2.3. HEAT EQUATION 47
A different derivation of the fundamental solution of the heat equation
appears in §4.3.1.
b. Initial-value problem. We now employ Φ to fashion a solution to the
initial-value (or Cauchy) problem
(8)
ut − Δu = 0 in
Rn
× (0, ∞)
u = g on
Rn
× {t = 0}.
Let us note that the function (x, t) → Φ(x, t) solves the heat equation
away from the singularity at (0, 0), and thus so does (x, t) → Φ(x − y, t) for
each fixed y ∈
Rn.
Consequently the convolution
(9)
u(x, t) =
Rn
Φ(x − y, t)g(y) dy
=
1
(4πt)n/2
Rn
e−
|x−y|
2
4t
g(y) dy (x ∈
Rn,
t 0)
should also be a solution.
THEOREM 1 (Solution of initial-value problem). Assume g ∈
C(Rn)
∩
L∞(Rn),
and define u by (9). Then
(i) u ∈
C∞(Rn
× (0, ∞)),
(ii) ut(x, t) − Δu(x, t) = 0 (x ∈
Rn,
t 0),
and
(iii) lim
(x,t)→(x0,0)
x∈Rn,
t0
u(x, t) =
g(x0)
for each point
x0
∈
Rn.
Proof. 1. Since the function
1
tn/2
e−
|x|
2
4t
is infinitely differentiable, with uni-
formly bounded derivatives of all orders, on
Rn
× [δ, ∞) for each δ 0, we
see that u ∈
C∞(Rn
× (0, ∞)). Furthermore
(10)
ut(x, t) − Δu(x, t) =
Rn
[(Φt − ΔxΦ)(x − y, t)]g(y) dy
= 0 (x ∈
Rn,
t 0),
since Φ itself solves the heat equation.
2. Fix
x0
∈
Rn,
ε 0. Choose δ 0 such that
(11) |g(y) −
g(x0)|
ε if |y −
x0|
δ, y ∈
Rn.
