ON LIOUVILLE’S THEOREM FOR BIHARMONIC FUNCTIONS
MINGFENG ZHAO
September 30, 2011
Definition 1. Let Ω be a domain in Rn , then we say that a function u on Ω is sub biharmonic in Ω if u ∈ C 4 (Ω), and
∆2 u(x) = ∆[∆u](x) ≥ 0,
for all x ∈ Ω.
we say that a function u on Ω is super biharmonic in Ω if u ∈ C 4 (Ω), and
∆2 u(x) ≤ 0,
for all x ∈ Ω.
we say that a function u on Ω is biharmonic in Ω if u ∈ C 4 (Ω), and
∆2 u(x) = 0,
for all x ∈ Ω.
Theorem 1 (The Mean Value Theorem for the Sub Biharmonic Functions, the equality (2), in Page 37, in Huilgol[3]).
Let u be a sub biharmonic function in BR (x0 ) for some x0 ∈ Rn and R > 0. Then for all 0 < r < R, we have
Z
1
r2
· ∆u(x0 ) ≤
·
u(x) dS(x)
u(x0 ) +
2n
|∂Br (x0 )| ∂Br (x0 )
Z
r2
1
u(x0 ) +
· ∆u(x0 ) ≤
·
u(x) dx.
2(n + 2)
|Br (x0 )| Br (x0 )
Proof. a. For any 0 < r < R, we consider the function:
r2
1
φ(r) = u(x ) +
· ∆u(x0 ) −
·
2n
|∂Br (x0 )|
0
Z
u(x) dS(x)
∂Br (x0 )
Z
r2
1
· ∆u(x0 ) −
·
u(x0 + ry) · rn−1 dS(y)
2n
nαn · rn−1 ∂B1 (0)
Z
1
r2
· ∆u(x0 ) −
·
u(x0 + ry) dS(y).
= u(x0 ) +
2n
nαn ∂B1 (0)
= u(x0 ) +
Also we know that
lim φ(r) = u(x0 ) + 0 −
r&0
1
· u(x0 ) · nαn = 0.
nαn
And
0
φ (r)
Z
=
r
r
· ∆u(x0 ) −
·
n
nαn
Z
=
1
1
· ∆u(x0 ) −
·
n
nαn
∂B1 (0)
d
u(x0 + ry) dS(y)
dr
∇u(x0 + ry) · y dS(y)
∂B1 (0)
1
Let x = x0 + ry
2
MINGFENG ZHAO
=
=
=
r
1
· ∆u(x0 ) −
·
n
nαn
Z
∇u(x) ·
∂Br
(x0 )
r
1
·
· ∆u(x0 ) −
n
nαn · rn−1
Z
r
1
·
· ∆u(x0 ) −
n
nαn · rn−1
Z
z − x0 1−n
·r
dS(x)
r
Let x = x0 + ry
∇u(x) · n(x) dS(x)
∂Br (x0 )
∂Br (x0 )
∂u(x)
dS(x)
∂n
Z
=
=
=
r
1
·
∆u(x) dx
· ∆u(x0 ) −
n
nαn · rn−1 Br (x0 )
#
"
Z
r
1
0
·
∆u(x) dx
· ∆u(x ) −
n
αn · rn Br (x0 )
#
"
Z
r
1
0
·
∆u(x) dx
· ∆u(x ) −
n
αn · rn Br (x0 )
By the Divergence Theorem
Since u is sub biharmonic in BR (x0 ), then
∆2 u(x) = ∆[∆u](x) ≥ 0,
for all x ∈ BR (x0 ).
That is, ∆u is a subharmonic function in BR (x0 ). Then by the Mean Value Theorem for the Subharmonic Function,
the Theorem 2.1, in Page 14, in Gilbarg and Trudinger[2], we know that for all 0 < r < R, we have
1
∆u(x ) ≤
·
αn · rn
0
Z
∆u(x) dx.
