Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 47, pp. 1–20.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

EXISTENCE AND ASYMPTOTIC BEHAVIOR OF POSITIVE

LEAST ENERGY SOLUTIONS FOR COUPLED NONLINEAR

CHOQUARD EQUATIONS

SONG YOU, PEIHAO ZHAO, QINGXUAN WANG

Abstract. In this article, we study the coupled nonlinear Schrödinger equa-
tions with Choquard type nonlinearities

−∆u+ ν1u = µ1(
1

|x|α
∗ u2)u+ β(

1

|x|α
∗ v2)u in RN ,

−∆v + ν2v = µ2(
1

|x|α
∗ v2)v + β(

1

|x|α
∗ u2)v in RN ,

u, v ≥ 0 in RN , u, v ∈ H1(RN ),

where ν1, ν2, µ1, µ2 are positive constants, β > 0 is a coupling constant, N ≥ 3,

α ∈ (0, N) ∩ (0, 4), and “∗” is the convolution operator. We show that the
nonlocal elliptic system has a positive least energy solution for positive small

β and positive large β via variational methods. For the case in which ν1 = ν2,

µ1 6= µ2, N = 3, 4, 5 and α = N − 2, we prove the uniqueness of positive least
energy solutions. Moreover, the asymptotic behaviors of the positive least

energy solutions as β → 0+ are studied.

1. Introduction

We consider the time-dependent coupled nonlinear Schrödinger equations with
Choquard type nonlinearities in the following form (see [12, 36]):

−i ∂
∂t

Φ1 = ∆Φ1 + µ1(V (x) ∗ |Φ1|2)Φ1 + β(V (x) ∗ |Φ2|2)Φ1 in RN ,

−i ∂
∂t

Φ2 = ∆Φ2 + µ2(V (x) ∗ |Φ2|2)Φ2 + β(V (x) ∗ |Φ1|2)Φ2 in RN ,

Φj = Φj(x, t) ∈ C, j = 1, 2,

Φj(x, t) = 0, x ∈ RN , t > 0, j = 1, 2,

(1.1)

where i is the imaginary unit, and “∗” is the convolution operator. System (1.1)
appears in many physical problem, especially in nonlinear optics. Physically, the
solution Φj denotes the j-th component of the beam in Kerr-like photorefractive
media (see [22, 23]). The positive constant µj indicate the self-focusing in the j-th
components of the beam. V (x) is the response function which possesses information
on the mutual interaction. The coupling constant β is the interaction between the

2010 Mathematics Subject Classification. 35B40, 35J47, 35J50.
Key words and phrases. Coupled Choquard equations; positive least energy solution;

asymptotic behavior; variational method.
c©2021 Texas State University.

Submitted July 17, 2019. Published May 28, 2021.

1



2 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

two components of the beam. The problem (1.1) also arises in the basic quantum
chemistry model of small number of electrons interacting with static nucleii which
can be approximated by Hartree or Hartree-Fock minimization problems (see [13,
14, 17]).

To obtain solitary wave solutions of system (1.1), we set Φ1(x, t) = eiν1tu(x) and
Φ2(x, t) = eiν2tv(x). Then system (1.1) turns into the elliptic system

−∆u+ ν1u = µ1(V (x) ∗ u2)u+ β(V (x) ∗ v2)u x ∈ RN ,

−∆v + ν2v = µ2(V (x) ∗ v2)v + β(V (x) ∗ u2)v x ∈ RN .
(1.2)

If the response function is a Dirac-delta function, i.e. V (x) = δ(x), then (1.2) turns
to be the following semilinear elliptic system with local nonlinearities:

−∆u+ ν1u = µ1u
3 + βuv2 x ∈ RN ,

−∆v + ν2v = µ2v
3 + βvu2 x ∈ RN .

(1.3)

Here, µ1, µ2 > 0 and β 6= 0 is a coupling constant. The existence and multiplicity of
solutions to (1.3) have been the subject of extensive mathematical studies in recent
years, see [3, 4, 5, 6, 7, 8, 19, 20, 26, 27, 28, 29, 34] and references therein.

In this paper we consider system (1.2) with a response function of Riesz potential,
i.e. V (x) = |x|−α, then (1.2) is reduced to the nonlocal elliptic system

−∆u+ ν1u = µ1(
1

|x|α
∗ u2)u+ β(

1

|x|α
∗ v2)u x ∈ RN ,

−∆v + ν2v = µ2(
1

|x|α
∗ v2)v + β(

1

|x|α
∗ u2)v x ∈ RN .

(1.4)

Here, α ∈ (0, N) ∩ (0, 4), ν1, ν2 > 0, µ1, µ2 > 0 and β 6= 0 is a coupling constant.
Before proceeding to state our resutls, we introduce the following classical Hardy-

Littlewood-Sobolev inequality (see [16]).

Lemma 1.1. Let p, r > 1 and 0 < α < N with 1
p + α

N + 1
r = 2, f ∈ Lp(RN ) and

h ∈ Lr(RN ). There exists a sharp constant C(N,α, p) such that∣∣ ∫
RN

∫
RN

f(x)h(y)

|x− y|α
dx dy

∣∣ ≤ C(N,α, p)|f |p|h|r, (1.5)

where | · |q is the Lq(RN )-norm with q ∈ [1,∞].

Assume that f, g ∈ L1
loc(RN ) and α ∈ (0, N), as [16] we define

D(f, g) :=

∫
RN

∫
RN

f(x)g(y)

|x− y|α
dx dy.

The following lemma is important for considering (1.4), and its proof is given in
[16, Theorem 9.8].

Lemma 1.2. If D(|f |, |f |) <∞, then

D(f, f) ≥ 0,

and there is equality if and only if f ≡ 0. Moreover, if D(|g|, |g|) <∞, then

|D(f, g)|2 ≤ D(f, f)D(g, g). (1.6)



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 3

Suppose that u, v ∈ H1(RN ) and α ∈ (0, N)∩ (0, 4), then by Lemma 1.1 we have

D(u2, u2) ≤ C|u2|2 2N
2N−α

= C|u|4 4N
2N−α

. (1.7)

It is well known that the solutions of (1.4) correspond to the critical points of
the C1 functional E : H → R given by

E(u, v) =
1

2

∫
RN

(|∇u|2 + ν1u
2 + |∇v|2 + ν2v

2)

− 1

4

∫
RN

µ1(
1

|x|α
∗ u2)u2 + 2β(

1

|x|α
∗ u2)v2 + µ2(

1

|x|α
∗ v2)v2,

(1.8)

where H := H1(RN ) ×H1(RN ). From (1.6) and (1.7), it is easy to check that E
is well defined in H. This allows to consider positive least energy solution, which is
defined as solution (u, v) of (1.4) with positive components and achieving the level

inf{E(u, v) : E′(u, v) = 0, (u, v) ∈ H, u > 0 and v > 0}.

We call a solution (u, v) semi-trivial if u = 0 or v = 0. A solution (u, v) nontrivial
if both u 6≡ 0 and v 6≡ 0. A nontrivial solution (u, v) positive if both u > 0 and
v > 0.

Note that system (1.4) admits a trivial solution (0, 0) and a pair of semi-trivial
solutions (ω1, 0) or (0, ω2), where ωi is the positive least energy solution of (see
[24])

−∆u+ λu = µ(
1

|x|α
∗ u2)u u ∈ H1(RN ), (1.9)

with (λ, µ) = (ν1, µ1) for ω1, and (λ, µ) = (ν2, µ2) for ω2 respectively. The existence
of solutions to (1.9) has received great interest recently, see [1, 2, 9, 10, 11, 15, 18,
21, 24, 25, 33] and references therein. Next, we will pay close attention to the
existence of nontrivial solutions to (1.4).

