# Positive Solutions for Resonant and Nonresonant Nonlinear Third-Order Multipoint Boundary Value Problems

Authors: Liu Yang; Chunfang Shen; Dapeng Xie
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Positive solutions for a kind of third-order multipoint boundary value problem under the nonresonant conditions and the resonant conditions are considered. In the nonresonant case, by using the Leggett-Williams fixed point theorem, the existence of at least three positive solutions is obtained. In the resonant case, by using the Leggett-Williams norm-type theorem due to O’Regan and Zima, the existence result of at least one positive solution is established. It is remarkable to point out that it is the first time that the positive solution is considered for the third-order boundary value problem at resonance. Some examples are given to demonstrate the main results of the paper.


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## Body

## 1. Introduction

We consider the existence of positive solutions for third-orderm-point boundary value problem:
(1)x′′′(t)+f(t,x(t))=0,t∈[0,1],x′′(0)=0,x′(0)=0,x(1)=∑i=1m-2‍βix(ξi),
where 0<ξ1<ξ2<⋯<ξm-2<1, 0≤βi≤1, i=1,2,…,m-2, ∑i=1m-2‍βi≤1, and f∈C([0,1]×[0,∞),R).If condition∑i=1m-2‍βi=1 holds, the problem is called resonant boundary value problem or boundary value problem at resonance. Otherwise, the associated problem is called nonresonant boundary value problem. Here, the condition ∑i=1m-2‍βi=1 is denoted by (1.2).Third-order differential equations arise in a variety of different areas of applied mathematics and physics, as the deflection of a curved beam having a constant or varying cross section, three-layer beam, and so on [1]. In recent years, the existence of positive solutions for nonresonant two-point or three-point boundary value problems (Bvp for short) for nonlinear third-order ordinary differential equations has been studied by several authors. For examples, Anderson [2] established the existence of at least three positive solutions to problem
(2)-x′′′(t)+f(x(t))=0,t∈(0,1),x(0)=x′(t2)=x′′(1)=0,
where f:R→[0,+∞) is continuous and 1/2≤t2<1.By using the well-known Guo-Krasnoselskii˘ fixed point theorem [3], Palamides and Smyrlis [4] proved that there exists at least one positive solution for third-order three-point problem:
(3)x′′′(t)=a(t)f(t,x(t)),t∈(0,1),x′′(η)=0,x(0)=x(1)=0,η∈(0,1).
For more existence results of positive solutions for boundary value problems of third-order ordinary differential equations, one can see [5–12] and references therein.For resonant problem of second-order or higher-order differential equations, many existence results of solutions have been established, see [13–25]. In [25], the authors considered the problem
(4)x′′′(t)=f(t,x,x′)+e(t),t∈(0,1),x′(0)=0,x(1)=βx(η),x(0)=∑i=1m-2‍αix(ξi).
By using the Mawhin continuation theorem, the existence results of solutions are established under the resonant condition β=1, ∑i=1m-2‍αi=1, ∑i=1m-2‍αiξi2=0 and β=1/η, ∑i=1m-2‍αi=1, ∑i=1m-2‍αiξi2=0, respectively.It is well known that the problem of existence for positive solution to nonlinear Bvp is very difficult when the resonant case is considered. Only few works gave the approach in this area for first- and second-order differential equations [26–31]. To our best knowledge, no paper deal with the existence result of positive solution for resonant third-order boundary value problems. Motivated by the approach in [27–29], we study the existence of positive solution for problem (1) under the nonresonant condition ∑i=1m-2‍βi<1 and resonant condition ∑i=1m-2‍βi=1, respectively. By using the Leggett-Williams fixed point theorem and its generalization [28, 30], we establish the existence results of positive solutions. The results obtained in this paper are interesting because(1)
the results obtained in the nonresonant case are more general than those established before;(2)
it is the first time that the positive solution is considered for third-order Bvp at resonance.The rest of the paper is organized as follows. Some definitions and lemmas are given in Section2. In Section 3, we consider the nonresonant case for problem (1). In Section 4, we discuss the existence of positive solution for problem (1) with resonant condition (1.2). Finally, in Section 5, we give some examples to illustrate the main results of the paper.


