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Research article

The existence of upper and lower solutions to second order random impulsive differential equation with boundary value problem

  • Received: 06 May 2020 Accepted: 06 July 2020 Published: 31 July 2020
  • MSC : 34A37, 34B, 34F05

  • In this article, we consider the existence of upper and lower solutions to a second-order random impulsive differential equation. We first study the solution form of the corresponding linear impulsive system of the second-order random impulsive differential equation. Based on the form of the solution, we define the resolvent operator. Then, we prove that the fixed point of this operator is the solution to the equation. Finally, we construct the sum of two monotonic iterative sequences and prove that they are convergent. Thus, we conclude that the system has upper and lower solutions.

    Citation: Zihan Li, Xiao-Bao Shu, Fei Xu. The existence of upper and lower solutions to second order random impulsive differential equation with boundary value problem[J]. AIMS Mathematics, 2020, 5(6): 6189-6210. doi: 10.3934/math.2020398

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  • In this article, we consider the existence of upper and lower solutions to a second-order random impulsive differential equation. We first study the solution form of the corresponding linear impulsive system of the second-order random impulsive differential equation. Based on the form of the solution, we define the resolvent operator. Then, we prove that the fixed point of this operator is the solution to the equation. Finally, we construct the sum of two monotonic iterative sequences and prove that they are convergent. Thus, we conclude that the system has upper and lower solutions.


    Impulsive differential equations have many applications in engineering, science and finance. As a ubiquitous phenomenon, pulses exist in mechanical systems with impacts, optimal control models in economics, and the transfers of satellite orbit. It is difficult to model such phenomena using continuous models or discrete models [1,2]. In the 1950s, an impulsive model was developed to describe such specific evolution of a dynamic system [1]. Impulsive differential systems describe the dynamic processes with discontinuous jump caused by sudden changes. A variety of impulsive systems were investigated in the literature [3,4,5,6,23,35,38,40].

    The characteristics of impulsive differential equations have attracted the attention of scholars [31,32]. In recent years, many scholars have studied the initial and boundary value problem of fixed impulse differential equations [16,17,22]. For example, the boundary value problem of impulsive equations have been examined in the literature [7,8,9,10,15,18,24]. The existence and uniqueness of solutions to the following impulsive equation with boundary value problems have been investigated in the literature [11].

    {u=f(t,u,u),tJ{t1,t2,,tm},Δu(tk)=Ik(u(tk)),k=1,2,,m,Δu(tk)=Nk(u(tk)),au(0)bu(0)=0,cu(1)+du(1)=0, (1.1)

    where a>0,b0,c>0,d0, 0=t0<t1<t2<<tm=1, J=[0,1].

    Some kinds of stochastic differential equation with fixed impulsive moments and Random impulsive differential equations also obtained considerable attention in the literature [12,13,14,20,21,25,26,36,37]. The Hyers-Ulam stability of random impulsive stochastic functional differential equations with finite delays has been studied in [19]. The authors considered the following system

    {d(x(t))=f(t,xt)dt+g(t,xt)dW(t),t0,tξk,x(ξk)=bk(τk)x(ξk),k=1,2,,xt0=ξ={ξ(θ):τθ0}, (1.2)

    where xt is Rd -valued stochastic process such that xtRd, xt={x(t+θ):τθ0}. Here, 0=ξ0<ξ1<ξ2<<ξk<<limkξk=+, and x(ξk)=limtξk0x(t). Note that {N(t),t0} is the simple counting process generated by ξk, and {W(t):t0} is a given m-dimensional Winer process.

    The fixed point method [28] had been used to study the random impulsive differential equations. Niu et al. [27] used the fixed point method to address the existence and Hyers-Ulam stability for the following differential equation

    {x(t)=f(t,x(t)),tJ, tξk,x(ξk)=bk(τk)x(ξk),k=1,2,,x0=x0,x(0)=x1. (1.3)

    Upper and lower solution method can be used to study fractional evolution equations [33] and impulsive differential equations [34].

    To the best of our knowledge, the boundary value problem of second order random impulsive differential equation has not been studied using the upper and lower solution method in the literature. In this paper, we use the upper and lower solution method to study the following second order random impulsive differential equation with boundary value problem.

    {x(t)=f(t,x(t),x(t)),tJ,x(ξ+k)=bk(τk)x(ξk),k=1,2,,α0x(0)α1x(0)=x0,β0x(1)+β1x(1)=x0. (1.4)

    Where f:J×R×RR is a continuous mapping. x(t) is a stochastic process taking values in the Euclidean space (R,). Then, we introduce τk to be a random variable defined from Ω to Ek=(0,dk), with 0<dk<1 for every kN+. We assume that τi and τj are independent of each other when ij for every i,jN+ and bk: EkR satisfies for every kN+, bk(τk)0. Set ξk=ξk1+τk. Obviously, {ξk} is a process with independent increments and the impulsive moments ξk form a strictly increasing sequence, i.e. 0=ξ0<ξ1<ξ2<<ξk<<1. We hold the opinion that x(ξk)=limtξk0x(t), x(ξ+k)=limtξk+0x(t). The convergence is under the meaning of the orbit. Since for a realization (sample) of random process, {ξk} will become a series of fixed time points. Under that sense, so we can define the limit as we would in general. We suppose that {N(t):t0} is the simple counting process generated by ξk. Let J=[0,1], R+=(0,+) and J=J{ξ1,ξ2,}. Here, α0,α1,β0,β1,x0,x0 are constants satisfying α0α10, β00, α0,α1,β0,β1,x0,x00.

