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Multiplicity of solutions for impulsive differential equation on the half-line via variational methods

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Abstract

In this paper, the existence of solutions for a second-order impulsive differential equation with two parameters on the half-line is investigated. Applying variational methods, we give some new criteria to guarantee that the impulsive problem has at least one classical solution, three classical solutions and infinitely many classical solutions, respectively. Some recent results are extended and significantly improved. Two examples are presented to demonstrate the application of our main results.

Introduction

In this paper, we consider the following boundary value problem with impulses:

$$ \begin{aligned} &{-}u^{\prime\prime}(t)+cu(t)=\lambda g \bigl(t,u(t) \bigr), \quad \mbox{a.e. } t\in[0,+\infty), \\ &\Delta u^{\prime}(t_{j})=I_{j} \bigl(u(t_{j}) \bigr), \quad j=1,2,\ldots,p, \\ &u^{\prime}\bigl(0^{+} \bigr)=h \bigl(u(0) \bigr), \quad u^{\prime}(+ \infty)=0, \end{aligned} $$
(1.1)

where c and λ are two positive parameters, \(0=t_{0}< t_{1}<\cdots <t_{p}<+\infty\), \(\Delta u^{\prime}(t_{j})=u^{\prime}(t_{j}^{+})-u^{\prime}(t_{j}^{-}) =\lim_{t\rightarrow t_{j}^{+}}u^{\prime}(t)-\lim_{t\rightarrow t_{j}^{-}}u^{\prime}(t)\), \(u^{\prime}(0^{+})=\lim_{t\rightarrow0^{+}}u^{\prime}(t)\), and \(u^{\prime}(+\infty)=\lim_{t\rightarrow+\infty}u^{\prime}(t)\), \(h,I_{j}\in C(\mathbb{R}, \mathbb{R})\), and \(g\in C([0,+\infty)\times \mathbb{R}, \mathbb{R})\).

Boundary value problems (BVPs) on the half-line occur in many applications; see [13]. Due to its significance, many researchers have studied BVPs for differential equations on the half-line, we refer the reader to [411].

On the other hand, impulsive differential equations have been widely applied in biology, control theory, industrial robotics, medicine, population dynamics and so on; see [1217]. Due to its significance, a lot of work has been done in the theory of impulsive differential equations, we refer the reader to [1824]. Some classical approaches and tools have been used to investigate BVPs for impulsive differential equations. These classical approaches and tools include the method of upper and lower solutions [23, 25], fixed point theorems [26] and topological degree theory [27, 28].

Recently, some researchers have used variational methods to investigate the existence and multiplicity of solutions for impulsive BVPs on the finite intervals [2937]. However, as far as we know, with the exception of [38, 39], the study of solutions of impulsive BVPs on the infinite intervals via variational methods has received considerably less attention. More precisely, in [38, 39], the authors studied the following BVP:

$$ \begin{aligned} &{-}u^{\prime\prime}(t)+u(t)=\lambda g \bigl(t,u(t) \bigr), \quad \mbox{a.e. } t\in[0,+\infty), \\ &\Delta u^{\prime}(t_{j})=I_{j} \bigl(u(t_{j}) \bigr), \quad j=1,2,\ldots,p, \\ &u^{\prime}\bigl(0^{+} \bigr)=h \bigl(u(0) \bigr), \quad u^{\prime}(+ \infty)=0, \end{aligned} $$
(1.2)

where λ is a positive parameter, \(h,I_{j}\in C(\mathbb{R}, \mathbb{R})\) and \(g\in C([0,+\infty)\times\mathbb{R}, \mathbb{R})\). They obtained the existence and multiplicity of solutions for (1.2) via variational methods.

Obviously, problem (1.1) is a generalization of problem (1.2). In fact, problem (1.2) follows from problem (1.1) by letting \(c=1\).

Motivated by the above facts, in this paper, we will improve and generalize some results in [38, 39].

In this paper, we need the following conditions.

