Computer Algebra Recipes

224 CHAPTER 5. LINEAR PDE MODELS. PART 1
but all 1 have failed to pull the sword out of the rock. Having taken a course
in thermodynamics as an engineering undergraduate, Russell speculates as to
whether the sword could be pulled out by cooling the ends of the sword with
large buckets of ice, thus causing heat to °ow out of the warmer interior of the
sword's blade to the ends. If the sword were cooled su±ciently, its diameter
might shrink slightly, and just possibly he could pull the sword out.
But remembering the thermodynamics course causes Russell's dream to alter
direction, and he then dreams of a related problem that appeared on that
course's ¯nal exam many years ago. A thin 1-m-long rod (the sword's shaft)
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(contactwiththe
buckets of ice). Taking one end of the rod to be at x = 0 and the other at x =
L = 1, the initial temperature (T ) 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 .
In the exam, he was asked to determine the temperature distribution T (x; t) for
any time t>0. Despite the passage of time, Russell remembers his approach to
solving this problem very well. His method of attack was to solve the di®usion
equation
@T(x; t)
@t
by the method of separation of variables, i.e., assume that T (x; t)=S(x) F (t).
Substituting this assumed form into (5.7) and dividing by T (x; t) yields
1 dF (t) d d
=
F (t) dt S(x)
2S(x) dx2 : (5.8)
Since the lhs of (5.8) involves a function of t alone and the rhs involves a function
of x alone, the only way that it can be generally true is for both sides to be
equal to a common constant, called the separation constant. The separated
ODE for S(x) is solved, subject to the boundary conditions, and the complete
product solution constructed, subject to the initial condition.
Russell's wild dream is interrupted by the roar of a big jet passing over his
house on its way into Sky Harbor Airport. Waking up, and unable to get back
to sleep, he decides to make a cup of instant decaf co®ee and implement an
animated Maple solution of the heat di®usion problem on his computer. To
carry out the animation, the plots library package is loaded. Noting that the
di®usion coe±cient d could be absorbed into either the spatial or time variable,
Russell sets d = 1 without loss of generality. He also speci¯es the rod's length,
L = 1, and enters the heat equation (5.7).
> restart: with(plots): d:=1: L:=1:
> heateq:=diff(T(x,t),t)=d*diff(T(x,t),x,x);
= d @2 T (x; t)
@x 2 ; (5.7)
heateq := @
@2
T (x; t) = T (x; t)
@t @x2 A general solution to the heat equation is sought using the PDE solve (pdsolve)
command, the HINT option explicitly telling Maple to separate variables.
1 King Arthur has not shown up yet.