# Positive Solutions for Neumann Boundary Value Problems of  Second-Order  Impulsive Differential Equations in Banach Spaces

Authors: Xiaoya Liu; Yongxiang Li
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The existence of positive solutions for Neumann boundary value problem of second-order impulsive differential equations−u″(t)+Mu(t)=f(t,u(t),
t∈J, t≠tk, -Δu′|t=tk=Ik(u(tk)), k=1,2,…,m, u′(0)=u′(1)=θ, in an ordered Banach space E was discussed by employing the fixed point index theory of condensing mapping, where M>0 is a constant, J=[0,1], f∈C(J×K,K), Ik∈C(K,K), k=1,2,…,m, and K is the cone of positive elements in E. Moreover, an application is given to illustrate the main result.


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## Body

## 1. Introduction

The theory of impulsive differential equations is a new and important branch of differential equation theory, which has an extensive physical, chemical, biological, engineering background and realistic mathematical model, and hence has been emerging as an important area of investigation in the last few decades; see [1]. Correspondingly, boundary value problems of second-order impulsive differential equations have been considered by many authors, and some basic results have been obtained; see [2–7]. But many of them obtained extremal solutions by monotone iterative technique coupled with the method of upper and lower solutions; see [2–4]. The research on positive solutions is seldom and most in real space ℝ; see [5–7].In this paper, we consider the existence of positive solutions to the second-order impulsive differential equation Neumann boundary value problem in an ordered Banach spaceE:(1.1)-u′′(t)+Mu(t)=f(t,u(t)),t∈J,t≠tk,-Δu′|t=tk=Ik(u(tk)),k=1,2,…,m,u′(0)=u′(1)=θ,
where M>0 is a constant, f∈C(J×E,E),  J=[0,1];0<t1<t2<···<tm<1;    Ik∈C(E,E) is an impulsive function, k=1,2,…,m.   Δu′|t=tk denotes the jump of u′(t) at t=tk,that is, Δu′|t=tk=u′(tk+)-u′(tk-), where u′(tk+) and u′(tk-) represent the right and left limits of u′(t) at t=tk, respectively.In the special case whereE=ℝ+=[0,+∞),    Ik=0,    k=1,2,…,m,  NBVP (1.1) has been proved to have positive solutions; see [8, 9]. Motivated by the aforementioned facts, our aim is to study the positive solutions for NBVP (1.1) in a Banach space by fixed point index theory of condensing mapping. Moreover, an application is given to illustrate the main result. As far as we know, no work has been done for the existence of positive solutions for NBVP (1.1) in Banach spaces.


## 2. Preliminaries

LetE be an ordered Banach space with the norm ∥·∥ and partial order ≤, whose positive cone K={x∈E∣x≥θ} is normal with normal constant N. Let J=[0,1];0=t0<t1<t2<···<tm<tm+1=1;    Jk=[tk-1,tk], k=1,2,…,m+1, J′=J∖{t1,t2,…,tm}. Let PC1(J,E)={u∈C(J,E)∣u′(t) is continuous at t≠tk, and left continuous at t=tk, and u′(tk+) exists, k=1,2,…,m}. Evidently, PC1(J,E) is a Banach space with the norm ∥u∥PC1=max{∥u(t)∥C,∥u′∥PC}, where ∥u∥C=supt∈J∥u(t)∥,∥u′∥PC=supt∈J∥u′(t)∥; see [2]. An abstract function u∈PC1(J,E)∩C2(J′,E) is called a solution of NBVP (1.1) if u(t) satisfies all the equalities of (1.1).LetC(J,E) denote the Banach space of all continuous E-value functions on interval J with the norm ∥u∥C=supt∈J∥u(t)∥. Let α(·) denote the Kuratowski measure of noncompactness of the bounded set. For the details of the definition and properties of the measure of noncompactness, see [10, 11]. For any B⊂C(J,E) and t∈J, set B(t)={u(t)∣u∈B}⊂E. If B is bounded in C(J,E), then B(t) is bounded in E, and α(B(t))≤α(B).Now, we first give the following lemmas in order to prove our main results.Lemma 2.1 (see [12]).