Recently, Wang and Shi [30] studied the existence and various qualitative prop-
erties of positive least energy solutions to system (1.4) with N = 3, α = 1. In [31]
the authors acquired the existence and multiplicity of nontrivial solutions of (1.4)
with perturbations. In [32] the authors studied the existence and nonexistence of
L2(RN )-normalized solutions of (1.4) with trapping potentials.

To the best of our knowledge, there are no papers considering system (1.4) with
α ∈ (0, N) ∩ (0, 4). In present paper, we will focus on providing conditions on the
coupling constant β that insures the existence of positive least energy solutions.
Moreover, we will investigate the asymptotic behaviors of those solutions.

We define

N =
{
u 6≡ 0, v 6≡ 0,

∫
RN
|∇u|2 + ν1u

2 =

∫
RN

µ1(
1

|x|α
∗ u2)u2 + β(

1

|x|α
∗ u2)v2,∫

RN
|∇v|2 + ν2v

2 =

∫
RN

µ2(
1

|x|α
∗ v2)v2 + β(

1

|x|α
∗ u2)v2

}
.

Then any nontrivial solution of (1.4) belongs to N . Let

A := inf
(u,v)∈N

E(u, v) = inf
(u,v)∈N

1

4

∫
RN
|∇u|2 + ν1u

2 + |∇v|2 + ν2v
2. (1.10)

Now, we list our main results. First, we consider the case ν1 = ν2 = ν. Let ϕ be
any a positive least energy solution of (1.9) with λ = ν and µ = 1. Then we have
the following two Theorems.



4 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

Theorem 1.3. Assume that N ≥ 3, α ∈ (0, N) ∩ (0, 4) and ν1 = ν2 = ν > 0.

(I) If 0 < β < min{µ1, µ2} or β > max{µ1, µ2}, then A is attained by

(
√
kϕ,
√
lϕ), where k, l > 0 satisfy

µ1k + βl = 1,

βk + µ2l = 1.
(1.11)

Therefore, (
√
kϕ,
√
lϕ) is a positive least energy solution of (1.4).

(II) If β ∈ [min{µ1, µ2},max{µ1, µ2}] and µ1 6= µ2, then (1.4) does not have a
nontrivial nonnegative solution.

Theorem 1.4. Assume that ν1 = ν2 = ν > 0, and let 0 < β < min{µ1, µ2} or
β > max{µ1, µ2}. Let (u, v) be any a least energy nontrivial solution of (1.4), then

(u, v) = (
√
kϕ,
√
lϕ), where (k, l) satisfies (1.11). In particular, when N = 3, 4, 5

and α = N − 2, (
√
kϕ,
√
lϕ) is a unique positive least energy solution of (1.4) up to

a translation.

For the general case in which ν1 6= ν2, we have the following theorem.

Theorem 1.5. Assume that N ≥ 3 and α ∈ (0, N) ∩ (0, 4).

(1) There exists β1 > 0 such that for any β ∈ (0, β1), (1.4) has a positive least
energy solution (u, v), which is radially symmetric.

(2) There exists β2 > 0 such that for any β ∈ (β2,+∞), (1.4) has a positive
least energy solution (u, v), which is radially symmetric.

(3) Assume that 0 < ν1 ≤ ν2 and µ2 < µ1. If µ2 ≤ β ≤ µ1, then (1.4) does not
have a nontrivial nonnegative solution.

In fact, we can give an accurate definition of β1 in Lemma 4.1 and β2 in Lemma
4.5, but do not give it here to avoid introducing heavy notation at this stage.

Remark 1.6. System (1.4) is critical when α = 4 in the sense of the Hardy-
Littlewood-Sobolev inequality, which leads to the lack of compactness. This will be
an interesting issue to be pursued in the future.

Finally, we study the asymptotic behavior of the positive least energy solutions
in the case β → 0+. Then we have the following result.

Theorem 1.7. Assume that N ≥ 3 and α ∈ (0, N)∩(0, 4). Let βn ∈ (0, β1), n ∈ N,
satisfy βn → 0 as n → +∞. Suppose that (un, vn) is the positive least energy
solutions of (1.4) with β = βn and (un, vn) is radially symmetric, which exists
by Theorem 1.5. Then passing to a subsequence, (un, vn) → (û, v̂) strongly in
H1(RN )×H1(RN ) as n→ +∞, where û is a positive least energy solution of

−∆u+ ν1u = µ1(
1

|x|α
∗ u2)u, u ∈ H1(RN ),

and v̂ is a positive least energy solution of

−∆v + ν2v = µ2(
1

|x|α
∗ v2)v, v ∈ H1(RN ).

The paper is organized as follows. Theorem 1.3 and Theorem 1.4 are proved
in Section 2 and Section 3, respectively. In Section 4, we use the Nehari manifold
approach and a mountain pass argument to prove Theorem 1.5. In Section 5, we
study the limit behavior of the positive least energy solutions as β → 0+.



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 5

We give some notation here. Throughout this paper, we denote the norm of Lq by

|u|q = (
∫
RN |u|

qdx)
1
q , the norm of H1(RN ) by ‖u‖2ν =

∫
RN (|∇u|2+νu2), the norm of

H by ‖(u, v)‖2H := ‖u‖2ν1 + ‖v‖2ν2 , Hr := {(u, v) ∈ H : u, v are radially symmetric},
and positive constants (possibly different in different places) by C,C1, C2.

2. Proof of Theorem 1.3

By [24] we know that∫
RN

(|∇u|2 + νu2) ≥ 2
√
B
(∫

RN
(

1

|x|α
∗ u2)u2

)1/2

, ∀u ∈ H1(RN ), (2.1)

where

B :=
1

4

∫
RN
|∇ϕ|2 + νϕ2 =

1

4

∫
RN

(
1

|x|α
∗ ϕ2)ϕ2, (2.2)

and ϕ is a positive least energy solution of (1.9) with λ = ν and µ = 1.

Conclusion of the proof of Theorem 1.3. Firstly, we prove (I) of Theorem 1.3. Since
0 < β < min{µ1, µ2} or β > max{µ1, µ2}, it follows that the equation (1.11) has

a solution (k, l) satisfying k > 0, l > 0. It is easy to see that (
√
kϕ,
√
lϕ) is a

nontrivial solution of (1.4). By (1.10) and (2.2) we have

A ≤ E(
√
kϕ,
√
lϕ) = (k + l)B. (2.3)

Let (un, vn) ⊂ N be a minimizing sequence for A, that is, E(un, vn) → A. For
simplicity of presentation, we set

αn =
(∫

RN
(

1

|x|α
∗ u2

n)u2
n

)1/2

, βn =
(∫

RN
(

1

|x|α
∗ v2

n)v2
n

)1/2

.

Then, by (1.6) and (2.1) we have

2
√
Bαn ≤

∫
RN
|∇un|2 + νu2

n

=

∫
RN

µ1(
1

|x|α
∗ u2

n)u2
n + β(

1

|x|α
∗ u2

n)v2
n

≤ µ1α
2
n + βαnβn,

(2.4)

2
√
Bβn ≤

∫
RN
|∇vn|2 + νv2

n

=

∫
RN

µ2(
1

|x|α
∗ v2

n)v2
n + β(

1

|x|α
∗ v2

n)u2
n

≤ µ2β
2
n + βαnβn.

