1 Introduction
In recent years, many authors have studied the existence of solutions of the following nonlinear Schrödinger system:
(1.1)
{
-
△
u
+
λ
1
u
=
μ
1
|
u
|
2
q
-
2
u
+
β
|
v
|
q
|
u
|
q
-
2
u
,
x
∈
Ω
,
-
△
v
+
λ
2
v
=
μ
2
|
v
|
2
q
-
2
v
+
β
|
u
|
q
|
v
|
q
-
2
v
,
x
∈
Ω
,
u
=
v
=
0
,
x
∈
∂
Ω
,
where λ1, λ2, μ1, μ2, β are constants, and Ω⊂ℝN is a bounded domain with smooth boundary or Ω=ℝN. System (1.1) arises in many physical problems, especially in nonlinear optics and Bose–Einstein condensation. We refer to [12, 14] and the references therein for experimental results and physical background of this system.
A solution (u0,v0)∈E:=H1(ℝN)×H1(ℝN) of (1.1) is called positive if u0>0, v0>0 and nontrivial if (u0,v0)≠(0,0). This solution is a ground state in the sense that (u0,v0)≠(0,0) and its energy is minimal among the energy of all nontrivial solutions. For the case q=2 and N≤3, if β>0 is small enough, existence of one positive solution of (1.1) was proved in [15], and if β∈(0,β1]∪[β2,+∞), where β1, β2 are positive constants determined in terms of λi, μi and N, existence of one positive solution of (1.1) was obtained in [1, 2, 4, 23]. For the general subcritical case 1<q<2*2:=N(N-2)+, N∈ℕ+ and β>β3, existence of one positive solution of (1.1) was shown in [20, 22]. The β3 mentioned previously is a nonnegative constant determined in terms of λi, μi, q and N. Moreover, β3>0 if q≥2 and β3=0 if 1<q<2. We refer the readers to [5, 8, 13, 18, 19, 21, 25, 24] for other results on the existence and multiplicity of solutions to (1.1) with subcritical exponent 1<q<2*2. For the critical case, where q=2*2 and Ω is a bounded smooth domain in ℝN, existence of a positive solution was studied in [11, 10, 27], and existence of a sign-changing solution was proved in [9].
Partially motivated by [1, 2, 4, 20, 22, 23], in this paper, we consider a more general system with critical exponent when β>0:
(1.2)
{
-
△
u
+
u
=
f
(
u
)
+
β
h
(
u
)
K
(
v
)
,
x
∈
ℝ
N
,
-
△
v
+
v
=
g
(
v
)
+
β
H
(
u
)
k
(
v
)
,
x
∈
ℝ
N
,
u
,
v
∈
H
1
(
ℝ
N
)
,
where f,g:ℝ→ℝ are C1 functions with critical growth; h,k are also C1 functions, and H(s):=∫0sh(t)𝑑t, K(s):=∫0sk(t)𝑑t. Our purpose is to search for a positive ground state solution of (1.2). Denote F(s):=∫0sf(t)𝑑t, G(s):=∫0sg(t)𝑑t, and the integral ∫ℝN⋅dx by ∫⋅. It is well known that a solution of (1.2) corresponds to a critical point of the energy functional I:E→ℝ defined by
I
(
u
,
v
)
:=
1
2
∫
(
|
∇
u
|
2
+
|
u
|
2
+
|
∇
v
|
2
+
|
v
|
2
)
-
∫
(
F
(
u
)
+
G
(
v
)
+
β
H
(
u
)
K
(
v
)
)
.
Set
𝒩
:=
{
(
u
,
v
)
∈
E
∖
{
(
0
,
0
)
}
:
〈
I
′
(
u
,
v
)
,
(
u
,
v
)
〉
=
0
}
.
Then a positive ground state of (1.2) is a positive solution of the minimization problem
m
:=
inf
𝒩
I
(
u
,
v
)
.
To find the minimizer, for the case N≥3, we assume that
lims→0+f(s)s=lims→0+g(s)s=0.
lim sups→+∞f(s)s2*-1≤1 and lim sups→+∞g(s)s2*-1≤1.
f′(s)s-f(s)>0 and g′(s)s-g(s)>0 if s>0.
There are λ>0 and 2<r<2* such that f(s)≥λsr-1 and g(s)≥λsr-1 for every s≥0.
|h(s)K(t)|≤|s|2*-1+|t|2*-1 and |H(s)k(t)|≤|s|2*-1+|t|2*-1 for |st|≥C0>0.
There exist C1>0,C2>0 and 1<p≤min{2*2,2} such that lims→0+h′(s)sp-2=C1 and lims→0+k′(s)sp-2=C2.
H(s)K(t)>0 and h(s)K(t)s+H(s)k(t)t-2H(s)K(t)>0 for s>0, t>0.
Let S and Cr be the best constants satisfying
S
(
∫
|
u
|
2
*
)
2
2
*
≤
∫
|
∇
u
|
2
for all
u
∈
D
1
,
2
(
ℝ
N
)
,
and
C
r
(
∫
|
u
|
r
)
2
r
≤
∫
(
|
∇
u
|
2
+
|
u
|
2
)
for all
u
∈
H
1
(
ℝ
N
)
.
The first result of the current paper is as follows.
Theorem 1.1.
Assume that (f1), (f2), (f3), (f4), (h1), (h2), and (h3) hold. If N≥3 and
λ
>
(
N
(
N
N
-
2
)
N
-
2
2
(
1
+
β
)
N
-
2
2
S
-
N
2
)
r
-
2
2
(
r
-
2
2
r
)
r
-
2
2
C
r
r
2
,
then problem (1.2) has a ground state solution (u,v) for all β>0. Moreover, if 1<p<2, then u>0,v>0; if p=2, then there is β*>0 such that u>0, v>0 if β>β*.
Remark 1.2.
Throughout this paper, the constants λ and r are defined as in condition (f4) and p is defined as in (h2).
For the case of dimension N=2, according to a Trudinger–Moser type inequality [7, Lemma 2.1], we replace the growth assumptions (f2) and (h1) with the following two conditions, respectively:
lims→+∞f(s)eαs2=lims→+∞g(s)eαs2=0(+∞) if α>4π(α<4π).
|h(s)K(t)|≤C(eαs2-1)+C(|t|(eαt2-1))r1-1r1 and |H(s)k(t)|≤C(eαt2-1)+C(|s|(eαs2-1))r2-1r2 if r1,r2>2 and α>4π.
