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

Authors: Fei Yang; Yuanjian Lin; Juan Zhang; Quanfu Lou
Journal: Journal of Applied Mathematics
 (2022)

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## Abstract

In this paper, by the use of a new fixed point theorem and the Green function of BVPs, the existence of at least one positive solution for the third-order boundary value problem with the integral boundary conditions is considered,where there is a nonnegative continuous function. Finally, an example which to illustrate the main conclusions of this paper is given.


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## Body

## 1. Introduction

Third-order boundary value problems are originated in a variety of different fields of applied mathematics and physics, for example, deflection of a buckling beam with a fixed or varying cross-section, three-layer beams, electromagnetic waves, and flood tides from gravitational blowing. In recent year, researches on third-order nonlinear boundary value problems have received widespread attention, and many excellent results have been obtained, in references [1–10].As we all know, boundary value problems with integral boundary conditions can describe many valuable phenomena more accurately. The study of many problems in the fields of heat conduction, chemical engineering, groundwater flow, thermoelasticity, plasma physics, etc. can be reduced to the study of boundary value problems with integral boundary conditions. However, in recent years, although the third-order boundary value problems with integral boundary conditions have received widespread attention, there are relatively few researches on the third-order boundary value problems with integral boundary conditions, in references [11–13].By using the Guo-Krasnoselskii fixed point theorem, Zhao and Cunchen [11] investigated the existence and nonexistence of at least one or two monotone positive solutions for the following third-order boundary value problem with integral boundary conditions:
(1)u′′′t+ft,ut,u′t=0,t∈0,1,u0=u′′1=0,u′0=∫01gtutdt.By using the mixed monotone operator method, He and Xiaoling [13] proved the existence and uniqueness of positive solutions for the following third-order ordinary differential equations with integral boundary conditions:
(2)−u′′′t=ft,ut,uξt+gt,ut,t∈0,1,ξ∈0,1,u0=u′′0=0,u′1=∫01qtu′tdt.All the above works were done under the assumption that 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+ft,x,x′+gt,x=0,0<t<1,x0=0,x′0=∫01qtx′tdt,x′′1=0.Throughout, we assumeH1f:0,1×0,+∞2⟶0,+∞andg:0,1×0,+∞⟶0,+∞ are continuous; H2μ=∫01qtdt,t∈0,1andμ≠1,σ=∫01tqtdt,σ∈0,1.


## 2. Preliminary

LetY=C0,1 be the Banach space equipped with the norm x0=maxt∈0,1xt.Lemma 1 (see [12]).
Letμ≠1. Then for any yt∈C0,1, the problem
(4)x′′′t+yt=0,0<t<1,x0=0,x′0=∫01qtx′tdt,x′′1=0,has a unique solution
(5)xt=∫01G1t,s+t1−μ∫01G2τ,sqτdτysds,t∈0,1,where,
(6)G1t,s=122ts−s2,0≤s≤t≤1,t2,0≤t≤s≤1,G2t,s=12s,0≤s≤t≤1,t,0≤t≤s≤1,Gt,s=G1t,s+t1−μ∫01G2τ,sqτdτ.
By (5), we get x′t=∫01G2t,s+1/1−μ∫01G2τ,sqτdτysds.
Letμ<1,γ∈0,1.Lemma 2 (see [11]).
For anyt,s∈0,1×0,1, G1t,sandG2t,s have
(i)
t2G11,s≤G1t,s≤2s−s2/2(ii)
ts≤G2t,s≤sLemma 3.
Ify∈C0,1,yt≥0, then the unique solution xt of problem (1) satisfies
(7)xt≥0,t∈0,1,mint∈γ,1xt≥γ2x0.Proof.
By Lemma1, we get
(8)xt=∫01G1t,s+t1−μ∫01G2τ,sqτdτysds,t∈0,1.
By Lemma2, we get
(9)0≤xt≤∫012s−s22+11−μ∫01G2τ,sqτdτysds.
So,(10)x0≤∫012s−s22+11−μ∫01G2τ,sqτdτysds.
Forγ∈0,1, we have
(11)mint∈γ,1xt=mint∈γ,1∫01G1t,s+t1−μ∫01G2τ,sqτdτysds≥mint∈γ,1∫01t2G1t,s+t1−μ∫01G2τ,sqτdτysds≥γ2∫012s−s22+11−μ∫01G2τ,sqτdτysds≥γ2x0.
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 αλx=λαx,βλx=λβx, for x∈X,λ∈R, and x≤Mmaxαx,βx, for x∈X and αx≤αy for x,y∈K,x≤y, where M>0 is a constant.Theorem 4 (see [14]).
Letr2>r1>0,L>0 be constants and Ωi=x∈X:αx<ri,βx<L,i=1,2,(13) two bounded open sets in X. Set
(12)Di=x∈X:αx=ri,i=1,2.AssumeT:K⟶K is a completely continuous operator satisfying the following:(A1)
α
T
x
<
r
1
,
x
∈
D
1
⋂
K
;
α
T
x
>
r
2
,
x
∈
D
2
⋂
K(A2)
β
T
x
<
L
,
x
∈
K(A3) There is a p∈Ω2⋂K\0 such that αp≠0 and αx+λp≥αx, for all x∈K and λ≥0Then,T has at least one fixed point in Ω2\Ω¯1⋂K.