Br (x0 )
Hence, we get that
φ0 (r) ≤ 0,
for all 0 < r < R.
That is, φ(r) is a decreasing function, which implies that
φ(r) ≤ lim φ(r) = 0,
r&0
for all 0 < r < R.
That is, we get
u(x0 ) +
r2
1
· ∆u(x0 ) ≤
·
2n
|∂Br (x0 )|
Z
u(x) dS(x).
∂Br (x0 )
b. Since, we have known that
u(x0 ) +
r2
· ∆u(x0 ) ≤
2n
=
Z
1
·
u(x) dS(x)
|∂Br (x0 )| ∂Br (x0 )
Z
1
·
u(x) dS(x),
nαn rn−1 ∂Br (x0 )
for all 0 < r < R.
Both sides are multiplied by n · rn−1 , we get
nrn−1 u(x0 ) +
rn+1
1
· ∆u(x0 ) ≤
·
2
αn
Z
u(x) dS(x).
∂Br (x0 )
ON LIOUVILLE’S THEOREM FOR BIHARMONIC FUNCTIONS
3
Then both sides integrate with respect to r from 0 to r, we get
Z r
Z r
Z rZ
∆u(x0 )
1
nu(x0 )
tn−1 dt +
·
tn+1 dr ≤
·
u(x) dS(x) dt.
2
αn 0 ∂Bt (x0 )
0
0
Then we can get
∆u(x0 ) rn+2
1
rn
+
·
≤
·
nu(x ) ·
n
2
n+2
αn
0
Z
u(x) dx.
Br (x0 )
Both sides are divided by rn , we get
r2
1
·
· ∆u(x0 ) ≤
2(n + 2)
αn · rn
Z
r2
1
· ∆u(x0 ) ≤
·
2(n + 2)
|Br (x0 )|
Z
u(x0 ) +
u(x) dx.
Br (x0 )
Therefore, we get
u(x0 ) +
u(x) dx.
Br (x0 )
As a direct consequence of the Theorem 1, we have the following:
Theorem 2. Let x0 ∈ Rn and R > 0. Then
a. If u is super biharmonic in BR (x0 ), then for all 0 < r < R, we have
Z
1
r2
· ∆u(x0 ) ≥
·
u(x) dS(x)
u(x0 ) +
2n
|∂Br (x0 )| ∂Br (x0 )
Z
r2
1
u(x0 ) +
· ∆u(x0 ) ≥
·
u(x) dx.
2(n + 2)
|Br (x0 )| Br (x0 )
b. If u is biharmonic in BR (x0 ), then for all 0 < r < R, we have
r2
u(x ) +
· ∆u(x0 )
2n
=
r2
· ∆u(x0 )
2(n + 2)
=
0
u(x0 ) +
Z
1
·
u(x) dS(x)
|∂Br (x0 )| ∂Br (x0 )
Z
1
·
u(x) dx.
|Br (x0 )| Br (x0 )
Theorem 3 (The Theorem 2, in Page 37, in Huilgol[3]). Let u be biharmonic in Rn , and
lim
|x|→∞
u(x)
= 0.
|x|2
Then u is a linear function on Rn . In particular, we know that if u is a bounded biharmonic function in Rn , then u
is a constant function on Rn .
Proof. For any x0 ∈ Rn and fix, by the Mean Value Theorem, the Theorem 2, we know that for all r > 0, we have
Z
r2
1
0
0
u(x ) +
· ∆u(x ) =
·
u(x) dS(x).
2n
|∂Br (x0 )| ∂Br (x0 )
That is, we get
(1)
u(x0 )
1
1
+
· ∆u(x0 ) =
·
2
r
2n
nαn · rn+1
Z
u(x) dS(x).
∂Br (x0 )
4
MINGFENG ZHAO
Since lim
|x|→∞
u(x)
= 0, then for any > 0, we know that there exists some constants R0 > 0 such that
|x|2
|u(x)| ≤ |x|2 ,
for all |x| ≥ R0 .
