BREZIS TYPE THEOREM ON UPPER HALF SPACE
MINGFENG ZHAO
April 06, 2014
Lemma 1 (Stampacchia’s Version of the Maximum Principle, Lemma 1 on Page 1 in Brezis[2]). Let Ω be any domain
in Rn with n ≥ 3, and u ∈ H 1 (Ω) satisfy

 −∆u(x) + c(x)u(x) ≤ 0,
(1)
 u(x) ≤ 0, ∀x ∈ ∂Ω.
∀x ∈ Ω,
Assume
kc− kL n2 (Ω) < SΩ ,
(2)
where SΩ is the Sobolev constant for Ω, i.e.,
SΩ =
k∇ϕk2L2 (Ω)
inf1
ϕ∈H0 (Ω)
kϕk2
.
2n (Ω)
L n−2
Then
u(x) ≤ 0,
∀x ∈ Ω.
Remark 1. The constant SΩ in Lemma 1 is well defined by the Sobolev embedding, Theorem 2.4.1 on Page 56 in
Ziemer[5].
Proof. Since u ∈ H 1 (Ω) and u(x) ≤ 0 for all x ∈ ∂Ω, by Corollary 2.1.8 on Page 47 in Ziemer[5], we know that
u+ ∈ H01 (Ω), and

 ∇u(x), if u(x) > 0,
∇u+ (x) =
 0, if u(x) ≤ 0.
Multiply u+ on the both sides of the first equation in (1), and integrative over Ω, use integration by parts, we have
Z
|∇u+ (x)|2 dx
Z
u+ (x)∆u(x) dx
= −
Ω
Ω
Z
≤
c(x)u(x)u+ (x) dx
−
Ω
Z
c+ (x)|u+ (x)|2 dx +
= −
Ω
Z
≤
−
Z
c− (x)|u+ (x)|2 dx
Ω
+
2
c (x)|u (x)| dx
Ω
1
Since c(x) = c+ (x) − c− (x)
2
MINGFENG ZHAO
kc− kL n2 (Ω) · ku+ k2
≤
2n
L n−2
Since
(Ω)
2
n−2
+
= 1, and by Holder’s inequality.
n
n
If u+ (x) ≡ 0 in Ω, that is, u(x) ≤ 0 in Ω, we are done. If u+ (x) 6≡ 0 in Ω, by the assumption on c− , we have
Z
|∇u+ (x)|2 dx
<
SΩ ku+ k2
2n
L n−2 (Ω)
Ω
R
|∇u+ (x)|2 dx
· ku+ k2 2n
ku+ k2 2n
L n−2 (Ω)
Ω
≤
Since u+ ∈ H01 (Ω), and by the definition of SΩ
L n−2 (Ω)
Z
=
|∇u+ (x)|2 dx.
Ω
We get a contradiction.
Remark 2. In order to satisfy (2), if we know that Ω is bounded and c− (x) is bounded in Rn , say 0 ≤ c− (x) ≤ L in
Ω, but
n
n
kc− kL n2 (Ω) ≤ kc− kL∞ (Rn ) · |Ω| 2 ≤ L · |Ω| 2 .
Recall that H01 (Ω) ⊂ H 1 (Rn ), which implies that
SΩ ≥ SRn = Sn :=
inf
∞
ϕ∈Cc (Rn )
k∇ϕk2L2 (Ω)
kϕk2
> 0.
2n (Ω)
L n−2
2
So if |Ω| is very small (e.g. |Ω| ≤ Sn · L−1 n ), Lemma 1 will hold.
Theorem 1 (Theorem 2 on Page 29 in Berestycki, Caffarelli and Nirenberg[1]). Let f be locally Lipschitz and u ∈
T
C 2 (Rn+1
C(Rn+1
+ ) satisfy
+ )


