Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 11, pp. 1–29.

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

LOCALIZED NODAL SOLUTIONS FOR SEMICLASSICAL

QUASILINEAR CHOQUARD EQUATIONS

WITH SUBCRITICAL GROWTH

BO ZHANG, XIANGQING LIU

Abstract. In this article, we study the existence of localized nodal solutions
for semiclassical quasilinear Choquard equations with subcritical growth

−εp∆pv + V (x)|v|p−2v = εα−N |v|q−2v

∫
RN

|v(y)|q

|x− y|α
dy, x ∈ RN ,

where N ≥ 3, 1 < p < N , 0 < α < min{2p,N−1}, p < q < p∗α, p∗α =
p(2N−α)
2(N−p) ,

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

invariant sets of descending flow, for small ε we establish 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 are interested in localized nodal solutions of the quasilinear
Choquard equation

−εp∆pv + V (x)|v|p−2v = εα−N |v|q−2v

∫
RN

|v(y)|q

|x− y|α
dy, x ∈ RN ,

v(x)→ 0 as |x| → ∞,
(1.1)

where ∆p = div(|∇v|p−2∇v) is the p-Laplacian operator, N ≥ 3, 1 < p < N ,

0 < α < min{2p,N − 1}, p < q < p∗α, p∗α = p(2N−α)
2(N−p) is the upper critical exponent

in the sense of Hardy-Littlewood-Sobolev inequality, ε > 0 is a small parameter.
The potential function V satisfies the 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, 35B40.
Key words and phrases. Quasilinear Choquard equation; nodal solutions;

perturbation method.
©2022. This work is licensed under a CC BY 4.0 license.

Submitted September 6, 2021. Published February 10, 2022.

1



2 B. ZHANG, X. LIU EJDE-2022/11

In the previous decades, the Choquard type equation has been widely studied,
we refer to [28, 11, 12, 16, 27, 20] and references therein. The Choquard equation

−∆u+ u = u

∫
R3

|u(y)|2

|x− y|
dy , x ∈ R3, (1.2)

which goes back to the description of the quantum theory of a polaron at rest
by Pekar [28], and which emerged in the work of Choquard on the modeling of an
electron trapped in its own hole, as a certain approximation to Hartree-Fock theory
of one-component plasma [11]. For the Choquard equation

−∆u+ V (x)u = |u|p−2u

∫
RN

|u(y)|p

|x− y|α
dy, x ∈ RN ,

u(x)→ 0 as |x| → ∞,
(1.3)

where 2N−α
N ≤ p ≤ 2N−α

N−2 . When the potential V is a positive constant, Lieb [12]

obtained the existence and uniqueness of positive radial ground states for (1.3), Li-
ons [16] extended Lieb’s results to a more general case and established the existence
of infinitely many radial solutions, Ma and Zhao [20] studied the radial symmetry
and uniqueness of positive ground states for (1.3) in higher dimension space via the
method of moving planes. For more related results, we refer to [23, 24, 22, 21, 25]
and references therein. For the semiclassical Choquard equation

−ε2∆u+ V (x)u = εα−Ng(x, u)

∫
RN

G(y, u(y))

|x− y|α
dy, x ∈ RN ,

u(x)→ 0 as |x| → ∞ .

(1.4)

Alves and Yang [3] proved the existence, multiplicity and concentration of solutions
for (1.4) by using the Lyapunov-Schmidt reduction method [10]. The existence and
qualitative properties of solutions for (1.4) have been also studied extensively for
recent decades by variational methods [30, 1, 2]. When p = 2, Moroz and Van
Schaftingen [26] constructed the single spike solution concentrating around the lo-
cal minimum of the potential V by using the nonlocal penalization method. Cassani
and Zhang [5] considered the existence of ground states for (1.4) involving a critical
nonlinearity in the sense of Hardy-Littlewood-Sobolev. Yang and Zhang [32] inves-
tigated the existence and concentration of solutions under the local potential well
condition. Under conditions (A1) and (/a2), Li and Ma [18] proved the existence
and concentration of infinite many solutions for the subcritical Choquard equation
by using the penalization method [4] and symmetric mountain pass lemma.

In this article, we consider the existence of localized nodal solutions for the
semiclassical Choquard equation (1.1) by using the method of invariant sets of
descending flow and the perturbation method. Byeon-Wang type penalization
method [4] can be used to deal with multiple localized nodal solutions for semi-
classical Schrödinger equations. Additional coercive term [17] can be used to make
the perturbed functional has necessary compactness properties in changed space.

Under the condition (A2), the critical set

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 denote

Bδ = {x ∈ RN : δx ∈ B},



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 3

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

Theorem 1.1. Assume that (A1) and (A2) hold. Then for any 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 any δ > 0 there exist
µ > 0, C = Ck > 0 and εk(δ) > 0 such that if 0 < ε < εk(δ), then it holds that

|vj,ε(x)| ≤ C exp{−µ
ε

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

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

−∆pu+ V (εx)|u|p−2u = |u|q−2u

∫
RN

|u(y)|q

|x− y|α
dy x ∈ RN . (1.5)

The energy functional associated with (1.5) is

Iε(u) =
1

p

∫
RN

(|∇u|p + V (εx)|u|p) dx− 1

2q

∫
RN

∫
RN

|u(x)|q|u(y)|q

|x− y|α
dx dy, (1.6)

for u ∈W 1,p(RN ).
To obtain multiple localized nodal solutions of (1.5), we use the penalization

method due to Byeon and Wang [4]. Let ζ ∈ C∞(RN ) 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. Define

χε(x) = ε−pζ(dist(x,Mε)).

Since the imbedding from W 1,p(RN ) to Ls(RN ) (p < s < p∗) is continuous but
not compact, we choose a suitable space as working space such that the functional Iε
recovers compactness on the changed space. We denoteXε = W 1,p(RN )

⋂
Lmε (RN ),

where Lmε (RN ) is a weighted space defined as

Lmε (RN ) = {u ∈ Lm(RN ) :

∫
RN

exp{(m− p) dist(εx,M)}|u|m dx < +∞}

endowed with the norm

‖u‖Lmε (RN ) =
(∫

RN
exp{(m− p) dist(εx,M)}|u|m dx

)1/m

.

We define

‖u‖W 1,p(RN ) =
(∫

RN
(|∇u|p + E(εx)|u|p) dx

)1/p

,

‖u‖Xε = ‖u‖W 1,p(RN ) + ‖u‖Lmε (RN ).

Meanwhile, we introduce one additional coercive term such that Iε has the nec-
essary compactness property on Xε. We need 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

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−p
|t|p−2t, Kε(x, t) =

∫ t

0

kε(x, τ) dτ,



4 B. ZHANG, X. LIU EJDE-2022/11

where p < m < min{2, q} if 1 < p < 2; p < m < q if p ≥ 2. We define

Γε(u) =
1

p

∫
RN

(|∇u|p + E(εx)|u|p) dx+ σ

∫
RN

Kε(x, u) dx

+
1

pβ

(∫
RN

χε(x)|u|p dx− 1
)β

+
− 1

2q

∫
RN

∫
RN

|u(x)|q|u(y)|q

|x− y|α
dx dy,

(1.7)

for u ∈ Xε, where p < pβ < q, E(x) = V (x) − σ, for small σ, E satisfies the
assumptions (A1) and (A2) (with a different constant a′ = a− σ > 0).

We will use the method of invariant sets of descending flow [19] to prove the
existence of sign-changing critical points of Γε,λ, but the method of invariant sets of
descending flow can not fit well for the functional (1.7). So we use the perturbation
method [9] 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,λ(u) =
1

p

∫
RN

(|∇u|p + E(εx)|u|p) dx+ σ

∫
RN

Kε(x, u) dx

+
1

pβ

(∫
RN

χε(x)|u|p dx− 1
)β

+
− 1

2q
gλ(ϕ1/2(u))ϕ(u), u ∈ Xε,

(1.8)

where

ϕ(u) =

∫
RN

∫
RN

|u(y)|q|u(x)|q

|x− y|α
dx dy.

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

ϕ1/2(u) ≤ C0‖u‖qW 1,p(RN )
.

Note that if

‖u‖W 1,p(RN ) ≤
( 1

C0λ

)1/q
,

|u(x)| ≤ 1

ε
exp{−dist(εx,M)} for x ∈ RN ,(∫
RN

χε(x)|u|p dx− 1
)

+
= 0,

for sufficiently small ε, λ, then Γε,λ(u) = Iε(u) and DΓε,λ(u) = DIε(u). Hence we
obtain solutions of (1.5).

In the following we use c to denote various constants, cε denotes constants de-
pending on ε and c, cε may be used from line to line for different constants but
independent of the arguments. This article is organized as follows. In Section 2 we
prove preliminary results and verify the Palais-Smale condition for the functional
Γε,λ. In Section 3 we construct a sequence of sign-changing critical points of Γε,λ
by using the method of invariant sets of descending flow. In Section 4 prove Theo-
rem 1.1. In Section 5 we prove the uniform bounds on the critical points obtained
in Section 3.



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 5

2. Palais-Smale condition

First collect some elementary results about the auxiliary functions and we prove
that Γε,λ satisfies the (PS) condition.