LetB⊂C(J,E) be equicontinuous. Then α(B(t)) is continuous on J, and
(2.1)α(B)=max⁡t∈Jα(B(t))=α(B(J)).Lemma 2.2 (see [13]).
LetB={un}⊂C(J,E) be a bounded and countable set. Then α(B(t)) is Lebesgue integral on J, and
(2.2)α({∫Jun(t)dt∣n∈N})≤2∫Jα(B(t))dt.Lemma 2.3 (see [14]).
LetD⊂E be bounded. Then there exists a countable set D0⊂D, such that α(D)≤2α(D0).To prove our main results, for anyh∈C(J,E), we consider the Neumann boundary value problem (NBVP) of linear impulsive differential equation in E:(2.3)-u′′(t)+Mu(t)=h(t),t∈J′,-Δu′|t=tk=yk,k=1,2,…,m,u′(0)=u′(1)=θ,
where M>0,yk∈E,k=1,2,…,m.Lemma 2.4.
For anyh∈C(J,E),M>0, and yk∈E,k=1,2,…,m, the linear NBVP (2.3) has a unique solution u∈PC1(J,E)∩C2(J′,E) given by
(2.4)u(t)=∫01G(t,s)h(s)ds+∑k=1mG(t,tk)yk,
where
(2.5)G(t,s)={cosh⁡M(1-t)cosh⁡MsMsinh⁡M,0≤s≤t≤1,cosh⁡Mtcosh⁡M(1-s)Msinh⁡M,0≤t<s≤1.Proof.
Suppose thatu(t) is a solution of (2.3); then
(2.6)u′′(t)-Mu(t)=-h(t),[e-2Mt(eMtu(t))′]′=-Me-Mtu(t)+e-Mtu′′(t)=-e-Mth(t).
Lety(t)=e-2Mt(eMtu(t))′; then
(2.7)y′(t)=-e-Mth(t),Δy|t=tk=-e-Mtkyk.
Integrating (2.7) from 0 to t1, we have
(2.8)y(t1)-y(0)=-∫0t1e-Msh(s)ds.
Again, integrating (2.7) from t1 to t, where t∈(t1,t2], then
(2.9)y(t)=y(t1+)-∫t1te-Msh(s)ds=y(0)-∫0te-Msh(s)ds-e-Mt1y1.
Repeating the aforementioned procession, for t∈J, we have
(2.10)y(t)=y(0)-∫0te-Msh(s)ds-∑0<tk<te-Mtkyk.
Hence,
(2.11)(eMtu(t))′=e2Mt(y(0)-∫0te-Msh(s)ds-∑0<tk<te-Mtkyk).
For t∈J, integrating (2.11) from 0 to t, we  have
(2.12)u(t)=e-Mt(∑0<tk<se-Mtkykdsu(0)+∫0te2Msy(0)ds-∫0te2Ms∫0se-Mτh(τ)dτds-∫0te2Ms∑0<tk<se-Mtkykds)=e-Mt{u(0)+12M[∑0<tk<se-Mtkykdsy(0)(e2Mt-1)-e2Mt∫0te-Msh(s)ds+∫0teMsh(s)ds-∑0<tk<t(e2Mt-e2Mtk)e-Mtkyk]}.
