# Positive Solutions for a Fractional Boundary Value Problem with Changing Sign Nonlinearity

Authors: Yongqing Wang; Lishan Liu; Yonghong Wu
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

We discuss the existence of positive solutions to the following fractionalm-point boundary value problem with changing sign nonlinearity D0+αu(t)+λf(t,u(t))=0,0<t<1,u(0)=0,D0+βu(1)=∑i=1m-2ηiD0+βu(ξi), where λ is a positive parameter, 1<α≤2, 0<β<α-1, 0<ξ1<⋯<ξm-2<1 with ∑i=1m-2ηiξiα-β-1<1, D0+α is the standard Riemann-Liouville derivative, f  and may be singular at t=0 and/or t=1 and also may change sign. The work improves and generalizes some previous results.


---

## Body

## 1. Introduction

In this paper, we consider the following fractional differential equation withm-point boundary conditions:
(1.1)D0+αu(t)+λf(t,u(t))=0,0<t<1,u(0)=0,D0+βu(1)=∑i=1m-2‍ηiD0+βu(ξi),
where 1<α≤2, λ>0 is a parameter, 0<β<α-1, 0<ξ1<⋯<ξm-2<1 with ∑i=1m-2‍ηiξiα-β-1<1, D0+α is the standard Riemann-Liouville derivative, and f∈C((0,1)×[0,+∞)→(-∞,+∞)) may be singular at t=0 and/or t=1 and also may change sign. In this paper, by a positive solution to (1.1), we mean a function u∈C[0,1] which is positive on (0,1] and satisfies (1.1).In recent years, great efforts have been made worldwide to study the existence of solutions for nonlinear fractional differential equations by using nonlinear analysis methods [1–24]. Fractional-order multipoint boundary value problems (BVP) have particularly attracted a great deal of attention (see, e.g., [13–19]). In [10], the authors discussed some properties of the Green function for the Direchlet-type BVP of nonlinear fractional differential equations
(1.2)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=0,u(1)=0,
where 1<α<2, D0+α is the standard Riemann-Liouville derivative and f:C([0,1]×[0,+∞)→[0,+∞)) is continuous. By using the Krasnosel'skii fixed-point theorem, the existence of positive solutions was obtained under some suitable conditions on f.In [14], the authors investigated the existence and multiplicity of positive solutions by using some fixed-point theorems for the fractional differential equation given by
(1.3)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=0,D0+βu(1)=aD0+βu(ξ),
where 1<α≤2, 0≤β≤1, 0<ξ<1, 0≤a≤1 with aξα-β-2<1-β, 0≤α-β-1, f is nonnegative.It should be noted that in most of the works in literature the nonlinearity needs to be nonnegative in order to establish positive solutions. As far as we know, semipositone fractional nonlocal boundary value problems with1<α≤2 have been seldom studied due to the difficulties in finding and analyzing the corresponding Green function.In [23], the authors investigated the following fractional differential equation with three-point boundary conditions:
(1.4)D0+αu(t)+f(t,u(t))+e(t)=0,0<t<1,u(0)=0,D0+βu(1)=aD0+βu(ξ),
where 1<α≤2, 0<β≤1, 0<ξ<1,  0≤a≤1,0≤α-β-1, e(t)∈L[0,1], and f satisfies the Caratheodory conditions. The authors obtained the properties of the Green function for (1.4) as follows:
(1.5)βtα-1s(1-s)α-β-1Γ(α)≤G(t,s)≤tα-1(1-s)α-β-1Γ(α)(1-aξα-β-1).
By using the Schauder fixed-point theorem, the authors obtained the existence of positive solution of (1.4) with the following assumptions:(A1)
for eachL>0, there exists a function ϕL≻0 such that f(t,tα-1x)≥ϕL(t) for a.e. t∈(0,1), for all x∈(0,L];(A2)
there existg(x),h(x), and k(t)≻0, such that
(1.6)0≤f(t,x)≤k(t){g(x)+h(x)},∀x∈(0,∞),a.e.t∈(0,1),here g:(0,∞)→[0,∞) is continuous and nonincreasing, h:[0,∞)→[0,∞) is continuous, and h/g is nondecreasing;(A3)
There exist two positive constantsR>r>0 such that
(1.7)R>ΦR1+γ*≥r>0,∫01‍k(s)g(rsα-1)ds<+∞,R≥(1+h(R)g(R))∫01‍(1-s)α-β-1Γ(α)(1-aξα-β-1)k(s)g(rsα-1)ds+γ*.Here
(1.8)ΦR1=∫01‍βs(1-s)α-β-1Γ(α)ϕR(s)ds.
The assumptions on nonlinearity are not suitable for frequently used conditions, such as superlinear or some sublinear. For instance, f(t,x)=xμ, μ>0, obviously, f does not satisfy (A1).Inspired by the previous work, the aim of this paper is to establish conditions for the existence of positive solutions of the more general BVP (1.1). Our work presented in this paper has the following new features. Firstly, we consider few cases of 1<α≤2 which has been studied before, and in dealing with the difficulties related to the Green function for this case, some new properties of the Green function have been discovered. Secondly, the BVP (1.1) possesses singularity; that is, f may be singular at t=0 and/or t=1. Thirdly, the nonlinearity f may change sign and may be unbounded from below. Finally, we impose weaker positivity conditions on the nonlocal boundary term; that is, some of the coefficients ηi may be negative.The rest of the paper is organized as follows. In Section2, we present some preliminaries and lemmas that are to be used to prove our main results. We also discover some new positive properties of the corresponding Green function. In Section 3, we discuss the existence of positive solutions of the semipositone BVP (1.1). In Section 4, we give an example to demonstrate the application of our theoretical results.


