# Existence Results for a Coupled System of Nonlinear Fractional Hybrid Differential Equations with Homogeneous Boundary Conditions

Authors: Josefa Caballero; Mohamed Abdalla Darwish; Kishin Sadarangani; Wafa M. Shammakh
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study an existence result for the following coupled system of nonlinear fractional hybrid differential equations with homogeneous boundary conditionsD0+α[x(t)/f(t,x(t),y(t))]=g(t,x(t),y(t)),D0+αy(t)/f(t,y(t),x(t))=g(t,y(t),x(t)),0<t<1, and x(0)=y(0)=0, where α∈(0,1) and D0+α denotes the Riemann-Liouville fractional derivative. The main tools in our study are the techniques associated to measures of noncompactness in the Banach algebras and a fixed point theorem of Darbo type.


---

## Body

## 1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modelling of a great number of processes which appear in physics, chemistry, aerodynamics, and so forth and involve also derivatives of fractional order. For details, see [1–5] and the references therein.On the other hand, about the theory of hybrid differential equations, we refer to the paper [6] where the authors studied the hybrid differential equation of first order:
(1)ddt[x(t)f(t,x(t))]=g(t,x(t)),t∈J=[0,T),x(t0)=x0∈R,
where f∈C(J×R,R∖{0}) and g∈C(J×R,R).In [7], the authors studied the fractional version of the abovementioned problem, that is,
(2)D0+α[x(t)f(t,x(t))]=g(t,x(t)),t∈J,0<α<1,x(0)=0,
under the same assumptions on f and g in [6].Recently, in [8], the authors studied the following fractional hybrid initial value problem with supremum:
(3)D0+α[x(t)f(t,x(t),max⁡0≤τ≤t|x(τ)|)]=g(t,x(t)),hhhhhhhhhhhhhhhhhhhhhhhhhhhh0<t<1,x(0)=0,
where 0<α<1, f∈C([0,1]×R×R,R∖{0}), and g∈C([0,1]×R,R).The coupled systems involving fractional differential equations are very important because they occur in numerous problems of applied nature; for instance, see [9–13].In this paper, we consider the following coupled system:(4)D0+α[x(t)f(t,x(t),y(t))]=g(t,x(t),y(t)),D0+α[y(t)f(t,y(t),x(t))]=g(t,y(t),x(t)),hhhhhhhhhhhhhhhhhhhhhhh0<t<1,x(0)=y(0)=0,
where α∈(0,1) and D0+α is the standard Riemann-Liouville fractional derivative.The main tool in our study is a fixed point theorem of Darbo type associated to measures of noncompactness.


## 2. Preliminaries

We begin this section with some definitions and results about fractional calculus.Letα>0 and n=[α]+1, where [α] denotes the integer part of α. For a function f:(0,∞)→R, the Riemann-Liouville fractional integral of order α>0 of f is defined as
(5)I0+αf(t)=1Γ(α)∫0tf(s)(t-s)1-αds,
provided that the right side is pointwise defined on (0,∞).The Riemann-Liouville fractional derivative of orderα of a continuous function f is defined by
(6)D0+αf(x)=1Γ(n-α)dndxn∫0xf(s)(x-s)α-n+1ds,
provided that the right side is pointwise defined on (0,∞).The following lemma will be useful for our study, [14].Lemma 1.
Leth∈L1(0,1) and 0<α<1. Then,(a)
(7)D0+αI0+αh(x)=h(x);(b)
(8)I0+αD0+αh(x)=h(x)-I0+1-αh(x)|x=0Γ(α)xα-1hhhhhhhhhhhhhhhhhhhha.e.on(0,1).Lemma 2.
Let0<α<1 and suppose that f∈C([0,1],R∖{0}) and y∈C[0,1]. Then, the unique solution of the fractional hybrid initial value problem
(9)D0+α[x(t)f(t)]=y(t),0<t<1x(0)=0,
is given by
(10)x(t)=f(t)Γ(α)∫0ty(s)(t-s)1-αds,t∈[0,1].Proof.
Suppose thatx(t) is a solution of problem (9). Using the operator I0+α and taking into account Lemma 1, we get
(11)I0+αD0+α[x(t)f(t)]=I0+αy(t),
or, equivalently,
(12)x(t)f(t)-I0+1-α(x(t)/f(t))|t=0Γ(α)tα-1=I0+αy(t).
