# Positive Solution of Fourth-Order Integral Boundary Value Problem

Authors: Guoqing Chai
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

The author investigates the fourth-order integral boundary value problem with two parametersu(4)(t)+βu′′(t)-αu(t)=f(t,u),t∈(0,1),u(0)=u(1)=0,u′′(0)=∫01u(s)ϕ1(s)ds,u′′(1)=∫01u(s)ϕ2(s)ds, where nonlinear term function f is allowed to change sign. Applying the fixed point index theorem on cone together with the operator spectrum theorem, some results on the existence of positive solution are obtained.


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## Body

## 1. Introduction

The theory of boundary value problems with integral boundary conditions for ordinary differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can all be reduced to nonlocal problems with integral boundary conditions (see, e.g., [1–3]). For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers by Gallardo [4], Karakostas and Tsamatos [5],  and Lomtatidze and Malaguti [6] and the references therein. For more information about the general theory of integral equations and their relation to boundary value problems, we refer to the books of Corduneanu [7] and Agarwal and O'Regan [8].Moreover, boundary value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoints and nonlocal boundary value problems as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention. To identify a few, we refer the reader to [9–15] and the references therein.In the recent literature, several sorts of boundary value problems with integral boundary conditions have been studied further, see [16–20]. Especially, Ruyun Ma and Yulian An [18] investigated the global structure of positive solutions for nonlocal boundary value problems(1.1)u′′(t)+λh(t)f(u(t))=0,0<t<1,u(0)=0,u(1)=∫01u(s)dA(s),
by using global bifurcation techniques, where f∈C([0,∞),[0,∞)),h∈C((0,1),[0,∞)). In [19], Jiqiang Jiang et al. investigated the existence of positive solution for second-order singular Sturm-Liouville integral boundary value problems(1.2)-u′′(t)=λh(t)f(t,u(t)),0<t<1,αu(0)-βu′(0)=∫01a(s)u(s)ds,γu(1)+δu′(1)=∫01b(s)u(s)ds,
by using the fixed point theory in cones, where f∈C([0,1]×(0,∞),[0,∞)).On the other hand, the fourth-order boundary value problem describe the deformations of an elastic beam in equilibrium state. Owing to its importance in physics, the existence of solutions to this problem has been studied by many authors; see, for example, [21–24] and references therein. Especially, in [22], Li studied existence of positive solution for fourth-order boundary value problem(1.3)u(4)(t)+βu′′(t)-αu(t)=f(t,u),t∈(0,1),u(0)=u(1)=0=u′′(0)==u′′(1)=0,
by using the fixed point index theorem, where f∈C([0,∞),[0,∞)).Motivated by the above-mentioned works [18, 19, 22], in this paper, we study the following fourth-order integral boundary value problem (for short BVP in the sequel) with two parameters:(1.4)u(4)(t)+βu′′(t)-αu(t)=f(t,u),t∈(0,1),u(0)=u(1)=0,u′′(0)=∫01u(s)ϕ1(s)ds,u′′(1)=∫01u(s)ϕ2(s)ds,
where nonlinear term function f is allowed to change sign. To the best of our knowledge, BVP has not been investigated up to now. In the literature such as above-mentioned paper [18, 19, 22], the nonnegativity on f is a usual assumption. In the present paper, since the function f is not assumed to be nonnegative, the corresponding integral operator doesn't map the cone into cone, and so, there exists difficulty in applying the cone fixed point theorem. On the other hand, owing to the occurrence of parameter α,β in this boundary value problem including integral boundary conditions, it is not easy to transform the BVP (1.4) into an integral equation directly. To overcome these difficulties, we first introduce operator spectrum method combined with some analysis technique, next apply the fixed point index theorem, and establish existence of positive solution to BVP (1.4).Let us begin with listing the following assumption conditions, which will be used in the sequel:LetI=[0,1],R=(-∞,+∞),R-=(-∞,0],R+=[0,+∞).(H1)
f∈C[I×R+,R]andexistsM∈L1(0,1)∩C[(0,1),R+]suchthat(1.5)f(t,u)+M(t)≥0,(t,u)∈(0,1)×R+.(H2)
α,β∈R,β<2π2,α≥-β2/4,α/π4+β/π2<1.Letλ1,λ2 be the roots of the polynomial p(λ)=λ2+βλ-α; namely,(1.6)λ1,λ2=-β±β2+4α2.