So when r > R0 + |x0 |, we know that for all x ∈ ∂Br (x0 ), we have |x| = |x − x0 + x0 | ≥ |x − x0 | − |x0 | = r − |x0 | ≥ R0 .
Then we have
1
·
nαn · rn+1
Z
≤
u(x) dS(x)
∂Br (x0 )
≤
=
≤
=
1
·
nαn · rn+1
Z
1
·
nαn · rn+1
Z
1
·
nαn · rn+1
Z
1
·
nαn · rn+1
Z
1
·
nαn · rn+1
Z
|u(x)| dS(x)
∂Br (x0 )
|x|2 dx
Br (x0 )
∂Br
|x − x0 + x0 |2 dx
(x0 )
[|x − x0 | + |x0 |]2 dx
∂Br (x0 )
[r + |x0 |]2 dx
∂Br (x0 )
[r + |x0 |]2
· nαn · rn−1
nαn · rn+1
=
= ·
[r + |x0 |]2
.
r2
By taking r % ∞, we get
lim sup
r%∞
1
·
nαn · rn+1
Z
1
·
nαn · rn+1
Z
u(x) dS(x) ≤ .
∂Br (x0 )
Then taking & 0, then we get
lim sup
r%∞
u(x) dS(x) = 0.
∂Br (x0 )
Therefore, by taking r % ∞ in (1), we can conclude that
∆u(x0 ) = 0.
Since x0 is arbitrary point in Rn , then we get
∆u(x) = 0,
for all x ∈ Rn .
For any x0 ∈ Rn and fix, by the Gradient Estimates for the Harmonic Function, the Theorem 2.10, in Page 23,
in Gilbarg and Trudinger[2], we know that for any r > R0 + |x0 |, we know that for all x ∈ ∂Br (x0 ), we have |x| =
|x − x0 + x0 | ≥ |x − x0 | − |x0 | = r − |x0 | ≥ R0 , we have
2
2n
2
0
|D u(x )| ≤
· sup |u(x)|
r
x∈Br (x0 )
=
4n2
· sup |u(x)|.
r2 x∈Br (x0 )
ON LIOUVILLE’S THEOREM FOR BIHARMONIC FUNCTIONS
5
By the Maximum Principle for the Harmonic Functions, the Theorem 2.2, in Page 15, in Gilbarg and Trudinger[2],
we know that
|u(x)| =
sup
x∈Br (x0 )
sup
|u(x)|.
x∈∂Br (x0 )
But for any x ∈ ∂Br (x0 ), we have
|u(x)| ≤ |x|2 = |x − x0 + x0 |2 ≤ · [|x − x0 | + |x0 |]2 = [r + |x0 |]2 .
Then we have
|D2 u(x0 )|
≤
=
4n2
· [r + |x0 |]2
r2
2
r + |x0 |
2
.
4n · ·
r
By taking r % ∞, we get
|D2 u(x0 )| ≤ 4n2 · .
By taking & 0, we get
D2 u(x0 ) = 0.
Since x0 is arbitrary in Rn , then
D2 u(x) = 0,
for all x ∈ Rn .
Therefore, u is a linear function on Rn . In particular, we know that if u is a bounded biharmonic function in Rn ,
then u is a constant function on Rn .
Remark 1. The Theorem 3 is sharp, since u(x) = |x|2 for all x ∈ Rn is biharmonic in Rn , but u is just bounded from
below in Rn , and unbounded from above in Rn .
