# Existence of the Solution for System of Coupled Hybrid Differential Equations with Fractional Order and Nonlocal Conditions

Authors: Khalid Hilal; Ahmed Kajouni
Journal: International Journal of Differential Equations
 (2016)

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## Abstract

This paper is motivated by some papers treating the fractional hybrid differential equations with nonlocal conditions and the system of coupled hybrid fractional differential equations; an existence theorem for fractional hybrid differential equations involving Caputo differential operators of order1<α≤2 is proved under mixed Lipschitz and Carathéodory conditions. The existence and uniqueness result is elaborated for the system of coupled hybrid fractional differential equations.


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## Body

## 1. Introduction

Our aim in this paper is to study the existence of solution for the boundary value problems for hybrid differential equations with fractional order1<α≤2 and nonlocal condition (BVPHDEFNL for short) of the form (1)Dαcxtft,xt=gt,xta.e.t∈J=0,T,1<α≤2,x0f0,x0=Lx,xTfT,xT=xT,where Dαc is the Caputo fractional derivative.f ∈ C ( J × R , R ∖ { 0 } ) ,   g ∈ C ( J × R , R ) ,   L : C ( J , R ) → R is a continuous function and xT∈R. And exploitation of results obtained to study the existence of solutions for a system of coupled hybrid fractional differential equations is as follows:(2)Dαcxtf1t,xt,yt=g1t,xt,yta.e.t∈0,1,1<α≤2,Dβcytf2t,xt,yt=g2t,xt,yta.e.t∈0,1,1<β≤2,x0f10,x0,y0=L1x,y;x1=0,y0f20,x0,y0=L2x,y;y1=0,where Dαc is the Caputo fractional derivative.f i ∈ C ( [ 0,1 ] × R × R , R ∖ { 0 } ), gi∈C([0,1]×R×R,R), and Li:C([0,1],R)×C([0,1],R)→R are continuous functions (i=1,2).Fractional differential equations are a generalization of ordinary differential equations and integration to arbitrary noninteger orders. The origin of fractional calculus goes back to Newton and Leibniz in the seventeenth century. It is widely and efficiently used to describe many phenomena arising in engineering, physics, economy, and science. There are several concepts of fractional derivatives, some classical, such as Riemann-Liouville or Caputo definitions. For noteworthy papers dealing with the integral operator and the arbitrary fractional order differential operator, see [1–7].The quadratic perturbations of nonlinear differential equations have attracted much attention. We call such fractional hybrid differential equations. There have been many works on the theory of hybrid differential equations, and we refer the readers to the articles [8–12].Dhage and Lakshmikantham [11] discussed the following first order hybrid differential equation (3)ddtxtft,xt=gt,xta.e.t∈J=0,T,xt0=x0∈R,where f∈C(J×R,R∖{0}) and g∈C(J×R,R). They established the existence, uniqueness results, and some fundamental differential inequalities for hybrid differential equations initiating the study of theory of such systems and proved, utilizing the theory of inequalities, the existence of extremal solutions and comparison results.Zhao et al. [13] have discussed the following fractional hybrid differential equations involving Riemann-Liouville differential operators:(4)Dqxtft,xt=gt,xta.e.t∈J=0,T,x0=0,where f∈C(J×R,R∖{0}) and g∈C(J×R,R). The authors of [13] established the existence theorem for fractional hybrid differential equation and some fundamental differential inequalities. They also established the existence of extremal solutions.Hilal and Kajouni [14] have studied boundary fractional hybrid differential equations involving Caputo differential operators of order 0<α<1 as follows: (5)Dαxtft,xt=gt,xta.e.t∈J=0,T,ax0f0,x0+bxTfT,xT=c,where f∈CJ×R,R∖0,g∈C(J×R,R) and a, b, and c are real constants with a+b≠0. They proved the existence result for boundary fractional hybrid differential equations under mixed Lipschitz and Carathéodory conditions. Some fundamental fractional differential inequalities are also established which are utilized to prove the existence of extremal solutions. Necessary tools are considered and the comparison principle is proved which will be useful for further study of qualitative behavior of solutions.The nonlocal condition is a condition attached to the main equation; it replaces the classic nonlocal condition in order to model physical phenomena of the fashion nearest from reality. The nonlocal condition involves the function(6)Lx=∑i=1pcixti,where ci,i=1,2,…,p, are given constants and 0<t1<t2<⋯<tp.Let us observe that Cauchy problems with nonlocal conditions were initiated by Byszewski and Lakshmikantham [2] and, since then, such problems have also attracted several authors including A. Aizicovici, K. Ezzinbi, Z. Fan, J. Liu, J. Liang, Y. Lin, T.-J. Xiao, G. N’Guérékata, E. Hernàndez, and H. Lee (see [2, 15]).


