Electronic Journal of Differential Equations, Vol. 2010(2010), No. 116, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ASYMPTOTIC BEHAVIOR OF GROUND STATE SOLUTION
FOR HÉNON TYPE SYSTEMS
YING WANG, JIANFU YANG
Abstract. In this article, we investigate the asymptotic behavior of positive
ground state solutions, as α → ∞, for the following Hénon type system
2p
2q
−∆u =
|x|α up−1 v q , −∆v =
|x|α up v q−1 , in B1 (0)
p+q
p+q
with zero boundary condition, where B1 (0) ⊂ RN (N ≥ 3) is the unit ball
centered at the origin, p, q > 1, p + q < 2∗ = 2N/(N − 2). We show that both
components of the ground solution pair (u, v) concentrate on the same point
on the boundary ∂B1 (0) as α → ∞.
1. Introduction
In this article, we investigate the asymptotic behavior of positive ground state
solution pairs of the following Hénon type system
2p
2q
−∆u =
|x|α up−1 v q , −∆v =
|x|α up v q−1 , in B1 (0)
(1.1)
p+q
p+q
with zero boundary condition, where B1 (0) ⊂ RN (N ≥ 3) is the unit ball centered
at the origin, α > 0, p, q > 1, p + q < 2∗ = 2N/(N − 2).
Hénon [6] considered the so called Hénon equation
−∆u = |x|α up−1 ,
x ∈ Ω,
x ∈ ∂Ω,
u = 0,
(1.2)
which stems from a research of rotating stellar structures. Such a problem enjoys
special features. As usual, for arbitrary bounded Ω, critical exponent for problem
(1.2) is 2∗ , while if Ω is a ball, it was shown in [8] that problem (1.2) has a radially
2(N +α)
+α)
symmetric solution for p ∈ (2, 2(N
N −2 ), the critical exponent N −2 is larger than
the critical Sobolev exponent 2∗ . Moreover, even in a ball, problem (1.2) possesses
non-radial solutions under some conditions, see [9] and references therein. This can
also be seen as in [3], where it was shown that the ground state solution of problem
(1.2) has a unique maximum point approaching to a point on ∂B1 (0) provided that
α > 0 fixed, p ∈ (2, 2∗ ) and p → 2∗ . Similar results for p ∈ (2, 2∗ ) fixed and α → ∞
can be found in [1, 2, 4].
2000 Mathematics Subject Classification. 35J50, 35J57, 35J47.
Key words and phrases. Asymptotic behavior; Hénon systems; ground state solution.
c
2010
Texas State University - San Marcos.
Submitted July 14, 2010. Published August 20, 2010.
1
2
Y. WANG, J. YANG
EJDE-2010/116
For system (1.1), we proved in [10] that there exists α∗ > 0 such that the ground
state solution of problem (1.1) is non-radial if α > α∗ , p, q > 1 and p + q < 2∗ ;
the maximum points of both components u and v of the ground state solution pair
(u, v) concentrate at the same point on the boundary ∂Ω as p + q → 2∗ .
In this paper, we investigate the asymptotic behavior of the ground state solution
pair of problem (1.1) as α → ∞. Our main result is as follows.
Theorem 1.1. Let (uα , vα ) be a positive ground state solution of (1.1) and denote x0 = (0, . . . , 0, 1). Suppose xα , yα ∈ B1 (0) is a maximum point of uα , vα
respectively. Then
xα , yα → x ∈ ∂B1 (0),
lim α(1 − |xα |),
Z
α→+∞
α−
lim α(1 − |yα |) ∈ (0, +∞),
α→+∞
(2−N )(p+q)+2N
p+q−2
2
(|∇(uα − α p+q−2 u(α(x − x0 ))|2
B1 (0)
+ |∇(vα − α
2
p+q−2
v(α(x − x0 ))|2 )dx → 0
as α → +∞, where (u, v) is a ground state solution of the system
2p xN p−1 q
2q xN p q−1
−∆u =
e u v , −∆v =
e u v
, in RN
−
p+q
p+q
(1.3)
with u = v = 0 on ∂RN
−.
The proof of Theorem 1.1 is inspired by that in [4]. In section 2, we prove that
(1.3) has a ground state solution pair. We establish in section 3 an asymptotic
estimate for Sα,p,q which is defined in the section 3. Then using the blow up
argument, we show in section 4 that the maximum points of both components of
the ground state solution of (1.1) concentrate on the same point of the boundary
of the domain. The proof of Theorem 1.1 is also given in section 4.