    The rest of the paper is organised as follows: In section 2, we introduce some notations and preliminaries. In section 3, we use the upper and lower solution method to study the existence of solutions to the second order random impulsive differential equations. In section 4, we give an example to show the application of the main result. Finally, conclusions are presented.

    Suppose (Ω,Γ,P) is a probability space. Let Lp(Ω,Rn) be the collection of all strongly measurable, pth integrable, and Γt-measurable with Rn-valued random variables x:ΩRn and norm Lp(Ω,Rn) for p1. Here, E(x)=ΩxdP< is the expectation of x, and Lp(Ω,Rn) is equipped with its natural norm xLp(Ω,Rn)=(ΩxpdP)1p=(Exp)1p.

    We introduce the space PC=PC(J,L2(Ω,Rn)): = {u(t) u(t)=u(t,ω) is a strongly measurable, square integrable, random process from J into L2(Ω,Rn), and u(t) is continuous when tJ and left continuous when tJJ}. We can prove that PC is a Banach space with norm

    uPC=(suptJEu(t)2)1/2. (2.1)

    Then, we consider the space PC1=PC1(J,L2(Ω,Rn)): = {u(t) u(t)=u(t,ω) is a strongly measurable, square integrable, random process from J into L2(Ω,Rn), u(t) is continuously differentiable when tJ and left continuous when tJJ, u(ξk) and u(ξ+k) exist for k=1,2,}. It is easy to see that PC1 is also a Banach space with norm

    uPC1=max{(suptJEu(t)2)1/2,(suptJEu(t)2)1/2}. (2.2)

    The functions in PC1 which satisfy the equation (1.4) are called the solutions of the equation (1.4).

    For convince, in the rest of the paper we write PC(J,L2(Ω,R)) as PC(J,R), write PC1(J,L2(Ω,R)) as PC1(J,R). In this paper, the upper and lower solutions of equation (1.4) are studied in PC1(J,R) space.

    Now we consider the equation (2.3)

    {u(t)=f(t,h(t),h(t))M[u(t)h(t)],tJ,u(ξ+k)=bk(τk)h(ξk),k=1,2,,α0u(0)α1u(0)=x0,β0u(1)+β1u(1)=x0, (2.3)

    where h(t)PC1(J,R) and M is a positive constant.

    Lemma 2.1. Equation (2.3) has one solution, given by

    u(t)=k=0[Ck1eMt+Ck2eMt+ˆh(t)]I(ξk,ξk+1](t), (2.4)
    {Ck1=C1,k=0,Ck1=δkC1e2Mξ1δkC2+kn=1Δn,k[bn(τn)h(ξn)ˆh(ξn)],k=1,2,,Ck2=C2,k=0,Ck2=e2Mξ1δ+kC1+δ+kC2+kn=1Δ+n,k[bn(τn)h(ξn)ˆh(ξn)],k=1,2,, (2.5)

    and

    δk=12k[1+1i1<j1ke2M(ξj1ξi1)+1i1<j1<i2<j2ke2M[(ξj2ξi2)+(ξj1ξi1)]++1i1<<jnkjn+1e2M[(ξjnξin)++(ξj1ξi1)]],δ+k=12k[1+1i1<j1ke2M(ξj1ξi1)+1i1<j1<i2<j2ke2M[(ξj2ξi2)+(ξj1ξi1)]++1i1<<jnkjn+1e2M[(ξjnξin)++(ξj1ξi1)]], (2.6)
    Δn,k=12kn+1eMξn[1+ni1<j1k(1)I{n}(i1)e2M(ξj1ξi1)+ni1<j1<i2<j2k(1)I{n}(i1)e2M[(ξj2ξi2)+(ξj1ξi1)]++ni1<<jlkjl+1(1)I{n}(i1)e2M[(ξjlξil)++(ξj1ξi1)]],Δ+n,k=12kn+1eMξn[1+ni1<j1k(1)I{n}(i1)e2M(ξj1ξi1)+ni1<j1<i2<j2k(1)I{n}(i1)e2M[(ξj2ξi2)+(ξj1ξi1)]++ni1<<jlkjl+1(1)I{n}(i1)e2M[(ξjlξil)++(ξj1ξi1)]], (2.7)
    C1=1|Q||x0α0+Mα1x0[eMB(1)(β0+Mβ1)e2Mξ1+MA(1)(β0+Mβ1)+eMB+(1)(β0Mβ1)+eMA+(1)(β0Mβ1)+β0ˆh(1)+β1ˆh(1)]|, (2.8)
    C2=1|Q||α0Mα1x0eMA(1)(β0+Mβ1)x0[eMB(1)(β0+Mβ1)e2Mξ1MA+(1)(β0Mβ1)+eMB+(1)(β0Mβ1)+β0ˆh(1)+β1ˆh(1)]|, (2.9)
    |Q|=|α0Mα1α0+Mα1eMA(1)(β0+Mβ1)e2Mξ1+MA(1)(β0+Mβ1)e2Mξ1MA+(1)(β0Mβ1)+eMA+(1)(β0Mβ1)|. (2.10)