Notice that y(0)=Mu(0)+u′(0); thus, for t∈J, we have
(2.13)u(t)=12M[∑0<tk<te-Mt2Mu(0)+(Mu(0)+u′(0))eMt-(Mu(0)+u′(0))e-Mt+e-Mt∫0teMsh(s)ds-eMt∫0te-Msh(s)ds-∑0<tk<t(e2Mt-e2Mtk)e-M(t+tk)yk]=12M[∑0<tk<t(Mu(0)-u′(0))e-Mt+(Mu(0)+u′(0))eMt+e-Mt∫0teMsh(s)ds-eMt∫0te-Msh(s)ds-∑0<tk<t(eM(t-tk)yk-eM(tk-t)yk)],(2.14)u′(t)=12[∑0<tk<t-(Mu(0)-u′(0))e-Mt+(Mu(0)+u′(0))eMt-e-Mt∫0teMsh(s)ds-eMt∫0te-Msh(s)ds-∑0<tk<t(eM(t-tk)yk+e-M(t-tk)yk)].
In view of that u′(0)=u′(1)=θ, we have
(2.15)u(0)=∫01eM(1-s)+e-M(1-s)M(eM-e-M)h(s)ds+∑k=1meM(1-tk)+e-M(1-tk)M(eM-e-M)=∫01cosh⁡M(1-s)Msinh⁡Mh(s)ds+∑k=1mcosh⁡M(1-tk)Msinh⁡Myk.
Substituting (2.15) into (2.13), for t∈J, we obtain
(2.16)u(t)=∫0t(eM(1-t)+e-M(1-t))(eMs+e-Ms)2M(eM-e-M)h(s)ds+∑0<tk<t(eM(1-t)+e-M(1-t))(eMtk+e-Mtk)2M(eM-e-M)yk+∫t1cosh⁡Mtcosh⁡M(1-s)Msinh⁡Mh(s)ds+∑t≤tk<1cosh⁡Mtcosh⁡M(1-tk)Msinh⁡Myk=∫01G(t,s)h(s)ds+∑k=1mG(t,tk)yk.
Inversely, we can verify directly that the functionu∈PC1(J,E)∩C2(J′,E) defined by (2.4) is a solution of the linear NBVP (2.3). Therefore, the conclusion of Lemma 2.4 holds.By (2.5), it is easy to verify that G(t,s) has the following property:(2.17)1Msinh⁡M≤G(t,s)≤cosh⁡2MMsinh⁡M.Evidently,C(J,E) is also an ordered Banach space with the partial order ≤ reduced by the positive cone C(J,K)={u∈C(J,E)∣u(t)≥θ,t∈J}. C(J,K) is also normal with the same normal constant N.Define an operatorA:C(J,K)→C(J,K) as follows:(2.18)Au(t)=∫01G(t,s)f(s,u(s))ds+∑k=1mG(t,tk)Ik(u(tk)).
Clearly, A:C(J,K)→C(J,K) is continuous, and the positive solution of NBVP (1.1) is the nontrivial fixed point of operator A. However, the integral operator A is noncompactness in general Banach space. In order to employ the topological degree theory and the fixed point theory of condensing mapping, there demands that the nonlinear f and impulsive function Ik satisfy some noncompactness measure condition. Thus, we suppose the following.(P0) For any R>0,f(J×KR) and  Ik(KR) are bounded and(2.19)α(f(t,D))≤Lα(D),α(Ik(D))≤Mkα(D),k=1,2,…,m,
where KR=K∩B(θ,R),D⊂K is arbitrarily countable set,  L>0  and  Mk≥0 are counsants and satisfy (4L/M)+(2cosh2M∑k=1mMk/MsinhM)<1.Lemma 2.5.
Suppose that condition(P0) is satisfied; then A:C(J,K)→C(J,K) is condensing.Proof.
SinceA(B) is bounded and equicontinuous for any bounded and nonrelative compact set B⊂C(J,K), by Lemma 2.3, there exists a countable set B1={un}⊂B, such that
(2.20)α(A(B))≤2α(A(B1)).