Since x(t)/f(t)|t=0=x(0)/f(0)=0/f(0)=0 (because f(0)≠0), we have
(13)x(t)=f(t)I0+αy(t).
This means that
(14)x(t)=f(t)Γ(α)∫0ty(s)(t-s)1-αds.
Conversely, suppose that x(t) is given by
(15)x(t)=f(t)Γ(α)∫0ty(s)(t-s)1-αds,t∈[0,1].
This means that
(16)x(t)=f(t)I0+αy(t),t∈[0,1].
Applying D0+α and taking into account Lemma 1 and that f(t)≠0 for t∈[0,1], we obtain
(17)D0+α[x(t)f(t)]=D0+αI0+αy(t)=y(t),0<t<1.
Moreover, for t=0 in (16), we have x(0)=f(0)·0=0. This completes the proof.In the sequel, we recall some definitions and basic facts about measures of noncompactness.Assume thatE is a real Banach space with norm ∥·∥ and zero element θ. By B(x,r) we denote the closed ball in E centered at x with radius r. By Br we denote the ball B(θ,r). If X is a nonempty subset of E, by the symbols X¯ and ConvX we denote the closure and the convex closure of X, respectively. By ∥X∥ we denote the quantity ∥X∥=sup⁡{∥x∥:x∈X}. Finally, by ME we will denote the family of all nonempty and bounded subsets of E and by NE we denote its subfamily consisting of all relatively compact subsets of E.Definition 3.
A mappingμ:ME→R+=[0,∞) is said to be a measure of noncompactness in E if it satisfies the following conditions.(a)
The familyker⁡μ={X∈ME:μ(X)=0} is nonempty and ker⁡⁡μ⊂NE.(b)
X⊂Y⇒μ(X)≤μ(Y).(c)
μ(X¯)=μ(ConvX)=μ(X).(d)
μ(λX+(1-λ)Y)≤λμ(X)+(1-λ)μ(Y) for λ∈[0,1].(e)
If(Xn) is a sequence of closed subsets from ME such that Xn+1⊂Xn(n≥1) and lim⁡n→∞μ(Xn)=0, then X∞=∩n=1∞Xn≠ϕ.The familyker⁡⁡μ appearing in (a) is called the kernel of the measure of noncompactness μ. Notice that the set X∞ appearing in (e) is an element of kerμ. Indeed, since μ(X∞)≤μ(Xn) for n=1,2,…, we infer that μ(X∞)=0 and this says that X∞∈ker⁡⁡μ.An important theorem about fixed point theorem in the context of measures of noncompactness is the following Darbo’s fixed point theorem [15].Theorem 4.
LetC be a nonempty, bounded, closed, and convex subset of a Banach space E and let T:C→C be a continuous mapping. Suppose that there exists a constant k∈[0,1) such that
(18)μ(T(X))≤kμ(X),
for any nonempty subset X of C.
Then,T has a fixed point.A generalization of Theorem4 which will be very useful in our study is the following theorem, due to Sadovskiǐ [16].Theorem 5.
LetC be a nonempty, bounded, closed, and convex subset of a Banach space E and let T:C→C be a continuous operator satisfying
(19)μ(T(X))<μ(X),
for any nonempty subset X of C with μ(X)>0.
Then,T has a fixed point.Next, we will assume that the spaceE has structure of Banach algebra. By xy we will denote the product of two elements x,y∈X and by XY we will denote the set defined by XY={xy:x∈X,y∈Y}.Definition 6.
LetE be a Banach algebra. We will say that a measure of noncompactness μ defined on E satisfies condition (m) if
(20)μ(XY)≤∥X∥μ(Y)+∥Y∥μ(X),
for any X, Y∈ME.This definition appears in [17].In this paper, we will work in the spaceC[0,1] consisting of all real functions defined and continuous on [0,1] with the standard supremum norm
(21)∥x∥=sup⁡{|x(t)|:t∈[0,1]},
for x∈C[0,1]. It is clear that (C[0,1],∥·∥) is a Banach algebra, where the multiplication is defined as the usual product of real functions.Next, we present the measure of noncompactness inC[0,1] which will be used later. Let us fix X∈MC[0,1] and ε>0. For x∈X, we denote by ω(x,ε) the modulus of continuity of x; that is,
(22)ω(x,ε)=sup⁡{|x(t)-x(s)|:t,s∈[0,1],|t-s|≤ε}.
