Differential transformation method for solving one-space-dimensional telegraph equation

Transcription

1 Volume 3, N 3, pp , 2 Copyright 2 SBMAC ISSN -825 wwwscielobr/cam Differential transformation method for solving one-space-dimensional telegraph equation B SOLTANALIZADEH Young Researchers Club, Sarab Branch, Islamic Azad University, Sarab, Iran babaksoltanalizadeh@gmailcom Abstract In this research, the Differential Transformation Method DTM has been utilized to solve the hyperbolic Telegraph equation This method can be used to obtain the exact solutions of this equation In the end, some numerical tests are presented to demonstrate the effectiveness and efficiency of the proposed method Mathematical subject classification: 35Lxx, 35Qxx Key words: differential transformation method, spectral method, telegraph equation, Taylor expansion Introduction In this paper we focus on the following one-space-dimensional Telegraph equation: 2 u t + α u 2 t + βu = 2 u + ψx, t, x 2 where α, β R, ψ : R R R are known and u : R R R is the unknown function This equation with initial and boundary conditions is considered by some authors Numerical solution of Telegraph equation with variable coefficient, was discussed in [6] Mohanty and his coworkers [7, 8], also developed new three-level alternative direction implicit schemes for the two and three-space-dimensional linear hyperbolic equations that these are unconditionally stable Saadatmandi and Dehghan [5], applied a shifted Chebyshev Tau #CAM-28/ Received: /XI/ Accepted: 2/IV/

2 64 METHOD FOR SOLVING ONE-SPACE-DIMENSIONAL TELEGRAPH EQUATION method Gao and Chi [22] solved Eq by a semi-discretization method They transformed Eq into a system consisting of ordinary differential equations with respect to time t and then found an exact solution containing an infinite matrix series Authors of [22], presented two unconditionally stable numerical schemes based on C 3 quartic splines with accuracy orders of Ok 5 + h 4 and Ok 7 + h 4 The concept of DTM was first proposed by Zhou [], who solved linear and nonlinear problems in electrical circuit problems Chen and Ho [2] developed this method for partial differential equations Ayaz [3] applied it to the system of differential equations During recent years, this method has been used for solving various types of equations by many authors For example, this method has been used for differential algebraic equations [8], partial differential equations [5, 6, 7], fractional differential equations [, ], Volterra integral equations [24] and Difference equations [9] Shahmorad et al [9, 2] developed DTM to fractional-order integro-differential equations with nonlocal boundary conditions and a class of two-dimensional Volterra integral equations Borhanifar and Abazari applied this method for Burgers and Schrödinger equations [4, 2, 23] In [4], this method has been utilized for the Kuramoto-Sivashinsky equation with an initial condition There exist similar problem For example, authors of [2, 3, 4] presented several matrix formulation method for solving some equation with a boundary integral condition The aim of this paper is to extend the differential transformation method to solve the hyperbolic Telegraph equation The method can be used to evaluate the approximating solution by the finite Taylor series and by an iteration procedure described by the transformed equations obtained from the original equation using the operations of differential transformation 2 The definitions and operations of DT 2 The one-dimensional differential transform The basic definitions and operations of one-dimensional DTM are introduced in [, 2, 3] as follows: Comp Appl Math, Vol 3, N 3, 2

3 B SOLTANALIZADEH 64 Definition 2 If ut is analytic in the time domain T then d k ut dt k = ϕt, k, t T 2 For t = t i, ϕt, k = ϕt i, k, where k belong to the non-negative integer, denoted as the K domain Therefore, Eq 2 can be rewritten as [ d k ] ut U i k = ϕt i, k =, t T, 3 dt k t=t i where U i k is called the spectrum of ut at t = t i in the K domain Definition 22 If ut is analytic, then it can be shown as t t i k ut = Uk 4 k= Equation 4 is known as the inverse transformation of Uk If Uk is defined as [ d k ] qtut Uk = Mk, k =,, 2,, 5 dt k t=t i then the function ut can be described as ut = qt t t i k k= Uk Mk, 6 where Mk =, qt = Mk is called the weighting factor and qt is regarded as a kernel corresponding to ut If Mk = and qt = then Eqs 4 and 6 are equivalent In this paper, the transformation with Mk = / and qt = is applied Thus from Eq 7, we have Uk = [ d k ] ut, k =,, 2, 7 dt k t=t i Using the differential transform, a differential equation in the domain of interest can be transformed to be an algebraic equation in the K domain and ut can be Comp Appl Math, Vol 3, N 3, 2

