# Existence Results for a Class of Coupled Hilfer Fractional Pantograph Differential Equations with Nonlocal Integral Boundary Value Conditions

Authors: Karim Guida; Khalid Hilal; Lahcen Ibnelazyz
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

This paper deals with the existence and uniqueness of solutions for a new class of coupled systems of Hilfer fractional pantograph differential equations with nonlocal integral boundary conditions. First of all, we are going to give some definitions that are necessary for the understanding of the manuscript; second of all, we are going to prove our main results using the fixed point theorems, namely, Banach’s contraction principle and Krasnoselskii’s fixed point theorem; in the end, we are giving two examples to illustrate our results.


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## Body

## 1. Introduction

Differential equations play a very important role in the understanding of qualitative features of many phenomenon and processes in different areas and practical fields. A lot of works have been done concerning these equations in the recent years for their importance in applied sciences; for more details about differential equations and their applications, we refer the readers to [1–7].A more general way to describe natural differential equations is through fractional calculus. Fractional calculus has attracted many researchers recently; this branch of mathematics is used in the modelling of many problems in various fields, like biology, physics, control theory, and economics; for more details, we give the following classical references [8–13].There are many different definitions of fractional integrals and derivatives in the literature [12]; the most popular definitions are the Riemann-Liouville and the Caputo fractional derivatives. A generalization of these derivatives was introduced by Hilfer in [14], known by the Hilfer fractional derivative of order α and type β∈0,1, and we can find the Riemann-Liouville fractional derivative when β=0, and the Caputo fractional derivative when β=1. Fractional differential equations involving the Hilfer fractional derivative have many applications, see [15–18] and the references therein.On the other hand, another important class of differential equations are called pantograph equations, which are a special class of delay differential equations arising in deterministic situations and are of the form(1)g′t=kgt+lgλt,t∈0,b,b>0,0<λ<1,g0=g0.They are also called equations with proportional delays. This class of differential equations was not properly investigated under fractional derivatives. Pantograph is a device used in drawing and scaling. But, recently, this device is being used in electric trains [19, 20]. Many researchers studied the pantograph differential equations and their applications in many sciences such as biology, physics, economics, and electrodynamics. For more details, please see [21, 22].In [23], the authors studied nonlocal boundary value problems for the Hilfer fractional derivative. Initial value problems involving Hilfer fractional derivatives were studied in [24–26]. Initial value problems for pantograph equations with the Hilfer fractional derivative were studied in [22, 27].To the best of our knowledge, there is no work involving systems of integral boundary value problems for pantograph equations with the Hilfer fractional derivative. Thus, the objective of this work is to introduce a new class of coupled systems of Hilfer fractional differential pantograph equations with nonlocal integral boundary conditions of the form(2)DHα1,β1xt=f1t,xt,xλ1t,ytt∈a,b,DHα2,β2yt=f2t,xt,yt,yλ2tt∈a,b,xa=0,A1xb+B1Iδ1xμ1=C1,μ1∈a,b,ya=0,A2yb+B2Iδ2yμ2=C2,μ2∈a,b,where HDα1,β1, HDα2,β2 are the Hilfer fractional derivatives of order α1 and α2, 1<α1,α2<2 and parameter β1, β2, 0<β1,β2<1, respectively, f1,f2:a,b×ℝ×ℝ×ℝ→ℝ are two continuous functions; Iδ1,Iδ2 are the Riemann-Liouville fractional integrals of order δ1 and δ2, respectively, a≥0, A1,A2,B1,B2,C1,C2∈ℝ, and 0<λ1,λ2<1.This paper is organized as follows: we first give some definitions and notions that will be used throughout the work, after that we will establish the existence and uniqueness results by means of the fixed point theorems, and last but not least, we will give some examples that illustrate the results.