2. A variational problem
We consider for γ > 0 the variational problem
R
(|∇u|2 + |∇v|2 )dx
RN
−
R
mγ,p,q =
inf
,
( RN eγxN |u|p |v|q dx)2/(p+q)
06=u,v∈D01,2 (RN
−)
(2.1)
−
where p + q ∈ (2, 2∗ ). We will show that mγ,p,q is achieved. First, we prove that
the problem is well defined. For any u ∈ C0∞ (RN
− ), by Hölder’s inequality,
Z 0 ∂u(x0 , t) 1/2
2 dt
.
|u(x0 , xN )| ≤ |xN |1/2
∂t
−∞
If p1 + q1 = 2 and u, v ∈ C0∞ (RN
− ), we have
Z
eγxN |u|p1 |v|q1 dx
RN
−
Z
∂u(x0 , t) 2 p21 0 ∂v(x0 , t) 2 q21 0
dt
dt
dx
≤
|xN |e
dxN
∂t
∂t
−∞
xN ≤0
RN −1
−∞
Z ∂u(x0 , x ) p21 Z ∂v(x0 , x ) q21
N 2
N 2
≤C
dx
dx
∂xN
∂xN
RN
RN
−
−
Z
γxN
Z
Z
0
EJDE-2010/116
Z
ASYMPTOTIC BEHAVIOR
3
(|∇u|2 + |∇v|2 ) dx.
≤C
RN
−
If p2 + q2 = 2∗ , again by Hölder’s inequality,
Z
Z
2∗ /2
(|∇u|2 + |∇v|2 ) dx
.
eγxN |u|p2 |v|q2 dx ≤ C
RN
−
RN
−
Using interpolation inequality for p + q ∈ (2, 2∗ ), we have
Z
Z
(p+q)/2
eγxN |u|p |v|q dx ≤ C
(|∇u|2 + |∇v|2 ) dx
.
RN
−
RN
−
This implies mγ,p,q > 0. Next, for every R > 0,
Z
Z
Z
γxN
p
q
−γR/2
e
|u| |v| dx ≤ Ce
xN ≤−R
(p+q)/2
(|∇u|2 + |∇v|2 ) dx
.
RN
−
RN −1
Hence, xN ≤−R eγxN |u|p |v|q dx is uniformly decay in the xN -direction. The variational problem mγ,p,q is compact in the xN -direction and it is translation invariant
in x1 , . . . , xN −1 . So we may prove as the proof of [11, Theorem 1.4] the following
result.
R
Proposition 2.1. Suppose p + q ∈ (2, 2∗ ), γ > 0, N ≥ 3. Then, mγ,p,q is achieved
by (u, v) with positive functions u, v ∈ D01,2 (RN
− ).
3. Estimate for Sα,p,q
It is known that the problem
(|∇u|2 + |∇v|2 ) dx
R B1 (0)
u,v∈H0 (B1 (0))\{0} ( B (0) |x|α |u|p |v|q dx)2/(p+q)
1
R
Sα,p,q =
inf
1
u,v∈H0 (B1 (0))\{0}
Jα (u, v) =
inf
1
is achieved and the minimizer is a solution of problem (1.1) up to a constant.
Furthermore, we have the following result.
Proposition 3.1. Let p + q ≥ 2. There is C > 0 such that
C≤
α
Sα,p,q
2N
2−N + p+q
≤ m1,p,q + o(1),
where o(1) → 0 as α → ∞.
Proof. We use the idea in [4]. We establish the upper bound first. For any ε > 0,
there exist wε , hε ∈ C0∞ (RN
− ), wε , hε 6= 0, such that
R
(|∇wε |2 + |∇hε |2 )dx
RN
−
R
Jε,RN
(w
,
h
)
=
< m1,p,q + ε.
ε
ε
−
( RN exN |wε |p |hε |q dx)2/(p+q)
−
Let
uα (x) = wε (αx0 , α(xN + (1 − |x0 |2 )1/2 )),
vα (x) = hε (αx0 , α(xN + (1 − |x0 |2 )1/2 )),
4
Y. WANG, J. YANG
EJDE-2010/116
where x0 = (x1 , x2 , . . . , xN −1 ). Then uα , vα ∈ H01 (B1 (0)) if α > 0 is large enough.
Denote B̃α = {y : α−1 y + x0 ∈ B1 (0)}, where x0 = (0, 0, . . . , 1). Then
Z
Z
N
−1 X
1
|∇uα |2 dx = α2−N
(
Di wε (x0 , xN + α(1 + (1 − 2 |x0 |2 )1/2 ))
α
B1 (0)
B̃α i=1
−1
α xi
1 0 2 1/2 2
0
+
|x | ) ))
D
w
(x
,
x
+
α(1
+
(1
−
N
ε
N
α2
(1 − α−2 |x0 |2 )1/2
1
+ |DN wε (x0 , xN + α(1 + (1 − 2 |x0 |2 )1/2 ))|2 ) dx.
