IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Extend Differential Transform Methods for Solving Differential Equations with Multiple Delay Gh. J. Mohammed, F.S. Fadhel Department of Mathematics, College of Education (Ibn Al-Haitham), University of Baghdad Department of Mathematics and Computer Applications, College of Science, Al-Nahrain University Received in : 29 November 2010 Accepted in :20 September 2011 Abstract In this paper, we present an approximate analytical and numerical solutions for the differential equations with multiple delay using the extend differential transform method (DTM). This method is used to solve many linear and non linear problems. Key words: (differential transform method, solving differential equation, multiple delay). Introduction In this paper, we extend the differential transform method (DTM) to solve the nth order differential equations with multiple delay of the form: y(n)(x) = f(x,y(x),y(x – r1), y(x – r2), …, y(x – rm)) , mℕ …(1) where y : I  ℝ, f: I ℝ2  ℝ, I  ℝ, r i > 0, i = 1,2, …, m The differential transform method was first applied in the engineering domain in [1]. In general, the DTM is applied to find the solution of electric circuit problems [2]. The DTM is numerical method based on Taylor series expansion, which is constructed as an analytical solution in the form of a polynomial. The traditional high order Taylor series method requires symbolic computation. However, the DTM obtains a polynomial series solution by means of an iterative procedure [3]. Recently the application of differential transform method is successfully extended to obtain approximate solutions to linear and nonlinear functional equations. Delay differential equations are observed in many fields of science and technology, such as biology engineering and physics. Many dynamic population first order nonlinear scalar equation of the form y'(x) = g(y(x)) – g(y(x – L)) …(2) may be used as a model for certain population growth if individuals have a constant life span L, where y(x) is the population size at time t and g(y) is the birth rate. Recently, various methods such as, monotone iterative technique, a domain decomposition method and the spline functions method have been considered for approximate solutions of DDE [4]. IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Hence, due to practical reasons and the papers mentioned above, we have been motivated to deal with DDE and develop DTM for both linear and nonlinear delay differential equations. According to the best of our knowledge, DTM has not been studied for DDE till now. With this technique, it is possible to obtain highly accurate numerical solution, analytical solution and as well as exact solutions. The aim this paper is to extend the method of differential transformation for solving differential equations with multiple delay of difference types as in equation (1). Differential Transform Method The differential transform of a function y(x) is defined as, [2]: 0 k x xk 1 d Y(k) [ y(x)] , k k! dx  � …(3) where Y(k) refers to the differential transform of a given function y(x), and 0x is the initial state. Throughout this paper, we use the small and capital letters to represent the original and transformed functions, respectively. The inverse of the differential transform Y(k) is defined by k 0 k 0 y(x) Y(k)(x x )     …(4) The automatic computation of DTM might be done. In this case, the following steps should be taken into consideration successively: i) The differential transform of each term in the DDE is computed ; ii) The recurrence equation is obtained ; iii)Y(0), Y(1), Y(2), Y(3),… are calculated