# Positive Solutions for Second-Order Three-Point Eigenvalue Problems

Authors: Chuanzhi Bai
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

With the help of the fixed point index theorem in cones, we get an existence theorem concerning the existence of positive solution for a second-order three-point eigenvalue problemx′′(t)+λf(t,x(t))=0,  0≤t≤1,x(0)=0,x(1)=x(η), where λ
is a parameter. An illustrative example is given to demonstrate the effectiveness of the obtained result.


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## Body

## 1. Introduction

Motivated by the work of Bitsadze and Samarskii [1] and Ilyin and Moiseev [2], much attention has been paid to the study of certain nonlocal boundary value problems (BVPs) in recent years.In the last twenty years, many mathematician, have considered the existence of positive solutions of nonlinear three-point boundary value problems; see, for example, Graef et al. [3] Webb [4], Gupta and Trofimchuk [5], Infante [6], Ehrke [7], Ma [8], Feng [9], He and Ge [10], Bai and Fang [11], and Guo [12]. Recently, by applying the Avery-Henderson [13] double fixed point theorem, Henderson [14] studied the existence of two positive solutions of the three-point boundary value problem for the second-order differential equation(1.1)y′′+f(y)=0,0≤t≤1,y(0)=0,y(p)-y(1)=0,
where 0<p<1 and f:R→[0,+∞) is continuous.In this paper, motivated and inspired by the above work and Wong [15], we apply a fixed point index theorem in cones to investigate the existence of positive solutions for nonlinear three-point eigenvalue problems(1.2)x′′(t)+λf(t,x(t))=0,0≤t≤1,x(0)=0,x(1)=x(η),
where 0<η<1 and f∈C([0,1]×[0,+∞),[0,∞)).We need the following well-known lemma. See [16] for a proof and further discussion of the fixed point index i(A,Kr,K).Lemma 1.1.
Assume thatE is a Banach space, and K⊂E is a cone in E. Let Kr={x∈K:∥x∥<r}. Furthermore, assume that A:K→K is a completely continuous map, and Ax≠x for x∈∂Kr={x∈K:∥x∥=r}. Then, one has the following conclusions: (1)
if∥x∥≤∥Ax∥ for x∈∂Kr, then i(A,Kr,K)=0;(2)
if∥x∥≥∥Ax∥ for x∈∂Kr, then i(A,Kr,K)=1.


