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√ √
by substituting q = x/ 4kt, dq = (dx)/ 4kt. Now look more carefully at the sketch of
(x, t) for a very small t. If we cut out the tall spike, the rest of (x, t) is very small. Thus

max S(x, t) → 0
|x|>δ

Notice that the value of the solution u(x, t) is a kind of weighted average of the initial
values around the point x. Indeed, letting z = x − y in the integral (7.6), we can write
Z ∞
X
u(x, t) = S(x − y, t)φ(y)dy = S(x − y i , t)φ(y i )∆y i
−∞ i
approximately. This is the average of the solutions S(x − y i , t) with the weights φ(y i ).
For very small t, the source function is a spike so that the formula exaggerates the values
of φ near x. For any t > 0 the solution is a spread-out version of the initial values at
t = 0.
Here’s the physical interpretation. Consider diffusion. S(x−y, t) represents the result
of a unit mass (say, 1 gram) of substance located at time zero exactly at the position
y which is diffusing (spreading out) as time advances. For any initial distribution of
concentration, the amount of substance initially in the interval ∆y spreads out in time
and contributes approximately the term S(x − y i , t)φ(y i )∆y i . All these contributions are
added up to get the whole distribution of matter. Now consider heat flow. S(x − y, t)
represents the result of a ”hot spot” at y at time 0. The hot spot is cooling off and
spreading its heat along the rod.
Another physical interpretation is brownian motion, where particles move randomly in
space. For simplicity, we assume that the motion is one-dimensional; that is, the particles
move along a tube. Then the probability that a particle which begins at position x ends
R b
up in the interval (a, b) at time t is precisely S(x − y, t)dy for some constant k. In
a
other words, if we let u(x, t) be the probability density (probability per unit length) and
if the initial probability density is φ(x), then the probability at all later times is given by
formula (7.6). That is, u(x, t) satisfies the diffusion equation.
It is usually impossible to evaluate the integral (7.8) completely in terms of elementary
functions. Answers to particular problems, that is, to particular initial data φ(x), are
sometimes expressible in terms of the error function of statistics,
Z x
2 −p 2
Erf(x) = √ e dp. (7.9)
π 0
Notice that Erf(0) = 0 and lim Erf(x) = 1.
x→+∞

Example 7.1 From (7.5) we can write Q(x, t) in terms of Erf as

1 1
Q(x, t) = + Erf( √ x ).
2 2 4kt
−x
Example 7.2 Solve the diffusion equation with the initial condition u(x, O) = e .
To do so, we simply plug this into the general formula (7.8):
Z ∞
1 −(x−y) /4kt −y
2
u(x, t) = √ e e dy.
4πkt −∞

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