We get the following result:
Theorem 1.3.
Assume that (f1), (f2’), (f3), (f4), (h1’), (h2) and (h3) hold. If N=2 and
λ
>
(
r
-
2
r
)
r
-
2
2
C
r
r
2
,
then problem (1.2) has a ground state solution (u,v) for all β>0. Moreover, if 1<p<2, then u>0, v>0; if p=2, then there is β**>0 such that u>0, v>0 if β>β**.
Our results complement those in [20, 22] in the sense that we address the critical exponent problem. Moreover, it is easy to verify that (1.1) is a special case of (1.2) if μ1>0,μ2>0 are large enough and 1<q<2*2. Our results indicates that (1.1) has a positive ground state for all β>0 if 1<q<2 and for β≥β* (or β**) if 2≤q<2*2, which is consistent with [20, Theorem 2.3] and [22, Corollary 1].
To prove Theorem 1.1 and Theorem 1.3, we first need to estimate certain upper bound for the minimum level m since we are dealing with a problem with critical exponent. However, it is too difficult to get this bound directly. Fortunately, when we consider the minimization problem
(1.3)
A
:=
inf
ℳ
1
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
,
where
ℳ
:=
{
{
(
u
,
v
)
∈
E
∖
{
(
0
,
0
)
}
:
∫
T
(
u
,
v
)
=
1
}
,
if
N
≥
3
,
{
(
u
,
v
)
∈
E
∖
{
(
0
,
0
)
}
:
∫
T
(
u
,
v
)
=
0
}
,
if
N
=
2
,
and
T
(
u
,
v
)
:=
F
(
u
)
+
G
(
v
)
+
β
H
(
u
)
K
(
v
)
-
|
u
|
2
2
-
|
v
|
2
2
,
we are able to estimate an upper bound of the minimum level A relating to the best constant of critical Sobolev imbedding, which allows us to prove that A is attained. Then by translation invariance we get that m is attained and (1.2) has a ground state solution. After that, we introduce the following single equations:
(1.4)
-
△
u
+
u
=
f
(
u
)
,
u
∈
H
1
(
ℝ
N
)
,
and
(1.5)
-
△
u
+
u
=
g
(
u
)
,
u
∈
H
1
(
ℝ
N
)
,
which play an important role in the proof of our results. Under the assumptions (f1)–(f4) for N≥3 and (f1), (f2’), (f3), (f4) for N=2, Alves and Souto [3] proved that both (1.4) and (1.5) have a positive ground state solution respectively if
λ
>
(
N
(
N
N
-
2
)
N
-
2
2
S
-
N
2
)
r
-
2
2
(
r
-
2
2
r
)
r
-
2
2
C
r
r
2
for
N
≥
3
,
and
λ
>
(
r
-
2
r
)
r
-
2
2
C
r
r
2
for
N
=
2
.
Denote the positive ground state solutions mentioned previously by U1 and U2, respectively. It is clear that (U1,0) and (0,U2) solve (1.2). However, we are going to find a solution with both components being positive. We need to compare the minimal level m with the energy of (U1,0) and (0,U2), which enables us to get sufficient conditions for existence of positive ground state solution of (1.2).
The rest of this paper is organized as follows. Section 2 is devoted to recall and fix some notations. The cases of dimension N≥3 and dimension N=2 are treated in Section 3 and Section 4, respectively.
3 The Case of Dimension N≥3
In this case, it is important to observe that the ℳ defined in Section 1 is a C1 manifold since we will use Ekeland’s Variational Principle in the following proof (f3), we have
f
(
s
)
s
-
2
F
(
s
)
=
∫
0
s
(
f
′
(
s
)
s
-
f
(
s
)
)
>
0
,
g
(
s
)
s
-
2
G
(
s
)
=
∫
0
s
(
g
′
(
s
)
s
-
g
(
s
)
)
>
0
for all s>0. Then for all s, it holds
(3.1)
f
(
s
)
s
-
2
F
(
s
)
≥
0
,
g
(
s
)
s
-
2
G
(
s
)
≥
0
.
Let J(u,v):=∫T(u,v), then from (3.1) and (h3),
〈
J
′
(
u
,
v
)
,
(
u
,
v
)
〉
=
∫
(
f
(
u
)
u
+
g
(
v
)
v
+
β
(
h
(
u
)
K
(
v
)
u
+
H
(
u
)
k
(
v
)
v
)
-
u
2
-
v
2
)
≥
∫
(
2
F
(
u
)
+
2
G
(
v
)
+
2
β
H
(
u
)
K
(
v
)
-
u
2
-
v
2
)
≥
2
J
(
u
,
v
)
=
2
≠
0
,
that is, J′(u,v)≠0 for every (u,v)∈ℳ. So the conclusion follows.
The following lemmas will be used in the sequel:
Lemma 3.1.
Any minimizing sequence {(un,vn)} of (1.3) is bounded in Er.
Proof.
If {(un,vn)} is a minimizing sequence of (1.3), we have
1
2
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
→
A
and
(3.2)
∫
(
F
(
u
n
)
+
G
(
v
n
)
+
β
H
(
u
n
)
K
(
v
n
)
)
=
∫
(
|
u
n
|
2
2
+
|
v
n
|
2
2
)
+
1
.
Using the growth assumptions on f, g, h and k, we know that there is C>0 such that
F
(
s
)
≤
1
8
s
2
+
C
s
2
*
,
G
(
s
)
≤
1
8
s
2
+
C
s
2
*
,
(3.3)
H
(
s
)
K
(
t
)
≤
C
(
s
2
*
+
t
2
*
)
+
1
8
β
(
s
2
+
t
2
)
.
From (3.2) and (3.3) we get
C
1
∫
(
|
u
n
|
2
*
+
|
v
n
|
2
*
)
≥
1
4
∫
(
|
u
n
|
2
+
|
v
n
|
2
)
+
1
for some C1>0. On the other hand, the definition of S indicates that
∫
(
|
u
n
|
2
*
+
|
v
n
|
2
*
)
≤
S
-
2
*
2
(
∫
|
∇
u
n
|
2
)
2
*
2
+
S
-
2
*
2
(
∫
|
∇
v
n
|
2
)
2
*
2
.