## 2. Preliminaries

In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper.ByX=C(J,R) we denote the Banach space of all continuous functions from J=[0,T] into R with the norm(7)y=sup⁡yt,t∈J.And let C(J×R,R) denote the class of functions g:J×R→R such that (i)
the mapt↦g(t,x) is measurable for each x∈R,(ii)
the mapx↦g(t,x) is continuous for each t∈J.The class C(J×R,R) is called the Carathéodory class of functions on J×R which are Lebesgue integrable when bounded by a Lebesgue integrable function on J.ByL1(J;R) we denote the space of Lebesgue integrable real-valued functions on J equipped with the norm ·L1 defined by(8)xL1=∫0Txsds.Definition 1.
The fractional integral of the functionh∈L1([a,b],R+) of order α∈R+ is defined by(9)Iaαht=1Γα∫att-sα-1hsds,where Γ is the gamma function.Definition 2.
For a functionh given on the interval [a,b], the Caputo fractional order derivative of h is defined by(10)Dca+αht=1Γn-α∫att-sn-α-1hnsds,where n=[α]+1 and [α] denotes the integer part of α.Lemma 3 (see [16]).
Letα>0. Then the fractional differential equation(11)Dαcht=0has solutions(12)ht=c0+c1t+⋯+cn-1tn-1,ci∈R,i=1,2,…,n-1,n=α+1.Lemma 4 (see [16]).
Letα>0. Then(13)IαDαht=ht+c0+c1t+⋯+cn-1tn-1,for some ci∈R,i=1,2,…,n-1,n=[α]+1.Definition 5.
By a solution of the BVPHDEFNL (1) we mean a function x∈C(J,R) such that (i)
d2u/dt2∈L1(J,R), where u:t↦x/f(t,x) for each x∈R,(ii)
x satisfies the equations in (1).


