# Positive Fixed Point of Strict Set Contraction Operators on Ordered Banach Spaces and Applications

Authors: Meiqiang Feng; Xuemei Zhang; Weigao Ge
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

The fixed point theorem of cone expansion and compression of norm type for a
strict set contraction operator is generalized by replacing the norms with a convex functional
satisfying certain conditions. We then show how to apply our theorem to prove the existence of
a positive solution to a second-order differential equation with integral boundary conditions
in an ordered Banach space. An example is worked out to demonstrate the main results.


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## Body

## 1. Introduction

The theory of integral and differential equations in Banach spaces, as two new branches of nonlinear functional analysis, has developed for no more than forty years, but it has extensive applications in such domains as the critical point theory, the theory of partial differential equations, and eigenvalue problems. For an introduction of the basic theory of integral and differential equations in Banach spaces, see Guo et al. [1], Guo and Lakshmikantham [2], Lakshmikantham and Leela [3], and Demling [4], and the references therein. In recent years, the theory of integral and differential equations in Banach spaces has become an important area of investigation in both pure and applied mathematics (see, for instance, [5–18] and references cited therein).On the other hand, the theory of fixed point is an important tool to study various boundary value problems of ordinary differential equations, difference differential equations, and dynamic equations on time scales. An overview of such results can be found in Guo et al. [1], in Guo and Lakshmikantham [2], and in Demling [4]. The Krasnoselskii's fixed point theorem concerning cone compression and expansion of norm type is worth mentioing here as follows (see [1, 2, 4]).Theorem 1.1.
LetΩ1 and Ω2 be two bounded open sets in Banach space E, such that 0∈Ω1 and Ω¯1⊂Ω2. Let P be a cone in E and let operator A:P∩(Ω¯2∖Ω1)→P be completely continuous. Suppose that one of the following two conditions is satisfied: (a)
∥Ax∥≥∥x∥,∀x∈P∩∂Ω1, and ∥Ax∥≤∥x∥,∀x∈P∩∂Ω2;(b)
∥Ax∥≤∥x∥,∀x∈P∩∂Ω1 and ∥Ax∥≥∥x∥,∀x∈P∩∂Ω2.Then, A has at least one fixed point in P∩(Ω2∖Ω¯1).To generalize Theorem1.1, one may consider the weakening of one or more of the following hypotheses: (i) the operator A, (ii) the norm.In [19], Sun generalized Theorem 1.1 for completely continuous operator to strict set contraction operator and obtain the following results.Theorem 1.2.
LetK be a cone of Banach space E and Kr={x∈K:∥x∥≤r},Kr,R={x∈K,r≤∥x∥≤R} with R>r>0. Suppose that A:KR→K is a strict set contraction such that one of the following two conditions is satisfied: (a)
∥Ax∥≥∥x∥,∀x∈K,∥x∥=r;∥Ax∥≤∥x∥,∀x∈K,∥x∥=R;(b)
∥Ax∥≤∥x∥,∀x∈K,∥x∥=r;∥Ax∥≥∥x∥,∀x∈K,∥x∥=R.Then, A has a fixed point x∈Kr,R.Recently, in [20], Anderson and Avery generalized the fixed point theorem of cone expansion and compression of norm type by replacing the norms with two functionals satisfying certain conditions to produce a fixed point theorem of cone expansion and compression of functional type. In [21], Guo and Ge extended Krasnoselskii's fixed point theorem by choosing two functionals that satisfy certain conditions which are used in place of the norm. In [22], Zhang and Sun generalized the classical Krasnoselskii's fixed point theorem concerning cone compression and expansion of norm type. The interesting point is that they took place norm by convex functional.In the past few years, we also notice a class of boundary value problems with integral boundary conditions appeared in heat conduction, chemical engineering, underground water flow, thermo elasticity, and plasma physics. Such problems include two, three, multi point and nonlocal boundary value problems as special cases and attracted the attention of Gallardo [23], Karakostas and Tsamatos [24], Lomtatidze and Malaguti [25], and others included in the references therein. On the other hand, we refer the reader to papers by Ahmad et al. [26], Feng et al. [27], Boucherif [28], Infante and Webb [29], Kang et al. [30], Ma [31], Webb [32], Webb and Infante [33, 34], Yang [35], Zhang et al.  [36–38], and Chang et al. [39] for other recent results on nonlinear boundary value problems with integral boundary conditions.Motivated by works mentioned above, we intend in this paper to generalize the fixed point theorem of cone expansion and compression of norm type for strict set contraction operator. The generalization allows the user to choose a convex functional that satisfies certain conditions which are used in place of the norm. In applications to boundary value problems, the functional will typically be maximum of the function over a specific interval. The flexibility of using functionals instead of norms allows the theorem to be used in a wider variety of situations. Our results either improve or generalize the corresponding results due to [19–22] and many of others. As an application of our main results, we consider the existence of positive solutions for second-order differential equations with integral boundary conditions in an ordered Banach space. On the other hand, our conditions are weaker than those of [22].The organization of this paper is as follows. We will introduce some lemmas and notations in the rest of this section. In Section2, the main results will be stated and proved. In Section 3, as an application of our main results, the existence of positive solutions for a second-order boundary value problem with integral boundary conditions in ordered Banach spaces is considered. Finally, in Section 4, one example is also included to illustrate the main results.Basic facts about ordered Banach spaceE can be found in [1–4]. Here we just recall a few of them. The cone P in E induces a partial order on E, that is, 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∥. Without loss of generality, suppose, in the present paper, the normal constant N=1.For a bounded setV in Banach space E, we denote α(V) the Kuratowski measure of noncompactness (see [1–4], for further understanding). The operator A:D→E(D⊂E) is said to be a k-set contraction if A:D→E is continuous and bounded and there is a constant k≥0 such that α(A(S))≤kα(S) for any bounded S⊂D; a k-set contraction with k<1 is called a strict set contraction.In the following, denote the Kuratowski's measure of noncompactness byα(·).For the application in the sequel, we first state the following definition and lemmas which can be found in [1], and some notation.Definition 1.3.
LetS be a bounded set of a real Banach space E. Let α(S)=inf{δ>0:S be  expressed  as  the  union of  a  finite  number  of  sets  such  that  the  diameter  of each set  does  not  exceed  δ, that is, S=⋃i=1mSi with  diam(Si)≤δ,i=1,2,…,m}. Clearly, 0≤α(S)<∞. α(S) is called the Kuratowski's measure of noncompactness.Definition 1.4.
LetP be a cone of a real Banach space E. If P*={Ψ∈E*∣Ψ(x)≥0,∀x∈P}, then P* is a dual cone of cone P.Definition 1.5.
LetP be a cone of real Banach space E. ρ:P→R is said to be a convex functional on P if ρ(tx+(1-t)y)≤tρ(x)+(1-t)ρ(y) for all x,y∈P and t∈[0,1].Definition 1.6.
A subsetX⊂E is said to be a retract of E if there exists a continuous mapping r:E→X satisfying r(x)=x,x∈X.Lemma 1.7.
LetD⊂E, D be a bounded set and f uniformly continuous and bounded from J×S into E; then
(1.1)α(f(J×S))=maxt∈Jα(f(t,S)),∀S⊂D.