Thus, ∫(|un|2+|vn|2) is bounded, and so {(un,vn)} is bounded in Er. ∎
Proof.
Notice that A≥0. To obtain a contradiction, assume that A=0 and let {(un,vn)} be a minimizing sequence of (1.3) in Er such that
1
2
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
→
0
,
and
∫
(
F
(
u
n
)
+
G
(
v
n
)
+
β
H
(
u
n
)
K
(
v
n
)
)
=
∫
(
|
u
n
|
2
2
+
|
v
n
|
2
2
)
+
1
.
Using a similar argument as in the proof of Lemma 3.1, it is not difficult to verify that
(3.4)
C
1
∫
(
|
u
n
|
2
*
+
|
v
n
|
2
*
)
≥
1
4
∫
(
|
u
n
|
2
+
|
v
n
|
2
)
+
1
≥
1
for some
C
1
>
0
.
However, since (un,vn)→0 in D1,2(ℝN)×D1,2(ℝN), it holds
∫
(
|
u
n
|
2
*
+
|
v
n
|
2
*
)
→
0
,
which contradicts (3.4). Thus, A>0. ∎
Lemma 3.3.
The following holds:
b
≥
1
N
(
N
-
2
2
N
)
N
-
2
2
(
2
A
)
N
2
.
Proof.
We first claim that γ([0,1])∩𝒫≠∅ for each γ∈Γ. Define
ϕ
(
u
,
v
)
:=
N
-
2
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
-
N
∫
T
(
u
,
v
)
=
N
I
(
u
,
v
)
-
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
.
Then there is ρ0>0 small enough such that ϕ(u,v)>0 for all (u,v) satisfying 0<∥(u,v)∥≤ρ0. For any γ∈Γ, it is clear that ϕ(γ(0))=0 and ϕ(γ(1))≤NI(γ(1))<0. Thus, there is t0∈(0,1) such that ∥γ(t0)∥>ρ0 and ϕ(γ(t0))=0, which implies that γ(t0)∈𝒫 and
d
≤
I
(
γ
(
t
0
)
)
≤
max
t
∈
[
0
,
1
]
I
(
γ
(
t
)
)
.
Hence,
d
≤
I
(
γ
(
t
0
)
)
≤
inf
γ
∈
Γ
max
t
∈
[
0
,
1
]
I
(
γ
(
t
)
)
=
b
.
On the other hand, there is a one-to-one correspondence Φ:ℳ→𝒫 such that
Φ
(
u
,
v
)
(
x
)
:=
(
u
(
x
t
u
)
,
v
(
x
t
u
)
)
,
t
u
:=
N
-
2
2
N
(
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
)
1
2
.
Then
d
=
inf
𝒫
I
(
u
,
v
)
=
inf
ℳ
I
(
Φ
(
u
,
v
)
(
x
)
)
=
inf
ℳ
1
2
t
u
N
-
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
-
t
u
N
∫
T
(
u
,
v
)
=
inf
ℳ
1
N
(
N
-
2
2
N
)
N
-
2
2
(
∫
|
∇
u
|
2
+
|
∇
v
|
2
)
N
2
=
1
N
(
N
-
2
2
N
)
N
-
2
2
(
2
A
)
N
2
.
Consequently,
b
≥
d
=
1
N
(
N
-
2
2
N
)
N
-
2
2
(
2
A
)
N
2
.
∎
Applying Ekeland’s Variational Principle [26], we obtain that there are sequences {(un,vn)}⊂ℳ∩Er and {λn}⊂ℝ such that
1
2
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
→
A
,
and
(3.5)
L
′
(
u
n
,
v
n
)
-
λ
n
J
′
(
u
n
,
v
n
)
→
0
in
H
-
1
(
ℝ
N
)
×
H
-
1
(
ℝ
N
)
,
where L(u,v):=12∫(|∇u|2+|∇v|2).
Lemma 3.4.
Let {λn} be the sequence obtained in (3.5), then lim supn→∞λn≤A.
Proof.
From (3.1), (3.5) and condition (h3), we get
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
=
λ
n
∫
(
f
(
u
n
)
u
n
+
g
(
v
n
)
v
n
+
β
h
(
u
n
)
K
(
v
n
)
u
n
+
β
H
(
u
n
)
k
(
v
n
)
v
n
-
u
n
2
-
v
n
2
)
+
o
(
1
)
≥
λ
n
∫
(
2
F
(
u
n
)
+
2
G
(
v
n
)
+
2
β
H
(
u
n
)
K
(
v
n
)
-
u
n
2
-
v
n
2
)
+
o
(
1
)
=
2
λ
n
+
o
(
1
)
.
Taking into account 12∫|∇un|2+|∇vn|2→A, we deduce that lim supn→∞λn≤A. ∎
According to the Concentration Compactness Principle of Lions [16, 17], for the minimizing sequence {(un,vn)} of (1.3), there are positive finite measures μ(1), μ(2), ν(1), ν(2) and sequences {μi(1)}, {νi(1)}, {μi(2)}, {νi(2)} such that
|
∇
u
n
|
2
⇀
μ
(
1
)
≥
|
∇
u
|
2
+
Σ
δ
x
i
μ
i
(
1
)
,
|
∇
v
n
|
2
⇀
μ
(
2
)
≥
|
∇
v
|
2
+
Σ
δ
x
i
μ
i
(
2
)
,
|
u
n
|
2
*
⇀
ν
(
1
)
=
|
u
|
2
*
+
Σ
δ
x
i
ν
i
(
1
)
,
|
v
n
|
2
*
⇀
ν
(
2
)
=
|
v
|
2
*
+
Σ
δ
x
i
ν
i
(
2
)
,
μ
i
(
1
)
≥
S
(
ν
i
(
1
)
)
2
2
*
,
μ
i
(
2
)
≥
S
(
ν
i
(
2
)
)
2
2
*
.
Let μi:=μi(1)+μi(2),νi:=νi(1)+νi(2),μ:=μ(1)+μ(2),ν:=ν(1)+ν(2). Then
(3.6)
|
∇
u
n
|
2
+
|
∇
v
n
|
2
⇀
μ
≥
|
∇
u
|
2
+
|
∇
v
|
2
+
Σ
δ
x
i
μ
i
,
|
u
n
|
2
*
+
|
v
n
|
2
*
⇀
ν
=
|
u
|
2
*
+
|
v
|
2
*
+
Σ
δ
x
i
ν
i
.
