# Positive Solutions for a Class of Third-Order Three-Point  Boundary Value Problem

Authors: Xiaojie Lin; Zhengmin Fu
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We investigate the problem of existence of positive solutions for the nonlinear third-order three-point boundary value problemu‴(t)+λa(t)f(u(t))=0, 0<t<1, u(0)=u′(0)=0, u″(1)=∝u″(η), where λ is a positive parameter, ∝∈(0,1), η∈(0,1), f:(0,∞)→(0,∞), a:(0,1)→(0,∞) are continuous. Using a specially constructed cone, the fixed point index theorems and Leray-Schauder degree, this work shows the existence and multiplicities of positive solutions
for the nonlinear third-order boundary value problem. Some examples
are given to demonstrate the main results.


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## Body

## 1. Introduction

This paper deals with the following third-order nonlinear boundary value problem:(1.1)u′′′(t)+λa(t)f(u(t))=0,0<t<1,u(0)=u′(0)=0,u′′(1)=αu′′(η).
Third-order boundary value problems arise in a variety of different areas of applied mathematics and physics. In the few years, there has been increasing interest in studying certain third-order boundary value problems for nonlinear differential equation and have received much attention. To identify a few, we refer the reader to [1–6].Recently, El-Shahed [1] discussed the following third-order two-point boundary value problem:
(1.2)u′′′(t)+λa(t)f(u(t))=0,0<t<1,u(0)=u′(0)=0,αu′(1)+βu′′(1)=0.
The methods employed in [1] are Kransnoselskii's fixed-point theorem of cone.In later work, by placing restrictions on the nonlinear termf, Sun [2] studied the following boundary value problems and obtained the three solution via leggett-williams fixed point theorem:
(1.3)u′′′(t)=a(t)f(t,u(t),u′(t),u′′(t)),0<t<1,u(0)=δu(η)=0,u′(η)=0,u′′(1)=0.The upper and lower solution is a powerful tool for proving existence for boundary value problems, Ma [7] studied the multiplicity of positive solutions of three-point boundary value problem of second-order ordinary differential equations. Du et al. [5] investigated a class of third-order nonlinear problem.Motivated by the work of the above papers, the purpose of this article is to study the existence of solution for boundary value problem (1.1) using a new technique (different from the proof of [1, 2, 7]) and we get a new existence result. The tools are based on the fixed point index theorems and Leray-Schauder degree.The paper is organized as follows: Section2 states some definitions and some lemmas which are important to obtain our main result. Section 3 is devoted to the existence result of BVP (1.1). Section 4 gives some examples to illustrate our main results.


