# Positive Solutions for a Class of Fourth-Order Boundary Value Problems in Banach Spaces

Authors: Jingjing Cai; Guilong Liu
Journal: Abstract and Applied Analysis
 (2011)

---

## Abstract

Using a specially constructed cone and the fixed point index theory, this work shows existence and
nonexistence results of positive solutions for fourth-order boundary value problem with two different parameters in Banach spaces.


---

## Body

## 1. Introduction

In this paper, we study the existence of positive solutions for the following fourth-order boundary value problem (BVP) with singularity in Banach space(E,∥·∥):(1.1)x(4)(t)+βx′′(t)=λf(t,x)+μg(t,x),t∈J,x(0)=x(1)=θ,x′′(0)=x′′(1)=θ,
where J=[0,1],  β<π2.g, f∈C(J,[0,+∞)), R+=[0,+∞),  and θ is the zero element of E.Fourth-order boundary value problems are studied not only in mathematics but also in physics and many other fields. For example, some models for bridge, underground water flow, and plasma physics can be reduced to fourth-order boundary value problems. In the recent years, some authors (cf. [1–7]) studied fourth-order boundary value problems, but these papers mainly dealt with the problems in real space or the problems without singularity. As far as we know, the fourth-order ordinary differential equation with singularity has been seldom studied in Banach spaces. Also, not so much is known about the case that the nonlinear term has two different parameters or has two parts with different properties.In [7], Yao studied the following two-point boundary value problem:(1.2)u(4)(t)-λh(t)f(u(t))=0,u(0)=u(1)=u′′(0)=u′′(1)=0,
where λ>0 and  f∈C[0,+∞). The author obtained positive solutions of BVP (1.2) in real space not abstract space.In [4], the authors studied the following boundary value problem in real space: (1.3)x(4)(t)+βx′′(t)=λf(t,x),x(0)=x(1)=x′′(0)=x′′(1)=0,
where f∈C([0,1]×[0,+∞)), β<π2, and λ>0. With the method of monotone iterative, the existence and uniqueness of positive solution of BVP (1.3) are obtained.Comparing with the results in [5–7], this paper has the following features. Firstly, we discuss positive solutions of BVP (1.1) in abstract space, not E=R as in [5, 7]. Secondly, BVP (1.1) has two parameters which may have different domains. Fourthly, f and g may have different properties, and the main tool we used is the fixed point theorem on cone. Thirdly, we talk about both the existence and nonexistence of positive solutions, but [5,9] only study the existence of positive solutions.Basic facts about ordered Banach spaceE can be found in [8]. Here we recall some of them. The cone P in E is said to be normal if there exists a positive constant N such that θ≤x≤y implies ∥x∥≤N∥y∥. In this paper, we always suppose that P is normal in E, and without loss of generality, suppose that the normal constant N=1. E* is dual space of E. Denote P*={ψ∈E*∣ψ(x)≥0,forallx∈P}, then P* is a dual cone of cone P. The noncompactness of Kuratowski is denoted by α(·).We study BVP (1.1) in C(J,E). Evidently, (C(J,E),∥·∥c) is a Banach space under the norm ∥x∥c=maxt∈J∥x(t)∥. x∈C(J,E) is a positive solution of BVP(1.1) if it satisfies BVP (1.1) and x(t)>θ, for all t∈J.


