# Positive Solutions for Fourth-Order Nonlinear Differential Equation with Integral Boundary Conditions

Authors: Qi Wang; Yanping Guo; Yude Ji
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

This paper investigates the existence and nonexistence of positive solutions for a class of fourth-order nonlinear differential equation with integral boundary conditions. The associated Green's function for the fourth-order boundary value problems is first given, and the arguments are based on Krasnoselskii's fixed point theorem for operators on a cone.


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## Body

## 1. Introduction

In this paper, we consider the following fourth-order boundary value problems (BVPs) with integral boundary conditions(1)x(4)(t)=w(t)f(t,x(t),x′′(t)),t∈(0,1),x(0)=∫01‍h1(s)x(s)ds,x(1)=∫01‍k1(s)x(s)ds,x′′(0)=∫01‍h2(s)x′′(s)ds,x′′(1)=∫01‍k2(s)x′′(s)ds,
where w may be singular at t=0 and(or) t=1, f:[0,1]×[0,+∞)×(-∞,0]→[0,+∞) is continuous, and h1,h2,k1,k2∈L1[0,1] are nonnegative.The existence of solutions for nonlinear higher-order nonlocal BVPs has been studied by several authors, for example, see [1–9] and the references therein. However, there are a few papers dealing with the existence of positive solutions for the fourth-order boundary value problems with integral boundary conditions, see [7–15]. For the case of h1,h2,k1,k2=0, BVP (1) reduces to the following two-point BVP
(2)x(4)(t)=w(t)f(t,x(t),x′′(t)),t∈(0,1),x(0)=x(1)=x′′(0)=x′′(1)=0.
A great deal of research has been devoted to the existence of solutions for problem (2) by using the Leray-Schauder continuation method, topological degree, and the method of lower and upper solutions.For the case ofh1=g,h2=h,k1=0,k2=0 or h1=0,h2=0,k1=g, and k2=h, BVP (1) reduces to the following BVP with integral boundary conditions
(3)x(4)(t)=w(t)f(t,x(t),x′′(t)),t∈(0,1),x(0)=∫01‍g(s)x(s)ds,x(1)=0,x′′(0)=∫01‍h(s)x′′(s)ds,x′′(1)=0,
or
(4)x(4)(t)=w(t)f(t,x(t),x′′(t)),t∈(0,1),x(0)=0,x(1)=∫01‍g(s)x(s)ds,x′′(0)=0,x′′(1)=∫01‍h(s)x′′(s)ds.
By using the fixed point theorem of cone expansion and compression of norm type, Zhang and Ge [11] have obtained the existence and multiplicity of positive solutions for BVP (3) and (4).The following assumptions will stand throughout this paper:(A1)
ω is nonnegative, and ω∈L1[0,1] may have singularities at t=0 and(or) t=1;(A2)
f∈C([0,1]×[0,+∞)×(-∞,0],[0,+∞));(A3)
h1,h2,k1,k2∈L1[0,1] are nonnegative and
(5)μ1=1-∫01‍h1(s)ds>0,ν1=1-∫01‍k1(s)ds>0,μ2=1-∫01‍h2(s)ds>0,ν2=1-∫01‍k2(s)ds>0.The paper is organized as follows. In Section2, we provide some necessary background material such as the Krasnoselskii’s fixed point theorem in cones. The associated Green’s function for the fourth-order boundary value problem is first given, and we also look at some properties of the Green’s function associated with the problem (1). In Section 3, the main results of problem (1) will be stated and proved. In Section 4, we give an example to illustrate how the main results can be used in practice.


## 2. Preliminary Results

In our main results, we will make use of the following lemmas.Lemma 1 (in [16]).
LetΩ1 and Ω2 be two bounded open sets in a real Banach space E, such that θ∈Ω1 and Ω1¯⊂Ω2. Let operator A:P∩(Ω2¯∖Ω1)→P be continuous, where θ denotes the zero element of E and P is a cone in E. Suppose that one of the two conditions (1)
∥Ax∥≤∥x∥, ∀x∈P∩∂Ω1 and ∥Ax∥≥∥x∥, ∀x∈P∩∂Ω2 or(2)
∥Ax∥≥∥x∥, ∀x∈P∩∂Ω1 and ∥Ax∥≤∥x∥, ∀x∈P∩∂Ω2 is satisfied. Then, A has at least one fixed point in P∩(Ω2¯∖Ω1).Lemma 2.