## 2. Basic Definitions and Preliminaries

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can also be found in the recent literature.Definition 2.1.
The fractional integral of orderα>0 of a function u:(0,+∞)→R is given by
(2.1)I0+αu(t)=1Γ(α)∫0t‍(t-s)α-1u(s)ds,
provided that the right-hand side is pointwisely defined on (0,+∞).Definition 2.2.
The Riemann-Liouville fractional derivative of orderα>0 of a function u:(0,+∞)→R is given by
(2.2)D0+αu(t)=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1u(s)ds,
where n=[α]+1 and [α] denotes the integer part of the number α, provided that the right-hand side is pointwisely defined on (0,+∞).Lemma 2.3 (see [3]).
Letα>0. Then the following equality holds for u∈L(0,1), D0+αu∈L(0,1);
(2.3)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋅⋅⋅+cntα-n,
where ci∈R, i=1,2,…,n, n-1<α≤n.Set(2.4)G0(t,s)=1Γ(α){tα-1(1-s)α-β-1,0≤t≤s≤1,tα-1(1-s)α-β-1-(t-s)α-1,0≤s≤t≤1,(2.5)p(s)=1-∑s≤ξi‍ηi(ξi-s1-s)α-β-1,(2.6)G(t,s)=G0(t,s)+q(s)tα-1,
where
(2.7)q(s)=p(s)-p(0)Γ(α)p(0)(1-s)α-β-1,p(0)=1-∑i=1m-2‍ηiξiα-β-1.
For convenience in presentation, we here list the assumption to be used throughout the paper.(H1)
p(0)>0, q(s)≥0 on [0,1].Remark 2.4.
Ifηi=0(i=1,…,m-2), we have p(0)=1 and q(s)≡0. If ηi≥0(i=1,…,m-2) and ∑i=1m-2‍ηiξiα-β-1<1, we have q(s)≥0 on [0,1].Lemma 2.5 (see [14]).
Assume thatg(t)∈L[0,1] and α>β>0. Then
(2.8)D0+β∫0t‍(t-s)α-1g(s)ds=Γ(α)Γ(α-β)∫0t‍(t-s)α-β-1g(s)ds.Lemma 2.6.
Assume(H1) holds, and y(t)∈L[0,1]. Then the unique solution of the problem
(2.9)D0+αu(t)+y(t)=0,0<t<1,u(0)=0,D0+βu(1)=∑i=1m-2‍ηiD0+βu(ξi)
is
(2.10)u(t)=∫01‍G(t,s)y(s)ds,
where G(t,s) is the Green function of the boundary value problem (2.9).Proof.
From Lemma2.3, the solution of (2.9) is
(2.11)u(t)=-I0+αy(t)+c1tα-1+c2tα-2.
Consequently,
(2.12)u(t)=-1Γ(α)∫0t‍(t-s)α-1y(s)ds+c1tα-1+c2tα-2.
From u(0)=0, we have c2=0.
By Lemma2.5, we have
(2.13)D0+βu(t)=-1Γ(α-β)∫0t‍(t-s)α-β-1y(s)ds+c1Γ(α)Γ(α-β)tα-β-1.
Therefore,
(2.14)D0+βu(1)=-1Γ(α-β)∫01‍(1-s)α-β-1y(s)ds+c1Γ(α)Γ(α-β),D0+βu(ξi)=-1Γ(α-β)∫0ξi‍(ξi-s)α-β-1y(s)ds+c1Γ(α)Γ(α-β)ξiα-β-1.
By D0+βu(1)=∑i=1m-2‍ηiD0+βu(ξi), we have
(2.15)c1=∫01‍(1-s)α-1y(s)ds-∑i=1m-2ηi∫0ξi‍(ξi-s)α-β-1y(s)dsΓ(α)p(0)=∫01‍(1-s)α-β-1p(s)y(s)dsΓ(α)p(0).
