# Existence of Solution to a Second-Order Boundary Value Problem via Noncompactness Measures

Authors: Wen-Xue Zhou; Jigen Peng
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

The existence and uniqueness of the solutions to the Dirichlet boundary value problem in the Banach spaces is discussed by using the fixed point theory of condensing mapping, doing precise computation of measure of noncompactness, and calculating the spectral radius of linear operator.


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## Body

## 1. Introduction

This paper is mainly concerned with the following second-order Dirichlet boundary value problem:(1.1)-u′′(t)=f(t,u(t)),t∈I=[0,1],u(0)=u(1)=θ,
in a Banach space E, where f(t,x)∈C(I×E,E), θ is the zero element of E.In the last several decades, there has been much attention focused on the boundary value problems for various nonlinear ordinary differential equations, difference equations, and functional differential equations, see [1–20] and the references therein. The existence of solutions for Neumann boundary value problems has been considerably investigated in many publications such as [2–5, 8–10]. Dirichlet boundary value problems have deserved the attention of many researchers, see [11–20] and the references therein.In particular, the authors in [11] have studied the following two-point boundary value problem:(1.2)x′′=H(t,x,x′),0<t<1,ax(0)-bx′(0)=x0,cx(1)+dx′(1)=x1,
where a,b,c,d≥0 and ad+bc>0. They obtained the existence of solutions by means of the Darbo fixed point theorem and properties of the measure of noncompactness.We would like to mention the results due to [11]. First, we point out that many authors applied the famous Sadovskii’s fixed point theorem to investigate similar problems and used the following hypothesis with respect to the Kuratowski measure of noncompactness α(·): there exists a constant k>0 such that for any bounded and equicontinuous set A,B⊂C(I,E) and t∈I, α(H(I×A×B))≤kmax{α(A),α(B)}. What is more, they required a stronger condition, that is, ∥H(t,x,y)∥≤L for (t,x,y)∈I×E×E and the constant k satisfies 0<k<1/2 (see Remarks 3.2–3.6).The authors in [15, 18] have studied the following boundary value problem:(1.3)x′′=f(t,x,x′),c1x(a)-d1x′(a)=x1,c2x(b)+d2x′(b)=x2,
where X is a real Banach space, J=[a,b]⊂ℝ, f:J×X2→X is continuous, ci,di∈ℝ, and xi∈X for i=1,2. They obtained the existence of solutions by means of Sadovskii’s fixed point theorem and properties of the measure of noncompactness.Motivated by the above-mentioned work [11, 15, 18], the main aim of this paper is to study the existence and uniqueness of solutions for the problem (1.1) under the new conditions. The main new features presented in this paper are as follows First, the existence and uniqueness of solutions to Banach space’s Dirichlet boundary value problem is proved precisely calculating the spectral radius of linear operation. Second, the conditions imposed on the BVP (1.1) are weak. Third, the main tools used in the analysis are Sadovskii’s fixed point theorem and precise computation of measure of noncompactess. Our results can be seen as a supplement of the results in [11] (see Remarks 3.2–3.6).This paper is organized as follows. In Section2, we provide some basic definitions, preliminaries facts, and various lemmas which will be used throughout this paper. In Section 3, we give main results in this paper.


## 2. Preliminaries and Lemmas

LetE be a real Banach space and P be a cone in E which defines a partial ordering in E by x≤y if and only if y-x∈P· P is said to be normal if there exists a positive constant N such that θ≤x≤y implies ∥x∥≤N∥y∥, where θ denotes the zero element of E, and the smallest N is called the normal constant of P (it is clear, N≥1). If x≤y and x≠y, we write x<y. For details on cone theory, see the monograph [7].LetI=[0,1]. By C(I,E) we denote the Banach space of all continuous functions from I into E with the norm(2.1)‖y‖c:=max⁡{|y(t)|:t∈I}.Definition 2.1 (see [7]).