## 2. Preliminary

Definition 2.1.
LetE be a real Banach space. A nonempty closed convex set K⊂E is called a cone of E if it satisfies the following two conditions:(1)
x∈K, λ≥0 implies λx∈K;(2)
x∈K, -x∈K implies x=0.Definition 2.2.
An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.Lemma 2.3.
Lety∈C[0,1], then the following boundary value problem:
(2.1)u′′′(t)+y(t)=0,0<t<1,(2.2)u(0)=u′(0)=0,u′′(1)=αu′′(η),
has the unique solution
(2.3)u(t)=∫01G(t,s)y(s)ds,
where
(2.4)G(t,s)={-12(t-s)2+t22,s≤η,s≤t,t22,t≤s≤η,-12(t-s)2+t22(1-α),η≤s≤t,t22(1-α),η≤s,t≤s.Proof.
From (2.1), we have
(2.5)u(t)=-12∫0t(t-s)2y(s)ds+At2+Bt+C.
In particular,
(2.6)u(t)=-12∫0t(t-s)2y(s)ds+At2+Bt+C,u′(t)=-t∫0ty(s)ds+∫0tsy(s)ds+2At+B,u′′(t)=-∫0ty(s)ds+2A.
Combining this with boundary conditions (2.2), we conclude that
(2.7)A=∫01y(s)ds2(1-α)-α∫0ηy(s)ds2(1-α),B=0,C=0.
Therefore, BVP (2.1)-(2.2) has a unique solution:
(2.8)u(t)=-12∫0t(t-s)2y(s)ds-αt2∫0ηy(s)ds2(1-α)+t2∫01y(s)ds2(1-α)={∫0t[-12(t-s)2+t22]y(s)ds+∫tηt22y(s)ds+∫η1t22(1-α)y(s)ds,t≤η,∫0η[-12(t-s)2+t22]y(s)ds+∫ηt[-12(t-s)2+t22(1-α)]y(s)ds+∫t1t22(1-α)y(s)ds,t≥η,=∫01G(t,s)y(s)ds.
The proof is completed.Lemma 2.4.
For all(t,s)∈[0,1]×[0,1], one has G(t,s)≥0.Lemma 2.5.
for all(t,s)∈[τ,1]×[0,1], one has
(2.9)γG(1,s)≤G(t,s)≤G(1,s),
where γ=ατ2/2, and τ statisfies ∫τ1G(t,s)a(s)ds>0.Proof.
Fors≤t,s≤η,
(2.10)G(t,s)=-12(t-s)2+t22=s(2t-s)2≤G(1,s),G(t,s)G(1,s)=2t-s2-s=t+t-s2-s≥t2.
For t≤s≤η,
(2.11)G(t,s)=t22≤G(1,s),G(t,s)G(1,s)=t2/21/2=t2.
For η≤s≤t,
(2.12)G(t,s)=-12(t-s)2+t22(1-α)=αt2+2ts(1-α)+s2(1-α)2(1-α)≤G(1,s),G(t,s)G(1,s)=αt2+2ts(1-α)+s2(1-α)α+2s(1-α)+s2(1-α)≥αt2.
For η≤s,t≤s,
(2.13)G(t,s)=t22(1-α)≤G(1,s),G(t,s)G(1,s)=t2.
Thus,
(2.14)αt22G(1,s)≤G(t,s)≤G(1,s),for(t,s)∈[0,1]×[0,1].
Therefore,
(2.15)γG(1,s)≤G(t,s)≤G(1,s),∀(t,s)∈[τ,1]×[0,1].
The proof is completed.Lemma 2.6.
Ify∈C[0,1] and y≥0, then the unique solutionu(t)of the BVP (2.1)-(2.2) is nonnegative and satisfies
(2.16)min⁡⁡u(t)t∈[τ,1]≥γ‖u‖.Proof.
Lety∈C+[0,1], it is obvious that it is nonnegative. For any t∈[0,1], by (2.3) and Lemma 2.5, it follows that
(2.17)u(t)=∫01G(t,s)y(s)ds≤∫01G(1,s)y(s)ds,
and thus,
(2.18)‖u‖≤∫01G(1,s)y(s)ds.
On the other hand, (2.3) and Lemma 2.5 imply, for any t∈[τ,1],
(2.19)u(t)=∫01G(t,s)y(s)ds≥γ∫01G(1,s)y(s)ds.
Therefore,
(2.20)min⁡t∈[τ,1]⁡u(t)≥γ‖u‖.
This completes the proof.LetE=C[0,1] with the usual normal ∥u∥=maxt∈[0,1]|u(t)|.Define the coneK by
(2.21)K={u∈C+[0,1]:min⁡t∈[τ,1]⁡u(t)≥γ‖u‖}.Define an operatorT by
(2.22)Tu(t)=λ∫01G(t,s)a(s)f(u(s))ds.By Lemma2.3, BVP (1.1) has a positive solution u=u(t) if and only if u is a fixed point of T.Lemma 2.7.
Assume that0<λ<∞. Then, T:K→K is completely continuous.Proof.
Firstly, it is easy to check thatT:K→K is well defined. By Lemma 2.6, we know that T(K)⊂K.
LetΩ be any boundary subset of K, then there exists r>0, ∥u∥≤r, for all u∈Ω. Therefore, we have
(2.23)|Tu|=λ|∫01G(t,s)a(s)f(u(s))ds|≤λ|∫01G(1,s)a(s)f(u(s))ds|.
So TΩ is boundary. Moreover, for any t1,t2∈[0,1],|t1-t2|≤δ,δ>0, we have
(2.24)|Tu(t1)-Tu(t2)|≤λ∫01|G(t1,s)-G(t2,s)|a(s)f(u(s))ds.
By the continuity of f and a, we have a(t) and f(u(t)) are boundary on u∈Ω,t∈[0,1], which means that there exists a constant Maf>0, depending only on Ω such that
(2.25)|a(t)f(u(t))|<Maf,
and thus for any ɛ>0,
(2.26)|G(t1,s)-G(t2,s)|≤ɛλMaf,|Tu(t1)-Tu(t2)|<ɛ.
Therefore, we can get TΩ is equicontinuity. Thirdly, we prove that T is continuous. Let un→u as n→∞, un∈K. Then, the continuity of f, we can get
(2.27)|Tun(t)-Tu(t)|=|λ∫01G(t,s)a(s)f(un(s))ds-λ∫01G(t,s)a(s)f(u(s))ds|=|λ∫01G(t,s)a(s)(f(un(s))-f(u(s)))ds|≤|λ∫01G(1,s)a(s)(f(un(s))-f(u(s)))ds|⟶0,n⟶∞.
Then, Tun(t)→Tu(t). Therefore, T is continuous.The operator T is completely continuous by an application of the Ascoli-Arzela theorem. This completes the proof.Lemma 2.8 (see [7, 8]).
LetE be a real Banach space and let K be a cone in E. For r≥0, define Kr={x∈K:∥x∥<r}. Assume T:K¯r→K is a completely continuous operator such that Tx≠x for x∈∂Kr={x∈K:∥x∥=r}.(1)
If∥Tx∥≥∥x∥ for x∈∂Kr, then
(2.28)i(T,Kr,K)=0.(2)
If∥Tx∥≤∥x∥ for x∈∂Kr, then
(2.29)i(T,Kr,K)=1.


