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.