(A1):

\(h(u)\), \(I_{j}(u)\) are nondecreasing, and \(h(u)u\geq 0\), \(I_{j}(u)u\geq0\) for any \(u\in\mathbb{R}\).

(A2):

\(h(u)u\geq0\), \(I_{j}(u)u\geq0\) for any \(u\in\mathbb {R}\) (\(j=1,2,\ldots,p\)) and there exist constants \(L,L_{j}\geq0\) such that

$$\bigl|h(u)-h(v) \bigr|\leq L|u-v|, \quad \bigl|I_{j}(u)-I_{j}(v) \bigr|\leq L_{j}|u-v|\quad \mbox{for any } u,v\in\mathbb{R}, $$

where L, \(L_{j}\) satisfy \(L+\sum_{j=1}^{p}L_{j}<\frac{1}{\beta^{2}}\), β will be given in (2.2).

(A3):

There exist \(d,q>0\) such that

$$\frac{d^{2}}{\beta^{2}}< \frac{(1+c)q^{2}}{2}+2\sum_{j=1}^{p} \int _{0}^{qe^{-t_{j}}}I_{j}(s)\,ds+2 \int_{0}^{q}h(s)\,ds $$

and

$$\alpha_{1}:=\frac{2\beta^{2}\int_{0}^{+\infty}\max_{|\xi|\leq d}G(t,\xi)\,dt}{d^{2}}< \alpha_{2}:=\frac{\int_{0}^{+\infty }G(t,qe^{-t})\,dt}{\frac{1+c}{4}q^{2}+\sum_{j=1}^{p}\int _{0}^{qe^{-t_{j}}}I_{j}(s)\,ds+\int_{0}^{q}h(s)\,ds}, $$

where \(G(t,u)=\int_{0}^{u}g(t,s)\,ds\), β will be given in (2.2).

Let \(|\cdot|_{k}\) denotes the usual norm on \(L^{k}[0,+\infty)\). Now, we state our main results.

Theorem 1.1

Assume that (A1) (or (A2)), (A3) hold and the following conditions are satisfied.

(A4):

There exist a positive constant \(\alpha\in(1, 2)\) and \(a_{1}, a_{2}, a_{3}\in L^{1}[0,+\infty)\) such that

$$\bigl|g(t,u) \bigr|\leq a_{1}(t)|u|+a_{2}(t)|u|^{\alpha-1}+a_{3}(t) $$

for a.e. \(t\in[0,+\infty)\) and all \(u\in\mathbb{R}\).

(A5):

There exists a constant m satisfying

$$\frac{(1+c)m^{2}}{2}+2\sum_{j=1}^{p} \int _{0}^{me^{-t_{j}}}I_{j}(s)\,ds+2 \int_{0}^{m}h(s)\,ds\leq\frac{d^{2}}{\beta^{2}} $$

such that

$$|a_{1}|_{1}\leq\frac{\int_{0}^{+\infty}G(t,me^{-t})\,dt}{d^{2}}. $$

Then, for each \(\lambda\in\,]\frac{1}{\alpha_{2}},\frac{1}{\alpha _{1}}[\), problem (1.1) has at least three classical solutions.

Remark 1.1

In (H2) of [38], \(l>1\) is needed; see (2.5) of [38].

Remark 1.2

Obviously, Theorem 1.1 generalizes Theorem 3.1 in [38]. Furthermore, the function

$$ g(t,u)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{\beta}e^{-t}, & u\leq\beta, \\ e^{-t} (\frac{u}{100}+600u^{\frac{1}{2}}-599\sqrt{\beta}-\frac {\beta}{100} ), & u>\beta, \end{array}\displaystyle \right . $$
(1.3)

does not satisfy (H2) in [38], while it satisfies (A4), and there are indeed many functions h and \(I_{j}\) not satisfying (H1) in [38], while they satisfy (A2), for example, \(h(u)=-\theta_{1}u\) and \(I_{j}(u)=\theta_{2}u(1+\sin u)\), where \(0<\theta_{1}<\frac{1}{2\beta^{2}}\) and \(0<\theta_{2}<\frac{1}{4p\beta^{2}}\).