 −∆u(z) = f (u(z)), ∀z := (x, y) ∈ Rn+1

+ ,


(3)
u(z) ≥ 0, ∀z ∈ Rn+1
+ ,




 u(x, 0) = 0, ∀x ∈ Rn .
Assume that
a. If n = 1, f (u) ≥ 0 for all u ∈ [0, ∞).
n+3
b. If n ≥ 2, the function f (u)u− n−1 is non-increasing in [0, ∞).
2
Then u(z) ≡ u(0, y)for all (x, y) ∈ Rn+1
+ . Moreover, if n = 1 and f (u) ≥ in [0, ∞), then uy (z) > 0 for all (x, y) ∈ R+ .
Proof. We follow the arguments in in Gidas and Spruck[3](Proof of Theorem 1.3 on Page 893 in Gidas and Spruck[3]).
Consider the spherical inversion with respect to B√2 (−en+1 ), that is,
J : z 7→ −en+1 +
2
· [z + en+1 ],
|z + en+1 |2
∀z ∈ Rn+1 .
BREZIS TYPE THEOREM ON UPPER HALF SPACE
For any z ∈ Rn+1
+ , let v(z) = u(z − en+1 ), w(z) = v(2z), p(z) =
J(z)
1
|z|n−1
·w
z
, and q(z) = p(z + en+1 ). Then
|z|2
= −en+1 +
2
· [z + en+1 ]
|z + en+1 |2
= −en+1 +
2(x, y + 1)
|z|2 + 2y + 1
=
(2x, 2y + 2 − |z|2 − 2y − 1)
|z|2 + 2y + 1
=
(2x, 1 − |z|2 )
.
|z|2 + 2y + 1
3
It’s easy to see that J(z) ∈ Rn+1
if and only if |z| < 1, and J(z) ∈ ∂Rn+1
if and only if |z| = 1. In summary, we
+
+
n+1
know that J is a 1 − 1 and onto map from B1 (0) to Rn+1
+ , and from ∂B1 (0)\{−en+1 } to R+ . Let v(z) = u(z − en+1 ),
1
z
, and q(z) = p(z + en+1 ), then
w(z) = v(2z), p(z) = n−1 · w
|z|
|z|2
q(z)
=
=
1
|z + en+1 |n−1
· u −en+1 +
1
[|x|2 + (y + 1)2 ]
n−1
2
2
· [z + en+1 ]
|z + en+1 |2
· u −en+1 +
!
2
[|x|2 + (y + 1)2 ]
n−1
2
· (x, y + 1) ,
∀z ∈ B1 (0).
By Problem 4.7 on Page 71 in Gilbarg and Trudinger[4], we know that
∆v(z)
=
∆u(z − en+1 )
∆w(z)
=
4 · ∆v (2z)
∆p(z)
= |z|−n−3 ∆w
∆q(z)
=
z
|z|2
∆p(z + en+1 )
z + en+1
|z + en+1 |2
2(z + en+1 )
−n−3
= 4|z + en+1 |
∆v
|z + en+1 |2
2
= 4|z + en+1 |−n−3 ∆u −en+1 +
·
[z
+
e
]
.
n+1
|z + en+1 |2
= |z + en+1 |−n−3 ∆w
Hence we get
−∆q(z)
2
·
[z
+
e
]
n+1
|z + en+1 |2
2
−n−3
= 4|z + en+1 |
f u −en+1 +
· [z + en+1 ]
|z + en+1 |2
= 4|z + en+1 |−n−3 f |z + en+1 |n−1 q(z) , ∀z ∈ B1 (0).
=
−4|z + en+1 |−n−3 ∆u −en+1 +
By (3)
4
MINGFENG ZHAO
So q satisfies



−∆q(z) = 4|z + en+1 |−n−3 f (|z + en+1 |n−1 q(z)),



q(z) > 0, ∀z ∈ B1 (0),




 q(z) = 0, ∀z ∈ ∂B1 (0)\{−en+1 }.
Let σ(z) = q(z − en+1 ), then






(4)