Ifh,k∈L1[0,1] are nonnegative, and μ=1-∫01‍h(s)ds>0,ν=1-∫01‍k(s)ds>0, then for any y∈C(0,1), the BVP
(6)-x′′(t)=y(t),t∈(0,1),x(0)=∫01‍h(s)x(s)ds,x(1)=∫01‍k(s)x(s)ds,
has a unique solution x which is given by
(7)x(t)=∫01‍H~(t,s)y(s)ds,
where
(8)H~(t,s)=G(t,s)+m+μtmν+nμ∫01‍k(τ)G(s,τ)dτ+n-νtmν+nμ∫01‍h(τ)G(s,τ)dτ,(9)G(t,s)={s(1-t),0≤s≤t≤1,t(1-s),0≤t≤s≤1,(10)m=∫01‍sh(s)ds,n=1-∫01‍sk(s)ds.Proof.
The general solution of-x′′(t)=y(t) can be written as
(11)x(t)=-∫0t‍(t-s)y(s)ds+At+B.
Now, we solve for A,B by x(0)=∫01‍h(s)x(s)ds and x(1)=∫01‍k(s)x(s)ds, it follows that
(12)B=-∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ+A∫01‍τh(τ)dτ+B∫01‍h(τ)dτ,-∫01‍(1-s)y(s)ds+A+B=-∫01‍k(τ)∫0τ‍(τ-s)y(s)dsdτ+A∫01‍τk(τ)dτ+B∫01‍k(τ)dτ,
that is,
(13)A∫01‍τh(τ)dτ-B(1-∫01‍h(τ)dτ)=∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ,A(1-∫01‍τk(τ)dτ)+B(1-∫01‍k(τ)dτ)=∫01‍(1-s)y(s)ds-∫01‍k(τ)∫0τ‍(τ-s)y(s)dsdτ.
Solving the above equations, we get
(14)A=1mν+nμ[ν∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ+μ(∫01‍(1-s)y(s)ds-∫01‍k(τ)∫0τ‍(τ-s)y(s)dsdτ)],B=1mν+nμ[m(∫01‍(1-s)y(s)ds-∫01‍k(τ)∫0τ‍(τ-s)y(s)dsdτ)-n∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ].
Therefore, (6) has a unique solution
(15)x(t)=-∫0t‍(t-s)y(s)ds+1mν+nμ×[νt∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ+μt(∫01‍(1-s)y(s)ds-∫01‍k(τ)×∫0τ‍(τ-s)y(s)dsdτ)+m(∫01‍(1-s)y(s)ds-∫01‍k(τ)×∫0τ‍(τ-s)y(s)dsdτ)-n∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ].
The unique solution of (6) be expressed as
(16)x(t)=∫0t‍s(1-t)y(s)ds+∫t1‍t(1-s)y(s)ds+1mν+nμ×[μt∫01‍τk(τ)dτ∫01‍(1-s)y(s)ds-μt×∫01‍(1-s)y(s)ds-νt×∫01‍τh(τ)dτ∫01‍(1-s)y(s)ds+νt∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ+μt×∫01‍(1-s)y(s)ds-μt×∫01‍k(τ)∫0τ‍(τ-s)y(s)dsdτ+∫01‍τh(τ)dτ∫01‍(1-s)y(s)ds-∫01‍τh(τ)dτ×∫01‍k(τ)∫0τ‍(τ-s)y(s)dsdτ-∫01‍h(τ)×∫0τ‍(τ-s)y(s)dsdτ+∫01‍τk(τ)dτ×∫01‍h(τ)∫0τ‍(τ-s)y(s)dsdτ-∫01‍τk(τ)dτ×∫01‍τh(τ)dτ∫01‍(1-s)y(s)ds+∫01‍τh(τ)dτ∫01‍τk(τ)dτ∫01‍(1-s)y(s)ds]=∫0t‍s(1-t)y(s)ds+∫t1‍t(1-s)y(s)ds+1mν+nμ×[μt(∫01‍k(τ)∫0τ‍s(1-τ)y(s)dsdτ+∫01‍k(τ)∫τ1‍τ(1-s)y(s)dsdτ)-νt(∫01‍h(τ)∫0τ‍s(1-τ)y(s)dsdτ+∫01‍h(τ)∫τ1‍τ(1-s)y(s)dsdτ)+(∫01‍h(τ)∫0τ‍s(1-τ)y(s)dsdτ+∫01‍h(τ)∫τ1‍τ(1-s)y(s)dsdτ)-∫01‍τk(τ)dτ×(∫01‍h(τ)∫0τ‍s(1-τ)y(s)dsdτ+∫01‍h(τ)∫τ1‍τ(1-s)y(s)dsdτ)+∫01‍τh(τ)dτ×(∫01‍k(τ)∫0τ‍s(1-τ)y(s)dsdτ+∫01‍k(τ)∫τ1‍τ(1-s)y(s)dsdτ)]=∫01‍G(t,s)y(s)ds+1mν+nμ×[μt∫01‍k(τ)∫01‍G(s,τ)y(s)dsdτ-νt∫01‍h(τ)∫01‍G(s,τ)y(s)dsdτ+∫01‍h(τ)∫01‍G(s,τ)y(s)dsdτ-∫01‍τk(τ)dτ∫01‍h(τ)×∫01‍G(s,τ)y(s)dsdτ+∫01‍τh(τ)dτ∫01‍k(τ)×∫01‍G(s,τ)y(s)dsdτ]=∫01‍G(t,s)y(s)ds+m+μtmν+nμ∫01‍y(s)∫01‍k(τ)G(s,τ)dτds+n-νtmν+nμ∫01‍y(s)∫01‍h(τ)G(s,τ)dτds.
