# On the Existence of Positive Solutions of Resonant and Nonresonant Multipoint Boundary Value Problems for Third-Order Nonlinear Differential Equations

Authors: Liu Yang; Chunfang Shen; Dapeng Xie; Xiping Liu
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 non-resonant conditions and the resonant conditions are considered. In the nonresonant case, by using Leggett-Williams fixed-point theorem, the existence of at least three positive solutions is obtained. In the resonant case, by using Leggett-Williams norm-type theorem due to O’Regan and Zima, existence result of at least one positive solution is established. The results obtained are valid and new for the problem discussed. Two examples are given to illustrate the main results.


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## Body

## 1. Introduction

We consider the third-orderm-point boundary value problem given by
(1)x′′′(t)+f(t,x(t))=0,t∈[0,1]x′′(0)=0,x′(0)=∑i=1m-2‍αix′(ξi),x(1)=∑i=1m-2‍βix(ξi),
where 0<ξ1<ξ2<⋯<ξm-2<1, 0≤αi, βi≤1, i=1,2,…,m-2, ∑i=1m-2‍αi<1, ∑i=1m-2‍βi≤1, and f∈C([0,1]×[0,∞),R). For the convenience of writing later, we denote ξ0=0,ξm-1=1, and α0=αm-1=β0=βm-1=0.If condition∑i=1m-2‍βi≠1 holds, the problem is nonresonant; that is, the associated linear problem
(1)x′′′(t)=0,t∈[0,1]x′′(0)=0,x′(0)=∑i=1m-2‍αix′(ξi),x(1)=∑i=1m-2‍βix(ξi)
has only zero solution, and the differential operator with boundary conditions is invertible. Otherwise, the problem is at resonance.Third-order differential equations arise in many different areas of applied mathematics and physics, as the varying cross-section or deflection of a curved beam having a constant, three-layer beam and so on [1]. Recently, there have been extensive studies on positive solutions for nonresonant two-point or three-point boundary value problems for nonlinear third-order ordinary differential equations. For examples, Anderson [2] established the existence of at least three positive solutions to problem
(3)-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-Krasnosel’skii fixed-point theorem [3], Palamides and Smyrlis [4] proved that there exists at least one positive solution for the third-order three-point problem:
(4)x′′′(t)=a(t)f(t,x(t)),t∈(0,1),x′′(η)=0,x(0)=x(1)=0,η∈(0,1).In another paper [5], Graef and Kong studied the existence of positive solutions for the third-order semipositone boundary value problem:
(5)u′′′(t)=λf(t,u(t))+e(t),t∈(0,1),u(0)=u′(p)=∫q1‍w(s)u′′(s)ds=0,
where 1/2<p<q<1 are constants and λ>0 is a parameter. Also f:(0,1)×[0,∞)→R,e:(0,1)→R, and e∈L(0,1). Moreover w:[q,1]→[0,∞) are continuous functions. By using the Guo-Krasnosel’skii fixed-point theorem, the author established the existence of positive solutions. For more existence results of positive solutions for boundary value problems of third-order ordinary differential equations, one can see [6–12] and references therein.For boundary value problems of second-order or higher-order differential equations at resonance, many existence results of solutions have been established; see [13–25]. In [25], the authors considered the following problem:
(6)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 Mawhin continuation theorem [26], the existence results of solutions were obtained under the resonant conditions β=1, ∑i=1m-2‍αi=1, and ∑i=1m-2‍αiξi2=0 and  β=1/η, ∑i=1m-2‍αi=1, and ∑i=1m-2‍αiξi2=0, respectively.It is well known that the problem of existence of positive solution for 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 [27–34]. To our best knowledge, few paper dealt with the existence result of positive solution for resonant third-order boundary value problems. Motivated by the approach in [28–30, 35], we study the positive solution for problem (1) under nonresonant condition ∑i=1m-2‍βi<1 and resonant condition ∑i=1m-2‍βi=1, respectively. By using Leggett-Williams fixed-point theorem and its generalization [27, 29], we establish the existence results of positive solutions. The results obtained in this paper are interesting in the following aspects.(1)
In the nonresonant case, Green’s function is established and the results obtained are more general than those of earlier work.(2)
It is the first time that the positive solution is considered for third-order boundary value problem at resonance.