Therefore, the solution of (2.9) is
(2.16)u(t)=c1tα-1-1Γ(α)∫0t‍(t-s)α-1y(s)ds=∫01‍G(t,s)y(s)ds.Lemma 2.7.
The functionG0(t,s) has the following properties:(1)
G0(t,s)>0, for  t,s∈(0,1);(2)
Γ(α)G0(t,s)≤tα-1, for  t,s∈[0,1];(3)
βtα-1h(s)≤Γ(α)G0(t,s)≤h(s)tα-2,  for  t,s∈(0,1),where
(2.17)h(s)=s(1-s)α-β-1.Proof.
(1) When0<t≤s<1, it is clear that
(2.18)G0(t,s)=1Γ(α)tα-1(1-s)α-β-1>0.
When 0<s≤t<1, we have
(2.19)tα-1(1-s)α-β-1-(t-s)α-1≥tα-1(1-s)α-β-1-tα-1(1-s)α-1=tα-1(1-s)α-β-1[1-(1-s)β]>0.
(2) By (2.4), for any t,s∈[0,1], we have
(2.20)Γ(α)G0(t,s)≤tα-1(1-s)α-β-1≤tα-1.
In the following, we will prove (3).(i)
When0<s≤t<1, noticing that 0<β<α-1≤1, we have
(2.21)∂∂β{tα-2s(1-s)α-β-1-tα-1(1-s)α-β-1}=tα-2(1-s)α-β-1(t-s)ln⁡(1-s)≤0.Therefore,
(2.22)tα-2s(1-s)α-β-1-(tα-1(1-s)α-β-1-(t-s)α-1)≥tα-2s-tα-1+(t-s)α-1=-tα-2(t-s)+(t-s)α-1≥0,
which implies
(2.23)Γ(α)G0(t,s)≤h(s)tα-2.
On the other hand, we have(2.24)dds{βs+(1-s)β}≤0,s∈[0,1).
Therefore, βs+(1-s)β≤1, which implies
(2.25)[1-(1-s)β]≥βs.
Then
(2.26)Γ(α)G0(t,s)=tα-1(1-s)α-β-1-(t-s)α-1≥tα-1(1-s)α-β-1-(t-s)β(t-ts)α-β-1=[1-(1-st)β]tα-1(1-s)α-β-1≥[1-(1-s)β]tα-1(1-s)α-β-1≥βtα-1h(s).(ii)
When0<t≤s<1, we have
(2.27)Γ(α)G0(t,s)=tα-1(1-s)α-β-1=tα-2t(1-s)α-β-1≤tα-2s(1-s)α-β-1=h(s)tα-2,On the other hand, clearly we have
(2.28)Γ(α)G0(t,s)=tα-1(1-s)α-β-1≥βtα-1h(s).
The inequalities (2.23)–(2.28) imply that (3) holds.By Lemma2.7, we have the following results.Lemma 2.8.
Assume(H1) holds, then the Green function defined by (2.6) satisfies(1)
G(t,s)>0, for all t,s∈(0,1);(2)
G(t,s)≤tα-1((1/(Γ(α)))+q(s)), for all  t,s∈[0,1];(3)
βtα-1Φ(s)≤G(t,s)≤tα-2Φ(s), for all t,s∈(0,1),where
(2.29)Φ(s)=(h(s)Γ(α)+q(s)).Lemma 2.9.
Assume(H1) holds, then the function G*(t,s)=:t2-αG(t,s) satisfies(1)
G*(t,s)>0, for all t,s∈(0,1);(2)
G*(t,s)≤t((1/(Γ(α)))+q(s)), for all t,s∈[0,1];(3)
βtΦ(s)≤G*(t,s)≤Φ(s), for all t,s∈[0,1].For convenience, we list here four more assumptions to be used later:(H2)
f∈C((0,1)×[0,+∞),(-∞,+∞)) satisfies
(2.30)f(t,x)≥-r(t),f(t,tα-2x)≤z(t)g(x),t∈(0,1),x∈[0,+∞),where r,z∈C((0,1),[0,+∞)), g∈C([0,+∞),[0,+∞)).(H3)
∫01‍r(s)ds<+∞, 0<∫01‍z(s)ds<+∞.(H4)
There exists[a,b]⊂(0,1) such that
(2.31)liminf⁡x→+∞min⁡t∈[a,b]f(t,x)x=+∞.(H5)
There exists[c,d]⊂(0,1) such that
