# Solution of Sequential Hadamard Fractional Differential Equations by Variation of Parameter Technique

Authors: Mohammed M. Matar
Journal: Abstract and Applied Analysis
 (2018)

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## Abstract

We obtain in this article a solution of sequential differential equation involving the Hadamard fractional derivative and focusing the orders in the intervals(1,2) and (2,3). Firstly, we obtain the solution of the linear equations using variation of parameter technique, and next we investigate the existence theorems of the corresponding nonlinear types using some fixed-point theorems. Finally, some examples are given to explain the theorems.


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## Body

## 1. Introduction

In recent years, there has been a great development in the study of fractional differential equations. This advancement is ranging from the theoretical analysis of the subject to analytical and numerical techniques (see [1–3] and the references cited therein). Among the theoretic approach, the existence theory of solutions for fractional differential models has gained attentions of many authors. Most of them have focused on using Riemann-Liouville and Caputo derivatives in representing the underlying fractional differential equation (see [4–10]). Another kind of fractional derivative is Hadamard type which was introduced in 1892 [11]. This derivative differs from aforementioned derivatives in the sense that the kernel of the integral in the definition of Hadamard derivative contains logarithmic function of arbitrary exponent. A detailed description of Hadamard fractional derivative and integral can be found in [12, 13]. Recently, the existence and uniqueness of solution for fractional differential equations in Hadamard sense were introduced in many faces by several authors ([6, 7] and references therein). We add in this article a new idea concerning the sequential definition of Hadamard fractional operator with constant coefficients of order less than three. For q∈(2,3), the Hadamard fractional differential equation can be transformed to classical Euler-Cauchy second-order nonhomogeneous differential equation that can be solved by variation of parameter technique.More precisely, we consider the nonlinear Hadamard fractional differential equations given by(1)DHa+qxt=f1t,xt,q∈0,1,xa+=0,(2)DHa+q+γDHa+q-1xt=f2t,xt,q∈1,2,xa+=x1a+=0,(3)DHa+q+λ1DHa+q-1+λ2DHa+q-2xt=f3t,xt,q∈2,3,xa+=x1a+=x2a+=0,where t∈J=[a,b], 1≤a<b<+∞, and fk:J×R→R is a given continuous function for each k=1,2,3. Here, DHa+q is the Hadamard fractional derivative, and γ, λ1, and λ2 are constants.


