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.