α
(3.1)
Let
1
yi = xi , i = 1, 2, . . . , N − 1; yN = xN + α(1 + (1 − 2 |x0 |2 )1/2 ),
α
∂y
then |det( ∂x
)| = 1. By (3.1),
Z
Z
N
−1
X
2
2−N
|Di wε + O(α−1 )DN wε |2 + |DN wε |2 ) dy
|∇uα | dx = α
(
B1 (0)
RN
−
i=1
(3.2)
Z
|∇wε |2 dy + O(α−1 )).
= α2−N (
RN
−
Similarly,
Z
Z
|∇vα |2 dx = α2−N (
|∇hε |2 dy + O(α−1 )).
RN
−
B1 (0)
(3.3)
For any x ∈ sptwε ∩ spthε , we have
|
−1
x
2xN
+ x0 |α = (1 +
+ O(α−2 ))α/2 = exN +O(α ) .
α
α
Therefore,
Z
|x|α |uα |p |vα |q dx
Z
x
1
| + x0 |α |wε (x0 , xN + α(1 + (1 − 2 0 2 )1/2 ))|p
= α−N
α |x |
eα α
B
1
× |hε (x0 , xN + α(1 + (1 − 2 0 2 )1/2 ))|q dx
α |x |
Z
1
x
−α(1−(1−
)1/2 )+O(α−1 )
α2 |x0 |2
= α−N
e N
|wε |p |hε |q dx
B1 (0)
RN
Z−
−N
=α (
exN |wε |p |hε |q dy + O(α−1 )).
RN
−
It follows from (3.2), (3.3) and (3.4) that
2N
Jα (uα , vα ) = α2−N + p+q (Jε,RN
(wε , hε ) + O(α−1 ))
−
2N
< α2−N + p+q (m1,p,q + ε + O(α−1 ))
and then,
α
Sα,p,q
2N
2−N + p+q
≤ m1,p,q + o(1).
(3.4)
EJDE-2010/116
ASYMPTOTIC BEHAVIOR
5
Next, we show the lower bound. Let r ∈ (0, 1], ω ∈ S N −1 . For any u, v ∈
\ {0}), we define ϕ(r, ω) = u(rβ , ω), ψ(r, ω) = v(rβ , ω), where β = NN+α .
Then
Z
Z 1Z
|ϕ(r, ω)|p |ψ(r, ω)|q rN −1 dr dω,
|x|α |u|p |v|q dx = β
(3.5)
H01 (B1 (0)
B1 (0)
and
Z
ω∈S N −1
0
|∇u|2 dx
B1 (0)
1
Z
Z
(
=β
ω∈S N −1
0
1
=
β
1
Z
Z
1
1
|ϕr (r, ω)|2 + 2β |∇ω ϕ(r, ω)|2 )rβ(N −1)+β−1 dr dω
r
β 2 r2(β−1)
(|ϕr (r, ω)|2 +
ω∈S N −1
0
β2
|∇ω ϕ(r, ω)|2 ))r(2−N )(1−β)+N −1 dr dω.
r2
(3.6)
Similarly,
Z
|∇v|2 dx
B1 (0)
1
=
β
1
Z
β2
(|ψr (r, ω)| + 2 |∇ω ψ(r, ω)|2 )r(2−N )(1−β)+N −1 dr dω.
r
ω∈S N −1
Z
0
(3.7)
2
Note that
|∇ω ϕ|2 =
N
−1
X
i=1
(
∂ϕ
∂ϕ 2
xi
−
)
0
2
1/2
∂xi
(1 − |x | ) ∂xN
(3.8)
and dω = (1 + |x0 |2 )−1/2 dx0 . Let ϕ̄(r, x0 ) = ϕ(r, βx0 ), ψ̄(r, x0 ) = ψ(r, βx0 ), Sβ =
{x ∈ S N −1 : |x0 | ≤ ηβ}, where η > 0 is small. Then, we may deduce as in [4] that
for β > 0 small,
Z 1Z
β2
(|ϕr (r, ω)|2 + 2 |∇ω ϕ(r, ω)|2 ))r(2−N )(1−β)+N −1 dr dω
r
0
Sβ
(3.9)
Z
≥ Cβ N −1
|∇ϕ̄(r, x0 )|2 r(2−N )(1−β) dr dx0
Bη
and
Z
0
1
Z
(|ψr (r, ω)|2 +
Sβ
≥ Cβ N −1
Z
β2
|∇ω ψ(r, ω)|2 ))r(2−N )(1−β)+N −1 dr dω
r2
(3.10)
|∇ψ̄(r, x0 )|2 r(2−N )(1−β) dr dx0 ,
Bη
where Bη = {x ∈ B1 (0) : |x0 | ≤ η}. Similarly,
Z 1Z
|ϕ(r, ω)|p |ψ(r, ω)|q rN −1 dr dω
0
Sβ
≤ Cβ N −1
Z
Bη
(3.11)
|ϕ̄(r, x0 )|p |ψ̄(r, x0 )|q rN −1 dr dx0 .