Theorem 1.2

Assume that the following conditions are satisfied.

(A6):

There exist positive constants \(c_{3}\), \(1<\sigma<2\), and \(c_{1},c_{2},c_{4},c_{5},c_{6}\in L^{1}[0,+\infty)\) such that

$$\bigl|G(t,u) \bigr|\leq c_{1}(t)|u|^{2}+c_{2}(t) \bigl(|u|^{\sigma}+c_{3} \bigr), \qquad \bigl|g(t,u) \bigr|\leq c_{4}(t)|u|+c_{5}(t)|u|^{\sigma-1}+c_{6}(t) $$

for a.e. \(t\in[0,+\infty)\) and all \(u\in\mathbb{R}\).

(A7):

\(h(u)u\geq0\), \(I_{j}(u)u\geq0\) for any \(u\in\mathbb {R}\) (\(j=1,2,\ldots,p\)).

Then, for each \(\lambda\in\,]0,\frac{1}{2\beta^{2}|c_{1}|_{1}}[\), problem (1.1) has at least one classical solution.

Remark 1.3

Let \(c=1\), it is clear that Theorem 1.2 improves Theorem 3.1 in [39]. In fact, there are many functions not satisfying the condition \(\mathrm{(S2)}\) in [39], while they satisfy (A6), for example, the function \(g(t,u)=e^{-t}(u+u^{\frac{1}{2}})\).

Theorem 1.3

Assume that the following conditions are satisfied.

(A8):

There exist constants \(c^{\prime}, c_{j}^{\prime}>0\) and \(\delta,\delta_{j}\in(0,1)\) such that

$$\bigl|I_{j}(u) \bigr|\leq c_{j}^{\prime}|u|^{\delta_{j}}, \qquad \bigl|h(u) \bigr|\leq c^{\prime}|u|^{\delta} \quad\textit{for any } u\in \mathbb{R}. $$
(A9):

There exist \(k_{1},k_{2}\in L^{1}[0,+\infty)\) and \(\gamma _{1}\in(0,1)\) such that

$$g(t,u)\leq k_{1}(t)|u|^{\gamma_{1}}+k_{2}(t), \quad \textit{for a.e. } t\in[0,+\infty) \textit{ and all } u\in\mathbb{R}. $$
(A10):

There exist an open set \(J\subset[0,+\infty)\) and constants \(T,\eta>0\) and \(\gamma_{2}\in(1,2)\) with \(\gamma_{2}<\min\{\min_{1\leq j\leq p}\{\delta_{j}\},\delta\}+1\) such that

$$G(t,u)\geq\eta|u|^{\gamma_{2}}, \quad\forall(t,u)\in J\times\mathbb {R}, |u| \leq T. $$

Furthermore, suppose that \(g(t,u)\), \(I_{j}(u)\), and \(h(u)\) are odd about u. Then problem (1.1) has infinitely many classical solutions for \(\lambda>0\).

Remark 1.4

By (S3) and (3.3) in [39], one has \(d\in L^{\frac{2}{2-\alpha}}([0,+\infty),[0,+\infty))\) (in (S3)). Let \(c=1\), it is clear that Theorem 1.1 generalizes Theorem 3.2 in [39]. Furthermore, there are many functions g, h, and \(I_{ij}\) satisfying our Theorem 1.3 and not satisfying Theorem 3.2 in [39]. For example, let \(I_{j}(u)=-u^{\frac{3}{5}}\), \(h(u)=-u^{\frac{3}{5}}\), and \(g(t,u)= (\frac{1}{(1+t^{2})^{2}}-\frac{1}{(1+t)^{2}} )u^{\frac{1}{3}}\).

The remainder of this paper is organized as follows. In Section 2, we present some preliminaries. In Section 3, we give the proof of Theorems 1.1-1.3. Finally, two examples are presented to illustrate the main results.