## 3. Main Results

Theorem 3.1.
Assume that(A1)
λ is a positive parameter,  η∈(0,1) and  α∈(0,1);(A2)
a:[0,1]→(0,∞) is continuous;(A3)
f:[0,∞)→(0,∞) is continuous;(A4)
f∞:=limu→∞(f(u)/u)=∞.When  λ is sufficiently small, (1.1) has at least one positive solution, whereas for  λ is sufficiently large, (1.1) has no positive solution.Proof.
Ifq>0, then
(3.1)β(q)=max⁡⁡u∈K,‖u‖=q[∫01G(t,s)a(s)f(u(s))ds]>0.
For any number 0<r1, let δ1=r1/β(r1), and set
(3.2)Kr1={u∈K:‖u‖<r1}.
Then, for λ∈(0,δ1) any u∈∂Kr1, we have
(3.3)Tu(t)<δ1[∫01G(t,s)f(u(s))ds]≤δ1β(r1)=r1.
Thus, Lemma 2.8 implies
(3.4)i(T,Kr1,K)=1.
Since f∞=∞, there is M>0, such that f(u)≥μu, for u>M, where μ is chosen so that
(3.5)λμγ∫τ1G(1,s)a(s)ds>1.
Let r2>M/γ, and set
(3.6)Kr2={u∈K:‖u‖<r2}.
If u∈∂Kr2, then
(3.7)min⁡t∈[τ,1]⁡u(t)≥γ‖u‖≥M.
Therefore,
(3.8)Tu(1)=λ∫01G(1,s)a(s)f(u(s))ds≥λ∫τ1G(1,s)a(s)f(u(s))ds≥λ∫τ1G(1,s)a(s)μu(s)ds≥λμ∫τ1G(1,s)a(s)dsγ‖u‖≥λμγ∫τ1G(1,s)a(s)ds‖u‖>‖u‖,
which implies that
(3.9)‖Tu‖≥‖u‖,
for u∈∂Kr2. An application of Lemma 2.8 again shows that
(3.10)i(T,Kr2,K)=0.
Since we can adjust r1,r2 so that r1<r2, it follows the additivity of the fixed-point index that
(3.11)i(T,Kr2∖K¯r1,K)=-1.
Thus, T has a fixed point in Kr2∖K¯r1 which is the desired positive solution of (1.1).
We verify that BVP of (1.1) has no positive solution for λ large enough.
Otherwise, there exist0<λ1<λ2<⋯<λn<⋯, with limn→∞λn=+∞, such that for any positive integer n, the BVP,
(3.12)u′′′(t)+λna(t)f(u(t))=0,0<t<1,u(0)=u′(0)=0,u′′(1)=αu′′(η),
has a positive solution un(t). By (2.22), we have
(3.13)un=λn∫01G(t,s)a(s)f(un(s))⟶+∞,(n⟶∞).
Thus,
(3.14)un⟶∞,(n⟶∞).
Since f∞, for c0>0, there exists r3>0, such that f(u)/u>c0, for u∈[r3,∞), which implies that
(3.15)f(u)>c0u,foru∈[r3,∞).
Let n be large enough that ∥un∥≥r3, then
(3.16)‖un‖≥un(1)=λn∫01G(1,s)a(s)f(un(s))ds≥λnγ∫01G(1,s)a(s)dsc0‖un‖>‖un‖.
Choose n so that c0λnγ∫01G(1,s)a(s)ds>1 which is a contradiction. The proof is completed.Theorem 3.2.
Assume that(B1)
λ is a positive parameter;  η∈(0,1) and  α∈(0,1);(B2)
a:[0,1]→(0,∞) is continuous and there exists  m>0 such that  a(t)≥m;(B3)
f:[0,∞)→(0,∞) is continuous;(B4)
f∞=limu→∞(f(u)/u)=0,f0=limu→0(f(u)/u)=0;(B5)
there existsσ>0, for  u≥σ, such that  f(u)≥β, where  β>0, then there exists  δ2>0, such that, for  λ>δ2, BVP (1.1) has at least two positive solutions  uλ1,uλ2 and  maxuλ1>σ.Proof.
Letδ2=(Mγmβ)-1σ, then for λ>δ2, Lemma 2.7 implies that T:K→K is completely continuous. Considering (B4), there exists 0<r<σ such that f(u)≤u/2Λλ, for 0≤u≤r, where Λ=∫01G(1,s)a(s)ds.
So, foru∈∂Kr, we have from (2.4)
(3.17)(Tu)(t)=λ[∫01G(t,s)a(s)f(s)ds]≤λ∫01G(1,s)a(s)f(u(s))ds≤λ[∫01G(1,s)a(s)ds]‖u‖2Λλ=‖u‖2<‖u‖=r.