Put
(23)ω(X,ε)=sup⁡{ω(x,ε):x∈X},ω0(X)=lim⁡ε→0ω(X,ε).
In [15], it is proved that ω0(X) is a measure of noncompactness in C[0,1].Proposition 7.
The measure of noncompactnessω0 on C[0,1] satisfies condition (m).Proof.
FixX, Y∈MC[0,1], ε>0, and t, s∈[0,1] with |t-s|≤ε. Then, we have
(24)|x(t)y(t)-x(s)y(s)|≤|x(t)y(t)-x(t)y(s)|+|x(t)y(s)-x(s)y(s)|=|x(t)||y(t)-y(s)|+|y(s)||x(t)-x(s)|≤∥x∥ω(y,ε)+∥y∥ω(x,ε).
This means that
(25)ω(xy,ε)≤∥x∥ω(y,ε)+∥y∥ω(x,ε),
and, therefore,
(26)ω(XY,ε)≤∥X∥ω(Y,ε)+∥Y∥ω(X,ε).
Taking ε→0, we get
(27)ω0(XY)≤∥X∥ω0(Y)+∥Y∥ω0(X).
This completes the proof.Proposition7 appears in [17] and we have given the proof for the paper is self-contained.


## 3. Main Results

We begin this section introducing the following classA of functions:
(28)A={lim⁡n→∞φn(t)φ:R+⟶R+:φisnondecreasingandlim⁡n→∞φn(t)=0foranyt>0},
where φn denotes the n-iteration of φ.Remark 8.
Notice that ifφ∈A, then φ(t)<t, for any t>0. Indeed, in contrary case, we can find t0>0 and t0≤φ(t0). Since φ is nondecreasing, we have
(29)0<t0≤φ(t0)≤φ2(t0)≤⋯≤φn(t0)≤⋯,
and, therefore, 0<t0≤lim⁡n→∞φn(t0) and this contradicts the fact that φ∈A.
Moreover, the fact thatφ(t)<t for any t>0 proves that if φ∈A, then φ is continuous at t0=0.Using Remark8 and Theorem 5, we have the following fixed point theorem.Theorem 9.
LetC be a nonempty, bounded, closed, and convex subset of a Banach space E and let T:C→C be a continuous operator satisfying
(30)μ(T(X))≤φ(μ(X)),
for any nonempty subset X of C, where φ∈A.
Then,T has a fixed point.Theorem9 appears in [18], where the authors present a proof without using Theorem 5.The following result which appears in [19] will be interesting in our study.Theorem 10.
Letμ1,μ2,…,μn be measures of noncompactness in the Banach spaces E1,E2,…,En, respectively. Suppose that F:[0,∞)n→[0,∞) is a convex function such that F(x1,x2,…,xn)=0 if and only if xi=0 for i=1,2,…,n.
Then,(31)μ~(X)=F(μ1(X1),μ2(X2),…,μn(Xn))
defines a measure compactness in E1×E2×⋯×En, where Xi denotes the natural projection of X into Ei, for i=1,2,…,n.Remark 11.
As a consequence of Theorem10, we have that if μ is a measure of noncompactness on a Banach space E and we consider the function F:[0,∞)×[0,∞)→[0,∞) defined by F(x,y)=max⁡(x,y), then, since F is convex and F(x,y)=0 if and only if x=y=0, μ~(X)=max⁡{μ(X1),μ(X2)} defines a measure of noncompactness in the space E×E.Next, we present the definition of a coupled fixed point.Definition 12.
An element(x,y)∈X×X is said to be a coupled fixed point of a mapping G:X×X→X if G(x,y)=x and G(y,x)=y.The following result is crucial for our study.Theorem 13.
LetΩ be a nonempty, bounded, closed, and convex subset of a Banach space E, and let μ be a measure of noncompactness in E. Suppose that G:Ω×Ω→Ω is a continuous operator satisfying
(32)μ(G(X1×X2))≤φ(max⁡(μ(X1),μ(X2)))
for all nonempty subsets X1 and X2 of Ω, where φ∈A.
Then,G has at least a coupled fixed point.Proof.
Notice that, by Remark11, μ~(X)=max⁡{μ(X1),μ(X2)} is a measure of noncompactness in the space E×E, where X1 and X2 are the projections of X into E.
Now, we consider the mappingG~:Ω×Ω→Ω×Ω defined by G~(x,y)=(G(x,y),G(y,x)). It is easily seen that Ω×Ω is a nonempty, bounded, closed, and convex subset of E×E. Since G is continuous, it is clear that G~ is also continuous.
