# Existence Results for a Coupled System of Nonlinear Fourth-Order Differential Equations

Authors: Bessem Samet
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Sufficient conditions are obtained for the existence of solutions to a coupled system of nonlinear fourth-order differential equations.


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## Body

## 1. Introduction

The purpose of this paper is to study the existence of solutions to the following coupled system of nonlinear fourth-order differential equations:(1)x′′′′(t)=f(t,x(t),y(t)),t∈(0,1),y′′′′(t)=g(t,x(t),y(t)),t∈(0,1),x(0)=x′′(0)=0,x′(1)=0,x′′′(1)=ξ(x(1)),y(0)=y′′(0)=0,y′(1)=0,y′′′(1)=ζ(y(1)),
where f,g:[0,1]×ℝ×ℝ→ℝ and ξ,ζ:ℝ→ℝ are given continuous functions.In recent years, boundary value problems for fourth-order ordinary differential equations have been extensively studied. Gupta [1] considered the boundary value problem
(2)x′′′′(t)+f(t)x(t)=g(t),t∈(0,π),x′(0)=x′′′(0)=x′′′(π)=x′(π)=0,
which describes the unstable static equilibrium of an elastic beam which is supported by sliding clamps at both ends, t=0 and t=π. Many authors (see [2–8] and references therein) studied the existence of solutions to the problem
(3)x′′′′(t)=f(t,x(t),x′′(t)),t∈(0,1),x(0)=x(1)=x′′(0)=x′′(1)=0,
which describes the deformation of the equilibrium state of an elastic beam with its two ends being simply supported. For other results on the study of fourth-order ordinary differential equations, we refer to [9–13] and references therein. On the other hand, to our knowledge, there are only few papers concerning existence results for systems of fourth-order ordinary differential equations. For example, in [14], via the fixed point index theory, the authors studied the existence of solutions to the system of bending elastic beam equations
(4)x′′′′(t)=f(t,x(t),y(t),x′′(t),y′′(t)),t∈(0,1),y′′′′(t)=g(t,x(t),y(t),x′′(t),y′′(t)),t∈(0,1),x(0)=x(1)=x′′(0)=x′′(1)=0,y(0)=y(1)=y′′(0)=y′′(1)=0.In this work, we provide four sufficient conditions for the existence of solutions to the problem (1). Some illustrative examples are presented.The paper is organized as follows. In Section2, we present the main results. In Section 3, we give some illustrative examples. In Section 4, we present some basic lemmas that will be used for the proofs. Finally, in Section 5, we prove our results.


## 2. Main Results

We obtained the following existence results.Theorem 1.
Suppose the following conditions hold:(i)
there existα,β,λ,μ>0 such that, for all t∈[0,1] and p,q,r,e∈ℝ, one has
(5)|f(t,p,q)-f(t,r,e)|≤α|p-r|+β|q-e|,|g(t,p,q)-g(t,r,e)|≤λ|p-r|+μ|q-e|,(ii)
there existγ,θ>0 such that, for all p,q∈ℝ, one has
(6)|ξ(p)-ξ(q)|≤γ|p-q|,|ζ(p)-ζ(q)|≤θ|p-q|,(iii)
ρ(A)<1, where A∈M2,2(ℝ+) is the matrix given by
(7)A=12(γ+αβθ+λμ)and ρ(A) denotes the spectral radius of A. Then(I)
equation (1) has a unique solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ),(II)
for anyu0,v0∈C([0,1];ℝ), the sequences {un},{vn}⊂C([0,1];ℝ) defined by
(8)un+1=F1(un,vn),vn+1=F2(un,vn)converge, respectively, to x* and y*; that is,
(9)limn→∞∥un-x*∥∞=limn→∞∥vn-y*∥∞=0.Theorem 2.
Suppose the following conditions hold:(i)
there exists a nonnegative functiona∈L1(0,1) such that, for all t∈[0,1] and p,q∈ℝ, one has
(10)|f(t,p,q)|≤a(t)+ϵ1|p|ρ1+ϵ2|q|ρ2,hhhhhhhhϵ1,ϵ2>0,0<ρ1,ρ2<1,(ii)
there exists a nonnegative functionb∈L1(0,1) such that, for all t∈[0,1] and p,q∈ℝ, one has
(11)|g(t,p,q)|≤b(t)+δ1|p|σ1+δ2|q|σ2,hhhhhhhlδ1,δ2>0,0<σ1,σ2<1,(iii)
for allp∈ℝ, one has
(12)max{|ξ(p)|,|ζ(p)|}≤|p|.Then (1) has at least one solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ).Theorem 3.
Suppose the following conditions hold:(i)
for allt∈[0,1] and p,q∈ℝ, one has
(13)|f(t,p,q)|≤a+ϵ|p|,a>0,0≤ϵ<1,(ii)
for allt∈[0,1] and p,q∈ℝ, one has
(14)|g(t,p,q)|≤b+δ|q|,b>0,0≤δ<1,(iii)
for allp∈ℝ, one has
(15)max{|ξ(p)|,|ζ(p)|}≤|p|.Then (1) has at least one solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ).Theorem 4.
Suppose the following conditions hold:(i)
for allt∈[0,1] and p,q∈ℝ, one has
(16)|f(t,p,q)|≤a+ϵ1|p|ρ1+ϵ2|q|ρ2,0≤2a<1,ϵ1,ϵ2>0,2(ϵ1+ϵ2)≤1,ρ1,ρ2≥1,(ii)
for allt∈[0,1] and p,q∈ℝ, one has
(17)|g(t,p,q)|≤b+δ1|p|σ1+δ2|q|σ2,0≤2b<1,δ1,δ2>0,2(δ1+δ2)≤1,σ1,σ2≥1,(iii)
for allp,q∈ℝ, one has
(18)max{|ξ(p)-ξ(q)|,|ζ(p)-ζ(q)|}≤k|p-q|,h0<k<2,(iv)
ξ(0)=ζ(0)=0.Then (1) has at least one solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ).