Therefore, the unique solution of (6) is x(t)=∫01‍H~(t,s)y(s)ds.Lemma 3 (in [11]).
Ifh,k∈L1[0,1] are nonnegative, then,(1)
n>ν,(2)
G(t,s)>0, for all t,s∈(0,1),G(t,s)≥0, for all t,s∈[0,1],(3)
Lettinge(t)=t(1-t), for all t,s∈[0,1], one has e(t)e(s)≤G(t,s)≤G(t,t)=e(t)≤1/4,(4)
Lettingδ=(0,1/2),Jδ=[δ,1-δ],for  allt∈Jδ,s∈[0,1], we have G(t,s)≥δG(s,s), where G(t,s) is defined by (9).Lemma 4.
Ifh,k∈L1[0,1] are nonnegative, and μ=1-∫01‍h(s)ds>0,ν=1-∫01‍k(s)ds>0, for function H~(t,s), we have the following properties:(1)
H~(t,s)>0,forallt,s∈(0,1),H~(t,s)≥0,forallt,s∈[0,1],(2)
ρe(t)e(s)≤H~(t,s)≤ηe(s),forallt,s∈[0,1],(3)
H~(t,s)≥δH~(s,s),t∈Jδ,s∈[0,1],where ρ=(1/(mν+nμ))(μ∫01‍e(τ)k(τ)dτ+ν∫01‍e(τ)h(τ)dτ), η=(m+n+μ(1-ν))/(mν+nμ).Proof.
(1)
From (8) and Lemma 3, we get (1).
(2)
For allt,s∈[0,1], by (8) and Lemma 3, we have
(17)H~(t,s)=G(t,s)+m+μtmν+nμ∫01‍k(τ)G(s,τ)dτ+n-νtmν+nμ∫01‍h(τ)G(s,τ)dτ≥m+μtmν+nμ∫01‍k(τ)G(s,τ)dτ+n-νtmν+nμ∫01‍h(τ)G(s,τ)dτ≥e(s)mν+nμ(μt∫01‍e(τ)k(τ)dτ+(1-t)ν∫01‍e(τ)h(τ)dτ)≥e(s)e(t)mν+nμ(μ∫01‍e(τ)k(τ)dτ+ν∫01‍e(τ)h(τ)dτ)=ρe(t)e(s),H~(t,s)=G(t,s)+m+μtmν+nμ∫01‍k(τ)G(s,τ)dτ+n-νtmν+nμ∫01‍h(τ)G(s,τ)dτ≤G(s,s)+m+μtmν+nμ∫01‍k(τ)G(s,s)dτ+n-νtmν+nμ∫01‍h(τ)G(s,s)dτ=e(s)(1+m+μtmν+nμ∫01‍k(τ)dτ+n-νtmν+nμ∫01‍h(τ)dτ)≤e(s)mν+nμ(∫01‍k(τ)mν+nμ+(m+μ)×∫01‍k(τ)dτ+n∫01‍h(τ)dτ)=ηe(s).
(
3
)
For allt∈Jδ,s∈[0,1], by (8) and Lemma 3, we have
(18)H~(t,s)=G(t,s)+m+μtmν+nμ∫01‍k(τ)G(s,τ)dτ+n-νtmν+nμ∫01‍h(τ)G(s,τ)dτ≥δG(s,s)+δ(m+μs)mν+nμ∫01‍k(τ)G(s,τ)dτ+δ(n-νs)mν+nμ∫01‍h(τ)G(s,τ)dτ=δH~(s,s).Lemma 5.