Assume thatS is a bounded set in E. Let α(S)=inf{δ>0:S be expressed as the union S=∪i=1mSi of a finite number of sets Si with diameter diam(Si)≤δ}.
α(S) is said to be the Kuratowski measure of noncompactness and is called the noncompactness measure for short. For details and properties of the noncompactness measure see [7].Definition 2.2 (see [7]).
The mappingA is said to be a condensing operator if A is continuous, bounded, and for any nonrelatively compact and bounded set S⊂D,
(2.2)α(A(S))<α(S).
The following lemmas are of great importance in the proof of our main results.Lemma 2.3.
Suppose thatM∉{-n2π2∣n=1,2,3,…}. Then for any h∈C(I,E), the linear boundary value problem
(2.3)-u′′(t)+Mu(t)=h(t),t∈I=[0,1],u(0)=u(1)=θ,
has a unique solution u:=TMh∈C2(I,E), and TM:C(I,E)↦C(I,E) is bounded linear operator.Proof.
Note thatM∉{-n2π2∣n=1,2,3,…}, which assures second-order boundary value problem
(2.4)-γ′′(t)+Mγ(t)=0,γ(0)=γ(1)=0,
has only a zero solution. To obtain a solution of the problem (2.3), we require a mapping whose kernel GM(t,s):I×I→ℝ is the Green’s function of the boundary value problem (2.4). Let β=|M|, we consider three cases.
Case 1.
ifM>0, we have
(2.5)GM(t,s)={sinh⁡(βt)sinh⁡(β(1-s))βsinh⁡β,if0≤t≤s≤1,sinh⁡(βs)sinh⁡(β(1-t))βsinh⁡β,if0≤s≤t≤1.Case 2.
ifM=0, we have
(2.6)G0(t,s)={t(1-s),if0≤t≤s≤1,s(1-t),if0≤s≤t≤1.Case 3.
ifM<0, M≠-n2π2,n=1,2,3,…, we have
(2.7)GM(t,s)={sin⁡(βt)sin⁡(β(1-s))βsin⁡β,if0≤t≤s≤1,sin⁡(βs)sin⁡(β(1-t))βsin⁡β,if0≤s≤t≤1.
After direct computations, it is easy to see that(2.8)u(t)=∫01GM(t,s)h(s)ds:=(TMh)(t)
is continuously differentiable, and u(t) is a solution of (2.3).
We now claim that solution of the boundary value problem (2.3) is unique. The proof is as follows. If possible, suppose that v(t)∈C2(I,E) is another solution of the problem (2.3). For any φ∈E*(E* denotes the dual space of E), let p(t)=φ(u(t)-v(t)), thus we obtain p(t)∈C2(I), p′′(t)=φ(u′′(t)-v′′(t)). By (2.3), we have
(2.9)-p′′(t)+Mp(t)=0,p(0)=p(1)=0,
that is, p(t) is a solution of the boundary value problem (2.4). However, on the other hand, problem (2.4) has only a zero solution, therefore we have φ(u(t)-v(t))=0. Thus, we get u(t)-v(t)=θ, hence u(t)≡v(t) in I, which implies that solution of the problem (2.3) is unique, say, u:=TMh, and TM:C(I,E)→C(I,E).
It is easy to see thatTM is bounded linear operator. This completes the proof.Remark 2.4.
IfM>-π2, it is easy to see that GM(t,s)≥0.Lemma 2.5.
Assume thatM>-π2, and TM:C(I,E)↦C(I,E) is given by (2.8). Then(1)
the spectral radiusr(TM)=1/(M+π2);(2)
IfE is an ordered Banach  space, then TM is a positive operator, that is, if h≥0, then TMh≥0.Proof.
(1) Define operatorℒ:𝒟(ℒ)↦C(I,E) by
(2.10)Lu:=-u′′+Mu,
where 𝒟(ℒ)={u∈C2(I,E)∣u(0)=u(1)=θ}. By Lemma 2.3, we have that TM is bounded invertible operator of ℒ, and if λ≠M+n2π2,n=1,2,3,…, then ℒ-λI has a bounded invertible operator, thus λ∈ρ(ℒ).
