AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 134691 10.1155/2013/134691 134691 Research Article Existence Result for Impulsive Differential Equations with Integral Boundary Conditions Ning Peipei Huan Qian Ding Wei Xia Yonghui Department of Mathematics Shanghai Normal University Shanghai 200234 China shnu.edu.cn 2013 2 4 2013 2013 05 12 2012 05 01 2013 2013 Copyright © 2013 Peipei Ning et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We investigate the following differential equations: -(y(x))'+q(x)y(x)=λf(x,y(x)), with impulsive and integral boundary conditions -Δ(y(xi))=Ii(y(xi)), i=1,2,,m, y(0)-ay(0)=0ωg0(s)y(s)ds, y(ω)-by(ω)=0ωg1(s)y(s)ds, where y(x)=p(x)y'(x). The expression of Green's function and the existence of positive solution for the system are obtained. Upper and lower bounds for positive solutions are also given. When p(t), I(·), g0(s), and g1(s) take different values, the system can be simplified to some forms which has been studied in the works by Guo and LakshmiKantham (1988), Guo et al. (1995), Boucherif (2009), He et al. (2011), and Atici and Guseinov (2001). Our discussion is based on the fixed point index theory in cones.

1. Introduction

The theory of impulsive differential equations in abstract spaces has become a new important branch and has developed rapidly (see ). As an important aspect, impulsive differential equations with boundary value problems have gained more attention. In recent years, experiments in a variety of different areas (especially in applied mathematics and physics) show that integral boundary conditions can represent the model more accurately. And researchers have obtained many good results in this field.

In this paper, we study the existence of positive solutions for the following system: (1)-(y(x))+q(x)y(x)=λf(x,y(x)),xxi,xJ-,-Δ(y(xi))=Ii(y(xi)),i=1,2,,m,y(0)-ay(0)=0ωg0(s)y(s)ds,y(ω)-by(ω)=0ωg1(s)y(s)ds, where y(x)=p(x)y(x), J-=J{x1,x2,,xm}, J=[0,ω], 0<x1<x2<<xm<ω, fC(J×R+,R+). y(x), y(x) are left continuous at x=xi, Δ(y(xi))=y(xi+)-y(xi-). IiC(R+,R+). And a>0, b<0, g0,g1:[0,1][0,) are continuous and positive functions.

When p(t), I(·), g0(s), and  g1(s) take different values, the system can be simplified to some forms which have been studied. For example,  discussed the existence of positive solution in case p(t)=1.

Let p(t)=1, g0,g1=0, [11, 12] investigated the system with only one impulse. Reference  studied the system when I(·)=0, g0,g1=0. Readers can read the papers in  for details.

Throughout the rest of the paper, we assume ω is a fixed positive number, and λ is a parameter. p(x), q(x) are real-valued measurable functions defined on J, and they satisfy the following condition:

(H1) p(x)>0, q(x)0, q(x)0 almost everywhere, and (2)0ω1p(x)dx<,0ωq(x)dx<.

This paper aims to obtain the positive solution for (1). In Section 2, we introduce some lemmas and notations. In particular, the expression and some properties of Green's functions are investigated. After the preparatory work, we draw the main results in Section 3.

2. Preliminaries Theorem 1 (Krasnoselskii's fixed point theorem).

Let E be a Banach space and CE. Assume Ω1, Ω2 are open sets in E with 0Ω1Ω-1Ω2, and S:C(Ω-2Ω1)C be a completely continuous operator such that either

s(y)y, yCΩ1, and s(y)y, yCΩ2; or

s(y)y, yCΩ1, and s(y)y, yCΩ2.

Then S has a fixed point in C(Ω-2Ω1).

Definition 2.

For two differential functions y and z, we defined their Wronskian by (3)Wx(y,z)=y(x)z(x)-y(x)z(x)=p(x)[y(x)z(x)-y(x)z(x)].

Consider the linear nonhomogeneous problem of the form (4)-(y(x))+q(x)y(x)=h(x),xJ. Its corresponding homogeneous equation is (5)-(y(x))+q(x)y(x)=0,xJ.

Lemma 3.

Suppose that y1 and y2 form a fundamental set of solutions for the homogeneous problem (5). Then the general solution of the nonhomogeneous problem (4) is given by (6)y(x)=c1y1(x)+c2y2(x)+0xy1(x)y2(s)-y1(s)y2(x)ws(y1,y2)h(s)ds, where c1 and c2 are arbitrary constants.

Proof.

