International Journal of Analysis and Applications Volume 18, Number 2 (2020), 243-253 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-18-2020-243 THE ANALYTICAL SOLUTION OF TELEGRAPH EQUATION OF SPACE-FRACTIONAL ORDER DERIVATIVE BY THE ABOODH TRANSFORM METHOD MOHAMED ELARBI BENATTIA1,∗, KACEM BELGHABA2 1High School of Economic, Laboratory of Mathematics and Its Applications (LAMAP), Oran, Algeria 2University of Oran 1, Laboratory of Mathematics and Its Applications (LAMAP), Oran, Algeria ∗Corresponding author: mohamed.benattia74@yahoo.com Abstract. 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. 1. 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 Received 2019-12-22; accepted 2020-01-20; published 2020-03-02. 2010 Mathematics Subject Classification. 65R10, 26A33. Key words and phrases. Aboodh Transform; fractional differential equation; Caputo fractional derivative; telegraph equation. c©2020 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 243 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-18-2020-243 Int. J. Anal. Appl. 18 (2) (2020) 244 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 Dαxu(x, t) = aut + utt + 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 frac- tional telegraphic equations using the Aboodh transformation method(ATM), where in [5] authors proposed a Sumudu transformation method (STM) which is used to solve this equation. 2. Preliminary 2.1. Fundamental Properties of Fractional Calculus. In this section we give definitions and some basic results. Definition 2.1. An Aboodh transform is defined for functions of exponential order. We consider functions in the set F defined by; F = { f(t) : ∣∣f(t)∣∣ < Me−vt, if t ∈[0;∞[, M,k1,k2 > 0; k1 ≤ v ≤ k2} (2.1) For a given function in the set F , M must be finite number and k1, k2 may be finite or infinite with variable v define as k1 ≤ v ≤ k2. Then, the Aboodh transform denoted by the operator A(:) is defined by the integral equation: T (v) = A [ f(t) ] = 1 v ∞∫ 0 f(t)e−vtdt, t ≥ 0, k1 ≤ v ≤ k2. (2.2) For a given function in the set F, M must be finite number and k1, k2 may be finite or infinite with variable v define as k1 ≤ v ≤ k2. Then, the Aboodh transform denoted by the operator A(:) is defined by the integral equation: T (v) = A [ f(t) ] = 1 v ∞∫ 0 f(t)e−vtdt, t ≥ 0, k1 ≤ v ≤ k2. (2.3) Standard Aboodh transform for some special functions found are given below in Table (2.1). Int. J. Anal. Appl. 18 (2) (2020) 245 f(t) T (v) = A [ f(t) ] 1 1 v2 t 1 v3 tn, n ≥ 1 n! vn+2 eat 1 v2 −av sin(at) 1 v(v2 + a2) cos(at) 1 v2 + a2 sinh(at) 1 v(v2 −a2) tcosh(at) 1 v2 −a2 TABLE(2.1): Aboodh transform of some functions. Definition 2.2. The Riemann-Liouville fractional integral of order α ∈ R+ is defined as D−αf (t) = Iαf(t) = 1 Γ (α) t∫ 0 f(x) (t−x)1−α dx, 0 < α ≤ 1 (2.4) I0f(t) = f(t) Properties of the operator Iα can be found in for α,β > 0, and γ > −1, we have: IαIβf(t) = IβIαf(t) = Iα+βf(t) Iαtγ = Γ (γ + 1) Γ (α + γ + 1) tα+γ Definition 2.3. The caputo fractional derivative (CFD) operator Dαt of order α is Dαt f(t) = I n−αDnf(t) = 1 Γ (n−α) t∫ 0 f(n)(x) (t−x)1+α−n dx, x > 0 (2.5) for n− 1 < α ≤ n, n ∈ N, t > 0. Definition 2.4. The Mittage Leffler function Eα (z) with α > 0, is definite by the following series: Eα (z) = zα Γ (nα + 1) , z ∈ C (2.6) where n ∈ Z+, α ∈ R+. Definition 2.5. The Aboodh transform A [Dαxf(x)] of the fractional derivative using the Caputo idea of the function is given by: A [Dαxf(x)] ] = vαT (v) − n−1∑ k=0 f(k)(0) v2−α+k (2.7) Int. J. Anal. Appl. 18 (2) (2020) 246 It is easy to understand that: A [Dαt f(x; t)] ] = vαA [f(x; t)] ] − n−1∑ k=0 f(k)(x; 0) v2−α+k , n− 1 < α ≤ n, (2.8) 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 A [ αf(t) ±βg(t) ] = αA [ f(t) ] ±βA [ g(t) ] . 