Preliminaries

In order to prove Theorem 1.1, we will need to the following critical points theorem.

Theorem 2.1

([40, 41])

Let X be a reflexive real Banach space, let \(\Phi:X\rightarrow\mathbb{R}\) be a sequentially weakly lower semicontinuous, coercive and continuously Gâteaux differentiable functional whose Gâteaux derivative admits a continuous inverse on \(X^{\ast}\), and let \(\Psi:X\rightarrow \mathbb{R}\) be a sequentially weakly upper semicontinuous and continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Assume that there exist \(r\in\mathbb{R}\) and \(x_{0},x_{1}\in X\), with \(\Phi(x_{0})< r<\Phi(x_{1})\) and \(\Psi(x_{0})=0\) such that

  1. (i)

    \(\sup_{\Phi(x)\leq r}\Psi(x)<(r-\Phi(x_{0}))\frac {\Psi(x_{1})}{\Phi(x_{1})-\Phi(x_{0})}\),

  2. (ii)

    for each \(\lambda\in\Lambda_{r}:=\,]\frac{\Phi (x_{1})-\Phi(x_{0})}{\Psi(x_{1})},\frac{r-\Phi(x_{0})}{\sup_{\Phi(x)\leq r}\Psi(x)}[\), the functional \(\Phi-\lambda\Psi\) is coercive.

Then for each \(\lambda\in\Lambda_{r}\), the functional \(\Phi-\lambda \Psi\) has at three distinct critical points in X.

In order to prove Theorem 1.3, we will need to the following definitions and theorems.

Let X be a Banach space, \(\varphi\in C^{1}(X,\mathbb{R})\) and \(e\in \mathbb{R}\). Let

$$\begin{aligned}& \Sigma:= \bigl\{ J\subset X-\{0\}:J \mbox{ is closed in } X \mbox{ and symmetric with respect to } 0 \bigr\} , \\& K_{e}:= \bigl\{ u\in X:\varphi(u)=e,\varphi^{\prime}(u)=0 \bigr\} , \quad \varphi^{e}:= \bigl\{ u\in X:\varphi(u)\leq e \bigr\} . \end{aligned}$$

Definition 2.1

([42])

For \(A\in\Sigma\), we say the genus of A is n (denoted by \(\gamma(A)=n\)) if there is an odd \(f\in C(A,\mathbb{R}^{n}\setminus\{0\})\) and n is the smallest integer with this property.

Definition 2.2

Suppose that X is a Banach space and \(\varphi\in C^{1}(X,\mathbb{R})\). If any sequence \(\{u_{n}\}\subset X\) for which \(\varphi(u_{n})\) is bounded and \(\varphi^{\prime}(u_{n})\rightarrow0\) as \(n \rightarrow\infty\) possesses a convergent subsequence in X, we say that φ satisfies the Palais-Smale condition.

Theorem 2.2

([43])

Let φ be an even \(C^{1}\) functional on X and satisfy the Palais-Smale condition. For any \(n\in\mathbb{N}\), set

$$\Sigma_{n}:= \bigl\{ A\in\Sigma:\gamma(A)\geq n \bigr\} ,\qquad d_{n}:=\inf_{A\in\Sigma_{n}}\sup_{u\in A} \varphi(u). $$
  1. (i)

    If \(\Sigma_{n}\neq\emptyset\) and \(d_{n}\in\mathbb{R}\), then \(d_{n}\) is a critical value of φ.

  2. (ii)

    If there exists \(k_{0}\in\mathbb{N}\) such that

    $$d_{n}=d_{n+1}=\cdots=d_{n+k_{0}}=e\in\mathbb{R}, $$

    and \(e\neq\varphi(0)\), then \(\gamma(K_{e})\geq k_{0}+1\).