## 3. Existence Result

In this section, we prove the existence results for the boundary value problems for hybrid differential equations with fractional order (1) on the closed and bounded interval J=[0,T] under mixed Lipschitz and Carathéodory conditions on the nonlinearities involved in it.We defined the multiplication inX by (14)xyt=xtyt,forx,y∈X.Clearly, X=C(J;R) is a Banach algebra with respect to above norm and multiplication in it.We prove the existence of solution for the BVPHDEFNL (1) by a fixed point theorem in Banach algebra due to Dhage [10].Lemma 6 (see [10]).
LetS be a nonempty, closed convex, and bounded subset of the Banach algebra X and let A:X→X and B:X→X be two operators such that(a)
A is Lipschitzian with a Lipschitz constant α,(b)
B is completely continuous,(c)
x=AxBy⇒x∈S for all y∈S,(d)
αM<1, where M=B(S)=sup⁡{B(x):x∈S}.Then the operator equation AxBx=x has a solution in S.We make the following assumptions:(H0):
the functionx↦x/f(t,x) is increasing in R almost everywhere for t∈J.(H1):
there exists a constantL>0 such that (15)ft,x-ft,y≤Lx-y,for allt∈J and x,y∈R.(H2):
there exists a functionh∈L1(J,R) such that (16)gt,x≤hta.e.t∈J,for allx∈R.(H3):
there exists a constantM>0 such that |L(y)|≤M, for each y∈C(J;R).As a consequence of Lemmas 3 and 4 we have the following result which is useful in what follows.Lemma 7.
Assume that hypothesis(H0) holds. Then for any h∈L1(J;R), the function x∈C(J;R) is a solution of the BVPHDEFNL: (17)Dcαxtft,xt=hta.e.t∈J=0,T,1<α≤2,x0f0,x0=Lx,xTfT,xT=xTif and only if x satisfies the hybrid integral equation(18)xt=ft,xt1Γα∫0tt-sα-1hsds-tTΓα∫0TT-sα-1hsds-tT-1Lx+tTxT.Proof.
Assume thatx is a solution of the problem (18). Applying the Caputo fractional operator of the order α, we obtain the first equation in (17). Again, substituting t=0 and t=T in (18) we will have the second equation in (17).
Conversely,Dcα(x(t)/f(t,x(t)))=h(t), so we get(19)xtft,xt+c0+c1t=Iαht.Then x(0)/f(0,x(0))+c0=0 and c0=-L(x), and even (20)xTfT,xT+c0+c1T=∫0TT-sα-1hsds.Thus,(21)c1=1TLx-xT+1Γα∫0TT-sα-1hsdsimplies that(22)xtft,xt=1Γα∫0tt-sα-1hsds-tTΓα∫0TT-sα-1hsds-tT-1Lx+tTxT.Theorem 8.
Assume hypotheses(H0)–(H3). Further, if(23)L2TαΓα+1hL1+M+xT<1,then the hybrid fractional order differential equation (1) has a solution defined on J.Proof.
We defined a subsetS of X by(24)S=x∈Xx≤N,where N=F0(2Tα/Γα+1hL1+M+|xT|)/(1-L((2Tα/Γ(α+1))hL1+M+|xT|)) and F0=supt∈J⁡|f(t,0)|.
It is clear thatS satisfies hypothesis of Lemma 6. By an application of Lemma 7, (1) is equivalent to the nonlinear hybrid integral equation(25)xt=ft,xt1Γα∫0tt-sα-1gs,xsds-tTΓα∫0TT-sα-1gs,xsds-tT-1Lx+tTxT.
