Least energy sign-changing solutions for Kirchhoff–Poisson systems

Abstract

The paper deals with the following Kirchhoff–Poisson systems:

$$ \textstyle\begin{cases} - ( {1+b\int _{{\mathbb{R}}^{3}} { \vert \nabla u \vert ^{2}\,dx} } ) \Delta u+u+k(x)\phi u+\lambda \vert u \vert ^{p-2}u=h(x) \vert u \vert ^{q-2}u, & x\in {\mathbb{R}}^{3}, \\ -\Delta \phi =k(x)u^{2}, & x\in {\mathbb{R}}^{3}, \end{cases} $$

(0.1)

where the functions k and h are nonnegative, $0\le \lambda , b$; $2\le p\le 4< q<6$. Via a constraint variational method combined with a quantitative lemma, some existence results on one least energy sign-changing solution with two nodal domains to the above systems are obtained. Moreover, the convergence property of $u_{b}$ as $b \searrow 0$ is established.

1 Introduction

Consider the following Kirchhoff–Poisson systems:

$$ \textstyle\begin{cases} - ( {1+b\int _{{\mathbb{R}}^{3}} { \vert \nabla u \vert ^{2}\,dx} } ) \Delta u+u+k(x)\phi u+\lambda \vert u \vert ^{p-2}u=h(x) \vert u \vert ^{q-2}u, & x\in {\mathbb{R}}^{3}, \\ -\Delta \phi =k(x)u^{2}, & x\in {\mathbb{R}}^{3}, \end{cases} $$

(1.1)

where k and h are nonnegative functions, $0\le \lambda , b$; $2 \le p\le 4< q<6$.

When $b=0$, systems (1.1) reduce to the following Schrödinger–Poisson systems:

$$ \textstyle\begin{cases} -\Delta u+u+k(x)\phi u+\lambda \vert u \vert ^{p-2}u=h(x) \vert u \vert ^{q-2}u,& x\in {\mathbb{R}}^{3}, \\ -\Delta \phi =k(x)u^{2}, & x\in {\mathbb{R}}^{3} , \end{cases} $$

(1.2)

which stem from quantum mechanics and have important applications in the semiconductor. From the physical viewpoint, the above systems have been introduced as a physical model describing a charged wave interacting with its own electrostatic field in quantum mechanics. The unknowns u and ϕ represent the wave functions associated to the particle and electric potential. For more details, one can refer to [1,2,3] and the references therein. In [4], Cerami and Vaira studied the following Schrödinger–Poisson systems:

$$ \textstyle\begin{cases} -\Delta u+u+k(x)\phi (x)u=a(x) \vert u \vert ^{p-1}u, & x\in {\mathbb{R}}^{3}, \\ -\Delta \phi =k(x)u^{2}, & x\in {\mathbb{R}}^{3}, \end{cases} $$

(1.3)

with $p\in (3,5)$. Under some suitable conditions, some existence results on positive solutions were obtained. Recently, Zhong and Tang [5] investigated the following Schrödinger–Poisson systems:

$$ \textstyle\begin{cases} -\Delta u+u+k(x)\phi (x)u=\lambda f(x)+ \vert u \vert ^{4}u, & x\in {\mathbb{R}}^{3}, \\ -\Delta \phi =k(x)u^{2}, & x\in {\mathbb{R}}^{3}, \end{cases} $$

where the functions k and f are nonnegative, $0<\lambda <\lambda _{1}$ and $\lambda _{1}$ is the eigenvalue of the problem $-\Delta u+u= \lambda f(x)u$ in ${H}^{1}({\mathbb{R}}^{3})$. Via the variational method, the authors obtained the existence results on the ground state sign-changing solution for $0<\lambda <\lambda _{1}$. Replacing u with $V(x)u$ in (1.3), Batista and Furtado in [6] studied the following systems:

$$ \textstyle\begin{cases} -\Delta u+V(x)u+k(x)\phi u=a(x) \vert u \vert ^{p-1}u,& x\in {\mathbb{R}} ^{3}, \\ -\Delta \phi =k(x)u^{2}, & x\in {\mathbb{R}}^{3}, \end{cases} $$

where $p\in (3,5)$ and $a(x)$ satisfies some mild conditions, especially, the potential function V can be nonconstant and indefinite in sign. By a variational approach, they also get some results of the existence of one nonnegative solution and one sign-changing solution. For the related research on this problem, the reader can refer to the literature [7,8,9,10,11,12,13,14].

On the other hand, if $k=0$ in (1.1), then (1.1) reduce to the following Kirchhoff-type problem:

$$ - \biggl( {1+b \int _{{\mathbb{R}}^{3}} { \vert \nabla u \vert ^{2}\,dx} } \biggr) \Delta u+u+\lambda \vert u \vert ^{p-2}u=h(x) \vert u \vert ^{q-2}u, \quad x\in {\mathbb{R}}^{3}. $$

(1.4)

Associated with the above problem, we have to mention the following Kirchhoff Dirichlet problem:

$$ \textstyle\begin{cases} - ( {a + b\int _{\varOmega }{ \vert \nabla u \vert ^{2} \,dx} } )\Delta u = f(x,u),& x \in \varOmega , \\ u = 0, & x \in \partial \varOmega , \end{cases} $$

which stems from the stationary analogue of the equation

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

proposed by Kirchhoff regarded as an extension of the classical D’Alembert wave equation on free vibrations of elastic strings. Due to its importance on the physical background, the Kirchhoff boundary problem received increasingly more attention. Recently, with the help of the variational methods, a number of results on the existence and multiplicity of solutions for the Kirchhoff problem

$$ \textstyle\begin{cases} - ( {a+b\int _{{\mathbb{R}}^{N}} { \vert \nabla u \vert ^{2}\,dx} } ) \Delta u=f(u),&x\in \varOmega , \\ u=0,&x\in \partial \varOmega , \end{cases} $$

(1.5)

have been established under various suitable conditions, where $\varOmega \subset {\mathbb{R}}^{N}$ is a bounded domain with a smooth boundary ∂Ω and f satisfies various suitable conditions; see, for example [15,16,17,18,19,20,21] and the references therein. Recently, Baraket and Molica Bisci in [22] studied the following Kirchhoff-type problem:

$$ \textstyle\begin{cases} - ( {a+b\int _{\varOmega }{ \vert \Delta u \vert ^{2}} } )\Delta u=\lambda f(x,u)+\mu g(x,u) & \text{in }\varOmega , \\ u=0 & \text{on }\partial \varOmega , \end{cases} $$

where $\varOmega \subset \mathbb{R}^{N}$ ($N\ge 3$) is a bounded open subset. Under some suitable conditions, the authors obtained multiplicity results via applying the three critical points theorem. Very recently, Xu and Chen in [23] investigated the following Kirchhoff-type problem:

$$ \textstyle\begin{cases} - ( {a+b\int _{\mathbb{R}^{3}} { \vert \Delta u \vert ^{2}} } ) \Delta u+Vu=f(u) & \text{in }\mathbb{R}^{3}, \\ u\in H^{1}(\mathbb{R}^{3}), \end{cases} $$

and established the existence of a ground state solution by applying a critical point theorem similar to the mountain pass lemma (see [24]). Moreover, for the related research on fractional Kirchhoff-type and Schrödinger-type problems, the reader can refer to [25, 26] and to the references therein and the monograph [27] published recently. However, regarding the existence of sign-changing solutions for the Kirchhoff problem, there are very few results in the literature. Recently, Shuai in [28] studied the existence of the least energy sign-changing solution of problem (1.5) and its convergence property on $\{ {u_{n} } \}$ as $b\searrow 0$. Later, under conditions different from [28], with the help of some analytical techniques and a non-Nehari manifold method, Tang and Cheng in [29] further studied problem (1.5) and obtained some existence results on a ground state sign-changing solution $u_{b}$ as well as its convergence property.

When $b\ne 0$, and $k\ne 0$, the systems (1.1) stand for Kirchhoff–Poisson systems. Because the nonlocal terms $b \int _{{\mathbb{R}}^{3}} {( \vert \nabla u \vert ^{2}\,dx)\Delta u}$ and $\phi _{u}$ are involved in the equation, the problem is totally different from the case $b=0$ and $k=0$. In [30], Zhang considered the following general singular Kirchhoff–Poisson systems:

$$ \textstyle\begin{cases} - ( {a + b\int _{\varOmega }{ \vert \Delta u \vert ^{2} } } )\Delta u + \phi u = \lambda hf(u) + g(u), &{\text{in }} \varOmega , \\ - \Delta \phi = u^{2} , & {\text{in }} \varOmega , \\ u > 0, & {\text{in }} \varOmega , \\ u = \phi = 0, & {\text{on }} \partial \varOmega , \end{cases} $$

where $\varOmega \subset \mathbb{R}^{3}$ is a smooth bounded domain with boundary ∂Ω, constants $a > 0$, $b \ge 0$ and $\lambda > 0$ is a parameter, functions f, g, h satisfy some conditions. By combining the variational method with a perturbation method, the author obtained the existence of two positive solutions if the parameter λ is small enough. In [31], Liu and Wang investigated the following Kirchhoff–Poisson systems:

$$ \textstyle\begin{cases} - ( {a + b\int _{ \mathbb{R}^{3} } { \vert \Delta u \vert ^{2} } } ) \Delta u + u - qK(x)\phi u = f(x) \vert u \vert ^{p - 1} u,&x \in \mathbb{R} ^{3} , \\ - \Delta \phi = qK(x)u^{2} , & x \in \mathbb{R}^{3}, \end{cases} $$

where the constants $a > 0$, $b \ge 0$, $1 < p < 5$, $q > 0$ and the functions $f,K: \mathbb{R}^{3} \to \mathbb{R}$ are nonnegative. Applying the critical point theorem with parameter λ (see [24]), the authors obtained the existence of a positive solution as well as a ground state solution with $q=1$ corresponding to its limit problem. Moreover, very recently, Wang, Rǎdulescu and Zhang in [32] studied a kind of fractional Kirchhoff–Poisson systems as follows:

$$ \textstyle\begin{cases} M([u]_{s}^{2} )( - \Delta )^{s} u + V(x)u + \phi (x)u = \lambda f(x,u) &{\text{in }} \mathbb{R}^{3} , \\ ( - \Delta )^{t} \phi (x) = u^{2} & {\text{in }} \mathbb{R}^{3} , \end{cases} $$

where $s,t \in (0,1)$ with $2t+4s>3$, $M:\mathbb{R}^{+} _{0} \rightarrow \mathbb{R}^{+}$ is a continuous function satisfying certain assumptions, the potential function $V: \mathbb{R}^{3} \rightarrow \mathbb{R}^{+}$ is continuous, f satisfies a Carathéodory condition, λ is a positive parameter. By applying the fountain theorem for the subcritical case and the symmetric mountain pass theorem for the critical case, respectively, the authors obtained infinitely many solutions for the system. Different from the works mentioned above, in the present paper, we shall combine a constraint variational method with quantitative deformation properties to establish the existence results as regards one least energy sigh-changing solution with two nodal domains to problem (1.1). Moreover, we also study the convergence property on $u_{b}$ as $b\searrow 0$.

2 Preliminaries

Throughout this paper, we always assume the following conditions hold.

$(l)$ :

$0\le \lambda , b$; $2\le p\le 4< q<6$.

$(k)$ :

$k\in L^{2}({\mathbb{R}}^{3})\cap L^{\infty }({\mathbb{R}} ^{3})\backslash \{ 0 \}$ and $k(x)\ge 0$ for a.e. $x\in {\mathbb{R}}^{3}$.

$(h)$ :

$h(x)>0$ for a.e. $x\in {\mathbb{R}}^{3}$ and there exists $q_{1}\in (q,6)$ such that $h\in L^{\frac{6}{6-q_{1}}}$.

In addition, ${\mathbb{R}}_{+} =[0,\infty )$, ${\mathbb{R}}_{+}^{0} =(0, \infty )$, $D^{1,2}({\mathbb{R}}^{3})$ is the Sobolev space equipped with the norm $\Vert u \Vert _{D^{1,2}} = ( {\int _{{\mathbb{R}} ^{3}} { \vert \nabla u \vert ^{2}\,dx} } )^{\frac{1}{2}}$ and $L^{s}$ is the Lebesgue space with norm $\vert u \vert _{s} = ( {\int _{{\mathbb{R}}^{3}} { \vert u \vert ^{s}\,dx} } )^{\frac{1}{s}}$ for $s\ge 1$. Also, $H^{1}({\mathbb{R}}^{3})$ is the Sobolev space equipped with the norm

$$ \Vert u \Vert = \biggl( { \int _{{\mathbb{R}}^{3}} { \bigl( \vert \nabla u \vert ^{2}+u^{2} \bigr)\,dx} } \biggr)^{\frac{1}{2}}. $$

C is for various positive constants, which can be different from one line to another line in the text.

