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

and Applied Analysis 3 ⋅ sλ1/2+1Jq−2 a+ g3 (s) ds, if λ21 > 4λ2 or x (t) = t−λ1/2 ∫ a sλ1/2+1 ln t sJq−2 a+ g3 (s) ds, if λ21 = 4λ2. (17) Proof. Applying the fractional integral operator Ja+ to (16), we have x (t) + b1 (ln t a) q−1 + b2 (ln t a) q−2 + b3 (ln t a) q−3 + λ1J1a+ (x (t) + c2 (ln t a) q−2 + c3 (ln t a) ) + λ2J2a+ (x (t) + d3 (ln t a) ) = Jqa+g3 (t) , (18) 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 x(1) (t) + b1 (q − 1) (ln t a) q−2 1t + b2 (q − 2) (ln t a) q−3 1t + λ1 t (x (t) + c2 (ln t a) q−2 + c3 (ln t a) ) + λ2 t ∫ t a (x (s) + d3 (ln s a) ) ds s = 1tJq−1 a+ g3 (t) . (19) The condition x(1)(a+) = 0 implies that b2 = c3 = 0. Multiplying by t and then the second derivative would imply tx(2) (t) + x(1) (t) + b1 (q − 1) (q − 2) (ln t a) q−3 1t + λ1 (x(1) (t) + c2 (q − 2) (ln t a) q−3 1t ) + λ2 t (x (t) + d3 (ln t a) ) = 1tJq−2 a+ g3 (t) . (20) For the last initial condition x(2)(a+) = 0, we get b1 = c2 = d3 = 0. Now, multiplying (20) by t, it follows that t2x(2) (t) + (λ1 + 1) tx(1) (t) + λ2x (t) = Jq−2 a+ g3 (t) , (21) 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 t2x(2) (t) + (λ1 + 1) tx(1) (t) + λ2x (t) = 0, (22) where m1 = (−λ1 − √λ21 − 4λ2)/2 and m2 = (−λ1 + √λ21 − 4λ2)/2 are the distinct real roots of the characteristic equation m2 + λ1m + λ2 = 0. (23) The complementary solution of the homogeneous equation (22) is xc (t) = a1tm1 + a2tm2 , (24) for some constants a1 and a2. These two constants can be evaluated by the initial conditionsx(a+) = x(1)(a+) = 0 given in (16), which both imply that a1am1 + a2am2 = 0, a1m1am1−1 + a2m2am2−1 = 0. (25) The only solution for these algebraic equations is a1 = a2 = 0, sincem1 ̸ = m2.TheWronskianW for the solutions x1 and x2 is W(x1, x2) = 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 tm1 tm2 m1tm1−1 m2tm2−1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 = (m2 − m1) tm1+m2−1 = √λ21 − 4λ2t−λ1−1 ̸ = 0, (26) since λ21 > 4λ2. Applying the variation of parameter technique we can get the particular solution xp (t) = ∫ a tm2sm1 − tm1sm2 (m2 − m1) sm1+m2−1 q−2 a+ g3 (s) ds. (27) Therefore, the general solution is x(t) = xp(t). If λ21 = 4λ2, we consider x1(t) = t−λ1/2 and x2(t) = t−λ1/2 ln t.Then, the complementary and the particular solutions are xc (t) = a1t−λ1/2 + a2t−λ1/2 ln t, xp (t) = t−λ1/2 ∫ a sλ1/2+1 ln t sJq−2 a+ g3 (s) ds, (28) where the Wronskian in this case is given by W(x1, x2) = 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 t−λ1/2 t−λ1/2 ln t −λ1 2 t−λ1/2−1 −λ1 2 t−λ1/2−1 ln t + t−λ1/2−1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 = −λ1 2 t−λ1−1 ln t + t−λ1−1 + λ1 2 t−λ1−1 ln t = t−λ1−1 ̸ = 0. (29) 4 Abstract and Applied Analysis The evaluation of the constants a1 and a2 in the complementary solution (28) leads to a1a−λ1/2 + a2a−λ1/2 ln a = 0, −λ1 2 a1a−λ1/2−1 − λ1 2 a2a−λ1/2−1 ln a + a2a−λ1/2−1 = 0, (30) 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. If λ21 < 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) = Jqa+g3(t), which is a trivial case and hereafter will be negligible. Remark 9. Any of initial conditions x(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 Lemmas 4, 5, and 7, we transform the initial value problems (1)–(3), respectively, into operator equations as Ψ1x (t) = Jqa+f1 (t, x (t)) , 0 < q < 1, Ψ2x (t) = t−γ ∫ a sγ−1Jq−1 a+ f2 (s, x (s)) ds, 1 < q < 2, (31) and for 2 < q < 3, we define the respective cases λ21 > 4λ2, and λ21 = 4λ2, as Ψ3x (t) = t−λ1/2 √λ21 − 4λ2 ⋅ ∫ a ((t s) (1/2)√λ2 1 −4λ2 − (st) (1/2)√λ2 1 ) ⋅ sλ1/2+1Jq−2 a+ g3 (s) ds, if λ21 > 4λ2 or Ψ3x (t) = t−λ1/2 ∫ a sλ1/2+1 ln t sJq−2 a+ g3 (s) ds,


