Now we give the proof of Theorem 1.1.
Proof
By Lemma 2.3, Φ is a sequentially weakly lower semicontinuous, continuously Gâteaux derivative and coercive functional whose Gâteaux derivative admits a continuous inverse on \(X^{\ast}\). By Lemma 2.5, Ψ is a sequentially weakly upper semicontinuous and continuously Gâteaux differentiable functional whose Gâteaux derivative is compact.
Let \(r=\frac{d^{2}}{2\beta^{2}}\), \(u_{0}(t)=0\), \(u_{1}(t)=qe^{-t}\) for any \(t\in[0,+\infty)\), one has \(u_{0},u_{1}\in X\), \(\Phi(u_{0})=\Psi(u_{0})=0\), \(\Phi(u_{1})=\frac{(1+c)q^{2}}{4}+\sum_{j=1}^{p}\int _{0}^{qe^{-t_{j}}}I_{j}(s)\,ds+\int_{0}^{q}h(s)\,ds\), \(\Psi(u_{1})=\int_{0}^{+\infty}G(t,qe^{-t})\,dt\). Therefore, we get
$$ \bigl(r-\Phi(u_{0}) \bigr)\frac{\Psi(u_{1})}{\Phi(u_{1})-\Phi(u_{0})}= \frac {d^{2}}{\beta^{2}}\frac{\int_{0}^{+\infty}G(t,qe^{-t})\,dt}{ \frac{(1+c)q^{2}}{2}+2\sum_{j=1}^{p}\int _{0}^{qe^{-t_{j}}}I_{j}(s)\,ds+2\int_{0}^{q}h(s)\,ds}, $$
(3.1)
and by (A3), we obtain \(\Phi(u_{0})< r<\Phi(u_{1})\).
On the other hand, for any \(u\in X\) such that \(\Phi(u)\leq r\), we have \(\|u\|_{X}\leq(2r)^{\frac{1}{2}}\). Owing to (2.2), one has \(\|u\|_{C}\leq\beta\|u\|_{X}\leq\beta(2r)^{\frac{1}{2}}=d\). Therefore,
$$ \sup_{\Phi(x)\leq r}\Psi(x)\leq \int_{0}^{+\infty}\max_{|\xi|\leq d}G(t,\xi) \,dt. $$
(3.2)
By (3.1), (3.2), and (A3), condition (i) in Theorem 2.1 is satisfied.
For any \(u\in X\), in view of (A1), (A4), and (2.2), we obtain
$$\begin{aligned} \Phi(u)-\lambda\Psi(u) =&\frac{1}{2}\|u\|_{X}^{2}+ \sum_{j=1}^{p} \int_{0}^{u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{u(0)}h(s)\,ds-\lambda \int _{0}^{+\infty}G \bigl(t,u(t) \bigr)\,dt \\ \geq& \biggl(\frac{1}{2}-\lambda\beta^{2}|a_{1}|_{1} \biggr)\|u\| _{X}^{2}-\lambda|a_{2}|_{1} \beta^{\alpha}\|u\|_{X}^{\alpha}-\lambda\beta |a_{3}|_{1}\|u\|_{X}. \end{aligned}$$
In view of \(\mathrm{(A_{5})}\), we get
$$ \frac{r-\Phi(u_{0})}{\sup_{\Phi(u)\leq r}\Psi(u)}=\frac{d^{2}}{2\beta ^{2}\sup_{\Phi(x)\leq r}\Psi(x)}\leq \frac{d^{2}}{2\beta^{2}\int_{0}^{+\infty}G(t,me^{-t})\,dt}\leq \frac {1}{2\beta^{2}|a_{1}|_{1}}. $$
(3.3)
Then, for any \(\lambda\in\,]0,\frac{1}{2\beta^{2}|a_{1}|_{1}}[\) (with the conventions \(\frac{1}{0}=+\infty\)), we get \(\lim_{\|u\|_{X}\rightarrow +\infty}(\Phi(u)-\lambda\Psi(u))=+\infty\). So condition \(\mathrm{(ii)}\) in Theorem 2.1 is satisfied. Hence, by Theorem 2.1, for each \(\lambda\in\,]\frac{1}{\alpha_{2}},\frac{1}{\alpha_{1}}[\), the functional \(\Phi-\lambda\Psi\) has at three distinct critical points in X. That is, for each \(\lambda\in\,]\frac{1}{\alpha_{2}},\frac{1}{\alpha_{1}}[\), problem (1.1) has at least three classical solutions. □
Now we give the proof of Theorem 1.2.