Lemma 2.1 (Hardy-Littlewood-Sobolev inequality [13]). 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, r, α,N), independently of g, h, such that∫

RN

∫
RN

g(x)h(y)

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

Lemma 2.2. For x ∈ RN and t ∈ R, it holds:

(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−p exp{(m− p) dist(εx,M)}|t|m−p)|t|p−2t ≤ kε(x, t)
≤ c2(1 + εm−p exp{(m− p) dist(εx,M)}|t|m−p)|t|p−2t;

(6) 1
m tkε(x, t) ≤ Kε(x, t) ≤ 1

p tkε(x, t);

(7) (kε(x, t1)−kε(x, t2))(t1−t2) ≥ cεm−p exp{(m−p) dist(εx,M)}|t1−t2|m(p ≥
2); (kε(x, t1) − kε(x, t2))(t1 − t2) ≥ cεm−p exp{(m − p) dist(εx,M)}|t1 −
t2|2(|t1|2−m − |t2|2−m)−1(1 < p < 2);

(8) |kε(x, t1)− kε(x, t2)| ≤ c(|t1|p−2 + |t2|p−2 + εm−p exp{(m− p) dist(εx,M)}
(|t1|m−2 + |t2|m−2))|t1 − t2|(p ≥ 2), |kε(x, t1)− kε(x, t2)| ≤ c(|t1 − t2|p−1 +
εm−p exp{(m− p) dist(εx,M)}|t1 − t2|m−1)(1 < p < 2).

Proof. The proof is straightforward. We only prove (6), (7) and (8).
(6) Let f(x, t) = Kε(x, t) − 1

p tkε(x, t), g(x, t) = Kε(x, t) − 1
m tkε(x, t), since

f(x, 0) = 0, ∂f(x,t)
∂t ≤ 0 for t ≥ 0; ∂f(x,t)

∂t ≥ 0 for t ≤ 0 and g(x, 0) = 0, ∂g(x,t)
∂t ≥ 0

for t ≥ 0; ∂g(x,t)
∂t ≤ 0 for t ≤ 0. So (6) holds.

We use the following elementary inequalities (see [8]). For p ≥ 2 and ξ, η ∈ RN ,

|ξ − η|p ≤ d1(|ξ|p−2ξ − |η|p−2η, ξ − η), (2.1)

| |ξ|p−2ξ − |η|p−2η| ≤ d2(|ξ|+ |η|)p−2|ξ − η|2. (2.2)

For 1 < p < 2 and ξ, η ∈ RN ,

|ξ − η|p ≤ d3

(
|ξ|p−2ξ − |η|p−2η, ξ − η

)p/2
(|ξ|p + |η|p)

2−p
2 , (2.3)

| |ξ|p−2ξ − |η|p−2η| ≤ d4|ξ − η|p−1. (2.4)

(7) For p ≥ 2, by (2.1) we have

(kε(x, t1)− kε(x, t2))(t1 − t2)

=

∫ 1

0

∂kε(x, θt1 + (1− θ)t2)

∂t
dθ(t1 − t2)2

≥ c
∫ 1

0

( θt1 + (1− θ)t2
mε(x, θt1 + (1− θ)t2)

)m−p
|θt1 + (1− θ)t2|p−2dθ(t1 − t2)2



6 B. ZHANG, X. LIU EJDE-2022/11

≥ cεm−p exp{(m− p) dist(εx,M)}
∫ 1

0

|θt1 + (1− θ)t2|m−2dθ(t1 − t2)2

≥ cεm−p exp{(m− p) dist(εx,M)}|t1 − t2|m.

For 1 < p < 2, by (2.3) we have

(kε(x, t1)− kε(x, t2))(t1 − t2)

≥ cεm−p exp{(m− p) dist(εx,M)}
∫ 1

0

|θt1 + (1− θ)t2|m−2dθ(t1 − t2)2

≥ cεm−p exp{(m− p) dist(εx,M)}|t1 − t2|2(|t1|2−m + |t2|2−m)−1.

(8) For p ≥ 2, by (2.2) we have

|kε(x, t1)− kε(x, t2)|

=
∣∣∣ ∫ 1

0

∂kε(x, θt1 + (1− θ)t2)

∂t
dθ(t1 − t2)

∣∣∣
≤ c

∫ 1

0

( θt1 + (1− θ)t2
mε(x, θt1 + (1− θ)t2)

)m−p
|θt1 + (1− θ)t2)|p−2dθ|t1 − t2|

≤ c
∫ 1

0

(1 + εm−p exp{(m− p) dist(εx,M)}|θt1 + (1− θ)t2|m−p)

× |θt1 + (1− θ)t2|p−2dθ|t1 − t2|

≤ c
∫ 1

0

|θt1 + (1− θ)t2|p−2dθ|t1 − t2|+ cεm−p exp{(m− p) dist(εx,M)}

×
∫ 1

0

|θt1 + (1− θ)t2|m−2dθ|t1 − t2|

≤ c
(
|t1|p−2 + |t1|p−2 + εm−p exp{(m− p) dist(εx,M)}

× (|t1|m−2 + |t2|m−2)
)
|t1 − t2|.

For 1 < p < 2, by (2.4) we have

|kε(x, t1)− kε(x, t2)|

≤ c
∫ 1

0

|θt1 + (1− θ)t2|p−2dθ|t1 − t2|+ cεm−p exp{(m− p) dist(εx,M)}

×
∫ 1

0

|θt1 + (1− θ)t2|m−2dθ|t1 − t2|

≤ c(|t1 − t2|p−1 + εm−p exp{(m− p) dist(εx,M)}|t1 − t2|m−1).

�

Lemma 2.3. 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 easy, we omit it.

Lemma 2.4. The imbedding Xε ↪→ Lr(RN )) (1 ≤ r < p∗) is compact.



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 7

Proof. Let {un} ⊂ Xε be bounded, then un ⇀ u in Xε, up to a subsequence if
necessary, un → u in Lrloc(RN )(1 ≤ r < p∗). We first prove un → u in L1(RN ). For
R > 0 we have∫

RN\BR(0)

|u| dx

≤
(∫

RN\BR(0)

e(m−p)dist(εx,M)|u|m dx
)1/m(∫

RN\BR(0)

e−
m−p
m−1 dist(εx,M) dx

)m−1
m

≤ ‖u‖Lmε (RN )

(∫
RN\BR(0)

e−
m−p
m−1 dist(εx,M) dx

)m−1
m

= oR(1).

Hence∫
RN
|un − u| dx =

∫
BR(0)

|un − u| dx+

∫
RN\BR(0)

|un − u| dx = on(1) + oR(1)

as n→∞. For 1 < r < p∗, we have∫
RN
|un − u|r dx

=

∫
RN
|un − u|rθ+(1−θ)r dx

≤
(∫

RN
|un − u|rθ

1
rθ dx

)rθ(∫
RN
|un − u|(1−θ)r

p∗
(1−θ)r dx

) (1−θ)r
p∗

≤ c
(∫

RN
|un − u| dx

)rθ
,

where 0 < θ < 1, 1
r = θ + 1−θ

p∗ . �

Lemma 2.5. Let {un} ⊂ Xε be a Palais-Smale sequence of the functional Γε,λ,
then {un} is bounded in Xε.

Proof. Since

〈DΓε,λ(u), v〉

=

∫
RN

(|∇u|p−2∇u∇v + E(εx)|u|p−2uv) dx+ σ

∫
RN

kε(x, u)v dx

+
(∫

RN
χε(x)|u|p dx− 1

)β−1

+

∫
RN

χε(x)|u|p−2uv dx

− 1

2
hλ(ϕ1/2(u))

∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)v(x)

|x− y|α
dx dy,

(2.5)

for any v ∈ Xε. By Lemma 2.2, we have

Γε,λ(u)− 1

q
〈DΓε,λ(u), u〉

=
(1

p
− 1

q

) ∫
RN

(|∇u|p + E(εx)|u|p) dx+ σ

∫
RN

Kε(x, u) dx

− σ

q

∫
RN

kε(x, u)u dx+
1

pβ

(∫
RN

χε(x)|u|2 dx− 1
)β

+



8 B. ZHANG, X. LIU EJDE-2022/11

− 1

q

(∫
RN

χε(x)|u|p dx− 1
)β−1

+

∫
RN

χε(x)|u|p dx

+
1

2q
hλ
(
ϕ1/2(u)

)
ϕ(u)− 1

2q
gλ
(
ϕ1/2(u)

)
ϕ(u)

≥
(1

p
− 1

q

) ∫
RN

(|∇u|p + E(εx)|u|p) dx+ σ
( 1

m
− 1

q

) ∫
RN

kε(x, u)u dx

+ c
(∫

RN
χε(x)|u|p dx− 1

)β
+
− c

≥ c
(
‖u‖p

W 1,p(RN )
+ ‖u‖mLmε (RN )

)
+ c
(∫

RN
χε(x)|u|p dx− 1

)β
+
− c.

Hence the Palais-Smale sequence {un} is bounded in Xε. �

Lemma 2.6. For every ε > 0, Γε,λ satisfies the Palais-Smale condition.

