IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3) 2010 Finite Difference Method for Solving Fractional Hyperbolic Partial Differential Equations G. J. Mohammed Department of Mathematics, College of Education -Ibn Al- Haitham, Baghdad University. Abstract In this paper, the finite difference method is used to solve fractional hyperbolic partial differential equations, by modifying the associated explicit and implicit difference methods used to solve fractional partial differential equation. A comparison with the exact solution is presented and the results are given in tabulated form in order to give a good comparison with the exact solution. Introduction An important type of differential equations which is called fractional differential equations in which the differintegration is of non-integer order [1]. Real life problems with fractional differential equations are of great importance, since fractional differential equations accumulate the whole information of the function in a weighted form. This has many applications in physics, chemistry, engineering ,etc. For that reason, we need a method for solving such equations, effectively, easy use and applied for different p roblems[2]. Consider the fractional order partial differential equation [3][4]: 2 2 u(x, t) t   c(x, t) q q u(x, t) x   +s(x, t), L  x  R, 0  t  T .............. (1) together with the initial and zero Dirichlet boundary conditions : tu(x,0) f (x), u (x,0)=h(x) , L x R u(L,t) 0, u(R,t) 0 for 0 t T           .………. (2) where q q u(x, t) x   denote the left-hand partial fractional derivative of order q of the function u with respect to x and 1 < q  2 . The left-handed shifted and the right-handed shifted Grünwald estimate to the left- handed and right-handed derivatives, are given by [1][5][6] : q q d f (x) d x  q 1 ( x) n k 0  gkf(x  (k  1)x) IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3) 2010 q q d f (x) d x  q 1 ( x) n k 0  gkf(x + (k  1) x) where n is the number of subdivisions of the interval [ L, R ] and q is a fractional number. Therefore: q i j q u(x , t ) x    q 1 ( x) i 1 k 0    gku(xi  (k  1)x, tj)  q 1 ( x) i 1 k 0    gkuik+1, ……………..(3) and q i j q u(x , t ) x    q 1 ( x)    1in 0k gku(xi + (k  1) x, tj)  q 1 ( x)    1in 0k gkui+k1,j ........................ (4) where g0 1 and gk  (1)k q(q 1)...(q k 1) k!    , k  1,2,… The Explicit Finite Difference Method for Solving Fractional Hyperbolic Partial Differential Equations The explicit finite difference method is improved to solve the initial-boundary value problem (1)-(2). To do this, we substitute t  tj, in eq. (1) and replace the partial derivative 2 2 u t   with its central difference approximation to get : q i, j 1 i, j i, j 1 i, j i, j i , j2 q u 2u u u c s ( t) x         ...................... (5) where tjjt, j0,1,…,m and m is the number of subdivisions of the interval [0, T], t R . Next, substitute equation (3) in equation (5) to obtain: i 1 i, j 1 i,j i, j 1 i, j k i k 1, j i, j2 q k 0 u 2u u c g u s , i 1,2,...,n 1; j 0,1, ...,m 1 ( t) ( x)                  …(6) On the other hand, the initial and boundary conditions given by eq.(2) becomes : i i,0 i i 0,j j n,j u(x ,0) u u(x ,0) f (x ), for i 0,1,...,n t u u(L,t ) 0, u u(R,t) 0 for j 0,1,...,m           IHJPAS IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3) 2010 and by using the central difference approximation to the initial derivative conditions ,one can get : i,1 i, 1 i 1 (u u ) h 2 t    , i0,1,…,n where i ih h(x ) for i0, 1,…, n. Hence : i,1 i, 1 iu u 2 th   , i0,1,…, n Moreover, equation (6) becomes: 2 i 1 i, j 2 i, j 1 i, j i, j 1 k i k 1,j i, jq k 0 ( t) c u 2u u g u s ( t) ( x)               ………….