The existence of solutions for Sturm–Liouville differential equation with random impulses and boundary value problems

Abstract

In this article, we consider the existence of solutions to the Sturm–Liouville differential equation with random impulses and boundary value problems. We first study the Green function of the Sturm–Liouville differential equation with random impulses. Then, we get the equivalent integral equation of the random impulsive differential equation. Based on this integral equation, we use Dhage’s fixed point theorem to prove the existence of solutions to the equation, and the theorem is extended to the general second order nonlinear random impulsive differential equations. Then we use the upper and lower solution method to give a monotonic iterative sequence of the generalized random impulsive Sturm–Liouville differential equations and prove that it is convergent. Finally, we give two concrete examples to verify the correctness of the results.

1 Introduction

Impulsive dynamical systems are an emerging field drawing attention from both theoretical and applied disciplines. They are often typically described by ordinary differential equations with instantaneous state jumps [24, 29]. And the impulsive differential equations serve as basic models to study the dynamics of processes that are subject to sudden changes in their states. Since many evolution processes, optimal control models in economics, mechanics, electricity, several fields in engineering stimulated neural networks, frequency modulated systems, and some motions of missiles or aircrafts are characterized by the impulsive dynamical behavior, the study of impulsive systems, especially the impulsive differential, is of great importance, for details, see [12, 17, 24, 27, 34]. But real systems are often subject to not only impulse effect but also noise perturbations. Taking into account the stochastic effects, the models are better described as random impulsive differential equations (RIDEs) rather than impulsive differential equations or stochastic differential equations. Hence the study of RIDEs has received some attention [2, 30]. Recently, a large number of important results about the impulsive differential equation have been reported in [3, 4, 6, 13–15, 18, 23, 24, 28–31, 35, 36, 41]. For example, in [6], Gowrisankar et al. investigated the existence and stability of mild solutions of the first order semilinear differential equation with random impulse (1.1) using the contraction principle.

$$ \textstyle\begin{cases} x'(t)=Ax(t)+f(t,x_{t}), \quad t\neq \xi _{k}, t\geq t_{0}, \\ x(\xi _{k})=b_{k}(\tau _{k})x(\xi _{k}^{-}), \quad k=1,2,\ldots, \\ x_{t_{0}}=\varphi , \end{cases} $$

(1.1)

where A is the infinitesimal generator of a strongly continuous semigroup of bounded linear operators $S(t)$ with domain $D(A)\subseteq X$. $x_{t}(s)=x(t+s)$ and $\xi _{k}$ is the random pulse time point. In addition, many scholars have also studied the properties of random impulsive differential equations. Radhakrishnan et al. [25] studied the existence of solutions for quasilinear random impulsive neutral differential evolution equation by using the analytic semigroup theory and Schauder fixed point theorem. Niu et al. [23] studied the existence and Hyers–Ulam stability of the random impulsive differential equation with the initial condition. Zhang et al. [40] studied the existence and exponential stability of random impulsive fractional differential equations using the Leray–Schauder fixed point theorem.

Moreover, Sturm–Liouville differential equations play an important role in the study of differential equations. For any second order homogeneous linear differential equation

$$ P(x)y''(x)+Q(x)y'(x)+R(x)y(x)=0, $$

we can multiply both sides of the equation by the integral factor $\mu (x)=\frac{1}{P(x)}e^{\int \frac{Q(x)}{P(x)}\,\mathrm{d}x}$, so that the equation becomes a Sturm–Liouville differential equation

$$ \bigl(\mu (x)P(x)y'(x) \bigr)'+\mu (x)R(x)y(x)=0. $$

Therefore, the study of Sturm–Liouville type differential equations is of great significance, and some scholars have conducted in-depth research [22, 37, 38].

Besides, in recent years, the boundary value problems of different order differential equations have emerged as an important area of research, since these problems have applications in various disciplines of science and engineering such as control theory, signal and image processing, polymer rheology, regular variation in thermodynamics, biophysics, aerodynamics, and damping [5, 36]. Many researchers studied the existence and stability theory for differential equations with a variety of boundary conditions, for instance, see the papers [1, 9–11, 16, 19–21, 26, 33, 42]. For example, Hua, Cong, and Cheng [8] studied equation (1.2) in 2012, which is the existence and uniqueness of solutions for the periodic-integrable boundary value problem of second order differential equations

$$ \textstyle\begin{cases} (p(t)x'(t))'+f(t,x(t))=0, \\ x(0)=x(T), \\ \int _{0}^{T}x(s)\,ds=0, \end{cases} $$

(1.2)

where $p(t)\in C(R,R)$ is a given T-periodic function in $t\in R$, and $p(t)>0$. $f\in C(R\times R,R)$ is T-periodic in t.

Finally, we found that the current main research results are focused on ordinary impulsive differential equations, and few scholars have studied random impulsive differential equations. Therefore, based on the importance of random impulsive differential equations, we have carried out research on them. Besides, we found that many researchers investigated the normal impulsive differential equations and the boundary value problem of the equations. But there are fewer people who studied the boundary value problem of Sturm–Liouville type differential equations with random impulses and the upper and lower solutions of this kind of equation. In this paper, we discuss the Sturm–Liouville type differential equations with random impulses and boundary value problems, and we derive the Green function (the researchers gave the Green function of the normal impulsive differential equations [4, 7]) of the equations with the random impulses which has never been studied in the past. At the same time, we use the upper and lower solution method to construct the monotone iterative sequence converging to the maximum and minimum solutions of the equation and prove their convergence.

The rest of the paper is organized as follows: in Sect. 2, we introduce some notations and necessary preliminaries to give an idea of some important definitions and lemmas. And the Green function of the random impulsive differential equations is derived. In Sect. 3, we use Dhage’s fixed point theorem to study the existence of the solutions of equation (2.1), and then the existence of solutions of general second order nonlinear random impulsive differential equations with boundary value problems is given. In Sect. 4, we use the upper and lower solution method to give the monotonic iterative convergent sequence of the generalized Sturm–Liouville differential equation with random impulses. Finally, two practical examples are given in Sect. 5 to verify the correctness of the theorem.

2 Preliminaries

In this article, we investigate the solution of the following equation:

$$ \textstyle\begin{cases} Lx=f(t,x(t)), \quad t\in J, t\neq \xi _{k}, k=1,2,3,\ldots, \\ x(\xi _{k}^{+})=b_{k}(\tau _{k})x(\xi _{k}^{-}), \quad k=1,2,3,\ldots, \\ a_{11}x(0)-a_{12}x'(0)=a_{21}x(1)+a_{22}x'(1)=0, \end{cases} $$

(2.1)

where L is the Sturm–Liouville operator defined as $Lu(t)=-(p(t)u'(t))'+q(t)u(t)$. Let X be a Banach space and Ω be a sample space. Assume that $\tau _{k}$ is a random variable defined from Ω to $D_{k}:=(0,d_{k})$ for $k=1,2,\ldots$ , where $0< d_{k}<\infty$. Furthermore, assume that $\tau _{i}$ and $\tau _{j}$ are independent from each other when $i\neq j$ for $i, j=1,2,\ldots$ $u(t)$ is a stochastic process taking values in X. For the sake of simplicity, we denote $J=[0,1]$. ${C}={C}(J,R)$ is the set of all the stochastic processes mapping J into R. $f:[J,{C}]\rightarrow R$ is a continuous function. $\xi _{k}=\xi _{k-1}+\tau _{k}$ for $k=1,2,\ldots$ , and $\xi _{0}=0$. Obviously, $0=\xi _{0}<\xi _{1}<\xi _{2}<\cdots<\xi _{k}<\cdots$ , i.e., $\xi _{k}$ forms a strictly increasing sequence. $b_{k}:D_{k}\rightarrow R$ for each $k=1,2,\ldots$ The convergence is under the meaning of the orbit, $u(\xi _{k}^{-})=\lim_{t\uparrow \xi _{k}}u(t)$. $p(t)$ and $q(t)$ are positive continuous functions mapping $J=[0,1]$ into $R^{+}=[0,+\infty ]$. $a_{ij}$ are positive constants for $i,j=1,2$.

Denote by $\{W_{t},t\in [0,1]\}$ the simple counting process generated by $\{\xi _{k}\}_{k\in N}$, it is to say that $\{W_{t}\geq n\}=\{\xi _{n}\leq t\}$ and denote by $\mathcal{F}$ the σ-algebra generated by $\{W_{t},t\in J\}$, then $\{\Omega ,\mathcal{F},P\}$ is a probability space. Define by $L_{p}=L_{p}(\Omega ,\mathcal{F},R)$ the Hilbert space of all $\mathcal{F}\textit{-measurable}$, pth integrable random variables with values in R.

$\Vert \cdot \Vert$ is any norm of R, and the expectation of the random variable x is defined as $\mathrm{E}(x)=\int _{\Omega }x\,\mathrm{d}P<\infty$. Then we introduce the space $\mathit{PC}=\mathit{PC}(J,L_{p}):=\{u(t):u(t)$ is strongly measurable, pth integrable random process from J into $L_{p}$, and $u(t)$ is continuously differentiable when $t\in J\setminus \{\xi _{1},\xi _{2},\ldots\}$ and left continuous when $t\in J\}$. It is easy to see that PC is a Banach space with the norm

$$ \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}= \Bigl(\sup_{t\in J} \mathrm{E} \bigl\Vert u(t) \bigr\Vert ^{2} \Bigr)^{\frac{1}{2}}. $$

We use the following notations: $P_{cl}(R)=\{Y\subseteq R$: Y is a closed set}, $P_{bd}(R)=\{Y\subseteq R$: Y is a bounded set}, $P_{cp}(R)=\{Y\subseteq R$: Y is a compact set}, $P_{cv}(R)= \{Y\subseteq R$: Y is a convex set}.

Definition 2.1

• The operator A is called upper semi-continuous (u.s.c.) on R if, for each open set V of R containing $A(x_{0})$, there exists an open neighborhood N of $x_{0}$ such that $A(N)\subseteq V$.

• A is closed graph if there exists a sequence $x_{n}\rightarrow x^{\ast }$, $y_{n}\rightarrow y^{\ast }$, $y_{n}=Ax_{n}$, then we can imply $y^{\ast }=Ax^{\ast }$.