## 2. Main Results

Lemma 2.1 (see [22]).
LetP be a cone in a real Banach space E. If ρ:P→[0,∞) is a uniformly continuous convex functional with ρ(θ)=0 and ρ(x)>0 for x≠θ, then ∀r>0, Dr={x∈P:ρ(x)≥r} is a retract of E.Lemma 2.2.
LetE be a real Banach space, ∥·∥ the norm in E, P a cone in E, and Ω={x∈E:∥x∥<R}, where R is a positive real number. Suppose that A:P∩Ω¯→P is a k-set contraction with k<1 and ρ:P→[0,∞) is a uniformly continuous convex functional with ρ(θ)=0, ρ(x)>0, ∀x≠θ and ρ(x)≤∥x∥. If (i)
infx∈P∩∂Ωρ(x)>0, and there exists δ>0 such that R/infx∈P∩∂Ωρ(x)≤1+δ/k and ρ(Ax)≥(k+δ)ρ(x),∀x∈P∩∂Ω;(ii)
Ax≠μx,μ∈(0,1], ∀x∈P∩∂Ω,hold, then the fixed point index i(A,P∩Ω,P)=0.Proof.
Without loss of generality, we supposek+δ<1 (If k+δ>1, then let N=k+δ. The proof is the same as the following process). Let N=1/(k+δ), then NA is a strict set contraction. Considering ht(x)=tAx+(1-t)NAx, t∈[0,1], x∈P∩∂Ω. If there exists t0∈[0,1], x0∈P∩∂Ω such that x0=t0Ax0+(1-t0)NAx0, then Ax0=(1/(t0+N(1-t0)))x0, which contradicts with (ii). Then by the homotopy invariance property of fixed point index, we have i(NA,P∩Ω,P)=i(A,P∩Ω,P).
Letr=infx∈P∩∂Ωρ(x). Define Dr={x∈P∣ρ(x)≥r}. It follows from θ∉Dr that d≜infx∈Dr∥x∥>0. From the fact that ρ(x)≤∥x∥, we have r≤d≤R. In fact, since (P∩∂)Ω⊂Dr, we have d≤R. On the other hand, for ∀x∈Dr,ρ(x)=r, combining this with ρ(x)≤∥x∥, we have r≤∥x∥,∀x∈Dr, then r≤infx∈Dr∥x∥=d. Let M=R/r, then by (i) we obtain Mk/(k+δ)<1 and Md>supx∈P∩Ω¯∥x∥, and MDr∩(P∩Ω¯)=∅, where MDr={Mx∣x∈Dr}.
LetH(t,x)=(1-t)NAx+tMNAx,∀(t,x)∈[0,1]×P∩Ω¯. Then, we have
(2.1)α(H(t,S))≤(1-t)α(NA(S))+tα(MNA(S))≤(1-t)kk+δα(S)+tMkk+δα(S)<Mkk+δα(S),∀S⊂P∩Ω,
and then we obtain that H(t,·):P∩Ω¯→P is the strict set contraction. In addition, it is obvious that H(t,x) is uniformly continuous about t for all x∈P∩Ω¯.
If there existsx1∈P∩∂Ω,t1∈[0,1] such that (1-t1)NAx1+t1MNAx1=x1, then Ax1=N(1-t1+t1M)-1x1, which contradicts with (ii). Thus by the homotopy invariance property of fixed point index, we have i(MNA,P∩Ω,P)=i(NA,P∩Ω,P).
SinceDr is a retract of E by Lemma 2.1, there exists a retraction r:E→Dr satisfying r(x)=x,x∈Dr. Let A1=NA,A¯1=r∘A1, then A¯1 is strict set contraction. From (i) and the definition of ρ, we have
(2.2)ρ(A1x)=ρ(NAx)≥Nρ(Ax)≥ρ(x)≥r,∀x∈P∩∂Ω.
Therefore, A1(∂Ω)⊂Dr, that is, A¯1x=A1x, ∀x∈P∩∂Ω. Then i(MA¯1,P∩Ω,P)=i(MA1,P∩Ω,P).
Ifi(A1,P∩Ω,P)≠0, then i(MA¯1,P∩Ω,P)≠0, which implies that MA¯1 has a fixed point x* in P∩Ω. Thus x*=MA¯1x*∈MDr. It is a paradox. The proof is complete.Lemma 2.3 (see [2]).
LetP be a cone and Ω a bounded open set in E with θ∈Ω. Suppose that A:P∩Ω¯→P is condensing and
