# Multiplicity Results for Positive Solutions to Differential Systems of Singular Coupled Integral Boundary Value Problems

Authors: Yujun Cui
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

By constructing a special cone and using a fixed-point theorem in cone, this paper investigates the existence of multiple solutions of coupled integral boundary value problems for a nonlinear singular differential system.


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## Body

## 1. Introduction

In recent years, singular uncoupled boundary value problems to differential systems have been studied widely and there are many excellent results (see [1–18] and references therein). Naturally we hope there are the same excellent results on singular uncoupled boundary value problems to differential systems with coupled boundary conditions. Many researchers put their efforts to study the existence of solutions for differential systems with coupled boundary conditions (see [19–30] and references therein).In [19], Asif and Khan addressed the question of the existence of coupled four-point boundary value conditions (1)-x′′t=f1t,xt,yt,t∈0,1,-y′′t=f2t,xt,yt,t∈0,1,x0=y0=0,x1=αyξ,y1=βxη,where the parameters α, β, ξ, and η satisfy ξ,η∈(0,1), 0<αβξη<1. The main tool in [19] is the Guo-Krasnosel’skiĭ fixed-point theorem.In [29], the authors studied the following nonlinear semipositone fractional differential equation with four-point coupled boundary value problem: (2)D0+αu+λft,u,v=0,t∈0,1,λ>0,D0+αv+λgt,u,v=0,ui0=vi0=0,0≤i≤n-2,u1=avξ,v1=buη,where λ is a parameter, a, b, ξ, and η satisfy ξ,η∈(0,1), 0<abξη<1, α∈(n-1,n] is a real number and n≥3, and D0+αu is Riemann-Liouville’s fractional derivative. The existence of positive solutions is established by using a nonlinear alternative of Leray-Schauder type and Guo-Krasnosel’skiĭ fixed-point theorem in a cone.In [20], Cui and Sun, using fixed-point index theory, studied the existence of positive solutions for superlinear differential system (3)-x′′t=f1t,xt,yt,t∈0,1,-y′′t=f2t,xt,yt,t∈0,1,x0=y0=0,x1=αy,y1=βx,where α[x] and β[x] are bounded linear functionals on C[0,1] given by (4)αx=∫01xtdAt,βx=∫01xtdBt,involving Stieltjes integrals; in particular, A and B are functions of bounded variation with positive measures.We should note that the nonlinear terms in two equations for the above problems have the same features. For instance, the nonlinear terms in two equations are both superlinear [20, 29] or both sublinear [19]. However, to the best of