# Existence Result for Impulsive Differential Equations with Integral Boundary Conditions

Authors: Peipei Ning; Qian Huan; Wei Ding
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We investigate the following differential equations:-(y[1](x))'+q(x)y(x)=λf(x,y(x)), with impulsive and integral boundary conditions -Δ(y[1](xi))=Ii(y(xi)), i=1,2,…,m, y(0)-ay[1](0)=∫0ωg0(s)y(s)ds, y(ω)-by[1](ω)=∫0ωg1(s)y(s)ds, where y[1](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.


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## Body

## 1. Introduction

The theory of impulsive differential equations in abstract spaces has become a new important branch and has developed rapidly (see [1–4]). 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[1](x))′+q(x)y(x)=λf(x,y(x)),x≠xi,x∈J-,-Δ(y[1](xi))=Ii(y(xi)),i=1,2,…,m,y(0)-ay[1](0)=∫0ωg0(s)y(s)ds,y(ω)-by[1](ω)=∫0ωg1(s)y(s)ds,
where y[1](x)=p(x)y′(x), J-=J∖{x1,x2,…,xm}, J=[0,ω], 0<x1<x2<⋯<xm<ω, f∈C(J×R+,R+). y(x), y[1](x) are left continuous at x=xi, Δ(y[1](xi))=y[1](xi+)-y[1](xi-). Ii∈C(R+,R+). And a>0, b<0, g0,g1:[0,1]→[0,∞) are continuous and positive functions.Whenp(t), I(·), g0(s), and  g1(s) take different values, the system can be simplified to some forms which have been studied. For example, [5–10] discussed the existence of positive solution in case p(t)=1.Letp(t)=1, g0,g1=0, [11, 12] investigated the system with only one impulse. Reference [13] studied the system when I(·)=0, g0,g1=0. Readers can read the papers in [13] for details.Throughout the rest of the paper, we assumeω 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).
LetE be a Banach space and C∈E. Assume Ω1, Ω2 are open sets in E with 0∈Ω1⊂Ω-1⊂Ω2, and S:C⋂(Ω-2∖Ω1)→C be a completely continuous operator such that either(i)
∥s(y)∥≤∥y∥, y∈C∩∂Ω1, and ∥s(y)∥≥∥y∥, y∈C∩∂Ω2; or(ii)
∥s(y)∥≥∥y∥, y∈C∩∂Ω1, and ∥s(y)∥≤∥y∥, y∈C∩∂Ω2.Then S has a fixed point in C⋂(Ω-2∖Ω1).Definition 2.
For two differential functionsy and z, we defined their Wronskian by
(3)Wx(y,z)=y(x)z[1](x)-y[1](x)z(x)=p(x)[y(x)z′(x)-y′(x)z(x)].
Consider the linear nonhomogeneous problem of the form(4)-(y[1](x))′+q(x)y(x)=h(x),x∈J.
Its corresponding homogeneous equation is
(5)-(y[1](x))′+q(x)y(x)=0,x∈J.Lemma 3.
Suppose thaty1 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[1](0)=0.
Thus, z(x) satisfies (4).Consider the following boundary value problem with integral boundary conditions:(11)-(y[1](x))′+q(x)y(x)=h(x),x∈J,y(0)-ay[1](0)=∫0ωg0(s)σ0(s)ds,y(ω)-by[1](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[1](0)=1,v(ω)=-b,v[1](ω)=-1.(H2) Letx,s∈J, denote a function
(13)ϕ(x,s)=u(x)u(ω)-bu[1](ω)g1(s)+v(x)v(0)-av[1](0)g0(s)
satisfies 0≤ϕ(x,s)<1/ω.For convenience, we denotem∶=min{ϕ(x,s);x,s∈J}, M∶=max{ϕ(x,s);x,s∈J}.Lemma 4.
LetK(x,s) be a nonnegative continuous function defined for -∞<x1≤x, s≤x2<∞ 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,x1≤x≤x2
has a unique solution y(x). Moreover, this solution is continuous and satisfied the inequality
(15)y(x)≥φ(x),x1≤x≤x2.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=0∞‍yn(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)maxx1⩽x⩽x2φ(x)=c,maxx1⩽x,s⩽x2K(x,s)=c1.
Then it is easy to get from (16) that
(18)0⩽yn(x)⩽cc1nn![∫x1xψ(s)ds]n,n=0,1,2,….
