Difference sche1nes as operator equations. General formulations 117- Difference schemes as operator equations. After replacing differential
equations by difference equations on a certain grid wh we obtain a system of
linear algebraic equations that can be written in matrix form. The outcome
of this is
(1)where U is a square N x N-matrix, Y = (y 1 , y 2 , .•. , yN) is the vector of
unknowns and <[> = ( <p 1 , <p 2 , ... , 'PN) is a known right-hand side including
the right-hand sides of boundary conditions. With every matrix U one
can associate a linear operator A acting from RN into RN. With this
correspondence in view, equation ( 1) takes the form(2) Ay=<p,
where the unknown vector y is sought, while the right-hand side 'P E RN is
a given vector. The operator A maps onto itself the space of grid functions
defined on wh and satisfying the h01nogeneous boundary conditions. Several
examples can add interest and aid in understanding.Example 1. The first boundary-value proble1n. Given on the segment
[O, l] an equidistant grid wh = {x; = ih, i = 0, 1, ... , N, h = 1/ N},
let us look for a solution of the first boundary-value problem(3) Ay=y_ xx =-f(x), O<x=ih<l,
or(3')i = 1, 2, ... , N - 1 ,
As the first step towards the solution of this problem, we form the vector
Y = (y 1 , y 2 , ..• , yN_ 1 ), making it possible to rewrite equation (3) in the
form (1) with the (N - 1) x (N - 1)-matrix
2 -1
-1 2
0 -10 00 0
-1 0
2 -10 00
0
00
0
0-1 2