Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 79, pp. 1–23.

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

DOI: 10.58997/ejde.2024.79

A GLOBAL COMPACTNESS RESULT FOR QUASILINEAR

ELLIPTIC PROBLEMS WITH CRITICAL SOBOLEV

NONLINEARITIES AND HARDY POTENTIALS ON RN

LINGYU JIN, SUTING WEI

Abstract. In this article, we study the elliptic equation with critical Sobolev

nonlinearity and Hardy potentials

(−∆)pu+ a(x)|u|p−1u− µ
|u|p−1u

|x|p
= |u|p

∗−2u+ f(x, u), u ∈ W 1,p(RN ),

where 0 < µ < min{ (N−p)p

pp
,
Np−1(N−p2)

pp
}, p∗ = Np

N−p
is the critical Sobolev

exponent. Through a compactness analysis of the associated functional oper-

ator, we obtain the existence of positive solutions under certain assumptions
on a(x) and f(x, u).

1. Introduction

For second-order semilinear elliptic differential equations on bounded domains,
Brezis and Nirenberg [3] obtained an existence result of solutions for a class of el-
liptic equations with critical Sobolev nonlinearities. by verifying a sub-level which
make the Palais-Smale conditions hold. A global compactness result for a semilin-
ear elliptic problem with critical Sobolev nonlinearities on the bounded domains
was obtained by Lions [19] and Struwe [27]. It was known that the sub-level which
makes the Palais-Smale conditions hold is determined by a compact result (refer to
[19, 27]). Alves [2] and Yan [29] generalized the result of Struwe [27] to the case of
p-Laplacian with critical Sobolev terms. Alves [2] also obtained the global compact-
ness result for the p-Laplace equation involving critical Sobolev terms on the whole
space. As for the case, the global compactness results for the p-Laplacian with crit-
ical Sobolev terms were obtained by Saintier [22] on a smooth Riemannian manifold
without boundary, and by Mercuri and Willem [20] on a smooth bounded domain
respectively. For the semilinear elliptic equation with Hardy potentials and critical
Sobolev terms, Cao and Peng [4] established global compactness results on bounded
domains, also demonstrating some new blow-up phenomena. On the whole space,
the global compactness result for the semilinear elliptic problem involving Hardy
potentials, and critical Sobolev terms was discussed in [7, 14, 25]. It is worth noting
that the equation discussed in [25] does not include sub-critical terms, whereas the

2020 Mathematics Subject Classification. 35J10, 35J20, 35J60.
Key words and phrases. p-Laplacian; compactness; positive solution; unbounded domain;

Sobolev nonlinearity.
©2024. This work is licensed under a CC BY 4.0 license.

Submitted April 5, 2024. Published December 3, 2024.

1



2 L. JIN, S. WEI EJDE-2024/79

equations discussed in [7, 14] include sub-critical terms caused new phenomena.
As for the p-Laplace equation with Hardy potentials and critical Sobolev terms on
bounded domains, the corresponding global compactness were proved in [13] and
[17]. Over the past two decades, the loss of compactness has led to numerous inter-
esting phenomena related to the existence and nonexistence of solutions for elliptic
equations (see, for example, [1, 2, 3, 4, 5, 13, 6, 7, 10, 12, 22, 23, 24, 25, 26] and
the references therein).

Motivated by [1, 7, 17, 20], we consider the nonlinear elliptic equation

(−∆)pu+ a(x)|u|p−1u− µ
|u|p−1u

|x|p
= |u|p

∗−2u+ f(x, u),

u ∈W 1,p(RN ),

(1.1)

where 0 < µ < min{ (N−p)p

pp , N
p−1(N−p2)

pp }, p∗ = Np
N−p is the critical Sobolev expo-

nent.
The main feature for this type of problems is the presence of the singular poten-

tial 1
|x|p related to the Sobolev-Hardy’s inequality. We recall the Sobolev-Hardy’s

inequality, ∫
RN

|u(x)|p

|x|p
dx ≤ c

∫
RN

|∇u(x)|p dx, ∀u ∈ D1,p(RN ) (1.2)

where c is a positive constant. The Sobolev embedding D1,p(RN ) ↪→ Lp(|x|−p,RN )
is not compact, even locally, in any neighborhood of zero. In addition to the inverse
square potential, another motivation for our investigation of problem (1.1) is the
presence of the critical Sobolev exponent and the unbounded domain, which result
in the loss of compactness of embeddings W 1,p(RN ) ↪→ Lp(RN ) and D1,p(RN ) ↪→
Lp∗

(RN ). Therefore, considering the noncompactness of embedding, we encounter
a triple loss of compactness, and their interaction introduces new challenges. To
address the challenges arising from the lack of compactness, we conduct a non-
compactness analysis, which allows us to distinctly identify and express all the ele-
ments responsible for non-compactness. To delve into more detail, in the context of
Palais-Smale sequences associated with the variational functional corresponding to
problem (1.1), we initially construct a comprehensive non-compact representation
encompassing all instances of singular behavior resulting from the critical Sobolev-
Hardy nonlinearity and the unbounded nature of the domain. Therefore, it can
determine the energy level intervals corresponding to the Palais-Smale sequence.
By leveraging the energy level intervals, we can more easily ascertain the existence
of both minimal energy solutions and high-energy solutions. In this paper we only
deduce the existence of minimal energy positive solutions for problem (1.1). Our
methods are based on techniques from [7, 14, 18, 21, 25, 27, 29].

This article is structured as follows. In Section 2, we present the main results of
the paper. In Section 3, we establish Theorem 2.1 through a meticulous analysis of
the characteristics of a positive Palais-Smale sequence for I. Section 4 is dedicated
to the proof of Theorem 2.3, achieved by employing both Theorem 2.1 and the
Mountain Pass Theorem. Finally, in the last section, we provide some preliminary
information as an appendix.



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 3

2. Main results

In this Section, we present the main results of this paper. For convenience, first
we provide some important notation and assumptions.

Let D1,p(RN ) be the homogeneous Sobolev space as the completion of C∞
0 (RN )

under the norm

∥u∥D1,p(RN ) = ∥∇u∥Lp(RN ), (2.1)

and denote by W 1,p(RN ) the usual nonhomogeneous Sobolev space with the norm

∥u∥W 1,p(RN ) = ∥∇u∥Lp(RN ) + ∥u∥Lp(RN ). (2.2)

Let u+ = max{u, 0}, u− = u+−u. Denote c and C as arbitrary constants which
may change from line to line. Let B(x, r) denote a ball centered at x with radius r
and B(x, r)C = RN \B(x, r).