(2.5)

Note that

E(un, vn) =
1

4

∫
RN
|∇un|2 + νu2

n + |∇vn|2 + νv2
n.

Combining this with (2.4) and (2.5) we have

2
√
B(αn + βn) ≤ 4E(un, vn) = 4A+ o(1) ≤ 4(k + l)B + o(1),

µ1αn + ββn ≥ 2
√
B,

µ2βn + βαn ≥ 2
√
B.



6 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

We deduce from (1.11) that the above three inequalities are equivalent to

(αn − 2k
√
B) + (βn − 2l

√
B) ≤ o(1),

µ1(αn − 2k
√
B) + β(βn − 2l

√
B) ≥ 0,

β(αn − 2k
√
B) + µ2(βn − 2l

√
B) ≥ 0.

Therefore, αn → 2k
√
B and βn → 2l

√
B as n→∞. Then

4A = lim
n→∞

4E(un, vn) ≥ lim
n→∞

2
√
B(αn + βn) = 4(k + l)B.

Combining this with (2.3) we have

A = (k + l)B = E(
√
kϕ,
√
lϕ). (2.6)

So (
√
kϕ,
√
lϕ) is a positive least energy solution of (1.4).

Now we prove (II) of Theorem 1.3. Suppose that (u, v) is a nontrivial solution
of (1.4) and satisfies u ≥ 0, v ≥ 0 in RN . By the strong maximum principle each of
the functions u, v is strictly positive in RN . Repeating the proof of [7, Proposition
4.1], we know that the solutions of (1.4) which are in H1(RN ) are also in C2(RN )
and tend to zero as |x| → ∞.

Next, we multiply the first equation in (1.4) by v, the second equation in (1.4)
by u, and integrate the resulting equations over RN . Then we obtain∫

RN
(∇u∇v + ν1uv) =

∫
RN

uv[µ1(
1

|x|α
∗ u2) + β(

1

|x|α
∗ v2)],∫

RN
(∇u∇v + ν2uv) =

∫
RN

uv[µ2(
1

|x|α
∗ v2) + β(

1

|x|α
∗ u2)].

Thus, ∫
RN

uv[(ν2 − ν1) + (µ1 − β)(
1

|x|α
∗ u2) + (β − µ2)(

1

|x|α
∗ v2)] = 0,

which is in a contradiction with the positivity of u and v as long as the three
constants (ν2− ν1), (µ1−β), (β−µ2) are of the same sign or zero, and one of them
is not zero. This implies that system (1.4) does not have a nontrivial solution with
nonnegative components if ν1 = ν2, µ1 6= µ2 and min{µ1, µ2} ≤ β ≤ max{µ1, µ2}.
The proof is complete. �

3. Proof of Theorem 1.4

Conclusion of the proof of Theorem 1.4. Firstly, we consider the case in which 0 <
β < min{µ1, µ2}. The following proof is inspired by [5]. Fix µ1 > 0, µ2 > 0 and
0 < β < min{µ1, µ2}. Let (u1, v1) be any a nontrivial least energy solution of (1.4),

then u1, v1 > 0 in RN by the strong maximum principle. Recalling (
√
kϕ,
√
lϕ) in

Theorem 1.3, first we claim that∫
RN

(
1

|x|α
∗ u2

1)u2
1 = k2

∫
RN

(
1

|x|α
∗ ϕ2)ϕ2. (3.1)

Observe that there exists δ > 0 such that 0 < β < min{µ, µ2} for any µ ∈ (µ1 −
δ, µ1 + δ). Then by Theorem 1.3, A is attained when µ1 is replaced by µ. Recall



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 7

the definition of E,N , A, they all depend on µ and we use notation Eµ,Nµ, A(µ)
in this proof. Recall (1.11) and (2.6), we have

A(µ) =
µ+ µ2 − 2β

µµ2 − β2
B,

so A′(µ1) := d
dµA(µ)|µ1

exists. We define

f(t, s, µ) := tµ

∫
RN

(
1

|x|α
∗ u2

1)u2
1 + s

∫
RN

β(
1

|x|α
∗ v2

1)u2
1 −

∫
RN

(|∇u1|2 + νu2
1),

g(t, s, µ) := s

∫
RN

µ2(
1

|x|α
∗ v2

1)v2
1 + t

∫
RN

β(
1

|x|α
∗ u2

1)v2
1 −

∫
RN

(|∇v1|2 + νv2
1).

It is easy to obtain that f(1, 1, µ1) = g(1, 1, µ1) = 0, and

∂f

∂t
(1, 1, µ1) = µ1D(u2

1, u
2
1),

∂f

∂s
(1, 1, µ1) = βD(u2

1, v
2
1),

∂g

∂t
(1, 1, µ1) = βD(u2

1, v
2
1),

∂g

∂s
(1, 1, µ1) = µ2D(v2

1 , v
2
1).

We define the matrix

G :=

(
∂f
∂t (1, 1, µ1) ∂f

∂s (1, 1, µ1)
∂g
∂t (1, 1, µ1) ∂g

∂s (1, 1, µ1)

)
,

then we see from (1.6) that

det(G) = µ1µ2D(u2
1, u

2
1)D(v2

1 , v
2
1)− β2D2(u2

1, v
2
1)

≥ (µ1µ2 − β2)D(u2
1, u

2
1)D(v2

1 , v
2
1) > 0.

(3.2)

Therefore, by the implicit function theorem, functions t(µ) and s(µ) are well defined
and class C1 on (µ1 − δ1, µ1 + δ1) for some δ1 ≤ δ. Moreover, t(µ1) = s(µ1) =
1, and so we may assume that t(µ), s(µ) > 0 for all µ ∈ (µ1 − δ1, µ1 + δ1) by
choosing a small δ1. Since f(t(µ), s(µ), µ) ≡ g(t(µ), s(µ), µ) ≡ 0, then we have

(
√
t(µ)u1,

√
s(µ)v1) ∈ Nµ. By a direct computation we see that

t′(µ1) = −µ2D(v2
1 , v

2
1)D(u2

1, u
2
1)

det(G)
, s′(µ1) =

βD(u2
1, v

2
1)D(u2

1, u
2
1)

det(G)
.

Note that t(µ) = 1 + t′(µ1)(µ− µ1) + o((µ− µ1)) and s(µ) = 1 + s′(µ1)(µ− µ1) +
o((µ− µ1)). Hence

A(µ) ≤ Eµ(
√
t(µ)u1,

√
s(µ)v1) = A(µ1) +

1

4
B(µ− µ1) + o((µ− µ1)),

where

B := t′(µ1)

∫
RN

(|∇u1|2 + νu2
1) + s′(µ1)

∫
RN

(|∇v1|2 + νv2
1) = −D(u2

1, u
2
1).

It follows that A(µ)−A(µ1)
µ−µ1

≥ B
4 + o(1), as µ ↗ µ1. So A′(µ1) ≥ B/4. Similarly,

we have A′(µ1) ≤ B/4. Therefore, A′(µ1) = B/4 = − 1
4

∫
RN ( 1

|x|α ∗ u
2
1)u2

1. By

Theorem 1.3, (
√
kϕ,
√
lϕ) is also a positive least energy solution of (1.4). Hence,

A′(µ1) = −k
2

4

∫
RN ( 1

|x|α ∗ ϕ
2)ϕ2, and so (3.1) holds.

