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630 CHAPTER 17 The Heat Equation

17.3.4 Solution by Fourier Sine Transform
We also can solve this problem on the half-line using the Fourier sine transform. Take this
transform of the heat equation with respect to x to get
∂u 2 2
ˆ u S (ω,t) =−ω k ˆu S (ω,t) + ωku(0,t) =−ω k ˆu S (ω,t).
∂t
This has general solution
2
ˆ u S (ω,t) = b ω e −ω kt .
Furthermore,
ˆ
ˆ u S (ω,0) = f S (ω) = b ω ,
so
2
ˆ u S (ω,t) = f S (ω)e −ω kt .
ˆ
Upon taking the inverse Fourier sine transform, we obtain the solution
2 ∞ −ω kt
2
ˆ
u(x,t) = f S (ω)e sin(ωx)dω.
π 0
It is a calculation along lines we have done previously to insert the integral for f S (ω) into this
ˆ
expression to obtain the solution by separation of variables.

SECTION 17.3 PROBLEMS

In each of Problems 1 through 4, solve the problem 5. f (x) = e −αx with α any positive constant.
2
∂u ∂ u 6. f (x) = xe −αx with α any positive constant.
= k for −∞ < x < ∞,t > 0
∂t ∂x 2
1 for 0 ≤ x ≤ h
and 7. f (x) = with h any positive number.
0 for x > h
u(x,0) = f (x) for −∞ < x < ∞.

x for 0 ≤ x ≤ 2
8. f (x) =
1. f (x) = e −4|x| 0 for x > 2

sin(x) for |x|≤ π
2. f (x) =
0 for |x| >π In each of Problems 9 and 10, use a Fourier transform on
the half-line to solve the problem.

x for 0 ≤ x ≤ 4
3. f (x) =
0 for x < 0and for x > 4
2
∂u ∂ u
9.
= 2 − tu for x > 0,t > 0,
e −x for |x|≤ 1 ∂t ∂x
4. f (x) =
0 for |x| > 1 u(0,t)= 0for t > 0,
u(x,0)= xe −x for x > 0.
In each of Problems 5 through 8, solve the problem
2
2
∂u ∂ u 10. ∂u ∂ u − u for x > 0,t > 0,
= k for x > 0,t > 0, = 2
∂t ∂x 2 ∂t ∂x
u(0,t)= 0for t > 0,
u(0,t) = 0for t > 0,
∂u
u(x,0) = f (x) for x > 0. (0,t)= f (t) for t > 0.
∂x
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October 14, 2010 15:25 THM/NEIL Page-630 27410_17_ch17_p611-640