Assume that (A1)–(A3) holds. Ifx(t)∈C2[0,1] is a solution of the following integral equation
(19)x(t)=(Tx)(t)≜∫01‍H(t,s)w(s)f(s,x(s),x′′(s))ds,
then x(t)∈C2[0,1]∩C4(0,1) is a solution of BVP (1), where
(20)H(t,s)=∫01‍H1(t,τ)H2(τ,s)dτ,(21)H1(t,τ)=G(t,τ)+m1+μ1tm1ν1+n1μ1∫01‍k1(ν)G(τ,ν)dν+n1-ν1tm1ν1+n1μ1∫01‍h1(ν)G(τ,ν)dν,(22)H2(τ,s)=G(τ,s)+m2+μ2τm2ν2+n2μ2∫01‍k2(ν)G(s,ν)dν+n2-ν2τm2ν2+n2μ2∫01‍h2(ν)G(s,ν)dν,(23)m1=∫01‍sh1(s)ds,n1=1-∫01‍sk1(s)ds,m2=∫01‍sh2(s)ds,n2=1-∫01‍sk2(s)ds.Proof.
By using Lemma2, the conclusion is obvious.Lemma 6.
If (A3) holds, then one has the following three properties:(1)
H(t,s)>0, for all t,s∈(0,1),(2)
0≤H(t,s)≤η1η2/16,t,s∈[0,1],(3)
H(1/2,s)≥(ρ1ρ2/120)e(s), for all s∈[0,1],where
(24)η1=m1+n1+μ1(1-ν1)m1ν1+n1μ1,η2=m2+n2+μ2(1-ν2)m2ν2+n2μ2,ρ1=1m1ν1+n1μ1×(μ1∫01‍e(τ)k1(τ)dτ+ν1∫01‍e(τ)h1(τ)dτ),ρ2=1m2ν2+n2μ2×(μ2∫01‍e(τ)k2(τ)dτ+ν2∫01‍e(τ)h2(τ)dτ).Proof.
By using Lemma3, the conclusions for (1) and the first part of (2) are obvious.
By using by (21) and Lemma 3, we have
(25)H1(t,τ)=G(t,τ)+m1+μ1tm1ν1+n1μ1∫01‍k1(ν)G(τ,ν)dν+n1-ν1tm1ν1+n1μ1∫01‍h1(ν)G(τ,ν)dν≤14+m1+μ1tm1ν1+n1μ1∫01‍14k1(s)ds+n1-ν1tm1ν1+n1μ1∫01‍14h1(s)ds≤η14.
Similarly, by (22), we get
(26)H2(τ,s)≤η24.
So, by (20), (25), and (26), we concluded
(27)H(t,s)=∫01‍H1(t,τ)H2(τ,s)dτ≤η1η216.
By using Lemmas 3 and 4, we have
(28)H(12,s)=∫01‍H1(12,τ)H2(τ,s)dτ≥ρ14∫01‍e(τ)H2(τ,s)dτ≥ρ1ρ24e(s)∫01‍e2(τ)dτ=ρ1ρ2120e(s).LetE=C2[0,1]. Then E is a Banach space with the norm ∥x∥2=∥x∥+∥x′′∥, where ∥x∥=maxt∈[0,1]|x(t)|,∥x′′∥=maxt∈[0,1]|x′′(t)|.For a fixedδ=(0,1/2),Jδ=[δ,1-δ], define the cone K⊂E by
(29)K≜{x∈E:x≥0,x′′≤0,mint∈Jδx(t)≥δ2∥x∥,maxt∈Jδx′′(t)≤-δ2∥x′′∥}.
It's easy to prove that K is a closed convex cone, and
(30)|x(t)|+|x′′(t)|≥δ2∥x∥2,t∈Jδ,x∈K.Lemma 7.
If (A1)–(A3) hold, thenT(K)⊂K, and T:K→K is completely continuous.Proof.
By the properties of Green's function, ifx∈C2[0,1] then Tx∈C2[0,1] and
(31)(Tx)′′(t)=-∫01‍H2(t,s)w(s)f(s,x(s),x′′(s))ds.
