@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Numerical Solutions Of The Nonlocal Problems For The Diffusion Partial Differential Equations Reem. M. Kubba Dept. of Mathematics/ College of Education for Pure Science( Ibn-Al-Haitham)/ University of Baghdad Received in:3 October 2012 , Accepted in:16 January 2013 Abstract In this work, we use the explicit and the implicit finite-difference methods to solve the nonlocal problem that consists of the diffusion equations together with nonlocal conditions. The nonlocal conditions for these partial differential equations are approximated by using the composite trapezoidal rule, the composite Simpson's 1/3 and 3/8 rules. Also, some numerical examples are presented to show the efficiency of these methods. Keywords: finite difference method, implicit finite difference method, explicit finite difference method, nonlocal condition, diffusion equation. 276 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Introduction An integral over the spatial domain is appeared in some part or in the whole boundary[1]. Such boundary value problems are known as nonlocal problems where nonlocal conditions appear when values of the function on the boundary are connected to values inside the domain, [2]. Many researchers studied the nonlocal problems say [3] studied the existence of the solutions for fractional ordinary differential equations with nonlocal initial condition. [4] used the finite-difference method for solving the diffusion equations together with homogeneous boundary conditions and initial condition of non-integral nonlocal type, [5] used the homotopy perturbation method to solve special types of nonlinear parabolic differential equations with the nonlocal boundary conditions. In this work we use the explicit and the implicit finite difference method to solve the following diffusion partial differential equation: 𝜕𝑢(𝑥, 𝑡) 𝜕𝑡 − 𝜕2𝑢(𝑥, 𝑡) 𝜕𝑥2 = 𝑓(𝑥, 𝑡) , 0 ≤ 𝑥 ≤ ℓ , 0 ≤ 𝑡 ≤ 𝑇 (1. 𝑎) together with the initial condition: u (x,0) = r(x) , 0 ≤ 𝑥 ≤ ℓ (1.b) and the Neumann condition : 𝜕𝑢(𝑥,𝑡) 𝜕𝑥 |𝑥=0 = 𝛼(𝑡) , 0 ≤ 𝑡 ≤ 𝑇 (1.c) and the nonlocal condition: ∫ 𝑢(𝑥, 𝑡)𝑑𝑥 = 𝛽(𝑡) ℓ 0 ,0 ≤ 𝑡 ≤ 𝑇 (1.d) where f, 𝛼, 𝛽, 𝑟 are known functions that must satisfy the compatibility conditions: ∫ 𝑟(𝑥)𝑑𝑥 = 𝛽(0) ℓ 0 , r (0) (0)′ = α The Explicit Finite Difference Method for Solving the Nonlocal Problem (1) It is well known that the explicit finite difference method is a numerical method that is used to solve the initial-boundary value problems for partial differential