## 2. Background Definitions and Lemmas

For the convenience of the reader, we present here the necessary definitions and two-fixed point theorems.LetX, Y be real Banach spaces. A nonempty convex closed set C⊂X is said to be a cone provided that(1)
ax∈C, for all x∈C, a≥0,(2)
x,-x∈C implies x=0.Definition 1.
The mapψ is said to be a nonnegative continuous concave functional on C provided that ψ:C→+∞ is continuous and
(5)ψ(tx+(1-t)y)≥tψ(x)+(1-t)ψ(y),
for all x,y∈C and t∈[0,1].Definition 2.
Let0<a<b be given and let ψ be a nonnegative continuous concave functional on the cone C. Define the convex sets Cr and C(ψ,a,b) by
(6)Cr={x∈C∣∥x∥<r},C(ψ,a,b)={x∈C∣a≤ψ(x),∥x∥≤b}.Lemma 3 (the Leggett-Williams fixed point theorem [32]).
LetT:C¯r→C¯r be a completely continuous operator and let ψ be a nonnegative continuous concave functional on C such that ψ(x)≤∥x∥ for all x∈C¯r. Suppose that there exist 0<a<b<d≤c such that(H1)
{x∈C(ψ,b,d)∣ψ(x)>b}≠⌀ and ψ(Tx)>b, for x∈C(ψ,b,d),(H2)
∥Tx∥<a for ∥x∥≤a,(H3)
ψ(Tx)>b for x∈C(ψ,b,c) with ∥Tx∥≥d.Then, T has at least three-fixed points x1,  x2, and x3 such that
(7)∥x1∥<a,b<ψ(x2),∥x3∥>a,ψ(x3)<b.
OperatorL:domL⊂X→Y is called a Fredholm operator with index zero, that is, ImL is closed and dimKerL=codimImL<∞, which implies that there exist continuous projections P:X→X and Q:Y→Y such that ImP=KerL and KerQ=ImL. Moreover, since dimImQ=codimImL, there exists an isomorphism J:ImQ→KerL. Denote by LP, the restriction of L to KerP∩domL to ImL and its inverse by KP, so KP:ImL→KerP∩domL and the coincidence equation Lx=Nx is equivalent to
(8)x=(P+JQN)x+KP(I-Q)Nx.
Denote γ:X→C to be a retraction, that is, a continuous mapping such that γx=x for all x∈C and
(9)Ψ:=P+JQN+KP(I-Q)N,Ψγ:=Ψ∘γ.Lemma 4 (the Leggett-Williams norm-type theorem [27]).
LetC be a cone in X,  and let Ω1, Ω2 be open bounded subsets of X with Ω¯1⊂Ω2, C∩(Ω¯2∖Ω1)≠∅. Assume that L:domL⊂X→Y is a Fredholm operator of index zero and(C1)
QN:X→Y is continuous and bounded, KP(I-Q)N:X→X is compact on every bounded subset of X,(C2)
Lx≠λNx for all x∈C∩∂Ω2∩domL and λ∈(0,1),(C3)
γ maps subsets of Ω¯2 into bounded subsets of C,(C4)
dB([I-(P+JQN)γ]|kerL,KerL∩Ω2,0)≠0, where dB stands for the Brouwer degree,(C5)
there existsu0∈C∖{0} such that ∥x∥≤σ(u0)∥Ψx∥ for x∈C(u0)∩∂Ω1, where C(u0)={x∈C:μu0≤x} for some μ>0 and σ(u0)  such that ∥x+u0∥≥σ(u0)∥x∥ for every x∈C,(C6)
(P+JQN)γ(∂Ω2)⊂C,(C7)
Ψγ(Ω¯2∖Ω1)⊂C,then the equationLx=Nx has a solution in the set C∩(Ω¯2∖Ω1).


