# Positive Solutions for a Nonlinear Higher Order Differential System with Coupled Integral Boundary Conditions

Authors: Yaohong Li; Haiyan Zhang
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We investigate the existence of positive solutions for a nonlinear higher order differential system, where the differential system is coupled not only in the differential system but also through the boundary conditions. By constructing a special cone and using the fixed point theorem of cone expansion and compression of norm type, the existence of single and multiple positive solutions is established. As an application, we give some examples to demonstrate our results.


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## Body

## 1. Introduction

In this paper, we consider the following nonlinear higher order differential system with coupled integral boundary conditions:(1)u(n)(t)+a1(t)f1(t,u(t),v(t))=0,t∈(0,1),v(n)(t)+a2(t)f2(t,u(t),v(t))=0,t∈(0,1),u(k)(0)=v(k)(0)=0,k=0,1,…,n-2,u(1)=α[v],v(1)=β[u],
where fi∈C([0,1]×[0,+∞)×[0,+∞),[0,+∞)), ai∈C([0,1],[0,+∞)), i=1,2, n≥3, α[v], β[u] are bounded linear functions on C[0,1] given by
(2)α[v]=∫01v(t)dA(t),β[u]=∫01u(t)dB(t),
involving Stieltjes integrals. In particular, A,B are functionals of  bounded variation with positive measures.In recent years, there were many works to be done for a variety of  nonlinear higher order ordinary differential system. However, most papers only focus on paying attention to the differential system with uncoupled boundary conditions (see [1–5] and the reference therein). Coupled boundary conditions arise in the study of reaction-diffusion equations and Sturm-Liouville problems (see [6]) and have wide applications in various fields of sciences and engineering, for example, the heat equation [7, 8].In a recent article [9], by applying a nonlinear alternative of  Leray-Schauder type and Guo-Krasnoselskii’s fixed point theorem on cone, the authors established the existence of multiple positive solutions of the following system with four-point coupled boundary conditions:
(3)D0+αu(t)+λf(t,u(t),v(t))=0,t∈(0,1),λ>0,D0+αu(t)+λh(t,u(t),v(t))=0,t∈(0,1),u(i)(0)=v(i)(0)=0,0≤i≤n-2,u(1)=av(ξ),v(1)=bu(η),
where D0+αu is the Riemann-Liouville’s fractional derivative.In [10], by using fixed point index theory, Yang studied the following system with uncoupled boundary conditions:
(4)u′′(t)+f1(t,u(t),v(t))=0,t∈(0,1),v′′(t)+f2(t,u(t),v(t))=0,t∈(0,1),u(0)=0,u(1)=β[u],v(0)=0,v(1)=β[v],
where β[·] are linear functionals defined by Stieltjes integrals.The work of above-mentioned papers and wide applications of coupled boundary value conditions motivate us to study the system (1). Further, the system is coupled not only in the differential system but also through the boundary conditions. By constructing a special cone and using the fixed point theorem on cone expansion and compression, the existence of single and multiple positive solutions is established.


## 2. Preliminaries

LetE=C[0,1]]; we write ∥u∥=max⁡{|u(t)|:t∈[0,1]}. Clearly, (E,∥·∥) is a Banach space. For each (u,v)∈E×E, we write ∥(u,v)∥=∥u∥+∥v∥. Define
(5)P={u∈E:u(t)≥0,t∈[0,1]},Q={(u,v)∈P×P:min⁡a≤t≤b(u,v)kkkkk=defmin⁡a≤t≤b(u(t)+v(t))≥γ∥(u,v)∥},
where [a,b] is some subset of (0,1),0<γ≤(an-1ν/ρ); consider
(6)ρ=max⁡⁡{α[tn-1]kβ[1]+1,β[tn-1]kα[1]+1,kkkkkkkkkβ[1]k,α[1]kβ[tn-1]k},ν=min⁡{α[tn-1]kβ[γn(t)],β[tn-1]kα[γn(t)],kkkkkkkkkkβ[γn(t)]k,α[γn(t)]kβ[tn-1]k},α[tn-1]=∫01tn-1dA(t)>0,β[tn-1]=∫01tn-1dB(t)>0,k=1-α[tn-1]β[tn-1]>0,
where γn(t) is defined by the following Lemma 2. Clearly, (E×E,∥·∥) is a Banach space and P is a cone of E.Lemma 1.