Hence it follows that (14) has a continuous solution
(19)y(x)=∑n=0∞yn(x)
and because y0=φ(x), yn≥0, 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 Lemma4 is also valid for the Volterra equation of the form
(20)y(x)=φ(x)+∫xx2K(x,s)ψ(s)y(s)ds,x1≤x≤x2.Lemma 6.
For the solutiony(x) of the BVP (11), the formula
(21)y(x)=w(x)+∫0ωG(x,s)h(s)ds,x∈J
holds, where
(22)w(x)=u(x)u(ω)-bu[1](ω)∫0ωg1(s)σ1(s)ds+v(x)v(0)-av[1](0)∫0ωg0(s)σ0(s)ds,G(x,s)=-1ws(u,v){u(s)v(x),0≤s≤x≤ω,u(x)v(s),0≤x≤s≤ω.Proof.
By Lemma3, 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[1](x)=c1u[1](x)+c2v[1](x)+∫0xu[1](x)v(s)-u(s)v[1](x)Ws(u,v)h(s)ds.
Consequently,
(25)y(0)=c1a+c2v(0),y[1](0)=c1+c2v[1](0).
Substituting these values of y(0) and y[1](0) into the first boundary condition of (11), we find
(26)c2=1v(0)-av[1](0)∫0ωg0(s)σ0(s)ds.
Similarly from the second boundary condition of (11), we can find
(27)c1=1u(ω)-bu[1](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 solutionu(x) and v(x), the inequality Wx(u,v)<0, x∈J holds.Proof.
Using the initial conditions (12), we can deduce from (5) for u(x) and v(x) the following equations:
(28)u[1](x)=1+∫0xq(s)u(s)ds,u(x)=a+∫0x1p(t)dt+∫0x[∫sxdtp(t)]q(s)u(s)ds,v[1](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[1](x)≥1>0,v(x)≥-b+∫xωdtp(t)>0,v[1](x)≤-1<0.
Now from (3), we get Wx(u,v)<0, x∈J. The proof is complete.From (21), (22), and Lemma 7, the following lemma follows.Lemma 8.
Under condition (H1) the Green's functionG(x,s) of the BVP (11) is positive. That is, G(x,s)>0 for x,s∈J.LetC(J) denote the Banach of all continuous functions y:I→ℝ equipped with the form ∥y∥=max{|y(x)|;x∈J}, for any y∈C(J). Denote P={y∈C(J);y(x)⩾0,y∈J}, then P is a positive cone in C(J).Let us setA=max0⩽x,s⩽ωG(x,s), B=min0⩽x,s⩽ωG(x,s), and by Lemma 8, obviously, A>B>0, x,s∈J.Define a mappingΦ in Banach space C(J) by
(30)(Φy)(x)=w(x)+λ∫0ωG(x,s)f(s,y(s))ds+∑i=0mG(x,xi)Ii(y(xi)),x∈J,
where
(31)w(x)=u(x)u(ω)-bu[1](ω)∫0ωg1(s)y(s)ds+v(x)v(0)-av[1](0)∫0ωg0(s)y(s)ds.Lemma 9.
The fixed point of the mappingΦ is a solution of (1).Proof.
Clearly,Φy is continuous in x for x∈J. For x≠xk,
(32)(Φy)′(x)=w′(x)+λ∫0ω∂G∂xf(s,y(s))ds+∑i=0m∂G(x,xi)∂xIi(y(xi)),
where
(33)w′(x)=u′(x)u(ω)-bu[1](ω)∫0ωg1(s)y(s)ds+v′(x)v(0)-av[1](0)∫0ωg0(s)y(s)ds.
We have
(34)(Φy)[1](x)=w[1](x)+λ∫0ωp(x)∂G∂xf(s,y(s))ds+∑i=0mp(x)∂G(x,xi)∂xIi(y(xi)),
where
(35)w[1](x)=u[1](x)u(ω)-bu[1](ω)∫0ωg1(s)y(s)ds+v[1](x)v(0)-av[1](0)∫0ωg0(s)y(s)ds.
We can easy get that
(36)(Φy)(0)-a(Φy)[1](0)=∫0ωg0(s)y(s)ds,(Φy)(ω)-b(Φy)[1](ω)=∫0ωg1(s)y(s)ds,Δ(Φy)[1](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+λ∫0ωp(x)∂G∂xf(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.
LetP0∶={y∈P;minx∈Jy(x)≥((1-Mω)B/(1-mω)A)∥y∥}, then P0 is a cone.Proof.