For all x∈K, we have by (19) and (31)
(32)(Tx)(t)≥0,(Tx)′′(t)≤0,(33)∥(Tx)∥≤∫01‍∫01‍[G(τ,τ)+m1+μ1m1ν1+n1μ1×∫01‍k1(ν)G(ν,ν)dν+n1m1ν1+n1μ1∫01‍h1(ν)G(ν,ν)dν]×H2(τ,s)ω(s)f(s,x(s),x′′(s))dτds,(34)∥(Tx)′′∥≤∫01‍[G(s,s)+m2+μ2m2ν2+n2μ2×∫01‍k2(τ)G(τ,τ)dτ+n2m2ν2+n2μ2∫01‍h2(τ)G(τ,τ)dτ]×ω(s)f(s,x(s),x′′(s))ds.
On the other hand, by (19), (33), and Lemma 3, we obtain
(35)mint∈Jδ(Tx)(t)=mint∈Jδ∫01‍∫01‍[G(t,τ)+m1+μ1tm1ν1+n1μ1×∫01‍k1(ν)G(τ,ν)dν+n1-ν1tm1ν1+n1μ1×∫01‍h1(ν)G(τ,ν)dν]×H2(τ,s)ω(s)f(s,x(s),x′′(s))dτds≥δ∫01‍∫01‍[G(τ,τ)+m1+μ1δm1ν1+n1μ1×∫01‍k1(ν)G(ν,ν)dν+n1-ν1(1-δ)m1ν1+n1μ1×∫01‍h1(ν)G(ν,ν)dν]×H2(τ,s)ω(s)f(s,x(s),x′′(s))dτds≥δ2∫01‍∫01‍[G(τ,τ)+m1+μ1m1ν1+n1μ1×∫01‍k1(ν)G(ν,ν)dν+n1m1ν1+n1μ1∫01‍h1(ν)G(ν,ν)dν]×H2(τ,s)ω(s)f(s,x(s),x′′(s))dτds≥δ2∥Tx∥.
Similarly, by (31), (34), and Lemma 3, we obtain
(36)maxt∈Jδ(Tx)′′(t)=-mint∈Jδ∫01‍H2(t,s)w(s)f(s,x(s),x′′(s))ds=-mint∈Jδ∫01‍[G(t,s)+m2+μ2tm2ν2+n2μ2×∫01‍k2(τ)G(s,τ)dτ+n2-ν2tm2ν2+n2μ2∫01‍h2(τ)G(s,τ)dτ]×ω(s)f(s,x(s),x′′(s))ds≤-δ2∫01‍[G(s,s)+m2+μ2m2ν2+n2μ2×∫01‍k2(τ)G(τ,τ)dτ+n2m2ν2+n2μ2∫01‍h2(τ)G(τ,τ)dτ]×ω(s)f(s,x(s),x′′(s))ds≤-δ2∥(Tx)′′∥.
So Tx⊂K, and then T(K)⊂K.
Clearly, the operatorT is continuous; thus by the Arzela-Ascoli theorem, it follows that T is completely continuous.


## 3. Main Results

For convenience, we introduce the following notations:(37)Kr={x∈K:∥x∥2≤r},∂Kr={x∈K:∥x∥2=r},K¯r,R={x∈K:r≤∥x∥2≤R},fβ=limsup|x|+|y|→βmaxt∈[0,1]f(t,x,y)|x|+|y|,fβ=liminf|x|+|y|→βmint∈[0,1]f(t,x,y)|x|+|y|,Mβ=maxt∈[0,1],0<|x|+|y|≤βf(t,x,y),L=(η1η216+η24)∫01‍w(s)ds,N=(ρ1ρ2120+ρ24)δ2∫δ1-δ‍e(s)w(s)ds.Theorem 8.
Assume that (A1)–(A3) hold. In addition, one supposes that one of the following conditions is satisfied(H1)
Lf(t,x,y)<|x|+|y|, for all |x|+|y|≠0,t∈[0,1],(H2)
Nf(t,x,y)>|x|+|y|, for all |x|+|y|≠0,t∈[0,1].
Then, the BVP (1) has no positive solution.Proof.
