Computer Algebra Recipes

228 CHAPTER 5. LINEAR PDE MODELS. PART 1
The answer will require that the two-dimensional form of Laplace's equation,
@ 2 T (x; y)
@x 2
+ @2 T (x; y)
@y 2 =0; (5.9)
be solved, subject to the four boundary conditions. After loading the plots
package, we enter Laplace's equation (LE).
> restart: with(plots):
> LE:=diff(T(x,y),x,x)+diff(T(x,y),y,y)=0;
μ μ
2
2
@ @
LE := T (x; y) + T (x; y) =0
@x2 @y2 Using the pdsolve commandwiththeHINT=f(x)*g(y),INTEGRATE, andbuild
options, we generate the general product solution of Laplace's equation.
> sol:=pdsolve(LE,HINT=f(x)*g(y),INTEGRATE,build);
sol := T (x; y) = C3 sin( p c1 y) C1 e (p c1 x) + C3 sin( p c1 y) C2
e (p c1 x)
+ C4 cos( p c1 y) C1 e (p c1 x) + C4 cos( p c1 y) C2
e (p c1 x)
The solution involves one separation constant c1 and four unknown coe±cients.
For notational convenience, we substitute p c1 = m in the rhs of sol.
> T:=subs(sqrt(_c[1])=m,rhs(sol));
T := C3 sin(my) C1 e (mx) +
+ C4 cos(my) C2
e (mx)
C3 sin(my) C2
e (mx)
+ C4 cos(my) C1 e (mx)
The coe±cients are determined from the four boundary conditions. To satisfy
T (x; 0) = 0, the cos(my) terms are removed since they don't vanish at y =0.
> T:=remove(has,T,cos(m*y)); #bc1
T := C3 sin(my) C1 e (mx) C3 sin(my) C2
+
e (mx)
Then T is converted completely to trigonometric form.
> T:=expand(convert(T,trig));
T := C3 sin(my) C1 cosh(mx)+ C3 sin(my) C1 sinh(mx)
C3 sin(my) C2
+
cosh(mx) + sinh(mx)
The second boundary condition is that T (0;y) = 0. Since cosh(mx)doesn't
vanish at x = 0, terms involving cosh(mx) are removed from T .
> T:=remove(has,T,cosh(m*x)); #bc2
T := C3 sin(my) C1 sinh(mx)
The third boundary condition, T (x; h) = 0, requires that sin(mh)=0,sothat
m = n¼=hwith n a positive integer. This relation is substituted into T ,and