Moreover, it is easy to verify that
(3.7)
μ
i
≥
S
ν
i
2
2
*
.
Indeed, let ψ be a smooth function with compact support satisfying 0≤ψ(x)≤1, ψ(x)=1 in B1(0) and ψ(x)=0 in ℝN∖B2(0). Set ψϵ(x):=ψ(x-xiϵ) for ϵ>0. Then
∫
|
∇
ψ
ϵ
u
n
|
2
+
∫
|
∇
ψ
ϵ
v
n
|
2
≥
S
(
∫
|
ψ
ϵ
u
n
|
2
*
)
2
2
*
+
S
(
∫
|
ψ
ϵ
v
n
|
2
*
)
2
2
*
≥
S
(
∫
|
ψ
ϵ
u
n
|
2
*
+
∫
|
ψ
ϵ
v
n
|
2
*
)
2
2
*
.
By letting ϵ→0, we obtain (3.7).
Lemma 3.5.
If νi>0 for some index i, then A≥2-2N(1+β)-N-2NS.
Proof.
Let {(un,vn)}⊂Er be a minimizing sequence of (1.3) to A. Then from (3.5) we infer that
∫
(
∇
u
n
∇
(
u
n
ψ
ϵ
)
+
∇
v
n
∇
(
v
n
ψ
ϵ
)
)
=
λ
n
∫
(
f
(
u
n
)
u
n
+
g
(
v
n
)
v
n
)
ψ
ϵ
(3.8)
+
λ
n
∫
(
β
h
(
u
n
)
K
(
v
n
)
u
n
+
β
H
(
u
n
)
k
(
v
n
)
v
n
-
u
n
2
-
v
n
2
)
ψ
ϵ
+
o
(
1
)
.
The growth assumptions on f, g, h and k imply that, for any η>1, there is C>0 such that
s
f
(
s
)
≤
η
s
2
*
+
s
2
4
+
C
|
s
|
q
1
,
s
g
(
s
)
≤
η
s
2
*
+
s
2
4
+
C
|
s
|
q
2
,
(3.9)
h
(
s
)
K
(
t
)
s
+
H
(
s
)
k
(
t
)
t
≤
η
(
s
2
*
+
t
2
*
)
+
C
(
|
s
|
q
3
+
|
t
|
q
4
)
+
1
4
β
(
s
2
+
t
2
)
for some 2<q1, q2, q3, q4<2*. From (3.8) and (3.9), we get
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
ψ
ϵ
+
∫
(
u
n
∇
u
n
+
v
n
∇
v
n
)
∇
ψ
ϵ
≤
η
(
1
+
β
)
λ
n
∫
(
|
u
n
|
2
*
+
|
v
n
|
2
*
)
ψ
ϵ
+
C
λ
n
∫
(
|
u
n
|
q
1
+
|
v
n
|
q
2
+
β
|
u
n
|
q
3
+
β
|
v
n
|
q
4
)
ψ
ϵ
+
λ
n
2
∫
(
|
u
n
|
2
+
|
v
n
|
2
)
ψ
ϵ
.
Letting ϵ→0 and using Lemma 3.4, we have μi≤Aη(1+β)νi for all η>1. So that μi≤A(1+β)νi, which together with (3.7) gives
S
ν
i
2
2
*
≤
μ
i
≤
A
(
1
+
β
)
ν
i
.
Then it follows
(3.10)
ν
i
≥
(
S
A
(
1
+
β
)
)
N
2
.
On the other hand, since
∫
(
|
u
n
|
2
*
+
|
v
n
|
2
*
)
≤
S
-
2
*
2
(
∫
|
∇
u
n
|
2
)
2
*
2
+
S
-
2
*
2
(
∫
|
∇
v
n
|
2
)
2
*
2
≤
S
-
2
*
2
(
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
)
2
*
2
,
we have
∫
(
|
u
n
|
2
*
+
|
v
n
|
2
*
)
ψ
ϵ
≤
S
-
2
*
2
(
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
)
2
*
2
.
Letting n→∞ and ϵ→0 gives
(3.11)
ν
i
≤
S
-
2
*
2
(
2
A
)
2
*
2
.
From (3.10) and (3.11), we deduce that
A
≥
2
-
2
N
(
1
+
β
)
-
N
-
2
N
S
.
∎
Lemma 3.6.
If
(3.12)
λ
>
(
N
(
N
N
-
2
)
N
-
2
2
(
1
+
β
)
N
-
2
2
S
-
N
2
)
r
-
2
2
(
r
-
2
2
r
)
r
-
2
2
C
r
r
2
,
then
(3.13)
b
<
1
N
(
N
-
2
N
)
N
-
2
2
(
1
+
β
)
-
N
-
2
2
S
N
2
.
Proof.
If we take h0∈Hr1(ℝN) such that
|
h
0
|
r
2
=
C
r
-
1
,
∥
h
0
∥
=
1
.
Then
(3.14)
b
≤
max
t
≥
0
I
(
t
h
0
,
0
)
≤
max
t
≥
0
(
t
2
2
∫
(
|
∇
h
0
|
2
+
|
h
0
|
2
)
-
λ
t
r
r
∫
|
h
0
|
r
)
=
r
-
2
2
r
λ
-
2
r
-
2
C
r
r
r
-
2
.
Substituting (3.12) into (3.14), we get (3.13). ∎
Lemma 3.7.
Let (u,v) be the weak limit of the minimizing sequence of (1.3), then (u,v)≠(0,0).
Proof.
Assume that (u,v)=(0,0). Notice that {(un,vn)}⊂Er. By a lemma due to Strauss [6, Radial Lemma A.II], (un,vn) is bounded in L∞(|x|≥R) for any R>0, which indicates that (un,vn) converge strongly to (u,v) in Lq(|x|>R) for all q>2. In particular,
(3.15)
(
u
n
,
v
n
)
→
(
0
,
0
)
in
L
2
*
(
|
x
|
>
R
)
×
L
2
*
(
|
x
|
>
R
)
for all
R
>
0
.