• A is called a completely continuous operator if A is a bounded linear operator and for every $x_{n}\rightharpoonup x^{\ast }$, we can get $Ax_{n}\rightarrow Ax^{\ast }$.

• A is called a compact operator if A is a linear operator and $\overline{A(V)}$ is compact for every $V\in P_{bd}(R)$.

Lemma 2.1

([32])

Suppose that $(\mathcal{X}, \Vert \cdot \Vert )$ is a normed linear space, then the set A is compact if and only if it is self-column compact.

Lemma 2.2

(Resonance theorem; [39])

Suppose that $\mathcal{X}$ is a Banach space, $\mathcal{Y}$ is a linear normed space ($B^{\ast }$ space), if W is a subset of all bounded linear operators from $\mathcal{X}$ to $\mathcal{Y}$ such that $\sup_{A\in W} \Vert Ax \Vert <\infty$, $\forall x\in \mathcal{X}$, then there exists a constant M such that $\Vert A \Vert \leq M$, $\forall A\in W$.

Theorem 2.1

If A is a compact operator, then A is a completely continuous operator.

Proof

Suppose $x_{n}\rightharpoonup x^{\ast }$, we use proof by contradiction. If $Ax_{n}$ does not converge to $y=Ax^{\ast }$, then there exist $\varepsilon _{0}>0$ and $\{n_{i}\}$ such that $\Vert Ax_{n_{i}}-Ax^{\ast } \Vert \geq \varepsilon _{0}$. From Lemma 2.2, we can know that $\{x_{n}\}$ is bounded. Combining that A is compact, we can get a subsequence from $\{x_{n_{i}}\}$, we write it as $\{x_{n_{i_{k}}}\}$ such that $Ax_{n_{i_{k}}}\rightarrow z$, but for every $y^{\ast }\in \mathcal{Y}^{\ast }$ ($\mathcal{Y}^{\ast }$ is the conjugate space of $\mathcal{Y}$), $\langle y^{\ast }, Ax_{n_{i_{k}}}-y\rangle =\langle A^{\ast }y^{\ast }, x_{n_{i_{k}}}-x^{ \ast }\rangle \rightarrow 0$, which implies $Ax_{n_{i_{k}}}\rightharpoonup y$, then $y=z$, which is a contradiction. The proof is completed. □

Definition 2.2

Suppose that T is a linear operator from $\mathcal{X}$ to $\mathcal{Y}$. $D(T)$ is the definitional domain of T, T is called closed if $x_{n}\in D(T)$, $x_{n}\rightarrow x$, and $Tx_{n}\rightarrow y$, then we can imply $x\in D(T)$ and $y=Tx$.

Remark 2.1

From the closed graph theorem, we can easily know that if A is a completely continuous operator, then A is u.s.c. if and only if A is a closed graph.

Theorem 2.2

(Dhage’s fixed point theorem)

Let X be a Banach space, $A: X\rightarrow P_{bd,cl,cv}(R)$, $B: X\rightarrow P_{cl,cv}(R)$ are two operators satisfying:

$(i)$:

A is contraction,

$(ii)$:

B is u.s.c. and completely continuous.

Then either

$(a)$:

the operator inclusion $x\in Ax+Bx$ has a solution for $\lambda =1$ or

$(b)$:

the set $U=\{u\in X: u\in \lambda Au+\lambda Bu, 0\leq \lambda \leq 1\}$ is unbounded.

Theorem 2.3

The solution of equation (2.1) is equivalent to the solution of the following integral equation:

$$ \begin{aligned}[b] u(t)={}& \int _{0}^{1}G(t,s) \bigl(-q(s)u(s)+f \bigl(s,u(s) \bigr) \bigr)\,\mathrm{d}s \\&{}+\sum_{k=1}^{ \infty } \bigl[W(t,k)u(\xi _{k}) \bigr]I_{A}(\xi _{1},\xi _{2},\ldots,\xi _{k},\ldots), \end{aligned} $$

(2.2)

where A is the set of all the sample orbits and $I_{A}(x)$ is the index function defined as

$$\begin{aligned}& I_{A}(t)= \textstyle\begin{cases} 1 &t\in A, \\ 0 &t\notin A, \end{cases}\displaystyle \end{aligned}$$

(2.3)

$$\begin{aligned}& G(t,s)= \textstyle\begin{cases} \frac{1}{ \vert Q \vert } (a_{12}+a_{11}p(0)\int _{0}^{s} \frac{1}{p(\tau )}\,\mathrm{d}\tau ) (a_{22}\frac{1}{p(1)}+a_{21} \int _{t}^{1}\frac{1}{p(\tau )}\,\mathrm{d}\tau ) &s< t, \\ \frac{1}{ \vert Q \vert } (a_{12}+a_{11}p(0)\int _{0}^{t} \frac{1}{p(\tau )}\,\mathrm{d}\tau ) (a_{22}\frac{1}{p(1)}+a_{21} \int _{s}^{1}\frac{1}{p(\tau )}\,\mathrm{d}\tau ) &s\geq t, \end{cases}\displaystyle \end{aligned}$$

(2.4)

$$\begin{aligned}& W(t,k)= \textstyle\begin{cases} \frac{1}{ \vert Q \vert }a_{11}p(0) (a_{21}\int _{t}^{1} \frac{1}{p(\tau )}\,\mathrm{d}\tau +a_{22}\frac{1}{p(1)} )(b_{k}( \tau _{k})-1) &0< \xi _{k}< t< 1, \\ -\frac{1}{ \vert Q \vert }a_{21}p(0) (a_{11}\int _{0}^{t} \frac{1}{p(\tau )}\,\mathrm{d}\tau +a_{12}\frac{1}{p(0)} )(b_{k}( \tau _{k})-1) &0< t\leq \xi _{k}< 1, \end{cases}\displaystyle \end{aligned}$$

(2.5)

and

$$ Q= \begin{bmatrix} a_{11} & -a_{12} \\ a_{21} & a_{21}\int _{0}^{1}\frac{p(0)}{p(s)}\,\mathrm{d}s+a_{22} \frac{p(0)}{p(1)} \end{bmatrix}. $$

(2.6)

Proof

Suppose that $\xi _{1}, \xi _{2},\ldots$ is a sample orbit. Thus, when $t\in (0,\xi _{1}]$, we have

$$\begin{aligned}& \bigl(p(t)u'(t) \bigr)'=q(t)u(t)-f \bigl(t,u(t) \bigr), \\& u'(t)=\frac{1}{p(t)}p(0)u'(0)+\frac{1}{p(t)} \int _{0}^{t}q(s)u(s) \,\mathrm{d}s- \frac{1}{p(t)} \int _{0}^{t}f \bigl(s,u(s) \bigr)\,\mathrm{d}s, \\& \begin{aligned} u(t)&=u(0)+p(0)u'(0) \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s+ \int _{0}^{t} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ &\quad {} - \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau . \end{aligned} \end{aligned}$$

When $t\in (\xi _{1},\xi _{2}]$, in the same way, we have

$$\begin{aligned} u'(t) =&\frac{1}{p(t)}p(\xi _{1})u'(\xi _{1})+\frac{1}{p(t)} \int _{ \xi _{1}}^{t}q(s)u(s)\,\mathrm{d}s - \frac{1}{p(t)} \int _{\xi _{1}}^{t}f \bigl(s,u(s) \bigr) \,\mathrm{d}s. \end{aligned}$$

The first derivative of this function has no impulses, so we can know

$$\begin{aligned} u(t) =&u \bigl(\xi _{1}^{+} \bigr)+p(\xi _{1})u'( \xi _{1}) \int _{\xi _{1}}^{t} \frac{1}{p(s)}\,\mathrm{d}s + \int _{\xi _{1}}^{t}\frac{1}{p(\tau )} \int _{\xi _{1}}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} - \int _{\xi _{1}}^{t}\frac{1}{p(\tau )} \int _{\xi _{1}}^{\tau }f \bigl(s,u(s) \bigr) \,\mathrm{d}s \, \mathrm{d}\tau \\ =&u \bigl(\xi _{1}^{+} \bigr)+ \biggl[p(0)u'(0)+ \int _{0}^{\xi _{1}}q(s)u(s) \,\mathrm{d}s - \int _{0}^{\xi _{1}}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \biggr] \int _{ \xi _{1}}^{t}\frac{1}{p(s)}\,\mathrm{d}s \\ & {} + \int _{\xi _{1}}^{t}\frac{1}{p(\tau )} \int _{\xi _{1}}^{\tau }q(s)u(s) \,\mathrm{d}s\,\mathrm{d}\tau - \int _{\xi _{1}}^{t}\frac{1}{p(\tau )} \int _{\xi _{1}}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau . \end{aligned}$$

Combining these two equations, we have

$$\begin{aligned} u(t) =&u(0)+u \bigl(\xi _{1}^{+} \bigr)-u(\xi _{1})+p(0)u'(0) \int _{0}^{t} \frac{1}{p(s)}\,\mathrm{d}s \\ & {} + \int _{\xi _{1}}^{t}\frac{1}{p(s)}\,\mathrm{d}s \int _{0}^{\xi _{1}}q(s)u(s) \,\mathrm{d}s + \int _{\xi _{1}}^{t}\frac{1}{p(\tau )} \int _{\xi _{1}}^{ \tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} + \int _{0}^{\xi _{1}}\frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s) \,\mathrm{d}s\,\mathrm{d}\tau - \biggl[ \int _{\xi _{1}}^{t}\frac{1}{p(s)} \,\mathrm{d}s \int _{0}^{\xi _{1}}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \\ & {} + \int _{\xi _{1}}^{t}\frac{1}{p(\tau )} \int _{\xi _{1}}^{\tau }f \bigl(s,u(s) \bigr) \,\mathrm{d}s \, \mathrm{d}\tau + \int _{0}^{\xi _{1}}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau \biggr]. \end{aligned}$$

