New solutions of hyperbolic telegraph equation

Mustafa Inc; Mohammad Partohaghighi; Mehmet Ali Akinlar; Gerhard-Wilhelm Weber

doi:10.3934/jdg.2020029

Article Contents

2021, Volume 8, Issue 2: 129-138. Doi: 10.3934/jdg.2020029

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New solutions of hyperbolic telegraph equation

1.
Department of Mathematics, Firat University, Elazig, Turkey
2.
Department of Mathematics, University of Bonab, Bonab, Iran
3.
Dept. of Mathematical Engineering, Yildiz Technical Univ., Istanbul, Turkey
4.
Poznan University of Technology, Poznan, Poland, IAM, Metu, Ankara, Turkey

^* Corresponding author: mehmetaliakinlar@gmail.com (Mehmet Ali Akinlar)
^* Corresponding author: mehmetaliakinlar@gmail.com (Mehmet Ali Akinlar)

Received: April 2020

Revised: July 2020

Early access: September 2020

Published: April 2021

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Abstract

We present a new method based on unification of fictitious time integration (FTI) and group preserving (GP) methods. The GP method is applied in numerically discretized ordinary differential equations obtained from application of FTI method to a given partial differential equation (PDE). The algorithm is applied to hyperbolic telegraph equation and utilizes the Cayley transformation and the Pade approximations in the Minkowski space. It avoids unauthentic solutions and ghost fixed points which is one of the advantages of the present method over other related numerical methods in the literature. The technique is tested on three specific examples for various parameter values appearing in the telegraph equation and discretization steps. Such solutions of the telegraph equation are obtained first time in this paper. Illustrative figures are provided. Efficiency of the method is determined by an error analysis which is achieved by comparing numerical solutions with exact solutions.

Keywords:

Mathematics Subject Classification: Primary:33E30, 34K28, 35G05, 35G15;Secondary:65M22, 65J15.

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Figure 1. Exact and numerical solutions and error for $ T = 0.05,\,\, m = n = 25,\,\, \zeta = 60000\,\,\kappa = 0.1 $ and $ u_i^j(0) = 0.1 $ and $ \Delta \xi = 3/10000000000 $ for Ex.1.

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Figure 2. Exact and numerical solutions and error for $ T = 0.05,\, m = m = 30,\, \zeta = 8000,\,\kappa = 0.1 $ and $ u_i^j(0) = 0.001 $ and $ \Delta \xi = 3/10000000000 $ for Ex.2.

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Figure 3. Exact and numerical solutions and error for $ T = 0.05,\, m = n = 40,\, \zeta = 7995,\,\kappa = 0.1 $ and $ u_i^j(0) = 0.001 $ and $ \Delta\xi = 3/10000000000 $ for Ex.2.

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Figure 4. Exact solution, numerical solution and error for $ T = 0.05,\, m = n = 25,\, \zeta = 5995,\,\kappa = 0.01 $ and $ u_i^j(0) = 0.01 $ and $ \Delta \xi = 3/10000000000 $ for Ex. 3.

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Table 1. Solution and error values for $ m=n=25 $ and $ u_i^j(0)=0.1 $ for Ex.1

(x, t)	Numerical	Exact	Error

(-2, 0)	0.1353	0.1353	8.0756e-14
(-1, 0.01)	0.3641	0.3641	2.5085e-13
(0, 0.02)	0.9773	0.9773	9.6537e-13
(1, 0.03)	2.6346	2.6346	7.8534e-12
(2, 0.04)	7.0875	7.0875	2.2148e-10

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Table 2. Numerical and exact solutions and error values for $ m=n=30 $, and $ u_i^j(0)=0 .001 $ for Ex.2

(x, t)	Numerical	Exact	Error

(0, 0)	0	0	0
(2, 0.01)	0.9289	0.9289	1.9657e-13
(4.02)	-0.6876	-0.6876	9.7778e-14
(6, 0.03)	-0.4197	-0.4197	3.7708e-14
(8, 0.04)	0.9628	0.9628	2.2468e-13
(10, 0.05)	3.6693e-16	3.6693e-16	2.9738e-44

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Table 3. Solutions with error values for $ m=n=40 $ and $ u_i^j(0)=0.001 $ for Ex. 2

(x, t)	Numerical	Exact	Error

(-10, 0)	-3.6739e-16	-3.6739e-16	2.9813e-44
(-5, 0.01)	0.9350	0.9350	4.9081e-12
(0.02)	-0.9924	-0.9924	1.3054e-11
(5, 0.03)	-0.9346	-0.9346	9.7228e-12
(10, 0.04)	3.6708e-16	3.6708e-16	2.9763e-44

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Table 4. Solutions with error values for $ m=n=25 $, $ u_i^j(0)=0.001 $ for Ex.3

(x, t)	Numerical	Exact	Error

(-2, 0)	-0.9093	-0.9093	2.6858e-13
(-1, 0.01)	-0.8328	-0.8328	6.0082e-12
(0, 0.02)	0	0	3.1758e-18
(1, 0.03)	0.8173	0.8173	1.3564e-11
(2, 0.04)	0.8740	0.8740	2.4814e-13

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