Now we show that
(3.16)
(
u
n
,
v
n
)
→
(
0
,
0
)
in
L
loc
2
*
(
ℝ
N
)
×
L
loc
2
*
(
ℝ
N
)
.
Otherwise, there is positive finite measure ν0 such that
|
u
n
|
2
*
+
|
v
n
|
2
*
⇀
ν
0
.
Then by Lemma 3.5 and Lemma 3.3, we get
b
>
1
N
(
N
-
2
N
)
N
-
2
2
(
1
+
β
)
-
N
-
2
2
S
N
2
,
which contradicts Lemma 3.6. Thus, ν0=0 and (3.16) holds. (3.15) and (3.16) tell us
(
u
n
,
v
n
)
→
(
0
,
0
)
in
L
2
*
(
ℝ
N
)
×
L
2
*
(
ℝ
N
)
,
which may lead to a contradiction by repeating the same arguments used in the proof of Lemma 3.2 (see inequality (3.4)). Thus, (u,v)≠(0,0). ∎
Lemma 3.8.
A is attained at some (u,v)≠(0,0), with u≥0, v≥0.
Proof.
Let {(un,vn)}⊂Er∩ℳ and 12∫(|∇un|2+|∇vn|)2→A. Applying Lemma 3.1, we may assume that (un,vn)⇀(u,v), then it follows
(3.17)
L
(
u
,
v
)
=
1
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
≤
lim inf
n
→
∞
1
2
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
→
A
.
Since {(un,vn)}⊂Er, by [6, Radial Lemma A.II], we know that {(un,vn)} is bounded in L∞(|x|>R), and then
F
(
u
n
)
→
F
(
u
)
,
G
(
v
n
)
→
G
(
v
)
,
H
(
u
n
)
K
(
v
n
)
→
H
(
u
)
K
(
v
)
in
L
1
(
|
x
|
≥
R
)
×
L
1
(
|
x
|
≥
R
)
.
On the other hand, from Lemma 3.3, Lemma 3.5 and Lemma 3.6, and an argument analogous to that used in the proof of Lemma 3.7, we obtain that νi=0 for every i, where νi is defined in (3.6). Then it follows
(
u
n
,
v
n
)
→
(
u
,
v
)
in
L
loc
2
*
(
ℝ
N
)
×
L
loc
2
*
(
ℝ
N
)
.
So we deduce that
F
(
u
n
)
→
F
(
u
)
,
G
(
v
n
)
→
G
(
v
)
,
H
(
u
n
)
K
(
v
n
)
→
H
(
u
)
K
(
v
)
in
L
1
(
B
R
(
0
)
)
×
L
1
(
B
R
(
0
)
)
.
Thus,
(3.18)
F
(
u
n
)
→
F
(
u
)
,
G
(
v
n
)
→
G
(
v
)
,
H
(
u
n
)
K
(
v
n
)
→
H
(
u
)
K
(
v
)
in
L
1
(
ℝ
N
)
×
L
1
(
ℝ
N
)
.
Notice that
∫
(
F
(
u
n
)
+
G
(
v
n
)
+
β
H
(
u
n
)
K
(
v
n
)
)
=
∫
(
|
u
n
|
2
2
+
|
v
n
|
2
2
)
+
1
,
which together with (3.17) and (3.18) gives
∫
(
F
(
u
)
+
G
(
v
)
+
β
H
(
u
)
K
(
v
)
)
≥
∫
(
|
u
|
2
2
+
|
v
|
2
2
)
+
1
,
that is, ∫T(u,v)≥1. If (u,v)∉ℳ, then ∫T(u,v)>1. Define a function l:[0,1]→ℝ by l(t)=∫T(tu,tv). It is clear that l(t)<0 for t>0 small enough and l(1)=∫T(u,v)>1. Hence, there is t0∈(0,1) such that l(t0)=0, which indicates t0(u,v)∈ℳ, and
(3.19)
t
0
2
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
≥
A
.
However, from (3.17) and t0∈(0,1), we know that
t
0
2
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
<
1
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
≤
A
,
which contradicts (3.19). Therefore, (u,v)∈ℳ and 12∫(|∇u|2+|∇v|2)=A>0.
It is obvious that (u,v)≠(0,0). Now it remains to prove that u≥0 and v≥0. Indeed, since
∫
(
|
∇
|
u
|
|
2
+
|
∇
|
v
|
|
2
)
≤
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
=
2
A
,
∫
T
(
|
u
|
,
|
v
|
)
=
∫
T
(
u
,
v
)
=
1
,
we may assume that u≥0 and v≥0. ∎
Lemma 3.9.
The following holds:
m
=
b
=
1
N
(
N
-
2
2
N
)
N
-
2
2
(
2
A
)
N
2
.
Proof.
For every (u,v)∈𝒩, it holds
b
≤
max
t
≥
0
I
(
t
u
,
t
v
)
=
I
(
u
,
v
)
.
Thus,
b
≤
inf
(
u
,
v
)
∈
𝒩
max
t
≥
0
I
(
t
u
,
t
v
)
=
inf
(
u
,
v
)
∈
𝒩
I
(
u
,
v
)
=
m
.
Then according to Lemma 3.3, we get
m
≥
b
≥
1
N
(
N
-
2
2
N
)
N
-
2
2
(
2
A
)
N
2
.
The proof will be accomplished once we show that
m
≤
1
N
(
N
-
2
2
N
)
N
-
2
2
(
2
A
)
N
2
.
Let (u,v) be such that
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
=
A
,
∫
T
(
u
,
v
)
=
1
.
It follows from the Lagrange multiplier rule that there is θ>0 such that (u,v) is a nontrivial solution of
{
-
△
u
=
θ
(
f
(
u
)
+
β
h
(
u
)
K
(
v
)
-
u
)
,
-
△
v
=
θ
(
g
(
v
)
+
β
h
(
u
)
K
(
v
)
-
v
)
.
Define (uθ,vθ):=(u(xθ),v(xθ)), then (uθ,vθ) solves (1.2) and
∫
(
|
∇
u
θ
|
2
+
|
∇
v
θ
|
2
)
=
2
θ
N
-
2
2
A
,
∫
T
(
u
θ
,
v
θ
)
=
θ
N
2
.
Moreover, the Pohozaev identity gives
θ
=
N
-
2
N
A
.
