Let A be a self-adjoint extension of the operator \(-(d^{2}/dt^{2})+L(t)\) with the domain \(\mathcal{D}(A)\subset L^{2}( R, R^{N})\). Let \(E=\mathcal{D}(\vert A\vert ^{1/2})\) be the domain of \(\vert A\vert ^{1/2}\) and define on E the inner product and the norm as \((u,w)_{0}=(\vert A\vert ^{1/2}u,\vert A\vert ^{1/2}w)_{2}+(u,w)_{2}\) and \(\Vert u\Vert _{0}=(u,u)^{1/2}\), respectively, where \((\cdot,\cdot)_{2}\) denotes the inner product of \(L^{2}\). Then E is a Hilbert space.
It is known that the spectrum \(\sigma (A)\) consists of eigenvalues numbered in \(\lambda_{1}\leq \lambda_{2} \leq \cdots \rightarrow \infty \), and a corresponding system of eigenfunctions \((e_{n})(Ae _{n} =\lambda_{n}e_{n})\) forms an orthogonal basis in \(L^{2}\). Let \(n^{-}=\#\{i\vert \lambda_{i}<0\}\), \(n^{0}=\#\{i\vert \lambda_{i}=0\}\), and \(\bar{n}=n^{-}+n^{0}\). Set \(E^{-}=\operatorname{span}\{e_{1},\ldots,e_{n^{-}}\}\), \(E^{0}=\operatorname{span}\{e_{n^{-}+1},\ldots,e_{\bar{n}}\}\), and \(E^{+}=\overline{\operatorname{span}\{e_{\bar{n}+1},\ldots \}}\). Then \(E=E^{-}\oplus E^{0}\oplus E ^{+}\). The inner product and the norm on E are introduced as
$$\begin{aligned} (u,w)=\bigl(\vert A\vert ^{1/2}u,\vert A\vert ^{1/2}w \bigr)_{2}+\bigl(u^{0},w^{0}\bigr)_{2},\qquad \Vert u\Vert =(u,u)^{1/2}, \end{aligned}$$
where \(u=u^{-}+u^{0}+u^{+}\) and \(w=w^{-}+w^{0}+w^{+}\in E=E^{-}\oplus E^{0}\oplus E^{+}\). Furthermore, let \(I: E \rightarrow R\) be the functional defined by
$$\begin{aligned} I(u) =& \int_{ R} \biggl( \frac{1}{2}\bigl\vert \dot{u}(t)\bigr\vert ^{2}+\frac{1}{2}\bigl(L(t)u(t),u(t)\bigr)-W\bigl(t,u(t) \bigr)+\bigl(f(t),u(t)\bigr) \biggr)\,dt { } \\ =& \frac{1}{2}\Vert u\Vert ^{2}- \int_{ R}W\bigl(t,u(t)\bigr)\,dt+ \int_{ R}\bigl(f(t),u(t)\bigr)\,dt. \end{aligned}$$
(4)
It is known that the critical points of I in E are the homoclinic solutions of (1). One can easily check that \(I\in C^{1}(E, R)\) and
$$\begin{aligned} \bigl\langle I'(u),v\bigr\rangle =& \int_{ R} \bigl( \bigl(\dot{u}(t),\dot{v}(t)\bigr)+ \bigl(L(t)u(t),v(t)\bigr)-\bigl( \nabla W\bigl(t,u(t)\bigr),v(t)\bigr) \bigr)\,dt \\ &{}+ \int_{ R}\bigl(f(t),v(t)\bigr)\,dt. \end{aligned}$$
(5)
By Lemma 2.2 in [6] we can conclude that E is compactly embedded in \(L^{p}\) for any \(p\in [1,+\infty ]\), which implies that there exists a constant \(C_{p}>0\) such that
$$\begin{aligned} \Vert u\Vert _{L^{p}}\leq C_{p}\Vert u\Vert \quad \mbox{for all } u\in E. \end{aligned}$$
(6)
Lemma 2.1
Suppose that the conditions of Theorem 1.1
hold, then there exist constants
α, \(\varrho >0\)
such that
\(I\vert_{S} \geq \alpha \), where
\(S=\{u\in E\vert \Vert u\Vert =\varrho \}\).