By assumption (P0) and Lemma 2.1,
(2.21)α(A(B1)(t))=α({∫01G(t,s)f(s,un(s))ds+∑k=1mG(t,tk)Ik(un(tk))})≤α({∫01G(t,s)f(s,un(s))ds})+α({∑k=1mG(t,tk)Ik(un(tk))})≤2∫01G(t,s)α(f(s,B1(s)))ds+∑k=1mG(t,tk)α(Ik(B1(tk)))≤2∫01G(t,s)Lα(B1(s))ds+∑k=1mG(t,tk)Mkα(B1(tk))≤2L∫01G(t,s)dsα(B1)+∑k=1mMkG(t,tk)α(B1)≤2LMα(B1)+cosh⁡2M∑k=1mMkMsinh⁡Mα(B1)=(2LM+cosh⁡2M∑k=1mMkMsinh⁡M)α(B1).
Since A(B1) is equicontinuous, by Lemma 2.1, we have
(2.22)α(A(B1))=max⁡t∈Jα(A(B1)(t))≤(2LM+cosh⁡2M∑k=1mMkMsinh⁡M)α(B1).
Combining (2.20) and (P0), we have
(2.23)α(A(B))≤2α(A(B1))≤(4LM+2cosh⁡2M∑k=1mMkMsinh⁡M)α(B).
Hence, A:C(J,K)→C(J,K) is condensing.LetP be a cone in C(J,K) defined by(2.24)P={u∈C(J,K)∣u(t)≥σu(τ),∀t,τ∈J},
where σ=1/cosh2M.Lemma 2.6.
For anyf(J,K)⊂K,A(C(J,K))⊂P.Proof.
For anyu∈C(J,K),t,τ∈J,  by (2.18) and the second inequality of (2.17), we have
(2.25)A(u(τ))=∫01G(τ,s)f(s,u(s))ds+∑k=1mG(τ,tk)Ik(u(tk))≤cosh⁡2MMsinh⁡M(∫01f(s,u(s))ds+∑k=1mIk(u(tk))).
By this, (2.18), and the first inequality of (2.17), we have
(2.26)A(u(t))=∫01G(t,s)f(s,u(s))ds+∑k=1mG(t,tk)Ik(u(tk))≥1Msinh⁡M(∫01f(s,u(s))ds+∑k=1mIk(u(tk)))≥σA(u(τ)).
Hence, A(C(J,K))⊂P.Thus, for anyf(J,K)⊂K,A:P→P is condensing mapping; the positive solution of NBVP (1.1) is equivalent to the nontrivial fixed point of A in P. For 0<r<R<∞, let Pr={u∈P∣∥u∥C<r}, and ∂Pr={u∈P∣∥u∥C=r}, which is the relative boundary bound of Pr in P. Denote that Pr,R=PR∖Pr¯; then the fixed point of A in Pr,R is the positive solution of NBVP (1.1). We will use the fixed point theory of condensing mapping to find the fixed point of A in Pr,R.LetX be a Banach space and let  P⊂X be a cone in X. Assume that Ω is a bounded open subset of X and let ∂Ω be its bound. Let Q:P∩Ω¯→P be a condensing mapping. If Qu≠u for every u∈P∩∂Ω, then the fixed point index i(Q,P∩Ω,P) is defined. If i(Q,P∩Ω,P)≠0, then Q has a fixed point in P∩Ω. As the fixed point index theory of completely continuous mapping, see [10, 11], we have the following lemmas that are needed in our argument for condensing mapping.Lemma 2.7.
LetQ:P→P be condensing mapping; if
(2.27)u≠λAu,∀u∈∂Pr,0<λ≤1,
then i(Q,Pr,P)=1.Lemma 2.8.
LetQ:P→P be condensing mapping; if there exists v0∈P,v0≠θ, such that
(2.28)u-Au≠τv0,∀u∈∂Pr,τ≥0,
then i(Q,Pr,P)=0.