    Denote

    ˆh(t)=eMt2Mt0eMsσ(s)ds+eMt2Mt0eMsσ(s)ds, (2.11)
    σ(s)=f(s,h(s),h(s))+Mh(s), (2.12)

    and the index function

    IA(t)={1tA,0tA, (2.13)
    I{n}(i1)={1n=i1,0ni1, (2.14)

    where

    A(t)=k=0δkI(ξk,ξk+1](t),A+(t)=k=0δ+kI(ξk,ξk+1](t), (2.15)

    and

    B(t)=k=0kn=1Δn,k[bn(τn)h(ξn)ˆh(ξn)]I(ξk,ξk+1](t),B+(t)=k=0kn=1Δ+n,k[bn(τn)h(ξn)ˆh(ξn)]I(ξk,ξk+1](t). (2.16)

    Proof. Suppose ξ1, ξ2, is a sample orbit. Thus, when t[0,ξ1], the solution of the equation (2.3) is

    u0(t)=C1eMt+C2eMt+ˆh(t). (2.17)

    When t(ξ1,ξ2], we assume that the solution of the equation (2.3) is

    u1(t)=C11eMt+C12eMt+ˆh(t). (2.18)

    Plug in the initial conditions

    u1(ξ+1)=b1(τ1)h(ξ1),u1(ξ+1)=u0(ξ1), (2.19)

    we can get

    C11=12eMξ1[C1eMξ1C2eMξ1+b1(τ1)h(ξ1)ˆh(ξ1)],C12=12eMξ1[C1eMξ1+C2eMξ1+b1(τ1)h(ξ1)ˆh(ξ1)]. (2.20)

    In the same way, we can get when t(ξk,ξk+1],

    Ck1=12eMξk[Ck11eMξkCk12eMξk+bk(τk)h(ξk)ˆh(ξk)],Ck2=12eMξk[Ck11eMξk+Ck12eMξk+bk(τk)h(ξk)ˆh(ξk)]. (2.21)

    Based on the above discussion, mathematical induction can be obtained as

    Ck1=δkC1e2Mξ1δkC2+kn=1Δn,k[bn(τn)h(ξn)ˆh(ξn)],Ck2=e2Mξ1δ+kC1+δ+kC2+kn=1Δ+n,k[bn(τn)h(ξn)ˆh(ξn)]. (2.22)

    Where δ+k,δk,Δ+k,Δk are defined as the equations (2.6) and (2.7).

    So, the solution of the equation (2.3) is

    u(t)=[k=0Ck1I(ξk,ξk+1](t)]eMt+[k=0Ck2I(ξk,ξk+1](t)]eMt+ˆh(t), (2.23)

    and

    k=0Ck1I(ξk,ξk+1](t)=[k=0δkI(ξk,ξk+1](t)]C1[k=0δkI(ξk,ξk+1](t)]e2Mξ1C2+k=0kn=1Δn,k[bn(τn)h(ξn)ˆh(ξn)]I(ξk,ξk+1](t), (2.24)
    k=0Ck2I(ξk,ξk+1](t)=[k=0δ+kI(ξk,ξk+1](t)]e2Mξ1C1+[k=0δ+kI(ξk,ξk+1](t)]C2+k=0kn=1Δ+n,k[bn(τn)h(ξn)ˆh(ξn)]I(ξk,ξk+1](t). (2.25)

    Therefore

    u(t)=[A(t)C1e2Mξ1A(t)C2+B(t)]eMt+[e2Mξ1A+(t)C1+A+(t)C2+B+(t)]eMt+ˆh(t), (2.26)
    u(t)=M[A(t)C1e2Mξ1A(t)C2+B(t)]eMtM[e2Mξ1A+(t)C1+A+(t)C2+B+(t)]eMt+ˆh(t). (2.27)

    Substituting these equations into the boundary value conditions of the equation (2.3) yields and using the Cramer's rule, it follows from (2.26) and (2.27) that (2.5)–(2.16) hold.

    Lemma 2.2. δk and δ+k are uniformly bounded series.

