First, we give some notations. A complete introduction to fractional Sobolev spaces can be found in [45]; we offer in the succeeding discussion a short review. We define the homogeneous fractional Sobolev space \(\mathcal{D}^{\alpha,2}(\mathbb{R}^{N})\) as follows:
$$\mathcal{D}^{\alpha,2}\bigl(\mathbb{R}^{N}\bigr) = \biggl\{ u \in L^{2^{\ast }_{\alpha}}\bigl(\mathbb{R}^{N}\bigr): \frac{ \vert u(x)-u(y) \vert }{ \vert x-y \vert ^{\frac{N}{2}+\alpha}} \in L^{2}\bigl(\mathbb{R}^{N}\times\mathbb {R}^{N} \bigr) \biggr\} , $$
which is the completion of \(\mathcal{C}^{\infty}_{0}(\mathbb {R}^{N})\) under the norm
$$\Vert u \Vert _{\mathcal{D}^{\alpha,2}(\mathbb {R}^{N})} = \biggl( \int_{\mathbb{R}^{N}} \bigl\vert (-\triangle)^{\frac {\alpha}{2}}u \bigr\vert ^{2}\,\mathrm{d}x \biggr)^{1/2} = \biggl( \int_{\mathbb{R}^{N}} \int_{\mathbb{R}^{N}}\frac{ \vert u(x)-u(y) \vert ^{2}}{ \vert x-y \vert ^{N+2\alpha }}\,\mathrm{d}x\,\mathrm{d}y \biggr)^{1/2}. $$
Moreover, the embedding \(\mathcal{D}^{\alpha,2}(\mathbb {R}^{N})\hookrightarrow L^{2^{\ast}_{\alpha}}(\mathbb{R}^{N})\) is continuous and for any \(\alpha \in (0,1)\), there exists a best constant \(S_{\alpha}\) such that
$$S_{\alpha} = \inf_{u\in\mathcal{D}^{\alpha,2}(\mathbb {R}^{N})}\frac{\int_{\mathbb{R}^{N}} \vert (-\triangle)^{\frac {\alpha}{2}}u \vert ^{2}\,\mathrm{d}x}{ (\int_{\mathbb{R}^{N}} \vert u^{2_{\alpha}^{\ast }} \vert \,\mathrm{d}x )^{\frac{2}{2_{\alpha}^{\ast}}}}. $$
The fractional Sobolev space \(H^{\alpha}(\mathbb{R}^{N})\) can be described by
$$H^{\alpha}\bigl(\mathbb{R}^{N}\bigr) = \biggl\{ u \in L^{2}\bigl(\mathbb{R}^{N}\bigr): \frac{ \vert u(x)-u(y) \vert }{ \vert x-y \vert ^{\frac{N}{2}+\alpha}} \in L^{2}\bigl(\mathbb{R}^{N}\times\mathbb {R}^{N} \bigr) \biggr\} , $$
endowed with the natural norm
$$\Vert u \Vert _{H^{\alpha}(\mathbb{R}^{N})} = \biggl( \int _{\mathbb{R}^{N}} \vert u \vert ^{2}\,\mathrm{d}x + \int_{\mathbb{R}^{N}} \int_{\mathbb{R}^{N}}\frac{ \vert u(x)-u(y) \vert ^{2}}{ \vert x-y \vert ^{N+2\alpha }}\,\mathrm{d}x\,\mathrm{d}y \biggr)^{1/2}. $$
It is well known that \(H^{\alpha}(\mathbb{R}^{N})\) is continuously embedded into \(L^{q}(\mathbb{R}^{N})\), and compactly embedded into \(L^{q}_{\mathrm{loc}}(\mathbb{R}^{N})\) for \(2\leq q \leq2_{\alpha }^{\ast} :=\frac{2N}{N-2\alpha}\).