4 642 METHOD FOR SOLVING ONE-SPACE-DIMENSIONAL TELEGRAPH EQUATION obtained by finite-term Taylor series plus a remainder, as ut = qt = n t t i k k= Uk Mk + R n+t n t t k Uk + R n+ t k= 8 In order to speed up the convergence rate and improve the accuracy of calculation, the entire domain of t needs to be split into sub-domains [9, ] 22 The two-dimensional differential transform Consider a function of two variables wx, t: R R R, and suppose that it can be represented as a product of two single-variable functions, ie, wx, t = uxvt Based on the properties of one-dimensional differential transform, function wx, t can be represented as wx, t = W i, jx i t j, 9 i= j= where W i, j is called the spectrum of wx, t Now we introduce the basic definitions and operations of two-dimensional DT as follows [] Definition 23 If wx, t is analytic and continuously differentiable with respect to time t in the domain of interest, then W h, k = [ k+h ] h! x k t wx, t, h x=x, t=t where the spectrum function W k, h is the transformed function, which is also called the T -function Let wx, t as the original function while the uppercase W k, h stands for the transformed function Now we define the differential inverse transform of W k, h as following: wx, t = k= h= W k, hx x k t t h Comp Appl Math, Vol 3, N 3, 2

5 B SOLTANALIZADEH 643 Using Eq in, we have wx, t = = k= h= k= h= [ k+h ] h! x k t wx, t x k t h h x=x, t=t W k, hx k t h 2 where x = and t = Now from the above definitions and Eqs and 2, we can obtain some of the fundamental mathematical operations performed by two-dimensional differential transform in Table Original function Transformed function wx, t = ux, t ± vx, t W k, h = Uk, h ± V k, h wx, t = cux, t W k, h = cuk, h wx, t = ux, t W k, h = k + Uk +, h x wx, t = r+s k + r!h + s! x r ux, t W k, h = Uk + r, h + s ts h! wx, t = ux, tvx, t W k, h = k h Ur, h sv k r, s wx, t = x ux, t t vx, t W h, k = k r= s= r= s= h k r + h s + Uk r +, sv r, h s + Table Operations of the two-dimensional differential transform 3 Application of the DTM In this section, we apply the DTM to solve the presented Telegraph equation Consider the equation, with the initial conditions 2 u t + α u 2 t + βu = 2 u + ψx, t, 3 x 2 ux, = f x, x L, 4 Comp Appl Math, Vol 3, N 3, 2

6 644 METHOD FOR SOLVING ONE-SPACE-DIMENSIONAL TELEGRAPH EQUATION and ux, t and boundary conditions = gx, x L, 5 u, t = rt, t T, 6 and u, t = st, t T 7 Let Uk, h as the differential transform of ux, t Applying Table, Eq 2 and Definition 23 when x = t =, we get the differential transform version of Eq 3 as following: h + h + 2Uk, h αh + Uk, h + + βuk, h = k + k + 2Uk + 2, h + ψk, h 8 By the first initial condition we get Uk, x k = k= k= f k x k, 9 which implies Uk, = f k, k =,, 2,, N, 2 and from the second initial condition and Table, we have which implies Uk, x k = k= k= g k x k, 2 Uk, = gk, k =,, 2,, N 22 Similarly, from the boundary conditions, we have U, h = r h, h = 2, 3,, N, 23 h! N k= Comp Appl Math, Vol 3, N 3, 2 Uh, k = sh, h = 2, 3,, N 24 h!

7 B SOLTANALIZADEH 645 Therefore, the values of Uk,, Uk,, and for U, h can be obtained from Eqs 2, 22 and 23 By using Eqs 8 and 24, the remainder values of U can be found as follows: Uk, h + 2 = βuk, h + k + k + 2Uk + 2, h h + h + 2 αh + Uk, h + + ψk, h, k =,,, N 2, h =,,, N 2 Example 3 Consider the Eqs 3-5 with the f x = x, rt =, gx = x st = exp t ψx, t = x exp t α = 2, β = 2 25 Applying Eqs 2, 22 and 23 in initial and boundary conditions of this problem, we get { {, for k =,, for k =, Uk, = and Uk, =, otherwise,, otherwise, and U, h =, h = 2, 3,, N From Eqs 24 and 25, we have U, 2 = 2U, + 2U2, 2 + U, + ϕ, 2 U, 2 = =, 2U, + 23U3, 2 + U, + ϕ, = 2 2!, U2, 2 = U3, 2 = =, U, 3 = U, 3 = 2U, + 2U2, 2 + U, 2 + ϕ, =, 2 3 2U, + 23U3, 2 + U, 2 + ϕ, = 2 3 3!, U2, 3 = U4, 3 = =, Comp Appl Math, Vol 3, N 3, 2