Proof
By \((W1)\), for any \(\varepsilon >0\), there exists \(\sigma >0\) such that
$$\begin{aligned} \bigl\vert \nabla W(t,x)\bigr\vert \leq \varepsilon \vert x\vert ,\qquad \vert x\vert \leq \sigma, \quad \forall t \in R. \end{aligned}$$
Then it follows from \((W1)\) and \((W2)\) that
$$\begin{aligned} W(t,x) =&\bigl\vert W(t,x)-W(t,0)\bigr\vert \\ =& \biggl\vert \int^{1}_{0}\bigl(\nabla W(t,sx),x\bigr)\,ds\biggr\vert \\ \leq & \int^{1}_{0}\bigl\vert \nabla W(t,sx)\bigr\vert \vert x\vert \,ds \\ \leq & \int^{1}_{0}\varepsilon \vert sx\vert \vert x \vert \,ds \\ \leq & \varepsilon \vert x\vert ^{2} \end{aligned}$$
(7)
for all \(t\in R\) and \(\vert x\vert \leq \sigma \). Let \(\varepsilon_{0}=\frac{1}{4C _{2}}\), then there exists \(\sigma_{0}>0\) such that (7) holds for all \(t\in R\) and \(\vert x\vert \leq \sigma_{0}\). Set
$$\begin{aligned} \varrho =\frac{\sigma_{0}}{C_{\infty }}, \qquad \alpha =\frac{1}{4}\varrho^{2}>0, \end{aligned}$$
which implies \(0 < \Vert u\Vert _{L^{\infty }} \leq \sigma_{0}\) for all \(u \in S\). Then it follows from the definition of I, (7) and (2) that
$$\begin{aligned} I(u) =& \frac{1}{2}\Vert u\Vert ^{2}- \int_{ R}W\bigl(t,u(t)\bigr)\,dt+ \int_{ R}\bigl(f(t),u(t)\bigr)\,dt { } \\ \geq & \frac{1}{2}\Vert u\Vert ^{2}- \frac{1}{4C_{2}} \int_{ R}\bigl\vert u(t)\bigr\vert ^{2}\,dt- \delta \int_{ R}\bigl\vert u(t)\bigr\vert \,dt{ } \\ \geq & \frac{1}{4}\Vert u\Vert ^{2}-\delta C_{1}\Vert u\Vert . \end{aligned}$$
By the definitions of ϱ and α, there exists \(\delta_{0}>0\) such that \(I\mid_{S} \geq \alpha \) for any f satisfying (2). □
Lemma 2.2
Suppose that the conditions of Theorem 1.1
hold, then there is
\(e\in E\)
such that
\(\Vert e\Vert > \varrho \)
and
\(I(e) \leq 0\), where
ϱ
is defined in Lemma 2.1.
Proof
It follows from \((W3)\) that there exist \(T>0\), \(\xi >0\), and \(\varepsilon_{1}>0\) such that
$$\begin{aligned} W(t,x)\geq \biggl( \frac{\pi^{2}}{2T^{2}}+\varepsilon_{1} \biggr) \vert x \vert ^{2} \end{aligned}$$
for all \(t\in [-T,T]\) and \(\vert x\vert > \xi \). Set \(\zeta =\max \{\vert W(t,x)\vert \vert t \in [-T,T], \vert x\vert \leq \xi \}\), hence we have
$$\begin{aligned} W(t,x)\geq \biggl( \frac{\pi^{2}}{2T^{2}}+\varepsilon_{1} \biggr) \bigl( \vert x\vert ^{2}- \xi^{2}\bigr)-\zeta. \end{aligned}$$
Set
$$\begin{aligned} Q_{1}(t)=\textstyle\begin{cases} \sin(\omega t)e,&t\in [-T,T], \\ 0,&t\in R\setminus [-T,T], \end{cases}\displaystyle \end{aligned}$$
where \(\omega =\frac{\pi }{T}\), \(e=(1,0,\ldots,0)\). It can be easily checked that \(( \frac{\pi^{2}}{2T^{2}}+\varepsilon_{1} ) m>M\), where