Letu,v∈E. Then differential system
(7)u(n)(t)+x(t)=0,t∈(0,1),v(n)(t)+y(t)=0,t∈(0,1),u(k)(0)=v(k)(0)=0,k=0,1,…,n-2,u(1)=α[v],v(1)=β[u],
has the following integral representation
(8)u(t)=∫01F1(t,s)x(s)ds+∫01G1(t,s)y(s)ds,v(t)=∫01F2(t,s)y(s)ds+∫01G2(t,s)x(s)ds,
where
(9)F1(t,s)=α[tn-1]tn-1k∫01Kn(ξ,s)dB(ξ)+Kn(t,s),G1(t,s)=tn-1k∫01Kn(ξ,s)dA(ξ),F2(t,s)=β[tn-1]tn-1k∫01Kn(ξ,s)dA(ξ)+Kn(t,s),G2(t,s)=tn-1k∫01Kn(ξ,s)dB(ξ),(10)Kn(t,s)=1(n-1)!{tn-1(1-s)n-1-(t-s)n-1,0≤s≤t≤1,tn-1(1-s)n-1,0≤t≤s≤1.Proof.
By Taylor’s formula, we have(11)u(t)=u(0)+tu′(0)+⋯+tn-1(n-1)!u(n-1)(0)+1(n-1)!∫0t(t-s)n-1u(n)(s)ds,v(t)=v(0)+tv′(0)+⋯+tn-1(n-1)!v(n-1)(0)+1(n-1)!∫0t(t-s)n-1v(n)(s)ds.
So, we reduce the equation of problems (7) to the following equivalent integral equation:
(12)u(t)=-1(n-1)!∫0t(t-s)n-1x(s)ds+tn-1(n-1)!u(n-1)(0),v(t)=-1(n-1)!∫0t(t-s)n-1y(s)ds+tn-1(n-1)!v(n-1)(0).
Let t=1; we have
(13)u(n-1)(0)=∫01(1-s)n-1x(s)ds+(n-1)!u(1),v(n-1)(0)=∫01‍(1-s)n-1y(s)ds+(n-1)!v(1).
By substituting u(n-1)(0) and v(n-1)(0) into (12), we have
(14)u(t)=-1(n-1)!∫0t(t-s)n-1x(s)ds+1(n-1)!∫01tn-1(1-s)n-1x(s)ds+tn-1u(1)lll=1(n-1)!∫0t[tn-1(1-s)n-1-(t-s)n-1]x(s)ds+1(n-1)!∫t1tn-1(1-s)n-1x(s)ds+tn-1u(1)lll=∫01Kn(t,s)x(s)ds+tn-1u(1),v(t)=∫01Kn(t,s)y(s)ds+tn-1v(1);
that is,
(15)u(t)=∫01Kn(t,s)x(s)ds+tn-1u(1),v(t)=∫01Kn(t,s)y(s)ds+tn-1v(1).
By applying β and α to (15), combined with the conditions u(1)=α[v], v(1)=β[u], respectively, we obtain
(16)u(1)=∫01v(t)dA(t)=∬01Kn(t,s)y(s)dsdA(t)+u(1)∫01tn-1dA(t),v(1)=∫01u(t)dB(t)=∬01Kn(t,s)x(s)dsdB(t)+u(1)∫01tn-1dB(t).
Therefore
(17)(-β[tn-1]11-α[tn-1])(u(1)v(1))=(∬01Kn(t,s)x(s)dsdB(t)∬01Kn(t,s)y(s)dsdA(t)),
and so
(18)(u(1)v(1))=1k(α[tn-1]11β[tn-1])×(∬01Kn(t,s)x(s)dsdB(t)∬01Kn(t,s)y(s)dsdA(t)).
