Initial Value Problem for Stochastic Hyprid Hadamard Fractional Differential Equation

In this paper, we discuss the existence of solutions for a stochastic initial value problem of Hyprid fractional differential equations of Hadamardtype given by D ( u(t) f(t, u(t)) ) = b(t, u(t)) +σ(t, u(t))Ẇ (t), 1 < t < T, 0 < α ≤ 1, J u(t) |t=1= ξ, where HDα is the Hadamard fractional derivative, f : [1, T ] × L2(Ω) → L2(Ω) and b, σ : [1, T ]×L2(Ω)→ L2(Ω), HJ (·) is the Hadamard fractional integral and ξ be such that (E(‖ ξ ‖)) < ∞, are investigated. The fractional calculus and stochastic analysis techniques are used to obtain the required results.

By hybrid differential equation, we mean that the terms in the equation are perturbed either linearly or quadratically or through the combination of first and second types. Perturbation taking place in form of the sum or difference of terms in an equation is called linear. On the other hand, if the equation is perturbed through the product or quotient of the terms in it, then it is called quadratic perturbation. So the study of hybrid differential equation is more general and covers several dynamic systems as particular cases ( [1]).
Fractional calculus and fractional-order differential equations have been applied in many fields of science and engineering, such as physics ([4]- [5]), chemical ([20]- [21]), etc. Actually, the concepts of fractional derivatives are not only generalization of the ordinary derivatives, but also it has been found that they can efficiently and properly describe the behavior of many physical systems (real-life phenomena) more accurately than integer order derivatives. Fractional differential equations involving Riemann-Liouville and Caputo-type fractional derivatives have extensively been studied by several researchers. However, the literature on Hadamard type fractional differential equations is not enriched yet. the fractional derivative due to Hadamard, introduce in 1892 [3], differs from the aforementioned derivatives in the sense that the kernel of the integral in the definition of the Hadamard derivative contains logarithmic function of arbitrary exponent. A detailed description of the Hadamard derivative and integral can be found in (e.g., [2]).

Material and Methods
This section contains some preliminary facts that we need in the sequel.
Let (Ω, , P) be a probability space, where Ω is a sample space, is a σalgebra and P is a probability measure. Let C = C([1, T ], L 2 (Ω)) be the space of all second order stochastic processes which is mean square continuous on [
satisfies the Volterra integral equation In particular, when 0 < α ≤ 1, problem (3) is equivalent to the integral equation For more details (see e.g., [2]). From Theorem (1) we have the following lemma.
= y(t), 0 < t < 1, is given by the following lemma (fixed point theorem due to [1]) is the fundamental in the proof of our main result. For t ∈ [1, T ], we define u r (t) = (ln t) r u(t), r ≥ 0. Let C r ([1, T ], L 2 (Ω)) be the space of all continuous processes u such that u r ∈ C([1, T ], L 2 (Ω)) which is indeed a Banach space endowed with the norm u C = max t {(ln t) r u(t) 2 }.
Proof. According to (Theorem 9.1, [1]) we split the proof into a sequence of steps.
Define a subset S of C as where ρ satisfies inequality (5).
Clearly, S is closed, convex and bounded subset of the Banach space C; by Lemma (2), the initial value problem (1) is equivalent to the integral equation for t ∈ [1, T ].
Define two operators A : C → C by and B : S → C by Then u = AuBu. We shall show that the operators A and B satisfy all the conditions of Lemma (2).
Step1. We first show that A is Lipschitz on C. Let u, v ∈ C. Then by (A1) we have for all t ∈ [1, T ]. Taking the maximum over the interval [1, T ], we obtain Step2. The operator B is a completely continuous on S. We show that B is continuous on S. Then after some simple calculations we get lim n→∞ (ln t) 1−α Bu n (t) = (ln t) 1−α Bu(t).
for all t ∈ [1, T ]. This shows that B is continuous on S. It sufficient to show that B(S) is a uniformly bounded and equi-continuous set in C. First we have, for all t ∈ [1, T ]. Taking the maximum over the interval [1, T ], the above inequality becomes This show that B is uniformly bounded on S.
Next, we show that B is an equi-continuous in C. Let τ 1 , τ 2 ∈ [1, T ] with τ 1 < τ 2 and u ∈ C. Then we have And after some calculations, we will see that the right hand side of the above inequality tends to zero independently of u ∈ S as τ 2 → τ 1 . Therefore, it follows from Arzelá − Ascoli theorem that B is a completely continuous operator on S.
Step3. Let u ∈ C and v ∈ S be arbitrary elements such that u = AuBv. Taking the maximum over [1, T ], we obtain Step4. According to lemma (2), now we show that M k < 1. This obvious by (A4), since we have Thus all the conditions of Lemma (2) are satisfied and hence the operator equation u = AuBu has a solution in S. In consequence, the problem (1) has a solution on [1, T ].