By (H2), it is to see that λ1≥λ2>-π2.LetΓ0=π4-βπ2-α. Then (H2) implies Γ0>0. Let X=C[0,1] be the real Banach space equipped with the norm ||u||=max0≤t≤1|u(t)|. Denote by P the set P={u∈X:u(t)≥0,t∈I} in X.


## 2. Preliminaries

In this section, we shall give some important preliminary lemmas, which will be used in proving of our main results.Lemma 2.1 (see [22, 23]).
Suppose that (H2) holds, then there exist uniqueφi,ψi,i=1,2 satisfying
(2.1)-φi′′(t)+λiφi(t)=0,t∈[0,1],φi(0)=0,φi(1)=1,-ψi′′(t)+λiψi(t)=0,t∈[0,1],ψi(0)=1,ψi(1)=0,
respectively, and φi≥0,ψi≥0 on [0,1], where λi is as in (1.6). Moreover, φi,ψi have the expression
(2.2)φi(t)={sinh⁡ωitsinh⁡ωi,λi>0,t,λi=0,sinωitsinωi,-π2<λi<0,ψi(t)={sinh⁡ωi(1-t)sinh⁡ωi,λi>0,1-t,λi=0,sinωi(1-t)sinωi,-π2<λi<0,
where ωi=|λi|,i=1,2.LetGi(t,s)(i=1,2) be the Green function of the linear boundary value problem(2.3)-u′′(t)+λiu(t)=0,t∈[0,1],u(0)=u(1)=0.
By [22, 23], Gi(t,s) can be expressed by the formula(2.4)Gi(t,s)=1σi{φi(t)ψi(s),0≤t≤s≤1,ψi(t)φi(s),0≤s≤t≤1,
where
(2.5)σi={ωisinh⁡ωi,ifλi>0,1,ifλi=0,i=1,2.ωisinωi,if-π2<λi<0,Lemma 2.2 (see [22, 23]).
Gi=Gi(t,s)(i=1,2) have the following properties: (i)
Gi(t,s)>0,∀t,s∈(0,1).(ii)
Gi(t,s)≤CiGi(s,s),∀t,s∈[0,1],φi≤Ci,ψi≤Ci,t∈[0,1].(iii)
Gi(t,s)≥δiGi(t,t)Gi(s,s),∀t,s∈[0,1],φi(t)≥δiGi(t,t),ψi(t)≥δiGi(t,t),t∈[0,1], where(2.6)Ci={1,ifλi≥0,1sinωi,if-π2<λi<0,δi={ωisinh⁡ωi,ifλi>0,1,ifλi=0,ωisinωi,if-π2<λi<0.PutDi=maxt∈I∫01Gi(t,s)ds,i=1,2. Set E21=D2C1,E12=D1C2, where Ci is described as before. We need also the following assumptions in the sequel.(H3) Functionsϕi∈C[I,R-],i=1,2, satisfy D=̇E12∫01|ϕ1(s)|ds+E21∫01|ϕ2(s)|ds<1.Leth∈C(0,1)∩L1(0,1), consider the following BVP:(2.7)u(4)(t)+βu′′(t)-αu(t)=h(t),t∈(0,1),u(0)=u(1)=u′′(0)=u′′(1)=0.
By papers [22, 23], BVP (2.7) has a unique solution u=Kh expressed by(2.8)Kh(t)=∫01∫01G1(t,s)G2(s,τ)h(τ)dτds=∫01∫01G2(t,s)G1(s,τ)h(τ)dτds,t∈[0,1].Letw=KM. Since M∈L1(0,1)∩C[(0,1),R+], by Lemma 2.2, it is easy to verify that w∈P.Let(2.9)g1(t)=-∫01G2(t,s)φ1(s)ds,t∈[0,1],
where φ1 is as in (2. 1). By Lemmas 2.1 and 2.2, we have g1∈C2([0,1],R-) and(2.10)-g1′′(t)+λ2g1(t)=-φ1(t),t∈[0,1],g1(0)=g1(1)=0.
On the other hand, φ1 satisfies the following relation:(2.11)-φ1′′(t)+λ1φ1(t)=0,t∈[0,1],φ1(0)=0,φ1(1)=1.
So, from (2.10)–(2.11), it follows that(2.12)g1′′(0)=λ2g1(0)+φ1(0)=0,g1′′(1)=λ2g1(1)+φ1(1)=1.