Letλ=λn=M+n2π2, λn is a eigenvalue of ℒ. For any x∈E,x≠θ, since sinM+n2π2t is eigenvector of λn, then the spectrum of operator ℒ is σ(ℒ)={M+n2π2∣n=1,2,3,…}. By the spectral mapping theorem [21], we get
(2.11)σ(TM)={θ}∪{1M+n2π2∣n∈N},
so r(TM)=1/(M+π2).
(2) Ifh≥0, by definition of GM(t,s) and M>-π2, then we get GM(t,s)≥0, so by (2.8), we have TMh≥0, that is, TM is a positive operator. This achieves the proof.Remark 2.6.
In particular. IfE=ℝ1, M=0, then by (1) of Lemma 2.5, we get r(T0)=1/π2, where T0 is an operator in C(I):
(2.12)(T0φ)(t)=∫01G0(t,s)φ(s)ds,
and ∥T0∥c=1/8. In fact:
(2.13)|(T0φ)(t)|=|∫01G0(t,s)φ(s)ds|≤∫01G0(t,s)ds⋅‖φ‖c=12t(t-1)⋅‖φ‖c≤18⋅‖φ‖c.
This means that ∥T0φ∥c≤(1/8)∥φ∥c, therefore ∥T0∥c≤1/8. However, on the other hand, ∥T0(1)∥c=1/8. As a result, we obtain ∥T0∥c=1/8.Lemma 2.7.
LetJ=[a,b], u∈C(J,E), φ∈C(J,ℝ+). Then
(2.14)∫abφ(s)u(s)ds∈(∫abφ(s)ds)⋅co⁡¯u(J),
where u(J)={u(t)∣t∈J}, co¯u(J) is closed convex hull of u(J).Proof.
If∫abφ(s)ds=0, then (2.14) is true. We suppose that ∫abφ(s)ds>0, and take a partition of [a,b]:
(2.15)Δn:a=t0(n)<t1(n)<t2(n)<⋯<tm(n)=b.
Let Δti(n)=ti(n)-ti-1(n), ∥Δn∥=max{Δti(n):1≤i≤mn}, by definition of Riemann integral, we get
(2.16)∫abφ(s)u(s)ds=lim⁡n→∞∑n=1mnφ(ti(n))u(ti(n))Δti(n),∫abφ(t)dt=lim⁡n→∞∑n=1mnφ(ti(n))Δti(n).
We taken sufficiently large, such that ∥Δn∥→0, then we get
(2.17)∫abφ(s)u(s)ds∫abφ(s)ds=lim⁡n→∞∑n=1mnφ(ti(n))u(ti(n))Δti(n)∑n=1mnφ(ti(n))Δti(n)∈co⁡¯u(J).
This finishes the proof.Lemma 2.8.
Suppose thatD is a bounded set in E, then there exists a countable subset D1 of D, such that
(2.18)α(D)≤2α(D1).Proof.
Letα(D)>0, D≠∅. For rn=α(D)(1-1/n), take x1(n)∈D, then D∖B(x1(n),rn/2) is not a cover of D. Take x2(n)∈D∖B(x2(n),rn/2), then B(x2(n),rn/2) is not a cover of D. Continuing this process, take xk+1(n)∈D∖∪i=1kB(xi(n),rn/2), then B(xi(n),rn/2)(i=1,2,3,…,k) is not a cover of D.
SetDn={xk(n)∣k=1,2,3,…}, then we get d(s,t)≥rn/2, where d(s,t) denote the distance between two points s and t of Dn. Thus it follows that α(Dn)≥rn/2.
SettingD1=∪i=1∞Dn, choose n sufficiently large such that α(D1)≥α(Dn)≥rn/2→α(D)/2, that is, α(D)≤2α(D1). The proof is completed.Lemma 2.9.