## 3. Illustrative Examples

Example 1.
Consider the initial value problem(19)x′′′′(t)=(12sin(x(t))+3316sin(y(t)))cos(t),12sin(x(t))+3316sin(y(t))lst∈(0,1),y′′′′(t)=(43|x(t)|+32|y(t)|)e-tsin(t),t∈(0,1),x(0)=x′′(0)=0,x′(1)=0,4x′′′(1)=sin(2x(1)),y(0)=y′′(0)=0,y′(1)=0,y′′′(1)=49sin(3(y(1))).
In this case, for all t∈[0,1] and p,q∈ℝ, we have
(20)f(t,p,q)=(12sin(p)+3316sin(q))cos(t),g(t,p,q)=(43|p|+32|q|)e-tsin(t),ξ(p)=14sin(2p),ζ(p)=49sin(3p).
Conditions (i) and (ii) of Theorem 1 are satisfied with
(21)α=12,β=3316,λ=43,μ=32,γ=12,θ=43.
Moreover, the matrix
(22)A=(α+γ2β2λ+θ2μ2)=(1233324334)
has the eigenvalues λ1=0 and λ2=0.9330. From Theorem 1, the considered problem has a unique solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ).Example 2.
Consider the initial value problem(23)x′′′′(t)=et+sin(|x(t)+y(t)|)cos(t),t∈(0,1),y′′′′(t)=3t2-5+e-t|x(t)|2/5+|y(t)|2(1+|y(t)|),hhhhhht∈(0,1),x(0)=x′′(0)=0,x′(1)=0,x′′′(1)=sin(x(1)),y(0)=y′′(0)=0,y′(1)=0,y′′′(1)=|y(1)|.
In this case, for all t∈[0,1] and p,q∈ℝ, we have
(24)f(t,p,q)=et+sin(|p+q|)cos(t),g(t,p,q)=3t2-5+e-t|p|2/5+|q|2(1+|q|),ξ(p)=sin(p),ζ(p)=|p|.
Conditions (i) and (ii) of Theorem 2 are satisfied with
(25)a(t)=et,ϵ1=ϵ2=1,ρ1=ρ2=12,b(t)=3t2-5,δ1=1,δ2=12,σ1=25,σ2=12.
From Theorem 2, the considered problem has at least one solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ).Example 3.
Consider the initial value problem(26)x′′′′(t)=e-t+12sin(x(t)1+y2(t)),t∈(0,1),y′′′′(t)=11+t2+x2(t)+y2(t),t∈(0,1),x(0)=x′′(0)=0,x′(1)=0,x′′′(1)=sin(x(1)1+x(1)2),y(0)=y′′(0)=0,y′(1)=0,y′′′(1)=y(1)1+y(1)4.
In this case, for all t∈[0,1] and p,q∈ℝ, we have
(27)f(t,p,q)=e-t+12sin(p1+q2),g(t,p,q)=11+t2+p2+q2,ξ(p)=sin(p1+p2),ζ(p)=p1+p4.
Conditions (i) and (ii) of Theorem 3 are satisfied with
(28)a=1,ϵ=12,b=1,δ=0.
From Theorem 3, the considered problem has at least one solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ).Example 4.
Consider the initial value problem(29)x′′′′(t)=t25+x(t)2+y(t)24cos(t),t∈(0,1),y′′′′(t)=tet-13+sin(x(t)+y(t))4,t∈(0,1),x(0)=x′′(0)=0,x′(1)=0,x′′′(1)=sin(3x(1)2),y(0)=y′′(0)=0,y′(1)=0,y′′′(1)=32|y(1)|.
In this case, for all t∈[0,1] and p,q∈ℝ, we have
(30)f(t,p,q)=t25+p2+q24cos(t),g(t,p,q)=tet-13+sin(p+q)4,ξ(p)=sin(3p2),ζ(p)=3|p|2.
Conditions (i), (ii), and (iii) of Theorem 4 are satisfied with
(31)a=15,ϵ1=ϵ2=14,ρ1=ρ2=2,b=13,δ1=1,δ2=14,σ1=σ2=1,k=32.
From Theorem 4, the considered problem has at least one solution (x*,y*)∈C([0,1];ℝ)×C([0,1];ℝ).