Case1. The condition (H1) holds. Assume towards a contraction that x*(t) is a positive solution of BVP (1), from the proof of Lemma 7 and (29), then x*∈K,x*(t)>0 for 0<t<1. By Lemma 6 and (29), we have
(38)∥x*∥=maxt∈[0,1]|x*(t)|=maxt∈[0,1]|Tx*(t)|=maxt∈[0,1]|∫01‍H(t,s)w(s)f(s,x*(s),x*′′(s))ds|<η1η216L∫01‍w(s)(|x*(s)|+|x*′′(s)|)ds≤η1η216L∥x*∥2∫01‍w(s)ds,∥x*′′∥=maxt∈[0,1]∥x*′′(t)∥=maxt∈[0,1]|(Tx*)′′(t)|=maxt∈[0,1]|∫01‍H2(t,s)w(s)f(s,x*(s),x*′′(s))ds|<η24L∫01‍w(s)(|x*(s)|+|x*′′(s)|)ds≤η24L∥x*∥2∫01‍w(s)ds.
So, ∥x*∥2=∥x*∥+∥x*′′∥<(1/L)(η1η2/16+η2/4)∥x*∥2∫01‍w(s)ds=∥x*∥2, which is a contradiction.
Case2. The condition (H2) holds. Assume towards a contraction that x**(t) is a positive solution of BVP (1), from the proof of Lemma 7 and (29), then x**∈K, x**(t)>0, for 0<t<1. By Lemmas 4 and 6 and (30), we have
(39)∥x**∥=maxt∈[0,1]|x**(t)|=maxt∈[0,1]|Tx**(t)|≥∫01‍H(12,s)w(s)f(s,x**(s),x**′′(s))ds>ρ1ρ2120N∫01‍e(s)w(s)(|x**(s)|+|x**′′(s)|)ds≥ρ1ρ2120N∫δ1-δ‍e(s)w(s)(|x**(s)|+|x**′′(s)|)ds≥ρ1ρ2120Nδ2∥x**∥2∫δ1-δ‍e(s)w(s)ds,∥x**′′∥=maxt∈[0,1]|x**′′(t)|=maxt∈[0,1]|(Tx**)′′(t)|=maxt∈[0,1]|∫01‍H2(t,s)w(s)f(s,x**(s),x**′′(s))ds|≥∫01‍H2(12,s)w(s)f(s,x**(s),x**′′(s))ds>ρ24N∫01‍e(s)w(s)(|x**(s)|+|x**′′(s)|)ds≥ρ24Nδ2∥x**∥2∫δ1-δ‍e(s)w(s)ds.
So, ∥x**∥2=∥x**∥+∥x**′′∥>(1/N)(ρ1ρ2/120+ρ2/4)δ2∥x**∥2∫δ1-δ‍e(s)w(s)ds=∥x**∥2, which is a contradiction, and this completes the proof.Theorem 9.
Assume that (A1)–(A3) hold. In addition, one supposes that one of the following conditions is satisfied(H3)
Lf0<1<Nf∞,(H4)
Lf∞<1<Nf0.
Then, the BVP (1) has at least one positive solution.Proof.
Case1. The condition (H3) holds.
ConsideringLf0<1, then there exists r1>0, such that f(t,x,y)≤(f0+ε1)(|x|+|y|), for t∈[0,1],(x,y)∈{(x,y):0<|x|+|y|≤r1}, where ε1>0, satisfies L(f0+ε1)≤1.
Then, fort∈[0,1],x∈∂Kr1, by (19), (26), and Lemma 6, we have
(40)∥Tx∥2=∥Tx∥+∥(Tx)′′∥≤(η1η216+η24)∫01‍w(s)f(s,x(s),s′′(s))ds≤(η1η216+η24)(f0+ε1)×∫01‍w(s)(|x(s)|+|x′′(s)|)ds≤(η1η216+η24)(f0+ε1)∥x∥2∫01‍w(s)ds≤∥x∥2,
which means that
(41)∥Tx∥2≤∥x∥2,forx∈∂Kr1.
Next, turning toNf∞>1, then there exists R¯1>0, such that f(t,x,y)≥(f∞-ε2)(|x|+|y|), for t∈[0,1],(x,y)∈{(x,y):0<|x|+|y|>R¯1}, where ε2>0, satisfies N(f∞-ε2)≥1.
LetR1=max{2r1,R¯1/δ2}, for all t∈Jδ,x∈∂KR1. By (30) and Lemma 6, we have
(42)∥Tx∥2=∥Tx∥+∥(Tx)′′∥≥|Tx(12)|+|(Tx)′′(12)|≥(ρ1ρ2120+ρ24)∫01‍e(s)w(s)f(s,x(s),s′′(s))ds≥(ρ1ρ2120+ρ24)(f∞-ε2)×∫01‍e(s)w(s)(|x(s)|+|x′′(s)|)ds≥(ρ1ρ2120+ρ24)(f∞-ε2)δ2∥x∥2×∫δ1-δ‍e(s)w(s)ds≥∥x∥2
which means that
(43)∥Tx∥2≥∥x∥2,forx∈∂KR1.