Thus,
m
≤
I
(
u
θ
,
v
θ
)
=
θ
N
-
2
2
A
-
θ
N
2
=
1
N
(
N
-
2
2
N
)
N
-
2
2
(
2
A
)
N
2
.
The conclusion follows. ∎
Lemma 3.10.
If 1<p<2, then m<min{B1,B2} for all β>0. Furthermore, if p=2, then there is β*>0 such that m<min{B1,B2} if β>β*.
Proof.
Let u be a positive ground state of (1.4) satisfying I1(u)=B1. Then there is t(s)>0 such that I(tu,tsu)=maxt≥0I(tu,tsu), that is,
t
(
1
+
s
2
)
∫
(
|
∇
u
|
2
+
|
u
|
2
)
=
∫
(
f
(
t
u
)
u
+
g
(
t
s
u
)
s
u
+
β
h
(
t
u
)
K
(
t
s
u
)
u
+
β
H
(
t
u
)
k
(
t
s
u
)
s
u
)
.
Denote
L
(
t
,
s
)
:=
t
(
1
+
s
2
)
∫
(
|
∇
u
|
2
+
|
u
|
2
)
-
∫
(
f
(
t
u
)
u
+
g
(
t
s
u
)
s
u
+
β
h
(
t
u
)
K
(
t
s
u
)
u
+
β
H
(
t
u
)
k
(
t
s
u
)
s
u
)
.
Calculating directly, we have
L
t
′
(
1
,
0
)
=
∫
(
f
(
u
)
u
-
f
′
(
u
)
u
2
)
>
0
.
Applying the Implicit Function Existence Theorem, we deduce that there is a C1 function t(s):(0,ϵ)→ℝ such that t(0)=1 and lims→0+t(s)=1. Moreover, if 1<p<2, we deduce that
lim
s
→
0
+
t
′
(
s
)
s
p
-
1
=
-
β
∫
(
f
′
(
u
)
u
2
-
f
(
u
)
u
)
L
1
:=
-
β
L
0
,
where
L
1
:=
lim
s
→
0
+
∫
(
h
(
t
u
)
k
(
t
s
u
)
u
2
+
H
(
t
u
)
k
′
(
t
s
u
)
t
s
u
2
+
H
(
t
u
)
k
(
t
s
u
)
u
)
s
p
-
1
.
According to (h1) and (h2), we know that L0 is a positive constant. Then
t
′
(
s
)
=
-
β
L
0
s
p
-
1
(
1
+
o
(
1
)
)
,
as
s
→
0
+
,
t
(
s
)
=
1
-
β
L
0
p
s
p
(
1
+
o
(
1
)
)
,
as
s
→
0
+
.
Therefore, by the Mean Value Theorem, we have for s>0 small enough,
m
≤
I
(
t
u
,
t
s
u
)
=
1
2
∫
(
f
(
t
u
)
t
u
-
2
F
(
t
u
)
+
g
(
t
s
u
)
t
s
u
-
2
G
(
t
s
u
)
)
+
1
2
β
∫
(
h
(
t
u
)
K
(
t
s
u
)
t
u
+
H
(
t
u
)
k
(
t
s
u
)
t
s
u
-
2
H
(
t
u
)
K
(
t
s
u
)
)
=
1
2
∫
(
f
(
u
)
u
-
2
F
(
u
)
-
(
f
′
(
u
)
u
2
-
f
(
u
)
u
)
β
L
0
p
s
p
)
+
o
(
s
p
)
+
o
(
s
2
)
+
1
2
β
∫
(
h
(
t
u
)
K
(
t
s
u
)
t
u
+
H
(
t
u
)
k
(
t
s
u
)
t
s
u
-
2
H
(
t
u
)
K
(
t
s
u
)
)
=
1
2
∫
(
f
(
u
)
u
-
2
F
(
u
)
)
+
o
(
s
p
)
+
o
(
s
2
)
-
1
2
β
(
L
1
p
s
p
-
∫
(
h
(
t
u
)
K
(
t
s
u
)
t
u
+
H
(
t
u
)
k
(
t
s
u
)
t
s
u
-
2
H
(
t
u
)
K
(
t
s
u
)
)
)
.
Let
(3.20)
L
2
:=
lim
s
→
0
+
∫
(
h
(
t
u
)
K
(
t
s
u
)
t
u
+
H
(
t
u
)
k
(
t
s
u
)
t
s
u
-
2
H
(
t
u
)
K
(
t
s
u
)
)
s
p
.
From (h2) and the continuity of k(s), we have
L
2
=
lim
s
→
0
+
∫
(
h
(
u
)
K
(
s
u
)
u
+
H
(
u
)
k
(
s
u
)
s
u
-
2
H
(
u
)
K
(
s
u
)
)
s
p
=
lim
s
→
0
+
∫
(
h
(
u
)
k
(
s
u
)
u
2
+
H
(
u
)
k
′
(
s
u
)
s
u
2
-
2
H
(
u
)
k
(
s
u
)
u
)
p
s
p
-
1
(3.21)
=
L
1
p
-
lim
s
→
0
+
3
∫
H
(
u
)
k
(
s
u
)
u
p
s
p
-
1
.
Denote
β
0
:=
lim
s
→
0
+
3
∫
H
(
u
)
k
(
s
u
)
u
p
s
p
-
1
>
0
,
then
∫
(
h
(
t
u
)
K
(
t
s
u
)
t
u
+
H
(
t
u
)
k
(
t
s
u
)
t
s
u
-
2
H
(
t
u
)
K
(
t
s
u
)
)
=
L
1
p
s
p
-
β
0
s
p
+
o
(
s
p
)
for s>0 small enough. Thus,
m
≤
1
2
∫
(
f
(
u
)
u
-
2
F
(
u
)
)
+
o
(
s
p
)
-
β
0
s
p
<
1
2
∫
(
f
(
u
)
u
-
2
F
(
u
)
)
=
I
(
u
,
0
)
=
B
1
for s>0 small enough. Analogously, if p=2,
lim
s
→
0
+
t
′
(
s
)
s
p
-
1
=
-
β
L
0
+
2
∫
(
|
∇
u
|
2
+
|
u
|
2
)
∫
(
f
′
(
u
)
u
2
-
f
(
u
)
u
)
:=
-
L
3
and
t
′
(
s
)
=
-
L
3
s
p
-
1
(
1
+
o
(
1
)
)
,
as
s
→
0
+
,
t
(
s
)
=
1
-
L
3
p
s
p
(
1
+
o
(
1
)
)
,
as
s
→
0
+
.