## 4. Preliminaries

In order to obtain an integral formulation of the problem (1), we need the following lemma. At first, we denote by C([0,1];ℝ) the set of continuous functions from [0,1] into ℝ.Lemma 5.
Letρ∈C([0,1];ℝ) and ω∈ℝ. The problem
(32)x′′′′(t)=ρ(t),t∈(0,1),x(0)=x′′(0)=0,x′(1)=0,x′′′(1)=ω
has a unique solution
(33)x(t)=∫01‍G(t,s)ρ(s)ds+φω(t),
where
(34)φω(t)=ω2t(t23-1),G(t,s)=16{t(6s-t2-3s2),0≤t≤s,s(6t-s2-3t2),s≤t≤1.Proof.
We omit the proof as it employs standard arguments.From Lemma5, problem (1) is equivalent to the following system of integral equations:
(35)x(t)=∫01‍G(t,s)f(s,x(s),y(s))ds+φξ(x(1))(t),y(t)=∫01‍G(t,s)g(s,x(s),y(s))ds+φζ(y(1))(t),y(t)=∫01‍G(t,s)g(s,x(s),y(s))ds+t∈[0,1].
Let X=C([0,1];ℝ). Define the mappings F1,F2:X×X→X by
(36)F1(u,v)(t)=∫01‍G(t,s)f(s,u(s),v(s))ds+φξ(u(1))(t),F2(u,v)(t)=∫01‍G(t,s)g(s,u(s),v(s))ds+φζ(v(1))(t)
for every t∈[0,1]. Let F:X×X→X×X be the mapping defined by
(37)F(u,v)=(F1(u,v),F2(u,v)).
Observe that (x,y)∈X×X is a solution to (35) if and only if (x,y) is a fixed point of F.Lemma 6.
One has(38)0≤t2s3≤G(t,s)≤12s(2t-t2)≤12,∀t,s∈[0,1].Proof.
For0≤t≤s≤1, we have
(39)∂G(t,s)∂t=(1-t)s-12(s-t)2≤(1-t)s.
Similarly, for s≤t≤1, we have
(40)∂G(t,s)∂t=(1-t)s.
Then, for all t,s∈[0,1], we have
(41)G(t,s)=∫0t‍∂G(r,s)∂rdr≤∫0t‍(1-r)sdr=12s(2t-t2)≤12.
Now, for 0≤t≤s≤1, we have
(42)G(t,s)=t6(6s-t2-3s2)≥t6(6s-4s2)≥t2s3.
Similarly, for s≤t≤1, we have
(43)G(t,s)=s6(6t-s2-3t2)≥s6(6t-4t2)≥t2s3.We endow the product setZ=X×X with the norm
(44)∥(x,y)∥Z=max{∥x∥∞,∥y∥∞},
where
(45)∥x∥∞=max{|x(t)|:t∈[0,1]}.Lemma 7.
The mappingF:Z→Z is continuous with respect to the norm ∥·∥Z.Proof.
Let(x,y)∈Z be an arbitrary point. Let R be a nonnegative real number such that ∥(x,y)∥Z≤R. To show the continuity of F at the point (x,y), let us consider a sequence {(xn,yn)}⊂Z converging to (x,y); that is,
(46)limn→∞∥xn-x∥∞=limn→∞∥yn-y∥∞=0.
Without restriction to the generality, we can suppose that ∥(xn,yn)∥Z≤R, for every n∈ℕ (ℕ is the set of natural numbers). Since f is continuous, we can define
(47)M=max{|f(t,p,q)|:0≤t≤1,-R≤p,q≤R}.
For all t∈[0,1], we have
(48)|F1(xn,yn)(t)-F1(x,y)(t)|≤∫01G(t,s)|f(s,xn(s),yn(s))-f(s,x(s),y(s))|ds+12|ξ(xn(1))-ξ(x(1))|.
Using Lemma 6, for all t∈[0,1], we have
(49)2|F1(xn,yn)(t)-F1(x,y)(t)|≤∫01‍|f(s,xn(s),yn(s))-f(s,x(s),y(s))|ds+|ξ(xn(1))-ξ(x(1))|,
which implies that
(50)2∥F1(xn,yn)-F1(x,y)∥∞≤∫01‍|f(s,xn(s),yn(s))-f(s,x(s),y(s))|ds+|ξ(xn(1))-ξ(x(1))|.
From the continuity of ξ, we have
(51)limn→∞|ξ(xn(1))-ξ(x(1))|=0.
The continuity of f implies that, for all s∈[0,1],
(52)limn→∞|f(s,xn(s),yn(s))-f(s,x(s),y(s))|=0.
On the other hand, for all n∈ℕ, we have
(53)∫01‍|f(s,xn(s),yn(s))-f(s,x(s),y(s))|ds≤2M.
Then, by the dominated convergence theorem, we have
(54)limn→∞∫01‍|f(s,xn(s),yn(s))-f(s,x(s),y(s))|ds=0.
Consequently, we have
(55)limn→∞∥F1(xn,yn)-F1(x,y)∥∞=0.
Similarly, we can show that
(56)limn→∞∥F2(xn,yn)-F2(x,y)∥∞=0.
We deduce that
(57)limn→∞∥F(xn,yn)-F(x,y)∥Z=0,
which yields the desired result.Lemma 8.
LetD be the subset of Z defined by
(58)D={(x,y)∈Z:∥(x,y)∥Z≤R},
for some R>0. Under condition (iii) of Theorem 2 (or conditions (iii)-(iv) of Theorem 4), F(D) is an equicontinuous set.Proof.
Suppose that (iii) of Theorem2 holds. Since f and g are continuous, we can define
(59)M=max{|f(t,p,q)|:0≤t≤1,-R≤p,q≤R},N=max{|g(t,p,q)|:0≤t≤1,-R≤p,q≤R}.
Let ɛ>0. Since G is continuous in [0,1]×[0,1], there exists α>0 such that
(60)|G(t,s)-G(t′,s)|<min{ɛ2M,ɛ2N}
whenever |t-t′|<α. Let κ=min{α,ɛ/2R}. For (x,y)∈D, for |t-t′|<κ, we have
(61)|F1(x,y)(t)-F1(x,y)(t′)|≤∫01‍|G(t,s)-G(t′,s)||f(s,x(s),y(s))|ds+|ξ(x(1))||t-t′|≤Mɛ2M+Rɛ2R(from(iii)ofTheorem2)=ɛ,|F2(x,y)(t)-F2(x,y)(t′)|≤∫01‍|G(t,s)-G(t′,s)||g(s,x(s),y(s))|ds+|ζ(y(1))||t-t′|≤Nɛ2N+Rɛ2R(from(iii)ofTheorem2)=ɛ.
Then, the equicontinuity of F(D) is proved. If conditions (iii)-(iv) of Theorem 4 hold, using the same technique with κ=min{α,ɛ/2kR}, we get the desired result.Now, we are ready to prove our main results.


