IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL. 24 (3) 2011 Extend Differential Transform Methods for Solving Differential Equations with Multiple Delay Gh. J. Mohammed, F.S. Fadhel Departme nt of Mathematics, College of Education (Ibn Al-Haitham), Unive rsity of Baghdad Departme nt of Mathematics and Computer Applications, College of Science, Al-Nahrain Unive rsity Received in : 29 November 2010 Accepte d in :20 September 2011 Abstract In this p ap er, we p resent an approximate analytical and numerical solutions for the differential equations with multip le delay using the extend differential transform method (DTM ). This method is used to solve many linear and non linear p roblems. Key words: (differential transform method, solvin g differ ential equation, multiple delay). Introduction In this p ap er, we extend the differ ential 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 ℝ 2  ℝ, I  ℝ, r i > 0, i = 1,2, …, m The differential transform method was first applied in the en gineering domain in [1]. I n general, the DTM is app lied to find the solution of electric circuit p roblems [2]. The DTM is numerical method based on Taylor series exp ansion, which is constructed as an analytical solution in the form of a polynomial. The traditional hi gh order Taylor series method requires sy mbolic comp utation. However, the DTM obtains a p oly nomial series solution by means of an iterative p rocedure [3]. Recently the app lication of differential transform method is successfully extended to obtain app roximate solutions to linear and non linear functional equations. Delay differential equations are observed in many fields of science and technology , such as biology engineering and p hy sics. M any dy namic p op ulation first order nonlinear scalar equ ation of the form y '(x) = g(y (x)) – g(y (x – L)) …(2) may be used as a model for certain p op ulation growt h if ind ividuals have a constant life sp an L, where y (x) is the pop ulation size at time t and g(y ) is the birth rate. Recently , various methods such as, monotone iterative technique, a do main decomposition method and the sp line functions method have been considered for app roximate solutions of DDE [4]. IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL. 24 (3) 2011 Hence, due to p ractical reasons and the p ap ers mentioned above, we have been motivated to deal with DDE and develop DTM for both linear and nonlinear delay differential equations. Accordin g to the best of our knowledge, DT M has not been st udied for DDE till now. With this technique, it is p ossible to obtain high ly accurate numer ical solution, analytical solution and as well as exact solutions. The aim this p ap er is to extend the method of differential transformation for solving differential equations with multiple delay of difference ty p es 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 st ate. Throughout this p ap er, we use the small and cap ital letters to represent the original and transformed functions, resp ectively. The inverse of the differential transform Y(k) is defin ed by k 0 k 0 y(x ) Y(k)(x x )     …(4) The automatic comp utation of DTM might be done. In this case, the following st eps should be taken into consideration successively : i) The differential transform of each term in the DDE is computed ; ii) The recurrence equation is obt ained ; iii)Y(0), Y(1), Y(2), Y(3),… are calculated by the recurrence equation and given initial condition ; iv)Finally , these values are subst itut ed back into eq. (4). Preliminaries of the DTM The following theorems give the p rop erties of the DTM which can be easily derived from equations (3) and (4), for their p roofs and more details (see [5], [6]). The orem (1):- If y (x) = f(x)  g( x), then Y(k) = F( x)  G(x). The orem (2):- If y (x) = c f(x), then Y(k) = c F( x) The orem (3):- If n n d f (x) (k n)! y(x) , then Y(k) [ ] F(k n ) k!dx     , k ℕ. The orem (4):- If 1 k 1 1 k 0 y( x) f (x ).g (x ), thenY(k ) F(k )G(k k )     , k ℕ. M oreover, we need the following theorem (see [4], [5]) : The orem (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. S CI. VOL. 24 (3) 2011 Main Result The following theorems are p roved 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 1 kh Y (k) ( 1) r F(h ), N     Proof: - Using the defin ition of the inverse DT M 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     The orem (7):- The differential transform of y (x) = f1( x – r1)f2( x – r2), p rovided 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), resp ectively. Using theorem (4), we hav e the differential transform of y (x) as: IBN AL- HAITHAM J. FOR PURE & APPL. S CI. 