Combining with the identity

$$\begin{aligned} \int _{\xi _{1}}^{t}\frac{1}{p(s)}\,\mathrm{d}s \int _{0}^{\xi _{1}}q(s)u(s) \,\mathrm{d}s= \int _{\xi _{1}}^{t}\frac{1}{p(\tau )} \int _{0}^{\xi _{1}}q(s)u(s) \,\mathrm{d}s\,\mathrm{d}\tau , \end{aligned}$$

we can get

$$\begin{aligned}& \begin{aligned} u(t)&=u(0)+p(0)u'(0) \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s + \int _{0}^{t} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ &\quad {} - \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +u \bigl(\xi _{1}^{+} \bigr)-u(\xi _{1}), \end{aligned} \\& u'(t)=\frac{1}{p(t)}p(0)u'(0)+\frac{1}{p(t)} \int _{0}^{t}q(s)u(s) \,\mathrm{d}s- \frac{1}{p(t)} \int _{0}^{t}f \bigl(s,u(s) \bigr)\,\mathrm{d}s. \end{aligned}$$

Suppose when $t\in (\xi _{k},\xi _{k+1}]$

$$\begin{aligned} u(t) =&u(0)+p(0)u'(0) \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s + \int _{0}^{t} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} - \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{n=1}^{k} \bigl[u \bigl( \xi _{n}^{+} \bigr)-u(\xi _{n}) \bigr]. \end{aligned}$$

Then, using the same way, we can get when $t\in (\xi _{k+1},\xi _{k+2}]$,

$$\begin{aligned} u(t) =&u(0)+p(0)u'(0) \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s + \int _{0}^{t} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} - \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{n=1}^{k+1} \bigl[u \bigl( \xi _{n}^{+} \bigr)-u(\xi _{n}) \bigr]. \end{aligned}$$

So, using the mathematical induction, we can get

$$\begin{aligned} u(t) =&\sum_{n=0}^{\infty } \biggl\{ u(0)+p(0)u'(0) \int _{0}^{t} \frac{1}{p(s)}\,\mathrm{d}s + \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{ \tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} - \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{0< \xi _{k}< t} \bigl[u \bigl(\xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \biggr\} I_{(\xi _{n},\xi _{n+1}]}(t) \end{aligned}$$

and

$$\begin{aligned} u'(t) =&\frac{1}{p(t)}p(0)u'(0)+\frac{1}{p(t)} \int _{0}^{t}q(s)u(s) \,\mathrm{d}s- \frac{1}{p(t)} \int _{0}^{t}f \bigl(s,u(s) \bigr)\,\mathrm{d}s. \end{aligned}$$

It is easy to see that

$$\begin{aligned}& \begin{aligned} u(1)&=u(0)+p(0)u'(0) \int _{0}^{1}\frac{1}{p(s)}\,\mathrm{d}s + \int _{0}^{1} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ &\quad {} - \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{k=1}^{\infty } \bigl[u \bigl( \xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr], \end{aligned} \\& u'(1)=\frac{1}{p(1)}p(0)u'(0)+\frac{1}{p(1)} \int _{0}^{1}q(s)u(s) \,\mathrm{d}s- \frac{1}{p(1)} \int _{0}^{1}f \bigl(s,u(s) \bigr)\,\mathrm{d}s. \end{aligned}$$

Plug in the boundary value conditions, we can get

$$\begin{aligned}& a_{11}u(0)-a_{12}u'(0)=0, \\& a_{21}u(0)+ \biggl[a_{21}p(0) \int _{0}^{1}\frac{1}{p(s)} \, \mathrm{d}s+a_{22} \frac{p(0)}{p(1)} \biggr]u'(0) \\& \quad = - \Biggl[a_{21} \Biggl( \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s) \,\mathrm{d}s\,\mathrm{d}\tau \\& \quad\quad {} - \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{k=1}^{\infty } \bigl[u \bigl( \xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \Biggr) \\& \quad\quad {} +a_{22} \biggl(\frac{1}{p(1)} \int _{0}^{1}q(s)u(s)\,\mathrm{d}s- \frac{1}{p(1)} \int _{0}^{1}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \biggr) \Biggr]. \end{aligned}$$

Then we define the matrix Q as

$$ Q= \begin{bmatrix} a_{11} & -a_{12} \\ a_{21} & a_{21}\int _{0}^{1}\frac{p(0)}{p(s)}\,\mathrm{d}s+a_{22} \frac{p(0)}{p(1)} \end{bmatrix}. $$

So, the solution of the above equation group is

$$ u(0)=\frac{1}{ \vert Q \vert } \begin{vmatrix} -a_{12} & 0 \\ a_{21} \int _{0}^{1}\frac{p(0)}{p(s)} \, \mathrm{d}s+a_{22} \frac{p(0)}{p(1)} & \zeta \end{vmatrix},\quad\quad u'(0)=\frac{1}{ \vert Q \vert } \begin{vmatrix} 0 & a_{11} \\ \zeta & a_{21} \end{vmatrix}, $$

where ζ is defined as

$$\begin{aligned} \zeta =& \Biggl[a_{21} \Biggl( \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{ \tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} - \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{k=1}^{\infty } \bigl[u \bigl( \xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \Biggr) \\ & {} +a_{22} \biggl(\frac{1}{p(1)} \int _{0}^{1}q(s)u(s)\,\mathrm{d}s- \frac{1}{p(1)} \int _{0}^{1}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \biggr) \Biggr]. \end{aligned}$$

So, we can write $u(t)$ as

$$\begin{aligned} u(t) =&\frac{1}{ \vert Q \vert } \Biggl\{ -a_{12}a_{21} \Biggl( \int _{0}^{1} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} - \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{k=1}^{\infty } \bigl[u \bigl( \xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \Biggr) \\ & {} -a_{12}a_{22} \biggl(\frac{1}{p(1)} \int _{0}^{1}q(s)u(s)\,\mathrm{d}s- \frac{1}{p(1)} \int _{0}^{1}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \biggr) \\ & {} - \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}sp(0) \Biggl[a_{11}a_{21} \Biggl( \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s) \,\mathrm{d}s\,\mathrm{d}\tau \\ & {} - \int _{0}^{1}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau +\sum_{k=1}^{\infty } \bigl[u \bigl( \xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \Biggr) \\ & {} +a_{11}a_{22} \biggl(\frac{1}{p(1)} \int _{0}^{1}q(s)u(s)\,\mathrm{d}s- \frac{1}{p(1)} \int _{0}^{1}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \biggr) \Biggr] \Biggr\} \\ & {} + \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s \,\mathrm{d}\tau - \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr) \,\mathrm{d}s \, \mathrm{d}\tau \\ & {} +\sum_{0< \xi _{k}< t} \bigl[u \bigl(\xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \\ =&\frac{1}{ \vert Q \vert } \biggl[-a_{12}a_{21} \int _{0}^{1} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} -a_{12}a_{22}\frac{1}{p(1)} \int _{0}^{1}q(s)u(s)\,\mathrm{d}s \\ & {} -a_{11}a_{21}p(0) \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s \int _{0}^{1} \frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s)\,\mathrm{d}s\,\mathrm{d}\tau \\ & {} -\frac{p(0)}{p(1)}a_{11}a_{22} \int _{0}^{1}q(s)u(s)\,\mathrm{d}s \\ & {} + \vert Q \vert \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }q(s)u(s) \,\mathrm{d}s\,\mathrm{d}\tau \biggr] \\ & {} +\frac{1}{ \vert Q \vert } \biggl[a_{12}a_{21} \int _{0}^{1} \frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau \\ & {} +a_{12}a_{22}\frac{1}{p(1)} \int _{0}^{1}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \\ & {} +a_{11}a_{21}p(0) \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s \int _{0}^{1} \frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr)\,\mathrm{d}s \, \mathrm{d}\tau \\ & {} +a_{11}a_{22}\frac{p(0)}{p(1)} \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s \int _{0}^{1}f \bigl(s,u(s) \bigr)\,\mathrm{d}s \\ & {} - \vert Q \vert \int _{0}^{t}\frac{1}{p(\tau )} \int _{0}^{\tau }f \bigl(s,u(s) \bigr) \,\mathrm{d}s \, \mathrm{d}\tau \biggr] \\ & {} +\frac{1}{ \vert Q \vert } \Biggl[-a_{12}a_{21}\sum _{k=1}^{\infty } \bigl[u \bigl(\xi _{k}^{+} \bigr)-u( \xi _{k}) \bigr] \\ & {} -a_{11}a_{21}p(0) \int _{0}^{t}\frac{1}{p(s)}\,\mathrm{d}s\sum _{k=1}^{ \infty } \bigl[u \bigl(\xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \\ & {} + \vert Q \vert \sum_{0< \xi _{k}< t} \bigl[u \bigl( \xi _{k}^{+} \bigr)-u(\xi _{k}) \bigr] \Biggr] \\ =&{-} \int _{0}^{1}G(t,s)q(s)u(s)\,\mathrm{d}s+ \int _{0}^{1}G(t,s)f \bigl(s,u(s) \bigr) \, \mathrm{d}s+ \sum_{k=1}^{\infty }W(t,k)u(\xi _{k}). \end{aligned}$$

And we have completed the proof of Theorem 2.3. □

Remark 2.2

We can see that $G(t,s)$ is a positive continuous function of t and s.