By substituting (18) into (15), we obtain
(19)u(t)=α[tn-1]tn-1k∬01Kn(t,s)x(s)dsdB(t)+tn-1k∬01Kn(t,s)y(s)dsdA(t)+∫01Kn(t,s)x(s)ds,v(t)=tn-1k∬01Kn(t,s)x(s)dsdB(t)+β[tn-1]tn-1k∬01Kn(t,s)y(s)dsdA(t)+∫01Kn(t,s)y(s)ds.
Therefore
(20)u(t)=∫01[α[tn-1]tn-1k∫01Kn(ξ,s)dB(ξ)]x(s)ds+∫01[tn-1k∫01Kn(ξ,s)dA(ξ)]y(s)ds+∫01Kn(t,s)x(s)ds,v(t)=∫01[tn-1k∫01Kn(ξ,s)dB(ξ)]x(s)ds+∫01[β[tn-1]tn-1k∫01Kn(ξ,s)dA(ξ)]y(s)ds+∫01Kn(t,s)y(s)ds.
which is equivalent to system (8).Lemma 2 (see [11]).
The continuous functionKn(t,s) has the following properties: (i)
0≤Kn(t,s)≤Kn(s), for all t,s∈[0,1], where Kn(s)=s(1-s)n-1/(n-2)!;(ii)
Kn(t,s)≥γn(t)Kn(s), for all t,s∈[0,1], where γn(t)=(1/(n-1))min⁡{tn-1,(1-t)tn-2}.Remark 3.
By combining (i) and (ii), we can easily see(21)Kn(s)≥Kn(t,s)≥γn(t)Kn(s),∀t,s∈[0,1].Remark 4.
From Remark3, and (9), for t,s∈[0,1], we have
(22)Fi(t,s)≤ρKn(s),Gi(t,s)≤ρKn(s),i=1,2,Fi(t,s)≥νtn-1Kn(s),Gi(t,s)≥νtn-1Kn(s),i=1,2.
Define the operatorT:Q→P×P by
(23)T(u,v)=(T1(u,v),T2(u,v)),
where operators T1,T2:Q→P are defined by
(24)T1(u,v)(t)=∫01F1(t,s)a1(s)f1(s,u(s),v(s))ds+∫01G1(t,s)a2(s)f2(s,u(s),v(s))ds,kkkkkkkkkkkkkkkkkkkkkkklllt∈[0,1],(25)T2(u,v)(t)=∫01F2(t,s)a2(s)f2(s,u(s),v(s))ds+∫01G2(t,s)a1(s)f1(s,u(s),v(s))ds,kkkkkkkkkkkkkkkkkkkkkkkkkt∈[0,1].
Moreover, by Lemma 1, if (u,v)∈Q is a fixed point of the operator T, then (u,v) is a solution of the system (1).Lemma 5.
The operatorT:Q→Q is completely continuous.Proof.
By Remark4, for s∈[0,1], we obtain
(26)min⁡t∈[a,b]Fi(t,s)≥νan-1Kn(s),min⁡t∈[a,b]Gi(t,s)≥νan-1Kn(s),i=1,2.
Therefore, By Lemma 2 and Remark 3, for (u,v)∈P, we have
(27)∥T1(u,v)∥≤ρ∫01Kn(s)a1(s)f1(s,u(s),v(s))ds+ρ∫01Kn(s)a2(s)f2(s,u(s),v(s))ds.
Moreover, we have
(28)min⁡t∈[a,b]T1(u,v)(t)=min⁡t∈[a,b][∫01F1(t,s)a1(s)f1(s,u(s),v(s))dskkkkkkkkk+∫01G1(t,s)a2(s)f2(s,u(s),v(s))ds]≥νan-1[∫01Kn(s)a1(s)f1(s,u(s),v(s))dskkkkkkkkkk+∫01Kn(s)a2(s)f2(s,u(s),v(s))ds]≥γ∥T1(u,v)∥.
In the same way, we can prove that
(29)min⁡t∈[a,b]T2(u,v)(t)≥γ∥T2(u,v)∥.