Then for s>0 small enough,
m
≤
I
(
t
u
,
t
s
u
)
=
1
2
∫
(
f
(
u
)
u
-
2
F
(
u
)
)
+
o
(
s
p
)
+
2
s
p
p
∫
(
|
∇
u
|
2
+
|
u
|
2
)
-
1
2
β
(
L
1
p
s
p
-
∫
(
h
(
t
u
)
K
(
t
s
u
)
t
u
+
H
(
t
u
)
k
(
t
s
u
)
t
s
u
-
2
H
(
t
u
)
K
(
t
s
u
)
)
)
.
Denote
β
1
*
:=
2
∫
|
∇
u
|
2
+
|
u
|
2
p
β
0
.
The previous inequality together with (3.20) and (3.21) implies that if β>β1*, then
m
≤
1
2
∫
(
f
(
u
)
u
-
2
F
(
u
)
)
-
C
s
p
+
o
(
s
p
)
<
I
(
u
,
0
)
=
B
1
for s>0 small enough. Similarly, we can prove that m<B2 for all β>0 if 1<p<2, and m<B2 for some β2*>0 and β>β2* if p=2. We finish the proof by letting β*:=max{β1*,β2*}. ∎
Proof of Theorem 1.1.
It is a direct result of Lemma 3.8, Lemma 3.9 and Lemma 3.10. ∎
4 The Case of Dimension N=2
In this case, the Pohozaev identity shows that any solution of (1.2) should satisfy ∫T(u,v)=0. Thus,
A
=
inf
{
1
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
:
∫
T
(
u
,
v
)
=
0
,
(
u
,
v
)
≠
(
0
,
0
)
}
=
inf
𝒫
1
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
.
Define the following minimax value which will be used in the sequel:
c
:=
inf
(
u
,
v
)
∈
E
∖
{
(
0
,
0
)
}
max
t
≥
0
I
(
t
u
,
t
v
)
.
Now we give a compactness lemma.
Lemma 4.1.
Assume that (f1), (f2’), (h1’), (h2) hold. Let {(un,vn)} be a sequence in Er such that
sup
n
∫
(
|
∇
u
n
2
+
|
∇
v
n
|
2
)
=
d
0
<
1
𝑎𝑛𝑑
sup
n
∫
(
|
u
n
|
2
+
|
v
n
|
2
)
=
a
0
<
∞
.
Then
∫
F
(
u
n
)
→
∫
F
(
u
)
,
∫
G
(
v
n
)
→
∫
G
(
v
)
,
∫
H
(
u
n
)
K
(
v
n
)
→
∫
H
(
u
)
K
(
v
)
when (un,vn)⇀(u,v) in Er.
Proof.
We can assume that there is (u,v)∈Er such that
(
u
n
,
v
n
)
⇀
(
u
,
v
)
in
E
,
(
u
n
,
v
n
)
→
(
u
,
v
)
on
ℝ
2
×
ℝ
2
,
lim
|
x
|
→
∞
(
u
n
(
x
)
+
v
n
(
x
)
)
=
0
.
Using a Trudinger–Moser type inequality [7, Lemma 2.1], we know that for d1∈(0,1) and d2>0, there is C(d1,d2) such that
sup
u
∈
Q
0
∫
(
e
4
π
u
2
-
1
)
≤
C
(
d
1
,
d
2
)
,
where Q0:={u∈H1(ℝ2):∫|∇u|2≤d1 and ∫|u|2≤d2}. Choose ϵ small enough such that d1:=d0(1-ϵ)2∈(0,1) and set α:=4π(1-ϵ)2>4π, then we have
(4.1)
∫
(
e
α
u
n
2
-
1
)
=
∫
(
e
4
π
(
u
n
2
(
1
-
ϵ
)
2
)
-
1
)
≤
C
(
d
1
,
d
2
)
.
Let Q(s):=eαs2-1,R1(s):=|s|(eα1s2-1)+|s|r1, and R2(s):=|s|(eα1s2-1)+|s|r2, where r1,r2>2. From (f1), (f2’), (h1’), (h2) and (4.1), we deduce that
lim
s
→
0
F
(
s
)
Q
(
s
)
=
lim
s
→
+
∞
F
(
s
)
Q
(
s
)
=
lim
s
→
0
G
(
s
)
Q
(
s
)
=
lim
s
→
+
∞
G
(
s
)
Q
(
s
)
=
0
,
lim
s
→
0
R
1
(
s
)
Q
(
s
)
=
lim
s
→
+
∞
R
1
(
s
)
Q
(
s
)
=
lim
s
→
0
R
2
(
s
)
Q
(
s
)
=
lim
s
→
+
∞
R
2
(
s
)
Q
(
s
)
=
0
,
for
α
1
<
α
,
sup
n
∫
|
Q
(
u
n
)
|
<
+
∞
,
sup
n
∫
|
Q
(
v
n
)
|
<
+
∞
,
and as n→∞,
F
(
u
n
)
→
F
(
u
)
,
G
(
u
n
)
→
G
(
u
)
,
R
1
(
u
n
)
→
R
1
(
u
)
,
R
2
(
v
n
)
→
R
2
(
v
)
a.e. on
ℝ
2
.
Using the Compactness Lemma of Strauss [6, Theorem A.I], we get
∫
F
(
u
n
)
→
∫
F
(
u
)
,
∫
G
(
v
n
)
→
∫
G
(
v
)
,
∫
R
1
(
u
n
)
→
∫
R
1
(
u
)
,
∫
R
2
(
v
n
)
→
∫
R
2
(
v
)
.
Choosing α1 such that 4π<α1<α and taking (h1’) and (h2) into account, we have that there are C1,C2>0, r1,r2>2 such that
H
(
u
n
)
K
(
v
n
)
≤
C
1
(
|
u
n
|
(
e
α
1
u
n
2
-
1
)
+
|
u
n
|
r
1
)
+
C
2
(
|
v
n
|
(
e
α
1
v
n
2
-
1
)
+
|
v
n
|
r
2
)
=
C
1
R
(
u
n
)
+
C
2
R
(
v
n
)
.
