Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 19, pp. 1–37.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: 10.58997/ejde.2024.19

LOCALIZED NODAL SOLUTIONS FOR SEMICLASSICAL

CHOQUARD EQUATIONS WITH CRITICAL GROWTH

BO ZHANG, WEI ZHANG

Abstract. In this article, we study the existence of localized nodal solutions

for semiclassical Choquard equation with critical growth

−ε2∆v + V (x)v = εα−N
(∫

RN

|v(y)|2∗α
|x− y|α

dy
)
|v|2
∗
α−2v + ϑ|v|q−2v, x ∈ RN ,

where ϑ > 0, N ≥ 3, 0 < α < min{4, N − 1},max{2, 2∗ − 1} < q < 2∗,

2∗α = 2N−α
N−2

, V is a bounded function. By the perturbation method and

the method of invariant sets of descending flow, we establish for small ε the

existence of a sequence of localized nodal solutions concentrating near a given
local minimum point of the potential function V .

1. Introduction

In this article, we study localized nodal solutions of the nonlinear Choquard
equation with critical exponent

−ε2∆v + V (x)v = εα−N
(∫

RN

|v(y)|2∗α
|x− y|α

dy
)
|v|2

∗
α−2v + ϑ|v|q−2v, x ∈ RN ,

v(x)→ 0 as |x| → ∞,
(1.1)

where ϑ > 0, N ≥ 3, 0 < α < min{4, N−1},max{2, 2∗−1} < q < 2∗, 2∗α = 2N−α
N−2 is

the upper critical exponent in the sense of the Hardy-Littlewood-Sobolev inequality,
ε > 0 is small parameter. The potential function V satisfies following assumptions:

(A1) V ∈ C1(RN , R) and there exist b > a > 0 such that

a ≤ V (x) ≤ b, ∀x ∈ RN .

(A2) There exists a bounded domain M ⊂ RN with the smooth boundary ∂M
such that

〈−→n (x), ∇V (x)〉 > 0 , ∀x ∈ ∂M ,

where −→n (x) is the outer normal of ∂M at x.

2020 Mathematics Subject Classification. 35B20, 35Q40.
Key words and phrases. Choquard equation; sign-changing solutions; nodal solutions;

variational perturbation method; semiclassical states.
©2024. This work is licensed under a CC BY 4.0 license.

Submitted October 17, 2023. Published February 16, 2024.

1



2 B. ZHANG, W. ZHANG EJDE-2024/19

Over the previous decades, the Choquard type equation has been widely studied.
The Choquard equation

−∆u+ u =
(∫

R3

|u(y)|2

|x− y|
dy
)
u , x ∈ R3, (1.2)

is the Choquard-Pekar equation which originated from the description of the quan-
tum theory of a polaron at rest by Pekar in 1954 [35]. Choquard also used (1.2) to
describe the Hartree-Fock [23] theory of one component plasma in 1976. If u is a
solution of (1.2), then ψ(t, x) = eitu(x) is a solitary wave solution of the Hartree
equation

iψt = −∆ψ −
(∫

R3

|ψ(y)|2

|x− y|
dy
)
ψ , x ∈ R3. (1.3)

In 1996, Penrose [29] proposed equation (1.2) as a model of self gravitation. Lieb
[21] proved the existence and uniqueness of the ground state solution for (1.2) by
using the symmetric rearrangement inequality. Lions [24] showed that the equation

−∆u+ λu =
(∫

R3

V (x− y)|u(y)|2 dy
)
u , x ∈ R3 (1.4)

has a positive radial symmetric solution and infinitely many radial symmetric so-
lutions, where λ > 0, V > 0 and V is radially symmetric. For semilinear Choquard
equation

−∆u+ V (x)u =
(∫

RN

F (y, u(y))

|x− y|α
dy
)
f(x, u) , x ∈ RN , (1.5)

where f(x, u) = ∂F (x,u)
∂u , a large number of research results have been obtained. We

refer the reader to [2, 11, 12, 13, 14, 15, 30, 31, 33, 39, 40], and references therein.
For the semiclassical Choquard equation with subcritical growth

− ε2∆u+ V (x)u = εα−N
(∫

RN

F (u(y))

|x− y|α
dy
)
f(u), x ∈ RN , (1.6)

Wei and Winter [41] proved the existence of solution by using the Lyapunov-Schmidt
reduction method [19] in 2009, where N = 3, α = 1, F (u) = |u|2, V satisfies

inf
x∈R3

V (x) > 0, V (x) ∈ C2(R3).

In 2015, Moroz and Van Schaftingen [32] constructed the single spike solution which
concentrating around the local minimum of potential V by using the nonlocal pe-
nalization method, where N ≥ 1, 0 < α < N , F (u) = |u|p, p ∈ [2, 2N−α

N−2 ),

V ∈ C(RN , [0,∞)). More results for the semiclassical Choquard equation with
subcritical growth we refer [8, 17, 18, 27, 37, 44, 45, 48] and references therein.

For the semiclassical Choquard equation with critical growth

− ε2∆u+ V (x)u = εα−N
(∫

RN

F (y, u(y))

|x− y|α
dy
)
f(x, u), x ∈ RN , (1.7)

Cassani and Zhang [5] proved the existence and decays exponentially of positive
solution for the semiclassical critical Choquard (1.7), where N = 3, 0 < α < 3, V, F
satisfies some suitable assumptions. In 2017, Alves and Gao [1] investigated the
existence of ground state solutions, multiplicity and concentration of semiclassical
solutions for the semiclassical Choquard (1.7) by using variational methods [3],
where N = 3, 0 < α < 3, V, F satisfy some suitable assumptions. In 2020, Gao,
Yang and Zhou [10] obtained existence and multiplicity of solutions for (1.7), where



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 3

N ≥ 3, 0 < α < min{4, N}, V and F satisfy some suitable assumptions. Qi and
Zou [36] investigated semiclassical critical Choquard equation

− ε2∆u+ V (x)u = εα−N
(∫

RN

|u(y)|2∗α
|x− y|α

dy
)
|u|2

∗
α−2u+ λg(u), x ∈ RN , (1.8)

where N ≥ 3, 0 < α < min{4, N}, 2∗α = 2N−α
N−2 , λ > 0, g ∈ C1(RN ,R) satisfies

(A3) limt→0+
g(t)
t = 0, and there exists p ∈ (1, N+2

N−2 ), such that limt→∞
g(t)
tp = 0;

(A4) There exists µ ∈ (2, 2(2N−α)
N−2 ), such that 0 < µG(t) ≤ g(t)t for all t ∈

(0,+∞);

(A5) g(t)
t is monotonic increasing on (0,∞).

and V satisfies:

There exists a bounded smooth domain M⊂ RN such that

m = min
x∈M

V (x) < min
x∈∂M

V (x).

They obtained a local solution concentrating around the local minimum of potential
V by using Byeon-Wang [4] type penalization method. More results for the equation
(1.7), one can see [9, 34, 42, 43, 46, 47] and references therein.

In recent years, there are have been many results of localized nodal solution
for the Choquard equation. In 2021, He and Liu [17] proved existence and con-
centration of infinitely many sign-changing solutions for the following Choquard
equation

− ε2∆u+ V (x)u = εα−N
(∫

RN

|u(y)|p

|x− y|α
dy
)
|u|p−2u, x ∈ RN , (1.9)

where N ≥ 3, 0 < α < min{4, N − 1}, the potential function V satisfies (A1)
and (A2). Zhang and Liu [45] investigated the semiclassical quasi-linear Choquard
equation with subcritical growth, and obtained a conclusion similar to that of [17]
in 2022.

For the semiclassical Choquard equation with critical growth

− ε2∆v+ V (x)v = εα−N
(∫

RN

|v(y)|2∗α
|x− y|α

dy
)
|v|2

∗
α−2v+ϑ|v|q−2v, x ∈ RN , (1.10)

where ϑ > 0, N ≥ 3, 0 < α < min{4, N − 1},max{2, 2∗ − 1} < q < 2∗, 2∗α = 2N−α
N−2 ,

there is no results on nodal solutions yet. Combining perturbation method, trun-
cation method and the method of invariant sets of descending flow, we prove (1.10)
possesses a sequence of localized nodal solutions. As for the method mentioned, we
refer [49, 45, 16] and the references therein.

Under the assumption (A2), the critical set safisfies

A = {x ∈M|∇V (x) = 0} 6= ∅,

and without loss of generality we assume 0 ∈ A. For any set B ⊂ RN and any
δ > 0, we set

Bδ = {x ∈ RN |δx ∈ B},

Bδ = {x ∈ RN |dist(x,B) := inf
y∈B
|x− y| < δ}.

The main result of this paper is as follows.



4 B. ZHANG, W. ZHANG EJDE-2024/19

Theorem 1.1. Assume that (A1) and (A2) hold. Then for each positive integer
k there exists εk > 0 such that if 0 < ε < εk, equation (1.1) has at least k pairs
of sign-changing solutions ±vj,ε, j = 1, . . . , k. Moreover, for each δ > 0 there exist
µ > 0, C = Ck > 0 and εk(δ) > 0 such that if 0 < ε < εk(δ); then

|vj,ε(x)| ≤ C exp{−µ
ε

dist(x,Aδ)}, ∀x ∈ RN , j = 1, . . . , k. (1.11)

Denoting u(x) = v(εx), equation (1.1) is equivalent to

−∆u+ V (εx)u =
(∫

RN

|u(y)|2∗α
|x− y|α

dy
)
|u|2

∗
α−2u+ ϑ|u|q−2u, x ∈ RN ,

u(x)→ 0 as |x| → ∞,
(1.12)

and the corresponding energy functional is

Iε(u) =
1

2

∫
RN

(|∇u|2 + V (εx)u2) dx− 1

2 · 2∗α

∫
RN

∫
RN

|u(y)|2∗α |u(x)|2∗α
|x− y|α

dx dy

− ϑ

q

∫
RN
|u|q dx, u ∈ H1(RN ).

To obtain multiple localized nodal solutions for Iε, we use the penalization
method due to Byeon and Wang [4]. Let ζ ∈ C∞ be a cut-off function, ζ(t) = 0 for
t ≤ 0; ζ(t) = 1 for t ≥ 1; 0 ≤ ζ ′(t) ≤ 2 and 0 ≤ ζ(t) ≤ 1. We define

χε(x) =

{
0, if x ∈Mε

ε−6ζ(dist(x,Mε)), if x /∈Mε.

We truncate the critical term to a subcritical term by truncation method. Now we
define some auxiliary functions. Let ξ(t) ∈ C∞(R, [0, 1]) be a smooth, even function
such that ξ(t) = 1 if |t| ≤ 1; ξ(t) = 0 if |t| ≥ 2; 0 ≤ ξ(t) ≤ 1 and ξ is decreasing in
[1, 2]. For ε ∈ (0, 1], x ∈ RN , t ∈ R, we define

bν(t) = ξ(νt), mν(t) =

∫ t

0

bν(τ) dτ,

fν(t) = |mν(t)|2
∗
α−r|t|r−2t, Fν(t) =

∫ t

0

fν(τ)dτ,

where 2 < r < 2∗α and r ≤ q. We now consider the equation

−∆u+ V (εx)u = 2∗α

(∫
RN

Fν(u(y))

|x− y|α
dy
)
fν(u(x)) + ϑ|u|q−2u, x ∈ RN ,

u(x)→ 0 as |x| → ∞,
(1.13)

and its corresponding energy functional

Iε,ν(u) =
1

2

∫
RN

(|∇u|2 + V (εx)u2) dx− 2∗α
2

∫
RN

∫
RN

Fν(u(y))Fν(u(x))

|x− y|α
dx dy

− ϑ

q

∫
RN
|u|q dx, u ∈ H1(RN ).

Since the imbedding from H1(RN ) to Lp(RN )(2 ≤ p ≤ 2∗) is continuous but
not compact, we need to choose a suitable function space as working space such



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 5

that the functional Iε,ν(u) recovers compactness. For this purpose, we denote
Xε = H1(RN ) ∩ Lmε (RN ), where Lmε (RN ) is a weighted space defined as

Lmε (RN ) =
{
u ∈ Lm(RN ) :

∫
RN

exp{(m− 2) dist(εx,M)}|u|m dx < +∞
}

endowed with the norm

‖u‖Lmε (RN ) =
(∫

RN
exp{(m− 2) dist(εx,M)}|u|m dx

)1/m

.

We define

‖u‖H1(RN ) =
(∫

RN
(|∇u|2 + E(εx)u2) dx

)1/2

,

‖u‖Xε = ‖u‖H1(RN ) + ‖u‖Lmε (RN ),

where E(x) = V (x) − σ, σ is small enough such that E satisfies the assumptions
(A1) and (A2) (with a different constant a′ = a− σ > 0).

Meanwhile, we introduce an additional coercive term such that Iε,ν has necessary
compactness property on Xε. For this purpose, we need some auxiliary functions.
For ε ∈ (0, 1] , x ∈ RN , t ∈ R, we define

bε(x, t) = ξ(ε exp{dist(εx,M)}t), mε(x, t) =

∫ t

0

bε(x, τ) dτ,

kε(x, t) = (
t

mε(x, t)
)m−2t, Kε(x, t) =

∫ t

0

kε(x, τ) dτ,

where 2 < m < r. We define the perturbed functional

Γε,ν(u) =
1

2

∫
RN

(|∇u|2 + E(εx)u2) dx+ σ

∫
RN

Kε(x, u) dx

+
1

2β

(∫
RN

χε(x)u2 dx− 1
)β

+
− 2∗α

2

∫
RN

∫
RN

Fν(u(y))Fν(u(x))

|x− y|α
dx dy

− ϑ

q

∫
RN
|u|q dx , u ∈ Xε,

where 2 < 2β < r. Note that the method of invariant sets of descending flow [28]
can not fit well for the functional Γε,ν , so we also use the perturbation method [17]
to overcome this difficulty. For t ∈ R+, we define

bλ(t) = ξ(λt), mλ(t) =

∫ t

0

bλ(τ) dτ,

gλ(t) =
mλ(t)

t
, hλ(t) = gλ(t) + bλ(t).

