# Positive Solutions for Third-Order Boundary-Value Problems with the Integral Boundary Conditions and Dependence on the First-Order Derivatives

Authors: Yanping Guo; Fei Yang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

By using a fixed point theorem in a cone and the nonlocal third-order BVP's Green function, the existence of at least one positive solution for the third-order boundary-value problem with the integral boundary conditionsx′′′(t)+f(t,x(t),x′(t))=0, t∈J, x(0)=0, x′′(0)=0, and x(1)=∫01g(t)x(t)dt is considered, where f is a nonnegative continuous function, J=[0,1], and g∈L[0,1]. The emphasis here is that f depends on the first-order derivatives.


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## Body

## 1. Introduction

Third-order boundary-value problems for differential equation play a very important role in a variety of different areas of applied mathematics and physics. Recently, third-order boundary-value problems have been many scholars' research object. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can produce boundary-value problems with integral boundary conditions [1–3]. For more information about the general theory of integral equations and their relation with boundary-value problems, we refer readers to the books of Corduneanu [4] and Agarwal and O'Regan [5].Moreover, boundary-value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoint, and nonlocal boundary-value problems as special cases. Such kind of BVPs in Banach space has been studied by some researchers [6–8].By the fixed point index theory in cones [9], Zhang et al. [10] investigated the multiplicity of positive solutions for a class of nonlinear boundary-value problems of fourth-order differential equations with integral boundary conditions in ordered Banach spaces. Feng et al. [11] investigated the existence and multiplicity of positive solutions for a class of nonlinear boundary-value problems of second-order differential equations with integral boundary conditions in ordered Banach spaces. Guo et al. [12] investigated the existence of positive solutions for the third-order boundary-value problems with integral boundary conditions and dependence on the second derivatives. In [13], by using the fixed point theorem of cone expansion and compression of norm type, Zhang and Ge proved the existence and multiplicity of symmetric positive solutions for the fourth-order boundary-value problems with integral boundary conditions. By using Krasnoselskii's fixed point theorem, Wang et al. [14] investigated the existence and nonexistence of positive solutions for a class of fourth-order nonlinear differential equation with integral boundary conditions
(1)x(4)(t)=ω(t)f(t,x(t),x′′(t)),0<t<1,x(0)=∫01‍h1(s)x(s)ds,x(1)=∫01‍k1(s)x(s)ds,x′′(0)=∫01‍h2(s)x′′(s)ds,x′′(1)=∫01‍k2(s)x′′(s)ds,
where the arguments are based on Krasnoselskii's fixed point theorem for operators on a cone.However, Zhao et al. [15] investigated the following third-order boundary-value problem with integral boundary conditions:
(2)x′′′(t)+f(t,x(t))=θ,t∈J,x(0)=θ,x′′(0)=θ,x(1)=∫01‍g(t)x(t)dt,
under the assumptions(1)
J=[0,1], and θ is the zero element of E,(2)
f:C([0,1]×P,P), and g∈L[0,1] is nonnegative,where P is a cone in the real Banach E.All the above works were done under the assumption that the first-order derivativex′ is not involved explicitly in the nonlinear term f. In this paper, we are concerned with the existence of positive solutions for the third-order boundary-value problem with the integral boundary conditions
(3)x′′′(t)+f(t,x(t),x′(t))=0,t∈J,x(0)=0,x′′(0)=0,x(1)=∫01g(t)x(t)dt.Throughout, we assume(H1)
J=[0,1], f:[0,1]×R2→R+ is continuous, g∈L[0,1], g(t)≥0, and σ∈[0,1), where σ=∫01‍sg(s)ds.To show the existence of positive solutions for (3), we define two positive continuous convex functionals. Then, by using the fixed point theorem [16] in a cone and the nonlocal third-order BVP's Green function, we give some new criteria for the existence of positive solutions for (3).