our knowledge, only a few papers discuss differential system under the case that the nonlinear terms of the system have different behaviors. Motivated by [19, 20, 29], the purpose of this paper is to establish the existence of multiple positive solutions for differential system with coupled integral boundary value problems (3) when f1 is superlinear in x and y and f2 is sublinear in x and y. Also suppose that f1(t,x,y) may be singular at t=0, t=1 and f2(t,x,y) may be singular at t=0, t=1, x=0, and y=0. Our main features are threefold. Firstly, our study is on singular nonlinear differential systems with general boundary value conditions. Secondly, f2 is allowed to be not only singular at t=0 and 1 but also singular at x=0 and y=0. Finally, a special cone is constructed to overcome difficulties due to singularities of nonlinear term.In the rest of this section, let us list the following assumptions:(
H
1
)
κ
1
>
0, κ2>0, and κ>0, where(5)κ1=αt=∫01tdAt,κ2=βt=∫01tdBt,κ=1-κ1κ2.(
H
2
)  f1∈C((0,1)×[0,∞)2,[0,∞)) satisfy (6)0<∫01s1-sf1s,1,1ds<+∞,and there exist constants λ1j, μ1j (0<λ1j≤μ1j<1, j=1,2, λ11+λ12>1) such that, for t∈(0,1), x,y∈(0,∞), (7)cμ11f1t,x,y≤f1t,cx,y≤cλ11f1t,x,y,cμ12f1t,x,y≤f1t,x,cy≤cλ12f1t,x,y,if0<c≤1.(
H
3
)  f2∈C((0,1)×(0,∞)2,[0,∞)) and there exist constants λ2j, μ2j  (-∞<λ2j≤μ2j<1,  j=1,2,  μ21+μ22<1) such that, for t∈(0,1), x,y∈(0,∞), (8)cμ21f2t,x,y≤f2t,cx,y≤cλ21f2t,x,y,cμ22f2t,x,y≤f2t,x,cy≤cλ22f2t,x,y,if0<c≤1.And they satisfy one of the following conditions:(
H
31
)  λ21>0, λ22>0, 0<∫01s(1-s)f2(s,1,1)ds<+∞, ρ∫01s(1-s)f1(s,1,1)ds+ρ∫01s(1-s)f2(s,1,1)ds<1.(
H
32
)  μ21<0, μ22<0, 0<∫01s(1-s)f2(s,s,s)ds<+∞, ρ∫01s(1-s)f1(s,1,1)ds+ργμ21+μ22∫01s(1-s)f2(s,s,s)ds<1.(
H
33
)  λ21>0, μ22<0, 0<∫01s(1-s)f2(s,1,s)ds<+∞, ρ∫01s(1-s)f1(s,1,1)ds+ργμ22∫01s(1-s)f2(s,1,s)ds<1.(
H
34
)  μ21<0, λ22>0, 0<∫01s(1-s)f2(s,s,1)ds<+∞.  ρ∫01s(1-s)f1(s,1,1)ds+ργμ21∫01s(1-s)f2(s,s,1)ds<1, where (9)0<γ=νρ<1,(10)ρ=maxαtκβ1+1,βtκα1+1,1κβ1,1κα1,ν=minαtκβt1-t,βtκαt1-t,1κβt1-t,1καt1-t.Remark 1.
Equations (7)-(8) imply (11)cλi1fit,x,y≤fit,cx,y≤cμi1fit,x,y,cλi2fit,x,y≤fit,x,cy≤cμi2fit,x,y,ifc≥1,i=1,2.Conversely, (11) implies (7)-(8).Remark 2.
Equation (7) implies (12)f1t,x1,y1≤f1t,x2,y2,if0<x1≤x2,0<y1≤y2.Remark 3.
(
H
31
) implies that f2(t,x,y) is nondecreasing in x and y.
(
H
32
) implies that f2(t,x,y) is nonincreasing in x and y.
(
H
33
) implies that f2(t,x,y) is nondecreasing in x and nonincreasing in y.
(
H
34
) implies that f2(t,x,y) is nonincreasing in x and nondecreasing in y.