∀z ∈ B1 (0),
σ satisfies
−∆σ(z) = 4|z|−n−3 f (|z|n−1 σ(z)),
σ(z) > 0,
∀z ∈ B1 (en+1 ),
σ(z) = 0,
∀z ∈ ∂B1 (en+1 )\{0}.
∀z ∈ B1 (en+1 ),
For any 0 < λ < 1, let
Σλ = {z ∈ B1 (en+1 ) : z1 > λ} .
For any z ∈ Σλ , we define
zλ
=
(2λ − z1 , z2 , · · · , zn+1 ),
σλ (z)
= σ(zλ ),
wλ (z)
= σλ (z) − σ(z).
For ∂Σλ ,we know that ∂Σλ consists two parts:
∂ + Σλ := ∂B1 (en+1 )
\
{z1 ≥ λ},
and ∂ 0 Σλ := B1 (en+1 )
\
It’s easy to see that wλ satisfies



−∆wλ (z) = 4 |zλ |−n−3 f (|zλ |n−1 σ(zλ )) − |z|−n−3 f (|z|n−1 σ(z)) ,



wλ (z) = σ(zλ ) > 0, ∀z ∈ ∂ + Σλ ,




 wλ (z) = 0, ∀z ∈ ∂ 0 Σλ ,
{z1 = λ}.
∀z ∈ Σλ ,
For all z ∈ Σλ , since z1 > λ, we get
2
2
|zλ |2 = (2λ − z1 )2 + z22 + · · · + zn+1
= 4λ2 − 4λz1 + z12 + · · · + zn+1
= 4λ(λ − z1 ) + |z|2 < |z|2 .
If n = 1 and f (u) ≥ 0 in [0, ∞), we have
|zλ |−n−3 f (|zλ |n−1 σ(zλ )) − |z|−n−3 f (|z|n−1 σ(z))
= |zλ |−4 f (σ(zλ )) − |z|−4 f (σ(z))
= |zλ |−4 [f (σλ (z)) − f (σ(z))] + [|zλ |−4 − |z|−4 ]f (σ(z))
≥
|zλ |−4 [f (σλ (z)) − f (σ(z))],
Since |zλ | < |z| and f (σ(z)) ≥ 0.
BREZIS TYPE THEOREM ON UPPER HALF SPACE
5
n+3
If n ≥ 2 and the function f (u)u− n−1 is non-increasing in [0, ∞), we know that
|zλ |−n−3 f (|zλ |n−1 σ(zλ )) − |z|−n−3 f (|z|n−1 σ(z))
= |zλ |−n−3 [f (|zλ |n−1 σ(zλ )) − f (|zλ |n−1 σ(z))] + [|zλ |−n−3 f (|zλ |n−1 σ(z)) − |z|−n−3 f (|z|n−1 σ(z))]
= |zλ |−n−3 [f (|zλ |n−1 σ(zλ )) − f (|zλ |n−1 σ(z))]
h
i
n+3
n+3
n+3
+σ n−1 (z) f (|zλ |n−1 σ(z))[|zλ |n−1 σ(z)]− n−1 − f (|z|n−1 σ(z))[|z|n−1 σ(z)]− n−1
≥
|zλ |−n−3 [f (|zλ |n−1 σ(zλ )) − f (|zλ |n−1 σ(z))]
Since |zλ | < |z| and σ(z) > 0
Let’s define
(5)

4|z |−n−3 [f (|zλ |n−1 σ(zλ )) − f (|zλ |n−1 σ(z))]


