In order to apply the variational method, we first recall some related preliminaries and establish a corresponding variational framework for our problem (1.1); then we give the proof of Theorem 1.1.
For \(1< s<+\infty\), define the Sobolev space
$$W^{m,s}\bigl(\mathbb{R}^{N}\bigr)=\bigl\{ u\in L^{s}\bigl(\mathbb{R}^{N}\bigr)\mid D^{\alpha}u\in L^{s}\bigl(\mathbb{R}^{N}\bigr), \vert \alpha \vert \leq m\bigr\} $$
equipped with the norm
$$\Vert u \Vert _{W^{m,s}(\mathbb{R}^{N})}= \biggl(\sum_{ \vert \alpha \vert \leq m} \int_{\mathbb{R}^{N}} \bigl\vert D^{\alpha}u \bigr\vert ^{s}\,dx \biggr)^{\frac{1}{s}}, $$
where \(\alpha=(\alpha_{1}, \alpha_{2},\ldots, \alpha_{N})\) with \(\alpha_{i}\in \mathbb{Z}^{+} \) (the set of all non-negative integers), \(i=1, 2, \ldots, N\), \(\vert \alpha \vert =\alpha_{1}+\alpha_{2}+\cdots+\alpha_{N}\) and
$$D^{\alpha}u=\frac{\partial^{ \vert \alpha \vert }u}{\partial x^{\alpha_{1}}_{1}\partial x^{\alpha_{2}}_{2}\cdots\partial x^{\alpha_{N}}_{N}}. $$
For \(s=2\), \(H^{m}(\mathbb{R}^{N})=W^{m,2}(\mathbb{R}^{N})\) is a Hilbert space equipped with the scalar product
$$\langle u,v\rangle_{H^{m}}=\sum_{ \vert \alpha \vert \leq m} \int_{\mathbb{R}^{N}}D^{\alpha}u D^{\alpha}v\,dx $$
and the norm
$$\Vert u \Vert _{H^{m}}=\langle u,u\rangle_{H^{m}}^{\frac{1}{2}}= \biggl(\sum_{ \vert \alpha \vert \leq m} \int_{\mathbb{R}^{N}} \bigl\vert D^{\alpha}u \bigr\vert ^{2}\,dx \biggr)^{\frac{1}{2}}. $$
Moreover, for \(m=2\) one has
$$\begin{aligned} &\langle u,v\rangle_{H^{2}}= \int_{\mathbb{R}^{N}}(\triangle u\triangle v+\nabla u \nabla v+uv)\,dx, \\ & \Vert u \Vert ^{2}_{H^{2}}=\langle u,v \rangle_{H^{2}}= \int_{\mathbb{R}^{N}}\bigl( \vert \triangle u \vert ^{2}+ \vert \nabla u \vert ^{2}+u^{2}\bigr)\,dx, \end{aligned}$$
whenever \(u,v \in H^{2}(\mathbb{R}^{N})\).
Under assumption \((V)\), we can find \(V_{0}\geq 0\) such that \(\widetilde{V}(x)=V(x)+V_{0}\geq 1\) for all \(x\in\mathbb{R}^{3}\). Then
$$E_{\lambda}= \biggl\{ u\in H^{2}\bigl(\mathbb{R}^{3} \bigr): \int_{\mathbb{R}^{3}}\bigl(a \vert \nabla u \vert ^{2}+ \lambda\widetilde{V}(x)u^{2}\bigr)\,dx< \infty \biggr\} $$
is a Hilbert space endowed with the norm
$$\Vert u \Vert _{\lambda}= \biggl( \int_{\mathbb{R}^{3}}\bigl( \vert \triangle u \vert ^{2}+a \vert \nabla u \vert ^{2}+\lambda\widetilde{V}(x)u^{2} \bigr)\,dx \biggr)^{\frac{1}{2}}. $$
Let
$$\begin{aligned} \Phi_{\lambda}(u) =&\frac{1}{2} \int_{\mathbb{R}^{3}}\bigl( \vert \triangle u \vert ^{2}+a \vert \nabla u \vert ^{2}+\lambda V(x)u^{2}\bigr)\,dx+ \frac{b}{4} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx \biggr)^{2} \\ &{}+\frac{1}{2} \int_{\mathbb{R}^{3}}u^{2} \vert \nabla u \vert ^{2}\,dx- \int_{\mathbb{R}^{3}}F(x,u)\,dx,\quad \forall u\in E_{\lambda}. \end{aligned}$$
(2.1)
By condition \((V)\), \((F_{1})\) and the fact \(\int_{\mathbb{R}^{3}}u^{2} \vert \nabla u \vert ^{2}\,dx<\infty\) (see Lemma 2.2 in [20]), \(\Phi_{\lambda}\) is a well-defined class \(C^{1}\) functional. For all \(u,v\in E_{\lambda}\)
$$\begin{aligned} \bigl\langle \Phi'_{\lambda}(u),v\bigr\rangle =& \int_{\mathbb{R}^{3}}\bigl(\triangle u\triangle v+a\nabla u\nabla v+\lambda V(x)uv\bigr)\,dx+b \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx \int_{\mathbb{R}^{3}}\nabla u\nabla v\,dx \\ &{}+ \int_{\mathbb{R}^{3}}\bigl(uv \vert \nabla u \vert ^{2}+u^{2} \nabla u \nabla v\bigr)\,dx- \int_{\mathbb{R}^{3}}f(x,u)v\,dx. \end{aligned}$$
(2.2)
Clearly, seeking a weak solution of problem (1.1) is equivalent to finding a critical point of the functional \(\Phi_{\lambda}\).