Remark 2.3

Using the same way, we can prove that the solution of the equation

$$ \textstyle\begin{cases} -x''(t)=f(t,x(t),x'(t)), \quad t\in J\setminus \{\xi _{1},\xi _{2},\ldots \}, \\ x(\xi _{k}^{+})=b_{k}(\tau _{k})x(\xi _{k}), \quad k=1,2,3,\ldots, \\ a_{11}x(0)-a_{12}x'(0)=a_{21}x(1)+a_{22}x'(1)=0, \end{cases} $$

(2.7)

is equivalent to the solution of the following integral equation:

$$ x(t)= \int _{0}^{1}G(t,s)f \bigl(s,x(s),x'(s) \bigr)\,\mathrm{d}s+\sum_{k=1}^{ \infty } \bigl[W(t,k)x(\xi _{k}) \bigr]I_{A}(\xi _{1},\xi _{2},\ldots,\xi _{k},\ldots), $$

(2.8)

where

$$\begin{aligned}& G(t,s)= \textstyle\begin{cases} \frac{1}{ \vert Q \vert }(a_{11}s+a_{12})(a_{21}(1-t)+a_{22}) &s< t, \\ \frac{1}{ \vert Q \vert }(a_{11}t+a_{12})(a_{21}(1-s)+a_{22}) &s\geq t, \end{cases}\displaystyle \end{aligned}$$

(2.9)

$$\begin{aligned}& W(t,k)= \textstyle\begin{cases} \frac{1}{ \vert Q \vert }a_{11}(a_{21}(1-t)+a_{22})(b_{k}( \tau _{k})-1) &0< \xi _{k}< t< 1, \\ -\frac{1}{ \vert Q \vert }a_{21}(a_{11}t+a_{12})(b_{k}(\tau _{k})-1) &0< t\leq \xi _{k}< 1, \end{cases}\displaystyle \end{aligned}$$

(2.10)

and

$$ \vert Q \vert = \begin{vmatrix} a_{11} & -a_{12} \\ a_{21} & a_{21}+a_{22} \end{vmatrix}=a_{11}a_{21}+a_{11}a_{22}+a_{12}a_{21}. $$

(2.11)

Definition 2.3

Define the operator $\Lambda : \mathit{PC}(J,L_{2})\rightarrow \mathit{PC}(J,L_{2})$ such that

$$ \Lambda u= \int _{0}^{1}G(t,s) \bigl[f \bigl(s,u(s) \bigr)-q(s)u(s) \bigr]\,\mathrm{d}s +\sum_{k=1}^{ \infty } \bigl[W(t,k)u(\xi _{k}) \bigr]I_{A}(\xi _{1},\xi _{2},\ldots,\xi _{k},\ldots). $$

(2.12)

Theorem 2.4

(Operator decomposition theorem)

Suppose that $\mathcal{X}$ and $\mathcal{Y}$ are two Banach spaces, T is an operator from $\mathcal{X}$ to $\mathcal{Y}$. $u(t):\Omega \rightarrow \mathcal{X}$ is a functional, $\{\Omega _{i}\}$ is a division of Ω, it is to say that $\bigcup_{i\in I}\Omega _{i}=\Omega$, and $\Omega _{i}$ and $\Omega _{j}$ have no element in common for every $i, j\in I$, $i\neq j$, where I is an arbitrary set. Then $u(t)$ is the fixed point of T if and only if, for every $i\in I$, $u_{i}(t)$ is the fixed point of T, where the definition domain of $u_{i}(t)$ is $\Omega _{i}$ and $u_{i}(t)\equiv u(t)$ when $t\in \Omega _{i}$.

3 The existence of solutions

In this section, we list the following basic assumptions of this paper and prove our main results.

$(H_{1})$::

There exists a constant M such that, for each $u_{1}, u_{2}\in \mathit{PC}(J,L_{2})$,

$$ \bigl\Vert f \bigl(t,u_{1}(t) \bigr)-f \bigl(t, u_{2}(t) \bigr) \bigr\Vert \leq M \bigl\Vert u_{1}(t)-u_{2}(t) \bigr\Vert . $$

$(H_{2})$::

There exists a constant $\eta _{2}$ such that

$$ \sup_{k,\omega } \frac{ \Vert b_{k}(\tau _{k})-1 \Vert }{E \Vert b_{k}(\tau _{k})-1 \Vert } \leq \eta _{2}< \infty . $$

$(H_{3})$::

$\{E \Vert b_{k}(\tau _{k})-1 \Vert \}$ is a convergent series, and

$$ \sum_{k=1}^{\infty }\sup_{\omega } \bigl\Vert b_{k}(\tau _{k})-1 \bigr\Vert \leq \eta _{3}< \infty , $$

where $\eta _{3}$ is a constant.

$(H_{4})$::

There exists a constant $M_{f}$ such that, for each $u\in \mathit{PC}(J,L_{2})$ and $t\in J$,

$$ \bigl\Vert f \bigl(t,u(t) \bigr) \bigr\Vert _{\mathit{PC}}\leq M_{f} \bigl( \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}+1 \bigr). $$

$(H_{5})$::

Suppose

$$\begin{aligned}& \eta _{0}=\frac{1}{ \vert Q \vert } \biggl(a_{12}+a_{11}p(0) \int _{0}^{1} \frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr) \biggl(a_{22}\frac{1}{p(1)}+a_{21} \int _{0}^{1}\frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr), \\& \begin{aligned} \mathcal{C}_{1}={}&\max \biggl\{ \frac{1}{ \vert Q \vert }a_{11}p(0) \biggl(a_{21} \int _{0}^{1}\frac{1}{p(\tau )}\,\mathrm{d}\tau +a_{22}\frac{1}{p(1)} \biggr), \\ &{}\frac{1}{ \vert Q \vert }a_{21}p(0) \biggl(a_{11} \int _{0}^{1} \frac{1}{p(\tau )}\,\mathrm{d}\tau +a_{12}\frac{1}{p(0)} \biggr) \biggr\} , \end{aligned} \\& M_{q}=\sup_{t\in J} \bigl\Vert q(t) \bigr\Vert , \end{aligned}$$

then they should satisfy the following equalities:

$$ 4\eta _{0}^{2}M_{q}^{2}+4\eta _{0}^{2}M_{f}^{2}+2 \mathcal{C}_{1}^{2} \eta _{2}\eta _{3}^{2}< 1 $$

and

$$ \eta _{0}M< 1. $$

Theorem 3.1

If conditions $(H_{1})\sim (H_{5})$ are met, equation (2.1) has a solution $u(t)$ in $\mathit{PC}(J,L_{2})$ which satisfies

$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}\leq \frac{4\eta _{0}^{2}M_{f}^{2} +2\eta _{0}M_{f}\sqrt{1-4\eta _{0}^{2}M_{q}^{2}-2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}}}{1-(4\eta _{0}^{2}M_{q}^{2}+4\eta _{0}^{2}M_{f}^{2}+2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2})}. \end{aligned}$$

Proof

First of all, we can decompose the operator Λ into A and B, that is to say $\Lambda u(t)=Au(t)+Bu(t)$ for every $u(t)\in \mathit{PC}(J,L_{2})$. The operators A and B are defined as

$$\begin{aligned}& Au= \int _{0}^{1}G(t,s)f \bigl(s,u(s) \bigr) \, \mathrm{d}s, \\& Bu=- \int _{0}^{1}G(t,s)q(s)u(s)\,\mathrm{d}s+\sum _{k=1}^{\infty } \bigl[W(t,k)x( \xi _{k}) \bigr]I_{A}(\xi _{1},\xi _{2}, \ldots,\xi _{k},\ldots). \end{aligned}$$

It is easy to see that B is a linear operator, and we can easily prove that the solution of equation (2.1) is equivalent to the fixed point of the operator $\Lambda =A+B$. Then we will prove Theorem 3.1 in six steps.

• Step (1):

A is a single-valued operator, so $Au\in P_{cl,cv}(R)$. Then we prove that, for every $u\in \mathcal{B}_{q}=\{u(t): \Vert u(t) \Vert _{\mathit{PC}}\leq q\}$, $\Vert Au(t) \Vert _{\mathit{PC}}\leq \eta _{1}$, where $\eta _{1}$ is a constant.

$$ G(t,s)\leq \frac{1}{ \vert Q \vert } \biggl(a_{12}+a_{11}p(0) \int _{0}^{1} \frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr) \biggl(a_{22}\frac{1}{p(1)}+a_{21} \int _{0}^{1}\frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr)=\eta _{0}. $$

Hence,

$$\begin{aligned} \mathrm{E} \Vert Au \Vert ^{2} \leq &\mathrm{E} \biggl( \int _{0}^{1} \bigl\Vert G(t,s)f \bigl(s,u(s) \bigr) \bigr\Vert \,\mathrm{d}s \biggr)^{2} \\ \leq &\mathrm{E} \biggl[ \int _{0}^{1} \bigl\Vert G(t,s) \bigr\Vert ^{2} \,\mathrm{d}s \int _{0}^{1} \bigl\Vert f \bigl(s,u(s) \bigr) \bigr\Vert ^{2}\,\mathrm{d}s \biggr] \\ \leq &\eta _{0}^{2}\mathrm{E} \biggl[ \int _{0}^{1} \bigl\Vert f \bigl(s,u(s) \bigr) \bigr\Vert ^{2}\,\mathrm{d}s \biggr] \\ \leq &\eta _{0}^{2} \int _{0}^{1} \bigl\Vert f \bigl(s,u(s) \bigr) \bigr\Vert _{\mathit{PC}}^{2} \,\mathrm{d}s \\ \leq &\eta _{0}^{2}M_{f}^{2}(q+1)^{2}= \eta _{1}^{2}. \end{aligned}$$

• Step (2):

We prove that A is a contraction.

$$\begin{aligned}& \begin{aligned} \Vert Au_{1}-Au_{2} \Vert ^{2}&= \biggl\Vert \int _{0}^{1}G(t,s) \bigl[f \bigl(s,u_{1}(s) \bigr)-f \bigl(s,u_{2}(s) \bigr) \bigr] \,\mathrm{d}s \biggr\Vert ^{2} \\ &\leq M^{2}\eta _{0}^{2} \int _{0}^{1} \bigl\Vert u_{1}(s)-u_{2}(s) \bigr\Vert ^{2}\,\mathrm{d}s, \end{aligned} \\& \begin{aligned} \mathrm{E} \Vert Au_{1}-Au_{2} \Vert ^{2}& \leq M^{2}\eta _{0}^{2} \int _{0}^{1} \int _{\Omega } \bigl\Vert u_{1}(s)-u_{2}(s) \bigr\Vert ^{2} \,\mathrm{d}\mu \,\mathrm{d}s \\ &\leq M^{2}\eta _{0}^{2}\sup _{t\in J}\mathrm{E} \bigl\Vert u_{1}(t)-u_{2}(t) \bigr\Vert ^{2}. \end{aligned} \end{aligned}$$

So, we get

$$\begin{aligned} \Vert Au_{1}-Au_{2} \Vert _{\mathit{PC}} \leq &M\eta _{0} \bigl\Vert u_{1}(t)-u_{2}(t) \bigr\Vert _{\mathit{PC}}, \end{aligned}$$

and we can easily know that A is a contraction.