We just need to show that the function (7)z(x)=0xy1(x)y2(s)-y1(s)y2(x)ws(y1,y2)h(s)ds is a particular solution of (4). From (7), we have for x[0,ω], (8)z(x)=0xy1(x)y2(s)-y1(s)y2(x)ws(y1,y2)h(s)ds,(9)[p(x)z(x)]=-h(x)+q(x)z(x). Besides, from (7) and (8), we have (10)z(0)=0,z(0)=0. Thus, z(x) satisfies (4).

Consider the following boundary value problem with integral boundary conditions: (11)-(y(x))+q(x)y(x)=h(x),xJ,y(0)-ay(0)=0ωg0(s)σ0(s)ds,y(ω)-by(ω)=0ωg1(s)σ1(s)ds. Denote by u(x) and v(x) the solutions of the homogenous equation (5) satisfying the initial conditions (12)u(0)=a,u(0)=1,v(ω)=-b,v(ω)=-1.

(H2) Let x,sJ, denote a function (13)ϕ(x,s)=u(x)u(ω)-bu(ω)g1(s)+v(x)v(0)-av(0)g0(s) satisfies 0ϕ(x,s)<1/ω.

For convenience, we denote m=min{ϕ(x,s);x,sJ}, M=max{ϕ(x,s);x,sJ}.

Lemma 4.

Let K(x,s) be a nonnegative continuous function defined for -<x1x, sx2< and ψ(x) a nonnegative integrable function on [x1,x2]. Then for arbitrary nonnegative continuous function φ(x) defined on [x1,x2], the Volterra integral equation (14)y(x)=φ(x)+x1xK(x,s)ψ(s)y(s)ds,x1xx2 has a unique solution y(x). Moreover, this solution is continuous and satisfied the inequality (15)y(x)φ(x),      x1xx2.

Proof.

We solve (14) by the method of successive approximations setting (16)y0(x)=φ(x),yn=x1xK(x,s)ψ(s)yn-1(s)ds,n=1,2,. If the series n=0yn(x) converges uniformly with respect to x[x1,x2], then its sum will be, obviously, a continuous solution of (14). To prove the uniform convergence of this series, we put (17)maxx1xx2φ(x)=c,maxx1x,sx2K(x,s)=c1. Then it is easy to get from (16) that (18)0yn(x)cc1nn![x1xψ(s)ds]n,n=0,1,2,. Hence it follows that (14) has a continuous solution (19)y(x)=n=0yn(x) and because y0=φ(x), yn0, n=1,2,, for this solution the inequality (15) holds. Uniqueness of the solution of (14) can be proved in a usual way. The proof is complete.

Remark 5.

Evidently, the statement of Lemma 4 is also valid for the Volterra equation of the form (20)y(x)=φ(x)+xx2K(x,s)ψ(s)y(s)ds,x1xx2.

Lemma 6.

For the solution y(x) of the BVP (11), the formula (21)y(x)=w(x)+0ωG(x,s)h(s)ds,xJ holds, where (22)w(x)=u(x)u(ω)-bu(ω)0ωg1(s)σ1(s)ds+v(x)v(0)-av(0)0ωg0(s)σ0(s)ds,G(x,s)=-1ws(u,v){u(s)v(x),0sxω,u(x)v(s),0xsω.

Proof.

By Lemma 3, the general solutions of the nonhomogeneous problem (4) has the form (23)y(x)=c1u(x)+c2v(x)+0xu(x)v(s)-u(s)v(x)Ws(u,v)h(s)ds, where c1 and c2 are arbitrary constants. Now we try to choose the constants c1 and c2 so that the function y(x) satisfies the boundary conditions of (11).

From (23), we have (24)y(x)=c1u(x)+c2v(x)+0xu(x)v(s)-u(s)v(x)Ws(u,v)h(s)ds. Consequently, (25)y(0)=c1a+c2v(0),y(0)=c1+c2v(0). Substituting these values of y(0) and y(0) into the first boundary condition of (11), we find (26)c2=1v(0)-av(0)0ωg0(s)σ0(s)ds. Similarly from the second boundary condition of (11), we can find (27)c1=1u(ω)-bu(ω)0ωg1(s)σ1(s)ds-0ωv(s)Ws(u,v)h(s)ds. Putting these values of c1 and c2 in (23), we get the formula (21), (22).

Lemma 7.

Let condition (H1) hold. Then for the Wronskian of solution u(x) and v(x), the inequality Wx(u,v)<0, xJ holds.

