24 2. FOUR IMPORTANT LINEAR PDE
As the expression on the right-hand side of (10) is continuous in the variable
x, we see u
C2(Rn).
2. Since Φ blows up at 0, we will need for subsequent calculations to
isolate this singularity inside a small ball. So fix ε 0. Then
(11)
Δu(x) =
B(0,ε)
Φ(y)Δxf(x y) dy +
Rn−B(0,ε)
Φ(y)Δxf(x y) dy
=: + Jε.
Now
(12) |Iε| C
D2f
L∞(Rn)
B(0,ε)
|Φ(y)| dy
Cε2|
log ε| (n = 2)
Cε2
(n 3).
An integration by parts (see §C.2) yields
(13)
=
Rn−B(0,ε)
Φ(y)Δyf(x y) dy
=
Rn−B(0,ε)
DΦ(y) · Dyf(x y) dy
+
∂B(0,ε)
Φ(y)
∂f
∂ν
(x y) dS(y)
=: + Lε,
ν
denoting the inward pointing unit normal along ∂B(0,ε). We readily check
(14) |Lε| Df
L∞(Rn)
∂B(0,ε)
|Φ(y)| dS(y)
Cε| log ε| (n = 2)
(n 3).
3. We continue by integrating by parts once again in the term Kε, to
discover
=
Rn−B(0,ε)
ΔΦ(y)f(x y) dy
∂B(0,ε)
∂Φ
∂ν
(y)f(x y) dS(y)
=
∂B(0,ε)
∂Φ
∂ν
(y)f(x y) dS(y),
since Φ is harmonic away from the origin. Now DΦ(y) =
−1
nα(n)
y
|y|n
(y = 0)
and
ν
=
−y
|y|
=
y
ε
on ∂B(0,ε). Consequently
∂Φ
∂ν
(y) =
ν
·DΦ(y) =
1
nα(n)εn−1
on ∂B(0,ε). Since
nα(n)εn−1
is the surface area of the sphere ∂B(0,ε), we
have
(15)
=
1
nα(n)εn−1
∂B(0,ε)
f(x y) dS(y)
= −−
∂B(x,ε)
f(y) dS(y) −f(x) as ε 0.
Previous Page Next Page