## 3. Positive Solution for the Nonresonant Problem

In this section, we suppose thatf∈C([0,1]×[0,+∞),[0,+∞)) and ∑i=1m-2‍βi<1. We begin with some preliminary results. Consider the problem
(10)x′′′(t)+y(t)=0,t∈[0,1],(11)x′′(0)=0,x′(0)=0,x(1)=∑i=1m-2‍βix(ξi).Lemma 5.
Denoteξ0=0, ξm-1=1, β0=βm-1=0, then for y(t)∈C[0,1], problems (10) and (11) have the unique solution
(12)x(t)=∫01‍G(t,s)∫0s‍y(τ)dτds,
where
(13)G(t,s)=11-∑k=0m-1βk×{(1-s)+∑k=im-1‍βk(s-ξk),t≤s,(1-t)+∑k=0i-1‍βk(t-s)+∑k=im-1‍βk(t-ξk),t≥s,
for ξi-1<s<ξi, i=1,2,…,m-1.Proof.
Integrating both sides of (10) and considering the boundary condition x′′(0)=0, we have
(14)-x′′(t)=∫0t‍y(s)ds.
Let G(t,s) be the Green function of problem
(15)-x′′(t)=0,(16)x′(0)=0,x(1)=∑i=0m-1‍βix(ξi).
From (15), we can suppose that
(17)G(t,s)={A+Btt≤s,ξi-1<s<ξi,C+Dtt≥s,ξi-1<s<ξi.
For the definition and properties of the Green function together with (16), we have
(18)A+Bs=C+Ds,B-D=1,B=0,C+D=∑k=0i-1‍βk(A+Bξk)+∑k=im-1‍βk(C+Dξk).
Hence,
(19)A=1-s+∑k=im-1‍βk(s-ξk)1-∑i=0m-1‍βi,C=1-∑k=0i-1‍βks-∑k=im-1‍βkξk1-∑i=0m-1‍βi,B=0,D=-1.
Thus,
(20)G(t,s)=11-∑k=0m-1βk×{(1-s)+∑k=im-1‍βk(s-ξk),t≤s,(1-t)+∑k=0i-1‍βk(t-s)+∑k=im-1‍βk(t-ξk),t≥s,
for ξi-1<s<ξi,  i=1,2,…,m-1.
Considering (14) together, we obtain that problems (10) and (11) have the unique solution
(21)x(t)=∫01‍G(t,s)∫0s‍y(τ)dτds.Lemma 6.
The functionG(t,s) established in Lemma 5 satisfies that G(t,s)≥0,  t,s∈[0,1].Proof.
Forξi-1≤s≤ξi, i=1,2,…,m-1 and t≤s,
(22)(1-s)+∑k=im-1‍βk(s-ξk)≥∑k=im-1‍βk(1-ξk)≥0.
For ξi-1≤s≤ξi, i=1,2,…,m-1 and t≥s,
(23)(1-t)+∑k=0i-1‍βk(t-s)+∑k=im-1‍βk(t-ξk)≥∑k=0i-1‍βk(1-s)+∑k=im-1‍βk(1-ξk)≥0.
These ensures that G(t,s)≥0, t,s∈[0,1].Lemma 7.
Ify∈C[0,1] and y≥0, then the unique solution x of problems (10) and (11) satisfy
(24)min0≤t≤1x(t)≥γmax0≤t≤1x(t),
where γ=(∑i=1m-2‍βi(1-ξi))/(1-∑i=1m-2‍βiξi)>0 is a constant.Proof.
Forx′′′(t)=-y(t)≤0, t∈[0,1], we get that x′′(t) is decreasing on [0,1]. Then, the condition x′′(0)=0 ensures that have x′′(t)≤0,t∈(0,1). This together with x′(0)=0x(t) is concave and decreasing on [0,1]. Thus,
(25)max0≤t≤1x(t)=x(0),min0≤t≤1x(t)=x(1).
