2.3. HEAT EQUATION 51

Now

(15) |Jε| ≤ ( ft

L∞

+

D2f

L∞

)

ε

0 Rn

Φ(y, s) dyds ≤ εC,

by the lemma. Integrating by parts, we also ﬁnd

(16)

Iε =

t

ε Rn

(

∂

∂s

− Δy)Φ(y, s) f(x − y, t − s) dyds

+

Rn

Φ(y, ε)f(x − y, t − ε) dy

−

Rn

Φ(y, t)f(x − y, 0) dy

=

Rn

Φ(y, ε)f(x − y, t − ε) dy − K,

since Φ solves the heat equation. Combining (14)–(16), we ascertain

ut(x, t) − Δu(x, t) = lim

ε→0

Rn

Φ(y, ε)f(x − y, t − ε) dy

= f(x, t) (x ∈

Rn,

t 0),

the limit as ε → 0 being computed as in the proof of Theorem 1. Finally

note u(·,t)

L∞

≤ t f

L∞

→ 0.

Solution of nonhomogeneous problem with general initial data. We

can of course combine Theorems 1 and 2 to discover that

(17) u(x, t) =

Rn

Φ(x − y, t)g(y) dy +

t

0

Rn

Φ(x − y, t − s)f(y, s) dyds

is, under the hypotheses on g and f as above, a solution of

(18)

ut − Δu = f in

Rn

× (0, ∞)

u = g on

Rn

× {t = 0}.

2.3.2. Mean-value formula.

First we recall some useful notation from §A.2. Assume U ⊂

Rn

is open

and bounded, and ﬁx a time T 0.