## 5. The Proofs

### 5.1. Proof of Theorem1

The proof is based on Perov fixed point theorem [15]. At first, let us recall some concepts.Letm≥1 be a positive integer. We endow the set ℝm with the partial order ⪯m defined by
(62)α=(α1,…,αm),β=(β1,…,βm)∈ℝm,hhhhα⪯mβ⟺αi≤βi,∀i=1,…,m.
We denote by 0ℝm the zero vector of ℝm.LetΣ be a nonempty set. A mapping d:Σ×Σ→Rm is called a vector valued metric on Σ if the following properties are satisfied:(d1)
d(x,y)=0ℝm⇔x=y, for all x,y∈Σ;(d2)
d(x,y)⪰m0ℝm, for all x,y∈Σ;(d3)
d(x,y)=d(y,x), for all x,y∈Σ;(d4)
d(x,y)⪯md(x,z)+d(z,y), for all x,y,z∈Σ.A set Σ equipped with a vector valued metric d is called a generalized metric space. We will denote such a space by (Σ,d). For the generalized metric spaces, the notions of convergent sequence, Cauchy sequence, completeness, open subset, and closed subset are similar to those for usual metric spaces.We denote byMm,m(ℝ+) the set of all m×m matrices with positive elements. For A∈Mm,m(ℝ+), we denote by ρ(A) the spectral radius of A, that is, the supremum among the absolute values of the elements in its spectrum.Lemma 9.
Let(M,d) be a complete generalized metric space and the mapping T:M→M with the property that there exists a matrix A∈Mm,m(ℝ+) such that
(63)d(Tx,Ty)⪯mAd(x,y),∀x,y∈M.
If ρ(A)<1, then T has a unique fixed point x*∈M. Moreover, for any x0∈M, the iterative sequence {xn}⊂M defined by xn+1=Txn converges to x*.Proof.
We endowZ=X×X with the vector valued metric d:Z×Z→ℝ2 defined by
(64)d((x,y),(u,v))=(∥x-u∥∞∥y-v∥∞),∀(x,y),(u,v)∈Z.
It is easy to show that (Z,d) is a complete generalized metric space. Let (x,y),(u,v)∈Z. For any t∈[0,1], we have
(65)F1(x,y)(t)-F1(u,v)(t)=∫01‍G(t,s)f(s,x(s),y(s))-f(s,u(s),v(s))ds+φξ(x(1))(t)-φξ(u(1))(t)=∫01‍G(t,s)f(s,x(s),y(s))-f(s,u(s),v(s))ds+t(t23-1)ξ(x(1))-ξ(u(1))2.
Using Lemma 6, for all t∈[0,1], we have
(66)2|F1(x,y)(t)-F1(u,v)(t)|≤∫01‍|f(s,x(s),y(s))-fs,u(s),v(s)|ds+|ξ(x(1))-ξ(u(1))|.
Using conditions (i) and (ii), we get that
(67)|F1(x,y)(t)-F1(u,v)(t)|≤(γ+α2)∥x-u∥∞+(β2)∥y-v∥∞,
for all t∈[0,1]. This implies that
(68)∥F1(x,y)-F1(u,v)∥∞≤(γ+α2)∥x-u∥∞+(β2)∥y-v∥∞.
Using similar calculations, we obtain that
(69)∥F2(x,y)-F2(u,v)∥∞≤(θ+λ2)∥x-u∥∞+(μ2)∥y-v∥∞.
It follows from (68) and (69) that