Now we define

Γε,ν,λ(u) =
1

2

∫
RN

(|∇u|2 + E(εx)u2) dx+ σ

∫
RN

Kε(x, u) dx

+
1

2β

(∫
RN

χε(x)u2 dx− 1
)β

+
− 2∗α

2
gλ(ϕ1/2(u))ϕ(u)

− ϑ

q

∫
RN
|u|q dx, u ∈ Xε,



6 B. ZHANG, W. ZHANG EJDE-2024/19

where

ϕ(u) =

∫
RN

∫
RN

Fν(u(x))Fν(u(y))

|x− y|α
dx dy.

By Hardy-Littlewood-Sobolev inequality and Sobolev inequality, we have

ϕ1/2(u) ≤ C0‖u‖
2∗α
H1(RN )

.

Note that if

|u(x)| ≤ 1

ε
exp{− dist(εx,M)} for x ∈ RN ,

|u(x)| ≤ 1

ν
,
(∫

RN
χε(x)u2 dx− 1

)
+

= 0, ‖u‖H1(RN ) ≤
( 1

C0λ

)1/2∗α

for sufficiently small ε, ν, λ, then Γε,ν,λ(u) = Iε(u), andDΓε,ν,λ(u) = DIε(u). Hence
we can obtain solutions of the equation (1.12) by researching Γε,ν,λ.

In the following c denotes various constants, cε denotes constants depending on
ε and c, cε may be used from line to line for different constants but independent of
the arguments.

This article organized as follows. In section 2 we prove preliminary results and
verify the Palais-Smale condition for the function Γε,ν,λ. In section 3 we construct
a sequence of nodal critical points of Γε,ν,λ by using the invariant sets method. In
Section 4, we prove uniform bound on the critical points obtained in Section 3.
Section 5 is devoted to the proof of Theorem 1.1.

2. Preliminaries and Palais-Smale condition for Γε,ν,λ

In this section, we first collect some elementary results about the auxiliary func-
tions involved in the perturbed functional Γε,ν,λ. Then, we prove that Γε,ν,λ satisfies
the (PS) condition.

Lemma 2.1 (Hardy-littlewood-Sobolev inequality [22]). Suppose α ∈ (0, N), and
s, r > 1 with 1

s + 1
r = 2N−α

N . Let g ∈ Ls(RN ), h ∈ Lr(RN ), there exists a sharp
constant C(s, α, r,N), independent of g, h, such that∫

RN

∫
RN

g(x)h(y)

|x− y|α
dx dy ≤ C(s, α, r,N)‖g‖Lr(RN )‖h‖Ls(RN ).

By the above lemma we have the following result.

Lemma 2.2. If v ∈ Ls(RN ) and s ∈ (1, N
N−α ), then

∫
RN

v(y)
|x−y|α dy ∈ L

Ns
N−Ns+αs (RN ),

and (∫
RN

∣∣∣ ∫
RN

v(y)

|x− y|α
dy
∣∣∣ Ns
N−Ns+αs

dx
)N−Ns+αs

Ns ≤ c(s,N, α)‖v‖Ls(RN ).

We denote

D(f, g) =

∫
RN

∫
RN

f(x)g(y)

|x− y|α
dx dy.

Lemma 2.3 ([22, Theorem 9.8]). Let N ≥ 3, 0 < α < N , and D(f, f), D(g, g) <
∞. Then

|D(f, g)|2 ≤ D(f, f)D(g, g)

with equality for g 6= 0 if and only if f = cg for some constant c.

Lemma 2.4. For x ∈ RN and t ∈ R the following holds;



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 7

(1) 0 ≤ bε(x, t) ≤ mε(x,t)
t ≤ 1;

(2) mε(x, t) = t ,if |t| < ε−1 exp{− dist(εx,M)};
(3) ε−1 exp{− dist(εx,M)} ≤ mε(x, t) ≤ Cε−1 exp{− dist(εx,M)}, if

ε−1 exp{− dist(εx,M)} ≤ t ≤ 2ε−1 exp{− dist(εx,M)};
(4) mε(x, t) = Cε−1 exp{− dist(εx,M)}, if t ≥ 2ε−1 exp{− dist(εx,M)}, where

C =
∫∞

0
ξ(τ) dτ ;

(5) C1(1 + εm−2 exp{(m− 2) dist(εx,M)}|t|m−2)t ≤ kε(x, t)
≤ C2(1 + εm−2 exp{(m− 2) dist(εx,M)}|t|m−2)t;

(6) 1
m tkε(x, t) ≤ Kε(x, t) ≤ 1

2 tkε(x, t);

(7) (kε(x, t1)− kε(x, t2))(t1 − t2) ≥ C(1 + εm−2 exp{(m− 2) dist(εx,M)}|t1 −
t2|m−2)|t1 − t2|2;

(8) |kε(x, t1) − kε(x, t2)| ≤ C(1 + εm−2 exp{(m − 2) dist(εx,M)}(|t1|m−2 +
|t2|m−2))|t1 − t2|;

(9) |Fν(t)| ≤ Cν |t|r;
(10) |fν(t)| ≤ Cν |t|r−1;
(11) 1

2∗α
tfν(t) ≤ Fν(t) ≤ 1

r tfν(t);

Proof. The proof is straightforward. We only prove (6). Let f(x, t) = Kε(x, t) −
1
2 tkε(x, t) and g(x, t) = Kε(x, t)− 1

m tkε(x, t). Since f(x, 0) = 0, we have ∂f(x,t)
∂t ≤ 0,

if t ≥ 0; and ∂f(x,t)
∂t ≥ 0, if t ≤ 0. g(x, 0) = 0; ∂g(x,t)

∂t ≥ 0, if t ≥ 0; ∂g(x,t)
∂t ≤ 0, if

t ≤ 0. So (6) holds. �

Lemma 2.5. For t ∈ R+ it holds

(1) gλ(t) = 1, g′λ(t) = 0 if 0 < t < 1
λ ;

(2) bλ(t)t ≤ gλ(t)t ≤ cλ, where cλ =
∫∞
0
ξ(τ) dτ

λ ;
(3) g′λ(t)t+ gλ(t) = bλ(t).

The proof of the above lemma is obviously, we omit it.

Lemma 2.6. The imbedding Xε = H1(RN ) ∩ Lmε (RN ) ↪→ Lp(RN ) is compact for
1 ≤ p < 2∗.

Proof. Let {un} is bounded in Xε and assume un ⇀ u in Xε and un → u in
Lploc(R

N ), 1 ≤ p < 2∗. We first prove un → u in L1(RN ). For R > 0, we have∫
RN\B(0,R)

|u| dx

≤
(∫

RN\B(0,R)

exp{(m− 2) dist(εx,M)}|u|m dx
)1/m

×
(∫

RN\B(0,R)

exp{−m− 2

m− 1
dist(εx,M)} dx

)m−1
m

≤ ‖u‖Lmε (RN )

(∫
RN\B(0,R)

exp{−m− 2

m− 1
dist(εx,M)} dx

)m−1
m

= oR(1).

Hence
∫
RN |un−u| dx =

∫
B(0,R)

|un−u| dx+
∫
RN\B(0,R)

|un−u| dx = on(1)+oR(1)→
0 as n→∞. For 1 < p < 2∗, we have∫

RN
|un − u|p dx



8 B. ZHANG, W. ZHANG EJDE-2024/19

=

∫
RN
|un − u|pθ+(1−θ)p dx

≤
(∫

RN
|un − u|pθ·

1
pθ dx

)pθ(∫
RN
|un − u|(1−θ)p·

2∗
(1−θ)p dx

) (1−θ)p
2∗

≤ c
(∫

RN
|un − u| dx

)pθ
,

where 0 < θ < 1, 1
p = θ + 1−θ

2∗ , so un → u in Lp(RN )(1 ≤ p < 2∗). �

Lemma 2.7. Let {un} be a (PS) sequence of the functional Γε,ν,λ, then {un} is
bounded in Xε.

Proof. A direct computation shows that

〈DΓε,ν,λ(u), v〉 =

∫
RN
∇u∇v + E(εx)uv dx+ σ

∫
RN

kε(x, u)v dx

+
(∫

RN
χε(x)u2 dx− 1

)β−1

+

∫
RN

χε(x)uv dx

− 2∗α
2
hλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u(x))v(x)

|x− y|α
dx dy

− ϑ
∫
RN
|u|q−2uv dx, for v ∈ Xε.

(2.1)

By Lemma 2.4, we have

Γε,ν,λ(un)− 1

r
〈DΓε,ν,λ(un), un〉

= (
1

2
− 1

r
)

∫
RN

(|∇un|2 + E(εx)u2
n) dx+ σ

∫
RN

Kε(x, un) dx

− σ

r

∫
RN

kε(x, un)un dx+
1

2β

(∫
RN

χε(x)u2
n dx− 1

)β
+

− 1

r

(∫
RN

χε(x)u2
n dx− 1

)β−1

+

∫
RN

χε(x)u2
n dx

+
2∗α
2r
hλ(ϕ1/2(un))

∫
RN

∫
RN

Fν(un(y))fν(un(x))un(x)

|x− y|α
dx dy

− 2∗α
2
gλ(ϕ1/2(un))ϕ(un) + ϑ(

1

r
− 1

q
)

∫
RN
|un|q dx

≥ (
1

2
− 1

r
)

∫
RN

(|∇un|2 + E(εx)u2
n) dx+ (

1

m
− 1

r
)σ

∫
RN

kε(x, un)un dx

+ c
(∫

RN
χε(x)u2

n dx− 1
)β

+
− c

≥ c
(
‖un‖2H1(RN ) + ‖un‖mLmε (RN )

)
+ c
(∫

RN
χε(x)u2

n dx− 1
)β

+
− c.

Combining |Γε,ν,λ(un)| ≤ C and DΓε,ν,λ(un)→ 0, it follows that {un} is bounded
in Xε. �

Lemma 2.8. For every ε, ν, λ > 0, Γε,ν,λ satisfies the (PS) condition.



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 9

Proof. Let {un} be a (PS) sequence of the functional Γε,ν,λ. By Lemma 2.7, {un}
is bounded in Xε. Up to a subsequence, we may assume un ⇀ u in Xε and un → u
in Lr(RN )(2 ≤ r < 2∗). By Lemma 2.1, we have

o(1) = 〈DΓε,ν,λ(un)−DΓε,ν,λ(u), un − u〉

=

∫
RN

(
|∇(un − u)|2 + E(εx)(un − u)2

)
dx

+ σ

∫
RN

(kε(x, un)− kε(x, u))(un − u) dx

+
(∫

RN
χε(x)u2

n dx− 1
)β−1

+

∫
RN

χε(x)un(un − u)dx

−
(∫

RN
χε(x)u2 dx− 1

)β−1

+

∫
RN

χε(x)u(un − u)dx

− 2∗α
2
hλ(ϕ1/2(un))

∫
RN

∫
RN

Fν(un(y))fν(un(x))(un(x)− u(x))

|x− y|α
dx dy

+
2∗α
2
hλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u(x))(un(x)− u(x))

|x− y|α
dx dy

− ϑ
∫
RN

(|un|q−2un − |u|q−2u)(un − u) dx

≥
∫
RN

(
|∇(un − u)|2 + E(εx)(un − u)2

)
dx

+ c(1 + εm−2 exp{(m− 2)dist(εx,M)}|un − u|m−2)|un − u|2

− c
∫
RN

(|un|q−1 + |u|q−1)|un − u| dx+ o(1)

≥ c
(
‖un − u‖2H1(RN ) + ‖un − u‖mLmε (RN )

)
+ o(1), as n→∞,

which implies ‖un − u‖Xε → 0. Therefore, Γε,ν,λ satisfies the (PS) condition. �

3. Existence of solutions for perturbed functions Γε,ν,λ

In this section, we construct a sequence of critical points of the functional Γε,ν,λ
by using the method of invariant sets with respect to a descending flow. Firstly we
define an operator A : X → X. The vector field u − Au will be used as pseudo-
gradient vector field of the functional Γε,ν,λ. To obtain multiple sign-changing
critical points of Γε,,ν,λ, we introduce the abstract critical point theorem [26, The-
orem 2.5], see also [7, Theorem 3.2].

Let X be a Banach space, f be an even C1-functional on X. Let P,Q be two
family of open convex sets of X,Q = −P . We set

W = P ∪Q, Σ = ∂P ∩ ∂Q .
Then we assume that

(A6) f satisfies the (PS) condition.
(A7) c∗ = infx∈Σ f(x) > 0 ,

and that there exists an odd continuous map A : X → X satisfying

(A8) For each c0, b0 > 0, there exists b = b(c0, b0) > 0 such that if ‖Df(x)‖ ≥
b0, |f(x)| ≤ c0, then

〈Df(x), x−Ax〉 ≥ b‖x−Ax‖X > 0 .