Now, we make the following decomposition:(2.13)g1(4)+βg1′′-αg1=(-d2dt2+λ1)(-d2dt2+λ2)g1=(-d2dt2+λ1)(-g1′′+λ2g1)=d2φ1dt2-λ1dφ1dt=0.
So by (2.10), (2.12)-(2.13), it follows that(2.14)g1(4)(t)+βg1′′(t)-αg1(t)=0,t∈[0,1],g1(0)=g1(1)=0,g1′′(0)=0,g1′′(1)=1,g1(t)≤0,t∈[0,1].
Similarly, by setting(2.15)g2(t)=-∫01G1(t,s)ψ2(s)ds,t∈[0,1],
we have(2.16)g2(4)(t)+βg2′′(t)-αg2(t)=0,t∈[0,1],g2(0)=g2(1)=0,g2′′(0)=1,g2′′(1)=0,g2(t)≤0,t∈[0,1].For anyu∈X, define u* as(2.17)u*(t)={u(t),ifu(t)≥0,0,ifu(t)<0.
Obviously, u*∈P for any u∈X.Leth∈L1(0,1)∩C(0,1); consider the BVP with integral boundary conditions(2.18)u(4)(t)+βu′′(t)-αu(t)=h(t),t∈(0,1),u(0)=u(1)=0,u′′(0)=∫01[u-w]*(s)ϕ1(s)ds,u′′(1)=∫01[u-w]*(s)ϕ2(s)ds.Denote operator B onC[0,1] by(2.19)Bu(t)=g2(t)∫01[u-w]*(s)ϕ1(s)ds+g1(t)∫01[u-w]*(s)ϕ2(s)ds.
It is easy to see that B maps C[0,1] into C[0,1].Define operator L:C4(0,1)→C(0,1) as follows:(2.20)Lu=u(4)+βu′′-αu.We need the following Lemma.Lemma 2.3.
Let (H2) holds. Assume thath∈L1(0,1)∩C(0,1) and  ϕi∈C[I,R-],i=1,2. Then  u¯∈C4(0,1)∩C2[0,1] is a solution of (2.18) if and only if  u¯ is a solution of operator equation  u=Kh+Bu in  C[0,1].Proof.
(1) Assumeu¯∈C4(0,1)∩C2[0,1] is a solution of (2.18). By (2.14)–(2.20), we have
(2.21)(Bu¯)(0)=(Bu¯)(1)=0,(Bu¯)′′(0)=∫01[u¯-w]*(s)ϕ1(s)ds,(Bu¯)′′(1)=∫01[u¯-w]*(s)ϕ2(s)ds,L(Bu¯)=(Lg2)∫01[u¯-w]*(s)ϕ1(s)ds+(Lg1)∫01[u¯-w]*(s)ϕ2(s)ds=0.
Let v¯=u¯-Bu¯. Then Lv¯(t)=Lu¯(t)-LBu¯(t)=Lu¯(t)=h(t),t∈(0,1);v¯(0)=u¯(0)-(Bu¯)(0)=0,v¯(1)=u¯(1)-(Bu¯)(1)=0;v¯′′(0)=u¯′′(0)-(Bu¯)′′(0)=0,v¯′′(1)=u¯′′(1)-(Bu¯)′′(1)=0. Thus, by (2.7)-(2.8), we have v¯=Kh,v¯∈C[0,1], and so u¯=Kh+Bu¯,u¯∈C[0,1].
(2) Inversely, assumeu¯∈C[0,1] satisfies u¯=Kh+Bu¯. Then u¯∈C4(0,1)∩C2[0,1]. By (2.7), (2.8),(2.14)–(2.20), we have
(2.22)LKh=h,LBu¯=0,(Kh)(0)=(Kh)(1)=(Kh)′′(0)=(Kh)′′(1)=0,(Bu¯)(0)=(Bu¯)(1)=0,(Bu¯)′′(0)=∫01[u¯-w]*(s)ϕ1(s)ds,(Bu¯)′′(1)=∫01[u¯-w]*(s)ϕ2(s)ds.
Consequently,
(2.23)Lu¯=LKh+LBu¯=h,u¯(0)=(Kh)(0)+(Bu¯)(0)=0,u¯(1)=(Kh)(1)+(Bu¯)(1)=0,u¯′′(0)=(Kh)′′(0)+(Bu¯)′′(0)=∫01[u¯-w]*(s)ϕ1(s)ds,u¯′′(1)=(Kh)′′(1)+(Bu¯)′′(1)=∫01[u¯-w]*(s)ϕ2(s)ds.