## 2. Background Definitions and Lemmas

LetX,Y be real Banach spaces. A nonempty closed convex set C⊂X is said to be a cone provided that ax∈C, if x∈C, a≥0 and x,-x∈C implies x=0.Definition 1.
The mapψ is a nonnegative continuous concave functional on C if ψ:C→+∞ is continuous and
(7)ψ(tx+(1-t)y)≥tψ(x)+(1-t)ψ(y),ψ(tx+(1-t)y)≥tψx,y∈C,t∈[0,1].Definition 2.
Let constants0<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) as follow:
(8)Cr={x∈C∣∥x∥<r},C(ψ,a,b)={x∈C∣a≤ψ(x),∥x∥≤b}.Lemma 3 (Leggett-Williams fixed-point theorem [35]).
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 operator T has at least triple fixed-points x1,x2, and x3 with ∥x1∥<a,b<ψ(x2),∥x3∥>a, and ψ(x3)<b.OperatorL:domL⊂X→Y is called a Fredholm operator with index zero, which means that ImL is closed and dim KerL = codim ImL<∞, and there exist continuous projections P:X→X and Q:Y→Y such that ImP=KerL and KerQ=ImL. Furthermore, for dim Im Q= codim ImL, 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 the operator equation:
(9)x=(P+JQN)x+KP(I-Q)Nx.
Let γ:X→C be a retraction, which means a continuous mapping such that γx=x for all x∈C and
(10)Ψ:=P+JQN+KP(I-Q)N,Ψγ:=Ψ∘γ.Lemma 4 (Leggett-Williams norm-type theorem [28]).
Assume thatC is a cone in X and that Ω1 and Ω2 are open bounded subsets of X with Ω¯1⊂Ω2 and C∩(Ω¯2∖Ω1)≠∅. Suppose that L:domL⊂X→Y is a Fredholm operator of index zero and that(C1)
QN:X→Y is bounded and continuous and 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) is 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 equation Lx=Nx has a solution in the set C∩(Ω¯2∖Ω1).


## 3. Main Results for Nonresonant Case

In this section we consider the positive solution for the nonresonant case with the condition0<∑i=0m-1‍βi<1 and we always suppose that f∈C([0,1]×[0,∞),[0,∞)).Firstly, we consider the third-orderm-point boundary value problem given by
(11)x′′′(t)+y(t)=0,t∈[0,1](12)x′′(0)=0,x′(0)=∑i=1m-2‍αix′(ξi),x(1)=∑i=1m-2‍βix(ξi).Lemma 5.
Supposey(t)∈C[0,1]. Then problem (11) and (12) is equivalent to
(13)x(t)=∫01‍G(t,s)y(s)ds,
where for i=1,2,…,m-1(14)G(t,s)={a1t+a0,t≤s,ξi-1≤s≤ξi,-12t2+b1t+b0,t≥s,ξi-1≤s≤ξi,a1=∑k=im-1‍αk(s-ξk)1-∑k=0m-1‍αk,b1=s-∑k=0i-1‍αks-∑k=im-1‍αkξk1-∑k=0m-1‍αk,a0=(∑k=0i-1‍βkξk∑k=im-1‍αk(s-ξk)1-∑k=0m-1‍αk000+(∑k=im-1‍βkξk-1)000×s-∑k=0i-1‍αks-∑k=im-1‍αkξk1-∑k=0m-1‍αk000+12s2+12-12∑k=im-1‍βk(ξk2+s2))×(1-∑k=0m-1‍βk)-1,b0=(∑k=0i-1‍βkξk∑k=im-1‍αk(s-ξk)1-∑k=0m-1‍αk+12∑k=0i-1‍βks2-12∑k=im-1‍βkξk2+12+(∑k=im-1‍βkξk-1)×s-∑k=0i-1‍αks-∑k=im-1‍αkξk1-∑k=0m-1‍αk)×(1-∑k=0m-1‍βk)-1.Proof.