Proof.

Using the initial conditions (12), we can deduce from (5) for u(x) and v(x) the following equations: (28)u(x)=1+0xq(s)u(s)ds,u(x)=a+0x1p(t)dt+0x[sxdtp(t)]q(s)u(s)ds,v(x)=-1-xωq(s)v(s)ds,v(x)=-b+xω1p(t)dt+xω[xsdtp(t)]q(s)v(s)ds. From (28), by condition (H1) and Lemma 4, it follows that (29)u(x)a+0xdtp(t)>0,u(x)1>0,v(x)-b+xωdtp(t)>0,v(x)-1<0. Now from (3), we get Wx(u,v)<0, xJ. The proof is complete.

From (21), (22), and Lemma 7, the following lemma follows.

Lemma 8.

Under condition (H1) the Green's function G(x,s) of the BVP (11) is positive. That is, G(x,s)>0 for x,sJ.

Let C(J) denote the Banach of all continuous functions y:I equipped with the form y=max{|y(x)|;xJ}, for any yC(J). Denote P={yC(J);y(x)0,yJ}, then P is a positive cone in C(J).

Let us set A=max0x,sωG(x,s), B=min0x,sωG(x,s), and by Lemma 8, obviously, A>B>0, x,sJ.

Define a mapping Φ in Banach space C(J) by (30)(Φy)(x)=w(x)+λ0ωG(x,s)f(s,y(s))ds+i=0mG(x,xi)Ii(y(xi)),xJ, where (31)w(x)=u(x)u(ω)-bu(ω)0ωg1(s)y(s)ds+v(x)v(0)-av(0)0ωg0(s)y(s)ds.

Lemma 9.

The fixed point of the mapping Φ is a solution of (1).

Proof.

Clearly, Φy is continuous in x for xJ. For xxk, (32)(Φy)(x)=w(x)+λ0ωGxf(s,y(s))ds+i=0mG(x,xi)xIi(y(xi)), where (33)w(x)=u(x)u(ω)-bu(ω)0ωg1(s)y(s)ds+v(x)v(0)-av(0)0ωg0(s)y(s)ds. We have (34)(Φy)(x)=w(x)+λ0ωp(x)Gxf(s,y(s))ds+i=0mp(x)G(x,xi)xIi(y(xi)), where (35)w(x)=u(x)u(ω)-bu(ω)0ωg1(s)y(s)ds+v(x)v(0)-av(0)0ωg0(s)y(s)ds. We can easy get that (36)(Φy)(0)-a(Φy)(0)=0ωg0(s)y(s)ds,(Φy)(ω)-b(Φy)(ω)=0ωg1(s)y(s)ds,Δ(Φy)(xk)=p(xk+)(Φy)(xk+)-p(xk-)(Φy)(xk-)=p(xk)[-u(xk)v(xk)Wxk(u,v)+u(xk)v(xk)Wtk(u,v)]×Ik(y(xk))=-Ik(y(xk)),(p(x)(Φy)(x))=[i=0mp(x)p(x)w(x)m+λ0ωp(x)Gxf(s,y(s))dsm+i=0mp(x)G(x,xi)xIi(y(xi))]=q(x)w(x)+λq(x)×0ωG(x,s)f(s,y(s))ds-λf(x,y(x))+q(x)×i=0mG(x,xi)Ii(y(xi))=q(x)(Φy)(x)-λf(x,y(x)), which implies that the fixed poind of Φ is a solution of (1). The proof is complete.

Lemma 10.

Let P0={yP;minxJy(x)((1-Mω)B/(1-mω)A)y}, then P0 is a cone.

Proof.

(i) For for all y1,y2P0 and for all α0, β0, we have (37)min(αy1)α·(1-Mω)B(1-mω)Ay1,min(βy2)β·(1-Mω)B(1-mω)Ay2. Moreover (38)min(αy1+βy2)(1-Mω)B(1-mω)A(αy1+βy2)(1-Mω)B(1-mω)Aαy1+βy2. Thus αy1+βy2P0.

(ii) If yP0 and -yP0, we have (39)minxJ(y(x))(1-Mω)B(1-mω)Ay,minxJ(-y(x))(1-Mω)B(1-mω)Ay. It implies that y=0. Hence P0 is a cone.

Defined a linear operator A:C(J)C(J) by (40)(Ay)(x)=0ωϕ(x,s)y(s)ds. Then we have the following lemma.

Lemma 11.