Let S̄ be the best Sobolev constant for the embedding of $D^{1,2}({\mathbb{R}}^{3})$ in $L^{6}({\mathbb{R}}^{3})$. That is,

$$ \bar{S}=\inf_{u\in D^{1,2}({\mathbb{R}}^{3})\backslash \{ 0 \} } \frac{ \Vert u \Vert _{D^{1,2}} }{ \vert u \vert _{6} }. $$

(2.1)

Similarly,

$$ S_{r} =\inf_{u\in H^{1}({\mathbb{R}}^{3})\backslash \{ 0 \} } \frac{ ( {\int _{{\mathbb{R}}^{3}} {( \vert \nabla u \vert ^{2}+u ^{2})\,dx} } )^{\frac{1}{2}}}{ \vert u \vert _{r} }, \quad r\in [1,6]. $$

(2.2)

For any fixed $u\in H^{1}({\mathbb{R}}^{3})$, from the Lax–Milgram theorem it follows that there exists an unique $\phi _{u} \in D^{1,2}( {\mathbb{R}}^{3})$ that satisfies $-\Delta \phi =k(x)u^{2}$ weakly, that is, for any $v\in D^{1,2}({\mathbb{R}}^{3})$,

$$ \int _{{\mathbb{R}}^{3}} {\nabla \phi _{u} \cdot \nabla v\,dx} = \int _{{\mathbb{R}}^{3}} {k(x)u^{2}v\,dx} . $$

Moreover,

$$ \phi _{u} (x)=\frac{1}{4\pi } \int _{{\mathbb{R}}^{3}} {\frac{k(y)u^{2}(y)}{ \vert x-y \vert }\,dy}. $$

(2.3)

Let

$$ L_{\phi _{u} } (v)= \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} v^{2}\,dx} , \quad v\in H^{1} \bigl({\mathbb{R}}^{3} \bigr), $$

then

$$ L_{\phi _{u} } (v)=\frac{1}{4\pi } \int _{{\mathbb{R}}^{3}} { \int _{{\mathbb{R}} ^{3}} {\frac{k(x)k(y)u^{2}(x)v^{2}(y)}{ \vert x-y \vert }\,dx\,dy} }. $$

(2.4)

Clearly, the energy functional associated with (1.1) can be expressed by

$$ \begin{aligned}[b] J_{b} (u)={}&\frac{1}{2} \Vert u \Vert ^{2}+\frac{b}{4} \vert \nabla u \vert _{2}^{4} +\frac{1}{4} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} (x)u^{2}\,dx} \\&{}+ \frac{\lambda }{p} \int _{{\mathbb{R}}^{3}} { \vert u \vert ^{p}\,dx} - \frac{1}{q} \int _{{\mathbb{R}}^{3}} {h(x) \vert u \vert ^{q}\,dx} . \end{aligned} $$

(2.5)

Lemma 2.1

([5])

Suppose that $k\in L^{\infty }( {\mathbb{R}}^{3})$. Then, for any $u\in H^{1}({\mathbb{R}}^{3})$, there exists $C>0$ such that

$$ L_{\phi _{u} } (u)= \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} u^{2}\,dx} = \int _{{\mathbb{R}}^{3}} { \vert \nabla \phi _{u} \vert ^{2}\,dx} \le C \Vert u \Vert ^{4}. $$

Lemma 2.2

Assume that condition $(k)$ holds. Then we have

• (i) $\phi _{u} \ge 0$, for any $u\in H^{1}({\mathbb{R}}^{3})$;

• (ii) for any $t\in R$, $\phi _{tu} =t^{2}\phi _{u}$;

• (iii) $\Vert \phi _{u} \Vert _{D^{1,2}} \le {\bar{S}} ^{-1}S_{6}^{-2} \vert k \vert _{2} \Vert u \Vert ^{2}$;

• (iv) $\vert \phi _{u} \vert _{6} \le \bar{S}^{-1} \Vert \phi _{u} \Vert _{D^{1,2}}$.

Proof

Under the condition $(k)$, the conclusions (i) and (ii) directly follow from Eq. (2.3). the conclusions (iii) and (iv) directly follow from (2.4) in [4]. □

For $R>0$, let $\varOmega _{R} =\{x\in {\mathbb{R}}^{3}: \vert x \vert \le R\}$, $\varOmega _{R}^{C} =\{x\in {\mathbb{R}}^{3}: \vert x \vert >R \}$. Denote

$$ u^{+}(x)=\max \bigl\{ u(x),0 \bigr\} , u^{-}(x)=\min \bigl\{ u(x),0 \bigr\} . $$

Lemma 2.3

Assume that conditions $(l)$, $(k)$ and $(h)$ hold. Then, for any $\{ {u_{n} } \} \subset H ^{1}({\mathbb{R}}^{3})$ with $u^{\pm }_{n} \rightharpoonup u^{\pm }$ weakly in $H^{1}({\mathbb{R}}^{3})$ and $u_{n}^{\pm }(x)\stackrel{ {}}{\longrightarrow }u^{\pm }(x)$ for a.e. $x\in {\mathbb{R}} ^{3}$, we have

$$\begin{aligned}& \begin{aligned} \mathrm{(i)} \quad \mathrm{(a)}&\quad \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{n} } u_{n} ^{2}\,dx} \to \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} u^{2}\,dx},\\& \quad \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{n}^{\pm }} \bigl(u_{n}^{ \pm } \bigr) ^{2}\,dx} \to \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{\pm }} \bigl(u ^{\pm } \bigr)^{2}\,dx}, \\ \mathrm{(b)}&\quad \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u _{n} } \bigl(u_{n}^{\pm } \bigr)^{2}\,dx} \to \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} \bigl(u^{\pm } \bigr)^{2}\,dx}, \\ \mathrm{(c)}&\quad \int _{{\mathbb{R}}^{3}} {k(x) \phi _{u_{n}^{+} } \bigl(u_{n}^{-} \bigr)^{2}\,dx} \to \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx}, \\ \mathrm{(d)}&\quad \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u _{n} } u_{n} \varphi \,dx} \to \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} u \varphi \,dx} ,\quad \forall \varphi \in H^{1} \bigl({\mathbb{R}}^{3} \bigr). \end{aligned} \\& \begin{aligned} \mathrm{(ii)}\quad \mathrm{(a)} &\quad \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{n} \vert ^{q} \,dx} \to \int _{{\mathbb{R}}^{3}} {h(x) \vert u \vert ^{q}\,dx}, \\ {\mathrm{(b)}} &\quad \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{n} \vert ^{q-2}u _{n} \varphi \,dx} \to \int _{{\mathbb{R}}^{3}} {h(x) \vert u \vert ^{q-2}u \varphi \,dx} , \quad \forall \varphi \in H^{1} \bigl({\mathbb{R}}^{3} \bigr), \\ {\mathrm{(c)}} &\quad \int _{{\mathbb{R}}^{3}} { \vert u_{n} \vert ^{p-2}u_{n} \varphi \,dx} \to \int _{{\mathbb{R}}^{3}} { \vert u \vert ^{p-2}u\varphi \,dx} , \quad \forall \varphi \in H^{1} \bigl({\mathbb{R}}^{3} \bigr). \end{aligned} \end{aligned}$$

Proof

Item (i) Conclusions (a) and (d) follow from Lemma 6 in [5] and Lemma 2.1 in [6] and Lemma 2.1 in [4]. Conclusion (b) and (c) can be similarly proved, we omit it.

Item (ii). We only prove that $\int _{{\mathbb{R}}^{3}} {h(x) \vert u _{n} \vert ^{q}\,dx} \to \int _{{\mathbb{R}}^{3}} {h(x) \vert u \vert ^{q}\,dx}$. The other relations can be obtained similarly.

In fact, the condition $q< q_{1}<6$ implies that $4<\frac{6q}{q_{1}}<6$. Let $r=\frac{6q}{q_{1}}$. The sequence $u_{n} \rightharpoonup u$ weakly in $H^{1}({\mathbb{R}}^{3})$ shows that $\{ {u_{n} } \}$ is bounded in $H^{1} ({\mathbb{R}}^{3})$. Therefore, by (2.2) there exists $M>0$ such that $\vert u_{n} \vert _{r} ^{q}\le M$, $\vert u \vert _{r}^{q} \le M$. Thus

$$\begin{aligned} \biggl\vert { \int _{\varOmega _{R}^{C} } {h(x) \bigl( \vert u_{n} \vert ^{q}- \vert u \vert ^{q} \bigr)\,dx} } \biggr\vert \le & \int _{\varOmega _{R}^{C} } {h(x) \bigl( \vert u _{n} \vert ^{q}+ \vert u \vert ^{q} \bigr)\,dx} \\ \le & \biggl( { \int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx} } \biggr) ^{\frac{6-q_{1}}{6}} \bigl( \vert u_{n} \vert _{r}^{q} + \vert u \vert _{r}^{q} \bigr) \\ \le & 2M \biggl( { \int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx} } \biggr) ^{\frac{6-q_{1}}{6}}. \end{aligned}$$

Because $h\in L_{\frac{6}{6-q_{1}}}$, we can choose $R>0$ large enough so that $2M ( {\int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx} } )^{\frac{6-q_{1}}{6}}<\varepsilon$, and therefore,

$$ \biggl\vert { \int _{\varOmega _{R}^{C} } {h(x) \bigl( \vert u_{n} \vert ^{q}- \vert u \vert ^{q} \bigr)\,dx} } \biggr\vert < \varepsilon . $$

On the other hand, by the absolute continuity on integral together with $h\in L_{\frac{6}{6-q_{1}}}$, for any $\eta >0$, there is $\delta >0$ such that $( {\int _{G} { \vert h(x) \vert ^{\frac{6}{6-q_{1}}}\,dx} } )^{\frac{6-q_{1}}{6}}<\eta$ for each $G\subset \varOmega _{R}$ with $\operatorname{mes} G<\delta$. Then

$$ \int _{G} {h(x) \vert u_{n} \vert ^{q} \,dx} \le \vert u_{n} \vert _{r}^{q} \biggl( { \int _{G} { \bigl\vert h(x) \bigr\vert ^{\frac{6}{6-q_{1}}} \,dx} } \biggr) ^{\frac{6-q_{1}}{6}}< M\eta . $$

So, we can apply the Vitali theorem to obtain

$$ \lim_{n\to \infty } \int _{\varOmega _{R} } {h(x) \vert u _{n} \vert ^{q} \,dx} = \int _{\varOmega _{R} } {h(x) \vert u \vert ^{q}\,dx} . $$

Hence,

$$\begin{aligned} \lim_{n\to \infty } \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{n} \vert ^{q} \,dx} =&\lim_{n\to \infty } \int _{\varOmega _{R}^{c} } {h(x) \vert u_{n} \vert ^{q} \,dx} +\lim_{n\to \infty } \int _{\varOmega _{R} } {h(x) \vert u_{n} \vert ^{q} \,dx} \\ =& \int _{\varOmega _{R}^{c} } {h(x) \vert u \vert ^{q}\,dx} + \int _{\varOmega _{R} } {h(x) \vert u \vert ^{q}\,dx} = \int _{{\mathbb{R}}^{3}} {h(x) \vert u \vert ^{q}\,dx} . \end{aligned}$$

The proof is complete. □

In terms of Lemmas 2.1–2.3, it can be verified that $J_{b}$ is well defined on $H^{1}({\mathbb{R}}^{3})$ and is of $C^{1}$ as well as

$$\begin{aligned} \bigl\langle {{J}'_{b} (u),v} \bigr\rangle =& \int _{{\mathbb{R}} ^{3}} {(\nabla u\cdot \nabla v+uv)\,dx} +b \vert \nabla u \vert _{2}^{2} \biggl( { \int _{{\mathbb{R}}^{3}} {\nabla u\cdot \nabla v\,dx} } \biggr) \\ &{}+ \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} uv \,dx} +\lambda \int _{{\mathbb{R}}^{3}} { \vert u \vert ^{p-2}uv \,dx} - \int _{{\mathbb{R}} ^{3}} {h(x) \vert u \vert ^{q-2}uv \,dx} . \end{aligned}$$

(2.6)

Obviously, $u\in H^{1}({\mathbb{R}}^{3})$ is a critical point of $J_{b}$ if and only if $(u,\phi _{u} )$ is a solution of systems (1.1). Noting that $\phi _{u}$ is nonnegative for any $u\in H^{1}({\mathbb{R}}^{3})$, $(u,\phi _{u} )$ is a sign-changing solution of system (1.1) if and only if u is a critical point of $J_{b}$ with $u^{\pm }\ne 0$.

By (2.3)–(2.4) and the Fubini theorem, we know that

$$ L_{\phi _{u^{+}} } \bigl(u^{-} \bigr)= \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u ^{-} \bigr)^{2}\,dx} = \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} \bigl(u^{+} \bigr)^{2}\,dx} =L_{\phi _{u^{-}} } \bigl(u^{+} \bigr) $$

(2.7)

and

$$\begin{aligned}& J_{b} (u)=J_{b} \bigl(u^{+} \bigr)+J_{b} \bigl(u^{-} \bigr)+\frac{b}{2} \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \frac{1}{2}L_{ \phi _{u^{+}} } \bigl(u^{-} \bigr), \end{aligned}$$

(2.8)

$$\begin{aligned}& \bigl\langle {{J}'_{b} (u),u^{+}} \bigr\rangle = \bigl\langle {{J}'_{b} \bigl(u^{+} \bigr),u^{+}} \bigr\rangle +b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} +L _{\phi _{u^{+}} } \bigl(u^{-} \bigr), \end{aligned}$$

(2.9)

$$\begin{aligned}& \bigl\langle {{J}'_{b} (u),u^{-}} \bigr\rangle = \bigl\langle {{J}'_{b} \bigl(u^{-} \bigr),u^{-}} \bigr\rangle +b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} +L _{\phi _{u^{+}} } \bigl(u^{-} \bigr). \end{aligned}$$

(2.10)

Let

$$ M_{b} = \bigl\{ {u\in H^{1} \bigl({\mathbb{R}}^{3} \bigr) : u^{\pm }\ne 0, \bigl\langle {{J}'_{b} (u),u^{+}} \bigr\rangle = \bigl\langle {{J}'_{b} (u),u^{-}} \bigr\rangle =0} \bigr\} . $$

In the following, we will look for the minimum point of the functional $J_{b}$ on $M_{b}$, which is the sigh-changing solution of systems (1.1).

The following lemma is crucial and plays an important role in obtaining our main results later.

Lemma 2.4

Assume $u\in H^{1}({\mathbb{R}}^{3})$ with $u^{\pm }\ne 0$. Then there is an unique pair $(s_{u} ,t_{u} ) \in {\mathbb{R}}_{+}^{0} \times {\mathbb{R}}_{+}^{0}$ satisfying that $s_{u} u^{+}+t_{u} u^{-}\in M_{b}$.