Define two operatorsA:X→X and B:S→X by (26)Axt=ft,xt,t∈J,Bxt=1Γα∫0tt-sα-1gs,xsds-tTΓα∫0TT-sα-1gs,xsds-tT-1Lx+tTxT.Then the hybrid integral equation (25) is transformed into the operator equation as (27)xt=AxtBxt,t∈J.We will show that the operators A and B satisfy all the conditions of Lemma 6.
Claim 1. Let x,y∈X. Then by hypothesis (H1), (28)Axt-Ayt=ft,xt-ft,yt≤Lxt-yt≤Lx-y,for all t∈J. Taking supremum over t, we obtain (29)Ax-Ay≤Lx-y,for all x,y∈X.
Claim 2 (we show that B is continuous in S ). Let (xn) be a sequence in S converging to a point x∈S. Then by Lebesgue dominated convergence theorem,(30)limn→∞⁡1Γα∫0tt-sα-1gs,xnsds=1Γα∫0tt-sα-1limn→∞⁡gs,xnsds,limn→∞bΓα∫0TT-sα-1gs,xnsds=bΓα∫0TT-sα-1limn→∞⁡gs,xnsds.And since L is a continuous function(31)limn→∞⁡Lxn=Lx,then(32)limn→∞⁡Bxnt=limn→∞⁡1Γα∫0tt-sα-1gs,xnsds-tTΓα∫0TT-sα-1gs,xnsds-tT-1Lxn+tTxT=limn→∞⁡1Γα∫0tt-sα-1gs,xnsds-limn→∞⁡tTΓα∫0TT-sα-1gs,xnsds-limn→∞⁡tT-1Lxn+tTxT=1Γα∫0tt-sα-1gs,xsds-tTΓα∫0TT-sα-1gs,xsds-tT-1Lx+tTxT=Bxt,for all t∈J. This shows that B is a continuous operator on S.
Claim 3 ( B is compact operator on S ). First, we show that B(S) is a uniformly bounded set in X.
Letx∈S. Then by hypothesis (H2), for all t∈J, (33)Bxt≤1Γα∫0tt-sα-1gs,xsds+1Γα∫0TT-sα-1gs,xsds+Lx+xT≤TααΓαhL1+TααΓαhL1+Lx+xT≤2TαΓα+1hL1+M+xT.Thus, Bx≤(2Tα/Γ(α+1))hL1+M+|xT|, for all x∈S.
This shows thatB is uniformly bounded on S.
Next, we show thatB(S) is an equicontinuous set on X.
We setp(t)=∫0th(s)ds.
Lett1,t2∈J. Then for any x∈S,(34)Bxt1-Bxt2=1Γα∫0t1t1-sα-1gs,xsds-1Γα∫0t2t2-sα-1gs,xsds-t1-t21TΓα∫0TT-sα-1gs,xsds-1Tt1-t2Lx+1Tt1-t2xT≤Tα-1Γα∫t1t2gs,xsds+Tα-1Γαt1-t2∫0Tgs,xsds+M+xTTt1-t2≤Tα-1Γαpt1-pt2+Tα-1Γα+M+xTTt1-t2.
Sincep is continuous on compact J, it is uniformly continuous. Hence,(35)t1-t2<η⟹Bxt1-Bxt2<ε,∀ε>0,∃η>0for all t1,t2∈J and for all x∈X.
This shows thatB(S) is an equicontinuous set in X.
Then by Arzelá-Ascoli theorem,B is a continuous and compact operator on S.
Claim 4 (hypothesis ( c ) of Lemma6 is satisfied). Let x∈X and y∈S be arbitrary such that x=AxBy. Then,(36)xt=AxtByt≤ft,xt1Γα∫0tt-sα-1gs,xsds-tTΓα∫0TT-sα-1gs,xsds-tT-1Lx+tTxT≤ft,xt-ft,0+ft,02TαΓα+1hL1+M+xT≤Lxt+F02TαΓα+1hL1+M+xT,and so,(37)xt-L2TαΓα+1hL1+M+xTxt≤F02Tαα+1hL1+M+xT,which implies(38)xt≤F02Tα/Γα+1hL1+M+xT1-L2Tα/Γα+1hL1+M+xT.Taking supremum over t,(39)x≤F02Tα/Γα+1hL1+M+xT1-L2Tα/Γα+1hL1+M+xT=N.Then x∈S, and hypothesis (c) of Lemma 6 is satisfied.
Finally, we have(40)M=BS=sup⁡Bx:x∈S≤2TαΓα+1hL1+M+xT.So, (41)αM≤L2TαΓα+1hL1+M+xT<1.Thus, all the conditions of Lemma 6 are satisfied and hence the operator equation AxBx=x has a solution in S. As a result, BVPHDEFNL (1) has a solution defined on J. This completes the proof.