Let us recall some basic concepts. Set

$$E= \bigl\{ u:[0,+\infty)\rightarrow\mathbb{R}\mid u \mbox{ is absolutely continuous}, u^{\prime}\in L^{2}[0,+\infty) \bigr\} . $$

Denote the Sobolev space by

$$X:= \biggl\{ u\in E\Bigm| \int_{0}^{+\infty} \bigl( \bigl|u^{\prime }(t) \bigr|^{2}+ \bigl|u(t) \bigr|^{2} \bigr)\,dt< \infty \biggr\} , $$

with the norm

$$ \|u\|_{X}= \biggl( \int_{0}^{+\infty} \bigl( \bigl|u^{\prime}(t) \bigr|^{2}+c \bigl|u(t) \bigr|^{2} \bigr)\,dt \biggr)^{\frac{1}{2}}, $$
(2.1)

this norm is equivalent to the usual norm. Hence, X is a reflexive Banach space.

Let \(C:=\{u\in C[0,+\infty)\mid\sup_{t\in[0,+\infty )}|u(t)|<+\infty\}\), with the norm \(\|u\|_{C}=\sup_{t\in[0,+\infty )}|u(t)|\). Then C is a Banach space. In addition, X is continuously embedded in C, thus, there exists a constant \(\beta>0\) such that

$$ \|u\|_{C}\leq\beta\|u\|_{X} \quad\mbox{for any } u\in X. $$
(2.2)

Suppose that \(u\in C[0,+\infty)\). Moreover, assume that for every \(j=0,1,2,\ldots,p-1\), \(u_{j}=u|_{(t_{j},t_{j+1})}\) satisfy \(u_{j}\in C^{2}(t_{j},t_{j+1})\) and \(u_{p}=u|_{(t_{p},+\infty)}\in C^{2}(t_{p},+\infty)\). We say u is a classical solution of problem (1.1) if it satisfies the equation in problem (1.1) a.e. on \([0,+\infty )\), the limits \(u^{\prime}(0^{+})\), \(u^{\prime}(+\infty)\), \(u^{\prime }(t_{j}^{+})\), \(u^{\prime}(t_{j}^{-})\) (\(j=1,2,\ldots,p\)) exist, and the impulsive conditions and boundary conditions in problem (1.1) hold.

For every \(u\in X\), put

$$ \Phi(u)=\frac{1}{2}\|u\|_{X}^{2} + \sum_{j=1}^{p} \int_{0}^{u(t_{j})}I_{j}(s)\,ds + \int_{0}^{u(0)}h(s)\,ds $$
(2.3)

and

$$ \Psi(u)= \int_{0}^{+\infty}G(t,u)\,dt. $$
(2.4)

It is clear that Φ is Gâteaux differentiable at any \(u\in X\) and

$$ \bigl\langle \Phi^{\prime}(u),v \bigr\rangle = \int_{0}^{+\infty} \bigl[u^{\prime}(t) v^{\prime}(t)+cu(t)v(t) \bigr]\,dt +\sum_{j=1}^{p}I_{j} \bigl(u(t_{j}) \bigr)v(t_{j})+h \bigl(u(0) \bigr)v(0) $$
(2.5)

for any \(v\in X\). Obviously, \(\Phi^{\prime}:X\rightarrow X^{\ast}\) is continuous.

Clearly, \(\Psi:X\rightarrow\mathbb{R}\) is continuously Gâteaux differentiable functional at any \(u\in X\) and

$$ \bigl\langle \Psi^{\prime}(u),v \bigr\rangle = \int_{0}^{+\infty}g \bigl(t,u(t) \bigr)v(t)\,dt $$
(2.6)

for any \(v\in X\).

Lemma 2.1

If \(u\in X\) is a critical point of \(\Phi-\lambda\Psi\), then u is a classical solution of problem (1.1).

Proof

The proof is similar to that of [38], and we omit it here. □

Lemma 2.2

Assume that \(\mathrm{(A_{2})}\) are satisfied, then Φ is sequentially weakly lower semicontinuous, coercive and its derivative admits a continuous inverse on \(X^{\ast}\).