A measurable function u : RN → R belongs to the Morrey space with p ∈ [1,∞)
and ν ∈ (0, N ], if

∥u∥p
Lp,ν(RN )

= sup
r>0,x̄∈RN

rν−N

∫
B(x̄,r)

|u(x)|p dx <∞.

By Hölder inequality, we can verify that

Lr,rN−p
p (RN ) ↪→ Lp∗

(RN ), for 1 ≤ r < p∗, 1 < p < N. (2.3)

Let X be a Banach space, Φ ∈ C1(X,R), c ∈ R, we call {un} ⊂ X is a Palais-
Smale sequence of Φ if

Φ(un) → c, Φ′(un) → 0 as n→ ∞. (2.4)

Next we establish specific assumptions regarding the functions a(x), f(x, u).

(A1) a(x) ∈ C(RN ), lim
x→∞

a(x) = ā > 0 and there exists a constant λ1 > 0 such

that∫
RN

[(
1−

( p

N − p

)p
µ
)
|∇u|p + a(x)|u|p

]
dx ⩾ λ1

∫
RN

(
ā− a(x)

)
|u|p dx, (2.5)

for all u ∈W 1,p(RN ). (Without loss of generality, we assume that ā = 1.)
(A2) f(x, t) is differentiable with respect to t ∈ [0,+∞) for all x ∈ RN and

continuous with respect to x ∈ RN for all t ∈ [0,+∞). Moreover, we
extend f(x, t) ≡ 0 for all t ∈ (−∞, 0), x ∈ RN .

(A3) There exists a constant q ∈ (p, Np
N−p ) such that lim

t→+∞
f(x,t)
tq−1 = 0 and

lim
t→0+

f(x,t)
tp−1 = 0 uniformly in x ∈ RN .

(A4) There exists a constant θ ∈ (0, p∗ − p) such that t ∂
∂tf(x, t) ⩾ (p − 1 +

θ)f(x, t) > 0, for all x ∈ RN , t > 0.
(A5) lim

|x|→+∞
f(x, t) = f̄(t) uniformly on any compact subset of [0,∞) and there

exists a constant σ > p( 1
p−1 )

1
p such that for any ε > 0 we can find Cε > 0

satisfying

f(x, t)− f̄(t) ⩾ −e−σ|x|(εtp−1 + Cεt
q−1) for all x ∈ RN , t ⩾ 0,

where q ∈ (p, Np
N−p ) is given by (A3).



4 L. JIN, S. WEI EJDE-2024/79

As in [8], assumption (A1) implies that(∫
RN

(|∇u|p + a(x)up − µ
up

|x|p
) dx

)1/p

is an equivalent norm of W 1,p(RN ). Also in Lemma 5.8, we give the proof of (2.5)
if a(x) satisfies some specific conditions.

As an example of a function that satisfies (A2)–(A5), we have

f(x, t) =

{
(1− e−σ|x|)tq, (p− 1 < q < p∗ − 1), for t ⩾ 0, x ∈ RN ,

0, for t < 0, x ∈ RN ,

In the following, we assume that a(x), f(x, u) satisfy (A1)–(A5).
The energy functional associated with problem (1.1) is

I(u) =
1

p

∫
RN

(
|∇u|p + a(x)|u|p − µ

|u|p

|x|p
)
dx

− 1

p∗

∫
RN

(
u+

)p∗

dx−
∫
RN

F (x, u)dx, ∀u ∈W 1,p(RN ),

(2.6)

with

F (x, u) =

∫ u

0

f(x, t) dt.

Next, we present some problems associated with problem (1.1). The limit equa-
tion of (1.1) involving sub-critical terms is

(−∆)pu+ ā|u|p−1u = f̄(u) + |u|p
∗−2u, u ∈W 1,p(RN ), (2.7)

and its corresponding variational functional is

I∞(u) =
1

p

∫
RN

(
|∇u|p + ā|u|p

)
dx− 1

p∗

∫
RN

(
u+

)p∗

dx−
∫
RN

F̄ (u) dx,

for all u ∈W 1,p(RN ), where F̄ (u) =
∫ u

0
f̄(t) dt.

The limit equation of (1.1) involving the Sobolev critical nonlinear term is

(−∆)pu = |u|p
∗−2u, u ∈ D1,p(RN ), (2.8)

and the corresponding variational functional is

I0(u) =
1

p

∫
RN

|∇u|p dx− 1

p∗

∫
RN

(
u+

)p∗

dx, ∀u ∈ D1,p(RN ).

The limit equation of (1.1) involving the Sobolev critical term and the Hardy
term is

(−∆)pu− µ
|u|p−1u

|x|p
= |u|p

∗−2u, u ∈ D1,p(RN ), (2.9)

and its corresponding variational functional is

Iµ(u) =
1

p

∫
RN

(
|∇u|p − µ

|u|p

|x|p
)
dx− 1

p∗

∫
RN

(
u+

)p∗

dx, ∀u ∈ D1,p(RN ).

Abdellaoui, Felli, and Peral [1] proved that all the positive solutions of problem
(2.9) take the form

Uε
µ(x) := ε

p−N
p Uµ(x/ε). (2.10)



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 5

Additionally, Uµ(r) and U
′
µ(r) have the following asymptotic properties

lim
r→0

ra(µ)Uµ(r) = c1 > 0,

lim
r→∞

rb(µ)Uµ(r) = c2 > 0,

lim
r→0

ra(µ)+1U ′
µ(r) = c1a(µ) > 0,

lim
r→∞

rb(µ)+1U ′
µ(r) = c2b(µ) > 0.

(2.11)

Here, c1 and c2 are positive constants depending only on N and p, while a(µ) and
b(µ) are the zeros of the function

g(t) = (p− 1)tp − (N − p)tp−1 + µ, t ⩾ 0 (0 < µ < ΛN,p =:
(N − p

p

)p
),

and satisfy 0 < a(µ) < b(µ).
We need further information on a(µ), b(µ), the two roots of g(t) = 0. After a

direct calculation, we infer that tmin = N−p
p is the only minimal point of g(t), t ⩾ 0,

and g(N−p
p ) = −ΛN,p + µ < 0 for 0 < µ < ΛN,p. Moreover, g′(t) < 0 for 0 < t <

tmin, g
′(t) > 0 for t > tmin. That is, g(t) is decreasing on the interval (0, tmin) and

increasing on the interval (tmin,∞). Thus,

a(µ) <
N − p

p
< b(µ) for 0 < µ < ΛN,p.

