Page 75 - 6099

Z Z ∞ ∂S Z ∞
t
= (x − y, t − s)f(y, s)dyds + lim S(x − y, t − s)f(y, s)dy. (7.27)
0 −∞ ∂t s→t −∞
taking special care due to the singularity of S(x − y, t − s) at t − s = 0. Using the fact
that S(x − y, t − s) satisfies the diffusion equation, we get

Z Z ∞ 2 Z ∞
t
∂u ∂ S
= k (x − y, t − s)f(y, s)dyds + lim S(x − y, ε)f(y, t)dy.
∂t 0 −∞ ∂x 2 ε→0 −∞
Pulling the spatial derivative outside the integral and using the initial condition satisfied
by S, we get

t
∂u ∂ 2 Z Z ∞
= k S(x − y, t − s)f(y, s)dyds + f(x, t) =
∂t ∂x 2 0 −∞
2
∂ u
= k + f(x, t).
∂x 2
This identity is exactly the PDE (7.19). Second, we verify the initial condition. Letting
t → 0, the first term in (7.20) tends to φ(x) because of the initial condition of S. The
second term is an integral from 0 to 0. Therefore,

0
Z
lim u(x, t) = φ(x) + . . . = φ(x).
t→0
0
This proves that (7.20) is the unique solution.
Remembering that S(x, t) is the gaussian distribution (7.7), the formula (7.20) takes
the explicit form

Z Z
t ∞
u(x, t) = S(x − y, t − s)f(y, s)dyds =
0 −∞
Z Z ∞ 1
t
2
= p e −(x−y) /4k(t−s) f(y, s)dyds. (7.28)
0 −∞ 4πk(t − s)
in the case that φ = 0.

7.4 Source on a half-line

For inhomogeneous diffusion on the half-line we can use the method of reflection.
Now consider the more complicated problem of a boundary source h(t) on the half-line;
that is,

v t − kv xx = f(x, t) for 0 < x < ∞, 0 < t < ∞,
v(0, t) = h(t), (7.29)
v(x, 0) = φ(x).

68