Proof

For any $u\in H^{1}({\mathbb{R}}^{3})$ with $u^{\pm }\ne 0$, clearly, by (2.6)–(2.10) we know that $su^{+}+tu^{-}\in M_{b}$ with $(s,t)\in {\mathbb{R}}_{+}^{0} \times {\mathbb{R}}_{+}^{0}$ if and only if the pair $(s,t)$ satisfies the following systems:

$$ \textstyle\begin{cases} s^{2} \Vert u^{+} \Vert ^{2}+bs^{4} \vert \nabla u^{+} \vert _{2}^{4} +s^{4}\int _{{\mathbb{R}}^{3}}{k(x)\phi _{u^{+}} (u^{+})^{2}\,dx}+ \lambda s^{p} \vert u^{+} \vert _{p}^{p} \\ \quad{} +s^{2}t^{2}[b \vert \nabla u^{+} \vert _{2}^{2} \vert \nabla u^{-} \vert _{2}^{2} +\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} (u^{-})^{2}\,dx}] -s^{q}\int _{{\mathbb{R}}^{3}} {h(x) \vert u^{+} \vert ^{q}\,dx} =0, \\ t^{2} \Vert u^{-} \Vert ^{2}+bt^{4} \vert \nabla u^{-} \vert _{2}^{4} +t^{4}\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} (u^{-})^{2}\,dx} +\lambda t^{p} \vert u^{-} \vert _{p}^{p} \\ \quad{} +s^{2}t^{2}[b \vert \nabla u^{+} \vert _{2}^{2} \vert \nabla u^{-} \vert _{2}^{2}+\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} (u^{-})^{2}\,dx}] -t^{q}\int _{{\mathbb{R}}^{3}} {h(x) \vert u^{-} \vert ^{q}\,dx} =0. \end{cases} $$

(2.11)

To study the solvability of systems (2.11), we investigate the following auxiliary systems with a parameter $\eta \in [0,1]$:

$$ \textstyle\begin{cases} s^{2} \Vert u^{+} \Vert ^{2}+bs^{4} \vert \nabla u^{+} \vert _{2}^{4} +s^{4}\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} (u^{+})^{2}\,dx}+ \lambda s^{p} \vert u^{+} \vert _{p}^{p} \\ \quad {}+\eta s^{2}t^{2} [ {b \vert \nabla u^{+} \vert _{2}^{2} \vert \nabla u^{-} \vert _{2}^{2} +\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u ^{+}} (u^{-})^{2}\,dx} } ] -s^{q}\int _{{\mathbb{R}}^{3}} {h(x) \vert u^{+} \vert ^{q}\,dx} =0, \\ t^{2} \Vert u^{-} \Vert ^{2}+bt^{4} \vert \nabla u^{-} \vert _{2}^{4} +t^{4}\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} (u^{-})^{2}\,dx}+ \lambda t^{p} \vert u^{-} \vert _{p}^{p} \\ \quad {}+\eta s^{2}t^{2} [ {b \vert \nabla u^{+} \vert _{2}^{2} \vert \nabla u^{-} \vert _{2}^{2} +\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u ^{+}} (u^{-})^{2}\,dx} } ] -t^{q}\int _{{\mathbb{R}}^{3}} {h(x) \vert u^{-} \vert ^{q}\,dx} =0. \end{cases} $$

(2.12)

Let

$$ E= \bigl\{ \eta| 0\le \eta \le 1 \text{ such that systems (2.12) have an unique solution in } {\mathbb{R}}_{+}^{0} \times { \mathbb{R}}_{+}^{0} \bigr\} . $$

Put

$$\begin{aligned} \textstyle\begin{cases} \varphi _{\eta }(s,t)=s^{2} \Vert u^{+} \Vert ^{2}+bs^{4} \vert \nabla u^{+} \vert _{2}^{4} +s^{4}\int _{{\mathbb{R}}^{3}} {k(x) \phi _{u^{+}} (u^{+})^{2}\,dx}+\lambda s^{p} \vert u^{+} \vert _{p}^{p} \\ \hphantom{\varphi _{\eta }(s,t)}\quad {} +\eta s^{2}t^{2} [ {b \vert \nabla u^{+} \vert _{2}^{2} \vert \nabla u^{-} \vert _{2}^{2} +\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} (u ^{-})^{2}\,dx} } ] -s^{q}\int _{{\mathbb{R}}^{3}} {h(x) \vert u ^{+} \vert ^{q}\,dx} , \\ \psi _{\eta }(s,t)=t^{2} \Vert u^{-} \Vert ^{2}+bt^{4} \vert \nabla u^{-} \vert _{2}^{4} +t^{4}\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u ^{-}} (u^{-})^{2}\,dx} +\lambda t^{p} \vert u^{-} \vert _{p}^{p} \\ \hphantom{\psi _{\eta }(s,t)}\quad{} +\eta s^{2}t^{2} [ {b \vert \nabla u^{+} \vert _{2}^{2} \vert \nabla u^{-} \vert _{2}^{2} +\int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} (u ^{-})^{2}\,dx} } ] -t^{q}\int _{{\mathbb{R}}^{3}} {h(x) \vert u ^{-} \vert ^{q}\,dx}. \end{cases}\displaystyle \end{aligned}$$

(2.13)

(1) In this part, we show that $0\in E$.

Since $\varphi _{0} (s,t)=\varphi _{0} (s,0)$, $\psi _{0} (s,t)=\psi _{0} (0,t)$, the solvability on $\varphi _{0} (s,t)=0$ and $\psi _{0} (s,t)=0$ is the same. So, we only show that there an unique $\bar{t}\in {\mathbb{R}} ^{0}_{+}$ such that $\psi _{0} (0,\bar{t})=0$.

In fact, owing to the fact that $2\le p\le 4< q<6$, $\lambda \ge 0$, and $h(x)>0$, a.e. $x\in {\mathbb{R}}$, we know that $\psi _{0} (0,t)>0$ as $t>0$ is small enough and $\psi _{0} (0,t)<0$ as $t>0$ is large enough. Hence, there is a $\bar{t}\in {\mathbb{R}}_{+}^{0}$ such that $\psi _{0} (0,\bar{t})=0$.

Now, we prove that such a number $\bar{t}\in {\mathbb{R}}_{+}^{0}$ is unique. To this end, let $g(t)=t^{-2}\psi _{0} (0,t)$, $t\geq 0$.

By

$$\begin{aligned} g(t) = &\bigl\Vert u^{-} \bigr\Vert ^{2}+bt^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{4} +t^{2} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} \bigl(u^{-} \bigr)^{2}\,dx} +\lambda t^{p-2} \bigl\vert u^{-} \bigr\vert _{p}^{p} \\ &{}-t^{q-2} \int _{{\mathbb{R}} ^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} , \quad t\ge 0, \end{aligned}$$

we have $g(0)>0$ and $g(\bar{t} )=0$. If there exists another number $t_{0}>0$ with $t_{0}\neq \bar{t}$ such that $\psi _{0}(0, t_{0})=0$, that is, $g(t_{0})=0$, then we will obtain a contradiction via the following argument:

Case 1. If $t_{0}< \bar{t}$, then

$$ \begin{aligned}[b] t_{0} {g}'(t_{0} )&=2bt_{0} ^{2} \bigl\vert u^{-} \bigr\vert _{2}^{4} +2t_{0} ^{2} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} \bigl(u^{-} \bigr)^{2}\,dx} +\lambda (p-2)t _{0} ^{p-2} \bigl\vert u^{-} \bigr\vert _{p}^{p} \\&\quad{} -(q-2)t_{0} ^{q-2} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx}. \end{aligned} $$

(2.14)

On the other hand, the fact that $g(t_{0} )=0$ implies that

$$ \begin{aligned}[b] & bt_{0} ^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{4} +t_{0} ^{2} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} \bigl(u^{-} \bigr)^{2}\,dx} \\ &\quad =- \bigl\Vert u^{-} \bigr\Vert ^{2}-\lambda t_{0} ^{p-2} \bigl\vert u^{-} \bigr\vert _{p} ^{p} +t_{0} ^{q-2} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} . \end{aligned} $$

(2.15)

Thus, substituting (2.15) into (2.14) and taking into account that $0< t_{0}$, $u^{-}\ne 0$, $2\le p\le 4< q<6$, $\lambda \geq 0$ and $h(x)>0$, a.e. $x\in {\mathbb{R}}^{3}$, we get

$$ t_{0} {g}'(t_{0} )=(p-4)\lambda t_{0} ^{p-2} \bigl\vert u^{-} \bigr\vert _{p}^{p} +(4-q)t _{0} ^{q-2} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} -2 \bigl\Vert u^{-} \bigr\Vert ^{2}< 0, $$

which together with $g(t_{0})=0$ implies that $g(t_{0}+\delta )<0$ as $0<\delta$ is small enough. Hence, there is a $t_{\ast }\in (t_{0}, \bar{t} )$ satisfying

$$ g(t_{\ast })=\min_{t\in [t_{0}, \bar{t}]} g(t)< 0, \quad\quad {g}''(t_{ \ast })\ge 0. $$

However, observing that

$$\begin{aligned} t_{\ast }^{2} {g}''(t_{\ast }) =&2bt_{\ast }^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{4} +2t_{\ast }^{2} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u ^{-}} \bigl(u^{-} \bigr)^{2}\,dx} +\lambda (p-2) (p-3)t_{\ast }^{p-2} \bigl\vert u^{-} \bigr\vert _{p}^{p} \\ & {} -(q-2) (q-3)t_{\ast }^{q-2} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} \end{aligned}$$

and

$$ bt_{\ast }^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{4} +t_{\ast }^{2} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} \bigl(u^{-} \bigr)^{2}\,dx} < -\lambda t _{\ast }^{p-2} \bigl\vert u^{-} \bigr\vert _{p}^{p} +t_{\ast }^{q-2} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} - \bigl\Vert u ^{-} \bigr\Vert ^{2} $$

(following from $g(t_{\ast })<0$), we have

$$ t_{\ast }^{2} {g}''(t_{\ast })< \lambda (p-1) (p-4)t_{\ast }^{p-2} \bigl\vert u^{-} \bigr\vert _{p}^{p} -(q-1) (q-4)t_{\ast }^{q-2} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} -2 \bigl\Vert u ^{-} \bigr\Vert ^{2}< 0 $$

(noting that $2\le p\le 4< q<6$, $\lambda \ge 0$ and $h(x)>0$, a.e. $x\in {\mathbb{R}}^{3}$), which contradicts the fact that ${g}''(t_{\ast })\ge 0$.

Case 2. If $\bar{t}< t_{0}$, the proof is the same. In fact, by only replacing $t_{0}$ with t̄ in the above argument on case 1, we also obtain a contradiction.

Hence, we have proved that $0\in E$.

(2) In this part, we show that the set E is open and closed in $[0,1]$.

• (i) E is an open set in $[0,1]$.

For any fixed $\eta _{0} \in E$ and $(s_{0} ,t_{0} )\in \mathbb{R}_{+} ^{0} \times \mathbb{R}_{+}^{0}$ is an unique solution of (2.12) associated with $\eta =\eta _{0}$. By calculation, from (2.13) we know that

$$\begin{aligned}& \begin{aligned}[b] \frac{\partial \varphi _{\eta _{0} } }{\partial s}\bigg| _{(s_{0} ,t_{0} )} &=2s_{0} \bigl\Vert u^{+} \bigr\Vert ^{2}+4bs_{0} ^{3} \bigl\vert \nabla u ^{+} \bigr\vert _{2}^{4} +4s_{0} ^{3} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u ^{+}} \bigl(u^{+} \bigr)^{2}\,dx} \\ &\quad {} +2\eta _{0} s_{0} t_{0} ^{2} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2} ^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x) \phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr] \\ &\quad {} +\lambda ps_{0} ^{p-1} \bigl\vert u^{+} \bigr\vert _{p}^{p} -qs_{0} ^{q-1} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx}, \end{aligned} \end{aligned}$$

(2.16)

$$\begin{aligned}& \frac{\partial \varphi _{\eta _{0} } }{\partial t}\bigg| _{(s_{0} ,t_{0} )} =2\eta _{0} s_{0} ^{2}t_{0} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr], \end{aligned}$$

(2.17)

$$\begin{aligned}& \frac{\partial \psi _{\eta _{0} } }{\partial s}\bigg| _{(s_{0} ,t_{0} )} =2 \eta _{0} s_{0} t_{0} ^{2} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr], \end{aligned}$$

(2.18)

$$\begin{aligned}& \begin{aligned}[b] \frac{\partial \psi _{\eta _{0} } }{\partial t}\bigg| _{(s_{0} ,t_{0} )} &= 2t_{0} \bigl\Vert u^{-} \bigr\Vert ^{2}+4bt_{0} ^{3} \bigl\vert \nabla u ^{-} \bigr\vert _{2}^{4} +4t_{0} ^{3} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u ^{-}} \bigl(u^{-} \bigr)^{2}\,dx} \\ &\quad {} +2\eta _{0} s_{0} ^{2}t_{0} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2} ^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x) \phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr] \\ &\quad {} +\lambda pt_{0} ^{p-1} \bigl\vert u^{-} \bigr\vert _{p}^{p} -qt_{0} ^{q-1} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx}. \end{aligned} \end{aligned}$$

(2.19)

Again, by $\varphi _{\eta _{0} } (s_{0} ,t_{0} )=0$, from (2.13) we have

$$\begin{aligned} \begin{aligned}[b] & bs_{0}^{3} \bigl\vert \nabla u^{+} \bigr\vert _{2}^{4} +s_{0}^{3} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{+} \bigr)^{2}\,dx} \\ &\quad =-\eta _{0} s_{0} t_{0}^{2} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u ^{-} \bigr)^{2}\,dx} } \biggr] \\ &\quad\quad {} -\lambda s_{0} ^{p-1} \bigl\vert u^{+} \bigr\vert _{p}^{p} +s_{0} ^{q-1} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx-s_{0} \bigl\Vert u^{+} \bigr\Vert ^{2}} . \end{aligned} \end{aligned}$$

(2.20)

By (2.16) combined with (2.20), we get

$$\begin{aligned} \frac{\partial \varphi _{\eta _{0} } }{\partial s}\bigg| _{(s_{0} ,t_{0} )} =&-2s_{0} \bigl\Vert u^{+} \bigr\Vert ^{2}-2\eta _{0} s_{0} t_{0} ^{2} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr] \\ & {} -\lambda (4-p)s_{0} ^{p-1} \bigl\vert u^{+} \bigr\vert _{p}^{p} -(q-4)s_{0} ^{q-1} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx} . \end{aligned}$$

(2.21)

Similarly, by $\psi _{\eta _{0} } (s_{0} ,t_{0} )=0$, we can deduce that

$$\begin{aligned} \frac{\partial \psi _{\eta _{0} } }{\partial t}\bigg| _{(s_{0} ,t_{0} )} =&-2t _{0} \bigl\Vert u^{-} \bigr\Vert ^{2}-2\eta _{0} s_{0} ^{2}t_{0} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr] \\ & {} -\lambda (4-p)t_{0} ^{p-1} \bigl\vert u^{-} \bigr\vert _{p}^{p} -(q-4)t_{0} ^{q-1} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} . \end{aligned}$$

(2.22)

Thus, by (2.17)–(2.18) and (2.21)–(2.22) as well as $\phi _{u^{+}} ,\phi _{u^{-}} \ge 0$, $\lambda \ge 0$, $2\le p\le 4< q$, $h(x)>0$, a.e. $x\in {\mathbb{R}} ^{3}$, we have the determinant

$$\begin{aligned} \begin{aligned} M &=\frac{\partial \varphi _{\eta _{0} } (s_{0} ,t_{0} )}{\partial s} \cdot \frac{\partial \psi _{\eta _{0} } (s_{0} ,t_{0} )}{\partial t}-\frac{ \partial \varphi _{\eta _{0} } (s_{0} ,t_{0} )}{\partial t}\cdot \frac{ \partial \psi _{\eta _{0} } (s_{0} ,t_{0} )}{\partial s} \\ &>AB-4\eta _{0} ^{2}s_{0} ^{3}t_{0} ^{3} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr]^{2}>0, \end{aligned} \end{aligned}$$

where

$$\begin{array}{c}A=2s_0{\parallel u^{+}\parallel }^2+2{\eta }_0{s}_0{t}_0^2\left(b\right|\mathrm{\nabla }{u}^{+}{|}_2^2|\mathrm{\nabla }{u}^{-}{|}_2^2+{\int }_{{\mathbb{R}}^3}k(x){\varphi }_{u^{+}}{\left(u^{-}\right)}^{2}dx),\hfill \\ B=2t_0{\parallel u^{-}\parallel }^2+2{\eta }_0{s}_0^2{t}_0\left(b\right|\mathrm{\nabla }{u}^{+}{|}_2^2|\mathrm{\nabla }{u}^{-}{|}_2^2+{\int }_{{\mathbb{R}}^3}k(x){\varphi }_{u^{+}}{\left(u^{-}\right)}^{2}dx),\hfill \\ M=\left|\begin{array}{cc}\frac{\partial {\phi }_{{\eta }_0}(s_0,t_0)}{\partial s}& \frac{\partial {\phi }_{{\eta }_0}(s_0,t_0)}{\partial t}\\ \frac{\partial {\psi }_{{\eta }_0}(s_0,t_0)}{\partial s}& \frac{\partial {\psi }_{{\eta }_0}(s_0,t_0)}{\partial t}\end{array}\right|.\hfill \end{array}$$

Hence, by the implicit function theorem, there exist an open neighborhood $V_{0}$ of $\eta _{0}$ and $\wedge _{0} \subset \mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0}$ of $(s_{0} ,t_{0} )$ such that the implicit function $s=s(\eta )$, $t=t(\eta )$ satisfies system (2.12) on $V_{0} \times \wedge _{0}$.