$$\begin{aligned} M=\frac{1}{2} \int^{T}_{-T}\bigl\vert \dot{Q}_{1}(t) \bigr\vert ^{2}\,dt,\qquad m= \int^{T}_{-T}\bigl\vert Q_{1}(t)\bigr\vert ^{2}\,dt. \end{aligned}$$
By (4), for every \(r \in R\setminus \{0\}\), the following inequality holds:
$$\begin{aligned} I(rQ_{1}) =& \frac{1}{2} \int^{T}_{-T}\bigl\vert r\dot{Q}_{1}(t) \bigr\vert ^{2}\,dt- \int ^{T}_{-T}W\bigl(t,rQ_{1}(t)\bigr)\,dt+ \int^{T}_{-T}\bigl(f(t),rQ_{1}(t)\bigr)\,dt{ } \\ \leq & \frac{\vert r\vert ^{2}}{2} \int^{T}_{-T}\bigl\vert \dot{Q}_{2}(t) \bigr\vert ^{2}\,dt- \biggl( \frac{\pi^{2}}{2T^{2}}+\varepsilon_{1} \biggr) \vert r\vert ^{2} \int^{T} _{-T}\bigl\vert Q_{1}(t)\bigr\vert ^{2}\,dt{ } \\ & {} +\vert r\vert \delta m^{1/2}+2T \biggl( \biggl( \frac{\pi^{2}}{2T^{2}}+ \varepsilon_{1} \biggr) \xi^{2}+\zeta \biggr) \\ =& - \biggl( \biggl( \frac{\pi^{2}}{2T^{2}}+\varepsilon_{1} \biggr) m-M \biggr) \vert r\vert ^{2}+\vert r\vert \delta m^{1/2} +2T \biggl( \biggl( \frac{\pi^{2}}{2T ^{2}}+\varepsilon_{1} \biggr) \xi^{2}+\zeta \biggr), \end{aligned}$$
which implies that there exists \(r \in R \setminus \{ 0\} \) such that \(\Vert r Q_{1}\Vert > \varrho \) and \(I(r Q_{1}) < 0\). Set \(e(t)=r Q_{1}(t)\). Then \(\Vert e\Vert > \varrho \) and \(I(e) < 0\). □
Lemma 2.3
Suppose that the conditions of Theorem 1.1
hold, then
I
satisfies the
\((C)\)
condition.
Proof
Assume that \(\{u_{n}\} \subset E\) is a sequence such that \(\{I(u_{n})\}\) is bounded and \(\Vert I^{\prime }(u_{n})\Vert (1+\Vert u_{n}\Vert ) \rightarrow 0\) as \(n \rightarrow \infty \). Then there exists a constant \(M_{1}> 0\) such that
$$\begin{aligned} \bigl\vert I(u_{n})\bigr\vert \leq M_{1},\qquad \bigl\Vert I^{\prime }(u_{n})\bigr\Vert \bigl(1+\Vert u_{n} \Vert \bigr) \leq M_{1}. \end{aligned}$$
(8)
Now we prove that \(\{u_{n}\}\) is bounded in E. Arguing in an indirect way, we assume that \(\Vert u_{n}\Vert \rightarrow +\infty \) as \(n\rightarrow \infty \). Set \(z_{n}=\frac{u_{n}}{\Vert u_{n}\Vert }\), then \(\Vert z_{n}\Vert =1\), which implies that there exists a subsequence of \(\{z_{n}\}\), still denoted by \(\{z_{n}\}\), such that \(z_{n}\rightharpoonup z_{0}\) in E. By (4) and (8), we get
$$\begin{aligned} \biggl\vert \int_{ R}\frac{W(t,u_{n}(t))}{\Vert u_{n}\Vert ^{2}}\,dt-\frac{1}{2}\biggr\vert =& \biggl\vert -\frac{I(u_{n})}{\Vert u_{n}\Vert ^{2}}+ \int_{ R}\frac{(f(t),u_{n}(t))}{ \Vert u_{n}\Vert ^{2}}\,dt\biggr\vert \\ \leq &\frac{M_{1}+\delta \Vert u_{n}\Vert }{\Vert u_{n}\Vert ^{2}}, \end{aligned}$$
(9)
which implies that
$$\begin{aligned} \biggl\vert \int_{ R}\frac{W(t,u_{n}(t))}{\Vert u_{n}\Vert ^{2}}\,dt\biggr\vert \leq 1 \end{aligned}$$
(10)
for n large enough. The following discussion is divided into two cases.