3. Procedure Solution Using ATM for Solving Linear TESFO We consider the following linear TESFO of the form: Dαxu(x, t) = aut + utt + bu(x, t) + g(x, t), 0 < x < 1, (3.1) t ≥ 0, 0 < α ≤ 2 where g(x, t) is the source term and a ,b are constants. With Initial Condition ∂(r)u(0, t) ∂xr = u(r)(0, t) ∣∣∣ t=0 = fr(t), r = 0, 1, 2, .......,n− 1. (3.2) Now applying the AT into Eq(3.1) we have: A [Dαxu(x, t)] = A [aut + utt + bu(x, t)] + A [g(x, t)] (3.3) Substituting Eq(2.8) into Eq(3.3) we get: vαA [u(x; t)] ] − m−1∑ k=0 u(k)(0; t) v2−α+k = A [aut + utt + bu(x, t)] + A [g(x, t)] (3.4) A [u(x; t)] = m−1∑ k=0 fk(t) v2+k + v−αA [aut + utt + bu(x, t)] + v −αA [g(x, t)] (3.5) So, according to Aboodh decomposition method (ADM) we can obtain the solution result u(x, t) as u(x, t) = ∞∑ n=0 un(x, t) (3.6) Now, substituting Eq(3.6) into Eq(3.5) gives A [ ∞∑ n=0 un(x, t) ] = m−1∑ k=0 fk(t) v2+k + v −α A [ a ( ∞∑ n=0 un(x, t) ) t + ( ∞∑ n=0 un(x, t) ) tt + b ∞∑ n=0 un(x, t) ] + v −α A [g(x, t)] (3.7) From Eq(3.7) we can define all the coefficients of un+1(x, t) So we get the zero coefficients u0(x, t) as: A [u0(x, t)] = m−1∑ k=0 fk(t) v2+k Int. J. Anal. Appl. 18 (2) (2020) 247 The first component u1(x, t) as: A [u1(x, t)] = v −αA [a (u0(x, t))t + (u0(x, t))tt + bu0(x, t) + g(x, t)] Finally the remaining coefficients of un+1(x, t)can be find in a way like each coefficients is found by using the coming before components. A [un+1(x, t)] = v −αA [a (un(x, t))t + (un(x, t))tt + bun(x, t) + g(x, t)] , n ≥ 0. Applying the Aboodh inverse to the above equations yields the following: u0(x, t) = A −1 [ m−1∑ k=0 fk(t) v2+k ] u1(x, t) = A −1 [v−αA [a (u0(x, t))t + (u0(x, t))tt + bu0(x, t) + g(x, t)]] ... un+1(x, t) = A −1 [v−αA [a (un(x, t))t + (un(x, t))tt + bun(x, t) + g(x, t)]] So that, the AS un(x, t) is given as: un(x, t) = n−1∑ j=0 uj(x, t) (3.8) Such that lim n→∞ un(x, t) = u(x, t) (3.9) 4. 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 Dαxu(x, t) = utt + ut + u, x, t ≥ 0, 0 < α ≤ 2, (4.1) with initial conditions   u(0, t) = e−t, t ≥ 0 ux(0, t) = e −t, t ≥ 0 (4.2) we appling the AT with (2.8) into (4.1) and (4.2) we get: vαA [u(x; t)] − 1∑ k=0 u(k)(0; t) v2−α+k = A [u(x, t)tt + u(x, t)t + u(x, t)] (4.3) So, we have A [u(x; t)] = e−t v2 + e−t v3 + v−αA [u(x, t)tt + u(x, t)t + u(x, t)] (4.4) Int. J. Anal. Appl. 18 (2) (2020) 248 So, according to ADM we can obtain the solution result u(x, t) as u(x, t) = ∞∑ n=0 un(x, t) substituting (3.6) into (4.4) gives A [ ∞∑ n=0 un(x, t) ] = e−t v2 + e−t v3 + v−αA [( ∞∑ n=0 un(x, t) ) )tt + ( ∞∑ n=0 un(x, t) ) )t + ( ∞∑ n=0 un(x, t) )] (4.5) according to equation (4.5), we can calculate the terms un+1(x, t) So, we get the coefficients of u0(x, t) as A [u0(x; t)] = e−t v2 + e−t v3 (4.6) So, we can use the Aboodh inverse in (4.6), we get u0(x; t) = A −1 [ e−t v2 + e−t v3 ] = e−t + xe−t and in the same way we calculate the coefficients of u1(x, t) A [u1(x, t)] = v −αA [(u0(x, t)) tt + (u0(x, t)) t + (u0(x, t))] (4.7) Also, we have u1(x, t) = A −1 [v−αA [(u0(x, t)) tt + (u0(x, t)) t + (u0(x, t))]] (4.8)   u1(x, t) = A −1 [v−αA [e−t + xe−t]] = A−1 [ e−t vα + e−t vα+3 ] = e−tA−1 [ 1 vα+2 + 1 vα+3 ] = e−t ( xα Γ(α + 1) + xα+1 Γ(α + 2) ) (4.9) We can find the coefficients of un(x, t) with the recurente relation as follows un+1(x, t) = A −1 [v−αA [(un(x, t)) tt + (un(x, t)) t + (un(x, t))]] , ∀n ≥ 0 (4.10) Also, we have u2(x, t) = e −t ( x2α Γ(2α + 1) + x2α+1 Γ(2α + 2) ) ... ... un(x, t) = e −t ( xnα Γ(nα + 1) + xnα+1 Γ(nα + 2) ) (4.11) Int. J. Anal. Appl. 18 (2) (2020) 249 Finally, we obtain the approximate solution un(x, t) = e −t ( 1 + x + xα Γ(α + 1) + xα+1 Γ(α + 2) + x2α Γ(2α + 1) + x2α+1 Γ(2α + 2) + .................. ) (4.12) If we put α = 1 in (4.12), we can conclude the exact solution u(x, t) = e−t ( 1 + x + x + x2 2! + x2 2! + x3 3! + x3 3! + .................. ) = 2e−t+x −e−t (a) (1.1) (b) (1.2) (c) (1.3) Figure 1. Comparison between (1.1) the exact solution for α = 1 and (1.2), (1.3) the approximative solutions using 4-term of the ATM for α = 1.7 and α = 1.9 respectively. Int. J. Anal. Appl. 18 (2) (2020) 250 Example 4.2. We consider a linear telegraph equation described by Dαxu(x, t) = utt + 2ut + u x, t ≥ 0, 0 < α ≤ 2, (4.13) with initial conditions   u(0, t) = e−3t, t ≥ 0 ux(0, t) = 2e −3t, t ≥ 0 (4.14) we appling the AT with (2.8) into (4.13) and (4.14) we get: vαA [u(x; t)] − 1∑ k=0 u(k)(0; t) v2−α+k = A [u(x, t)tt + 2u(x, t)t + u(x, t)] (4.15) So, we have A [u(x; t)] = e−3t v2 + 2 e−3t v3 + v−αA [u(x, t)tt + 2u(x, t)t + u(x, t)] (4.16) So, according to ADM we can obtain the solution result u(x; t) as u(x, t) = ∞∑ n=0 un(x, t) (4.17) substituting (3.6) into (4.16) gives A [ ∞∑ n=0 un(x, t) ] = e−3t v2 + 2 e−3t v3 + v −α A [( ∞∑ n=0 un(x, t) ) )tt + 2 ( ∞∑ n=0 un(x, t) ) )t + ( ∞∑ n=0 un(x, t) )] (4.18) according to equation (4.18), we can calculate the terms un+1(x, t). So, we get the coefficients of u0(x, t) as A [u0(x; t)] = e−3t v2 + 2 e−3t v3 (4.19) we use the Aboodh inverse in (4.19), we obtain u0(x; t) = e −3t + 2xe−3t and in the same way we calculate the coefficients of u1(x, t) A [u1(x, t)] = v −αA [(u0(x, t)) tt + 2 (u0(x, t)) t + (u0(x, t))] (4.20) Also, we have u1(x, t) = A −1 [v−αA [(u0(x, t)) tt + 2 (u0(x, t)) t + (u0(x, t))]] (4.21) Int. J. Anal. Appl. 18 (2) (2020) 251   u1(x, t) = A −1 [ v−αA [ 4e−3t + 8xe−3t ]] = e−3tA−1 [ 4 vα+2 + 8 vα+3 ] = 4e−3t ( xα Γ(α + 1) + 2xα+1 Γ(α + 2) ) (4.22) We can find the coefficients of un(x, t) with the recurrence relation as follows un+1(x, t) = A −1 [v−αA [(un(x, t)) tt + 2 (un(x, t)) t + (un(x, t))]] (4.23) Also, we have   u2(x, t) = A −1 [v−αA [(u1(x, t)) tt + 2 (u1(x, t)) t + (u1(x, t))]] = 4A−1 [ v−αA [ e−3t ( 4xα Γ(α + 1) + 8xα+1 Γ(α + 2) )]] = 4e−3tA−1 [ 4 v2α+2 + 8 v2α+3 ] = 16e−3t ( x2α Γ(2α + 1) + 2x2α+1 Γ(2α + 2) ) (4.24) we can give the general solution as follow un(x, t) = 4 ne−3t ( xnα Γ(nα + 1) + 2xnα+1 Γ(nα + 2) ) (4.25) Finally, we obtain the approximate solution un(x, t) = e −3t ( 1 + 2x + 4xα Γ(α + 1) + 8xα+1 Γ(α + 2) + 16x2α Γ(2α + 1) + 32x2α+1 Γ(2α + 2) + ....... 4nxnα Γ(nα + 1) + 2 4nxnα+1 Γ(nα + 2) + ..... ) (4.26) If we put α = 2 in (4.26), we can conclude the exact solution u(x, t) = e−3t ( 1 + 2x + 4x2 2! + 8x3 3! + 16x4 4! + ..... ) = e−3te2x = e−3t+2x (4.27) Int. 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