Proof
First of all, we will show that \(\Phi-\lambda\Psi \) is weakly lower semicontinuous. Let \(\{u_{n}\}\subset X\), \(u_{n}\rightharpoonup u\) in X, we see that \(\{u_{n}\}\) converges uniformly to u on \([0,M]\) with \(M\in(0,+\infty)\) an arbitrary constant and \(\liminf_{n\rightarrow\infty}\|u_{n}\|_{X}\geq\|u\|_{X}\). By Lemma 2.4, we have
$$\begin{aligned} \liminf_{n\rightarrow\infty} \bigl(\Phi(u_{n})-\lambda\Psi (u_{n}) \bigr) \geq&\liminf_{n\rightarrow\infty} \Biggl( \frac {1}{2}\|u_{n}\|_{X}^{2} +\sum _{j=1}^{p} \int_{0}^{u_{n}(t_{j})}I_{j}(s)\,ds+ \int _{0}^{u_{n}(0)}h(s)\,ds \Biggr) \\ &{}-\lambda\limsup_{n\rightarrow\infty}\Psi(u_{n}) \\ \geq&\frac{1}{2}\|u\|_{X}^{2}+\sum _{j=1}^{p} \int _{0}^{u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{u(0)}h(s)\,ds \\ &{}-\lambda\limsup_{n\rightarrow+\infty}\Psi(u_{n}) \\ \geq&\frac{1}{2}\|u\|_{X}^{2}+\sum _{j=1}^{p} \int _{0}^{u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{u(0)}h(s)\,ds \\ &{}-\lambda \int_{0}^{+\infty}G(t,u)\,dt \\ =&\Phi(u)-\lambda\Psi(u). \end{aligned}$$
Then \(\Phi-\lambda\Psi\) is sequentially weakly lower semicontinuous.
Second, we will show that \(\Phi-\lambda\Psi\) is coercive. By (A6), (A7), and (2.2), we obtain
$$\begin{aligned} \Phi(u)-\lambda\Psi(u) =&\frac{1}{2}\|u\|_{X}^{2}+ \sum_{j=1}^{p} \int_{0}^{u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{u(0)}h(s)\,ds-\lambda \int _{0}^{+\infty}G \bigl(t,u(t) \bigr)\,dt \\ \geq& \biggl(\frac{1}{2}-\lambda\beta^{2}|c_{1}|_{1} \biggr)\|u\| _{X}^{2}-\lambda|c_{2}|_{1} \bigl(\beta^{\sigma}\|u\|_{X}^{\sigma}+c_{3} \bigr), \end{aligned}$$
for any \(u\in X\). Since \(0<\sigma<2\), for any \(\lambda\in\,]0,\frac{1}{2\beta^{2}|c_{1}|_{1}}[\) (with the conventions \(\frac{1}{0}=+\infty\)), we obtain \(\lim_{\|u\|\rightarrow\infty}(\Phi(u)-\lambda\Psi(u)) = +\infty\), that is, \(\Phi-\lambda\Psi\) is coercive. Hence, \(\Phi-\lambda\Psi\) has a minimum (Theorem 1.1 of [45]), which is a critical point of \(\Phi-\lambda\Psi\). Thus, for each \(\lambda\in\,]0,\frac{1}{2\beta^{2}|c_{1}|_{1}}[\), problem (1.1) has at least one classical solution. □
Now we give the proof of Theorem 1.3.