Definition 2.1
A sequence \(\{u_{n}\}\subset E_{\lambda}\) is said to be a \((C)_{c}\) sequence if
$$\Phi_{\lambda}(u_{n})\rightarrow c,\qquad \bigl\Vert \Phi'_{\lambda}(u_{n}) \bigr\Vert _{\lambda} \bigl(1+ \Vert u_{n} \Vert _{\lambda}\bigr)\rightarrow 0. $$
\(\Phi_{\lambda}\) is said to satisfy the \((C)_{c}\) condition if any \((C)_{c}\) sequence possesses a convergent subsequence.
Let \(E'_{\lambda}= \{u\in H^{2}(\mathbb{R}^{N}):\int_{\mathbb{R}^{N}}(a \vert \nabla u \vert ^{2}+\lambda\widetilde{V}(x)u^{2})\,dx<\infty \}\).
Lemma 2.2
Under assumption
\((V)\), the embedding
\(E'_{\lambda}\hookrightarrow L^{s}(\mathbb{R}^{N})\)
is compact for
\(2\leq s<2_{\ast}\), where
\(2_{\ast}=\frac{2N}{N-4}\), if
\(N>4\); \(2_{\ast}=+\infty\), if
\(N\leq4\).
Proof
Define
$$E= \biggl\{ u\in H^{1}\bigl(\mathbb{R}^{N}\bigr): \int_{\mathbb{R}^{N}}\bigl(a \vert \nabla u \vert ^{2}+ \lambda\widetilde{V}(x)u^{2}\bigr)\,dx< \infty \biggr\} . $$
By Propositions 3.1 and 3.3 in [13], we know that the embedding \(E\hookrightarrow L^{s}(\mathbb{R}^{N})\) is compact for \(2\leq s<2_{\ast}\) due to the condition \((V)\), and the embedding \(E'_{\lambda}\hookrightarrow E\) is continuous, therefore, the embedding \(E'_{\lambda}\hookrightarrow L^{s}(\mathbb{R}^{N})\) is compact for \(2\leq s<2_{\ast}\). □
Lemma 2.3
Under assumptions
\((V)\), \((F_{1})\), any bounded
\((C)_{c}\)
sequence of
\(\Phi_{\lambda}\)
has a strongly convergent subsequence in
\(E_{\lambda}\).
Proof
Let \(\{u_{n}\}\subset E_{\lambda}\) hold with
$$ \sup_{n} \Vert u_{n} \Vert _{\lambda}< +\infty. $$
(2.3)
Then up to a subsequence, there exists a constant \(c\in\mathbb{R}\) such that
$$ \Phi_{\lambda}(u_{n})\rightarrow c,\qquad \Phi'_{\lambda}(u_{n})\rightarrow 0. $$
(2.4)
According to Lemma 2.2, going if necessary to a subsequence, we can assume that there exists \(u\in E_{\lambda}\) such that
$$\begin{aligned} &u_{n}\rightharpoonup u\quad \mbox{in }E_{\lambda}, \\ &u_{n}\rightarrow u \quad \mbox{in }L^{s}\bigl( \mathbb{R}^{3}\bigr)\ (2\leq s< +\infty), \\ &u_{n}\rightarrow u\quad \mbox{a.e. in } \mathbb{R}^{3}. \end{aligned}$$
(2.5)
By an elementary computation,
$$\begin{aligned} &\bigl\langle \Phi'_{\lambda}(u_{n})- \Phi'(u),u_{n}-u\bigr\rangle \\ &\quad \geq \Vert u_{n}-u \Vert ^{2}_{\lambda}- \lambda V_{0} \int_{\mathbb{R}^{3}} \vert u_{n}-u \vert ^{2}\,dx \\ &\qquad {}+b\biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{n} \vert ^{2}\,dx- \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx\biggr) \int_{\mathbb{R}^{3}}\nabla u_{n} \nabla (u_{n}-u) \,dx \\ &\qquad {}+ \int_{\mathbb{R}^{3}}\bigl(u_{n} \vert \nabla u_{n} \vert ^{2}-u \vert \nabla u \vert ^{2}\bigr) (u_{n}-u)\,dx + \int_{\mathbb{R}^{3}}\bigl(u^{2}_{n}-u^{2} \bigr)\nabla u\nabla (u_{n}-u)\,dx \\ &\qquad {}+ \int_{\mathbb{R}^{3}}\bigl(f(x,u)-f(x,u_{n})\bigr) (u_{n}-u)\,dx. \end{aligned}$$