From the concavity of x(t), we have
(26)ξi(x(1)-x(0))≤x(ξi)-x(0).
Multiplying both sides with βi and considering x(1)=∑i=1m-2‍βix(ξi), we have
(27)(1-∑i=1m-2‍βiξi)x(1)≥∑i=1m-2‍βi(1-ξi)x(0).
This completes the proof of Lemma 7.Let Banach spaceE=C[0,1] be endowed with the maximum norm. We define the cone C⊂E by
(28)C={∑i=1m-2‍x∈E∣x(t)≥0,x′′(0)=0,x′(0)=0,x(1)=∑i=1m-2‍βix(ξi),x(t)isconcaveon[0,1]}.Define the nonnegative continuous concave functionalψ:C→[0,∞) by
(29)ψ(x)=min0≤t≤1x(t),x∈C.
Define the constants σ, δ by
(30)σ=1max0≤t≤1∫01‍sG(t,s)ds,δ=∫01‍sG(1,s)ds.Theorem 8.
Suppose that there exist constants0<a<b<b/γ≤c such that (A1)
f(t,x)<σa, (t,x)∈[0,1]×[0,a],(A2)
f(t,x)>b/δ, (t,x)∈[0,1]×[b,(b/γ)],(A3)
f(t,x)<σc, (t,x)∈[0,1]×[0,c],then problem (1) has at least three positive solutions x1, x2, and x3 satisfying
(31)∥x1∥≤a,b<min0≤t≤1x2,∥x3∥>awithmin0≤t≤1x3<b.Proof.
We define operatorT:C→E by
(32)Tx(t)=∫01‍G(t,s)∫0s‍f(τ,x(τ))dτds.
It is easy to check that T:C→C and is completely continuous.
Next, the conditions of Lemma3 are checked. If x∈C¯c, then ∥x∥≤c, and condition (A3) implies that
(33)f(t,x)≤σc,0≤t≤1,∥T(x)∥=max0≤t≤1∫01‍G(t,s)∫0s‍f(τ,x(τ))dτds≤max0≤t≤1∫01‍sG(t,s)ds×σc≤c.
Thus, T:C¯c→C¯c. In the same way, we see that T:C¯a→C¯a. Hence, condition (H2) of Lemma 3 is satisfied.
The fact that constant functionx(t)=(b/γ)∈{C(ψ,b,(b/γ))∣ψ(x)>b} implies that {C(ψ,b,(b/γ))∣ψ(x)>b}≠Ø. If x∈C(ψ,b,(b/γ)), from the assumption (A2), f(t,x)≥b/δ. Thus,
(34)ψ(Tx)=min0≤t≤1∫01‍G(t,s)∫0s‍f(τ,x(τ))dτds=∫01‍G(1,s)∫0s‍f(τ,x(τ))dτds≥∫01‍sG(1,s)ds×bδ≥b,
which ensures that condition (H1) of Lemma 3 is satisfied. Finally, we show that condition (H3) of Lemma 3 also holds. Suppose that x∈C(ψ,b,c) with ∥Tx∥>b/γ,
(35)ψ(Tx)=min0≤t≤1Tx(t)≥γ×∥Tx∥>γ×bγ=b.
So, condition (H3) of Lemma 3 is satisfied. Thus, an application of Lemma 3 implies that the nonresonant third-order boundary value problem (1) has at least three positive solutions x1, x2, and x3 satisfying (31).


## 4. Positive Solution for Resonant Problem

In this section, the condition∑i=1m-2‍βi=1 is considered. Obviously, problem (1) is at resonance under this condition. The norm-type Leggett-Williams fixed point theorem will be used to establish the existence results of positive solution. We define the Banach spaces X=Y=C[0,1] endowed with the maximum norm.Define linear operatorL:domL⊂X→Y, Lx=-x′′′(t), t∈[0,1], where
(36)domL={∑i=1m-2‍x∈X∣x′′′∈C[0,1],x′′(0)=0,x′(0)=0,x(1)=∑i=1m-2‍βix(ξi)},
and N:X→Y with
(37)(Nx)(t)=f(t,x(t)),t∈[0,1].