## 2. Preliminaries and Notations

In this section, we introduce some notations and definitions related to fractional calculus that we will use throughout this paper.We first define the following spaces:Ca,b,ℝ with a≥0 is the Banach space of all continuous functions from a,b to ℝ, La,b is the space of Lebesgue integrable functions on a finite closed interval a,bb>a of the real line ℝ, and ACka,b is the space of real-valued functions ft which have continuous derivatives up to order k−1 on a,b such that fk−1t belongs to the space of absolutely continuous functions ACa,b.Definition 1. (see [8, 11]).
The Riemann-Liouville fractional integral of orderα>0 of a continuous function f:a,∞→ℝ, is defined by
(3)Iα=1Γα∫att−sα−1fsds,provided the right-hand side exists on a,∞.Definition 2 (see [8, 11]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function f, is defined by
(4)RLDαft≔DnIn−αft=1Γn−αddtn∫att−sn−α−1fsds,where n=α+1, α denotes the integer part of the real number α, provided the right-hand side is pointwise defined on a,∞.Definition 3 (see [8, 11]).
The Caputo fractional derivative of orderα>0 of a continuous function f, is defined by
(5)CDαft≔In−αDnft=1Γn−α∫att−sn−α−1ddtnfsds,n−1<α<n,provided the right-hand side is pointwise defined on a,∞.Definition 4 (see [14]).
The Hilfer fractional derivative of orderα and parameter β of a function f is given by
(6)HDα,βft=Iβn−αDnI1−βn−αft,where n−1<α<n, 0≤β≤1, t>a, and D=d/dt.Remark 5.
Whenβ=0, the Hilfer fractional derivative becomes the Riemann-Liouville fractional derivative, while when β=1, the Hilfer fractional derivative becomes the Caputo fractional derivative.The following lemma gives a composition between the Riemann-Liouville fractional integral operator and the Hilfer fractional derivative operator.Lemma 6 (see [15]).
Letf∈La,b, n−1<α<n, n∈ℕ, 0<β<1, In−α1−βf∈ACka,b; then, we have
(7)IαHDα,βft=ft−∑k=0n−1t−ak−n−α1−βΓk−n−α1−β+1limt→a+dkdtkI1−βn−αft.
Now, we give a lemma which is the solution of a variant of the integral boundary value coupled systems (2).Lemma 7.
Leta≥0, for i=1,2, 1<αi<2, γi=αi+2βi−αiβi, hi∈Ca,b,ℝ, and
(8)Λi=Aib−aγi−1Γγi+Biμi−aγi+δi−1Γγi+δi≠0.
Then, the following problem(9)DHα1,β1xt=h1t,t∈a,b,DHα2,β2yt=h2t,t∈a,b,xa=0,A1xb+B1Iδ1xμ1=C1,μ1∈a,b,ya=0,A2yb+B2Iδ2yμ2=C2,μ2∈a,b,is equivalent to the system of equations:
(10)xt=Iα1h1t+t−aγ1−1Λ1Γγ1C1−A1Iα1h1b−B1Iα1+δ1h1μ1,yt=Iα2h2t+t−aγ2−1Λ2Γγ2C2−A2Iα2h2b−B2Iα2+δ2h2μ2.Proof.
Let us assume thatx,y is a solution of problem (9). Applying the fractional integrals Iα1 and Iα2 on both sides of the equations in (9) and using Lemma 6, we obtain
(11)xt=c01t−a−2−α11−β1Γ1−2−α11−β1+c11t−a1−2−α11−β1Γ2−2−α11−β1+Iα1h1t,yt=c02t−a−2−α21−β2Γ1−2−α21−β2+c12t−a1−2−α21−β2Γ2−2−α21−β2+Iα2h2t.
Since fori=1,2, 1−βi2−α1=2−γi, we obtain
(12)xt=c01t−aγ1−2Γγ1−1+c11t−aγ1−1Γγ1+Iα1h1t,yt=c02t−aγ2−2Γγ2−1+c12t−aγ2−1Γγ2+Iα2h2t,where for i=1,2, c0i, c1i are real constants.
Since we havex0=y0=0 and limt→at−aγ1−2=limt→at−aγ2−2=∞, we can obtain that c01=c02=0.
Then, we get(13)xt=c11t−aγ1−1Γγ1+Iα1h1t,yt=c12t−aγ2−1Γγ2+Iα2h2t,from the conditions: A1xb+B1Iδ1xμ1=C1 and A2yb+B2Iδ2yμ2=C2; we can find that
(14)c11=1Λ1C1−A1Iα1h1b−B1Iα1+δ1h1μ1,c12=1Λ2C2−A2Iα2h2b−B2Iα2+δ2h2μ2.
By substituting the values above in (13), we obtain
(15)xt=Iα1h1t+t−aγ1−1Λ1Γγ1C1−A1Iα1h1b−B1Iα1+δ1h1μ1,yt=Iα2h2t+t−aγ2−1Λ2Γγ2C2−A2Iα2h2b−B2Iα2+δ2h2μ2,which is the solution to problem (9).
We get the converse by direct computations. This ends the proof.