by the recurrence equation and given initial condition ; iv)Finally, these values are substituted back into eq. (4). Preliminaries of the DTM The following theorems give the properties of the DTM which can be easily derived from equations (3) and (4), for their proofs and more details (see [5], [6]). Theorem (1):- If y(x) = f(x)  g(x), then Y(k) = F(x)  G(x). Theorem (2):- If y(x) = c f(x), then Y(k) = c F(x) Theorem (3):- If n n d f (x) (k n)! y(x) , then Y(k) [ ] F(k n) k!dx     , k ℕ. Theorem (4):- If 1 k 1 1 k 0 y(x) f (x).g(x), thenY(k) F(k )G(k k )     , k ℕ. Moreover, we need the following theorem (see [4], [5]) : Theorem (5):- If y(x) = f1(x)  f2(x) … fn – 1 (x)  fn(x), then n 1 n 2 2 1 k 3n 1 2 1 2 2 1 n 1 n 1 n 2 n 2 1 k 0 k 0 k 0 k 0 kk k Y(k) ... F (k) F (k k )...F (k k ) F (k k )                  IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Main Result The following theorems are proved to give the differential transform of given functions with constant delay. Theorem (6):- The differential transform of y(x) = f(x – r), r  1, is  1 1 1 N hh k h k1 1kh Y(k) ( 1) r F(h ), N     Proof: - Using the definition of the inverse DTM where the definition given by eq. (4), we get: k 0 k 0 y(x) F(k)(x r x )      0 1 2 3 0 0 0 0F(0)(x r x ) F(1)(x r x ) F(2)(x r x ) F(3)(x r x ) ...             2 3 0 2 3 0 1 2 3 2 0 0 [F(0) rF(1) r F(2) r F(3) ...](x x ) [F(1) 2F(2) 3r F(3) 4r F(4) ...] (x x ) [F(2) 3F(3) 6r F(4) 10r F(5) ...](x x ) ...                            1 1 1 1 1 1 1 1 1 1 1 1 h hh h 0 h 1 h 1 11 1 1 0 1 0h 0 h 10 1 h hh 2 h 2 h 3 h 32 31 1 1 0 1 0h 2 h 32 3 ( 1) r F(h )(x x ) ( 1) r F(h )(x x ) ( 1) r F(h )(x x ) ( 1) r F(h )(x x ) ...                              1 1 1 hh k h k k1 h1 0kk 0 h k y(x) ( 1) r F( )(x x )           …(5) Taking in to account eq. (4) and eq. (5), we have Y(k) as:  1 1 1 N hh k h k1 1kh Y(k) ( 1) r F(h ), N     Theorem (7):- The differential transform of y(x) = f1(x – r1)f2(x – r2), provided that r1 > 0 and r2 > 0 is: 1 1 2 1 2 1 2 1 1 11 1 2 1 h h hk N h h k h k h k k h h1 1 2 2 k k k kk h h k k Y(x) ( 1) r r F ( )F ( ), for                        N Proof: let the differential transforms of f1(x – r1) and f2(x – r2) at x = x0 be G1(k) and G2(k), respectively. Using theorem (4), we have the differential transform of y(x) as: IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 1 k 1 2 1 k 0 Y(x) G(k)G (k k) ...(6)    From theorem (6), we get: 1 1 1 1 11 1 hN h k h k h1 kh k G (k) ( 1) r F( ), forN            and 2 2 1 2 1 2 11 1 hN h k k h k k h2 1 k kh k k G (k k ) ( 1) r F( ), forN                Substituting these values in to equation (6), we obtain : 1 2 1 2 1 2 1 1 11 1 2 1 h hk N 1h h k h k h k k 1 1 2 2 k k kk h h k k h Y(x) ( 1) ( ) r r F (h )F (h ), N k                       The next remark shows that the DTM in DDES is generalization to the DTM in ODES. Remark:- If r1 = 0 and r2 = 0, then theorem (7) reduces to theorem (4). Similarly, as in theorem (7) we may generalize the result for the n-delays as in the next theorem: Theorem (8):- The differential transform of y(x) = f1(x – r1)f2(x – r2)… fn(x – rn), provided that ri 0, i = 1, 2, …, n is: h h hk kk k N N N N 1 2 n 13n 1 2 h h ... h k1 2 nY(k) ..... ..... ( 1) .. k k k k kk 0h 0 k 0k 0h k h k k h k k h k k 1 2 1 n 1 n 2n 1 n 2 2 1 1 1 2 2 1 n 1 n 1 n 2 n n 1 hn h k h k k h1 2 1r r ....r1 2 n 1 k kn 1                                                            k k h k kn 1 n 1 n 2 n n 1r F (h )F (h )...F (h )F (h ), for Nn 1 1 2 2 n 1 n 1 n n           …(7) Illustrative Examples In this section, some liner and nonlinear differential equations with multiple delay are considered .By using DTM , we obtain an approximate and exact solutions when the original problem has an exact solution in polynomial form. Example 1:- Let us consider the following initial value problem: dy y(x 1) y(x 2) 2x 2,0 x 1 ...