Proof

Let \(\{u_{n}\}\subset X\), \(u_{n}\rightharpoonup u\) in X, we see that \(\{u_{n}\}\) converges uniformly to u on \([0,M]\) for any \(M\in(0,+\infty)\) and \(\liminf_{n\rightarrow\infty}\|u_{n}\| _{X}\geq\|u\|_{X}\). Thus

$$\begin{aligned} \liminf_{n\rightarrow\infty}\Phi(u_{n}) =&\liminf _{n\rightarrow\infty} \Biggl(\frac{1}{2}\|u_{n} \|_{X}^{2} +\sum_{j=1}^{p} \int_{0}^{u_{n}(t_{j})}I_{j}(s)\,ds+ \int _{0}^{u_{n}(0)}h(s)\,ds \Biggr) \\ \geq& \frac{1}{2}\|u\|_{X}^{2}+\sum _{j=1}^{p} \int _{0}^{u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{u(0)}h(s)\,ds=\Phi(u). \end{aligned}$$

So Φ is sequentially weakly lower semicontinuous. Furthermore, in view of (2.3) and \(\mathrm{(A_{2})}\), we have

$$\begin{aligned} \Phi(u) =&\frac{1}{2}\|u\|_{X}^{2}+\sum _{j=1}^{p} \int _{0}^{u(t_{j})}I_{j}(s)\,ds+ \int_{0}^{u(0)}h(s)\,ds \geq\frac{1}{2}\|u\|_{X}^{2}. \end{aligned}$$

Thus, Φ is coercive.

Next we will show that \(\Phi^{\prime}\) admits a continuous inverse on \(X^{\ast}\). For each \(u\in X\backslash\{0\}\), by (2.5) and (A2), we have

$$\begin{aligned} \bigl\langle \Phi^{\prime}(u),u \bigr\rangle =& \int_{0}^{+\infty} \bigl[ \bigl|u^{\prime}(t) \bigr|^{2}+c \bigl|u(t) \bigr|^{2} \bigr]\,dt+\sum _{j=1}^{p}I_{j} \bigl(u(t_{j}) \bigr)u(t_{j})+h \bigl(u(0) \bigr)u(0)\geq\|u \|_{X}^{2}. \end{aligned}$$

So \(\lim_{\|u\|_{X}\rightarrow+\infty}\langle\Phi^{\prime }(u),u\rangle/\|u\|_{X}=+\infty\), that is, \(\Phi^{\prime}\) is coercive.

For any \(u, v\in X\), in view of (A2) and (2.2), we have

$$\begin{aligned} \bigl\langle \Phi^{\prime}(u)-\Phi^{\prime}(v),u-v \bigr\rangle =& \|u-v\| _{X}^{2}+\sum_{j=1}^{p} \bigl[I_{j} \bigl(u(t_{j}) \bigr)-I_{j} \bigl(v(t_{j}) \bigr) \bigr] \bigl(u(t_{j})-v(t_{j}) \bigr) \\ &{}+ \bigl(h \bigl(u(0) \bigr)-h \bigl(v(0) \bigr) \bigr) \bigl(u(0)-v(0) \bigr) \\ \geq& \Biggl[1-\beta^{2} \Biggl(L+\sum_{j=1}^{p}L_{j} \Biggr) \Biggr]\|u-v\|_{X}^{2}. \end{aligned}$$

Since \(L+\sum_{j=1}^{p}L_{j}<\frac{1}{\beta^{2}}\), so \(\Phi^{\prime}\) is uniformly monotone. By [44], Theorem 26.A(d), we see that \(\Phi^{\prime}\) admits a continuous inverse on \(X^{\ast}\). □

Lemma 2.3

Assume that \(\mathrm{(A_{1})}\) holds, then Φ is sequentially weakly lower semicontinuous, coercive and its derivative admits a continuous inverse on \(X^{\ast}\).

Proof

The proof is similar to the proof of Lemma 2.2, and we omit it here. □

Lemma 2.4

Suppose that (A4) is satisfied. If \(u_{n}\rightharpoonup u\) in E, then \(g(t,u_{n})\rightarrow g(t,u)\) in \(L^{1}[0,+\infty)\).