10 B. ZHANG, W. ZHANG EJDE-2024/19

(A9) A(∂P ) ⊂ P and A(∂Q) ⊂ Q.

We define

Γj = {E ⊂ X : E is compact, −E = E, γ(E ∩ η−1(Σ)) ≥ j for η ∈ Λ},
Λ = {η ∈ C(X, X) : η is odd, η(P ) ⊂ P, η(Q) ⊂ Q, η(x) = x if f(x) < 0}

where γ is the genus of symmetric sets, defined as

γ(E) = inf
{
n : there exists an odd map η : E → Rn\{0}

}
.

Now we assume

(A10) Γj is nonempty.

We define

cj = inf
E∈Γj

sup
x∈E\W

f(x), j = 1, 2, . . . ,

Kc = {x : Df(x) = 0, f(x) = c}, K∗c = Kc \W .

Theorem 3.1 ([26, Theorem 3.1]). Assume (A6)–(A10) hold. Then

(1) cj ≥ c∗, K∗cj 6= ∅.
(2) cj →∞ as j →∞.
(3) If cj = cj+1 = · · · = cj+k−1 = c, then γ(K∗c ) ≥ k.

We prove the existence of critical points of Γε,,ν,λ by using the method of invari-
ant sets of descending flow. First, we need to define the operator A.

Definition 3.2. Given u ∈ Xε define v = Au by the equation∫
RN

(∇v∇η + E(εx)vη)dx+ σ

∫
RN

kε(x, v)η dx

+
(∫

RN
χε(x)u2 dx− 1

)β−1

+

∫
RN

χε(x)vη dx

=
2∗α
2
hλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u(x))η(x)

|x− y|α
dx dy

+ ϑ

∫
RN
|u|q−2uη dx, for η ∈ Xε .

(3.1)

Lemma 3.3 (Brezis-Libe type lemma [6]). Assume 0 < α < min{4, N − 1} and
that f satisfies

(1) there exists a constant C > 0 such that

|f(t)| ≤ C(|t|
N−α
N + |t|

N+2−α
N−2 ), ∀t ∈ R.

(2) limt→0+
f(t)
t = 0, limt→∞

f(t)

t
N+2−α
N−2

= 1.

Let {un} ⊂ H1(RN ) be such that un ⇀ u in H1(RN ) and un(x) → u(x) a.e.
x ∈ RN , then up to a subsequence if necessary, it holds∫

RN

∫
RN

F (un(x))F (un(y))

|x− y|α
dx dy

=

∫
RN

∫
RN

F (un(x)− u(x))F (un(y)− u(y))

|x− y|α
dx dy

+

∫
RN

∫
RN

F (u(x))F (u(y))

|x− y|α
dx dy + on(1),

(3.2)



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 11∫
RN

∫
RN

F (un(x))f(un(y))v(y)

|x− y|α
dx dy

=

∫
RN

∫
RN

F (un(x)− u(x))f(un(y)− u(y))v(y)

|x− y|α
dx dy

+

∫
RN

∫
RN

F (u(x))f(u(y))v(y)

|x− y|α
dx dy + on(1),

(3.3)

where on(1)→ 0 uniformly as n→∞ for any v ∈ C∞0 (RN ).

Lemma 3.4. Function A is well defined, odd and continuous on Xε.

Proof. For simplicity, we denote

ψε(u) =
(∫

RN
χε(x)u2 dx− 1

)β−1

+
,

and define

B(v) =
1

2

∫
RN

(|∇v|2 + E(εx)v2) dx+ σ

∫
RN

Kε(x, v) dx+
1

2
ψε(u)

∫
RN

χε(x)v2 dx.

Since

〈DB(v1)−DB(v2), v1 − v2〉 ≥ c(‖v1 − v2‖2H1(RN ) + ‖v1 − v2‖mLmε (RN )), (3.4)

for all v1, v2 ∈ Xε, it follows that DB is strongly monotone. Then problem (3.1)
has a unique solution v = Au, which can be obtained by solving the minimization
problem

inf{B(v)− F (v)
∣∣v ∈ Xε},

where

F (v) =
2∗α
2
hλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u(x))v(x)

|x− y|α
dx dy + ϑ

∫
RN
|u|q−2uv dx.

So A is well defined. Moreover, it is easy to check the operator A is odd. Finally,
let un → u in Xε, and denote vn = Aun, v = Au. By choosing η = vn − v in (3.1),
we have∫

RN
(|∇(vn − v)|2 + E(εx)|vn − v|2) dx+ σ

∫
RN

(kε(x, vn)− kε(x, v))(vn − v) dx

+ ψε(u)

∫
RN

χε(x)(vn − v)2 dx

= (ψε(un))− ψε(u))

∫
RN

χε(x)vn(v − vn) dx+
2∗α
2
hλ(ϕ1/2(un))

×
∫
RN

∫
RN

(
Fν(un(y))fν(un(x))− Fν(u(y))fν(u(x))

)
(vn(x)− v(x))

|x− y|α
dx dy

+
2∗α
2

(hλ(ϕ1/2(un))− hλ(ϕ1/2(u)))

×
∫
RN

∫
RN

Fν(u(y))fν(u(x))(vn(x)− v(x))

|x− y|α
dx dy

+ ϑ

∫
RN

(|un|q−2un − |u|q−2u)(vn − v) dx.

(3.5)



12 B. ZHANG, W. ZHANG EJDE-2024/19

By Lemma 2.4, can estimate the two side of (3.5):

LHS ≥ c(‖vn − v‖2H1(RN ) + ‖vn − v‖mLmε (RN )),

RHS = on(1)‖vn − v‖H1(RN ).

So we obtain ‖Aun −Au‖Xε → 0. This means A is continuous. �

Lemma 3.5. Let u ∈ Xε, v = Au. Then it holds:

(1) 〈DΓε,ν,λ(u), u− v〉 ≥ c(‖u− v‖2H1(RN ) + ‖u− v‖mLmε (RN ));

(2) ‖DΓε,ν,λ(u)‖ ≤ c(1 + |Γε,ν,λ(u)|+ ‖u− v‖Xε)γ‖u− v‖Xε(γ > 1).

Proof. (1) By (3.1), for η ∈ Xε, we have

〈DΓε,ν,λ(u), η〉

=

∫
RN

(∇u∇η + E(εx)uη)dx+ σ

∫
RN

kε(x, u)η dx

+
(∫

RN
χε(x)u2 dx− 1

)β−1

+

∫
RN

χε(x)uη dx

− 2∗α
2
hλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u(x))η(x)

|x− y|α
dx dy − ϑ

∫
RN
|u|q−2uη dx

=

∫
RN

(
∇(u− v)∇η + E(εx)(u− v)η

)
dx+ σ

∫
RN

(
kε(x, u)− kε(x, v)

)
η dx

+
(∫

RN
χε(x)u2 dx− 1

)β−1

+

∫
RN

χε(x)(u− v)η dx.

Hence

〈DΓε,ν,λ(u), u− v〉

=

∫
RN

(|∇(u− v)|2 + E(εx)(u− v)2) dx+ σ

∫
RN

(
kε(x, u)− kε(x, v)

)
(u− v) dx

+
(∫

RN
χε(x)u2 dx− 1

)β−1

+

∫
RN

χε(x)(u− v)2 dx

≥ c
(
‖u− v‖2H1(RN ) + ‖u− v‖mLmε (RN )

)
.

(2) We define

Jε(u) =
1

2

∫
RN

(|∇u|2 + E(εx)u2) dx+ σ

∫
RN

Kε(x, u) dx.

On the one hand,

Γε,ν,λ(u)− 1

r
〈DJε(u)−DJε(v), u〉

= (
1

2
− 1

r
)

∫
RN

(|∇u|2 + E(εx)u2) dx+ σ

∫
RN

Kε(x, u) dx− σ

r

∫
RN

kε(x, u)u dx

+
1

2β

(∫
RN

χε(x)u2 dx− 1
)β

+
− 1

r

(∫
RN

χε(x)u2 dx− 1
)β−1

+

∫
RN

χε(x)uv dx

+
2∗α
2r
hλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u(x))u(x)

|x− y|α
dx dy − 2∗α

2
gλ(ϕ1/2(u))ϕ(u)

+ ϑ(
1

r
− 1

q
)

∫
RN
|u|q dx.

(3.6)



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 13

By Hölder’s inequality and the Young’s inequality, we have

1

2β

(∫
RN

χε(x)u2 dx− 1
)β

+
− 1

r

(∫
RN

χε(x)u2 dx− 1
)β−1

+

∫
RN

χε(x)uv dx

=
1

2β

(∫
RN

χε(x)u2 dx− 1
)β

+
− 1

r

(∫
RN

χε(x)u2 dx− 1
)β−1

+

∫
RN

χε(x)u2 dx

+
1

r

(∫
RN

χε(x)u2 dx− 1
)β−1

+

∫
RN

χε(x)u(u− v) dx

≥ c
(∫

RN
χε(x)u2 dx− 1

)β
+
− c
(∫

RN
χε(x)|u− v||u| dx

)β
− c

≥ c
(∫

RN
χε(x)u2 dx− 1

)β
+
− c
(∫

RN
χε(x)|u− v|2 dx

)β
− c.

Consequently,

Γε,ν,λ(u)− 1

r
〈DJε(u)−DJε(v), u〉

≥ C
(
‖u‖2H1(RN ) + ‖u‖mLmε (RN )

)
+ C

( ∫
RN

χε(x)u2dx− 1
)β

+

− c
( ∫

RN
χε(x)(u− v)2dx

)β − c.
On the other hand,

Γε,ν,λ(u)− 1

r
〈DJε(u)−DJε(v), u〉

≤ |Γε,ν,λ(u)|+ c‖u‖H1(RN )‖u− v‖H1(RN )

+ c(‖u‖m−2
Lmε (RN )

+ ‖v‖m−2
Lmε (RN )

)‖u− v‖Lmε (RN )‖u‖Lmε (RN ).

(3.7)

By (3.6), (3.7) and Young’s inequality, we have

‖u‖2H1(RN ) + ‖u‖mLmε (RN ) +
(∫

RN
χε(x)u2 dx− 1

)β
+

≤ c(1 + |Γε,ν,λ(u)|+ ‖u− v‖2H1(RN ) + ‖u− v‖mLmε (RN ) + ‖u− v‖2β
H1(RN )

).

By (3.1), we have

|〈DΓε,ν,λ(u), η〉|

≤ c‖u− v‖H1(RN )‖η‖H1(RN ) +
(∫

RN
χε(x)u2 dx− 1

)β−1

+
‖u− v‖H1(RN )‖η‖H1(RN )

+ c

∫
RN

εm−2 exp{(m− 2) dist(εx,M)}(|u|m−2 + |u|m−2)|u− v||η| dx

≤ c(‖u‖m−2
Lmε (RN )

+ ‖v‖m−2
Lmε (RN )

)‖u− v‖Lmε (RN )‖η‖Lmε (RN )

+ c
(

1 +
( ∫

RN
χε(x)u2 dx− 1

)β−1

+

)
‖u− v‖H1(RN )‖η‖H1(RN ).

This implies

‖DΓε,ν,λ(u)‖
≤ c (‖u‖m−2

Lmε (RN )
+ ‖v‖m−2

Lmε (RN )
)‖u− v‖Lmε (RN )

+ c
(

1 +
( ∫

RN
χε(x)u2 dx− 1

)β−1

+

)
‖u− v‖H1(RN )



14 B. ZHANG, W. ZHANG EJDE-2024/19

≤ c
(

1 + ‖u‖mLmε (RN ) + ‖u− v‖mLmε (RN ) +
( ∫

RN
χε(x)u2 dx− 1

)β
+

)
‖u− v‖Xε

≤ c(1 + |Γε,ν,λ(u)|+ ‖u− v‖2H1(RN ) + ‖u− v‖mLmε (RN ) + ‖u− v‖2β
H1(RN )

)‖u− v‖Xε
≤ c(1 + |Γε,ν,λ(u)|+ ‖u− v‖Xε)γ‖u− v‖Xε ,

where γ > 1. �

Corollary 3.6. Given b0, c0, there exist b = b(b0, c0) such that if |Γε,ν,λ(u)| ≤ c0
and ‖DΓε,ν,λ(u)‖ ≥ b0, then u−Au 6= 0 and

〈DΓε,ν,λ(u), u−Au〉 ≥ b‖u−Au‖ > 0.

For δ > 0, we define the convex open sets

P = {u
∣∣u ∈ Xε(RN ), ‖u−‖H1(RN ) < δ},

Q = {u
∣∣u ∈ Xε(RN ), ‖u+‖H1(RN ) < δ}.

