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 De partment of Mathe matics, College of Education -Ibn Al- Haitham, Bag hdad University. Abstract In this p aper, the finite difference method is used to solve fractional hy p erbolic p artial differential equations, by modify ing the associated exp licit and imp licit difference methods used to solve fractional p artial differential equation. A comp arison with the exact solution is p resented and the results are given in tabulated form in order to give a good comp arison with the exact solution. Introduction An imp ortant ty p e of differential equations which is called fractional differential equations in which the differintegration is of non-integer order [1]. Real life p roblems with fractional differential equations are of great imp ortance, since fractional differential equations accumulate the whole information of the function in a weighted form. This has many app lications in p hy sics, chemist ry , engineering ,etc. For that reason, we need a method for solving such equations, effectively , easy use and app lied for different p roblems[2]. Consider the fractional order p artial 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 resp ect 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 exp licit finite difference method is imp roved to solve the initial-boundary value p roblem (1)-(2). To do this, we substitut e t  tj , in eq. (1) and replace the p artial derivative 2 2 u t   with its central difference ap p roximation 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 t he number of subdivisions of t he interval [0, T], t R . Next, substitut e 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 t he 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 app roximation 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 M oreover, 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 substitut ing i, 1 i,1 iu u 2 th    back into eq.(8) one can show t hat 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 t he numerical solution of eq.(1). Then the resulting equation can be exp licitly 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 q k 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. S CI. VOL.23 (3) 2010 Let the error Ei,j=u(hi, kj ) ui,j then the st ability condition under which the finite difference eq. (9) is st able, to find the stability conditions under which the error Ei,j is bounded . Smith [7] shows t hat t he error Ei,j can be writt en in the form : Ei,j= ih j e   , where s= k e  γ = 1 ………………….(11) Where α is a comp lex constant, one can substitut e eqs .(10), (11) into (9), to get: 1 h(1 w ) 2 rqe 0         Assuming that, h(1 w )   ,then it is easily known t hat t he 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 t ime if 1    , for all real β ……………… (12) Equation (12) is called the Von-Neumann’s condition for st ability .Thus we will use eq. (12) to find the st ability condition of the finite difference equation. For stability ; as r, q and β are real and when giving st ability while 2 gives inst ability . When −1 ≤ A ≤ 1 , we get 1 2and  are comp lex 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 p rove 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 t he 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 imp rove and introduce similar app roach for the imp licit finite difference method to solved the one–sided fractional hy p erbolic p artial 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 subst itut ing eqs. (10) and (11) into eq. (13), one can get: 1 2    < 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 st ability 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 st ability for this p roblem . 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 st ability 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 illust rative numerical examp le is considered: Example:- IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.23 (3) 2010 Consider the fractional order p artial 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 examp le has the exact solution as: 2 2 u( x, t) x (x 2)t ,  [8]. which is considered for the comp arison p urp ose. Here; we use the exp licit and imp licit finite difference methods to solve this examp le 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, t he 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 app roximation 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. M oreover, 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 . 2 5 ( 4 x 2 x 2 . 5 4 6 x t 2 . 5 4 6 x t )    IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.23 (3) 2010 IHJPAS By substitut ing 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 t he 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    fo r i  0,1,…,10. ju(0, t ) 0 , ju(1, t ) 0 , f or 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 est imate for the fractional derivatives . 2. The st ability results in the finite p artial differential equation case as generalization and unification for the corresp onding result in the classical hy p erbolic p artial differential equation. 3. Similar to this work, the exp licit finite difference method can be also used to solve the initial-boundary value p roblems of the two-sided fractional hy p erbolic p artial 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 p artial fractional derivatives of order q of the function u with resp ect 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 imp licit finite difference method can be also used to solve the initial-boundary value p roblems of the two-sided fractional hy p erbolic p artial 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. Re ferences 1. Nishimoto, K. (1983) , "Fractional Calculus: Integrations and Differentiations of Arbitrary Order", Descartes Press Comp any Koriy ama 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) "anly sis of fractional differential Equations",Dep artment of M athematics , University of M anchester England. 4. M eerschaert, M . and Tadjeran, C.(2006) ”finite difference app roximation for two –sided sp ace fractional p artial differential equations”, Ap p lied Numerical M athematics , 56: 80-90. 5. Ames, W. F. (1992) "Numerical M ethods for Partial Differential Equations", 3 rd Edition, Academic Press, Inc. 6. Samko, S.; Kilbas, A. and M aricllev,O.(1993)T heory and Ap p lications. Gordon and Breach, London. 7. Smith, G. (1978), "Numerical Solution of Partial Differential Equations: Finite Difference M ethods", 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 soluti ons for nm2 of example. xi tj Numerical sol uti on ui,j Exact soluti on u(xi,tj) Explicit me thod Implicit me thod 1 0.5 0.25 0.25 0.25 1 1 0.9472 0.9684 1 IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.23 (3) 2010 IHJPAS Table (2) Represents the numerical and the exact soluti ons for nm10 of example. xi tj Numerical sol uti on ui,j Exact soluti on u(xi,tj) Explicit me thod Implicit me thod 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