Definition 2. Let k ∈ N, and Ω be a domain in Rn , then we say that a function u on Ω is sub k-harmonic in Ω if
u ∈ C 2k (Ω), and
∆k u(x) = ∆∆
· · · ∆} u(x) ≥ 0,
| {z
for all x ∈ Ω.
k many times
we say that a function u on Ω is super k-harmonic in Ω if u ∈ C 2k (Ω), and
∆k u(x) ≤ 0,
for all x ∈ Ω.
we say that a function u on Ω is biharmonic in Ω if u ∈ C 2k (Ω), and
∆k u(x) = 0,
Theorem 4. Let k ∈ N, x0 ∈ Rn and R > 0. Then
for all x ∈ Ω.
6
MINGFENG ZHAO
a. If u is sub k-harmonic in BR (x0 ), then for all 0 < r < R, we have
0
u(x ) +
k−1
X
i=1
0
u(x ) +
n2i · i! ·
k−1
X
i=1
2i · i! ·
r2i
Qi−1
i
j=1 (n
r2i
Qi
+ 2j)
· ∆ u(x ) ≤
i
j=1 (n
+ 2j)
0
0
· ∆ u(x ) ≤
Z
1
·
|∂Br (x0 )|
1
·
|Br (x0 )|
u(x) dS(x)
∂Br (x0 )
Z
u(x) dx.
Br (x0 )
b. If u is super k-harmonic in BR (x0 ), then for all 0 < r < R, we have
u(x0 ) +
k−1
X
i=1
0
u(x ) +
n2i · i! ·
k−1
X
i=1
2i · i! ·
r2i
Qi−1
j=1 (n
r2i
Qi
+ 2j)
· ∆i u(x0 ) ≥
i
j=1 (n
+ 2j)
0
· ∆ u(x ) ≥
Z
1
·
|∂Br (x0 )|
1
·
|Br (x0 )|
u(x) dS(x)
∂Br (x0 )
Z
u(x) dx.
Br (x0 )
c. If u is k-harmonic in BR (x0 ), then for all 0 < r < R, we have
u(x0 ) +
k−1
X
i=1
u(x0 ) +
n2i · i! ·
k−1
X
i=1
r2i
Qi−1
j=1 (n
r2i
Qi
2i · i! ·
j=1 (n
+ 2i)
+ 2j)
· ∆i u(x0 )
· ∆i u(x0 )
=
=
Z
1
·
|∂Br (x0 )|
1
·
|Br (x0 )|
u(x) dS(x)
∂Br (x0 )
Z
u(x) dx.
Br (x0 )
Proof. We just prove for the case of sub k-harmonic functions, that is, let u be a sub k-harmonic function in BR (x0 ).
Claim I: For any 0 < r < R, we have
u(x0 ) +
k−1
X
i=1
0
u(x ) +
n2i · i! ·
k−1
X
i=1
2i · i! ·
r2i
Qi−1
j=1 (n
r2i
Qi
+ 2j)
· ∆i u(x0 ) ≤
i
j=1 (n
+ 2j)
0
· ∆ u(x ) ≤
Z
1
·
|∂Br (x0 )|
1
·
|Br (x0 )|
u(x) dS(x)
∂Br (x0 )
Z
u(x) dx.
Br (x0 )
We prove the Claim I by induction. When k = 1, that is, u is a subharmonic function in BR (x0 ). We know that
u(x0 ) +
1−1
X
i=1
u(x0 ) +
n2i · i! ·
1−1
X
i=1
2i · i! ·
r2i
Qi−1
· ∆i u(x0 )
= u(x0 )
r2i
Qi
· ∆i u(x0 )
= u(x0 ).
j=1 (n + 2j)
j=1 (n + 2j)
By the Mean Value Theorem for the Subharmonic Function, the Theorem 2.1, in Page 14, in Gilbarg and Trudinger[2],
we know that for all 0 < r < R, we have
u(x0 ) ≤
u(x0 ) ≤
Z
1
·
u(x) dS(x)
|∂Br (x0 )| ∂Br (x0 )
Z
1
·
u(x) dx.