(2.6)
Clearly, \(\lambda V_{0}\int_{\mathbb{R}^{3}} \vert u_{n}-u \vert ^{2}\,dx\rightarrow 0\), and \(\langle\Phi'_{\lambda}(u_{n})-\Phi'(u),u_{n}-u\rangle\rightarrow 0\). Then, since \(\{u_{n}\}\subset E_{\lambda}\) is bounded, we have
$$\begin{aligned} & \biggl\vert b\biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{n} \vert ^{2}\,dx- \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx\biggr) \int_{\mathbb{R}^{3}}\nabla u_{n}\nabla (u_{n}-u)\,dx \biggr\vert \\ &\quad \leq \biggl\vert b\biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{n} \vert ^{2}\,dx- \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx\biggr) \int_{\mathbb{R}^{3}}\nabla u\nabla (u_{n}-u)\,dx \biggr\vert \\ &\qquad {}+ \biggl\vert b\biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{n} \vert ^{2}\,dx- \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx\biggr) \int_{\mathbb{R}^{3}} \bigl\vert \nabla (u_{n}-u) \bigr\vert ^{2}\,dx \biggr\vert \\ &\quad \rightarrow 0. \end{aligned}$$
(2.7)
Note that \(E_{\lambda}\hookrightarrow H^{2}(\mathbb{R}^{3})\hookrightarrow W^{1,s}(\mathbb{R}^{3})\) for \(2\leq s \leq+\infty\),
$$\begin{aligned} \int_{\mathbb{R}^{3}} \vert \nabla u_{n} \vert ^{3}\,dx &\leq \int_{\mathbb{R}^{3}} \Biggl( \vert u_{n} \vert ^{2}+\sum^{3}_{i=1} \biggl\vert \frac{\partial u_{n}}{\partial x_{i}} \biggr\vert ^{2} \Biggr)^{\frac{3}{2}}\,dx \\ &\leq \int_{\mathbb{R}^{3}} \Biggl( \vert u_{n} \vert +\sum ^{3}_{i=1} \biggl\vert \frac{\partial u_{n}}{\partial x_{i}} \biggr\vert \Biggr)^{3}\,dx \\ &\leq \int_{\mathbb{R}^{3}} \biggl[4\max \biggl\{ \vert u_{n} \vert , \biggl\vert \frac{\partial u_{n}}{\partial x_{1}} \biggr\vert , \biggl\vert \frac{\partial u_{n}}{\partial x_{2}} \biggr\vert , \biggl\vert \frac{\partial u_{n}}{\partial x_{3}} \biggr\vert \biggr\} \biggr]^{3}\,dx \\ &\leq4^{3} \int_{\mathbb{R}^{3}} \Biggl( \vert u_{n} \vert ^{3}+\sum^{3}_{i=1} \biggl\vert \frac{\partial u_{n}}{\partial x_{i}} \biggr\vert ^{3} \Biggr)\,dx \\ &=4^{3} \Vert u_{n} \Vert ^{3}_{W^{1,3}(\mathbb{R}^{3})} \\ &\leq4^{3}S^{3}_{3} \Vert u_{n} \Vert ^{3}_{\lambda}, \end{aligned}$$
(2.8)
where
$$S_{s}=\sup_{u\in E_{\lambda}, \Vert u \Vert _{\lambda}=1} \Vert u \Vert _{W^{1,s}}, \quad \forall 2\leq s\leq +\infty. $$
Applying (2.3)–(2.5) and (2.8), there exist constants \(C_{1}>0\) such that