(2.3)Ax≠μx,∀x∈P∩∂Ω,μ≥1.
Then i(A,P∩Ω,P)=1.Lemma 2.4.
LetP be a cone and Ω a bounded open set in E. Suppose that A:P∩Ω¯→P is a k-set contraction with k<1 and ρ:P→[0,∞) is a uniformly continuous convex functional with ρ(θ)=0 and ρ(x)>0 for x≠θ. If ρ(Ax)≤ρ(x) and Ax≠x for x∈P∩∂Ω, then i(A,P∩Ω,P)=1.Proof.
If there existx1∈P∩∂Ω and μ1≥1 such that Ax1=μ1x1, then μ1>1. Therefore,
(2.4)ρ(x1)=ρ(1μ1Ax1)≤1μ1ρ(Ax1)≤1μ1ρ(x1)<ρ(x1).
It is a paradox. From Lemma 2.3, it follows that i(A,P∩Ω,P)=1. The proof is complete.Theorem 2.5.
LetΩ1 be a bounded open set in E such that θ∈Ω1, and Ω2={x∈E∣∥x∥<R} and Ω¯1⊂Ω2. Suppose that A:P∩(Ω¯2∖Ω1)→P is a k-set contraction with k<1 and ρ:P→[0,∞) is a uniformly continuous convex functional with ρ(θ)=0 and ρ(x)>0,∀x≠θ and ρ(x)≤∥x∥. If (a)
ρ(Ax)≤ρ(x), ∀x∈P∩∂Ω1;(b)
infx∈P∩∂Ω2ρ(x)>0, and there exists δ>0 such that R/infx∈P∩∂Ω2ρ(x)≤1+δ/k and ρ(Ax)≥(k+δ)ρ(x), and Ax≠μx,μ∈(0,1], ∀x∈P∩∂Ω2hold, then A has at least one fixed point in P∩(Ω2∖Ω¯1).Proof.
It is easy to obtain the results by Lemmas2.2 and 2.4. So we omit it.Theorem 2.6.
LetΩ1={x∈E∣∥x∥<R} and Ω2 a bounded open set in E such that Ω¯1⊂Ω2. Suppose that A:P∩(Ω¯2∖Ω1)→P is a k-set contraction with k<1, and ρ:P→[0,∞) is a uniformly continuous convex functional with ρ(θ)=0 and ρ(x)>0,∀x≠θ and ρ(x)≤∥x∥. If (a)
infx∈P∩∂Ω1ρ(x)>0, and there exists δ>0 such that R/infx∈P∩∂Ω1ρ(x)≤1+δ/k and ρ(Ax)≥(k+δ)ρ(x), and Ax≠μx,μ∈(0,1],∀x∈P∩∂Ω1;(b)
ρ(Ax)≤ρ(x),∀x∈P∩∂Ω2are satisfied, then A has at least one fixed point in P∩(Ω2∖Ω¯1).Proof.
It is easy to obtain the results by Lemmas2.2 and 2.4. So we omit it.Remark 2.7.
If we letk=0, then A is completely continuous. Comparing with Corollary 2.1 of [22], our conditions are weaker.Corollary 2.8.
LetΩ1 be a bounded open set in E such that θ∈Ω1, Ω2={x∈E∣∥x∥<R}, and Ω¯1⊂Ω2. Suppose that A:P∩(Ω¯2∖Ω1)→P is a k-set contraction with k<1 and ρ:P→[0,∞) is a uniformly continuous convex functional with ρ(θ)=0 and ρ(x)>0,∀x≠θ and ρ(x)≤∥x∥. If (a)
ρ(Ax)≤ρ(x), ∀x∈P∩∂Ω1;(b)
infx∈P∩∂Ω2ρ(x)>0 with R/infx∈P∩∂Ω2ρ(x)≤1/k,ρ(Ax)≥ρ(x) and Ax≠μx,μ∈(0,1),∀x∈P∩∂Ω2hold, then A has at least one fixed point in P∩(Ω2∖Ω¯1).Proof.
It follows by takingk+δ=1.Corollary 2.9.
LetΩ1={x∈E∣∥x∥<R} and Ω2 a bounded open set in E such that Ω¯1⊂Ω2. Suppose that A:P∩(Ω¯2∖Ω1)→P is a k-set contraction with k<1 and ρ:P→[0,∞) is a uniformly continuous convex functional with ρ(θ)=0 and ρ(x)>0,∀x≠θ and ρ(x)≤∥x∥. If (a)
infx∈P∩∂Ω1ρ(x)>0 with R/infx∈P∩∂Ω1ρ(x)≤1/k,ρ(Ax)≥ρ(x), and Ax≠μx,μ∈(0,1],∀x∈P∩∂Ω1;(b)
ρ(Ax)≤ρ(x), ∀x∈P∩∂Ω2hold, then A has at least one fixed point in P∩(Ω2∖Ω¯1).Proof.
It follows by takingk+δ=1.