Thus, we obtain ∫H(un)K(vn)→∫H(u)K(v). ∎
Proof.
For each (u,v)∈E∖{(0,0)}, we set
l
(
t
)
:=
∫
T
(
t
u
,
t
v
)
=
∫
(
F
(
t
u
)
+
G
(
t
v
)
+
β
H
(
t
u
)
K
(
t
v
)
-
t
2
u
2
2
-
t
2
u
2
2
)
.
It is clear that l(t)<0 for t>0 small and l(t)>0 for t>0 large enough. Thus, there is t0>0 such that l(t0)=0, which indicates t0(u,v)∈ℳ. Then
A
≤
I
(
t
0
u
,
t
0
v
)
≤
max
t
≥
0
I
(
t
u
,
t
v
)
,
which implies A≤c. ∎
Proof.
To obtain a contradiction, suppose that A=0. Let {(un,vn)} be such that
1
2
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
→
A
=
0
,
∫
T
(
u
n
,
v
n
)
=
0
.
For each θn>0, let (ξn(x),ζn(x))=(un(xθn),vn(xθn)), then
∫
(
|
∇
ξ
n
|
2
+
|
∇
ζ
n
|
2
)
=
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
,
∫
T
(
ξ
n
,
ζ
n
)
=
0
,
and
∫
(
|
ξ
n
|
2
+
|
ζ
n
|
2
)
=
θ
n
2
∫
(
|
u
n
|
2
+
|
v
n
|
2
)
.
Choose
θ
n
2
=
1
∫
(
|
u
n
|
2
+
|
v
n
|
2
)
,
then we have
1
2
∫
(
|
∇
ξ
n
|
2
+
|
∇
ζ
n
|
2
)
→
A
=
0
,
∫
(
|
ξ
n
|
2
+
|
ζ
n
|
2
)
=
1
,
∫
T
(
ξ
n
,
ζ
n
)
=
0
.
According to Lemma 4.1, we get
∫
F
(
ξ
n
)
→
∫
F
(
ξ
)
,
∫
G
(
ζ
n
)
→
∫
G
(
ζ
)
,
∫
H
(
ξ
n
)
K
(
ξ
n
)
→
∫
H
(
ξ
)
K
(
ζ
)
.
Note that ∫T(ξn,ζn)=0, which implies
∫
(
F
(
ξ
n
)
+
G
(
ζ
n
)
+
β
H
(
ξ
n
)
K
(
ζ
n
)
)
=
1
2
.
Thus,
∫
(
F
(
ξ
)
+
G
(
ζ
)
+
β
H
(
ξ
)
K
(
ζ
)
)
=
1
2
,
and (ξ,ζ)≠(0,0). However,
∫
(
|
∇
ξ
|
2
+
|
∇
ζ
|
2
)
≤
lim inf
n
→
∞
∫
(
|
∇
ξ
n
|
2
+
|
∇
ζ
n
|
2
)
=
A
=
0
,
which is a contradiction. Thus, A>0. ∎
In much the same way as in the proof of Lemma 3.6 and Lemma 3.9, we deduce the following two lemmas:
Lemma 4.4.
If
λ
>
(
r
-
2
r
)
r
-
2
2
C
r
r
2
,
then c<12.
Proof of Theorem 1.3.
Let {(un,vn)}⊂Er be a minimizing sequence such that
(4.2)
1
2
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
→
A
,
∫
T
(
u
n
,
v
n
)
→
0
.
Arguing as in the proof of Lemma 4.3, we may assume that
(4.3)
∫
(
|
u
n
|
2
+
|
v
n
|
2
)
=
1
.
From (4.2), Lemma 4.2 and Lemma 4.4, we have
lim sup
n
→
∞
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
≤
2
A
≤
2
c
<
1
.
Then using Lemma 4.1, we get
(4.4)
∫
F
(
u
n
)
→
∫
F
(
u
)
,
∫
G
(
v
n
)
→
∫
G
(
v
)
,
∫
H
(
u
n
)
K
(
v
n
)
→
∫
H
(
u
)
K
(
v
)
,
where (u,v) is the weak limit of (un,vn). Comparing (4.4) together with (4.2) and (4.3) indicates
∫
(
F
(
u
)
+
G
(
v
)
+
β
H
(
u
)
K
(
v
)
)
=
1
2
.
Thus, (u,v)≠(0,0). Moreover,
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
≤
lim inf
n
→
∞
∫
(
|
∇
u
n
|
2
+
|
∇
v
n
|
2
)
≤
A
.
It remains to show that ∫T(u,v)=0. Since
∫
(
|
u
|
2
+
|
v
|
2
)
≤
lim inf
n
→
∞
(
|
u
n
|
2
+
|
v
n
|
2
)
=
1
,
we get
∫
T
(
u
,
v
)
≥
0
.
If ∫T(u,v)>0, let us consider the function l(t) defined in Lemma 4.2. Observe that l(t)<0 for t>0 small enough, and l(1)=∫T(u,v)>0. Then there is t0∈(0,1) such that l(t0)=0, which is equivalent to
∫
T
(
t
0
u
,
t
0
v
)
=
0
.
Hence, (t0u,t0v)∈ℳ and
(4.5)
1
2
∫
(
|
∇
(
t
0
u
)
|
2
+
|
∇
(
t
0
v
)
|
2
)
≥
A
.
However, since t0∈(0,1), it is clear that
1
2
∫
(
|
∇
(
t
0
u
)
|
2
+
|
∇
(
t
0
v
)
|
2
)
<
1
2
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
≤
A
,
which conflicts with (4.5). So ∫T(u,v)=0, and A is attained. Furthermore, since
∫
(
|
∇
|
u
|
|
2
+
|
∇
|
v
|
|
2
)
≤
∫
(
|
∇
u
|
2
+
|
∇
v
|
2
)
=
2
A
,
∫
T
(
|
u
|
,
|
v
|
)
=
∫
T
(
u
,
v
)
=
0
,
we may assume that u≥0 and v≥0. Using Lemma 4.5, we know that (u,v) is a ground state solution of (1.2). Arguing analogously as in the proof of Lemma 3.10, we get that if 1<p<2, then u>0, v>0 for all β>0, and if p=2, then there exists β**>0 such that u>0, v>0 for β>β**. ∎
and
Yongqing Li