IfB is a bounded set in C(I,E), B(I)={u(t)∣u∈B,t∈I}⊂E. Then
(2.19)α(B(I))≤2α(B).Proof.
For anyɛ>0, there exists a partition B=∪i=1nBi such that
(2.20)diam⁡(Bi)<α(B)+ε,
for i=1,2,3,…,n. Choose ui∈Bi(i=1,2,3,…,n). Since ui is uniformly continuous on I, there exists δ>0, such that t′,t′′∈I, and |t′-t′′|<δ, we have
(2.21)‖ui(t′)-ui(t′′)‖<ε.
LetΔ:0=t0<t1<t2<t3<⋯<tm=1 be a partition of I, and ∥Δ∥<δ. Set Ij=[tj-1,tj], Dij=Bi(Ij)={u(t)∣u∈Bi,t∈Ij}. Clearly, we have
(2.22)D=B(I)=⋃i=1nBi(I)=⋃j=1m⋃i=1nBi(Ij).
For anyu,v∈Bi and t,s∈Ij, it follows from (2.20) and (2.21) that
(2.23)‖u(t)-v(s)‖≤‖u(t)-ui(t)‖+‖ui(t)-ui(s)‖+‖v(s)-ui(s)‖≤‖u-ui‖c+ε+‖v-ui‖c≤2diam⁡(Bi)+ε<2α(B)+3ɛ.
So,
(2.24)diam⁡(Dij)=sup⁡u,v∈Bi;t,s∈Ij‖u(t)-v(s)‖≤2α(B)+3ɛ.
Thus it follows that α(D)≤2α(B)+3ε. Therefore, by using the arbitrariness of ɛ, we have α(B(I))≤2α(B). The lemma is proved.Lemma 2.10 (see [7]).
Assume thatH⊂C[J,E] is bounded and equicontinuous. Then α(H(t)) is continuous on J and
(2.25)α({∫Jx(t)dt:x∈H})≤∫Jα(H(t))dt.Lemma 2.11 (see [7]).
Suppose thatH is a countable family of strongly measurable functions x:J↦E. If there exists a function M∈L[J,ℝ+] such that ∥x(t)∥≤M(t) for a.e. t∈J, then α(H(t))∈L[J,ℝ+] and
(2.26)α({∫Jx(t)dt:x∈H})≤2∫Jα(H(t))dt.Lemma 2.12.
Assume thatΩ1 is equicontinuous in C(I,E). Then co¯(Ω1) is equicontinuous.Proof.
For anyɛ>0, it follows from the equicontinuity of Ω1 that there exists δ>0 such that |t1-t2|<δ implies ∥u(t1)-u(t2)∥<ε/3 for all t1,t2∈I and u∈Ω1.
For anyh∈co¯(Ω1), by virtue of definition of co¯(Ω1), we have
(2.27)∀u1,u2,…,un∈Ω1,λ1,λ2,…,λn>0,∑i=1nλi=1,‖∑i=1nλiui-h‖c<ε3.
Thus, we get(2.28)‖h(t1)-h(t2)‖≤‖h(t1)-∑i=1nλiui(t1)‖+‖∑i=1nλiui(t2)-h(t2)‖+‖∑i=1nλi[ui(t1)-ui(t2)]‖≤2‖∑i=1nλiui-h‖c+∑i=1nλi‖ui(t1)-ui(t2)‖<23ε+∑i=1nλi⋅13ε=ε.
Hence, co¯(Ω1) is equicontinuous. This finishes the proof.Lemma 2.13 (see [7]).
IfH⊂C[I,E] is bounded and equicontinuous, then α(B(t)) is continuous in I and α(B)=α(B(I))=maxx∈Iα(B(t)).Lemma 2.14 (see [7] (Sadovskii’s Theorem)).
Assume thatD is a nonempty bounded, closed, and convex set. If a mapping A:D↦D is condensing, then A has a fixed point in D.


## 3. Main Results

In this section, we present and prove our main results.Theorem 3.1.