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                  Subst itut ing 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 t he DTM in DDES is generalization to the DTM in ODES. Remark:- If r1 = 0 and r2 = 0, then theorem (7) reduces to theorem (4). Similar ly, as in theorem (7) we may generalize the r esult for t he n-delays as in the next theorem: The orem (8):- The differential transform of y (x) = f1( x – r1)f2( x – r2)… fn( x – rn), p rovided 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 0 h 0 k 0 k 0 h 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 h n h k h k k h1 2 1r r ....r1 2 n 1 k k n 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 differ ential equations with multiple delay are considered .By using DTM , we obtain an approximate and exact solutions when the original p roblem has an exact solution in p olynomial form. Exam ple 1:- Let us consider the following initial value p roblem: dy y(x 1) y(x 2) 2x 2,0 x 1 ...(8) dx         IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL. 24 (3) 2011 with the initial condition y (0) = 0 …(9) By app lying the DTM, we can get t he exact solution for eq. (8) and eq. (9). Indeed, usin g theorems (2), (3) and (6) the differ ential 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 x n at x0 = 0 and it is easily show that: 1,k n (k n) ...(11) 0,k n,n 0,2       Considerin g, the differential transform of y (x) at x0 = 0, the initial conditions in equation (9) are transformed into Y(0) = 0 , resp ectively. 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. Exam ple 2:In this examp le, 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 followin g initial condition: y (0) = 0, y '(0) = 0 …(13) Using theorem (3) and (6), the diff erential 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. S CI. VOL. 24 (3) 2011 Taking N = 3 ,we obtain the followin g sy st em of linear algebraic equations by usin g equati on (14) and eq.(15) for k = 0, 1. 7Y(2) – 9Y(3) = 7 – 6Y(2) + 21Y(3) = – 6 Solvin g this sy stem, we obtain: Y(2) = 1, Y(3) = 0 Similar ly, we have Y(k) = 0, for k  4. Then, by using equation (4), we obtain the exact solution y (x) = x 2 . Exam ple 3:- consider the linear diff erential equation of third order 3 x 0.3x 0.3 3 dy y(x) y( ) ,0 x 1 ...(16)e dx      and the followin g initial conditions y (0) = 1, y '(0) = – 1, y ''(0) = 1 …(17) Using equation (3), the differential transform of x 0.3e   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!                   Considerin g, the differential transform of y(x) at x0 = 0 the in itial conditions in equ ation (17) are transformed into: 1Y(0) 0,Y(1) 1,Y(2) , ...(19) 2    Taking N = 6, we obtain the following lin ear algebraic equations sy stem by using equati on (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 Solvin g this sy stem, we obtain: Y(3) 0.1666666589, Y(4) 0.04166662168 Y(5) 0.008333153417, Y(6) 0.001388386354       Subsist ing these value of Y(K) for k = 0, 1, 2, 3 into equation (4) we obtain the followin g app roximate solution : IBN AL- HAITHAM J. FOR PURE & APPL. S CI. 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 app roximate 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 r esults for N = 6, N = 8 and the exact solution xy( x) e (see[3]) , are given in table (1). Exam ple 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 cond itions y (0) = 1, y '(0) = 1, …(21) Using theorem (3) and (7) the diff erential 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!           Subst itut ing these values in to eq. (4), we obtain the following analytical solution, 2 4 6 8 10 121 1 1 1 1 1y(x) 1 x x x x x x ... 2! 4! 6! 8! 10! 12!         IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL. 24 (3) 2011 Which is formally the same as M aclaurin Series of cos x. In fact y (x) = cos x is the exact solution for eq. (20) and (21). Conclusions In this p ap er the differential transform method is developed for solving d ifferential equations with multiple constant delay s. First , some new t heorems are p rovided and then used to solve linear and nonlinear DDES. The obtained results are found to be very accurate in comp arison with t he exact solution. Re ferences 1. Kurnaz ,A. and Ot uranç, G. (2005)T he differential transform app roximation for the sy st em of ordinary differential equations,Int . J. Comp uter M ath. 82:709–719. 2. Zhou,J.K. (1986)Differential Transformation and its App lication for Electrical C ircuit, 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 M ath. 82 : 49–54. 4. Arikoglu and Öz kol, I. (2005) Solution of boundary value p roblems for integro-d ifferential equations by using differential transform method, appl. M ath. Comp ut 168: 1145-1158. 5. Ay az, F. (2004) Solutions of the sy stem of differential equations by differential transform method, App l. M ath. Comput. 147:547–567. 6. Cooke ,K.L. and Yorke,J.A. (1973) So me equations modellin g growt h p rocesses and gonorrhea ep idemics, M ath. Biosci.16: 75–101. IBN AL- HAITHAM J. FOR PURE & APPL. S CI. 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