Proof. Let {un} ⊂ Xε be a Palais-Smale sequence of the functional Γε,λ. By
Lemma 2.5, {un} is bounded in Xε. By Lemma 2.4, we can assume un → u in
Lr(RN ), 1 ≤ r < p∗. By Lemma 2.1 and Lemma 2.4, we obtain

o(1) =〈DΓε,λ(uk)−DΓε,λ(ul), uk − ul〉

=

∫
RN

(|∇uk|p−2∇uk − |∇ul|p−2∇ul,∇uk −∇ul) dx

+

∫
RN

E(εx)(|uk|p−2uk − |ul|p−2ul)(uk − ul) dx

+ σ

∫
RN

(kε(x, uk)− kε(x, ul))(uk − ul) dx

+
(∫

RN
χε(x)|uk|p dx− 1

)β−1

+

∫
RN

χε(x)|uk|p−2uk(uk − ul)dx

+
(∫

RN
χε(x)|ul|p dx− 1

)β−1

+

∫
RN

χε(x)|ul|p−2ul(uk − ul)dx

− 1

2
hλ(ϕ(uk))

∫
RN

∫
RN

|uk(y)|q|uk(x)|q−2uk(x)(uk(x)− ul(x))

|x− y|α
dx dy

+
1

2
hλ(ϕ(ul))

∫
RN

∫
RN

|ul(y)|q|ul(x)|q−2ul(x)(uk(x)− ul(x))

|x− y|α
dx dy

≥
∫
RN

(|∇uk|p−2∇uk − |∇ul|p−2∇ul)∇(uk − ul) dx

+

∫
RN

E(εx)(|uk|p−2uk − |ul|p−2ul)(uk − ul) dx

+ σ

∫
RN

(kε(x, uk)− kε(x, ul))(uk − ul) dx+ o(1).

So ∫
RN

(|∇uk|p−2∇uk − |∇ul|p−2∇ul,∇uk −∇ul) dx→ 0, as k, l→∞,∫
RN

(|uk|p−2uk − |ul|p−2ul)(uk − ul) dx→ 0, as k, l→∞,∫
RN

(kε(x, uk)− kε(x, ul))(uk − ul) dx→ 0, as k, l→∞.



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 9

For p ≥ 2, by (2.1) we have∫
RN
|∇(uk − ul)|p dx

≤ c
∫
RN

(|∇uk|p−2∇uk − |∇ul|p−2∇ul,∇uk −∇ul) dx→ 0, as k, l→∞,∫
RN
|uk − ul|p dx

≤ c
∫
RN

(|uk|p−2uk − |ul|p−2ul)(uk − ul) dx→ 0, as k, l→∞,∫
RN

exp{(m− p) dist(εx,M)}|uk − ul|m dx

≤ c
∫
RN

(kε(x, uk)− kε(x, ul))(uk − ul) dx→ 0, as k, l→∞.

For 1 < p < 2, by (2.3) we have∫
RN
|∇(uk − ul)|p dx

≤ c
(∫

RN
(|∇uk|p−2∇uk − |∇ul|p−2∇ul,∇uk −∇ul) dx

)p/2
×
(∫

RN
(|∇uk|p + |∇ul|p) dx

) 2−p
2

≤ c
(∫

RN
(|∇uk|p−2∇uk − |∇ul|p−2∇ul,∇uk −∇ul) dx

)p/2
→ 0, as k, l→∞,

∫
RN

E(εx)|uk − ul|p dx

≤ c
(∫

RN
(|uk|p−2uk − |ul|p−2ul)(uk − ul) dx

)p/2
×
(∫

RN
(|uk|p + |ul|p) dx

) 2−p
2

≤ c
(∫

RN
(|uk|p−2uk − |ul|p−2ul)(uk − ul) dx

)p/2
→ 0, as k, l→∞

and ∫
RN

exp{(m− p) dist(εx,M)}|(uk − ul)|m dx

≤ c
(∫

RN
(kε(x, uk)− kε(x, ul))(uk − ul) dx

)m/2
×
(∫

RN
εm−p exp{(m− p) dist(εx,M)}(|uk|m + |ul|m) dx

) 2−m
2

≤ c
( ∫

RN
(kε(x, uk)− kε(x, ul))(uk − ul) dx

)m/2 → 0, as k, l→∞.

So {un} is a Cauchy sequence in Xε. �



10 B. ZHANG, X. LIU EJDE-2022/11

3. Existence of solutions

In this section we construct a sequence of critical points of the functional Γε,λ
by using the method of invariant sets of 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,λ. In order to obtain multiple sign-changing critical
points of Γε,λ, we introduce the abstract critical point theorem [14, Theorem 2.5],
see also [6, Theorem 3.2].

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

W = P ∪Q, Σ = ∂P ∩ ∂Q.

Assume

(A3) f satisfies the (PS) condition.
(A4) c∗ = infx∈Σ f(x) > 0.

Assume there exists an odd continuous map A : X → X satisfying

(A5) Given 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 .

(A6) A(∂P ) ⊂ P , 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,

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

}
.

Assume

(A7) 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. Assume (A3)–(A7) 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.

Lemma 3.2. For any v ∈ Ls(RN ), s ∈ (1, N
N−α ),

∫
RN

v(y)
|x−y|α dy ∈ L

Ns
N−Ns+αs (RN ).

Moreover(∫
RN

∣∣ ∫
RN

v(y)

|x− y|α
dy
∣∣ Ns
N−Ns+αs dx

)N−Ns+αs
Ns

dx ≤ c(s,N, α)‖v‖Ls(RN ).

By Hardy-Littlewood-Sobolev inequality, the proof of Lemma 3.2 is straightfor-
ward, we omit it.



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 11

Lemma 3.3 ([31, Theorem 4.2.7]). Let Ω ⊆ RN be a domain and let {un} be
bounded in Lq(Ω) for some q > 1, If un(x) → u(x) a.e. in Ω as n → ∞, then
un ⇀ u in Lq(Ω) as n→∞.

Lemma 3.4. If {un} is bounded in W 1,p(RN ), un ⇀ u in W 1,p(RN ) and un(x)→
u(x) a.e. in RN , then for p < q < p∗α,

(1)
∫
RN
∣∣|un|q − |un − u|q − |u|q∣∣ 2N

2N−α dy → 0 as n→∞.

(2)
∫
RN
∣∣|un|q + |un − u|q − |u|q

∣∣ 2N
2N−α dy → 0 as n→∞.

(3)
∫
RN
∣∣|un|q−2un − |un − u|q−2(un − u) − |u|q−2u

∣∣ 2Np
2Np−αp−2N+2p dx → 0 as

n→∞.

The proof of the above lemma is similar to that of [33, Theorem 2.5], we omit
it.

Lemma 3.5 ([7, Theorem 2.6]). Let α ∈ (0, N), s ∈ (1, N
N−α ) and let {un} ⊂

L1(RN )
⋂
Ls(RN ) be bounded and such that, up to a subsequence, for any bounded

domain Ω ⊂ RN , un → 0 in Ls(RN ) as n → ∞. Then, up to a subsequence if

necessary,
∫
RN

un(y)
|x−y|α dy → 0 a.e. in RN as n→∞.

Lemma 3.6. Let {un} ⊂ W 1,p(RN ) be such that un ⇀ u in W 1,p(RN ) and
un(x)→ u(x) a.e. in RN as n→∞. Then, up to a subsequence if necessary,∫

RN

∫
RN

|un(y)|q|un(x)|q

|x− y|α
dx dy

=

∫
RN

∫
RN

|un(y)− u(y)|q|un(x)− u(x)|q

|x− y|α
dx dy

+

∫
RN

∫
RN

|u(y)|q|u(x)|q

|x− y|α
dx dy + on(1).

(3.1)

∫
RN

∫
RN

|un(y)|q|un(x)|q−2un(x)φ(x)

|x− y|α
dx dy

=

∫
RN

∫
RN

|un(y)− u(y)|q|un(x)− u(x)|q−2(un(x)− u(x))φ(x)

|x− y|α
dx dy

+

∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)φ(x)

|x− y|α
dx dy + on(1)‖φ‖W 1,p(RN ),

(3.2)

where 0 < α < min{2p,N}, p < q < p∗α, φ(x) ∈ C∞0 (RN ) and on(1) → 0 as
n→∞.

From Lemmas 3.2-3.5, we can prove this lemma according to [7, Lemma 2.2,
Lemma 2.4], we omit it. We define

Jε(u) =
1

p

∫
RN

(|∇u|p + E(εx)|u|p) dx+ σ

∫
RN

Kε(x, u) dx.

Definition 3.7. Given u ∈ Xε, define v = Au by the equation

1

2
〈DJε(u) +DJε(v)−DJε(u− v), η〉

+
(∫

RN
χε(x)|u|p dx− 1

)β−1

+

∫
RN

χε(x)|v|p−2vη dx

=
1

2
hλ
(
ϕ1/2(u)

) ∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)η(x)

|x− y|α
dx dy, η ∈ Xε.

(3.3)



12 B. ZHANG, X. LIU EJDE-2022/11

Lemma 3.8. If ‖u‖Xε is bounded, then ‖Au‖Xε is bounded.

