European Journal of Scientific Research ISSN 1450-216X Vol.80 No.4 (2012), pp.552-559 © EuroJournals Publishing, Inc. 2012 http://www.europeanjournalofscientificresearch.com

A Numerical Approach to Solve Integro-Differential Equations System Mustafa Haluk Çelik Kırgızistan–Türkiye Manas Üniversitesi Fen Bilimleri Enstitüsü, Bişkek, Kırgızistan E-mail: [email protected] Tel: +996 557 110011 İsa Muslu Kırgızistan–Türkiye Manas Üniversitesi Fen Bilimleri Enstitüsü, Bişkek, Kırgızistan E-mail: [email protected] Tel: +996 771 111000 Bülent Bayraktar Kırgızistan–Türkiye Manas Üniversitesi Tradional and Voc.College, Bişkek, Kırgızistan E-mail: [email protected] Tel: +996 700103680 Abstract In this paper, we represent a numerical method to solve linear and nonlinear integro-differential equations system. By this method, we can find the solution of the system by power series and reproduce the analytical solution if the exact solutions are polynomial, otherwise we find their's Taylor series. The comparison of the approximate solution and exact solution shows that the used method is easy and practical for linear and nonlinear integro-differential equations system. The package Maple 13 is used for computations. Keywords: Integro-differential equation, Power series, Approximate solution,Exact solution.

1. Introduction Integro-differential equations arise in a lot of physical process such as nano-hydrodynamics [1], glassforming process [2], drop wise condensation [3], and wind ripple in the desert [4]. There are many numerical methods to solve system of linear and nonlinear integro-differential equations, like the Adomian decomposition methods [5], homotopy perturbation method [6] and [7], Galerkin method [8] and variational iteration method [9]. Nowadays, the Power series method (PSM) has used to solve stiff ordinary differential equations system [10], linear Volterra integral equations system of the second kind and system of integro-differential equations, [12], [13]. In this paper Power series method in which the Taylor exansion of the exact solution of linear or nonlinear integro-differential equations

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553

system is obtained by recursive procedure is used. To show main logic of this method, let’s consider the following integro-differential equations system x

U ' ( x ) = H ( x, U ( x ) ) + ∫ K ( x, t , U ( t ) , U ' ( t ) ) dt

(1.1)

0

along with initial condition F(0) = b, where T U = [u1 , u2 ,..., un ] , H = [ h1 , h2 ,..., hn ] , T

K = ⎡⎣ kij ⎤⎦ , i, j = 1, 2,..., n, b = [b1 , b2 ,..., bn ] . In the equation (1.1), H and K are given analytic functions and without any loss of generality, assume that they are polynomials, if they are not polynomials then they can be substituted by their's Taylor expansion. Also, b is a fixed constant vector and the vector function U is the solution of the equation (1.1), which will be determined. T

2. The Method of Power Series Let us search the solution of the integro-differential equations system (1.1) as follow n

ui = ∑ cij x j , i = 1, 2,..., n.

(2.1)

j =0

By initial conditions, we have ci 0 = ui ( 0 ) ,

i = 1, 2,.., n. We can calculate the coefficients of

(2.1). So, we think the solution of problem (1.1) as U ( x ) = c0 + c1 x,

(2.2)

where c j = ( cij ) , i = 1, 2,..., n and c1 is unknown.Substituting (2.2) into (1.1), we get the following

( B1c1 − a1 ) + P1 ( x ) = 0 where B 1 is n x n constant matrix, a 1 is n x 1 constant vector, P1 ( x ) = ⎡⎣ pi1 ( x ) , i = 1, 2,..., n ⎤⎦ and pi1 ( x ) are polynomials of order equal or bigger than 1. By ignoring P1 ( x ) , we have an algebraic linear equations system of c1 . If we solve this system, the coefficient of x

system

in (2.2) can be determined. For next step, we suppose that U ( x ) = c0 + c1 x + c2 x 2

(2.3) where c0 , c1 are known and c2 is unknown.Substituting (2.3) into (1.1), we get the following system

( B1c2 − a2 ) + P2 ( x ) = 0 where B 2 and a 2 are similar to B 1 and a 1 , respectively, P2 ( x ) = ⎡⎣ pi 2 ( x ) , i = 1, 2,..., n ⎤⎦ and pi 2 ( x ) are polynomials of order greater than 1. By ignoring P2 ( x ) ,

we have again an algebraic system of linear equations of c2 and by solving this system, coefficients of x 2 in (2.3) can be determined. This process is repeated till the arbitrary order coefficients of Power series of the solution for the problem be obtained. We can investigate the convergence of the spoken method by the following theorem. Without loss of generality, we can prove it for n = 1. Theorem 2.1. Let u = f(x) be the exact solution of the following integro-differential equation, x

u ' ( x ) = h ( x, u ( x ) ) + ∫ K ( x, t , u ( t ) , u ' ( t ) ) dt , u ( 0 ) = b

(2.4)

0

And, assume that f(x) has a power series representation. Then, the spoken method obtains it (the Taylor expansion of f(x)).