Next, we take a nonemptyX of Ω×Ω. Then,
(33)μ~(G~(X))=μ~(G~(X1×X2))=μ~(G(X1×X2)×G(X2×X1))=max⁡{μ(G(X1×X2)),μ(G(X2×X1))}≤max⁡{φ(max⁡(μ(X1),μ(X2))),φ(max⁡(μ(X2),μ(X1)))}=φ(max⁡(μ(X1),μ(X2)))=φ(μ~(X1×X2)).
Since φ∈A, by Theorem 9, the mapping G~ has at least one fixed point. This means that there exists (x0,y0)∈Ω×Ω such that G~(x0,y0)=(x0,y0) or, equivalently, G(x0,y0)=x0 and G(y0,x0)=y0. This proves that G has at least a coupled fixed point.Now, we consider the coupled system of integral equations:(34)x(t)=f(t,x(t),y(t))Γ(α)∫0tg(s,x(s),y(s))(t-s)1-αds,y(t)=f(t,y(t),x(t))Γ(α)∫0t‍g(s,y(s),x(s))(t-s)1-αds,t∈[0,1].Lemma 14.
Assume thatf∈C([0,1]×R×R,R∖{0}) and g∈C([0,1]×R×R,R). Then, (x,y)∈C[0,1]×C[0,1] is a solution of (34) if and only if (x,y)∈C[0,1]×C[0,1] is a solution of (4).Proof.
The proof is an immediate consequence of Lemma2, so we omit it.Next, we will study problem (4) under the following assumptions.(H1)
f∈C([0,1]×R×R,R∖{0}) and g∈C([0,1]×R×R,R).(H2)
The functionsf and g satisfy
(35)|f(t,x1,y1)-f(t,x2,y2)|≤φ1(max⁡⁡(|x1-x2|,|y1-y2|)),|g(t,x1,y2)-g(t,x2,y2)|≤φ2(max⁡(|x1-y1|,|x2-y2|)),respectively, for anyt∈[0,1] and x1, x2, y1, y2∈R, where φ1, φ2∈A and φ1 is continuous.Notice that assumption(H1) gives us the existence of two nonnegative constants k1 and k2 such that |f(t,0,0)|≤k1 and |g(t,0,0)|≤k2, for any t∈[0,1].(H3)
There existsr0>0 satisfying the inequalities
(36)(φ1(r)+k1)·(φ2(r)+k2)≤rΓ(α+1),φ2(r)+k2≤Γ(α+1).Theorem 15.
Under assumptions(H1)–(H3), problem (4) has at least one solution in C[0,1]×C[0,1].Proof.
In virtue of Lemma14, a solution (x,y)∈C[0,1]×C[0,1] of problem (4) satisfies
(37)x(t)=f(t,x(t),y(t))Γ(α)∫0tg(s,x(s),y(s))(t-s)1-αds,y(t)=f(t,y(t),x(t))Γ(α)∫0tg(s,y(s),x(s))(t-s)1-αds,hhhhhhhhhhhhhhhhhhhhhhhhhht∈[0,1].
We consider the space C[0,1]×C[0,1] equipped with the norm ∥(x,y)∥C[0,1]×C[0,1]=max⁡{∥x∥,∥y∥}, for any (x,y)∈C[0,1]×C[0,1].
InC[0,1]×C[0,1], we define the operator
(38)G(x,y)(t)=f(t,x(t),y(t))Γ(α)∫0t‍g(s,x(s),y(s))(t-s)1-αds,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhht∈[0,1].
Let F and G be the operators given by
(39)F(x,y)(t)=f(t,x(t),y(t)),G(x,y)(t)=1Γ(α)∫0tg(s,x(s),y(s))(t-s)1-αds
for any (x,y)∈C[0,1]×C[0,1] and any t∈[0,1]. Then,
(40)G(x,y)=F(x,y)·G(x,y).
Firstly, we will prove that G applies C[0,1]×C[0,1] into C[0,1]. To do this, it is sufficient to prove that F(x,y), G(x,y)∈C[0,1] for any (x,y)∈C[0,1]×C[0,1] since the product of continuous functions is continuous.