## 2. Preliminaries and Lemmas

In this paper, we make the following conditions.(H1)
f,g∈C(J′×P,P), f(t,Pr)={f(t,u),u∈Pr} is relatively compact and g(t,Pr) is also relatively compact, where Pr={u∈P:∥u∥≤r}. For any u∈P, there exist ai(t), bi(t)∈L(J,R+), w1(x), w2(x)∈C(J,R+) such that(2.1)‖f(t,u)‖≤a1(t)+b1(t)w1(‖u‖),‖g(t,u)‖≤a2(t)+b2(t)w2(‖u‖),a.e.t∈J.(H2)
There exist a functionm∈L(J,(0,+∞)), such that ψ(f(t,u))≥m(t)∥u∥ or ψ(g(t,u))≥m(t)∥u∥, for t∈J,u∈P, where ψ∈P*, ∥ψ∥=1.Lemma 2.1 (see [8]).
LetH={x∣x:J→E,x is strong measurable} be countable, and there exists M∈L[J,R+] such that ∥x(t)∥≤M(t),a.e.t∈J,x∈H, then α(H(t))∈L[J,R+] and
(2.2)α({∫Jx(t)dt:x∈H})≤2∫Jα(H(t))dt.Lemma 2.2 (see [8]).
LetP be a cone in Banach space E. For r>0, define Pr={x∈P:∥x∥<r}. Assume that T:P¯r→P is a completely continuous map such that x≠Tx for x∈∂Pr.(i)
If∥x∥≤∥Tx∥ for x∈∂Pr, then i(T,Pr,P)=0.(ii)
If∥x∥≥∥Tx∥ for x∈∂Pr, then i(T,Pr,P)=1.Lemma 2.3 (see [4]).
For anyφ∈C(J,P),  the following fourth-order boundary value problem:
(2.3)x(4)(t)+βx′′(t)=φ(t),x(0)=x(1)=x′′(0)=x′′(1)=θ
has the solution
(2.4)x(t)=∫01∫01G1(t,τ)G2(τ,s)φ(s)dsdτ,
where G2(t,s) is defined in Proposition 2.4 and
(2.5)G1(t,s)={t(1-s),0≤t≤s≤1,s(1-t),0≤s≤t≤1.Proposition 2.4.
For allt,s∈[0,1]×[0,1] one has the following properties:
(2.6)t(1-t)G1(s,s)≤G1(t,s)≤G1(s,s)=s(1-s),G1(t,s)≤t(1-t);G1(t,s)≤14.
Let ω=β. G2(t,s) is explicitly given by the following
Ifβ<0, then
(2.7)G2(t,s)={sinh⁡ωtsinh⁡ω(1-s)ωsinh⁡ω,0≤t≤s≤1,sinh⁡ωssinh⁡ω(1-t)ωsinh⁡ω,0≤s≤t≤1.
Ifβ=0, then G2(t,s)=G1(t,s).
If0<β<π2, then
(2.8)G2(t,s)={sinωtsinω(1-s)ωsinω,0≤t≤s≤1,sinωssinω(1-t)ωsinω0≤s≤t≤1.Proposition 2.5.
For anyt,s∈(0,1), one has G1(t,s)>0, G2(t,s)>0, and G1(t,s)≥δG1(u,s), for t∈Jδ,s,u∈[0,1], where δ∈(0,1/2).Proposition 2.6.
It is easy to see that(2.9)G2(t,s)≤{14,β=0,e2ω4,β<0,sinωω,β>0.Letd=max{1/4,e2ω/4,sinω/ω}, Q={x∈C(J,E):x(t)≥θ,t∈J}, and K={x∈Q:x(t)≥δx(s), t∈Jδ,s∈[0,1]}. It is easy to know that K is a cone in C(J,E). Let Kr={x∈K:∥x∥≤r}⊂K, K⊂Q. Define T as (2.10)(Tx)(t)=∫01∫01G1(t,τ)G2(τ,s)(λf(s,x(s))+μg(s,x(s)))dsdτ,
then x is the solution of BVP(1.1) if and only if x is the fixed point of T.Lemma 2.7.
Suppose that condition (H1) holds, then T:K→K is completely continuous.Proof.
By the continuity ofG1(t,s) and (H1), we have Tx∈Q. And for any t∈Jδ, x∈K, by Proposition 2.5, we get
(2.11)(Tx)(t)=δ(Tx)(u).
Next we prove thatT is compact. Let V={xn}n=1∞⊂K be any bounded set, and we suppose that ∥xn∥≤r, for some r>0. Let Mr=max0≤v≤r{wi(v),i=1,2}, then by condition (H1),
(2.12)f(t,xn(t))≤a1(t)+b1(t)Mr,g(t,xn(t))≤a2(t)+b2(t)Mr,a.e.t∈J,xn∈V.
We obtain from Lemma 2.1(2.13)α{(Txn)(t):xn∈V}=α{∬01G1(t,τ)G2(τ,s)(λf(s,xn(s))+μg(s,xn(s)))dsdτ:xn∈V}≤2d∫01α((λf(s,xn(s))+μg(s,xn(s))):xn∈V)ds=0.
Hence, TV is relatively compact, so there exists a subsequence {xni} of {xn} such that {Txni} converges to some v∈C(J,E). So T:K→K is completely continuous.In this paper, foru∈P we denote (2.14)(ψf)ν=liminf⁡‖u‖→νmin⁡t∈Jψ(f(t,u))‖u‖,(ψg)ν=liminf⁡‖u‖→νmin⁡t∈Jψ(g(t,u))‖u‖,
where ν denotes 0 or ∞, ψ∈P*, ∥ψ∥=1.


