Difference approximation of elementary differential operators 51
We would like to discuss the questions raised above in more detail.
Obviously, in numerical solution of mathematical problems it is unrealistic
to reproduce a difference solution for all the values of the argument varying
in a certain domain of a prescribed Euclidean space. The traditional way
of covering this is to select some finite set of points in this domain and look
for an approximate solution only at those points. Any such set of points is
called a grid and the isolated points are termed the grid nodes.
Any such function defined at nodal points is called a grid function.
Thus, the first step in this direction is to replace the domain of continuous
variation of an argument by a grid, that is, by the domain of discrete
variation of the same argument. In other words, we have approximated the
space of solutions to a differential equation by the space of grid functions.
For this reason the properties of a difference solution and, in particular, its
proximity to an exact solution depend essentially on a proper choice of the
grid. It seems worthwhile giving several simple grids to help motivate what
is done.
Example 1. An equidistant grid on a segment. The segment [O, l]
of unit length is splitted into N equal intervals. The spacing between the
adjacent nodes xi - xi-i = h = "Jv is termed a grid step or simply step.
The splitting points xi = ih constitute what is called the set of grid nodes
wh = {xi = ih, i = 1, ... , N - l} and generate one possible grid on this
segment (see Fig. 1):
x
Figure 1.
When the boundary points x 0 = 0 and x N = 1 are put together with
the grid wh, it will be denoted by wh ={xi= ih, i = 0, 1, ... , N - l, N}.
On the segment [O, l] we are working with a new function Yh (xi) of
the discrete argument instead of a given function y( x) of the continuous
argument. The values of such a function are calculated at the grid nodes
xi and the function itself depends on the step h as on the parameter.
Example 2. An equidistant grid in a plane. We now consider a
set of functions u(x, t) of two arguments in the rectangle i5 = {O < x <
1, 0 < t < T} and split up the segment [O, l] on the x-axis and the segment
[O, T] on the t-axis into N 1 and N2 parts with steps h = l/N 1 and T =