equations as well as, ordinary differential equations. Here we use it to solve the nonlocal problem given by equations (1). To do this, we divide the x-interval [0,ℓ] and the t-interval [0,T] into m and n subintervals i i 1[x , x ], i=0,1,...,m 1+ − and j j 1[t , t ], j=0,1,...,n 1+ − respectively such that ix ih,i=0,1,...,m 1= − ; jt jk, j=0,1,...,n 1= − ; where ℎ = ℓ 𝑚 and T k n = . So the problem here is to find the solution of the nonlocal problem given by equations (1) at the mesh points i j(x , t ), i=0,1,...,m, j=0,1,...,n. Let i, ju denote the numerical solution of the nonlocal problem given by equations (1) at each mesh point i j(x , t ), i=0,1,...,m, j=0,1,...,n. By 277 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 replacing the partial derivatives u t ∂ ∂ and 2 2 u x ∂ ∂ by the forward difference approximation and central difference approximation respectively, equation (1.a) becomes: i, j 1 i 1, j i, j i 1, j i ju ru (1 2r)u ru kf (x , t ), i 1, 2,..., m 1, j=0,1,...,n 1+ + −= + − + + = − − (2) where 2 k r h = . From the initial condition given by equation (1.b) one can have: 𝑢𝑖,0 = 𝑟(𝑥𝑖), 𝑖 = 0,1, … , 𝑚 (3) By evaluating equation (2) at each i 1, 2,..., m 1,= − j=0 and by using equation (3) one can get the numerical solutions i,1u , i=1,2,...,m 1− . Therefore it remains to find the numerical solutions 0,1 m,1u , u . To do this, we use the forward finite difference approximation for the Neumann condition given by equation (1.e) to get: 𝑢0,𝑗 = 𝑢1,𝑗 − ℎ𝛼�𝑡𝑗� , j=1,2,…,n (4) By evaluating equation (4) at j=1 one can get the value of 𝑢0,1.Then we approximate the integral that appeared in the right hand side of the nonlocal condition given by equation (1.d) with any suitable method of quadrature methods, say the composite trapezoid rule to get: 𝑢𝑚,𝑗 = 2 ℎ 𝛽�𝑡𝑗� − 𝑢0,𝑗 − 2 � 𝑢𝑖,𝑗 , 𝑗 = 1,2, … , 𝑛 (5) 𝑚−1 𝑖=1 Therefore by evaluating equation (5) at j=1 one can get the value of 𝑢𝑚,1.Hence the numerical solutions of the nonlocal problem given by equation (1),𝑢𝑖,1, 𝑖 = 0,1, … , 𝑚 are obtained. Next, we evaluate equation (2) at each i=1,2,…,m-1, j=1 and by using the numerical solutions 𝑢𝑖,1, i=0,1,…,m one can get 𝑢𝑖,2 , i=1,2,…,m-1 and by following the same previous steps one can get the value of 𝑢0,2, 𝑢𝑚,2 .By continuing in this manner, one can get the numerical solutions 𝑢𝑖,𝑗 , i=0,1,…,m , j=0,1,…,n of the nonlocal problem given by equations (1). The Implicit Finite Difference Method for Solving the Nonlocal Problem (1) It is well known that the implicit finite difference method