## 3. Main Results

The spaceX=x:xt∈Ca,b,ℝ endowed with the norm x=supxt,t∈a,b and the space Y=y:yt∈Ca,b,ℝ endowed with the norm y=supyt,t∈a,b are two Banach spaces. Moreover, the product space X×Y,x,y is a Banach space with the norm x,y=x+y.In view of Lemma7, we define the operator T:X×Y→X×Y by
(16)Tx,yt=Ux,yt,Vx,yt,where
(17)Ux,yt=t−aγ1−1Λ1Γγ1C1−A1Iα1f1s,xs,xλ1s,ysb−B1Iα1+δ1f1s,xs,xλ1s,ysμ1+Iα1f1s,xs,xλ1s,yst,Vx,yt=t−aγ2−1Λ2Γγ2C2−A2Iα2f2s,xs,ys,yλ2sb−B2Iα2+δ2f2s,xs,ys,yλ2sμ2+Iα2f2s,xs,ys,yλ2st.We should note that problem (2) has a solution x,y if and only if the operator T has a fixed point.In what is coming, for convenience, we set the following:Fori=1,2,
(18)Ωi=b−aγi−1ΛiΓγiAib−aαiΓαi+1+Biμi−aαi+δiΓαi+δi+1+b−aαiΓαi+1.We are going to prove the existence and uniqueness as well as the existence results for problem (2) by using the Banach contraction principle and Krasnoselskii’s fixed point theorem.The first result is based on Banach’s fixed point theorem.Theorem 8.
Assume that
H1 For all x,x¯,y,y¯∈ℝ,t∈a,b there exist Lf1,Lf2>0 such that
(19)f1t,xt,xλ1t,yt−f1t,x¯t,x¯λ1t,y¯t≤Lf12x−x¯+y−y¯,f2t,xt,yt,yλ2t−f1t,x¯t,y¯t,y¯λ2t≤Lf2x−x¯+2y−y¯,in addition, if we have
(20)3Lf1Ω1+Lf2Ω2<1,where Ω1,Ω2 are defined by (18); then, the boundary coupled systems (2) has a unique solution x∗,y∗ on a,b.Proof.
We transform the boundary value coupled systems (2) into a fixed point problem. Applying the Banach contraction mapping principle, we show that T defined by (16) and (17) has a unique fixed point. We let f1t,0,0,0=M1t∈a,bsup and f2t,0,0,0=M2t∈a,bsup and choose
(21)r0≥M1Ω1+M2Ω2+C1b−aγ1−1/Λ1Γγ1+C2b−aγ2−1/Λ2Γγ21−3Lf1Ω1+Lf2Ω2.
We first show thatTBr0⊂Br0, where Br0=x,y∈X×Y:x,y≤r0.
For anyx,y∈Br0, we have
(22)Ux,yt≤supt∈a,bt−aγ1−1Λ1Γγ1C1+A1Iα1f1s,xs,xλ1s,ysb+B1Iα1+δ1f1s,xs,xλ1s,ysμ1+Iα1f1s,xs,xλ1s,yst≤b−aγ1−1Λ1Γγ1C1+A1Iα1f1s,xs,xλ1s,ys−f1s,0,0,0+f1s,0,0,0b+B1Iα1+δ1f1s,xs,xλ1s,ys−f1s,0,0,0+f1s,0,0,0μ1+Iα1f1s,xs,xλ1s,ys−f1s,0,0,0+f1s,0,0,0b≤Lf12x+y+M1b−aγ1−1Λ1Γγ1A1b−aα1Γα1+1+B1μ1−aα1+δ1Γα1+δ1+1+b−aα1Γα1+1+C1b−aγ1−1Λ1Γγ1≤3r0Lf1+M1Ω1+C1b−aγ1−1Λ1Γγ1.
Similarly, we get(23)Vx,yt≤3r0Lf2+M2Ω2+C2b−aγ2−1Λ2Γγ2.
Finally,(24)Tx,yt≤3r0Lf1+M1Ω1+3r0Lf2+M2Ω2+C1b−aγ1−1Λ1Γγ1+C2b−aγ2−1Λ2Γγ2≤r0,which implies that TBr0⊂Br0.
Next, we show that the operatorT is a contraction; we let x,y,x¯,y¯∈X×Y; then for t∈a,b, we have
(25)Ux,yt−Ux¯,y¯t≤b−aγ1−1Λ1Γγ1A1Iα1f1s,xs,xλ1s,ys−f1s,x¯s,x¯λ1s,y¯sb+B1Iα1+δ1f1s,xs,xλ1s,ys−f1s,x¯s,x¯λ1s,y¯sμ1+Iα1f1s,xs,xλ1s,ys−f1s,x¯s,x¯λ1s,y¯sb≤Lf1b−aγ1−1Λ1Γγ1A1b−aα1Γα1+1+B1μ1−aα1+δ1Γα1+δ1+1+b−aα1Γα1+12x−x¯+y−y¯≤Lf1Ω12x−x¯+y−y¯≤3Lf1Ω1x−x¯+y−y¯,with a similar method, we also get
(26)Vx,yt−Vx¯,y¯t≤3Lf2Ω2x−x¯+y−y¯.
Finally, we can obtain(27)Tx,yt−Tx¯,y¯t≤3Lf1Ω1+Lf2Ω2x−x¯+y−y¯.
And since,3Lf1Ω1+Lf2Ω2<1, then the operator T is a contraction.