Similarly, by computing A′(µ2) and A′(β), respectively, we see that∫
RN

(
1

|x|α
∗ v2

1)v2
1 = l2

∫
RN

(
1

|x|α
∗ϕ2)ϕ2,

∫
RN

(
1

|x|α
∗ u2

1)v2
1 = kl

∫
RN

(
1

|x|α
∗ϕ2)ϕ2.



8 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

Therefore,∫
RN

(
1

|x|α
∗ u2

1)v2
1 =

l

k

∫
RN

(
1

|x|α
∗ u2

1)u2
1 =

k

l

∫
RN

(
1

|x|α
∗ v2

1)v2
1 .

We define (u, v) := ( 1√
k
u1,

1√
l
v1). Combining these with (1.11) and (u1, v1) ∈ N ,

we obtain∫
RN

(|∇u|2 + νu2) =

∫
RN

(
1

|x|α
∗ u2)u2,

∫
RN

(|∇v|2 + νv2) =

∫
RN

(
1

|x|α
∗ v2)v2.

(3.3)
Then, by (2.1) we have

1

4

∫
RN

(|∇u|2 + νu2) ≥ B, 1

4

∫
RN

(|∇v|2 + νv2) ≥ B.

Therefore,

A = (k + l)B =
1

4

∫
RN

(|∇u1|2 + νu2
1 + |∇v1|2 + νv2

1)

=
1

4
k

∫
RN

(|∇u|2 + νu2) +
1

4
l

∫
RN

(|∇v|2 + νv2)

≥ (k + l)B.

This implies that

1

4

∫
RN

(|∇u|2 + νu2) = B,
1

4

∫
RN

(|∇v|2 + νv2) = B.

Combining this with (3.3), it is easy to see that u and v are both positive least
energy solutions of (1.9) with λ = ν and µ = 1. Since (u1, v1) satisfies (1.4), then
we know that

−∆u+ νu = µ1k(
1

|x|α
∗ u2)u+ βl(

1

|x|α
∗ v2)u = (

1

|x|α
∗ u2)u,

that is, 1
|x|α ∗ (u2 − v2) ≡ 0. It follows from lemma 1.2 that u = v. Denote ϕ = u,

then (u1, v1) = (
√
kϕ,
√
lϕ), where ϕ is a positive least energy solution of (1.9) with

λ = ν and µ = 1.
Next, we consider the case β > max{µ1, µ2}. The following proof is inspired by

[34]. First, we claim that if (u2, v2) is a least energy nontrivial solution of (1.4),

then we obtain v2(x) = au2(x), where a =
√

(β − µ1)/(β − µ2) is a constant.
In fact, if this claim holds, it is easy to see that u2 is a positive least energy

solution of the equation

−∆u+ νu =
β2 − µ1µ2

β − µ2
(

1

|x|α
∗ u2)u, (3.4)

and so u2 =
√
kϕ, where ϕ is a positive least energy solution of (1.9) with λ = ν

and µ = 1.
It suffices to prove this claim. Set û2(x) = a−1v2(x), then (u2, û2) satisfies

−∆u2 + νu2 = µ1(
1

|x|α
∗ u2

2)u2 + βa2(
1

|x|α
∗ û2

2)u2, x ∈ RN ,

−∆û2 + νû2 = µ2a
2(

1

|x|α
∗ û2

2)û2 + β(
1

|x|α
∗ u2

2)û2, x ∈ RN ,

u2, û2 ≥ 0, u, v ∈ H1(RN ).

(3.5)



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 9

By the proof of Theorem 1.3(II), we know that u2, û2 ∈ C2(RN ) and tend to zero
as |x| → ∞. Let Ω+ ≡ {x ∈ RN |u2(x) − û2(x) > 0}. Then Ω+ is a piecewise
C1 smooth domain. Multiplying the first equation in (3.5) by û2 and the second
equation in (3.5) by u2, then integrating by parts on Ω+ and subtracting together,
we obtain the following integral identity∫

∂Ω+

(û2
∂u2

∂n
− u2

∂û2

∂n
) +

∫
Ω+

(µ1 − β)u2û2(
1

|x|α
∗ (u2

2 − û2
2)) = 0, (3.6)

where n denotes the unit outward normal to ∂Ω+.
On the one hand, since u2(x)− û2(x) > 0 in Ω+ and u2(x)− û2(x) = 0 on ∂Ω+,

then we know that∫
∂Ω+

(û2
∂u2

∂n
− u2

∂û2

∂n
) =

∫
∂Ω+

u2
∂(u2 − û2)

∂n
≤ 0. (3.7)

On the other hand, since µ1 − β < 0 and 1
|x|α ∗ (u2

2 − û2
2) ≥ 0 in Ω+, then we

have ∫
Ω+

(µ1 − β)u2û2(
1

|x|α
∗ (u2

2 − û2
2)) ≤ 0. (3.8)

Therefore, from (3.6)-(3.8) we have Ω+ = ∅. Similarly, we may prove that the
set Ω− ≡ {x ∈ RN |u2(x) − û2(x) < 0} is also an empty set. It follows that

u2(x) = û2(x) in RN . Therefore, v2(x) = au2(x), where a =
√

(β − µ1)/(β − µ2).
Finally, based on the above arguments, we can obtain the uniqueness of pos-

itive least energy solutions of system (1.4) when 0 < β < min{µ1, µ2} or β >
max{µ1, µ2} due to the uniqueness of positive solutions to (1.9) for N = 3, 4, 5,
α = N − 2 (see [33]). We completed the proof. �

4. Proof of Theorem 1.5

Multiply the equation for u in (1.4) by v, the equation for v by u, and integrate
over RN , which yields∫

RN
uv[(ν2 − ν1) + (µ1 − β)(

1

|x|α
∗ u2) + (β − µ2)(

1

|x|α
∗ v2)] = 0.

Hence, (3) of Theorem 1.5 holds.
Firstly, we show the proof of (1) in Theorem 1.5. Similarly to (2.1), we get that∫

RN
(|∇u|2 + νiu

2) ≥ 2
√
µiBi

(∫
RN

(
1

|x|α
∗ u2)u2

)1/2

, ∀u ∈ H1(RN ), (4.1)

where

Bi :=
1

2

∫
RN

(|∇ωi|2 + νiω
2
i )− 1

4

∫
RN

µi(
1

|x|α
∗ ω2

i )ω2
i , (4.2)

and ωi is a positive least energy solution of (1.9) with λ = νi and µ = µi, i = 1, 2.
We define

β3 := min
{√

µ1µ2
B1

B2
,

√
µ1µ2

B2

B1

}
. (4.3)

Then we have the following estimate.

Lemma 4.1. For any β ∈ (0, β3), it holds

A < B1 +B2. (4.4)



10 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

Proof. Note that (
√
t1ω1,

√
t2ω2) ∈ N for some t1, t2 > 0 is equivalent to t1, t2 > 0

satisfying∫
RN

µ1(
1

|x|α
∗ ω2

1)ω2
1 =

∫
RN
|∇ω1|2 + ν1ω

2
1

= t1

∫
RN

µ1(
1

|x|α
∗ ω2

1)ω2
1 + t2

∫
RN

β(
1

|x|α
∗ ω2

1)ω2
2 ,∫

RN
µ2(

1

|x|α
∗ ω2

2)ω2
2 =

∫
RN
|∇ω2|2 + ν2ω

2
2

= t2

∫
RN

µ2(
1

|x|α
∗ ω2

2)ω2
2 + t1

∫
RN

β(
1

|x|α
∗ ω2

1)ω2
2 .