$$\begin{aligned} & \biggl\vert \int_{\mathbb{R}^{3}}\bigl(u_{n} \vert \nabla u_{n} \vert ^{2}-u \vert \nabla u \vert ^{2}\bigr) (u_{n}-u)\,dx \biggr\vert \\ &\quad \leq \int_{\mathbb{R}^{3}} \vert u_{n} \vert \vert \nabla u_{n} \vert ^{2} \vert u_{n}-u \vert \,dx+ \int_{\mathbb{R}^{3}} \vert u \vert \vert \nabla u \vert ^{2} \vert u_{n}-u \vert \,dx \\ &\quad \leq \biggl( \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{6}\,dx \biggr)^{\frac{1}{6}} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{n} \vert ^{3}\,dx \biggr)^{\frac{2}{3}} \biggl( \int_{\mathbb{R}^{3}} \vert u_{n}-u \vert ^{6} \,dx \biggr)^{\frac{1}{6}} \\ &\qquad {}+ \biggl( \int_{\mathbb{R}^{3}} \vert u \vert ^{6}\,dx \biggr)^{\frac{1}{6}} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{3}\,dx \biggr)^{\frac{2}{3}} \biggl( \int_{\mathbb{R}^{3}} \vert u_{n}-u \vert ^{6} \,dx \biggr)^{\frac{1}{6}} \\ &\quad \leq C_{1} \Vert u_{n}-u \Vert _{L^{6}} \rightarrow 0, \quad \mbox{as } n\rightarrow\infty, \end{aligned}$$
(2.9)
and \(C'_{1}>0\) such that
$$\begin{aligned} & \biggl\vert \int_{\mathbb{R}^{3}}\bigl(u_{n}^{2}-u^{2} \bigr)\nabla u\nabla(u_{n}-u)\,dx \biggr\vert \\ &\quad \leq \int_{\mathbb{R}^{3}} \vert u_{n}-u \vert \vert u_{n}+u \vert \vert \nabla u \vert \bigl\vert \nabla(u_{n}-u) \bigr\vert \,dx \\ &\quad \leq \biggl( \int_{\mathbb{R}^{3}} \vert u_{n}-u \vert ^{6} \biggr)^{\frac{1}{6}} \biggl( \int_{\mathbb{R}^{3}} \vert u_{n}+u \vert ^{6} \biggr)^{\frac{1}{6}} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{3} \biggr)^{\frac{1}{3}} \biggl( \int_{\mathbb{R}^{3}} \bigl\vert \nabla(u_{n}-u) \bigr\vert ^{3} \biggr)^{\frac{1}{3}} \\ &\quad \leq C'_{1} \Vert u_{n}-u \Vert _{L^{6}}\rightarrow 0, \quad \mbox{as }n\rightarrow\infty. \end{aligned}$$
(2.10)
By \((F_{1})\) and the Hölder inequality,
$$\begin{aligned} & \biggl\vert \int_{\mathbb{R}^{3}}\bigl(f(x,u)-f(x,u_{n})\bigr) (u_{n}-u)\,dx \biggr\vert \\ &\qquad \leq C_{0} \int_{\mathbb{R}^{3}}\bigl[ \vert u \vert + \vert u \vert ^{p-1}+ \vert u_{n} \vert + \vert u_{n} \vert ^{p-1}\bigr] \vert u_{n}-u \vert \,dx \\ &\qquad \leq C_{0}\bigl[\bigl( \Vert u_{n} \Vert _{L^{2}}+ \Vert u \Vert _{L^{2}}\bigr) \Vert u_{n}-u \Vert _{L^{2}}+\bigl( \Vert u_{n} \Vert ^{p-1}_{L^{p}}+ \Vert u \Vert ^{p-1}_{L^{p}} \bigr) \Vert u_{n}-u \Vert _{L^{p}}\bigr]. \end{aligned}$$
Then, combining the last inequality with (2.5), we get
$$ \int_{\mathbb{R}^{3}}\bigl(f(x,u)-f(x,u_{n})\bigr) (u_{n}-u)\,dx\rightarrow 0,\quad \mbox{as }n\rightarrow\infty. $$
(2.11)
Hence, the combination of (2.7) and (2.9)–(2.11) implies that
$$u_{n}\rightarrow u \quad \mbox{in }E_{\lambda}. $$
Therefore, the proof is complete. □
Lemma 2.4
Assume that
\((V)\)
and
\((F_{1})\)–\((F_{3})\)
hold, then
\(\Phi_{\lambda}\)
satisfies the
\((C)_{c}\)
condition.
