Lecture 10
Interior C 2,α estimate for Newtonian potential (continued)
Proposition 1 Consider
B1 = BR (x0 ) ⊂ B2R (x0 ) = B2 , f ∈ C α (B2R ), where 0 <
�
α < 1. Let ω(x) = B2 Γ(x − y)f (y)dy, the Newtonian Potential of f in B2 . Then
ω ∈ C 2,α (BR (x0 )) and we have estimate
�D2 ω�0;BR + Rα |D2 ω |α;BR ≤ C(�f �0;B2R + Rα |f |α;B2R ),
where C = C(n, α) is constant.
Proof:(continued)
�
(V I) = (f (x) − f (x))
Dij Γ(x − y)dy
B2 \Bδ (ξ)
�
≤ |f |C α (x) |x − x|α |
Di Γ(x − y)νj dsy |
∂(B2 \Bδ (ξ))
�
�
α
Di Γ(x − y)νj dsy |)
≤ |f |C α (x) δ (|
Di Γ(x − y)νj dsy | + |
∂Bδ (ξ)
∂B2
�
�
1
1
δ
1
α
≤ c|f |C α (x) δ (
dsy +
dsy )
( ≤ |y − ξ | ≤ |y − x|)
n−1
n−1
2
2
∂B2 |x − y |
∂Bδ (ξ) |x − y|
1
1
nωn (δ)n−1 )
≤ c|f |C α (x) (δ α n−1 nωn (2R)n−1 +
R
(δ/2)n−1
≤ c|f |C α (x) δ α .
�
(Dij Γ(x − y) − Dij Γ(x − y))(f (y) − f (x))dy
(V ) =
B2 \Bδ (ξ)
�
|DDij Γ(ˆ
x − y)||x − x||f |C α (x) |x − y |α
dy
�
1
≤ cδ |f |C α (x)
|x − y|α dy
x − y |n+1
|y−ξ|≥δ |ˆ
≤
B2 \Bδ (ξ)
Since
|x − y | ≤ |x − ξ | + |ξ − y | ≤
δ
3
+ |ξ − y | ≤ |ξ − y |
2
2
and
|y − ξ | ≤ |y − x|
ˆ + |x
ˆ − ξ | ≤ |y − x|
ˆ +
1
ˆ
=⇒ |y − ξ | ≤ |y − x|,
2
1
δ
1
≤ |y − x|
ˆ + |y − ξ |
2
2
We thus get
�
dy
≤ cδ|f |C α (x)
|y − ξ|
n+1−α
(V ) ≤ cδ |f |C α (x)
|y−ξ|≥δ
3R
α−2
�
≤ cδ |f |C α (x)
r
dr ≤ cδ |f |C α (x)
δ
�
δ
3R
1
rn−1 dr
rn+1−α
1
((3R)α−1 − δ α−1 )
α−1
c
≤
|f | α δ α .
1 − α C (x)
Combine all the results, we have shown
|Dij ω(x) − Dij ω(x)| ≤ C(
thus
Rα
|f (x)|
+ |f |C α (x) + |f |C α (x) )|x − x|α ,
Rα
|Dij ω(x) − Dij ω(x)|
≤ C(|f (x)| + Rα |f |C α (x) + Rα |f |C α (x) ),
|x − x|α
i.e.
Rα |D2 ω |α;BR ≤ C(�f �C 0 ;B2R + Rα |f |C α (B2R ) ).
Since we have already known that
|D2 ω|0 ≤ c(�f �C 0 + Rα |f |α ),
we finally get
�D2 ω�0;BR + Rα |D2 ω |α;BR ≤ C(�f �0;B2R + Rα |f |α;B2R ).
�
Exercise:
1) Find a continuous function f s.t. Δu = f does not have a C 2 solution.
2) Find g ∈ C 1 and Δu = g but u is not C 2,1 .
Interior C 2,α estimates for Poisson’s equation.
Application: u ∈ C 2 (B2R (x0 )), Δu = f, f ∈ C α (B2R (x0 )).
Theorem 1
�u�C 2,α (BR ) ≤
C
(�u�C 0 (B2R ) + �f �C α (B2R ) ).
Rα
Proof: Since Δ(u − N f ) = 0, where N f is the Newtonian potential of f , thus from
the C 2,α estimate of N f we can get the C 2,α of u.
�
2
Theorem 2 Let u ∈ C 2 (Ω), Δu = f, f ∈ C α (Ω), then for any Ω� ⊂⊂ Ω, we have
�u�C 2,α (Ω� ) ≤ C(�u�C 0 (Ω) + �f �C α (Ω) ).
Proof: Take x ∈ Ω� , choose R s.t.B2R ⊂ Ω, then
|u|C 2,α (x) ≤ |u|C 2,α (BR (x)) ≤
≤
C
(�u�C 0 ;B2R + |f |C α (B2R ) )
Rα
C
(�u�C 0 ;Ω + |f |C α (Ω) ).
Rα
Taking superior over all x and using previous estimate, we get
�u�C 2,α (Ω� ) ≤ C(�u�C 0 (Ω) + �f �C α (Ω) ).
�
Boundary estimate on Newtonian potential: C 2,α estimate up to the bound­
ary for domain with flat boundary portion.
Suppose BR ⊂ B2R , with flat boundary portion B1+ ⊂ B2+ .
�
Lemma 1 Let f ∈ C α (B2+ ), ω(x) = B + Γ(x − y)f (y)dy be the Newtonian potential of
2
f in B2+ . Then ω ∈ C 2,alpha (B1+ ) and
|D2 ω |0;B1 + Rα |D2 ω |α;B + ≤ C(|f |0;B2 + Rα |f |α;B + ).
2
1
Proof: Examine the proof form last time. From C 2 estimate have
�
�
Dij ω(x) =
Dij Γ(x − y)(f (y) − f (x))dy − f (x)
Di Γ(x − y)νj (y)dsy .
B2+
(Note
�
∂B2+
∂B2+
Di Γ(x − y)νj (y)dsy =
�
∂B2+
Dj Γ(x − y)νi (y)dsy .)
So if either i �= n or j =
� n, the integral on the lower boundary portion of B2+ vanishes.
(Since ν = (0, 0, · · · , −1).)
If i = j = n, then
�
Dnn ω = f (x)
(Dn Γ(x − y) − Dn Γ(x − y))(−1)dσ
∂B2+
�
≤ |f (x)|
+
∂B2
|DDn Γ(ˆ
x − y)||x − x|dσ
�
≤ |f (x)|δ
∂B2+
1
dσ
|ˆ
x − y |n
1
nωRn−1
Rn
≤ c|f (x)|δ α .
≤ |f (x)|δ
Since we know Δω = f , thus ωnn = f − ω11 − ω22 − · · · , so we see that ωnn ∈ C α ,
and we can get estimate for ωnn from estimates for ωii , i < n.
�
3