THE ANALYTICAL SOLUTION OF TELEGRAPH EQUATION OF SPACE-FRACTIONAL ORDER DERIVATIVE BY THE ABOODH TRANSFORM METHOD

In this article, an analytical solution based on the series expansion method is proposed to solve the telegraph equation of space fractional order (TESFO), namely the Aboodh transformation method (ATM) subjected to the appropriate initial condition. Using ATM, it is possible to find exact solution or a closed approximate solution of a differential equation. Finally, several numerical examples are given to illustrate the accuracy and stability of this method.


Introduction
In the last few decades, fractional calculus found many applications in various fields of physical sciences such as viscoelasticity, diffusion, control, relaxation processes and so on [1]. Suspension flows are traditionally modeled by parabolic partial differential equations. Sometimes they can be better modeled by hyperbolic equations such as the telegraph equation, which have parabolic asymptotic. In particular the experimental data described in [1] seem to be better modeled by the telegraph equation than by the heat equation. The telegraph equation is used in signal analysis for transmission and propagation of electrical signals and also used modeling reaction diffusion. The different type solutions of the fractional telegraph equations have been discussed by Momani [2] by using decomposition method, Yildirim [3] by homotopy perturbation method.
Our concern in this work is to consider the space-fractional telegraph equations as D α x u(x, t) = au t + u tt + bu(x, t) + g(x, t), 0 < x < 1 where t ≥ 0, 0 < α ≤ 2, a, b are given constants, g(x, t) is given function.
The main objective of this paper is to introduce a new analytical and approximate solution of spatial fractional telegraphic equations using the Aboodh transformation method(ATM), where in [5] For a given function in the set F , M must be finite number and k 1 , k 2 may be finite or infinite with variable v define as k 1 ≤ v ≤ k 2 . Then, the Aboodh transform denoted by the operator A(:) is defined by the integral equation: For a given function in the set F , M must be finite number and k 1 , k 2 may be finite or infinite with variable v define as k 1 ≤ v ≤ k 2 . Then, the Aboodh transform denoted by the operator A(:) is defined by the integral equation: Standard Aboodh transform for some special functions found are given below in Table (2.1).
Properties of the operator I α can be found in for α, β 0, and γ −1, we have: Definition 2.4. The Mittage Leffler function E α (z) with α > 0, is definite by the following series: where n ∈ Z + , α ∈ R + .
] of the fractional derivative using the Caputo idea of the function is given by: It is easy to understand that: Remark 2.1. The Aboodh transform is linear, i.e., if α and β are any constants and f (t) and g(t) are functions defined over the set F above, then

Procedure Solution Using ATM for Solving Linear TESFO
We consider the following linear TESFO of the form: where g(x, t) is the source term and a ,b are constants.
With Initial Condition Now applying the AT into Eq(3.1) we have: So, according to Aboodh decomposition method (ADM) we can obtain the solution result u(x, t) as Now, substituting Eq(3.6) into Eq(3.5) gives From Eq(3.7) we can define all the coefficients of u n+1 (x, t) So we get the zero coefficients u 0 (x, t) as: The first component u 1 (x, t) as: Finally the remaining coefficients of u n+1 (x, t)can be find in a way like each coefficients is found by using the coming before components.
Applying the Aboodh inverse to the above equations yields the following: So that, the AS u n (x, t) is given as: Such that lim n→∞ u n (x, t) = u(x, t) (3.9)

Illustrative Examples
In this section we shall test two examples using the ATM to solve the TESFO and the solutions we got it by using the present procedure will be comparing with original ES.
Example 4.1. consider the following homogeneous TESFO we appling the AT with (2.8) into (4.1) and (4.2) we get:

So, we have
So, according to ADM we can obtain the solution result u(x, t) as according to equation (4.5), we can calculate the terms u n+1 (x, t) So, we get the coefficients of u 0 (x, t) as So, we can use the Aboodh inverse in (4.6), we get and in the same way we calculate the coefficients of u 1 (x, t) Also, we have We can find the coefficients of u n (x, t) with the recurente relation as follows x, t)) tt + (u n (x, t)) t + (u n (x, t))] , ∀n ≥ 0 (4.10) Also, we have x nα Γ(nα + 1) + x nα+1 Γ(nα + 2) (4.11) Finally, we obtain the approximate solution If we put α = 1 in (4.12), we can conclude the exact solution D α x u(x, t) = u tt + 2u t + u x, t ≥ 0, 0 < α ≤ 2, (4.13) we appling the AT with (2.8) into (4.13) and (4.14) we get: So, according to ADM we can obtain the solution result u(x; t) as substituting (3.6) into (4.16) gives un(x, t) (4.18) according to equation (4.18), we can calculate the terms u n+1 (x, t).