Proof. By (3.3)

‖Au‖p
W 1,p(RN )

+ ‖Au‖mLmε (RN )

≤ c〈DJε(v), v〉

≤ chλ(ϕ1/2(u))

∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)v(x)

|x− y|α
dx dy

≤ c‖u‖2q−1
W 1,p(RN )

‖Au‖W 1,p(RN )

≤ c‖Au‖W 1,p(RN ),

so ‖Au‖Xε is bounded. �

Lemma 3.9. For u, v ∈ Xε, the following holds: (1) For p ≥ 2,

〈DJε(u)−DJε(v), φ〉

≤ c
(
‖u‖p−2

W 1,p(RN )
+ ‖v‖p−2

W 1,p(RN )

)
‖u− v‖W 1,p(RN )‖φ‖W 1,p(RN )

+ c
(
‖u‖m−2

Lmε (RN )
+ ‖v‖m−2

Lmε (RN )

)
‖u− v‖Lmε (RN )‖φ‖Lmε (RN ).

(2) For 1 < p < 2,

〈DJε(u)−DJε(v), φ〉

≤ c
(
‖u− v‖p−1

W 1,p(RN )
‖φ‖W 1,p(RN ) + ‖u− v‖m−1

Lmε (RN )
‖φ‖Lmε (RN )

)
.

(3) For p > 1,

〈DJε(u− v), φ〉 ≤ c
(
‖u− v‖p−1

W 1,p(RN )
‖φ‖W 1,p(RN ) + ‖u− v‖m−1

Lmε (RN )
‖φ‖Lmε (RN )

)
.

Proof. We only verify (1). By (2.2) and the Hölder inequality, for p ≥ 2, we have

〈DJε(u)−DJε(v), φ〉

≤ c
∫
RN

(|∇u|p−2∇u− |∇v|p−2∇v,∇φ) dx+ c

∫
RN

(|u|p−2u− |v|p−2v)φdx

+ c

∫
RN

(kε(x, u)− kε(x, v))φdx

≤ c
∫
RN

(|∇u|p−2 + |∇v|p−2)|∇(u− v)||∇φ| dx

+ c

∫
RN

(|u|p−2 + |v|p−2)|u− v||φ| dx

+ c

∫
RN

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

≤ c
(
‖u‖p−2

W 1,p(RN )
+ ‖v‖p−2

W 1,p(RN )

)
‖u− v‖W 1,p(RN )‖φ‖W 1,p(RN )

+ c
(
‖u‖m−2

Lmε (RN )
+ ‖v‖m−2

Lmε (RN )

)
‖u− v‖Lmε (RN )‖φ‖Lmε (RN ).

�

Lemma 3.10. A is odd, well defined, and continuous on Xε.

Proof. It is easy to see that A is odd. We define

G(v) =
1

2
〈DJε(u), v〉+

1

2
Jε(v) +

1

2
Jε(u− v)



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 13

+
1

p

(∫
RN

χε(x)|u|p dx− 1
)β−1

+

∫
RN

χε(x)|v|p dx

− 1

2
hλ
(
ϕ1/2(u)

) ∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)v(x)

|x− y|α
dx dy, v ∈ Xε.

Equation (3.3) has a unique solution v = Au, which can be obtained by solving the
minimization problem inf{G(v) : v ∈ Xε}. Since

G(v) ≥ 1

2
〈DJε(u), v〉+

1

2
Jε(v)− cλ ≥ c1‖v‖pW 1,p(RN )

− c2‖v‖W 1,p(RN ) − cλ,

G is coercive.
Let {vn} ⊂ Xε be a minimizing sequence for the functional G, vn ⇀ v in Xε.

By the lower semicontinuity

G(v) = lim inf
n→∞

G(vn) = inf{G(v)
∣∣v ∈ Xε},

so v is a solution of (3.3). Assume v1, v2 are solutions of (3.3), then taking (v1−v2)
as the test function, we have

1

2
〈DJε(v1)−DJε(v2), v1 − v2〉 −

1

2
〈DJε(u− v1)−DJε(u− v2), v1 − v2〉

+
(∫

RN
χε(x)|u|p dx− 1

)β−1

+

∫
RN

χε(x)(|v1|p−2v1 − |v2|p−2v2)(v1 − v2) dx = 0.

Hence

〈DJε(v1)−DJε(v2), v1 − v2〉 = 0.

For p ≥ 2, by (2.1) we have

〈DJε(v1)−DJε(v2), v1 − v2〉 ≥ c(‖v1 − v2‖pW 1,p(RN )
+ ‖v1 − v2‖mLmε (RN )).

For 1 < p < 2, by (2.3) we have

〈DJε(v1)−DJε(v2), v1 − v2〉

≥ c
(∫

RN
|∇(v1 − v2)|p dx

)2/p(∫
RN

(|∇v1|p + |∇v2|p) dx
) p−2

p

+ c
(∫

RN
|v1 − v2|p dx

)2/p(∫
RN

(|v1|p + |v2|p) dx
) p−2

p

+ c
(∫

RN
exp{(m− p) dist(εx,M)}|v1 − v2|m dx

) 2
m

×
(∫

RN
exp{(m− p) dist(εx,M)}(|v1|m + |v2|m) dx

)m−2
m

≥ c‖v1 − v2‖2W 1,p(RN )

(
‖v1‖2−pW 1,p(RN )

+ ‖v2‖2−pW 1,p(RN )

)−1

+ ‖v1 − v2‖2Lmε (RN )

(
‖v1‖2−mLmε (RN )

+ ‖v2‖2−mLmε (RN )

)−1
.

Then we have v1 = v2 in Xε. So (3.3) has a unique solution v = Au. Denoting

ψε(u) =
(∫

RN
χε(x)|u|p dx− 1

)β−1

+
,



14 B. ZHANG, X. LIU EJDE-2022/11

and taking η = vn − v in (3.3), we have

1

2
〈DJε(u− v)−DJε(un − vn), (u− v)− (un − vn)〉

+
1

2
〈DJ

ε
(vn)−DJε(v), vn − v〉

+ ψε(un)

∫
RN

χε(x)(|vn|p−2vn − |v|p−2v)(vn − v) dx

=
1

2
〈DJε(u− v)−DJε(un − vn), u− un〉+

1

2
〈DJε(un)−DJε(u), v − vn〉

+
(
ψε(un)− ψε(u)

) ∫
RN

χε(x)|v|p−2v(v − vn) dx

+
1

2
hλ(ϕ1/2(un))t

∫
RN

∫
RN

((
|un(y)|q|un(x)|q−2un(x)− |u(y)|q|u(x)|q−2u(x)

)
×
(
vn(x)− v(x)

))
/|x− y|α dx dy

+
1

2
(hλ(ϕ1/2(un))− hλ(ϕ1/2(u)))

×
∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)(vn(x)− v(x))

|x− y|α
dx dy.

(3.4)
Now we estimate the right-hand side of (3.4). By Lemma 3.8 and Lemma 3.9,
suppose un → u in Xε, for p ≥ 2, we have

〈DJε(u− v)−DJε(un − vn), u− un〉+ 〈DJε(un)−DJε(u), v − vn〉

≤ c
(
‖u− v‖p−2

W 1,p(RN )
+ ‖un − vn‖p−2

W 1,p(RN )

)
‖u− v − un + vn‖W 1,p(RN )

× ‖u− un‖W 1,p(RN )

+ c
(
‖u− v‖m−2

Lmε (RN )
+ ‖un − vn‖m−2

Lmε (RN )

)
‖u− v − un + vn‖Lmε (RN )

× ‖u− un‖Lmε (RN )

+ c
(
‖un‖p−2

W 1,p(RN )
+ ‖u‖p−2

W 1,p(RN )

)
‖un − u‖W 1,p(RN )‖v − vn‖W 1,p(RN )

+ c
(
‖un‖m−2

Lmε (RN )
+ ‖u‖m−2

Lmε (RN )

)
‖un − u‖Lmε (RN )‖v − vn‖Lmε (RN )

≤ c‖u− un‖Xε = on(1).

(3.5)

For 1 < p < 2, we have

〈DJε(u− v)−DJε(un − vn), u− un〉+ 〈DJε(un)−DJε(u), v − vn〉

≤ c
(
‖u− v − un + vn‖p−1

W 1,p(RN )
+ ‖u− v − un + vn‖m−1

Lmε (RN )

)
‖u− un‖Xε

+ c
(
‖un − u‖p−1

W 1,p(RN )
+ ‖un − u‖m−1

Lmε (RN )

)
‖v − vn‖Xε = on(1).

(3.6)

By Lemmas 2.1and 3.8 and the continuity of ψε(u) and hλ
(
ϕ1/2(u)

)
for u, we have(

ψε(un)− ψε(u)
) ∫

RN
χε(x)|v|p−2v(v − vn) dx

+
1

2

(
hλ(ϕ1/2(un))− hλ(ϕ1/2(u))

)
×
∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)(vn(x)− v(x))

|x− y|α
dx dy = on(1).

(3.7)



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 15

By Lemmas 2.1,3.6 and 3.8, we have

1

2
hλ
(
ϕ1/2(un)

) ∫
RN

∫
RN

((
|un(y)|q|un(x)|q−2un(x)

− |u(y)|q|u(x)|q−2u(x)
)(
vn(x)− v(x)

))
/|x− y|α dx dy

=
1

2
hλ(ϕ1/2(un))

∫
RN

∫
RN

(
|un(y)− u(y)|q|un(x)− u(x)|q−2

× (un(x)− u(x))(vn(x)− v(x))
)
/|x− y|α dx dy + on(1) = on(1).