It is obvious that KerL={x∈domL:x(t)≡c,t∈[0,1]}. Denote the function G(s),s∈[0,1] as follows:
(38)G(s)=(1-s)2-∑i=km-1‍βi(ξi-s)2,ξk-1≤s≤ξk,k=1,2,…,m-1.
Note that G(s)≥0, s∈[0,1].Denote the functionU(t,s) and positive number κ as follows:
(39)U(t,s)={-16s3+12s2-12s+16+4t3+2324∫01‍G(s)dsG(s),16+4t3+2324∫01‍G(s)dsG(s),0≤t≤s≤1,-16s3-12s+st-12t2+16+4t3+2324∫01‍G(s)dsG(s),t-12t2+16+4t3+2324∫01‍G(s)dsG(s),0≤s≤t≤1,κ:=min{1,mins∈[0,1]∫01‍G(τ)dτG(s),mint,s∈[0,1]1U(t,s)}.Theorem 9.
Assume that there exists positive constantR∈(0,∞) such that f:[0,1]×[0,R]→(-∞,+∞) is continuous and satisfies the following conditions:(S1)
f(t,x)≥-κx, for (t,x)∈[0,1]×[0,R],(S2)
f(t,x)<0 for [t,x]∈[0,1]×[(1-(κ/2))R,R],(S3)
there existsr∈(0,R),t0∈[0,1], a∈(0,1],M1<M∈(0,1) and continuous functions g:[0,1]→[0,+∞), h:(0,r]→[0,+∞) such that f(t,x)≥g(t)h(x),  [t,x]∈[0,1]×(0,r] and h(x)/xa is nonincreasing on (0,r] with
(40)h(r)ra∫01‍U(t0,s)g(s)ds≥1-MMa,then problem (1) with resonant condition (1.2) has at least one positive solution.Proof.
Firstly, we claim that(41)ImL={y∈Y∣∫01‍G(s)y(s)ds=0}.
In fact, for each y∈{y∈Y∣∫01‍G(s)y(s)ds=0}, we take
(42)x(t)=-12∫0t‍(t-s)2y(s)ds.
It is easy to check that -x′′′(t)=y(t),x′′(0)=0,x′(0)=0, and x(1)=∑i=1m-2‍βix(ξi), which means x(t)∈domL. Thus,
(43){y∈Y∣∫01‍G(s)y(s)ds=0}⊂ImL.
On the other hand, for each y(t)∈ImL, there exists x(t)∈domL,
(44)-x′′′(t)=y(t),x′′(0)=0,x′(0)=0,x(1)=∑i=1m-2‍βix(ξi).
Integrating both sides on [0,t], we have
(45)x(t)=-12∫0t‍(t-s)2y(s)ds+12x′′(0)t2+x′(0)t+x(0).
Considering condition x′′(0)=0,x′(0)=0,x(1)=∑i=1m-2‍βix(ξi), and ∑i=0m-1‍βi=1, we conclude that
(46)∫01‍(1-s)2y(s)ds-∑i=0m-1‍βi∫0ξi‍(ξi-s)2y(s)ds=0,
which equivalents to the conclusion that ∫01‍G(s)y(s)ds=0. So, we have
(47)ImL⊂{y∈Y∣∫01‍G(s)y(s)ds=0}.
Thus,
(48)ImL={y∈Y∣∫01‍G(s)y(s)ds=0}.
Clearly, dimKerL=1 and ImL are closed. Next, we see that Y=Y1⊕ImL, where
(49)Y1={y1∣y1=1∫01‍G(s)ds∫01‍G(s)y(s)ds,y∈Y}.