## 3. The Main Results

LetX=C10,1 be the Banach space equipped with the norm x=maxt∈0,1xt+maxt∈0,1x′t and K=x∈X:xt≥0,mint∈γ,1xt≥γ2x0 is a cone in X.Define two continuous convex functionalsαx=maxt∈0,1xt and βx=maxt∈0,1x′t, for each x∈X, and then x≤2maxαx,βx and αλx=λαx,βλx=λβx, for x∈X,λ∈R;αx≤αy for x,y∈K,x≤y.In the following, we denote(13)η0=13+11−μ∫01∫01G2τ,sqτdτds,η1=∫γ12s−s22+11−μ∫01G2τ,sqτdτds,η2=12+11−μ∫01∫01G2τ,sqτdτds.We will suppose that there areL>b>γ2b>c>0 such that ft,x,y+gt,x satisfies the following growth conditions: H3ft,x,y+gt,x<c/η0, for t,x,y∈0,1×0,c×−L,L,t,x∈0,1×0,c; H4ft,x,y+gt,x≥b/γ2η1, for t,x,y∈γ,1×γ2b,b×−L,L,t,x∈γ,1×γ2b,b; and H5ft,x,y+gt,s<L/η2, for t,x,y∈0,1×0,b×−L,L,t,x∈0,1×0,b.Let(14)f∗t,x,y=ft,x,y,t,x,y∈0,1×0,b×−∞,∞,ft,b,y,t,x,y∈0,1×b,∞×−∞,∞,f1t,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,∞.Let(15)g∗t,x=gt,x,t,x∈0,1×0,b,gt,b,t,x∈0,1×b,∞,g1t,x=g∗t,x,t,x∈0,1×0,∞,g∗t,x,t,x∈0,1×0,∞,g∗t,x,t,x∈0,1×0,∞.We denote(16)Txt=∫01G1t,s+t1−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds,Tx′t=∫01G2t,s+11−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds.Lemma 5.
SupposeH1 hold. Then, T:K⟶K is completely continuous.Proof.
Forx∈K, by Lemma 3, we have Tx≥0.
So, we can getTK⊂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⟶∞; then, there is a constant M0>0, such that maxt∈0,1xnt,x∗t,xn′t,x∗′t≤M0, then
(17)Txnt−Tx∗t=∫01Gt,sf1s,xn,xn′+g1s,xnds−∫01Gt,sf1s,x∗,x∗′+g1s,x∗ds≤∫01Gt,sf1s,xn,xn′+g1s,xn−f1s,x∗,x∗′+g1s,x∗ds,Txn′t−Tx∗t=∫01G2t,s+11−μ∫01G2τ,sqτdτf1s,x,xn′+g1s,xnds−∫01G2t,s+11−μ∫01G2τ,sqτdτ×f1s,x∗,x∗′+g1s,x∗ds≤∫01G2t,s+11−μ∫01G2τ,sqτdτf1s,x,xn′+g1s,xn−f1s,x∗,x∗′+g1s,x∗ds<∫01s+11−μ∫01G2τ,sqτdτf1s,x,xn′+g1s,xn−f1s,x∗,x∗′+g1s,x∗ds.
Iff is uniformly continuous on 0,1×−M0,M0×−M0,M0, we get
(18)Txn−Tx∗⟶0,n⟶∞.
Next, we show thatT:K⟶K is compact.
LetB⊂K be bounded, and then, there is M>0, such that x≤M. For x∈B, we have
(19)Txt=∫01G1t,s+t1−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds<∫012s−s22+11−μ∫01G2τ,sqτdτds×C∗,where C∗=maxf1t,x,x′+g1t,x;t∈0,1,x∈B.(20)Tx′t=∫01G2t,s+11−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds<∫01s+11−μ∫01G2τ,sqτdτds×C∗.
It is clear thatTB is a bounded set in K; because Gt,s is uniformly continuous on 0,1×0,1, for ε>0, there exists δ∈0,ε, such that G1t1,s−G1t2,s<ε,G2t1,s−G2t2,s<ε for t1,t2∈0,1,t1−t2<δ.
Forx∈B, we have
(21)Tx′t1−Tx′t2=∫01∂Gt,s∂tt=t1f1s,x,x′+g1s,xds−∫01∂Gt,s∂tt=t2f1s,x,x′+g1s,xds=∫01G2t1,s+11−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds−∫01G2t2,s+11−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds≤∫0t1s+11−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds−∫0t2s+11−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds≤12t1−t2t1+t2×C∗≤εC∗.
ThereforeTB is equicontinuous. Using the Arzela-Ascoli theorem, a standard proof yields T:K⟶K is completely continuous.Theorem 6.
SupposeH1-H5 hold. Then, BVP (1) has at least one positive solution xt satisfying
(22)c<αx<b,βx<L.Proof.
TakeΩ1=x∈X:xt<c,x′t<LandΩ2=x∈X:xt<b,x′t<L, two bounded open sets in X, and D1=x∈X:αx=candD2=x∈X:αx=b.
By Lemma5, T:K⟶K is completely continuous, and there is a p∈Ω2⋂K\0 such that αp≠0, for all u∈K and λ≥0.
By Lemma2 and H3, for x∈D1∩K,αx=c, we get
(23)αTx=maxt∈0,1∫01G1t,s+t1−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds≤∫012s−s22+11−σ∫01Gτ,sgτdτf1s,x,x′+g1s,xds=13+11−μ∫01∫01G2τ,sgτdτds×cη0=c.
By Lemma2, for x∈D2∩K,αx=b, there is xt≥γ2αx=γ2b,t∈γ,1.
So, byH4, we get
(24)αTx=maxt∈0,1∫01G1t,s+t1−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds≥∫01t2G11,s+t1−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds>γ2∫γ12s−s22+11−μ∫01G2τ,sqτdτds×bγ2η1=b.
ByH5, for x∈K, we have
(25)βTx=maxt∈0,1∫01G2t,s+t1−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds≤∫01s+11−μ∫01G2τ,sqτdτf1s,x,x′+g1s,xds<12+11−μ∫01∫01G2τ,sgτdτds×Lη2=L.
Theorem4 implies there is x∈Ω2\Ω¯1⋂K such that x=Tx. So, xt is a positive solution for BVP (1) satisfying
(26)c<αx<b,βx<L.
Thus, Theorem6 is completed.