(3.8)

So the right-hand side of (3.4) satisfies

RHS = on(1). (3.9)

Next, we estimate the left-hand side of (3.4), for p ≥ 2, by (2.1),

LHS ≥ 〈DJε(vn)−DJε(v), vn − v〉
≥ c
(
‖vn − v‖pW 1,p(RN )

+ ‖vn − v‖mLmε (RN )

)
.

(3.10)

For 1 < p < 2, by (2.3),

LHS ≥〈DJε(vn)−DJε(v), vn − v〉

≥c‖vn − v‖2W 1,p(RN )

(
‖vn‖2−pW 1,p(RN )

+ ‖v‖2−p
W 1,p(RN )

)−1

+ ‖vn − v‖2Lmε (RN )

(
‖vn‖2−mLmε (RN )

+ ‖v‖2−m
Lmε (RN )

)−1
.

(3.11)

By (3.9)-(3.11), for p > 1, we obtain ‖vn − v‖Xε → 0 as n→∞. �

Lemma 3.11. Let u ∈ Xε, v = A(u), then the following holds:

(1) 〈DΓε,λ(u), u− v〉 ≥ c(‖u− v‖p
W 1,p(RN )

+ ‖u− v‖mLmε (RN ));

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

Proof. (1) We denote v = Au. For η ∈ Xε, we have

〈DΓε,λ(u), η〉

=
1

2
〈DJε(u)−DJε(v), η〉+

1

2
〈DJε(u− v), η〉

+
(∫

RN
χε(x)|u|p dx− 1

)β−1

+

∫
RN

χε(x)(|u|p−2u− |v|p−2v)η dx .

(3.12)

Hence

〈DΓε,λ(u), u− v〉

=
1

2
〈DJε(u)−DJε(v), u− v〉+

1

2
〈DJε(u− v), u− v〉

+
(∫

RN
χε(x)|u|p dx− 1

)β−1

+

∫
RN

χε(x)(|u|p−2u− |v|p−2v)(u− v) dx

≥ c
(
‖u− v‖p

W 1,p(RN )
+ ‖u− v‖mLmε (RN )

)
.



16 B. ZHANG, X. LIU EJDE-2022/11

(2) By (3.3) and Lemma 3.9, we have

Γε,λ(u)− 1

2q
〈DJε(u)−DJε(v), u〉

=
(1

p
− 1

q

) ∫
RN

(|∇u|p + E(εx)|u|p) dx+ σ

∫
RN

Kε(x, u) dx

− σ

q

∫
RN

kε(x, u)u dx+
1

pβ

(∫
RN

χε(x)|u|p dx− 1
)β

+

− 1

q

(∫
RN

χε(x)|u|p dx− 1
)β−1

+

∫
RN

χε(x)|v|p−2vu dx

+
1

2q
bλ
(
ϕ1/2(u)

)
ϕ(u) +

1

2q
〈DJε(u− v), u〉

≥ c
(
‖u‖p

W 1,p(RN )
+ ‖u‖mLmε (RN )

)
+ c
(∫

RN
χε(x)|u|p dx− 1

)β
+

− c
(∫

RN
χε(x)|u− v|p dx

)β
− c
(
‖u‖p−2

W 1,p(RN )
+ ‖v‖p−2

W 1,p(RN )

)
‖u− v‖W 1,p(RN )‖u‖W 1,p(RN )

− c
(
‖u‖m−2

Lmε (RN )
+ ‖v‖m−2

Lmε (RN )

)
‖u− v‖Lmε (RN )‖u‖Lmε (RN ) − c,

(3.13)

where we have used the estimate
1

pβ

(∫
RN

χε(x)|u|p dx− 1
)β

+
− 1

q

(∫
RN

χε(x)|u|p dx− 1
)β−1

+

∫
RN

χε(x)|v|p−2vu dx

=
1

pβ

(∫
RN

χε(x)|u|p dx− 1
)β

+
− 1

q

(∫
RN

χε(x)|u|p dx− 1
)β−1

+

∫
RN

χε(x)|u|p dx

+
1

q

(∫
RN

χε(x)|u|p dx− 1
)β−1

+

∫
RN

χε(x)(|u|p−2u− |v|p−2v)u dx

≥ c
(∫

RN
χε(x)|u|p dx− 1

)β
+
− c
(∫

RN
χε(x)(|u|p−2 + |v|p−2)|u− v||u| dx

)β
− c

≥ c
(∫

RN
χε(x)|u|p dx− 1

)β
+
− c
(∫

RN
χε(x)|u− v|p dx

)β
− c.

On the other hand, by Lemma 3.9,

Γε,λ(u)− 1

2q
〈DJε(u)−DJε(v), u〉

≤ |Γε,λ(u)|+ c
(
‖u‖p−2

W 1,p(RN )
+ ‖v‖p−2

W 1,p(RN )

)
‖u− v‖W 1,p(RN )‖u‖W 1,p(RN )

+ c
(
‖u‖m−2

Lmε (RN )
+ ‖v‖m−2

Lmε (RN )

)
‖u− v‖Lmε (RN )‖u‖Lmε (RN ).

(3.14)

By (3.13), (3.14) and the Young’s inequality, we have

‖u‖p
W 1,p(RN )

+ ‖u‖mLmε (RN ) +
(∫

RN
χε(x)|u|p dx− 1

)β
+

≤ c(1 + |Γε,λ(u)|+ ‖u− v‖p
W 1,p(RN )

+ ‖u− v‖mLmε (RN ) + ‖u− v‖pβ
W 1,p(RN )

).

(3.15)

By Lemma 3.9, (3.12), (3.15) and Young’s inequality, we have

‖DΓε,λ(u)‖

≤ c
((∫

RN
χε(x)|u|p dx− 1

)β−1

+
+ 1
)

(‖u‖p−2
W 1,p(RN )

+ ‖v‖p−2
W 1,p(RN )

)‖u− v‖Xε



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 17

+ c
(
‖u‖m−2

Lmε (RN )
+ ‖v‖m−2

Lmε (RN )

)
‖u− v‖Xε

≤ c
(
1 + |Γε,λ(u)|+ ‖u− v‖p

W 1,p(RN )
+ ‖u− v‖mLmε (RN ) + ‖u− v‖pβ

W 1,p(RN )

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

�

Corollary 3.12. For all b0, c0 > 0, there exists b = b(b0, c0) > 0 such that if
|Γε,λ(u)| ≤ c0 and ‖DΓε,λ(u)‖ ≥ b0, then u−Au 6= 0 and

〈DΓε,λ(u), u−Au〉 ≥ b‖u−Au‖Xε > 0.

Now we define the convex open sets

P = {u
∣∣u ∈ Xε, ‖u−‖W 1,p(RN ) < δ},

Q = {u
∣∣u ∈ Xε, ‖u+‖W 1,p(RN ) < δ},

where δ is a positive constant, u− = min{u, 0} and u+ = max{u, 0}. We denote

D(f, g) =

∫
RN

∫
RN

f(x)g(y)

|x− y|α
dx dy.

Lemma 3.13 ([13, 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 only when f = cg for some constant c.

Lemma 3.14. There exists δλ > 0 such that for 0 < δ < δλ,

A(∂P ) ⊂ P, A(∂Q) ⊂ Q.

Proof. We only prove A(∂Q) ⊂ Q. For u ∈ ∂Q, let v = Au. By Lemma s2.3 and
3.13, we have

‖v+‖pW 1,p(RN )

≤ c〈Jε(v), v+〉

≤ chλ
(
ϕ1/2(u)

) ∫
RN

∫
RN

|(u(y)|q|u(x)|q−2u(x)v+(x)

|x− y|α
dx dy

≤ chλ
(
ϕ1/2(u)

) ∫
RN

∫
RN

|(u(y)|q|u+(x)|q−2u+(x)v+(x)

|x− y|α
dx dy

≤ chλ
(
ϕ1/2(u)

)
ϕ1/2(u)

(∫
RN

∫
RN

|u+(x)|q−1|v+(x)||u+(y)|q−1|v+(y)|
|x− y|α

dx dy
)1/2

≤ cλ‖u+‖q−1
W 1,p(RN )

‖v+‖W 1,p(RN ),

taking δλ ≤ c
− 1
q−p

λ the conclusion follows. �

Lemma 3.15. There exist δ0 > 0 and c∗ = c∗(δ), such that for any 0 < δ < δ0,

Γε,λ(u) ≥ c∗ > 0 for all u ∈ ∂P ∩ ∂Q.