is a numerical method that is used to solve the initial-boundary value problems for partial differential equations. Here we shall use this method to solve the nonlocal problem given by equations (1). To do this, we replace the partial derivatives u t ∂ ∂ and 2 2 u x ∂ ∂ by the backward difference approximation and central difference approximation respectively in equation (1.a) to get: 278 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 (1 + 𝑟)𝑢𝑖,𝑗+1 − 𝑟 2 𝑢𝑖+1,𝑗+1 − 𝑟 2 𝑢𝑖−1,𝑗+ = 𝑟 2 𝑢𝑖+1,𝑗 − 𝑟𝑢𝑖,𝑗 + 𝑟 2 𝑢𝑖−1,𝑗 + 𝑘𝑓�𝑥𝑖, 𝑡𝑗� (6) Where i=1,2,…,m-1 ,j=0,1,…,n-1 From the initial condition given by equation (1.b) we get: 𝑢𝑖,0 = 𝑟(𝑥𝑖), 𝑖 = 0,1, … , 𝑚 (7) By evaluating equation (6 ) at each i=1,2,…,m-1, j=0 one can get the following equations: (1 + 𝑟)𝑢𝑖,1 − 𝑟 2 𝑢𝑖+1,1 − 𝑟 2 𝑢𝑖−1,1 = 𝑟 2 𝑢𝑖+1,0 − 𝑟𝑢𝑖,𝑗 + 𝑟 2 𝑢𝑖−1,0 + 𝑘𝑓(𝑥𝑖, 𝑡0) , i=1,2,…,m-1 From equations (4)-(5) one can have: 𝑢1,1 − 𝑢0,1 = ℎ𝛼(𝑡1) 𝑢0,1 + 2∑ − = 1 1 1. m i iu + 𝑢𝑚,1 = 2 ℎ 𝛽(𝑡1) So the above m+1 equations with m+1 unknowns that can be written in matrix form as: Au=g (8) where A= ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ −1 1 0 0 0 0 0 0 − 𝑟 2 1 + 𝑟 − 𝑟 2 0 0 0 0 0 0 − 𝑟 2 1 + 𝑟 − 𝑟 2 0 0 0 0 0 0 − 𝑟 2 1 + 𝑟 − 𝑟 2 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 0 − 𝑟 2 1 + 𝑟 − 𝑟 2 1 2 2 2 2 2 2 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ , u= ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 𝑢0,1 𝑢1,1 𝑢2,1 𝑢3,1 . . . 𝑢𝑚−1,1 𝑢𝑚,1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ and g= ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ ℎ𝛼(𝑡1) 𝑟 2 𝑢2,0 − 𝑟𝑢1,0 + 𝑟 2 𝑢0,0 + 𝑘𝑓(𝑥1, 𝑡0) 𝑟 2 𝑢3,0 − 𝑟𝑢2,0 + 𝑟 2 𝑢1,0 + 𝑘𝑓(𝑥2, 𝑡0) 𝑟 2 𝑢3,0 − 𝑟𝑢2,0 + 𝑟 2 𝑢1,0 + 𝑘𝑓(𝑥3, 𝑡0) . . . 𝑟 2 𝑢𝑚,0 − 𝑟𝑢𝑚−1,0 + 𝑟 2 𝑢𝑚−2,0 + 𝑘𝑓(𝑥𝑚−1, 𝑡𝑜) 2 ℎ 𝛽(𝑡1) ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ . The above system can be solved to get the values of 𝑢𝑖,1 , 𝑖 = 0,1, … , 𝑚. By repeating the same previous steps one can the values of 𝑢𝑖,2 , 𝑖 = 0,1, … , 𝑚. By continuing in this manner, 279 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 one can get the numerical solutions 𝑢𝑖,𝑗 , 𝑖 = 0,1, … , 𝑚 , 𝑗 = 0,1, … , 𝑛 of the nonlocal problem given by equations (1). Remarks (1) If the number of x-subintervals m is an even integer, one can approximate the integral that appeared in the right hand side of the nonlocal condition given by equation (1.d) with the composite Simpson's 1/3 rule to get: ℎ 3 �𝑢0,𝑗 + 4 ∑ 𝑥2𝑖−1𝑢2𝑖−1,𝑗 + 2 ∑ 𝑥2𝑖𝑢2𝑖,𝑗 + 𝑢𝑚,𝑗 𝑚 2 −1 𝑖=1 𝑚 2 𝑖=1 � = 𝛽�𝑡𝑗�, 𝑗 = 1,2, … , 𝑛 (2) If the number of