## 2.  Preliminaries

LetY=C[0,1] be the Banach space equipped with the norm ∥x∥0=maxt∈[0,1]|x(t)|.Lemma 1 (see [15]).
Suppose(H1) holds. Then for any y(t)∈C[0,1], the problem
(4)x′′′(t)+y(t)=0,t∈J,x(0)=0,x′′(0)=0,x(1)=∫01g(t)x(t)dt
has a unique solution
(5)x(t)=∫01H(t,s)y(s)ds,
where
(6)H(t,s)=G(t,s)+t1-σ∫01‍G(τ,s)g(τ)dτ,G(t,s)={12t(1-s)2-12(t-s)2,0≤s≤t≤1,12t(1-s)2,0≤t≤s≤1.Lemma 2 (see [15]).
Fort,s∈[0,1], one has 0≤G(t,s)≤max0≤t,s≤1G(t,s)≤1/8.Remark 3.
Whent, s∈(0,1), it is easy to check that G(t,s)>0.
In addition, for0≤s≤t≤1, the maximum of G(t,s) occurs at t=(1+s2)/2.Lemma 4 (see [15]).
Chooseδ∈(0,1/2) and Jδ=[δ,1-δ]; then for all t∈Jδ, v, s∈[0,1], one has
(7)G(t,s)≥ρG(v,s),
where ρ=4δ2(1-δ).Remark 5.
For0≤s≤t≤1, denote G(t,s)=G1(t,s). Notice that G1(t,s) is concave with respect to t; we have
(8)mint∈Jδ,0≤s≤tG1(t,s)=min{G1(δ,s),G1(1-δ,s)}=12δ2(1-δ).Lemma 6 (see [15]).
Assume that(H1) holds; then (i)
H(t,s)≤(1/2)γ, t∈[0,1],(ii)
H(t,s)≥ρH(v,s), t∈Jδ, v,s∈[0,1],where γ=(1+∫01‍(1-s)g(s)ds)/(1-σ).Lemma 7.
Ify∈C[0,1], y(t)≥0, then the unique solution x(t) of problem (4) satisfies
(9)mint∈Jδx(t)≥ρ∥x∥0.Proof.
By Lemmas4 and 6 and (5), we get
(10)mint∈Jδx(t)=mint∈Jδ∫01‍H(t,s)y(s)ds≥ρ∫01‍H(v,s)y(s)ds≥ρx(v).
Forv∈[0,1], we have
(11)mint∈Jδx(t)≥ρx(v).
So,(12)mint∈Jδx(t)≥ρmaxv∈[0,1]x(v)=ρmaxv∈[0,1]|x(v)|=ρ∥x∥0.
The proof is completed.LetX be a Banach space and K⊂X a cone. Suppose α, β:X→R+ are two continuous convex functionals satisfying α(λx)=|λ|α(x), β(λx)=|λ|β(x), for x∈X, λ∈R, ∥x∥≤Mmax{α(x),β(x)}, for x∈X, and α(x)≤α(y) for x, y∈K, x≤y, where M>0 is a constant.Theorem 8 (see [16]).
Letr2>r1>0, L>0 be constants and
(13)Ωi={x∈X:α(x)<ri,β(x)<L},i=1,2,
two bounded open sets inX. Set
(14)Di={x∈X:α(x)=ri},i=1,2.
AssumeT:K→K is a completely continuous operator satisfying (A1)
α(Tx)<r1,  x∈D1∩K; α(Tx)>r2, x∈D2∩K;(A2)
β(Tx)<L, x∈K;(A3)
there is ap∈(Ω2∩K)∖{0} such that α(p)≠0 and α(x+λp)≥α(x), for all x∈K and λ≥0.
ThenT has at least one fixed point in (Ω2∖Ω¯1)∩K.