## 2. Linear Fractional Differential Equations

We introduce some basic ideas of fractional calculus that may be used in the sequel of this article [12].Definition 1.
The Hadamard fractional integral of orderq>0 is defined as (4)Ja+qht=1Γq∫atln⁡tsq-1hssds,provided that the integral exists.Definition 2.
The Hadamard derivative of fractional orderq>0 is defined as (5)DHa+qht=δnJa+qht,n-1<q<n,n=q+1,δ=tddt,where [q] denotes the integer part of the real number q.LetC(J,R) denote the Banach space of all real valued continuous functions defined on J, and Cδn(J,R) denotes the Banach space of all real valued functions f such that δnf∈C(J,R).Lemma 3 (see [12]).
Letx∈Cδn(J,R), h∈C(J,R), and ck∈R; then,(6)Ja+qJa+pht=Ja+q+pht,DHa+qJa+qxt=xt,Ja+qDHa+qxt=xt+c1ln⁡taq-1+c2ln⁡taq-2+⋯+cnln⁡taq-n.The following result is obvious by Lemma 3.Lemma 4.
Letx∈Cδ1(J,R), and g1∈C(J,R). The fractional differential equation(7)DHa+qxt=g1t,t∈J,0<q<1,xa+=0,has a solution given by(8)xt=Ja+qg1t.Lemma 5.
Letx∈Cδ2(J,R), and g2∈C(J,R). The fractional differential equation(9)DHa+q+γDHa+q-1xt=g2t,t∈J,1<q<2,xa+=x1a+=0,has a solution given by(10)xt=t-γ∫atsγ-1Ja+q-1g2sds.Proof.
Applying the fractional integral operatorJa+q to (9), we have(11)Ja+qg2t=xt+b1ln⁡taq-1+b2ln⁡taq-2+γJa+1xt+c1ln⁡taq-2,for some constants b1, b2, and c1. The initial condition x(a+)=0 implies that b2=0. Taking the first derivative of (11), it follows that(12)1tJa+q-1g2t=x1t+b1q-1ln⁡taq-21t+γtxt+c1lntaq-2.The condition x(1)(a+)=0 implies that b1=0=c1. Multiplying (12) by the integrating factor given by (13)ζt=e∫γ/tdt=tγ,we get (14)ddttγxt=tγ-1Ja+q-1g2t.Integrating (14), and using again x(a+)=0, we conclude that (15)xt=t-γ∫atsγ-1Ja+q-1g2sds.This finishes the proof.Remark 6.
Forγ=0, the solution is still valid, since Ja+qg2(t)=Ja+1Ja+q-1g2(t)=∫atJa+q-1g2(s)/sds. This trivial case will be negligible hereafter.Lemma 7.
Letλ12≥4λ2, x∈Cδ3(J,R), and g3∈C(J,R). The linear system (16)DHa+q+λ1DHa+q-1+λ2DHa+q-2xt=g3t,t∈J,2<q<3,xa+=x1a+=x2a+=0,has a solution (17)xt=t-λ1/2λ12-4λ2∫atts1/2λ12-4λ2-st1/2λ12-4λ2sλ1/2+1Ja+q-2g3sds,ifλ12>4λ2orxt=t-λ1/2∫atsλ1/2+1ln⁡tsJa+q-2g3sds,ifλ12=4λ2.Proof.
Applying the fractional integral operatorJa+q to (16), we have(18)xt+b1ln⁡taq-1+b2ln⁡taq-2+b3ln⁡taq-3+λ1Ja+1xt+c2ln⁡taq-2+c3ln⁡taq-3+λ2Ja+2xt+d3ln⁡taq-3=Ja+qg3t,for some constants b1, b2, b3, c2, c3, and d3. The initial condition x(a+)=0 implies that b3=0. Taking the first derivative of (18), and using Leibniz’s rule, it follows that(19)x1t+b1q-1ln⁡taq-21t+b2q-2ln⁡taq-31t+λ1txt+c2ln⁡taq-2+c3ln⁡taq-3+λ2t∫atxs+d3ln⁡saq-3dss=1tJa+q-1g3t.The condition x(1)(a+)=0 implies that b2=c3=0. Multiplying by t and then the second derivative would imply (20)tx2t+x1t+b1q-1q-2ln⁡taq-31t+λ1x1t+c2q-2ln⁡taq-31t+λ2txt+d3ln⁡taq-3=1tJa+q-2g3t.For the last initial condition x(2)(a+)=0, we get b1=c2=d3=0. Now, multiplying (20) by t, it follows that (21)t2x2t+λ1+1tx1t+λ2xt=Ja+q-2g3t,which is a second-order Cauchy-Euler differential equation that has a general solution x(t)=xc(t)+xp(t), where xc(t) and xp(t) are the complementary and particular solution of (14), respectively. To find the complementary solution of (21), consider the solutions x1(t)=tm1 and x2(t)=tm2 for the homogeneous equation (22)t2x2t+λ1+1tx1t+λ2xt=0,where m1=-λ1-λ12-4λ2/2 and m2=-λ1+λ12-4λ2/2 are the distinct real roots of the characteristic equation(23)m2+λ1m+λ2=0.The complementary solution of the homogeneous equation (22) is (24)xct=a1tm1+a2tm2,for some constants a1 and a2. These two constants can be evaluated by the initial conditions x(a+)=x(1)(a+)=0 given in (16), which both imply that (25)a1am1+a2am2=0,a1m1am1-1+a2m2am2-1=0.The only solution for these algebraic equations is a1=a2=0, since m1≠m2. The Wronskian W for the solutions x1 and x2 is (26)Wx1,x2=tm1tm2m1tm1-1m2tm2-1=m2-m1tm1+m2-1=λ12-4λ2t-λ1-1≠0,since λ12>4λ2. Applying the variation of parameter technique we can get the particular solution (27)xpt=∫attm2sm1-tm1sm2m2-m1sm1+m2-1Ja+q-2g3sds.Therefore, the general solution is x(t)=xp(t).
Ifλ12=4λ2, we consider x1(t)=t-λ1/2 and x2(t)=t-λ1/2ln⁡t. Then, the complementary and the particular solutions are (28)xct=a1t-λ1/2+a2t-λ1/2ln⁡t,xpt=t-λ1/2∫atsλ1/2+1ln⁡tsJa+q-2g3sds,where the Wronskian in this case is given by(29)Wx1,x2=t-λ1/2t-λ1/2ln⁡t-λ12t-λ1/2-1-λ12t-λ1/2-1ln⁡t+t-λ1/2-1=-λ12t-λ1-1ln⁡t+t-λ1-1+λ12t-λ1-1ln⁡t=t-λ1-1≠0.The evaluation of the constants a1 and a2 in the complementary solution (28) leads to(30)a1a-λ1/2+a2a-λ1/2ln⁡a=0,-λ12a1a-λ1/2-1-λ12a2a-λ1/2-1ln⁡a+a2a-λ1/2-1=0,which imply a1=a2=0. Therefore, the general solution becomes again x(t)=xp(t). The two cases together constitute the required solution. This finishes the proof.Remark 8.
Ifλ12<4λ2, then (23) has two complex conjugate roots that will not be considered in this article. On the other hand, if λ1=λ2=0, the solution is x(t)=Ja+qg3(t), which is a trivial case and hereafter will be negligible.Remark 9.
Any of initial conditionsx(a+)=x(1)(a+)=x(2)(a+)=constant≠0 should not be used because it would imply ill-posed systems.


