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17.2 The Heat Equation on [0, L] 615

EXAMPLE 17.2
Suppose the ends of the bar are insulated and the left half of the bar is initially at constant
temperature A while the right half is initially at temperature zero. Then

A for 0 ≤ x ≤ L/2
f (x) =
0 for L/2 < x ≤ L.
Compute
2 L/2
c 0 = Adξ = A
L 0
and, for n = 1,2,···,
2 L/2 nπξ 2A
c n = A cos dξ = sin(nπ/2).
L 0 L nπ
The solution is
∞
A 2A 1 nπ nπx 2 2 2
u(x,t) = + sin cos e −n π kt/L .
2 π n 2 L
n=1
Since sin(nπ/2)=0if n is even, we can retain only odd n in this summation to write this solution
as
∞
A 2A 1 (2n − 1)πx −(2n−1) π kt/L 2
2 2
u(x,t) = + cos e .
2 π 2n − 1 L
n=1
17.2.3 Radiating End
Suppose the left end of the bar is maintained at temperature zero, while the right end radiates
energy into the surrounding medium, which is kept at temperature zero. If the initial temperature
function is f , then the temperature distribution is modeled by the initial-boundary value problem
2
∂u ∂ u
= k for 0 < x < L,t > 0,
∂t ∂x 2
∂u
u(0,t) = 0, (L,t) =−Au(L,t) for t > 0,
∂x
and
u(x,0) = f (x) for 0 ≤ x ≤ L.
A is a positive constant called the transfer coefficient.Let u(x,t) = X(x)T (t) to obtain

X + λX = 0, T + λkT = 0.

Because u(0,t) = X(0)T (t) = 0, then X(0) = 0. From the radiation condition at the right end,

X (L)T (t) =−AX(L)T (t) for t > 0,
so

X (L) + AX(L) = 0.
The problem for X is

X + λX = 0; X(0) = 0, X (L) + AX(L) = 0.

This is a regular Sturm-Liouville problem. To solve for the eigenvalues and eigenfunctions,
consider cases on λ.

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October 14, 2010 15:25 THM/NEIL Page-615 27410_17_ch17_p611-640