Proof
Let \(\{u_{n}\}\subset E_{\lambda}\) be such that
$$ \Phi_{\lambda}(u_{n})\rightarrow c,\qquad \bigl\Vert \Phi'_{\lambda}(u_{n}) \bigr\Vert _{\lambda}\bigl(1+ \Vert u_{n} \Vert _{\lambda}\bigr) \rightarrow 0. $$
(2.12)
First, we prove that \(\{u_{n}\}\) is bounded in \(E_{\lambda}\). By \((F_{3})\), (2.1), (2.2) and (2.12), one has
$$\begin{aligned} c+o(1) =&\Phi_{\lambda}(u_{n})-\frac{1}{4} \bigl\langle \Phi'_{\lambda}(u_{n}),u_{n} \bigr\rangle \\ =&\frac{1}{4} \int_{\mathbb{R}^{3}}\bigl( \vert \triangle u_{n} \vert ^{2}+a \vert \nabla u_{n} \vert ^{2}+\lambda \widetilde{V}(x)u^{2}_{n}\bigr)\,dx \\ &{}+ \int_{\mathbb{R}^{3}} \biggl[\frac{1}{4}f(x,u_{n})u_{n}-F(x,u_{n})- \frac{\lambda}{4}V_{0}u^{2}_{n} \biggr]\,dx \\ \geq&\frac{1}{4} \Vert u_{n} \Vert ^{2}_{\lambda}- \frac{\alpha+\lambda V_{0}}{4} \int_{\mathbb{R}^{3}}u^{2}_{n}\,dx. \end{aligned}$$
(2.13)
Thus, it remains to show that \(\{u_{n}\}\) is bounded in \(L^{2}(\mathbb{R}^{3})\). Otherwise, suppose that \(\Vert u_{n} \Vert _{2}\rightarrow\infty\) and then \(\Vert u_{n} \Vert _{\lambda}\rightarrow\infty\). Let \(\omega_{n}=\frac{u_{n}}{ \Vert u_{n} \Vert _{\lambda}}\), then \(\Vert \omega_{n} \Vert _{\lambda}=1\). According to Lemma 2.2, up to a subsequence, for some \(\omega\in E_{\lambda}\), we obtain
$$\begin{aligned} &\omega_{n} \rightharpoonup\omega \quad \mbox{in } E_{\lambda}, \\ &\omega_{n} \rightarrow\omega \quad \mbox{in } L^{2}\bigl( \mathbb{R}^{3}\bigr), \\ &\omega_{n} \rightarrow\omega \quad \mbox{a.e. in } \mathbb{R}^{3}. \end{aligned}$$
Clearly, we deduce that \(\omega\neq 0\) from (2.13). Then, for \(x\in\{y\in\mathbb{R}^{3}:\omega(y)\neq 0\}\), we have \(\vert u_{n}(x) \vert \rightarrow\infty\) as \(n\rightarrow\infty\). For any given \(u\in H^{2}(\mathbb{R}^{3})\backslash\{0\}\), define
$$\begin{aligned} g(t)&= \bigl\Vert t^{-1}u(tx) \bigr\Vert ^{2}_{H^{2}}-1 \\ &=\frac{1}{t} \int_{\mathbb{R}^{3}} \vert \triangle u \vert ^{2}\,dx+ \frac{1}{t^{3}} \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{t^{5}} \int_{\mathbb{R}^{3}}u^{2}\,dx-1, \quad \forall t>0. \end{aligned}$$
By an elementary computation, there exists a unique \(T=\widetilde{t}(u)>0\) such that
$$g(T)=0, \quad \forall u\in H^{2}\bigl(\mathbb{R}^{3}\bigr) \backslash\{0\}. $$
This implies that \(g(t)=0\) defines a functional \(T=\widetilde{t}(u)\) on \(H^{2}(\mathbb{R}^{3})\backslash\{0\}\). We define \(\widetilde{t}(0)=0\). It is easy to verify that \(T=\widetilde{t}(u)\) is continuous and \(\widetilde{t}(u)\rightarrow\infty\) as \(\Vert u \Vert _{H^{2}}\rightarrow\infty\).