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Mustafa Haluk Çelik, İsa Muslu and Bülent Bayraktar

Proof. By the spoken method, we assume that the approximate solution to Eqn.(2.4) be as follows, if ( x ) = c + c x + c x 2 + ⋅⋅⋅. (2.5) 0 1 2 Hence, it is enough that we only prove, m f ( ) ( 0) cm = , m = 1, 2,3,..., m! Note that for m = 0, the initial condition gives, c0 = f ( 0 ) = b.

(2.6) (2.7)

Then, for m = 1, if we place u = f(x) and x = 0 in Eqn.(2.4), we get f ' ( 0 ) = h ( 0, f ( 0 ) ) + 0

(2.8)

On the other hand, from (2.5) and (2.7), we have if ( x ) = c + c x 0 1

(2.9)

by substituting (2.9) into Eqn.(2.4) and placing s = 0, we get c1 = g ( 0, f ( 0 ) ) + 0 = f ' ( 0 ) .

(2.10)

For m = 2, differentiating Eqn.(2.4) with respect to x, we have x ∂ ∂ ∂ f '' ( x) = h( x, f ( x) ) + h( x, f ( x) ) f ' ( x) + K ( x, f ( x) , f ' ( x) ) + ∫ K ( x, t, f ( t ) , f ' ( t ) ) dt ∂x ∂u ∂x 0 placing s = 0 in (2.11) and we get ∂ ∂ f '' ( 0 ) = h ( 0, f ( 0 ) ) + h ( 0, f ( 0 ) ) f ' ( 0 ) + K ( 0, f ( 0 ) , f ' ( 0 ) ) ∂x ∂u According to (2.5),(2.7) and (2.10), let if ( x ) = f ( 0 ) + f ' ( 0 ) x + c x 2 2 by substituting (2.13) into (2.11), and placing x = 0, we get ∂ ∂ 2c2 = h ( 0, c0 ) + h ( 0, c0 ) c1 + K ( 0, c0 , c1 ) ∂x ∂u Hence, with comparison (2.12) and (2.14), we conclude that 2c2 = f '' ( 0 ) , or c2 =

(2.11)

(2.12) (2.13) (2.14) f '' ( 0 ) , 2

If we continue the above procedure, we can easily prove (2.6) for m = 3,4,… Corollary 2.2. If the exact solution to Eqn.(2.4) be a polynomial, then the discussed method will be gotten the real solution.

3. Numerical Test of the Method To make clear of the method, we consider three examples of integro-differential equations and then we will compare the obtained results with the exact solutions or the other methods. Example 3.1. Consider the following system of linear Volterra integro-differential equations, x ⎧ ' 2 u = 2 + x + 2 x − u x − ⎪ 1 2( ) ∫0 ( u1 ( s ) + u2 ( s ) )ds, ⎪ x ⎪⎪ ' (3.1) ⎨u2 = −1 − 3 x + u1 ( x ) − ∫ ( u1 ( s ) + u2 ( s ) )ds, 0 ⎪ ⎪u ( 0 ) = −1 , u ( 0 ) = 1. 2 ⎪ 1 ⎪⎩

A Numerical Approach to Solve Integro-Differential Equations System

555

This problem has the exact solution u1* ( x ) = x + e x , u2* ( x ) = x − e x .By using the initial conditions, e0 = [ −1,1] .Let the solution of (3.1) be T

⎧⎪u1 ( x ) = e10 + e11 x = −1 + e11 x, ⎨ ⎪⎩u2 ( x ) = e20 + e21 x = 1 + e21 x. To find e11 , e21 we substitute (3.2) into (3.1) then we have 11 ( x ) ⎧

p  ⎪ 1 1 ⎛ ⎞ ⎪( e11 − 1) = ⎜ x + 2 x 2 − e21 x − e11 x 2 − e21 x 2 ⎟ , 2 2 ⎪ ⎝ ⎠ ⎨ ⎪( e + 2 ) = ⎛ −3 x + e x − 1 e x 2 − 1 e x 2 ⎞ , 11 11 21 ⎜ ⎟ ⎪ 21 2 ⎝2  ⎠