Therefore, we conclude by Banach’s contraction mapping principle thatT has a fixed point which is the unique solution x∗,y∗ of problem (2). The proof is completed.Next, we present a result based on Krasnoselskii’s fixed point theorem.Theorem 9.
Letf:a,b×ℝ×ℝ×ℝ→ℝ be a continuous function that satisfies H1.
In addition, we assume that
H2f1t,xt,xλ1t,yt≤2Cf1x+Df1y+Mf1,(28)f2t,xt,yt,yλ2t≤Cf2x+2Df2y+Mf2,for all x,y∈ℝ, 0<λ1,λ2<1 and t∈a,b, with Cf1,Cf2,Df1,Df2,Mf1,Mf2 are positive real numbers.
Then, problem (2) has at least a solution x,y on a,b, provided
(29)∑i=123Lfib−aγi−1ΛiΓγiAib−aαiΓαi+1+Biμi−aαi+δiΓαi+δi+1<1.Proof.
We setε0≥Mf1Ω1+C1b−aγ1−1/Λ1Γγ1/1−2Cf1−Df1+Mf2Ω2+C2b−aγ2−1/Λ2Γγ2/1−Cf2−2Df2, where Ω1,Ω2 are defined in (18), and we consider Bε0=x,y∈X×Y:x,y≤ε0.
We define the operatorT by Tx,yt=Ux,yt,Vx,yt, for any x,y∈Bε0, and t∈a,b where
(30)Ux,yt=t-aγ1-1Λ1Γγ1C1-A1Iα1f1s,xs,xλ1s,ysb-B1Iα1+δ1f1s,xs,xλ1s,ysμ1+Iα1f1s,xs,xλ1s,yst,Vx,yt=t-aγ2-1Λ2Γγ2C2-A2Iα2f2s,xs,ys,yλ2sb-B2Iα2+δ2f2s,xs,ys,yλ2sμ2+Iα2f2s,xs,ys,yλ2st,by splitting the two operators above, we have
(31)U1x,yt=Iα1f1s,xs,xλ1s,yst,U2x,yt=t−aγ1−1Λ1Γγ1C1−A1Iα1f1s,xs,xλ1s,ysb−B1Iα1+δ1f1s,xs,xλ1s,ysμ1,V1x,yt=Iα2f2s,xs,ys,yλ2st,V2x,yt=t−aγ2−1Λ2Γγ2C2−A2Iα2f2s,xs,ys,yλ2sb−B2Iα2+δ2f2s,xs,ys,yλ2sμ2.
This upcoming part of the proof requires us to rewrite the operatorT as
(32)Tx,yt=T1x,yt+T2x,yt,where
(33)T1x,yt=U1x,yt,V1x,yt,T2x,yt=U2x,yt,V2x,yt.
For anyx,y∈Bε0, we have
(34)T1x,yt≤supt∈a,bt−aγ1−1Λ1Γγ1C1+A1Iα1f1s,xs,xλ1s,ysb+B1Iα1+δ1f1s,xs,xλ1s,ysμ1+Iα1f1s,xs,xλ1s,yst≤2Cf1x+Df1y+Mf1Ω1+C1b−aγ1−1Λ1Γγ1,
Similarly, we have(35)T2x,yt≤Cf2x+2Df2y+Mf2Ω2+C2b−aγ2−1Λ2Γγ2,
Sinceε0≥Mf1Ω1+C1b−aγ1−1/Λ1Γγ1/1−2Cf1−Df1+Mf2Ω2+C2b−aγ2−1/Λ2Γγ2/1−Cf2−2Df2.
Then, this shows thatTx,y∈Bε0.
Next, we show thatT2 is a contraction mapping.
For allx,y,x¯,y¯∈X×Y, and for t∈a,b, we have
(36)U2x,yt−U2x¯,y¯t≤3Lf1b−aγ1−1Λ1Γγ1A1b−aα1Γα1+1+B1μ1−aα1+δ1Γα1+δ1+1x−x¯+y−y¯.
Similarly,(37)V2x,yt−V2x¯,y¯t≤3Lf2b−aγ2−1Λ2Γγ2A2b−aα2Γα2+1+B2μ2−aα2+δ2Γα2+δ2+1x−x¯+y−y¯.
It is easy to see, using (29), that T2 is a contraction mapping.
The continuity of the functionsf1 and f2 implies the continuity of the operator T1. In addition, T1 is uniformly bounded on Bε0 as
(38)T1x,yt≤U1x,yt+V1x,yt≤b−aα1Γα1+12Cf1x+Df1y+Mf1+b−aα2Γα2+1Cf2x+2Df2y+Mf2≤ε0.
Now we prove the compactness of the operatorT1.
For anyt1,t2∈a,b, with t1<t2, we have
(39)U1x,yt2−U1x,yt1=1Γα1∫at1t2−sα1−1−t1−sα1−1f1t,xs,xλ1s,ysds+∫t1t2t2−sα1−1f1t,xs,xλ1s,ysds≤2Cf1x+Df1y+Mf1Γα1+12t2−t1α1+t2−aα1−t1−aα1,which tends to zero as t2−t1→0 independently of x,y.Similarly, we haveV1x,yt2−V2x,yt1→0 as t2−t1→0.Finally, the operatorT1 is equicontinuous, which means that T1 is relatively compact on Bε0. Hence, by Arzela-Ascoli theorem, T1 is compact on Bε0. Thus, all the assumptions of Krasnoselskii’s fixed point theorem are satisfied. So, the boundary value coupled systems (2) have at least a solution x,y on a,b.