LetG(t,s) be Green’s function of problem -x′′′(t)=0 with boundary condition (12). We can suppose
(15)G(t,s)={a2t2+a1t+a0t≤s,ξi-1≤s≤ξi,i=1,2,…,m-1b2t2+b1t+b0t≥s,ξi-1≤s≤ξi,i=1,2,…,m-1,
where a0,a1,a2,b0,b1, and b2 are undetermined coefficients.
Considering the properties of Green's function together with boundary condition (12), we have
(16)a2s2+a1s+a0=b2s2+b1s+b0,2a2s+a1=2b2s+b1,2a2-2b2=1,a2=0,a1=∑k=0i-1‍αk(2a2ξk+a1)+∑k=im-1‍αk(2b2ξk+b1),b2+b1+b0=∑k=0i-1‍βk(a2ξk2+a1ξk+a0)+∑k=im-1‍βk(b2ξk2+b1ξk+b0).
The explicit expression of Green’s function is obtained by solving the linear function systems.Lemma 6.
Green’s functionG(t,s) satisfies that G(t,s)≥0,t,s∈[0,1].Proof.
Forξi-1≤s≤ξi, i=1,2,…,m-1, and t≤s,
(17)∂G(t,s)∂t=a1<0,
which means that G(t,s) is decreasing on variable t. Thus
(18)G(t,s)≥G(s,s),0≤t≤s.
On the other hand, for ξi-1≤s≤ξi, i=1,2,…,m-1, and t≥s,
(19)∂G(t,s)∂t=-t+b1=-t+s-∑k=0i-1‍αks-∑k=im-1‍αkξk1-∑k=0m-1‍αk≤∑k=im-1‍αk(s-ξk)1-∑k=0i-1‍αk<0.
Thus,
(20)G(t,s)≥G(1,s),s≤t≤1.
Furthermore,
(21)∂G(1,s)∂s≤-(1-∑k=0i-1‍αk)∑k=0m-1‍βk(1-ξk)<0.
Thus, for t,s∈[0,1],
(22)G(t,s)≥G(1,1)=-12+b1+b0=-12+1-∑k=0m-1‍αk1-∑k=0m-1‍αk+((1-∑k=0m-1‍βk)-1(12∑k=0m-1‍βk+12-1-∑k=0m-1‍αk1-∑k=0m-1‍αk)×(1-∑k=0m-1‍βk)-1)=0.
This gives that G(t,s)≥0,t,s∈[0,1].Lemma 7.
Ify(t)≥0, t∈[0,1], and x(t) is the solution of problem (11) and (12), then
(23)min0≤t≤1|x(t)|≥δmax0≤t≤1|x(t)|,
where δ=(∑i=1m-2‍βi(1-ξi))/(1-∑i=1m-2‍βiξi) is a positive constant.Proof.
From(24)x′′′(t)=-y(t)≤0,t∈[0,1],
we see that x′′(t) is decreasing on [0,1]. Considering x′′(0)=0, we have x′′(t)≤0,t∈(0,1). Next we claim that x′(0)≤0. Suppose that, on the contrary, x′(0)>0. We have
(25)0=x′(0)-x′(0)=x′(0)-∑i=1m-2‍αix′(ξi)>∑i=1m-2‍αi(x′(0)-x′(ξi))≥0,acontradiction.
Thus,
(26)max0≤t≤1x(t)=x(0),min0≤t≤1x(t)=x(1).
From the concavity of x(t), we have
(27)ξi(x(1)-x(0))≤x(ξi)-x(0).
Multiplying left and right sides by βi and considering x(1)=∑i=1m-2‍βix(ξi), we have
(28)(1-∑i=1m-2‍βiξi)x(1)≥∑i=1m-2‍βi(1-ξi)x(0).
This completes the proof of Lemma 7.Let the Banach spaceE=C[0,1] be endowed with the maximum norm. We define the cone C⊂E by
(29)C={∑i=1m-2‍x∈E∣x(t)≥0,x′′(0)=0,00x′(0)=∑i=1m-2‍αix′(ξi),00x(1)=∑i=1m-2‍βix(ξi),x(t)isconcaveon[0,1]}.Define the continuous nonnegative concave functionalψ:C→[0,∞) by
(30)ψ(x)=min0≤t≤1x(t),x∈C.
