50 2. FOUR IMPORTANT LINEAR PDE
THEOREM 2 (Solution of nonhomogeneous problem). Define u by (13).
Then
(i) u ∈ C1
2(Rn
× (0, ∞)),
(ii) ut(x, t) − Δu(x, t) = f(x, t) (x ∈
Rn,
t 0),
and
(iii) lim
(x,t)→(x0,0)
x∈Rn,
t0
u(x, t) = 0 for each point
x0
∈
Rn.
Proof. 1. Since Φ has a singularity at (0, 0), we cannot directly justify
differentiating under the integral sign. We instead proceed somewhat as in
the proof of Theorem 1 in §2.2.1.
First we change variables, to write
u(x, t) =
t
0 Rn
Φ(y, s)f(x − y, t − s) dyds.
As f ∈ C1
2(Rn
× [0, ∞)) has compact support and Φ = Φ(y, s) is smooth
near s = t 0, we compute
ut(x, t) =
t
0 Rn
Φ(y, s)ft(x − y, t − s) dyds
+
Rn
Φ(y, t)f(x − y, 0) dy
and
uxixj (x, t) =
t
0 Rn
Φ(y, s)fxixj (x − y, t − s) dyds (i, j = 1,...,n).
Thus ut,Dxu,
2
and likewise u, Dxu, belong to
C(Rn
× (0, ∞)).
2. We then calculate
(14)
ut(x, t) − Δu(x, t) =
t
0
Rn
Φ(y, s)[(
∂
∂t
− Δx)f(x − y, t − s)] dyds
+
Rn
Φ(y, t)f(x − y, 0) dy
=
t
ε Rn
Φ(y, s)[(−
∂
∂s
− Δy)f(x − y, t − s)] dyds
+
ε
0
Rn
Φ(y, s)[(−
∂
∂s
− Δy)f(x − y, t − s)] dyds
+
Rn
Φ(y, t)f(x − y, 0) dy.
=: Iε + Jε + K.