(2.32)liminf⁡x→+∞min⁡t∈[c,d]f(t,x)=+∞,lim⁡x→+∞g(x)x=0.Remark 2.10.
The second limit of(H5) implies
(2.33)lim⁡u→+∞g*(u)u=0,
where
(2.34)g*(u)=max⁡x∈[0,u]g(x).Proof.
Bylimu→+∞(g(u)/u)=0, for any ϵ>0, there exists N1>0, such that for any u>N1 we have
(2.35)0≤g(u)<ϵu.
Let  N=max{N1,((g*(N1))/ϵ)}, for any u>N we have
(2.36)0≤g*(u)<ϵu+g*(N1)<2ϵu.
Therefore, limu→+∞((g*(u))/u)=0.Lemma 2.11.
Assume(H1) holds and r(t)∈C(0,1)∩L[0,1] is nonnegative, then the BVP
(2.37)D0+αu(t)+r(t)=0,0<t<1,u(0)=0,D0+βu(1)=∑i=1m-2‍ηiD0+βu(ξi)
has a unique solution ω(t)=∫01‍G(t,s)r(s)ds with ω(t)≤ktα-1, where
(2.38)k=∫01‍(1Γ(α)+q(s))r(s)ds,t∈[0,1].Proof.
By Lemma2.6, ω(t)=∫01‍G(t,s)r(s)ds is the unique solution of (2.37). By (2) of Lemma 2.8, we have
(2.39)ω(t)=∫01‍G(t,s)r(s)ds≤tα-1∫01‍(1Γ(α)+q(s))r(s)ds.LetE=C[0,1] be endowed with the maximum norm ∥u∥=max0≤t≤1|u(t)| and define a cone P by
(2.40)P={u(t)∈E:thereexistslu>0suchthatβt‖u‖≤u(t)≤lut},
and then set Br={u(t)∈E:∥u∥<r}, Pr=P∩Br, ∂Pr=P∩∂Br.Next we consider the following singular nonlinear BVP:(2.41)D0+αu(t)+λ[f(t,[u(t)-λω(t)]+)+r(t)]=0,0<t<1,u(0)=0,D0+βu(1)=∑i=1m-2‍ηiD0+βu(ξi),
where λ>0, [v(t)]+=max{v(t),0}, ω(t) is defined in Lemma 2.11.Let(2.42)Tu(t)=λ∫01‍G*(t,s)[f(s,[sα-2u(s)-λω(s)]+)+r(s)]ds.Clearly, ifu(t)∈P is a fixed point of T, then y(t)=tα-2u(t) is a positive solution of (2.41).Lemma 2.12.
Suppose that (H1)–(H3) hold. Then T:P→P is a completely continuous operator.Proof.
It is clear thatT is well defined on P. For any u∈P, Lemma 2.9 implies
(2.43)Tu(t)≥βtλ∫01‍Φ(s)[f(s,[sα-2u(s)-λω(s)]+)+r(s)]ds.
On the other hand,
(2.44)Tu(t)≤λ∫01‍Φ(s)[f(s,[sα-2u(s)-λω(s)]+)+r(s)]ds.
Therefore, Tu(t)≥βt∥Tu∥. Noticing that
(2.45)Tu(t)≤λt∫01‍(1Γ(α)+q(s))[f(s,[sα-2u(s)-λω(s)]+)+r(s)]ds,
we have T:P→P.
Using the Ascoli-Arzela theorem, we can then get thatT:P→P is a completely continuous operator.Lemma 2.13 (see [25]).
LetE be a real Banach space and let P⊂E be a cone. Assume that Ω1 and Ω2 are two-bounded open subsets of E with θ∈Ω1,Ω¯1⊂Ω2, T:P∩(Ω¯2∖Ω1)→P a completely continuous operator such that either(1)
∥Tu∥≤∥u∥, u∈P∩∂Ω1 and ∥Tu∥≥∥u∥, u∈P∩∂Ω2, or(2)
∥Tu∥≥∥u∥, u∈P∩∂Ω1 and ∥Tu∥≤∥u∥, u∈P∩∂Ω2.Then T has a fixed point in P∩(Ω¯2∖Ω1).


