Homogeneous difference schernes for the heat conduction 489
Let yh 71 (x;, t 1 ) and Yh 72 (x;, t 1 ) be solutions of the same difference
problem with different steps r 1 = T and r 2 = T. Via the linear transforn1
with expression (77) standing for y 1 , 71 and Yh 72 both, it is necessary to
reduce to zero the coefficient at the member rn^1. All this enables us to find
with reasonable accuracy
the values
A similar procedure works or: two different grids wh and w 0 5 h for fixed T,
thus causing a grid solution y such that
When providing current manipulations, the solution u = u(x, t) and the
available data oft.he original problen1 are preassumecl to be smooth enough
and sufficient for the existence of the asyn1ptotic expansion
- The third boundary-value problem. For the moment, the statement of
the problem is
OU
Ft= Lu+f(x,t), Lu= o ( ou)
0
xk(x,t)
0
x, O<x<l,
' ou(O, t)
k(O,t) ox =(3 1 (t)u(O,t)-μ 1 (t), (3 1 >O,
ou(l,t)
-k(l, t) ch: = /32(t) u(l, t) - μ2(t) I f32 > 0.
In Chapter 3, Section 5 we have formed a difference equation of the
third kind for the stationary equation Lu+ f = 0. The formal passage
from the stationary equation to the nonstationary one is provided by the
replacement off by f - OU I ot. This trick has been already encountered in