Furthermore, we obtain that

N

p
< b(µ) ⇐⇒ −N

p−1(N − p2)

pp
+ µ = g

(N
p

)
< g

(
b(µ)

)
= 0

⇐⇒ 0 < µ <
Np−1(N − p2)

pp
(N > p2).

Moreover, Uε
µ(x) are also minimizers for the quotient

Sµ = inf
u∈D1,p(RN )\{0}

∫
RN

(
|∇u|p − µ |u|p

|x|p
)
dx( ∫

RN |u|p∗ dx
)p/p∗ . (2.12)

For the case that µ = 0,

U0 =
1

(1 + |x|
p

p−1 )
N−p

p

. (2.13)

We can define

J∞ = inf
u∈N

I∞(u), (2.14)

with

N =
{
u ∈W 1,p(RN ) \ {0} :

∫
RN

(
|∇u|p + ā|u|p − (u+)p

∗
− F̄ (u)

)
dx = 0

}
. (2.15)

It is well known that N ̸= ∅ since problem (2.7) has at least one positive solution
if N > p2 (see [15]). Moreover, the authors in [15] proved that J∞ can be achieved
by a function w(x) ∈ N satisfies following properties

c1(1 + |x|)−
N−1

p(p−1) e−( ā
p−1 )

1/p|x| ≤ w(x) ≤ c2(1 + |x|)−
N−1

p(p−1) e−( ā
p−1 )

1/p|x|. (2.16)



6 L. JIN, S. WEI EJDE-2024/79

For convenience, we define the quantities

D0 =

∫
RN

(1
p
|∇U0|p −

1

p∗
|U0|p

∗
)
dx =

1

N
S
N/p
0 , (2.17)

Dµ =

∫
RN

[1
p

(
|∇Uµ|p − µ

|Uµ|p

|x|p
)
− 1

p∗
|Uµ|p

∗
]
dx =

1

N
SN/p
µ . (2.18)

The main result of our paper reads as follows.

Theorem 2.1. Suppose a(x), f(x, u) satisfy (A1)–(A5), N > p2, and

0 < µ < min
{ (N − p)p

pp
,
Np−1(N − p2)

pp
}
.

Also assume that {un} is a positive Palais-Smale sequence of I at level d ≥ 0. Then
there exist sequences {ykn} ⊂ RN (1 ≤ k ≤ l1), {R̄i

n} ⊂ R+(1 ≤ i ≤ l2), {Rj
n} ⊂

R+, {xjn} ⊂ RN (1 ≤ j ≤ l3) and uk ∈ W 1,p(RN )(1 ≤ k ≤ l1), 0 ≤ u ∈ W 1,p(RN )
(l1, l2, l3 ∈ N+) such that up to a subsequence:

d = I(u) +

l1∑
k=1

I∞(uk) + l2Dµ + l3D0 + o(1)

and ∥∥un − u−
l1∑

k=1

uk(x− ykn)−
l2∑
i=1

U R̄i
n −

l3∑
j=1

U
Rj

n,x
j
n

0

∥∥
W 1,p(RN )

= o(1) (2.19)

as n→ ∞, where u and uk(1 ≤ k ≤ l1) satisfy

I ′(u) = 0, I∞′(uk) = 0,

R̄i
n → 0, Rj

n → 0,
|xjn|
Rj

n

→ ∞.

In particular, if u ̸≡ 0, then u is a weakly solution of (1.1). Note that the corre-
sponding sum in (2.19) will be treated as zero if li = 0 (i = 1, 2, 3).

Remark 2.2. (1) Similar to [25, Corollary 3.3], one can demonstrate that any
Palais-Smale sequence for I at a level that does not have the form m1Dµ+m2J

∞+
m3D0, m1,m2,m3 ∈ N

⋃
{0}, gives rise to a non-trivial weak solution of (1.1).

(2) To account for the lower-order terms in problem (1.1), it becomes necessary
to impose the condition that u ∈ W 1,p(RN ) in order to ensure the well-defined
nature of the functional I(u). Specifically, when u ∈ W 1,p(RN ), the Sobolev in-
equality implies that u ∈ Lq(RN ) for p ≤ q < p∗. It is worth highlighting that
the quantities ∥u∥Lp(RN ) and ∥u∥Lq(RN ) are influenced solely by translation invari-

ance, while the integral
∫
RN

|u|p
|x|p dx is affected by scaling invariance. Consequently,

these considerations give rise to three limiting equations introducing intriguing new
structures.

Using the compactness results and the Mountain Pass Theorem [3] we prove the
following existence result.

Theorem 2.3. Assume that p < q < p∗, 0 < µ < min{ (N−p)p

pp , N
p−1(N−p2)

pp } and

N > p2. If a(x), f(x, u) satisfy (A1)–(A5), then problem (1.1) has a nontrivial



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 7

solution u ∈W 1,p(RN ) which satisfies

I(u) < min
{ 1

N
SN/p
µ , J∞

}
.

3. Non-compactness analysis

In this section, we prove Theorem 2.1 by using the Concentration-Compactness
Principle and a delicate analysis of the Palais-Smale sequences of I. Firstly, we give
the following Lemmas.

Lemma 3.1. Let {un} ⊂ D1,p(RN ) be a bounded sequence such that

inf
n∈N+

∫
RN

|un|p
∗
dx ≥ c > 0. (3.1)

Then, up to subsequence, there exist two sequences {rn} ⊂ R+ and {xn} ⊂ RN

such that

ūn ⇀ ū0 ̸≡ 0 in D1,p(RN ), (3.2)

where

ūn =

r
N−p

p
n un(rnx) if |xn|

rn
is bounded,

r
N−p

p
n un(rnx+ xn) if |xn|

rn
→ ∞.

(3.3)

Proof. By [21, Theorem 2], we have

∥un∥Lp∗ (RN ) ≤ C∥un∥θD1,p(RN )∥un∥
1−θ
Lp,N−p(RN )

, (3.4)

where p
p∗ ≤ θ < 1.

