Computer Algebra Recipes

6.3. NUMERICAL SIMULATION OF PDES 275
For example, in the forward di®erence approximation, the ¯rst spatial and
time derivatives at P are written as
μ
@Ã
=
@x P
(Ãi+1;j
μ
¡ Ãi;j) @Ã
; =
h
@t P
(Ãi;j+1 ¡ Ãi;j)
; (6.16)
k
while the \standard" CDAs for the second derivatives at P are
μ 2 @ Ã
@x2
=
P
(Ãi+1;j ¡ 2 Ãi;j + Ãi¡1;j)
h2 ; (6.17)
μ 2 @ Ã
@t2
=
P
(Ãi;j+1 ¡ 2 Ãi;j + Ãi;j¡1)
k2 : (6.18)
Although rectangular meshes are most commonly employed, diamond-shaped
grids can also prove to be useful in certain numerical schemes used to model
wave equations.
6.3.1 Freeing Excalibur the Numerical Way
\Are ¯ve nights warmer than one night, then?"
Alice ventured to ask. \Five times as warm of course."
\But they could be ¯ve times as cold, by the same rule|"
\Just so!" cried the Red Queen.
Lewis Carroll (Charles Lutwidge Dodgson), English writer (1832{1898)
Recall that Russell, the aerospace engineer, was having a wild dream about
freeing the sword Excalibur from its stony tomb by cooling its ends with buckets
of ice water when his dream took a sudden detour and he recalled the
following related problem from his undergraduate thermodynamics course.
A thin 1-meter-long rod (the shaft of the sword), whose lateral surface is
insulated to prevent heat °ow through that surface, has its ends suddenly held
at the freezing point of water, 0 ± C, by placing them in contact with buckets of
ice. Taking one end of the rod to be at x = 0 and the other at x = 1, the initial
temperature distribution was T (x; t =0)=100x (1 ¡ x), a parabolic pro¯le
with a maximum temperature of 25 ± at the midpoint x = 1
2 .
Russell was able to determine the analytic solution to this problem, using
the separation of variables technique with the aid of Maple. Since it is now
necessary in his work to numerically solve a system of nonlinear PDEs, and he
hasn't done any numerical work for a while, he decides to tackle the Excalibur
problem ¯rst, using an explicit ¯nite-di®erence scheme.
In the linear di®usion equation (with di®usion coe±cient d =1), Russell
replaces the time derivative with the forward-di®erence approximation and the
second spatial derivative with the standard CDA, so that
@T
@t = @2T @x2 ) (Ti;j+1 ¡ Ti;j)
=
k
(Ti+1;j ¡ 2 Ti;j + Ti¡1;j)
h2 : (6.19)