• Step (3):

It is easy to see that B is a single-valued operator, so for each $u(t)\in \mathit{PC}(J,L_{2})$, $Bu\in P_{cl,cv}(R)$. Next, we prove, for every $u(t)\in \mathcal{B}_{q}$, that Bu is bounded.

$$\begin{aligned}& Bu(t)=- \int _{0}^{1}G(t,s)q(s)u(s)\,\mathrm{d}s+\sum _{k=1}^{\infty } \bigl[W(t,k)x(\xi _{k}) \bigr]I_{A}(\xi _{1},\xi _{2}, \ldots,\xi _{k},\ldots), \\& \begin{aligned} \Vert Bu \Vert ^{2}&\leq 2 \int _{0}^{1} \bigl\Vert G(t,s)q(s)u(s) \bigr\Vert ^{2}\,\mathrm{d}s+2 \Biggl\Vert \sum _{k=1}^{\infty }W(t,k)u(\xi _{k}) \Biggr\Vert ^{2} \\ &\leq 2\eta _{0}^{2} \int _{0}^{1} \bigl\Vert q(s)u(s) \bigr\Vert ^{2} \,\mathrm{d}s+2 \Biggl\Vert \sum_{k=1}^{\infty }W(t,k)u( \xi _{k}) \Biggr\Vert ^{2}. \end{aligned} \end{aligned}$$

And based on the second mean value theorem of integrals, we know, for each k, that there exists $\omega _{k}\in \Omega$ such that

$$\begin{aligned}& \mathrm{E} \Biggl\Vert \sum_{k=1}^{\infty }W(t,k)u( \xi _{k}) \Biggr\Vert ^{2} \\& \quad = \int _{\Omega } \Biggl\Vert \sum_{k=1}^{\infty }W(t,k) u(\xi _{k}) \Biggr\Vert ^{2} \,\mathrm{d}\mu \\& \quad = \int _{\Omega } \Biggl\Vert \sum_{k=1}^{\infty }W(t,k)u( \xi _{k}) \Biggr\Vert \,\mathrm{d}\mu \Biggl\Vert \sum _{k=1}^{\infty }W(t,k) (\omega _{k})u \bigl(\xi _{k}( \omega _{k}) \bigr) \Biggr\Vert \\& \quad \leq \mathcal{C}_{1}^{2} \int _{\Omega } \Biggl\Vert \sum_{k=1}^{\infty } \bigl(b_{k}( \tau _{k})-1 \bigr)u(\xi _{k}) \Biggr\Vert \,\mathrm{d}\mu \Biggl\Vert \sum_{k=1}^{ \infty } \bigl(b_{k}(\tau _{k})-1 \bigr)u \bigl(\xi _{k}( \omega _{k}) \bigr) \Biggr\Vert \\& \quad \leq \mathcal{C}_{1}^{2}\eta _{3} \Biggl[ \sum_{k=1}^{\infty } \frac{\sup_{\omega _{k}} \Vert b_{k}(\tau _{k}(\omega _{k}))-1 \Vert }{\int _{\Omega } \Vert b_{k}(\tau _{k})-1 \Vert \,\mathrm{d}\mu } \int _{\Omega } \bigl\Vert b_{k}( \tau _{k})-1 \bigr\Vert \,\mathrm{d}\mu \int _{\Omega } \bigl\Vert u(\xi _{k}) \bigr\Vert \, \mathrm{d} \mu \Biggr] \\& \quad\quad{}\times \max_{k} \bigl\Vert u \bigl(\xi _{k}(\omega _{k}) \bigr) \bigr\Vert \\& \quad \leq \mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}\max_{k} \int _{ \Omega } \bigl\Vert u(\xi _{k}) \bigr\Vert \, \mathrm{d}\mu \bigl\Vert u \bigl(\xi _{k}(\omega _{k}) \bigr) \bigr\Vert \\& \quad \leq \mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2} \max_{k} \int _{ \Omega } \bigl\Vert u(\xi _{k}) \bigr\Vert ^{2}\,\mathrm{d}\mu \\& \quad \leq \mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2} \bigl\Vert u(\xi _{k}) \bigr\Vert _{\mathit{PC}}^{2}. \end{aligned}$$

Based on the above equation, we can easily know that Bu is bounded for each $u(t)\in \mathcal{B}_{q}$.

• Step (4):

Define $B(\mathcal{B}_{q})=\{Bu(t):u(t)\in \mathcal{B}_{q}\}$, then we prove that $B(\mathcal{B}_{q})$ is equicontinuous. Based on Theorem 2.4, we only need to prove for each $(\xi _{k},\xi _{k+1}]$, $k=0,1,2, \ldots$ and every $\varepsilon >0$ that there exists $\delta >0$ such that, for any $t_{1}, t_{2}\in (\xi _{k},\xi _{k+1}]$, $\Vert t_{1}-t_{2} \Vert <\delta$, we have $\Vert Bu(t_{1})-Bu(t_{2}) \Vert _{\mathit{PC}}<\varepsilon$ for every $u(t)\in \mathcal{B}_{q}$.

$$\begin{aligned}& \bigl\Vert (Bu) (t_{1})-(Bu) (t_{2}) \bigr\Vert ^{2} \\& \quad = \Biggl\Vert \int _{0}^{1} \bigl[G(t_{1},s)-G(t_{2},s) \bigr] \bigl(-q(s)u(s) \bigr)\,\mathrm{d}s \\& \quad \quad {} + \sum_{n=1}^{\infty } \bigl[W(t_{1},n)-W(t_{2},n) \bigr]u(\xi _{n})I_{A}( \cdot ) \Biggr\Vert ^{2} \\& \quad \leq 2 \biggl( \int _{0}^{1} \bigl\Vert G(t_{1},s)-G(t_{2},s) \bigr\Vert \bigl\Vert q(s)u(s) \bigr\Vert \,\mathrm{d}s \biggr)^{2} \\& \quad \quad {} + 2 \Biggl\Vert \sum_{n=1}^{\infty } \bigl[W(t_{1},n)-W(t_{2},n) \bigr]u(\xi _{n}) \Biggr\Vert ^{2}, \end{aligned}$$

so we have

$$\begin{aligned}& \mathrm{E} \bigl\Vert (Bu) (t_{1})-(Bu) (t_{2}) \bigr\Vert ^{2} \\& \quad \leq 2M_{q}^{2}q^{2} \int _{0}^{1} \bigl\Vert G(t_{1},s)-G(t_{2},s) \bigr\Vert ^{2}\,\mathrm{d}s \\& \quad\quad {} + 2\sup_{n,\omega } \frac{ \Vert W(t_{1},n)-W(t_{2},n) \Vert }{\int _{\Omega } \Vert W(t_{1},n)-W(t_{2},n) \Vert \,\mathrm{d}\mu } \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}^{2} \\& \quad\quad {} \times \Biggl(\sum_{n=1}^{\infty } \mathrm{E} \bigl\Vert W(t_{1},n)-W(t_{2},n) \bigr\Vert \Biggr) \Biggl(\sum_{n=1}^{\infty }\sup _{\omega } \bigl\Vert W(t_{1},n)-W(t_{2},n) \bigr\Vert \Biggr), \end{aligned}$$

where

$$\begin{aligned}& \sum_{n=1}^{\infty } \bigl\Vert W(t_{1},k)-W(t_{2},k) \bigr\Vert \\& \quad \leq \Biggl\Vert \frac{1}{ \vert Q \vert }a_{11}a_{21}p(0) \int _{t_{1}}^{t_{2}} \frac{1}{p(\tau )}\,\mathrm{d}\tau \sum_{n=1}^{k} \bigl(b_{n}(\tau _{n})-1 \bigr) \Biggr\Vert \\& \quad\quad {} + \Biggl\Vert \frac{1}{ \vert Q \vert }a_{11}a_{21}p(0) \int _{t_{1}}^{t_{2}} \frac{1}{p(\tau )}\,\mathrm{d}\tau \sum_{n=k+1}^{\infty } \bigl(b_{n}(\tau _{n})-1 \bigr) \Biggr\Vert \\& \quad \leq \biggl\Vert \frac{1}{ \vert Q \vert }a_{11}a_{21}p(0) \int _{t_{1}}^{t_{2}} \frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr\Vert \sum_{n=1}^{\infty } \bigl\Vert b_{n}(\tau _{n})-1 \bigr\Vert , \end{aligned}$$

hence

$$\begin{aligned} \sum_{n=1}^{\infty }\mathrm{E} \bigl\Vert W(t_{1},n)-W(t_{2},n) \bigr\Vert \leq &\eta _{3} \biggl\Vert \frac{1}{ \vert Q \vert }a_{11}a_{21}p(0) \int _{t_{1}}^{t_{2}}\frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr\Vert . \end{aligned}$$

So, combining with the above equations, we have

$$\begin{aligned}& \mathrm{E} \bigl\Vert (Bu) (t_{1})-(Bu) (t_{2}) \bigr\Vert ^{2} \\& \quad \leq 2M_{q}^{2}q^{2} \int _{0}^{1} \bigl\Vert G(t_{1},s)-G(t_{2},s) \bigr\Vert ^{2}\,\mathrm{d}s \\& \quad \quad {} + 2\frac{1}{ \vert Q \vert ^{2}}a_{11}^{2}a_{21}^{2}p^{2}(0) \sup_{k, \omega } \frac{ \Vert b_{k}(\tau _{k})-1 \Vert }{\int _{\Omega } \Vert b_{k}(\tau _{k})-1 \Vert \,\mathrm{d}\mu } \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}^{2} \biggl( \int _{t_{1}}^{t_{2}} \frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr)^{2} \\& \quad \quad {} \times \Biggl(\sum_{k=1}^{\infty }E \bigl\Vert b_{k}(\tau _{k})-1 \bigr\Vert \,\mathrm{d} \mu \Biggr) \Biggl(\sum_{n=1}^{\infty }\sup _{\omega } \bigl\Vert b_{k}(\tau _{k})-1 \bigr\Vert \Biggr) \\& \quad \leq 2M_{q}^{2}q^{2} \int _{0}^{1} \bigl\Vert G(t_{1},s)-G(t_{2},s) \bigr\Vert ^{2}\,\mathrm{d}s \\& \quad \quad {} + 2\eta _{2}\eta _{3}^{2} \frac{1}{ \vert Q \vert ^{2}}a_{11}^{2}a_{21}^{2}p^{2}(0) \biggl( \int _{t_{1}}^{t_{2}}\frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr)^{2} \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}^{2}. \end{aligned}$$

It is easy to see that $\mathrm{E} \Vert (Bu)(t_{1})-(Bu)(t_{2}) \Vert ^{2} \rightarrow 0$ as $\vert t_{1}-t_{2} \vert \rightarrow 0$, so we have proved that the set is equicontinuous. From step (3) to step (4), combining with the Arzela–Ascoli theorem, we can easily know that $B(\mathcal{B}_{q})$ is sequentially compact and $\overline{B(\mathcal{B}_{q})}$ is self-listed. Using Lemma 2.1, we can know that $\overline{B(\mathcal{B}_{q})}$ is compact. Combining with Theorem 2.1, we have proved that B is completely continuous.