Then there exists a constant c > 0 such that

∥un∥pLp,N−p(RN )
= sup

x̄∈RN , R∈R+

R−p

∫
B(x̄,R)

|un|p dx ≥ c > 0. (3.5)

From (3.5), we may find rn > 0 and xn ∈ RN such that for n large enough,

r−p
n

∫
B(xn,rn)

|un|p dx ≥ ∥un∥pLp,N−p(RN )
− c

2n
≥ c

2
> 0. (3.6)

We define

ūn =

r
N−p

p
n un(rnx) when |xn|

rn
is bounded,

r
N−p

p
n un(rnx+ xn) when |xn|

rn
→ ∞.

(3.7)

Since {un} is bounded in D1,p(RN ), from the scaling and translation invariance
of D1,p(RN ), it follows that {ūn} is also bounded in D1,p(RN ), therefore, up to a
subsequence (still denoted by ūn),

ūn ⇀ ū0 in D1,p(RN ) and ūn → ū0 in Lp
loc(R

N ), as n→ ∞.

If |xn|/rn is bounded, there exists a constant R > 1 such that B(xn

rn
, 1) ⊂ B(0, R),

then
c

2
<

∫
B( xn

rn
,1)

|ūn|p dx ≤
∫
B(0,R)

|ūn|p dx→
∫
B(0,R)

|ū0(x)|p dx. (3.8)

If |xn|/rn → ∞, then

c

2
<

∫
B(0,1)

|ūn|p dx ≤
∫
B(0,R)

|ūn|p dx→
∫
B(0,R)

|ū0(x)|p dx, (3.9)



8 L. JIN, S. WEI EJDE-2024/79

where R > 1. Obviously we have ū0 ̸≡ 0. From (3.8) and (3.9), the proof is
complete. □

Lemma 3.2. Let {vn} ⊂W 1,p(RN ) be a Palais-Smale sequence of I at level d and
vn ⇀ 0 in W 1,p(RN ), ∥vn∥Lq(RN ) → 0 for all 1 < q < p∗, as n→ ∞. If there exist

sequences {rn} ⊂ R+, {xn} ⊂ RN with rn → 0, |xn|/rn → ∞ as n→ ∞ such that

v̄n(x) := r
N−p

p
n vn(rnx+xn) converges weakly in D1,p(RN ) and almost everywhere to

some 0 ̸= v0 ∈ D1,p(RN ) as n→ ∞, then v0 solves problem (2.8) and the sequence

zn := vn − r
p−N

p
n v0(

x−xn

rn
) is a Palais-Smale sequence of I at level d− I0(v0).

Proof. First, we prove that v0 solves problem (2.8). Fix a ball B(0, r) and a test
function ϕ ∈ C∞

0 (B(0, r)). Since

vn ⇀ 0, v̄n ⇀ v0 in D1,p(RN ), ∥vn∥Lq(RN ) → 0, and
|xn|
rn

→ ∞,

it follows that∫
RN

a(x)|vn|p−2
vnϕn dx = o(1),

∫
RN

f(x, vn)vnϕn dx = o(1),

µ

∫
RN

|v̄n|p−2v̄nϕ

|x+ xn

rn
|p

dx = o(1),

where ϕn = r
p−N

p
n ϕ(x−xn

rn
). It implies

⟨I ′0(v0), ϕ⟩

=

∫
RN

|∇v0|p−2∇v0∇ϕdx−
∫
RN

(
v+0

)p∗−1
ϕdx

=

∫
RN

|∇v̄n|p−2∇v̄n∇ϕdx− µ

∫
RN

|v̄n|p−2v̄nϕ

|x+ xn

rn
|p

dx−
∫
RN

(
v̄+n

)p∗−1
ϕdx+ o(1)

=

∫
RN

|∇vn|p−2∇vn∇ϕn dx− µ

∫
RN

|vn|p−2vnϕn
|x|p

dx−
∫
RN

(
v+n

)p∗−1
ϕn dx

+

∫
RN

a(x)ϕn|vn|p−2
vn dx−

∫
RN

f(x, vn)vnϕn dx+ o(1) = o(1)

(3.10)

as n→ ∞. The last equality in (3.10) holds since∫
RN

|ϕn|p dx = rpn

∫
RN

|ϕ|p dx = o(1),

and

∥ϕ∥D1,p(RN ) = ∥ϕn∥W 1,p(RN ) + o(1) as n→ ∞.

Thus v0 solves problem (2.8). From Lemma 5.6, (2.13) and N > p2, it follows that∫
RN

|v0(x)|q dx ≤ c

∫
RN

1

(1 + |x|
p

p−1 )
q
p (N−p)

dx ≤ c, ∀q ≥ p, (3.11)

which implies that v0 ∈ Lp(RN ).
Let

zn(x) = vn(x)− r
p−N

p
n v0

(x− xn
rn

)
∈W 1,p(RN ).



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 9

Obviously zn ⇀ 0 in W 1,p(RN ) as n → ∞. Now we prove that {zn} is a Palais-
Smale sequence of I at level d− I0(v0). From (3.11) it follows∫

RN

∣∣r p−N
p

n v0

(x− xn
rn

)∣∣p dx = rpn∥v0∥
p
Lp(RN )

→ 0, as n→ ∞, (3.12)

by Brézis-Lieb Lemma and the weak convergence, similar to Lemma 5.7, we can
prove that

I(zn) = I(vn)− I0(v0),

and ⟨I ′(zn), ϕ⟩ = o(1) as n→ ∞. This completes the proof. □

Lemma 3.3. Assume 0 < µ < min{ (N−p)p

pp , N
p−1(N−p2)

pp }. Let {vn} ⊂ W 1,p(RN )

be a Palais-Smale sequence of I at level d and vn ⇀ 0 inW 1,p(RN ), ∥vn∥Lq(RN ) → 0

for all 1 < q < p∗, as n→ ∞. If there exists a sequence {rn} ⊂ R+, with rn → 0 as

n→ ∞ such that v̄n(x) := r
N−p

p
n vn(rnx) converges weakly in D1,p(RN ) and almost

everywhere to some 0 ̸= v0 ∈ D1,p(RN ) as n → ∞, then v0 solves problem (2.9)

and the sequence zn := vn − r
p−N

p
n v0(

x
rn
) is a Palais-Smale sequence of I at level

d− Iµ(v0).