    Proof. We firstly consider δk,

    e2M[(ξjmξim)++(ξj1ξi1)]1. (2.28)

    So, for every nN+, we have

    1i1<<jnkjn+1e2M[(ξjnξin)++(ξj1ξi1)](k2n), (2.29)
    δk=12k[1+1i<jke2M(ξjξi)+1i1<j1<i2<j2ke2M[(ξj2ξi2)+(ξj1ξi1)]++1i1<<jnkjn+1e2M[(ξjnξin)++(ξj1ξi1)]]12k[1+(k2)+(k4)++(k2n)]=12. (2.30)

    In the same way, we can prove that δ+k is uniformly bounded and

    δ+k12e2M. (2.31)

    Lemma 2.3. For every nN+, Δn,k and Δ+n,k are uniformly bounded series.

    Proof. We can easily prove that

    2n1δkΔn,k2n1δk,2n1eMδ+kΔ+n,k2n1eMδ+k. (2.32)

    So, we have

    |Δn,k|2n1δk2n2,|Δ+n,k|2n1eMδ+k2n2e3M, (2.33)

    and we have proved the lemma.

    Definition 2.1. Define the operator Λ:PC1[J,R]PC1[J,R] such that

    Λh=k=0[Ck1eMt+Ck2eMt+ˆh(t)]I(ξk,ξk+1](t). (2.34)

    Definition 2.2. v0(t)PC1[J,R] is called a lower solution of equation (1.4) if v0(t) satisfies the inequality group

    {v0(t)f(t,v0(t),v0(t)),v0(ξ+k)bk(τk)v0(ξk),α0v0(0)α1v0(0)x0,β0v0(1)+β1v0(1)x0. (2.35)

    Definition 2.3. ω0(t)PC1[J,R] is called an upper solution of equation (1.4) if ω0(t) satisfies the inequality group

    {ω0(t)f(t,ω0(t),ω0(t)),ω0(ξ+k)bk(τk)ω0(ξk),α0ω0(0)α1ω0(0)x0,β0ω0(1)+β1ω0(1)x0. (2.36)

    Lemma 2.4. h(t)PC1[J,R] is the solution of equation (1.4) if and only if h(t)PC1[J,R] is the fix point of the operator Λ.

    Proof. If h(t) is the fix point of the operator Λ, it is to say that h(t) satisfies the equation Λh(t)=h(t), then, in the equation (2.3), we have u(t)=Λh(t)=h(t), so, we can replace u(t) with h(t) in the equation (2.3), and we have

    {h(t)=f(t,h(t),h(t)),h(ξ+k)=bk(τk)h(ξk),α0h(0)α1h(0)=x0,β0h(1)+β1h(1)=x0, (2.37)

    and we have proved that h(t) is a solution of the equation (1.4).

    If h(t) is the solution of the equation (1.4), then using the same method, we can easily know that it is also the fix point of the operator Λ, and we have proved the lemma.

    Lemma 2.5.(The Arzela-Ascoli Theorem)([29]) The set MC2[J,Rn] is column compact tight if and only if

    (i) The functions in the set M are uniformly bounded, that is to say, there exists a fixed constant K for all u(t)M, where u(t)K.

    (ii) Functions in the set M are equally continuous, that is to say, for all ϵ>0, there exists δ=δ(ϵ) such that when t1,t2J and t1t2∥<δ for all u(t)M, there is u(t1)u(t2)∥<ϵ.

    Lemma 2.6.([30]) Suppose E is a semi-ordered Banach space. For x0,y0E, x0y0, and D=[x0(t),y0(t)], A:DE is an operator. Assuming that the following conditions are satisfied

    (i) A is an increasing operator,

    (ii) x0 is the lower solution of A and y0 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 x0 and y0 be the initial conditions. We then have the iteration sequences

    xn=Axn1,yn=Ayn1,n=1,2,. (2.38)

    Thus,

    x0x1xnyny1y0, (2.39)

    and

    xnx,yny. (2.40)

    (H1) The equation (1.4) has the lower solution v0(t) and the upper solution ω0(t) and they meet the inequality

    v0(t)ω0(t), (3.1)

    for any tJ.

    (H2) There exists a constant M>0, such that

    f(t,x1,y)f(t,x2,y)M(x1x2), (3.2)

    for any tJ, yPC1(J,R) and v0(t)x2(t)x1(t)ω0(t).

    (H3) There exist constants B1 and B2 such that

    {k=0{kn=1Δn,kbn(τn)}I(ξk,ξk+1](t)B1,k=0{kn=1Δ+n,kbn(τn)}I(ξk,ξk+1](t)B2. (3.3)

    (H4) There exists an increasing continuous function Θ(x) satisfies that Θ(x)x is a decreasing function and

    Ef(t,x,y)2Θ(xPC1), (3.4)

    for every tJ, x1,x2D=[v0(t),ω0(t)], and yPC1(J,R).

    (H5) There exist a function Ψ(t,x,y) and a constant K such that

    (i) For each tJ, the function Ψ(t,,):R×RR is continuous and Ψ(t,0,0)=0. For every x,yR, the function Ψ(,x,y):JR is measurable;

    (ii)

    Ef(t,x1,y1)f(t,x2,y2)2KΨ(t,Ex1x22,Ex1x22), (3.5)

    for every tJ, x(t)D=[v0(t),ω0(t)].