## 3. Main Results

(P1)
(i) There existδ>0,   a,ak>0,  such that for all x∈Pδ and t∈J,  f(t,x)≤ax,  Ik(x)≤akx, and a+(∑k=1mak/σ2)<M.(ii) There exist b,bk>0,  h0∈C(J,K),  and yk∈K, such that for all x∈P and t∈J,  f(t,x)≥bx-h0(t),  Ik(x)≥bkx-yk, and b+σ2∑k=1mbk>M.(P2)
(i) There existδ>0,  b,bk>0,  such that for all x∈Pδ and t∈J,  f(t,x)≥bx,  Ik(x)≥bkx, and b+σ2∑k=1mbk>M.(ii) There exist a,ak>0,  h0∈C(J,K), and yk∈K, such that for all x∈P and t∈J,  f(t,x)≤ax+h0(t),  Ik(x)≤akx+yk, and a+(∑k=1mak/σ2)<M.Theorem 3.1.
LetE be an ordered Banach space, whose positive cone K is normal, f∈C(J×K,K), and Ik∈C(K,K),k=1,2,…,m. Suppose that conditions (P0) and (P1) or (P2)  are satisfied; then the NBVP (1.1) has at least one positive solution.Proof.
We show, respectively, that the operatorA defined by (2.18) has a nontrivial fixed point in two cases that (P1) is satisfied and (P2) is satisfied.
Case 1.
Assume that(P1) is satisfied; let 0<r<δ, where δ is the constant in condition (P1), to prove that A satisfies
(3.1)u≠λAu,∀u∈∂Pr,0<λ≤1.
If (3.1) is not true, then there exist u0∈∂Pr and 0<λ0≤1, such that u0=λ0Au0; by the definition of A, u0(t) satisfies
(3.2)-u0′′(t)+Mu0(t)=λ0f(t,u0(t)),t∈J,t≠tk,-Δu0′|t=tk=λ0Ik(u0(tk)),k=1,2,…,m,u0′(0)=u0′(1)=θ.
Integrating (3.2) from 0 to 1, using (i) of assumption (P1), we have
(3.3)(M-a)∫01u0(t)dt≤∑k=1maku0(tk).
Since u0∈P,  for any   t,s∈J, by the definition of P, we have u0(t)≥σu0(s), u0(tk)≤(1/σ)u0(s), and thus
(3.4)σ(M-a)u0(s)≤∑k=1makσu0(s),
that is; (M-(a+(∑k=1mak/σ2)))u0(s)≤θ. So we obtain that u0(s)≤θ in J, which contracts with u0∈∂Pr. Hence (3.1) is satisfied; by Lemma 2.7, we have
(3.5)i(A,Pr,P)=1.
Let e∈C(J,K),∥e∥=1,  v0(t)≡e, and obviously v0∈P. We show that if R is large enough, then
(3.6)u-Au≠τv0,∀u∈∂PR,τ≥0.
In fact, if there exist u0∈∂PR,  τ0≥0 such that u0-Au0=τ0v0, then Au0=u0-τ0v0; by the definition of A,u0(t) satisfies
(3.7)-u0′′(t)+Mu0(t)-Mτ0v0=f(t,u0(t)),t∈J,t≠tk,-Δu0′|t=tk=Ik(u0(tk)),k=1,2,…,m,u0′(0)=u0′(1)=θ.
By  (ii) of assumption (P1), we have
(3.8)-u0′′(t)+Mu0(t)=f(t,u0(t))+Mτ0v0≥bu0(t)-h0(t),t∈J′.
Integrating on J and using (ii) of assumption (P1), we have
(3.9)(b-M)∫01u0(t)dt+∑k=1mbku0(tk)≤∫01h0(t)dt+∑k=1myk.
If b>M, for any t,s∈J, by the definition of P, we have u0(t)≥σu0(s);  u0(tk)≥σu0(s); thus
(3.10)(σ(b-M)+σ∑k=1mbk)u0(s)≤∫01h0(t)dt+∑k=1myk.
By σ(b-M)+σ∑k=1mbk>0 and the normality of cone K, we have
(3.11)‖u0‖C≤N(‖h0‖C+∑k=1m∥yk∥)σ(b-M)+σ∑k=1mbk≜R1.