 − λ
,
wλ (z)
cλ (z) =



0, if wλ (z) = 0.
if wλ (z) 6= 0,
Then wλ satisfies



−∆wλ (z) + cλ (z)wλ (z) ≥ 0, ∀z ∈ Σλ ,



wλ (z) = σ(zλ ) > 0, ∀z ∈ ∂ + Σλ ,




 wλ (z) = 0, ∀z ∈ ∂ 0 Σλ ,
For any 0 < λ0 < 1 and fix, it’s easy to see that there exists some δλ0 > 0 such that for any λ ∈ (λ0 , 1), for any
z ∈ Σλ , we have z, zλ ∈
/ Bδλ0 (0). Since f is Locally Lipschitz, then f is Lipschitz on [0, kukL∞ (Ωλ0 ) ], let Lλ0 be the
Lipschitz constant, where Ωλ0 = B1 (en+1 )\Bδλ0 (0), which implies that
(6)
−n−3
.
kc−
· |zλ |n−1 = 4Lλ0 |zλ |−4 ≤ 4Lλ0 δλ−4
λ kL∞ (Σλ ) ≤ Lλ0 4|zλ |
0
By Remark 2, we know that there exists some λ1 ∈ (λ0 , 1) such that λ ∈ [λ0 , 1), we have
wλ (z) ≥ 0,
∀z ∈ Σλ .
Now we can define the nonempty set
Λ := {λ ∈ (0, 1) : wµ (z) ≥ 0 in Σµ , ∀µ ∈ [λ, 1)} =
6 ∅.
Claim I: Λ is open in (0, 1).
In fact, for any λ ∈ Λ, we have wλ (z) ≥ 0 for all z ∈ Σλ , by the Strong Maximum Principle or Harnack inequality,
we know that
wλ (z) > 0,
∀z ∈ Σλ .
Take λ2 ∈ (0, λ) and fix, let δλ2 and Lλ2 be numbers defined as before, now we can take a compact subset K ⊂ Σλ
such that
|Σλ \K| <
i2
1h
−1 n
Sn · (4Lλ2 δλ2 )
.
2
6
MINGFENG ZHAO
Since K ⊂ Σλ and K is compact, then
δ := inf wλ (z) > 0.
z∈K
By the uniform continuity, there exists some λ3 ∈ (0, λ) such that
wµ (z) ≥
δ
,
2
∀z ∈ K, ∀µ ∈ [λ3 , λ].
For any µ ∈ [λ3 , λ), we know that Σλ ⊂ Σµ . Let
Σµ,λ,K := Σµ \K = (Σλ \K)
[
(Σµ \Σλ )
[
∂ 0 Σλ .
So there exists some λ4 ∈ (λ3 , λ) such that
h
i2
−1 n
|Σµ,λ,K | < Sn · (4Lλ2 δλ2 )
,
∀µ ∈ [λ4 , λ].
By Remark 2, for all µ ∈ (λ4 , λ], we have
wµ (z) ≥ 0,
∀z ∈ Σµ,λ,K .
Hence, for all µ ∈ (λ4 , λ], we get
wµ (z) ≥ 0,
∀z ∈ Σµ .
In summary, we know that wµ (z) ≥ 0 in Σµ for all µ ∈ (λ4 , 1), that is, λ is an interior point of Λ, that is, Λ is open
in (0, 1). By taking λ & 0, we get
σ(−z1 , z2 , · · · , zn+1 ) ≥ σ(z),
∀z1 > 0.
Let σ(z) = σ(−z1 , z2 , · · · , zn+1 ), then σ is also a solution to (4). By the previous argument, we get
σ(z) = σ(−z1 , z2 , · · · , zn+1 ) ≥ σ(z) = σ(−z1 , z2 , · · · , zn+1 ),
∀z1 > 0.
Hence we know that
σ(z) = σ(−z1 , z2 , · · · , zn+1 ),
∀z ∈ B1 (en+1 ).
For any A ∈ O(n), let σ(z) = σ(Ax, y), then σ is also a solution to (4). By the previous argument, we know that
σ(·, y) is radially symmetric in B1 (0) ⊂ Rn , which implies that u(·, y) is radially symmetric in Rn . For any x0 ∈ Rn ,
let U (z) = u(x0 + x, y) in Rn+1
+ , then U is a solution to (3). By the previous argument, we have U (·, y) is radially
symmetric in Rn , that is, u(·, y) is symmetric with respect to x0 . Since x0 ∈ Rn is arbitrary, then u(·, y) ≡ constant in
Rn , that is, u(x, y) ≡ U (y) for all y ≥ 0. Hence U satisfies



−U 00 (y) = f (U (y)),



U (y) > 0, ∀y > 0,




 U (0) = 0.
∀y > 0,
BREZIS TYPE THEOREM ON UPPER HALF SPACE
7
In the case n = 1 and f (u) ≥ 0 for u ≥ 0. If U 0 (y0 ) < 0 for some y0 > 0, since −U 00 (y) = f (U (y)) ≥ 0 for y > 0, then
U 0 (y) ≤ U 0 (y0 ) < 0 for all y ≥ y0 , which implies that U (y) ≤ U (y0 ) + U 0 (y0 )(y − y0 ) for all y ≥ y0 . But U 0 (y0 ) < 0, then
U (y) → −∞ as y → ∞, contradiction. So we must have U 0 (y) ≥ 0 for all y > 0. Since U 0 satisfies −U 000 = f 0 (U (y))U 0 (y)
for y > 0. By the Strong Maximum Principle, since U (y) > 0 = U (0) for y > 0, then we know that U 0 (y) > 0 for all
y > 0.
Corollary 1. Let u satisfy