    (H6) Define P=12e2MM, P=12e2M+ln3(m1+m2), m1=suptECk1eMt2, and m2=suptECk2eMt2.

    Then we define the sequences vn and ωn as

    vn(t)=k=0[Ck1(vn1)eMt+Ck2(vn1)eMt]I(ξk,ξk+1](t)eMt2Mt0eMs[Mvn1(s)+f(s,vn1(s),vn1(s))]ds+eMt2Mt0eMs[Mvn1(s)+f(s,vn1(s),vn1(s))]ds, (3.6)
    ωn(t)=k=0[Ck1(ωn1)eMt+Ck2(ωn1)eMt]I(ξk,ξk+1](t)eMt2Mt0eMs[Mωn1(s)+f(s,ωn1(s),ωn1(s))]ds+eMt2Mt0eMs[Mωn1(s)+f(s,ωn1(s),ωn1(s))]ds. (3.7)

    Theorem 3.1.

    If conditions (H1)(H6) are met, the equation (1.4) has the maximum solution x(t) and the minimum solution x(t) in [v0(t),ω0(t)]PC1[J,R]. And there exist ωn(t)=Λωn1(t) uniformly convergent to x(t), vn(t)=Λvn1(t) uniformly convergent to x(t), n=1,2,. And if x(t) is the solution of the equation (1.4), it satisfies

    ln(Ex(t)2)P10Θ(Ex(t)2)Ex(t)2dt+P. (3.8)

    Proof. We will prove this theorem in five steps.

    Step(1). We prove that v0(t) and ω0(t) are the lower and upper solutions of the operator Λ, i.e., we should prove v0(t)Λv0(t) and ω0(t)Λω0(t).

    When there is no random impulsive, we set v1(t)=Λv0(t). Now, we only need to prove that v0(t)v1(t). Here, we use proof by contradiction. If it is not true, then there exist t0J and ε>0 such that

    {v0(t0)=v1(t0)+ε,v0(t)v1(t)+ε, (3.9)

    for every tJ. If t0J({0}{1}), it is easy to see that

    {v0(t0)v1(t0)=0,v0(t0)v1(t0)0. (3.10)

    However,

    v1(t)=f(t,v0(t),v0(t))M[v1(t)v0(t)]v0(t)M[v1(t)v0(t)],v0(t0)v1(t0)M[v0(t0)v1(t0)]=Mε>0, (3.11)

    which is a contradiction to the inequality v0(t0)v1(t0)0. Thus, our hypothesis does not work.

    When t0=0 or t0=1, we assume that t0=0. Therefore,

    m(t)=v1(t)v0(t),mintJ{m(t)}=m(0), (3.12)

    assuming that v1(0)+ε=v0(0), it is easy to see that

    v1(0)v0(0)0. (3.13)

    By the boundary value conditions, we can get

    α0v0(0)α1v0(0)x0,α0v1(0)α1v1(0)=x0. (3.14)

    Thus, we have

    α1v1(0)+α0εα1v0(0)0,α1[v1(0)v0(0)]<0, (3.15)

    which is a contradiction to the hypothesis. Therefore, we have proved that v(t)Λv(t).

    When the equation has the random pulses, we have

    {v1(t)=f(t,v0(t),v0(t))M[v1(t)v0(t)],v1(ξ+k)=bk(τk)v0(ξk), (3.16)

    and v0(t) is the lower solution of the equation (1.4). Thus, according to the second inequality of (2.35), for every t{ξk}kN+, we have

    v0(ξ+k)bk(τk)v0(ξk)=v1(ξ+k). (3.17)

    Thus, based on our discussion, we conclude that for every t(ξk,ξk+1], k=1,2,,

    v1(t)v0(t). (3.18)

    Hence, we have proved that v1(t)v0(t) for every tJ.

    In the same way, we can prove that ω0(t)Λω0(t) for every tJ.

    Step(2). We prove that Λ is an increasing operator.

    First of all, we take any h1(t) and h2(t), h1(t),h2(t)PC1[J,R]. Suppose h1(t)h2(t) for any tJ. Then, we prove that Λh1(t)Λh2(t). Here, we use proof by contradiction. Let h1(t)=Λh1(t), h2(t)=Λh2(t). Then, we need to prove h1(t)h2(t).

    When there is no random pulse, if the hypothesis is not true, then there must exist t0J and ε>0 such that h1(t0)+ε=h2(t0) and h1(t)+εh2(t) for every tJ. If t0J({0}{1}), then we have

    h1(t0)h2(t0)=0,h1(t0)h2(t0)0, (3.19)

    and

    h1(t)=f(t,h1(t),h1(t))M[h1(t)h1(t)],h2(t)=f(t,h2(t),h2(t))M[h2(t)h2(t)]. (3.20)

    Thus

    h1(t0)h2(t0)=f(t0,h2(t0),h2(t0))f(t0,h1(t0),h1(t0))+M[h1(t0)h2(t0)]+M[h2(t0)h1(t0)]M[h1(t0)h2(t0)]+M[h1(t0)h2(t0)]+M[h2(t0)h1(t0)]Mε<0. (3.21)

    Which is a contradiction. When t0=0 or t0=1, we assume that t0=0 and h1(0)+ε=h2(0).