In virtue of assumption(H1), it is clear that F(x,y)∈C[0,1] for (x,y)∈C[0,1]×C[0,1]. In order to prove that G(x,y)∈C[0,1] for (x,y)∈C[0,1]×C[0,1], we fix t0∈[0,1] and we consider a sequence (tn)∈[0,1] such that tn→t0, and we have to prove that G(x,y)(tn)→G(x,y)(t0). Without loss of generality, we can suppose that tn>t0. Then, we have
(41)|G(x,y)(tn)-G(x,y)(t0)|=1Γ(α)|∫0tn‍g(s,x(s),y(s))(tn-s)1-αds-∫0t0‍g(s,x(s),y(s))(t0-s)1-αds|≤1Γ(α)|∫0tn‍g(s,x(s),y(s))(tn-s)1-αds-∫0tn‍g(s,x(s),y(s))(t0-s)1-αds|+1Γ(α)|∫0tn‍g(s,x(s),y(s))(t0-s)1-αds-∫0t0‍g(s,x(s),y(s))(t0-s)1-αds|≤1Γ(α)∫0tn‍|(tn-s)α-1-(t0-s)α-1|×|g(s,x(s),y(s))|ds+1Γ(α)∫t0tn‍|(t0-s)α-1||g(s,x(s),y(s))|ds.
By assumption (H1), since g∈C([0,1]×R×R,R), g is bounded on the compact set [0,1]×[-∥x∥,∥x∥]×[-∥y∥,∥y∥]. Denote by
(42)M=sup⁡{|g(s,x1,y1)|:s∈[0,1],x1∈[-∥x∥,∥x∥],hhhhhhhhhhhhhhhhhhhhhhhhy1∈[-∥y∥,∥y∥]}.
From the last estimate, we obtain
(43)|G(x,y)(tn)-G(x,y)(t0)|≤MΓ(α)∫0tn|(tn-s)α-1-(t0-s)α-1|ds+MΓ(α)∫t0tn|(t0-s)α-1|ds.
As 0<α<1 and tn>t0, we infer that
(44)|G(x,y)(tn)-G(x,y)(t0)|≤MΓ(α)[∫0t0|(tn-s)α-1-(t0-s)α-1|ds+∫t0tn|(tn-s)α-1-(t0-s)α-1|ds]+MΓ(α)∫t0tn‍1(s-t0)1-αds=MΓ(α)[∫t0tnds(tn-s)1-α+∫t0tn‍ds(s-t0)1-α∫0t0[(t0-s)α-1-(tn-s)α-1]ds+∫t0tnds(tn-s)1-α+∫t0tn‍ds(s-t0)1-α]+MΓ(α)∫t0tn‍1(s-t0)1-αds≤MΓ(α+1)[(tn-t0)α+t0α-tnα+(tn-t0)α+(tn-t0)α]+MΓ(α+1)(tn-t0)α=4MΓ(α+1)(tn-t0)α+MΓ(α+1)(t0α-tnα)<4MΓ(α+1)(tn-t0)α,
where the last inequality has been obtained by using the fact that t0α-tnα<0.
Therefore, sincetn→t0, from the last estimate, we deduce that G(x,y)(tn)→G(x,y)(t0). This proves that G(x,y)∈C[0,1]. Consequently, G:C[0,1]×C[0,1]→C[0,1]. On the other hand, for (x,y)∈C[0,1]×C[0,1] and t∈C[0,1], we have
(45)|G(x,y)(t)|=|F(x,y)(t)·G(x,y)(t)|=|f(t,x(t),y(t))||1Γ(α)∫0tg(s,x(s),y(s))(t-s)1-αds|≤[|f(t,x(t),y(t))-f(t,0,0)|+|f(t,0,0)|]×[1Γ(α)|∫0tg(s,x(s),y(s))-g(s,0,0)(t-s)1-αds+∫0tg(s,0,0)(t-s)1-αds|]≤1Γ(α)[φ1(max⁡⁡(|x(t)|,|y(t)|))+k1]×[∫0t‍|g(s,x(s),y(s))-g(s,0,0)|(t-s)1-αds+∫0t‍|g(s,0,0)|(t-s)1-αds]≤1Γ(α)[φ1(max⁡⁡(∥x∥,∥y∥))+k1]×[∫0t‍φ2(max⁡(|x(s)|,|y(s)|))(t-s)1-αds+k2∫0tds(t-s)1-α]≤1Γ(α)[φ1(max⁡⁡(∥x∥,∥y∥))+k1]·[φ2(max⁡⁡(∥x∥,∥y∥))+k2]∫0t‍ds(t-s)1-α≤1Γ(α+1)(φ1(∥(x,y)∥)+k1)·(φ2(∥(x,y)∥)+k2).