LetE be a Banach space. Suppose that f(t,x)∈C(I×E,E) and the following conditions hold:(H1)
there exist two positive numbersc0 and c1, such that(3.1)‖f(t,x)‖≤c0+c1‖x‖,∀t∈I,x∈E,(H2)
for any a bounded setD in E, there exists a constant L>0 such that(3.2)α(f(I×D))≤Lα(D),(H3)
there exist two positive numbersc1 and L with c1<π2, L<4. Then problem (1.1) has at least one solution.
Proof.
Define the integral operatorA:C(I,E)↦C(I,E)(3.3)(Au)(t)=∫01G0(t,s)f(s,u(s))ds=T0(f(⋅,u)).
Then A:C(I,E)↦C(I,E) is continuous, and it is clear that u is a solution of the problem (1.1) if and only if u is a fixed point of A.
We now show thatA is a condensing operator. Let B be bounded in C(I,E), by (H1), we claim that {-(Au)′′∣u∈B} is bounded. Since (Au)(0)=0, we know that {(Au)′∣u∈B} is bounded, this means that A(B) is equicontinuous. Therefore, it follows from Lemma 2.13 that α(A(B))=maxx∈Iα(A(B)(t)).
For anyu∈B, t∈I, from Lemma 2.7, we have
(3.4)(Au)(t)=∫01G0(t,s)f(s,u(s))ds∈(∫01G0(t,s)ds)⋅co⁡¯{f(s,u(s))∣s∈I}⊂(∫01G0(t,s)ds)⋅co⁡¯{f(I×B(I))}.
Hence,
(3.5)(A(B))(t)⊂(∫01G0(t,s)ds)⋅co⁡¯{f(I×B(I))}.
Using the properties of the noncompactness measure together with(H2), we obtain
(3.6)α(A(B)(t))≤∫01G0(t,s)ds⋅α(f(I×B(I)))≤Lα(B(I))⋅(∫01G0(t,s)ds)=12t(1-t)⋅Lα(B(I))≤18Lα(B(I)).
By Lemma2.9, we have
(3.7)α(A(B)(t))≤18L⋅α(B(I))≤18L⋅2α(B)=14L⋅α(B).
Hence,(3.8)α(A(B))≤14L⋅α(B).
By (H3), we get 0<L/4<1, therefore A is condensing.
LetΩ:=Bc(θ,R)={u∈C(I,E):∥u∥<R}, we will prove that u-λAu≠θ, 0<λ≤1,u∈∂Ω for R sufficiently large. By means of the homotopy invariance theorem, we have deg(I-A,Ω,θ)=1. By virtue of the solvability of Kronecker  [6], we know that A has a fixed point in Ω, and the fixed point of A is a solution of the problem (1.1).
Indeed. If there exists a constantλ0∈(0,1], λ0∈∂Ω such that u0-λ0Au0=θ, then u0 satisfies
(3.9)u0=λ0∫01G0(t,s)f(s,u0(s))ds.
Letφ0(t)=∥u0(t)∥, and T0 is an operator in C(I) defined by
(3.10)(T0φ)(t)=∫01G0(t,s)φ(s)ds.
Then by Lemma 2.5 we have r(T0)=1/π2.
By (3.9) and (H1), we have
(3.11)φ0(t)≤∫01G0(t,s)(c0+c1φ0(s))ds=c0T0(1)+c1(T0φ0)(t)≤c0‖T0‖+c1(T0φ0)(t)≤c0+c1(T0φ0)(t).
Thus,(3.12)φ0≤c0‖T0‖+c1T0φ0≤c0+c1T0φ0,
continuing this process, by induction, we obtain
(3.13)φ0≤c0+c1T0(c0+c1T0φ0)=c0+c0c1‖T0‖+c12T02φ0≤⋯≤c0(1+c1‖T0‖+c12‖T02‖+⋯+c1n-1‖T0n-1‖+c1n‖T0n‖φ0),
which implies that
(3.14)φ0≤c0∑n=0n-1c1n‖T0n‖+c1nT0nφ0.