If b≤M, then for any t,s∈J, by the definition of P, we have u0(t)≤(1/σ)u0(s),  u0(tk)≥σu0(s); thus
(3.12)((b-M)σ+σ∑k=1mbk)u0(s)≤∫01h0(t)dt+∑k=1myk.
By b+σ2∑k=1mbk>M, and the normality of K, we have
(3.13)‖u0‖C≤N(‖h0‖C+∑k=1m∥yk∥)((b-M)/σ)+σ∑k=1mbk≜R2.
LetR>max{R1,R2,r}; then (3.6) is satisfied; by Lemma 2.8, we have
(3.14)i(A,PR,P)=0.
Combining (3.5), (3.14), and the additivity of fixed point index, we have
(3.15)i(A,Pr,R,P)=i(A,PR,P)-i(A,Pr,P)=-1≠0.
Therefore A has a fixed point in Pr,R, which is the positive solution of NBVP (1.1).
Case 2.
Assume that(P2) is satisfied; let 0<r<δ, where δ is the constant in condition (P2), to proof that A satisfies
(3.16)u-Au≠τv0,∀u∈∂Pr,τ≥0,
where v0(t)=e∈P,e≠θ. In fact, if there exists u0∈∂Pr and τ0≥0, such that u0-Au0=τ0v0, then u0 satisfies (3.7) and (i) of condition (P2), and we have
(3.17)-u0′′(t)+Mu0(t)=f(t,u0(t))+Mτ0v0≥bu0(t),t∈J′.
Integrating on J and using (i) of assumption (P2), we have
(3.18)(b-M)∫01u0(t)dt+∑k=1mbku0(tk)≤θ.
Ifb>M, for any t,s∈J, by the definition of P, we have u0(t)≥σu0(s),u0(tk)≥σu0(s), for allt,s∈J, and thus
(3.19)(σ(b-M)+σ∑k=1mbk)u0(s)≤θ.
By σ(b-M)+σ∑k=1mbk>0, we obtain that u0(s)≤θ, which contracts with u0∈∂Pr. Hence (3.16) is satisfied.
Ifb≤M, for any t,s∈J, by the definition of P, we have u0(t)≤(1/σ)u0(s),  u0(tk)≥σu0(s), for all t,s∈J,  and thus
(3.20)(1σ(b-M)+σ∑k=1mbk)u0(s)≤θ.
By b+σ2∑k=1mbk>M, we obtain that u0(s)≤θ, which contracts with u0∈∂Pr. Hence (3.16) is satisfied.
Hence, by Lemma2.8, we have
(3.21)i(A,Pr,P)=0.
Next, we show that ifR is large enough, then
(3.22)u≠λAu,∀u∈∂PR,0<λ≤1.
In fact, if there exists u0∈∂PR and 0<λ0≤1 such that u0=λ0Au0, then u0 satisfies (3.2). Integrating (3.2) on J, and using (ii) of (P2),  we have
(3.23)(M-a)∫01u0(t)dt-∑k=1maku0(tk)≤∫01h0(t)dt+∑k=1myk.
For any t,s∈J, by the definition of P, we have u0(t)≥σu0(s),  u0(tk)≤(1/σ)u0(s), for allt,s∈J, and thus
(3.24)(σ(M-a)-1σ∑k=1mak)u0(s)≤∫01h0(t)dt+∑k=1myk.
By a+(∑k=1mak/σ2)<M, we have
(3.25)u0(s)≤∫01h0(t)dt+∑k=1mykσ(M-a)-(1/σ)∑k=1mak.
By the normality of K, we have
(3.26)‖u0(s)‖≤N‖∫01h0(t)dt+∑k=1myk‖σ(M-a)-(1/σ)∑k=1mak.
Thus
(3.27)‖u0‖C≤N(‖h0‖C+∑k=1m‖yk‖)σ(M-a)-(1/σ)∑k=1mak≜R3.