• Step (5):

We prove that B is u.s.c. Based on Remark 2.1, we only need to prove that for each $u_{n}(t)\rightarrow u^{\ast }(t)$ we have $(Bu_{n})(t)\rightarrow (Bu^{\ast })(t)$.

$$\begin{aligned}& \bigl\Vert (Bu_{n}) (t)- \bigl(Bu^{\ast } \bigr) (t) \bigr\Vert ^{2} \\& \quad = \Biggl\Vert \int _{0}^{1}G(t,s) \bigl[-q(s) \bigl(u_{n}(s)-u^{\ast }(s) \bigr) \bigr]\,\mathrm{d}s + \sum _{k=1}^{\infty }W(t,k) \bigl[u_{n}(\xi _{k})-u^{\ast }(\xi _{k}) \bigr]I_{A}( \cdot ) \Biggr\Vert ^{2} \\& \quad \leq 2\eta _{0}^{2}M_{q}^{2} \int _{0}^{1} \bigl\Vert u_{n}(s)-u^{\ast }(s) \bigr\Vert ^{2}\,\mathrm{d}s +2\mathcal{C}_{1}^{2} \Biggl\Vert \sum_{k=1}^{ \infty } \bigl(b_{k}(\tau _{k})-1 \bigr) \bigl[u_{n}(\xi _{k})-u^{\ast }(\xi _{k}) \bigr] \Biggr\Vert ^{2}. \end{aligned}$$

Hence,

$$\begin{aligned}& \mathrm{E} \bigl\Vert (Bu_{n}) (t)- \bigl(Bu^{\ast } \bigr) (t) \bigr\Vert ^{2} \\& \quad \leq 2\eta _{0}^{2}M_{q}^{2} \bigl\Vert u_{n}(s)-u^{\ast }(s) \bigr\Vert _{\mathit{PC}}^{2} \\& \quad\quad {} + 2\mathcal{C}_{1}^{2}\sup _{k,\omega } \frac{ \Vert b_{k}(\tau _{k}(\omega _{k}))-1 \Vert }{\int _{\Omega } \Vert b_{k}(\tau _{k})-1 \Vert \,\mathrm{d}\mu } \sum_{k=1}^{\infty } \int _{\Omega } \bigl\Vert b_{k}(\tau _{k})-1 \bigr\Vert \,\mathrm{d}\mu \int _{\Omega } \bigl\Vert u_{n}(\xi _{k})-u^{ \ast }(\xi _{k}) \bigr\Vert \,\mathrm{d} \mu \\& \quad \quad {} \times \max_{k} \bigl\Vert u_{n} \bigl(\xi _{k}(\omega _{k}) \bigr)-u^{\ast } \bigl( \xi _{k}( \omega _{k}) \bigr) \bigr\Vert \sum _{k=1}^{\infty }\sup_{\omega } \bigl\Vert b_{k}(\tau _{k})-1 \bigr\Vert \\& \quad \leq 2\eta _{0}^{2}M_{q}^{2} \bigl\Vert u_{n}(s)-u^{\ast }(s) \bigr\Vert _{\mathit{PC}}^{2}+2 \mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2} \bigl\Vert u_{n}(s)-u^{\ast }(s) \bigr\Vert _{\mathit{PC}}^{2}. \end{aligned}$$

So, we have proved that B is u.s.c.

• Step (6):

We prove that the set $U=\{u(t): u(t)\in \lambda (Au)(t)+\lambda (Bu)(t), 0\leq \lambda \leq 1\}$ is bounded. If $u(t)\in U$, then we have

$$\begin{aligned}& u(t)=\lambda \int _{0}^{1}G(t,s) \bigl[-q(s)u(s)+f \bigl(s,u(s) \bigr) \bigr]\,\mathrm{d}s + \lambda \sum_{k=1}^{\infty } \bigl[W(t,k)u(\xi _{k}) \bigr]I_{A}(\xi _{1}, \xi _{2},\ldots,\xi _{k},\ldots), \\& \begin{aligned} \bigl\Vert u(t) \bigr\Vert ^{2}&\leq 2\lambda ^{2} \biggl( \int _{0}^{1} \bigl\Vert G(t,s) \bigl[-q(s)u(s)+f \bigl(s,u(s) \bigr) \bigr] \bigr\Vert \,\mathrm{d}s \biggr)^{2} \\ &\quad {} +2\lambda ^{2} \Biggl(\sum_{k=1}^{\infty } \bigl\Vert W(t,k)u(\xi _{k}) \bigr\Vert \Biggr)^{2} \\ &\leq 4\lambda ^{2}\eta _{0}^{2}M_{q}^{2} \int _{0}^{1} \bigl\Vert u(s) \bigr\Vert ^{2}\,\mathrm{d}s+4\lambda ^{2}\eta _{0}^{2} \int _{0}^{1} \bigl\Vert f \bigl(s,u(s) \bigr) \bigr\Vert ^{2}\,\mathrm{d}s \\ &\quad {} +2\lambda ^{2}\mathcal{C}_{1}^{2} \Biggl\Vert \sum_{k=1}^{\infty } \bigl(b_{k}( \tau _{k})-1 \bigr)u(\xi _{k}) \Biggr\Vert ^{2}, \end{aligned} \end{aligned}$$

hence

$$\begin{aligned} \mathrm{E} \bigl\Vert u(t) \bigr\Vert ^{2} \leq &4\lambda ^{2}\eta _{0}^{2}M_{q}^{2} \bigl\Vert u(s) \bigr\Vert _{\mathit{PC}}^{2}+4\lambda ^{2}\eta _{0}^{2} \int _{0}^{1} \mathrm{E} \bigl\Vert f \bigl(s,u(s) \bigr) \bigr\Vert ^{2}\,\mathrm{d}s \\ & {} +2\lambda ^{2}\mathcal{C}_{1}^{2}\eta _{2} \bigl\Vert u(\xi _{k}) \bigr\Vert _{\mathit{PC}}^{2}\sum_{k=1}^{\infty } \int _{\Omega } \bigl\Vert b_{k}( \tau _{k})-1 \bigr\Vert \,\mathrm{d}\mu \sum _{k=1}^{\infty } \sup_{\omega } \bigl\Vert b_{k} \bigl(\tau _{k}(\omega ) \bigr)-1 \bigr\Vert \\ \leq &4\lambda ^{2}\eta _{0}^{2}M_{q}^{2} \bigl\Vert u(s) \bigr\Vert _{\mathit{PC}}^{2}+4 \lambda ^{2}\eta _{0}^{2}M_{f}^{2} \bigl( \bigl\Vert u(s) \bigr\Vert _{\mathit{PC}}+1 \bigr)^{2} \\ & {} +2\lambda ^{2}\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2} \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}^{2}. \end{aligned}$$

By simplifying, we can get

$$\begin{aligned}& \bigl[1- \bigl(4\lambda ^{2}\eta _{0}^{2}M_{q}^{2}+4 \lambda ^{2}\eta _{0}^{2}M_{f}^{2}+2 \lambda ^{2}\mathcal{C}_{1}^{2}\eta _{2} \eta _{3}^{2} \bigr) \bigr] \Vert u \Vert _{\mathit{PC}}^{2} \\& \quad{} -8\lambda ^{2}\eta _{0}^{2}M_{f}^{2} \Vert u \Vert _{\mathit{PC}}-4\lambda ^{2}\eta _{0}^{2}M_{f}^{2} \leq 0. \end{aligned}$$

This is a quadratic function, and

$$\begin{aligned} \Delta = 16\lambda ^{2}\eta _{0}^{2}M_{f}^{2} \bigl(1-4\lambda ^{2}\eta _{0}^{2}M_{q}^{2}-2 \lambda ^{2}\mathcal{C}_{1}^{2}\eta _{2} \eta _{3}^{2} \bigr)\geq 0. \end{aligned}$$

Then we can get

$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}} \leq & \frac{4\lambda ^{2}\eta _{0}^{2}M_{f}^{2} +2\lambda \eta _{0}M_{f}\sqrt{1-4\lambda ^{2}\eta _{0}^{2}M_{q}^{2}-2\lambda ^{2}\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}}}{1-(4\lambda ^{2}\eta _{0}^{2}M_{q}^{2}+4\lambda ^{2}\eta _{0}^{2}M_{f}^{2}+2\lambda ^{2}\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2})} \\ =&\frac{4\eta _{0}^{2}M_{f}^{2} +2\eta _{0}M_{f}\sqrt{\frac{1}{\lambda ^{2}}-4\eta _{0}^{2}M_{q}^{2}-2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}}}{\frac{1}{\lambda ^{2}}-(4\eta _{0}^{2}M_{q}^{2}+4\eta _{0}^{2}M_{f}^{2}+2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2})} \\ \leq &\frac{4\eta _{0}^{2}M_{f}^{2} +2\eta _{0}M_{f}\sqrt{1-4\eta _{0}^{2}M_{q}^{2}-2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}}}{1-(4\eta _{0}^{2}M_{q}^{2}+4\eta _{0}^{2}M_{f}^{2}+2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2})}. \end{aligned}$$

So, the set U is bounded. As a consequence of Theorem 2.2, we deduce that $A+B$ has a fixed point $u(t)$ which is a solution of equation (2.1), and we have completed the proof of Theorem 3.1. □

Using the same way, we can prove the following theorem.

