Page 641 - Advanced_Engineering_Mathematics o'neil

17.2 The Heat Equation on [0, L] 621

Substitute this into equation (17.9) to get
2
2
n π k

T =− T n (t) + B n (t).
n 2
L
For n = 1,2,···, this is a first-order differential equation for T n (t), which we write as
2
2
n π k

T (t) + T n (t) = B n (t).
n 2
L
Next, use equation (17.5) to obtain the condition
2 L nπξ 2 L nπξ
T n (0) = u(ξ,0)sin dξ = f (ξ)sin dξ = b n
L 0 L L 0 L
which is the nth coefficient in the Fourier sine expansion of f on [0, L]. Solve the differential
equation for T n (t) subject to this boundary condition to get
t

2 2
2 2
T n (t) = e −n π k(t−τ)/L 2 B n (τ)dτ + b n e −n π kt/L 2 .
0
Finally, substitute this into equation (17.4) to obtain the solution
∞ t
2 2 2 nπx
u(x,t) = e −n π k(t−τ)/L B n (τ)dτ sin
L
0
n=1
2 L nπξ nπx 2 2 2
∞
+ f (ξ)sin dξ sin e −n π kt/L . (17.10)
L 0 L L
n=1
Notice that the last term is the solution of the problem if F(x,t) = 0, while the first term is
the effect of this source term on the solution.
EXAMPLE 17.4
We will solve the problem
2
∂u ∂ u
= 4 + xt for 0 < x <π,t > 0,
∂t ∂x 2
u(0,t) = u(π,t) = 0for t ≥ 0,
and

20 for 0 ≤ x ≤ π/4
u(x,0) = f (x) =
0 for π/4 < x ≤ π.
Since we have a formula (17.10) for the solution, we need only carry out the integrations. First
compute
2 π 2(−1) n+1
B n (t) = ξt sin(nξ)dξ = t.
π 0 n
This enables us to evaluate
t 2 t 2(−1) n+1 2
e −4n (t−τ) B n (τ)dτ = τe −4n (t−τ) dτ
n
0 0
2
2
1 −1 + 4n t + e −4n t
= (−1) n+1 .
8 n 5

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

October 14, 2010 15:25 THM/NEIL Page-621 27410_17_ch17_p611-640