⎪ p21 ( x ) ⎩ p11 ( x ) , p21 ( x ) are O(x) and Where

(3.2)

(3.2)

by

ignoring

B1e1 = a1 , Where B = ⎡1 0 ⎤ , a = ⎡ 2⎤ , e = ⎡e11 ⎤ . thus, e ⎡ 1⎤ and then 1 = ⎢ ⎢0 1⎥ 1 ⎢ 0 ⎥ 1 ⎢e ⎥ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ −2 ⎦ ⎣ 21 ⎦ Let’s continue to next step. Let ⎧⎪u1 ( x ) = −1 + x + e12 x 2 , ⎨ 2 ⎪⎩u2 ( x ) = 1 − 2 x + e22 x . Like previous step, by substitute (3.3) p12 ( x ) ⎧

 ⎪ 1 1 ⎛5 2 ⎞ ⎪( 2e12 − 3) x = ⎜ x − e22 x 2 − e12 x3 − e22 x3 ⎟ , 3 3 ⎪ ⎝2 ⎠ have ⎨ ⎪( 2e + 2 ) x = ⎛ e x 2 − 1 e x3 − 1 e x3 + 1 x 2 ⎞ , 12 22 ⎜ 12 ⎟ ⎪ 22 3 3 2 ⎠ ⎝ 

⎪ p22 ( x ) ⎩

we

have

⎧⎪u1 ( x ) = −1 + x, ⎨ ⎪⎩u2 ( x ) = 1 − 2 x. (3.3) into

(3.1),

we

⎡ 3⎤ By ignoring p12 ( x ) , p22 ( x ) which are O ( x ) and solve system B2 e2 = a2 , we get c2 = ⎢ 2 ⎥ ⎢ ⎥ ⎣ −1 ⎦ 1 2 ⎧ 3 2 ⎧ u x = 1 + 2 x + x + e13 x3 , ( ) 1 ⎪ ⎪u1 ( x ) = −1 + x + x , ⎪ 2 2 then .We go to next step ⎨ And ⎨ ⎪u2 ( x ) = 1 − 2 x − x 2 . ⎪u ( x ) = −1 − 1 x 2 + e x3 . 23 ⎩ ⎪⎩ 2 2 2

and

them,

13 ( x ) ⎧

 p  ⎪⎛ 7⎞ 1 1 1 ⎛ ⎞ ⎪⎜ 3e13 − ⎟ x 2 = ⎜ −e23 x 3 − x 3 − e13 x 4 − e23 x 4 ⎟ = 0, 2⎠ 6 4 4 ⎪⎝ ⎝ ⎠ ⎨ ⎪( 3e − 2 ) x 2 = ⎛ e x3 − 1 x3 − 1 e x 4 − 1 e x 4 ⎞ = 0, 13 23 ⎜ 13 ⎟ ⎪ 23 6 4 4 ⎝ 

⎠ ⎪ p23 ( x ) ⎩

556

Mustafa Haluk Çelik, İsa Muslu and Bülent Bayraktar 3 2 7 3 ⎡ 7⎤ ⎧ ⎢ 6⎥ ⎪⎪u1 ( x ) = −1 + x + 2 x + 6 x , So c2 = ⎢ ⎥ and then ⎨ ⎢ 2⎥ ⎪u ( x ) = 1 − 2 x − x 2 + 2 x 3 . ⎪⎩ 2 3 ⎣⎢ 3 ⎦⎥ The rest of components of the solution by iteration method can be found by a similar way. 3 2 7 3 5 4 17 5 1 6 ⎧ 7 ⎪⎪u1 ( x ) = −1 + x + 2 x + 6 x − 24 x − 120 x + 48 x + O ( x ) , ⎨ ⎪u ( x ) = 1 − 2 x − x 2 + 2 x 3 + 1 x 4 − 2 x 5 − 1 x 6 + O ( x 7 ) . ⎪⎩ 2 3 4 15 40 We can see that these solutions yield to the exact solutions, or simply, the exact solutions are