Hence, u¯ is a solution of (2.18). The proof is complete.
We have also the following lemma.Lemma 2.4.
Suppose (H3) holds. ThenB:X→X is a bounded operator with  ||B||≤D(<1) and  BX⊂P.Proof.
In view of Lemma2.2 (ii), by (2.9),(2.15),(2.19) and (H3), noticing that w∈P, for any u∈X and t∈I, we have
(2.24)|(Bu)(t)|≤|g2(t)|∫01[u-w]*ϕ1(s)|ds+|g1(t)|∫01[u-w]*ϕ2(s)|ds≤E12∫01|u(s)||ϕ1(s)|ds+E21∫01|u(s)||ϕ2(s)|ds≤D‖u‖.
Thus, ||Bu||≤D||u||, and so ||B||≤D(<1).
On the other hand, fromgi(t)≤0,ϕi(t)≤0,t∈I,i=1,2, we have BX⊂P. So, Lemma 2.4 is true.
By (2.7)-(2.8), it follows from w=KM that
(2.25)w(4)(t)+βw′′(t)-αw(t)=M(t),t∈(0,1),w(0)=w(1)=w′′(0)=w′′(1)=0.
For anyu∈X, let f¯u(t)=f(t,[u-w]*(t)),t∈[0,1] and Gu(t)=f¯u(t)+M(t),t∈(0,1). Under conditions (H1)–(H3), consider the following auxiliary BVP:
(2.26)u(4)(t)+βu′′(t)-αu(t)=Gu(t),t∈(0,1),u(0)=u(1)=0,u′′(0)=∫01[u-w]*(s)ϕ1(s)ds,u′′(1)=∫01[u-w]*(s)ϕ2(s)ds.
Notice that w(t) satisfies (2.25), it is easy to see that u¯∈C4(0,1)∩C2[0,1] is a solution of (2.26) if and only if u¯-w∈C4(0,1)∩C2[0,1] is a solution of the following BVP:
(2.27)u(4)(t)+βu′′(t)-αu(t)=f(t,u*(t)),t∈(0,1),u(0)=u(1)=0,u′′(0)=∫01u*(s)ϕ1(s)ds,u′′(1)=∫01u*(s)ϕ2(s)ds.
Thus, if and only if u¯(t)≥w(t),t∈[0,1], then u¯-w is a solution of BVP (1.4).
Now, by Lemma2.3, u¯∈C4(0,1)∩C2[0,1] is a solution of (2.26) if u¯∈X is a fixed point of the operator KG+B. So, we only need focusing our attention on the existence of the fixed point of KG+B.
For the remainder of this section, we give the definition of positive solution.
By a positive solution of BVP (1.4), we mean a function u∈C4(0,1)∩C2[0,1] such that u(t)≥0,t∈[0,1],  u(t)>0,t∈(0,1), and u satisfies (1.4).


## 3. Main Results

We introduce now some notations, which will be used in the sequel.LetC1, δ1, and D be as described in Lemma 2.2 and (H3), respectively. We also set(3.1)d1=∫01∫01G2(s,τ)M(τ)dτds,b0=C12d1δ1(1-D),f0=lim⁡u→+0¯inf⁡t∈If(t,u)u,f∞=lim⁡¯u→+∞max⁡t∈If(t,u)u.
We also need the following assumption.(H4) There exists a numberr0∈(b0,+∞), and Γ1≥r0/Γ0(r0-b0) such that(3.2)f(t,u)+M(t)≥Γ1u,∀(t,u)∈(0,1)×[0,r0].We are now in a position to state and prove our main results on the existence.Theorem 3.1.
Suppose that (H1)–(H4) hold. Iff∞=0, then BVP (1.4) has a positive solution.Proof.
By Lemma2.4 together with (H3), we have ||B||≤D(<1). By operator spectrum theorem, we know that (I-B)-1 exists and is bounded. Furthermore, by Neumann expression, (I-B)-1can be expressed by
(3.3)(I-B)-1=∑n=0∞Bn.
Noticing that BP⊂P and from (3.3), we have
(3.4)(I-B)-1u=∑n=0∞Bnu≥u,∀u∈P,(3.5)‖(I-B)-1‖≤∑n=0∞‖B‖n=11-‖B‖≤11-D.