## 3. Applications

Throughout the remainder of this paper, we apply the above results to a second-order differential equation in Banach spaces:(3.1)x′′+f(t,x)=θ,0<t<1,
subject to the following integral boundary conditions: (3.2)x(0)=∫01g(t)x(t)dt,x(1)=θ,
where f∈C([0,1]×P,P), θ is the zero element of E, and g∈L1[0,1] is nonnegative.We consider problem (3.1)-(3.2) in C(J,E), in which J=[0,1]. Evidently, (C(J,E),∥·∥c) is a Banach space with norm ∥x∥c=maxt∈J∥x(t)∥ for x∈C(J,E).To establish the existence of positive solutions inC(J,E) of (3.1)-(3.2), let us list the following assumptions.(ℍ)  f∈C(J×P,P), and for any l>0,f is uniformly continuous on J×(P∩𝒯l). Further suppose that g∈L1[0,1] is nonnegative, σ∈[0,1), and there exist nonnegative constants ηl with (3.3)12γηl<1
such that (3.4)α(f(t,S))≤ηlα(S),t∈J,S⊂P∩𝒯l,
where 𝒯l={x∈E:∥x∥≤l}, γ=(1+∫01sg(s)ds)/(1-σ), and σ=∫01(1-s)g(s)ds.It is easy to see that the problem (3.1)-(3.2) has a solution x=x(t) if and only if x is a solution of the operator equation (3.5)(Tx)(t)=∫01H(t,s)f(s,x(s))ds,
where (3.6)H(t,s)=G(t,s)+1-t1-σ∫01G(s,τ)g(τ)dτ,(3.7)G(t,s)={t(1-s),0≤t≤s≤1,s(1-t),0≤s≤t≤1.From (3.6) and (3.7), we can prove that H(t,s),G(t,s) have the following properties.Proposition 3.1.
Assume that(ℍ) holds. Then for t,s∈[0,1] we have
(3.8)H(t,s)≥0,G(t,s)≥0.Proposition 3.2.
Fort,s∈[0,1], we have
(3.9)e(t)e(s)≤G(t,s)≤G(t,t)=t(1-t)=e(t)≤e¯=maxt∈[0,1]e(t)=14.Proposition 3.3.
Letδ¯∈(0,1/2), Jδ¯=[δ¯,1-δ¯]. Then for all t∈Jδ¯, s,u∈[0,1], we have
(3.10)G(t,s)≥δ¯G(u,s).Proposition 3.4.
Assume that(ℍ) holds. Then for t,s∈[0,1], we have
(3.11)ρe(t)e(s)≤H(t,s)≤γt(1-t)=γe(t),
where γ is defined in (ℍ), and
(3.12)ρ=∫01e(τ)g(τ)dτ1-σ.Proof.
By (3.6) and (3.9), we have
(3.13)H(t,s)=G(t,s)+1-t1-σ∫01G(s,τ)g(τ)dτ≥1-t1-σ∫01G(s,τ)g(τ)dτ≥∫01G(s,τ)g(τ)dτ1-σt(1-t)≥∫01e(τ)g(τ)dτ1-σt(1-t)s(1-s)=ρe(t)e(s),t∈[0,1].
On the other hand, noticingG(t,s)≤s(1-s), we obtain
(3.14)H(t,s)=G(t,s)+1-t1-σ∫01G(s,τ)g(τ)dτ≤s(1-s)+1-t1-σ∫01s(1-s)g(τ)dτ≤s(1-s)[1+11-σ∫01g(τ)dτ]≤s(1-s)1+∫01sg(s)ds1-σ=γe(s),t∈[0,1].Proposition 3.5.
Assume that(ℍ) holds. Then for all t∈Jδ¯,s,u∈[0,1], we have
(3.15)H(t,s)≥δ¯H(u,s).Proof.
By (3.10), we have
(3.16)H(t,s)=G(t,s)+1-t1-σ∫01G(s,τ)g(τ)dτ≥δ¯G(u,s)+δ¯1-σ∫01G(s,τ)g(τ)dτ≥δ¯G(u,s)+δ¯(1-u)1-σ∫01G(s,τ)g(τ)dτ=δ¯H(u,s),s,u∈[0,1].We construct a coneK by (3.17)K={x∈Q:x(t)≥δ¯x(s),t∈Jδ¯,s∈[0,1]},
where (3.18)Q={x∈C(J,E):x(t)≥θ,t∈J}.It is easy to see thatK is a cone of C(J,E).We will make use of the following lemmas.Lemma 3.6.
Suppose that(ℍ) holds. Then for each l>0,T is strict set contraction on Q∩Bl, that is, there exists a constant 0≤kl<1 such that α(T(S))≤klα(S) for any S⊂Q∩Bl, where Bl={x∈C[J,E],∥x∥c≤l}.Proof.
By(ℍ), we know that f is uniformly continuous on J×(P∩𝒯l). Hence, f is bounded on J×(P∩𝒯l). This together with (3.4) and Lemma 1.7 implies that