(8) dx         IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 with the initial condition y(0) = 0 …(9) By applying the DTM, we can get the exact solution for eq. (8) and eq. (9). Indeed, using theorems (2), (3) and (6) the differential transform for equation (8) is found as h hN N1 2h k h kh k h k1 21 2(k 1)Y(k 1) ( 1) 1 Y(h ) ( 1) 2 Y(h ) 2 (K 1) 2 (K)1 2 k kh k h k1 11 1 2                             …(10) where (k – n) is the differential transform of xn at x0 = 0 and it is easily show that: 1,k n (k n) ...(11) 0,k n,n 0,2       Considering, the differential transform of y(x) at x0 = 0, the initial conditions in equation (9) are transformed into Y(0) = 0 , respectively. Form equation(10), we obtain: Y(1) = 1, Y(k) = 0, for k  2 Then, by using the inverse transform defined by equation (4), we obtain the exact solution y(x) = x. Example 2:In this example, we consider the second order linear differential equation with multiple delay: 2 2 2 d y y(x 1) y(x 2) 2x 6x 7,0 x 1 ...(12) dx          and the following initial condition: y(0) = 0, y'(0) = 0 …(13) Using theorem (3) and (6), the differential transform for eq. (12) is found as: h hN N1 2h k h kh k h k1 21 2(k 1)(k 2)Y(k 2) ( 1) 1 Y(h ) ( 1) 2 Y(h ) 2 (K 2) 6 (K 1) 7 (K)1 2 k kh k h k1 2                                 …(14) where (k – n) is defined in equation (11) Considering, the differential transform of y(x) at x0 = 0, the initial conditions in equation (13) are transformed into: Y(0) = 0, Y(1) = 0 …(15) IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Taking N = 3 ,we obtain the following system of linear algebraic equations by using equation (14) and eq.(15) for k = 0, 1. 7Y(2) – 9Y(3) = 7 – 6Y(2) + 21Y(3) = – 6 Solving this system, we obtain: Y(2) = 1, Y(3) = 0 Similarly, we have Y(k) = 0, for k  4. Then, by using equation (4), we obtain the exact solution y(x) = x2. Example 3:- consider the linear differential equation of third order 3 x 0.3x 0.3 3 dy y(x) y( ) ,0 x 1 ...(16)e dx      and the following initial conditions y(0) = 1, y'(0) = – 1, y''(0) = 1 …(17) Using equation (3), the differential transform of x 0.3 e   at x0 = 0 are obtained to be 1 k 0.3( )( 1) k! e .then because of theorems (3) and (6) the differential transform of equation (16) is: 1 hN 1h k h k k 0.31 kh k1 1 (k 1)(k 2)(k 3)Y(k 3) Y(k) ( 1) Y(h ) ( 1) e ...(18)(0.3) 1 k!                   Considering, the differential transform of y(x) at x0 = 0 the initial conditions in equation (17) are transformed into: 1Y(0) 0,Y(1) 1,Y(2) , ...(19) 2    Taking N = 6, we obtain the following linear algebraic equations system by using equation (18) and equation (19) for k = 0, 1, 2, 3. 5.973Y(3) 0.0081Y(4) 0.00243Y(5) 0.000729Y(6) 0.995141192 0.27Y(3) 23.892Y(4) 0.0405Y(5) 0.01458Y(6) 0.950141192 0.9Y(3) 0.54Y(4) 59.73Y(5) 0.1215Y(6) 0.325070596 2Y(3) 1.2Y(4) 0.9Y(5) 119.46Y(6) 0.2                     24976468 Solving this system, we obtain: Y(3) 0.1666666589, Y(4) 0.04166662168 Y(5) 0.008333153417, Y(6) 0.001388386354       Subsisting these value of Y(K) for k = 0, 1, 2, 3 into equation (4) we obtain the following approximate solution : IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 2 3 4 5y(x) 1 x 0.5x 0.1666666589x 0.04166662168x 0.008333153417x 60.001388386354x ...         For N=8, the approximate solution is given as: 2 3 4 5y(x) 1 x 0.5x 0.1666666666x 0.04166666657x 0.0083333329417x 6 4 7 5 8 6 90.001388887387x 1.98084138 10 x 2.479261343 10 x 2.742273669 10 x 7 10 8 112.61716267 10 x 1.626927858 10 x ...                      