## 4. Example

Example 1.
Consider the following boundary value problem(27)x′′′t+ft,x,x′+gt,x=0,0<t<1,x0=0,x′0=∫01qtx′tdt,x′′1=0,where,
(28)ft,x,y+gt,x=t3x+x+cosy,t,x,y∈0,1×0,2.2×−60306,60306,t,x∈0,1×0,2.22280t3x−2.2+420001x−2.2+11t15+2.2+cosy,t,x,y∈0,1×2.2,2.3×−60306,60306,t,x∈0,1×2.2,2.3t3231−x+180.039x+231+cosy,t,x,y∈0,1×2.3,231×−60306,60306,t,x∈0,1×2.3,231.
In this problem, we know thatqt=t2; then, we can get μ=∫01t2dt=1/3. Choose γ=1/10∈0,1/2, then ρ=4δ21−δ=7/128.Furthermore, we obtain(29)η0=67120,η1≈0.551,η2=2940.If we takec=2.2,b=231,andL=60306, then we get γ2b≈2.31>2.3.
(30)ft,x,y+gt,x=t3x+x+cosy≤2.17<cη0≈3.94,fort,x,y∈0,1×0,2.2×−60306,60306,t,x∈0,1×0,2.2,ft,x,y+gt,x=t3231−x+180.039x+231+cosy>42011.7>bγ2η1≈42000,fort,x,y∈0,1×2.3,231×−60306,60306,t,x∈0,1×2.3,231,ft,x,y+gt,xt3231−x+180.039x+231+cosy<83179.02<Lη2≈83180.69,fort,x,y∈0,1×2.3,231×−60306,60306,t,x∈0,1×2.3,231.Then, all the conditions of Theorem6 are satisfied. Therefore, by Theorem 6 we know that boundary value problem (1) has at least one positive solution xt satisfying
(31)2.2<αx<231,βx<60306.



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