If (H2) is satisfied, then

A is a bounded linear operator, A(P)P;

(I-A) is invertible;

(I-A)-11/(1-Mω).

Proof.

(i) (41)A(αy1(x)+βy2(x))=0ωϕ(x,s)[αy1(s)+βy2(s)]ds=α(Ay1)(x)+β(Ay2)(x), for all α,β, y1,y2C(J).

Using ϕ(x,s)M, it is easy to see that |(Ay)(t)|Mωy.

Let yP. Then y(s)0 for all sJ. Since ϕ(t,s)m0, it follows that (Ay)(x)0 for each xJ. So A(P)P.

(ii) We want to show that (I-A) is invertible, or equivalently 1 is not an eigenvalue of A.

Since M<1/ω, it follows from condition (H2) that AyMωy<y.

So (42)A=supy0AyyMω<1. On the other hand, we suppose 1 is an eigenvalue of A, then there exists a yC(J) such that Ay=y. Moreover, we can obtain that Ay/y=1. So A1. Thus this assumption is false.

Conversely, 1 is not an eigenvalue of A. Equivalently, (I-A) is invertible.

(iii) We use the theory of Fredholm integral equations to find the expression for (I-A)-1.

Obviously, for each xJ, y(x)=(I-A)-1z(x)y(x)=z(x)+(Ay)(x).

By (40), we can get (43)y(x)=z(x)+0ωϕ(x,s)y(s)ds. The condition M<1/ω implies that 1 is not an eigenvalue of the kernel ϕ(x,s). So (43) has a unique continuous solution y for every continuous function z.

By successive substitutions in (43), we obtain (44)y(x)=z(x)+0ωR(x,s)z(s)ds, where the resolvent kernel R(x,s) is given by (45)R(x,s)=j=1ϕj(x,s). Here ϕj(x,s)=0ωϕ(x,τ)ϕj-1(τ,s)ds, j=2, and ϕ1(x,s)=ϕ(x,s).

The series on the right in (45) is convergent because |ϕ(x,s)|M<1/ω.

It can be easily verified that R(x,s)M/(1-Mω).

So we can get (46)(I-A)-1z(x)=z(x)+0ωR(x,s)z(s)ds. Therefore (47)(I-A)-1z(x)z(x)+M1-Mω0ωz(s)dsz(1+Mω1-Mω)=11-Mωz. So (48)(I-A)-1zz11-Mω. Thus (I-A)-11/(1-Mω). This completes the proof of the lemma.

Remark 12.

Since ϕ(x,s)m for each (x,s)J, it is easy to prove that R(x,s)m/(1-mω).

3. Main Results

Consider the following boundary value problem (BVP) with impulses: (49)-(y(x))+q(x)y(x)=λf(x,y(x)),mmmmmmmmmmmmmxxi,xJ,-Δ(y(xi))=Ii(y(xi)),i=1,2,,m,y(0)-ay(0)=0ωg0(s)y(s)ds,y(ω)-by(ω)=0ωg1(s)y(s)ds. Denote a nonlinear operator T:PC(J)PC(J) by (50)(Ty)(x)=λ0ωG(x,s)f(s,y(s))ds+i=0mG(x,xi)Ii(y(xi)). It is easy to see that solutions of (49) are solutions of the following equation: (51)y(x)=Ty(x)+Ay(x),xJ-1. According to Lemma 11, y is a solution of (51) if and only if it is a solution of (52)y(x)=(I-A)-1Ty(x). It follows from (46) that y is a solution of (52) if and only if (53)y(x)=(Ty)(x)+0ωR(x,s)(Ty)(s)ds. So, the operator Φ can be written as (54)(Φy)(x)=(Ty)(x)+0ωR(x,s)(Ty)(s)ds. It satisfies the conditions of Theorem 1 with E=C(J) and the cone C=P0.

Let us list some marks and conditions for convenience.

The nonlinearity f:J×[0,)[0,) is continuous and satisfies the following.

(H3) There exist L1>0 and α(x)P, r1 with r1i=0mIi(y(xi))/λ0ωα(s)ds  such that (55)f(x,y)α(x)[y(1-Mω)-r1] for all y(0,L1], xJ.

(H4) There exist L2>L1 and β(x)P, p1 with p1i=0mIi(y(xi))/λ0ωβ(s)ds such that (56)f(x,y)β(x)[y(1-mω)-p1] for all y(L2,], xJ.

Then, we can get the following theorem.

Theorem 13.