## 3. Existence of Positive Solutions

Theorem 3.1.
Suppose that (H1)–(H4) hold. Then there exists λ*>0 such that the BVP (1.1) has at least one positive solution for any λ∈(0,λ*).Proof .
Chooser1>kβ-1. Let
(3.1)λ*=min⁡{1,r1(g*(r1)+1)∫01‍Φ(s)(z(s)+r(s))ds},
where
(3.2)g*(r)=max⁡x∈[0,r]g(x).
In the following of the proof, we suppose λ∈(0,λ*).
For anyu∈∂Pr1, noticing u(t)≥βtr1 and Lemma 2.11, we have
(3.3)tα-2u(t)-λω(t)≥(βr1-λk)tα-1≥(βr1-k)tα-1≥0,(3.4)r1≥u(t)-λt2-αω(t)≥(βr1-k)t≥0.
Therefore,
(3.5)Tu(t)=λ∫01‍G*(t,s)(f(s,[sα-2u(s)-λω(s)]+)+r(s))ds≤λ∫01‍Φ(s)(z(s)g([u(s)-λs2-αω(s)]+)+r(s))ds≤λ(g*(r1)+1)∫01‍Φ(s)(z(s)+r(s))ds<λ*(g*(r1)+1)∫01‍Φ(s)(z(s)+r(s))ds≤r1.
Thus,
(3.6)‖Tu‖≤‖u‖,∀u∈∂Pr1.
Now choose a real number(3.7)L>2λβ2∫ab‍Φ(s)sα-1ds.
By (H4), there exists a constant N>0 such that
(3.8)f(t,x)>Lx,foranyt∈[a,b],x≥N.
Let
(3.9)r2=r1+2kβ+2Nβaα-1.
Then for any u∈∂Pr2, we have
(3.10)tα-2u(t)-λω(t)≥(βr2-k)tα-1≥βr22tα-1,∀t∈(0,1].
Thus, for any t∈[a,b], we have tα-2u(t)-λω(t)>N. Hence, we get
(3.11)‖Tu‖=max⁡t∈[0,1]λ∫01‍G*(t,s)[f(s,[sα-2u(s)-λω(s)]+)+r(s)]ds≥max⁡t∈[0,1]λ∫ab‍G*(t,s)f(s,[sα-2u(s)-λω(s)])ds≥max⁡t∈[0,1]λL∫ab‍G*(t,s)(sα-2u(s)-λω(s))ds≥max⁡t∈[0,1]λL∫ab‍G*(t,s)βr22sα-1ds≥max⁡t∈[0,1]λLβ2r22t∫ab‍Φ(s)sα-1ds=λLβ2r22∫ab‍Φ(s)sα-1ds≥r2.
Therefore,
(3.12)‖Tu‖≥‖u‖,∀u∈∂Pr2.
By Lemma 2.13, T has a fixed point u∈P such that r1≤∥u∥≤r2. Let u¯(t)=tα-2u(t)-λω(t). Since ∥u∥≥r1, by (3.3) we have u¯(t)≥0 on (0,1] and limt→0+tα-2u(t)=0. Notice that ω(t) is the solution of (2.37) and tα-2u(t) is the solution of (2.41). Thus, u¯(t) is a positive solution of the BVP (1.1).Theorem 3.2.
Suppose that(H1)–(H3) and (H5) hold. Then there exists λ*>0 such that the BVP (1.1) has at least one positive solution for any λ∈(λ*,+∞).Proof.
By the first limit of(H5), there exists N>0 such that
(3.13)f(t,x)≥2kβ2∫cd‍Φ(s)ds,foranyt∈[c,d],x≥N.
Let(3.14)λ*=Nkcα-1.
In the following part of the proof, we suppose λ>λ*.
Let(3.15)R1=2λkβ.
Then for any u∈∂PR1, we have
(3.16)tα-2u(t)-λω(t)≥(βR1-λk)tα-1=λktα-1≥λ*ktα-1,∀t∈(0,1].
Therefore, tα-2u(t)-λω(t)≥N, for any t∈[c,d] and u∈∂PR1. Then
(3.17)Tu(t)=λ∫01‍G*(t,s)[f(s,[sα-2u(s)-λω(s)]+)+r(s)]ds≥λ∫cd‍G*(t,s)f(s,[sα-2u(s)-λω(s)]+)ds≥2λkβ2∫cd‍Φ(s)ds∫cd‍G*(t,s)ds≥2λktβ∫cd‍Φ(s)ds∫cd‍Φ(s)ds=R1t.
This implies
(3.18)‖Tu‖≥‖u‖,∀u∈∂PR1.
On the other hand,g(x) is continuous on [0,+∞), and thus from the second limit of (H5), we have
(3.19)lim⁡x→+∞g*(x)x=0,
where g*(x) is defined by (3.2). For
(3.20)ϵ=(2λ∫01‍Φ(s)z(s)ds)-1,
there exists X0>0 such that g(u)≤ɛx for any x≥X0 and u∈[0,x].
Let(3.21)R2=X0+R1+2λ∫01‍Φ(s)r(s)ds.
For any u∈∂PR2, by (3.16) we can get R2≥u(t)-λt2-αω(t)≥0, for all t∈[0,1]. Therefore,
(3.22)‖Tu‖≤λ∫01‍Φ(s)[z(s)g([u(s)-λs2-αω(s)]+)+r(s)]ds≤λɛR2∫01‍Φ(s)z(s)ds+λ∫01‍Φ(s)r(s)ds≤R22+R22=R2.
Thus,
(3.23)‖Tu‖≤‖u‖,∀u∈∂PR2.
By Lemma 2.13, T has a fixed point u∈P such that R1≤∥u∥≤R2. Let u¯(t)=tα-2u(t)-λω(t). Since ∥u∥≥R1, by (3.16) we have u¯(t)≥0 on (0,1] and limt→0+tα-2u(t)=0. Notice that ω(t) is a solution of (2.37) and tα-2u(t) is a solution of (2.41). Thus, u¯(t) is a positive solution of the BVP (1.1).By the proof of Theorem3.2, we have the following corollary.Corollary 3.3.
The conclusion of Theorem3.2 is valid if (H5) is replaced by (H5*). There exist [c,d]⊂(0,1) and N>0 such that for any t∈[c,d] and x≥N,
(3.24)f(t,x)≥2kβ2∫cd‍Φ(s)ds,lim⁡x→+∞g(x)x=0.