Thus, from the reversibility of I-B, we have
(3.6)u=KGu+Bu,u∈X⟺u=(I-B)-1KGu,u∈X.
The following proof will be divided into five steps.Step 1.
We will show that(I-B)-1KG:P→P is completely continuous.(1)
KG maps P into P.For any u∈P, it follows from (H1) that f(t,[u-w]*(t))∈X, and so (Gu)(t)≥0,t∈(0,1),Gu∈C(0,1)∩L1(0,1). By (H1)-(H2) together with Lemma 2.2, for any t∈[0,1], we have
(3.7)0≤(KGu)(t)=∫01∫01G1(t,s)G2(s,τ)[f(τ,[u-w]*(τ))+M(τ))dτds≤ρη+ρ∫01M(τ)dτ<+∞,
where η=maxt∈[0,1]|f(t,[u-w]*(t))|<∞,ρ=maxt,s,τ∈[0,1]G1(t,s)G2(s,τ)<∞.From the continuity ofG1(t,s), it is easy to see that KGu∈X, and so KGu∈P.(2)
KG is a compact operator on P.Assume that U is a arbitrary bounded set in P. Then there exists a L0>0 such that ∥u∥≤L0 for all u∈U. Also, we have ∥[u-w]*∥≤L0 for all u∈U since w∈P. Consequently,
(3.8)0≤(KGu)(t)≤ρb+ρ∫01M(τ)dτ<+∞,∀w∈P,
where b=max(t,u)∈I×[0,L0]|f(t,u)|,ρ=maxt,s,τ∈[0,1]G1(t,s)G2(s,τ). That means {KGu∣u∈U} is a uniformly bounded set in P.On the other hand, the continuity ofG1 on I×I yields that for every ɛ>0, there exists δ>0 such that for any t1,t2∈I with |t1-t2|<δ, the following inequality(3.9)|G1(t2,s)-G1(t1,s)|<ɛ
holds for all s∈I, and so,
(3.10)|(KGu)(t2)-(KGu)(t1)|≤ɛ∫01∫01G2(s,τ)(Gu)(τ)dτds≤ɛ[be2+e2∫01M(τ)dτ],
for any u∈U, where e2=maxs,τ∈[0,1]G2(s,τ)<+∞,b=max(t,u)∈I×[0,L0]|f(t,u)|<+∞. That is, {KGu∣u∈U}is equicontinuous.
Hence, in view of Arzela-Ascoli theorem, we know that the operatorKG is compact on P.(3)
Now, we show that the operatorKG is continuous.Indeed, for any sequence {un} in P with un→u and any t∈I, we have
(3.11)|[un-w]*(t)-[u-w]*(t)|=12|[|un(t)-w(t)|+(un(t)-w(t))]-[|u(t)-w(t)|+u(t)-w(t)]|=12|[|un(t)-w(t)|-|u(t)-w(t)|+un(t)-u(t)]|≤12{|[|un(t)-w(t)|-|u(t)-w(t)|]|+|un(t)-u(t)|}≤|un(t)-u(t)|.
Thus, ∥[un-w]*-[u-w]*∥→0, and, by Lemma 2.2, it follows from the continuity of f that
(3.12)‖KGun-KGu‖≤C1∫01∫01G1(s,s)G2(s,τ)|f(τ,[un-w]*(τ))-f(τ,[u-w]*(τ))|dτds⟶0.
By (1)–(3) we obtain that KG:P→P is completely continuous.
Now, from (3.4), we have (I-B)-1:P→P is continuous, and so, (I-B)-1KG:P→P is completely continuous.
Now we set(3.13)Q=(I-B)-1KG,q1(t)=δ1C1G1(t,t),t∈I,
where δ1,C1 are described in Lemma 2.2. Set
(3.14)P0={u∈P:u(t)≥(1-‖B‖)q1(t)‖u‖,t∈I}.
Obviously, P0 is a cone in X.Step 2.
Q:P→P0.
In fact, for anyu∈P and every t,σ in I, by Lemma 2.2, we have
(3.15)(KGu)(t)=∫01∫01G1(t,s)G2(s,τ)(Gu)(τ)dτds≥δ1G1(t,t)∫01∫01G1(s,s)G2(s,τ)(Gu)(τ)dτds≥δ1C1G1(t,t)∫01∫01G1(σ,s)G2(s,τ)(Gu)(τ)dτds=q1(t)(KGu)(σ).