LetR>max{R3,r}; then  (3.22)  is satisfied; by  Lemma 2.7, we have
(3.28)i(A,PR,P)=1.
Combining (3.21), (3.28), and the additivity of fixed index, we have
(3.29)i(A,Pr,R,P)=i(A,PR,P)-i(A,Pr,P)=1.
ThereforeA has a fixed point in Pr,R, which is the positive solution of NBVP (1.1).Remark 3.2.
The conditions(P1) and (P2) are a natural extension of suplinear condition and sublinear condition in Banach space E. Hence if Ik=θ, then Theorem 3.1 improves and extends the main results in [8, 9].


## 4. Example

We provide an example to illustrate our main result.Example 4.1.
Consider the following problem:(4.1)-∂2∂t2w(t,x)+w(t,x)=∫01e(t-s)w2(s,x)ds+1100w(t,x),t∈J′,x∈I,-Δ∂∂tw(t,x)|t=1/2=1100w(12,x),x∈I,∂∂tw(0,x)=∂∂tw(1,x)=0,x∈I,
where J=[0,1],J′=J∖{1/2},I=[0,T], and   T>0 is a constant.Conclusion
Problem (4.1) has at least one positive solution.Proof.
LetE=C(I),and K={w∈C(I)∣w(x)≥0,x∈I}; then E is a Banach space with norm ∥w∥=maxt∈I|w(x)|, and  K is a positive cone of E. Let u(t)=w(t,·); then the problem (4.1) can be transformed into the form of NBVP (1.1), where M=1,  f(t,u)=∫01e(t-s)u2(s)ds+(1/100)u(t), and  I1(u(1/2))=(1/100)u(1/2). Evidently C(J,E) is a Banach space with norm ∥u∥C=maxt∈J∥u(t)∥, and C(J,K) is positive cone of C(J,E). Let P={u∈C(J,K)∣u(t)≥σu(s),t,s∈J}, where σ=1/cosh21;  then P is a cone in C(J,K), and for any u∈P,t∈J, we have σ∥u∥C≤u(t)≤∥u∥C.
Next, we will verify that the conditions(P0) and (P1) in Theorem 3.1 are satisfied.
It is easy to verify that for anyR>0,   f(J×KR) and  I1(KR) are bounded. Let g(t,u)=∫01e(t-s)u2(s)ds; then g is completely continuous. So for any countable bounded set D⊂K, we have
(4.2)α(f(t,D))≤α(g(t,D))+1100α(D)=1100α(D),α(I1(D))=1100α(D),
and for L=1/100,  and  M1=1/100, by simple calculations, (4L/M)+(2cosh2MM1/MsinhM)<1. So condition (P0) is satisfied.
Letδ=1/100; then for u∈Pδ, we have
(4.3)f(t,u)≤δ∫01e(t-s)dsσu(t)+1100u(t)≤(δ(e-1)σ+1100)u(t),I1(u(12))=1100u(12)≤1100σu(t).
Let a=(δ(e-1)/σ)+(1/100), a1=(1/100)σ; by simple calculations, we have a+(a1/σ2)<1. So (i) of condition (P1) is satisfied.
LetR=100,  for u∈P,∥u∥C≥R; we have f(t,u)≥(1/2σ2R+(1/100))u(t). For u∈P,  0≤∥u∥C≤R, we have f(t,u)≤R2et+1. Hence, let h0(t)=10000et+1,   y1=0; then for any u∈P, we have
(4.4)f(t,u)≥(12σ2R+1100)u(t)-h0(t)I1(u(12))≥1100σu(t).
Let b=(1/2)σ2R+(1/100), and   b1=(1/100)σ; by simple calculations, we have b+σ2b1>1,  so (ii) of condition (P1) is satisfied.
By Theorem3.1, Problem (4.1) has at least one positive solution.



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*Source: 401923-2012-03-15.xml*