….(7) where i1,2,…,n1, j=0,1,…,m1. Therefore: 2 i 1 i,0 2 i,1 i,0 i, 1 k i k 1,0 i,0q k 0 ( t) c u 2u u g u s ( t) ( x)              ……..………(8) By substituting i, 1 i,1 iu u 2 th    back into eq.(8) one can show that 1,iu can be calculated from the following equation: 2 2i 1 i,0 i,1 i k i k 1 i,0 iq k 0 ( t) c ( t) u f g f s tg 22( x)              , i=0,1,……..n −1 where i0, 1,….. n1. By evaluating the above equation for each i0,1,…,n1, one can get the values of i,1u , i 1,2,..., n 1  . Then by evaluating equation (7) at each i1,2,…,n1 and j2,3,…,m1 one can get the numerical solution of eq.(1). Then the resulting equation can be explicitly solved to give: i 1 i, j 1 i, j i, j 1 w i w 1, j w 0 u 2u u r g u           …………..……..(9) Where r = 2 qk h .The resulting difference equation is stable since we let g0=1 and gw = wq(q-1)k(q-w-1) (-1) w , w = 1,2,….. 1 ≤ q ≤ 2 , i≠1 ,hence gi ≥ 0 for all value of i. Therefore: i 1 w w 0 g     g1  (q)  q ………………(10) The difference between the analytical and numerical solutions of the difference equation remains bounded as j increases. IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (3) 2010 Let the error Ei,j=u(hi, kj) ui,j then the stability condition under which the finite difference eq. (9) is stable, to find the stability conditions under which the error Ei,j is bounded . Smith [7] shows that the error Ei,j can be written in the form : Ei,j= ih je  , where s= ke γ = 1 ………………….(11) Where α is a complex constant, one can substitute eqs .(10), (11) into (9), to get: 1 h(1 w)2 rqe 0        Assuming that, h(1 w)   ,then it is easily known that the equation for R is: 2 (2 rqe ) 1 0      Let A 2 rqe  , where e 1  Hence the values of  are: 2 1 A A 4 2     and 2 2 A A 4 2     From eq.(11), the error will not grow with time if 1  , for all real β ……………… (12) Equation (12) is called the Von-Neumann’s condition for stability .Thus we will use eq. (12) to find the stability condition of the finite difference equation. For stability; as r, q and β are real and when giving stability while 2 gives instability. When −1 ≤ A ≤ 1 , we get 1 2and  are complex number, hence: 2 1 A y 4 A 2     and 2 1 A y 4 A 2     Then using Von-Neumann’s condition (12) to prove that eq.(9) is stable For 1 A 1   , the only useful inequality is A 1  , hence 2 rqe 1  , where e 1  .Therefore; 1 r q   , where 1 ≤ q ≤ 2. Hence, 1 r 2  , which is the stability condition. IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3) 2010 The Implicit Finite Difference Method for Solving Fractional Hyperbolic Partial Differential Equations IHJPAS Now, we can improve and introduce similar approach for the implicit finite difference method to solved the one–sided fractional hyperbolic partial differential equations. The resulting discretization takes the following form: i 1 i, j 1 i, j i, j 1 i, j w i w 1, j2 q w 0 u 2u u c g u k h           Where i 1,2,...,n 1; j 0,1,...,m 1    . Then to get i 1 i, j 1 i, j i, j 1 w i w 1, j 1 w 0 u 2u u r g u            …………………..(13) In the above equation and under the same conditions of eq.