By the Gelfand  theorem [22], we have
(3.15)r(T0)=lim⁡n→∞‖T0n‖n=1π2.
Setc2=(c1+π2)/2∈(c1,π2), then we have 1/c2>1/π2. By (3.15), there exists an integer N0, such that ∥T0n∥n<1/c2 as n≥N0, that is, ∥T0n∥<1/c2n, this means that c1n∥T0n∥<(c1/c2)n. In view of series ∑n=0∞(c1/c2)n converges, we know that ∑n=0∞c1n∥T0n∥ also converges.
DenoteR0=c0∑n=0∞c1n∥T0n∥. By (3.14), we get φ0(t)≤R0, which implies that ∥u0(t)∥≤R0, hence ∥u0∥c≤R0.
TakeR>R0, then we have u-λAu≠θ, forallλ∈(0,1],u∈∂Ω. Thus problem (1.1) has at least one solution. This proves the theorem.Remark 3.2.
In [11], the nonlinear term f(t,u,u′) is bounded, if f(t,u,u′)=f(t,u), in our result, the nonlinear term f(t,u) may no more than a linear growth.Remark 3.3.
In [11], if f(t,u,u′)=f(t,u), the growth restriction of L for (H2) is 0<L<1/2. However, in our result, L satisfies 0<L<4.Theorem 3.4.
LetE be a Banach space, and f(t,x)∈C(I×E,E). Assume that condition (H1) and the following conditions hold:(H2)′
for allt∈I, for any a bounded set D in E, there exists a constant L>0 such that(3.16)α(f(t,D))≤Lα(D),(H3)′
there exist two positive numbersc1 and L with 0<c1<π2, L<2. Then problem (1.1) has at least one solution.Proof.
Assume that the operatorA is defined the same as in Theorem 3.1. We show that the operator A is condensing. In fact, for a bounded set B∈C(I,E), there exists a countable subset B1={un}, such that α(A(B))≤2α(A(B1)). However, on the other hand, we have
(3.17)α(A(B1))=max⁡t∈Iα(A(B1)(t)).
By Lemma2.11 and (H2)′, we obtain
(3.18)α(A(B1))=α({∫01G0(t,s)f(s,un(s))ds∣n∈N})≤2∫01G0(t,s)α({f(s,un(s))∣n∈N})ds≤2∫01G0(t,s)Lα(B1(s))ds≤2L∫01G0(t,s)ds⋅α(B1)≤14Lα(B1).
Thus,
(3.19)α(A(B))≤12Lα(B).
By (H3)′, we get 0<L/2<1, therefore A is condensing. By using the same arguments of Theorem 3.1, we can obtain the conclusion of Theorem 3.4. The detailed proof is omitted here. The proof is achieved.Next, we establish a uniqueness of solution for the problem (1.1).Theorem 3.5.
LetE be a Banach space. Suppose that f(t,x)∈C(I×E,E) and that there exists a constant L with 0<L<π2 such that
(3.20)‖f(t,u2)-f(t,u1)‖≤L‖u2-u1‖,∀u1,u2∈E.
Then problem (1.1) has a unique solution.Proof.
Assume that operatorA is defined the same as in Theorem 3.1, and the fixed point of A is a solution of the problem (1.1).
We will prove that for sufficiently largen the operator An is a contraction operator. Indeed, by the definition of A and (3.20), we have the estimate
(3.21)‖(Anu2)(t)-(Anu1)(t)‖=‖∫01G0(t,s)[f(s,(An-1u2)(s))-f(s,(An-1u1)(s))]ds‖≤L∫01G0(t,s)‖(An-1u2)(s)-(An-1u1)(s)‖ds=LT0(‖(An-1u2)(s)-(An-1u1)(s)‖).
By induction, we have(3.22)‖(Anu2)(t)-(Anu1)(t)‖≤Ln⋅T0n⋅‖u2(s)-u1(s)‖≤Ln⋅‖T0n‖⋅‖u2-u1‖c.