## 2. Main Results

In the following, we will denote byC[0,1] the space of all continuous functions x:[0,1]→R. This is a Banach space when it is furnished with usual sup-norm ∥x∥:=sup{|x(s)|:s∈[0,1]}.By [14], the Green's function for the three-point boundary-value problem (2.1)-x′′=0,x(0)=0,x(1)=x(η)
is given by(2.2)G(t,s)={t,t≤s≤η,s,s≤t,s≤η,1-s1-ηt,t≤s,s≥η,s+η-s1-ηt,η≤s≤t.From the Green's functionG(t,s), we have that a function x is a solution of the boundary value problem (1.2) if and only if it satisfies(2.3)x(t)=λ∫01G(t,s)f(s,x(s))ds,t∈[0,1].Lemma 2.1.
Suppose thatG(t,s) is defined as above. Then we have the following results:  (1)
0≤G(t,s)≤G(s,s),0≤t,s≤1,(2)
G(t,s)≥ηtG(s,s),0≤t,s≤1.Proof.
It is easy to see that (1) holds. To show that (2) holds, we distinguish four cases.(i)
Ift≤s≤η, then
(2.4)G(t,s)=t≥ηts=ηtG(s,s).(ii)
Ifs≤t and s≤η, then
(2.5)G(t,s)=s≥ηts=ηtG(s,s).(iii)
Ift≤s and s≥η, then
(2.6)G(t,s)=1-s1-ηt≥ηst1-s1-η=ηtG(s,s).(iv)
Finally, ifη≤s≤t, then
(2.7)G(t,s)=s-s-η1-ηt≥s-s-η1-η=η(1-s)1-η≥tsη(1-s)1-η=ηts(1-s)1-η=ηtG(s,s).Remark 2.2.
Ifs≤η and s≥η, then G(s,s)=s and G(s,s)=s(1-s)/(1-η), respectively.
Define(2.8)P={u∈C[0,1]:u(t)≥0,mint∈[η/2,1]u(t)≥η22∥u∥}.
Obviously, P is a cone in the Banach space C[0,1].
Define an operatorA:P→C[0,1] as follows:(2.9)(Ax)(t):=λ∫01G(t,s)f(s,x(s))ds,t∈[0,1].
It is easy to know that fixed points ofA are solutions of the BVP (1.2).
Now, we can state and prove our main results.Lemma 2.3.
A:P→P is completely continuous.Proof.
For anyx∈P, by Lemma 2.1 (1), we have (Ax)(t)≥0, for each t∈[0,1]. It follows from Lemma 2.1 that
(2.10)mint∈[η/2,1](Ax)(t)=λmint∈[η/2,1]∫01G(t,s)f(s,x(s))ds≥λη22∫01G(s,s)f(s,x(s))ds≥λη22∫01G(t,s)f(s,x(s))ds.
Hence, mint∈[η/2,1](Ax)(t)≥(η2/2)∥Ax∥, which implies AP⊂P. Moreover, it is easy to check that A:P→P is completely continuous.By simple calculation, we obtain that(2.11)∫η/21G(η2,s)ds=∫η/2ηη2ds+∫η11-s1-ηη2ds=η24+η(1-η)4=η4.Lemma 2.4.
Suppose that there exists a positive constantr>0 such that
(2.12)(H1)f(t,x)≥4rηon[η2,1]×[η22r,r]
holds. If λ>1, then
(2.13)i(A,Pr,P)=0.Proof.
Forx∈∂Pr, it follows from the definition of the cone that
(2.14)η22r=η22∥x∥≤mint∈[η/2,1]x(t)≤x(t)≤∥x∥=r,t∈[η2,1],
which implies
(2.15)η22r≤x≤r,t∈[η2,1].
Thus, we have by  (H1) and (2.11) that
(2.16)(Ax)(η2)=λ∫01G(η2,s)f(s,x(s))ds>∫η/21G(η2,s)f(s,x(s))ds≥4rη∫η/21G(η2,s)ds=r=∥x∥
since x∈∂Pr. This shows that
(2.17)∥Ax∥>∥x∥,∀x∈∂Pr.
It is obvious that Ax≠x for x∈∂Pr. Therefore, by Lemma 1.1 (1), we conclude that i(A,Pr,P)=0.Lemma 2.5.
Suppose that there exists a positive constantm>0 such that
(2.18)(H2)f(t,x)≤p(t)g(x)on[0,1]×[0,m],
where p∈C([0,1],[0,+∞)) and g∈C([0,+∞),[0,+∞)). If
(2.19)λ<12p1(∫0mdsH(m)-H(s))2,
where H(u):=∫0ug(s)ds and p1=maxt∈[0,1]p(t)>0, then
(2.20)i(A,Pm,P)=1.Proof.
First, we claim that(2.21)Ax≠μx,forx∈∂Pm,μ≥1.
Suppose to the contrary that there exist x∈∂Pm and μ0≥1 such that
(2.22)Ax=μ0x,forx∈∂Pm.
It is clear that (2.22) is equivalent to
(2.23)x′′(t)+λμ0f(t,x)=0.
Sincex∈C[0,1] and x(η)=x(1), it follows that there exists a ξ∈(η,1) such that x'(ξ)=0. From x′′≤0 on (0,1), we see that x(ξ)=∥x∥:=m>0, x'(t)≥0 on (0,ξ), and x'(t)≤0 on (ξ,1). By (2.18) and (2.23), we have(2.24)x′′(t)=-λμ0f(t,x(t))≥-λμ0p(t)g(x(t)),t∈[0,1].
Multiplying (2.24) by x' and then integrating from t to ξ (t∈[0,ξ)), we get from x'(ξ)=0 that
(2.25)-12(x′(t))2≥-λμ0∫tξp(s)g(x(s))x'(s)ds,t∈[0,ξ),
that is
(2.26)(x′(t))2≤2λμ0∫tξp(s)g(x(s))x'(s)ds≤2λp1μ0∫x(t)x(ξ)g(u)du,t∈[0,ξ),
which implies that
(2.27)0≤x'(t)≤2λp1μ0(H(m)-H(x(t))),t∈[0,ξ).
Thus,
(2.28)∫0mds(2λp1/μ0)(H(m)-H(s))≤∫0ξdt=ξ≤1.
Hence, we obtain from (2.19) and (2.28) that
(2.29)1≥μ02λp1∫0mdsH(m)-H(s)≥12λp1∫0mdsH(m)-H(s)>1.
This contradiction implies that (2.21) holds. By (2.21), we have ∥Ax∥≤∥x∥ for x∈∂Pm, and Ax≠x for x∈∂Pm. Thus, by Lemma 1.1 (2), we obtain
(2.30)i(A,Pm,P)=1.For convenience, let(2.31)(H3)2p1<(∫0mdsH(m)-H(s))2.Theorem 2.6.
Assume that there exist two distinct positive constantsr,m such that  (H1)–(H3) hold. If
(2.32)1<λ<12p1(∫0mdsH(m)-H(s))2,
then BVP (1.2) has at least one positive solution.Proof.
From Lemmas2.4 and 2.5, and the property of the fixed point index, we can easily get that the operator A has a fixed point in Pm¯∖Pr (m>r) or in Pr¯∖Pm (r>m). Therefore, BVP (1.2) has at least one positive solution.


## 3. An Example

To illustrate our results we present the following example.Example 3.1.
Consider the following boundary value problem(3.1)x′′(t)+λ2et(1+|sin(xt)|)x2(t)=0,0<t<1,x(0)=0,x(1)=x(12).
Let f(t,x)=(1/2)et(1+|sin(xt)|)x2 and η=1/2. Choosing m=1/3, r=800, we have
(3.2)f(t,x)≤p(t)g(x),on[0,1]×[0,13],
where p(t)=et and g(x)=x2. Thus, p1=e, H(u)=(1/3)u3, and
(3.3)(∫0mdsH(m)-H(s))2=(∫01/3ds1/81-s3/3)2=6.3701>5.4366=2p1.
Hence,  (H2) and  (H3) hold. Moreover, we get
(3.4)f(t,x)≥12etx2(on[η2,1]×[η22r,r])≥12e1/41002(on[14,1]×[100,800])=6420>6400=8r=4rη(on[14,1]×[100,800]),
which implies  that (H1) holds. Therefore, it follows from Theorem 2.6 that BVP (3.1) has at least one positive solution if
(3.5)1<λ<12p1(∫0mdsH(m)-H(s))2=6.37015.4366=1.1717.



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*Source: 236826-2010-08-30.xml*