6
Y. WANG, J. YANG
EJDE-2010/116
Since ϕ̄, ψ̄ = 0 on S N −1 , there exists a constant C > 0 such that
Z
Z
0 p+q N −1
0 2/(p+q)
(
|ϕ̄(r, x )| r
dr dx )
≤C
|∇ϕ̄(r, x0 )|2 r(2−N )(1−β) dr dx0
Bη
Bη
(3.12)
and
Z
|ψ̄(r, x0 )|p+q rN −1 dr dx0
2/(p+q)
Z
|∇ψ̄(r, x0 )|2 r(2−N )(1−β) dr dx0 .
≤C
Bη
Bη
(3.13)
Therefore, by (3.12) and (3.13),
Z
2/(p+q)
|ϕ̄(r, x0 )|p |ψ̄(r, x0 )|q rN −1 dr dx0
Bη
Z
(|ϕ̄(r, x0 )|p+q rN −1 dr dx0 )2/(p+q) + C
≤C
Bη
Z
Z
(|ψ̄(r, x0 )|p+q rN −1 dr dx0 )2/(p+q)
Bη
(|∇ϕ̄(r, x0 )|2 + |∇ψ̄(r, x0 )|2 )r(2−N )(1−β) dr dx0 .
≤C
Bη
(3.14)
We derive from (3.9)-(3.14) that
Z 1 Z
2/(p+q)
|ϕ(r, ω)|p |ψ(r, ω)|q rN −1 dr dω
0
Sβ
≤ Cβ
2(N −1)
p+q
Z
(
2(N −1)
p+q
Z
2
|ϕ̄(r, x0 )|p |ψ̄(r, x0 )|q rN −1 dr dx0 ) p+q
Bη
≤ Cβ
(|∇ϕ̄(r, x0 )|2 + |∇ψ̄(r, x0 )|2 )r(2−N )(1−β) dr dx0
(3.15)
Bη
≤ Cβ
1−N +
2(N −1)
p+q
Z
1
Z
0
(|ϕr (r, ω)|2 +
Sβ
β2
|∇ω ϕ(r, ω)|2
r2
β2
+ |ψr (r, ω)|2 + 2 |∇ω ψ(r, ω)|2 )r(2−N )(1−β)+N −1 dr dω
r
Since (1.1) is rotation invariant, we may choose β > 0 so that S N −1 can be covered
by finite number Sβ up to a rotation, that is S N −1 ⊂ ∪Sβ . Then
Z 1 Z
2/(p+q)
|ϕ(r, ω)|p |ψ(r, ω)|q rN −1 dr dω
0
≤
ω∈S N −1
1Z
XZ
0
≤ Cβ
|ϕ(r, ω)|p |ψ(r, ω)|q rN −1 dr dω
2/(p+q)
Sβ
2(N −1)
1−N + p+q
1
Z
0
2
Z
Sβ
(|ϕr (r, ω)|2 +
β2
|∇ω ϕ(r, ω)|2
r2
β
+ |ψr (r, ω)|2 + 2 |∇ω ψ(r, ω)|2 )r(2−N )(1−β)+N −1 dr dω
r
Z 1Z
2(N −1)
β2
(|ϕr (r, ω)|2 + 2 |∇ω ϕ(r, ω)|2
≤ Cβ 1−N + p+q
r
0
S N −1
2
β
+ |ψr (r, ω)|2 + 2 |∇ω ψ(r, ω)|2 )r(2−N )(1−β)+N −1 dr dω.
r
(3.16)
EJDE-2010/116
ASYMPTOTIC BEHAVIOR
7
Hence, we deduce from (3.6)-(3.7) and (3.16) that
R
(|∇u|2 + |∇v|2 ) dx
2N
2N
R B1 (0)
≥ Cβ N −2− p+q = Cα2−N + p+q .
α
p
q
2/(p+q)
( B1 (0) |x| |u| |v| dx)
(3.17)
It yields
(|∇u|2 + |∇v|2 ) dx
R B1 (0)
≥ C.
( B1 (0) |x|α |u|p |v|q dx)2/(p+q)
R
α
2N
N −2− p+q
(3.18)
The proof is complete since u and v are arbitrary.
4. Asymptotic behavior of ground state solution
1
2
Let (Uα , Vα ) be a minimizer of Sα,p,q . Choosing λα = ( Sα,p,q
) 2−(p+q) and defining
uα = λα Uα , vα = λα Vα , we see that (uα , vα ) is a solution pair of (1.1), which is
also a minimizer of Sα,p,q . That is,
Z
Z
(|∇uα |2 + |∇vα |2 ) dx = 2
|x|α |uα |p |vα |q dx
(4.1)
B1 (0)
B1 (0)
and
(|∇uα |2 + |∇vα |2 ) dx
B1 (0)
R
.