|Br (x0 )| Br (x0 )
So the Claim I is true for k = 1. When k = 2, that is, u is a sub biharmonic function in BR (x0 ). We know that
0
u(x ) +
2−1
X
i=1
n2i · i! ·
r2i
Qi−1
j=1 (n
+ 2j)
· ∆i u(x0 )
= u(x0 ) +
r2
· ∆u(x0 )
2n
ON LIOUVILLE’S THEOREM FOR BIHARMONIC FUNCTIONS
u(x0 ) +
2−1
X
i=1
2i · i! ·
r2i
Qi
j=1 (n
+ 2j)
· ∆i u(x0 )
= u(x0 ) +
7
r2
· ∆u(x0 ).
2(n + 2)
By the Theorem 1, we know that
u(x0 ) +
u(x0 ) +
r2
· ∆u(x0 ) ≤
2n
r2
· ∆u(x0 ) ≤
2(n + 2)
Z
1
·
u(x) dS(x)
|∂Br (x0 )| ∂Br (x0 )
Z
1
·
u(x) dx.
|Br (x0 )| Br (x0 )
So the Claim I is true for k = 2. Now we assume for k = m ≥ 2, the Claim I is true. Look at k = m + 1, that is, u is
(
a sub m + 1-harmonic function in BR x0 ), so
∆m+1 u(x) = ∆m [∆u](x) ≤ 0,
for all x ∈ BR (x0 ).
That is, ∆u is sub m-harmonic function in BR (x0 ). Then by the assumption for k = m, we have
0
∆u(x ) +
m−1
X
i=1
∆u(x0 ) +
(2)
n2i · i! ·
m−1
X
i=1
2i · i! ·
r2i
Qi−1
j=1 (n
r2i
Qi
j=1 (n
i
+ 2j)
+ 2j)
0
· ∆ [∆u](x ) ≤
· ∆i [∆u](x0 ) ≤
Z
1
·
|∂Br (x0 )|
1
·
|Br (x0 )|
∆u(x) dS(x)
∂Br (x0 )
Z
∆u(x) dx.
Br (x0 )
Now let
φ(r)
=
u(x0 ) +
m
X
i=1
=
0
u(x ) +
n2i · i! ·
r2i
Qi−1
n2i · i! ·
r2i
Qi−1
m
X
i=1
j=1 (n
+ 2j)
· ∆i u(x0 ) −
1
·
|∂Br (x0 )|
Z
1
· ∆ u(x ) −
·
n−1
nα
·
n r
j=1 (n + 2j)
i
0
u(x) dS(x)
∂Br (x0 )
Z
u(x0 + ry) · rn−1 dS(y)
∂B1 (0)
Let x = x0 + ry
=
0
u(x ) +
m
X
i=1
n2i · i! ·
r2i
Qi−1
1
·
· ∆ u(x ) −
nα
n
j=1 (n + 2j)
i
0
Z
u(x0 + ry) dS(y).
∂B1 (0)
So we can get
lim φ(r) = 0.