Proof. First, we prove that v0 solves problem (2.9). Fix a ball B(0, r) and a test
function ϕ ∈ C∞

0 (B(0, r)). Since

vn ⇀ 0, v̄n ⇀ v0 in D1,p(RN ), ∥vn∥Lq(RN ) → 0,

it follows that∫
RN

a(x)|vn|p−2
vnϕn dx = o(1),

∫
RN

f(x, vn)vnϕn dx = o(1).

So, we obtain that

⟨I ′µ(v0), ϕ⟩

=

∫
RN

|∇v0|p−2∇v0∇ϕdx− µ

∫
RN

|v0|p−2v0ϕ

|x|p
dx−

∫
RN

(
v+0

)p∗−1
ϕdx

=

∫
RN

|∇v̄n|p−2∇v̄n∇ϕdx− µ

∫
RN

|v̄n|p−2v̄nϕ

|x|p
dx−

∫
RN

(
v̄+n

)p∗−1
ϕdx+ o(1)

=

∫
RN

|∇vn|p−2∇vn∇ϕn dx− µ

∫
RN

|vn|p−2vnϕn
|x|p

dx−
∫
RN

(
v+n

)p∗−1
ϕn dx

+

∫
RN

a(x)|vn|p−2
vnϕn dx−

∫
RN

f(x, vn)vnϕn dx+ o(1)

= o(1) as n→ ∞,

(3.13)

where ϕn = r
p−N

p
n ϕ( x

rn
). The last equality in (3.13) holds since∫

RN

|ϕn|p dx = rpn

∫
RN

|ϕ|p dx = o(1),

∥ϕ∥D1,p(RN ) = ∥ϕn∥W 1,p(RN ) + o(1) as n→ ∞.

Thus v0 solves (2.9). From (2.11) and µ < Np−1(N−p2)
pp , it follows that∫

RN

|v0(x)|p dx ≤ c

∫
|x|≤1

1

|x|a(µ)p
dx+ c

∫
|x|≥1

1

|x|b(µ)p
dx ≤ c, (3.14)



10 L. JIN, S. WEI EJDE-2024/79

which implies that v0 ∈ Lp(RN ).
Let

zn(x) = vn(x)− r
p−N

p
n v0

( x
rn

)
∈W 1,p(RN ).

Obviously zn ⇀ 0 in W 1,p(RN ) as n → ∞. Now, we prove that {zn} is a Palais-
Smale sequence of I at level d− Iµ(v0). From (3.14) it follows that∫

RN

∣∣r p−N
p

n v0

( x
rn

)∣∣p dx = rpn∥v0∥
p
Lp(RN )

→ 0, as n→ ∞. (3.15)

By the Brézis-Lieb Lemma and the weak convergence, as in Lemma 5.7, we can
prove that

I(zn) = I(vn)− Iµ(v0),

⟨I ′(zn), ϕ⟩ = o(1)

as n→ ∞. This completes the proof. □

Lemma 3.4. Let ν be a unit vector of RN and w be that in (2.16). There exist
some constants C1 > 0 and C2 > 0 independent of R ⩾ 1 such that: (1)∫

|x|⩽1

(w(x−Rν))p dx ⩾ C1R
− (N−1)

p−1 e−p( 1
p−1 )

1
p R, for R ⩾ 1,

and (2)∫
RN

e−σ|x|(w(x−Rν))q dx ⩽ C2R
− q(N−1)

p(p−1) e−min{σ,q( 1
p−1 )

1
p R}, for R ⩾ 1.

The above lemma can be proved by the similar arguments as that of [5, Lemma
3.6]. We omit its proof.

Proof of Theorem 2.1. By Lemma 5.4 in the appendix, we can assume that {un}
is bounded in W 1,p(RN ). Up to a subsequence, as n→ ∞, we assume that

un ⇀ u in W 1,p(RN ),

un → u in Lq
loc(R

N ) for 1 < q < p∗,

un → u a.e. in RN .

We denote vn(x) = un(x)− u(x), then {vn} is a Palais-Smale sequence of I and

vn ⇀ 0 in W 1,p(RN ), (3.16)

vn → 0 in Lq
loc(R

N ) for 1 < q < p∗, (3.17)

vn → 0 a.e. in RN . (3.18)

Then by Lemma 5.7, we know that

I(vn) = I(un)− I(u) + o(1), as n→ ∞, (3.19)

I ′(vn) = o(1), as n→ ∞, (3.20)

∥vn∥W 1,p(RN ) = ∥un∥W 1,p(RN ) − ∥u∥W 1,p(RN ) + o(1), as n→ ∞. (3.21)

Without loss of generality, we may assume that

∥vn∥pW 1,p(RN )
→ l > 0 as n→ ∞.

In fact if l = 0, Theorem 2.1 is proved for l1 = 0, l2 = 0, l3 = 0.



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 11

Step 1: Getting rid of the blowing up bubbles caused by unbounded domains.
Suppose there exists a constant 0 < δ <∞ such that

∥vn∥Lp(RN ) ≥ δ > 0. (3.22)

By Lemma 5.1, there exists a subsequence still denoted by {vn}, such that one of
the following two cases occurs.

(i) Vanishing occurs: for all 0 < R <∞,

sup
y∈RN

∫
B(y,R)

(
|∇vn|p + |vn|p

)
dx→ 0 as n→ ∞.

By the Sobolev inequality, for 0 < R <∞, we have

sup
y∈RN

∫
B(y,R)

|vn|r dx ≤ sup
y∈RN

c

∫
B(y,R)

(|∇vn|p + |vn|p) dx→ 0 as n→ ∞, (3.23)

where 1 < r < p∗. Since vn is bounded in W 1,p(RN ), from (3.23) and Lemma 5.2
it follows that ∫

RN

|vn|q dx→ 0 as n→ ∞, ∀1 < q < p∗,

which contradicts (3.22).
(ii) Nonvanishing occurs: There exist β > 0, 0 < R̄ < ∞, and {yn} ⊂ RN such

that

lim inf
n→∞

∫
yn+BR̄

(
|∇vn|p + |vn|p

)
dx ≥ β > 0. (3.24)

We claim that there exists at least one |yn| → ∞ as n → ∞. Otherwise, if any
{yn} satisfying (3.24) is bounded, then there exists a R > 0 large enough such that

∥vn∥W 1,p(B(0,R)C) → 0 as n→ ∞. (3.25)

From the fact

vn → 0 in Lq
loc(R

N ) for 1 < q < p∗,

(3.25), and the Sobolev inequality, it follows that ∥vn∥Lp(RN ) → 0 as n→ ∞. This
contradicts (3.22).