(70)d(F(x,y),F(u,v))⪯2Ad((x,y),(u,v)),∀(x,y),(u,v)∈Z,
where
(71)A=12(γ+αβθ+λμ).
Since ρ(A)<1, the desired result follows from Lemma 9.


### 5.2. Proof of Theorem2

The proof is based on Schauder fixed point theorem [16].Lemma 10.
LetD be a closed bounded convex subset of a normed space E. If T:D→E is a continuous compact map such that T(D)⊆D, then T has a fixed point.Proof.
LetD⊂Z be the set defined by
(72)D={(x,y)∈Z:∥(x,y)∥Z≤R},
where
(73)R≥max{3∫01‍a(s)ds,3∫01‍b(s)ds,(3ϵ1)1/(1-ρ1),max{(3ϵ2)1/(1-ρ2),(3δ1)1/(1-σ1),(3δ2)1/(1-σ2)∫01‍}.
We will prove that(74)F(D)⊆D.
Let (x,y)∈D; that is, ∥x∥∞≤R and ∥y∥∞≤R. For all t∈[0,1], we have
(75)F1(x,y)(t)=∫01‍G(t,s)f(s,x(s),y(s))ds+φξ(x(1))(t).
Using Lemma 6 and conditions (i) and (iii), we obtain that
(76)|F1(x,y)(t)|≤12(∫01‍a(s)ds)+ϵ12Rρ1+ϵ22Rρ2+R2≤R6+R6+R6+R2=R.
Thus, we have
(77)∥F1(x,y)∥∞≤R.
Similarly, for all t∈[0,1], we have
(78)F2(x,y)(t)=∫01‍G(t,s)g(s,x(s),y(s))ds+φζ(y(1))(t).
Using Lemma 6 and conditions (ii) and (iii), we obtain that
(79)|F2(x,y)(t)|≤12(∫01‍b(s)ds)+δ12Rσ1+δ22Rσ2+R2≤R6+R6+R6+R2=R.
Thus, we have
(80)∥F2(x,y)∥∞≤R.
Then, (74) follows from (77) and (80).
Now, using Lemmas7 and 8, Arzelá-Ascoli theorem, and Lemma 10, we obtain the desired result.


### 5.3. Proof of Theorem3

The proof is based on Schaefer fixed point theorem [17].Lemma 11.
LetE be a Banach space. Assume that T:E→E is a continuous compact operator and the set
(81)V={u∈E:u=λTu,0≤λ≤1}
is bounded. Then T has a fixed point in E.Proof.
The continuity and the compactness of the mappingF:Z→Z follow from Lemmas 7 and 8. We have just to show that the set
(82)V={(x,y)∈Z:(x,y)=(λF1(x,y),λF2(x,y)),hh0≤λ≤1(λF1(x,y),λF2(x,y))}
is bounded with respect to the norm ∥·∥Z. Then, the result will be obtained by applying Lemma 11.
Let(x,y)∈Z such that x=λF1(x,y) and y=λF2(x,y) for some λ∈[0,1]. For all t∈[0,1], using (i), (iii), and Lemma 6, we obtain that
(83)|x(t)|=λ|∫01‍G(t,s)f(s,x(s),y(s))ds+ξ(x(1))2t(t23-1)|≤a2+ϵ2∥x∥∞+∥x∥∞2,
which implies that
(84)∥x∥∞≤a1-ϵ.
Similarly, using (ii), (iii), and Lemma 6, for all t∈[0,1], we have
(85)|y(t)|=λ|∫01‍G(t,s)g(s,x(s),y(s))ds+ζ(y(1))2t(t23-1)|≤b2+δ2∥y∥∞+∥y∥∞2,
which implies that
(86)∥y∥∞≤b1-δ.
Consequently, V is a bounded set.


