International Mathematical Forum, 2, 2007, no. 45, 2231 - 2236
An Approximation to the Solution of Telegraph Equation by Adomian Decomposition Method J. Biazar
1
and H. Ebrahimi
Department of Mathematics, Faculty of Science University of Guilan, P.O. Box 1914, Rasht, Iran Abstract Adomian decomposition method has been applied to solve many functional equations so far. In this article, this method is applied to solve a famous partial differential equation, called Telegraph equation.
Keywords: Adomian decomposition method; Telegraph equation; Adomian polynomials
1
Introduction
Consider an infinitesimal piece of telegraph wire as an electrical circuit, which consists of resistor of resistance Rdx and a coil of inductance Ldx. If i(x, t) is the current through the wire, the voltage across the resistor is iRdx while that across the coil is it Ldx. Denoting by u(x, t) the voltage at position x and time t. We have that the change in voltage between the ends of the piece of wire is du = −iRdx − it Ldx
(1)
Suppose further that current can escape from the wire to ground, either through a resistor of conductance Gdx or through a capacitor of capacitance Cdx. The amount that escapes through the resistor is uGdx. Because the charge on the capacitor is q = uCdx, the amount that escapes from the capacitor ut Cdx. In total di = −uGdx − ut Cdx (2) With dividing by dx both sides of (1) and (2) and taking the limit dx −→ 0, we get the differential equations ut + Ri + Lit = 0 1
Corresponding author. e-mail:
[email protected]
(3)
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J. Biazar and H. Ebrahimi
Cut + Gu + ix = 0 By From
∂ ∂x
of (3) and
∂ ∂t
(4)
of (4), we have: uxx + Rix + Litx = 0
(5)
Cutt + Gut + ixt = 0
(6)
From (4) and (6), we obtain: ix = −Cut − Gu
(7)
ixt = −Cutt − Gut
(8)
Substituting (7) and (8) into (5) gives: utt + ( Let α =
G , C
β=
R L
and c2 =
1 G R GR + )ut + u= uxx C L CL LC 1 . LC
(9)
Thus
utt + (α + β)ut + αβu = c2 uxx Where u(x, t) is voltage at position x and time t. Also by using ∂ of (4), we have: ∂x itt + (α + β)it + αβi = c2 ixx
(10) ∂ ∂t
of (3) and (11)
Where i(x, t) is current through the wire. Equations (10) and (11) are called telegraph equation.
2
Adomian decomposition method applied to telegraph equation
Consider the following telegraph equation with the indicated initial conditions: utt + (α + β)ut + αβu = c2 uxx
(12)
u(x, 0) = f (x)
(13)
∂u(x, 0) = g(x) (14) ∂t Where u can be voltage or current through the wire. For solving by Adomian decomposition method, pay attention to initial conditions consider operator −1 ∂2 Ltt = ∂t = 0t 0t (.)dtdt and by applying 2 . The inverse operator of Ltt is Ltt the inverse operator to both sides of (12), we get: ∂u(x, 0) t+ u(x, t) = u(x, 0) + ∂t
t t 0
0
(c2
∂2u ∂u − αβu)dtdt − (α + β) 2 ∂x ∂t
(15)
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Solution of telegraph equation
Substituting (13) and (14) into (15), we have: u(x, t) = f (x) + g(x)t +
t t 0
0
(c2
∂2u ∂u − αβu)dtdt − (α + β) 2 ∂x ∂t
(16)
To solve equation (16) by Adomian decomposition method let, as usual in this method, the solution u as the sum of a series u = Σ∞ n=0 un
(17)
So that the components un will be determined recursive and the integrand on the right side as the sum of a series as: c2
∂2u ∂u − αβu = Σ∞ − (α + β) n=0 An (u0 , u1 , . . . , un ) 2 ∂x ∂t
(18)
Where An (u0 , u1, . . . , un ) are called Adomian polynomials and should be computed. Adomian polynomials have been calculated by using an alternate algorithm for calculating Adomian polynomials [4]. Therefore we have: An (u0 , u1 , . . . , un ) = c2
∂un ∂ 2 un − (α + β) − αβun 2 ∂x ∂t
n = 0, 1, 2, . . .
(19)
Substituting (17) and (19) into (16) yields to: ∞ Σ∞ n=0 un = f (x) + g(x)t + Σn=0
t t 0
0
(c2
∂ 2 un ∂un − αβun )dtdt − (α + β) 2 ∂x ∂t
To determine the components un (n ≥ 0) we first identify u0 by terms that arise from the initial conditions at t = 0. Second, the remaining components of u(x, t) can be determined by procedure that each component is determined by using the preceding component. In other words, the method introduces the following recursive relation: u0 = f (x) + g(x)t t t
∂ 2 un ∂un − αβun )dtdt − (α + β) ∂x2 ∂t 0 0 For the first few n, we have: un+1 =
(c2
n = 0, 1, 2, . . .