Under assumption (V), we see that
$$E= \biggl\{ u\in H^{\alpha}\bigl(\mathbb{R}^{N}\bigr): \int_{\mathbb{R}^{N}} \bigl(a \bigl\vert (-\triangle)^{\frac{\alpha}{2}}u \bigr\vert ^{2}+V(x)u^{2} \bigr)\,\mathrm{d}x< +\infty \biggr\} $$
is a Hilbert space equipped with the norm
$$\Vert u \Vert = \biggl( \int_{\mathbb{R}^{N}}\bigl(a \bigl\vert (-\triangle)^{\frac{\alpha}{2}}u \bigr\vert ^{2}+V(x)u^{2} \bigr)\,\mathrm{d}x \biggr)^{1/2} . $$
Let
$$ \begin{aligned}[b]\Phi(u)&=\frac{1}{2} \int_{\mathbb{R}^{N}} \bigl(a \bigl\vert (-\triangle)^{\frac{\alpha}{2}}u \bigr\vert ^{2}+V(x)u^{2} \bigr)\,\mathrm{d}x\\ &\quad {} + \frac{b}{4} \biggl( \int_{\mathbb{R}^{N}} \bigl\vert (-\triangle )^{\frac{\alpha}{2}}u \bigr\vert ^{2}\,\mathrm{d}x \biggr)^{2} - \int_{\mathbb{R}^{N}}F(x,u)\,\mathrm{d}x,\quad\forall u\in E. \end{aligned} $$
(2.1)
From (F1) and (F2), it is easy to see that \(\Phi \in \mathcal {C}^{1}(E,\mathbb{R})\) as a functional, and that
$$\begin{aligned} \bigl\langle \Phi'(u),v \bigr\rangle =& \int_{\mathbb{R}^{N}} \bigl(a(-\triangle)^{\frac{\alpha }{2}}u(- \triangle)^{\frac{\alpha}{2}}v +V(x)uv \bigr)\,\mathrm{d}x \\ &{} +b \int_{\mathbb{R}^{N}} \bigl\vert (-\triangle)^{\frac{\alpha }{2}}u \bigr\vert ^{2}\,\mathrm{d}x \int_{\mathbb{R}^{N}}(-\triangle)^{\frac{\alpha}{2}}u(-\triangle )^{\frac{\alpha}{2}}v\,\mathrm{d}x \\ &{}- \int_{\mathbb{R}^{N}}f(x,u)v\,\mathrm{d}x,\quad\forall u,v \in E. \end{aligned}$$
(2.2)
Clearly, any critical point of Φ is a weak solution of (1.1). Set
$$\begin{aligned} \mathcal{N} := \bigl\{ u \in E : \bigl\langle \Phi'(u),u \bigr\rangle =0, u\neq0\bigr\} . \end{aligned}$$
(2.3)
Lemma 2.1
Assume that (V), (F1), (F2) and (F4) hold. Then
$$\begin{aligned} \Phi(u) \geq & \Phi(tu)+\frac{1-t^{4}}{4}\bigl\langle \Phi^{'}(u),u \bigr\rangle +\frac{(1-\theta_{0})(1-t^{2})^{2}}{4} \Vert u \Vert ^{2}, \quad\forall u \in E, t\geq0. \end{aligned}$$
(2.4)
Proof
For any \(x \in \mathbb{R}^{N}\), \(t\geq0\), \(\tau \in \mathbb{R}\setminus\{0\}\), (F4) yields
$$ \begin{aligned}[b] &\frac{1-t^{4}}{4}\tau f(x,\tau)+F(x,t\tau)-F(x,\tau)+ \frac {\theta_{0}V(x)}{4}\bigl(1-t^{2}\bigr)^{2}\tau^{2} \\ &\quad = \int^{1}_{t} \biggl[ \frac{f(x,\tau)}{\tau^{3}}- \frac{f(x,\xi \tau)}{(\xi\tau)^{3}} +\theta_{0}V(x)\frac{1-\xi^{2}}{(\xi\tau)^{2}} \biggr] \xi^{3}\tau ^{4}\,\mathrm{d}\xi\geq0. \end{aligned} $$
(2.5)
By (2.1), (2.2) and (2.5), we have