⎧u1* ( x ) = u1 ( x ) + O ( x 7 ) , ⎪ ⎨ * 7 ⎪⎩u2 ( x ) = u2 ( x ) + O ( x ) . Here O ( x 7 ) is the reminder, which define the error between the exact solution and the Taylor polynomial solution. Example 3.2. Consider the following nonlinear integro-differential equation, x ⎧ 2 ⎪u ' ( x ) = 1 + ∫ u ( s ) u ' ( s ) ds, ⎨ 0 ⎪u 0 = −1. ⎩ ( )

(3.4)

We use the Power series method to get the solution of the problem. From the initial condition, e0 = −1 . Let the solution of (3.4) is in the form of

u ( x ) = e0 + e1 ( x ) = −1 + e1 ( x )

(3.5)

1( x)

p  1 ⎛ ⎞ To find e1 , we substitute (3.5) into (3.4), we have ( e1 − 3) = ⎜ x + e13 x 3 − e12 x 2 + e1 x ⎟ 3 ⎝ ⎠ p ( x) By ignoring 1 which is O(x), we obtain e1 = 3 and then u(x) = x. For the next step, we assume that u ( x ) = −1 + 3x + e2 x 2 (3.6)

p2 ( x )

 1 ⎛1 ⎞ By substituting (3.6) into (3.4), we have ( 2e2 − 4 ) x = ⎜ e23 x 6 + 3e23 x 5 + ... + ( 2e2 − 18 ) x 2 ⎟ 2 ⎝3 ⎠ From the relation above and by ignoring p2 ( x ) , we have e2 = 2 and u(x) is the same as in

previous step . By repeating this progress, we can calculate more coefficients of the solution. We have e8 and the result is calculated these coefficients till 7 4 16 41 403 7 1489 8 u ( x ) = −1 + 3 x + 2 x 2 − x 3 − x 4 + x 5 + x 6 − x − x + O ( x 9 ) , It is easy to verify that the 3 3 3 18 42 1008 ⎛ 2 ⎞ exact solution is u * ( x ) = 2 tan ⎜⎜ x ⎟⎟ = u ( x ) + O ( x 9 ) , where O ( x9 ) is the reminder of the Taylor ⎝ 2 ⎠ polynomial solution. Example 3.3. Let’s consider the following nonlinear integro-differential equation,

A Numerical Approach to Solve Integro-Differential Equations System

557

x ⎧ 5 3x 1 3 x ⎪u ' ( x ) = e − e + + ∫ u ( s ) ds, 6 3 0 ⎨ ⎪u 0 = −1, ⎩ ( )

(3.7)

u ( x ) = e0 + e1 x = −1 + e1 x

(3.8)

u* ( x ) = e x .In this example, in m th step, we use m+1 terms of Taylor expansion with the exact solution x 3x of e and e . Again, we use the Power series method for obtaining the solution of the problem. From the initial condition, e0 = 1 . Assume, the solution of (3.7) is the form 1( x)

 p  4 1⎞ ⎛ 5 1 ( −1 + e1 x ) 1 ⎞ ⎛ By substituting (3.8) into (3.7), we get, ⎜ e1 − ⎟ = ⎜ e x − e3 x + − ⎟. 3 ⎠ ⎜⎝ 6 4 4e1 ⎟⎠ e1 ⎝ 1 1 By ignoring p1 ( x ) , we get e1 = and then u ( x ) = −1 + x : For the next step, we assume that 3 3 1 u ( x ) = −1 + x + e2 x 2 (3.9) 3 and by substituting it into (3.7), we have 12 ( x )

 p  ⎛ x 5 3x 1 3 7 1 2 6 x2 ⎞ ( 2e2 + 1) x = ⎜ e − e + e2 x + e2 x + ... + ⎟ = 0, 6 7 6 2⎠ ⎝ 1 and we found From the above relation and by ignoring p2 ( x ) , we have e2 = − 2 1 1 u ( x ) = −1 + x − x 2 By going on this procedure, more coefficients of the solution can be calculated. 3 2 These coefficients till c6 have been computed and the result is 1 1 1 11 4 83 5 41 6 23 7 373 8 u ( x ) = −1 + x − x 2 + x 3 − x + x − x + x − x , 3 2 6 72 1080 720 720 17280 This function is exactly the first seven terms of Taylor expansion of the analytic solution. So, u * ( x ) = u ( x ) + O ( x 9 ) and O ( x 9 ) is the error between the exact solution and the Taylor