## 4. Examples

Example 1.
We first consider the following problem:(40)DH3/2,2/5xt=t3+sinxt+cosyt90+sinxt/3100,t∈12,52,DH3/2,2/5yt=cosxt+yt50+yt/3t3+54,t∈12,52,x12=0,23x52+34I1/2x32=45,y12=0,23y52+34I1/2y32=45,where, α1=α2=3/2, β1=β2=2/5, λ1=λ2=1/3, a=1/2, b=5/2, A1=A2=2/3, B1=B2=3/4, δ1=δ2=1/2, μ1=μ2=3/2, and C1=C2=4/5.
The setting yieldsγ1=γ2=17/10, Λ1=Λ2=1.872599119, and Ω1=Ω2=4.129461300.
Now, we put(41)f1t,xt,xλ1t,yt=t3+sinxt+cosyt90+sinxt/3100,f2t,xt,yt,yλ2t=cosxt+yt50+yt/3t3+54,
which satisfiesH1 as
(42)f1t,xt,xλ1t,yt−f1t,x¯t,x¯λ1t,y¯t≤0.01112x−x¯+y−y¯,f2t,xt,yt,yλ2t−f1t,x¯t,y¯,y¯λ2t≤0.020x−x¯+2y−y¯,
Setting:Lf1=0.0111, Lf2=0.020, we obtain 3Lf1Ω1+Lf2Ω2≃0.3852787392<1, which shows that inequality (20) is verified. Then, by Theorem 8, we can conclude that problem (40) has a unique solution x∗,y∗ on 1/2,5/2.
We now consider the following problem:(43)DH5/3,1/2xt=sinxt+sinxt/4100+sinyt90+e1−3t,t∈13,53,DH5/3,1/2yt=sinxt+sinyt90+sinyt/4100+e1−3t,t∈13,52,x13=0,35x53+14I3/2x23=34,y13=0,35y53+14I3/2y23=34,where, α1=α2=5/3, β1=β2=1/2, λ1=λ2=1/4, a=1/3, b=5/3, A1=A2=3/5, B1=B2=1/4, δ1=δ2=3/2, μ1=μ2=2/3, and C1=C2=3/4.
The setting yieldsγ1=γ2=11/6, Λ1=Λ2=0.8175877260.
The functionf1,f2 verify H1 as
(44)f1t,xt,xλ1t,yt−f1t,x¯t,x¯λ1t,y¯t≤1902x−x¯+y−y¯,f2t,xt,yt,yλ2t−f1t,x¯t,y¯,y¯λ2t≤190x−x¯+2y−y¯.
Hence,Lf1=Lf2=1/90, by the definitions of f1 and f2 the conditon H2 is also satisfied and we have
(45)∑i=123Lfib−aγi−1ΛiΓγiAib−aαiΓαi+1+Biμi−aαi+δiΓαi+δi+1≃0.07108034<1,
which shows that problem (43) has at least a solution x,y on 1/3,5/3.



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