1
u1 = (
1 t3 t2 t2 Ci1 c2i (−αβ)1−i g (2i)(x)) −(α+β)g(x) +( Ci1 c2i (−αβ)1−i f (2i) (x)) 3! 2! i=0 2! i=0
u2 = (
2 i=0
Ci2 c2i (−αβ)2−i g (2i) (x))
1 t5 t4 − 2(α + β)( Ci1 c2i (−αβ)1−i g (2i) (x)) 5! 4! i=0
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J. Biazar and H. Ebrahimi
2 1 t3 t4 t3 +(α+β)2 g(x) +( Ci2 c2i (−αβ)2−i f (2i) (x)) −(α+β)( Ci1 c2i (−αβ)1−i f (2i) (x)) 3! i=0 4! 3! i=0
u3 = (
3
Ci3 c2i (−αβ)3−i g (2i) (x))
i=0
+3(α + β)2 (
1 i=0
+(
3 i=0
2 t7 t6 − 3(α + β)( Ci2 c2i (−αβ)2−i g (2i) (x)) 7! 6! i=0
Ci1 c2i (−αβ)1−i g (2i) (x))
Ci3 c2i (−αβ)3−i f (2i) (x)) +(α + β)2 (
t5 t4 − (α + β)3g(x) 5! 4!
2 t6 t5 − 2(α + β)( Ci2 c2i (−αβ)2−i f (2i) (x)) 6! 5! i=0
1 i=0
Ci1 c2i (−αβ)1−i f (2i) (x))
t4 4!
.. . Therefore the general term will be as: n−k+1 n un = Σn+1 Ck−1 (α+β)n−k+1( k=1 ((−1)
n−1 +Σnk=1 ((−1)n−k Ck−1 (α+β)n−k (
k i=0
k−1 i=0
Cik−1c2i (−αβ)k−i−1g (2i) (x))
Cik c2i (−αβ)k−i f (2i) (x))
tn+k ) (n + k)!
tn+k ) (n + k)! n = 0, 1, 2, . . .
And the solution is: u(x, t) =
∞ n+1 n=0 k=1
+
n ∞
n ((−1)n−k+1Ck−1 (α+β)n−k+1(
n−k
((−1)
n=0 k=1
n−1 Ck−1 (α
+ β)
n−k
(
k i=0
k−1 i=0
Cik−1 c2i (−αβ)k−i−1g (2i) (x))
Cik c2i (−αβ)k−i f (2i) (x))
tn+k ) (k + n)!
tn+k ) (20) (k + n)!
The exact solution of equation (12) is now entirely determined. However, in practice all terms of the series Σ∞ n=0 un can not be determined; thus we take an approximation of the solution form the truncated series ϕn = Σn−1 i=0 ui with limn→∞ ϕn = u.
3
Numerical results
example 1: Consider the telegraph equation with the initial conditions: utt + (α + β)ut + αβu = c2 uxx
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Solution of telegraph equation
u(x, 0) = sin x ∂u(x, 0) =0 ∂t By using Adomian decomposition method we have: u0 = sin x un = (−1)n sin x
n
n−1 (Ck−1 (α + β)n−k (c2 + αβ)k
k=1
tn+k ) (n + k)!
n = 1, 2, 3, . . .
Thus, the solution is: u(x, t) = sin x(
n ∞
n−1 ((−1)n Ck−1 (α + β)n−k (c2 + αβ)k
n=0 k=1
tn+k )) (n + k)!
For α = β = c = 1 seven-terms approximation to the solution will be as follows: 2 1 1 1 7 u(x, t) ≈ sin x(1 − t2 + t3 − t4 + t6 − t 3 6 90 315 1 8 4 9 19 10 1 1 t + t + t + t11 + t12 ) + 280 2835 113400 124740 7484400 example 2: Let u(x, 0) = ex ∂u(x, 0) =x ∂t α=2 β=3 c=4 By using (20) seven-terms approximation to the solution would be as: 625 x 12 3125 x 11 3 369 3 xt13 + ( x+ e )t + ( x− e )t u(x, t) ≈ 400400 6160 299376 30800 49896 1625 x 10 18125 x 9 375 x 8 47 879 57 x+ e )t + ( x− e )t + ( x + e )t +( 336 2268 1120 4536 32 32 1375 x 7 1475 x 6 75 2059 133 211 x− e )t + (− x+ e )t + ( x − ex )t5 +( 5040 72 144 72 120 4 175 x 4 25 65 19 −5 e )t + ( x − ex )t3 + ( x + 5ex )t2 + xt + ex +(− x + 24 12 6 3 2
4
Conclusion and Discussion
The Adomian decomposition method is a powerful method, which has provided an efficient potential for the solution of physical applications modeled by nonlinear differential equations [1, 2, 3]. The main goal of article has been to derive an approximation to the solution of telegraph equation. We have achieved this goal by applying Adomian decomposition method. In Examples, the approximation can be obtained to any desired number of terms to increase the level of accuracy. Also For computations we used the package Maple 9.
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J. Biazar and H. Ebrahimi
Reference
[1] Adomian, G., Nonlinear Stochastic Systems Theory and Applications to Physics, Kluwer Academic Press, (1989). [2] Adomian, G., Solving Frontier Problems of Physics The Decomposition Method , Kluwer Academic Press, (1994). [3] Adomian, G., Bellman, R., Partial Differential Equations, Reidel Publishing, (1985). [4] Biazar, J., Babolian, E., Nouri, A., Islam, R., An Alternate Algorithm for Computing Adomian Polynomial in Special Cases, Applied Mathematics and Computing, 28(2-3),PP.523-529,(2003) Received: June 1, 2006