Now, we show that, for any $\eta \in V_{0}$, the system (2.12) has no solution in $(\mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0} ) \backslash \wedge _{0}$.

Suppose by contradiction that there exists $\eta _{1} \in V_{0}$ such that system (2.12) has another solution $(\bar{s},\bar{t})$ in $(\mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0} )\backslash \wedge _{0}$ associated with $\eta _{1}$ apart from the solution $(s,t)$ in ∧₀ associated with $\eta _{1}$. Then, by the implicit function theorem again, we can find a solution function $\bar{s}= \bar{s}(\eta )$, $\bar{t}=\bar{t}(\eta )$ in $(\eta _{1} -\varepsilon , \eta _{1} +\varepsilon )$ for some $\varepsilon >0$, which satisfies (2.12) and goes through $( {\eta _{1} ,(\bar{s},\bar{t})} )$.

• 1. If $\eta _{0} <\eta _{1}$, then consider the saturated solution $\bar{s}=\bar{s}(\eta )$, $\bar{t}=\bar{t}(\eta )$ on its saturated interval. Since it cannot be defined at $\eta _{0}$ and cannot enter $V_{0} \times \wedge _{0}$, there exists a point $\eta _{2} \in [\eta _{0} ,\eta _{1} )$ such that the solution $\bar{s}=\bar{s}(\eta )$, $\bar{t}=\bar{t}(\eta )$ in $(\eta _{2} ,\eta _{1} )$ and $\bar{s}^{2}( \eta )+\bar{t}^{2}(\eta )\to \infty$ as $\eta \to \eta _{2} ^{+}$, which contradicts systems (2.12) noting that $2\le p\le 4< q<6$, $\lambda \ge 0$, and $h(x)>0$, a.e. $x\in {\mathbb{R}}^{3}$. Hence $V_{0} \subset E$.

• 2. If $\eta _{0} >\eta _{1}$, the proof is similar.

• (ii) E is a closed set in $[0,1]$.

In fact, let $\{ {\eta _{n} } \} \subset E$ be a sequence with $\eta _{n} \to \eta _{0} \in [0,1]$ and $(s_{n} ,t_{n} )\in \mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0}$ be the unique solution of (2.12) associated with $\eta _{n}$. Because the sequence $\{\eta _{n} \}$ is bounded, from (2.12) it follows that $\{ {(s_{n} ,t_{n} )} \}$ is bounded. Therefore, there exists a subsequence of $\{ {(s_{n} ,t_{n} )} \}$, still denoted by $\{ {(s_{n} ,t_{n} )} \}$, such that $(s_{n} ,t_{n} )\to (s_{0} ,t_{0} )$. Of course, $(s_{0} ,t_{0} )$ satisfies systems (2.12) for $\eta =\eta _{0}$. Furthermore, by (2.12), we have

$$ \bigl\Vert u^{+} \bigr\Vert ^{2}\le s_{n} ^{q-2} \int _{{\mathbb{R}} ^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx} , $$

which implies that $s_{n} \ge c_{1} >0$ for some $c_{1} >0$ because $2\le p\le 4< q<6$, $h(x)>0$, a.e. $x\in {\mathbb{R}}^{3}$ and $u^{+}\ne 0$. Similarly, there exists $c_{2} >0$ such that $t_{n} \ge c_{2} >0$. Thus, $(s_{0} ,t_{0} )\in \mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0}$ is a solution of (2.12). Also, the implicit function theorem ensures that $(s_{0} ,t_{0} )$ is the unique solution of (2.12) for $\eta =\eta _{0}$ again. Hence, E is closed in $[0,1]$.

Summing up the above arguments (i) and (ii), we get $E=[0,1]$, and therefore, the conclusion of Lemma 2.4 is true. □

Lemma 2.5

Assume that $u\in H^{1}({\mathbb{R}} ^{3})$ with $u^{\pm }\ne 0$ and $\varphi _{1} (1,1)\le 0$, $\psi _{1} (1,1) \le 0$, where $\varphi _{1}$, $\psi _{1}$ are given as in (2.13) with $\eta =1$. Then the unique pair $(s_{u} ,t_{u} )\in \mathbb{R}_{+} ^{0} \times \mathbb{R}_{+}^{0}$ given in Lemma 2.4 satisfies $0< s_{u} ,t_{u} \le 1$.

Proof

If $s_{u} \ge t_{u} >0$, then, by $s_{u} u^{+}+t_{u} u^{-}\in M_{b}$ together with (2.11), we have

$$\begin{aligned}& s_{u} ^{2} \bigl\Vert u^{+} \bigr\Vert ^{2}+s^{4}b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{4} +s_{u}^{4} \int _{{\mathbb{R}}^{3}} k(x)\phi _{u^{+}} \bigl(u ^{+} \bigr)^{2}\,dx \\& \quad\quad{} +s_{u}^{4} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr] \\& \quad \ge -\lambda s_{u}^{p} \bigl\vert u^{+} \bigr\vert _{p}^{p} +s_{u}^{q} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx} . \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned}[b] & s_{u} ^{-2} \bigl\Vert u^{+} \bigr\Vert ^{2}+b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{4} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{+} \bigr)^{2}\,dx+} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2} ^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr] \\ &\quad \ge -\lambda s_{u}^{p-4} \bigl\vert u^{+} \bigr\vert _{p}^{p} +s_{u}^{q-4} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx} . \end{aligned} \end{aligned}$$

(2.23)

On the other hand, the assumption $\varphi _{1} (1,1)\le 0$ implies that

$$\begin{aligned} \begin{aligned}[b] & \bigl\Vert u^{+} \bigr\Vert ^{2}+b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{4} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{+} \bigr)^{2}\,dx+} \biggl[ {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr] \\ &\quad \le -\lambda \bigl\vert u^{+} \bigr\vert _{p}^{p} + \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx} . \end{aligned} \end{aligned}$$

(2.24)

By (2.23)–(2.24), we have

$$ \bigl( {s_{u} ^{-2}-1} \bigr) \bigl\Vert u^{+} \bigr\Vert ^{2}+ \lambda \bigl( {s_{u}^{p-4} -1} \bigr) \bigl\vert u^{+} \bigr\vert _{p}^{p} \ge \bigl( {s_{u} ^{q-4}-1} \bigr) \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx} . $$

Because $u^{+}\ne 0$, $0\le \lambda$, $2\le p\le 4< q<6$ and $h(x)>0$, for a.e. $x\in {\mathbb{R}}^{3}$, the above inequality implies that $0< s_{u} \le 1$, and so, $0< t_{u} \le 1$.

For the case $0< t_{u} \le s_{u}$, the proof is similar. □

Let $m_{b} :=\inf \{ {J_{b} (u)| u\in M_{b} } \}$. We have the following lemma.

Lemma 2.6

Assume that $(l)$, $(k)$, $(h)$ hold. Then $m_{b} >0$ can be achieved at some point $u_{b} \in M_{b}$.

Proof

For any $u\in M_{b}$, by $\langle {J_{b} ^{ \prime }(u),u} \rangle =0$ and (2.2), we have

$$\begin{aligned} \Vert u \Vert ^{2} \le & \Vert u \Vert ^{2}+b \vert \nabla u \vert _{2}^{4} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u} u ^{2}\,dx+ \lambda \int _{{\mathbb{R}}^{3}} { \vert u \vert ^{p}\,dx} } \\ =& \int _{{\mathbb{R}}^{3}} {h(x) \vert u \vert ^{q}\,dx} \le \vert h \vert _{\frac{6}{6-q_{1}}} \vert u \vert _{r}^{q} \le \vert h \vert _{\frac{6}{6-q_{1}}} S_{r}^{-q} \Vert u \Vert ^{q}, \end{aligned}$$

where $r=\frac{6q}{q_{1}}$ satisfying $4< r<6$. Hence, $\Vert u \Vert \ge c_{0} >0$, where $c_{0} = ( {\frac{S_{r}^{q} }{ \vert h \vert _{\frac{6}{6-q_{1}}} }} )^{\frac{1}{q-2}}$. Also, by

$$\begin{aligned} J_{b} (u) =&J_{b} (u)-\frac{1}{q} \bigl\langle {J_{b} ^{\prime }(u),u} \bigr\rangle \\ =& \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \Vert u \Vert ^{2}+ \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr)b \vert \nabla u \vert _{2}^{4} + \biggl( {\frac{1}{4}- \frac{1}{q}} \biggr) \int _{{\mathbb{R}}^{3}} k(x)\phi _{u} u^{2}\,dx \\ &{}+ \lambda \biggl( {\frac{1}{p}- \frac{1}{q}} \biggr) \vert u \vert _{p}^{p} \\ \ge & \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \Vert u \Vert ^{2}\ge \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr)c_{0}^{2}>0, \end{aligned}$$

we have $m_{b} \ge ( {\frac{1}{2}-\frac{1}{q}} )c_{0} ^{2}>0$.

Let $\{ {u_{n} } \} \subset M_{b}$ with $J_{b} (u_{n} )\to m_{b}$ as $n\rightarrow \infty$. By

$$ J_{b} (u_{n} )=J_{b} (u_{n} )- \frac{1}{q} \bigl\langle {J_{b} ^{ \prime }(u_{n} ),u_{n} } \bigr\rangle \ge \biggl( {\frac{1}{2}- \frac{1}{q}} \biggr) \Vert u_{n} \Vert ^{2}, $$

and $J_{b} (u_{n} )\to m_{b}$, we know that $\{ {u_{n} } \}$ is bounded in $H^{1}({\mathbb{R}}^{3})$, which implies that there exist $u_{b} \in H^{1}({\mathbb{R}}^{3})$ and a subsequence, still denoted by $\{u_{n}\}$, such that $u_{n}\rightharpoonup u_{b}$ as well as $u_{n} ^{\pm }\rightharpoonup u_{b} ^{\pm }$ weakly in $H^{1}( {\mathbb{R}}^{3})$.

By $\{ {u_{n} } \} \subset M_{b}$, it follows from $\langle {J_{b} ^{\prime }(u_{n} ),u_{n} ^{\pm }} \rangle =0$ that

$$\begin{aligned} \bigl\Vert u_{n} ^{\pm } \bigr\Vert ^{2} \le & \bigl\Vert u_{n} ^{ \pm } \bigr\Vert ^{2}+b \vert \nabla u_{n} \vert _{2}^{2} \bigl\vert \nabla u_{n}^{\pm } \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{n} } \bigl(u_{n} ^{\pm } \bigr)^{2}\,dx+} \lambda \bigl\vert u_{n} ^{\pm } \bigr\vert _{p}^{p} \\ =& \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u_{n}^{\pm } \bigr\vert ^{q}\,dx} \le \vert h \vert _{\frac{6}{6-q_{1}}} S_{r}^{-q} \bigl\Vert u_{n}^{\pm } \bigr\Vert ^{q}, \end{aligned}$$

(2.25)

where $r=\frac{6q}{q_{1}}$. Therefore,

$$ \bigl\Vert u_{n} ^{\pm } \bigr\Vert \ge c_{0} >0. $$

(2.26)

Thus, by (2.25)–(2.26) and Lemma 2.3, we obtain $0< c_{0}^{2} \le \lim_{n\to \infty } \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{n}^{ \pm } \vert ^{q}\,dx} =\int _{{\mathbb{R}}^{3}} h(x) \vert u_{b}^{\pm } \vert ^{q}\,dx$, which implies that $u_{b}^{\pm }\ne 0$ taking into account the assumption $h(x)>0$, a.e. $x\in {\mathbb{R}}^{3}$.

