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We may use the following subtraction device to reduce (7.29) to a simpler problem. Let
V (x, t) = v(x, t) − h(t). Then V (x, t) will satisfy

0
V t − kV xx = f(x, t) − h (t) for 0 < x < ∞, 0 < t < ∞,
V (0, t) = 0, (7.30)
V (x, 0) = φ(x) − h(0). (7.31)

To verify (7.30), just subtract! This new problem has a homogeneous boundary condition
to which we can apply the method of reflection. Once we find V , we recover v by
v(x, t) = V (x, t) + h(t). This simple subtraction device is often used to reduce one linear
problem to another.The domain of independent variables (x, t) in this case is a quarter-
plane with specified conditions on both of its half-lines. If they do not agree at the corner
[i.e., if φ(0) 6= h(0)], then the solution is discontinuous there (but continuous everywhere
else). This is physically sensible . Think for instance, of suddenly at t = 0 sticking a hot
iron bar into a cold bath.
For the inhomogeneous Neumann problem on the half-line,

w t − kw xx = f(x, t) for 0 < x < ∞, 0 < t < ∞,
w x (0, t) = h(t), (7.32)
w(x, 0) = φ(x), (7.33)

we would subtract off the function xh(t). That is, W(x, t) = w(x, t) − xh(t). Differen-
tiation implies that W x (0, t) = 0.

7.5 Diffusion revisited

In this section we make a careful mathematical analysis of the solution of the diffusion
equation that we found in previous sections. (On the other hand, the formula for the
solution of the wave equation is so much simpler that it does not require a special justi-
fication.)
The solution formula for the diffusion equation is an example of a convolution, the
convolution of φ with S (at a fixed t). It is
Z ∞ Z ∞
u(x, t) = S(x − y, t)φ(y)dy = S(z, t)φ(x − z)dz, (7.34)
−∞ −∞
√ 2 √
where S(z, t) = 1/ 4πkte −z /4kt . If we introduce the variable p = z/ kt, it takes the
equivalent form
Z ∞
1 −p /4 √
2
u(x, t) = √ e φ(x − p kt)dp. (7.35)
4π −∞
Now we are prepared to state a precise theorem.

Theorem 7.5.1 Let φ(x) be a bounded continuous function for−∞ <
x < ∞. Then the formula (7.35) defines an infinitely differentiable
function u(x, t) for −∞ < x < ∞, 0 < t < ∞, which satisfies the

equation u = ku xx and lim t→+0 u(x, t) = φ(x) for each x.
t

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