Now, taking into account assumption (H3), we infer that the operator G applies Br0×Br0 into Br0. Moreover, from the last estimates, it follows that
(46)∥F(Br0×Br0)∥≤φ1(r0)+k1,∥G(Br0×Br0)∥≤φ2(r0)+k2Γ(α+1).
Next, we will prove that the operatorsF and G are continuous on the ball Br0×Br0 and, consequently, G will be also continuous.
In fact, we fixε>0 and we take (x0,y0), (x,y)∈Br0×Br0 with ∥(x,y)-(x0,y0)∥ = ∥(x-x0,y-y0)∥ = max⁡{∥x-x0∥,∥y-y0∥}≤ε. Then, for t∈[0,1], we have
(47)|F(x,y)(t)-F(x0,y0)(t)|=|f(t,x(t),y(t))-f(t,x0(t),y0(t))|≤φ1(max⁡⁡(|x(t)-x0(t)|,|y(t),y0(t)|))≤φ1(max⁡(∥x-x0∥,∥y-y0∥))≤φ1(ε)<ε,
where we have used Remark 8. This proves the continuity of F on Br0×Br0.
In order to prove the continuity ofG on Br0×Br0, we have
(48)|G(x,y)(t)-G(x0,y0)(t)|=1Γ(α)|∫0t‍g(s,x(s),y(s))(t-s)1-αds-∫0t‍g(s,x0(s),y0(s))(t-s)1-αds|≤1Γ(α)∫0t‍|g(s,x(s),y(s))-g(s,x0(s),y0(s))|(t-s)1-αds≤1Γ(α)∫0t‍φ2(max⁡⁡(|x(s)-x0(s)|,|y(s)-y0(s)|))(t-s)1-αds≤1Γ(α)φ2(max⁡(∥x-x0∥,∥y-y0∥))∫0t‍ds(t-s)1-α≤1Γ(α+1)φ2(ε)<εΓ(α+1).
Therefore,
(49)∥G(x,y)-G(x0,y0)∥<εΓ(α+1)
and, consequently, G is a continuous operator on Br0×Br0.
In order to prove thatG satisfies assumptions of Theorem 13, only we have to check the condition
(50)μ(G(X1×X2))≤φ(max⁡(μ(X1),μ(X2)))
for any subsets X1 and X2 of Br0.
To do this, we fixε>0, t1, t2∈[0,1] with |t1-t2|≤ε and (x,y)∈X1×X2; then, we have
(51)|F(x,y)(t1)-F(x,y)(t2)|=|f(t1,x(t1),y(t1))-f(t2,x(t2),y(t2))|≤|f(t1,x(t1),y(t1))-f(t1,x(t2),y(t2))|+|f(t1,x(t2),y(t2))-f(t2,x(t2),y(t2))|≤φ1(max⁡(|x(t1)-x(t2)|,|y(t1)-y(t2)|))+ω(f,ε)≤φ1(max⁡(ω(x,ε),ω(y,ε)))+ω(f,ε),
where ω(f,ε) denotes the quantity
(52)ω(f,ε)=sup⁡{|f(t,x,y)-f(s,x,y)|:t,s∈[0,1],|t-s|≤ε,x,y∈[-r0,r0]}.
From the last estimate, we infer that
(53)ω(F(X1×X2),ε)≤φ1(max⁡⁡(ω(X1,ε),ω(X2,ε)))+ω(f,ε).
Since f(t,x,y) is uniformly continuous on bounded subsets of [0,1]×R×R, we deduce that ω(f,ε)→0 as ε→0 and, therefore,
(54)ω0(F(X1×X2))≤lim⁡ε→0φ1(max⁡(ω(X1,ε),ω(X2,ε))).
By assumption (H2), since φ1 is continuous, we infer
(55)ω0(F(X1×X2))≤φ1(max⁡(ω0(X1),ω0(X2))).
Now, we estimate the quantity ω0(G(X1×X2)).