=
( B1 (0) |x|α |uα |p |vα |q dx)2/(p+q)
R
Sα,p,q
It yields
Z
(|∇uα |2 + |∇vα |2 ) dx = 2
B1 (0)
Z
(4.2)
2
p+q
p+q−2
.
|x|α |uα |p |vα |q dx = 2− p+q−2 Sα,p,q
(4.3)
B1 (0)
By Proposition 3.1
Cα
(2−N )(p+q)+2N
p+q−2
Z
(|∇uα |2 + |∇vα |2 ) dx ≤ C 0 α
≤
(2−N )(p+q)+2N
p+q−2
.
(4.4)
B1 (0)
Let
Then
2
2
x
x
ūα (x) = α− p+q−2 uα ( ), v̄α (x) = α− p+q−2 vα ( ), x ∈ Bα (0).
α
α
Z
C≤
(|∇ūα |2 + |∇v̄α |2 ) dx ≤ C 0 .
(4.5)
Bα (0)
Choose p1 , q1 > 0 such that p > p1 , q > q1 and p1 + q1 = 2.
Lemma 4.1. As α → +∞, we have
0 < C ≤ C max |ūα |p−p1 max |v̄α |q−q1 .
x∈Bα (0)
x∈Bα (0)
Proof. By Proposition 3.1,
R
2
Cα ≤ Sα,p1 ,q1
B1 (0)
= R
(|∇uα |2 + |∇vα |2 ) dx
B1 (0)
|x|α |uα |p1 |vα |q1 dx
.
Equation (4.5) implies
Z
Z
x
| |α |ūα |p1 |v̄α |q1 dx ≤ C
(|∇ūα |2 + |∇v̄α |2 ) dx ≤ C.
Bα (0) α
Bα (0)
Hence, by (4.3) and (4.4),
Z
x
0<C≤
| |α |ūα |p |v̄α |q dx
α
Bα (0)
(4.6)
8
Y. WANG, J. YANG
≤ max (|ūα |p−p1 |v̄α |q−q1 )
EJDE-2010/116
Z
x∈Bα (0)
x
| |α |ūα |p1 |v̄α |q1 dx
Bα (0) α
≤ C max |ūα |p−p1 max |v̄α |q−q1 .
x∈Bα (0)
x∈Bα (0)
The assertion follows.
Lemma 4.2. There is C > 0 such that |ūα (x)| ≤ C, |v̄α (x)| ≤ C for x ∈ Bα (0).
Proof. Since (uα , vα ) is a solution pair of (1.1), then for x ∈ Bα (0) we have
−∆ūα (x) =
2p x α p−1
2p p−1
| | ūα (x)v̄αq (x) ≤
ū (x)v̄αq (x)
p+q α
p+q α
−∆v̄α (x) =
2q p
2q x α p
| | ūα (x)v̄αq−1 (x) ≤
ū (x)v̄αq−1 (x).
p+q α
p+q α
(4.7)
and
Now we use the Moser iteration to prove the result. Without confusion, we use
(u, v) to denote (ūα , v̄α ). Let s ≥ 1. Multiplying (4.7) by u2s and integrating by
parts, we obtain
Z
Z
2p
−2
s 2
s (2s − 1)
|∇u | dx ≤
up−1+2s v q dx.
p + q Bα (0)
Bα (0)
Since s−2 (2s − 1) ≥ s−1 if s ≥ 1,
Z
Z
2sp
s 2
up−1+2s v q dx.
|∇u | dx ≤
p + q Bα (0)
Bα (0)
By Sobolev inequality and Hölder’s inequality, we deduce
Z
2/2∗
∗
u2 s dx
Bα (0)
Z
2sp
≤
up−1+2s v q dx
p + q Bα (0)
Z
p−1+2s Z
q
p+q−1+2s
p+q−1+2s
2sp ≤
up+q−1+2s dx
v p+q−1+2s dx
p + q Bα (0)
Bα (0)
Z
sp
p+q−1+2s
p+q−1+2s
≤
(u
+v
) dx.
p + q Bα (0)
Similarly, we have
Z
2/2∗
∗
v 2 s dx
≤
Bα (0)
Therefore,
Z
Bα (0)
(u2
∗
s
sq
p+q
Z
(up+q−1+2s + v p+q−1+2s ) dx.
(4.8)
(4.9)
Bα (0)
2/2∗ Z
2/2∗ Z
2/2∗
∗
∗
∗
+ v 2 s ) dx
≤
u2 s dx
+
v 2 s dx
B (0)
Bα (0)
Z α
≤s
(up+q−1+2s + v p+q−1+2s ) dx.