Thus,
(3.23)‖Anu2-Anu1‖c≤Ln⋅‖T0n‖⋅‖u2-u1‖c.
Moreover, we can choosen to be sufficiently large such that ∥T0n∥n tends to r(T0)=1/π2.
Further, takeL1∈(L,π2), there exists an integer N0, such that ∥T0∥≤1/L1n as n≥N0. By (3.23), we obtain
(3.24)‖Anu2-Anu1‖c≤Ln⋅‖T0n‖⋅‖u2-u1‖c≤(LL1)n⋅‖u2-u1‖c,
which implies that An is a contraction mapping by L/L1<1. By the contraction mapping principle, we conclude that there exists a unique fixed point for A, this proves that problem (1.1) has a unique solution. This completes the proof.Remark 3.6.
By the direct application of the Banach contraction mapping principle, the conclusion of Theorem3.5 holds true under the condition 0<L<8. However, we require the condition 0<L<π2, here π2 is optimum.
The following theorem is concerned with the existence of positive solutions for problem (1.1).Theorem 3.7.
LetE be an ordered Banach space, K be a normal cone with positive elements. Suppose that f(t,x)∈C(I×E,E) satisfy the following conditions:(P1)
there exists a constantc with 0<c<π2, and h0∈C(I,K), such that(3.25)θ≤f(t,x)≤cx+h0(t),∀x≥θ,(P2)
for any a bounded setD in E, there exists a constant L with 0<L<8 such that(3.26)α(f(I×D))≤Lα(D).
Then problem (1.1) has at least one positive solution.Proof.
Consider the linear boundary value problem(3.27)-u′′-cu=h0,u(0)=u(1)=θ.
By Lemma 2.3, the linear boundary value problem (3.27) has unique a positive solution u0∈C(I,K).
SetD=[θ,u0]⊂C(I,E), then D is bounded and convex closed set in C(I,E). For any u∈D, by (P1), we have
(3.28)θ≤f(t,u(t))≤cu(t)+h0(t)≤cu0(t)+h0(t).
Multiply byG0(t,s) and integrate from 0 to 1, we obtain
(3.29)θ≤Au(t)≤∫01G0(t,s)[cu0(s)+h0(s)]ds=u0(t),
that is, Au∈D, so A(D)⊂D. By the proof of Theorem 3.1, it follows that A is equicontinuous. Thus, by Lemma 2.12, we know that  Ω0=co¯(A(D)) is equicontinuous.
Next we show thatA:Ω0→Ω0 is condensing. For any B⊂Ω0, then B is bounded and equicontinuous, therefore A(B)⊂Ω0 is bounded and equicontinuous. By Lemma 2.13, we have
(3.30)α(A(B))=max⁡t∈Iα(A(B(t))).
However, on the other hand, we have(3.31)α(A(B(t)))=α({∫01G0(t,s)f(s,u(s))ds∣u∈B}).
Thus, For anyu∈B, t∈I, we acquire
(3.32)Au(t)=∫01G0(t,s)f(s,u(s))ds∈(∫01G0(t,s)ds)⋅co⁡¯{f(s,u(s))∣s∈I}⊂(∫01G0(t,s)ds)⋅co⁡¯{f(I×B(I))}.
Hence,(3.33)A(B)(t)⊂(∫01G0(t,s)ds)⋅co⁡¯{f(I×B(I))}.
Further, we obtain(3.34)α(A(B)(t))≤(∫01G0(t,s)ds)⋅α(f(I×B(I)))≤L2t(1-t)⋅α(B(I))≤L8⋅α(B).
This means that(3.35)α(A(B))≤L8⋅α(B).
By(P2), we have 0<L/8<1, so A is condensing. Applying Lemma 2.14, we conclude that A has a fixed point which is a solution of problem (1.1). The proof of the theorem is completed.



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*Source: 786404-2012-05-08.xml*