## 2. Main Results

For eachu∈E≔C[0,1], we write u=max{ut:t∈[0,1]}. Clearly, (E,·) is a Banach space. Similarly, for each (x,y)∈E×E, we write x,y1=max{x,y}. For any real constant r>0, define Ωr={(x,y)∈E×E:x,y1<r}. Define (13)P=x,y∈E×E:xt≥γtx,y1,yt≥γtx,y1,t∈0,1,where γ is given by (9). Clearly, (E×E,·1) is a Banach space and P is a cone of E×E.Remark 4.
The coneP defined by (13) is completely different from the cone used in the uncoupled boundary value problems. This means that the cone P has the following property:(14)ifx,y∈P∖θ,thenγtx,y1≤xt,yt≤x,y1,fort∈0,1,which is crucial in the definition of T and in the proof of Lemma 9.Our main result is the following theorems.Theorem 5.
Assume that(H1), (H2), and (H3) are satisfied. Then differential system (3) has at least two positive solutions (x1,y1),(x2,y2)∈C[0,1]×C[0,1] such that 0<x1,y11<1<x2,y21.Note. We need only to prove this theorem under condition (H33), since the proof is similar when (H31) or (H32) or (H34) is satisfied.The proof of Theorem5 is based on the following theorem in [31].Lemma 6.
LetE be a Banach space and P a cone in E. Suppose that Ω1 and Ω2 are two bounded open subsets of E with θ∈Ω1, Ω1¯⊂Ω2. Let operator T:P∩(Ω2¯∖Ω1)→P be completely continuous. Suppose that one of the two conditions (15)iTx≤x,∀x∈P∩∂Ω1,Tx≥x,∀x∈P∩∂Ω2,iiTx≥x,∀x∈P∩∂Ω1,Tx≤x,∀x∈P∩∂Ω2is satisfied. Then T has a fixed point in P∩(Ω2¯∖Ω1).Lemma 7 (see [20]).
Assume that(H1) holds. Let u,v∈E; then the system of BVPs (16)-x′′t=ut,-y′′t=vt,t∈0,1,x0=y0=0,x1=αy,y1=βxhas integral representation(17)xt=∫01G1t,susds+∫01H1t,svsds,yt=∫01G2t,svsds+∫01H2t,susds,where (18)kt,s=t1-s,0≤t≤s≤1,s1-t,0≤s≤t≤1,G1t,s=k1tκ∫01ks,τdBτ+kt,s,H1t,s=tκ∫01ks,τdAτ,G2t,s=k2tκ∫01ks,τdAτ+kt,s,H2t,s=tκ∫01ks,τdBτ.Remark 8 (see [20]).
From (13) and (H1), for t∈[0,1], we have(19)Git,s≤ρs1-s,Hit,s≤ρs1-s,Git,s≥νts1-s,Hit,s≥νts1-s,i=1,2.
Define an operatorT:P∖{θ}→P by (20)Tx,y=T1x,y,T2x,y,where operators T1,T2:P∖{θ}→Q={u∈E∣u(t)≥0,t∈[0,1]} are defined by(21)T1x,yt=∫01G1t,sf1s,xs,ysds+∫01H1t,sf2s,xs,ysds,T2x,yt=∫01G2t,sf2s,xs,ysds+∫01H2t,sf1s,xs,ysds,t∈0,1.Lemma 9.
Assume that(H1), (H2), and (H3) hold. Then, for any 0<a<b<+∞, T:P∩(Ωb¯∖Ωa)→P is a completely continuous operator.Proof.
For(x,y)∈P∩(Ωb¯∖Ωa), let c be a positive number such that x,y1/c<1 and c>1. From (H2), (H3), and Remark 2, we have(22)f1t,xt,yt≤f1t,c,c≤cμ11+μ12f1t,1,1,f2t,xt,yt≤f2t,c,γx,y1t≤cμ21f2t,1,min⁡γx,y1,1t≤cμ21min⁡γa,1λ22f2t,1,t.Hence for any t∈[0,1], by Remark 8, we get(23)Tix,yt≤ρ∫01s1-sf1s,xs,ysds+ρ∫01s1-sf2s,xs,ysds≤ρ∫01s1-sf1s,c,cds+ρ∫01s1-sf2s,c,γsx,y1ds≤ρcμ11+μ12∫01s1-sf1s,1,1ds+ρcμ21min⁡γa,1λ22∫01s1-sf2s,1,sds,i=1,2.Thus, T is well defined on P∩(Ωb¯∖Ωa).