## 4. An Example

In this section we give an example to illustrate the usefulness of our main results. Let us consider the following fractional boundary value problem:(42)Dcα2+ln⁡t+1xt+x2te1-t=e-x2tx2t+t2+1a.e.t∈J=0,1,1<α≤2,x0f0,x0=∑i=1ncixti;x1f1,x1=1,where 0<t1<t2<⋯<tn<1,ci,i=1,2,…,n, are given positive constants.And∑i=1nci<(1-π)/2M¯, where M¯=max1≤i≤n⁡L(ti).We set(43)ft,x=et-12+ln⁡1+t1+x,t,x∈0,1×0,+∞,gt,x=e-x2x2+t2+1,Lxt=∑i=1ncixti.Let x,y∈[0,+∞) and t∈J. We have(44)ft,x-ft,y=et-12+ln⁡1+t11+x-11+y≤12x-y1+x1+y≤12x-y.Hence, condition (H1) holds with L=1/2. Also we have(45)gt,x=e-x2x2+t2+1≤ht,where h(t)=1/(1+t2). We have (46)∫01htdt=∫0111+t2=π4.Then condition (H2) holds.Furthermore, sinceL∈C(J,R), then we set M¯=max1≤i≤n⁡L(ti) and we have(47)Lti=∑i=1nciLti≤M¯∑i=1nci.We will check that condition (23) is satisfied with T=1.Since∑i=1nci<(1-π)/2M¯, then π+2M¯∑i=1nci<1.Thus,(48)πΓα+1+2M¯∑i=1nci<1,which is satisfied for each α∈(1,2]. Then by Theorem 8 problem (42) has a solution on [0,1].