$$\begin{aligned} \Phi(u)-\Phi(tu) =& \frac{1-t^{2}}{2} \Vert u \Vert ^{2}+\frac{1-t^{4}}{4}b \bigl\Vert (-\triangle)^{\frac{ \alpha}{2}} \bigr\Vert ^{4}_{2}+ \int_{\mathbb {R}^{N}}\bigl[F(x,tu)-F(x,u)\bigr]\,\mathrm{d}x \\ =& \frac{1-t^{4}}{4}\bigl\langle \Phi^{'}(u),u \bigr\rangle + \frac {(1-t^{2})^{2}}{2} \Vert u \Vert ^{2} \\ &{}+ \int_{\mathbb{R}^{N}} \biggl[\frac {1-t^{4}}{4}f(x,u)u+F(x,tu)-F(x,u) \biggr]\,\mathrm{d}x \\ \geq&\frac{1-t^{4}}{4}\bigl\langle \Phi^{'}(u),u \bigr\rangle + \frac{(1-\theta _{0})(1-t^{2})^{2}}{4} \Vert u \Vert ^{2} \\ &{}+ \int_{\mathbb{R}^{N}} \biggl[\frac {1-t^{4}}{4}f(x,u)u+F(x,tu)-F(x,u)+ \frac{\theta _{0}V(x)}{4}\bigl(1-t^{2}\bigr)^{2}u^{2} \biggr]\,\mathrm{d}x \\ \geq&\frac{1-t^{4}}{4}\bigl\langle \Phi^{'}(u),u \bigr\rangle + \frac{(1-\theta _{0})(1-t^{2})^{2}}{4} \Vert u \Vert ^{2}. \end{aligned}$$
This shows that (2.4) holds. □
Corollary 2.2
Assume that (V), (F1), (F2) and (F4) hold. Then, for an
\(u\in\mathcal{N}\),
$$\begin{aligned} \Phi(u) \geq & \Phi(tu)+\frac{(1-\theta_{0})(1-t^{2})^{2}}{4} \Vert u \Vert ^{2}, \quad\forall t\geq0. \end{aligned}$$
(2.6)
Corollary 2.3
Assume that (V), (F1), (F2) and (F4) hold. Then, for any
\(u\in\mathcal{N}\),
$$\begin{aligned} \Phi(u)=\max_{t\geq0}\Phi(tu). \end{aligned}$$
(2.7)
Under (F3), to show \(\mathcal{N}\neq\emptyset\) in our situation, we have to overcome the competing effect of \(\int_{\mathbb{R}^{N}} \vert (-\triangle)^{\frac{\alpha}{2}}u \vert ^{2}\,\mathrm{d}x\). To this end, we define a set \(\mathcal{E}\) as follows:
$$\begin{aligned} \mathcal{E}= \biggl\{ u \in E : b \bigl\Vert (- \triangle)^{\frac {\alpha}{2}}u \bigr\Vert ^{4}_{2}+ \int_{\mathbb{R}^{N}} \bigl[V(x)u^{2} -f(x,u)u \bigr] \,\mathrm{d}x< 0 \biggr\} . \end{aligned}$$
(2.8)
Lemma 2.4
Assume that (V) and (F1)-(F4) hold. Then
\(\mathcal{E}\neq\emptyset\). Moreover, \(\mathcal{N} \subset \mathcal{E}\).
Proof
For any fixed \(u \in E\) with \(u\neq0\), set \(u_{t}(x)=tu(t^{-1} x)\) for \(t>0\). By (V), one has
$$ \begin{aligned}[b] &b \bigl\Vert (-\triangle)^{\frac{\alpha}{2}}u_{t} \bigr\Vert ^{4}_{2}+ \int_{\mathbb{R}^{N}} \bigl[V(x)u_{t}^{2}-f(x,u_{t})u_{t} \bigr]\,\mathrm{d}x \\ &\quad = t^{4+2N-4\alpha}b \bigl\Vert (-\triangle)^{\frac{\alpha }{2}}u \bigr\Vert ^{4}_{2} +t^{2+N} \int_{\mathbb{R}^{N}}V(tx)u^{2}\,\mathrm{d}x -t^{N} \int_{\mathbb{R}^{N}}f(tx,tu)tu\,\mathrm{d}x \\ &\quad \leq t^{2+N} \biggl[t^{2+N-4\alpha}b \bigl\Vert (- \triangle)^{\frac {\alpha}{2}}u \bigr\Vert ^{4}_{2} + \Vert V \Vert _{\infty} \Vert u \Vert ^{2}_{2} - \int_{\mathbb{R}^{N}}\frac{f(tx,tu)tu}{t^{2}}\,\mathrm{d}x \biggr] \\ &\quad =t^{2+N} \biggl\{ t^{2+N-4\alpha} \biggl[b \bigl\Vert (-\triangle )^{\frac{\alpha}{2}}u \bigr\Vert ^{4}_{2} - \int_{\mathbb{R}^{N}}\frac{f(tx,tu)tu}{t^{4+N-4\alpha}}\,\mathrm{d}x \biggr] + \Vert V \Vert _{\infty} \Vert u \Vert ^{2}_{2} \biggr\} . \end{aligned} $$
(2.9)
Note that, for \(u\neq0\), \(F(tx,tu)/ \vert tu \vert ^{4+N-4\alpha}\rightarrow+\infty\) as \(t\rightarrow+\infty\) by (F3), where \(F(x,t)=\int_{0}^{t}f(x,s)\,\mathrm{d}s\). From (2.5) with \(t=0\), one has