In fact, for each y(t)∈Y, we have
(50)∫01‍G(s)[y(s)-y1]ds=0.
This shows that y-y1∈ImL. Since Y1∩ImL={0}, we have Y=Y1⊕ImL. Thus, L is a Fredholm operator with index zero.
Then, define the projectionsP:X→X, Q:Y→Y by
(51)Px=∫01‍x(s)ds,Qy=1∫01‍G(s)ds∫01‍G(s)y(s)ds.
Clearly, ImP=KerL, KerQ=ImL and KerP={x∈X:∫01‍x(s)ds=0}. Note that for y∈ImL, the inverse KP of LP is given by
(52)(KP)(y)=∫01‍k(t,s)y(s)ds,
where
(53)k(t,s)={-16s3+12s2-12s+16,0≤t≤s≤1,-16s3-12s+st-12t2+16,0≤s≤t≤1.
In fact, it is easy to check that
(54)L(KP)(y)=(-∫01‍k(t,s)y(s)ds)′′′=y(t),(55)KP(L)(x)=∫0t‍(-16s3-12s+st-12t2+16)(-x′′′(s))ds+∫t1‍(-16s3+12s2-12s+16)(-x′′′(s))ds=x(t).
Considering that f can be extended continuously on [0,1]×(-∞,+∞); condition (C1) of Lemma 4 is fulfilled.
Define the cone of nonnegative functionsC and subsets of XΩ1,Ω2 by
(56)C={x∈X:x(t)≥0,t∈[0,1]},Ω1={x∈X:r>|x(t)|>M∥x∥,t∈[0,1]},Ω2={x∈X:|x(t)|<R,t∈[0,1]}.
Clearly, Ω1 and Ω2 are bounded and open sets, furthermore
(57)Ω¯1={x∈X:r≥|x(t)|≥M∥x∥,t∈[0,1]}⊂Ω2,C∩Ω¯2∖Ω1≠Ø.
Let the isomorphismJ=I and (γx)(t)=|x(t)| for x∈X. Then, it is easy to check that γ is a retraction and maps subsets of Ω¯2 into bounded subsets of C, which means that condition (C3) of Lemma 4 is satisfied.
Next, we confirm that (C2) of Lemma4 holds. For this purpose, suppose that there exists x0∈C∩∂Ω2∩domL and λ0∈(0,1) such that Lx0=λ0Nx0. Then,
(58)-x0′′′(t)=λ0f(t,x0),
for all t∈[0,1]. Thus,
(59)x0′′(t)=-λ0∫0t‍f(s,x0(s))ds,x0′(t)=-λ0∫0t‍(t-s)f(s,x0(s))ds.
Let x0(t0)=∥x0∥=R. We verify that t0≠0 and t0≠1. The step is divided into three cases:(1)
we show thatt0≠1. Suppose, on the contrary, that x0(t) achieves maximum value R only at t=1. Then, x(1)=∑i=1m-2‍βix0(ξi) in combination with ∑i=1m-2‍βi=1 yields that max1≤i≤m-2x0(ξi)≥R, which is a contradiction,(2)
we show thatt0≠0. Suppose, on the contrary, that x0(t) achieves maximum value R at t=0. Then,
(60)x0′′′(0)=-λ0f(0,R)>0which together with the condition x′′(0)=x′(0)=0 yield that x(t) is increasing near the point t=0. This contradicts to the fact that x0(t) achieves maximum value R at t=0.
Thus, there existst0∈(0,1) such that x0(t0)=R=max0≤t≤1x0(t). We may choose η<t0 nearest to t0 with x0′′(η)=0. From the mean value theory, we claim that there exists ξ∈(η,t0) such that
(61)x0(η)=x0(t0)-x0′(ξ)(t0-η).
Here,
(62)x0′(ξ)=-λ0∫0ξ‍(ξ-s)f(s,x0)ds≤λ0κ∫0ξ‍(ξ-s)x0(s)ds≤λ0κR∫0ξ‍(ξ-s)ds=12ξ2λ0κR.
