Section 2.6: Moving Plane Method
2.6 Moving Plane Method
In this section we will use the moving plane method to discuss the symmetry of
solutions. The following result was rst proved by Gidas, Ni, and Nirenberg.
Theorem 2.34. Suppose u ∈ C(B
1
) ∩ C
2
(B
1
) is a positive solution of
∆u + f(u) = 0 in B
1
,
u = 0 on ∂B
1
,
where f is locally Lipschitz in R. Then u is radially symmetric in B
1
and
∂u
∂r
(x) < 0 for
x 6= 0.
The original proof requires that solutions be C
2
up to the boundary. Here we give
a method that does not depend on the smoothness of domains nor the smoothness of
solutions up to the boundary.
Lemma 2.35. Suppose that Ω is a bounded domain that is convex in the x
1
-direction
and symmetric with respect to the plane {x
1
= 0}. Suppose u ∈ C(Ω) ∩ C
2
(Ω) is a
positive solution of
∆u + f(u) = 0 in Ω,
δu = 0 on ∂Ω,
where
f
is locally Lipschitz in
R
. Then
u
is symmetric with respect to
x
1
and
D
x
1
u(x) <
0 for any x ∈ Ω with x
1
> 0.
Proof of Lemma 2.35. Write x = (x
1
, y) ∈ Ω for y ∈ R
n−1
. We will prove
u(x
1
, y) < u(x
∗
1
, y) (2.4)
for any x
1
> 0 and x
∗
1
< x
1
with x
∗
1
+ x
1
> 0. Then by letting x
∗
1
→ − x
1
, we get
u(x
1
, y) ≤ u(−x
1
, y) for any x
1
. Then by changing the direction x
1
→ −x
1
, we get the
symmetry. Let a = sup x
1
for (x
1
, y) ∈ Ω. For 0 < λ < a, dene
Σ
λ
= {x ∈ Ω : x
1
> λ},
T
λ
= {x
1
= λ},
Σ
′
λ
= reection of Σ
λ
with respect to T
λ
,
x
λ
= (2λ − x
1
, . . . , x
n
) for x = (x
1
, . . . , x
n
).
In Σ
λ
we dene
w
λ
(x) = u(x) − u(x
λ
) for x ∈ Σ
λ
.
57
Chapter : Maximum Principles
Then we have by the mean value theorem
∆w
λ
+ c(x, λ)w
λ
= 0 in Σ
λ
,
w
λ
≤ 0 and w
λ
6≡ 0 on ∂Σ
λ
,
where c(x, λ) is a bounded function in Σ
λ
. We need to show w
λ
< 0 in Σ
λ
for any
λ ∈ (0, a). This implies in particular that w
λ
assumes along ∂Σ
λ
∩ Ω its maximum in
Σ
λ
. By Theorem 2.5 (the Hopf lemma) we have for any such λ ∈ (0, a)
D
x
1
w
λ
x
1
=λ
= 2D
x
1
u
x
1
=λ
< 0.
For any λ close to a, we have w
λ
< 0 by Proposition 2.13 (the maximum principle for a
narrow domain) or Theorem 2.32. Let ( λ
0
, a) be the largest interval of values of λ such
that w
λ
< 0 in Σ
λ
. We want to show λ
0
= 0. If λ
0
> 0, by continuity, w
λ
0
≤ 0 in
Σ
λ
0
and w
λ
0
6≡ 0 on ∂Σ
λ
0
. Then Theorem 2.7 (the strong maximum principle) implies
w
λ
0
< 0 in Σ
λ
0
. We will show that for any small ε > 0
w
λ
0
−ε
< 0 in Σ
λ
0
−ε
.
Fix δ > 0 (to be determined). Let K be a closed subset in Σ
λ
0
such that | Σ
λ
0
\K| <
δ
2
.
The fact that w
λ
0
< 0 in Σ
λ
0
implies
w
λ
0
(x) ≤ −η < 0 for any x ∈ K.
By continuity we have
w
λ
0
−ε
< 0 in K.
For ε > 0 small, |Σ
λ
0
−ε
\ K| < δ. We choose δ in such a way that we may apply
Theorem 2.32 (the maximum principle for a domain with small volume) to w
λ
0
−ε
in
Σ
λ
0
−ε
\ K. Hence we get
w
λ
0
−ε
(x) ≤ 0 in Σ
λ
0
−ε
\ K
and then by Theorem 2.10
w
λ
0
−ε
(x) < 0 in Σ
λ
0
−ε
\ K.
Therefore we obtain for any small ε > 0
w
λ
0
−ε
(x) < 0 in Σ
λ
0
−ε
.
This contradicts the choice of λ
0
.
58