Consequently, foru∈∂Kr, we have ∥Tu∥<∥u∥, by Lemma 2.8,
(3.18)i(T,Kr,K)=1.
Now considering (B4), there exists h>0, for u>h, such that f(u)≤u/2Λλ.  Letting ρ=max0≤u≤hf(u), then
(3.19)0≤f(u)≤u2Λλ+ρ.
Choose
(3.20)R>max⁡⁡{r,2Λρλ}.
So for u∈∂KR, from (3.18) and (3.19), we have
(3.21)(Tu)(t)=λ[∫01G(t,s)a(s)f(u)ds]≤λ[∫01G(1,s)a(s)f(u)ds]≤λ[∫01G(1,s)a(s)ds](12Λλ‖u‖+ρ)<‖u‖2+R2=‖u‖,
That is, by Lemma 2.8,
(3.22)i(T,KR,K)=1.
On the other hand, for u∈K¯σR={u∈K:∥u∥≤R,mint∈Jθu(t)≥σ,θ∈(0,1/2),Jθ=[θ,1-θ]}, (2.3) and (2.4) yield that
(3.23)‖Tu‖≤λ[∫01G(t,s)a(s)ds](12Λλ‖u‖+ρ)<R.
Furthermore, for u∈K¯σR, from (2.3) and (2.4), we obtain
(3.24)min⁡t∈Jθ⁡(Tu)(t)=min⁡t∈Jθ⁡⁡λ[∫01G(1,s)a(s)f(u(s))ds]≥⁡min⁡t∈Jθ⁡λ∫θ1-θG(t,s)a(s)f(u(s))ds≥λγ∫θ1-θG(1,s)a(s)f(u(s))ds≥λMγmβ>δ2Mγmβ=σ,
where M=∫θ1-θG(1,s)ds. Let u0≡(σ+R)/2 and H(t,u)=(1-t)Tu+tu0, then H:[0,1]×K¯σR→K is continuous, and from the analysis above, we obtain for (t,u)∈[0,1]×K¯σR:
(3.25)H(t,u)∈KσR.
Therefore, for u∈∂KσR, we have H(t,u)≠u. Hence, by the normality property and the homotopy invariance property of the fixed point index, we obtain
(3.26)i(T,KσR,K)=i(u0,KσR,K)=1.
Consequently, by the solution property of the fixed point index, T has a fixed point uλ1and uλ1∈KσR. By Lemma 2.4, it follows that u1λ is a solution to BVP (1.1), and
(3.27)max⁡t∈[0,1]⁡⁡uλ1≥min⁡t∈Jθ⁡⁡uλ1>γ.
On the other hand, from (3.18) and (3.19) together with the additivity of the fixed point index, we get
(3.28)i(T,KR∖(K¯r∪K¯σR))=i(T,KR,K)-i(T,KσR,K)-i(T,Kr,K)=1-1-1=-1.
Hence, by the solution property of the fixed point index,T has a fixed point uλ2 and uλ2∈KR∖(K¯r∪K¯σR). By Lemma 2.3, it follows that uλ2 is also a solution to BVP (1.1), and uλ1≠uλ2. The proof is completed.


## 4. Examples

Example 4.1.
We consider the following third-order boundary value problems:(4.1)u′′′(t)+λ(2t+1)eu=0,u(0)=u′(0)=0,u′′(1)=34u′′(14),
here η=1/4,α=3/4,f(u(t))=eu,a(t)=2t+1,f∞=limu→∞(f(u)/u)=∞,f is continuous, a(t) is continuous. By direct calculations, we obtain that λ<r1(1-α),forr1>0. Therefore, by Theorem 3.1, there exists at least one solution u(t) for BVP (4.1), whereas for λ large enough, (4.1) has no solution.Example 4.2.
Consider the following third-order ordinary differential equation:(4.2)u′′′+λ(2t+1)f(u(t))=0,u(0)=u′(0)=0,u′′(1)=14u′′(12),
where
(4.3)f(u(t))={u2e-u,ifu≤a,a3/2ue-a,ifu>a,f is continuous, a(t) is continuous. Here, m=1, α=1/4,β=a2e-a,σ=a,a>0. Choose δ2=6a/(2θ3-3θ2+3θ-1),θ∈(120,1/212),τ∈(0,1), when λ>δ2, by Theorem 3.2, there exist at least two solutions uλ1(t),uλ2(t) for BVP (4.1).



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