Case 1: \(z_{0}\equiv 0\). Let \(s>\frac{\delta C_{1}}{2}\). From \((W1)\) and \((W4)\), we can deduce that there exists \(M_{2}>0\) such that
$$\begin{aligned} \bigl\vert W(t,x)\bigr\vert \leq M_{2}\bigl(\vert x\vert ^{2}+\vert x\vert ^{\tau }\bigr) \quad \forall t \in R, \end{aligned}$$
(11)
and
$$\begin{aligned} \bigl\vert \bigl(\nabla W(t,x),x\bigr)\bigr\vert \leq M_{2}\bigl( \vert x\vert ^{2}+\vert x\vert ^{\tau }\bigr) \quad \forall t \in R. \end{aligned}$$
(12)
By the compactness of the embedding, one can obtain
$$\begin{aligned} \limsup_{n\rightarrow \infty } \int_{ R}\bigl\vert W\bigl(t,sz_{n}(t)\bigr)\bigr\vert \,dt\leq M_{2} \limsup_{n\rightarrow \infty } \int_{ R}\bigl( s^{2}\vert z_{n}\vert ^{2}+s^{\tau }\vert z_{n}\vert ^{\tau } \bigr)\,dt=0. \end{aligned}$$
(13)
Set \(\lambda_{n}=\frac{s}{\Vert u_{n}\Vert }\). It follows from (8)–(13), \((W2)\), and \((W5)\) that
$$\begin{aligned} M_{1} \geq & I(u_{n}) \\ =&I(\lambda_{n}u_{n})+\frac{1-\lambda_{n}^{2}}{2}\Vert u_{n}\Vert ^{2} \\ &{}+ \int _{ R} \bigl( W\bigl(t,\lambda_{n}u_{n}(t) \bigr)-W\bigl(t,u_{n}(t)\bigr) \bigr)\,dt+(1-\lambda _{n}) \int_{ R}\bigl(f(t),u_{n}(t)\bigr)\,dt \\ =&I(\lambda_{n}u_{n})+ \biggl( \frac{1-\lambda_{n}^{2}}{2}- \lambda_{n} \biggr) \bigl\langle I'(u_{n}),u_{n} \bigr\rangle \\ &{}+ \biggl( \lambda_{n}\Vert u_{n}\Vert ^{2}+ \biggl( \frac{(1- \lambda_{n})^{2}}{2}+\lambda_{n} \biggr) \int_{ R}\bigl(f(t),u_{n}(t)\bigr)\,dt \biggr) \\ &+ \int_{ R} \biggl( \biggl( \frac{1-\lambda_{n}^{2}}{2}- \lambda_{n} \biggr) \bigl( \nabla W\bigl(t,u_{n}(t) \bigr),u_{n}(t)\bigr)+W\bigl(t,\lambda_{n}u_{n}(t) \bigr)-W\bigl(t,u_{n}(t)\bigr) \biggr)\,dt \\ \geq & I(sz_{n})+o(1)+ \biggl( s-\delta C_{1} \biggl( \frac{(1-\lambda_{n})^{2}}{2}+ \lambda_{n} \biggr) \biggr) \Vert u_{n}\Vert \\ &{}+ \int_{\vert u_{n}\vert \geq r_{\infty }} \biggl( \biggl( \frac{1-\lambda_{n}^{2}}{2}- \lambda_{n} \biggr) \bigl(\nabla W\bigl(t,u_{n}(t) \bigr),u_{n}(t)\bigr) \\ &{}+W\bigl(t,\lambda_{n}u _{n}(t)\bigr)-W\bigl(t,u_{n}(t)\bigr) \biggr)\,dt{ } \\ &{}+ \int_{\vert u_{n}\vert \leq r_{\infty }} \biggl( \biggl( \frac{1-\lambda_{n}^{2}}{2}- \lambda_{n} \biggr) \bigl(\nabla W\bigl(t,u_{n}(t) \bigr),u_{n}(t)\bigr) \\ &{}+W\bigl(t,\lambda_{n}u _{n}(t)\bigr)-W\bigl(t,u_{n}(t)\bigr) \biggr)\,dt \\ \geq & \frac{s^{2}}{2}- \int_{ R}W\bigl(t,sz_{n}(t)\bigr)\,dt+ \int_{ R}\bigl(f(t),sz _{n}(t)\bigr)\,dt+o(1) \\ &{}-d_{2} \int_{\vert u_{n}\vert \geq r_{\infty }}\lambda_{n}^{\mu }\bigl\vert u_{n}(t)\bigr\vert ^{ \mu }\,dt+ \int_{\vert