## 3. Existence Theorems

We establish sufficient conditions for existence of solutions to problems (1)–(3) using different types of fixed-point theorems.In view of Lemmas4, 5, and 7, we transform the initial value problems (1)–(3), respectively, into operator equations as(31)Ψ1xt=Ja+qf1t,xt,0<q<1,Ψ2xt=t-γ∫atsγ-1Ja+q-1f2s,xsds,1<q<2,and for 2<q<3, we define the respective cases λ12>4λ2, and λ12=4λ2, as (32)Ψ3xt=t-λ1/2λ12-4λ2∫atts1/2λ12-4λ2-st1/2λ12-4λ2sλ1/2+1Ja+q-2g3sds,ifλ12>4λ2orΨ3xt=t-λ1/2∫atsλ1/2+1ln⁡tsJa+q-2g3sds,ifλ12=4λ2.where t∈J. If the operator Ψk, k=1,2,3, has a fixed point, the corresponding problem in systems (1)–(3) has this fixed point as a solution.Lemma 10.
The operatorΨk:C(J,R)→C(J,R), k=1,2,3, is completely continuous.Proof.
Obviously, the continuity of the operatorΨk follows from the continuity of the function fk. Let U be a bounded proper subset of C(J,R); then for any t∈J, and x∈U, there exists a positive constant Lk such that fk(t,x(t))≤Lk. Accordingly, for any t∈J, and x∈U, it follows that (33)Ψ1xt≤L1ln⁡t/aqΓq+1,0<q<1,Ψ2xt≤L2ln⁡t/aq-11-a/tγγΓq,1<q<2,γ≠0,and for 2<q<3, we have (34)Ψ3xt≤L3a2ln⁡t/aq-2λ12-4λ2Γq-1×t/a2-t/a-λ1/2+1/2λ12-4λ22+λ1/2-1/2λ12-4λ2-t/a2-a/tλ1/2+1/2λ12-4λ22+λ1/2+1/2λ12-4λ2,ifλ12>4λ2,Ψ3xt≤L3a2ln⁡t/aq-1Γq-1t/a2-a/tλ1/2λ1/2+2,ifλ12=4λ2,λ1≠-4.Consequently, for any k=1,2,3, the operator Ψk is bounded on U⊂C(J,R). Furthermore, if t1,t2∈J, such that a≤t1<t2≤b, then (35)Ψ1xt2-Ψ1xt1≤L1ln⁡t2/aq-ln⁡t1/aqΓq+1,for 0<q<1. Let 1<q<2, and then (36)Ψ2xt2-Ψ2xt1≤L2γΓq×t1γ-aγln⁡t1aq-1+t2γln⁡t2aq-1t2γ-t1γ,γ≠0.For 2<q<3, let λ12>4λ2, and then (37)Ψ3xt2-Ψ3xt1≤t2-λ1/2+1/2λ12-4λ2-t1-λ1/2+1/2λ12-4λ2λ12-4λ2×∫at1sλ1/2+1-1/2λ12-4λ2Ja+q-2f3s,xsds+t2-λ1/2-1/2λ12-4λ2-t1-λ1/2-1/2λ12-4λ2λ12-4λ2×∫at1sλ1/2+1+1/2λ12-4λ2Ja+q-2f3s,xsds+t2-λ1/2+1/2λ12-4λ2λ12-4λ2×∫t1t2sλ1/2+1-1/2λ12-4λ2Ja+q-2f3s,xsds+t2-λ1/2-1/2λ12-4λ2λ12-4λ2×∫t1t2sλ1/2+1+1/2λ12-4λ2Ja+q-2f3s,xsds≤L3t2-λ1/2+1/2λ12-4λ2-t1-λ1/2+1/2λ12-4λ2ln⁡t1/aq-2λ12-4λ2Γq-1×t1λ1/2-1/2λ12-4λ2+2-aλ1/2-1/2λ12-4λ2+2λ1/2-1/2λ12-4λ2+2+L3t2-λ1/2-1/2λ12-4λ2-t1-λ1/2-1/2λ12-4λ2ln⁡t1/aq-2λ12-4λ2Γq-1×t1λ1/2+1/2λ12-4λ2+2-aλ1/2+1/2λ12-4λ2+2λ1/2+1/2λ12-4λ2+2+L3t2-λ1/2+1/2λ12-4λ2ln⁡t2/aq-2λ12-4λ2Γq-1×t2λ1/2-1/2λ12-4λ2+2-t1λ1/2-1/2λ12-4λ2+2λ1/2-1/2λ12-4λ2+2+L3t2-λ1/2-1/2λ12-4λ2ln⁡t2/aq-2λ12-4λ2Γq-1×t2λ1/2+1/2λ12-4λ2+2-t1λ1/2+1/2λ12-4λ2+2λ1/2+1/2λ12-4λ2+2.Now, let λ12=4λ2, and