x-subintervals m is a multiple of three, one can approximate the integral that appeared in the right hand side of the nonlocal condition given by equation (1.d) with the composite Simpson's 3/8 rule to get: 3ℎ 8 �𝑢0,𝑗 + 3 �∑ 𝑥3𝑖−1𝑢3𝑖−1,𝑗 + ∑ 𝑥3𝑖−2𝑢3𝑖−2,𝑗 𝑚 3 𝑖=1 𝑚 3 𝑖=1 � +2 ∑ 𝑥3𝑖𝑢3𝑖,𝑗 𝑚 3 −1 𝑖=1 + 𝑢𝑚,𝑗� = 𝛽�𝑡𝑗� (9) 𝑗 = 1,2, … , 𝑛 Numerical Examples In this section we present three examples of the nonlocal problems that are solved by using the explicit and the implicit finite difference methods. Example (1) Consider the following diffusion partial differential equation: 𝜕𝑢(𝑥,𝑡) 𝜕𝑡 − 𝜕 2𝑢(𝑥,𝑡) 𝜕𝑥2 = 1 0 ≤ 𝑥 ≤ 1, 𝑡 ≥ 0 with the initial condition: u (x,0) = 3x , 0 ≤ 𝑥 ≤ 1 and the Neumann condition: 𝜕𝑢(𝑥,𝑡) 𝜕𝑥 |𝑥=0 = 3 , 𝑡 ≥ 0 and the nonlocal condition: ∫ 𝑢(𝑥, 𝑡)𝑑𝑥 = 3 2 + 𝑡 1 0 , 𝑡 ≥ 0 This example is constructed such that the exact solution is u(x,t)=3x+t. We shall use the explicit finite difference method to solve this example. To do this, let m=10 and n=100, then𝑥𝑖 = 𝑖 10 , 𝑖 = 0,1, … ,10, 𝑡𝑗 = 𝑗 100 , 𝑗 = 0,1, … ,100 and r=1. In this case, equations (2)-(3) becomes: 𝑢𝑖,𝑗+1 = 𝑢𝑖+1,𝑗 − 𝑢𝑖,𝑗 + 𝑢𝑖−1,𝑗 + 1 100 , 𝑖 = 0,1, … ,10, 𝑗 = 0,1, … ,100 (10) and 𝑢𝑖,0 = 3𝑥𝑖 , 𝑖 = 0,1, … ,10 280 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 By evaluating equation (10) at each i=1,2,…,9 and j=0 one can get 𝑢𝑖,1, i=1,2,…,9 and by substituting j=1 at equation (4) we get: 𝑢0,1 = 𝑢1,1 − 1 10 (3 2 + 𝑡1) and this implies that 𝑢𝑚,1 = 20 � 3 2 + 𝑡1� − 𝑢0,1 − 2 ∑ = 9 1 1, i iu Next, we evaluate equation (10) at each i=1,2,…,9, j=1, one can get the values of 𝑢𝑖,2, i =1,2,…,9 and by substituting j=2 at equation (4) we get: 𝑢0,2 = 𝑢1,2 − 1 10 (3 2 + 𝑡2) and this implies that 𝑢𝑚,2 = 20 � 3 2 + 𝑡2� − 𝑢0,2 − 2∑ = 9 1 2, i iu By continuing in this manner one can get the numerical solutions 𝑢𝑖,𝑗 , i=0,1,…,10, j= 0,1,…,100, of the above nonlocal problem. Some of these results can be seen in table (1) So the errors of this example via trapezoidal rule are zero. Example (2) Consider the following diffusion partial differential equation: 𝜕𝑢(𝑥,𝑡) 𝜕𝑡 − 𝜕 2𝑢(𝑥,𝑡) 𝜕𝑥2 = 2𝑥2𝑡 + 2𝑥 − 2𝑡2 , 0 ≤ 𝑥 ≤ 1, 𝑡 ≥ 0 with the initial condition: u (x,0) = 0 , 0 ≤ 𝑥 ≤ 1 and the Neumann condition : 𝜕𝑢(𝑥,𝑡) 𝜕𝑥 |𝑥=0 = 2𝑡 , , 𝑡 ≥ 0 and the nonlocal condition : ∫ 𝑢(𝑥, 𝑡)𝑑𝑥 = 1 3 𝑡2 + 𝑡 1 0 , 𝑡 ≥ 0 We use the explicit finite difference method to solve this example. This example is constructed such that the exact solution is 𝑢(𝑥, 𝑡) = 𝑥2𝑡2 + 2𝑡𝑥. To do this, first, let m=15 281 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 and n=200, then 𝑥𝑖 = 𝑖 15 , 𝑖 = 0,1, … ,15, 𝑡𝑗 = 𝑗 200 , 𝑗 = 0,1, … ,200 and. 