8 646 METHOD FOR SOLVING ONE-SPACE-DIMENSIONAL TELEGRAPH EQUATION By continuing this process, we obtain which implies that which is the Taylor expansion of the k \ h ! 3! 4! Table 2 Different obtained values of Uk, h ux, t x t + 2! t 2 3! t 3 + ux, t = x exp t x L, which is the exact solution of the Example 3 The computational results of Example 3 are presented in Table 3, and the plot of corresponding exact and approximate functions are shown in Figs and 2 x, t n = n = 2,, , , , , , , , , 7 422, Table 3 Numerical results of Example 3 Comp Appl Math, Vol 3, N 3, 2

9 B SOLTANALIZADEH ux,t t x Figure Plot of exact function from Example Ux,t t x Figure 2 Plot of approximate function from Example 3 for n = 2 Comp Appl Math, Vol 3, N 3, 2

10 648 METHOD FOR SOLVING ONE-SPACE-DIMENSIONAL TELEGRAPH EQUATION Example 32 [] Consider the Eqs 3-5 with the f x = sinx, rt =, gx = sinx st = sin exp t ψx, t = α =, β = From the above initial and boundary conditions and Eqs 2, 22 and 23, we obtain the corresponding spectra as follows:, for k is even, Uk, = k 2, for k is odd, and and, for k is even, Uk, = k+ 2, for k is odd, U, h = r h, h = 2, 3,, N h! From Eqs 24 and 25, we have U, 2 = U, 2 = U2, 2 = U3, 2 = U4, 2 = U5, 2 = U, + 2U2, + U, =, 2 U, + 2U3, + U, = 2 2!, U2, + 2U4, + U2, =, 2 U3, + 2U5, + U3, = 2 2!3!, U4, + 2U6, + U4, =, 2 U5, + 2U7, + U5, = 2 2!5!, Comp Appl Math, Vol 3, N 3, 2

11 B SOLTANALIZADEH 649 U, 3 = U, 3 = 2U, + 2U2, 22U, 2 =, 2 3 2U, + 2U3, 22U, 2 = 2 3 3!, Thus we get k \ h ! 3! 4! 2 3 3! 3! 2!3! 3!3! 4!3! 4 5 5! 5! 2!5! 3!5! 4!5! Table 4 Different obtained values of Uk, h which implies that ux, t x t + 2! t 2 3! t x 3 2! + 2! t 2!3! t 2 + 2!5! t 3 + +, which is the Taylor expansion of the ux, t = exp t sinx, which is the exact solution of the Example 32 The computational results of Example 42 are presented in Table 5, and the plot of corresponding exact and approximate functions are shown in Figs 3 and 4 4 Conclusions In this paper, we solved the Telegraph equation with initial conditions by the Differential Transformation method By using this method, Numerical/analytical results obtained by a simple iterative process The numerical results prove Comp Appl Math, Vol 3, N 3, 2

12 65 METHOD FOR SOLVING ONE-SPACE-DIMENSIONAL TELEGRAPH EQUATION x, t n = n = 2,, , , , , , , , , , Table 5 Numerical results of Example 32 ux,t t x Figure 3 Plot of exact function from Example 32 that this method is a powerful techniques for this case of problems Consequently, it is seen that this method can be an alternative way for the solution of partial differential equations that have no analytic solutions Comp Appl Math, Vol 3, N 3, 2

13 B SOLTANALIZADEH 65 Ux,t t x Figure 4 Plot of approximate function from Example 32 for n = 2 Acknowledgments The author would like to thank the reviewers for carefully reading the paper and for their constructive comments and suggestions that have improved the paper The author is deeply grateful to the Young Researchers Club, Sarab Branch, Islamic Azad University, for the financial supports REFERENCES [] JK Zhou, Differential Transformation and Its Applications for Electrical Circuits Huazhong University Press, Wuhan, China, 986 in Chinese [2] CK Chen and SH Ho, Solving partial differential equations by two dimensional differential transform method Appl Math Comput, 6 999, 7 79 [3] F Ayaz, Solutions of the systems of differential equations by differential transform method Appl Math Comput, 47 24, [4] R Abazari and A Borhanifar, Numerical study of the solution of the Burgers and coupled Burgers equations by a differential transformation method Comput Math Appl, 59 2, Comp Appl Math, Vol 3, N 3, 2

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