Define the constants m*,m* by
(31)m*=max0≤t≤1∫01‍G(t,s)ds,m*=min0≤t≤1∫01‍G(t,s)ds.Theorem 8.
Suppose that there exist constants0<a<b<b/δ≤c such that (A1)
f(t,x)<a/m*, (t,x)∈[0,1]×[0,a],(A2)
f(t,x)>b/m*, (t,x)∈[0,1]×[b,b/δ],(A3)
f(t,x)<c/m*, (t,x)∈[0,1]×[0,c].Then problem (1) has at least three positive solutions x1,x2, and x3 satisfying
(32)∥x1∥≤a,b<min0≤t≤1x2,∥x3∥>a,min0≤t≤1x3<b.Proof.
The operatorT:C→E is defined by
(33)Tx(t)=∫01‍G(t,s)f(s,x(s))ds.
It is clear that T:C→C and it is completely continuous.
Next, the conditions of Lemma3 are checked. If x∈C¯c, then ∥x∥≤c and condition (A3) implies that
(34)f(t,x)≤cm*,0≤t≤1.
Then
(35)∥T(x)∥=max0≤t≤1∫01‍G(t,s)f(s,x(s))ds≤max0≤t≤1∫01‍G(t,s)ds×cm*≤c.
Thus, T:C¯c→C¯c.Similar to the proof above, we obtain thatT:C¯a→C¯a. Hence, condition (H2) of Lemma 3 is satisfied.The fact that the constant functionx(t)=b/δ∈{P(ψ,b,b/δ)∣ψ(x)>b} implies that {P(ψ,b,b/δ)∣ψ(x)>b}≠Ø. If x∈P(ψ,b,b/δ), from assumption (A2),
(36)f(t,x)≥bm*.
Thus
(37)ψ(Tx)=min0≤t≤1∫01‍G(t,s)f(s,x(s))ds≥min0≤t≤1∫01‍G(t,s)ds×bm*=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∈P(ψ,b,c) with ∥Tx∥>b/δ. Then
(38)ψ(Tx)=min0≤t≤1Tx(t)≥δ×∥Tx∥>δ×bδ=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
(39)∥x1∥≤a,b<min0≤t≤1x2,∥x3∥>a,min0≤t≤1x3<b.Here an example is given to illustrate the main results of this section. We consider the following nonresonant three-point boundary value problem:(40)x′′(t)+f(t,x)=0,t∈(0,1)x′′(0)=0,x′(0)=12x′(13),x(1)=12x(13),
where
(41)f(t,x)={110et+x3+5π,0<x<6110et+221π,x≥6.
Here α1=1/2,  β1=1/2,  ξ1=1/3, and
(42)G(t,s)={12s2+st-103s-13t+32,0≤s≤13,t≤s,-12t2+2st-13t-103s+32,0≤s≤13,t≥s,(s-1)2,13≤s≤1,t≤s,-12t2+12s2+st+1-2s,13≤s≤1,t≥s.
By a simple computation, we can get that
(43)m*=max0≤t≤1∫01‍G(t,s)ds=3281,m*=min0≤t≤1∫01‍G(t,s)ds=1681.
We choose a=1,b=4, and c=32. It is easy to check that (1)
f(t,x)<81/32, [t,x]∈[0,1]×[0,1],(2)
f(t,x)>81/4, [t,x]∈[0,1]×[4,10],(3)
f(t,x)<81, [t,x]∈[0,1]×[0,32].Thus all conditions of Theorem8 hold. This ensures that problem (40) has at least three positive solutions x1,x2, and x3 satisfying
(44)max0≤t≤1x1≤1,min0≤t≤1x2>4,max0≤t≤1x3>1,min0≤t≤1x3<4.