Case2. The condition (H4) holds.
ConsideringNf0>1, then there exists r2>0, such that f(t,x,y)≥(f0-ε3)(|x|+|y|), for t∈[0,1],(x,y)∈{(x,y):0<|x|+|y|≤r2}, where ε3>0, satisfies N(f0-ε3)≥1.
Then, fort∈Jδ,x∈∂Kr2, by (30) and Lemma 6, we have
(44)∥Tx∥2≥|Tx(12)|+|(Tx)′′(12)|≥(ρ1ρ2120+ρ24)(f0-ε3)×∫01‍e(s)w(s)(|x(s)|+|x′′(s)|)ds≥(ρ1ρ2120+ρ24)(f0-ε3)δ2∥x∥2×∫δ1-δ‍e(s)w(s)ds≥∥x∥2,
which means that
(45)∥Tx∥2≥∥x∥2,forx∈∂Kr2.
Next, turning toLf∞<1, then there exists R¯2>0, such that f(t,x,y)≤(f∞+ε4)(|x|+|y|), for t∈[0,1],(x,y)∈{(x,y):|x|+|y|>R¯2}, where ε4>0, satisfies L(f∞+ε4)≤1.
Let(46)MR¯2=maxt∈[0,1],0<|x|+|y|≤R¯2f(t,x,y),
for t∈[0,1],f(t,x,y)≤MR¯2+(f∞+ε4)(|x|+|y|).
LetR2=max{2r2,R¯2,LMR¯2/(1-L(f∞+ε4))}, for t∈[0,1],x∈∂KR2. By (26) and Lemma 6, we have
(47)∥Tx∥2≤(η1η216+η24)∫01‍w(s)f(s,x(s),s′′(s))ds≤(η1η216+η24)×(MR¯2∫01‍w(s)ds+(f∞+ε4)×∫01‍w(s)(|x(s)|+|x′′(s)|)ds)≤LMR¯2+∥x∥2Ł(f∞+ε4)≤R2-L(f∞+ε4)R2+∥x∥2L(f∞+ε4)=∥x∥2,
which means that
(48)∥Tx∥2≤∥x∥2,forx∈∂KR2.
Applying Lemma1 to (41) and (43), or (45), and (48) yields that T has a fixed point x*∈K¯ri,Ri(i=1,2) with x*(t)≥δ∥x*∥2, t∈(0,1). Thus, it follows that BVP (1) has at least one positive solution, and the theorem is proved.Theorem 10.
Assume that (A1)–(A3) hold, as do the following two conditions:(H5)
Nf0>1 and Nf∞>1,(H6)
there existsb>0, such that maxt∈[0,1],0<|x|+|y|≤bf(t,x,y)<b/L.
Then, the BVP (1) has at least two positive solutions x*(t),x**(t), which satisfy
(49)0<∥x*∥2<b<∥x**∥2.Proof.
We chooser,R with 0<r<b<R.
IfNf0>1, then by the proof of (45), we have
(50)∥Tx∥2≥∥x∥2,for∥x∥2=r.
IfNf∞>1, then by the proof of (43), we have
(51)∥Tx∥2≥∥x∥2,for∥x∥2=R.
On the other hand, by (H6), for x∈∂Kb, we have
(52)∥Tx∥2≤(η1η216+η24)∫01‍w(s)f(s,x(s),s′′(s))ds≤(η1η216+η24)Mb∫01‍w(s)ds=LMb,
where Mb=maxt∈[0,1],0<|x|+|y|≤bf(t,x,y)<b/L.
By (52), we have
(53)∥Tx∥2<b=∥x∥2.
Applying Lemma1 to (50), (51), and (53) yields that T has a fixed point x*∈K¯r,b, and a fixed point x**∈K¯b,R. Thus it follows that BVP (1) has at least two positive solutions x* and x**. Noticing (53), we have ∥x*∥2≠b and ∥x**∥2≠b. Therefore, (49) holds, and the proof is complete.Theorem 11.
Assume that (A1)–(A3) hold, as do the following two conditions:(H7)
Lf0<1 and Lf∞<1,(H8)
there existδ∈(0,1/2) and B>0 such that f(t,x,y)>δ2B/N for all t∈Jδ,x∈[δ2B,B] and y∈[-B,-δ2B].
