Mathematical Physics - Volume II - Numerical Methods

1.1 Finite difference method

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Let ( x n , t j ) = ( nh , jk ) ( n = 0 , 1 , . . . , N ; j = 0 , 1 , 2 . . . ) with Nh = 1, and let the vector column U j be defined as: U j = [ U 1 j , . . . , U N − 1 , j ] T . Explicit method can be expressed as the following matrix equations: U j + 1 = CU j ( j = 0 , 1 , 2 , . . . ) , (1.31) where U 0 = [ f 1 , f 2 , . . . , f N − 1 ] T , and C is the three-diagonal square matrix of the N ˘1 order.

    r

    (1.32)

1 − 2 r

r

0

0 0 0 0 0 0

1 − 2 r

r

r

1 − 2 r

r

0 0

. . . . . .

r

1 − 2 r . . .

0

0

r

r

1 − 2 r

Let us assume that error E n j was introduced at x n in time step j , (( n = 1 , . . . , N − 1 ) , such that the solution of equation (1.31) becomes U j + E j , where E j is the vector-column whose n -th component is E n j . In this case, the following holds: U j + 1 + E j + 1 = CU j + CE j or E j + 1 = CE j (1.33) (1.34) λ 1 , . . . , λ N − 1 and V 1 , . . . , V N − 1 be the eigenvalues and their respective linearly independent eigenvectors of a symmetrical matrix C . If E j is written as a linear combination of V k : E j = N − 1 ∑ k = 1 a k V k , (1.35) and if the following is used CV k = λ k V k (1.36) we obtain: E j + m = N − 1 ∑ k = 1 λ m k a k V k (1.37) Equation (1.37) indicates that the error E n j remains limited if and only if | λ k | ≤ 1 for k = 1 , . . . , N − 1. Knowing that the eigenvalues of a real symmetrical i.e., after a total of m steps E j + m = C m E j