Lemma 3.7. There exists δλ > 0 such that for 0 < δ < δλ,

A(∂P ) ⊂ P, A(∂Q) ⊂ Q.
Proof. We only prove A(∂Q) ⊂ Q. Similarly, A(∂P ) ⊂ P . For u ∈ ∂Q, let v = Au.
By Lemma 2.3 and Lemma 2.5, we have

‖v+‖2H1(RN )

≤ c
∫
RN

(∇v∇v+ + E(εx)vv+) dx+ c

∫
RN

kε(x, v)v+ dx

≤ chλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u(x))v+(x)

|x− y|α
dx dy

+ c

∫
RN
|u|q−2uv+ dx

≤ chλ(ϕ1/2(u))

∫
RN

∫
RN

Fν(u(y))fν(u+(x))v+(x)

|x− y|α
dx dy + c

∫
RN
|u+|q−1v+ dx

≤ chλ(ϕ1/2(u))ϕ1/2(u)
(∫

RN

∫
RN

fν(u+(x))v+(x)fν(u+(y))v+(y)

|x− y|α
dx dy

)1/2

+ c‖u+‖q−1
H1(RN )

‖v+‖H1(RN )

≤ cλ
(∫

RN

∫
RN

|u+(x)|2∗α−1v+(x)|u+(y)|2∗α−1v+(y)

|x− y|α
dx dy

)1/2

+ c‖u+‖q−1
H1(RN )

‖v+‖H1(RN )

≤ cλ(‖u+‖
2∗α−1

H1(RN )
+ ‖u+‖q−1

H1(RN )
)‖v+‖H1(RN ).

Taking δλ = min{ 1
2c
− 1

2∗α−2

λ , 1
2c
− 1
q−2

λ }, it is easy to get that ‖v+‖H1(RN ) < δ for
0 < δ < δλ. Consequently, the conclusion follows. �

Lemma 3.8. There exist δ0 > 0 and 0 < c∗ = c∗(δ), such that for 0 < δ < δ0 and
u ∈ ∂P ∩ ∂Q, we have Γε,ν,λ(u) ≥ c∗.
Proof. For u ∈ ∂P∩∂Q, we have ‖u‖H1(RN ) ≥ δ, ‖u‖L2∗ (RN ) ≤ cδ, and ‖u‖Lq(RN ) ≤
cδ. Hence

Γε,ν,λ(u) ≥ 1

2

∫
RN

(|∇u|2 + E(εx)u2) dx− 2∗α
2
gλ(ϕ1/2(u))ϕ(u)− ϑ

q

∫
RN
|u|q dx



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 15

≥ c1‖u‖2H1(RN ) − c2
(
‖u‖2·2

∗
α

L2∗ (RN )
+ ‖u‖q

Lq(RN )

)
≥ c1‖u‖2H1(RN ) − c2(δ2·2∗α−2 + δq−2)‖u‖2H1(RN ).

Taking δ0 = min{( c14c2
)

1
2·2∗α−2 , ( c14c2

)
1
q−2 }, then for 0 < δ < δ0, we have

Γε,ν,λ(u) ≥ c1
2
‖u‖2H1(RN ) ≥

c1
2
δ2 := c∗. �

Assume B(0, R) ⊂M. Let {en}∞n=1 be a family of linearly independent functions
in C∞0 (B(0, R)). There exists an increasing sequence Rn such that

J0(u) < 0, ∀u ∈ Hn, ‖u‖ ≥ Rn.

where Hn := span{e1, . . . , en} and

J0(u) =
1

2

∫
RN

(|∇u|2 + bu2) dx+ σ

∫
RN

e(m−2)|x||u|m dx− ϑ

q

∫
RN
|u|q dx.

We define ϕn ∈ C(Bn, C
∞
0 (B(0, R))),

ϕn(t) = Rn

n∑
i=1

tiei, t = (t1, . . . , tn) ∈ Bn = {t ∈ Rn : |t| ≤ 1} .

Let

Γj = {E ⊂ Xε : E is compact, E = E, γ(E ∩ η−1(Σ)) ≥ j for η ∈ Λ},
Λ = {η ∈ C(Xε, Xε) : ηis odd, η(P ) ⊂ P, η(Q) ⊂ Q, η(u) = u if Γε(u) ≤ 0}.

Similarly from [25, Lemma 5.6], we obtain the following Lemma.

Lemma 3.9. Γj is nonempty, for j = 1, 2, . . . .

Theorem 3.10. Assume that conditions (A1) and (A2) hold, then there exist 0 <

ε̃ < 1, 0 < ν̃ < 1, and 0 < λ̃ < 1, such that if 0 < ε < ε̃, 0 < ν < ν̃, and
0 < λ < λ̃, then the functional Γε,ν,λ has infinitely many sign-changing critical
points; the corresponding critical values are

cj(ε, ν, λ) = inf
E∈Γj

sup
u∈E\W

Γε,ν,λ(u), j = 1, 2, . . . . (3.8)

Moreover

(1) there exist mj, j = 1, . . . , independent of ε, ν, λ such that

cj(ε, ν, λ) ≤ mj , j = 1, 2, . . . . (3.9)

(2) If cj(ε, ν, λ) = · · · = cj+k(ε, ν, λ) = c, then γ(K∗c ) ≥ k + 1, where

K∗c = Kc \W, Kc = {x : DΓε,ν,λ(u) = 0, Γε,ν,λ(u) = c}.

Proof. For the functional Γε,µ,λ, it is easy to check that Γε,ν,λ satisfies the as-
sumptions of Theorem 3.1. Therefore, we only need to prove (3.9). Note that
Ej = ϕj+1(Bj+1) ∈ Γj . It is easy to know that there exist 0 < ε̃ < 1, 0 < ν̃ < 1,

and 0 < λ̃ < 1, such that if 0 < ε < ε̃, 0 < ν < ν̃, and 0 < λ < λ̃, then

Γε,ν,λ(u) ≤ J0(u) for u ∈ ϕj+1(Bj+1) and
(∫

RN χε(x)|u|pdx− 1
)β

+
= 0. Hence

cj(ε, ν, λ) ≤ mj := sup
u∈Ej

J0(u) . �



16 B. ZHANG, W. ZHANG EJDE-2024/19

4. Uniform bounds

In this section, we the following theorem that gives uniform bounds needed for
proving Theorem 1.1.

Theorem 4.1. (1) Assume Γε,ν,λ(u) ≤ L and DΓε,ν,λ(u) = 0. Then there exists a
constant H = H(L) such that

‖u‖H1(RN ) ≤ H.

(2) Assume Γε,ν(u) ≤ L and DΓε,ν(u) = 0. Then there exist constants µ > 0,
C = C(L) such that, for any δ > 0, there exists ε = ε(δ) > 0, for 0 < ε < ε(δ),

|u(x)| ≤ C exp{−µdist(x, (Aδ)ε)} for x ∈ RN .

(3) Assume Iε,ν(u) ≤ L and DIε,ν(u) = 0. Then there exists a positive constant
M = M(L) such that |u(x)| ≤M for x ∈ RN .

Similar to Lemma 2.7, it is easy to obtain Theorem 4.1(1). Before proving parts
(2) and (3), we need establish some preliminary lemmas.

Lemma 4.2. Assume DΓε,ν(u) = 0 and Γε,ν(u) ≤ L. Then
(1) there exist cν,L, such that |u(x)| ≤ cν,L for x ∈ RN ;

(2) there exist bν,L, such that
∫
RN

Fν(u(y))
|x−y|α dy ≤ bν,L for x ∈ RN ;

(3) for any δ > 0 there exist c = c(δ, ν, L) such that |u(x)| ≤ cε3 for x ∈ RN\(Mε)
δ.

Proof. (1) Assume DΓε,ν(u) = 0 and Γε,ν(u) ≤ L, it is easy to show that u is

bounded in H1(RN ) and (
∫
RN χε(x)u2 dx− 1)β+ is bounded. Choose φ = |uT |2k−2u

as test function in 〈DΓε,ν(u), φ〉 = 0, where k ≥ 1, T > 0 and uT (x) = ±T if
±u(x) ≥ T , uT (x) = u(x) if |u(x)| ≤ T . By 〈DΓε,ν(u), φ〉 = 0, it is easy to obtain
the inequality∫

RN
(∇u∇φ+ E(εx)uφ) dx ≤ cν

∫
RN

∫
RN

|u(y)|r|u(x)|r−1φ(x)

|x− y|α
dx dy

+ τ

∫
RN

uφ dx+ cτ

∫
RN
|u|

2Nr
2N−α−2uφ dx,

(4.1)

where τ ≤ infx∈RN E(x). Hence∫
RN
∇u∇φdx

≤ cν
∫
RN

(

∫
RN

|u(y)|r

|x− y|α
dy + |u|

αr
2N−α )|u|r−2uφ dx

≤ cν
(∫

RN
(

∫
RN

|u(y)|r

|x− y|α
dy + |u|

αr
2N−α )2N/α dx

) α
2N

×
(∫

RN
(|u|r|uT |2k−2)

2N
2N−α dx

) 2N−α
2N

≤ cν
(∫

RN
(|u||uT |k−1)

2Nr
2N−α dx

) 2N−α
Nr

(∫
RN
|u|

2Nr
2N−α dx

) (r−2)(2N−α)
2Nr

≤ cν
(∫

RN
(|u||uT |k−1)

2Nr
2N−α dx

) 2N−α
Nr

.

(4.2)



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 17

The left-hand side of (4.1) satisfies

LHS ≥
∫
RN
|∇u|2|uT |2k−2 dx

≥ c

k2

∫
RN
|∇(|u||uT |k−1)|2 dx

≥ c

k2

(∫
RN

(|u||uT |k−1)2∗ dx
)2/2∗

.

(4.3)

Combining (4.2) and (4.3), we have

(∫
RN

(|u||uT |k−1)2∗ dx
)2/2∗

≤ cνk2
(∫

RN
(|u||uT |k−1)

2Nr
2N−α dxBig)

2N−α
Nr . (4.4)

Letting T →∞ in (4.4) we obtain

(∫
RN
|u|2

∗k dx
)2/2∗

≤ cνk2
(∫

RN
|u|

2Nrk
2N−α dx

) 2N−α
Nr

. (4.5)

We denote χ = 2N−α
r(N−2) > 1, k1 = χ, by iterations, we obtain

(∫
RN
|u|2

∗χn dx
) 1

2∗χn ≤
(
Cνχ

2n
) 1

2χn
(∫

RN
|u|2

∗χn−1

dx
) 1

2∗χn−1

n = 1, 2, . . . .

(4.6)
Hence

‖u‖L∞(RN ) ≤ cν‖u‖L2∗ (RN ) ≤ cν,L. (4.7)

(2) In view of 1 < α < N − 1, for x ∈ RN we have∫
RN

Fν(u(y))

|x− y|α
dy ≤ bν

(∫
|x−y|≥1

|u(y)|r

|x− y|α
dx+

∫
|x−y|<1

|u(y)|r

|x− y|α
dx
)

≤ bν
(
‖u‖rLr(RN ) +

∫
|x−y|<1

1

|x− y|α
dx‖u‖rL∞(RN )

)
≤ bν(‖u‖rLr(RN ) + ‖u‖rL∞(RN ))

≤ bν,L.

(4.8)

(3) For y ∈ RN , 0 < ρ < R ≤ 1. We choose η ∈ C∞0 (RN , [0, 1]) such that
η(x) = 0 for x /∈ B(y,R); η(x) = 1 for x ∈ B(y, ρ) and |∇η| ≤ c

R−ρ . Setting

ϕ = u|u|2k−2ηm, k ≥ 1 as test function in 〈DΓε,ν(u), ϕ〉 = 0, we have∫
RN
∇u∇ϕdx

≤ c
∫
RN

∫
RN

Fν(u(y))|u(x)|2∗α−1|ϕ(x)|
|x− y|α

dx dy + c

∫
RN
|u|q−2uϕdx.

(4.9)



18 B. ZHANG, W. ZHANG EJDE-2024/19

The left-hand side of (4.9) satisfies

LHS ≥
∫
RN
|∇u|2|u|2k−2η2 dx− c

∫
RN
|∇u||∇η||u|2k−1η dx

≥ 1

2

∫
RN
|∇u|2|u|2k−2η2 dx− c

∫
RN
|u|2k|∇η|2η2 dx

≥ c

k2

∫
RN
|∇(|u|kη)|2 dx− c

∫
RN
|u|2k|∇η|2 dx

≥ c

k2
(

∫
B(y,ρ)

|u|2
∗k dx)2/2∗ − c

(R− ρ)2

∫
B(y,R)

|u|2k dx,

(4.10)

By (4.7) and (4.8) we obtain

RHS ≤ cν,L
∫
RN
|u|2

∗
α−2|u|2kη2 dx+ c

∫
RN
|u|q−2|u|2kη2 dx

≤ cν,L
∫
B(y,R)

|u|2k dx.
(4.11)

With the above estimations, we have(∫
B(y,ρ)

|u|2
∗k dx

)2/2∗

≤ cν,Lk
2

(R− ρ)2

∫
B(y,R)

|u|2k dx, for k ≥ 1.

By iteration again, we obtain

‖u‖L∞(B(y,R2 )) ≤ cν,L‖u‖L2(B(y,R)).

Since ∫
RN\(Mε)δ

u2 dx ≤ cδε6,

it follows that |u(x)| ≤ cδ,ν,Lε3 for all x ∈ RN\(Mε)
δ. �

For ν fixed, let εn → 0, and assume un ∈ H1(RN ) is such that DΓεn,ν(un) = 0
and Γεn,ν(un) ≤ L. It is easy to show that {un} is bounded in H1(RN ), hence we
have the following profile decomposition [38],

un =
∑
k∈Λ

Uk(· − yn,k) + rn, (4.12)

where Λ is an index set, yn,k ∈ RN . Moreover,

(1) un(·+ yn,k) ⇀ Uk in H1(RN ) as n→∞.
(2) |yn,k − yn,l| → ∞ as n→∞ for k 6= l.
(3) ‖un‖2H1(RN ) =

∑
k∈Λ ‖Uk‖2H1(RN ) + ‖rn‖2H1(RN ) + o(1) as n→∞.