Thus, we have
(3.16)(KGu)(t)≥q1(t)‖KGu‖,t∈I.
Since KGP⊂P, by (3.4) together with (3.16) for every t∈I, we have
(3.17)((I-B)-1KGu)(t)≥(KGu)(t)≥q1(t)‖KGu‖.
On the other hand, since ∥(I-B)-1(KGu)∥≤∥(I-B)-1∥·∥KGu∥, by (3.5), we have
(3.18)‖KGu‖≥1‖(I-B)-1‖‖(I-B)-1KGu‖≥(1-‖B‖)‖(I-B)-1KGu‖.
Inequality (3.17) together with (3.18) implies for every t∈I(3.19)((I-B)-1KGu)(t)≥(1-‖B‖)q1(t)‖(I-B)-1KGu‖,
namely, (Qu)(t)≥(1-∥B∥)q1(t)∥Qu∥,t∈I. Thus, we obtain that Q maps P into P0.Step 3.
We shall deduce that for anyu∈P0 and t∈I, the following inequality holds:
(3.20)u(t)-w(t)≥(1-b0‖u‖)u(t),
where b0=C12d1/δ1(1-D).
In fact, in view of Lemma2.2 and the symmetry of G1(t,s), we have
(3.21)G1(t,s)=G1(s,t)≤C1G1(t,t),∀t,s∈I.
Thus, keeping in mind that d1=∫01∫01G2(s,τ)M(τ)dτds, it follows from w=KM that
(3.22)w(t)=∫01∫01G1(t,s)G2(s,τ)M(τ)dτds≤C1G1(t,t)∫01∫01G2(s,τ)M(τ)dτds=C12δ1d1q1(t),t∈I.
On the other hand, from u∈P0, it follows that
(3.23)u(t)≥q1(t)(1-‖B‖)‖u‖≥q1(t)(1-D)‖u‖,t∈I.
Thus, by (3.22)-(3.23), we have
(3.24)w(t)≤C12d1δ1(1-D)‖u‖,
and so,
(3.25)u(t)-w(t)≥(1-C12d1δ1(1-D)‖u‖)u(t)=(1-b0‖u‖)u(t),t∈I,
where b0=C12d1/(δ1(1-D)).Step 4.
By (H4), we have(3.26)f(t,u)+M(t)≥Γ1u,(t,u)∈(0,1)×[0,r0].
Let ϕ0=sinπt. By (2.7)-(2.8), we easily know that sinπt is a positive eigenfunction of operator K with respect to positive eigenvalue Γ0, that is, Kϕ0=Γ0ϕ0.
Now, we show thatϕ0∈P0, that is, ϕ0(t)≥(1-∥B∥)q1(t)∥ϕ0∥,t∈I. We discuss it in three different cases.(1)
λ1=0. In this case, G1(t,t)=t(1-t),t∈I, and C1=δ1=1.
(i)
Ift∈[0,1/2], then πt∈[0,π/2]. By Jordan's inequality, we have
(3.27)sinπt≥2π⋅πt=2t,t∈[0,12].(ii)
Ift∈[1/2,1], by setting x=1-t, we have x∈[0,1/2]. Then from (3. 12), it follows that(3.28)sinπt=sinπ(1-x)=sinπx≥2x=2(1-t),t∈[12,1].
Thus, by (i)-(ii) above, we have
(3.29)sinπt≥2t(1-t)=2G1(t,t)=2q1(t),t∈I.(2)
λ1>0. In this case, G1(t,t)=(sinhω1t·sinhω1(1-t))/ω1sinhω1,t∈I, and C1=1,δ1=ω1/sinhω1.
(i)
Ift∈[0,1/2], by setting φ(t)=(cosh(ω1/2))t-(sinhω1t/ω1),t∈[0,1/2], we have(3.30)φ′(t)=cosh⁡ω12-cosh⁡ω1t≥cosh⁡ω12-cosh⁡ω12=0,t∈[0,12].
From ϕ(0)=0, it follows that 0≤sinhω1t/ω1≤(cosh(ω1/2))t,t∈[0,1/2]. Keeping in mind that 0<sinhω1(1-t)/sinhω1≤1 for all t∈[0,1/2], it follows immediately that
(3.31)G1(t,t)≤(cosh⁡ω12)t,t∈[0,12].(ii)
Ift∈[1/2,1], by setting x=1-t, we have x∈[0,1/2]. From (2)(i) above, it follows that(3.32)G1(t,t)=G1(1-x,1-x)=G1(x,x)≤(cosh⁡⁡(ω1/2))x=(cosh⁡⁡(ω1/2))(1-t),t∈[12,1].
