Section 1.5: Energy Method
1.5 Energy Method
In this section we discuss harmonic functions by using the energy method. In general
we assume throughout this section that a
ij
∈ C(B
1
) satises
λ|ξ|
2
≤ a
ij
(x)ξ
i
ξ
j
≤ Λ|ξ|
2
for any x ∈ B
1
and ξ ∈ R
n
for some positive constants λ and Λ. We consider the function u ∈ C
1
(B
1
) satisfying
Z
B
1
a
ij
D
i
uD
j
φ = 0 for any φ ∈ C
1
0
(B
1
).
It is easy to check by integration by parts that the harmonic functions satisfy the
above equation for a
ij
= δ
ij
.
Lemma 1.36 (Caccioppolli’s Inequality). Suppose u ∈ C
1
(B
1
) satises
Z
B
1
a
ij
D
i
uD
j
φ = 0 for any φ ∈ C
1
0
(B
1
).
Then for any function η ∈ C
1
0
(B
1
), we have
Z
B
1
η
2
|Du|
2
≤ C
Z
B
1
|Dη|
2
u
2
where C is a positive constant depending only on λ and Λ.
Proof. For any η ∈ C
1
0
(B
1
) set φ = η
2
u. Then we have
λ
Z
B
1
η
2
|Du|
2
≤ Λ
Z
B
1
η|u||Dη||Du|.
We obtain the result by the Hölder inequality.
Corollary 1.37. Let u be as in Lemma 1.36. Then for any 0 ≤ r < R ≤ 1 there holds
Z
B
r
|Du|
2
≤
C
(R − r)
2
Z
B
R
u
2
where C is a positive constant depending only on λ and Λ.
Proof. Take η such that η = 1 on B
r
, η = 0 outside B
R
, and |Dη| ≤ 2(R − r)
−1
.
Corollary 1.38. Let u be as in Lemma 1.36. Then for any 0 < R ≤ 1 there hold
Z
B
R/2
u
2
≤ θ
Z
B
R
u
2
and
Z
B
R/2
|Du|
2
≤ θ
Z
B
R
|Du|
2
where θ = θ(n, λ, Λ) ∈ (0, 1).
29
Chapter 1: Harmonic Functions
Proof. Take η ∈ C
1
0
(B
R
) with η = 1 on B
R/2
and |Dη| ≤ 2R
−1
. Then Lemma 1.36
yields
Z
B
R
|D(ηu)|
2
≤ C
Z
B
R
|Dη|
2
u
2
≤
C
R
2
Z
B
R
\B
R
/2
u
2
by noting Dη = 0 in B
R/2
. Hence by the Poincaré inequality we get
Z
B
R
(ηu)
2
≤ c(n)R
2
Z
B
R
|D(ηu)|
2
.
Therefore we obtain
Z
B
R/2
u
2
≤ C
Z
B
R
\B
R/2
u
2
, which implies (C + 1)
Z
B
R/2
u
2
≤ C
Z
B
R
u
2
.
For the second inequality, observe that Lemma 1.36 holds for u−a for arbitrary constant
a. Then as before we have
Z
B
R
η
2
|Du|
2
≤ C
Z
B
R
|Dη|
2
(u − a)
2
≤
C
R
2
Z
B
R
\B
R/2
(u − a)
2
.
The Poincaré inequality implies with a = |B
R
\ B
R/2
|
−1
R
B
R
\B
R/2
u
Z
B
R
\B
R/2
(u − a)
2
≤ c(n)R
2
Z
B
R
\B
R/2
|Du|
2
.
Hence we obtain
Z
B
R/2
|Du|
2
≤ C
Z
B
R
\B
R/2
|Du|
2
;
in particular,
(C + 1)
Z
B
R/2
|Du|
2
≤ C
Z
B
R
|Du|
2
.
Remark 1.39. Corollary 1.38 implies, in particular, that a harmonic function in R
n
with nite L
2
-norm is identically 0 and that a harmonic function in R
n
with nite
Dirichlet integral is constant.
Remark 1.40. By iterating the result in Corollary 1.38, we have the following estimates.