Again, by the weak semi-continuity of the norm and Lemma 2.3, we get

$$\begin{aligned}& \bigl\Vert u_{b} ^{\pm } \bigr\Vert ^{2}+b \vert \nabla u_{b} \vert _{2}^{2} \bigl\vert \nabla u_{b} ^{\pm } \bigr\vert _{2}^{2} + \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u_{b} } \bigl(u_{b} ^{\pm } \bigr)^{2}\,dx+} \lambda \bigl\vert u_{b} ^{\pm } \bigr\vert _{p}^{p} \\& \quad \le \lim_{n\to \infty } \inf \biggl\{ { \bigl\Vert u_{n} ^{\pm } \bigr\Vert ^{2}+b \vert \nabla u_{n} \vert _{2}^{2} \bigl\vert \nabla u_{n} ^{\pm } \bigr\vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x) \phi _{u_{n} } \bigl(u_{n} ^{\pm } \bigr)^{2}\,dx+} \lambda \bigl\vert u_{n} ^{\pm } \bigr\vert _{p}^{p} } \biggr\} \\& \quad =\lim_{n\to \infty } \inf \biggl\{ { \int _{{\mathbb{R}} ^{3}} {h(x) \bigl\vert u_{n} ^{\pm } \bigr\vert ^{q}\,dx} } \biggr\} = \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u_{b} ^{\pm } \bigr\vert ^{q}\,dx} , \end{aligned}$$

which means that $\varphi _{1} (1,1)\le 0$ and $\psi _{1} (1,1)\le 0$ in (2.13) corresponding to $u_{b}$. Hence, by Lemma 2.5, there exists an unique $(s,t)\in (0,1]\times (0,1]$ with $u_{b}^{\ast }=su_{b}^{+} +tu _{b}^{-} \in M_{b}$. Moreover, by the weak semi-continuity of the norm and Lemmas 2.2–2.3 and (2.7), we have

$$\begin{aligned} m_{b} &\le J_{b} \bigl(u_{b}^{\ast } \bigr)-\frac{1}{q} \bigl\langle {J_{b} ^{ \prime } \bigl(u_{b}^{\ast } \bigr),u_{b}^{\ast }} \bigr\rangle \\ &= \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \bigl\Vert u_{b}^{\ast } \bigr\Vert ^{2}+ \biggl( { \frac{1}{4}-\frac{1}{q}} \biggr)b \bigl\vert \nabla u_{b}^{\ast } \bigr\vert _{2}^{4} + \biggl( {\frac{1}{4}- \frac{1}{q}} \biggr) \int _{{\mathbb{R}}^{3}} k(x)\phi _{u_{b}^{\ast }} \bigl(u_{b}^{\ast } \bigr)^{2}\,dx \\ &\quad{} + \lambda \biggl( {\frac{1}{p}-\frac{1}{q}} \biggr) \bigl\vert u_{b}^{\ast } \bigr\vert _{p}^{p} \\ &= \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \bigl[s^{2} \bigl\Vert u_{b} ^{+} \bigr\Vert ^{2}+t^{2} \bigl\Vert u_{b}^{-} \bigr\Vert ^{2} \bigr] \\ &\quad{}+ \biggl( { \frac{1}{4}-\frac{1}{q}} \biggr)b \bigl[ {s^{4} \bigl\vert \nabla u_{b}^{+} \bigr\vert _{2}^{4} +t^{4} \bigl\vert \nabla u_{b}^{-} \bigr\vert _{2}^{4} +2s ^{2}t^{2} \bigl\vert \nabla u_{b}^{+} \bigr\vert _{2}^{4} \bigl\vert \nabla u_{b} ^{-} \bigr\vert _{2}^{4} } \bigr] \\ & \quad {} + \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr) \biggl[ s^{4} \int _{{\mathbb{R}}^{3}} k(x)\phi _{u_{b}^{+} } \bigl(u_{b}^{+} \bigr)^{2}\,dx+t ^{4} \int _{{\mathbb{R}}^{3}} k(x)\phi _{u_{b}^{-} } \bigl(u_{b}^{-} \bigr)^{2}\,dx \\ &\quad{}+2s ^{2}t^{2} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{b}^{+} } \bigl(u_{b}^{-} \bigr)^{2}\,dx} \biggr] \\ & \quad {} +\lambda \biggl( {\frac{1}{p}-\frac{1}{q}} \biggr) \bigl( {s^{p} \bigl\vert u_{b}^{+} \bigr\vert _{p}^{p} +t^{p} \bigl\vert u_{b}^{-} \bigr\vert _{p} ^{p} } \bigr) \\ &\le \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \bigl( { \bigl\Vert u _{b}^{+} \bigr\Vert ^{2}+ \bigl\Vert u_{b}^{-} \bigr\Vert ^{2}} \bigr)+ \biggl( { \frac{1}{4}-\frac{1}{q}} \biggr)b \bigl[ { \bigl\vert \nabla u _{b}^{+} \bigr\vert _{2}^{4} + \bigl\vert \nabla u_{b}^{-} \bigr\vert _{2}^{4} +2 \bigl\vert \nabla u_{b}^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u_{b}^{-} \bigr\vert _{2}^{2} } \bigr] \\ & \quad {} + \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr) \biggl[ { \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u_{b}^{+} } \bigl(u_{b}^{+} \bigr)^{2}\,dx+ \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u_{b}^{-} } \bigl(u_{b}^{-} \bigr)^{2}\,dx+2 \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u_{b}^{+} } \bigl(u_{b}^{\_} \bigr)^{2}\,dx} } } } \biggr] \\ & \quad {} +\lambda \biggl( {\frac{1}{p}-\frac{1}{q}} \biggr) \bigl( { \bigl\vert u_{b}^{+} \bigr\vert _{p}^{p} + \bigl\vert u_{b}^{-} \bigr\vert _{p}^{p} } \bigr) \\ &\le \lim_{n\to \infty } \inf \biggl\{ { \biggl( {\frac{1}{2}- \frac{1}{q}} \biggr) \bigl( { \bigl\Vert u_{n}^{+} \bigr\Vert ^{2}+ \bigl\Vert u_{n}^{-} \bigr\Vert ^{2}} \bigr)} \\ &\quad{}+ \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr)b \bigl[ { \bigl\vert \nabla u_{n}^{+} \bigr\vert _{2}^{4} + \bigl\vert \nabla u_{n}^{-} \bigr\vert _{2}^{4} +2 \bigl\vert \nabla u_{n}^{+} \bigr\vert _{2}^{2} \bigl\vert \nabla u_{n}^{-} \bigr\vert _{2}^{2} } \bigr] \\ & \quad {} + \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr) \biggl[ { \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u_{n}^{+} } \bigl(u_{n} ^{+} \bigr)^{2}\,dx+ \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u_{n}^{-} } \bigl(u_{n} ^{-} \bigr)^{2}\,dx+2 \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u_{n}^{+} } \bigl(u_{n} ^{\_} \bigr)^{2}\,dx} } } } \biggr] \\ & \quad {} + {\lambda \biggl( {\frac{1}{p}-\frac{1}{q}} \biggr) \bigl( { \bigl\vert u_{n}^{+} \bigr\vert _{p}^{p} + \bigl\vert u_{n}^{-} \bigr\vert _{p}^{p} } \bigr)} \biggr\} \\ &=\lim_{n\to \infty } \inf \biggl[ {J_{b} (u_{n} )- \frac{1}{q} \bigl\langle {J_{b} ^{\prime }(u_{n} ),u_{n} } \bigr\rangle } \biggr] =\lim_{n\to \infty } \inf \bigl[ {J_{b} (u_{n} )} \bigr]=m_{b} . \end{aligned}$$

Thus, $s=t=1$, and therefore, $u_{b} \in M_{b}$, $J_{b} (u_{b} )=m_{b}$. □

Lemma 2.7

For any given $u\in H^{1}({\mathbb{R}} ^{3})$ with $u^{\pm }\ne 0$. Let $g (s,t)=J_{b} (su^{+}+tu^{-})$, $(s,t) \in \mathbb{R}_{+} \times \mathbb{R}_{+}$, then there exists an unique pair $(s_{u} ,t_{u} )\in \mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0}$ such that g attains its global maximum on $\mathbb{R}_{+} \times \mathbb{R}_{+}$ at the point $(s_{u} ,t_{u} )$, which is exactly obtained in Lemma 2.4.

Proof

Because $2\le p\le 4< q<6$, $\lambda \ge 0$ and $h(x)>0$, a.e. $x\in {\mathbb{R}}^{3}$, we can know that $g (s,t)\to -\infty$ as $\vert (s,t) \vert \to \infty$. Hence, we only need to prove that g cannot achieve its maximum on the boundary of $\mathbb{R}_{+} \times \mathbb{R}_{+}$. In fact, if not—if g achieved its maximum at point $(0,\bar{t})$, $\bar{t}>0$, then, by

$$\begin{aligned} g (s,\bar{t}) =&J_{b} \bigl(su^{+}+\bar{t}u^{-} \bigr) \\ =& \frac{s^{2}}{2} \bigl\Vert u^{+} \bigr\Vert ^{2}+ \frac{s^{4}}{4}b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{4} +\frac{s^{4}}{4} \int _{{\mathbb{R}} ^{3}} {k(x)\phi _{u^{+}} \bigl(u^{+} \bigr)^{2}\,dx} +\frac{\lambda }{p}s^{p} \bigl\vert u^{+} \bigr\vert _{p}^{p} \\ &{}-\frac{s^{q}}{q} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{+} \bigr\vert ^{q} \,dx} \\ & {} + \frac{\bar{t}^{2}}{2} \bigl\Vert u^{-} \bigr\Vert ^{2}+ \frac{ \bar{t}^{4}}{4}b \bigl\vert \nabla u^{-} \bigr\vert _{2}^{4} + \frac{\bar{t}^{4}}{4} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{-}} \bigl(u^{-} \bigr)^{2}\,dx} +\frac{\lambda }{p}\bar{t}^{p} \bigl\vert u^{-} \bigr\vert _{p}^{p} \\ &{} -\frac{\bar{t}^{q}}{q} \int _{{\mathbb{R}}^{3}} {h(x) \bigl\vert u^{-} \bigr\vert ^{q} \,dx} \\ & {} +2s^{2}\bar{t}^{2} \biggl( {b \bigl\vert \nabla u^{+} \bigr\vert _{2}^{4} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{4} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u^{+}} \bigl(u^{-} \bigr)^{2}\,dx} } \biggr) \ge g (s,0)+g (0,\bar{t}), \end{aligned}$$

and the fact that $g (s,0)>0$ as $s>0$ is small enough, we know that $g (s,\bar{t})>g (0,\bar{t})$ as $s>0$ is small enough, which contradicts the assumption that g achieved its maximum value at $(0,\bar{t})$ on $\mathbb{R}_{+} \times \mathbb{R}_{+}$. So, g achieved its maximum in some point $(s_{u} ,t_{u} )\in \mathbb{R}_{+} ^{0} \times \mathbb{R}_{+}^{0}$ on $\mathbb{R}_{+} \times \mathbb{R} _{+}$. Of course, $(s_{u} ,t_{u} )$ is a critical point of g on $\mathbb{R}_{+} \times \mathbb{R}_{+}$, and therefore it follows from the proof of Lemma 2.4 that the pair $(s_{u} ,t_{u} )$ is the unique solution of systems (2.11) in $\mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0}$. □

3 Main results

We are now in a position to give our main results in this paper.

Theorem 3.1

Assume that conditions $(l)$, $(k)$, $(h)$ hold, then problem (1.1) possesses one least energy sign-changing solution $u_{b}$, which has exactly two nodal domains, where $u_{b}$ is given in Lemma 2.6.

Proof

We apply the quantitative deformation lemma to prove that $J_{b} ^{\prime }(u_{b} )=0$.

Owing to the fact that $u_{b} \in M_{b}$ and $J_{b} (u_{b} )=m_{b}$, in terms of Lemma 2.7, for any $(s,t)\in \mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0}$ with $(s,t)\ne (1,1)$, we immediately obtain

$$ J_{b} \bigl(su_{b}^{+} +tu_{b}^{-} \bigr)< J_{b} \bigl(u_{b}^{+} +u_{b}^{-} \bigr)=m_{b} . $$

(3.1)

If $J_{b} ^{\prime }(u_{b} )\ne 0$, then there exist $r>0$ and $\tau >0$ satisfying

$$\begin{aligned}& \bigl\Vert J_{b} ^{\prime }(u) \bigr\Vert >\tau \quad \text{as } \Vert u-u _{b} \Vert < 3r. \end{aligned}$$

Set $U=(1-\sigma _{0} ,1+\sigma _{0} )\times (1-\sigma _{0} ,1+\sigma _{0} )$, where $0<\sigma _{0} <\frac{1}{2}$, and $h(s,t)=su_{b}^{+} +tu _{b}^{-}$, $(s,t)\in \bar{U}$. It follows from (3.1) that

$$ m^{\ast }:=\max_{\partial U} J_{b} \circ h=\max_{\partial U} J_{b} \bigl(su_{b}^{+} +tu_{b}^{-} \bigr)< m_{b} . $$

(3.2)

Take $0<\varepsilon <\min \{ {(m_{b} -m^{\ast })/2,\tau r/8} \}$, $S= \{ {u\in H^{1}({\mathbb{R}}^{3}): \Vert u-u_{b} \Vert < r} \}$.

Let $S_{2r} = \{ {u\in H^{1}({\mathbb{R}}^{3}): \operatorname{dist}(u,S) \le 2r} \}$. Applying Lemma 2.3 in [33], there exists a deformation η satisfying that

• (i) $\eta (1,u)=u$, if $u\notin J_{b}^{-1} (m_{b} -2\varepsilon ,m _{b} +2\varepsilon )\cap S_{2r}$;

• (ii) $\eta (1,J_{b}^{m_{b} +\varepsilon } \cap S)\subset J_{b}^{m_{b} -\varepsilon }$;

• (iii) $J_{b} (\eta (1,u))\le J_{b} (u)$ for any $u\in H^{1}({\mathbb{R}} ^{3})$.

Now, we show that

$$ \eta \bigl(1,h(U) \bigr)\cap M_{b} \ne \emptyset , $$

which will contradict the definition of $m_{b}$.

First, we claim that

$$ \max_{(s,t)\in \bar{U}} J_{b} \bigl(\eta \bigl(1,h(s,t) \bigr) \bigr)< m_{b} . $$

(3.3)

In fact, for any $(s,t)\in \bar{U}$, we consider the following two cases:

• (a) If $(s,t)\ne (1,1)$, then, by (3.3) together with (iii) above, we have

  $$ J_{b} \bigl(\eta \bigl(1,h(s,t) \bigr) \bigr)\le J_{b} \bigl(h(s,t) \bigr)=J_{b} \bigl(su_{b}^{+} +tu_{b} ^{-} \bigr)< m_{b} . $$

• (b) If $(s,t)=(1,1)$, then, by $J_{b} (h(1,1))=J_{b} (u_{b} )=m_{b} < m_{b} +\varepsilon$ and $h(1,1)=u_{b} \in S$, it follows from (ii) that

  $$\begin{aligned}& J_{b} \bigl(\eta \bigl(1,h(1,1) \bigr) \bigr)\le m_{b} - \varepsilon < m_{b} . \end{aligned}$$

Hence, from the above arguments (a) and (b), we know that (3.3) is true.