18 B. ZHANG, X. LIU EJDE-2022/11

Proof. For u ∈ ∂P ∩ ∂Q, we have

Γε,λ(u) ≥ 1

p

∫
RN

(|∇u|p + E(εx)|u|p) dx− 1

2q
gλ(ϕ(u))

∫
RN

∫
RN

|u(x)|q|u(y)|q

|x− y|α
dx dy

≥ c1‖u‖pW 1,p(RN )
− c2‖u‖2q

L
2Nq

2N−α (RN )

≥ c1p‖u‖pW 1,p(RN )
− c2δ2q−p‖u‖p

W 1,p(RN )

≥ c1
2
‖u‖p

W 1,p(RN )
≥ c1

2
δp := c∗,

where δ0 = ( c12c2
)

1
2q−p . �

Assume B = {x ∈ RN
∣∣|x| ≤ R} ⊂ M. Let {en}∞n=1 be a family of linearly

independent functions in C∞0 (B). There exists an increasing sequence Rn such
that

J0(u) < 0, ∀u ∈ Hn, ‖u‖Xε ≥ Rn
where Hn := span{e1, . . . , en} and

J0(u) =
1

p

∫
RN

(|∇u|p + b|u|p) dx+ σ

∫
RN

e(m−p)|x||u|m dx− 1

2q
gλ(ϕ1/2(u))ϕ(u).

We define ϕn ∈ C(Bn, C
∞
0 (B)) as

ϕn(t) = Rn

n∑
i=1

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

where Rn is also chosen such that ‖ϕn(t)‖Xε ≥ Rn for t ∈ ∂Bn. Let

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

Lemma 3.16. The set Γj is nonempty.

For the proof of the above lemma, we refer to [15, Lemma 5.6].

Theorem 3.17. Assume that conditions (A1) and (A2) hold. Then there exist
0 < ε < 1 and 0 < λ < 1, such that if 0 < ε < ε, 0 < λ < λ, then the functional
Γε,λ has infinitely many sign-changing critical points, the corresponding critical
values are

cj(ε, λ) = inf
E∈Γj

sup
u∈E\W

Γε,λ(u), j = 1, 2, . . . . (3.16)

Moreover

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

cj(ε, λ) ≤ mj , j = 1, 2, . . . . (3.17)

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

Proof. All the assumptions of Theorem 3.1 are satisfied, so we only need to prove
estimate (3.17). Since Ej = ϕj+1(Bj+1) ∈ Γj , for t ∈ Bj+1, u = ϕj+1(t), there

exist 0 < ε < 1 and 0 < λ < 1, such that
( ∫

RN χε(x)|u|pdx − 1
)β

+
= 0, and

Γε,λ(u) ≤ J0(u), for u ∈ ϕj+1(Bj+1), if 0 < ε < ε, 0 < λ < λ. Hence

cj(ε, λ) ≤ mj := sup
u∈Ej

J0(u) .

�



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 19

4. Proof of Theorem 1.1

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

‖u‖W 1,p(RN ) ≤ H.
(2) Assume Γε(u) ≤ L, DΓε(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 .

The proof of the above theorem is given in Section 5.

Corollary 4.2. (1) Assume Γε,λ(u) ≤ L, DΓε,λ(u) = 0. Then there exists λ =

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

Γε(u) = Iε(u) and DIε(u) = 0 if 0 < ε ≤ ε.

Proof. (1) By Theorem 4.1 (1), if 0 < λ < λ(L) = 1
C0Hq

, then

‖u‖W 1,p(RN ) ≤ (
1

C0λ
)1/q.

It follows that Γε,λ(u) = Γε(u) and DΓε(u) = 0.
(2) By Theorem 4.1 (2), there exist constants µ, 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 . Let ε = ε(L) ≤ min{µ, 1

c}, then for 0 < ε ≤ ε,

|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). 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|,

hence

|u(x)| ≤ c exp{−cdist(x, (Aδ)ε)}

≤ c exp{−c
l
|x|}, for x /∈Mε.

We have ∫
RN

χε(x)|u|p dx ≤ cε−p
∫
|x|≥cε−1

exp{−c
l
|x|} dx

≤ cε−N−p+1 exp{− c
ε
} < 1

and (∫
RN

χε(x)|u|p dx− 1
)

+
= 0



20 B. ZHANG, X. LIU EJDE-2022/11

for 0 < ε < ε(δ) sufficiently small. It follows that Γε(u) = Iε(u) and DIε(u) =
0. �

The proof of Theorem 1.1. For each positive integer k, by Theorem 3.17, there exist
0 < ε̃ < 1 and 0 < λ̃ < 1, such that if 0 < ε < ε̃ and 0 < λ < λ̃, the functional Γε,λ
has k pairs of sign-changing critical points ±uj , j = 1, . . . , k. The corresponding
critical values satisfy

0 < c1(ε, λ) ≤ . . . ≤ ck(ε, λ) ≤ mk.

By Corollary 4.2 (2), there exists εk = εk(mk), such that if

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

then

Γε(u) = Iε(u), DIε(u) = 0.

Fixed ε ∈ (0, ε̃k). By Corollary 4.2 (1), there exists λk = λk(mk), such that if

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

then

Γε,λ(u) = Γε(u), DΓε(u) = 0.

Now for 0 < ε < ε̃k, 0 < λ < λ̃k, uj,ε = uj(ε, 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)}, x ∈ RN ,

hence

|vj,ε| ≤ c exp{−µ
ε

dist(x,Aδ)}, x ∈ RN . �

5. Uniform bound

In this section we prove Theorem 4.1. It is easy to obtain part (1). So we only
prove part (2).

Lemma 5.1. Assume Γε(u) < L,DΓε(u) = 0. Then

(1) there exists cL, such that |u(x)| ≤ cL for x ∈ RN ;

(2) there exists d, such that
∫
RN
|u(y)|q
|x−y|α dy ≤ d for x ∈ RN ;

(3) for any δ > 0 there exists c = c(δ, L) such that |u(x)| ≤ cε for x ∈
RN\(Mε)

δ.

Proof. (1) It is easy to show that u is bounded in W 1,p(RN ) and (
∫
RN χε(x)|u|p dx−

1)β+ is bounded. Choose φ = |uT |p(k−1)u as the test function in 〈DΓε(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(u), φ〉 = 0, it is easy to obtain the inequality∫

RN
|∇u|p−2∇u∇φdx ≤ c

∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)φ(x)

|x− y|α
dx dy. (5.1)



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 21

First, let us estimate the right-hand side of the inequality (5.1). According to
Lemma 2.1 and Hölder’s inequality, we obtain∫

RN

∫
RN

|u(y)|q)|u(x)|q−2u(x)φ(x)

|x− y|α
dx dy

≤ c
(∫

RN
|u|

2Nq
2N−α dx

) (2N−α)(2q−p)
2Nq

(∫
RN

(|u||uT |k−1)
2Nq

2N−α dx
) p(2N−α)

2Nq

≤ c
(∫

RN
(|u||uT |k−1)

2Nq
2N−α dx

) p(2N−α)
2Nq

.

(5.2)

The left-hand side of (5.1) satisfies∫
RN
|∇u|p−2∇u∇φdx ≥

∫
RN
|∇u|p|uT |p(k−1) dx

≥ c

kp

∫
RN
|∇(|u||uT |k−1)|p dx

≥ c

kp
( ∫

RN
(|u||uT |k−1)p

∗
dx
) p
p∗ .

(5.3)

By (5.2) and (5.3), we have(∫
RN

(|u||uT |k−1)p
∗
dx
) p
p∗ ≤ ckp

(∫
RN

(|u||uT |k−1)
2Nq

2N−α dx
) p(2N−α)

2Nq

. (5.4)

Letting T →∞ in (5.4), we obtain(∫
RN
|u|p

∗k dx
) p
p∗ ≤ ckp

(∫
RN
|u|

2Nqk
2N−α dx

) p(2N−α)
2Nq

.

We write χ = p(2N−α)
2q(N−p) > 1. By using iteration, starting from k1 = p(2N−α)

2q(N−p) > 1,

we have (∫
RN
|u|p

∗χn dx
) 1
p∗χn ≤ (cχpn)

1
pχn

(∫
RN
|u|p

∗χn−1

dx
) 1

p∗χn−1

, (5.5)

for n = 1, 2, . . . . Hence by (5.5) we have

‖u‖L∞(RN ) ≤ c‖u‖Lp∗ (RN ) ≤ cL. (5.6)

(2) In view of 0 < α < N − 1, for x ∈ RN we have∫
RN

|u(y)|q

|x− y|α
dy ≤ c

(∫
|x−y|≥1

|u(y)|q

|x− y|α
dy +

∫
|x−y|<1

|u(y)|q

|x− y|α
dy
)

≤ c
(
‖u‖q

Lq(RN )
+

∫
|x−y|<1

1

|x− y|α
dx‖u‖q

L∞(RN )

)
≤ c
(
‖u‖q

Lq(RN )
+ ‖u‖q

L∞(RN )

)
≤ C.

(3) For x0 ∈ RN , 0 < ρ < R ≤ 1. Choose η ∈ C∞0 (RN , [0, 1]) such that
η(x) = 0 for x /∈ BR(x0); η(x) = 1 for x ∈ Bρ = Bρ(x0) and |∇η| ≤ c

R−ρ . Take

ϕ = u|u|p(k−1)ηp, p ≥ 1 as the test function in 〈DΓε,λ(u), ϕ〉 = 0, we have∫
RN
|∇u|p−2∇u∇ϕdx ≤ c

∫
RN

∫
RN

|u(y)|q|u(x)|q−2u(x)ϕ(x)

|x− y|α
dx dy. (5.7)



22 B. ZHANG, X. LIU EJDE-2022/11

The left-hand side of (5.7) satisfies

LHS

≥
∫
RN
|∇u|p|u|p(k−1)ηp dx+ p

∫
RN
|∇u|p−2∇u|u|p(k−1)uηp−1∇η dx

≥
∫
RN
|∇u|p|u|p(k−1)ηp dx− p

∫
RN
|∇u|p−1|u|(p−1)(k−1)|u|kηp−1|∇η| dx

≥ c
∫
RN
|∇u|p|u|p(k−1)ηp dx− c

∫
RN
|u|pk|∇η|p dx

≥ c

kp

∫
RN
|∇(|u|kη)|p dx− c

∫
RN
|u|pk|∇η|p dx

≥ c

kp
(

∫
Bρ

|u|p
∗k dx)

p
p∗ − c

(R− ρ)p

∫
BR

|u|pk dx.