Assume (H1), (H2), (H3), and (H4) are satisfied. And (57)(1-mω)A20ωα(s)ds(1-Mω)B20ωβ(s)ds, then, if λ satisfies (58)(1-mω)A(1-Mω)B20ωβ(s)dsλ1A0ωα(s)ds. The problem (49) has at least one positive solution.

Proof.

First of all, we show that operator Φ is defined by (54) maps P0 into itself. Let yP0.

Then (Φy)(x)0 for all that tJ-1, and (59)(Φy)(x)λA1-Mω0ωf(s,y(s))ds+A1-Mωi=0mIi(y(xi)). Because from the formula (54), we have (60)(Φy)(x)=(Ty)(x)+0ωR(x,s)(Ty)(s)ds=λ0ωG(x,s)f(s,y(s))ds+i=0mG(x,xi)Ii(y(xi))+λ0ωR(x,s)0ωG(x,τ)f(τ,y(τ))dτds+0ωR(x,s)i=0mG(x,xi)Ii(y(xi))dsλ(1+Mω1-Mω)0ωG(x,s)f(s,y(s))ds+i=0mG(x,xi)Ii(y(xi))+Mω1-Mωi=0mG(x,xi)Ii(y(xi))λA1-Mω0ωf(s,y(s))ds+A1-Mωi=0mIi(y(xi)). Hence, inequality (59) is established.

This implies that (61)ΦyλA1-Mω0ωf(s,y(s))ds+A1-Mωi=0mIi(y(xi)), or equivalently (62)0ωf(s,y(s))ds1-MωλAΦy-1λi=0mIi(y(xi)). On the other hand, it follows that (63)(Φy)(x)λB1-mω0ωf(s,y(s))ds+B1-mωi=0mIi(y(xi)). In fact, we have (64)(Φy)(x)=λ0ωG(x,s)f(s,y(s))ds+i=0mG(x,xi)Ii(y(xi))+λ0ωR(x,s)0ωG(x,τ)f(τ,y(τ))dτds+0ωR(x,s)i=0mG(x,xi)Ii(y(xi))dsλ(1+mω1-mω)0ωG(x,s)f(s,y(s))ds+i=0mG(x,xi)Ii(y(xi))+mω1-mωi=0mG(x,xi)Ii(y(xi))λB1-mω0ωf(s,y(s))ds+B1-mωi=0mIi(y(xi)). It follows from (62) that (65)(Φy)(x)λB1-mω·[1-MωλAΦy-1λi=0mIi(y(xi))]+B1-mωi=0mIi(y(xi))=(1-Mω)B(1-mω)AΦy-B1-mωi=0mIi(y(xi))+B1-mωi=0mIi(y(xi))=(1-Mω)B(1-mω)AΦy. So, we get (66)(Φy)(x)(1-Mω)B(1-mω)AΦy. This show that ΦyP0.

It is easy to see that Φ is the complete continuity.

We now proceed with the construction of the open sets Ω1 and Ω2.

First, let yP0 with y=L1. Inequality (59) implies (67)(Φy)(x)λA1-Mω0ωf(s,y(s))ds+A1-Mωi=0mIi(y(xi))λA1-Mω0ωα(s)[y(s)(1-Mω)-r1]ds+A1-Mωi=0mIi(y(xi))=λA0ωα(s)y(s)ds-λA1-Mωr1×0ωα(s)ds+A1-Mωi=0mIi(y(xi))=λA0ωα(s)y(s)ds+A1-Mω×[i=0mIi(y(xi))-λr10ωα(s)ds]. By condition (H3) and (58), we obtain (68)i=0mIi(y(xi))-λr10ωα(s)ds0,λA0ωα(s)ds1. So (69)(Φy)(x)λA0ωα(s)dsyy. Consequently, Φyy.

Let Ω1={yC(J);y<L1}. Then, we have Φyy for yP0Ω1.

Next, let L2~=max{2L1,((1-mω)A/(1-Mω)B)L2} and set Ω2={yC(J);y<L2~}.