Secondly, for $(s,t)\in \bar{U}$, let $\varphi (s,t)=\eta (1,h(s,t))$ and

$$ \phi (s,t)= \bigl(\phi _{1} (s,t),\phi _{2} (s,t) \bigr),\quad\quad \psi (s,t)= \bigl(\psi _{1} (s,t), \psi _{2} (s,t) \bigr), $$

where

$$\begin{aligned}& \phi _{1} (s,t)= \bigl\langle J_{b} ^{\prime } \bigl(h(s,t) \bigr), u_{b}^{+} \bigr\rangle ,\quad\quad \phi _{2} (s,t)= \bigl\langle J_{b} ^{\prime } \bigl(h(s,t) \bigr), u_{b}^{-} \bigr\rangle , \\& \psi _{1} (s,t)=\frac{1}{s} \bigl\langle J_{b} ^{\prime } \bigl(\varphi (s,t) \bigr), \varphi ^{+}(s,t) \bigr\rangle ,\quad\quad \psi _{2} (s,t)=\frac{1}{t} \bigl\langle J_{b} ^{\prime } \bigl(\varphi (s,t) \bigr), \varphi ^{-}(s,t) \bigr\rangle . \end{aligned}$$

Since $u_{b} \in M_{b}$, by Lemma 2.4, $\phi (s,t)=0\Leftrightarrow (s,t)=(1,1)$. Again, by the proof of Lemma 2.4, we know that $\frac{\partial (\phi _{1} ,\phi _{2} )}{\partial (s,t)}| _{(1,1)} >0$, and therefore, the degree theory yields

$$ \deg (\phi ,U,\theta )=1, $$

where $\theta =(0,0)$. On the other hand, for any $(s,t)\in \partial U$, by (3.2) and $0<\varepsilon <\frac{m_{b} -m^{\ast }}{2}$, we have

$$ J_{b} \bigl(h(s,t) \bigr)\le m^{\ast }< m_{b} -2 \varepsilon . $$

Thus, $h(s,t)\notin J_{b} ^{-1}(m_{b} -2\varepsilon ,m_{b} +2\varepsilon )\cap S_{2r}$. So, the conclusion (i) above implies that

$$ \varphi (s,t)=\eta \bigl(1,h(s,t) \bigr)=h(s,t), \quad \text{for any } (s,t) \in \partial U, $$

which means that $\phi (s,t)=\psi (s,t)$, for any $(s,t)\in \partial U$. Hence, $\deg (\psi ,U,\theta )=\deg (\phi ,U,\theta )=1$. Therefore, there exists $(s_{0} ,t_{0} )\in U$ such that $\psi (s_{0} ,t_{0} )=0$. That means that $\eta (1,h(s_{0} ,t_{0} ))=\varphi (s_{0} ,t_{0} ) \in M_{b}$. Hence, $J_{b} (1,h(s_{0} ,t_{0} ))\ge m_{b}$, which contradicts (3.3). Thus, we have deduced that $J_{b} ^{\prime }(u_{b} )=0$, $u_{b} \in M_{b}$, and $u_{b}$ is a least energy sign-changing solution of problem (1.1).

Finally, we show that $u_{b}$ has exactly two nodal domains. To this end, let $u_{b} =u_{1} +u_{2} +u_{3}$, satisfying that $u_{1} (x) \ge 0$, $u_{2} (x)\le 0$, for any $x\in {\mathbb{R}}^{3}$; $u_{1} (x)=u _{2} (x)=0$, for any $x\in {\mathbb{R}}^{3}\backslash (\varOmega _{1} \cup \varOmega _{2} )$; $u_{3} (x)=0$, for any $x\in \varOmega _{1} \cup \varOmega _{2}$, where $\varOmega _{1} = \{ {x\in {\mathbb{R}}^{3}| u_{1} (x)>0} \}$, $\varOmega _{2} = \{ {x\in {\mathbb{R}} ^{3}| u_{2} (x)<0} \}$ are connected open subsets of Ω.

Set $u=u_{1} +u_{2}$. Then $u^{+}=u_{1}$ and $u^{-}=u_{2}$ with $u^{\pm }\ne 0$.

In the following, we deduce that $u_{3} =0$.

Suppose by contradiction that $u_{3} \ne 0$. Then, by $J_{b} ^{\prime }(u_{b} )u_{1} =0$, we can deduce that $\varphi _{1} (1,1)\le 0$. Similarly, by $J_{b} ^{\prime }(u_{b} )u_{2} =0$, we can deduce that $\psi _{1} (1,1)\le 0$, where $\varphi _{1}$, $\psi _{1}$ are given by (2.13) with $u^{+}=u_{1}$, $u^{-}=u_{2}$. So, by Lemmas 2.4–2.5, there exists an unique pair $(s_{u} ,t_{u} )\in (0,1]\times (0,1]$ such that $s_{u} u^{+}+t_{u} u^{-}\in M_{b}$. Then

$$ m_{b} \le J_{b} (s_{u} u_{1} +t_{u} u_{2} ), \quad\quad \bigl\langle {J_{b} ^{ \prime }(s_{u} u_{1} +t_{u} u_{2} ),s_{u} u_{1} +t_{u} u_{2} } \bigr\rangle =0. $$

Consequently,

$$\begin{aligned} m_{b} \le & J_{b} (s_{u} u_{1} +t_{u} u_{2} )=J_{b} (s_{u} u_{1} +t _{u} u_{2} )- \frac{1}{q} \bigl\langle {J_{b} ^{\prime }(s_{u} u_{1} +t _{u} u_{2} ),s_{u} u_{1} +t_{u} u_{2} } \bigr\rangle \\ =& \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \bigl(s_{u}^{2} \Vert u _{1} \Vert ^{2}+t_{u}^{2} \Vert u_{2} \Vert ^{2} \bigr)+ \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr)b \bigl[ {s_{u}^{2} \vert \nabla u_{1} \vert _{2}^{2} +t_{u}^{2} \vert \nabla u_{2} \vert _{2} ^{2}} \bigr]^{2} \\ & {} + \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr) { \int _{{\mathbb{R}} ^{3}} {k(x) \bigl[s_{u}^{4}\phi _{u_{1} } (u_{1} )^{2}+t_{u}^{4} \phi _{u_{2}}(u _{2})^{2}+2s_{u}^{2}t_{u}^{2} \phi _{u_{1}}(u_{2})^{2} \bigr]\,dx}} \\ & {} +\lambda \biggl( {\frac{1}{p}-\frac{1}{q}} \biggr) \bigl(s_{u}^{p} \vert u_{1} \vert _{p}^{p} +t_{u}^{p} \vert u_{2} \vert _{p}^{p} \bigr) \\ \le & \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \bigl( \Vert u_{1} \Vert ^{2}+ \Vert u_{2} \Vert ^{2} \bigr)+ \biggl( {\frac{1}{4}- \frac{1}{q}} \biggr)b \bigl[ \vert \nabla u_{1} \vert _{2}^{2} + \vert \nabla u_{2} \vert _{2}^{2} \bigr]^{2} \\ & {} + \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr) { \int _{{\mathbb{R}} ^{3}} {k(x) \bigl[\phi _{u_{1} } (u_{1} )^{2}+\phi _{u_{2}}(u_{2})^{2}+2 \phi _{u_{1}}(u_{2})^{2} \bigr]\,dx}} \\ &{}+\lambda \biggl( { \frac{1}{p}-\frac{1}{q}} \biggr) \bigl( \vert u_{1} \vert _{p}^{p}+ \vert u_{2} \vert _{p}^{p} \bigr) \\ =&J_{b} (u_{1}+u_{2} )-\frac{1}{q} \bigl\langle {J_{b} ^{\prime }(u _{1} +u_{2} ),u_{1} +u_{2} } \bigr\rangle \\ =&J_{b}(u_{b})-\frac{1}{q} \bigl\langle {J_{b} ^{\prime }(u_{b} ),u _{1} +u_{2} } \bigr\rangle \\ & {} - \biggl[J_{b}(u_{3})+ \biggl(\frac{1}{2}- \frac{1}{q} \biggr) \biggl(b \bigl\vert \nabla (u_{1}+u_{2}) \bigr\vert _{2}^{2} \vert \nabla u_{3} \vert _{2}^{2}+ \int _{\mathbb{R} ^{3} } {k(x)\phi _{u_{1} + u_{2} } (u_{3} )^{2} \,dx} \biggr) \biggr] \\ =&J_{b}(u_{b}) - \biggl[J_{b}(u_{3})+ \biggl(\frac{1}{2}-\frac{1}{q} \biggr) \biggl(b \bigl\vert \nabla (u_{1}+u_{2}) \bigr\vert _{2}^{2} \vert \nabla u_{3} \vert _{2}^{2} \\ &{}+ \int _{\mathbb{R}^{3} } {k(x)\phi _{u_{1} + u_{2} } (u_{3} )^{2} \,dx} \biggr) \biggr]. \end{aligned}$$

(3.4)

On the other hand, by the fact that $u_{3} \ne 0$, $2\le p\le 4< q<6$, $\lambda \ge 0$, we have

$$\begin{aligned} J_{b} (u_{3} ) =&J_{b} (u_{3} )- \frac{1}{q} \bigl\langle {J_{b} ^{ \prime }(u_{b} ),u_{3} } \bigr\rangle \\ =& \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \Vert u_{3} \Vert ^{2}+ \biggl( {\frac{1}{4}- \frac{1}{q}} \biggr)b \vert \nabla u _{3} \vert _{2}^{4} + \biggl( { \frac{1}{4}-\frac{1}{q}} \biggr) \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{3} } (u_{3} )^{2}\,dx} \\ & {} +\lambda \biggl( {\frac{1}{p}-\frac{1}{q}} \biggr) \vert u_{3} \vert _{p}^{p} -\frac{b}{q} \bigl\vert \nabla ( u_{1}+ u_{2}) \bigr\vert _{2}^{2} \vert \nabla u_{3} \vert _{2}^{2} -\frac{1}{q} \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{1}+u_{2} }(u_{3} )^{2}\,dx} \\ >&-\frac{b}{q} \bigl\vert \nabla ( u_{1}+ u_{2}) \bigr\vert _{2}^{2} \vert \nabla u_{3} \vert _{2}^{2} -\frac{1}{q} \int _{{\mathbb{R}}^{3}} {k(x) \phi _{u_{1}+u_{2} } (u_{3})^{2} \,dx}. \end{aligned}$$

Thus,

$$ J_{b} (u_{3} )+\frac{1}{q} \biggl(b \bigl\vert \nabla ( u_{1}+ u_{2}) \bigr\vert _{2}^{2} \vert \nabla u_{3} \vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{1}+u_{2} } (u_{3})^{2} \,dx} \biggr) >0. $$

The condition $q>4$ shows that $\frac{1}{2}-\frac{1}{q}>\frac{1}{q}$, and therefore, it follows from the above inequality that

$$ J_{b} (u_{3} )+ \biggl(\frac{1}{2}- \frac{1}{q} \biggr) \biggl(b \bigl\vert \nabla ( u_{1}+ u_{2}) \bigr\vert _{2}^{2} \vert \nabla u_{3} \vert _{2}^{2} + \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{1}+u_{2} } (u_{3})^{2} \,dx} \biggr)>0. $$

(3.5)

By (3.4)–(3.5), we have

$$ m_{b}< J_{b}(u_{b})=m_{b}, $$

which is a contradiction. Hence, $u_{3}=0$, and therefore, $u_{b}$ has exactly two nodal domains. □

In the following, we always assume that $b>0$ in problem (1.1). We will investigate the convergence of $u_{b}$ as $b\searrow 0$.

Theorem 3.2

Assume that the conditions $(l)$, $(k)$, $(h)$ hold. Then, for any sequence $\{ {b_{n} } \}$ with $b_{n} \searrow 0$ as $n\to \infty$, there exists a subsequence, still denoted by $\{ {b_{n} } \}$ such that $u_{b_{n} } \to u_{0} \in H^{1}({\mathbb{R}}^{3})$ in $H^{1}({\mathbb{R}}^{3})$ as $n\to \infty$. Moreover, $u_{0}$ is a least energy sign-changing solution of problem (1.2).

Proof

For any sequence $\{ {b_{n} } \}$ with $b_{n}\searrow 0$ as $n\to \infty$, $u_{b_{n} }$ is one least energy sign-changing solution of problem (1.1) corresponding to $b=b_{n}$.

(1) Firstly, we show that $\{ {u_{b_{n} } } \}$ is bounded in $H^{1}({\mathbb{R}}^{3})$.

Take a nonzero function $g\in C_{0}^{\infty }({\mathbb{R}}^{3})$ with $g^{\pm }\ne 0$. Because $2\le p\le 4< q<6$, $\lambda \ge 0$, and $h(x)>0$, a.e. $x\in {\mathbb{R}}^{3}$, we can choose an appropriate positive number $\tau >0$ such that

$$ \varphi (1,1)\le 0, \quad\quad \psi (1,1)\le 0 $$

holds for all $b\in [0,1]$ corresponding to $u^{+}=\tau g^{+}$, $u^{-}=\tau g^{-}$ in (2.13). Thus, by Lemma 2.5, for each $b\in (0,1]$, there exists a pair $(s_{b} ,t_{b} )\in (0,1]\times (0,1]$ such that $s_{b} u^{+}+t_{b} u^{-}\in M_{b}$. Let $\bar{u}:=s_{b} u ^{+}+t_{b} u^{-}$ and $u_{b}$ be a least energy sign-changing solution of problem (1.1), then

$$\begin{aligned} J_{b} (u_{b} )&\le J_{b} (\bar{u})=J_{b} (\bar{u})-\frac{1}{q} \bigl\langle {J _{b} ^{\prime }( \bar{u}),\bar{u}} \bigr\rangle \\ &= \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \bigl(s_{b} \bigl\Vert u^{+} \bigr\Vert ^{2}+t_{b} \bigl\Vert u^{-} \bigr\Vert ^{2} \bigr) + \biggl( { \frac{1}{4}- \frac{1}{q}} \biggr) b \bigl[ {s_{b}^{2} \bigl\vert \nabla u ^{+} \bigr\vert _{2}^{2} +t_{b}^{2} \bigl\vert \nabla u^{-} \bigr\vert _{2}^{2}} \bigr] ^{2} \\ & \quad {} + \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr) \int _{{\mathbb{R}} ^{3}} {k(x) \bigl[s_{b}^{4} \phi _{u^{+}} \bigl(u^{+} \bigr)^{2}+t_{b}^{4} \phi _{u^{-}} \bigl(u ^{-} \bigr)^{2}+2s_{b}^{2} t_{b}^{2}\phi _{u^{+}} \bigl(u^{-} \bigr)^{2} \bigr]\,dx} \\ & \quad {} +\lambda \biggl( {\frac{1}{p}-\frac{1}{q}} \biggr) \bigl(s_{b}^{p} \bigl\vert u^{+} \bigr\vert _{p}^{p} +t_{b}^{p} \bigl\vert u^{-} \bigr\vert _{p}^{p} \bigr) \\ &\le \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \bigl( \bigl\Vert u^{+} \bigr\Vert ^{2}+ \bigl\Vert u^{-} \bigr\Vert ^{2} \bigr)+ \biggl( {\frac{1}{4}- \frac{1}{q}} \biggr) \bigl[ \bigl\vert \nabla u^{+} \bigr\vert _{2}^{2} + \bigl\vert \nabla u ^{-} \bigr\vert _{2}^{2} \bigr]^{2} \\&\quad{} +\lambda \biggl( { \frac{1}{p}-\frac{1}{q}} \biggr) \bigl( \bigl\vert u^{+} \bigr\vert _{p}^{p} + \bigl\vert u^{-} \bigr\vert _{p}^{p} \bigr) \\ & \quad {} + \biggl( {\frac{1}{4}-\frac{1}{q}} \biggr) \biggl[ { \int _{{\mathbb{R}} ^{3}} {k(x) \bigl( {\phi _{u^{+}} \bigl(u^{+} \bigr)^{2}+\phi _{u^{-}} \bigl(u^{-} \bigr)^{2}+2 \phi _{u^{+}} \bigl(u^{-} \bigr)^{2}} \bigr)\,dx} } \biggr] :=M_{0}. \end{aligned}$$

Thus, $J_{b} (u_{b} )\le M_{0}$ for all $b\in [0,1]$.