Thus
(30)min⁡t∈[a,b]T(u,v)=min⁡t∈[a,b](T1(u,v)(t)+T2(u,v)(t))≥γ∥T1(u,v)∥+γ∥T2(u,v)∥=γ∥T(u,v)∥.
Then operator T:Q→Q is continuous since Kn(t,s), f1(t,u,v), f2(t,u,v), a1(t), a2(t) are continuous. Standard applications of Arzelà-Ascoli theorem; it is easy to prove that operator T:Q→Q is completely continuous.Lemma 6 (see [12]).
SupposeE is a real Banach space and P is cone in E, and let Ω1, Ω2 be bounded open sets in E such that θ∈Ω1, Ω¯1⊂Ω2. Let operator T:P∩(Ω¯2∖Ω1)→P be completely continuous. Suppose that one of the following two conditions holds:(i)
∥Tu∥≤∥u∥, for all u∈P∩∂Ω1; ∥Tu∥≥∥u∥, for all u∈P∩∂Ω2,(ii)
∥Tu∥≥∥u∥, for all u∈P∩∂Ω1; ∥Tu∥≤∥u∥, for all u∈P∩∂Ω2,then operator T has at least one fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Results

In this section, we show the existence of positive solutions to the system (1). For convenience, we first introduce the following notations:
(31)f1λ=liminf⁡u+v→λmin⁡t∈[0,1]f1(t,u,v)u+v,f1λ=limsup⁡u+v→λmax⁡t∈[0,1]f1(t,u,v)u+v,f2λ=liminf⁡u+v→λmin⁡t∈[0,1]f2(t,u,v)u+v,f2λ=limsup⁡u+v→λmax⁡t∈[0,1]f2(t,u,v)u+v,
where λ=0 or ∞. Let r=min⁡{r1,r2}, R=max⁡{R1,R2}, where
(32)ri=(4ρ∫01ai(s)Kn(s)ds)-1,Ri=(4γνan-1∫abai(s)Kn(s)ds)-1,i=1,2.Theorem 7.
Iff10,f20∈[0,r) and f1∞,f2∞∈(R,+∞], then system (1) has at least one positive solution.Proof.
At first, it follows from the assumptionf10,f20∈[0,r) that there exists μ1>0 and a sufficiently small ε1>0 such that
(33)f1(t,u,v)≤(f10+ε1)(u+v),∀t∈[0,1],u+v≤μ1,f2(t,u,v)≤(f20+ε1)(u+v),∀t∈[0,1],u+v≤μ1,
where ε1 satisfies f10+ε1≤r and f20+ε1≤r.
SetΩ1={(u,v)∈P×P:∥(u,v)∥<μ1}. For any (u,v)∈∂Ω1∩Q, by (24), (25), and (33), we have
(34)∥T1(u,v)∥≤ρ∫01Kn(s)a1(s)f1(s,u(s),v(s))ds+ρ∫01Kn(s)a2(s)f2(s,u(s),v(s))ds≤ρ[(f10+ε1)∫01Kn(s)a1(s)dskkk+(f20+ε1)∫01Kn(s)a2(s)ds]·∥(u,v)∥≤ρ[14ρ+14ρ]∥(u,v)∥≤12∥(u,v)∥,∥T2(u,v)∥≤ρ∫01Kn(s)a2(s)f2(s,u(s),v(s))ds+ρ∫01Kn(s)a1(s)f1(s,u(s),v(s))ds≤ρ[(f20+ε1)∫01Kn(s)a2(s)dskkkk+(f10+ε1)∫01Kn(s)a1(s)ds]·∥(u,v)∥≤ρ[14ρ+14ρ]∥(u,v)∥≤12∥(u,v)∥.
Therefore
(35)∥T(u,v)∥=∥T1(u,v)∥+∥T2(u,v)∥≤∥(u,v)∥,kkkkkkkkkkkkkfor(u,v)∈∂Ω1∩Q.
