Page 659 - Advanced_Engineering_Mathematics o'neil

17.6 Heat Conduction in a Rectangular Plate 639

satisfying the heat equation and the boundary conditions. Reasoning as we did with the two-
dimensional wave equation, we get
1
1
2
c nm = 4 ξ(1 − ξ )η(1 − η)sin(nπξ)sin(mπη)dξ dη
0 0
(−1) (−1) − 1
n m
= 48 .
3
3
n π 3 m π 3
The solution is
u(x, y,t) =
∞
m
∞
48 (−1) n (−1) − 1 2 2 2
sin(nπx)sin(mπy)e −(n +m )π kt .
π 6 n 3 m 3
n=1 m=1
SECTION 17.6 PROBLEMS

1. Write the solution for the general problem and
2
2
∂u ∂ u ∂ u u(x, y,0) = f (x, y) for 0 ≤ x ≤ L,0 ≤ y ≤ K.
= k + for 0 < x < L,
∂t ∂x 2 ∂y 2
2. Solve this problem when k = 4, L = 2, K = 3, and
2
0 < y < K,t > 0, f (x, y) = x (L − x)sin(y)(K − y).
3. Solve this problem when k = 1, L = π, K = π,and
u(x,0,t) = u(x, K,t) = 0for 0 < x < L,t > 0,
f (x, y) = sin(x)cos(y/2).
u(0, y,t) = u(L, y,t) = 0for0 < y < K,t > 0,

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October 14, 2010 15:25 THM/NEIL Page-639 27410_17_ch17_p611-640