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Its restriction,
v(x, t) = u(x, t) for x > 0, (7.14)

will be the unique solution of our new problem (7.10). There is no difference at all between
v and u except that the negative values of x are not considered when discussing v.
Why is v(x, t) the solution of (7.10)? Notice first that u(x, t) must also be an odd
function of x. That is, u(−x, t) = −u(x, t). Putting x = 0, it is dear that u(0, t) = 0.
So the boundary condition v(0, t) = 0 is automatically satisfied! Furthermore, v solves
the PDE as well as the initial condition for x > 0, simply because it is equal to u for
x > 0 and u satisfies the same PDE for all x and the same initial condition for x > 0.
The explicit formula for v(x, t) is easily deduced from (7.13) and (7.14). From (7.13) and
(7.11) we have

∞ 0
Z Z
u(x, t) = S(x − y, t)φ(y)dy − S(x − y, t)φ(−y)dy.
0 −∞
Changing the variable −y to +y in the second integral, we get

Z ∞
u(x, t) = S(x − y, t) − S(x + y, t)φ(y)dy.
0
(Notice the change in the limits of integration.) Hence for 0 < x < ∞, 0 < t < ∞, we
have Z
1 ∞ 2 2
v(x, t) = √ (e −(x−y) /4kt − e −(x+y) /4kt )φ(y)dy. (7.15)
4πkt 0

This is the complete solution formula for (7.10). We have just carried out the method
of odd extensions or reflection method, so called because the graph of φ odd (x) is the
reflection of the graph of φ(x) across the origin.

Example 7.3 Solve (7.10) with φ(x) ≡ 1. The solution is given by formula (7.15).
√
This case can be simplified as follows. Let p = (x − y)/ 4kt in the first integral and
√
q = (x + y)/ 4kt in the second integral. Then
√
Z x/ 4kt √ Z +∞ √
u(x, t) = e −p 2 dp/ π − e −q 2 dq/ π =
√
−∞ x/ 4kt

1 1 1 1
= + Erf( √ x ) − − Erf( √ x ) = Erf( √ x ).
2 2 4kt 2 2 4kt 4kt

Now let’s play the same game with the Neumann problem

w t − kw xx = 0 for 0 < x < ∞, 0 < t < ∞,
w(x, 0) = φ(x), (7.16)
w x (0, t) = 0.

In this case the reflection method is to use even, rather than odd, extensions. An even
function is a function φ such that φ(−x) = +φ(x). If φ is an even function, then
differentiation shows that its derivative is an odd function. So automatically its slope at

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