∆u(x, y) = 0, ∀(x, y) ∈ Rn+1

+ ,


u(x, y) > 0, ∀(x, y) ∈ Rn+1
+ ,




 u(x, 0) = 0, ∀x ∈ Rn .
Then there exists some a > 0 such that u(x, y) = ay for all (x, y) ∈ Rn+1
+ .
Proof. If u(z) ≡ 0 in R2+ , we are done. If u(z) 6≡ 0 in R2+ , by the Strong Maximum Principle or Harnack inequality, we
know that u(z) > 0 for all z ∈ Rn+1
+ . By Theorem 1, we have u(x, y) ≡ U (y) for all y ≥ 0. Since ∆u(x, y) = 0 for all
and u(x, 0) = 0 for all x ∈ Rn , then there exists some a > 0 such that u(x, y) ≡ ay for all (x, y) ∈ Rn+1 .
(x, y) ∈ Rn+1
+
Corollary 2. Let p ≥ 1 and u satisfy



−∆u(z) = up (z), ∀z := (x, y) ∈ R2+ ,



u(z) ≥ 0, ∀z ∈ R2+ ,




 u(x, 0) = 0, ∀x ∈ R2 .
Then u(z) ≡ 0 in R2+ .
Proof. If u(z) ≡ 0 in R2+ , we are done. If u(z) 6≡ 0 in R2+ , by the Strong Maximum Principle or Harnack inequality, we
n+1
know that u(z) > 0 for all z ∈ Rn+1
+ . By Theorem 1, we know that u(z) = U (y) for all (z ∈ R+ , and U satisfies



−U 00 (y) = U p (y), ∀y > 0,






U (y) > 0, ∀y > 0,

 U 0 (y) > 0, ∀y > 0,





 U (0) = 0.
Then we know that U 00 (y) = −U p (y) ≤ −U p (1) = −α for all y ≥ 1 with α > 0, which implies that U 0 (y) ≤
−α(y − 1) + U 0 (1) for all y ≥ 1. Hence U 0 (y) → −∞ as y → ∞, contradiction.
8
MINGFENG ZHAO
Corollary 3 (Theorem 1.3 on Page 886 and Remarks on Page 895 in Gidas and Spruck[3]). Let n ≥ 2, 1 ≤ p ≤
n+3
n−1
and u satisfy



−∆u(z) = up (z), ∀z := (x, y) ∈ Rn+1

+ ,


n+1
u(z) ≥ 0, ∀z ∈ R+ ,




 u(x, 0) = 0, ∀x ∈ Rn .
Then u(z) ≡ 0 in Rn+1
+ .
n+1
Proof. If u(z) ≡ 0 in Rn+1
+ , we are done. If u(z) 6≡ 0 in R+ , by the Strong Maximum Principle or Harnack inequality,
n+3
n+3
we know that u(z) > 0 for all z ∈ Rn+1
, then α :=
− p > 0 and
+ . Since 1 ≤ p ≤
n−1
n−1
n+3
up u− n−1 = u−α is decreasing.
By Theorem 1, we know that u(z) = U (y) for all (z ∈ Rn+1
+ , and U satisfies


 −U 00 (y) = U p (y), ∀y > 0,





 U (y) > 0, ∀y > 0,


U 0 (y) > 0, ∀y > 0,





 U (0) = 0.
By the proof of Corollary 2, we will get a contradiction.
Corollary 4 (Corollary on Page 29 in Berestycki, Caffarelli and Nirenberg[1]). Let n ≥ 2, p ≥