(4) ‖rn‖Ls(RN ) → 0 as n→∞, 2 < s < 2∗,

‖un‖sLs(RN ) =
∑
k∈Λ

‖Uk‖sLs(RN ) + o(1)

as n→∞.

By Lemma 4.2 (3) we have

lim
n→∞

dist(yn,k,Mεn) < +∞.

We denote y∗k = limn→∞ εnyn,k. Since dist(yn,k,Mεn) = ε−1
n dist(εnyn,k,M), we

have
dist(y∗k,M) = 0, i.e. y∗k ∈M. (4.13)



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 19

Moreover, we obtain the following properties of {un}.

Lemma 4.3. If ũn = un(·+yn) ⇀ U in H1(RN ) for yn ∈ RN , and limn→∞ εnyn =
y∗, then Z = |U | satisfies∫

RN
∇Z∇ϕdx+

∫
RN

Zϕdx

≤ Cν
∫
RN

∫
RN

|Z(y)|r|Z(x)|r−1ϕ(x)

|x− y|α
dx dy + c

∫
RN

Zq−1ϕdx,

(4.14)

for ϕ ∈ H1(RN ) and ϕ ≥ 0.

Proof. Let ϕ ∈ C∞0 (RN ), R > 0, such that ϕ(x) = 1 for |x| ≤ R; ϕ(x) = 0
for |x| ≥ 2R and |∇ϕ| ≤ c

R . Choosing ϕn = ϕ(· − yn) as the test function in
〈DΓεn,ν(un), ϕn〉 = 0, we deduce∫

RN
(∇ũn∇ϕdx+ E(εn(x+ yn))ũnϕ) dx+ σ

∫
RN

kεn(x+ yn, ũn)ϕdx

+
(∫

RN
χεn(x)u2

n dx− 1
)β−1

+

∫
RN

χεn(x+ yn)ũnϕdx

= 2∗α

∫
RN

∫
RN

Fν(ũn(y))fν(ũn(x))ϕ(x)

|x− y|α
dx dy + ϑ

∫
RN
|ũn|q−2ũnϕdx.

(4.15)

By Rellich’s imbedding theorem, we have ũn → U in Lsloc(RN )(1 ≤ s < 2∗). By
Lemma 2.1, we have∫

RN
(|∇(ũk − ũl)|2ϕdx

= −
∫
RN

(∇(ũk − ũl),∇ϕ)(ũk − ũl) dx

−
∫
RN

(E(εk(x+ yk))ũk − E(εl(x+ yl))ũl)(ũk − ũl)ϕdx

− σ
∫
RN

(kεk(x+ yk, ũk)− kεl(x+ yl, ũl))(ũk − ũl)ϕdx

−
∫
RN

(
ψεk(uk)χεk(x+ yk)ũk − ψεl(ul)χεl(x+ yl)ũl

)
(ũk − ũl)ϕdx

+ 2∗α

∫
RN

∫
RN

Fν(ũk(y))fν(ũk(x))(ũk(x)− ũl(x))ϕ(x)

|x− y|α
dx dy

− 2∗α

∫
RN

∫
RN

Fν(ũk(y))fν(ũk(x))(ũk(x)− ũl(x))ϕ(x)

|x− y|α
dx dy

+ ϑ

∫
RN

(|ũk|q−2ũk − |ũl|q−2ũl)(ũk − ũk)ϕdx

≤ c‖ũk − ũl‖L2(B(0,2R)) + c‖ũk − ũl‖Lm(B(0,2R)) + c‖ũk − ũl‖Lq(B(0,2R))

+ c‖ũk − ũl‖
L

2Nr
2N−α (B(0,2R))

→ 0, as k, l→∞.

Since ϕ = 1 in B(0, R) and ϕ ≥ 0, ũn → U inH1
loc(RN ). Let zn = |ũn|, wn,δ = (ũ2

n+

δ2)1/2 − δ. Then from Lebesgue’s controlled convergence theorem it follows that
wn,δ ∈ H1(RN ), and wn,δ → zn in H1(RN ) as δ → 0. Now for any ϕ ∈ C∞0 (RN ),



20 B. ZHANG, W. ZHANG EJDE-2024/19

ϕ ≥ 0, we have ϕδ = ϕũn(ũ2
n + δ2)−1/2 ∈ H1

loc(RN ), and∫
RN

(∇wn,δ∇ϕ+ E(εn(x+ yn))wn,δϕ) dx

=

∫
RN

(ũn∇ũn∇ϕ(ũ2
n + δ2)−1/2 + E(εn(x+ yn))((ũ2

n + δ2)1/2 − δ)ϕ) dx

=

∫
RN

(∇ũn∇ϕδ − |∇ũn|pϕ(ũ2
n + δ2)−

3
2 δ2 + E(εn(x+ yn))((ũ2

n + δ2)1/2 − δ)ϕ) dx

≤
∫
RN

(∇ũn∇ϕδ + E(εn(x+ yn))ũnϕδ) dx

≤ c
∫
RN

∫
RN

|zn(y)|r|zn(x)|r−1|ϕδ(x)|
|x− y|α

dx dy + c

∫
RN
|zn|q−1|ϕδ| dx.

Letting δ → 0 in the above inequality we obtain∫
RN

(∇zn∇ϕ+ znϕ) dx

≤ Cν
∫
RN

∫
RN

|zn(y)|q|zn(x)|q−1ϕ(x)

|x− y|α
dx dy + c

∫
RN
|zn|q−1ϕdx.

(4.16)

for ϕ ∈ C∞0 (RN ), ϕ ≥ 0. By ũn → U in H1
loc(RN ) as n → ∞, we have zn → Z in

W 1,2
loc (RN ) as n→∞. Finally, by a density argument we complete the proof. �

Corollary 4.4. Λ is a finite set.

Proof. Zk = |Uk| satisfies (4.14) and taking ϕ = Zk in (4.14), we have

‖Zk‖2H1(RN ) ≤ Cν
∫
RN

∫
RN

|Zk(y)|r|Zk(x)|r

|x− y|α
dx dy + c

∫
RN
|Zk|q dx

≤ Cν‖Zk‖2rH1(RN ) + c‖Zk‖qLq(RN )
.

(4.17)

So there exists m > 0 such that ‖Uk‖W 1,p(RN ) ≥ m. By property (3) of the profile
decomposition (4.12), we know that Λ is a finite set. �

Assume that the sequence {un} has the profile decomposition (4.12). By Corol-
lary 4.4, we can assume that Λ = {1, . . . , k}. Meanwhile, we denote

Ω
(n)
R = RN\{∪k∈ΛB(yn,k, R) ∪B(0, R)},

for the above {un}, we have the following statements.

Lemma 4.5. Assume Γεn,ν(un) ≤ L, DΓεn,ν(un) = 0, then there exist c, µ, inde-
pendent of n, such that∫

Ω
(n)
R

Gεn(x, un,∇un) dx ≤ c exp{−µR},

where

Gεn(x, un,∇un)

= |∇un|2 + E(εnx)u2
n + σkεn(x, un)un +

(∫
RN

χε(x)u2
n dx− 1

)β−1

+
χε(x)u2

n.

Moreover,

|un(x)| ≤ c exp{−µR}, x ∈ Ω
(n)
R .



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 21

Proof. By the decomposition (4.12) we have

‖un‖Ls(Ω(n)
R )

= oR(1), 2 < s < 2∗

where oR(1)→ 0 as R→ +∞. By Moser’s iteration we have

‖un‖L∞(Ω
(n)
R )

= oR(1). (4.18)

Let η ∈ C∞(RN ) such that η(x) = 0 for x /∈ Ω
(n)
R , and η(x) = 1 for x ∈ Ω

(n)
R+1,

|∇η| ≤ 2. Take ϕn = unη
2 as a test function in 〈DΓεn(un), ϕ〉 = 0, we have∫

Ω
(n)
R

(|∇un|2 + E(εnx)|un|2)η2 dx+ σ

∫
Ω

(n)
R

kεn(x, un)unη
2 dx

+
(∫

RN
χε(x)u2

n dx− 1
)β−1

+

∫
Ω

(n)
R

χε(x)u2
nη

2 dx

=
2∗α
2

∫
Ω

(n)
R

∫
RN

F (un(y))f(un(x))un(x)η2(x)

|x− y|α
dx dy + ϑ

∫
Ω

(n)
R

|un|qη2 dx

− 2

∫
Ω

(n)
R \Ω

(n)
R+1

∇un∇ηunη dx.

(4.19)

By Lemma 4.2(2) and (4.18), for n large enough, we have

2∗α
2

∫
Ω

(n)
R

∫
RN

F (un(y))f(un(x))un(x)η2(x)

|x− y|α
dx dy + ϑ

∫
Ω

(n)
R

|un|qη2 dx

≤ 1

2

∫
Ω

(n)
R

E(εnx)|un|pηp dx.
(4.20)

Also

2
∣∣ ∫

Ω
(n)
R \Ω

(n)
R+1

∇un∇ηunη dx
∣∣ ≤ c∫

Ω
(n)
R \Ω

(n)
R+1

(|∇un|2 + |un|2) dx. (4.21)

By (4.19)-(4.21), we have∫
Ω

(n)
R+1

Gεn(x, un,∇un) dx ≤ c
∫

Ω
(n)
R \Ω

(n)
R+1

Gεn(x, un,∇un) dx.

Consequently,∫
Ω

(n)
R+1

Gεn(x, un,∇un) dx ≤ θ
∫

Ω
(n)
R

Gεn(x, un,∇un) dx,

where θ = c
c+1 < 1. Finally∫

Ω
(n)
R

Gεn(x, un,∇un) dx ≤ c exp{−µR},

where µ = − ln θ > 0 . And by Moser’s iteration, we have

|un(x)| ≤ c exp{−µR}, x ∈ Ω
(n)
R . �

Lemma 4.6. For every k ∈ Λ it holds k∗ = limn→∞ εny
k
n ∈ Ā.



22 B. ZHANG, W. ZHANG EJDE-2024/19

Proof. If the lemma does not hold, we assume that there exist k ∈ Λ and εn > 0
such that limn→∞ εn = 0 and dist(y∗k,A) > 0. Let tk = ∇V (y∗k) 6= 0, by (A2) we
deduce that there exists δ1 > 0 such that

(tk,∇V (x)) ≥ 1

2
|tk|2 > 0, (tk,∇ dist(x,M)) ≥ 0 for x ∈ Bδ1(y∗k). (4.22)

Let

δ2 = min{|y∗k − y∗l |
∣∣∣y∗k 6= y∗l , k, l = 1, 2, . . . , k}, 0 < δ < min{1

2
δ1,

1

100
δ2}.

From 〈DΓεn,ν(un), ϕ〉 = 0, we have∫
RN

(∇un∇ϕ+ E(εnx)unϕ) dx+ σ

∫
RN

kεn(x, un)ϕdx

+
(∫

RN
χεn(x)u2

n dx− 1
)β−1

+

∫
RN

χεn(x)unϕdx

= 2∗α

∫
RN

∫
RN

Fν(un(y))fν(un(x))ϕ(x)

|x− y|α
dx dy + ϑ

∫
RN
|un|q−2unϕdx,

(4.23)

for all ϕ ∈ H1(RN ). Let η ∈ C∞0 (RN ) be such that η(x) = 0 if |x− yn,k| ≥ 2δε−1
n ;

η(x) = 1 if |x − yn,k| ≤ δε−1
n and |∇η| ≤ 2

δ εn(≤ 1). Choosing ϕ = (tk,∇un)η as
test function in (4.23), we obtain the local Pohoẑaev identity

εn
2

∫
RN

(tk,∇E(εnx))u2
nη dx+ σ

∫
RN

(tk,∇xKεn(x, un))η dx

+
1

2

(∫
RN

χεn(x)u2
n dx− 1

)β−1

+

∫
RN

(∇χεn(x), tk)u2
nη dx

=

∫
RN

(∇un,∇η)(tk,∇η) dx− 1

2

∫
RN

(|∇un|2 + E(εnx))u2
n)(tk,∇η) dx

− 1

2

(∫
RN

χεn(x)u2
n dx− 1

)β−1

+

∫
RN

χεn(x)u2
n(tk,∇η) dx

− α2∗α

∫
RN

∫
RN

(tk, x− y)
Fν(un(y))Fν(un(x))η(x)

|x− y|α+2
dx dy

+ 2∗α

∫
RN

∫
RN

(tk,∇η)
Fν(un(y))Fν(un(x))

|x− y|α
dx dy

+ σ

∫
RN

Kεn(x, un)(tk,∇η) dx+
ϑ

q

∫
RN
|un|q(tk,∇η) dx.

(4.24)

We denote

Bn = B(yn,k, 2δε
−1
n ), Tn = B(yn,k, 2δε

−1
n )\B(yn,k, δε

−1
n ),

T̃n = B(yn,k, 3δε
−1
n )\B(yn,k, δε

−1
n ).

Next, we estimate equation (4.24). By (4.22), we have

εn

∫
Bn

(tk,∇E(εnx))u2
nη dx ≥ cεn,

(tk,∇xKεn(x, un)) = c(tk,∇ dist(εnx,M)) ≥ 0, ∀x ∈ Bn,(∫
RN

χεn(x)u2
n dx− 1

)β−1

+

∫
Bn

(∇χεn(x), tk)u2
nη dx ≥ 0.