That is,

t1 =
µ2D(ω2

2 , ω
2
2)[µ1D(ω2

1 , ω
2
1)− βD(ω2

1 , ω
2
2)]

µ1µ2D(ω2
1 , ω

2
1)D(ω2

2 , ω
2
2)− β2D2(ω2

1 , ω
2
2)
,

t2 =
µ1D(ω2

1 , ω
2
1)[µ2D(ω2

2 , ω
2
2)− βD(ω2

1 , ω
2
2)]

µ1µ2D(ω2
1 , ω

2
1)D(ω2

2 , ω
2
2)− β2D2(ω2

1 , ω
2
2)
.

Meanwhile, we deduce from (1.6) and 0 < β < β3 ≤
√
µ1µ2 that

βD(ω2
1 , ω

2
2) <

√
µ1µ2

B1

B2
D1/2(ω2

1 , ω
2
1)D1/2(ω2

2 , ω
2
2)

= µ1D(ω2
1 , ω

2
1).

Similarly, we have

βD(ω2
1 , ω

2
2) < µ2D(ω2

2 , ω
2
2), µ1µ2D(ω2

1 , ω
2
1)D(ω2

2 , ω
2
2)− β2D2(ω2

1 , ω
2
2) > 0.

So t1, t2 > 0 and (
√
t1ω1,

√
t2ω2) ∈ N . Then

A ≤ E(
√
t1ω1,

√
t2ω2)

=
t1
4

∫
RN

(|∇ω1|2 + ν1ω
2
1) +

t2
4

∫
RN

(|∇ω2|2 + ν2ω
2
2)

=
t1
4

∫
RN

µ1(
1

|x|α
∗ ω2

1)ω2
1 +

t2
4

∫
RN

µ2(
1

|x|α
∗ ω2

2)ω2
2

<
t1
4

∫
RN

µ1(
1

|x|α
∗ ω2

1)ω2
1 + β(

1

|x|α
∗ ω2

1)ω2
2

+
t2
4

∫
RN

µ2(
1

|x|α
∗ ω2

2)ω2
2 + β(

1

|x|α
∗ ω2

1)ω2
2

=
1

4

∫
RN

(|∇ω1|2 + ν1ω
2
1) +

1

4

∫
RN

(|∇ω2|2 + ν2ω
2
2)

= B1 +B2.

�

The following lemma plays a crucial role in the proof of our main results.

Lemma 4.2. Let α ∈ (0, N) ∩ (0, 4). If (uj , vj) ⊂ Hr be a sequence converging
weakly to some (u, v) ∈ Hr as j →∞, then∫

RN
(

1

|x|α
∗ u2

j )v
2
j →

∫
RN

(
1

|x|α
∗ u2)v2 as j →∞, (4.5)



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 11∫
RN

(
1

|x|α
∗ u2

j )u
2
j →

∫
RN

(
1

|x|α
∗ u2)u2 as j →∞. (4.6)

Proof. Note that∫
RN

(
1

|x|α
∗u2

j )v
2
j =

∫
RN

(
1

|x|α
∗u2)v2+

∫
RN

(
1

|x|α
∗u2

j )(v
2
j−v2)+

∫
RN

(
1

|x|α
∗(u2

j−u2))v2,

and

u2
j → u2 in L

2N
2N−α (RN ) as j →∞, v2

j → v2 in L
2N

2N−α (RN ) as j →∞.
Combining these with (1.7), we know that (4.5) and (4.6) hold. �

The following proposition shows the role of A.

Proposition 4.3. If A is attained by a couple (u, v) ∈ N , then this couple is a
solution of (1.4), provided 0 < β <

√
µ1µ2.

Based upon (1.6), the proof of the above proposition is similar to that of [28,
Proposition 1.1], and so we omit it. Before proceeding, we recall some facts about
spherical rearrangement (see [16]).

Proposition 4.4. Assume N ≥ 3 and α ∈ (0, N) ∩ (0, 4). Suppose that u1, u2 ∈
H1(RN ) and let u∗1, u

∗
2 be the symmetric-decreasing rearrangement of u1, u2. Then

‖u∗i ‖νi ≤ ‖ui‖νi ,
∫
RN

(
1

|x|α
∗ (u∗1)2)(u∗2)2 ≥

∫
RN

(
1

|x|α
∗ u2

1)u2
2.

Let κ1 be the smaller root of the equation

β2 −√µ1µ2

(√B1

B2
+ 2

√
B2

B1

)
β + µ1µ2 = 0,

and κ2 be the smaller root of the equation

β2 −√µ1µ2

(√B2

B1
+ 2

√
B1

B2

)
β + µ1µ2 = 0.

Set

β1 := min
{
β3,

√
µ1µ2B1B2

B1 +B2
, κ1, κ2

}
, (4.7)

where β3 is defined in (4.3).

The proof of (1) in Theorem 1.5. Assume that β ∈ (0, β1). The ideas of the fol-
lowing proof mainly come from [28]. Take a minimizing sequence {(un, vn)} ⊂ N
for A, then {(un, vn)} is bounded in H. By Proposition 4.4 the sequence of rear-
rangements {(u∗n, v∗n)} is bounded in H. Up to a subsequence, we may assume that

(u∗n, v
∗
n) → (u∗, v∗) weakly in H and strongly in L

4N
2N−α (RN ) × L

4N
2N−α (RN ). The

proof is divided into three steps.

Step 1. We show that u∗ 6≡ 0, v∗ 6≡ 0. Define

an = D1/2((u∗n)2, (u∗n)2), bn = D1/2((v∗n)2, (v∗n)2).

By (1.6), (4.1) and Proposition 4.4, we have

2
√
µ1B1an ≤

∫
RN
|∇u∗n|2 + ν1(u∗n)2

≤
∫
RN
|∇un|2 + ν1u

2
n



12 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

=

∫
RN

µ1(
1

|x|α
∗ u2

n)u2
n + β(

1

|x|α
∗ u2

n)v2
n

≤
∫
RN

µ1(
1

|x|α
∗ (u∗n)2)(u∗n)2 + β(

1

|x|α
∗ (u∗n)2)(v∗n)2

≤ µ1a
2
n + βanbn,

and

2
√
µ2B2bn ≤

∫
RN
|∇v∗n|2 + ν2(v∗n)2

≤
∫
RN
|∇vn|2 + ν2v

2
n

=

∫
RN

µ2(
1

|x|α
∗ v2

n)v2
n + β(

1

|x|α
∗ v2

n)u2
n

≤
∫
RN

µ2(
1

|x|α
∗ (v∗n)2)(v∗n)2 + β(

1

|x|α
∗ (v∗n)2)(u∗n)2

≤ µ2b
2
n + βanbn.

Note that

E(un, vn) =
1

4

∫
RN
|∇un|2 + ν1u

2
n + |∇vn|2 + ν2v

2
n.

We deduce from lemma 4.4 that

µ1an + βbn ≥ 2
√
µ1B1,

βan + µ2bn ≥ 2
√
µ2B2,√

µ1B1an +
√
µ2B2bn ≤ 2(B1 +B2) + o(1).

(4.8)

We would like to infer from (4.8) that there exists C2 > C1 > 0 such that

C1 < an, bn < C2.

