Computer Algebra Recipes

252 CHAPTER 6. LINEAR PDE MODELS. PART 2
> F[m,n]:=subs(fp=m*Pi/a,q=n*Pi/(2*a),_C1=1,_C2=1g,sol2);;
Fm; n := sin
à r
³
m¼x
´ ³
n¼y
´
2 2 m ¼
sin cos c
a 2 a
a2 + n2 ¼2 !
t
4 a2 Since she will need it for determining the coe±cients, Vectoria evaluates Fm;n
at t = 0, labeling the result gm;n.
> g[m,n]:=eval(F[m,n],t=0);
³
m¼x
´ ³
n¼y
´
gm;n := sin sin
a 2 a
The initial pro¯le of the rectangular membrane is entered.
> f:=4*h*x^2*(a-x)*y^3*(2*a-y)/a^7;
f := 4 hx2 (a ¡ x) y3 (2 a ¡ y)
a7 Now, the general form of the transverse displacement at time t will be given by
the following linear superposition,
1X 1X
1X 1X
Ã(x; y; t) = Am;n Fm;n ´ Ãm;n; (6.3)
m=1 n=1
m=1 n=1
with the coe±cients Am;n determined by the initial pro¯le f. Setting t =0in
(6.3) yields
X1
1X
Am;n gm;n = f:
m=1 n=1
Multiplying both sides by sin(m 0¼x=a)sin(n0¼y=(2 a)), integrating over x from
0toa and over y from0to2a, and taking the double sum will yield zero unless
m = m 0 =andn = n 0 . Removing the primes and solving for Am;n yields
Z 2 a Z a
Á Z 2 a Z a
Am;n = fgm;n dx dy
g
0 0
0 0
2 m;n dx dy:
Using this relation, we calculate Am;n in the next command line.
> A[m,n]:=int(int(f*g[m,n],x=0..a),y=0..2*a)/int(int(g[m,n]^2,
x=0..a),y=0..2*a):
Then, the general Fourier term Ãm;n = Am;n Fm;n is determined, the result
being simpli¯ed assuming that m and n are integers and a>0.
> psi[m,n]:=simplify(A[m,n]*F[m,n]) assuming m::integer,
n::integer,a>0;
Ãm; n := 3072 h ¡ 4+8(¡1) m + ¼2 n2 (¡1) n +2(¡1) (m+n) ¼2 n2 +8 (¡1) (1+m+n) +4(¡1) (1+n)¢ ³
m¼x
´ ³
n¼y
´
sin sin
Ã
a 2 a
c¼
cos
p 4 m2 + n2 !
t
.
(m
2 a
3 ¼8 n5 )
Given the mathematical form of f, Vectoria is not surprised at how complicated
the coe±cient turns out to be in Ãm;n.