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 23

Hence the left-hand side of (4.24) satisfies

LHS ≥ cεn. (4.25)

We estimate the right-hand side of (4.24). Since∫
RN

∫
RN

(tk, x− y)
Fν(un(y))Fν(un(x))η(x)η(y)

|x− y|α+2
dx dy = 0,

and∫
RN

∫
RN

(tk, x− y)
Fν(un(y))Fν(un(x))η(x)

|x− y|α+2
dx dy

=

∫
RN

∫
RN

(tk, x− y)
Fν(un(y))Fν(un(x))(1− η(y))

|x− y|α+2
dx dy

≤ c
∫
Bn

∫
(RN\Bn)∪Tn

|un(y)|r|un(x)|r

|x− y|α+1
dx dy

≤ c
∫
Bn

∫
T̃n

|un(y)|r|un(x)|r

|x− y|α+1
dx dy + c

∫
Bn

∫
RN\(T̃n∪Bn)

|un(y)|r|un(x)|r

|x− y|α+1
dx dy

=: I + II.

Then

II ≤ c
∫
Bn

∫
RN\(T̃n∪Bn)

|un(y)|r|un(x)|r 1

δα+1
εα+1
n dx dy ≤ cεα+1

n .

By Lemma 4.5, we have

|un(y)| ≤ c exp{−µδε−1
n }, ∀y ∈ T̃n.

Consequently, for n large enough, we have

I ≤ c exp{−rµδε−1
n }

∫
Bn

∫
T̃n

|un(x)|r

|x− y|α+1
dx dy

≤ c exp{−rµδε−1
n }

∫
Bn

|un(x)|r dx
∫
|x−y|≤5δε−1

n

1

|x− y|α+1
dy

≤ c exp{−rµδε−1
n }ε−N+α+1

n ≤ cεα+1
n .

By Lemma 4.5, for n large enough, the right hand side of (4.24) satisfies

RHS ≤ c
∫
Tn

Gεn(x, un,∇un) dx+ c exp{−µδ
2
ε−1
n }+ c exp{−qµδε−1

n }ε−Nn + cεα+1
n

≤ c exp{−µδε−1
n }+ c exp{−µδ

2
ε−1
n }+ c exp{−qµδε−1

n }ε−Nn + cεα+1
n ≤ cεα+1

n .

Hence
εn ≤ cεα+1

n .

Since 0 < α < min{N − 1, 4}, we arrive at a contradiction as n→∞ and complete
the proof. �

Proof of Theorem 4.1 part 2. By Lemma 4.5,

|un(x)| ≤ c exp{−µR} for x ∈ Ω
(n)
R .

Let Rn(x) = min{|x− yn,k|
∣∣∣k ∈ Λ}, then

|un(x)| ≤ c exp{−µRn(x)} for x ∈ Ω
(n)
Rn
.



24 B. ZHANG, W. ZHANG EJDE-2024/19

Since εnyn,k → y∗k ∈ A, for any δ > 0, there exists ε(δ) such that for εn ≤ ε(δ),
εnyn,k ∈ Aδ, hence

|un(x)| ≤ c exp{−µRn} ≤ c exp{−µdist(x, (Aδ)εn)}, x ∈ RN . �

In the following , we assume un ∈ H1(RN ), L > 0, Iεn,νn(un) ≤ L, DIεn,νn(un) =
0, νn → 0,and εn → ε∗ ∈ (0, 1). The case νn → ν∗ ∈ (0, 1) is easier, since in that
case we need only to deal with subcritical problems. It is easy to show that {un}
is bounded in H1(RN ), we have the following profile decomposition [38].

un =
∑
k∈Λ1

Uk(· − yn,k) +
∑
k∈Λ∞

σ
N−2

2

n,k Uk(σn,k(· − yn,k)) + rn, (4.26)

where yn,k ∈ RN , σn,k ∈ R+, Λ is an index set, Uk ∈ H1(RN ) for k ∈ Λ1, Uk ∈
D1,2(RN ) for k ∈ Λ∞, rn ∈ D1,2(RN ) satisfying

(1) For k ∈ Λ1, un(· + yn,k) ⇀ Uk in H1(RN ) as n → ∞ . For k ∈ Λ∞,

σ
−N−2

2

n,k un(σ−1
n,k(·+ yn,k)) ⇀ Uk in D1,2(RN ) as n→∞ .

(2) For k ∈ Λ1, k 6= l, |yn,k − yn,l| → ∞ as n → ∞. For k ∈ Λ∞, k 6= l,
σn,k
σn,l

+
σn,l
σn,k

+ σn,kσn,l|yn,k − yn,l|2 →∞ as n→∞.

(3) ‖un‖2D1,2(RN ) =
∑
k∈Λ ‖Uk‖2D1,2(RN ) + ‖rn‖2D1,2(RN ) + o(1) as n → ∞, Λ =

Λ1∪Λ∞. ‖un‖2
∗

L2∗ (RN )
=
∑
k∈Λ ‖Uk‖2

∗

L2∗ (RN )
+o(1) as n→∞, Λ = Λ1∪Λ∞.

(4) rn = un −
∑
k∈Λ1

Uk(· − yn,k) −
∑
k∈Λ∞

σ
N−2

2

n,k Uk(σn,k(· − yn,k)) → 0 in

L2∗(RN ) as n→∞.

Similar to Lemma 4.3 and Corollary 4.4, we have the following lemma.

Lemma 4.7. (1) Assume yn ∈ RN and set ũn = un(· + yn) ⇀ U in H1(RN ).
Then Z = |U | satisfies∫

RN
∇Z∇ϕdx+

∫
RN

Zϕdx

≤ c
∫
RN

∫
RN

|Z(y)|2∗α |Z(x)|2∗α−1ϕ(x)

|x− y|α
dx dy + c

∫
RN

Zq−1ϕdx,

(4.27)

for ϕ ∈ H1(RN ), ϕ ≥ 0.
(2) The index sets Λ1, Λ∞ in the profile decomposition (4.26) are infinite.

Lemma 4.8. Assume yn ∈ RN , σn →∞. Set ũn = σ
−N−2

2
n un(σ−1

n ·+yn) ⇀ U in
D1,2(RN ). Then Z = |U | satisfies∫

RN
∇Z∇ϕdx ≤ c

∫
RN

∫
RN

|Z(y)|2∗α |Z(x)|2∗α−1ϕ(x)

|x− y|α
dx dy, (4.28)

for ϕ ∈ H1(RN ), ϕ ≥ 0.

Proof. Let ϕ ∈ C∞0 (RN ), R > 0, such that ϕ(x) = 1 for |x| ≤ R; ϕ(x) = 0 for

|x| ≥ 2R, and |∇ϕ| ≤ c
R . Select ϕn = σ

N−2
2

n ϕ(σn(· − yn)) as the test function in



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 25

〈DIεn,νn(un), ϕn〉 = 0, we have∫
RN

(∇ũn∇ϕdx+ σ−2
n V (εn(x+ yn))ũnϕ) dx

= 2∗α

∫
RN

∫
RN

F̃νn(ũn(y))f̃νn(ũn(x))ϕ(x)

|x− y|α
dx dy

+ ϑσ
qN−2

2 −N
n

∫
RN
|ũn|q−2ũnϕdx,

(4.29)

where F̃νn(t) = σ
α
2−N
n Fνn(σ

N−2
N

n t), f̃νn(t) = σ
α−N

2 −1
n fνn(σ

N−2
N

n t). By Rellich’s
imbedding theorem, we have ũn → U in Lsloc(RN )(1 ≤ s < 2∗). By (4.29) and
Lemma 2.1, we have∫

RN
(|∇(ũk − ũl)|2ϕdx

= −
∫
RN

(∇(ũk − ũl),∇ϕ)(ũk − ũl) dx

−
∫
RN

(V (εk(x+ yk))ũk − V (εl(x+ yl))ũl)(ũk − ũl)ϕdx

+ 2∗α

∫
RN

∫
RN

Fνn(ũk(y))fνn(ũk(x))(ũk(x)− ũl(x))ϕ(x)

|x− y|α
dx dy

− 2∗α

∫
RN

∫
RN

Fνn(ũk(y))fνn(ũk(x))(ũk(x)− ũl(x))ϕ(x)

|x− y|α
dx dy

+ ϑ

∫
RN

(|ũk|q−2ũk − |ũl|q−2ũl)(ũk − ũk)ϕdx

≤ c‖ũk − ũl‖L2(B(0,2R)) + c‖ũk − ũl‖Lq(B(0,2R)) + c‖ũk − ũl‖
L

2Nr
2N−α (B(0,2R))

→ 0, as k, l→∞.

Since ϕ = 1 in B(0, R) and ϕ ≥ 0, ũn → U in D1,2
loc(RN ). Let σn → ∞ in (4.29),

we have∫
RN
∇ũn∇ϕdx = 2∗α

∫
RN

∫
RN

F̃νn(ũn(y))f̃νn(ũn(x))ϕ(x)

|x− y|α
dx dy (4.30)

Let zn = |ũn|, wn,δ = (ũ2
n + δ2)1/2 − δ, then it follows from Lebesgue’s controlled

convergence theorem that wn,δ ∈ D1,2
loc(RN ), and wn,δ → zn in D1,2

loc(RN ) as δ → 0.

Now for any ϕ ∈ C∞0 (RN ), ϕ ≥ 0, we have ϕδ = ϕũn(ũ2
n + δ2)−1/2 ∈ D1,2

loc(RN ),
and ∫

RN
∇wn,δ∇ϕdx =

∫
RN

ũn∇ũn∇ϕ(ũ2
n + δ2)−1/2 dx

=

∫
RN

(
∇ũn∇ϕδ − |∇ũn|pϕ(ũ2

n + δ2)−
3
2 δ2
)
dx

≤
∫
RN
∇ũn∇ϕδ dx

≤ c
∫
RN

∫
RN

|zn(y)|2∗α |zn(x)|2∗α−1|ϕδ(x)|
|x− y|α

dx dy.

(4.31)



26 B. ZHANG, W. ZHANG EJDE-2024/19

Letting δ → 0 in the above inequality, we obtain∫
RN
∇zn∇ϕdx ≤ c

∫
RN

∫
RN

|zn(y)|2∗α |zn(x)|2∗α−1ϕ(x)

|x− y|α
dx dy. (4.32)

for ϕ ∈ C∞0 (RN ), ϕ ≥ 0. By ũn → U in D1,2
loc(RN ) as n → ∞, we have zn → Z in

D1,2
loc(RN ) as n→∞. Hence by a density argument we complete the proof. �

Lemma 4.9 ([49]). Let w ≥ 0, w ∈ D1,2(RN ) be a solution of −∆w ≤ av, where

a ∈ LN/2(RN ), v ≥ 0, v ∈ L2∗(RN ) ∩ Ls(RN ), s > 2∗

2 . Then

‖w‖Ls(RN ) ≤ c‖a‖LN/2(RN )‖v‖Ls(RN ).

Lemma 4.10. Let w ≥ 0, w ∈ D1,2(RN ) be a solution of −∆w ≤ aw2∗α−1, where

a(x) =
∫
RN

w2∗α(y)

|x−y|α dy + w2∗−2∗α(x), a ∈ L2N/α(RN ), and ‖a‖L2N/α(RN ) ≤ d1. As-

sume
∫
B(y,2ρ)

w2∗ dx ≤ d2 := ( S
2∗d1

)
2∗κ

2∗−2κ , ll y ∈ RN and 0 < ρ < 1, where

κ = 2N
2N−α , S is optimal constant of the embedding D1,2(RN ) ↪→ L2∗(RN ).

S = inf
u∈D1,2(RN )

∫
RN |∇u|

2dx( ∫
RN |u|2

∗dx
)2/2∗ ,

then there exist constants c, c∗, depending on ‖w‖L2∗ (RN ) and ρ such that

‖w‖L∞(B(y, ρ2 )) ≤ c‖w‖L2∗ (B(y,ρ)) ≤ c∗,
‖w‖L∞(B(y, ρ2 )) ≤ c‖w‖Lτ (B(y,ρ)), ∀τ ∈ (0, 2∗].

Proof. For every ϕ ∈ H1(RN ) with ϕ ≥ 0, we have∫
RN
∇w∇ϕdx ≤ ‖a‖L2N/α(RN )

(∫
RN

w2∗−κϕκ dx
)1/κ

≤ d1

(∫
RN

w2∗−κϕκ dx
)1/κ

.

(4.33)

Let p > 1, ψ ∈ C∞0 (RN ), ψ(x) = 1 if x ∈ B(y, ρ); ψ(x) = 0 if x /∈ B(y, 2ρ), and
|∇ψ| ≤ 2

ρ . By choosing the test function ϕ = w2p−1ψ2 in (4.33), we deduce that∫
RN
|∇(wpψ)|2 dx ≤ pd1

(∫
RN

w2∗−2κ(w2pψ2)κ dx
)1/κ

+

∫
RN

w2p|∇ψ|2 dx. (4.34)

Taking p = 1 + δ, and δ > 0 small, we obtain∫
RN
|∇(w1+δψ)|2 dx

≤ pd1

(∫
RN

w2∗−2κ(w1+δψ)2κ dx
)1/κ

+

∫
RN

w2(1+δ)|∇ψ|2 dx.
(4.35)

The left-hand side of (4.35) satisfies

LHS ≥ S
(∫

RN
(w1+δψ)2∗ dx

)2/2∗

, (4.36)



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 27

where S is the Sobolev constant. The first term of right-hand side of (4.35),

pd1

(∫
RN

w2∗−2κ(u1+δψ)2κ dx
)1/κ

≤ pd1

(∫
B(y,2ρ)

w2∗ dx
) 2∗−2κ

2∗·κ
(∫

RN
(w1+δψ)2∗ dx

)2/2∗

.