𝑟 = 9 8 In this case, equation (2)-(3) becomes: 𝑢𝑖,𝑗+1 = 9 8 𝑢𝑖+1,𝑗 − 5 4 𝑢𝑖,𝑗 + 9 8 𝑢𝑖−1,𝑗 + 1 200 �2𝑥𝑖 2𝑡𝑗 + 2𝑥𝑖 − 2𝑡𝑗 2� (11) and 𝑢𝑖,0 = 0 , 𝑖 = 0,1, … ,15 (12) Respectively, by evaluating equation (11) at each i=1,2,…,14 and j=0 one can get the values of 𝑢𝑖,1, i=1,2,…,14 and by substituting j=1 in equation (4) we get: 𝑢0,1 = 𝑢1,1 − 2 15 𝑡1 and from equation (5) we get: 𝑢𝑚,1 = 30 � 1 3 𝑡12 + 𝑡1� − 𝑢0,1 − 2∑ = 14 1 1, i iu Now, we evaluate equation (11) at each i=1,…,14, j=1,one can get the values of 𝑢𝑖,2, i =1,2,…,14 and by substituting j=2 at equation (4) we get: 𝑢0,2 = 𝑢1,2 − 2 15 𝑡2 If we use equation (5), then : 𝑢𝑚,2 = 30 � 1 3 𝑡22 + 𝑡2� − 𝑢0,2 − 2∑ = 14 1 2, i iu If we use equation (9) then 𝑢𝑚,2 =0.02 By continuing in this manner one can get the numerical solutions 𝑢𝑖,𝑗, i=0,1,…,15, j= 0,1,…,200, of the above nonlocal problem. Some of these results can be seen in table (2) On the other hand, since m=15 is a multiple of 3, one can use either equation (5) or equation (9) to get the values of 𝑢𝑚,1. If we use equation (5), then : 𝑢𝑚,1=0.01 On the other hand ,if we use equation (9) then: 𝑢𝑚,1 =0.01 Second, let m=16 and n=200, then𝑥𝑖 = 𝑖 16 , 𝑖 = 0,1, … ,16, 𝑡𝑗 = 𝑗 200 , 𝑗 = 0,1, . . ,200 and 𝑟 = 32 25 Similar to the previous, one can get the values of the numerical solutions 𝑢𝑖,𝑗, 𝑖 = 0,1, … ,16, 𝑗 = 0,1, … ,200.These values can be seen in tables (6)-(7). Example (3) Consider the following diffusion partial differential equation: 282 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 𝜕𝑢(𝑥,𝑡) 𝜕𝑡 − 𝜕 2𝑢(𝑥,𝑡) 𝜕𝑥2 = 6𝑥2𝑡 + 2𝑥 − 6𝑡2 0 ≤ 𝑥 ≤ 1, 𝑡 ≥ 0 with the initial condition: u (x,0) = 0, 0 ≤ 𝑥 ≤ 1 and the Neumann condition: 𝜕𝑢(𝑥,𝑡) 𝜕𝑥 |𝑥=0 = 2𝑡 , 𝑡 ≥ 0 and the non- local condition: ∫ 𝑢(𝑥, 𝑡)𝑑𝑥 = 𝑡2 + 𝑡 1 0 , 𝑡 ≥ 0 This example is constructed such that the exact solution isu(x, t) = 3𝑥2𝑡2 + 2𝑥𝑡. We shall use the implicit finite difference method to solve this example. To do this, let m=10 and n=1000, then 𝑥𝑖 = 𝑖 10 , 𝑖 = 0,1, … ,10, 𝑡𝑗 = 𝑗 1000 , 𝑗 = 0,1, … ,1000 and. 𝑟 = 1 10 In this case equation (8) becomes ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ −1 1 0 0 0 0 0 0 −0.05 2 −0.05 0 0 0 0 0 0 −0.05 2 −0.05 0 0 0 0 0 0 −0.05 2 −0.05 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 0 −0.05 2 −0.05 1 2 2 2 2 2 2 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 𝑢0,1 𝑢1,1 𝑢2,1 𝑢3,1 . . . 