Then, the BVP (1) has at least two positive solutions x*(t),x**(t), which satisfy
(54)0<∥x*∥2<B<∥x**∥2.Proof.
We chooser¯,R¯ with 0<r¯<B<R¯.
IfLf0<1, then by the proof of (41), we have
(55)∥Tx∥2≤∥x∥2,for∥x∥2=r¯.
IfLf∞<1, then by the proof of (48), we have
(56)∥Tx∥2≤∥x∥2,for∥x∥2=R¯.
On the other hand, by (H8), for x∈∂KB, we have
(57)∥Tx∥2≥(ρ1ρ2120+ρ24)∫01‍e(s)w(s)×f(s,x(s),s′′(s))ds>B=∥x∥2.
Applying Lemma1 to (55), (56), and (57) yields that T has a fixed point x*∈K¯r¯,B, and a fixed point x**∈K¯B,R¯. Thus it follows that BVP (1) has at least two positive solutions x* and x**. Noticing (57), we have ∥x*∥2≠B and ∥x**∥2≠B. Therefore, (54) holds, and the proof is complete.


## 4. Example

Example 12.
Consider the following fourth-order BVP(58)x(4)(t)=1tf(t,x(t),x′′(t)),t∈(0,1),x(0)=∫01‍sx(s)ds,x(1)=∫01‍sx(s)ds,x′′(0)=∫01‍s2x′′(s)ds,x′′(1)=∫01‍s2x′′(s)ds.
By calculation, we obtain
(59)μ1=1-∫01‍sds=12,ν1=1-∫01‍sds=12,m1=∫01‍s2ds=13,n1=1-∫01‍s2ds=23,μ2=1-∫01‍s2ds=23,ν2=1-∫01‍s2ds=23,m2=∫01‍s3ds=14,n2=1-∫01‍s3ds=34,η1=52,η2=116,ρ1=16,ρ2=110,L=14396,δ=13,N=18132187000(32-2)≈0.00032.(1)
About the nonexistence of positive solution, we consider BVP (58) with
(60)f(t,x,y)=k1et(|x|+|y|)+k2|sin(|x|+|y|)|,where k1 and k2 are two positive real numbers, then
(61)f(t,x,y)|x|+|y|<ek1+k2,f(t,x,y)|x|+|y|>k1.Ifek1+k2<1/L or k1>1/N, by Theorem 8, BVP (58) has no positive solution.(2)
About the existence of positive solution, we consider BVP (58) with
(62)f(t,x,y)=k1et-1(|x|+|y|)+k2|sin(|x|+|y|)|,where k1 and k2 are two positive real numbers, then
(63)f0=liminf|x|+|y|→0mint∈[0,1]f(t,x,y)|x|+|y|=k1e+k2,f∞=limsup|x|+|y|→∞maxt∈[0,1]f(t,x,y)|x|+|y|=k1.Ifk1<1/L and k2>1/N, then we have Lf∞<1<Nf0. By Theorem 9, BVP (58) has at least one positive solution.(3)
About the multiplicity of positive solution, we consider BVP (58) with
(64)f(t,x,y)={et|sink1(|x|+|y|)|,|x|+|y|≤b,et|sink1(|x|+|y|)|+k2(1+t)(|x|+|y|-b)2|x|+|y|+1,|x|+|y|>b,where k1 and k2 are two positive real numbers, then
(65)f0=liminf|x|+|y|→0mint∈[0,1]f(t,x,y)|x|+|y|=k1,f∞=limsup|x|+|y|→∞maxt∈[0,1]f(t,x,y)|x|+|y|=2k2.
If k1>1/N and k2>1/2N, then we have Nf0>1 and Nf∞>1.
Furthermore, lettingb>LMb>143e/96, then we have Mb<b/L. By Theorem 10, BVP (58) has at least two positive solutions.Remark 13.
In [4, 7–10], the existence of solutions for fourth-order ordinary differential equations BVP has been treated but did not discuss problems with singularities. Although, [11] seems to have considered the existence of positive solutions for the fourth-order BVP with integral boundary conditions for the singularity allowed to appear at t=0 and/or t=1. However, in [11], only the boundary conditions x(0)=0,x′′(0)=0, or x(1)=0,x′′(1)=0 have been considered. It is clear that the boundary conditions of Example 12 are x(0)≠0,x′′(0)≠0, and x(1)≠0,x′′(1)≠0. Hence, we generalize the fourth-order ordinary differential equations BVP.



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*Source: 684962-2013-02-19.xml*