla f,x)\vert u\vert ^{p+1}. \end{aligned}

From ($$f_{1}$$), ($$f_{2}$$), and ($$K_{2}$$)-($$K_{4}$$) we infer

$$I_{1}(u)\geqslant\frac{a}{4} \int_{\mathbb {R}^{3}}\vert \nabla u\vert ^{2}.$$
(5.24)

Therefore, we obtain that $$m\geqslant0$$. In the following, we will prove that $$m>0$$. By contradiction, let $$\{u_{n}\}$$ be a $$(\mathrm{PS})_{0}$$ sequence of $$I_{1}$$. Inequality (5.24) implies that $$\lim _{n\rightarrow \infty} \vert \nabla u_{n}\vert ^{2}=0$$. This conclusion, combined with $$\langle I'_{1}(u),u \rangle=0$$, implies that $$\lim _{n\rightarrow\infty} u_{n}^{2}=0$$. Therefore, we obtain $$\lim _{n\rightarrow\infty} \Vert u_{n}\Vert =0$$, a contradiction with $$\Vert u_{n}\Vert \geqslant\delta>0$$ for all n.

Step 3. m is is attained at some $$u\in\mathcal{S}$$.

Let $$\{u_{n}\}\subset\mathcal{S}$$ satisfy $$I_{1}(u_{n})\rightarrow m$$. Using the same arguments as in Step 1, we can deduce that $$\{u_{n}\}$$ is bounded in $$H^{1}(\mathbb {R}^{3})$$, that is, $$\{u_{n}\}$$ is a bounded $$(\mathrm{PS})_{m}$$ sequence of $$I_{1}$$. Similarly to the arguments in Lemma 5.10, there exists a nontrivial point $$u\in H^{1}(\mathbb {R}^{3})$$ such that $$I_{1}(u)=m$$ and $$I_{1}'(u)=0$$. By the standard regularity arguments as in the proof of Theorem 1.4 of  we see that u is a positive ground state solution for problem (1.4). □

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## Acknowledgements

The authors would like to thank the referee for his/her valuable observations, which greatly improved the paper. SW is supported by the Science Council of Shanxi Province (No. 201601D102001).

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Authors

### Contributions

FL participated in the design of the study and drafted the manuscript. SW carried out the theoretical studies and helped to draft the manuscript. Both authors read and approved the final manuscript.

### Corresponding author

Correspondence to Shuli Wang.

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Fengxia Liu and Shuli Wang contributed equally to this work.

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