Due to the definition of g, for any \(u\in H^{2}(\mathbb{R}^{3})\backslash\{0\}\), there exists
$$v(x)=T^{-1}u(Tx)\in H^{2}\bigl(\mathbb{R}^{3} \bigr) $$
such that
$$\Vert v \Vert _{H^{2}}=1. $$
Note that \(u_{n}\neq 0\) for large \(n\in\mathbb{N}\), then there exist
$$v_{n}(x)=T_{n}^{-1}u_{n}(T_{n}x) \in H^{2}\bigl(\mathbb{R}^{3}\bigr) $$
such that
$$\Vert v_{n} \Vert _{H^{2}}=1. $$
That is,
$$u_{n}(x)=T_{n} v_{n}\bigl(T^{-1}_{n}x \bigr), $$
with \(\Vert v_{n} \Vert _{H^{2}}=1\) for large \(n\in\mathbb{N}\). Moreover, we have
$$T_{n}=\widetilde{t}(u_{n})\rightarrow\infty\quad \mbox{as }n\rightarrow\infty $$
and
$$\bigl\{ x\in\mathbb{R}^{3}:v_{n}(x)\neq0\bigr\} \neq\emptyset \quad \mbox{for large }n\in\mathbb{N}. $$
From \((F_{1})\)–\((F_{3})\), there are \(R_{0}>0\) and \(C_{2}>0\) such that, for all \(x\in\mathbb{R}^{3}\),
$$ f(x,u)u+\alpha u^{2}\geq 4F(x,u)\geq 0, \quad \forall \vert u \vert \geq R_{0}, $$
(2.14)
and
$$ \bigl\vert f(x,u)u \bigr\vert \leq C_{2}u^{2}, \quad \forall \vert u \vert \leq R_{0}. $$
(2.15)
Thus, by \((F_{3})\), (2.1), (2.2), (2.12)–(2.15) and \(\Vert v_{n} \Vert _{H^{2}}=1\),
$$\begin{aligned} c+o(1) =&\Phi_{\lambda}(u_{n})-\frac{1}{2} \bigl\langle \Phi'_{\lambda}(u_{n}),u_{n} \bigr\rangle \\ \geq&-\frac{b}{4} \Vert \nabla u_{n} \Vert ^{4}_{2}-\frac{\alpha}{4} \int_{\mathbb{R}^{3}}u^{2}_{n}\,dx-\frac{1}{2} \int_{\mathbb{R}^{3}}u_{n}^{2} \vert \nabla u_{n} \vert ^{2}\,dx+\frac{1}{4} \int_{\mathbb{R}^{3}}f(x,u_{n})u_{n}\,dx \\ =&-\frac{bT^{6}_{n}}{4} \Vert \nabla v_{n} \Vert ^{4}_{2}-\frac{\alpha T^{5}_{n}}{4} \int_{\mathbb{R}^{3}}v^{2}_{n}\,dx-\frac{T^{5}_{n}}{2} \int_{\mathbb{R}^{3}}v_{n}^{2} \vert \nabla v_{n} \vert ^{2}\,dx \\ &{}+\frac{T^{3}_{n}}{4} \int_{ \vert T_{n}v_{n} \vert \leq R_{0}}f(T_{n}x,T_{n}v_{n})T_{n}v_{n} \,dx +\frac{T^{6}_{n}}{4} \int_{ \vert T_{n}v_{n} \vert \geq R_{0}}\frac{f(T_{n}x,T_{n}v_{n})T_{n}v_{n}}{T^{3}_{n}}\,dx \\ \geq&\frac{T^{6}_{n}}{4} \biggl\{ -b-\frac{\alpha+C_{2}}{T_{n}}+ \int_{ \vert T_{n}v_{n} \vert \geq R_{0}}\frac{f(T_{n}x,T_{n}v_{n})T_{n}v_{n}}{T^{3}_{n}}\,dx \\ \ &{} -\frac{2\int_{\mathbb{R}^{3}}v_{n}^{2} \vert \nabla v_{n} \vert ^{2}\,dx}{T_{n}} \biggr\} . \end{aligned}$$
(2.16)
By the Hölder inequality and the Sobolev embedding inequality, we see that the sequence of integrals \(\int_{\mathbb{R}^{3}}v_{n}^{2} \vert \nabla v_{n} \vert ^{2}\,dx<\infty\), since \(\Vert v_{n} \Vert _{H^{2}}=1\); on the other hand, by \((F_{2})\) and (2.14), we have
$$\int_{ \vert T_{n}v_{n} \vert \geq R_{0}}\frac{f(T_{n}x,T_{n}v_{n})T_{n}v_{n}}{T^{3}_{n}}\,dx\rightarrow+\infty\quad \mbox{as }n\rightarrow+\infty, $$
which contradicts (2.16). Hence, \(\{u_{n}\}\) is bounded in \(L^{2}(\mathbb{R}^{3})\). This shows that \(\{u_{n}\}\) is bounded in \(E_{\lambda}\) due to (2.13). By Lemma 2.3, \(\{u_{n}\}\) contains a convergent subsequence. □
Next, we define
$$X_{j}=\mathbb{R}e_{j}, \qquad Y_{k}= \bigoplus_{j=1}^{k}X_{j}, \qquad Z_{k}= \overline{\bigoplus_{j=k+1}^{\infty}X_{j}}, \quad k\in \mathbb{Z}, $$
where \(\{e_{j}\}\) is an orthonormal basis of \(E_{\lambda}\).
Lemma 2.5
Assume that
\((V)\)
holds, then, for
\(2\leq s<2_{\ast}\),
$$\beta_{k}(s)=\sup_{u\in Z_{k}, \Vert u \Vert _{\lambda}=1} \Vert u \Vert _{s}\rightarrow 0,\quad k\rightarrow\infty. $$
Proof
By virtue of Lemma 2.2, we can prove the conclusion in a similar way to [16, Lemma 3.8] and [17, Corollary 8.18]. □
Lemma 2.6
Assume that
\((V)\)
and
\((F_{1})\)
hold, then there exist constants
\(\rho, \alpha>0\)
such that
\(\Phi|_{\partial B_{\rho}\cap Z_{m}}\geq\alpha\).
