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A more elaborate analogy is the following. Let’s suppose that φ is an n-vector, u(t) is an
n-vector function of time, and A is a fixed n×n matrix. Then (7.21) is a coupled system
of n linear ODEs. In case f(t) ≡ 0, the solution of (7.21) is given as u(t) = S(t)φ, where
S(t) is the matrix S(t) = e −tA . So in case f(t) 6= 0, an integrating factor for (7.21) is
tA
S(−t) = e . Now we multiply (7.21) on the left by this integrating factor to get

d du
(S(−t)u(t)) = S(−t) + S(−t)Au(t) = S(−t)f(t).
dt dt
Integrating from 0 to t, we get

Z
t
S(−t)u(t) − φ = S(−s)f(s)ds.
0

Multiplying this by S(t), we end up with the solution formula

Z t
u(t) = S(t)φ + S(t − s)f(s)ds. (7.23)
0

The first term in (7.23) represents the solution of the homogeneous equation, the second
the effect of the source f(t). For a single equation, of course, (7.23) reduces to (7.22).
Now let’s return to the original diffusion problem (7.19). There is an analogy between
(7.20) and (7.23) which we now explain. The solution of (7.19) will have two terms.
The first one will be the solution of the homogeneous problem, already solved in previous
section, namely
Z
∞
S(x − y, t)φ(y)dy = (L(t)φ)(x). (7.24)
−∞
S(x − y, t) is the source function given by the formula (7.7). Here we are using L(t) to
denote the source operator, which transforms any function φ to the new function given
by the integral in (7.23). (Remember: Operators transform functions into functions.) We
can now guess what the whole solution to (7.19) must be. In analogy to formula ((7.21),
we guess that the solution of (7.19) is

t
Z
u(t) = L(t)φ + L(t − s)f(s)ds. (7.25)
0

Formula (7.25) is exactly the same as (7.20). The method we have just used to find
formula (7.20) is the operator method.
Proof of (7.20). All we have to do is verify that the function u(x, t), which is defined
by (7.20), in fact satisfies the PDE and IC (7.19). Since the solution of (7.19) is unique,
we would then know that u(x, t) is that unique solution. For simplicity, we may as well
let φ ≡ 0, since we understand the φ term already.
We first verify the PDE. Differentiating (7.20), assuming φ ≡ 0 and using the rule for
differentiating integrals, we have

Z Z ∞
t
∂u ∂
= S(x − y, t − s)f(y, s)dyds = (7.26)
∂t ∂t 0 −∞

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