Bα (0)
(4.10)
EJDE-2010/116
ASYMPTOTIC BEHAVIOR
9
Now we define {sj } by induction. Let p+q−1+2s0 = 2∗ and p+q−1+2sj+1 = 2∗ sj ,
2∗
j = 0, 1, 2, . . . . We also define M0 = 1, Mj+1 = (sj Mj ) 2 , j = 0, 1, 2, . . . . We claim
that for all j ≥ 0,
Z
(4.11)
(up+q−1+2sj + v p+q−1+2sj ) dx ≤ CMj
Bα (0)
and
Mj ≤ emsj−1 ,
(4.12)
where C, m > 0. (4.11) and (4.12) imply
1
Z
2∗1s
msj−1
∗
∗
2∗ s
∗
j
(u2 sj + v 2 sj ) dx
≤ CMj j ≤ e 2 sj ≤ C
Bα (0)
for all j. The assertion then follows. Now, we show (4.11). Obviously, if j = 0,
(4.11) holds. Suppose it holds for j, we deduce it holds for j + 1. Indeed,
Z
(up+q−1+2sj+1 + v p+q−1+2sj+1 ) dx
Bα (0)
Z
∗
∗
=
(u2 sj + v 2 sj ) dx
Bα (0)
≤
2∗ /2
sj
2∗ /2
(up+q−1+2sj + v p+q−1+2sj ) dx
Z
Bα (0)
≤ sj Mj
2∗ /2
= Mj+1 .
Inequality (4.12) can be proved, as in [7].
Let Mα = maxx∈B1 (0) uα , Nα = maxx∈B1 (0) vα .
Lemma 4.3. There holds
2
2
C2 α p+q−2 ≤ Mα , Nα ≤ C3 α p+q−2 .
Proof. By Lemmas 4.1 and 4.2, we have
0 < C1 ≤ C max |ūα |p−p1
x∈Bα (0)
This yields the result.
and 0 < C1 ≤ C max |v̄α |q−q1 .
x∈Bα (0)
Let xα ∈ B1 (0) be a maximum point of uα and yα ∈ B1 (0) be a maximum point
of vα .
Lemma 4.4. The following hold limα→+∞ α(1 − |xα |) and limα→+∞ α(1 − |yα |)
are in ∈ (0, +∞).
Proof. We only prove limα→+∞ α(1 − |xα |) = L and L ∈ (0, +∞). The other case
can be done in the same way. Let Bα (−xα ) = {x : αx + xα ∈ B1 (0)} and define
2
2
x
x
ũα (x) = α− p+q−2 uα ( + xα ), ṽα (x) = α− p+q−2 vα ( + xα ).
α
α
Then, for x ∈ Bα (−xα ),
α
2p x
q
−∆ũα (x) =
+ xα ũp−1
α (x)ṽα (x).
p+q α
10
Y. WANG, J. YANG
EJDE-2010/116
By Lemma 4.3,
ũα , ṽα ≤ C,
and
ũα (0) =
Z
max
x∈Bα (−xα )
ũα (x) ≥ C1 > 0
(|∇ũα |2 + |∇ṽα |2 ) dx ≤ C.
Bα (−xα )
Suppose that α(1 − |xα |) → +∞, we assume that there are ũ, ṽ ∈ D1,2 (RN ) such
that
ũα * ũ,
ṽα * ṽ,
in D1,2 (RN )
ũα → ũ,
ṽα → ṽ,
1
in Cloc
(RN ).
Now, we distinguish two cases: (i) |xα | ≤ l < 1; (ii) |xα | → 1 as α → +∞. For any
x with |x| ≤ C, in case (i), we have
C
C
x
| + xα |α ≤ ( + |xα |)α ≤ ( + l)α ≤ (l + ε)α → 0, as α → +∞.
α
α
α
In case (ii), since α(1 − |xα |) → +∞,
|x|
|x|
x
| + xα |α ≤ eα ln( α +|xα |−1+1) = O(eα( α +|xα |−1) ) = O(e|x|+α(|xα |−1) ) → 0.
α
So ũ satisfies
−∆ũ = 0, ũ ∈ D1,2 (RN ).
This implies ũ = 0, a contradiction to ũ(0) = limα→+∞ ũα (0) ≥ C > 0. Therefore,
α(1 − |xα |) → L < +∞.
Now, we claim L > 0. Indeed, we have ũ(0) = limα→+∞ ũα (0) > 0. Since (1.1)
is invariant under the rotations. After suitably rotating the coordinate system, we
α
may assume that xα = (0, . . . , 0, xα
N ), where xN → 1, as α → +∞. Then (ũ, ṽ) is a
N
positive solution pair of (1.3) in Ω = R− + (0, . . . , 0, L) with ũ = ṽ = 0 on ∂Ω. If
L = 0, we would have Ω = RN
− , and then we obtain ũ(0) = 0, a contradiction. The
proof is complete.