$(H_{6})$::

There exists a constant M such that, for each $x_{1}, x_{2}\in \mathit{PC}(J,L_{2})$,

$$\begin{aligned} \bigl\Vert f \bigl(t,x_{1},x'_{1} \bigr)-f \bigl(t,x_{2},x'_{2} \bigr) \bigr\Vert \leq M \Vert x_{1}-x_{2} \Vert . \end{aligned}$$

$(H_{7})$::

There exists a constant $M_{f}$ such that, for each $u\in \mathit{PC}(J,L_{2})$ and $t\in J$,

$$\begin{aligned} \bigl\Vert f \bigl(t,x(t),x'(t) \bigr) \bigr\Vert \leq M_{f} \bigl( \bigl\Vert x(t) \bigr\Vert _{\mathit{PC}}+1 \bigr). \end{aligned}$$

$(H_{8})$::

Suppose

$$\begin{aligned}& \eta _{0}=\frac{(a_{11}+a_{12})(a_{21}+a_{22})}{a_{11}a_{21}+a_{12}a_{21}+a_{11}a_{22}}, \\& \mathcal{C}_{1}=\frac{\max \{a_{11}(a_{21}+a_{22}),a_{21}(a_{11}+a_{12})\}}{a_{11}a_{21}+a_{12}a_{21}+a_{11}a_{22}}, \end{aligned}$$

then they should satisfy the following equalities:

$$\begin{aligned} 4\eta _{0}^{2}M_{f}^{2}+2 \mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}< 1, \end{aligned}$$

and

$$\begin{aligned} \eta _{0}M< 1. \end{aligned}$$

Theorem 3.2

If conditions $(H_{2})\sim (H_{3})$ and $(H_{6})\sim (H_{8})$ are met, then equation (2.7) has a solution $x(t)$ in $\mathit{PC}(J,L_{2})$ which satisfies

$$\begin{aligned} \bigl\Vert x(t) \bigr\Vert _{\mathit{PC}}\leq \frac{4\eta _{0}^{2}M_{f}^{2} +2\eta _{0}M_{f}\sqrt{1-2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}}}{1-(+4\eta _{0}^{2}M_{f}^{2}+2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2})}. \end{aligned}$$

4 The upper and lower solutions

In this section, we consider the upper and lower solutions of the following generalized Sturm–Liouville differential equation with random impulses:

$$ \textstyle\begin{cases} -u''(t)=f(t,u(t),u'(t)), \quad t\in J', \\ u(\xi _{k}^{+})=b_{k}(\tau _{k})u(\xi _{k}), \quad k=1,2,3,\ldots, \\ a_{11}u(0)-a_{12}u'(0)=a_{21}u(1)+a_{22}u'(1)=0. \end{cases} $$

(4.1)

The notation in this equation is the same as the previous definition. First of all, we consider the following linear random impulsive differential equation:

$$ \textstyle\begin{cases} -u''(t)=f(t,h(t),h'(t))-M(u(t)-h(t)), \quad t\in J', \\ u(\xi _{k}^{+})-u(\xi _{k})=(b_{k}(\tau _{k})-1)h(\xi _{k}), \quad k=1,2,3,\ldots, \\ a_{11}u(0)-a_{12}u'(0)=a_{21}u(1)+a_{22}u'(1)=0, \end{cases} $$

(4.2)

where $h(t)\in \mathit{PC}(J,L_{2})$ is a stochastic process. Based on Theorem 2.3, we can get that the solution of equation (4.2) is equivalent to the solution of the following integral equation:

$$\begin{aligned} u(t) =&{-}M \int _{0}^{1}G(t,s)u(s)\,\mathrm{d}s+ \int _{0}^{1}G(t,s) \bigl[f \bigl(s,h(s),h'(s) \bigr)+Mh(s) \bigr] \,\mathrm{d}s \\ & {} +\sum_{k=1}^{\infty } \bigl[W(t,k)h(\xi _{k}) \bigr]I_{A}(\xi _{1},\xi _{2}, \ldots, \xi _{k},\ldots), \end{aligned}$$

where $G(t,s)$ and $W(t,k)$ are defined as (2.9) and (2.10).

Then we define the operator $\Lambda : \mathit{PC}(J,L_{2})\rightarrow \mathit{PC}(J,L_{2})$ as

$$\begin{aligned} \Lambda h =&{-}M \int _{0}^{1}G(t,s)\Lambda h(s)\,\mathrm{d}s+ \int _{0}^{1}G(t,s) \bigl[f \bigl(s,h(s),h'(s) \bigr)+Mh(s) \bigr] \,\mathrm{d}s \\ & {} +\sum_{k=1}^{\infty } \bigl[W(t,k)h(\xi _{k}) \bigr] I_{A}(\xi _{1}, \xi _{2}, \ldots,\xi _{k},\ldots), \end{aligned}$$

we can easily prove that $h(t)$ is the solution of the equation if and only if it is the fixed point of the operator Λ.

Definition 4.1

$\omega _{0}(t)\in \mathit{PC}(J,L_{2})$ is called an upper solution of equation (4.1) if $\omega _{0}(t)$ satisfies the following inequality:

$$\begin{aligned} \textstyle\begin{cases} -\omega _{0}''(t)\geq f(t,\omega _{0}(t),\omega _{0}'(t)), \\ \omega _{0}(\xi _{k}^{+})\geq b_{k}(\tau _{k})\omega _{0}(\xi _{k}), \\ a_{11}\omega _{0}(0)-a_{12}\omega _{0}'(0)\geq x_{0}, \\ a_{21}\omega _{0}(1)+a_{22}\omega _{0}'(1)\geq x^{*}_{0}. \end{cases}\displaystyle \end{aligned}$$

If the above inequalities are reversed, we call it a lower solution of equation (4.1).

Lemma 4.1

([32])

Suppose that E is a semi-ordered Banach space. For $x_{0}, y_{0}\in E$, $x_{0}\leq y_{0}$, and $D=[x_{0}(t),y_{0}(t)]$, $A:D\rightarrow E$ is an operator. Assume that the following conditions are satisfied:

• (i) A is an increasing operator,

• (ii) $x_{0}$ is the lower solution of A and $y_{0}$ is the upper solution of A,

• (iii) A is a continuous operator,

• (iv) $A(D)$ is a relatively compact set of columns in E.

Then A has a maximum fixed point and a minimum fixed point in D. Let $x_{0}$ and $y_{0}$ be the initial conditions. We then have the iteration sequences

$$ x_{n}=Ax_{n-1}, \quad\quad y_{n}=Ay_{n-1}, \quad n=1,2,\ldots. $$

Thus,

$$ x_{0}\leq x_{1}\leq \cdots\leq x_{n}\leq \cdots \leq y_{n}\leq \cdots\leq y_{1} \leq y_{0}, $$

and

$$ x_{n}\rightarrow x^{*}, \quad\quad y_{n}\rightarrow y^{*}. $$

$(H_{9})$ :

$v_{0}(t)$ is the lower solution of equation (4.1) and $\omega _{0}(t)$ is the upper solution of equation (4.1), and they meet the following inequality:

$$ v_{0}(t)\leq \omega _{0}(t) $$

for any $t\in J$.

$(H_{10})$ :

$$ \inf_{k} \bigl\{ b_{k}(\tau _{k})-1:k \in N \bigr\} >0. $$

Theorem 4.1

If conditions $(H_{2})\sim (H_{3})$, $(H_{6})\sim (H_{10})$ are met, then equation (4.1) has the maximum solution $u^{*}(t)$ and the minimum solution $u_{*}(t)$ in $[v_{0}(t),\omega _{0}(t)]\cap \mathit{PC}(J,L_{2})$. $\omega _{n}(t)=\Lambda \omega _{n-1}(t)$ uniformly converges to $u^{*}(t)$, $v_{n}(t)=\Lambda v_{n-1}(t)$ uniformly converges to $u_{*}(t)$, where $n=1,2,\ldots$ .

Proof

First of all, we prove that $v_{0}$ is the lower solution of Λ. It is to say that we should prove that $v_{0}(t)\leq \Lambda v_{0}(t)=v_{1}(t)$. We can easily prove this when there is no pulse (for more details, see [34]). When the equation is equipped with the random impulses, we have

$$\begin{aligned} v_{0} \bigl(\xi _{k}^{+} \bigr)-v_{0}( \xi _{k})\leq \bigl(b_{k}(\tau _{k})-1 \bigr)v_{0}( \xi _{k})=v_{1} \bigl(\xi _{k}^{+} \bigr)-v_{1}(\xi _{k}), \end{aligned}$$

combining with $v_{1}(\xi _{k})\geq v_{0}(\xi _{k})$, we can get $v_{1}(\xi _{k}^{+})\geq v_{0}(\xi _{k}^{+})$. So, we have proved $v_{0}(t)\leq \Lambda v_{0}(t)$.

Then we prove that it is an increasing operator. It is to say that, for any $h_{1}(t)\leq h_{2}(t)$, we have $\Lambda h_{1}(t)\leq \Lambda h_{2}(t)$. When there is no pulse, we can easily prove this conclusion (for more details, see [13]). When there are random impulses, we have

$$\begin{aligned} \Lambda h_{1} \bigl(\xi _{k}^{+} \bigr)-\Lambda h_{1}(\xi _{k}) =& \bigl(b_{k}(\tau _{k})-1 \bigr)h_{1}( \xi _{k}) \\ \leq & \bigl(b_{k}(\tau _{k})-1 \bigr)h_{2}(\xi _{k})= \Lambda h_{2} \bigl(\xi _{k}^{+} \bigr)- \Lambda h_{2}(\xi _{k}), \end{aligned}$$

so we have proved that $\Lambda h_{1}(t)\leq \Lambda h_{2}(t)$.

Based on the proof of Theorem 3.1, we can easily prove that Λ is a continuous operator and $\Lambda ([v_{0}, \omega _{0}])$ is a relatively compact set of columns. Combining with Lemma 4.1, we complete the proof of Theorem 4.1. □

In this section, we give some examples to illustrate our main result.