r&0
And
φ0 (r)
=
m
X
i=1
=
m
X
i=1
=
m
X
i=1
1
2i · r2i−1
· ∆i u(x0 ) −
·
Qi−1
nαn
n2i · i! · j=1 (n + 2j)
Z
2i · r2i−1
1
· ∆i u(x0 ) −
·
Q
i−1
nαn
n2i · i! · j=1 (n + 2j)
Z
2i · r2i−1
1
· ∆i u(x0 ) −
·
Qi−1
i
nαn
n2 · i! · j=1 (n + 2j)
Z
Let x = x0 + ry
∂B1 (0)
d
u(x0 + ry) dS(y)
dr
∇u(x0 + ry) · y dS(y)
∂B1 (0)
∇u(x) ·
∂Br (x0 )
x − x0 n−1
·r
dS(x)
r
8
MINGFENG ZHAO
m
X
=
i=1
m
X
=
i=1
m
X
=
i=1
2i · r2i−1
1
·
· ∆i u(x0 ) −
Qi−1
i
nαn · rn−1
n2 · i! · j=1 (n + 2j)
Z
1
2i · r2i−1
· ∆i u(x0 ) −
·
Qi−1
n−1
i
nα
·
n r
n2 · i! · j=1 (n + 2j)
Z
1
2i · r2i−1
· ∆i u(x0 ) −
·
Qi−1
n−1
i
nα
·
n r
n2 · i! · j=1 (n + 2j)
Z
∇u(x) · n(x) dS(x)
∂Br (x0 )
∂Br
(x0 )
∂u(x)
dS(x)
∂n
u(x) dx
Br (x0 )
By the Divergence Theorem
r
·
n
=
"
m
X
i=1
2i · r2i−2
1
·
· ∆i u(x0 ) −
Qi−1
nαn · rn−1
2i · i! · j=1 (n + 2j)
#
Z
∆u(x) dx
Br (x0 )
"
#
Z
m
X
r
1
2i · r2i−2
0
i
0
·
· ∆u(x ) +
· ∆ u(x ) −
∆u(x) dx
Qi−1
i
n
nαn · rn−1 Br (x0 )
j=1 (n + 2j)
i=2 2 · i! ·
"
#
Z
m−1
2i
X
2(i
+
1)
·
r
r
1
∆u(x) dx
· ∆u(x0 ) +
·
· ∆i+1 u(x0 ) −
Qi
i+1 · (i + 1)! ·
n
nαn · rn−1 Br (x0 )
j=1 (n + 2j)
i=1 2
"
#
Z
m−1
2i
X
r
r
1
· ∆u(x0 ) +
· ∆i [∆u](x0 ) −
·
∆u(x) dx
Qi
i · i! ·
n
nαn · rn−1 Br (x0 )
2
(n
+
2j)
j=1
i=1
=
=
=
≤ 0,
By (2)
That is, φ(r) is decreasing function on 0 < r < R, in particular, we get
φ(r) ≤ lim φ(r) = 0.
r&0
That is, we get
m
X
u(x0 ) +
n2i · i! ·
r2i
Qi−1
n2i · i! ·
r2i
Qi−1
i=1
1
·
|∂Br (x0 )|
Z
1
· ∆ u(x ) ≤
·
n−1
nα
·
n r
j=1 (n + 2j)
Z
j=1 (n
+ 2j)
· ∆i u(x0 ) ≤
u(x) dS(x).
∂Br (x0 )
That is, we get
0
u(x ) +
m
X
i=1
i
0
u(x) dS(x).
∂Br (x0 )
Both sides are multiplied by n · rn−1 , and then integrate with respect to r from 0 to r, we can get
n · u(x0 ) ·
Z
r
tn−1 dt + n ·
m
X
0
i=1
n2i · i! ·
1
Qi−1
j=1 (n
+ 2j)
· ∆i u(x0 ) ·
Z
r
t2i+n−1 ≤
0
1
·
αn
Z
r
Z
u(x) dS(x)dt.
0
∂Bt (x0 )
That is, we have
u(x0 ) · rn +
m
X
r2i+n
1
≤
·
n + 2i
αn
Z
r2i
1
· ∆ u(x ) ·
≤
·
Q
i−1
n
i
n
+
2i
α
n·r
2 · i! · j=1 (n + 2j)
Z
i=1
1
2i · i! ·
Qi−1
j=1 (n
+ 2j)
· ∆i u(x0 ) ·
u(x) dx.
Bt (x0 )
Both sides are divided by rn , we can get
0
u(x ) +
m
X
i=1
1
i
0
u(x) dx.
Bt (x0 )
ON LIOUVILLE’S THEOREM FOR BIHARMONIC FUNCTIONS
9
That is, we have
u(x0 ) +
m
X
i=1
2i · i! ·
r2i
Qi
j=1 (n
+ 2j)
· ∆i u(x0 ) ≤
1
·
|Br (x0 )|
Z
u(x) dx.