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][2][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][5][6][7][8][9][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  ∈ (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.

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 order  > 0 is defined as provided that the integral exists.
Definition 2. The Hadamard derivative of fractional order  > 0 is defined as where [] denotes the integer part of the real number .
Let (, R) denote the Banach space of all real valued continuous functions defined on , and    (, R) denotes the Banach space of all real valued functions  such that    ∈ (, R).
Multiplying (12) by the integrating factor given by we get Integrating (14), and using again (+) = 0, we conclude that This finishes the proof.
Multiplying by  and then the second derivative would imply  (2) () +  (1) For the last initial condition  (2) (+) = 0, we get  1 =  2 =  3 = 0. Now, multiplying (20) by , it follows that ) which is a second-order Cauchy-Euler differential equation that has a general solution () =   () +   (), where   () and   () are the complementary and particular solution of (14), respectively.To find the complementary solution of (21), consider the solutions  1 () =   1 and  2 () =   2 for the homogeneous equation where 1 − 4 2 )/2 are the distinct real roots of the characteristic equation The complementary solution of the homogeneous equation ( 22) is for some constants  1 and  2 .These two constants can be evaluated by the initial conditions (+) =  (1) (+) = 0 given in ( 16), which both imply that The only solution for these algebraic equations is The Wronskian  for the solutions  1 and  2 is Applying the variation of parameter technique we can get the particular solution Therefore, the general solution is () =   ().

Abstract and Applied Analysis
The evaluation of the constants  1 and  2 in the complementary solution (28) leads to which imply  1 =  2 = 0. Therefore, the general solution becomes again () =   ().The two cases together constitute the required solution.This finishes the proof.
Remark 8.If  2 1 < 4 2 , then (23) has two complex conjugate roots that will not be considered in this article.On the other hand, if  1 =  2 = 0, the solution is () = J  +  3 (), which is a trivial case and hereafter will be negligible.

Existence Theorems
We establish sufficient conditions for existence of solutions to problems (1)-(3) using different types of fixed-point theorems.
We need the following Schauder's fixed point theorem ( [14]).Theorem 11.If U 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.
As   < 1, Ψ  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.
If the operator Ψ  ,  = 1, 2, 3, has a fixed point, the corresponding problem in systems (1)-(3) has this fixed point as a solution.R); then for any  ∈ , and  ∈ U, there exists a positive constant   such that |  (, ())| ≤   .Accordingly, for any  ∈ , and  ∈ U, it follows that Lemma 10.The operator Ψ  : (, R) → (, R),  = 1, 2, 3, is completely continuous.Proof.Obviously, the continuity of the operator Ψ  follows from the continuity of the function   .Let U be a bounded proper subset of (, for any  ∈ .On the other hand, the operator Ψ  : U  → U  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.