### 5.4. Proof of Theorem4

The following fixed point theorem due to Agarwal et al. [18] will be used in the proof.Lemma 12.
LetE be a Banach space, let D be a bounded closed convex subset of E, and let A,B:D→E such that Au+Bv∈D for every (u,v)∈D. If A is a contraction and B is continuous and compact, then the equation Aw+Bw=w has a solution w∈D.Proof.
We define the operatorsA,B:Z→Z as follows: for all (x,y)∈Z,
(87)A(x,y)=(A1(x,y),A2(x,y)),B(x,y)=(B1(x,y),B2(x,y)),
where, for every t∈[0,1],
(88)A1(x,y)(t)=ξ(x(1))2t(t33-1),A2(x,y)(t)=ζ(y(1))2t(t33-1),B1(x,y)(t)=∫01‍G(t,s)f(s,x(s),y(s))ds,B2(x,y)(t)=∫01‍G(t,s)g(s,x(s),y(s))ds.
Clearly, for every (x,y)∈Z, we have
(89)F(x,y)=A(x,y)+B(x,y).
Let 2max{a,b}<R<1. We define the set D⊂Z as follows:
(90)D={(x,y)∈Z:∥(x,y)∥Z≤R}.
Let (x,y),(u,v)∈D. For every t∈[0,1], we have
(91)|A1(x,y)(t)+B1(u,v)(t)|≤∫01‍G(t,s)|f(s,u(s),v(s))|ds+|ξ(x(1))|2(fromLemma6andconditions(i),(iii),and(iv))≤12(a+ϵ1Rρ1+ϵ2Rρ2)+R2≤a+(ϵ1+ϵ2+1)2R≤1/2+(ϵ1+ϵ2+1)2R≤R.
Similarly, for every t∈[0,1], we have
(92)|A2(x,y)(t)+B2(u,v)(t)|≤∫01‍G(t,s)|g(s,u(s),v(s))|ds+|ζ(u(1))|2(fromLemma6andconditions(ii),(iii),and(iv))≤12(b+δ1Rσ1+δ2Rσ2)+R2≤b+(δ1+δ2+1)2R≤1/2+(δ1+δ2+1)2R≤R.
Thus, we proved that, for every (x,y),(u,v)∈D, we have
(93)A(x,y)+B(u,v)∈D.
Now, we will prove that A:D→Z is a contraction. Let (x,y),(u,v)∈D. For every t∈[0,1], we have
(94)|A1(x,y)(t)-A1(u,v)(t)|≤|ξ(x(1))-ξ(u(1))|2(fromcondition(iii))≤k2|x(1)-u(1)|(fromcondition(iii))≤k2∥x-u∥∞.
Similarly, for every t∈[0,1], we have
(95)|A2(x,y)(t)-A2(u,v)(t)|≤|ζ(y(1))-ζ(v(1))|2(fromcondition(iii))≤k2|y(1)-v(1)|(fromcondition(iii))≤k2∥y-v∥∞.
Since 0<k<2, then A is a contraction (with respect to the norm ∥·∥Z). Finally, the continuity and the compactness of the operator B follow by using the same arguments of the proof of Theorem 2. Applying Lemma 12, we obtain the existence of (x,y)∈D such that (x,y) is a fixed point of A+B=F.