By Lemma 4.4, we know that xα → x0 ∈ ∂B1 (0), yα → y0 ∈ ∂B1 (0) if α → +∞.
In the following, we show that x0 = y0 .
Lemma 4.5. Both xα and yα converge to a point x0 ∈ ∂B1 (0) as α → +∞.
Proof. We argue by contradiction. Suppose x0 6= y0 , then there is a δ > 0 such
that Bδ (x0 ) ∩ Bδ (y0 ) = ∅. After suitably rotating the coordinate system, we may
assume that x0 = (0, . . . , 0, 1). Applying the blow up argument for
2
2
x
x
ũα (x) = α− p+q−2 uα ( + x0 ), ṽα (x) = α− p+q−2 vα ( + x0 )
α
α
in B1 (0)∩Bδ (x0 ), since ũα , ṽα are bounded in D01,2 (RN
− ), we may assume that there
)
such
that
are ũ, ṽ ∈ D01,2 (RN
−
ũα * ũ,
ṽα * ṽ,
in D01,2 (RN
− ),
ũα → ũ,
ṽα → ṽ,
1
in Cloc
(RN
− ).
Moreover, (ũ, ṽ) with ũ, ṽ ∈ D01,2 (RN
− ) is a positive solution of (1.3). In the same
way, we may assume y0 = (0, . . . , 0, 1). Define
2
2
x
x
ūα (x) = α− p+q−2 uα ( + y0 ), v̄α (x) = α− p+q−2 vα ( + y0 ).
α
α
EJDE-2010/116
ASYMPTOTIC BEHAVIOR
11
Then
ūα * ū,
v̄α * v̄,
in D01,2 (RN
− ),
ūα → ū,
v̄α → v̄,
1
in Cloc
(RN
− ),
and (ū, v̄) is a positive solution of (1.3). It implies
Z
Z
p+q
2
p+q−2
(|∇ũ|2 + |∇ṽ|2 )dx,
(|∇ū|2 + |∇v̄|2 )dx ≥ 2− p+q−2 m1,p,q
RN
−
RN
−
By Proposition 3.1,
Z
1
1
)
(|∇uα |2 + |∇vα |2 )dx
I(uα , vα ) = ( −
2 p + q B1 (0)
p+q
2
1
1
p+q−2
=( −
)2− p+q−2 Sα,p,q
2 p+q
p+q
(2−N )(p+q)+2N
2
1
1
p+q−2
p+q−2
)2− p+q−2 α
+ o(1)).
≤( −
(m1,p,q
2 p+q
On the other hand,
Z
1
1
I(uα , vα ) ≥ ( −
)
(|∇uα |2 + |∇vα |2 )dx
2 p + q B1 (0)∩Bδ (x0 )
Z
1
1
)
(|∇uα |2 + |∇vα |2 )dx
+( −
2 p + q B1 (0)∩Bδ (y0 )
Z
(2−N )(p+q)+2N 1
1
p+q−2
≥α
( −
)
(|∇ũα |2 + |∇ṽα |2 )dx
2 p + q Bα (−x0 )∩Bαδ (0)
Z
(2−N )(p+q)+2N 1
1
p+q−2
+α
( −
)
(|∇ūα |2 + |∇v̄α |2 )dx.
2 p + q Bα (−y0 )∩Bαδ (0)
So we obtain
Z
(|∇ũα |2 + |∇ṽα |2 )dx +
Z
Bα (−x0 )∩Bαδ (0)
2
− p+q−2
≤2
(|∇ūα |2 + |∇v̄α |2 )dx
Bα (−θy0 )∩Bαδ (0)
p+q
p+q−2
(m1,p,q + o(1)).
Therefore,
2
− p+q−2
2
p+q
p+q−2
Z
(m1,p,q + o(1)) ≥
2
2
Z
(|∇ũ| + |∇ṽ| )dx +
RN
−
(|∇ū|2 + |∇v̄|2 )dx
RN
+
2
p+q
p+q
p+q−2
p+q−2
≥ 2− p+q−2 (m1,p,q
+ M1,p,q
),
which is impossible. The proof is complete.
Now, we may assume that x0 = (0, . . . , 0, 1). Let
2
2
x
x
ûα (x) = α− p+q−2 uα ( + x0 ), v̂α (x) = α− p+q−2 vα ( + x0 ),
α
α
which, as before, satisfies
ûα * û,
ûα → û,
v̂α * v̂,
v̂α → v̂,
in D01,2 (RN
− ),
1
in Cloc
(RN
−)
and (û, v̂) 6= (0, 0) is a positive solution of (1.3).