Bt (x0 )
Hence, we get that the Claim I is true for k = m + 1. Therefore, by induction, we can conclude that if u is a sub
k-harmonic function in BR (x0 ), then for any 0 < r < R, we have
u(x0 ) +
k−1
X
i=1
u(x0 ) +
r2i
Qi−1
n2i · i! ·
j=1 (n
k−1
X
i=1
r2i
Qi
2i · i! ·
j=1 (n
+ 2j)
+ 2j)
· ∆i u(x0 ) ≤
· ∆i u(x0 ) ≤
Z
1
·
|∂Br (x0 )|
1
·
|Br (x0 )|
u(x) dS(x)
∂Br (x0 )
Z
u(x) dx.
Br (x0 )
Theorem 5 (The Remark 2, in Page 38, in Huilgol[3]). Let k ∈ N, and u be a bounded k-harmonic in Rn . Then u is
a constant function on Rn .
Proof. We prove this Theorem by induction. When k = 1, that is, u is a bounded harmonic function in Rn , then by the
Usual Liouville Theorem for the Harmonic Functions, Theorem 8, in Page 30, in Evans[1], we know that u is a constant
function on Rn . So the Theorem is true for k = 1.
When k = 2, that is, u is a bounded biharmonic function in Rn , then by the Theorem 3, we know that u is a constant
function on Rn . So the Theorem is true for k = 2.
Now we assume the Theorem is true for k = m, that is, if u is a bounded m-harmonic function, then u is a constant
function on Rn .
Look at k = m + 1, assume that u is a m + 1-harmonic function in Rn , then by the Theorem 4, we know that for
any x0 ∈ Rn and fix, then for all r > 0, we have
u(x0 ) +
m
X
i=1
2i · i! ·
r2i
Qi
j=1 (n
+ 2j)
· ∆i u(x0 ) =
1
·
|Br (x0 )|
Z
u(x) dx.
Br (x0 )
Both sides are divided by r2m , we get
m
(3)
u(x0 ) X
r2i−2m
1
+
· ∆i u(x0 ) =
·
Q
i
2m
n+2m
i · i! ·
r
α
·
r
n
(n
+
2j)
2
j=1
i=1
Z
u(x) dx.
Br (x0 )
And we note that
1
·
αn · rn+2m
Z
u(x) dx
≤
Br (x0 )
≤
=
=
1
·
αn · rn+2m
Z
1
·
αn · rn+2m
Z
|u(x)| dx
Br (x0 )
kukL∞ (Rn ) dx
Br (x0 )
kukL∞ (Rn )
· αn · r n
αn · rn+2m
kukL∞ (Rn )
.
r2m
10
MINGFENG ZHAO
By taking r % ∞, we get
1
·
r%∞ αn · r n+2m
Z
lim
u(x) dx = 0.
Br (x0 )
So for (3), we take r % ∞, we can get
2m · m! ·
1
Qm
j=1 (n
+ 2j)
· ∆m u(x0 ) = 0.
By the arbitrary of x0 ∈ Rn , so we get
∆m u(x) = 0,
for all x ∈ Rn .
By the assumption for k = m, we know that u is a constant function on Rn . That is, the Theorem is true for
k = m + 1.
Therefore, by induction, we know that the Theorem is true for all k ∈ N.
ON LIOUVILLE’S THEOREM FOR BIHARMONIC FUNCTIONS
References
[1] Lawrence C. Evans. Partial differential equations. American Mathematical Society, 1998.
[2] David Gilbarg and Neil S. Trudinger. Elliptic partial differential equations of second order. Springer, 2001.
[3] R. R. Huilgol. On liouville’s theorem for biharmonic functions. SIAM J. Appl. Math., 20:37–39, 1971.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
E-mail address: mingfeng.zhao@uconn.edu
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