$$\begin{aligned} \frac{1}{4}\tau f(x,\tau)-F(x,\tau)+\frac{\theta_{0}V(x)}{4}\tau ^{2}\geq0, \quad\forall x \in \mathbb{R}^{N}, \tau \in \mathbb{R}, \end{aligned}$$
(2.10)
then we have
$$\begin{aligned} \frac{f(tx,tu)tu}{ \vert tu \vert ^{4+N-4\alpha }}\rightarrow+\infty, \quad\mbox{as } t \rightarrow+\infty\mbox{ uniformly in } x \in \mathbb{R}^{N}. \end{aligned}$$
(2.11)
For \(N\in(2\alpha,4\alpha)\), thus from (2.9) and (2.11), one has
$$\begin{aligned} b \bigl\Vert (-\triangle)^{\frac{\alpha}{2}}u_{t} \bigr\Vert ^{4}_{2}+ \int_{\mathbb{R}^{3}} \bigl[V(x)u_{t}^{2}-f(x,u_{t})u_{t} \bigr]\,\mathrm{d}x\rightarrow-\infty, \quad\mbox{as } t\rightarrow+\infty. \end{aligned}$$
Thus, taking \(v=u_{T}\) for T large, we have \(v \in \mathcal{E}\). From (2.2), it is easy to see that and \(\mathcal{N}\subset \mathcal{E}\). □
Lemma 2.5
Assume that (V) and (F1)-(F4) hold. If
\(u \in \mathcal{E}\), then there exists a unique
\(t(u)>0\)
such that
\(t(u) \in \mathcal{N}\).
Proof
First, we prove the existence of \(t(u)\). In view of Lemma 2.4, let \(u\in\mathcal{E}\) be fixed and define a function \(g(t)=\langle\Phi'(tu), tu \rangle\) on \([0,+\infty)\). By (F4), one has
$$\begin{aligned} \theta_{0}V(x) (t\tau)^{2}-f(x,t\tau)t\tau \leq\bigl[\theta_{0}V(x)\tau ^{2}-f(x,\tau)\tau \bigr]t^{4},\quad\forall x \in \mathbb{R}^{N}, t \geq 1, \tau \in \mathbb{R}, \end{aligned}$$
(2.12)
Since \(u \in \mathcal{E}\), (2.12) yields
$$ \begin{aligned}[b] &bt^{4} \bigl\Vert (-\triangle)^{\frac{\alpha}{2}}u \bigr\Vert ^{4}_{2}+ \int_{\mathbb{R}^{N}}\bigl[\theta_{0}V(x) (tu)^{2}-f(x,tu)tu \bigr]\,\mathrm{d}x \\ &\quad \leq t^{4} \biggl(b \bigl\Vert (-\triangle)^{\frac{\alpha }{2}}u \bigr\Vert ^{4}_{2}+ \int_{\mathbb{R}^{N}}\bigl[\theta _{0}V(x)u^{2}-f(x,u)u \bigr]\,\mathrm{d}x \biggr)< 0,\quad\forall t \geq 1. \end{aligned} $$
(2.13)
It follows from (2.1) and (2.13) that
$$\begin{aligned} g(t) =&t^{2} \biggl[a \bigl\Vert (- \triangle)^{\frac{\alpha}{2}}u \bigr\Vert ^{2}_{2}+(1- \theta_{0}) \int_{\mathbb{R}^{N}}V(x)u^{2}\,\mathrm{d}x \biggr] +bt^{4} \bigl\Vert (-\triangle)^{\frac{\alpha}{2}}u \bigr\Vert ^{4}_{2} \\ &{}+ \int_{\mathbb{R}^{N}}\bigl[\theta_{0}V(x) (tu)^{2}-f(x,tu)tu \bigr]\,\mathrm{d}x \\ \leq&t^{2} \biggl[a \bigl\Vert (-\triangle)^{\frac{\alpha}{2}}u \bigr\Vert ^{2}_{2}+(1-\theta_{0}) \int_{\mathbb{R}^{N}}V(x)u^{2}\,\mathrm{d}x \biggr] \\ &{}+t^{4} \biggl(b \bigl\Vert (-\triangle)^{\frac{\alpha}{2}}u \bigr\Vert ^{4}_{2}+ \int_{\mathbb{R}^{N}}\bigl[\theta_{0}V(x)u^{2}-f(x,u)u \bigr]\,\mathrm{d}x \biggr). \end{aligned}$$
(2.14)
For (F1), (F2) and (2.14), it is easy to verify that \(g(0)=0\), \(g(t)>0\) for \(t>0\) small and \(g(t)<0\) for t large because of \(u\in \mathcal{E}\). Therefore, there exists \(t_{0}=t(u)>0\) so that \(g(t_{0})=0\) and \(t(u)u \in \mathcal{N}\).