## 5.1. Proof of Theorem1

The proof is based on Perov fixed point theorem [15]. At first, let us recall some concepts.Letm≥1 be a positive integer. We endow the set ℝm with the partial order ⪯m defined by
(62)α=(α1,…,αm),β=(β1,…,βm)∈ℝm,hhhhα⪯mβ⟺αi≤βi,∀i=1,…,m.
We denote by 0ℝm the zero vector of ℝm.LetΣ be a nonempty set. A mapping d:Σ×Σ→Rm is called a vector valued metric on Σ if the following properties are satisfied:(d1)
d(x,y)=0ℝm⇔x=y, for all x,y∈Σ;(d2)
d(x,y)⪰m0ℝm, for all x,y∈Σ;(d3)
d(x,y)=d(y,x), for all x,y∈Σ;(d4)
d(x,y)⪯md(x,z)+d(z,y), for all x,y,z∈Σ.A set Σ equipped with a vector valued metric d is called a generalized metric space. We will denote such a space by (Σ,d). For the generalized metric spaces, the notions of convergent sequence, Cauchy sequence, completeness, open subset, and closed subset are similar to those for usual metric spaces.We denote byMm,m(ℝ+) the set of all m×m matrices with positive elements. For A∈Mm,m(ℝ+), we denote by ρ(A) the spectral radius of A, that is, the supremum among the absolute values of the elements in its spectrum.Lemma 9.
Let(M,d) be a complete generalized metric space and the mapping T:M→M with the property that there exists a matrix A∈Mm,m(ℝ+) such that
(63)d(Tx,Ty)⪯mAd(x,y),∀x,y∈M.
If ρ(A)<1, then T has a unique fixed point x*∈M. Moreover, for any x0∈M, the iterative sequence {xn}⊂M defined by xn+1=Txn converges to x*.Proof.
We endowZ=X×X with the vector valued metric d:Z×Z→ℝ2 defined by
(64)d((x,y),(u,v))=(∥x-u∥∞∥y-v∥∞),∀(x,y),(u,v)∈Z.
It is easy to show that (Z,d) is a complete generalized metric space. Let (x,y),(u,v)∈Z. For any t∈[0,1], we have
(65)F1(x,y)(t)-F1(u,v)(t)=∫01‍G(t,s)f(s,x(s),y(s))-f(s,u(s),v(s))ds+φξ(x(1))(t)-φξ(u(1))(t)=∫01‍G(t,s)f(s,x(s),y(s))-f(s,u(s),v(s))ds+t(t23-1)ξ(x(1))-ξ(u(1))2.
Using Lemma 6, for all t∈[0,1], we have
(66)2|F1(x,y)(t)-F1(u,v)(t)|≤∫01‍|f(s,x(s),y(s))-fs,u(s),v(s)|ds+|ξ(x(1))-ξ(u(1))|.
Using conditions (i) and (ii), we get that
(67)|F1(x,y)(t)-F1(u,v)(t)|≤(γ+α2)∥x-u∥∞+(β2)∥y-v∥∞,
for all t∈[0,1]. This implies that
(68)∥F1(x,y)-F1(u,v)∥∞≤(γ+α2)∥x-u∥∞+(β2)∥y-v∥∞.
Using similar calculations, we obtain that
(69)∥F2(x,y)-F2(u,v)∥∞≤(θ+λ2)∥x-u∥∞+(μ2)∥y-v∥∞.
It follows from (68) and (69) that
(70)d(F(x,y),F(u,v))⪯2Ad((x,y),(u,v)),∀(x,y),(u,v)∈Z,
where
(71)A=12(γ+αβθ+λμ).
Since ρ(A)<1, the desired result follows from Lemma 9.


## 5.2. Proof of Theorem2

The proof is based on Schauder fixed point theorem [16].Lemma 10.
LetD be a closed bounded convex subset of a normed space E. If T:D→E is a continuous compact map such that T(D)⊆D, then T has a fixed point.Proof.
LetD⊂Z be the set defined by
(72)D={(x,y)∈Z:∥(x,y)∥Z≤R},
where
(73)R≥max{3∫01‍a(s)ds,3∫01‍b(s)ds,(3ϵ1)1/(1-ρ1),max{(3ϵ2)1/(1-ρ2),(3δ1)1/(1-σ1),(3δ2)1/(1-σ2)∫01‍}.
We will prove that(74)F(D)⊆D.
Let (x,y)∈D; that is, ∥x∥∞≤R and ∥y∥∞≤R. For all t∈[0,1], we have
(75)F1(x,y)(t)=∫01‍G(t,s)f(s,x(s),y(s))ds+φξ(x(1))(t).
Using Lemma 6 and conditions (i) and (iii), we obtain that
(76)|F1(x,y)(t)|≤12(∫01‍a(s)ds)+ϵ12Rρ1+ϵ22Rρ2+R2≤R6+R6+R6+R2=R.
Thus, we have
(77)∥F1(x,y)∥∞≤R.
Similarly, for all t∈[0,1], we have
(78)F2(x,y)(t)=∫01‍G(t,s)g(s,x(s),y(s))ds+φζ(y(1))(t).
Using Lemma 6 and conditions (ii) and (iii), we obtain that
(79)|F2(x,y)(t)|≤12(∫01‍b(s)ds)+δ12Rσ1+δ22Rσ2+R2≤R6+R6+R6+R2=R.
Thus, we have
(80)∥F2(x,y)∥∞≤R.
Then, (74) follows from (77) and (80).
Now, using Lemmas7 and 8, Arzelá-Ascoli theorem, and Lemma 10, we obtain the desired result.