Proof

Assume that \(u_{n}\rightharpoonup u\). In view of (A4) and (2.2), we have

$$\begin{aligned} \bigl|g(t,u_{n})-g(t,u) \bigr| \leq& \bigl(a_{1}(t)|u_{n}|+a_{2}(t)|u_{n}|^{\alpha-1}+a_{3}(t) \bigr)+ \bigl(a_{1}(t)|u|+a_{2}(t)|u|^{\alpha-1}+a_{3}(t) \bigr) \\ \leq& \bigl(\|u_{n}\|_{C}+\|u\|_{C} \bigr)a_{1}(t)+ \bigl(\|u_{n}\|_{C}^{\alpha-1}+\|u \|_{C}^{\alpha -1} \bigr)a_{2}(t)+2a_{3}(t) \\ \leq&\beta \bigl(\|u_{n}\|_{X}+\|u\|_{X} \bigr)a_{1}(t)+ \beta^{\alpha-1} \bigl(\|u_{n}\| _{X}^{\alpha-1}+\|u \|_{X}^{\alpha-1} \bigr)a_{2}(t)+2a_{3}(t). \end{aligned}$$

Applying the Lebesgue dominated convergence theorem, the lemma is proved. □

Lemma 2.5

The functional Ψ is a sequentially weakly upper semicontinuous and its derivative is compact.

Proof

Let \(\{u_{n}\}\subset X\), \(u_{n}\rightharpoonup u\) in X, we see that \(\{u_{n}\}\) converges uniformly to u on \([0,M]\) for any \(M\in(0,+\infty)\). It follows from the reverse Fatou lemma that

$$\begin{aligned} \limsup_{n\rightarrow+\infty}\Psi(u_{n}) =&\limsup _{n\rightarrow +\infty}\lim_{M\rightarrow+\infty} \int_{0}^{M}G(t,u_{n})\,dt \\ \leq&\lim_{M\rightarrow+\infty} \int_{0}^{M}\limsup_{n\rightarrow +\infty}G(t,u_{n}) \,dt \\ =& \int_{0}^{+\infty}G(t,u)\,dt=\Psi(u). \end{aligned}$$

So Ψ is sequentially weakly upper semicontinuous.

Next we will show that \(\Psi^{\prime}\) is compact. Let \(\{u_{n}\} \subset X\), \(u_{n}\rightharpoonup u\) in X. By Lemma 2.4, we get

$$\begin{aligned} \bigl\| \Psi^{\prime}(u_{n})-\Psi^{\prime}(u) \bigr\| _{X^{\ast}} =&\sup_{\|v\| _{X}=1} \bigl\| \bigl(\Psi^{\prime}(u_{n})- \Psi^{\prime}(u) \bigr)v \bigr\| \\ =&\sup_{\|v\|_{X}=1} \biggl\vert \int_{0}^{+\infty } \bigl(g(t,u_{n})-g(t,u) \bigr)v\,dt \biggr\vert \\ \leq&\|v\|_{C}\sup_{\|v\|_{X}=1} \int_{0}^{+\infty } \bigl|g(t,u_{n})-g(t,u) \bigr|\,dt \\ \leq&\beta \int_{0}^{+\infty} \bigl|g(t,u_{n})-g(t,u) \bigr|\,dt \rightarrow0 \end{aligned}$$

as \(k\rightarrow\infty\), for any \(u\in X\). Thus, \(\Psi^{\prime}\) is strongly continuous on X, which implies that \(\Psi^{\prime}\) is a compact operator by [44], Proposition 26.2. □

Proof of Theorems 1.1-1.3

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. □

Examples

In order to illustrate our results, we give two examples.