## 3. Main Results

LetX=C1[0,1] be the Banach space equipped with the norm ∥x∥=maxt∈[0,1]|x(t)|+maxt∈[0,1]|x′(t)|, and K={x∈X:x(t)≥0, mint∈Jδx(t)≥ρ∥x∥0} is a cone in X.Define two continuous convex functionalsα(x)=maxt∈[0,1]|x(t)| and β(x)=maxt∈[0,1]|x′(t)|, for each x∈X; then ∥x∥≤2max{α(x),β(x)} and α(λx)=|λ|α(x), β(λx)=|λ|β(x), for x∈X, λ∈R; α(x)≤α(y) for x, y∈K, x≤y.In the following, we denote(15)η0=18+∫01‍[11-σ∫01‍G(τ,s)g(τ)dτ]ds,η1=maxv∈[0,1]∫δ1-δ‍H(v,s)ds,η2=23+∫01‍[11-σ∫01‍G(τ,s)g(τ)dτ]ds.We will suppose that there areL>b>ρb>c>0 such that f(t,x,y) satisfies the following growth conditions: (H2)
f(t,x,y)<c/η0, for (t,x,y)∈[0,1]×[0,c]×[-L,L],(H3)
f(t,x,y)≥b/ρη1, for (t,x,y)∈[δ,1-δ]×[ρb,b]×[-L,L],(H4)
f(t,x,y)<L/η2, for (t,x,y)∈[0,1]×[0,b]×[-L,L].Let(16)f*(t,x,y)={f(t,x,y),(t,x,y)∈[0,1]×[0,b]×(-∞,∞),f(t,b,y),(t,x,y)∈[0,1]×(b,∞)×(-∞,∞),f1(t,x,y)={f*(t,x,y),(t,x,y)∈[0,1]×[0,∞)×[-L,L],f*(t,x,-L),(t,x,y)∈[0,1]×[0,∞)×(-∞,-L],f*(t,x,L),(t,x,y)∈[0,1]×[0,∞)×[L,∞).
We denote
(17)(Tx)(t)=∫01‍H(t,s)f1(s,x,x′)ds,(Tx)′(t)=∫01‍∂H(t,s)∂tf1(s,x,x′)ds.Lemma 9.
Suppose(H1) holds. Then T:K→K is completely continuous.Proof.
Forx∈K, by Lemmas 2 and 4, it is obviously that Tx≥0.
By Lemma7, we have
(18)mint∈JδTx(t)≥ρ∥Tx∥0.
So, we can get T(K)⊂K.
In the following, we will show thatT:K→K is completely continuous.
At first we show thatT:K→K is continuous.
Letxn, x*∈K, it satisfies ∥xn-x*∥→0, (n→∞), and then there is a constant M0>0, such that maxt∈[0,1]{|xn(t)|,|x*(t)|,|xn′(t)|,|x*′(t)|}≤M0; then
(19)|(Txn)(t)-(Tx*)(t)|=|∫01‍H(t,s)f1(s,xn,xn′)ds-∫01‍H(t,s)f1(s,x*,x*′)ds|≤∫01‍H(t,s)|f1(s,xn,xn′)-f1(s,x*,x*′)|ds,|(Txn)′(t)-(Tx*)′(t)|=|∫01‍∂H(t,s)∂tf1(s,x,xn′)ds-∫01‍∂H(t,s)∂tf1(s,x*,x*′)ds|≤∫01‍|∂H(t,s)∂t||f1(s,x,xn′)-f1(s,x*,x*′)|ds<∫01‍[12(1-s)2+(1-s)]×|f1(s,x,xn′)-f1(s,x*,x*′)|ds+∫01‍[11-σ∫01‍G(τ,s)g(τ)dτ]×|f1(s,x,xn′)-f1(s,x*,x*′)|ds.
Byf which is uniformly continuous on [0,1]×[-M0,M0]×[-M0,M0], we get
(20)∥Txn-Tx*∥⟶0,(n⟶∞).
Next we show thatT:K→K is compact.
LetB⊂K be bounded; then there is M>0, such that ∥x∥≤M. For x∈B, we have
(21)|(Tx)(t)|=|∫01‍H(t,s)f1(s,x,x′)ds|≤∫01‍12γf1(s,x,x′)ds=12∫01‍1+∫01‍(1-s)g(s)ds1-σds×C*,
where C*=max{|f1(t,x,x′)|;t∈[0,1],x∈B}.
Consider(22)|(Tx)′(t)|=|∫01‍∂H(t,s)∂tf1(s,x,x′)ds|=|∫01‍[∂G(t,s)∂t+11-σ∫01‍G(τ,s)g(τ)dτ]×f1(s,x,x′)ds∫01‍|<|∫01‍[12(1-s)2+(1-s)]ds+∫01‍[11-σ∫01‍G(τ,s)g(τ)dτ]ds|×C*=|23+∫01‍[11-σ∫01‍G(τ,s)g(τ)dτ]ds|×C*.
It is clear thatT(B) is a bounded set in K, because H(t,s) is uniformly continuous on [0,1]×[0,1], for ε>0, there exists δ∈(0,ε), such that |H(t1,s)-H(t2,s)|<ε, and for t1, t2∈[0,1], |t1-t2|<δ.
Forx∈B, we have
(23)|(Tx)(t1)-(Tx)(t2)|=|∫01‍H(t1,s)f1(s,x,x′)ds-∫01‍H(t2,s)f1(s,x,x′)ds|≤∫01‍|H(t1,s)-H(t2,s)|ds×C*≤εC*,|(Tx)′(t1)-(Tx)′(t2)|=|∫01‍∂H(t,s)∂t|t=t1f1(s,x,x′)ds-∫01‍∂H(t,s)∂t|t=t2f1(s,x,x′)ds|=|∫01‍∂G(t,s)∂t|t=t1f1(s,x,x′)ds-∫01‍∂G(t,s)∂t|t=t2f1(s,x,x′)ds|=|∫0t1‍(s-t1)f1(s,x,x′)ds+∫0t2‍(t2-s)f1(s,x,x′)ds|≤12|(t1-t2)(t1+t2)|×C*≤εC*.
ThereforeT(B) is equicontinuous. Using the Arzela-Ascoli theorem, a standard proof yields T:K→K which is completely continuous.Theorem 10.
Suppose(H1)–(H4) hold. Then BVP (3) has at least one positive solution x(t) satisfying
(24)c<α(x)<b,β(x)<L.Proof.
Take(25)Ω1={x∈X:|x(t)|<c,|x(t)′|<L},Ω2={x∈X:|x(t)|<b,|x(t)′|<L},
two bounded open sets in X, and
(26)D1={x∈X:α(x)=c},D2={x∈X:α(x)=b}.
By Lemma9, T:K→K is completely continuous, and there is a p∈(Ω2∩K)∖{0} such that α(p)≠0 and α(x+λp)≥α(x) for all u∈K and λ≥0.
By(H2), for x∈D1∩K and α(x)=c, we get
(27)α(Tx)=maxt∈[0,1]|∫01‍H(t,s)f1(s,x,x′)ds|=maxt∈[0,1]|∫01‍[G(t,s)+t1-σ∫01‍G(τ,s)g(τ)dτ]×f1(s,x,x′)ds∫01‍|<∫01‍[maxt∈[0,1]G(t,s)+t1-σ∫01‍G(τ,s)g(τ)dτ]×f1(s,x,x′)ds<[∫01‍18ds+∫01‍(t1-σ∫01‍G(τ,s)g(τ)dτ)ds]×cη0=[18+∫01‍(t1-σ∫01‍G(τ,s)g(τ)dτ)ds]×cη0=c.
By Lemma7, for x∈D2∩K and α(x)=b, there is x(t)≥ρα(x)=ρb, t∈Jδ.
So, by(H3), we get
(28)α(Tx)=maxt∈[0,1]|∫01‍H(t,s)f1(s,x,x′)ds|>∫δ1-δ‍H(t,s)f1(s,x,x′)ds>ρ∫δ1-δ‍H(v,s)ds×bρη1.
Forv∈[0,1], we have
(29)α(Tx)>ρ∫δ1-δ‍H(v,s)ds×bρη1.
So,(30)α(Tx)>ρmaxv∈[0,1]∫δ1-δ‍H(v,s)ds×bρη1=b.
By(H4), for x∈K, we have
(31)β(Tx)=maxt∈[0,1]|∫01‍∂H(t,s)∂tf1(s,x,x′)ds|<|∫01[∂G(t,s)∂t+11-σ∫01‍G(τ,s)g(τ)dτ]×f1(s,x,x′)ds=|∫0t(12(1-s)2-(t-s))f1(s,x,x′)ds+∫t112(1-s)2f1(s,x,x′)ds+∫01‍[11-σ∫01‍G(τ,s)g(τ)dτ]×f1(s,x,x′)ds∫01‍|<[∫01‍(12(1-s)2+(1-s))ds+∫01‍(11-σ∫01‍G(τ,s)g(τ)dτ)ds]×Lη2=[23+∫01‍(11-σ∫01‍G(τ,s)g(τ)dτ)ds]×Lη2=L.
Theorem8 implies there is x∈(Ω2∖Ω¯1)∩K such that x=Tx. So, x(t) is a positive solution for BVP (3) satisfying
(32)c<α(x)<b,β(x)<L.
Thus, Theorem10 is completed.