Let u be in Lemma 1.36. Then for any 0 < ρ < r ≤ 1 there hold
Z
B
ρ
u
2
≤ C
ρ
r
µ
Z
B
r
u
2
and
Z
B
ρ
|Du|
2
≤ C
ρ
r
µ
Z
B
r
|Du|
2
for some positive constant µ = µ(n, λ, Λ). Later on we will prove that we can take
µ ∈ (n − 2, n) in the second inequality. For harmonic functions we have better results.
30
Section 1.5: Energy Method
Lemma 1.41. Suppose {a
ij
} is a constant positive denite matrix with
λ|ξ|
2
≤ a
ij
ξ
i
ξ
j
≤ Λ|ξ|
2
for any ξ ∈ R
n
for some constants 0 < λ ≤ Λ. Suppose u ∈ C
1
(B
1
) satises
Z
B
1
a
ij
D
i
uD
j
φ = 0 for any φ ∈ C
1
0
(B
1
). (1.8)
Then for any 0 < ρ ≤ r, there hold
Z
B
ρ
|u|
2
≤ c
ρ
r
n
Z
B
r
|u|
2
, (1.9)
Z
B
ρ
|u − u
ρ
|
2
≤ c
ρ
r
n+2
Z
B
r
|u − u
r
|
2
, (1.10)
where c = c(λ, Λ) is a positive constant and u
r
denotes the average of u in B
r
.
Proof. By dilation, consider r = 1. We restrict our consideration to the range ρ ∈ (0,
1
2
],
since (1.9) and (1.10) are trivial for ρ ∈ (
1
2
, 1].
CLAIM.
kuk
2
L
∞
(B
1/2
)
+ kDuk
2
L
∞
(B
1/2
)
≤ c(λ, Λ)
Z
B
1
|u|
2
.
Therefore for ρ ∈ (0,
1
2
]
Z
B
ρ
|u|
2
≤ ρ
n
kuk
2
L
∞
(B
1/2
)
≤ cρ
n
Z
B
1
|u|
2
and
Z
B
ρ
|u − u
ρ
|
2
≤
Z
B
ρ
|u − u(0)|
2
≤ ρ
n+2
kDuk
2
L
∞
(B
1/2
)
≤ cρ
n+2
Z
B
1
|u|
2
.
If u is a solution of (1.8), so is u − u
1
. With u replaced by u − u
1
in the above
inequality, there holds
Z
B
ρ
|u − u
ρ
|
2
≤ cρ
n+2
Z
B
1
|u − u
1
|
2
.
Proof. We present two methods.
METHOD 1. By rotation, we may assume {a
ij
} is a diagonal matrix. Hence (1.8)
becomes
n
X
i=1
λ
i
D
ii
u = 0
31
Chapter 1: Harmonic Functions
with 0 < λ ≤ λ
i
≤ Λ for i = 1, . . . , n. It is easy to see there exists an r
0
= r
0
(λ, Λ) ∈
(0,
1
2
) such that for any x
0
∈ B
1/2
the rectangle
x :
|x
i
− x
0i
|
√
λ
i
< r
0
is contained in B
1
. Change the coordinate
x
i
7−→ y
i
=
x
i
√
λ
i
and set v(y) = u( x) .
Then v is harmonic in
y :
P
n
i=1
λ
i
y
2
i
< 1
. In the ball {y : |y − y
0
| < r
0
} use the
interior estimates to yield
|v(y
0
)|
2
+ |Dv(y
0
)|
2
≤ c(λ, Λ)
Z
B
r
0
(y
0
)
v
2
≤ c(λ, Λ)
Z
{
∑
n
i=1
λ
i
y
2
i
<1
}
v
2
.
Transform back to u to get
|u(x
0
)|
2
+ |Du(x
0
)|
2
≤ c(λ, Λ)
Z
|x|<1
u
2
.
METHOD 2. If u is a solution to (1.8), so are any derivatives of u. By applying
Corollary 1.37 to derivatives of u we conclude that for any positive integer k
kuk
H
k
(B
1/2
)
≤ c(k, λ, Λ)kuk
L
2
(B
1
)
.
If we x a value of k suciently large with respect to n, H
k
(B
1/2
) is continuously
embedded into C
1
(B
1/2
) and therefore
|u|
L
∞
(B
1/2
)
+ |Du|
L
∞
(B
1/2
)
≤ c(λ, Λ)kuk
L
2
(B
1
)
.
This nishes the proof.
32