Mathematical Physics - Volume II - Numerical Methods

24 Chapter 1. Finite difference method and Finite element method

(4) If the contour problem has a unique solution, and h is sufficiently small, A is a non-singular matrix, thus the system of linear algebraic equations has a unique solution. The rounding error in the case the central differences are used to approximate a Laplacian, u xx + u yy on a rectangular mesh, ( x m , y n ) = ( mh , nk ) , can be expressed as (assuming that u xxxx and u yyyy : − h 2 12 U xxxx ( ˜ x , y n ) − k 2 12 U yyyy ( x m , ˜ y ) = O ( h 2 + k 2 ) (1.59) It follows from the above that if the solution of a contour problem for the Poisson equation has fourth-order derivatives equal to ≡ 0, as in the case of u = xy , then the solution obtained by finite differences is exact.

1.1.9 Neumann problem

Let us now also consider the Neumann problem, given as:

u xx + u yy = f ( x , y ) on the domain Ω

(1.60)

and

∂ u ∂ n

= g ( x , y ) on the boundary S ,

(1.61)

where Ω represents a rectangle 0 < x < a , 0 < y < b . Mesh nodes will be selected so that Mh = a and Nh = b . Based on (1.61) it follows that:

− U m − 1 , n − U m , n − 1 + 4 U mn − U m + 1 , n − U m , n + 1 = − h 2 f mn ,

(1.62)

wherein the rounding error is O ( h 2 ) , m = 0 , 1 , . . . , M and n = 0 , 1 , . . . , N , and f ( x , y ) is defined in S as well. The main characteristic of Neumann problem solution is that the derivative representation via central differences is achieved by introducing primary nodes (Figure 1.1), which are not located in the mesh nodes (domain): ∂ u ∂ n

U M + 1 , n − U M − 1 , n = 2 hg Mn ( n = 1 , 2 , . . . , N − 1 ) , U m , N + 1 − U m , N − 1 = 2 hg mN ( m = 1 , 2 , . . . , M − 1 ) , U − 1 , n − U 1 n = 2 hg 0 n ( n = 1 , 2 , . . . , N − 1 ) U m , − 1 − U m 1 = 2 hg m 0 ( m = 1 , 2 , . . . , M − 1 ) .

(1.63) (1.64) (1.65) (1.66)

In mesh nodes, where the normal is not determined, we will adopt the mean value of the two nearest normal, which provides additional four boundary conditions:

U − 1 , 0 + U 0 , − 1 = U 10 + U 01 + 4 hg 00 , U M , − 1 + U M + 1 , 0 = U M 1 + U M − 1 , 0 + 4 hg M 0 , U M + 1 , N + U M , N + 1 = U M − 1 , N + U M , N − 1 + 4 hg MN , U 0 , N + 1 + U − 1 , N = U 0 , N − 1 + U 1 , N + 4 hg 0 N .

(1.67) (1.68) (1.69) (1.70)