Thus,
(63)x0(η)=x0(t0)-x0′(ξ)(t0-η)≥R-12ξ2λ0κ(t0-η)R≥(1-κ2)R.
Then,
(64)0≥x0′′(t0)-x0′′(η)=-λ0∫ηt0‍f(s,x0(s))ds,
which contradict to condition (S2). Thus, (C2) holds.Remark 10.
The sign of third-order derivative of a functiony(t) at point t0 cannot be confirmed even if t0 is a maximal value of y(t). Thus, the method in [29] is not applicable directly to problem (1). In our opinion, it is the key that the conditions (S2) in this paper are stronger than that in [29].
Forx∈KerL∩Ω2, define
(65)H(x,λ)=x-λ|x|-λ∫01‍G(s)ds∫01‍G(s)f(s,|x|)ds,
where x=c∈KerL∩Ω2 and λ∈[0,1]. Suppose that H(x,λ)=0. In view of (S1), we obtain
(66)c=λ|c|+λ∫01‍G(s)ds∫01‍G(s)f(s,|c|)ds≥λ|c|-λ∫01‍G(s)ds∫01‍G(s)κ|c|ds=λ|c|(1-κ)≥0.
Hence, H(x,λ)=0 implies c≥0. Furthermore, if H(R,λ)=0, we get
(67)0≤R(1-λ)∫01‍G(s)ds=λ∫01‍G(s)f(s,R)ds,
contradicting to (S2). Thus, H(x,λ)≠0 for x∈∂Ω2 and λ∈[0,1]. Therefore,
(68)dB(H(x,0),KerL∩Ω2,0)=dB(H(x,1),KerL∩Ω2,0)=dB(I,KerL∩Ω2,0)=1.
This ensures
(69)dB([I-(P+JQN)γ]|KerL,KerL∩Ω2,0)=dB(H(x,1),KerL∩Ω2,0)≠0.
Let x∈Ω¯2∖Ω1 and t∈[0,1]. From condition (S1), we see
(70)(Ψγx)(t)=∫01‍|x(t)|dt+1∫01‍G(s)ds∫01‍G(s)f(s,|x(s)|)ds+∫01‍k(t,s)[f(s,|x(s)|)-1∫01‍G(s)ds+∫01‍k(t,s)×∫01‍G(τ)f(τ,|x(τ)|)dτ1∫01‍G(s)ds]ds=∫01‍|x(t)|dt+∫01‍U(t,s)f(s,|x(s)|)ds≥∫01‍|x(s)|ds-κ∫01‍U(t,s)|x(s)|ds=∫01‍(1-κU(t,s))|x(s)|ds≥0.
Hence, Ψγ(Ω¯2∖Ω1)⊂C. Moreover, for x∈∂Ω2, we have
(71)(P+JQN)γx=∫01‍|x(s)|ds+1∫01‍G(s)ds∫01‍G(s)f(s,|x(s)|)ds≥∫01‍(1-κ∫01‍G(s)dsG(s))|x(s)|ds≥0,
which means (P+JQN)γ(∂Ω2)⊂C. These ensure that (C6) and (C7) of Lemma 4 hold.
At last, we confirm that (C5) is satisfied. Takingu0(t)≡1 on [0,1], we see
(72)u0∈C∖{0},C(u0)={x∈Cx(t)>0on[0,1]},
and we can take σ(u0)=1. Let x∈C(u0)∩∂Ω1, we have
(73)x(t)>0,t∈[0,1],0<∥x∥≤r,x(t)≥M∥x∥on[0,1].
Therefore, in view of (S3), we obtain, for all x∈C(u0)∩∂Ω1,
(74)(Ψx)(t0)=∫01‍x(s)ds+∫01‍U(t0,s)f(s,x(s))ds≥M∥x∥+∫01‍U(t0,s)g(s)h(x(s))ds=M∥x∥+∫01‍U(t0,s)g(s)h(x(s))xa(s)xa(s)ds≥M∥x∥+h(r)ra∫01‍U(t0,s)g(s)Ma∥x∥ads≥M∥x∥+(1-M)∥x∥=∥x∥.