(i) For for ally1,y2∈P0 and for all α≥0, β≥0, we have
(37)min(αy1)⩾α·(1-Mω)B(1-mω)A∥y1∥,min(βy2)⩾β·(1-Mω)B(1-mω)A∥y2∥.
Moreover
(38)min(αy1+βy2)⩾(1-Mω)B(1-mω)A(α∥y1∥+β∥y2∥)⩾(1-Mω)B(1-mω)A∥αy1+βy2∥.
Thus αy1+βy2∈P0.
(ii) Ify∈P0 and -y∈P0, we have
(39)minx∈J(y(x))⩾(1-Mω)B(1-mω)A∥y∥,minx∈J(-y(x))⩾(1-Mω)B(1-mω)A∥y∥.
It implies that y=0. Hence P0 is a cone.Defined a linear operatorA: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(i)
A is a bounded linear operator, A(P)⊂P;(ii)
(I-A) is invertible;(iii)
∥(I-A)-1∥≤1/(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∈ℝ, y1,y2∈C(J).
Usingϕ(x,s)≤M, it is easy to see that |(Ay)(t)|≤Mω∥y∥.
Lety∈P. Then y(s)≥0 for all s∈J. Since ϕ(t,s)≥m≥0, it follows that (Ay)(x)≥0 for each x∈J. So A(P)⊂P.
(ii) We want to show that(I-A) is invertible, or equivalently 1 is not an eigenvalue of A.
SinceM<1/ω, it follows from condition (H2) that ∥Ay∥≤Mω∥y∥<∥y∥.
So(42)∥A∥=supy≠0∥Ay∥∥y∥⩽Mω<1.
On the other hand, we suppose 1 is an eigenvalue of A, then there exists a y∈C(J) such that Ay=y. Moreover, we can obtain that ∥Ay∥/∥y∥=1. So ∥A∥⩾1. Thus this assumption is false.
Conversely, 1 is not an eigenvalue ofA. Equivalently, (I-A) is invertible.
(iii) We use the theory of Fredholm integral equations to find the expression for(I-A)-1.
Obviously, for eachx∈J, 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=1∞ϕj(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 thatR(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-Mω∫0ωz(s)ds≤∥z∥(1+Mω1-Mω)=11-Mω∥z∥.
So
(48)∥(I-A)-1z∥∥z∥≤11-Mω.
Thus ∥(I-A)-1∥≤1/(1-Mω). This completes the proof of the lemma.Remark 12.
Sinceϕ(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[1](x))′+q(x)y(x)=λf(x,y(x)),mmmmmmmmmmmmmx≠xi,x∈J,-Δ(y[1](xi))=Ii(y(xi)),i=1,2,…,m,y(0)-ay[1](0)=∫0ωg0(s)y(s)ds,y(ω)-by[1](ω)=∫0ωg1(s)y(s)ds.
Denote a nonlinear operator T:PC(J)→PC(J) by
(50)(Ty)(x)=λ∫0ωG(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),x∈J-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 nonlinearityf:J×[0,∞)→[0,∞) is continuous and satisfies the following.(H3) There existL1>0 and α(x)∈P, r1∈ℝ with r1⩾∑i=0mIi(y(xi))/λ∫0ω‍α(s)dssuch that
(55)f(x,y)≤α(x)[y(1-Mω)-r1]
for all y∈(0,L1], x∈J.(H4) There existL2>L1 and β(x)∈P, p1∈ℝ with p1⩽∑i=0mIi(y(xi))/λ∫0ω‍β(s)ds such that
(56)f(x,y)≥β(x)[y(1-mω)-p1]
for all y∈(L2,∞], x∈J.Then, we can get the following theorem.Theorem 13.
Assume (H1), (H2), (H3), and (H4) are satisfied. And(57)(1-mω)A2∫0ωα(s)ds⩽(1-Mω)B2∫0ωβ(s)ds,
then, if λ satisfies
(58)(1-mω)A(1-Mω)B2∫0ωβ(s)ds⩽λ⩽1A∫0ωα(s)ds.
The problem (49) has at least one positive solution.Proof.
First of all, we show that operatorΦ is defined by (54) maps P0 into itself. Let y∈P0.
Then(Φy)(x)≥0 for all that t∈J-1, and
(59)(Φy)(x)⩽λA1-Mω∫0ωf(s,y(s))ds+A1-Mω∑i=0mIi(y(xi)).