Further, it follows from the the assumption f1∞,f2∞∈(R,+∞] that there exists l>μ1>0 and a sufficiently small ε2>0 such that
(36)f1(t,u,v)≥(f1∞-ε2)(u+v),∀t∈[0,1],u+v≥l,f2(t,u,v)≥(f2∞-ε2)(u+v),∀t∈[0,1],u+v≥l,
where ε2 satisfies f1∞-ε2≥R and f2∞-ε2≥R. Let μ2=max⁡{2μ1,l/γ}}; set Ω2={(u,v)∈P×P:∥(u,v)∥<μ2}. Then (u,v)∈∂Ω2∩Q implies that min⁡t∈[a,b]⁡(u,v)≥γ∥(u,v)∥=γμ2≥l. So, by (24), (25), and (36), we have
(37)min⁡t∈[a,b]T1(u,v)(t)≥νan-1[∫abKn(s)a1(s)f1(s,u(s),v(s))dskkkkkkkkkkk+∫abKn(s)a2(s)f2(s,u(s),v(s))ds]≥γνan-1[(f1∞-ε2)∫abKn(s)a1(s)dskkkkkkkkk+(f2∞-ε2)∫abKn(s)a2(s)ds]·∥(u,v)∥≥γνan-1[14γνan-1+14γνan-1]∥(u,v)∥≥12∥(u,v)∥,min⁡t∈[a,b]T2(u,v)(t)≥νan-1[∫abKn(s)a2(s)f2(s,u(s),v(s))dskkkkkkkkkk+∫abKn(s)a1(s)f1(s,u(s),v(s))ds]≥γνan-1[(f2∞-ε2)∫abKn(s)a2(s)dskkkkkkkkkk+(f1∞-ε1)∫abKn(s)a1(s)ds]·∥(u,v)∥≥12∥(u,v)∥.
Therefore,
(38)∥T(u,v)∥=∥T1(u,v)∥+∥T2(u,v)∥≥min⁡t∈[a,b]T1(u,v)(t)+min⁡t∈[a,b]T2(u,v)(t)≥∥(u,v)∥,(u,v)∈∂Ω2∩Q.
By applying Lemmas 5 and 6 to (35) and (38), it follows that operator T has at least one fixed point (u,v) in Q∩(Ω¯2∖Ω1). This means that system (1) has at least one positive solution (u,v).Using similar arguments as those used in the proof of Theorem7, we can also obtain the following result.Theorem 8.
Iff1∞,f2∞∈[0,r) and f10,f20∈(R,+∞], then system (1) has at least one positive solution.Next we discuss the multiplicity of positive solutions for system (1).Theorem 9.
Iff10,f2∞∈(4R,+∞], and there exist an m>0 such that f1(t,u,v),f2(t,u,v)∈(0,mr), for for all t∈[0,1],(u,v)∈∂Ω3∩Q, where Ω3={(u,v)∈P×P,∥(u,v)∥<m}, then system (1) has at least two positive solutions.Proof.
At first, it follows from the assumptionf10∈(4R,+∞],+∞] that there exists an 0<m1<m and a sufficiently small ε4>0 such that
(39)f1(t,u,v)≥(f10-ε4)(u+v),∀t∈[0,1],u+v≤m1,
where ε4 satisfies f10-ε4≥4R.
SetΩ4={(u,v)∈P×P:∥(u,v)∥<m1} and (u,v)∈∂Ω4∩Q. By (24), we have
(40)∥T(u,v)∥≥min⁡t∈[a,b]T(u,v)(t)≥min⁡t∈[a,b]T1(u,v)(t)≥νan-1∫abKn(s)a1(s)f1(s,u(s),v(s))ds≥γνan-1(f10-ε4)∫abKn(s)a1(s)ds·∥(u,v)∥≥∥(u,v)∥.
Further, by using f2∞∈(4R,+∞], there exists m2>m>0 and a sufficiently small ε5>0 such that
(41)f2(t,u,v)≥(f2∞-ε5)(u+v),∀t∈[0,1],u+v≥m2,
where ε5 satisfies f2∞-ε5≥4R. Set Ω5={(u,v)∈P×P:∥(u,v)∥<m3}, where m3>m2. For all (u,v)∈∂Ω5∩Q, by (24), we have
(42)∥T(u,v)∥≥min⁡t∈[a,b]T(u,v)(t)≥min⁡t∈[a,b]T1(u,v)(t)≥νan-1∫abKn(s)a2(s)f2(s,u(s),v(s))ds≥γνan-1(f2∞-ε5)∫abKn(s)a2(s)ds·∥(u,v)∥≥∥(u,v)∥.
