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the origin is zero: φ (0) = 0. If φ(x) is defined only on the half-line, its even extension is
defined to be
(
φ(x), for x ≥ 0,
φ even (x) = (7.17)
+φ(−x), for x ≥ 0.

By the same reasoning as we used above, we end up with an explicit formula for w(x, t).
It is Z
1 ∞ 2 2
w(x, t) = √ (e −(x−y) /4kt + e −(x+y) /4kt )φ(y)dy. (7.18)
4πkt 0
Notice that the only difference between (7.15) and (7.18) is a single minus sign!

Example 7.4 Solve (7.16) with φ(x) ≡ 1. This is the same as Example 7.3 except
for the single sign. So we can copy from that example:

1 1 1 1
u(x, t) = + Erf( √ x ) + − Erf( √ x ) = 1
2 2 4kt 2 2 4kt

(That was stupid: We could have guessed it!)

7.3 Diffusion with a source

In this section we solve the inhomogeneous diffusion equation on the whole line

u t − ku xx = f(x, t) (−∞ < x < ∞, 0 < t < ∞), (7.19)

u(x, 0) = φ(x)

with f(x, t) and φ(x) arbitrary given functions. For instance, if u(x, t) represents the
temperature of a rod, then φ(x) is the initial temperature distribution and f(x, t) is a
source (or sink) of heat provided to the rod at later times. We will show that the solution
of (7.19) is

∞ t ∞
Z Z Z
u(x, t) = S(x − y, t)φ(y)dy + S(x − y, t − s)f(y, s)dyds. (7.20)
−∞ 0 −∞

Notice that there is the usual term involving the initial data φ and another term involving
the source f. Both terms involve the source function S. Let’s begin by explaining
where (7.20) comes from. Later we will actually prove the validity of the formula. Our
explanation is an analogy. The simplest analogy is the ODE

du
+ Au(t) = f(t), u(0) = φ, (7.21)
dt

tA
where A is a constant. Using the integrating factor e , the solution is
Z t
u(t) = e −tA φ + e (s−t)A f(s)ds. (7.22)
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