For this it is sufficient to show that each two of the lines

l1 =
{
z = (x, y) ∈ R2 :

√
µ1B1x+

√
µ2B2y = 2(B1 +B2)

}
,

l2 =
{
z ∈ R2 : µ1x+ βy = 2

√
µ1B1

}
,

l3 =
{
z ∈ R2 : βx+ µ2y = 2

√
µ2B2

}
,

meet, and their crossing points have strictly positive coordinates (these lines are
determined by the parameters in (4.8)). Indeed, for large n the point (an, bn) is
arbitrarily close to the triangle (or segment, or point) between these crossing points.
Let (x0, y0) be the crossing points of l2 and l3, by direct computation we have

x0 =
2µ2

√
µ1B1 − 2β

√
µ2B2

µ1µ2 − β2
, y0 =

2µ1

√
µ2B2 − 2β

√
µ1B1

µ1µ2 − β2
.

Since β < β1 ≤
√
µ1µ2, we see that we have to verify the following inequalities√

B1

µ1
<
B1 +B2√
µ1B1

<

√
µ2B2

β
, (4.9)√

B2

µ2
<
B1 +B2√
µ2B2

<

√
µ1B1

β
, (4.10)



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 13√
µ1B1x0 +

√
µ2B2y0 ≤ 2(B1 +B2). (4.11)

Inequalities (4.9) and (4.10) can be recast as β <
√
µ1µ2B1B2

B1+B2
, which is true by the

definition of β1. Elementary computations show that (4.11) also holds. Finally,
from lemma 4.2 and (4.8) we infer that there exists C1 > 0 such that

C1 ≤
(∫

RN
(

1

|x|α
∗ (u∗)2)(u∗)2

)1/2

≤
2
√
µ2B1√

µ1µ2B1 − β
√
B2

<
2
√
µ2B2

β
, (4.12)

C1 ≤
(∫

RN
(

1

|x|α
∗ (v∗)2)(v∗)2

)1/2

≤
2
√
µ1B2√

µ1µ2B2 − β
√
B1

<
2
√
µ1B1

β
, (4.13)

where C1 depends on µ1, µ2, β, B1, B2. Therefore u∗ 6≡ 0 and v∗ 6≡ 0.

Step 2. We show that there exist t1 > 0 and t2 > 0 such that (
√
t1u
∗,
√
t2v
∗) ∈ N .

By Lemma 4.2 and Proposition 4.4, we know that

‖u∗‖2ν1 ≤ lim inf
n→∞

‖u∗n‖ν1 ≤ lim inf
n→∞

‖un‖ν1

= lim inf
n→∞

∫
RN

µ1(
1

|x|α
∗ u2

n)u2
n + β(

1

|x|α
∗ u2

n)v2
n

≤ lim inf
n→∞

∫
RN

µ1(
1

|x|α
∗ (u∗n)2)(u∗n)2 + β(

1

|x|α
∗ (u∗n)2)(v∗n)2

=

∫
RN

µ1(
1

|x|α
∗ (u∗)2)(u∗)2 + β(

1

|x|α
∗ (u∗)2)(v∗)2,

that is,

‖u∗‖2ν1 ≤
∫
RN

µ1(
1

|x|α
∗ (u∗)2)(u∗)2 + β(

1

|x|α
∗ (u∗)2)(v∗)2. (4.14)

Similarly, we have

‖v∗‖2ν2 ≤
∫
RN

β(
1

|x|α
∗ (u∗)2)(v∗)2 + µ2(

1

|x|α
∗ (v∗)2)(v∗)2, (4.15)

E(u∗, v∗) ≤ lim inf
n→∞

E(u∗n, v
∗
n) ≤ lim inf

n→∞
E(un, vn) = A. (4.16)

Let t1, t2 be the solutions of the linear system(∫
RN

µ1(
1

|x|α
∗ (u∗)2)(u∗)2

)
t1 +

(∫
RN

β(
1

|x|α
∗ (u∗)2)(v∗)2

)
t2 = ‖u∗‖2ν1 ,(∫

RN
β(

1

|x|α
∗ (u∗)2)(v∗)2

)
t1 +

(∫
RN

µ2(
1

|x|α
∗ (v∗)2)(v∗)2

)
t2 = ‖v∗‖2ν2 .

We claim that the solution of the above equations satisfies t1 > 0, t2 > 0 if β ∈
(0, β1). In fact, by a direct computation we see that

t1 =
‖u∗‖2ν1

∫
RN µ2( 1

|x|α ∗ (v∗)2)(v∗)2 − ‖v∗‖ν2
∫
RN β( 1

|x|α ∗ (u∗)2)(v∗)2

µ1µ2D((u∗)2, (u∗)2)D((v∗)2, (v∗)2)− β2D2((u∗)2, (v∗)2)
,

t2 =
‖v∗‖2ν2

∫
RN µ1( 1

|x|α ∗ (u∗)2)(u∗)2 − ‖u∗‖ν1
∫
RN β( 1

|x|α ∗ (u∗)2)(v∗)2

µ1µ2D((u∗)2, (u∗)2)D((v∗)2, (v∗)2)− β2D2((u∗)2, (v∗)2)
.

Next, we prove that t1 > 0. We need to show that

‖u∗‖2ν1

∫
RN

µ2(
1

|x|α
∗ (v∗)2)(v∗)2 > ‖v∗‖2ν2

∫
RN

β(
1

|x|α
∗ (u∗)2)(v∗)2. (4.17)



14 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

By (1.6), (4.1), (4.12), (4.13) and (4.15), inequality (4.17) is implied by

µ2‖u∗‖2ν1D
1/2((v∗)2, (v∗)2) > β‖v∗‖2ν2D

1/2((u∗)2, (u∗)2)

⇐ µ2D
1/2((v∗)2, (v∗)2) >

β

2
√
µ1B1

‖v∗‖2ν2

⇐ 1 >
β

2
√
µ1B1

( β
µ2
D1/2((u∗)2, (u∗)2) +D1/2((v∗)2, (v∗)2)

)
⇐ 1 >

β

2
√
µ1B1

( β
µ2

2
√
µ2B2

β
+

2
√
µ1B2√

µ1µ2B2 − β
√
B1

)
.

Note that the last inequality above can be recast as√
B1B2

µ2
β2 − (

√
µ1B1 + 2

√
µ1B2)β + µ1

√
µ2B1B2 > 0. (4.18)

Therefore, by the definition of β1, we know the inequality (4.17) is true, and so
t1 > 0. Similarly, we can prove that t2 > 0. Moreover, we have (

√
t1u
∗,
√
t2v
∗) ∈ N .

Step 3. We show that (|u∗|, |v∗|) is a positive solution of (1.4) and E(|u∗|, |v∗|) =
A. Note that (

√
t1u
∗,
√
t2v
∗) ∈ N . Then

A ≤ E(
√
t1u
∗,
√
t2v
∗) =

t1
4
‖u∗‖2ν1 +

t2
4
‖v∗‖2ν2

≤ t1
4

∫
RN

µ1(
1

|x|α
∗ (u∗)2)(u∗)2 + β(

1

|x|α
∗ (u∗)2)(v∗)2

+
t2
4

∫
RN

β(
1

|x|α
∗ (u∗)2)(v∗)2 + µ2(

1

|x|α
∗ (v∗)2)(v∗)2

=
1

4
‖u∗‖2ν1 +

1

4
‖v∗‖2ν2

≤ 1

4
lim inf
n→∞

(‖un‖2ν1 + ‖vn‖2ν2) = A,

which implies that (4.14) and (4.15) are equalities. Thus, (u∗, v∗) ∈ N . Combining
this with (4.16), one has that A = E(u∗, v∗). Therefore, (|u∗|, |v∗|) ∈ N and
A = E(|u∗|, |v∗|). By Proposition 4.3 and the maximum principle, we see that
(|u∗|, |v∗|) is a positive least energy solution of (1.4). �

It remains to prove (2) of Theorem 1.5. Assume that β > 0. Without loss of
generality, we may assume that ν1 ≤ ν2. We define

A := inf
h∈Γ

max
t∈[0,1]

E(h(t)), (4.19)

where Γ = {h ∈ C([0, 1], H) : h(0) = (0, 0), E(h(1)) < 0}. By (1.8), we know that
for any (u, v) ∈ H, and (u, v) 6= (0, 0),

max
t>0

E(tu, tv) = E(tu,vu, tu,vv)

=
1

4
t2u,v

∫
RN

(|∇u|2 + ν1u
2 + |∇v|2 + ν2v

2),
(4.20)

where tu,v > 0 satisfies

t2u,v =

∫
RN (|∇u|2 + ν1u

2 + |∇v|2 + ν2v
2)∫

RN µ1( 1
|x|α ∗ u2)u2 + 2β( 1

|x|α ∗ u2)v2 + µ2( 1
|x|α ∗ v2)v2

.