Owing to the fact that $b_{n} \searrow 0$, we can assume $b_{n} \in (0,1]$, and therefore, $J_{b_{n} } (u_{b_{n} } )\le M_{0}$ for all $n\ge 1$. On the other hand, for all $n\ge 1$,

$$ M_{0} \ge J_{b_{n} } (u_{b_{n} } )=J_{b_{n} } (u_{b_{n} } )- \frac{1}{q} \bigl\langle {J_{b_{n} } ^{\prime }(u_{b_{n} } ),u_{b_{n} } } \bigr\rangle \ge \biggl( {\frac{1}{2}-\frac{1}{q}} \biggr) \Vert u_{b_{n} } \Vert ^{2}. $$

Hence, $\{ {u_{b_{n} } } \}$ is bounded in $H^{1}( {\mathbb{R}}^{3})$.

(2) Because $\{ {u_{b_{n} } } \}$ is bounded in $H^{1}({\mathbb{R}}^{3})$, there exists a subsequence of $\{ {b_{n} } \}$, still denoted by $\{ {b_{n} } \}$, such that $u_{b_{n} } \rightharpoonup u_{0}$ as well as $u_{b_{n} }^{\pm }\rightharpoonup u_{0}^{\pm }$ weakly in $H^{1}( {\mathbb{R}}^{3})$. Then $u_{b_{n} } \to u_{0}$ in $L_{\mathrm{loc}} ^{r} ({\mathbb{R}}^{3})$ for $r\in [1,6)$ and $u_{b_{n} } (x)\to u _{0} (x)$ a.e. $x\in {\mathbb{R}}^{3}$.

Now, we show that $u_{b_{n} } \to u_{0}$ in $H^{1}({\mathbb{R}}^{3})$.

In fact, from $u_{b_{n} } \rightharpoonup u$ weakly in $H^{1}({\mathbb{R}} ^{3})$ and $J_{b_{n} } ^{\prime }(u_{b_{n} } )=0$, it follows that

$$\begin{aligned}& \bigl\langle {J_{b_{n} } ^{\prime }(u_{b_{n} } )-J_{0} ^{\prime }(u _{0} ),u_{b_{n} } -u_{0} } \bigr\rangle =- \bigl\langle {J_{0} ^{ \prime }(u_{0} ),u_{b_{n} } -u_{0} } \bigr\rangle \to 0\quad \text{as } n\to \infty . \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \Vert u_{b_{n} } -u_{0} \Vert ^{2} =& \bigl\langle {J_{b _{n} } ^{\prime }(u_{b_{n} } )-J_{0} ^{\prime }(u_{0} ),u_{b_{n} } -u _{0} } \bigr\rangle -d_{n} \\ & - \int _{{\mathbb{R}}^{3}} {k(x)\phi _{u_{b_{n} } } (u_{b_{n} } -u _{0} )^{2}\,dx} -\lambda \int _{{\mathbb{R}}^{3}} { \vert u_{b_{n} } \vert ^{p-2}(u_{b_{n} } -u_{0} )^{2}\,dx} \\ \le & \bigl\langle {J_{b_{n} } ^{\prime }(u_{b_{n} } )-J_{0} ^{ \prime }(u_{0} ),u_{b_{n} } -u_{0} } \bigr\rangle -d_{n} , \end{aligned}$$

where

$$\begin{aligned} d_{n} =&b_{n} \vert \nabla u_{b_{n} } \vert _{2}^{2} \int _{{\mathbb{R}}^{3}} {\nabla u_{b_{n} } \cdot (} \nabla u_{b_{n} } -\nabla u_{0} )\,dx+ \int _{{\mathbb{R}}^{3}} {k(x) (\phi _{u_{b_{n} } } - \phi _{u_{0} } )u_{0} (u_{b_{n} } -u_{0} )\,dx} \\ & {} +\lambda \int _{{\mathbb{R}}^{3}} { \bigl( { \vert u_{b_{n} } \vert ^{p-2}- \vert u_{0} \vert ^{p-2}} \bigr)u_{0}(u_{b_{n} }-u_{0} )\,dx} \\ & {} - \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{b_{n} } \vert ^{q-2}u_{b_{n} }(u_{b_{n} }-u_{0} )\,dx} + \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{0} \vert ^{q-2}u_{0}(u_{b_{n} }-u_{0} )\,dx}. \end{aligned}$$

In the following, we deduce that $d_{n} \to 0$ as $n\to \infty$.

Owing to the fact that $\{ {u_{b_{n} } } \}$ is bounded in $H^{1}({\mathbb{R}}^{3})$, it is easy to verify that the sequence $\{ { \vert \nabla u_{b_{n} } \vert _{2}^{2} \int _{{\mathbb{R}} ^{3}} {\nabla u_{b_{n} } \cdot (} \nabla u_{b_{n} } -\nabla u_{0} )\,dx} \}$ is bounded, and therefore

$$ b_{n} \vert \nabla u_{b_{n} } \vert _{2}^{2} \int _{{\mathbb{R}}^{3}} {\nabla u_{b_{n} } \cdot (} \nabla u_{b_{n} } -\nabla u_{0} )\,dx\to 0 $$

(3.6)

as $n\to \infty$ noting that $b_{n} \to 0$ as $n\to \infty$.

Now, we prove that

$$ \int _{{\mathbb{R}}^{3}} { \bigl( { \vert u_{b_{n} } \vert ^{p-2}- \vert u_{0} \vert ^{p-2}} \bigr)u_{0} (u_{b_{n} } -u_{0} )\,dx} \to 0 \quad \text{as } n\to \infty . $$

(3.7)

For $R>0$, let $\varOmega _{R} = \{ {x\in {\mathbb{R}}^{3}: \Vert x \Vert < R} \}$, $\varOmega _{R}^{C} = \{ {x \in {\mathbb{R}}^{3}: \Vert x \Vert \ge R} \}$, we have

$$\begin{aligned} A_{n} :=& \biggl\vert { \int _{\varOmega _{R}^{C} } { \bigl( { \vert u_{b_{n} } \vert ^{p-2}- \vert u_{0} \vert ^{p-2}} \bigr)u_{0} (u_{b_{n} } -u_{0} )\,dx} } \biggr\vert \\ \le & \int _{\varOmega _{R}^{C} } { \bigl( { \vert u_{b_{n} } \vert + \vert u_{0} \vert } \bigr)^{p-1} \vert u_{0} \vert \,dx} \\ \le & \biggl( { \int _{\varOmega _{R}^{C} } { \bigl( { \vert u_{b_{n} } \vert + \vert u_{0} \vert } \bigr)^{2(p-1)}\,dx} } \biggr)^{ \frac{1}{2}} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{2} \,dx} } \biggr)^{\frac{1}{2}} \\ \le & \biggl( { \int _{{\mathbb{R}}^{3}} { \bigl( { \vert u_{b_{n}} \vert + \vert u_{0} \vert } \bigr)^{2(p-1)}\,dx} } \biggr)^{ \frac{1}{2}} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{2} \,dx} } \biggr)^{\frac{1}{2}} \\ \le & 4^{p-1} \biggl( { \int _{{\mathbb{R}}^{3}} { \bigl( { \vert u_{b _{n}} \vert ^{2(p-1)}+ \vert u_{0} \vert ^{2(p-1)}} \bigr)\,dx} } \biggr) ^{\frac{1}{2}} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{2} \,dx} } \biggr)^{\frac{1}{2}} \\ =&4^{p-1} \bigl( { \vert u_{b_{n}} \vert _{2(p-1)}^{2(p-1)} + \vert u _{0} \vert _{2(p-1)}^{2(p-1)} } \bigr)^{\frac{1}{2}} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{2} \,dx} } \biggr)^{\frac{1}{2}}. \end{aligned}$$

Because of the boundedness of $\{ {u_{b_{n} } } \}$ combined with (2.2), there exists $M_{1} >0$ such that

$$ 4^{p-1} \bigl( { \vert u_{n} \vert _{2(p-1)}^{2(p-1)} + \vert u_{0} \vert _{2(p-1)}^{2(p-1)} } \bigr)^{\frac{1}{2}}\le M_{1}. $$

On the other hand, the fact that $u_{0} \in L^{2} ({\mathbb{R}}^{3})$ implies that $\forall \varepsilon >0$, we can choose $R>0$ large enough so that

$$ \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{2} \,dx} } \biggr) ^{\frac{1}{2}}< \frac{\varepsilon }{2M_{1} }. $$

Then

$$ \biggl\vert { \int _{\varOmega _{R}^{C} } { \bigl( { \vert u_{b_{n} } \vert ^{p-2}- \vert u_{0} \vert ^{p-2}} \bigr)u_{0} (u_{b_{n} } -u_{0} )\,dx} } \biggr\vert < \frac{ \varepsilon }{2}. $$

(3.8)

Similarly, for given R above, there exists $M_{2}>0$ such that

$$\begin{aligned} B_{n} :=& \biggl\vert { \int _{\varOmega _{R} } { \bigl( { \vert u_{b_{n} } \vert ^{p-2}- \vert u_{0} \vert ^{p-2}} \bigr)u_{0} (u_{b_{n} } -u_{0} )\,dx} } \biggr\vert \\ \le & \int _{\varOmega _{R} } { \bigl( { \vert u_{b_{n} } \vert ^{p-2}+ \vert u_{0} \vert ^{p-2}} \bigr) \vert u_{0} \vert \vert u_{b_{n} } -u_{0} \vert \,dx} \\ \le & \int _{\varOmega _{R} } { \bigl( { \vert u_{b_{n} } \vert + \vert u _{0} \vert } \bigr)^{p-2} \vert u_{0} \vert \vert u_{b_{n} } -u _{0} \vert \,dx} \\ \le & \int _{\varOmega _{R} } { \bigl( { \vert u_{b_{n} } \vert + \vert u _{0} \vert } \bigr)^{p-1} \vert u_{b_{n} } -u_{0} \vert \,dx} \\ \le & \biggl( { \int _{\varOmega _{R} } { \bigl( { \vert u_{b_{n} } \vert + \vert u_{0} \vert } \bigr)^{2(p-1)}\,dx} } \biggr)^{\frac{1}{2}} \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u_{0} \vert ^{2}\,dx} } \biggr) ^{\frac{1}{2}} \\ \le & M_{2} \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u_{0} \vert ^{2}\,dx} } \biggr)^{\frac{1}{2}}. \end{aligned}$$

By the fact that $u_{b_{n} } \to u_{0}$ in $L_{\mathrm{loc}}^{r} ( {\mathbb{R}}^{3})$, $r\in [1, 6)$, there exists $N_{1} \ge 1$, such that

$$ M_{2} \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u_{0} \vert ^{2}\,dx} } \biggr)^{\frac{1}{2}}< \frac{\varepsilon }{2} $$

as $n\ge N_{1}$. Thus,

$$ \biggl\vert { \int _{\varOmega _{R} } { \bigl( { \vert u_{b_{n} } \vert ^{p-2}- \vert u_{0} \vert ^{p-2}} \bigr)u_{0}(u_{b_{n} } -u_{0} )\,dx} } \biggr\vert < \frac{ \varepsilon }{2} $$

(3.9)

as $n\ge N_{1}$.

Hence, (3.8)–(3.9) shows that (3.7) holds.

Now, we show that

$$ \begin{aligned}[b] & \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{b_{n} } \vert ^{q-2}u_{b_{n} }(u _{b_{n} } -u_{0} )\,dx} \\ &\quad{} - \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{0} \vert ^{q-2}u_{0}(u_{b_{n} } -u_{0} )\,dx} \to 0, \quad \text{as }n\to \infty . \end{aligned} $$

(3.10)

We only prove that $\int _{{\mathbb{R}}^{3}} {h(x) \vert u_{b_{n} } \vert ^{q-2}u_{b_{n} }(u_{b_{n} }-u_{0} )\,dx}\rightarrow 0$ as $n \to \infty$, because the proof on $\int _{{\mathbb{R}}^{3}} {h(x) \vert u_{0} \vert ^{q-2}u_{0}(u_{b_{n} } -u_{0} )\,dx} \to 0$ is similar.

We make an argument similar to (3.7) as follows:

For $R>0$, we have

$$\begin{aligned}& \biggl\vert { \int _{{\mathbb{R}}^{3}} {h(x) \vert u_{b_{n} } \vert ^{q-2}u _{b_{n} } (u_{b_{n} } -u_{0} )\,dx} } \biggr\vert \\& \quad \le \int _{{\mathbb{R}} ^{3}} {h(x) \vert u_{b_{n} } \vert ^{q-1} \vert u_{b_{n} } -u_{0} \vert \,dx} \\& \quad = \int _{\varOmega _{R} } {h(x) \vert u_{b_{n} } \vert ^{q-1} \vert u _{b_{n} } -u_{0} \vert \,dx} + \int _{\varOmega _{R}^{C} } {h(x) \vert u_{b _{n} } \vert ^{q-1} \vert u_{b_{n} } -u_{0} \vert \,dx} . \end{aligned}$$

Take $r_{1} =\frac{6}{q-1}$, $r_{2} =\frac{6}{1+(q_{1}-q)}$. Then the condition $(l)$ ensures that $1< r_{1}< 2<r_{2}<6$. By the boundedness of $\{ {u_{b_{n} } } \}$ combined with (2.2) and the fact that $u_{b_{n} } \to u_{0}$ in $L_{\mathrm{loc}}^{r} ({\mathbb{R}} ^{3})$, $r\in [1,6)$, there exists $M_{1} >0$ such that

$$\begin{aligned} A_{n} :=& \int _{\varOmega _{R} } {h(x) \vert u_{b_{n} } \vert ^{q-1} \vert u _{b_{n} } -u_{0} \vert \,dx} \\ \le & \biggl( { \int _{\varOmega _{R} } {h^{\frac{6}{6-q_{1}}}\,dx} } \biggr) ^{\frac{6-q_{1}}{6}} \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } \vert ^{(q-1)r_{1} } \,dx} } \biggr)^{\frac{1}{r_{1} }} \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u_{0} \vert ^{r_{2} }\,dx} } \biggr)^{\frac{1}{r _{2} }} \\ \le & \vert h \vert _{\frac{6}{6-q_{1}}} \vert u_{b_{n} } \vert _{6} ^{q-1} \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u_{0} \vert ^{r _{2} }\,dx} } \biggr)^{\frac{1}{r_{2} }} \\ \le & M_{1} \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u_{0} \vert ^{r_{2} }\,dx} } \biggr)^{\frac{1}{r_{2} }}\to 0 \end{aligned}$$

(3.11)

as $n\to \infty$.