Example 4.1

Consider the following problem:

$$ \begin{aligned} &{-}u^{\prime\prime}(t)+u(t)=\lambda g \bigl(t,u(t) \bigr), \quad \mbox{a.e. } t\in[0,+\infty), \\ &\Delta u^{\prime}(t_{j})=I_{j} \bigl(u(t_{j}) \bigr), \quad j=1, \\ &u^{\prime}\bigl(0^{+} \bigr)=h \bigl(u(0) \bigr), \qquad u^{\prime}(+\infty)=0, \end{aligned} $$
(4.1)

where \(h(u)=u\), \(I_{j}(u)=u\).

Compared to problem (1.1), \(c=1\). It is clear that (A1) is satisfied. β is defined in (2.2). When β lies in different intervals, we can choose different g satisfies the conditions. So we only consider one case. If \(\beta <\frac{\sqrt{10}}{12}\), we take

$$g(t,u)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{\beta}e^{-t}, & u\leq\beta, \\ e^{-t} (\frac{u}{100}+600u^{\frac{1}{2}}-599\sqrt{\beta}-\frac {\beta}{100} ), & u>\beta. \end{array}\displaystyle \right . $$

Then

$$G(t,u)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{\beta}e^{-t}u, & u\leq\beta, \\ e^{-t} [\frac{u^{2}}{200}+400u^{\frac{3}{2}}- (599\sqrt {\beta}+\frac{\beta}{100} )u+200\beta^{\frac{3}{2}}+\frac {\beta^{2}}{200} ], & u>\beta. \end{array}\displaystyle \right . $$

Take \(t_{1}=\ln\sqrt{2}\), \(a_{1}(t)=\frac{e^{-t}}{200}\), \(\alpha=\frac{3}{2}\), \(a_{2}(t)=400e^{-t}\), \(a_{3}=\frac{1}{\sqrt{\beta}}\), \(b_{3}(t)=\frac{e^{-t}}{100}\), \(b_{4}(t)=600e^{-t}\), \(b_{5}(t)=\sqrt{\beta}e^{-t}\), and choose constants \(d,q>0\) and m satisfying \(6\beta^{2}q< d<\min\{\beta, \frac{\sqrt{10}}{2}\beta q\}\) and \(\frac{5}{2}m^{2}=\frac{d^{2}}{\beta^{2}}\). A simple calculation shows that (A3), (A4), and (A5) are satisfied. Applying Theorem 1.1, then, for each \(\lambda\in\, ]\frac{1}{\alpha_{2}},\frac{1}{\alpha_{1}}[\), problem (4.1) has at least three classical solutions.

Example 4.2

Consider the following problem:

$$ \begin{aligned} &{-}u^{\prime\prime}(t)+u(t)=\lambda g \bigl(t,u(t)\bigr), \quad \mbox{a.e. } t\in[0,+\infty), \\ &\Delta u^{\prime}(t_{j})=I_{j}\bigl(u(t_{j}) \bigr), \quad j=1, \\ &u^{\prime}\bigl(0^{+}\bigr)=h\bigl(u(0)\bigr), \qquad u^{\prime}(+ \infty)=0, \end{aligned} $$
(4.2)

where \(\lambda>0\), \(I_{j}(u)=-u^{\frac{3}{5}}\), \(h(u)=-u^{\frac{3}{5}}\), and \(g(t,u)= (\frac{1}{(1+t^{2})^{2}}-\frac{1}{(1+t)^{2}} )u^{\frac{1}{3}}\).

Compared to problem (1.1), \(c=1\). By simple calculations, all conditions in Theorem 1.3 are satisfied. Applying Theorem 1.3, then (4.2) has infinitely many classical solutions.

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Acknowledgements

The first author was supported by the Doctor Priming Fund Project of University of South China (2014XQD13) and the Tianyuan Fund for Mathematics of NSFC (No. 11526111). The third author was supported by the Scientific Research Fund of Hunan Provincial Education Department (No. 14A098).

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Correspondence to Zhimin He.

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MSC

  • 34B08
  • 34B15
  • 34B37
  • 58E30

Keywords

  • impulsive differential equation
  • variational methods
  • critical points
  • half-line