(9) and substituting eqs. (10) and (11) into eq. (13), one can get: 12    < rqe , where h(1 w)   . Hence the values of  are : 1 2 1 1 (1 A) A     and 1 2 2 1 (1 A) A     where A 1 rqe  . To discuss the stability of eq. (13); by using Von-Neumann’s condition (12). When A < 1 , we get real the roots, 1 also, which gives instability while 2 gives stability for this problem . Now, when 1 A 1   , we get complex number, which are 1 2 1 1 (1 A) A      and 1 2 1 1 (1 A) A      .and the condition of the stability leads to r ≥ 1 when 1 ≤ q ≤ 2 and e 1  Therefore; the finite difference eq. (13) is instable for r ≤ 2 q , 1 ≤ q ≤ 2. Illustrative Example To illustrate the methods of the solution, an illustrative numerical example is considered: Example:- IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (3) 2010 Consider the fractional order partial differential equation : IHJPAS 1t0 ,2x0 ,xt546.2tx546.2x2x4 x u x )5.0( 1 t u 22232 5.1 5.1 2 1 2 2        Together with initial and zero Dirichlet boundary conditions: u(x, 0)=0, u(x,0) 0 t    for 0 x 2  .. u(0, t)0, u(1, t)0 for 0 t 1  . This example has the exact solution as: 2 2u(x,t) x (x 2)t ,  [8]. which is considered for the comparison purpose. Here; we use the explicit and implicit finite difference methods to solve this example numerically. To do this, first we divide the x-interval into 2 subintervals such that ix i, i0,1,2 and the t-interval into 2 subintervals such that j j t , 2  j0,1,2. Thus, the initial and zero Dirichlet boundary conditions become: iu(x ,0) 0 for i0,1,2. iu(x ,0) 0 t    for i0,1,2. ju(0,t ) 0 for j0,1,2. ju(1, t ) 0 for j0,1,2. By using the central difference approximation to the initial derivative condition one can get: i,1 i, 1 1 (u u ) 0 2 t    ; hence i,1 i, 1u u  for i0,1,2. Moreover, equation (7) becomes: 1 i 1 2 i, j 1 i, j i, j 1 i k i k 1,j k 0 u 2u u 0.25x g u           2 3 2 2 2 i i i j i j0 .2 5 ( 4 x 2 x 2 .5 4 6 x t 2 .5 4 6 x t )     where i1 and j0, 1. Therefore 1 i 1 2 i,1 i,0 i, 1 i k i k 1,0 k 0 u 2u u 0.25x g u          2 3 2 2 2 i i i 0 00 .25( 4 x 2 x 2.54 6x t 2.546 x t )    IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (3) 2010 IHJPAS By substituting i, 1 i,1u u  in the above equation one can show that i,1u can be calculated from the equation 1 i 1 2 i,1 i,0 i k i k 1,0 k 0 u u 0.125x g u        2 3 2 2 2 i i i 0 00.125( 4x 2x 2.546x t 2.546xt )    Thus 1,1u  2 3 1 10.125( 4x 2x ) 0.25   . Then 1 2 2 1,2 1,1 1,0 1 k 2 k,1 k 0 u 2u u 0.25x g u       2 3 2 2 2 1 1 1 1 1 10 .2 5 ( 4 x 2 x 2 .5 4 6 x t 2 .5 4 6 x t )    0.947.  These values are tabulated down with the comparison with the exact solution. See table (1) Second, we divide the x-interval into 10 and the t-interval into 10 subinterval. Thus, the initial and zero Dirichlet boundary conditions become: iu(x ,0) 0 , iu(x ,0) 0 t    for i  0,1,…,10. ju(0, t ) 0 , ju(1, t ) 0 , for j  0,1,…,10. The results are presented in table (2). Conclusions 1. The finite difference method gave the numerical solution of the fractional differential equations and it depended on the Grunwald estimate for the fractional derivatives . 2. The stability results in the finite partial differential equation case as generalization and unification for the corresponding result in the classical hyperbolic partial differential equation. 