𝑢9,1 𝑢10,1⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 2 10 𝑡1 1 20 𝑢2,0 − 1 10 𝑢1,0 + 1 20 𝑢0,0 + 1 1000 (6𝑥12𝑡02 + 2𝑥1 − 6𝑡02) 1 20 𝑢3,0 − 1 10 𝑢2,0 + 1 20 𝑢1,0 + 1 1000 (6𝑥22𝑡02 + 2𝑥2 − 6𝑡02) 1 20 𝑢3,0 − 1 10 𝑢2,0 + 1 20 𝑢1,0 + 1 1000 (6𝑥32𝑡02 + 2𝑥3 − 6𝑡02) . . . 1 20 𝑢10,0 − 1 10 𝑢9,0 + 1 20 𝑢8,0 + 1 1000 (6𝑥92𝑡02 + 2𝑥9 − 6𝑡02) 20(𝑡12 + 𝑡1) ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ By solving the above system we get the following tabulated results which can seen in tables (8) - (9) 283 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Conclusions In this work the explicit and the implicit finite difference methods are powerful numerical methods that gave satisfactory results for solving the nonlocal problem for the parabolic partial differential equations and can be also used to solve the nonlocal problems for the hyperbolic and elliptic partial differential equations. References 1. Gulin, A.V., Lonkin N.I. and .Morozova, V.A., ( 2005) , "Stability of Nonlocal Finite- Difference Problems" Computational Mathematics and Modeling, 16,( 3), 248-256. 2. Hayder, D. M., (2011),"Some Analytical Methods for Solving Some Types of Nonlocal Problems", M.Sc. Thesis, College of Science, Al-Nahrain University. 3. El-Sayed A. M., and Bin-Taher E.O., (2010), "Anonlocal Problem of an Arbitrary (Fractional) Orders Differential Equation", Alexandria Journal of Mathematics, 1, (2), 1-7, 4. Karatay I., Bayramoglu S. R., Yildiz B. and Kokluce B., (2011) , "Matrix Stability of the Difference Schemes for Nonlocal Boundary Value Problems for Parabolic Differential Equations", International Journal of the Physical Sciences, 6, (4), 819-827. 5. GhoreishM. I., Ismail A.I., (2011), "The Homotopy Perturbation Method (HPM) for Nonlinear Parabolic Equation with Nonlocal Boundary Conditions", Applied Mathematical Sciences, 5, (3), 113-123. Table (1): represents the numerical and the exact solutions of example (1) via trapezoid rule for r=1 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 3 0 0 0.01 0.02 0.03 1 0.3 0.31 0.32 0.33 2 0.6 0.61 0.62 0.63 3 0.9 0.91 0.92 0.93 4 1.2 1.21 1.22 1.23 5 1.5 1.51 1.52 1.53 6 1.8 1.81 1.82 1.83 7 2.1 2.11 2.12 2.13 8 2.4 2.41 2.42 2.43 9 2.7 2.71 2.72 2.73 10 3 3.01 3.02 3.03 284 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Table (2): represents the numerical solutions of example (2) by using trapezoid rule for r==9 8 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 3 0 0 0 -2.778 × 10−8 1.667 × 10−7 1 0 6.667 × 10−4 1.333 × 10−3 2 × 10−3 2 0 1.333 × 10−3 2.667 × 10−3 4.002 × 10−3 3 0 2 × 10−3 4.002 × 10−3 6.005 × 10−3 4 0 2.667 × 10−3 5.337 × 10−3 