## 4. Main Results for Resonant Case

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 spacesX=Y=C[0,1] endowed with the maximum norm. It is well known that X and Y are the Banach spaces.Define the linear operatorL:domL⊂X→Y and
(45)(Lx)(t)=-x′′′(t),t∈[0,1],
where
(46)domL={∑i=1m-2‍x∈X∣x′′′∈C[0,1],x′′(0)=0,0x′(0)=∑i=1m-2‍αix′(ξi),x(1)=∑i=1m-2‍βix(ξi)}
and the nonlinear operator N:X→Y with
(47)(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 follow:
(48)G(s)=(1-s)2-∑k=im-2‍βk(ξk-s)2+21-∑k=im-2‍βkξk1-∑k=1m-2‍αk∑k=im-2‍αk(ξk-s),0000ξi-1≤s≤ξi,i=1,2,…,m-2.
Define the function k(t,s) as follow:(49)k(t,s)={16(1-s)3-12(t-s)2-st+12s-k0-∑k=0i-1‍βk((1/2)s2-ξks+(1/2)ξk2)(1/2-t)1-∑i=0m-1‍βiξi,t≥s,ξi-1≤s≤ξi,16(1-s)3-st+12s-k0-∑k=0i-1‍βk((1/2)s2-ξks+(1/2)ξk2)(1/2-t)1-∑i=0m-1‍βiξi,t≤s,ξi-1≤s≤ξi,for i=1,2,…,m-1 and k0=(1/2-t)[1-(1/2)(1-∑i=0m-1‍βiξi)].The functionU(t,s) and positive number κ are given by
(50)U(t,s)=k(t,s)+G(s)∫01‍G(s)ds(1-∫01‍k(t,s)ds),(t,s)+G(s)∫01‍G(s)ds(1-∫01‍k(t,s)ds)t,s∈[0,1],κ≔min{1,mins∈[0,1]∫01‍G(s)dsG(s),mint,s∈[0,1]1U(t,s)},σ=1+∑i=1m-2‍αiξi21-∑i=1m-2‍αi.Theorem 9.
Suppose 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-(κσ/2))R,R],(S3)
there existr∈(0,R),M∈(0,1),t0∈[0,1],a∈(0,1], and continuous functions
(51)g:[0,1]⟶[0,+∞),h:(0,r]⟶[0,+∞)such that
(52)f(t,x)≥g(t)h(x),[t,x]∈[0,1]×(0,r]
and h(x)/xa is nonincreasing on (0,r] with
(53)h(r)ra∫01‍U(t0,s)g(s)ds≥1-MMa.
Then resonant problem (1) has at least one positive solution.Proof.
Firstly we prove that(54)ImL={y∈Y∣∫01‍G(s)y(s)ds=0}.
Indeed, for each y∈{y∈Y∣∫01‍G(s)y(s)ds=0}, we choose
(55)x(t)=-12∫0t‍(t-s)2y(s)ds-t1-∑i=1m-2‍αi∑i=1m-2‍αi∫0ξi‍(ξi-s)y(s)ds.
We can check that
(56)-x′′′(t)=y(t),x′′(0)=0,x′′(0)=0,x′(0)=∑i=1m-2‍αix′(ξi),x(1)=∑i=1m-2‍βix(ξi),
which means x(t)∈domL. Thus
(57){y∈Y∣∫01‍G(s)y(s)ds=0}⊂ImL.
On the other hand, for each y(t)∈ImL, there exists x(t)∈domL such that
(58)-x′′′(t)=y(t),x′′(0)=0,x′′(0)=0,x′(0)=∑i=1m-2‍αix′(ξi),x(1)=∑i=1m-2‍βix(ξi).
Integrating both sides on [0, t], we have
(59)x(t)=-12∫0t‍(t-s)2y(s)ds+12x′′(0)t2+x′(0)t+x(0).
Considering the boundary condition together with the resonant condition ∑i=0m-1‍βi=1, we have
(60)∫01‍(1-s)2y(s)ds-∑i=1m-2‍βi∫0ξi‍(ξi-s)2y(s)ds+21-∑i=1m-2‍βiξi1-∑i=1m-2‍αi∑i=1m-2‍αi∫0ξi‍(ξi-s)y(s)ds=0.
Thus,
(61)ImL={y∈Y∣∫01‍G(s)y(s)ds=0}.