    Then

    h1(0)h2(0)0. (3.22)

    Thus

    α0h1(0)α1h1(0)=x0,α0h2(0)α1h2(0)=x0. (3.23)

    Take the difference of these equation yields

    α1[h1(0)h2(0)]+α0ε=0. (3.24)

    That is to say h1(0)h2(0)<0, which is a contradiction.

    When there exits random pulses, we have

    Λh2(ξ+k)=h2(ξ+k)=bk(τk)h2(ξk)bk(τk)h1(ξk)=Λh1(ξ+k). (3.25)

    Then, for every t(ξk,ξk+1], k=1,2,, we have

    Λh1(t)Λh2(t). (3.26)

    Thus, for every tJ, the inequality Λh1(t)Λh2(t) is true.

    Step(3). We prove that Λ is a continuous operator. That is to say, we should prove that for any ε>0, there exists δ(ε)>0 such that when h1(t)h2(t)PC1<δ, Λh1(t)Λh2(t)PC1<ε.

    We assume that

    Λh1=k=0[Ck1eMt+Ck2eMt]I(ξk,ξk+1](t)+ˆh1(t), (3.27)
    Λh2=k=0[˜Ck1eMt+˜Ck2eMt]I(ξk,ξk+1](t)+ˆh2(t), (3.28)
    EΛh1Λh223EeMtk=0(Ck1˜Ck1)I(ξk,ξk+1](t)2+3EeMtk=0(Ck2˜Ck2)I(ξk,ξk+1](t)2+3Eˆh1(t)ˆh2(t)2. (3.29)

    Among them,

    Ek=0eMt(Ck1˜Ck1)I(ξk,ξk+1](t)2=Ek=0eMt[δk(C1˜C1)e2Mξ1δk(C2˜C2)+kn=1Δn,kbn(τn)[(h1(ξn)h2(ξn))(ˆh1(ξn)ˆh2(ξn))]]I(ξk,ξk+1](t)234EeMt(C1˜C1)2+34EeMte2Mξ1(C2˜C2)2+3EeMtk=0kn=1Δn,kbn(τn)[(h1(ξn)h2(ξn))(ˆh1(ξn)ˆh2(ξn))]I(ξk,ξk+1](t)2, (3.30)

    and we have

    A+(t)=˜A+(t), (3.31)
    A(t)=˜A(t). (3.32)
    EB(t)˜B(t)2=Ek=0kn=1Δn,kbn(τn)[(h1(ξn)h2(ξn))(ˆh1(ξn)ˆh2(ξn))]I(ξk,ξk+1](t)2suptJ[Eh1(t)h2(t)2+Eˆh1(t)ˆh2(t)2]Ek=0kn=1Δn,kbn(τn)I(ξk,ξk+1](t)2B21suptJ[Eh1(t)h2(t)2+Eˆh1(t)ˆh2(t)2], (3.33)
    EB+(t)˜B+(t)2=Ek=0kn=1Δ+n,kbn(τn)[(h1(ξn)h2(ξn))(ˆh1(ξn)ˆh2(ξn))]I(ξk,ξk+1](t)2suptJ[Eh1(t)h2(t)2+Eˆh1(t)ˆh2(t)2]Ek=0kn=1Δ+n,kbn(τn)I(ξk,ξk+1](t)2B22suptJ[Eh1(t)h2(t)2+Eˆh1(t)ˆh2(t)2], (3.34)

    and

    Eˆh1(t)ˆh2(t)22E(eMt2M2t0eMs[σ1(s)σ2(s)]ds2)+2E(eMt2M2t0eMs[σ1(s)σ2(s)]ds2)2E(eMt2M2t0eMs[f(s,h1(s),h1(s))f(s,h2(s),h2(s))+M(h1(s)h2(s))]ds2)+2E(eMt2M2t0eMs[f(s,h1(s),h1(s))f(s,h2(s),h2(s))+M(h1(s)h2(s))]ds2), (3.35)

    combing with (H5), we have

    Et0eMs[f(s,h1(s),h1(s))f(s,h2(s),h2(s))+M(h1(s)h2(s))]ds22t0KΨ(s,Eh1(s)h2(s)2,Eh1(s)h2(s)2)ds+2t0MEh1(s)h2(s)2ds. (3.36)

    From (3.31) and (3.32), we can easily know that |Q| is dependent with h(t). So, based on the above discussion, we can know when h1(t)h2(t)∥→0, ˆh1(t)ˆh2(t)0, C1˜C10 and C2˜C20. And then, we can get when h1(t)h2(t)∥→0, EΛh1(t)Λh2(t)20.