u_{n}\vert \leq r_{\infty }} \biggl( -\frac{\lambda_{n}^{2}+ \lambda_{n}}{2}\bigl(\nabla W \bigl(t,u_{n}(t)\bigr),u_{n}(t)\bigr) \biggr)\,dt \\ \geq & \frac{s^{2}}{2}-\delta C_{1} \int_{ R}\bigl\vert z_{n}(t)\bigr\vert \,dt-d _{2}s^{\mu } \int_{\vert u_{n}\vert \geq r_{\infty }}\bigl\vert z_{n}(t)\bigr\vert ^{\mu }\,dt+o(1) \\ &{}+M_{2} \int_{\vert u_{n}\vert \leq r_{\infty }} \biggl( -\frac{\lambda_{n}^{2}+ \lambda_{n}}{2}\bigl(\bigl\vert u_{n}(t)\bigr\vert ^{2}+\bigl\vert u_{n}(t) \bigr\vert ^{\tau }\bigr) \biggr)\,dt \\ \geq & \frac{s^{2}}{2}-\frac{M_{2}}{2} \int_{\vert u_{n}\vert \leq r_{\infty }} \bigl( \lambda_{n}^{2}\bigl\vert u_{n}(t)\bigr\vert ^{2}+ \lambda_{n} \bigl\vert u_{n}(t)\bigr\vert ^{2}+\lambda_{n}^{2} \bigl\vert u_{n}(t)\bigr\vert ^{\tau }+\lambda _{n}\bigl\vert u_{n}(t)\bigr\vert ^{\tau } \bigr)\,dt+o(1) \\ \geq & \frac{s^{2}}{2}-\frac{M_{2}}{2} \int_{\vert u_{n}\vert \leq r_{\infty }} \bigl( s^{2}\bigl\vert z_{n}(t)\bigr\vert ^{2}+r_{\infty }s\bigl\vert z_{n}(t)\bigr\vert +r_{\infty }^{\tau -2}s^{2} \bigl\vert z_{n}(t)\bigr\vert ^{2}+r_{\infty }^{\tau -1}s \bigl\vert z_{n}(t)\bigr\vert \bigr)\,dt+o(1) \\ \geq &\frac{s^{2}}{2}-o(1) \end{aligned}$$
for s and n large enough, which is a contradiction. Hence \(\Vert u_{n}\Vert \) is still bounded in this case, which implies that \(\{u_{n}\}\) is bounded in E.
Case 2: \(z_{0}\not \equiv 0\). Let \(\Omega =\{t\in R\vert \vert z_{0}(t)\vert >0 \}\). Then we can see that \(\operatorname{meas}(\Omega)>0\), where meas denotes the Lebesgue measure. Since \(\Vert u_{n}\Vert \rightarrow +\infty \) as \(n\rightarrow \infty \) and \(\vert u_{n}(t)\vert =\vert z_{n}(t)\vert \cdot \Vert u_{n}\Vert \), then we have \(\vert u_{n}(t)\vert \rightarrow +\infty \) as \(n\rightarrow \infty \) for a.e. \(t\in \Omega \). By \((W2)\), \((W3)\), and Fatou’s lemma, we can obtain
$$\begin{aligned} \liminf_{n\rightarrow \infty } \int_{ R}\frac{W(t,u_{n}(t))}{\Vert u_{n}\Vert ^{2}}\,dt \geq &\liminf _{n\rightarrow \infty } \int_{\Omega }\frac{W(t,u _{n}(t))}{\Vert u_{n}\Vert ^{2}}\,dt{ } \\ =& \liminf_{n\rightarrow \infty } \int_{\Omega }\frac{W(t,u _{n}(t))}{\vert u_{n}\vert ^{2}}\bigl\vert z_{n}(t) \bigr\vert ^{2}\,dt \\ =& +\infty, \end{aligned}$$
which contradicts (10). So \(\Vert u_{n}\Vert \) is bounded in this case. □
By a standard argument, we see that \(\{u_{n}\}\) has a convergent subsequence in E. Hence I satisfies the \((C)\) condition.
Proof of Theorem 1.1
The proof of this theorem is divided into two steps.