It is obvious that dim KerL=1 and ImL is closed.
Secondly we seeY=Y1⊕ImL, where
(62)Y1={y1∣y1=1∫01‍G(s)ds∫01‍G(s)y(s)ds,y∈Y}.
In fact, for each y(t)∈Y, we have
(63)∫01‍G(s)[y(s)-y1]ds=0.
This induces that y-y1∈ImL. Since Y1∩ImL={0}, we have Y=Y1⨁ImL. Thus L is a Fredholm operator with index zero.
Define two projectionsP:X→X and Q:Y→Y by
(64)Px=∫01‍x(s)ds,Qy=1∫01‍G(s)ds∫01‍G(s)y(s)ds.
Clearly, ImP=KerL and KerQ=ImL. Note that, for y∈ImL, the inverse KP of LP is given by
(65)(KP)y=∫0t‍k(t,s)y(s)ds.
In fact, It is easy to check that
(66)L(KP)(y)=(-∫01‍k(t,s)y(s)ds)′′′=y(t)KP(L)(x)=∫01‍k(t,s)(-x′′′(s))ds=x(t).
Next we will check that every condition of Lemma4 is fulfilled. Remark that f can be extended continuously on [0,1]×(-∞,+∞) and condition (C1) of Lemma 4 is fulfilled.
Define the set of nonnegative functionsC and subsets of X Ω1,Ω2 by
(67)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]}.
Remark that Ω1 and Ω2 are open and bounded sets. Furthermore
(68)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 γ is a retraction and maps subsets of Ω¯2 into bounded subsets of C, which ensures that condition (C3) of Lemma 4 is fulfilled.
Then we prove that (C2) of Lemma 4 is fulfilled. For this purpose, suppose that there exist x0∈C∩∂Ω2∩domL and λ0∈(0,1) such that Lx0=λ0Nx0. Then
(69)-x0′′′(t)=λ0f(t,x0)
for all t∈[0,1]. Thus
(70)x0′′(t)=-λ0∫0t‍f(s,x0(s))ds,(71)x0′(t)=-λ0∫0t‍(t-s)f(s,x0(s))ds-λ011-∑i=1m-2‍αi∑i=1m-2‍αi∫0ξi‍(ξi-s)f(s,x0(s))ds.
Let x0(t0)=∥x0∥=R. The proof is divided into three cases.(1)
We show thatt0≠1. Suppose, on the contrary, that x0(t) achieves maximum value R only at t0=1. Then the boundary condition x(1)=∑i=1m-2‍βix0(ξi) in combination with the resonant condition ∑i=1m-2‍βi=1 yields that max1≤i≤m-2x0(ξi)≥R, which is a contradiction.(2)
We claim thatt0≠0. Suppose, on the contrary, that x0(t) achieves maximum value R at t0=0. From condition (S2), there exists t1>0 near to zero such that
(72)f(t,x0(t))<0,t∈[0,t1].
Then
(73)x0′(t)=-λ0∫0t‍(t-s)f(s,x0(s))ds-λ011-∑i=1m-2‍αi∑i=1m-2‍αi∫0ξi‍(ξi-s)f(s,x)ds>0,0000000000000000000000000000000t∈[0,t1],
which contradicts the fact that x0(t) achieves maximum value at t0=0.(3)
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
(74)x0(η)=x0(t0)-x0′(ξ)(t0-η).However,
(75)x0′(ξ)=-λ0∫0ξ‍(ξ-s)f(s,x0)ds-λ011-∑i=1m-2‍αi∑i=1m-2‍αi∫0ξi‍(ξi-s)f(s,x)ds≤λ0κ∫0ξ‍(ξ-s)x0(s)ds+λ0κ11-∑i=1m-2‍αi∑i=1m-2‍αi∫0ξi‍(ξi-s)x0ds≤λ0κR∫0ξ‍(ξ-s)ds+λ0κR11-∑i=1m-2‍αi∑i=1m-2‍αi∫0ξi‍(ξi-s)ds=12λ0κR(ξ2+∑i=1m-2‍αiξi21-∑i=1m-2‍αi).
