Computer Algebra Recipes

6.2. SEMI-INFINITE AND INFINITE DOMAINS 267
where K is a positive constant with units in kg/(m 2 ¢ s). The coe±cients can be
eliminated and the two equations cast into dimensionless form by introducing
the new variables t ´ ¸T, x ´ p ¸=d X, and c = ( p ¸d=K) C. Then the
concentration equation becomes
@c(x; t)
@t
= @2c(x; t)
¡ c(x; t); (6.10)
@x2 with @c=@x = ¡1 as the boundary condition at x =0.
Assuming that initially c =0forx¸ 0, we want to determine the distribution
of radioactive gas in the atmosphere for times t ¸ 0 and animate the
solution. The method of attack will be to make use of the Laplace transform.
The Laplace transform of a function f(t) isde¯nedas
Z 1
L(f(t)) ´ F (s) = f(t) e ¡st dt: (6.11)
Integrating by parts and assuming that e ¡stf(t)!0 ast!1,then
μ
μ 2 df
d f
L = sF(s) ¡ f(0); and L
dt
dt2
= s 2 df (0)
F (s) ¡ sf(0) ¡ : (6.12)
dt
To solve the concentration equation (6.10), subject to the boundary and
initial conditions, the Laplace transform can be applied to the time part of the
equation, the resultant second-order ODE in x solved, and the inverse Laplace
transform performed to regain the time dependence.
This program is now carried out using the laplace command contained in
the integral transform library package.
> restart: with(plots): with(inttrans):
The dimensionless concentration equation is entered,
> CE:=diff(c(x,t),t)=diff(c(x,t),x,x)-c(x,t);
CE := @ μ 2
@
c(x; t) = c(x; t) ¡ c(x; t)
@t @x2 and the initial concentration speci¯ed.
> c(x,0):=0:
The Laplace transform of CE is taken with respect to the time variable, and
the function laplace(c(x; t); t;s) then replaced with f (x) inLT .
> LT:=laplace(CE,t,s);
> LT:=subs(laplace(c(x,t),t,s)=f(x),LT);
μ 2 d
LT := s f(x) = f (x) ¡ f (x)
dx2 This second-order ODE is solved for f(x), the following DEtools command line
being used to obtain exponential solutions, which are convenient here.
> sol:=DEtools[expsols](LT,f(x));
sol := [e (p s+1 x) ;e (¡ p s+1 x) ]
0