Hence, by (2)(i)-(ii) above, we have
(3.33)G(t,t)≤cosh⁡1ω1⋅{t,t∈[0,12],1-t,t∈[12,1].
On the other hand, by (3.27)-(3.28), we have
(3.34)sinπt≥2⋅{t,t∈[0,12],1-t,t∈[12,1].
Thus, we have immediately
(3.35)sinπt≥2cosh⁡(ω1/2)G1(t,t)=2C1δ1cosh⁡(ω1/2)q1(t)=2sinh⁡ω1ω1cosh⁡(ω1/2)q1(t)=2sinh⁡(ω1/2)ω1/2q1(t),t∈I.
It is easy to verity that sinh(ω12)/ω1/2≥1. Hence, sinπt≥2q1(t),t∈I.
(3)
-π2<λ1<0. In this case, G1(t,t)=sinω1t·sinω1(1-t)/ω1sinω1,t∈I, and C1=1/sinω1, δ1=ω1sinω1.
(i)
Ift∈[0,1/2], then 0≤sinω1t≤sinπt,0<sinω1(1-t)≤1. Thus,
(3.36)G1(t,t)≤sinπtω1sinω1,t∈[0,12].(ii)
Ift∈[1/2,1],  from (i), by letting x=1-t, then we have x∈[0,1/2], and
(3.37)G1(t,t)=G1(1-x,1-x)=G1(x,x)≤sinπxω1sinω1=sinπ(1-t)ω1sinω1=sinπtω1sinω1,t∈[12,1].Thus, (3)(i)-(ii) above implies that
(3.38)sinπt≥ω1sinω1G1(t,t)=ω1sinω1C1δ1q1(t)=1sinω1q1(t)≥q1(t),t∈I.
Summing up (1)–(3) keeping in mind that ∥ϕ0∥=1, we have
(3.39)ϕ0(t)=sinπt≥q1(t)=q1(t)‖ϕ0‖≥(1-‖B‖)q1(t)‖ϕ0‖,t∈I,
that is, ϕ∈P0.
Now, setΩr0={u∈P0:||u0||<r0}. We claim that
(3.40)u≠Qu+λϕ0,∀λ≥0,u∈∂Ωr0.Indeed, if not, then exists a u0∈∂Ωr0 and λ0≥0 with u0=Qu0+λ0ϕ0. Without loss of generality, assume that λ0>0 (otherwise, by proving later on, we will know that the theorem is true). By u0∈∂Ωr0, we have ||u0||=r0, and so, it follows from (3.25) that
(3.41)r0≥u0(t)≥u0(t)-w(t)≥(1-b0r0)u0(t)≥0,t∈I,
since r0>b0.
Thus, by (3.26) and (3.41), we have
(3.42)f(t,[u0-w]*(t))+M(t)=f(t,(u0-w)(t))+M(t)≥Γ1(1-b0r0)u0(t),t∈(0,1).
Therefore, by (3.4), (3.16), we have
(3.43)u0=Qu0+λ0ϕ0=(I-B)-1KGu0+λ0ϕ0≥KGu0+λ0ϕ0=∫01∫01G1(t,s)G2(s,τ)[f(τ,[u-w]*(τ))+M(τ)]dτds+λ0ϕ0≥Γ1(1-b0r0)∫01∫01G1(t,s)G2(s,τ)u0(τ)dτds+λ0ϕ0=Γ1(1-b0r0)Ku0+λ0ϕ0.
Thus, u0≥λ0ϕ0. Let λ*=sup{λ∣u0≥λϕ0}. Then λ*≥λ0, and u0≥λ*ϕ0. By KP⊂P and Kϕ0=Γ0ϕ0, it follows that
(3.44)Ku0≥λ*Kϕ0=λ*Γ0ϕ0.
Thus, by (3.43), we have
(3.45)u0≥Γ1(1-b0r0)Γ0λ*ϕ0+λ0ϕ.