Thus
(76)x0(η)=x0(t0)-x0′(ξ)(t0-η)≥R-12λ0κ(ξ2+∑i=1m-2‍αiξi21-∑i=1m-2‍αi)(t0-η)R≥[1-κ2(1+∑i=1m-2‍αiξi21-∑i=1m-2‍αi)]R=(1-κσ2)R.
Then, considering assumption (S2), we have
(77)0≥x0′′(t0)-x0′′(η)=-λ0∫ηt0‍f(s,x0(s))ds>0,
which is a contradiction. Thus (C2) of Lemma 4 is fulfilled.Remark 10.
The sign of third-order derivative of a functionh(t) at point t0 cannot be confirmed when t0 is a maximal value of h(t). Thus the methods in [28] are not applicable directly to this problem.Forx∈KerL∩Ω2, define the projection H(x,λ) as follows:
(78)H(x,λ)=x-λ|x|-λ∫01‍G(s)ds∫01‍G(s)f(s,|x|)ds,
where λ∈[0,1] and x∈KerL∩Ω2. Suppose H(x,λ)=0. In view of (S1) we obtain
(79)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. Hence, if H(R,λ)=0, we get
(80)0≤R(1-λ)∫01‍G(s)ds=λ∫01‍G(s)f(s,R)ds,
contradicting (S2). Thus H(x,λ)≠0 for x∈∂Ω2 and λ∈[0,1]. Therefore
(81)dB(H(x,0),KerL∩Ω2,0)=dB(H(x,1),KerL∩Ω2,0)=dB(I,KerL∩Ω2,0)=1.
This ensures
(82)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
(83)(Ψγx)(t)=∫01‍|x(t)|dt+1∫01‍G(s)ds∫01‍G(s)f(s,|x(s)|)ds+∫01‍k(t,s)000000×[f(s,|x(s)|)-1∫01‍G(s)ds0000000000×∫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
(84)(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. Taking u0(t)≡1 on [0,1], we see
(85)u0∈C∖{0},C(u0)={x∈C∣x(t)>0on[0,1]}
and we can choose σ(u0)=1. For x∈C(u0)∩∂Ω1, we have
(86)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,
(87)(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 with the application of Lemma4, we confirm that the equation Lx=Nx has a solution x∈C∩(Ω¯2∖Ω1), which implies that the resonant problem (1) has at least one positive solution.Finally an example is given to illustrate the main results of the resonance case. We investigate the resonant third-order three-point boundary value problem:(88)x′′′(t)+(-12t2+12t+516)(x2-4x+115)×x2-6x+10=0,t∈[0,1]x′′(0)=0,x′(0)=12x′(23),x(0)=x(23),
where α1=1/2,  β=1,  ξ=2/3, and
(89)f(t,x)=(-12t2+12t+516)(x2-4x+115)×x2-6x+10.
Here
(4)G(s)={1-43s,0≤s≤23(1-s)2,23≤s≤1,k(t,s)={16(1-s)3-12(t-s)2-st+12s-56(12-t),t≥s,0≤s≤2316(1-s)3-st+12s-56(12-t),t≤s,0≤s≤23.16(1-s)3-12(t-s)2-st+12s-(12-t)×(32s2-2s+32),t≥s,23≤s≤116(1-s)3-st+12s-(12-t)×(32s2-2s+32),t≤s,23≤s≤1.
By a simple computation, we have
(91)∫01‍G(s)ds=3181,σ=139,κ=3181,∫01‍U(0,s)ds=1.
Choose R=1, r=1/4, t0=0,a=1, and M=1/2.We take(92)g(t)=-12t2+13t+516,t∈[0,1],h(x)=x2-6x+10,x∈[0,14].
Then,
(93)748≤g(t)≤53144<3181,t∈[0,1],x2-4x+115≥-x,x∈[0,1].
It is easy to check that (1)
f(t,x)>-(31/81)x, for all (t,x)∈[0,1]×[0,1],(2)
f(t,x)<0, for all (t,x)∈[0,1]×[1055/1458,1],(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/x is nonincreasing on (0,1/4] with
(94)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].



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*Source: 687595-2013-12-15.xml*