∆u(z) = up (z), ∀z := (x, y) ∈ Rn+1

+ ,


u(z) ≥ 0, ∀z ∈ Rn+1
+ ,




 u(x, 0) = 0, ∀x ∈ Rn .
Then u(z) ≡ 0 in Rn+1
+ .
Proof. Since p ≥
n+3
n−1 ,
then α =
n+3
n−1
− p > 0, and
n+3
−up u− n−1 = −uα is decreasing.
By Theorem 1, we know that u(z) = U (y) for all z ∈ Rn+1
+ , and U satisfies



U 00 (y) = U p (y), ∀y > 0,



U (y) > 0, ∀y > 0,




 U (0) = 0.
n+3
and u satisfies
n−1
BREZIS TYPE THEOREM ON UPPER HALF SPACE
9
By the uniqueness theorem of ODEs, we know that U 0 (0) > 0. Since U 00 (y) = U p (y) > 0 for y > 0, then U 0 (y) ≥
U 0 (0) > 0 for all y ≥ 0, which implies that U (y) ≥ U 0 (0)y for all y ≥ 0. Since U 0 (0) > 0, then lim U (y) = ∞. Since
y→∞
U 00 (y) = U p (y) for y > 0, then
[U 0 (y)]2 = [U 0 (0)]2 +
2
[U (y)]p+1 ,
p+1
∀y ≥ 0.
So we get
U 0 (y)
q
[U 0 (0)]2 +
2
p+1
p+1 [U (y)]
∀y ≥ 0.
= 1,
Then we get
Z
U (y)
dt
q
0
[U 0 (0)]2 +
2
p+1
= y,
∀y > 0.
· tp+1
By taking y → ∞, we get
Z
∞
0
dt
q
0
2
[U (0)] +
2
p+1
· tp+1
dt
q
0
2
[U (0)] +
2
p+1
· tp+1
= ∞.
On the other hand, sine p > 1, then
Z
0
∞
< ∞,
we get a contradiction.
Corollary 5. Let n ≥ 1, p > 1 and u satisfies



∆u(z) = up (z), ∀z := (x, y) ∈ Rn+1

+ ,


u(z) ≥ 0, ∀z ∈ Rn+1
+ ,





u(x, 0) = 0, ∀x ∈ Rn .
Then u(z) ≡ 0 in Rn+1
+ .
Proof. Since p > 1, then there exists some m ∈ N such that
p≥
n+m+3
.
n+m−1
Let v(w, z) = u(z) for all (w, z) ∈ Rm+n+1
, then v satisfies
+


 ∆v = v p , ∀(w, z) ∈ Rm+n+1
,

+


v(w, z) ≥ 0, ∀(w, z) ∈ Rn+1
+ ,





v(w, x, 0) = 0, ∀(w, x) ∈ Rm+n .
, which implies that Then u(z) ≡ 0 in Rn+1
By Corollary 4, we get u(z) = v(w, z) ≡ 0 in Rm+n+1
+
+ .
10
MINGFENG ZHAO
Remark 3. Let u(x, y) = yex1 > 0 in Rn+1
+ , then u satisfies



∆u(z) = u(z), ∀z := (x, y) ∈ Rn+1

+ ,


n+1
u(z) ≥ 0, ∀z ∈ R+ ,




 u(x, 0) = 0, ∀x ∈ Rn .
References
[1] H. Berestycki, L. A. Caffarelli, and L. Nirenberg. Symmetry for elliptic equations in a half space. Boundary Value Problems for Partial
Differential Equations and Applications, 29:27–42, 1993.
[2] Haı̈m Brezis. Symmetry in nonlinear PDE’s. Proceedings of Symposia in Pure Mathematics, 65:1–12, 1999.
[3] B. Gidas and J. Spruck. A priori bounds for positive solutions of nonlinear elliptic equations. Comm. in Partial Differential Equations,
6(8):883–901, 1981.
[4] David Gilbarg and Neil S. Trudinger. Elliptic partial differential equations of second order. Springer, 2001.
[5] William P. Ziemer. Weakly Differentiable Functions. Sobolev Spaces and Functions of Bounded Variations, volume 120 of Graduate Texts
in Mathematics. Springer, 1989.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
E-mail address: mingfeng.zhao@uconn.edu