To proceed, we first construct the Palais-Smale sequences of I∞. We denote v̄n =
vn(x+ yn). Since ∥v̄n∥W 1,p(RN ) = ∥vn∥W 1,p(RN ) ≤ C, without loss of generality, we
assume that as n→ ∞,

v̄n ⇀ v0 in W 1,p(RN ),

v̄n → v0 in Lq
loc(R

N ), ∀1 < q < p∗.

Since for all ϕ ∈ C∞
0 (RN ), for n large enough,∫

RN

|v̄n|p−2v̄nϕ

|x+ yn|p
dx ≤ 2

|yn|p

∫
RN

|vn|p−2vnϕn dx

≤ 2

|yn|p
(∫

RN

|vn|p dx
) p−1

p
(∫

RN

|ϕn|p dx
)1/p

(3.26)

where ϕn = ϕ(x− yn). Obviously∫
RN

|ϕn|p dx =

∫
RN

|ϕ|p dx ≤ c,

∫
RN

|vn|p dx ≤ c. (3.27)



12 L. JIN, S. WEI EJDE-2024/79

Let |yn| → ∞, from (3.26) and (3.27), we have∫
RN

|v̄n|p−2v̄nϕ

|x+ yn|p
dx = o(1) as n→ ∞. (3.28)

Since vn ⇀ 0 weakly in W 1,p(RN ) and limn→∞ a(x + yn) = ā, by the Lebesgue
convergence Theorem, as n→ ∞, we have∫

RN

a(x)|vn|p−2
vnϕn dx

=

∫
RN

ā|vn|p−2
vnϕdx+

∫
RN

[
a(x+ yn)− ā

]
|vn|p−2

vnϕdx

=

∫
RN

ā|v̄n|p−2v̄nϕdx+ o(1).

(3.29)

Similarly, we have∫
RN

f(x, vn)vnϕn dx =

∫
RN

f̄(v̄n)v̄nϕdx+

∫
RN

[
f(x+ yn, v̄n)− f̄(v̄n)

]
v̄nϕdx

=

∫
RN

f̄(v̄n)v̄nϕdx+ o(1).

(3.30)
Recall that vn is a Palais-Smale sequence of I, by (3.28)-(3.30) we have

⟨I ′(vn), ϕn⟩+ o(1) = ⟨I∞′(v̄n), ϕ⟩ = o(1), as n→ ∞. (3.31)

This shows that v̄n is a Palais-Smale sequence of I∞(u), and v0 is a weak solution
of (2.7).

We claim that v0 ̸≡ 0. From (3.22), we may assume there exists a sequence {yn}
satisfying (3.24) and∫

B(yn,R)

|vn(x)|p dx = b+ o(1) > 0, as n→ ∞, (3.32)

where b > 0 is a constant. If v0 ≡ 0, we have∫
B(R)

|v̄n|p dx =

∫
B(yn,R)

|vn|p dx = o(1) as n→ ∞, 0 < R <∞,

which contradicts (3.32).
We denote zn = vn − v0(x− yn); therefore, as n→ ∞,

∥zn∥W 1,p(RN ) = ∥vn∥W 1,p(RN ) − ∥v0∥W 1,p(RN ) + o(1), (3.33)

I(zn) = I(vn)− I∞(v0) + o(1). (3.34)

Hence zn ⇀ 0 in W 1,p(RN ) as n → ∞, and zn is a Palais-Smale sequence of I.
Then by Brézis-Lieb Lemma, we have∫

Rn

|zn|p dx =

∫
Rn

|vn − v0|p dx+ o(1)

=

∫
Rn

|vn|p dx−
∫
Rn

|v0|p dx+ o(1)

≤
∫
Rn

|vn|p dx− c,

(3.35)

where the last inequality follows from the fact v0 ̸≡ 0. If ∥zn∥Lp(RN ) → δ2 > 0 as
n→ ∞, from (3.35) and the boundedness of ∥vn∥Lp(RN ), then one can repeat Step



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 13

1 for finite times (l1 times) since the amount of sequences satisfying (3.22) is finite.

Step 2: Getting rid of the blowing up bubbles caused by the Sobolev term. Suppose
there exists 0 < δ <∞ such that

inf
n∈N+

∫
RN

(
v+n

)p∗

dx ≥ δ > 0. (3.36)

It follows from Lemma 3.1 that there exist two sequences {rn} ⊂ R+ and {xn} ⊂
RN , such that

v̄n ⇀ v0 ̸= 0 in D1,p(RN ), (3.37)

where

v̄n =

r
N−p

p
n vn(rnx) if |xn|/rn is bounded,

r
N−p

p
n vn(rnx+ xn) if |xn|/rn → ∞.

(3.38)

Now we claim that rn → 0 as n→ ∞. In fact, there exists a R1 > 0 such that∫
B(0,R1)

|v0|p dx = δ1 > 0. (3.39)

From the Sobolev compact embedding, (3.16)-(3.18), (3.37)-(3.39), for all r > 0 we
have

vn → 0 in Lp(B(0, r)), v̄n → v0 in Lp(B(0, r)),

0 ̸= ∥v0∥pLp(B(0,R1))
+ o(1)

=

∫
B(0,R1)

|v̄n|p dx

=

{
r−p
n

∫
B(0,rnR1)

|vn|p dx, if |xn|/rn is bounded,

r−p
n

∫
B(xn,rnR1)

|vn|p dx, if |xn|/rn → ∞.

(3.40)

From ∥vn∥Lp(RN ) = o(1), (3.39) and (3.40), it follows that rn → 0.