Proof
From (2.1) and \((F_{1})\), for all \(u\in E_{\lambda}\) we have
$$\begin{aligned} \Phi_{\lambda}(u) =&\frac{1}{2} \int_{\mathbb{R}^{3}}\bigl( \vert \triangle u \vert ^{2}+a \vert \nabla u \vert ^{2}+\lambda V(x)u^{2}\bigr)\,dx + \frac{b}{4} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx \biggr)^{2} \\ &{}+\frac{1}{2} \int_{\mathbb{R}^{3}}u^{2} \vert \nabla u \vert ^{2}\,dx- \int_{\mathbb{R}^{3}}F(x,u)\,dx \\ \geq&\frac{1}{2} \Vert u \Vert ^{2}_{\lambda}- \biggl(\frac{\lambda V_{0}+C_{0}}{2} \Vert u \Vert ^{2}_{2}+ \frac{C_{0}}{p} \Vert u \Vert ^{p}_{p} \biggr). \end{aligned}$$
(2.17)
By virtue of Lemma 2.5, we can choose an integer \(m\geq 1\), for all \(u\in Z_{m}\), satisfying
$$\begin{aligned} & \Vert u \Vert ^{2}_{2}\leq\frac{1}{2(\lambda V_{0}+C_{0})} \Vert u \Vert ^{2}_{\lambda}, \\ & \Vert u \Vert ^{p}_{p}\leq\frac{p}{4C_{0}} \Vert u \Vert ^{p}_{\lambda}. \end{aligned}$$
Combining this with (2.17), one has
$$\Phi_{\lambda}(u)\geq\frac{1}{4} \Vert u \Vert ^{2}_{\lambda}\bigl(1- \Vert u \Vert ^{p-2}_{\lambda} \bigr). $$
Note that, if we let \(\rho= \Vert u \Vert _{\lambda}>0\) be sufficiently small, then \(\Phi_{\lambda}(u)\geq\frac{1}{8}\rho^{2}>0\). □
Lemma 2.7
Assume that
\((V)\), \((F_{1})\)
and
\((F_{2})\)
hold, then, for any finite dimensional subspace
\(E\subset E_{\lambda}\), there exists
\(R=R(E)>0\)
such that
\(\Phi_{\lambda}|_{E\backslash B_{\rho}}<0\).
Proof
According to the proof of Lemma 2.4, we know that, for any \(u\in E\backslash \{0\}\), there exists a unique \(T=\widetilde{t}(u)>0\) such that
$$v(x)=T^{-1}u(Tx)\in H^{2}\bigl(\mathbb{R}^{3} \bigr) \quad \mbox{and}\quad \Vert v \Vert _{H^{2}}=1. $$
Hence
$$u(x)=Tv\bigl(T^{-1}x\bigr) \quad \mbox{with } \Vert v \Vert _{H^{2}}=1\mbox{ and } T>0. $$
By the equivalence of norms in the finite dimensional space E, there exists \(C_{3}>0\) such that
$$\min\{a,1\} \Vert u \Vert ^{2}_{H^{2}}\leq \Vert u \Vert ^{2}_{\lambda}\leq C_{3} \Vert u \Vert ^{2}_{2}. $$
Combining this with
$$T=\widetilde{t}(u)\rightarrow\infty\quad \mbox{as } \Vert u \Vert _{\lambda}\rightarrow\infty\mbox{ uniformly in }E, $$
we find that, for any \(\delta>0\), there exists a large \(R=R(E,\delta)>0\) such that
$$T=\widetilde{t}(u)\geq\delta\quad \mbox{for all }u\in E\mbox{ with } \Vert u \Vert _{\lambda}\geq R. $$
By \((F_{1})\), there exists \(C_{4}>0\), for all \(x\in\mathbb{R}^{N}\), \(\vert u \vert \leq R_{0}\) such that
$$\bigl\vert F(x,u) \bigr\vert \leq C_{4}u^{2}, $$
where \(R_{0}\) is given by (2.15). Combining (2.1) with \(\Vert v \Vert _{H^{2}}=1\), it follows that for all \(u\in E\backslash\{0\}\)