Finally, we have following result.
(4.13)
12
Y. WANG, J. YANG
EJDE-2010/116
Proposition 4.6. The pair (û, v̂) is a minimizer of m1,p,q , which satisfies
Z
(|∇(ûα − û)|2 + |∇(v̂α − v̂)|2 )dx → 0, as α → +∞.
N
R−
Proof. By (1.3), we have
Z
Z
(|∇û|2 + |∇v̂|2 )dx = 2
RN
−
and
exN ûp v̂ q dx,
RN
−
(|∇u|2 + |∇v|2 )dx
RN
.
m1,p,q ≤ R − x
( RN e N |u|p |v|q dx)2/(p+q)
R
−
So we obtain
Z
p+q
2
RN
−
p+q−2
(|∇u|2 + |∇v|2 )dx ≥ 2− p+q−2 m1,p,q
.
For R > 0 define
x
x
+ x0 ∈ B R (x0 ) ∩ B1 (0)}, Ωα = {x : + x0 ∈ B1 (0)}.
α
α
α
By Proposition 3.1, (uα , vα ) is a minimizer of Sα,p,q and satisfies (1.1), then
Z
p+q
(2−N )(p+q)+2N
2
p+q−2
p+q−2
(|∇uα |2 + |∇vα |2 )dx ≤ 2− p+q−2 α
+ o(1)).
(4.14)
(m1,p,q
BR,α = {x :
B1 (0)
Moreover,
Z
(|∇uα |2 + |∇vα |2 )dx
B1 (0)
Z
Z
2
2
=
(|∇uα | + |∇vα | )dx +
B R (x
0
α
=α
∩B1 (0)
)
(2−N )(p+q)+2N
p+q−2
Z
(|∇ûα |2 + |∇v̂α |2 )dx +
≥α
Zα
0)
(|∇ûα |2 + |∇v̂α |2 )dx
Ωα /BR,α
BR,α
(2−N )(p+q)+2N
p+q−2
(|∇uα |2 + |∇vα |2 )dx
B1 (0)/B R (x
Z
(|∇ûα |2 + |∇v̂α |2 )dx + o(1)
RN
− ∩BR (0)
≥α
Z
(2−N )(p+q)+2N
(|∇u|2 + |∇v|2 )dx + o(1)
p+q−2
RN
− ∩BR (0)
≥α
(2−N )(p+q)+2N
p+q−2
2
= 2− p+q−2 α
2
p+q
p+q−2
(2− p+q−2 m1,p,q
+ o(1))
(2−N )(p+q)+2N
p+q−2
p+q
p+q−2
((m1,p,q
+ o(1)).
(4.15)
By (4.9) and (4.15),
Z
(|∇ûα |2 + |∇v̂α |2 )dx = o(1) + oR (1),
Ωα /BR,α
Z
2
2
Z
(|∇u|2 + |∇v|2 )dx + o(1)
(|∇ûα | + |∇v̂α | )dx =
BR,α
RN
− ∩BR (0)
2
p+q
p+q−2
= 2− p+q−2 m1,p,q
+ o(1) + oR (1).
EJDE-2010/116
ASYMPTOTIC BEHAVIOR
Let R → +∞, the above equation yields
Z
p+q
2
p+q−2
(|∇u|2 + |∇v|2 )dx = 2− p+q−2 m1,p,q
.
13
(4.16)
RN
−
An application of the Brezis-Lieb’s Lemma gives
Z
(|∇(ûα − u)|2 + |∇(v̂α − v)|2 )dx → 0
RN
−
as α → +∞. On the other hand, by (1.3) and (4.16),
R
Z
(|∇u|2 + |∇v|2 )dx
p+q−2
RN
−
2/(p+q)
R
=2
(
(|∇u|2 + |∇v|2 )dx) p+q = m1,p,q .
p
q
2/(p+q)
x
N
( RN e |u| |v| dx)
RN
−
−
This implies that (u, v) achieves m1,p,q . As a consequence, we have
Z
(2−N )(p+q)+2N
2
−
p+q−2
(|∇(uα − α p+q−2 u(α(x − x0 ))|2
α
B1 (0)
+ |∇(vα − α
2
p+q−2
v(α(x − x0 ))|2 )dx → 0,
as α → +∞.
Now the the proof of Theorem 1.1 is completed by Lemmas 4.4, 4.5 and Proposition 4.6.
Acknowledgements. This research was partially supported by grants N10961016
and 10631030 from the NNSF of China, and 2009GZS0011 from NSF of Jiangxi.
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Ying Wang
Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022,
China
E-mail address: yingwang00@126.com
14
Y. WANG, J. YANG
EJDE-2010/116
Jianfu Yang
Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022,
China
E-mail address: jfyang 2000@yahoo.com