For |xn|/rn → ∞, we define zn = vn − r
p−N

p
n v0

(
x−xn

rn

)
. Then zn ⇀ 0 in

W 1,p(RN ). It follows from Lemma 3.3 that {zn} is a Palais-Smale sequence of
I satisfying

I(zn) = I(vn)− I0(v0) + o(1), as n→ ∞. (3.41)

Since v0 satisfies (2.8), from Lemma 3.1, (2.10) and (2.17) there exists ε1 > 0 such
that

v0 = ε
p−N

p

1 U0

(x− x̄1
ε1

)
, I0(v0) = D0. (3.42)

Let R1
n = rnε1, x

1
n = rnx̄1 + xn, it follows that

r
p−N

p
n v0

(x− xn
rn

)
= (R1

n)
p−N

p U0

(x− x1n
R1

n

)
= U

R1
n,x

1
n

0 , (3.43)

with R1
n → 0, |x1n|/R1

n → ∞. Then from (3.19) it follows that

zn = vn − U
R1

n,x
1
n

0 = un − u− U
R1

n,x
1
n

0 ,

I(zn) = I(vn)−D0 + o(1) = I(un)− I(u)−D0 + o(1)

with R1
n → 0, |x1n|/R1

n → ∞. Obviously

∥zn∥Lp∗ (RN ) = ∥vn∥Lp∗ (RN ) − ∥U0∥Lp∗ (RN ) + o(1).



14 L. JIN, S. WEI EJDE-2024/79

For |xn|/rn bounded, we define zn = vn − r
p−N

p
n v0(

x
rn
). Then zn ⇀ 0 in

W 1,p(RN ). It follows from Lemma 3.3 that {zn} is a Palais-Smale sequence of
I satisfying

I(zn) = I(vn)− Iµ(v0) + o(1), as n→ ∞. (3.44)

Since v0 satisfies (2.9), from (2.10) and (2.18) there exists ε1 > 0 such that

v0 = ε
p−N

p

1 Uµ

( x
ε1

)
, Iµ(v0) = Dµ. (3.45)

Let R̄1
n = rnε1, from (3.45), it follows that

r
p−N

p
n v0

( x
rn

)
= (R̄1

n)
p−N

p Uµ

( x

R̄1
n

)
= U

R̄1
n

µ , (3.46)

with R̄1
n → 0. Then from (3.19) it follows that

zn = vn − U
R̄1

n
µ = un − u− U

R̄1
n

µ ,

I(zn) = I(vn)−Dµ + o(1) = I(un)− I(u)−Dµ + o(1)
(3.47)

with R̄1
n → 0. Obviously

∥zn∥Lp∗ (RN ) = ∥vn∥Lp∗ (RN ) − ∥Uµ∥Lp∗ (RN ) + o(1). (3.48)

If still there exists a δ̄ > 0 such that∫
RN

(
z+n

)p∗

dx ≥ δ̄ > 0,

then we repeat the previous argument. From (3.48) and that∫
RN

(
z+n

)p∗

dx ≤ ∥zn∥p
∗

W 1,p(RN )
≤ c,

we deduce that the iteration must stop after finite times. That is, from step 1 and
step 2, there exist constants l1, l2, l3 and a new Palais-Smale sequence of I, (without
loss of generality) denoted by {vn}, such that as n→ ∞,

d = I(vn) + I(u) +

l1∑
k=1

I∞(uk) + l2Dµ + l3D0 + o(1), (3.49)

vn = un − u−
l1∑

k=1

uk(x− ykn)−
l2∑
i=1

U R̄i
n −

l3∑
j=1

U
Rj

n,x
j
n

0 ,

with R̄i
n, Rj

n → 0,
|xjn|
Rj

n

→ ∞, (3.50)

∥vn∥Lq(RN ) → 0,

∫
RN

(v+n )
p∗
dx→ 0 (3.51)

as n→ ∞. Then from ⟨I ′(vn), vn⟩ = o(1), it follows that

∥vn∥W 1,p(RN ) ≤ c

∫
RN

(∣∣∇vn|p + a(x)|vn|p − µ
|vn|p

|x|p
)
dx

= c
(∫

RN

f(x, vn)vn dx+

∫
RN

(
v+n

)p∗

dx
)
→ 0

(3.52)

as n→ ∞. From (3.51) and (3.52), it gives that

I(vn) = o(1). (3.53)



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 15

From (3.49)-(3.53), the proof of Theorem 2.1 is complete. □

4. Proof of Theorem 2.3

For this proof we use Mountain Pass Theorem [3] and Theorem 2.1. From

I(tu) =
tp

p

[ ∫
RN

(
|∇u|p+a(x)|u|p−µ |u|

p

|x|p
)
dx

]
− t

p∗

p∗

∫
RN

(u+)
p∗

dx−
∫
RN

F (x, tu) dx,

we deduce that for a fixed u ̸≡ 0 in W 1,p(RN ), I(tu) → −∞ if t→ +∞. Since∫
RN

F (x, u)dx ≤ C∥u∥q
W 1,p(RN )

+ ε∥u∥p
W 1,p(RN )

,

∫
RN

|u|p
∗
dx ≤ C∥u∥p

∗

W 1,p(RN )
,

we have

I(u) ≥ c∥u∥p
W 1,p(RN )

− C
(
∥u∥q

W 1,p(RN )
+ ∥u∥p

∗

W 1,p(RN )

)
.

Hence, there exists r0 > 0 small such that I(u)
∣∣
∂B(0,r0)

≥ ρ > 0 for q, p∗ > p.

As a consequence, I(u) satisfies the geometry structure of Mountain-Pass The-
orem. Now define

c∗ =: inf
γ∈Γ

sup
t∈[0,1]

I(γ(t)),

where Γ = {γ ∈ C([0, 1],W 1,p(RN )) : γ(0) = 0, γ(1) = ψ0 ∈ W 1,p(RN )} with
I(tψ0) ≤ 0 for all t ≥ 1.

To complete the proof of Theorem 2.3, we need to verify that I(u) satisfies the
local Palais-Smale conditions. According to Remarks 2.2(1), we only need to verify
that

c∗ < min
{ 1

N
SN/p
µ ,

1

N
S
N/p
0 , J∞}

= min
{ 1

N
SN/p
µ , J∞}

. (4.1)

Let vε =
Uε

µ

(
∫
RN |Uε

µ|p
∗ d)1/p∗

, we claim that

max
t>0

I(tvε) <
1

N
SN/p
µ . (4.2)

Since, Uε
µ are the minimizers of Sµ, we have∫

RN

|∇vε|p dx−
∫
RN

µ
|vε|p

|x|p
dx = Sµ. (4.3)

From (2.11) (also refer to [6]), and a(µ) < N−p
p , µ < Np−1(N−p2)

pp , it is easy to

calculate the estimate∫
RN

|vε|p dx ≤ cεp
∫
RN

|Uµ|p dx

≤ cεp
∫
|x|≤1

1

|x|a(µ)p
dx+ cεp

∫
|x|≥1

1

|x|b(µ)p
dx = O(εp).