5 Examples

Example 5.1

Consider the following equation:

$$ \textstyle\begin{cases} -(\frac{9}{1.5-t}u'(t))'+\frac{9t}{t^{2}+30}u(t)=\frac{1}{8}te^{t} \sin u(t), \quad t\in J\setminus \{\xi _{1},\xi _{2},\ldots\}, \\ x(\xi _{k}^{+})=(\sqrt{\tau _{k}}+1)x(\xi _{k}), \quad k=1,2,\ldots, \\ x(0)-\frac{1}{2}x'(0)=\frac{1}{3}x(1)+x'(1)=0, \end{cases} $$

(5.1)

where $\{\tau _{k}\}$ is a variable sequence, $\tau _{i}$ and $\tau _{j}$ are independent from each other for each $i\neq j$. $\tau _{k}\sim U(0,\frac{1}{4^{k}})$, it is to say that the probability density function of $\tau _{k}$ is

$$\begin{aligned} p(x)= \textstyle\begin{cases} 4^{k} &x\in (0,\frac{1}{4^{k}}), \\ 0 &\text{others}. \end{cases}\displaystyle \end{aligned}$$

Then it is easy to see that $\{\sqrt{\tau _{k}}\}$ is also a variable sequence, and for every $i\neq j$, $\sqrt{\tau _{i}}$ and $\sqrt{\tau _{j}}$ are independent. We can easily get the probability density function of $\sqrt{\tau _{k}}$:

$$\begin{aligned} p(x)= \textstyle\begin{cases} 2^{2k+1}x &x\in (0,\frac{1}{2^{k}}), \\ 0 &\text{others}. \end{cases}\displaystyle \end{aligned}$$

Set $\xi _{0}=0$, $\xi _{k+1}=\xi _{k}+\tau _{k+1}$, obviously, $\{\xi _{k}\}$ is a process with independent increments, and the impulsive moments $\{\xi _{k}\}$ form a strictly increasing sequence. And we have

$$\begin{aligned} \xi _{k}\leq \frac{1}{4}+\frac{1}{4^{2}}+\cdots+ \frac{1}{4^{k}}= \frac{1}{3} \biggl(1-\frac{1}{4^{k}} \biggr)< 1. \end{aligned}$$

In this example, we define the norm $\Vert x \Vert = \vert x \vert$. Then we have

$$\begin{aligned}& \eta _{3}=\sum_{k=1}^{\infty }\sup _{\omega _{k}} \bigl\Vert b_{k}( \tau _{k})-1 \bigr\Vert =\sum_{k=1}^{\infty } \frac{1}{2^{k}}=1, \\& \eta _{2}=\sup_{k,\omega } \frac{ \Vert b_{k}(\tau _{k})-1 \Vert }{\mathrm{E} \Vert b_{k}(\tau _{k})-1 \Vert }= \frac{\frac{1}{2^{k}}}{\frac{1}{3}\frac{1}{2^{k-1}}}=\frac{3}{2}, \\& \vert Q \vert = \begin{vmatrix} a_{11} & -a_{12} \\ a_{21} & a_{21} \int _{0}^{1}\frac{p(0)}{p(s)} \, \mathrm{d}s+a_{22} \frac{p(0)}{p(1)} \end{vmatrix}=\frac{13}{18}, \\& \eta _{0}=\frac{1}{ \vert Q \vert } \biggl(a_{12}+a_{11}p(0) \int _{0}^{1} \frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr) \biggl(a_{22}\frac{1}{p(1)}+a_{21} \int _{0}^{1}\frac{1}{p(\tau )}\,\mathrm{d}\tau \biggr)=\frac{35}{26}, \end{aligned}$$

and

$$\begin{aligned} \mathcal{C}_{1} =&\max \biggl\{ \frac{1}{ \vert Q \vert }a_{11}p(0) \biggl(a_{21} \int _{0}^{1}\frac{1}{p(\tau )}\,\mathrm{d}\tau +a_{22}\frac{1}{p(1)} \biggr), \\ &\frac{1}{ \vert Q \vert }a_{21}p(0) \biggl(a_{11} \int _{0}^{1} \frac{1}{p(\tau )}\,\mathrm{d}\tau +a_{12}\frac{1}{p(0)} \biggr) \biggr\} = \frac{7}{13}. \end{aligned}$$

So we have

$$\begin{aligned}& \eta _{0}M=\frac{35}{1872}\approx 0.019< 1, \\& 4\eta _{0}^{2}M_{q}^{2}+4\eta _{0}^{2}M_{f}^{2}+2 \mathcal{C}_{1}^{2} \eta _{2}\eta _{3}^{2}=\frac{739{,}855{,}753}{841{,}928{,}256}\approx 0.879< 1. \end{aligned}$$

Based on Theorem 3.1, we can know that equation (5.1) has a solution $u(t)$ which satisfies

$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}\leq \frac{4\eta _{0}^{2}M_{f}^{2} +2\eta _{0}M_{f}\sqrt{1-4\eta _{0}^{2}M_{q}^{2}-2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}}}{1-(4\eta _{0}^{2}M_{q}^{2}+4\eta _{0}^{2}M_{f}^{2}+2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2})} \approx 0.119. \end{aligned}$$

We have completed the proof of the example.

Example 5.2

Now we consider the following second order random impulsive differential equation with boundary value problems:

$$ \textstyle\begin{cases} -33u''(t)=e^{t^{2}}u(t)\sin u'(t)+t^{2}+\frac{1}{2}t, \quad t\in J \setminus \{\xi _{1},\xi _{2},\ldots\}, \\ x(\xi _{k}^{+})=(\tau _{k}+1)x(\xi _{k}), \quad k=1,2,\ldots, \\ x(0)-x'(0)=\frac{1}{7}x(1)+\frac{1}{4}x'(1)=0. \end{cases} $$

(5.2)

Here we define the norm $\Vert u \Vert _{\mathit{PC}}=\sup_{t\in J}\mathrm{E} \vert u(t) \vert ^{2}$. $\{\tau _{k}\}$ is a variable sequence and $\tau _{i}$ and $\tau _{j}$ are independent from each other when $i\neq j$. The probability density function of $\tau _{k}$ is

$$\begin{aligned} p(x)= \textstyle\begin{cases} 4^{k+2}x &x\in [0,\frac{1}{2^{k+2}}), \\ 2^{k+3}-4^{k+2}x &x\in [\frac{1}{2^{k+2}},\frac{1}{2^{k+1}}), \\ 0 &\text{others}. \end{cases}\displaystyle \end{aligned}$$

Suppose $\xi _{k+1}=\xi _{k}+\tau _{k+1}$, $k=0,1,2,\ldots$ . Obviously, we have

$$\begin{aligned} \xi _{k}< \sum_{n=1}^{\infty } \frac{1}{2^{n+1}}=\frac{1}{2}< 1. \end{aligned}$$

Then we have

$$\begin{aligned}& \eta _{3}=\sum_{k=1}^{\infty } \frac{1}{2^{k+1}}=\frac{1}{2}, \\& \eta _{2}=\sup_{k,\omega } \frac{ \Vert b_{k}(\tau _{k})-1 \Vert }{E \Vert b_{k}(\tau _{k})-1 \Vert }= \frac{\frac{1}{2^{k+1}}}{\frac{1}{2^{k+2}}}=2. \end{aligned}$$

And we can easily get $M_{f}=\frac{1}{22}$, $M=\frac{1}{33}$,

$$\begin{aligned}& \eta _{0}= \frac{(a_{11}+a_{12})(a_{21}+a_{22})}{a_{11}a_{21}+a_{12}a_{21}+a_{11}a_{22}}= \frac{22}{15}, \\& \mathcal{C}_{1}= \frac{\max \{a_{11}(a_{21}+a_{22}),a_{21}(a_{11}+a_{12})\}}{a_{11}a_{21}+a_{12}a_{21}+a_{11}a_{22}}= \frac{11}{15}. \end{aligned}$$

Hence, $\eta _{0}M=\frac{2}{45}\approx 0.044<1$ and $4\eta _{0}^{2}M_{f}^{2}+2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}= \frac{5}{9}\approx 0.556<1$. So, this equation satisfies all the conditions of Theorem 3.2, which shows that the equation has a solution satisfying

$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert _{\mathit{PC}}\leq \frac{4\eta _{0}^{2}M_{f}^{2} +2\eta _{0}M_{f}\sqrt{1-2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2}}}{1-(+4\eta _{0}^{2}M_{f}^{2}+2\mathcal{C}_{1}^{2}\eta _{2}\eta _{3}^{2})} = \frac{9(1+\sqrt{26})}{225}\approx 0.244. \end{aligned}$$

We can easily prove that $\omega _{0}(t)=0$ is an upper solution and $v_{0}(t)=-\frac{1}{1+\sin t}$ is a lower solution of equation (5.2) and $\omega _{0}(t)\geq v_{0}(t)$. So, equation (5.2) satisfies all the conditions of Theorem 4.1. Hence, we can get the extremal solutions of problem (5.2) between $v_{0}$ and $\omega _{0}$ by constructing iterative sequences starting from $v_{0}$ and $\omega _{0}$:

$$\begin{aligned}& \begin{aligned} v_{n}(t)&=-M \int _{0}^{1}G(t,s)v_{n}(s)\, \mathrm{d}s+ \int _{0}^{1}G(t,s) \bigl[f \bigl(s,v_{n-1}(s),v_{n-1}'(s) \bigr)+Mv_{n-1}(s) \bigr] \,\mathrm{d}s \\ &\quad {} +\sum_{k=1}^{\infty } \bigl[W(t,k)v_{n-1}(\xi _{k}) \bigr]I_{A}(\xi _{1}, \xi _{2},\ldots,\xi _{k},\ldots), \end{aligned} \\& \begin{aligned} \omega _{n}(t)&=-M \int _{0}^{1}G(t,s)\omega _{n}(s) \, \mathrm{d}s+ \int _{0}^{1}G(t,s) \bigl[f \bigl(s,\omega _{n-1}(s),\omega _{n-1}'(s) \bigr)+M\omega _{n-1}(s) \bigr] \,\mathrm{d}s \\ &\quad {} +\sum_{k=1}^{\infty } \bigl[W(t,k) \omega _{n-1}(\xi _{k}) \bigr] I_{A}( \xi _{1},\xi _{2},\ldots,\xi _{k},\ldots). \end{aligned} \end{aligned}$$

We have completed the proof of the example.