Step 1: We show that there exists a function \(u_{0}\in E\) such that \(I'(u_{0})=0\) and \(I(u_{0})<0\). Let \(f(t)=(f_{1}(t), f_{2}(t),\ldots,f_{N}(t))\), where \(f_{i}(t)\in C( R, R)\) (\(i=1,2,\ldots,N\)). Since \(f\not \equiv 0\), there exists \(i_{0}\in [1,N]\cap Z\) such that \(f_{i_{0}}(t)\not \equiv 0\). Without loss of generalization, we assume that there exist an interval \((a,b)\subset R\) and a constant \(A>0\) such that
$$\begin{aligned} f_{i_{0}}(t)\geq A\quad \mbox{for all } t\in (a,b). \end{aligned}$$
We choose a function \(\psi_{0}\in C_{0}^{\infty }(a,b)\) satisfying
$$\begin{aligned} \textstyle\begin{cases} \psi_{0}(t)=-f_{i_{0}}(t),&t\in ( \frac{3a+b}{4}, \frac{a+3b}{4} ), \\ \psi_{0}(t)\leq 0,&t\in ( a,b) \setminus ( \frac{3a+b}{4}, \frac{a+3b}{4} ), \\ \vert \psi_{0}'(t)\vert \leq 2,&t\in ( a,b). \end{cases}\displaystyle \end{aligned}$$
Set \(\psi (t)=(\psi_{1}(t),\psi_{2}(t),\ldots,\psi_{N}(t))\), where \(\psi_{j}(t)=0\) for all \(j\in [1,N]\cap Z\setminus \{i_{0}\}\) and \(\psi_{i}(t)=\psi_{0}(t)\) for \(j=i_{0}\). Therefore, \(\psi \in E\) and we can deduce that
$$\begin{aligned} \int_{ R}\bigl(f(t),\psi (t)\bigr)\,dt= \int_{a}^{b}f_{i_{0}}(t) \psi_{0}(t)\,dt \leq - \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}}f_{i_{0}}^{2}(t)\,dt\leq -\frac{A ^{2}}{2}(b-a)< 0. \end{aligned}$$
Hence we have
$$\begin{aligned} I(r\psi) =& \frac{1}{2} \int_{ R}\bigl\vert r\dot{\psi }(t)\bigr\vert ^{2}\,dt- \int_{ R}W\bigl(t,r \psi (t)\bigr)\,dt+ \int_{ R}\bigl(f(t),r\psi (t)\bigr)\,dt{ } \\ \leq & \frac{r^{2}}{2} \int_{ R}\bigl\vert \dot{\psi }(t)\bigr\vert ^{2}\,dt-M_{2} \biggl( r^{2} \int_{ R}\bigl\vert \psi (t)\bigr\vert ^{2}\,dt+r^{\tau } \int_{ R}\bigl\vert \psi (t)\bigr\vert ^{ \tau } \biggr)\,dt+r \int_{ R}\bigl(f(t),\psi (t)\bigr)\,dt \\ < &0 \end{aligned}$$
for \(r>0\) small enough. Then we obtain
$$\begin{aligned} c_{0}=\inf \bigl\{ I(u): u\in B_{\varrho }\bigr\} < 0, \end{aligned}$$
where ϱ is defined in Lemma 2.1 and \(B_{\varrho }=\{u\in E\vert \Vert u\Vert \leq \varrho \}\). By Ekeland’s variational principle, there exists a sequence \(\{u_{n}\}\subset B_{\varrho }\) such that
$$\begin{aligned} c_{0}\leq I(u_{n})\leq c_{0}+ \frac{1}{n}, \end{aligned}$$
and
$$\begin{aligned} I(w)\leq I(u_{n})-\frac{1}{n}\Vert w-u_{n}\Vert \end{aligned}$$
for all \(w\in B_{\varrho }\). Then, by a standard procedure, we can show that \(\{u_{n}\}\) is a \((C)\) sequence of I. Therefore, it follows from Lemma 2.3 that there exists a function \(u_{0}\in E\) such that \(I'(u_{0})=0\) and \(I(u_{0})<0\).
Step 2: By the mountain pass theorem and Lemmas 2.1–2.3, there exists \({\widetilde{u}_{0}}\in E\) such that \(I'(\widetilde{u}_{0})=0\) and \(I(\widetilde{u}_{0})>0\).
Then we finish the proof of Theorem 1.1. □