(5.8)

By (5.6), the right-hand side of (5.7) satisfies

RHS ≤ c
∫
RN
|u|q−p|u|pkηp dx ≤ cL

∫
BR

|u|pk dx. (5.9)

By (5.8) and (5.9), we have(∫
Bρ

|u|p
∗k dx

)p/p∗
≤ cLk

p

(R− ρ)p

∫
BR

|u|pk dx for k ≥ 1.

By iteration we obtain

‖u‖L∞(B(x,R/2)) ≤ cL‖u‖Lp(B(x,R)). (5.10)

Since ∫
RN\(Mε)δ

|u|p dx ≤ cδεp,

by (5.10), we have |u(x)| ≤ c(δ, L)ε for x ∈ RN\(Mε)
δ. �

Let εn → 0, assume un ∈W 1,p(RN ), DΓεn(un) = 0, Γεn(un) ≤ L. Since {un} is
bounded in W 1,p(RN ), we have the following profile decomposition [29],

un =
∑
k∈Λ

Uk(· − yn,k) + rn, (5.11)

where Λ is an index set, yn,k ∈ RN ,

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

=
∑
k∈Λ ‖Uk‖

p
W 1,p(RN )

+ ‖rn‖pW 1,p(RN )
+ o(1) as n→∞ .

(4) ‖rn‖Ls(RN ) → 0 asn→∞, p < s < p∗, ‖un‖sLs(RN ) =
∑
k∈Λ ‖Uk‖sLs(RN ) +

o(1) as n→∞.

By Lemma 5.1 (3) we have

lim
n→∞

dist(yn,k,Mεn) < +∞.

We denote

y∗k = lim
n→∞

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. (5.12)



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 23

Lemma 5.2. Assume εn → 0, DΓεn(un) = 0, Γεn(un) ≤ L. Let ũn = un(·+yn) ⇀
U in W 1,p(RN ), yn ∈ RN , limn→∞ εnyn = y∗. Then Z = |U | satisfies∫

RN
|∇Z|p−2∇Z∇ϕdx+

∫
RN
|Z|p−1ϕdx

≤ c
∫
RN

∫
RN

|Z(y)|q|Z(x)|q−1ϕ(x)

|x− y|α
dx dy,

(5.13)

for ϕ ∈W 1,p(RN ), ϕ ≥ 0.

Proof. Let ϕ ∈ C∞0 (RN ). Selecting ϕn(x) = ϕ(x − yn) as a test function in
〈DΓεn(un), ϕn〉 = 0, we have∫

RN

(
|∇ũn|p−2∇ũn∇ϕ+ E(εn(x+ yn))|ũn|p−2ũnϕ

)
dx

+ σ

∫
RN

kεn(x+ yn, ũn)ϕdx+ ψεn(un)

∫
RN

χεn(x+ yn)|ũn|p−2ũnϕdx

=

∫
RN

∫
RN

|ũn(y)|q|ũn(x)|q−2ũn(x)ϕ(x)

|x− y|α
dx dy.

(5.14)

Let R > 0, such that ϕ(x) = 1 for |x| ≤ R and ϕ(x) = 0 for |x| ≥ 2R. The sequence
{ũn} converges in Lqloc(RN ), p < q < p∗. By Lemma 3.6, we have∫

RN
(|∇ũk|p−2∇ũk − |∇ũl|p−2∇ũl,∇ũk −∇ũl)ϕdx

≤ c
(∫

RN
(|∇ũk|p + |∇ũl|p) dx

) p−1
p ·

(∫
B2R(0)

|ũk − ũl|p dx
)1/p

+ c
(∫

RN
(|ũk|p + |ũl|p) dx

) p−1
p
(∫

B2R(0)

|ũk − ũl|p dx
)1/p

+ c
(∫

RN
(|ũk|m + |ũl|m) dx

)m−1
m
(∫

B2R(0)

|ũk − ũl|m dx
)1/m

+ c
(∫

RN
(|ũk|

2Nq
2N−α + |ũl|

2Nq
2N−α ) dx

) 2N−α
2N

×
(∫

B2R(0)

|ũk − ũl|
2Nq

2N−α dx
) 2N−α

2N

+ o(1)

≤ c‖ũk − ũl‖Lp(B2R(0)) + c‖ũk − ũl‖Lm(B2R(0))

+ c‖ũk − ũl‖
L

2Nq
2N−α (B2R(0))

+ o(1)

→ 0 as k, l→∞ .

(5.15)

For p ≥ 2, by (2.1) and (5.15) we have∫
BR(0)

|∇(ũk − ũl)|p dx

≤ c
∫
RN

(|∇ũk|p−2∇ũk − |∇ũl|p−2∇ũl,∇(uk − ul))ϕdx→ 0,

as k, l→∞.



24 B. ZHANG, X. LIU EJDE-2022/11

For 1 < p < 2, by (2.3) and (5.15) we have∫
BR(0)

|∇(ũk − ũl)|p dx

≤ c
(∫

RN
(|∇ũk|p−2∇ũk − |∇ũl|p−2∇ũl,∇(ũk − ũl))ϕdx

)p/2
×
(∫

RN
(|∇ũk|p + |∇ũl|p)ϕdx

) 2−p
2

≤ c
(∫

RN
(|∇ũk|p−2∇ũk − |∇ũl|p−2∇ũl)∇(ũk − ũl)ϕdx

)p/2
→ 0 as k, l→∞.

Since ϕ(x) = 1 in BR(0) and ϕ ≥ 0. Hence ũn → u in W 1,p
loc (RN ). Let zn = |ũn|,

wn,δ = (ũ2
n + δ2)1/2 − δ, then it follows from Lebesgue dominated convergence

theorem that wn,δ ∈ W 1,p(RN ), and wn,δ → zn in W 1,p(RN ) as δ → 0. Now for

any ϕ ∈ C∞0 (RN ), ϕ ≥ 0, we have ϕδ = ϕũn(ũ2
n + δ2)−

1
2 ∈W 1,p

loc (RN ), and∫
RN

(
|∇ũn|p−2∇wn,δ∇ϕ+ E(εn(x+ yn))|ũn|p−2wn,δϕ

)
dx

=

∫
RN
|∇ũn|p−2ũn∇ũn∇ϕ(ũ2

n + δ2)−
1
2 dx

+

∫
RN

E(εn(x+ yn))|ũn|p−2
(
(ũ2
n + δ2)1/2 − δ

)
ϕdx

=

∫
RN

(|∇ũn|p−2∇ũn∇ϕδ − |∇ũn|pϕ(ũ2
n + δ2)−

3
2 δ2

+ E(εn(x+ yn))|ũn|p−2((ũ2
n + δ2)1/2 − δ)ϕ) dx

≤
∫
RN

(|∇ũn|p−2∇ũn∇ϕδ + E(εn(x+ yn))|ũn|p−2ũnϕδ) dx

≤
∫
RN

∫
RN

|zn(y)|q|zn(x)|q−1|ϕδ(x)|
|x− y|α

dx dy.

(5.16)

Let δ → 0 in (5.16), for ϕ ∈ C∞0 (RN ), ϕ ≥ 0, we obtain∫
RN

(|∇zn|p−2∇zn∇ϕ+ |zn|p−1ϕ) dx

≤ c
∫
RN

∫
RN

|zn(y)|q|zn(x)|q−1ϕ(x)

|x− y|α
dx dy.

(5.17)

By ũn → u in W 1,p
loc (RN ) as n → ∞, we have zn → Z in W 1,p

loc (RN ) as n → ∞.
Hence we complete the proof by a denseness argument. �

Lemma 5.3. Λ is a finite set.

Proof. Zk = |Uk| satisfies (5.13) and take ϕ = Zk in (5.13), we have

‖Zk‖pW 1,p(RN )
≤ c

∫
RN

∫
RN

|Zk(y)|q|Zk(x)|q

|x− y|α
dx dy ≤ c‖Zk‖2qW 1,p(RN )

. (5.18)

So there exists m > 0 such that ‖Uk‖W 1,p(RN ) ≥ m. By the property (3) of the
profile decomposition (5.11), we know Λ is a finite set. �



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 25

Assume that the sequence {un} has the profile decomposition (5.11). Define

Λ = {1, . . . , k}, Ω
(n)
R = RN\{∪k∈ΛB(yn,k, R) ∪B(0, R)}.

