280 CHAPTER 11. INTEGRATION ON MANIFOLDS
=∫
∂Ug
∂v∂n−=0
v∂u∂n
dσ − limε→0
∫∂Bε
(u
∂v∂n− v
∂u∂n
)dσ
0 =∫
∂Ug
∂v∂n
dσ − limε→0
∫∂Bε
(u
∂ rn
∂n− rn
∂u∂n
)dσ
Now rx ∂u∂n is bounded uniformly on ∂Bε for small ε and ∂ rx
∂n = −(n−2)|y−x|n−1
y−x|y−x| ·
y−x|y−x| =
−(n−2)|y−x|n−1
0 =∫
∂Ug
∂v∂n
dσ + limε→0
ωn−1
∫∂Bε
u(n−2)
ωn−1εn−1 dσ
=∫
∂Ug
∂v∂n
dσ +(n−2)ωn−1u(x) (11.15)
u(x) =1
−(n−2)ωn−1
∫∂U
g(y)(n−2)
r|x|2− r2
|y−x|ndσ =
∫∂U
g(y)1r
r2−|x|2
|y−x|ndσ
In case n = 2, u(x) =∫
∂U g(y) 1r
r2−|x|2
|y−x|2dσ .
Theorem 11.7.4 Let u ∈ C2(
B(0,r))
satisfy ∆u = 0 and u = g on ∂B(0,r), and
also B(0,r)⊆ Rn. If n≥ 2, u(x) = 1ωn−1
∫∂B(0,r) g(y) 1
rr2−|x|2|y−x|n dσ (y) . If u(x) is given by
this formula, then in fact ∆u = 0 and u = g on ∂B(0,r) in the sense that limx→x0 u(x) =g(x0) and u is infinitely differentiable since all partial derivatives exist.
Proof: I know a solution to ∆u = 0,u = 1 on B(0,r), namely u = 1. It follows from
what was just shown that if n > 2,1 = 1ωn−1
∫∂B(0,r)
1r
r2−|x|2|y−x|n dσ (y) and if n = 2, then it
follows that 1 = 12π
∫∂B(0,r)
1r
r2−|x|2
|y−x|2dσ (y) .
Let n > 2. It is similar if n = 2. Let x0 ∈ ∂B(0,r) and x will be close to x0 and it isdesired to show that |g(x0)−u(x)| is small.
|g(x0)−u(x)| ≤ 1ωn−1r
∫∂B(0,r)
|g(y)−g(x0)|(
r2−|x|2
|y−x|n
)dσ (y)
≤ 1ωn−1r
∫[|y−x0|<δ ]
|g(y)−g(x0)|(
r2−|x|2
|y−x|n
)dσ (y)+
1ωn−1r
∫[|y−x0|≥δ ]
|g(y)−g(x0)|(
r2−|x|2
|y−x|n
)dσ (y) (11.16)
Now since g is continuous, there is a constant M such that |g(y)|< M for all y. Therefore,
|g(x0)−u(x)| <2M
ωn−1r
∫[|y−x0|≥δ ]
(r2−|x|2
|y−x|n
)dσ (y)+
1ωn−1r
∫[|y−x0|<δ ]
|g(y)−g(x0)|(
r2−|x|2
|y−x|n
)dσ (y)