3. Similar to this work, the explicit finite difference method can be also used to solve the initial-boundary value problems of the two-sided fractional hyperbolic partial differential equations given by, u(x, t) t    c(x, t) q q u(x, t) x   + d(x, t) q q u(x, t) x   + s(x, t) together with the initial and zero Dirichlet boundary conditions: u(x,0) f (x) ,u(L, t) 0,u(R,t) 0 for L x R     where L  x  R, 0  t  T, q q u(x, t) x   and q q u(x, t) x   denote the left-handed and the right- handed partial fractional derivatives of order q of the function u with respect to x and 1 < q  2. IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VO L.23 (3) 2010 IHJPAS In this case equation (4) becomes i 1 n i 1 i, j 1 i, j 1 i, j 1 i, j i, j 1 w i w 1, j w i w 1, j i, jq q w 0 w 0 c d u 2u u g u g u s ( x) ( x)                       4. In a similar manner, the implicit finite difference method can be also used to solve the initial-boundary value problems of the two-sided fractional hyperbolic partial differential equations given by equations:- u(x, t) t    c(x, t) q q u(x, t) x   + d(x, t) q q u(x, t) x   + s(x, t) In this case eq.(4)becomes: i 1 n i 1 i, j 1 i,j 1 i, j 1 i, j i, j 1 w i w 1,j 1 w i w 1,j 1 i, j 1q q w 0 w 0 c d u 2u u g u g u s ( x) ( x)                          where i1,2,…, n1; j0,1,…,m1. References 1. Nishimoto, K. (1983) , "Fractional Calculus: Integrations and Differentiations of Arbitrary Order", Descartes Press Company Koriyama Japan. 2. Al-Rahhal, D. (2005),"Numerical Solution for Fractional Integro-Differential Equations", Ph.D. Thesis, Colle ge of Science, University of Baghdad. 3. Diethelm, K. (1999) "anlysis of fractional differential Equations",Department of Mathematics , University of Manchester England. 4. Meerschaert, M. and Tadjeran, C.(2006) ”finite difference approximation for two –sided space fractional partial differential equations”, Applied Numerical Mathematics , 56: 80-90. 5. Ames, W. F. (1992) "Numerical Methods for Partial Differential Equations", 3 rd Edition, Academic Press, Inc. 6. Samko, S.; Kilbas, A. and Maricllev,O.(1993)Theory and Applications. Gordon and Breach, London. 7. Smith, G. (1978), "Numerical Solution of Partial Differential Equations: Finite Difference Methods", Oxford University Press 8. Sidiqi, L. (2007), ”some finite difference method for solving fractional differential equations”, M.Sc. Thesis, college of Science , Al-Nahrain University . Table (1) Represents the numerical and the exact solutions for nm2 of example. xi tj Numerical solution ui,j Exact solution u(xi,tj) Explicit method Implicit method 1 0.5 0.25 0.25 0.25 1 1 0.9472 0.9684 1 IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (3) 2010 IHJPAS Table (2) Represents the numerical and the exact solutions for nm10 of example. xi tj Numerical solution ui,j Exact solution u(xi,tj) Explicit method Implicit method 1 0.5 0.25 0.25 0.25 1 1 0.994 0.995 1 0.8 0.2 0.0398 0.0389 0.031 0.2 0.7 0.0326 0.0393 0.035 0.4 0.9 0.2129 0.2026 0.207 0.6 1 0. 5096 0. 5063 0. 504 1.2 0.7 0.5655 0.5641 0.564 1.4 0.3 0.1014 0.1042 0.106 1.6 0.8 0.6568 0.6551 0.655 1.8 1 0.6466 0.6467 0.648 IHJPAS 2010) 3( 23مجلة ابن الھیثم للعلوم الصرفة والتطبیقیة المجلد ق الفروقات المنتھیة لحل المعادالت التفاضلیة ائطر الجزئیة الكسریة من نوع القطع الزائد غدیر جاسم محمد .جامعة بغداد، ابن الھیثم -قسم الریاضیات، كلیة التربیة لخالصةا لح�ل ) finite difference method( في ھذا البحث ، استخدمت طریق�ة الفروق�ات المنتھی�ة Hyperbolic partial ( الكس�ریة م�ن ن�وع القط�ع الزائ�د الرت�ب يت التفاض�لیة الجزئی�ة ذالمع�ادال differational equation ( ھیة الصریحة والض�منیة ،بتطویر طریقة الفروقات المنت)Explicit and Implicit method . ( وأعطیت النت�ائج ف�ي ج�داول للحص�ول عل�ى أفض�ل مقارن�ة حیحالنتائج العددیة مع الحل الص ورنتق . حیحمع الحل الص IHJPAS