Proof
Let \(\varphi=\Phi-\lambda\Psi\). Obviously, \(\varphi\in C^{1}(X,\mathbb{R})\). In the following, we first show that φ is bounded from below. By (A8), (A9), and (2.2), we have
$$\begin{aligned} \varphi(u) =& \frac{1}{2}\|u\|_{X}^{2}+ \sum_{j=1}^{p} \int _{0}^{u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{u(0)}h(s)\,ds - \lambda \int _{0}^{+\infty}G \bigl(t,u(t) \bigr)\,dt \\ \geq& \frac{1}{2}\|u\|_{X}^{2}- \sum _{j=1}^{p}c_{j}^{\prime}\beta^{\delta_{j}+1}\|u\|_{X}^{\delta_{j}+1}-c^{\prime}\beta^{\delta+1}\|u\|_{X}^{\delta+1} \\ &{}- \lambda \int_{0}^{+\infty} \bigl(k_{1}(t)|u|^{\gamma _{1}+1}+k_{2}(t)|u| \bigr)\,dt \\ \geq& \frac{1}{2}\|u\|_{X}^{2}- \sum _{j=1}^{p}c_{j}^{\prime}\beta^{\delta_{j}+1}\|u\|_{X}^{\delta_{j}+1}-c^{\prime}\beta^{\delta+1}\|u\|_{X}^{\delta+1} \\ &{}-\lambda\beta^{\gamma_{1}+1}|k_{1}|_{1}\|u \|_{X}^{\gamma_{1}+1}-\lambda \beta|k_{2}|_{1}\|u \|_{X}. \end{aligned}$$
(3.4)
Since \(\delta_{j},\delta\in(0,1)\) and \(\gamma_{1}\in(0,1)\), (3.4) implies that \(\varphi(u)\rightarrow\infty\) as \(\| u\| _{X}\rightarrow\infty\). Consequently, φ is bounded from below.
Next, we prove that φ satisfies the Palais-Smale condition. Suppose that \(\{u_{n}\}\subset X\) such that \(\{\varphi(u_{n})\}\) be a bounded sequence and \(\varphi^{\prime}(u_{n})\rightarrow0\) as \(n\rightarrow \infty\), it follows from (3.4) that \(\{u_{n}\}\) is bounded in X. From the reflexivity of X, we may extract a weakly convergent subsequence, which, for simplicity, we call \(\{u_{n}\}\), \(u_{n}\rightharpoonup u\) in X. Next we will prove that \(u_{n}\rightarrow u\) in X. By (2.5) and (2.6), we have
$$\begin{aligned} \bigl(\varphi^{\prime}(u_{n})- \varphi^{\prime}(u) \bigr) (u_{n}-u) =&\|u_{n}-u\| _{X}^{2}+ \bigl[h(u_{n}(0)-h \bigl(u(0) \bigr) \bigr] \bigl(u_{n}(0)-u(0) \bigr) \\ &{}+ \sum_{j=1}^{p} \bigl(I_{j} \bigl(u_{n}(t_{j}) \bigr)-I_{j} \bigl(u(t_{j}) \bigr) \bigr) \bigl(u_{n}(t_{j})-u(t_{j}) \bigr) \\ &{}- \lambda \int_{0}^{+\infty} \bigl(g \bigl(t,u_{n}(t) \bigr)-g \bigl(t,u(t) \bigr) \bigr) \bigl(u_{n}(t)-u(t) \bigr)\,dt. \end{aligned}$$
(3.5)
Obviously,
$$ \bigl(\varphi^{\prime}(u_{n})- \varphi^{\prime}(u) \bigr) (u_{n}-u)\rightarrow0. $$
(3.6)
We claim that if \(u_{k}\rightharpoonup u\) in E, then \(g(t,u_{k})\rightarrow g(t,u)\) in \(L^{1}[0,+\infty)\). The proof is similar to that of Lemma 2.4, and we omit it here. By (2.2), we obtain
$$\begin{aligned} & \int_{0}^{+\infty} \bigl(g \bigl(t,u_{n}(t) \bigr)-g \bigl(t,u(t) \bigr) \bigr) \bigl(u_{n}(t)-u(t) \bigr)\,dt \\ &\quad\leq \bigl(\|u_{n}\|_{C}+\|u\|_{C} \bigr) \int_{0}^{+\infty } \bigl|g \bigl(t,u_{n}(t) \bigr)-g \bigl(t,u(t) \bigr) \bigr|\,dt \\ &\quad\leq\beta \bigl(\|u_{n}\|_{X}+\|u\|_{X} \bigr) \int_{0}^{+\infty } \bigl|g \bigl(t,u_{n}(t) \bigr)-g \bigl(t,u(t) \bigr) \bigr|\,dt\rightarrow0 \end{aligned}$$
(3.7)
as \(n\rightarrow\infty\). Since \(u_{n}\rightharpoonup u\) in X, for any \(M>0\), we get \(u_{n}\rightarrow u\) in \(C[0,M]\). So
$$ \begin{aligned} &\sum_{j=1}^{p} \bigl(I_{j} \bigl(u_{n}(t_{j}) \bigr)-I_{j} \bigl(u(t_{j}) \bigr) \bigr) \bigl(u_{n}(t_{j})-u(t_{j}) \bigr) \rightarrow0, \\ & \bigl[h \bigl(u_{n}(0) \bigr)-h \bigl(u(0) \bigr) \bigr] \bigl(u_{n}(0)-u(0) \bigr)\rightarrow0. \end{aligned} $$
(3.8)
In view of (3.5)-(3.8), we obtain \(\|u_{n}-u\| _{X}\rightarrow0\) as \(n\rightarrow\infty\). Then φ satisfies the Palais-Smale condition.