Next, we prove the uniqueness. For any given \(u \in \mathcal{E}\), let \(t_{1}, t_{2}>0\) such that \(g(t_{1})={g(t_{2})=0}\). Jointly with (2.6), one has
$$\begin{aligned} \Phi(t_{1}u) \geq\Phi(t_{2}u)+ \frac{(1-\theta _{0})(t_{1}^{2}-t_{2}^{2})^{2}}{4t_{1}^{2}} \Vert u \Vert ^{2} \end{aligned}$$
(2.15)
and
$$\begin{aligned} \Phi(t_{2}u) \geq\Phi(t_{1}u)+ \frac{(1-\theta _{0})(t_{2}^{2}-t_{1}^{2})^{2}}{4t_{2}^{2}} \Vert u \Vert ^{2}. \end{aligned}$$
(2.16)
Both (2.15) and (2.16) imply \(t_{1}=t_{2}\). Hence, \(t(u)>0\) is unique for any \(u \in \mathcal{E}\). □
Lemma 2.6
Assume that (V) and (F1)-(F4) hold. Then
$$\inf_{u\in\mathcal{N}}\Phi(u) = c = \inf_{u\in\mathcal{E},u\neq0} \max _{t\geq0}\Phi(t u)>0. $$
Proof
Corollary 2.3 and Lemma 2.5 imply that \(c = \inf_{u\in\mathcal{E},u\neq0} \max_{t\geq0}\Phi(t u)\). From Lemma 2.1, it is easy to see that \(c> 0\). □
Lemma 2.7
Assume that (V) and (F1)-(F4) hold. Then there exist a constant
\(c_{\ast} \in (0,c]\)
and a sequence
\({u_{n}}\subset E\)
satisfying
$$\begin{aligned} \Phi(u_{n}) \rightarrow c_{\ast},\quad \bigl\Vert \Phi'(u_{n}) \bigr\Vert \bigl(1+ \Vert u_{n} \Vert \bigr)\rightarrow0. \end{aligned}$$
(2.17)
Proof
We use the non-Nehari manifold approach developed in [42, 43] to show (2.17). From (F1), (F2) and (2.1), we know that there exist \(\delta_{0}>0\) and \(\rho_{0}>0\) such that
$$\begin{aligned} \Phi(u)\geq\rho_{0},\quad \Vert u \Vert = \delta_{0}. \end{aligned}$$
(2.18)
Choose \(v_{k}\in\mathcal{N}\subset\mathcal{E}\) such that
$$\begin{aligned} c\leq\Phi(v_{k})< c+\frac{1}{k},\quad k \in \mathbb{N}. \end{aligned}$$
(2.19)
For Lemma 2.1, it is easy to see that \(\Phi(tv_{k})<0\) for large \(t>0\). In fact, if \(\Phi(tv_{k})\geq0\) for large \(t>0\). By (2.4) and \(v_{k} \in \mathcal{N}\), we have
$$\begin{aligned} \Phi(v_{k})\geq\Phi(tv_{k})+ \frac{(1-\theta_{0})(1-t^{2})^{2}}{4} \Vert v_{k} \Vert ^{2}, \end{aligned}$$
(2.20)
which contradicts (2.19). From the mountain pass lemma, there exists a sequence \(\{u_{k,n}\}_{n\in\mathbb{N}}\subset E\) satisfying
$$\begin{aligned} \Phi(u_{k,n}) \rightarrow c_{k},\qquad \bigl\Vert \Phi '(u_{k,n}) \bigr\Vert \bigl(1+ \Vert u_{k,n} \Vert \bigr)\rightarrow 0,\quad k \in \mathbb{N}, \end{aligned}$$
(2.21)
where \(c_{k} \in [\rho_{0},\sup_{t\geq0}\Phi(tv_{k})]\). By (2.7) and (2.19), one has
$$\begin{aligned} \rho_{0}\leq c_{k}\leq\sup _{t\geq0}\Phi(tv_{k})=\Phi(v_{k})< c+ \frac {1}{k},\quad k \in \mathbb{N}. \end{aligned}$$