Similarly, let $r=\frac{6q}{q_{1}}$, $r_{1} =\frac{6}{q-1}$, $r_{2} =\frac{6}{1+(q _{1}-q)}$, then $4< r<6$ and there exists $M_{2} >0$ such that

$$\begin{aligned} B_{n} :=& \int _{\varOmega _{R}^{C} } {h(x) \vert u_{b_{n} } \vert ^{q-1} \vert u_{b_{n} } -u_{0} \vert \,dx} \\ \le & \int _{\varOmega _{R}^{C} } {h(x) \bigl( { \vert u_{b_{n} } \vert ^{q}+ \vert u_{b_{n} } \vert ^{q-1} \vert u_{0} \vert } \bigr)\,dx} \\ \le & \biggl( { \int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx} } \biggr) ^{\frac{6-q_{1}}{6}} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{b_{n} } \vert ^{r} \,dx} } \biggr)^{\frac{q}{r}} \\ & {} + \biggl( { \int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx} } \biggr) ^{\frac{6-q_{1}}{6}} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{b_{n} } \vert ^{r_{1} (q-1)} \,dx} } \biggr)^{\frac{1}{r_{1} }} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{r_{2} } \,dx} } \biggr)^{ \frac{1}{r_{2} }} \\ \le & \biggl( { \int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx} } \biggr) ^{\frac{6-q_{1}}{6}} \bigl[ { \vert u_{b_{n} } \vert _{r}^{q} + \vert u _{b_{n} } \vert _{6}^{q-1} \vert u_{0} \vert {^{r_{2} }} } \bigr] \le M_{2} \biggl( { \int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx} } \biggr)^{\frac{6-q_{1}}{6}}. \end{aligned}$$

Because $h\in L^{\frac{6}{6-q_{1}}}({\mathbb{R}}^{3})$, for any $\varepsilon >0$, we can choose $R>0$ large enough so that

$$ \biggl({ \int _{\varOmega _{R}^{C} } {h^{\frac{6}{6-q_{1}}}\,dx}} \biggr)^{\frac{6-q _{1}}{6}} < \frac{\varepsilon }{M_{2} } $$

and therefore

$$ \int _{\varOmega _{R}^{C} } {h(x) \vert u_{b_{n} } \vert ^{q-1} \vert u_{b _{n} } -u_{0} \vert \,dx} < \varepsilon . $$

(3.12)

Hence, by (3.11)–(3.12), we conclude that $\int _{{\mathbb{R}} ^{3}} {h(x) \vert u_{b_{n} } \vert ^{q-2}u_{b_{n} } (u_{b_{n} } -u_{0} )\,dx} \to 0$ as $n\to \infty$.

Finally, we show that

$$ \int _{{\mathbb{R}}^{3}} {k(x) ( {\phi _{u_{b_{n} }} -\phi _{u_{0} } } )u_{0} (u_{b_{n} } -u_{0} )\,dx} \to 0\quad \text{as }n\to \infty . $$

(3.13)

In fact, by Lemma 2.2(iii)–(iv) together with the boundedness of $\{ {u_{b_{n} } } \}$ in $H^{1}({\mathbb{R}}^{3})$, we have

$$\begin{aligned}& \biggl\vert { \int _{{\mathbb{R}}^{3}} {k(x) ( {\phi _{u_{b_{n} }} - \phi _{u_{0} } } )u_{0} (u_{b_{n} } -u_{0} )\,dx} } \biggr\vert \\& \quad \le \int _{{\mathbb{R}}^{3}} {k(x) \vert \phi _{u_{b_{n} }} -\phi _{u_{0} } \vert \bigl\vert u_{0} (u_{b_{n} } -u_{0} ) \bigr\vert \,dx} \\& \quad \le \vert k \vert _{\infty } \biggl( { \int _{{\mathbb{R}}^{3}} { \vert \phi _{u_{b_{n} }} -\phi _{u_{0} } \vert ^{6}} } \biggr)^{\frac{1}{6}} \biggl( { \int _{{\mathbb{R}}^{3}} { \bigl\vert u_{0} (u_{b_{n} } -u_{0} ) \bigr\vert ^{\frac{6}{5}}\,dx} } \biggr)^{\frac{5}{6}} \\& \quad \le \vert k \vert _{\infty }\bigl( \vert \phi _{u_{b_{n} }} \vert _{6} + \vert \phi _{u_{0} } \vert _{6} \bigr) \biggl( { \int _{{\mathbb{R}}^{3}} { \bigl\vert u _{0} (u_{b_{n} } -u_{0} ) \bigr\vert ^{\frac{6}{5}}\,dx} } \biggr)^{ \frac{5}{6}} \\& \quad \le \vert k \vert _{\infty }\bar{S}^{-2}S_{6}^{-2} \vert k \vert _{2} \bigl( \Vert u_{b_{n} } \Vert ^{2}+ \Vert u_{0} \Vert ^{2} \bigr) \biggl( { \int _{{\mathbb{R}}^{3}} { \bigl\vert u_{0} (u_{b_{n} } -u _{0} ) \bigr\vert ^{\frac{6}{5}}\,dx} } \biggr)^{\frac{5}{6}} \\& \quad \le M_{1} \biggl( { \int _{{\mathbb{R}}^{3}} { \bigl\vert u_{0} (u_{b_{n} } -u _{0} ) \bigr\vert ^{\frac{6}{5}}\,dx} } \biggr)^{\frac{5}{6}}. \end{aligned}$$

(3.14)

For any $R>0$, we know that

$$ \int _{{\mathbb{R}}^{3}} { \bigl\vert u_{0} (u_{b_{n} } -u_{0} ) \bigr\vert ^{\frac{6}{5}}\,dx} = \int _{\varOmega _{R} } { \bigl\vert u_{0} (u_{b_{n} } -u_{0} ) \bigr\vert ^{\frac{6}{5}}\,dx} + \int _{\varOmega _{R}^{C} } { \bigl\vert u_{0} (u _{b_{n} } -u_{0} ) \bigr\vert ^{\frac{6}{5}}\,dx}. $$

Because

$$\begin{aligned} \biggl( { \int _{\varOmega _{R}^{C} } { \bigl\vert u_{0} (u_{b_{n} } -u_{0} ) \bigr\vert ^{\frac{6}{5}}\,dx} } \biggr) \le & \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{\frac{3}{2}} \,dx} } \biggr)^{\frac{4}{5}} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{b_{n} } -u_{0} \biggr) \vert ^{6}\,dx} } ) ^{\frac{1}{5}} \\ \le & \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{ \frac{3}{2}} \,dx} } \biggr)^{\frac{4}{5}} \biggl( { \int _{\varOmega _{R}^{C} } {\bigl( \vert u_{b_{n} } \vert + \vert u_{0} \vert \bigr)^{6}\,dx} } \biggr)^{ \frac{1}{5}} \\ \le & \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{ \frac{3}{2}} \,dx} } \biggr)^{\frac{4}{5}}\bigl( \vert u_{b_{n} } \vert _{6} + \vert u_{0} \vert _{6} \bigr)^{\frac{6}{5}} \\ \le & M_{2} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{\frac{3}{2}} \,dx} } \biggr)^{\frac{4}{5}}, \end{aligned}$$

(3.15)

and $u_{0} \in L^{\frac{3}{2}}({\mathbb{R}}^{3})$, for any given $\varepsilon >0$, we can choose $R>0$ large enough so that

$$ M_{2} \biggl( { \int _{\varOmega _{R}^{C} } { \vert u_{0} \vert ^{\frac{3}{2}} \,dx} } \biggr)^{\frac{4}{5}}< \varepsilon /2. $$

(3.16)

On the other hand, since $u_{b_{n} } \to u_{0}$ in $L_{\mathrm{loc}} ^{r}({\mathbb{R}}^{3})$, $r\in [1,6)$, there exists $N>0$ such that

$$\begin{aligned} \int _{\varOmega _{R} } { \bigl\vert u_{0} (u_{b_{n} } -u_{0} ) \bigr\vert ^{ \frac{6}{5}}\,dx} =& \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u _{0} \biggr) \vert ^{\frac{3}{2}}\,dx} } )^{\frac{4}{5}} \biggl( { \int _{\varOmega _{R} } { \vert u_{0} \vert ^{6} \,dx} } \biggr)^{\frac{1}{5}} \\ \le & \biggl( { \int _{\varOmega _{R} } { \vert u_{b_{n} } -u_{0} \biggr) \vert ^{\frac{3}{2}}\,dx} } )^{\frac{4}{5}} \biggl( { \int _{{\mathbb{R}} ^{3}} { \vert u_{0} \vert ^{6} \,dx} } \biggr)^{\frac{1}{5}}< \varepsilon /2 \end{aligned}$$

(3.17)

as $n>N$. Thus, by (3.14)–(3.17), we find that (3.13) holds. Consequently, by (3.6), (3.7), (3.10) and (3.13), we obtain $d_{n} \to 0$ as $n\to \infty$, and therefore, $u_{b_{n} } \to u_{0}$ as $n\to \infty$ in $H^{1}({\mathbb{R}}^{3})$.

Now, by

$$\begin{aligned} \bigl\langle J_{0} ^{\prime }(u_{0} ),u_{b_{n} }^{\pm } \bigr\rangle =& \bigl\langle J'_{b_{n} } (u_{b_{n} } ),u_{b_{n} }^{\pm } \bigr\rangle - \bigl\langle J'_{b _{n} } (u_{b_{n} } )- J'_{0} (u_{b_{n} } ),u_{b_{n} }^{\pm } \bigr\rangle - \bigl\langle J'_{0} (u_{b_{n} } )-J'_{0} (u_{0} ), u_{b_{n} }^{\pm } \bigr\rangle \\ =&- \bigl\langle J'_{b_{n} } (u_{b_{n} } )- J'_{0} (u_{b_{n} } ),u_{b_{n} }^{\pm } \bigr\rangle - \bigl\langle J'_{0} (u_{b_{n} } )-J'_{0} (u_{0} ), u_{b _{n} }^{\pm } \bigr\rangle \end{aligned}$$

combined with the fact that $u_{b_{n}}\rightarrow u_{0}$, $u_{b_{n}} ^{\pm }\rightharpoonup u_{0}^{\pm }$ in $H^{1}({\mathbb{R}}^{3})$ and $b_{n}\rightarrow 0$ and that $J_{b}$ is a $C^{1}$ functional in $H^{1}({\mathbb{R}}^{3})$, we have $J_{0}'(u_{0})=0$ and

$$ \bigl\langle J_{0} ^{\prime }(u_{0} ),u_{0}^{\pm } \bigr\rangle =\lim_{n \to \infty } \bigl\langle J_{0} ^{\prime }(u_{0} ),u_{b_{n} } ^{\pm } \bigr\rangle =0. $$

On the other hand, by an argument similar to (2.26) and a subsequent derivation on $u_{b}^{\pm }\neq 0$, we know that $u_{0}^{\pm }\neq 0$. So, $u_{0} \in M_{0}$. That is, $u_{0}$ is a sign-changing weak solution of problem (1.2)

Now, we prove that $u_{0}$ is also a least energy sign-changing solution of problem (1.2). In fact, assume that $v_{0} \in M_{0}$ is a least energy sign-changing solution of problem (1.2). Then, by Lemma 2.4, there is a sequence of pairs $(s_{b_{n} } ,t_{b_{n} } ) \in \mathbb{R}_{+}^{0} \times \mathbb{R}_{+}^{0}$ with $s_{b_{n} } v _{0}^{+} +t_{b_{n} } v_{0}^{-} \in M_{b_{n} }$. Because $b_{n} \to 0$ as $n\to \infty$, observing that (2.11) holds corresponding to $b=b_{n}$, $u=v_{0}$ and taking into account that $2\le p\le 4< q<6$, $\lambda \ge 0$, and $h(x)>0$, a.e. $x\in {\mathbb{R}} ^{3}$, it is easy to check that $\{ {(s_{b_{n} }, t_{b_{n} } )} \}$ is bounded. Without loss of generality, we can assume that $s_{b_{n} } \to s_{0} \in \mathbb{R}_{+}$, $t_{b_{n} } \to t_{0} \in \mathbb{R}_{+}$. Moreover, by an argument similar to that in (2)(ii) of the proof of Lemma 2.4, it can be verified that $s_{0} >0$, $t_{0} >0$. Consequently, we have

$$ s_{0} v_{0}^{+} +t_{0} v_{0}^{-} =\lim_{n\to \infty } \bigl(s_{b_{n} } v_{0}^{+} +t_{b_{n} } v_{0}^{-} \bigr)\in M_{0} . $$

Applying Lemma 2.4 for $b=0$, it follows from $v_{0} \in M_{0}$ that $s_{0} =t_{0} =1$. Thus, we have

$$ J_{0} (v_{0} )\le J_{0} (u_{0} )= \lim_{n\to \infty } J_{b_{n} } (u_{b_{n} } )\le \lim _{n\to \infty } J _{b_{n} } \bigl(s_{b_{n} } v_{0}^{+} +t_{b_{n} } v_{0}^{-} \bigr) =J_{0} \bigl(v_{0} ^{+} +v_{0}^{-} \bigr)=J_{0} (v_{0} ). $$

Hence, $J_{0} (v_{0} )=J_{0} (u_{0} )$. That is, $u_{0}$ is also a least energy sign-changing solution of problem (1.2). The proof of Theorem 3.2 is complete. □

4 Conclusion

In this paper, with the help of the constraint variational method combined with a quantitative lemma, Kirchhoff–Poisson systems (1.1) are investigated and the existence result on the least energy sign-changing solution with two nodal domains to the problem is established. Moreover, the convergence property of $u_{b}$ as $b \searrow 0$ is also obtained. It should be pointed out that, because the nonlocal terms $b\int _{{\mathbb{R}}^{3}} {( \vert \nabla u \vert ^{2}\,dx)\Delta u}$ and $\phi _{u}$ are involved here, the above Kirchhoff–Poisson systems are totally different from the case $b=0$ and $k=0$ and there are more difficulties we need to overcome in the proof.