Look at the condition u(x,y,0) =
T o xy (x - a) (y - b)
Then:
å¥ n=1 å¥ m= 1 a nm sin (np/ a) x sin (mp/ b)
Suppose a function f(x, y) which is periodic of period 2p
Consider y fixed and suppose the function has the Fourier Series expansion;
f(x, y) = å¥ n=o [a (y) cos nx + b(y) sin nx]
Suppose in addition that a(y) and b(y) have Fourier series expansions, e.g.
a(y) = å¥ n=o [a nm cos ny + b nm sin ny]
And:
b(y) = å¥ m=o (g nm cos my + d nm sin ny]
Substitution of a n (y), b n (y) into series for f(x,y) results in terms of the form:
cos nx cos my, cos nx sin my, sin nx cos my, sin nx sin my
Hence:
f(x, y) = å¥ n=o å¥ m=o a nm (cos nx cos my) + b nm (cos nx sin my) +
c nm (sin nx cos my) + d nm (sin nx sin my)
Note that the preceding is a double Fourier series.
This brief diversion enables us to continue our treatment as follows, so writing:
ò p -p
f(x,y) sin ax dx = p å¥ m=1 c am cos my + p å¥ m=1 d am sin my
Next, multiply both sides by sin by and integrate, i.e.
Which tells us
ò p -p [ò p -p f(x,y) sin ax dx] sin by dy
Þ
å m ò p -p d am sin by sin my dy = p d ab
Which tells us:
d ab = 1/ p 2 ò p -p [ò p -p f(x,y) sin ax sin by dx dy
Similarly:
a ab = 1/ p ò p -p [ò p -p f(x,y) cos ax cos by dx dy
b ab = 1/ p ò p -p [ò p -p f(x,y) cos ax sin by dx dy
c ab = 1/ p ò p -p [ò p -p f(x,y) sin ax cos by dx dy
And further:
d n0 = d 0m = c 0m = b n0
Now take:
ò p -p [ò p -p f(x,y) sin ax dx dy = p (2 p c a0)
c a0 = 1/ 2p 2 ò p -p [ò p -p f(x,y) sin ax dx dy
And:
b 0b = 1/ 2p 2 ò p -p [ò p -p f(x,y) sin by dx dy
Using the preceding we may write:
1/ 2p 2 ò p -p [ò p -p f(x,y) cos ax dx dy = a a0
Similarly:
a ab = 1/ 2p 2 ò p -p [ò p -p f(x,y) cos by dx dy
To get a a0: Take: ò p -p f(x,y) dx =
2p å¥ m=0 a 0m cos my + 2p å¥ m=0 b 0m sin my etc.
If f(x,y) is an odd function in x and y then:
a oo = a 0m = a nm = b 0m = b nm= c no = c nm = 0
The series then reduces to:
å¥ n=1 å¥ m= 1 d nm sin nx sin my
And:
d nm = 4/p 2 ò p -p [ò p -p f(x,y) sin nx sin my dx dy {if f(x,y) odd