(2.22)
Therefore, by (2.21) and (2.22), for \(k \in \mathbb{N}\), one has
$$\begin{aligned} \Phi(u_{k,n}) \rightarrow c_{k} \in \biggl[ \rho_{0},c+\frac {1}{k} \biggr),\qquad \bigl\Vert \Phi'(u_{k,n}) \bigr\Vert \bigl(1+ \Vert u_{k,n} \Vert \bigr)\rightarrow0,\quad\mbox{as }n\rightarrow \infty. \end{aligned}$$
(2.23)
In view of (2.23), for \(k=1\), there exists \(n_{1}>0\) large enough such that
$$\begin{aligned} \rho_{0}\leq\Phi(u_{1,n_{1}})< c_{1}+1, \qquad \bigl\Vert \Phi '(u_{1,n_{1}}) \bigr\Vert \bigl(1+ \Vert u_{1,n_{1}} \Vert \bigr)< 1; \end{aligned}$$
(2.24)
for \(k=2\), there exists \(n_{2}>n_{1}>0\) large enough such that
$$\begin{aligned} \rho_{0}\leq\Phi(u_{2,n_{2}})< c_{2}+1/2, \qquad \bigl\Vert \Phi '(u_{2,n_{2}}) \bigr\Vert \bigl(1+ \Vert u_{2,n_{2}} \Vert \bigr)< 1/2. \end{aligned}$$
(2.25)
In this way, we can choose a sequence \(\{n_{k}\}\subset\mathbb{N}\) with \(n_{k}\rightarrow\infty\) as \(k\rightarrow\infty\) such that
$$\begin{aligned} \Phi(u_{k,n_{k}}) \in \biggl[\rho_{0},c+ \frac{1}{k} \biggr),\qquad \bigl\Vert \Phi'(u_{k,n_{k}}) \bigr\Vert \bigl(1+ \Vert u_{k,n_{k}} \Vert \bigr)< \frac{1}{k}, \quad k\in\mathbb{N}. \end{aligned}$$
(2.26)
Let \(u_{k}=u_{k,n_{k}}\), \(k \in \mathbb{N}\). Then, going if necessary to a subsequence, we conclude from (2.26) that
$$\begin{aligned} \Phi(u_{n}) \rightarrow c_{\ast},\qquad \bigl\Vert \Phi'(u_{n}) \bigr\Vert \bigl(1+ \Vert u_{n} \Vert \bigr)\rightarrow0. \end{aligned}$$
(2.27)
□
Lemma 2.8
Assume that (V) and (F1)-(F4) hold. Then any sequence
\(\{u_{n}\}\subset E\)
satisfying (2.17) is bounded in
E.
Proof
By (2.2), (2.10) and (2.17), one has
$$ \begin{aligned}[b] c_{\ast}+o(1) &=\Phi(u_{n})-\frac{1}{4}\bigl\langle \Phi'(u_{n}), u_{n} \bigr\rangle \\ &\geq\frac{1-\theta_{0}}{4} \Vert u_{n} \Vert ^{2} + \int_{\mathbb{R}^{N}} \biggl(\frac{1}{4}f(x,u)u-F(x,u)+ \frac{\theta _{0}V(x)}{4}u^{2} \biggr)\,\mathrm{d}x \\ &\geq\frac{1-\theta_{0}}{4} \Vert u_{n} \Vert ^{2}. \end{aligned} $$
(2.28)
This shows that the sequence \(\{u_{n}\}\) is bounded in E. □
Next, we prove the minimizer of the constrained problem is a critical point, which plays a crucial role in the asymptotically periodic case.
Lemma 2.9
Assume that (V) and (F1)-(F4) hold. If
\(u_{0}\in \mathcal{N}\)
and
\(\Phi(u_{0})=c\), then
\(u_{0}\)
is a critical point of Φ.
Proof
Analogous to the proof of [21], it is easy to show this lemma by combining the quantitative deformation lemma and degree theory. □