Fixε>0, t1, t2∈[0,1] with |t1-t2|≤ε and (x,y)∈X1×X2. Without loss of generality, we can suppose that t1<t2; then, we have
(56)|G(x,y)(t2)-G(x,y)(t1)|=1Γ(α)|∫0t2‍g(s,x(s),y(s))(t2-s)1-αds-∫0t1‍g(s,x(s),y(s))(t1-s)1-αds|≤1Γ(α)[∫0t1‍|(t2-s)α-1-(t1-s)α-1|×|g(s,x(s),x(σs))|ds+∫t1t2‍(t2-s)α-1|g(s,x(s),y(s))|ds]≤1Γ(α)[∫0t1‍[(t1-s)α-1-(t2-s)α-1]+∫t1t2‍(t2-s)α-1|g(s,x(s),y(s))|ds×|g(s,x(s),x(σs))|ds+∫t1t2‍(t2-s)α-1|g(s,x(s),y(s))|ds].
Since g∈C([0,1]×R×R,×R) is bounded on the compact subsets of [0,1]×R×R, particularly, on [0,1]×[-r0,r0]×[-r0,r0]. Put L=sup⁡{|g(t,x,y)|:t∈[0,1], x, y∈[-r0,r0]}. Then, from the last inequality, we infer that
(57)|G(x,y)(t2)-G(x,y)(t1)|≤LΓ(α)[∫0t1‍[(t1-s)α-1-(t2-s)α-1]ds+∫t1t2‍(t2-s)α-1ds]≤LΓ(α+1)[(t2-t1)α+t1α-t2α+(t2-t1)α]≤2LΓ(α+1)(t2-t1)α≤2LΓ(α+1)εα,
where we have used the fact that t1α-t2α≤0. Therefore,
(58)ω(G(X1×X2),ε)≤2LΓ(α+1)εα.
From this, it follows that
(59)ω0(G(X1×X2))=0.
Next, by Proposition7, (46), (55), and (59), we have
(60)ω0(G(X1×X2))=ω0(F(X1×X2)·G(X1×X2))≤∥F(X1×X2)∥ω0(G(X1×X2))+∥G(X1×X2)∥ω0(F(X1×X2))≤∥F(Br0×Br0)∥ω0(G(X1×X2))+∥G(Br0×Br0)∥ω0(F(X1×X2))≤φ2(r0)+k2Γ(α+1)φ1(max⁡(ω0(X1),ω0(X2))).
By assumption (H3), since φ2(r0)+k2≤Γ(α+1) and since it is easily proved that if α∈[0,1] and φ∈A, then αφ∈A, we deduce that
(61)ω0(G(X1×X2))≤φ(max⁡(ω0(X1),ω0(X2))),
where φ∈A.
Finally, by Theorem13, the operator G has at least a coupled fixed point and this is the desired result. This completes the proof.The nonoscillary character of the solutions of problem (4) seems to be an interesting question from the practical point of view. This means that the solutions of problem (4) have a constant sign on the interval (0,1). In connection with this question, we notice that if f(t,x,y) and g(t,x,y) have constant sign and are equal (this means that f(t,x,y)>0 and g(t,x,y)≥0 or f(t,x,y)<0 and g(t,x,y)≤0 for any t∈[0,1] and x, y∈R) and under assumptions of Theorem 15, then the solution (x,y)∈C[0,1]×C[0,1] of problem (4) given by Theorem 15 satisfies x(t)≥0 and y(t)≥0 for t∈[0,1], since the solution (x,y) satisfies the system of integral equations
(62)x(t)=f(t,x(t),y(t))Γ(α)∫0t‍g(s,x(s),y(s))(t-s)1-αds,y(t)=f(t,y(t),x(t))Γ(α)∫0t‍g(s,y(s),x(s))(t-s)1-αds,0≤t≤1.
On the other hand, if we perturb the data function in problem (4) of the following manner:
(63)D0+α[x(t)f(t,x(t),y(t))]=g(t,x(t),y(t))+η(t),D0+α[y(t)f(t,y(t),x(t))]=g(t,y(t),x(t))+η(t),hhhhhhhhhhhhhhhhhhhhhhhhhhhhh0<t<1,
where 0<α<1, f∈C([0,1]×R×R,R∖{0}), g∈C([0,1]×R×R,R), and η∈C[0,1], then, under assumptions of Theorem 15, problem (63) can be studied by using Theorem 15, where assumptions (H1) and (H2) are automatically satisfied and we only have to check assumption (H3). This fact gives a great applicability to Theorem 15.Before presenting an example illustrating our results, we need some facts about the functions involving this example. The following lemma appears in [18].Lemma 16.