The hypothesis in (H4) yields Γ1(1-b0/r0)Γ0≥1, and so u0≥(λ*+λ0)ϕ0, which contradicts to the definition of λ*( noticing that λ0>0). This shows that (3.40) fulfils. Therefore, in terms of the fixed point index theorem on cone ([25]), we have
(3.46)i(Q,Ωr0,P0)=0.
Step 5.
Letd2=maxt∈[0,1]∫01∫01G1(t,s)G2(s,τ)dτds,Γ2=(1-D)/d2. By hypothesis f∞=0, we have f∞<(Γ2-ɛ0) for a fixed ɛ0∈(0,Γ2), and so, there exists R1>0 such that
(3.47)f(t,u)<(Γ2-ɛ0)u,t∈I
holds when u≥R1.Let C=max{|f(t,u)|:(t,u)∈I×[0,R1]}. Then
(3.48)f(t,u)≤(Γ2-ɛ0)u+C,(t,u)∈I×R+.
Let d3=maxt∈[0,1]∫01∫01G1(t,s)G2(s,τ)M(τ)dτds,letE0=(Cd2+d3)/ɛ0d2. Take R0>max{r0,E0}. Set ΩR0={u∈P0:∥u∥<R0}. We shall show that
(3.49)λu≠Qu,u∈∂ΩR0,λ≥1.
Suppose on the contradiction that there exist u0∈∂ΩR0 and λ0≥1 with λ0u0=Qu0. Then ∥u0∥=R0>r0>b0. By (3.48), we have
(3.50)f(t,[u0-w]*(t))≤(Γ2-ɛ0)[u0-w]*(t)+C≤(Γ2-ɛ0)u0(t)+C,t∈I.
Hence,
(3.51)(KGu0)(t)=∫01∫01G1(t,s)G2(s,τ)(f(τ,[u0-w]*(τ))+M(τ))dτds≤∫01∫01G1(t,s)G2(s,τ)((Γ2-ɛ0)u0(τ)+C+M(τ))dτds≤(Γ2-ɛ0)d2‖u0‖+Cd2+d3,t∈I.
So, ∥KGu0∥≤(Γ2-ɛ0)d2∥u0∥+Cd2+d3. Thus, from (3.5) and Γ2=(1-D)/d2, it follows that
(3.52)‖u0‖≤‖λ0u0‖=‖(I-B)-1KGu0‖≤11-D‖KGu0‖≤d2(Γ2-ɛ0)1-D‖u0‖+Cd2+d31-D.
Then, R=∥u0∥≤E0, which contradicts to the choice of R0. Hence, (3.49) holds. Therefore, the fixed point index theorem ([25]) implies
(3.53)i(Q,ΩR0,P0)=1.
By (3.46)–(3.53), applying additivity of fixed point index [25], we have
(3.54)i(Q,ΩR0∖Ω¯r0,P0)=i(Q,ΩR0,P0)-i(Q,Ωr0,P0)=1.
Therefore, Q has a fixed point u¯∈ΩR0∖Ω¯r0. Hence, v¯=u¯-w is a solution of BVP (1.4).
Now, fromu¯∉Ω¯r0, we have ∥u¯∥>r0(>b0), and so, (3.20) together with the fact that u¯∈P0 gives
(3.55)v¯(t)=u¯(t)-w(t)≥(1-b0‖u¯‖)u¯(t)≥(1-b0r0)u¯(t)≥(1-b0r0)(1-‖B‖)q1(t)r0,t∈I.
Thus, v¯(t)≥0,t∈[0,1]. Moreover, v¯(t)>0,t∈(0,1) from q1(t)>0,t∈(0,1). That means that v¯ is a positive solution of BVP (1.4). The proof is completed.Corollary 3.2.
Let(H2),(H3) hold. Assume that  f∈C[I×R+,R+]. If  f0>1/Γ0,f∞=0, then BVP (1.4) has a positive solution.Proof.
Let us takeM(t)=0 in Theorem 3.1. Then d1=0, and so b0=0. By f0>1/Γ0, we can take a Γ1∈(1/Γ0,∞) such that f0>Γ1. Then there exists a r0∈(0,+∞) such that
(3.56)f(t,u)≥Γ1u,(t,u)∈I×[0,r0].
Hence, all hypotheses in Theorem 3.1 are satisfied, and the conclusion of Corollary 3.2 follows. This completes the proof.Remark 3.3.
Even in the case thatM(t)=0, the conclusion of Corollary 3.2 is still new.



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*Source: 859497-2011-07-26.xml*