By assumption, for for all (u,v)∈∂Ω3∩Q, we have
(43)∥T(u,v)∥=∥T1(u,v)∥+∥T2(u,v)∥≤2rmρ[∫01Kn(s)a1(s)ds+∫01Kn(s)a2(s)ds]≤m=∥(u,v)∥.
From (40)–(43), it is easy to know that two conditions of Lemma 6 are both satisfied. By applying Lemmas 5 and 6 to (40)–(43), it follows that operator T has at least a fixed point (u1,v1)∈Q∩(Ω¯3∖Ω4) and a fixed point (u2,v2)∈Q∩(Ω¯5∖Ω3). Both are positive solutions of system (1) and satisfy m1≤∥(u1,v1)∥<m<∥(u2,v2)∥≤m3. This means system (1) has at least two positive solutions.Similarly, we have the following results.Theorem 10.
Iff10,f1∞∈(4R,+∞], and there exist an m>0 such that f1(t,u,v),f2(t,u,v)∈(0,mr), for for all t∈[0,1],(u,v)∈∂Ω6∩Q, where Ω6={(u,v)∈P×P,∥(u,v)∥<m}, then system (1) has at least two positive solutions.Theorem 11.
Iff20,f1∞∈(4R,+∞], and there exist an m>0 such that f1(t,u,v),f2(t,u,v)∈(0,mr), for all t∈[0,1],(u,v)∈∂Ω7∩Q, where Ω7={(u,v)∈P×P,∥(u,v)∥<m}. Then system (1) has at least two positive solutions.Theorem 12.
Iff20,f2∞∈(4R,+∞], and there exist an m>0 such that f1(t,u,v),f2(t,u,v)∈(0,mr), for all t∈[0,1],(u,v)∈∂Ω8∩Q, where Ω8={(u,v)∈P×P,∥(u,v)∥<m}. Then system (1) has at least two positive solutions.


## 4. Some Examples

In order to illustrate our result, we consider some examples.Example 1.
Consider the following system(44)-u(3)(t)=1+t8[(u2+v2)+λsin(u+v)],t∈(0,1),-v(3)(t)=14[(u2+v2)2+e-(u2+v2)],t∈(0,1),u(0)=u′(0)=0,u(1)=2∫01v(t)dt,v(0)=v′(0)=0,v(1)=∫01u(t)dt,
where n=3, a1(t)=(1+t)/8, a2(t)=1/4, f1(t,u,v)=(u2+v2)+λsin(u+v), f2(t,u,v)=(u2+v2)2+e-(u2+v2), A(t)=2t, B(t)=t, Kn(s)=s(1-s)2, γn(t)=(1/2)min⁡{t2,(1-t)t}. By direct calculation, we can obtain that ρ=18/7, ν=3/56, r=14/15. Choose λ∈[0,14/15)); then conditions of  Theorem 7 are satisfied. This means that system (44) has at least one position solution.Example 2.
Consider the following system(45)-u(3)(t)=1+t16(u2+v2)(1/3),t∈(0,1),-v(3)(t)=132(u2+v2),t∈(0,1),u(0)=u′(0)=0,u(1)=2∫01v(t)dt,v(0)=v′(0)=0,v(1)=∫01u(t)dt,
where n=3, a1(t)=(1+t)/16, a2(t)=1/32, f1(t,u,v)=(u2+v2)1/3, f2(t,u,v)=(u2+v2), A(t)=2t,  B(t)=t,  Kn(s)=s(1-s)2, γn(t)=(1/2)min⁡{t2,(1-t)t}. By direct calculation, we can obtain that ρ=(18/7), υ=(3/56), r=(112/15). Choose m=2; then conditions of Theorem 9 are satisfied. This means that system (45) has at least two position solutions.



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