So, ∥x∥≤σ(u0)∥Ψx∥, for all x∈C(u0)∩∂Ω1, which means (C5) of Lemma 4 holds.
Thus, by Lemma4, we confirm that the equation Lx=Nx has a solution x∈C∩(Ω¯2∖Ω1), which implies that nonlinear third-order multipoint boundary value problem (1) with resonance condition (1.2) has at least one positive solution.


## 5. Examples

In this section, we give two examples to illustrate the main results of the paper. First, we consider the nonresonant four-point boundary value problem(75)x′′′(t)+f(t,x)=0,t∈(0,1),x′′(0)=0,x′(0)=0,x(1)=12x(13)+13x(23),
where
(76)f(t,x)={1100et+x3π,0<x<6,1100et+216π,x≥6.
Here, β1=1/2, β2=1/3, ξ1=1/3, ξ2=2/3, and
(77)G(t,s)={-s+113,t≤s,0≤s≤13,-t+113,t≥s,0≤s≤13,-4s+143,t≤s,13≤s≤23,-3s-2t+143,t≥s,13≤s≤23,6-6s,t≤s,23≤s≤1,6-t-5s,t≥s,23≤s≤1.
By a simply computation, we can get that
(78)m=143162,δ=89162,γ=811.We choosea=1, b=4,  and c=162. It is easy to check that (1)
f(t,x)<143/162, [t,x]∈[0,1]×[0,1],(2)
f(t,x)>648/89, [t,x]∈[0,1]×[4,11/2],(3)
f(t,x)<143,[t,x]∈[0,1]×[0,162]. Thus, all the conditions of Theorem 8 are satisfied. This ensures that problem (75) has at least three positive solutions x1, x2, and x3 satisfying
(79)max0≤t≤1x1≤1,min0≤t≤1x2>4,max0≤t≤1x3>1,min0≤t≤1x3<4.
Next, we consider the resonant third-order four-point boundary value problem
(80)x′′′(t)+(-12t2+12t+516)(x2-4x+115)x2-6x+10=0,t∈[0,1],x′′(0)=0,x′(0)=0,x(0)=x(23),
where β=1, ξ=2/3 and
(81)f(t,x)=(-12t2+12t+516)×(x2-4x+115)x2-6x+10.
Here,
(82)G(s)={59-23s,0≤s≤23,(1-s)2,23≤s≤1.
By a simple computation, we have
(83)∫01‍G(s)ds=1981,κ=1945,∫01‍U(0,s)ds=1.
Choose R=1, r=1/4, t0=0, a=1, and M=1/2.We take(84)g(t)=-12t2+13t+516,t∈[0,1],h(x)=x2-6x+10,x∈[0,14].
Then,
(85)748≤g(t)≤53144<1945,t∈[0,1],x2-4x+115≥-x,x∈[0,1].
It is easy to check that (1)
f(t,x)>-(19/45)x, for all (t,x)∈[0,1]×[0,1],(2)
f(t,x)<0, for all (t,x)∈[0,1]×[71/90,1],(3)
f(t,x)≥(101/80)(-(1/2)t2+(1/3)t+(5/16))x2-6x+10≥g(t)h(x), [t,x]∈[0,1]×(0,1/4] and h(x)/x=x2-6x+10/xis nonincreasing on (0,1/4] with
(86)h(r)ra∫01‍U(0,s)g(s)ds>713748>1=1-MMa.
Then, all conditions of Theorem 9 are satisfied. This ensures that the resonant problem has at least one solution, positive on [0,1].Remark 11.
The established existence results of positive solutions for third-order boundary value problems in [2, 3, 5–12], for examples, are not applicable to the problem (75) or (5.2).



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*Source: 519346-2013-03-24.xml*