Letφ:R+→R+ be a nondecreasing and upper semicontinuous function. Then, the following conditions are equivalent:(i)
lim⁡n→∞φn(t)=0 for any t≥0;(ii)
φ(t)<t for any t>0.Particularly, the functionsα1, α2:R+→R+ given by α1(t)=arctant and α2(t)=t/(1+t) belong to the class A, since they are nondecreasing and continuous, and, as it is easily seen, they satisfy (ii) of Lemma 16.On the other hand, since the functionα1(t)=arctant is concave (because α′′(t)≤0) and α1(0)=0, we infer that α1 is subadditive and, therefore, for any t,t′∈R+, we have
(64)|α1(t)-α1(t′)|=|arctant-arctant′|≤arctan|t-t′|.
Moreover, it is easily seen that max⁡(α1,α2) is a nondecreasing and continuous function because α1 and α2 are nondecreasing and continuous and max⁡(α1,α2) satisfies (ii) of Lemma 16. Therefore, max⁡(α1,α2)∈A.Now, we are ready to present an example where our results can be applied.Example 17.
Consider the following coupled system of fractional hybrid differential equations:(65)D0+1/2[(1/20)(|y(t)|/(1+|y(t)|)))-1x(t)×((1/20)(|y(t)|/(1+|y(t)|))14+(110)arctan|x(t)|(120)(|y(t)|(1+|y(t)|)))-1+(120)(|y(t)|(1+|y(t)|)))-1]=17+19x(t)+110y(t),D0+1/2[+(1/20)(|x(t)|/(1+|x(t)|)))-1y(t)×(14+(110)arctan|y(t)|(120)(|x(t)|(1+|x(t)|))+(120)(|x(t)|(1+|x(t)|)))-1+(120)(|x(t)|(1+|x(t)|)))-1]=17+19y(t)+110x(t),hhhhhhhhhhhhh0<t<1,x(0)=y(0)=0.Notice that problem (65) is a particular case of problem (4), where α=1/2, f(t,x,y)=1/4+(1/10)arctan|x|+(1/20)(|y|/(1+|y|)), and g(t,x,y)=1/7+(1/9)x+(1/10)y.It is clear thatf∈C([0,1]×R×R,R∖{0}) and g∈C([0,1]×R×R,R) and, moreover, k1=sup⁡{|f(t,0,0)|:t∈[0,1]}=1/4 and k2=sup⁡{|g(t,0,0)|:t∈[0,1]}=1/7. Therefore, assumption (H1) of Theorem 15 is satisfied.Moreover, fort∈[0,1] and x1, x2, y1, y2∈R, we have
(66)|f(t,x1,y1)-f(t,x2,y2)|≤110|arctan|x1|-arctan|x2||+120||y1|1+|y1|-|y2|1+|y2||≤110arctan||x1|-|x2||+120||y1|-|y2|(1+|y1|)(1+|y2|)|≤110arctan⁡(|x1-x2|)+120|y1-y2|1+|y1-y2|=110α1(|x1-x2|)+120α2(|y1-y2|)≤110max⁡⁡(α1,α2)(|x1-x2|)+110max⁡⁡(α1,α2)(|y1-y2|)≤110[2max⁡(α1,α2)×max⁡(|x1-x2|,|y1-y2|)]=15max⁡(α1,α2)(max⁡(|x1-x2|,|y1-y2|)).
Therefore, φ1(t)=(1/5)max⁡(α1(t),α2(t)) and φ1∈A.On the other hand,(67)|g(t,x1,y1)-g(t,x2,y2)|≤19|x1-x2|+110|x2-y2|≤19(|x1-y1|+|y2-y2|)≤19(2max⁡(|x1-y1|,|y2-y2|))=29max⁡(|x1-y1|,|y2-y2|),
and φ2(t)=(2/9)t. It is clear that φ2∈A. Therefore, assumption (H2) of Theorem 15 is satisfied.In our case, the inequality appearing in assumption(H3) of Theorem 15 has the expression
(68)[15max⁡(arctanr,r1+r)+14][29r+17]≤rΓ(32).
It is easily seen that r0=1 satisfies the last inequality. Moreover,
(69)29r0+17=29+17≤Γ(32)≅0.88623.
Finally, Theorem 15 says that problem (65) has at least one solution (x,y)∈C[0,1] such that max⁡(∥x∥,∥y∥)≤1.



---

*Source: 672167-2014-07-14.xml*