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 15

Note that (tu,vu, tu,vv) ∈ N ′, where

N ′ :=
{

(u, v) ∈ H\{(0, 0)}, F (u, v) :=

∫
RN

(|∇u|2 + ν1u
2 + |∇v|2 + ν2v

2)

−
∫
RN

µ1(
1

|x|α
∗ u2)u2 + 2β(

1

|x|α
∗ u2)v2 + µ2(

1

|x|α
∗ v2)v2 = 0

}
,

(4.21)

it is standard to see that

A = inf
H3(u,v) 6=(0,0)

max
t>0

E(tu, tv)

= inf
(u,v)∈N ′

E(u, v)

= inf
H\{(0,0)}

(
‖u‖2ν1 + ‖v‖2ν2

)2
4
∫
RN µ1( 1

|x|α ∗ u2)u2 + 2β( 1
|x|α ∗ u2)v2 + µ2( 1

|x|α ∗ v2)v2
.

(4.22)

Note that N ⊆ N ′, one has that A ≤ A. Denote

β2 := max
{µ1(2 + ν2−ν1

ν1
)2 − (µ1 + µ2)

2
,

3µ2 − µ1

2
, 0
}
. (4.23)

Then we have the following lemma.

Lemma 4.5. Assume that N ≥ 3 and α ∈ (0, N) ∩ (0, 4). Then for any β > β2 it
holds A < min{B1, B2}.

Proof. It follows from (4.22) and (4.23) that

A ≤
(
‖ω1‖2ν1 + ‖ω1‖2ν2

)2
4
∫
RN µ1( 1

|x|α ∗ ω
2
1)ω2

1 + 2β( 1
|x|α ∗ ω

2
1)ω2

1 + µ2( 1
|x|α ∗ ω

2
1)ω2

1

≤ µ1(2 + ν2 − ν1)2

µ1 + µ2 + 2β
B1 < B1.

(4.24)

Similarly, we see that

A ≤
(
‖ω2‖2ν1 + ‖ω2‖2ν2

)2
4
∫
RN µ1( 1

|x|α ∗ ω
2
2)ω2

2 + 2β( 1
|x|α ∗ ω

2
2)ω2

2 + µ2( 1
|x|α ∗ ω

2
2)ω2

2

≤ 4µ2

µ1 + µ2 + 2β
B2 < B2.

(4.25)

By (4.24) and (4.25), we know that A < min{B1, B2}. �

Proof of (2) in Theorem 1.5. Assume that β > β2. Since the functional E has a
mountain pass structure, then by the mountain pass theorem (see [35]) there exists
{(un, vn)} such that

lim
n→∞

E(un, vn) = A, lim
n→∞

E′(un, vn) = 0.

It is standard to see that {(un, vn)} is bounded in H. By Proposition 4.4 the
sequence of rearrangements {(u∗n, v∗n)} is bounded in H. Up to a subsequence, we

may assume that (u∗n, v
∗
n) → (u∗, v∗) weakly in H and strongly in L

4N
2N−α (RN ) ×



16 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

L
4N

2N−α (RN ). Similarly to the proof of (1) in Theorem 1.5, we know that

‖(u∗, v∗)‖2H

≤
∫
RN

µ1(
1

|x|α
∗ (u∗)2)(u∗)2 + 2β(

1

|x|α
∗ (u∗)2)(v∗)2 + µ2(

1

|x|α
∗ (v∗)2)(v∗)2,

E(u∗, v∗) ≤ lim inf
n→∞

E(u∗n, v
∗
n) ≤ lim inf

n→∞
E(un, vn) = A.

(4.26)

Step 1. We show that (u∗, v∗) 6= (0, 0). We define

cn =
(∫

RN
(

1

|x|α
∗ (u∗n)2)(u∗n)2

)1/2

, dn =
(∫

RN
(

1

|x|α
∗ (v∗n)2)(v∗n)2

)1/2

.

It follows from (4.1) and (4.26) that

2
√
B1cn + 2

√
B2dn ≤ µ1c

2
n + 2βcndn + µ2d

2
n.

By lemma 4.2 we have∫
RN

(
1

|x|α
∗ (u∗)2)(u∗)2 +

∫
RN

(
1

|x|α
∗ (v∗)2)(v∗)2 ≥ C > 0,

which implies (u∗, v∗) 6= (0, 0).

Step 2. We show that (|u∗|, |v∗|) is a positive solution of (1.4) and E(|u∗|, |v∗|) =
A. If inequality (4.26) is an equality, then (u∗, v∗) ∈ N ′ and A is attained by
(u∗, v∗). If not, that is,

‖(u∗, v∗)‖2H <

∫
RN

µ1(
1

|x|α
∗ (u∗)2)(u∗)2 + 2β(

1

|x|α
∗ (u∗)2)(v∗)2

+ µ2(
1

|x|α
∗ (v∗)2)(v∗)2,

take s ∈ (0, 1) such that (u1, v1) = s(u∗, v∗) ∈ N ′. Therefore,

A ≤ 1

4
‖(u1, v1)‖2H <

1

4
‖(u∗, v∗)‖2H ≤

1

4
lim inf
n→∞

‖(u∗n, v∗n)‖2H

≤ lim inf
n→∞

‖(un, vn)‖2H = A,

which is a contradiction. Thus (u∗, v∗) ∈ N ′ and E(u∗, v∗) = A. Therefore,
E(|u∗|, |v∗|) = A and (|u∗|, |v∗|) ∈ N ′. So there exists a Lagrange multiplier L ∈ R
such that

E′(|u∗|, |v∗|)− LG′(|u∗|, |v∗|) = 0,

where

G(u, v) = ‖(u, v)‖2H −
∫
RN

µ1(
1

|x|α
∗ u2)u2

+ 2β(
1

|x|α
∗ u2)v2 + µ2(

1

|x|α
∗ v2)v2.

Since E′(|u∗|, |v∗|)(|u∗|, |v∗|) = G(|u∗|, |v∗|) = 0 and

G′(|u∗|, |v∗|)(|u∗|, |v∗|)

= −2

∫
RN

µ1(
1

|x|α
∗ (u∗)2)(u∗)2 + 2β(

1

|x|α
∗ (u∗)2)(v∗)2 + µ2(

1

|x|α
∗ (v∗)2)(v∗)2 6= 0,



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 17

then we get that L = 0 and E′(|u∗|, |v∗|) = 0, that is (|u∗|, |v∗|) is a solution
of (1.4). It follows from lemma 4.5 that |u∗| 6≡ 0 and |v∗| 6≡ 0. This means
(|u∗|, |v∗|) ∈ N ⊂ N ′, and so E(|u∗|, |v∗|) = A = A. Then using the strong
maximum principle, we see that |u∗|, |v∗| > 0 in RN . Therefore, (|u∗|, |v∗|) is a
positive least energy solution of (1.4). The proof is complete. �

5. Proof of Theorem 1.7

Recall the definitions of E,N , A, they all depend on β, and we use notation
Eβ ,Nβ , Aβ in this section.