It is easy to see that φ is even and \(\varphi(0)=0\). In order to apply Theorem 2.2, we prove now that
$$ \mbox{for each } n\in\mathbb{N}, \mbox{ there exists } \varepsilon>0 \mbox{ such that } \gamma \bigl(\varphi^{-\varepsilon } \bigr)\geq n. $$
(3.9)
For each \(n\in\mathbb{N}\), we take n disjoint open sets \(B_{i}\) such that
$$\bigcup_{i=1}^{n}B_{i}\subset J. $$
For \(i=1,2,\ldots,n\), let \(u_{i}\in(W_{0}^{1,2}(B_{i})\cap X)\) and \(\| u_{i}\|_{X}=1\), and
$$E_{n}= \operatorname{span}\{u_{1},u_{2}, \ldots,u_{n}\}, \qquad J_{n}= \bigl\{ u\in E_{n}:\|u \|_{X}=1 \bigr\} . $$
For any \(u\in E_{n}\), there exist \(\lambda_{i}\in\mathbb{R}\), \(i=1,2,\ldots,n\), such that
$$ u(t)=\sum_{i=1}^{n} \lambda_{i}u_{i}(t) \quad\mbox{for } t\in[0,+\infty). $$
(3.10)
Then
$$ |u|_{\gamma_{2}}= \biggl( \int_{0}^{+\infty} \bigl|u(t) \bigr|^{\gamma_{2}}\,dt \biggr)^{\frac{1}{\gamma_{2}}}= \Biggl(\sum_{i=1}^{n}| \lambda _{i}|^{\gamma_{2}} \int_{B_{i}} \bigl|u_{i}(t) \bigr|^{\gamma_{2}}\,dt \Biggr)^{\frac{1}{\gamma_{2}}} $$
(3.11)
and
$$\begin{aligned} \|u\|_{X}^{2} =& \int_{0}^{+\infty} \bigl( \bigl|u_{i}^{\prime}(t) \bigr|^{2}+c \bigl|u(t) \bigr|^{2} \bigr)\,dt \\ =& \sum_{i=1}^{n}\lambda_{i}^{2} \int_{B_{i}} \bigl( \bigl|u_{i}^{\prime}(t) \bigr|^{2}+c \bigl|u_{i}(t) \bigr|^{2} \bigr)\,dt \\ =& \sum_{i=1}^{n}\lambda_{i}^{2} \int_{0}^{+\infty} \bigl( \bigl|u_{i}^{\prime}(t) \bigr|^{2}+c \bigl|u_{i}(t) \bigr|^{2} \bigr)\,dt \\ =& \sum_{i=1}^{n}\lambda_{i}^{2} \|u_{i}\|_{X}^{2} \\ =& \sum_{i=1}^{n}\lambda_{i}^{2}. \end{aligned}$$
(3.12)
Since all norms of any finite dimensional normed space are equivalent, so there exists \(M_{0}>0\) such that
$$ M_{0}\|u\|_{X}\leq|u|_{\gamma_{2}} \quad \mbox{for } u\in E_{n}. $$
(3.13)
In view of (A8), (A10), (2.2), (3.11), (3.12), and (3.13), we get
$$\begin{aligned} \varphi(\rho u) =& \frac{\rho^{2}}{2}\|u\|_{X}^{2}+ \sum_{j=1}^{p} \int_{0}^{\rho u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{\rho u(0)}h(s)\,ds - \lambda \int_{0}^{+\infty}G \bigl(t,\rho u(t) \bigr)\,dt \\ =& \frac{\rho^{2}}{2}\|u\|_{X}^{2}+\sum _{j=1}^{p} \int _{0}^{\rho u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{\rho u(0)}h(s)\,ds \\ &{}- \lambda\sum _{i=1}^{n} \int_{B_{i}}G \bigl(t,\rho u(t) \bigr)\,dt \\ \leq& \frac{\rho^{2}}{2}\|u\|_{X}^{2}+ \sum _{j=1}^{p}c_{j}^{\prime}(\rho \beta)^{\delta_{j}+1}\|u\|_{X}^{\delta_{j}+1} +c^{\prime}(\rho \beta)^{\delta+1}\|u\|_{X}^{\delta+1} \\ &{}-\lambda\eta\rho^{\gamma_{2}} \sum_{i=1}^{n}| \lambda _{i}|^{\gamma_{2}} \int_{B_{i}} \bigl|u_{i}(t) \bigr|^{\gamma_{2}}\,dt \\ =& \frac{\rho^{2}}{2}\|u\|_{X}^{2}+ \sum _{j=1}^{p}c_{j}^{\prime}(\rho \beta)^{\delta_{j}+1}\|u\|_{X}^{\delta_{j}+1} +c^{\prime}(\rho \beta)^{\delta+1}\|u\|_{X}^{\delta+1}-\lambda\eta \rho^{\gamma_{2}}|u|_{\gamma_{2}}^{\gamma_{2}} \\ \leq& \frac{\rho^{2}}{2}\|u\|_{X}^{2}+ \sum _{j=1}^{p}c_{j}^{\prime}(\rho \beta)^{\delta_{j}+1}\|u\|_{X}^{\delta_{j}+1} +c^{\prime}(\rho \beta)^{\delta+1}\|u\|_{X}^{\delta+1}-\lambda\eta (M_{0}\rho)^{\gamma_{2}}\|u\|_{X}^{\gamma_{2}} \\ =& \frac{\rho^{2}}{2}+ \sum_{j=1}^{p}c_{j}^{\prime}(\rho\beta)^{\delta_{j}+1} +c^{\prime}(\rho\beta)^{\delta+1}-\lambda \eta(M_{0}\rho)^{\gamma_{2}}, \end{aligned}$$
(3.14)
for \(\forall u\in J_{n}\), \(0<\rho\leq\frac{T}{\beta}\).
Since \(\gamma_{2}\in(1,2)\) with \(\gamma_{2}<\min\{\min_{1\leq j\leq p}\{\delta_{j}\},\delta\}+1\), there exist \(\varepsilon>0\) and \(\delta>0\) such that
$$ \varphi(\delta u)< -\varepsilon\quad\mbox{for } u\in J_{n}. $$
(3.15)
Let
$$J_{n}^{\delta}=\{\delta u:u\in J_{n}\}, \qquad\Omega= \Biggl\{ (\lambda _{1},\lambda_{2},\ldots,\lambda_{n}) \in\mathbb{R}:\sum_{l=1}^{n}\lambda _{l}^{2}< \delta^{2} \Biggr\} , $$
then it follows from (3.13) that
$$\varphi(u)< -\varepsilon\quad\mbox{for } u\in J_{n}^{\delta}. $$
Together with the fact that \(\varphi\in C^{1}(X,\mathbb{R})\) and is even, it implies that
$$ J_{n}^{\delta}\subset\varphi^{-\varepsilon}\in \Sigma. $$
(3.16)
By virtue of (3.10) and (3.12), there exists an odd homeomorphism mapping \(f\in C(J_{n}^{\delta},\partial\Omega )\). By some properties of the genus (see 3∘ of Propositions 7.5 and 7.7 in [42]), one has
$$ \gamma \bigl(\varphi^{-\varepsilon} \bigr)\geq\gamma \bigl(J_{n}^{\delta}\bigr)=n, $$
(3.17)
so the proof of (3.9) follows. Let
$$d_{n}:=\inf_{J\in\Sigma_{n}}\sup_{u\in J} \varphi(u). $$
It follows from (3.17) and the fact that φ is bounded from below on X that \(-\infty< d_{n}\leq-\varepsilon<0\), that is, for any \(n\in\mathbb{N}\), \(d_{n}\) is a real negative number. By Theorem 2.2, φ has infinitely many critical points, and so problem (1.1) has infinitely many solutions. □