(3.19)α(f(J×S))=maxt∈Jα(f(t,S))≤ηlα(S),forS⊂Q∩Bl.
From f being uniformly continuous and bounded on S⊂Q∩Bl, we can obtain that T is continuous and bounded from Q∩Bl into Q.
On the other hand, it is clear that0≤H(t,s)≤(1/4)γ and using a similar method as in the proof of Lemma 2 in [40], we can get that
(3.20)α(T(S))≤214γηlα(S).
Therefore,(3.21)α(T(S))≤klα(S),S⊂Q∩Bl,
where kl=(1/2)γηl, 0≤kl<1. The proof is complete.Lemma 3.7.
Suppose that(ℍ) holds. Then T(K)⊂K and T:K→K is a strict set contraction.Proof.
From (3.5) and (3.15), we obtain
(3.22)mint∈Jδ¯(Tx)(t)=mint∈Jδ¯∫01H(t,s)f(s,x(s))ds≥δ¯∫01H(u,s)f(s,x(s))ds≥δ¯(Tx)(u),u∈J.
Therefore, T(x)∈K, that is, T(K)⊂K.
Next by Lemma3.6, one can prove that T:K→K is a strict set contraction. So it is omitted.Let(3.23)Δ=∫δ¯1-δ¯H(12,s)ds.Theorem 3.8.
Assume that(ℍ) holds and P is normal. If there exist a,b with 0<a<b such that Ψ(f(t,x))≥Δ-1a for t∈Jδ¯,∥x∥≤a, and ∥f(t,x)∥≤6γ-1b for t∈Jδ¯,∥x∥≤δ¯-1b, where Ψ∈P*,∥Ψ∥=1, then problem (3.1)-(3.2) has at least one positive solution.Proof.
LetT be the cone preserving, strict set contraction that was defined by (3.5).
Letρ(x)=supt∈Jδ¯∥x(t)∥. Then ρ:K→[0,+∞) is a uniformly continuous convex functional with ρ(θ)=0 and ρ(x)>0 for x≠θ. Let
(3.24)Ω1={x∈C[J,E]∣∥x∥c<a},Ω2={x∈C[J,E]∣ρ(x)<b}.
It is clear thatΩ1 and Ω2 are open sets in C[J,E] with θ∈Ω1 and Ω¯1⊂Ω2. If x∈K∩Ω2, we have ∥x∥c≤δ¯-1b which implies that K∩Ω2 is bounded.
Ifx∈K∩∂Ω1, then ∥x∥c=a.
(3.25)ρ(Tx)=supt∈Jδ¯∥(Tx)(t)∥≥∥(Tx)(12)∥≥Ψ((Tx)(12))=∫01H(12,s)Ψ(f(s,x(s)))ds≥∫δ¯1-δ¯H(12,s)Ψ(f(s,x(s)))ds≥∫δ¯1-δ¯H(12,s)dsΔ-1a=a=∥x∥c≥ρ(x).
Ifx∈K∩∂Ω2, then ρ(x)=b and ∥x∥c≤δ¯-1b.
(3.26)ρ(Tx)=supt∈Jδ¯∥(Tx)(t)∥≤supt∈Jδ¯∫01H(t,s)∥f(s,x(s))∥ds≤6γ-1bsupt∈Jδ¯∫01H(t,s)ds=b=ρ(x).
Hence, the proof is finished by Corollary2.9.


## 4. Example

Example 4.1.
To illustrate how our main results can be used in practice, we present an example. For the convenience of computation, we study a two-point boundary value problem. Now we consider the following boundary value problem:(4.1)-x′′=f(t,x),0<t<1,x(0)=0,x(1)=0,
where g(t)=0, and
(4.2)f(t,x)={10te-x+25,0<t<1,0≤x≤10,1324(x-100)2+10te-10,0<t<1,x≥10.
Hence σ=0,  γ=1. In this case E=R, P=R+,  and k=0. Let Ψ≡1,  δ¯=1/10; then Δ=2/5.
Selecta=10,b=20; then we can see that
(4.3)|f(t,x)|≥10×110×e-10+25>25=Δ-1a,∀t∈[110,910],x≤10,|f(t,x)|≤1324×1002+10×110×e-10<60=6γ-1a,∀t∈[110,910],x≤200.
Hence, the conditions of the Theorem 3.8 are satisfied. Then problem (4.1) has at least one positive solution.Remark 4.2.
Example4.1 implies that there is a large number of functions that satisfy the conditions of Theorem 3.8. In addition, the conditions of Theorem 3.8 are also easy to check.



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*Source: 439137-2010-08-18.xml*