    Then

    Λh(t)=k=0(MCk1eMtMCk2eMt)I(ξk,ξk+1](t)+12eMtt0eMsσ(s)ds+12eMtt0eMsσ(s)ds, (3.37)

    hence,

    EΛh1(t)Λh2(t)24MEk=0(Ck1˜Ck1)eMtI(ξk,ξk+1](t)2+4MEk=0(Ck2˜Ck2)eMtI(ξk,ξk+1](t)2+2EeMtt0eMs[σ1(s)σ2(s)]ds2+2EeMtt0eMs[σ1(s)σ2(s)]ds2, (3.38)

    which implies that

    limδ0Λh1(t)Λh2(t)PC1=0. (3.39)

    Thus, we have proved that Λ is a continuous operator.

    Step(4). We prove that the functions in the set {uPC1(J,R)uΛ(D)} are uniformly bounded.

    Because uΛ(D), for any u{uPC1(J,R)uΛ(D)}, there exists h(t)D such that u=Λh(t),

    EΛh(t)23Ek=0Ck1eMtI(ξk,ξk+1](t)2+3Ek=0Ck2eMtI(ξk,ξk+1](t)2+3Eˆh(t)2, (3.40)

    suppose r1=1M, r2=e2MM and we have

    Eˆh(t)2r1t0Mr2Ef(s,h(s),h(s))+Mh(s)2ds+r2t0Mr1Ef(s,h(s),h(s))+Mh(s)2ds4Mr1r2t0[Ef(s,h(s),h(s))2+MEh(s)2]ds4Mr1r2t0[Θ(Eh(s)2)+MEh(s)2]ds, (3.41)

    so, if h(t) is the solution of the equation (1.4), we have

    Eh(t)23(m1+m2)+12Mr1r2t0Θ(Eh(s)2)+MEh(s)2ds. (3.42)

    Next, define ϕ(t)=Eh(t)2 and we have the inequality

    ϕ(t)3(m1+m2)+12Mr1r2t0Θ(ϕ(s))+Mϕ(s)ds. (3.43)

    Define the right of the inequality (3.43) as the function φ(t), we can get

    ϕ(t)φ(t),tJ, (3.44)

    so,

    φ(t)12Mr1r2[Θ(φ(t))+Mφ(t)], (3.45)
    lnφ(t)lnφ(0)12Mr1r2t0Θ(φ(s))φ(s)+Mds12Mr1r2[10Θ(ϕ(s))ϕ(s)ds+M], (3.46)

    then we can easily get

    lnϕ(t)12Mr1r2[10Θ(ϕ(s))ϕ(s)ds+M]+ln(3m1+3m2)P10Θ(ϕ(t))ϕ(t)dt+P. (3.47)

    For Δn,k, Δ+n,k, δk and δk are bounded, combining with

    EB(t)2Ek=0kn=1Δn,kbn(τn)[h(ξn)ˆh(ξn)]I(ξk,ξk+1](t)2suptJ[Eh(t)2+Eˆh(t)2]Ek=0kn=1Δn,kbn(τn)I(ξk,ξk+1](t)2, (3.48)

    and consider that Θ(s) satisfies the condition (H4) and D=[v0(t),ω0(t)], where v0(t), ω0(t) are all square integrable, so, EΛh(t)2 is bounded.

    Then,

    ˆh(t)=eMt2t0eMsσ(s)ds+eMt2t0eMsσ(s)ds, (3.49)

    and

    EΛh(t)24Ek=0MCk1eMtI(ξk,ξk+1](t)2+4Ek=0MCk2eMtI(ξk,ξk+1](t)2+2EeMtt0eMsσ(s)ds2+2EeMtt0eMsσ(s)ds2. (3.50)

    Using the same way, we can prove that EΛh(t)2 is bounded. Thus, we have proved that the functions in the set {u(t)C2(J,R)u(t)Λ(D)} are uniformly bounded.

    Step(5). We prove that the set {u(t)u(t)=Λh(t)} is equicontinuous.

    For every u(t){u(t)u(t)=Λh(t)}, and every t1,t2J, t1t2<δ,

    u(t)=k=0[Ck1eMt+Ck2eMt+ˆh(t)]I(ξk,ξk+1](t), (3.51)

    and,

    Eu(t1)u(t2)23Ek=0Ck1[eMt1I(ξk,ξk+1](t1)eMt2I(ξk,ξk+1](t2)]2+3Ek=0Ck2[eMt1I(ξk,ξk+1](t1)eMt2I(ξk,ξk+1](t2)]2+3Eˆh(t1)ˆh(t2)23Ek=0Ck1[eMt1I(ξk,ξk+1](t1)eMt2I(ξk,ξk+1](t2)]2+3Ek=0Ck2[eMt1I(ξk,ξk+1](t1)eMt2I(ξk,ξk+1](t2)]2+6E(eMt2M2t2t1eMsσ(s)ds2)+6E(eMt2M2t2t1eMsσ(s)ds2). (3.52)

