Use the equation for u (0, y, t) to obtain the value of A, and thence find X(x):
u(0, y, t) = X(0) Y(y) T(t)
Þ
X(0) = 0 Þ A = 0
∴ X(x) = B sin b x
Since: X(x) = A cos bx + B sin b x
Further: u(a, y, t) = X(a) Y(y) T(t) = 0
Þ X(a) = 0 Þ B sin b a
Here: ba = np Or B = np /a
n = 1, 2, 3......
Similarly: u(x, 0, t) = X(x) Y(0) T(t) = 0
Þ Y(0) = 0 Þ C = 0
Then: Y(y) = D sin a y
And:
u(x, b, t) = X(x) Y(b) T(t) = 0
Þ Y(b) = 0 Þ D sin a b = 0
Or: a b = mp
Or: a = mp /b, m = 1,2, 3......
Further: X n (x) = B n sin (np/ a) x, n = 1,2 ...
Y n (y) = D n sin (mp/ b) y, m = 1,2 ....
For T(t):
T' = ¶ T / ¶ t = - Eg sin g t + F g cos gt
Then:
¶ u/ ¶ t = X(x) Y(y) T'(t)
And:
¶ u(x,y,t)/ ¶ t = X(x) Y(y) T'(0) = 0
Þ T'(0) = 0 Þ F g= 0
Þ F = 0
Therefore:
T nm (t) = E nm cos [cÖ ((mp/ b) 2 + (np/ a) 2 t]
Where the subscript nm denotes that we have a two fold, infinite solution for the above equation.
∴ u nm (x, y, t) =
B n D n E nm sin (np/ a) x sin (mp/ b) y cos [cÖ ((mp/ b) 2 + (np/ a) 2 t]
Finally,
u(x,y,t) =
å¥ n=1 å¥ m=-oo a nm sin (np/ a) x sin (mp/ b) y cos [cÖ ((mp/ b) 2 + (np/ a) 2 t]
To be continued..
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