## 5. System of Coupled Hybrid Fractional Differential Equations

The aim of this section is to obtain the existence results, by means of Banach’s fixed point theorem, for the problem of coupled hybrid fractional differential equations for (1). Consider (49)Dcαxtf1t,xt,yt=g1t,xt,yta.e.t∈0,1,1<α≤2,Dcβytf2t,xt,yt=g2t,xt,yta.e.t∈0,1,1<β≤2,x0f10,x0,y0=L1x,y,x1=0,y0f20,x0,y0=L2x,y,y1=0,where Dαc is the Caputo fractional derivative.f i ∈ C ( J × R × R , R ∖ { 0 } ), gi∈C([0,1]×R×R,R), and Li:C([0,1],R)×C([0,1],R)→R are continuous functions (i=1,2).Main Results. Let Ω={ω(t)∖ω(t)∈C1([0,1])} denote a Banach space equipped with the norm ω=sup⁡{|ω(t)|,t∈[0,1]}, where Ω=K×R. Notice that the product space (K×R,(x,y)) with the norm (x,y)=x+y, (x,y)∈K×R is also a Banach space.In view of Lemma7, we define an operator Φ:K×R→K×R by (50)Φx,yt=Φ1x,yt,Φ2x,yt,where(51)Φ1x,yt=f1t,xt,yt1Γα∫0tt-sα-1g1s,xs,ysds-tΓα∫011-sα-1g1s,xs,ysds-t-1L1x,y,Φ2x,yt=f2t,xt,yt1Γβ∫0tt-sβ-1g2s,xs,ysds-tΓβ∫011-sβ-1g2s,xs,ysds-t-1L2x,y.In the sequel, we need the following assumptions: (H1′):
the functionsfi are continuous and bounded; that is, there exist positive numbers Li such that |fi(t,u,v)|≤Li for all (t,u,v)∈[0,1]×R×R(i=1,2).(H2′):
there exist real constantsρ0,δ0>0 and ρi,δi≥0(i=1,2) such that |g1(t,x,y)|≤ρ0+ρ1|x|+ρ2|y| and |g2(t,x,y)|≤δ0+δ1|x|+δ2|y| for all x,y∈R(i=1,2).(H3′):
there exist real constantsM1,M2>0|L1(x,y)|≤M1 and |L2(x,y)|≤M2 for each x,y∈C([0,1]).(H4′):
there exist real constantsγ1,γ,γ1′,γ2′ such that |L1(x1,y1)-L1(x2,y2)|≤γ1|x1-x2|+γ2|y1-y2| and |L2(x1,y1)-L2(x2,y2)|≤γ1′|x1-x2|+γ2′|y1-y2|.For brevity, let us set(52)μ1=2L1Γα+1,μ2=2L2Γβ+1,(53)μ0=min⁡1-μ1ρ1+μ2δ1,1-μ1ρ2+μ2δ2.Now we present our result for the existence and uniqueness of solutions for problem (49). This result is based on Banach’s contraction mapping principle.Theorem 9.
Suppose that conditions(H1′), (H3′), and (H4′) hold and that g1,g2:[0,1]×R2→R are continuous functions. In addition, there exist positive constants ηi,ξi,i=1,2, such that(54)g1t,x1,y1-g1t,x2,y2=η1x1-x2+η2y1-y2,g2t,x1,y1-g2t,x2,y2=ξ1x1-x2+ξ2y1-y2,∀t∈0,1,x1,x2,y1,y2∈R.If μ1(η1+η2)+μ2(ξ1+ξ2)+L1(γ1+γ2)+L2(γ1′+γ2′)<1, then problem (49) has a unique solution.Proof.
Let us setsupt∈[0,1]⁡g1(t,0,0)=κ1<∞ and supt∈[0,1]g2(t,0,0)=κ2<∞ and define a closed ball: Br={(x,y)∈K×R:(x,y)≤r}, where(55)r≥μ1κ1+μ2κ2+M1L1+M2L21-μ1η1+η2-μ2ξ1+ξ2.
Claim 5 (we show that Φ B r ⊂ B r ). Let (x,y)∈Br. We have (56)Φ1x,yt≤M1supt∈0,11Γα∫0tt-sα-1g1s,xs,ysds+1Γα∫011-sα-1g1s,xs,ysds+L1x,y=M1supt∈0,11Γα∫0tt-sα-1g1s,xs,ys-g1s,0,0+g1s,0,0ds+1Γα∫01t-sα-1g1s,xs,ys-g1s,0,0+g1s,0,0ds+L1x,y≤M12Γα+1η1x+η2y+κ1+L1≤M12Γα+1η1+η2r+κ1+L1≤μ1η1+η2r+κ1+M1L1.Hence,(57)Φ1x,y≤μ1η1+η2r+κ1+M1L1,Φ2x,y≤μ2ξ1+ξ2r+κ2+M2L2.From (57), it follows that Φ(x,y)≤r.