Because from the formula (54), we have
(60)(Φy)(x)=(Ty)(x)+∫0ωR(x,s)(Ty)(s)ds=λ∫0ωG(x,s)f(s,y(s))ds+∑i=0mG(x,xi)Ii(y(xi))+λ∫0ωR(x,s)∫0ωG(x,τ)f(τ,y(τ))dτds+∫0ωR(x,s)∑i=0mG(x,xi)Ii(y(xi))ds≤λ(1+Mω1-Mω)∫0ωG(x,s)f(s,y(s))ds+∑i=0mG(x,xi)Ii(y(xi))+Mω1-Mω∑i=0mG(x,xi)Ii(y(xi))≤λA1-Mω∫0ωf(s,y(s))ds+A1-Mω∑i=0mIi(y(xi)).
Hence, inequality (59) is established.
This implies that(61)∥Φy∥⩽λA1-Mω∫0ωf(s,y(s))ds+A1-Mω∑i=0mIi(y(xi)),
or equivalently
(62)∫0ωf(s,y(s))ds⩾1-MωλA∥Φy∥-1λ∑i=0mIi(y(xi)).
On the other hand, it follows that
(63)(Φy)(x)⩾λB1-mω∫0ωf(s,y(s))ds+B1-mω∑i=0mIi(y(xi)).
In fact, we have
(64)(Φy)(x)=λ∫0ωG(x,s)f(s,y(s))ds+∑i=0mG(x,xi)Ii(y(xi))+λ∫0ωR(x,s)∫0ωG(x,τ)f(τ,y(τ))dτds+∫0ωR(x,s)∑i=0mG(x,xi)Ii(y(xi))ds≥λ(1+mω1-mω)∫0ωG(x,s)f(s,y(s))ds+∑i=0mG(x,xi)Ii(y(xi))+mω1-mω∑i=0mG(x,xi)Ii(y(xi))≥λB1-mω∫0ωf(s,y(s))ds+B1-mω∑i=0mIi(y(xi)).
It follows from (62) that
(65)(Φy)(x)≥λB1-mω·[1-MωλA∥Φy∥-1λ∑i=0mIi(y(xi))]+B1-mω∑i=0mIi(y(xi))=(1-Mω)B(1-mω)A∥Φy∥-B1-mω∑i=0mIi(y(xi))+B1-mω∑i=0mIi(y(xi))=(1-Mω)B(1-mω)A∥Φy∥.
So, we get
(66)(Φy)(x)⩾(1-Mω)B(1-mω)A∥Φy∥.
This show that Φy∈P0.
It is easy to see thatΦ is the complete continuity.
We now proceed with the construction of the open setsΩ1 and Ω2.
First, lety∈P0 with ∥y∥=L1. Inequality (59) implies
(67)(Φy)(x)≤λA1-Mω∫0ωf(s,y(s))ds+A1-Mω∑i=0mIi(y(xi))≤λA1-Mω∫0ωα(s)[y(s)(1-Mω)-r1]ds+A1-Mω∑i=0mIi(y(xi))=λA∫0ωα(s)y(s)ds-λA1-Mωr1×∫0ωα(s)ds+A1-Mω∑i=0mIi(y(xi))=λA∫0ωα(s)y(s)ds+A1-Mω×[∑i=0mIi(y(xi))-λr1∫0ωα(s)ds].
By condition (H3) and (58), we obtain
(68)∑i=0mIi(y(xi))-λr1∫0ωα(s)ds⩽0,λA∫0ωα(s)ds⩽1.
So
(69)(Φy)(x)⩽λA∫0ωα(s)ds∥y∥⩽∥y∥.
Consequently, ∥Φy∥⩽∥y∥.
LetΩ1∶={y∈C(J);∥y∥<L1}. Then, we have ∥Φy∥⩽∥y∥ for y∈P0∩∂Ω1.
Next, letL2~=max{2L1,((1-mω)A/(1-Mω)B)L2} and set Ω2∶={y∈C(J);∥y∥<L2~}.
Fory∈P0 with ∥y∥=L2~, we have
(70)minx∈Jy(x)≥(1-Mω)B(1-mω)A∥y∥=(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-mω∫0ωf(s,y(s))ds+B1-mω∑i=0mIi(y(xi))≥λB1-mω∫0ωβ(s)[y(s)(1-mω)-p1]ds+B1-mω∑i=0mIi(y(xi))=λB∫0ωβ(s)y(s)ds+B1-mω×(∑i=0m‍Ii(y(xi))-λp1∫0ωβ(s)ds).