Lemma 5.4. Assume DΓεn(un) = 0,Γεn(un) < L. Then there exist c = c(L), µ,
independent of n, such that∫

Ω
(n)
R

Gεn(x, un,∇un) dx ≤ ce−µR for x ∈ Ω
(n)
R ,

where

Gεn(x, un,∇un)

= |∇un|p + |un|p + kεn(x, un)un +
(∫

RN
χεn(x)|un|p dx− 1

)β−1

+
χεn(x)|un|p.

Moreover, we have

|un(x)| ≤ ce−µR for x ∈ Ω
(n)
R .

Proof. By the decomposition (5.11) we have

‖un‖Lq(Ω(n)
R )

= oR(1), p < q < p∗,

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

‖un‖L∞(Ω
(n)
R )

= oR(1).

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

(n)
R+1 and

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

Ω
(n)
R

(
|∇un|p + E(εnx)|un|p

)
ηp dx+ σ

∫
Ω

(n)
R

kεn(x, un)unη
p dx

+
(∫

RN
χεn(x)|un|p dx− 1

)β−1

+

∫
Ω

(n)
R

χεn(x)|un|pηp dx

=

∫
Ω

(n)
R

∫
RN

|un(y)|q|un(x)|qηp(x)

|x− y|α
dx dy

− p
∫

Ω
(n)
R \Ω

(n)
R+1

|∇un|p−2∇ununηp−1∇η dx.

By Lemma 5.1 (2) and ‖un‖L∞(Ω
(n)
R )

= oR(1) we obtain∫
Ω

(n)
R

dx

∫
RN

|un(y)|q|un(x)|qηp(x)

|x− y|α
dy ≤ 1

2

∫
Ω

(n)
R

E(εnx)|un|pηp dx.

Also ∣∣ ∫
Ω

(n)
R \Ω

(n)
R+1

|∇un|p−2∇ununηp−1∇η dx
∣∣

≤ τ
∫

Ω
(n)
R

|∇un|pηp + cτ

∫
Ω

(n)
R \Ω

(n)
R+1

|un|pηp dx.

So, we have∫
Ω

(n)
R+1

Gεn(x, un,∇un) dx ≤ c
∫

Ω
(n)
R \Ω

(n)
R+1

Gεn(x, un,∇un) dx.



26 B. ZHANG, X. LIU EJDE-2022/11

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 ≤ ce−µR,

where µ = − ln θ > 0. And by Moser’s iteration, we have |un(x)| ≤ ce−µR for

x ∈ Ω
(n)
R . �

Lemma 5.5. For every k ∈ Λ, it holds y∗k = limn→∞ εnyn,k ∈ Ā.

Proof. If not, we assume that there exist k ∈ Λ, εn > 0, εn → 0 as n → ∞ and
dist(y∗k, Ā) > 0. Let tk = ∇V (y∗k) 6= 0. Then by condition (A2) there exists δ1 > 0
such that(

tk,∇V (x)
)
≥ 1

2
|tk|2 > 0,

(
tk,∇dist(x,M)

)
≥ 0 for x ∈ Bδ1(y∗k). (5.19)

Set
δ2 = min{|y∗k − y∗l |y∗k 6= y∗l , k, l = 0, 1, . . . , k0, y∗0 = 0}.

Let

0 < δ < min{1

2
δ1,

1

100
δ2}.

Denote

Bn = {x||x− yn,k| ≤ 2δε−1
n },

Tn = x|δε−1
n ≤ |x− yn,k| ≤ 2δε−1

n .

Choose η ∈ C∞0 (RN ) 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). By 〈DΓεn(un), ϕ〉 = 0 for ϕ ∈W 1,p(RN ), we have∫
RN

(
|∇un|p−2∇un∇ϕ+ E(εnx)|un|p−2unϕ

)
dx+ σ

∫
RN

kεn(x, un)ϕdx

+
(∫

RN
χεn(x)|un|p dx− 1

)β−1

+

∫
RN

χεn(x)|un|p−2unϕdx

=

∫
RN

∫
RN

|un(y)|q|un(x)|q−2un(x)ϕ(x)

|x− y|α
dx dy.

(5.20)

Choosing ϕ = (tk,∇un)η as test function in (5.20), we obtain the local Pohožaev
identity

εn
p

∫
RN

(
tk,∇E(εnx)

)
|un|pη dx+ σ

∫
RN

(
tk,∇xkεn(x, un)

)
η dx

+
1

p

(∫
RN

χεn(x)|un|p dx− 1
)β−1

+

∫
RN

(
tk,∇χεn(x)

)
|un|pη dx

=

∫
RN
|∇un|p−2(∇un,∇η)(tk,∇un) dx

− 1

p

∫
RN

(|∇un|p + E(εnx))|un|p)(tk,∇η) dx− σ
∫
RN

kεn(x, un)un(tk,∇η) dx

− 1

p

(∫
RN

χεn(x)|un|p dx− 1
)β−1

+

∫
RN

χεn(x)|un|p(tk,∇η) dx



EJDE-2022/11 SEMICLASSICAL CHOQUARD EQUATIONS 27

− α

q

∫
RN

∫
RN

(tk, x− y)
|un(y)|q|un(x)|qη(x)

|x− y|α+2
dx dy

+
1

q

∫
RN

∫
RN

(tk,∇η)
|un(y)|q|un(x)|q

|x− y|α
dx dy. (5.21)

Next, we estimate all terms of the above inequality. By (5.19), we have

εn

∫
RN

(
tk,∇E(εnx)

)
|un|pη dx ≥ cεn,

and (∫
RN

χεn(x)|un|p dx− 1
)β−1

+

∫
RN

(
∇χεn(x), tk

)
|un|pη dx ≥ 0.

Hence the left-hand side of (5.21), satisfies

LHS ≥ cεn. (5.22)

We estimate the right-hand side of (5.21), by∫
RN

∫
RN

(tk, x− y)
|un(y)|q|un(x)|qη(x)η(y)

|x− y|α+2
dx dy = 0,

then ∫
RN

∫
RN

(tk, x− y)
|un(y)|q|un(x)|qη(x)

|x− y|α+2
dx dy

=

∫
RN

∫
RN

(tk, x− y)
|un(y)|q|un(x)|qη(x)(1− η(y))

|x− y|α+2
dx dy

≤ c
∫∫
|y−yn,k|≥δε−1

n

|x−yn,k|≤2δε−1
n

|un(x)|q|un(y)|q

|x− y|α+1
dx dy

≤ c
∫∫

δε−1
n ≤|y−yn,k|≤3δε−1

n

|x−yn,k|≤2δε−1
n

|un(x)|q|un(y)|q

|x− y|α+1
dx dy

+ c

∫∫
|y−yn,k|≥3δε−1

n

|x−yn,k|≤2δε−1
n

|un(x)|q|un(y)|q

|x− y|α+1
dx dy

=: I + II,

where

II ≤ c
∫∫
|y−yn,k|≥3δε−1

n

|x−yn,k|≤2δε−1
n

|un(y)|q|un(x)|q 1

δα+1
εα+1
n dx dy ≤ cεα+1

n .

The region T̃n = {y|δε−1
n ≤ |y−yn,k| ≤ 3δε−1

n } is contained in Ω
(n)

δε−1
n

, by Lemma 5.4,

we have
|un(y)| ≤ ce−µδε

−1
n , y ∈ T̃n.

Then

I ≤ ce−qµδε
−1
n

∫∫
δε−1
n ≤|y−yn,k|≤3δε−1

n

|x−yn,k|≤2δε−1
n

|un(x)|q

|x− y|α+1
dx dy

≤ ce−qµδε
−1
n

∫∫
|x−y|≤5δε−1

n

|x−yn,k|≤2δε−1
n

1

|x− y|α+1
|un(x)|q dy dx

≤ ce−qµδε
−1
n ε−N+α+1

n ≤ cεα+1
n .



28 B. ZHANG, X. LIU EJDE-2022/11

By Lemmas 5.1 and 5.4, we obtain that the right-hand side of (5.21), satisfies

RHS ≤ c
∫
Tn

Gεn(x, un,∇un) dx+ cεα+1
n

≤ ce−µδε
−1
n + cεα+1

n ≤ cεα+1
n .

Therefore cεn ≤ cεα+1
n . Since 0 < α < min{N −1, 2p}, we arrive at a contradiction

as n→∞. The proof is complete. �

The proof of Theorem 4.1 (2). By Lemma 5.4,

|un(x)| ≤ ce−µR for x ∈ Ω
(n)
R .

Let Rn(x) = min{|x− yn,k|k ∈ Λ}. Then

|un(x)| ≤ ce−µRn(x) for x ∈ Ω
(n)
Rn
.

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

|un(x)| ≤ ce−µRn ≤ ce−µ dist(x,(Aδ)εn ), x ∈ RN .

�

Acknowledgments. This work was supported by NSFC Grant Nos. 11761082 and
12161093, and by the Calculus of Variations and its Applications Innovation Team
in Universities of Yunnan Province.

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Bo Zhang
Department of Mathematics, Yunnan Normal University, Kunming, Yunnan 650500, China

Email address: zhangbo371013@163.com

Xiangqing Liu

Department of Mathematics, Yunnan Normal University, Kunming, Yunnan 650500, China

Email address: lxq8u8@163.com


	1. Introduction
	2. Palais-Smale condition
	3. Existence of solutions
	4. Proof of Theorem ??
	5. Uniform bound
	Acknowledgments

	References