$$\begin{aligned} \Phi_{\lambda}(u) =&\frac{1}{2} \Vert u \Vert ^{2}_{\lambda}+\frac{b}{4} \Vert \nabla u \Vert ^{4}_{2}+\frac{1}{2} \int_{\mathbb{R}^{3}}u^{2} \vert \nabla u \vert ^{2}\,dx- \int_{\mathbb{R}^{3}}\biggl[\frac{\lambda V_{0}}{2}u^{2}+F(x,u)\biggr] \,dx \\ \leq&\frac{C_{3}}{2} \Vert u \Vert ^{2}_{2}+ \frac{b}{4} \Vert \nabla u \Vert ^{4}_{2}+ \frac{1}{2} \int_{\mathbb{R}^{3}}u^{2} \vert \nabla u \vert ^{2}\,dx- \int_{\mathbb{R}^{3}}\biggl[\frac{\lambda V_{0}}{2}u^{2}+F(x,u)\biggr] \,dx \\ =&\frac{C_{3}-\lambda V_{0}}{2}T^{5} \Vert v \Vert ^{2}_{2}+ \frac{bT^{6}}{4} \Vert \nabla v \Vert ^{4}_{2}-T^{3} \int_{\mathbb{R}^{3}}F(Tx,Tv)\,dx+\frac{1}{2} \int_{\mathbb{R}^{3}}u^{2} \vert \nabla u \vert ^{2}\,dx \\ \leq& T^{6} \biggl(\frac{b}{4}+\frac{C_{3}+\lambda V_{0}+2C_{4}}{2T}- \int_{ \vert Tv \vert \geq R_{0}}\frac{F(Tx,Tv)}{T^{3}}\,dx \biggr)+\frac{1}{2} \int_{\mathbb{R}^{3}}u^{2} \vert \nabla u \vert ^{2}\,dx \\ =&\Psi(T). \end{aligned}$$
(2.18)
Note that \(v\neq0\), then it follows from \((F_{2})\) that
$$\frac{F(Tx,Tv)}{ \vert Tv \vert ^{3}}\rightarrow+\infty\quad \mbox{as }T\rightarrow+\infty. $$
Thus
$$\int_{ \vert Tv \vert \geq R_{0}}\frac{F(Tx,Tv)}{T^{3}}\rightarrow+\infty\quad \mbox{as }T \rightarrow+\infty. $$
Combining this with (2.18), we obtain
$$\Psi(T)\rightarrow-\infty\quad \mbox{as }T\rightarrow+\infty. $$
Thus, there exists a large \(T_{0}>0\) such that
for all \(T\geq T_{0}\). Taking \(\delta=T_{0}\), then there exists a large \(R=R(E)>0\) such that
$$T=\widetilde{t}(u)\geq T_{0} $$
for all \(u\in E\) with \(\Vert u \Vert _{\lambda}\geq R\).
Hence, \(\Phi_{\lambda}(u)<0\) for all \(u\in E\) with \(\Vert u \Vert _{\lambda}\geq R\). □
Proof of Theorem 1.1
Let \(X=E_{\lambda}\), \(Y=Y_{m}\) and \(Z=\overline{Z_{m}}\). Clearly, \(\Phi(0)=0\) and \(\Phi(u)=\Phi(-u)\) due to \((F_{4})\). By virtue of Lemma 2.4, Lemma 2.6, Lemma 2.7 and the fountain theorem (Theorem 3.6 [16]), problem (1.1) possesses infinitely many high energy solutions. □
Proof of Corollary 1.2
Let us consider the Hilbert space
$$H= \biggl\{ u\in H^{1}\bigl(\mathbb{R}^{3}\bigr): \int_{\mathbb{R}^{3}}\bigl(a \vert \nabla u \vert ^{2}+ \lambda\widetilde{V}(x)u^{2}\bigr)\,dx< \infty \biggr\} $$
endowed with the norm
$$\Vert u \Vert = \biggl( \int_{\mathbb{R}^{3}}\bigl(a \vert \nabla u \vert ^{2}+ \lambda\widetilde{V}(x)u^{2}\bigr)\,dx \biggr)^{\frac{1}{2}}. $$
Let
$$\Phi(u)=\frac{1}{2} \int_{\mathbb{R}^{3}}\bigl(a \vert \nabla u \vert ^{2}+ \lambda V(x)u^{2}\bigr)\,dx+\frac{b}{4} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx \biggr)^{2} - \int_{\mathbb{R}^{3}}F(x,u)\,dx,\quad \forall u\in H. $$
Obviously, Φ is a well-defined class \(C^{1}\) functional, and the embedding \(H\hookrightarrow L^{s}\) is compact for \(2\leq s<6\) (see the proof of Lemma 2.2). By Lemma 2.4, Lemma 2.6, Lemma 2.7 and the fountain theorem (Theorem 3.6 [16]), problem (1.3) possesses infinitely many high energy solutions. □
Remark 2.8
In the next paper, we wish to consider the sign-changing solutions for the biharmonic problem like in [19, 21] and so on.