## 3. Main Results

Theorem 3.1.
Assume that (H1) holds and the following conditions are satisfied:
(3.1)(ψf)∞=∞,(ψg)0=∞.
Then BVP(1.1) has at least two positive solutions when λ and μ are sufficiently small.Proof.
T is completely continuous by Lemma 2.7. For any l>0, set
(3.2)F(l)=d4max⁡x∈K,‖x‖c=l∫01‖f(s,x(s))‖ds,G(l)=d4max⁡x∈K,‖x‖c=l∫01‖g(s,x(s))‖ds.
Since (ψf)∞=∞, there exists r1>0 such that F(r1)>0. If G(r1)=0, let λ1=r1/F(r1); then for any λ∈(0,λ1) and x∈K,∥x∥c=r1, we have
(3.3)‖(Tx)(t)‖≤d4∫01‖λf(s,x(s))+μg(s,x(s))‖ds≤d4∫01‖λf(s,x(s))‖ds+d4∫01‖μg(s,x(s))‖ds≤λ1F(r1)=r1=‖x‖c.
IfG(r1)≠0, let μ1=r1/2G(r1) and λ1=r1/2F(r1). For λ∈(0,λ1),μ∈(0,μ1), and x∈K,∥x∥c=r1, we get
(3.4)‖(Tx)(t)‖≤d4∫01‖λf(s,x(s))‖ds+d4∫01‖μg(s,x(s))‖ds≤d4λ1∫01‖λf(s,x(s))‖ds+d4μ1∫01‖g(s,x(s))‖ds≤r1F(r1)+μ1G(r1)≤r12+r12=r1=‖x‖c,
which means that ∥Tx∥c<r1=∥x∥c, for x∈K, ∥x∥c=r1, and Lemma 2.2 implies that i(T,Kr1,K)=1.
Since(ψf)∞=∞, there exists r¯2>0 such that ψ(f(t,u))≥ɛ1∥u∥, u∈P,t∈J,∥u∥≥r¯2, where ɛ1>0 satisfies ɛ1λδ∬δ1-δG1(1/2,τ)G2(τ,s)dsdτ>1. Let r2=max{2r1,r¯2/δ}, then for any t∈Jδ,x∈K, ∥x∥c=r2, by the definition of K we get ∥x(t)∥≥δ∥x∥c≥r¯2, and we have
(3.5)‖(Tx)(12)‖≥ψ((Tx)(12))≥λ∫01∫01G1(12,τ)G2(τ,s)ψ(f(s,x(s)))dsdτ≥λ∫01∫δ1-δG1(12,τ)G2(τ,s)ψ(f(s,x(s)))dsdτ≥λɛ1∫01∫δ1-δG1(12,τ)G2(τ,s)‖x(s)‖dsdτ≥λɛ1δ‖x‖c∫01∫δ1-δG1(12,τ)G2(τ,s)dsdτ>‖x‖c,
which implies that ∥Tx∥c>r2, x∈Kr2. By Lemma 2.2 we have i(T,Kr2,K)=0.
On the other hand, since(ψg)0=∞, there exists r¯3:0<r¯3<r1 such that ψ(g(t,u))≥ɛ2∥u∥, for t∈J,u∈P,∥u∥≤r¯3, where ɛ2>0 and satisfies
(3.6)ɛ2μδ∫01∫δ1-δG1(12,τ)G2(τ,s)dsdτ>1.Accordingly,0<r3<r¯3. Then for x∈K,∥x∥c=r3, similar to (3.5) we get ∥Tx∥c>r3. By Lemma 2.2 we have i(T,Kr3,K)=0. Hence, from the additivity of fixed point index, we obtain
(3.7)i(T,Kr2K¯r1,K)=-1,i(T,Kr1K¯r3,K)=1.
Hence, T has two fixed points x1∈Kr2/K¯r1 and x2∈Kr1/K¯r3. Therefore, BVP(1.1) has at least two fixed points x1 and x2.Corollary 3.2.
If (H1),  (ψg)∞=∞, and (ψf)0=∞ hold, then BVP(1.1) has at least two positive solutions when λ and μ are sufficiently small.Theorem 3.3.
Suppose that (H1) and (H2) hold, and if λ is sufficiently large, then BVP(1.1) has no positive solutions.Proof.
First by (H2) we can suppose that ψ(f(t,u))≥m(t)∥u∥, for t∈J. suppose that BVP(1.1) has a positive solution x, and choose λ sufficiently large satisfying λδ∬δ1-δG1(1/2,τ)G2(τ,s)m(s)dsdτ>1. Then we have
(3.8)‖x‖c≥‖x(12)‖≥ψ(x(12))≥λ∫01∫01G1(12,τ)G2(τ,s)ψ(f(s,x(s)))dsdτ≥λ∫01∫δ1-δG1(12,τ)G2(τ,s)m(s)‖x(s)‖dsdτ≥λδ∫01∫δ1-δG1(12,τ)G2(τ,s)m(s)dsdτ‖x‖c>‖x‖c,
which is a contradiction; so BVP(1.1) has no positive solutions. Similarly, if ψ(g(t,u))≥m(t)∥u∥, for t∈J, and μ is sufficiently large, then BVP(1.1) has no positive solutions.



---

*Source: 831730-2011-02-20.xml*