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This is one of the few fortunate examples that can be integrated. The exponent is
2
2
x − 2xy + y + 4kty
− .
4kt
Completing the square in the y variable, it is
(y + 2kt − x) 2
− + kt − x.
4kt
√ √
We let p = (y + 2kt − x)/ 4kt so that dp = dy/ 4kt. Then
2 dp
u(x, t) = e kt−x ]int ∞ e −p √ = e kt−x .
−∞
π
7.2 Diffusion on the half-line

Let’s take the domain to be D = the half-line (0, ∞) and take the Dirichlet boundary
condition at the single endpoint x = 0. So the problem is

v t − kv xx = 0 in {0 < x < ∞, 0 < t < ∞}, (7.10)
v(x, 0) = φ(x) for t = 0,
v(0, t) = 0 for x = 0.

The PDE is supposed to be satisfied in the open region {0 < x < ∞, 0 < t < ∞}. If it
exists, we know that the solution v(x, t) of this problem is unique. It can be interpreted,
for instance, as the temperature in a very long rod with one end immersed in a reservoir
of temperature zero and with insulated sides.
We are looking for a solution formula analogous to (7.8). In fact, we shall reduce our
new problem to our old one. Our method uses the idea of an odd function. Any function
φ(x) that satisfies φ(−x) = −φ(+x) is called an odd function. Its graph y = φ(x) is
symmetric with respect to the origin. Automatically (by putting x = 0 in the definition),
φ(0) = 0.
Now the initial datum φ(x) of our problem is defined only for x ≥ 0. Let φ odd be the
unique odd extension of φ to the whole line. That is


φ(x), for x > 0,

φ odd (x) = −φ(−x), for x < 0, (7.11)


0, for x = 0.
Let u(x, t) be the solution of

u t − ku xx = 0, (7.12)
u(x, 0) = φ odd (x)

for the whole line −∞ < x < ∞, 0 < t < ∞. It is given by the formula
∞
Z
u(x, t) = S(x − y, t)φ odd (y)dy. (7.13)
−∞

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