then(38)Ψ3xt2-Ψ3xt1≤t2-λ1/2ln⁡t2a-t1-λ1/2ln⁡t1a∫at1sλ1/2+1Ja+q-2f3s,xsds+t2-λ1/2ln⁡t2a∫t1t2sλ1/2+1Ja+q-2f3s,xsds≤L3t2-λ1/2ln⁡t2a-t1-λ1/2ln⁡t1at1λ1/2+2-aλ1/2+2λ1/2+2ln⁡t1/aq-2Γq-1+L3t2-λ1/2ln⁡t2at2λ1/2+2-t1λ1/2+2λ1/2+2ln⁡t2/aq-2Γq-1.As t2→t1, (Ψkx)(t2)-(Ψkx)(t1)→0, for k=1,2,3. This implies that Ψk is equicontinuous on J. In consequence, it follows by the Arzela-Ascoli theorem that the operator Ψk is completely continuous. This finishes the proof.We need the following Schauder’s fixed point theorem ([14]).Theorem 11.
IfU is a closed, bounded, convex subset of a Banach space X and the mapping Δ:U→U is completely continuous, then Δ has a fixed point in U.Accordingly, if we define a closed, bounded, convex subsetU of C(J,R) on which Ψk′s, as defined by (31)-(32), are completely continuous, then problems (1)–(3) have the respective solution.Theorem 12.
Assume that(39)limx→0fkt,xx≤ξk<∞,t∈J;then, each problem of (1)–(3) has a solution.Proof.
For anyk, the given condition implies that fk(t,x(t))≤1+ξkx(t), whenever x(t)<δk, for a fixed number δk>0. Therefore, define a subset Uk as(40)Uk=x∈CJ,R:xt≤δk,t∈J.Hence, Uk is a closed, bounded, and convex subset of C(J,R). If x∈Uk, then fk(t,x(t))<δk1+ξk, for any t∈J. On the other hand, the operator Ψk:Uk→Uk is completely continuous by Lemma 10; then by Schauder’s fixed-point Theorem 11, each problem of (1)–(3) has a solution. This finishes the proof.Our next result deals with the uniqueness of solution for problems (1)–(3) and is based on the Banach contraction theorem. The following assumptions are essential for proving the next theorem.(H1)
For anyt∈J, x,y∈R, k=1,2,3,(41)fkt,x-fkt,y≤Bkx-y,supt∈Jfkt,0=Ak,Ck=max⁡Ak,Bk.(H2)
Letηk<1, such that (42)η1=C1ln⁡b/aqΓq+1,η2=C2ln⁡b/aq-1maxt∈J1-a/tγγΓq,γ≠0,η3=C3ln⁡b/aq-2LΓq-1.HereL=M, if λ12>4λ2, or L=N, if λ12≠4λ2, where(43)N=maxt∈Jt2-a2a/tλ1/2λ1/2+2ln⁡ta,M=1λ12-4λ2maxt∈Jt2-1/a2t/a-λ1/2+1/2λ12-4λ22+λ1/2-1/2λ12-4λ2-t2-1/a2a/tλ1/2+1/2λ12-4λ22+λ1/2+1/2λ12-4λ2.Theorem 13.
Assume that (H1) and (H2) are satisfied. Then there exists a unique solution for each of the problems (1)–(3) on J.Proof.
Letrk≥1-ηk-1ηk, for k=1,2,3. We show that ΨkBrk⊂Brk, where Brk={x∈C(J,R):x≤rk}. This follows by the following estimates: (44)Ψ1xt≤B1ln⁡b/aqΓq+1x+A1ln⁡b/aqΓq+1≤1-η1r1+η1r1=r1,Ψ2xt≤B2ln⁡b/aq-11-a/tγγΓqx+A2ln⁡b/aq-11-a/tγγΓq≤r2,Ψ3xt≤C3ln⁡b/aq-2MΓq-1A3+B3x,ifλ12>4λ2,C3ln⁡t/aq-2NΓq-1A3+B3x,ifλ12=4λ2≤r3.for any x∈Brk. Moreover, for x,y∈C(J,R), k=1,2,3, and for each t∈J, we obtain (45)Ψkxt-Ψkyt≤ηkx-y.As ηk<1, Ψk is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle. This completes the proof.We close the existence theorems by applying Leray-Schauder degree theorem [14]. Therefore, we need the following assumption.(H3)