Next, for(x1,y1),(x2,y2)∈K×R and for any t∈[0,1], we have (58)Φ1x2,y2t-Φ1x1,y1t≤L1∫0tt-sα-1Γαg1s,x2s,y2s-g1s,x1s,y1sds+∫011-sα-1Γαg1s,x2s,y2s-g1s,x1s,y1sds+L1x2,y2-L1x1,y1≤L12Γα+1η1x2t-x1t+η2y2t-y1t+γ1x2t-x1t+γ2y2t-y1t≤μ1η1x2-x1+η2y2-y1+L1γ1x2-x1+γ2y2-y1≤μ1η1+η2+L1γ1+γ2x2-x1+y2-y1which yields(59)Φ1x2,y2-Φ1x1,y1≤μ1η1+η2+L1γ1+γ2x2-x1+y2-y1.Working in a similar manner, one can find that(60)Φ2x2,y2-Φ2x1,y1≤μ2ξ1+ξ2+L2γ1′+γ2′x2-x1+y2-y1.We deduce that (61)Φx2,y2-Φx1,y1≤μ1η1+η2+μ2ξ1+ξ2+L1γ1+γ2+L2γ1′+γ2′x2-x1+y2-y1.In view of condition μ1(η1+η2)+μ2(ξ1+ξ2)+L1(γ1+γ2)+L2(γ1′+γ2′)<1, it follows that Φ is a contraction. So Φ has a unique fixed point. This implies that problem (49) has a unique solution on [0,1]. This completes the proof.In our second result, we discuss the existence of solutions for problem (49) by means of Leray-Schauder alternative.Lemma 10 (see [17]).
LetF:J→J be a completely continuous operator (i.e., a map that is restricted to any bounded set in G is compact). Let P(F)={x∈J:x=λFxforsome0<λ<1}. Then either the set P(F) is unbounded or F has at least one fixed point.Theorem 11.
Assume that conditions(H1′)–(H3′) hold. Furthermore, it is assumed that μ1ρ1+μ2δ1<1 and μ1ρ2+μ2δ2<1, where μ1 and μ2 are given by (52). Then the boundary value problem (49) has at least one solution.Proof.
We will show that the operatorΦ:K×R→K×R satisfies all the assumptions of Lemma 10. In the first step, we prove that the operator Φ is completely continuous. Clearly, it follows by the continuity of functions f1, f2, g1, and g2 that the operator Φ is continuous.
LetS⊂K×R be bounded. Then we can find positive constants N1 and N2 such that (62)g1t,xt,yt≤N1g2t,xt,yt≤N2,∀x,y∈S.Thus, for any x,y∈S, we can get (63)Φ1x,yt≤L1∫0tt-sα-1Γαg1s,xs,ysds+t∫011-sβ-1Γβg1s,xs,ysds+1-tL1x,y≤L1N12Γα+1+M1which yields(64)Φ1x,y=N1μ1+M1.In a similar manner,(65)Φ2x,y=N2μ2+M2.We deduce that the operator Φ is uniformly bounded.
Now we show that the operatorΦ is equicontinuous.
We takeτ1,τ2∈[0,1] with τ1<τ2 and obtain (66)Φ1xτ2,yτ2-Φ1xτ1,yτ1≤L1N1∫0τ2τ2-sα-1Γαds-∫0τ1τ1-sα-1Γαds+L1N1τ2-τ1∫011-sα-1Γαds+M1τ2-τ1≤L1N1∫0τ1τ1-sα-1-τ2-sα-1Γαds-∫τ1τ2τ2-sα-1Γαds+L1N1τ2-τ1∫011-sα-1Γαds+M1τ2-τ1⟶τ1→τ20,Φ2xτ2,yτ2-Φ2xτ1,yτ1≤L2N2∫0τ2τ2-sβ-1Γβds-∫0τ1τ1-sβ-1Γβds+L2N2τ2-τ1∫011-sβ-1Γβds+M2τ2-τ1≤L2N2∫0τ1τ1-sβ-1-τ2-sβ-1Γβds-∫τ1τ2τ2-sβ-1Γβds+L2N2τ2-τ1∫011-sβ-1Γβds+M2τ2-τ1⟶τ1→τ20,which tend to 0 independently of (x,y). This implies that the operator Φ(x,y) is equicontinuous. Thus, by the above findings, the operator Φ(x,y) is completely continuous.
In the next step, it will be established that the setP={(x,y)∈K×R/(x,y)=λΦ(x,y),0≤λ≤1} is bounded.
Let(x,y)∈P. Then we have (x,y)=λΦ(x,y). Thus, for any t∈[0,1], we can write (67)xt=λΦ1x,yt,yt=λΦ2x,yt.Then,(68)xt≤2L1Γα+1ρ0+ρ1x+ρ2y+M1,yt≤2L2Γβ+1δ0+δ1x+δ2y+M2,which imply that(69)x≤μ1ρ0+ρ1x+ρ2y+M1,y≤μ2δ0+δ1x+δ2y+M2.Thus,(70)x+y≤μ1ρ0+μ2δ0+M1+M2+μ1ρ1+μ2δ1x+μ1ρ2+μ2δ2y,which, in view of (55), gives(71)x,y≤μ1ρ0+μ2δ0+M1+M2μ0.This shows that the set is bounded. Hence, all the conditions of Lemma 10 are satisfied and consequently the operator Φ has at least one fixed point, which corresponds to a solution of problem (49). This completes the proof.



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*Source: 4726526-2016-06-05.xml*