## 4. Example

Example 1.
Consider the following boundary-value problem(33)x′′′(t)+f(t,x(t),x′(t))=0,0<t<1,x(0)=0,x′′(0)=0,x(1)=∫01‍x(t)dt,
where
(34)f(t,x,y)={t3x+2x+|cosy|,(t,x,y)∈[0,1]×[0,0.5]×[-3667,3667],109t3(x-0.5)+25742(x-0.5)+t6+1+|cosy|,(t,x,y)∈[0,1]×[0.5,0.6]×[-3667,3667],t3(11-x)+222(x+11)+|cosy|,(t,x,y)∈[0,1]×[0.6,11]×[-3667,3667].
In this problem, we know thatg(t)=1; then we can get σ=∫01‍sg(s)ds=1/2. Choose δ=1/8∈(0,1/2); then ρ=4δ2(1-δ)=7/128.
Furthermore, we obtain(35)η0=524,ρη1=358192,η2=34.
If we takec=0.5, b=11, and L=3667, then we get ρb≈0.601>0.6:
(36)f(t,x,y)=t3x+2x+|cosy|≤2.17<cη0≈2.4,
for (t,x,y)∈[0,1]×[0,0.5]×[-3667,3667],
(37)f(t,x,y)=t3(11-x)+222(x+11)+|cosy|≥2575.2>bρη1≈2574.1,
for (t,x,y)∈[δ,1-δ]×[δb,11]×[-3667,3667],
(38)f(t,x,y)≤4888.8<Lη2≈4889.3,
for (t,x,y)∈[0,1]×[0,11]×[-3667,3667].
Then all the conditions of Theorem10 are satisfied. Therefore, by Theorem 10 we know that boundary-value problem (33) has at least one positive solution x(t) satisfying
(39)0.5<α(x)<11,β(x)<3667.



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