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