Lemma 5.1. Assume that N ≥ 3 and α ∈ (0, N) ∩ (0, 4). Let β ∈ (0, β1) and
(uβ , vβ) be the positive least energy solution of (1.4) which exists by Theorem 1.5.
Then it holds∫

RN

( 1

|x|α
∗ u2

β

)
u2
β ≥

(2
√
µ1µ2B1B2 − 2β(B1 +B2)

µ1

√
µ2B2 − β

√
µ1B1

)2

, (5.1)∫
RN

( 1

|x|α
∗ v2

β

)
v2
β ≥

(2
√
µ1µ2B1B2 − 2β(B1 +B2)

µ2

√
µ1B1 − β

√
µ2B2

)2

. (5.2)

Proof. Note that (uβ , vβ) ∈ Nβ with E(uβ , vβ) ≤ 4(B1 +B2). We denote

D1 =
(∫

RN
(

1

|x|α
∗ u2

β)u2
β

)1/2

, D2 =
(∫

RN
(

1

|x|α
∗ v2

β)v2
β

)1/2

.

Similarly to the proof of (1) in Theorem 1.5, we obtain

µ1D1 + βD2 ≥ 2
√
µ1B1, (5.3)

βD1 + µ2D2 ≥ 2
√
µ2B2, (5.4)√

µ1B1D1 +
√
µ2B2D2 ≤ 2(B1 +B2). (5.5)

Therefore, we deduce from (5.3)-(5.5) that∫
RN

(
1

|x|α
∗ u2

β)u2
β ≥

(2
√
µ1µ2B1B2 − 2β(B1 +B2)

µ1

√
µ2B2 − β

√
µ1B1

)2

,∫
RN

(
1

|x|α
∗ v2

β)v2
β ≥

(2
√
µ1µ2B1B2 − 2β(B1 +B2)

µ2

√
µ1B1 − β

√
µ2B2

)2

.

�

Conclusion of the proof of Theorem 1.7. Let 0 < βn < β1, n ∈ N, satisfy βn → 0
as n → +∞. Suppose that (un, vn) is the positive least energy solutions of (1.4)
with β = βn, and (un, vn) is radially symmetric. By Lemma 4.1, we know that
{(un, vn)} is uniformly bounded in H. Passing to a subsequence, we may assume
that

un ⇀ û, vn ⇀ v̂ weakly in H1(RN ),

un → û, vn → v̂ almost everywhere in RN .

Combining these with lemma 4.2 and lemma 5.1 we obtain∫
RN

(
1

|x|α
∗ û2)û2 ≥ 2

√
B1

µ1
,

∫
RN

(
1

|x|α
∗ v̂2)v̂2 ≥ 2

√
B2

µ2
.



18 S. YOU, P. ZHAO, Q. WANG EJDE-2021/47

Hence, û 6≡ 0, v̂ 6≡ 0 and û(x), v̂(x) ≥ 0 a.e. x ∈ RN . We multiply the equation for
u in (1.4) by û and integrate over RN , which implies∫

RN
∇un∇û+ ν1unû =

∫
RN

µ1(
1

|x|α
∗ u2

n)unû+ βn(
1

|x|α
∗ v2

n)unû. (5.6)

We claim that∫
RN

(|x|−α ∗ u2
n)unû→

∫
RN

(|x|−α ∗ û2)û2, as n→∞. (5.7)

Note that
u2
n → û2 strongly in L

2N
2N−α (RN ), as n→∞.

By the Hardy-Littlewood-Sobolev inequality, the Riesz potential defines a linear

continuous map from L
2N

2N−α (RN ) to L
2N
α (RN ). Then we obtain

|x|−α ∗ u2
n → |x|−αû2 strongly in L

2N
α (RN ), as n→∞.

Combining this with the fact that

un → û strongly in L
4N

2N−α (RN ), as n→∞,
we have

(|x|−α ∗ u2
n)un → (|x|−α ∗ û2)û strongly in L

4N
2N+α (RN ), as n→∞.

Therefore, (5.7) holds. It follows from (4.1), (5.6) and (5.7) that∫
Ω

(|∇û|2 + ν1û
2) ≤

∫
RN

µ1(
1

|x|α
∗ û2)û2 ≤ (4B1)−1

(∫
RN

(|∇û|2 + ν1û
2)
)2

, (5.8)

and so ∫
RN

(|∇û|2 + ν1û
2) ≥ 4B1. (5.9)

Similarly we have ∫
RN

(|∇v̂|2 + ν2v̂
2) ≥ 4B2. (5.10)

Combining these with lemma 4.1 we know that

B1 +B2

≥ lim
n→∞

Aβn

=
1

4
lim
n→∞

∫
RN
|∇un|2 + ν1u

2
n + |∇vn|2 + ν2v

2
n

=
1

4

∫
RN
|∇û|2 + ν1û

2 +
1

4

∫
RN
|∇v̂|2 + ν2v̂

2

+
1

4
lim
n→∞

∫
RN
|∇(un − û)|2 + ν1|un − û|2 + |∇(vn − v̂)|2 + ν2|vn − v̂|2

≥ B1 +B2 +
1

4
lim
n→∞

∫
RN
|∇(un − û)|2 + ν1|un − û|2

+
1

4
lim
n→∞

∫
RN
|∇(vn − v̂)|2 + ν2|vn − v̂|2

≥ B1 +B2.

(5.11)

This means that

lim
n→∞

∫
RN
|∇(un − û)|2 + ν1|un − û|2 + |∇(vn − v̂)|2 + ν2|vn − v̂|2 = 0,



EJDE-2021/47 POSITIVE LEAST ENERGY SOLUTION FOR CHOQUARD SYSTEMS 19

and so

(un, vn)→ (û, v̂) strongly in H1(RN )×H1(RN ) as n→ +∞.
Moreover,

∫
RN (|∇û|2 + ν1û

2) = 4B1, and so we see from (5.8) that∫
RN

(|∇û|2 + ν1û
2) =

∫
RN

µ1(
1

|x|α
∗ û2)û2 = 4B1.

Therefore, û is a positive least energy solution of

−∆u+ ν1u = µ1(
1

|x|α
∗ u2)u, u ∈ H1(RN ).

Similarly, we know that v̂ is a positive least energy solution of

−∆v + ν2v = µ2(
1

|x|α
∗ v2)v, v ∈ H1(RN ).

The proof is complete �

Acknowledgments. This research was partially supported by the National Nat-
ural Science Foundation of China (NSFC 11471147).

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Song You
Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China

Email address: yousong@mail.tsinghua.edu.cn

Peihao Zhao
School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000,

China

Email address: zhaoph@lzu.edu.cn

Qingxuan Wang

Department of Mathematics, Zhejiang Normal University, Jinhua, Zhejiang 321004, China
Email address: wangqx@zjnu.edu.cn


	1. Introduction
	2. Proof of Theorem ??
	3. Proof of Theorem ??
	4. Proof of Theorem ??
	5. Proof of Theorem ??
	Acknowledgments

	References