Next we show thatT(P∩(Ωb¯∖Ωa))⊂P. By Remark 8, for τ,t,s∈[0,1], we obtain (24)Git,s≥γtGiτ,s,Hit,s≥γtHiτ,s,i=1,2,H1t,s≥γtG2τ,s,G1t,s≥γtH2τ,s,H2t,s≥γtG1τ,s,G2t,s≥γtH1τ,s.Hence, for (x,y)∈P∩(Ωb¯∖Ωa), t∈[0,1], we have(25)T1x,yt=∫01G1t,sf1s,xs,ysds+∫01H1t,sf2s,xs,ysds≥γt∫01G1τ,sf1s,xs,ysds+γt∫01H1τ,sf2s,xs,ysds=γtT1x,yτ,T1x,yt=∫01G1t,sf1s,xs,ysds+∫01H1t,sf2s,xs,ysds≥γt∫01H2τ,sf1s,xs,ysds+γt∫01G2τ,sf2s,xs,ysds=γtT2x,yτ.Then T1(x,y)(t)≥γtT1(x,y) and T1(x,y)(t)≥γtT2(x,y), that is, T1(x,y)(t)≥γtT1x,y,T2x,y1. In the same way, we can prove that T2(x,y)(t)≥γtT1x,y,T2x,y1. Therefore, T(P∩(Ωb¯∖Ωa))⊂P.
Moreover,T:P∩(Ωb¯∖Ωa)→P is a completely continuous operator. This is a standard textbook result using Ascoli-Arzela theorem (see, e.g., [31]) and is omitted.Proof of Theorem5.
By(H2) and (H3), we can get (26)tμ11+μ12f1t,1,1≤f1t,t,t≤f1t,1,1,tμ21-λ22f2t,1,t≤f2t,t,1≤f2t,1,t,t∈0,1.This implies that (27)0<∫01s1-sf1s,s,sds<+∞,0<∫01s1-sf2s,s,1ds<+∞.Choose constants r and R such that(28)R>max⁡γ-1,1,ν4γλ11+λ12∫01s1-sf1s,s,sds-1/λ11+λ12-1,0<r<min⁡12,ν4γμ21∫01s1-sf2s,s,1ds1/1-μ21-μ22.
It follows from Lemma9 that T:P∩(ΩR¯∖Ωr)→P is a completely continuous operator. Moreover, by Lemma 7, if (x,y)∈P∩(ΩR¯∖Ωr) is a fixed point of T, then (x,y) is a solution of differential system (3).
For anyt∈[0,1], (x,y)∈P∩∂Ωr, it follows from the definition of cone P that (29)γrt=γtx,y1≤xt,yt≤r<1. Thus for any (x,y)∈P∩∂Ωr, by (H33) and Remarks 2 and 8, we have(30)Tix,yt≥ν4∫01s1-sf2s,γrs,rds≥ν4γμ21rμ21+μ22∫01s1-sf2s,s,1ds≥r=x,y1,i=1,2,t∈14,1.Consequently, (31)Tx,y1≥x,y1,∀x,y∈P∩∂Ωr.
Again, for anyt∈[0,1], (x,y)∈P∩∂ΩR, we have (32)γRt=γtx,y1≤xt,yt≤R.Thus for any (x,y)∈P∩∂ΩR, noting that γR>1, we have (33)Tix,yt≥ν4∫01s1-sf1s,γRs,γRsds≥ν4γRλ11+λ12∫01s1-sf1s,s,sds≥R=x,y1,i=1,2,t∈14,1.This guarantees (34)Tx,y1≥x,y1,∀x,y∈P∩∂ΩR.
On the other hand, for anyt∈[0,1], (x,y)∈P∩∂Ω1, it follows from the definition of cone P that (35)γt≤xt,yt≤1. Thus for any (x,y)∈P∩∂Ω1, by (H33) and Remarks 2 and 8, we have(36)Tix,yt≤ρ∫01s1-sf1s,1,1ds+ρ∫01s1-sf2s,1,γsds≤ρ∫01s1-sf1s,1,1ds+ργμ22∫01s1-sf2s,1,sds<1=x,y1,i=1,2.That is, (37)Tx,y1<x,y1,∀x,y∈P∩∂Ω1.Therefore, from (31), (37), and Lemma 6, it follows that differential system (3) has one positive solution (x1,y1)∈P with r≤x1,y11<1. In the same way, from (34), (37), and Lemma 6, it follows that differential system (3) has one positive solution (x2,y2)∈P with 1<x1,y11≤R.


## 3. An Example

In this section we give an example to illustrate the usefulness of our main results. Let us consider the singular differential system with coupled boundary value problem(38)-x′′=xp1yq1πt1-t,-y′′=xp24yt,x0=y0=0,x1=y14+y12,y1=∫01xtdt,where 0≤p1,q1,p2<+∞, p1+q1>1, p2<1.Let(39)At=0,t∈0,14,1,t∈14,12,2,t∈12,1,Bt=t,λ11=μ11=p1,λ12=μ12=q1,λ21=μ21=p2,λ22=μ22=-12,f1t,x,y=xp1yq1πt1-t,f2t,x,y=xp24yt;then(40)κ1=34,κ2=12,κ=1-κ1κ2=58,ρ=165,ν=15,γ=116,ρ∫01s1-sf1s,1,1ds+ργμ22∫01s1-sf2s,1,sds=16518+435=134175<1.So all conditions of Theorem 5 are satisfied for (38), and our conclusion follows from Theorem 5.



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*Source: 3608352-2017-06-05.xml*