    We suppose t1(ξk1,ξk1+1), t2(ξk2,ξk2+1), so when t1t2<δ, ξk1+1ξk2<δ

    Ek=0Ck1[eMt1I(ξk,ξk+1](t1)eMt2I(ξk,ξk+1](t2)]2=ECk11eMt1Ck21eMt22. (3.53)

    So, it is easy to see that when δ0, Eu(t1)u(t2)20. Then, we consider

    u(t)=k=0(MCk1eMtMCk2eMt)I(ξk,ξk+1](t1)+12eMtt0eMsσ(s)ds+12eMtt0eMsσ(s)ds. (3.54)

    Therefore,

    Eu(t1)u(t2)24Ek=0MCk1[eMt1I(ξk,ξk+1](t1)eMt2I(ξk,ξk+1](t2)]2+4Ek=0MCk2[eMt1I(ξk,ξk+1](t1)eMt2I(ξk,ξk+1](t2)]2+2E(eMt2t2t1eMsσ(s)ds2)+2E(eMt2t2t1eMsσ(s)ds2). (3.55)

    Using the same method, we can prove that when δ0, Eu(t1)u(t2)20. We have already proved that when t1t20, Eu(t1)u(t2)20. So

    u(t1)u(t2)PC10. (3.56)

    That is to say, the set {u(t)u(t)=Λh(t)} is equicontinuous.

    Using Lemma (2.5), we know that the set {u(t)u(t)=Λh(t)} is a column compact set. It follows from Lemma (2.6) that the equation (1.4) has a solution in D=[v0(t),ω0(t)], where t[0,1]. Thus, theorem (3.1) is established.

    The main result could have many applications, now, we give an example to illustrate this theorem. We consider the following second order random impulsive differential equation with boundary value problems.

    {x(t)=(x(t)sin(t)+t)3,tJ,x(ξ+k)=k3kτkx(ξk),k=1,2,,x(0)2x(0)=1,2x(1)+x(1)=1. (4.1)

    Let τkU(0,12k), then the probability density function of τk is

    p(x)={2kx(0,12k),0x(0,12k). (4.2)

    Set ξ0=0, ξk+1=ξk+τk+1. Obviously, {ξk} is a process with independent increments and the impulsive moments ξk form a strictly increasing sequence. And for every kN,

    ξk<ξk+112+122++12k+1<1. (4.3)

    So in this example, bk(τk)=k3kτk, τk is a random variable defined from Ω to Ek=(0,dk)=(0,12k). Suppose τi and τj are independent of each other when ij, x(ξ+k)=limtξk+0x(t) and x(ξk)=limtξk0x(t).

    Taking v0(t)=0, ω0(t)=6cost, we can easily prove that v0(t) is the lower solution and ω0(t) is the upper solution of the equation (4.1). And for every v0(t)x2(t)<x1(t)ω0(t), we have

    [x1(t)sin(t)+t]3[x2(t)sin(t)+t]3=sint[x1(t)x2(t)][(x1(t)sin(t)+t)2+(x1(t)sin(t)+t)(x2(t)sin(t)+t)+(x2(t)sin(t)+t)2]3[x1(t)x2(t)][ω0(t)+1]2147[x1(t)x2(t)]. (4.4)

    So, we can easily know that M=147.

    kn=1Δn,kbn(τn)kn=12n1n3nτn=3[1(23)k]k(23)k+1. (4.5)

    So, we have proved that k=0{kn=1Δn,kbn(τn)}I(ξk,ξk+1](t)3. That is to say B1=3. In the same way we can prove that B2=3e3147.

    For every v0(t)x2(t)<x1(t)ω0(t), we have

    E[x(t)sint+t]32Eω0(t)3+3ω0(t)2+3ω0(t)+12<, (4.6)

    and

    E[x1(t)sint+t]3[x2(t)sint+t]32E[(x1(t)x2(t)][(x1(t)+1)2+(x1(t)+1)(x2(t)+1)+(x2(t)+1)2]2. (4.7)

    So, the equation (4.1) meets all the conditions of the theorem (3.1). We can get the solution of the equation of the equation (4.1) between v0(t)=0 and ω0(t)=6cost by constructing iterative sequences starting from v0 an ω0 respectively.

    In this article, we study the existence of upper and lower solutions of second order random impulse equation (1.4). First, we study the solution form of the corresponding linear impulsive system (2.3) induced by system (1.4). Based on the form of the solution, we define the solution operator. Secondly, we prove that the fixed point of this operator is the solution of equation (1.4). Finally, we construct two monotone iterative sequences by the solution to (2.3). We then prove that they converge. Thus, it is concluded that there exists upper and lower solution to system (1.4). Impulsive differential equations have been studied in literature [7,8,9,10]. Random impulsive differential equations have also been discussed in the literature [12,13,14,19,27,39]. In this paper, we extend the form of solutions to initial value problems of random impulsive differential equations to more general boundary value problems. The upper and lower methods are applied to Random impulsive differential equations and the related conclusions are generalized.

    The authors would like to thank the editor and the reviewers for their helpful comments and suggestions.

    No potential conflict of interest was reported by the authors.



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