48 2. FOUR IMPORTANT LINEAR PDE
Then if |x
x0|
δ
2
, we have, according to the lemma,
|u(x, t)
g(x0)|
=
Rn
Φ(x y, t)[g(y)
g(x0)]
dy

B(x0,δ)
Φ(x y, t)|g(y)
g(x0)|
dy
+
Rn−B(x0,δ)
Φ(x y, t)|g(y)
g(x0)|
dy
=: I + J.
Now
I ε
Rn
Φ(x y, t) dy = ε,
owing to (11) and the lemma. Furthermore, if |x
x0|

δ
2
and |y
x0|
δ,
then
|y
x0|
|y x| +
δ
2
|y x| +
1
2
|y
x0|.
Thus |y x|
1
2
|y
x0|.
Consequently
J 2 g
L∞
Rn−B(x0,δ)
Φ(x y, t) dy

C
tn/2
Rn−B(x0,δ)
e−
|x−y|2
4t
dy

C
tn/2
Rn−B(x0,δ)
e−
|y−x0|2
16t
dy
= C
Rn−B(0,δ/

t)
e−
|z|2
16
dz 0 as t
0+.
Hence if |x
x0| δ
2
and t 0 is small enough, |u(x, t)
g(x0)|
2ε.
Interpretation of fundamental solution. In view of Theorem 1 we
sometimes write
Φt ΔΦ = 0 in
Rn
× (0, ∞)
Φ = δ0 on
Rn
× {t = 0},
δ0 denoting the Dirac measure on
Rn
giving unit mass to the point 0.
Infinite propagation speed. Notice that if g is bounded, continuous,
g 0, g 0, then
u(x, t) =
1
(4πt)n/2
Rn
e−
|x−y|2
4t
g(y) dy
Previous Page Next Page