Computer Algebra Recipes

262 CHAPTER 6. LINEAR PDE MODELS. PART 2
that vanishes at x = 1 is de¯ned as
r Z 1
2
FST (f(x)) ´ F (s) = f(x)sin(sx) dx; (6.5)
¼ 0
and the inverse Fourier sine transform of F (s) by
r Z 1
2
f(x) = F (s)sin(sx) ds: (6.6)
¼ 0
For the present problem, the temperature distribution will satisfy the onedimensional
heat di®usion equation (with d the heat di®usion coe±cient),
@T(x; t)
= d
@t
@2T (x; t)
@x2 : (6.7)
Taking the FST of (6.7) with respect to x and integrating the rhs twice by parts,
and assuming f 0 (x !1) ! 0, will convert the PDE into a ¯rst-order ODE for
the transformed quantity F (s; t). The ODE will depend on the boundary condition
T (0;t). The ODE is then solved for F (s; t) making use of the Fourier sine
transform of the initial condition. Finally, on performing the inverse transform
with respect to s, the temperature distribution T (x; t) is found.
Once again, Vectoria disdains doing the problem \the old-fashioned way"
favored by some instructors, but instead intends to let the computer assist her.
To this end, she loads the plots and integral transform packages,
> restart: with(plots): with(inttrans):
and enters the di®usion equation DE,
> DE:=diff(T(x,t),t)=d*diff(T(x,t),x,x);
DE := @ T (x; t) =d
@t
μ
@ 2
2 T (x; t)
@x
and the boundary condition T (0;t)=T0.
> T(0,t):=T0:
To keep the recipe general for the moment, Vectoria has not yet speci¯ed the
value of T0 . She then takes the FST of DE with respect to x, the integration
by parts and boundary condition substitution being automatically done.
> FST:=fouriersin(DE,x,s);
FST := @
@t fouriersin(T (x; t);x;s)=ds(p 2 T0 ¡ s fouriersin(T (x; t);x;s) p ¼)
p
¼
To simplify the notation and facilitate solving the above ODE, she replaces
fouriersin(T (x; t);x;s)withF (t), temporarily suppressing the argument s.
> FST:=subs(fouriersin(T(x,t),x,s)=F(t),FST);
FST := d
dt F (t) = ds(p2 T0 ¡ s F (t) p ¼)
p
¼
The Fourier sine transform of the initial condition is zero, so Vectoria solves
the ¯rst-order linear di®erential equation FST for F (t), subject to F (0) = 0.
> eq:=dsolve(fFST,F(0)=0g,F(t));