## 4. Example

Example 4.1 (a 4-point BVP with coefficients of both signs).
Consider the following problem:(4.1)D0+7/4u(t)+λf(t,u(t))=0,t∈(0,1),u(0)=0,D0+1/4u(1)=D0+1/4u(14)-12D0+1/4u(49),
where
(4.2)f(t,x)=x2+ln⁡t.
We have(4.3)G0(t,s)=1Γ(7/4){t3/4(1-s)1/2,0≤t≤s≤1,t3/4(1-s)1/2-(t-s)3/4,0≤s≤t≤1,(4.4)p(s)={1-((1/4)-s1-s)1/2-12((4/9)-s1-s)1/2,0≤s≤14,1-12((4/9)-s1-s)1/2,14<s≤49,1,49<s≤1.
By direct calculations, we have p(0)=(1/6) and q(s)≥0, which implies that (H1) holds.
Letr(t)=-lnt, z(t)=t-1/2, g(x)=x2. It is easy to see that (H2) and (H3) hold. Moreover,
(4.5)liminf⁡x→+∞min⁡t∈[(1/4),(3/4)]f(t,x)x=+∞.
Therefore, the assumptions of Theorem 3.1 are satisfied. Thus, Theorem 3.1 ensures that there exists λ*>0 such that the BVP (4.1) has at least one positive solution for any λ∈(0,λ*).Remark 4.2.
Noticing thatλx2 does not satisfy (A1), therefore, the work in the present paper improves and generalizes the main results of [23].



---

*Source: 214042-2012-07-19.xml*