(4.37)

From (4.35)-(4.37), we have

S
(∫

RN
(w1+δψ)2∗ dx

)2/2∗

≤ pd1

(∫
B(y,2ρ)

w2∗ dx
) 2∗−2r

2∗r
(∫

RN
(w1+δψ)2∗ dx

)2/2∗

+

∫
RN

w2(1+δ)|∇ψ|2 dx.

(4.38)
Taking 1 + δ = 2∗/2, q = (1 + δ)2∗ > 2∗, and denoting

∫
B(y,2ρ)

w2∗ dx ≤ d2, we

have

S

2

(∫
B(y,ρ)

wq dx
)2/2∗

≤
∫
RN

w2∗ |∇ψ|2 dx ≤ 4

ρ2

∫
B(y,2ρ)

w2∗ dx ≤ 4d2

ρ2
.

hence ∫
B(y,ρ)

wq dx ≤ d3 :=
( 8d2

Sρ2

) 2∗
2 .

Let 0 < r < R < ρ < 1, ψ ∈ C∞0 (RN ), ψ(x) = 1 if x ∈ B(y, r); ψ(x) = 0 if
x /∈ B(y,R), and |∇ψ| ≤ 2

R−r . Then the left-hand side of (4.34) satisfies

LHS ≥ S
(∫

RN
(wpψ)2∗ dx

)2/2∗ ≥ S(∫
B(y,r)

(w2∗·p dx
)2/2∗

. (4.39)

The right-hand side of (4.34) satisfies

RHS ≤
(∫

B(y,ρ)

wq dx
) 2∗−2κ

qκ
(∫

RN
(w2pψ2)

qκ
q−2∗+2κ dx

) q−2∗+2κ
qκ

+
c

(R− r)2

(∫
B(y,R)

w
2∗·p
d dx

)2d/2∗

≤ (c+
c

(R− r)2
)
(∫

B(y,R)

w
2∗·p
d dx

)2d/2∗

,

(4.40)

where d = 2∗(q−2∗+2κ)
2qκ . By (4.34), (4.39) and (4.40), we have(∫
B(y,r)

w2∗·p dx
) 1

2∗·p ≤ (
c

R− r
)

1
p

(∫
B(y,R)

w
2∗·p
d dx

) d
2∗·p

, (4.41)

Taking p = pk = dk, R = rk = ρ
2 + ρ

2k
, r = rk+1, k = 1, 2, . . . . By (4.41), we have(∫

B(y,rk+1)

w2∗dk dx
) 1

2∗dk ≤ (
c2k+1

ρ
)1/dk

(∫
B(y,rk)

w2∗dk−1

dx
) 1

2∗dk−1

,

Letting k →∞, we obtain

‖w‖L∞(B(y, ρ2 )) ≤ c‖w‖L2∗ (B(y,ρ)) ≤ c∗.



28 B. ZHANG, W. ZHANG EJDE-2024/19

Taking p = pk = dk, rk = r + R−r
2k−1 , k = 1, 2, . . . . By (4.41), we obtain(∫

B(y,rk+1)

w2∗dk dx
) 1

2∗dk ≤
( c2k

R− r
)1/dk(∫

B(y,rk)

w2∗dk−1

dx
) 1

2∗dk−1

,

Letting k →∞, we have

‖w‖L∞(B(y,r)) ≤ c(
1

R− r
)

1
d−1 ‖w‖L2∗ (B(y,R)).

hence

‖w‖L∞(B(y,r)) ≤ c(
1

R− r
)

1
d−1 ‖w‖L2∗ (B(y,R))

≤ c( 1

R− r
)

1
d−1 ‖w‖

2∗−τ
2∗

L∞(B(y,R))‖w‖
τ
2∗
Lτ (B(y,R))

≤ 1

2
‖w‖L∞(B(y,R)) + c(

1

R− r
)

2∗
τ(d−1) ‖w‖Lτ (B(y,R)).

(4.42)

By iteration, we have

‖w‖L∞(B(y, ρ2 )) ≤ c‖w‖Lτ (B(y,ρ)). �

from the above lemma, we have the following Lemma.

Lemma 4.11. Let w ≥ 0, w ∈ D1,2(RN ) be a solution of

−∆w ≤ aw2∗α−1,

where a(x) =
∫
RN

w2∗α (y)
|x−y|α dy + w2∗−2∗α(x), and a ∈ L2N/α(RN ). Then there exist

R > 0 and a constant c depending on ‖w‖L2∗ (RN ), such that

|w(x)| ≤ c
(∫
|x|≥R2

w2∗ dx
)1/2∗

, x ∈ RN , |x| ≥ R.

Lemma 4.12. There exist positive constants c, µ such that

|Uk(x)| ≤ c(1 + |x|2)
2−N

2 for k ∈ Λ∞,

|Uk(x)| ≤ c exp{−µ|x|} for k ∈ Λ1.

Proof. Let w ≥ 0, w ∈ D1,2(RN ) be a solution of

−∆w ≤ aw2∗α−1,

where a ∈ L2N/α(RN ). Using the Kelvin transformation v(x) = |x|2−Nw( x
|x|2 ), we

know that v satisfies

−∆v ≤ ãv2∗α−1, ∀|x| ≤ 1,

where ã(x) = |x|α−Na( x
|x|2 ). We also have

∫
Bρ(0)

|v|2∗ dx = oρ(1), and

‖ã‖L2N/α(RN ) = ‖a‖L2N/α(RN ) ≤ b.

Choose ρ small enough, By Lemma 4.10, we obtain |v(x)| ≤ c in B ρ
2 (0). This shows

that

|w(x)| ≤ c

|x|N−2
, |x| ≥ 2ρ.

Similarly, we obtain that |w(x)| ≤ c, |x| ≤ 2ρ. So we have

w(x) ≤ c(1 + |x|2)
2−N

2 .



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 29

We use the hole-filling technique. For

−∆w + w ≤ aw2∗α−1,

we also have
w(x) ≤ c(1 + |x|2)

2−N
2 .

So there exists R0 > 0 such that

w(x)2∗−2 ≤ 1

2
, for x ∈ RN , |x| ≥ R0.

Without loss of generality, we assume that

−∆w + w ≤ 0, for x ∈ RN , |x| ≥ R0.

For R > R0, we choose ψ ∈ C∞(RN ) such that ψ(x) = 1, |x| ≥ R + 1; ψ(x) = 0,
|x| ≤ R, |∇ψ| ≤ 2. Taking ϕ = wψ2 as test function we have∫

RN
|∇w|2ψ2 dx+

∫
RN

w2ψ2 dx ≤ 2

∫
RN
∇wψw∇ψ dx

≤ 1

2

∫
RN
|∇w|2ψ2 dx+ c

∫
RN

w2|∇ψ|2 dx.
(4.43)

Therefore,∫
|x|≥R+1

w2 dx ≤
∫
RN

w2ψ2 dx ≤ c
∫
RN

w2|∇ψ|2 dx ≤
∫
R≤|x|≤R+1

w2 dx,∫
|x|≥R+1

w2 dx ≤ c

c+ 1

∫
|x|≥R

w2 dx.

It follows that there exist c, µ such that∫
|x|≥R2

w2 dx ≤ c exp{−µR}.

By Lemma 4.11, there exist c′, µ′ such that

w(x) ≤ c
(∫
|x|≥R2

w2 dx
)1/2∗

≤ c′ exp{−µ′R}, x ∈ RN , |x| ≥ R.

�

Suppose 1 ≤ p2 < 2∗ < p1, σ ≥ 1 and m > 0. Consider the system of inequalities

‖u1‖Lp1 (RN ) ≤ m,

‖u2‖Lp2 (RN ) ≤ mσ
N
2∗−

N
p2 .

(4.44)

We define the norm

‖u‖p1,p2,σ = inf{m : there exists u1, u2 satisfying |u| ≤ u1 + u2 and (4.44) holds}.

Lemma 4.13. Assume the profile decomposition (4.26) holds. Without loss of
generality assume k∞ ∈ Λ∞, σn = σn,k∞ = min{σn,k

∣∣k ∈ Λ∞}. N
N−2 < p2 < 2∗ <

p1, then there exists a constant c, depending on p1, p2 such that ‖u‖p1,p2,σ ≤ c.

Proof. We divide un into three parts

un = zn + wn + rn,

where

zn =
∑
k∈Λ1

Uk(· − yn,k), wn =
∑
k∈Λ∞

σ
N−2

2

n,k Uk(σn,k(· − yn,k)).



30 B. ZHANG, W. ZHANG EJDE-2024/19

By Lemma 4.12, for k ∈ Λ1, Uk decays exponentially. Hence |zn|p1 ≤ c for all

p1 ≥ 1. For k ∈ Λ∞, Uk decays polynomially. Hence for N
N−2 < p2 < 2∗,

‖σ
N−2

2

n,k Uk(σn,k(· − yn,k))‖Lp2 (RN ) = σ
N−2

2 −
N
p2

n,k ‖Uk‖Lp2 (RN )

≤ cσ
N
2∗−

N
p2

n,k

( ∫
RN

(1 + |x|2)
2−N

2 p2 dx
) 1
p2

≤ cσ
N
2∗−

N
p2

n,k ≤ cσ
N
2∗−

N
p2

n

(4.45)

and ‖wn‖Lp2 (RN ) ≤ cσ
N
2∗−

N
p2

n . We define Zn,Wn, Rn ∈ D1,2(RN ) as follows:

−∆Zn = an(x)|zn|2
∗
α−1, in RN ,

−∆Wn = an(x)|wn|2
∗
α−1, in RN ,

−∆Rn = an(x)|rn|2
∗
α−1, in RN .

Here an(x) =
∫
RN
|un(y)|2

∗
α

|x−y|α dy + |un(x)|
α

N−2 . By Lemma 4.9, we have

‖Zn‖Lp1 (RN ) ≤ c‖an|zn|2
∗
α−2‖LN/2(RN )‖zn‖Lp1 (RN )

≤ c‖an‖L2N/α(RN )‖|zn|2
∗
α−2‖

L
2N
4−α (RN )

‖zn‖Lp1 (RN )

≤ c‖zn‖Lp1 (RN ) ≤ c.

(4.46)

Similar to (4.46), we have

‖Wn‖Lp2 (RN ) ≤ c‖wn‖Lp2 (RN ) ≤ cσ
N
2∗−

N
p2

n ,

‖Rn‖p1,p2,σn ≤ o(1)‖rn‖p1,p2,σn .
Let un satisfy the equation

−∆un + V (εnx)un = 2∗α

(∫
RN

Fνn(un(y))

|x− y|α
dy
)
fνn(un) + ϑ|un|q−2un, (4.47)

where x ∈ RN .
By Lemma 2.4 and the inequality |t| ≤ ε|t| + cε|t|2

∗
, the function vn = |un|

satisfies
−∆vn ≤ can(x)|un|2

∗
α−1 ≤ −c∆(Zn +Wn +Rn).

By the maximum principle, we obtain

|un| = vn ≤ c(Zn +Wn +Rn).

Now we have the estimate

‖un‖p1,p2,σn ≤ c(‖Zn‖p1,p2,σn + ‖Wn‖p1,p2,σn + ‖Rn‖p1,p2,σn)

≤ c+ o(1)‖rn‖p1,p2,σn
≤ c+ o(1)(‖un‖p1,p2,σn + ‖zn‖p1,p2,σn + ‖wn‖p1,p2,σn)

≤ c+ o(1)‖un‖p1,p2,σn ,

(4.48)

and ‖un‖p1,p2,σn ≤ c. �

Assume the profile decomposition (4.26) holds. Let k∞ ∈ Λ∞ be such that
σn = σn,k∞ = min{σn,k

∣∣k ∈ Λ∞}. We denote yn = yn,k∞ . Since the index set Λ∞
is finite, we can find a constant c > 0 such that the region

A1
n = B(yn, 7cσ

−1/2
n )\B(yn, cσ

−1/2
n )



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 31

does not contain any concentration points, corresponding to the index set Λ∞,

A1
n ∩ {yn,k

∣∣k ∈ Λ∞} = ∅.

Lemma 4.14 ([20]). Let u ≥ 0, u ∈ H1(RN ) and satisfy

−∆u ≤ f, in RN ,
where f ≥ 0, f ∈ L1

loc(RN ). Then for γ ∈ (1, N
N−1 ), there exists a positive constant

c = c(N, γ) such that for x0 ∈ RN , r ∈ (0, 1)(
r−N

∫
B(x0,r)

|u|γ dx
)1/γ

≤ c
(

1 +

∫ 1

r

(t1−N
∫
B(x0,t)

f dx)dt
)
.

Lemma 4.15. Assume the profile decomposition (4.26) holds. Then there exists a
constant c > 0, independent of n, such that |un(x)| ≤ c for x ∈ A2

n, and∫
A3
n

|∇un|2 dx ≤ cσ
2−N

2
n ,

where

A2
n = B(yn, 6cσ

−1/2
n )\B(yn, 2cσ

−1/2
n ), A3

n = B(yn, 5cσ
−1/2
n )\B(yn, 3cσ

−1/2
n ).

Proof. It is easy to show that

−∆vn ≤ cwnv
2∗α−1
n ,

where

wn(x) =

∫
RN

|vn(y)|2∗α
|x− y|α

dy + |vn(x)|2
∗−2∗α .