For yP0 with y=L2~, we have (70)minxJy(x)(1-Mω)B(1-mω)Ay=(1-Mω)B(1-mω)AL2~(1-Mω)B(1-mω)A·(1-mω)A(1-Mω)BL2=L2. It follows from (63) that (71)(Φy)(x)λB1-mω0ωf(s,y(s))ds+B1-mωi=0mIi(y(xi))λB1-mω0ωβ(s)[y(s)(1-mω)-p1]ds+B1-mωi=0mIi(y(xi))=λB0ωβ(s)y(s)ds+B1-mω×(i=0mIi(y(xi))-λp10ωβ(s)ds). By condition (H4) and (58), we obtain (72)i=0mIi(y(xi))-λp10ωβ(s)ds0,λB(1-mω)A(1-Mω)B0ωβ(s)ds. Since yP0 we have y(x)((1-Mω)B/(1-mω)A)y for all xJ. It follows from the above inequality that (73)(Φy)(x)(1-mω)A(1-Mω)B0ωβ(s)ds0ωβ(s)ds·(1-Mω)B(1-mω)Ay=y. Hence Φyy for yP0Ω2.

It follows from (i) of Theorem 1 that Φ has a fixed point in P0(Ω-2Ω1), and this fixed point is a solution of (49).

This completes the proof.

Next, with L1 and L2 as above, we assume that f satisfied the following.

(H5) There exist α*(x)P, r1* with r1*i=0mIi(y(xi))/λ0ωα*(s)ds such that (74)f(x,y)α*(x)[y(1-mω)-r1*] for all y(0,L1], xJ.

(H6) There exist β*(x)P, p1* with p1*i=0mIi(y(xi))/λ0ωβ*(s)ds such that (75)f(x,y)β*(x)[y(1-Mω)-p1*] for all y(L2,], xJ.

Theorem 14.

Assume (H1), (H2), (H5), and (H6) are satisfied. And (76)(1-mω)A20ωβ*(s)ds(1-Mω)B20ωα*(s)ds, then, if λ satisfies (77)(1-mω)A(1-Mω)B20ωα*(s)dsλ1A0ωβ*(s)ds. The problem (49) has at least one positive solution.

Proof.

Let Φ be a completely continuous operator defined by (54). Then Φ maps the cone P0 into itself.

First, let yP0 with y=L1. Inequality (63) implies (78)(Φy)(x)λB1-mω0ωf(s,y(s))ds+B1-mωi=0mIi(y(xi))λB1-mω0ωα*(s)[y(s)(1-mω)-r1*]ds+B1-mωi=0mIi(y(xi))=λB0ωα*(s)y(s)ds+B1-mω×(i=0mIi(y(xi))-λr1*0ωα*(s)ds). By condition (H5) and (77), we obtain (79)i=0mIi(y(xi))-λr1*0ωα*(s)ds0,λB(1-mω)A(1-Mω)B0ωα*(s)ds. Hence (80)(Φy)(x)(1-mω)A(1-Mω)B0ωα*(s)ds0ωα*(s)y(s)ds. Since yP0, we have y(x)((1-Mω)B/(1-mω)A)y for all xJ. It follows from the above inequality that (81)(Φy)(x)(1-mω)A(1-Mω)B0ωα*(s)ds0ωα*(s)ds·(1-Mω)B(1-mω)Ay=y. Let Ω1={yC(J);y<L1}. Then, we have Φyy for yP0Ω1.

Next, let L2~=max{2L1,((1-mω)A/(1-Mω)B)L2} and set Ω2={yC(J);y<L2~}.

Then for yP0 with y=L2~ for all xJ, we have minxJy(x)L2. Inequality (59) implies (82)(Φy)(x)λA1-Mω0ωf(s,y(s))ds+A1-Mωi=0mIi(y(xi))λA1-Mω0ωβ*(s)[y(s)(1-Mω)-p1*]ds+A1-Mωi=0mIi(y(xi))=λA0ωβ*(s)y(s)ds+A1-Mω[i=0mIi(y(xi))-λp1*0ωβ*(s)ds]. By condition (H6) and (77), we obtain (83)i=0mIi(y(xi))-λp1*0ωβ*(s)ds0,λA0ωβ*(s)ds1. So (84)(Φy)(x)λA0ωβ*(s)dsyy1. Therefore Φyy with y=L2~.

Then, we have Φyy for yP0Ω2.

We see the case (ii) of Theorem 1 is met. It follows that Φ has a fixed point in P0(Ω-2Ω1), and this fixed point is a solution of (49).

This completes the proof.

Acknowledgments

This work was supported by the NNSF of China under Grant no. 11271261, Natural Science Foundation of Shanghai (no. 12ZR1421600), Shanghai Municipal Education Commission (no. 10YZ74), the Slovenian Research Agency, and a Marie Curie International Research Staff Exchange Scheme Fellowship within the 7th European Community Programme (FP7-PEOPLE-2012-IRSES-316338).

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