## 5.3. Proof of Theorem3

The proof is based on Schaefer fixed point theorem [17].Lemma 11.
LetE be a Banach space. Assume that T:E→E is a continuous compact operator and the set
(81)V={u∈E:u=λTu,0≤λ≤1}
is bounded. Then T has a fixed point in E.Proof.
The continuity and the compactness of the mappingF:Z→Z follow from Lemmas 7 and 8. We have just to show that the set
(82)V={(x,y)∈Z:(x,y)=(λF1(x,y),λF2(x,y)),hh0≤λ≤1(λF1(x,y),λF2(x,y))}
is bounded with respect to the norm ∥·∥Z. Then, the result will be obtained by applying Lemma 11.
Let(x,y)∈Z such that x=λF1(x,y) and y=λF2(x,y) for some λ∈[0,1]. For all t∈[0,1], using (i), (iii), and Lemma 6, we obtain that
(83)|x(t)|=λ|∫01‍G(t,s)f(s,x(s),y(s))ds+ξ(x(1))2t(t23-1)|≤a2+ϵ2∥x∥∞+∥x∥∞2,
which implies that
(84)∥x∥∞≤a1-ϵ.
Similarly, using (ii), (iii), and Lemma 6, for all t∈[0,1], we have
(85)|y(t)|=λ|∫01‍G(t,s)g(s,x(s),y(s))ds+ζ(y(1))2t(t23-1)|≤b2+δ2∥y∥∞+∥y∥∞2,
which implies that
(86)∥y∥∞≤b1-δ.
Consequently, V is a bounded set.


## 5.4. Proof of Theorem4

The following fixed point theorem due to Agarwal et al. [18] will be used in the proof.Lemma 12.
LetE be a Banach space, let D be a bounded closed convex subset of E, and let A,B:D→E such that Au+Bv∈D for every (u,v)∈D. If A is a contraction and B is continuous and compact, then the equation Aw+Bw=w has a solution w∈D.Proof.
We define the operatorsA,B:Z→Z as follows: for all (x,y)∈Z,
(87)A(x,y)=(A1(x,y),A2(x,y)),B(x,y)=(B1(x,y),B2(x,y)),
where, for every t∈[0,1],
(88)A1(x,y)(t)=ξ(x(1))2t(t33-1),A2(x,y)(t)=ζ(y(1))2t(t33-1),B1(x,y)(t)=∫01‍G(t,s)f(s,x(s),y(s))ds,B2(x,y)(t)=∫01‍G(t,s)g(s,x(s),y(s))ds.
Clearly, for every (x,y)∈Z, we have
(89)F(x,y)=A(x,y)+B(x,y).
Let 2max{a,b}<R<1. We define the set D⊂Z as follows:
(90)D={(x,y)∈Z:∥(x,y)∥Z≤R}.
Let (x,y),(u,v)∈D. For every t∈[0,1], we have
(91)|A1(x,y)(t)+B1(u,v)(t)|≤∫01‍G(t,s)|f(s,u(s),v(s))|ds+|ξ(x(1))|2(fromLemma6andconditions(i),(iii),and(iv))≤12(a+ϵ1Rρ1+ϵ2Rρ2)+R2≤a+(ϵ1+ϵ2+1)2R≤1/2+(ϵ1+ϵ2+1)2R≤R.
Similarly, for every t∈[0,1], we have
(92)|A2(x,y)(t)+B2(u,v)(t)|≤∫01‍G(t,s)|g(s,u(s),v(s))|ds+|ζ(u(1))|2(fromLemma6andconditions(ii),(iii),and(iv))≤12(b+δ1Rσ1+δ2Rσ2)+R2≤b+(δ1+δ2+1)2R≤1/2+(δ1+δ2+1)2R≤R.
Thus, we proved that, for every (x,y),(u,v)∈D, we have
(93)A(x,y)+B(u,v)∈D.
Now, we will prove that A:D→Z is a contraction. Let (x,y),(u,v)∈D. For every t∈[0,1], we have
(94)|A1(x,y)(t)-A1(u,v)(t)|≤|ξ(x(1))-ξ(u(1))|2(fromcondition(iii))≤k2|x(1)-u(1)|(fromcondition(iii))≤k2∥x-u∥∞.
Similarly, for every t∈[0,1], we have
(95)|A2(x,y)(t)-A2(u,v)(t)|≤|ζ(y(1))-ζ(v(1))|2(fromcondition(iii))≤k2|y(1)-v(1)|(fromcondition(iii))≤k2∥y-v∥∞.
Since 0<k<2, then A is a contraction (with respect to the norm ∥·∥Z). Finally, the continuity and the compactness of the operator B follow by using the same arguments of the proof of Theorem 2. Applying Lemma 12, we obtain the existence of (x,y)∈D such that (x,y) is a fixed point of A+B=F.



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*Source: 324848-2013-12-23.xml*