8.01 × 10−3 5 0 3.333 × 10−3 6.672 × 10−3 0.01 6 0 4 × 10−3 8.008 × 10−3 0.012 7 0 4.667 × 10−3 9.344 × 10−3 0.014 8 0 5.333 × 10−3 0.011 0.016 9 0 6 × 10−3 0.012 0.018 10 0 6.667 × 10−3 0.013 0.02 Table (3): represents the absolute errors of example (2) by using trapezoid rule for r==9 8 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 3 0 0 0 2.778 × 10−8 1.667 × 10−7 1 0 1.111 × 10−7 4.722 × 10−7 8.333 × 10−7 2 0 4.444 × 10−7 1.139 × 10−6 2.083 × 10−6 3 0 1 × 10−6 2.25 × 10−6 3.75 × 10−6 4 0 1.778 × 10−6 3.806 × 10−6 6.083 × 10−6 5 0 2.778 × 10−6 5.806 × 10−6 9.083 × 10−6 6 0 4 × 10−6 8.25 × 10−6 1.275 × 10−5 7 0 5.444 × 10−6 1.114 × 10−5 1.708 × 10−5 8 0 7.111 × 10−6 1.447 × 10−5 2.208 × 10−5 9 0 9 × 10−6 1.825 × 10−5 2.775 × 10−5 10 0 1.111 × 10−5 2.247 × 10−5 3.408 × 10−5 285 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Table(4): represents the numerical solutions of example (2) by using Simpson's 3\8 rule for r==9 8 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 3 6 0 4 × 10−3 8.008 × 10−3 0.012 7 0 4.667 × 10−3 9.344 × 10−3 0.014 8 0 5.333 × 10−3 0.011 0.016 9 0 6 × 10−3 0.012 0.018 10 0 6.667 × 10−3 0.013 0.02 11 0 7.333 × 10−3 0.015 0.022 12 0 8 × 10−3 0.016 0.024 13 0 8.667 × 10−3 0.017 0.027 14 0 9.333 × 10−3 0.019 0.027 15 0 0.01 0.02 0.033 Table (5): represents the absolute errors of example (2) by using Simpson's 3\8 rule for r=9 8 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 3 6 0 4 × 10−6 8.25 × 10−6 1.275 × 10−5 7 0 5.444 × 10−6 1.114 × 10−5 1.708 × 10−5 8 0 7.111 × 10−6 1.447 × 10−5 2.208 × 10−5 9 0 9 × 10−6 1.825 × 10−5 2.775 × 10−5 10 0 1.111 × 10−5 2.247 × 10−5 3.408 × 10−5 11 0 1.344 × 10−5 2.714 × 10−5 4.108 × 10−5 12 0 1.6 × 10−5 3.225 × 10−5 4.875 × 10−5 13 0 1.878 × 10−5 3.781 × 10−5 3.648 × 10−4 14 0 2.178 × 10−5 3.312 × 10−4 1.039 × 10−3 15 0 3.083 × 10−4 4.988 × 10−4 2.607 × 10−3 286 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 able(6): represents the numerical solutions of example (2) by using Simpson's 1/3 rule for r==32 25 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 3 0 0 0 5.469 × 10−8 8.594 × 10−8 1 0 6.25 × 10−4 1.25 × 10−3 1.875 × 10−3 2 0 1.25 × 10−3 2.501 × 10−3 3.752 × 10−3 3 0 1.875 × 10−3 3.752 × 10−3 5.63 × 10−3 4 0 2.5 × 10−3 5.003 × 10−3 7.509 × 10−3 5 0 3.125 × 10−3 6.255 × 10−3 9.389 × 10−3 6 0 3.75 × 10−3 7.507 × 10−3 0.011 7 0 4.375 × 10−3 8.759 × 10−3 0.013 8 0 5 × 10−3 0.01 0.015 9 0 5.625 × 10−3 0.011 0.017 10 0 6.25 × 10−3 0.013 0.019 Table (7): represents the absolute errors of example (2) by using Simpson's 1/3 rule for r=32 25 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 3 0 0 0 5.469 × 10−8 8.594 × 10−8 1 0 9.766 × 10−8 4.453 × 10−7 7.93 × 10−7 2 0 3.906 × 10−7 1.031 × 10−6 1.922 × 10−6 3 0 8.789 × 10−7 2.008 × 10−6 3.387 × 10−6 4 0 1.562 × 10−6 3.375 × 10−6 5.438 × 10−6 5 0 2.441 × 10−6 5.133 × 10−6 8.074 × 10−6 6 0 3.516 × 10−6 7.281 × 10−6 1.13 × 10−5 7 0 4.785 × 10−6 9.82 × 10−6 1.511 × 10−5 8 0 6.25 × 10−6 1.275 × 10−5 1.95 × 10−5 9 0 7.91 × 10−6 1.607 × 10−5 2.448 × 10−5 10 0 9.766 × 10−5 1.978 × 10−5 3.005 × 10−5 287 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Table (8): represents the numerical solutions of example (3) by using implicit finite difference method for r= 1 10 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 0 9.717 × 10−5 2.898 × 10−4 4.758 × 10−4 1 1.028 × 10−4 1.102 × 10−4 1.242 × 10−4 2 2.105 × 10−4 2.107 × 10−4 2.114 × 10−4 3 3.158 × 10−4 3.161 × 10−4 3.164 × 10−4 4 4.211 × 10−4 4.216 × 10−4 4.221 × 10−4 5 5.263 × 10−4 5.271 × 10−4 5.279 × 10−4 6 6.316 × 10−4 6.327 × 10−4 6.339 × 10−4 7 7.37 × 10−4 7.392 × 10−4 7.422 × 10−4 8 8.478 × 10−4 8.721 × 10−4 9.118 × 10−4 9 1.176 × 10−3 1.852 × 10−3 2.729 × 10−3 10 0.01 0.028 0.046 Table(9): represents the absolute errors of example (3) by using the implicit finite difference method for r= 1 10 i j 𝑢�𝑥𝑖, 𝑡𝑗� = 𝑢𝑖,𝑗 0 1 2 0 0 2.898 × 10−4 4.758 × 10−4 1 2 × 10−4 2.899 × 10−4 4.76 × 10−4 2 4.001 × 10−4 5.898 × 10−4 9.897 × 10−4 3 6.003 × 10−4 8.85 × 10−4 1.486 × 10−3 4 8.005 × 10−4 1.18 × 10−3 1.982 × 10−3 5 1.001 × 10−3 1.476 × 10−3 2.479 × 10−3 6 1.201 × 10−3 1.772 × 10−3 2.976 × 10−3 7 1.401 × 10−3 2.067 × 10−3 3.471 × 10−3 8 1.602 × 10−3 2.336 × 10−3 3.905 × 10−3 9 1.802 × 10−3 1.758 × 10−3 2.693 × 10−3 10 2.003 × 10−3 0.024 0.04 288 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 الحلول العددیة للمسائل الالمحلیة لمعادالت االنتشار التفاضلیة الجزئیة ریم منذر كبة جامعة بغداد ) / ابن الھیثم( للعلوم الصرفة كلیة التربیة/ الریاضیات علوم قسم 2013كانون الثاني 16 : قبل البحث في ، 2012تشرین االول 3 : استلم البحث في الخالصة في ھذا العمل استعملنا طریقة الفروقات المنتھیة الصریحة والضمنیة لحل المعادالت التفاضلیة الجزئیة مع الشروط لھذه المعادالت التفاضلیة الجزئیة قربت باستخدام قاعدة شبھ المنحرف المركبة وقاعدة .الالمحلیة الشروط الالمحلیة .بعض االمثلة العددیة لتبیان دقة ھذه الطرائق قدرتسمبسون ثلث وثالثة اثمان المركبة كذلك ،روقات المنتھیة الضمنیةطریقة الفروقات المنتھیة، طریقة الفروقات المنتھیة الصریحة، طریقة الف :الكلمات المفتاحیة ..الشروط الالمحلیة، معادلة االنتشار 289 | Mathematics