By condition (H4) and (58), we obtain
(72)∑i=0mIi(y(xi))-λp1∫0ωβ(s)ds⩾0,λB⩾(1-mω)A(1-Mω)B∫0ωβ(s)ds.
Since y∈P0 we have y(x)⩾((1-Mω)B/(1-mω)A)∥y∥ for all x∈J. It follows from the above inequality that
(73)(Φy)(x)⩾(1-mω)A(1-Mω)B∫0ωβ(s)ds∫0ωβ(s)ds·(1-Mω)B(1-mω)A∥y∥=∥y∥.
Hence ∥Φy∥⩾∥y∥ for y∈P0∩∂Ω2.
It follows from (i) of Theorem1 that Φ has a fixed point in P0∩(Ω-2∖Ω1), and this fixed point is a solution of (49).
This completes the proof.Next, withL1 and L2 as above, we assume that f satisfied the following.(H5) There existα*(x)∈P, r1*∈ℝ with r1*⩽∑i=0mIi(y(xi))/λ∫0ωα*(s)ds such that
(74)f(x,y)≥α*(x)[y(1-mω)-r1*]
for all y∈(0,L1], x∈J.(H6) There existβ*(x)∈P, p1*∈ℝ with p1*⩾∑i=0mIi(y(xi))/λ∫0ωβ*(s)ds such that
(75)f(x,y)≤β*(x)[y(1-Mω)-p1*]
for all y∈(L2,∞], x∈J.Theorem 14.
Assume (H1), (H2), (H5), and (H6) are satisfied. And(76)(1-mω)A2∫0ωβ*(s)ds⩽(1-Mω)B2∫0ωα*(s)ds,
then, if λ satisfies
(77)(1-mω)A(1-Mω)B2∫0ωα*(s)ds⩽λ⩽1A∫0ωβ*(s)ds.
The problem (49) has at least one positive solution.Proof.
LetΦ be a completely continuous operator defined by (54). Then Φ maps the cone P0 into itself.
First, lety∈P0 with ∥y∥=L1. Inequality (63) implies
(78)(Φy)(x)≥λB1-mω∫0ωf(s,y(s))ds+B1-mω∑i=0mIi(y(xi))≥λB1-mω∫0ωα*(s)[y(s)(1-mω)-r1*]ds+B1-mω∑i=0mIi(y(xi))=λB∫0ωα*(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)ds⩾0,λB⩾(1-mω)A(1-Mω)B∫0ωα*(s)ds.
Hence
(80)(Φy)(x)⩾(1-mω)A(1-Mω)B∫0ωα*(s)ds∫0ωα*(s)y(s)ds.
Since y∈P0, we have y(x)⩾((1-Mω)B/(1-mω)A)∥y∥ for all x∈J. It follows from the above inequality that
(81)(Φy)(x)⩾(1-mω)A(1-Mω)B∫0ωα*(s)ds∫0ωα*(s)ds·(1-Mω)B(1-mω)A∥y∥=∥y∥.
Let Ω1∶={y∈C(J);∥y∥<L1}. Then, we have ∥Φy∥⩾∥y∥ for y∈P0∩∂Ω1.
Next, letL2~=max{2L1,((1-mω)A/(1-Mω)B)L2} and set Ω2∶={y∈C(J);∥y∥<L2~}.
Then fory∈P0 with ∥y∥=L2~ for all x∈J, we have minx∈Jy(x)⩾L2. Inequality (59) implies
(82)(Φy)(x)≤λA1-Mω∫0ωf(s,y(s))ds+A1-Mω∑i=0mIi(y(xi))≤λA1-Mω∫0ωβ*(s)[y(s)(1-Mω)-p1*]ds+A1-Mω∑i=0mIi(y(xi))=λA∫0ωβ*(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)ds⩽0,λA∫0ωβ*(s)ds⩽1.
So
(84)(Φy)(x)⩽λA∫0ωβ*(s)ds∥y∥⩽∥y∥⩽1.
Therefore ∥Φy∥⩽∥y∥ with ∥y∥=L2~.
Then, we have∥Φy∥⩽∥y∥ for y∈P0∩∂Ω2.
We see the case (ii) of Theorem1 is met. It follows that Φ has a fixed point in P0∩(Ω-2∖Ω1), and this fixed point is a solution of (49).
This completes the proof.



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*Source: 134691-2013-04-02.xml*