Fort∈J, x∈R, and k=1,2,3,(46)fkt,x≤Dkx+1,where Dk<∞, satisfies (47)F1=D1ln⁡b/aqΓq+1<1,F2=D2ln⁡b/aq-1maxt∈J1-a/tγγΓq<1,γ≠0,F3=D3ln⁡b/aq-2LΓq-1<1,where L is defined as in (H2).Theorem 14.
Assume that (H3) is satisfied. Then every problem of (1)–(3) has a solution.Proof.
Define a ballBRk={x∈C(J,R):x<Rk}, k=1,2,3, for some positive real number Rk which will be determined later. We show that Ψk:BRk¯→C(J,R) satisfies (48)0∉I-λkΨk∂BRk,for any x∈∂BRk, and λk∈[0,1]; here ∂BRk denotes the boundary set of BRk. Define the homotopy (49)hλkx=Hλk,x=x-λkΨkx,x∈CJ,R,λk∈0,1.Then, by Lemma 10, hλk is completely continuous. Let I denote the identity operator; then the homotopy invariance and normalization properties of topological degrees imply that (50)deg⁡hλk,BRk,0=deg⁡I-λkΨk,BRk,0=deg⁡h1,BRk,0=deg⁡h0,BRk,0=deg⁡I,BRk,0=1,since 0∈BRk. By the nonzero property of the Leray-Schauder degree, h1(x)=x-Ψkx=0 for at least one x∈BRk. To find Rk, we assume that x(t)=λkΨkx(t) for some λk∈[0,1] and for all t∈J. Then, using (H3), we have (51)xt=λkΨkxt≤Fk1+x.Therefore,(52)x≤Fk1-Fk.The value of Rk=1/1-Fk>x is sufficient for applicability of Leray-Schauder degree theorem. This completes the proof.Example 1.
Consider the following nonlinear fractional differential equation:(53)DH2+1.2-4DH2+0.2xt=tsin⁡xt,t∈2,3,x2=x′2=0.Here q=1.2, f2(t,x(t))=t/3sin⁡x(t), and γ=-4. Simple calculations lead to ξ2=C2=D2=1, and (54)η2=C2ln⁡b/aq-11-a/tγγΓq=ln⁡3/20.2maxt∈J1-2/t-440.9=0.95.Therefore, all previous existence theorems ensure the existence and uniqueness of the solution in C[2,3],R for problem (53).Example 2.
Consider the following nonlinear fractional differential equation:(55)DH1+2.6-2DH1+1.6-DH1+0.6xt=e-3.2xt1+xt,t∈1,e,x1=x′1=x′′1=0.Here q=2.6, f3(t,x(t))=e-3.2|x(t)|/1+|x(t)|, λ1=-2, and λ2=-1. Simple calculations lead to (56)M=122maxt∈Jt2-t1+21-2-t2-t1-21+2≤20.4,η3=C3ln⁡b/aq-2LΓq-1=20.4e-3.2Γ1.6=0.94.We apply any of the following:(1)
Theorem12, with ξ3=e-3.2(2)
Theorem13, with C3=e-3.2, and η3<1(3)
Theorem14, with D3≤e-3.2, and F3=0.94<1
Problem (55) has a solution in C[1,e],R. The uniqueness of this solution follows from Theorem 13.



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