polynomial expansion. Example 3.4. Finally, let’s consider the following nonlinear integro-differential equation, t ⎧ 2 2 2 ⎪ x '' ( t ) + 2tx ' ( t ) = ∫ ( ts ln x ' ( s ) + y '' ( s ) ) ds, 0 ⎪ t ⎪ ⎪ y '' t = 2 − t 2 + ln y '' s − x ' s + x ' s y ' s ds, (3.10) ⎨ ( ) ∫0 ( ( ) ( ) ) ( ) ( ) ⎪ ⎪ x ( 0 ) = 0, x ' ( 0 ) = 1, ⎪ ⎪⎩ y ( 0 ) = y ' ( 0 ) = 0. This system has the exact solution of x* ( t ) = t and y* ( t ) = t 2 . Let u1 ( t ) = x ( t ) , u2 ( t ) = x ' ( t ) , u3 ( t ) = y ( t ) , u4 ( t ) = y ' ( t ) , so, the system (3.10) convert to

558

Mustafa Haluk Çelik, İsa Muslu and Bülent Bayraktar

⎧u1' ( t ) = u2 ( t ) ⎪ t ⎪ ' 2 2 2 1 ⎪u2 ( t ) + 2tu2 ( t ) = ∫ ( ts ln u2 ( s ) + u4 ( s ) ) ds, 0 ⎪ ⎪ ' ⎨u3 ( t ) = u4 ( t ) , ⎪ t ⎪u ' ( t ) = 2 − t 2 + ln ( u ' ( s ) − u ( s ) ) + u ( s ) u ( s ) ds, 4 2 2 4 ∫0 ⎪ 4 ⎪ ⎪⎩u1 ( 0 ) = 0, u2 ( 0 ) = 1, u3 ( 0 ) = 0, u4 ( 0 ) = 0.

(

(3.11)

)

Let U ( t ) = ⎡⎣u1 ( t ) , u2 ( t ) , u3 ( t ) , u4 ( t ) ⎤⎦

T

and ei = [ e1i , e2i , e3i , e4i ] . From the initial conditions, T

e0 = [ 0,1, 0, 0 ] and thus, U ( t ) = e0 . Assume, the solution of (3.11) is in the form of T

U ( t ) = e0 + te1 = [ e11t ,1 + e21t , e31t , e41t ]

T

(3.12)

By using the Taylor series of ln u2 about 1 and ln ( u4' ( s ) − u2 ( s ) ) about e41 − 1 and replacing (3.12) into (3.11), we get, 11 ( t ) ⎧

p  ⎪( e − 1) + ( −e t ) = 0, 21 ⎪ 11 21 ( t ) ⎪

p  ⎪e + ( −e t ) = 0, 41 ⎪ 31 p31 ( t ) ⎪

 ⎨ ⎪e + ⎡( 2 − e ) t + 4e t 2 + 2e2 t 3 − 2 e t 5 + 1 e2 t 6 − 2 e3 t 7 + 1 e4 t 8 ⎤ = 0, 41 21 21 21 21 21 21 ⎥ ⎪ 21 ⎢⎣ 3 3 9 9 ⎦ ⎪ p41 ( t )

 ⎪ ⎪ ⎡⎛ 1 1 ⎞ 2 ⎛1 2 1 ⎞ 3 1 3 4 1 4 5⎤ ⎪( e41 − 2 ) + ⎢⎜1 + e21 − e41 ⎟ t + ⎜ e21 − e21c41 ⎟ t + e21t + e21t ⎥ = 0, 20 2 ⎠ 3 12 ⎝6 ⎠ ⎣⎝ 2 ⎦ ⎩ By ignoring pi1 ( t ) , i = 1, 2,3, 4, , we find the following U ( t ) = [t ,1, 0, 2t ]

(3.13)

For the next step, we assume that T U ( t ) = [t ,1, 0, 2t ] + e2t 2

(3.14)

T

⎧ 2e12t + O ( t 2 ) = 0, ⎪ ⎪( 2e32 − 2 ) t + O ( t 2 ) = 0, ⎪ and by replacing it into (3.11), we have ⎨ 2 ⎪ 2e22 + O ( t ) = 0, ⎪ 2 ⎪⎩ 2e42t + O ( t ) = 0,

From the relation above , we have e2 = [ 0, 0,1, 0] and we find U ( t ) = ⎡⎣t ,1, t 2 , 2t ⎤⎦ . By going on T

T

this procedure, we will find zero for all of the remain coefficients. So we have been calculated the exact solution for this example.

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4. Conclusions In this study, we used the method of Power series to find numerical solution of linear and nonlinear Volterra Integro-differential equations system. As shown in the examples, the discussed method is a powerful progress for solving the problems.

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