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17.3 Solutions in an Infinite Medium 629

17.3.3 Problems on the Half-Line
We will solve the problem

2
∂u ∂ u
= k for 0 < x < ∞,t > 0,
∂t ∂x 2
u(0,t) = 0for t > 0,
and

u(x,0) = f (x) for 0 < x < ∞.
for the half-line, taking the case that the left end at x = 0 is kept at zero temperature. Putting
u(x,t) = X(x)T (t) leads in the usual way to
X + λX = 0; X(0) = 0

and
T + kλT = 0.

2
The eigenvalues are λ = ω with ω> 0 and eigenfunctions b ω sin(ωx). With these values of λ,
2
T (t) is a constant multiple of e −ω kt . For each ω> 0, the function
2
u ω (x,t) = b ω sin(ωx)e −ω kt
satisfies the heat equation and the boundary condition. For the initial condition, write
∞ 2

u(x,t) = b ω sin(ωx)e −ω kt dω.
0
Then
∞

u(x,0) = f (x) = b ω sin(ωx)dω
0
requires that we choose the coefficients as the Fourier integral sine coefficients of f on [0,∞):

2 ∞
b ω = f (ξ)sin(ωξ)dξ.
π 0

EXAMPLE 17.8
Suppose the initial temperature function is

π − x for 0 ≤ x ≤ π
f (x) =
0 for x >π.
Compute the coefficients

2 π 2 πω − sin(πω)
b ω = (π − ξ)sin(ωξ)dξ = .
π 0 ω ω 2
The solution is
2 ∞ πω − sin(πω) 2
u(x,t) = sin(ωx)e −ω kt dω.
π 0 ω 2

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October 14, 2010 15:25 THM/NEIL Page-629 27410_17_ch17_p611-640