Comparison between the numerical results for N = 6, N = 8 and the exact solution xy(x) e (see[3]) , are given in table (1). Example 4:- Consider the nonlinear DDE with multiple delays: 2 x x x 2 d y 1 y(x ).y(x ) sin .cos cos ,0 x 1 ...(20) 4 4 2dx           and the initial conditions y(0) = 1, y'(0) = 1, …(21) Using theorem (3) and (7) the differential transform of equation (20) is obtained as 1 2 1 1 1 2 1 k N N h h k k h k h k k (k 1)(k 2)Y(k 2) Y(k) ( 1)              h h h k h k k1 2 1 1 1 h h1 2 k k k1 Y( )Y( )( ) ( ) 4 4                   k k x x x k k 1 1 1d d[ (sin cos )] [ (cos )] (k) ...(22) k! k! 2dx dx      where (k – n) is defined in eq. (11), with n = 0. The initial conditions in eq. (21) are transformed in to: Y(0) = 1, Y(1) = 0 From eq. (22), we obtain: 1 1 Y(2) , Y(3) 0, Y(4) , Y(5) 0 2! 4!      1 1 Y(6) , Y(3) 0,Y(8) , Y(9) 0 6! 8! 1 1 Y(10) , Y(11) 0, Y(12) , Y(13) 0,... 10! 12!           Substituting these values in to eq. (4), we obtain the following analytical solution, 2 4 6 8 10 121 1 1 1 1 1 y(x) 1 x x x x x x ... 2! 4! 6! 8! 10! 12!         IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Which is formally the same as Maclaurin Series of cos x. In fact y(x) = cos x is the exact solution for eq. (20) and (21). Conclusions In this paper the differential transform method is developed for solving differential equations with multiple constant delays. First, some new theorems are provided and then used to solve linear and nonlinear DDES. The obtained results are found to be very accurate in comparison with the exact solution. References 1. Kurnaz ,A. and Oturanç, G. (2005)The differential transform approximation for the system of ordinary differential equations,Int. J. Computer Math. 82:709–719. 2. Zhou,J.K. (1986)Differential Transformation and its Application for Electrical Circuit, Huazhong University Press,Wuhan,China, 3. Evans ,D.J. and Raslan,K.R. (2005)The Adomian decomposition method for solving delay differential equation, Int. J.Computer Math. 82 : 49–54. 4. Arikoglu and Özkol, I. (2005) Solution of boundary value problems for integro-differential equations by using differential transform method, appl. Math. Comput 168: 1145-1158. 5. Ayaz, F. (2004) Solutions of the system of differential equations by differential transform method, Appl. Math. Comput. 147:547–567. 6. Cooke ,K.L. and Yorke,J.A. (1973) Some equations modelling growth processes and gonorrhea epidemics, M ath. Biosci.16: 75–101. IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Table (1):Comparison of numerical results x N = 6 N = 8 Exact 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 0.90483734 0.81873067 0.74081810 0.67320002 0.60653066 0.54881137 0.49658521 0.49328974 0.40656969 0.36787959 1 0.90483739 0.81873073 0.74081814 0.67320001 0.60653065 0.54881161 0.49653034 0.44932890 0.40656966 0.36787946 1 0.90483741 0.81873075 0.74081822 0.67032004 0.60653065 0.54881163 0.49658530 0.44932896 0.40656965 0.36787944 2011) 3( 24المجلد مجلة ابن الهیثم للعلوم الصرفة والتطبیقیة التباطؤ ي توسیع طریقة التحویالت التفاضلیة لحل المعادالت التفاضلیة ذ المتعدد فاضل صبحي فاضل ،غدیر جاسم محمد ابن الهیثم ، جامعة بغداد –قسم الریاضیات ، كلیة التربیة لعلوم ، جامعة النهرین قسم الریاضیات وتطبیقات الحاسوب ، كلیة ا 2010تشرین الثاني 29: استلم البحث في 2011أیلول 20 : قبل البحث في ةصالخال توسـیع ال المعالتباطؤ المتعدد بأست يلیة ذضق عددیة لحل المعادالت التفاائق تقریبیة وطر ائا طر نقدم ،في هذا البحث .خطیة الت هذه الطریقة لحل العدید من المسائل الخطیة وغیر ملعستأ .ةلطریقة التحویالت التفاضلی