By Lemma 4.14, for all y ∈ RN and r ∈ (0, 1), we have(
r−N

∫
B(y,r)

|un|γ dx
)1/γ

=
(
r−N

∫
B(y,r)

vγn dx
)1/γ

≤ c
(

1 +

∫ 1

r

(t1−N
∫
B(y,t)

wnv
2∗α−1
n dx)dt

)
.

(4.49)

By Lemma 4.13, ‖vn‖p1,p2,σn ≤ c for N
N−2 < p2 < 2∗ < p1. There exist functions

v
(1)
n , v

(2)
n such that vn ≤ v(1)

n + v
(2)
n , ‖v(1)

n ‖Lp1 (RN ) ≤ c, and

‖v(2)
n ‖Lp2 (RN ) ≤ cσ

N
2∗−

N
p2

n .

We choose p1 = N(N+2)
N−2 and p2 = 2N(N+2−α)

(N−2)(2N−α) . We have the estimate(
r−N

∫
B(y,r)

|un|γ dx
)1/γ

≤ c+ c

∫ 1

r

(
t1−N

∫
B(y,t)

w(1)
n |v(1)

n |2
∗
α−1 dx

)
dt

+ c

∫ 1

r

(
t1−N

∫
B(y,t)

w(2)
n |v(2)

n |2
∗
α−1 dx

)
dt,

where∫ 1

r

(
t1−N

∫
B(y,t)

w(1)
n |v(1)

n |2
∗
α−1 dx

)
dt

≤
(∫ 1

r

t1−Ndt
)(∫

B(y,t)

|w(1)
n |N |v(1)

n |N(2∗α−1) dx
)1/N(∫

B(y,t)

dt
)N−1

N



32 B. ZHANG, W. ZHANG EJDE-2024/19

≤ c
(∫

B(y,1)

|w(1)
n |N |v(1)

n |N(2∗α−1) dx
)1/N

≤ c
(∫

B(y,1)

|w(1)
n |

N(N+2)
α dx

) α
N+2 ·

(∫
B(y,1)

|v(1)
n |N(2∗α−1) N+2

N+2−α dx
)N+2−α

N+2

≤ c
(∫

B(y,1)

|v(1)
n |

N(N+2)
N−2 dx

)N+2−α
N+2 ≤ c,

and∫ 1

r

(
t1−N

∫
B(y,t)

w(2)
n |v(2)

n |2
∗
α−1 dx

)
dt

≤
(∫ 1

r

t1−Ndt
)(∫

B(y,t)

|w(2)
n |2N/α dx

) α
2N
(∫

B(y,t)

|v(2)
n |(2

∗
α−1) 2N

2N−α dx
) 2N−α

2N

≤ c
(∫ 1

r

t1−Ndt
)(∫

B(y,t)

|v(2)
n |p2 dx

) 2∗α−1

p2

≤ c
(∫ 1

r

t1−Ndt
)
σ

2−N
2

n

≤ c(rσ1/2
n )2−N ≤ c, provided r ≥ c

4
σ−1/2
n .

Therefore (
σN/2n

∫
B(y, c4σ

−1/2
n )

|un|γ dx
)1/γ

≤ c, y ∈ A2
n.

In (4.26), with un = zn + wn + rn, zn ∈ L∞(RN ), |rn|2∗ = o(1), by (4) of the

profile decomposition (4.26) becomes wn =
∑
k∈Λ∞

σ
N−2

2

n,k Uk(σn,k(· − xn,k)). For

y ∈ A2
n, x ∈ B(y, c2σ

−1/2
n ), by our choice A1

n, |y − yn,k| ≥ cσ
−1/2
n , |x − yn,k| ≥

|yn,k − y| − |x− y| ≥ c
2σ
−1/2
n ≥ c

2σ
−1/2
n,k . Hence for n is large enough,∫

B(y, c2σ
−1/2
n )

|un|2
∗
dx =

∫
B(y, c2σ

−1/2
n )

|σ
N−2

2

n,k Uk(σn,k(· − yn,k))|2
∗
dx+ o(1)

≤
∫
|x−xn,k|≥ c2σ

−1/2
n,k

|σ
N−2

2

n,k Uk(σn,k(· − yn,k))|2
∗
dx+ o(1)

=

∫
|x|≥ c2σ

−1/2
n,k

|Uk(x)|2
∗
dx+ o(1) = o(1), ∀y ∈ A2

n.

By Lemma 4.10, we have

|un(x)| ≤ c
(
σN/2n

∫
B(y, c4σ

−1/2
n )

|un|γ dx
)1/γ

≤ c, ∀x ∈ B(y,
c

8
σ−1/2
n ), y ∈ A2

n.

Hence |un(x)| ≤ c for x ∈ A2
n.

Let ϕ ∈ C∞0 (RN ) be such that ϕ(x) = 1 for x ∈ A3
n; ϕ(x) = 0 for x /∈ A2

n and

|∇ϕ| ≤ 2c−1σ
1/2
n . Taking φ = unϕ

2 as test function in 〈DIεn,νn(un), φ〉 = 0, we
obtain∫

RN
(|∇un|2ϕ2 + V (εnx)u2

nϕ
2) dx



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 33

= −2

∫
RN

un∇unϕ∇ϕdx+ 2∗α

∫
RN

∫
RN

Fνn(un(y))

|x− y|α
dyfνn(un)unϕ

2 dx

+ ϑ

∫
RN
|un|qϕ2 dx

≤ 1

2

∫
RN
|∇un|2ϕ2 dx+ c

∫
RN
|un|2|∇ϕ|2 dx

+ c

∫
RN

∫
RN

|un(y)|2∗α
|x− y|α

dy|un|2
∗
αϕ2 dx+ c

∫
RN
|un|qϕ2 dx.

Since |un(x)| ≤ c for x ∈ A2
n, we have∫

A3
n

|∇un|2 dx

≤
∫
RN
|∇un|2ϕ2 dx

≤ cσn
∫
A2
n

|un|2 dx+ c

∫
A2
n

∫
RN

|un(y)|2∗α
|x− y|α

dy|un|2
∗
αϕ2 dx+ c

∫
A2
n

|un|q dx

≤ c
(∫
A2
n

|
∫
RN

|un(y)|2∗α
|x− y|α

dy|2N/α dx
) α

2N
(∫
A2
n

|un|2
∗
dx
) 2N−α

2N

+ cσ
1−N2
n

≤ cσ
2−N

2
n ,

for n large enough. �

Similar to [16, Lemma 4.3], we can prove that the index set Λ∞ in the profile
decomposition (4.26) is empty.

The proof part 3 of Theorem 4.1. Note that Λ∞ is empty, the profile decomposition
in (4.26) reduces to

un =
∑
k∈Λ1

Uk(· − yn,k) + rn, (4.50)

where rn → 0 in L2∗(RN ) as n → ∞. By Lemma 4.12, there exist c, µ, such
that |Uk(x)| ≤ c exp{−µ|x|} for k ∈ Λ1. Hence,

∑
k∈Λ1

Uk(· − yn,k) are uniformly

bounded. It follows that {un} are uniformly bounded, that is there exists a constant
M depending on L, but not on n, such that

|un(x)| ≤M, for x ∈ RN , n = 1, 2, . . . .

�

5. Proof of Theorem 1.1

From Theorem 4.1, we deduce the following corollary.

Corollary 5.1. (1) Assume Γε,ν,λ(u) ≤ L and DΓε,ν,λ(u) = 0. Then there exists

λ = λ(L) such that Γε,ν,λ(u) = Γε,ν(u) and DΓε,ν(u) = 0 if 0 < λ ≤ λ.
(2) Assume Γε,ν(u) ≤ L and DΓε,ν(u) = 0. Then there exists ε = ε(L) such that

Γε,ν(u) = Iε,ν(u) and DIε,ν(u) = 0 if 0 < ε ≤ ε.
(3) Assume Iε,ν(u) ≤ L and DIε,ν(u) = 0. Then there exists ν = ν(L) such that

Iε,ν(u) = Iε(u) and DIε(u) = 0 if 0 < ν ≤ ν.



34 B. ZHANG, W. ZHANG EJDE-2024/19

Proof. (1) By Theorem 4.1(1), if 0 < λ < λ(L) = 1
CH2∗α

, then

‖u‖H1(RN ) ≤ (
1

Cλ
)

1
2∗α .

It follows that Γε,ν,λ(u) = Γε,ν(u) and DΓε,ν(u) = 0.
(2) By Theorem 4.1(2), there exist constants µ, c = c(L) such that, for every

δ > 0, there exists ε = ε(δ) > 0, for 0 < ε < ε(δ) and x ∈ RN

|u(x)| ≤ c exp{−µdist(x, (Aδ)ε)}.
Let ε(L) ≤ min{µ, 1

c}, then for 0 < ε ≤ ε and x ∈ RN , it holds

|u(x)| ≤ c exp{−µdist(x, (Aδ)ε)}

≤ 1

ε
exp{−εdist(x, (Aδ)ε)}

≤ 1

ε
exp{−dist(εx,M)}.

Hence mε(x, u) = u, for x ∈ RN . Moreover we denote D = max{|y|
∣∣∣y ∈ M},

d = dist(Aδ, ∂M). We choose an integer l > 1 such that ld ≥ D. Then for x /∈Mε,

l dist(x, (Aδ)ε) ≥ l dist((Aδ)ε, ∂Mε) + dist(x, ∂Mε)

≥ l

ε
d+ |x| − D

ε
≥ |x|,

and hence

|u(x)| ≤ c exp{−cdist(x, (Aδ)ε)} ≤ c exp{−c
l
|x|}, ∀x /∈Mε.

As a consequence, for 0 < ε < ε(δ) sufficiently small, one has∫
RN

χε(x)u2 dx ≤ cε−6

∫
|x|≥cε−1

exp{−c
l
|x|} dx ≤ cε−N−5 exp{− c

ε
} < 1

and (∫
RN

χε(x)|u|p dx− 1
)

+
= 0.

It follows that Γε(u) = Iε(u) and DIε(u) = 0.
(3) By Theorem 4.1(3), if 0 < ν < ν(L) = 1

M , then

|u(x)| ≤ 1

ν
, ∀x ∈ RN .

Hence, Iε,ν(u) = Iε(u) and DIε(u) = 0. �

Proof of Theorem 1.1. Given a positive integer k, by Theorem 3.10, there exist
0 < ε̃ < 1, 0 < ν̃ < 1, and 0 < λ̃ < 1, such that if 0 < ε < ε̃, 0 < ν < ν̃,
0 < λ < λ̃, then the functional Γε,ν,λ has k pairs of sign-changing critical points
±uj , j = 1, . . . , k, and the corresponding critical values satisfy

0 < c1(ε, ν, λ) ≤ · · · ≤ ck(ε, ν, λ) ≤ mk.

By Corollary 5.1(3), there exists νk = νk(mk), such that if

0 < ν < ν̃k = min{νk, ν̃}, Iε,ν(u) ≤ mk, DIε,ν(u) = 0,

then

Iε,ν(u) = Iε(u), DIε(u) = 0.



EJDE-2024/19 SEMICLASSICAL CHOQUARD EQUATIONS 35

Fixed ν ∈ (0, νk). By Corollary 5.1(2), there exists εk = εk(mk), such that if

0 < ε < ε̃k = min{εk, ε̃}, Γε,ν(u) ≤ mk, DΓε,ν(u) = 0,

then

Γε,ν(u) = Iε,ν(u), DIε,ν(u) = 0.

We fix ν ∈ (0, νk) and ε ∈ (0, εk). By Corollary 5.1(1), there exists λk = λk(mk),
such that if

0 < λ < λ̃k = min{λk, λ̃}, Γε,ν,λ(u) ≤ mk, DIε,ν,λ(u) = 0,

then

Γε,ν,λ(u) = Γε,ν(u), DΓε,ν(u) = 0.

Now for 0 < ν < ν̃k, 0 < ε < ε̃k, 0 < λ < λ̃k, we have that uj,ε = uj(ε, ν, λ),
j = 1, . . . , k are critical points of the functional Iε. Moreover, by Theorem 4.1,
there exist constants µ > 0, c = c(mk), such that for any δ > 0, there exists εk(δ)
such that for 0 < ε < εk(δ) it holds

|uj,ε| ≤ c exp{−µdist(x, (Aδ)ε)}, for x ∈ RN ,
hence

|vj,ε| ≤ c exp{−µ
ε

dist(x,Aδ)}, for x ∈ RN . �

Acknowledgments. This research was partially supported by the Sichuan Uni-
versity of Arts Science Research Project (No. 2022QD070), by the Natural Science
Foundation of Sichuan Province (No. 2023NSFSC0077), by the Yunnan Funda-
mental Research Projects (No. 202101AU070083), and by the Yunnan University
of Finance and Economics Science Research Project (No. 2023D39).

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Bo Zhang

School of Mathematics, Sichuan University of Arts and Science, Dazhou 635000, China

Email address: zhangbo371013@163.com

Wei Zhang (corresponding author)

School of Statistics and Mathematics, Yunnan University of Finance and Economics,
Kunming 650221, China

Email address: weizyn@163.com


	1. Introduction
	2. Preliminaries and Palais-Smale condition for ,,
	3. Existence of solutions for perturbed functions ,,
	4. Uniform bounds
	5. Proof of Theorem ??
	Acknowledgments

	References