(4.4)

Similarly, ∫
RN

|vε|q dx = O(ε
(p−N)q

p +N ). (4.5)

Since p∗ > q, we have

O(εp) = o(ε
(p−N)q

p +N ). (4.6)



16 L. JIN, S. WEI EJDE-2024/79

We denote by tε the attaining point of maxt>0 I(tvε), similar to the proof of [6,
Lemma 3.5] we can prove that tε is uniformly bounded. In fact, we consider the
function

h(t) = I(tvε)

=
tp

p

[
∥∇vε∥pLp(RN )

+

∫
RN

(
a(x)|vε|p − µ

|vε|p

|x|p
)
dx

]
− tp

∗

p∗

∫
RN

|vε|p
∗
dx−

∫
RN

F (x, tvε)dx

≥ ctp

p
∥vε∥pW 1,p(RN )

− ctp
∗

p∗
∥vε∥p

∗

W 1,p(RN )
− δtp∥vε∥pW 1,p(RN )

− ctq∥vε∥qW 1,p(RN )

≥ (c− δp)tp

p
∥vε∥pW 1,p(RN )

− ctp
∗

p∗
∥vε∥p

∗

W 1,p(RN )
− ctq∥vε∥qW 1,p(RN )

,

(4.7)

where δ > 0 small enough. Then h(t) > 0 when t is closed to 0, it follows that
maxt>0 h(t) is attained for tε > 0. From

∫
RN |vε|p

∗
dx = 1, it follows that

0 = h′(tε)

= tp−1
ε

[
∥∇vε∥pLp(RN )

+

∫
RN

(
a(x)|vε|p − µ

|vε|p

|x|p
)
dx

]
− tp

∗−1
ε −

∫
RN

f(x, tvε)vε dx.

(4.8)

Since f(x, vε) > 0, from (4.3) and (4.4), for ε sufficiently small, we have

tp
∗−p

ε ≤ ∥∇vε∥pLp(RN )
+

∫
RN

(
a(x)|vε|p − µ

|vε|p

|x|p
)
dx < 2Sµ. (4.9)

Then
1

2
Sµ < ∥∇vε∥pLp(RN )

+

∫
RN

(
a(x)|vε|p − µ

|vε|p

|x|p
)
dx

= tp
∗−p

ε + t−p+1
ε

∫
RN

f(x, tvε)vε dx

≤ tp
∗−p

ε +O(εN−qN−p
p ).

(4.10)

Choosing ε > 0 small enough, by (4.3)-(4.5), there exists a constant γ > 0 such
that tε > γ > 0. Combining this with (4.9), it implies that tε is bounded for ε > 0
small enough. Hence, for ε > 0 small,

max
t>0

I(tvε) = I(tεvε)

≤ max
t>0

{ tp
p

∫
RN

(
|∇vε|p − µ

|vε|p

|x|p
)
dx− tp

∗

p∗

∫
RN

|vε|p
∗
dx

}
−O(ε

(p−N)q
p +N ) +O(εp)

<
1

N
SN/p
µ (by (4.6)).

This completes the proof of (4.2). By the definition of c∗, we have c∗ < 1
N S

N/p
µ .

Next we verify that
c∗ < J∞. (4.11)



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 17

We only need to verify that

sup
0≤t≤t̄

I(twR) < J∞

for R large enough. Here wR = w(x−Rν), ν is a unit vector of RN and w be that
in (2.16). Since a(x) ∈ C(RN ), we can choose a small τ ∈ (0, 1) such that

1− a(x) +
µ

|x|p
⩾

µ

2|x|p
, ∀|x| ⩽ τ.

Then, we find that∫
RN

(1− a(x) +
µ

|x|p
)wp

R dx

⩾
∫
|x|⩽τ

µcp1
2τp

(|x−Rν|+ 1)
−N−1

p−1
e−p( 1

p−1 )
1
p |x−Rν| dx

⩾ c
µ

τp
(R+ 1)−

(N−1)
p−1 e−p( 1

p−1 )
1
p (R+1)

∫
|x|⩽τ

dx

⩾ cτN−pR− (N−1)
p−1 e−p( 1

p−1 )
1
p R = c̄R− (N−1)

p−1 e−p( 1
p−1 )

1
p R,

(4.12)

where c̄, c are positive constants. On the other hand, it follows from (A5) and
Lemma 3.4 that∫

RN

[
F̄ (twR)− F (x, twR)

]
dx

=

∫
RN

∫ twR

0

[
f̄(s)− f(x, s)

]
ds dx

⩽ ε
tp

p

∫
RN

e−σ|x|wp
R dx+ Cε

tq

q

∫
RN

e−σ|x|wq
R dx

⩽ εcR− (N−1)
p−1 e−p( 1

p−1 )
1
p R + CεcR

− (N−1)q
p(p−1) e−min{σ,q( 1

p−1 )
1
p R}

(4.13)

where c, Cε are positive constants. Hence, noting σ > p( 1
p−1 )

1
p , we see that for R

large enough,

I(twR)

⩽ I∞(twR)−
tp

p

∫
RN

(1− a(x) +
µ

|x|p
)wp

R dx+

∫
RN

(F̄ (twR)− F (x, twR)) dx

⩽ J∞ − cR− (N−1)
p−1 e−p( 1

p−1 )
R/p

+ cεR− (N−1)
p−1 e−p( 1

p−1 )
1
p R + CεcR

− (N−1)q
(p−1)p e−min{σ,q( 1

p−1 )
R/p} < J∞.

5. Appendix

In this section, we give detailed proofs some lemmas used above.

Lemma 5.1 ([28, Lemma 2.1]). Let {ρn}n≥1 be a sequence in L1(RN ) satisfying

ρn ≥ 0 on RN , lim
n→∞

∫
RN

ρn dx = λ > 0, (5.1)



18 L. JIN, S. WEI EJDE-2024/79

where λ > 0 is fixed. Then there exists a subsequence {ρnk
} satisfying one of the

following two possibilities: (i) Vanishing:

lim
k→∞

sup
y∈RN

∫
B(y,R)

ρnk
dx = 0, for all R < +∞. (5.2)

(ii) Nonvanishing: there exists α > 0, R < +∞ and {yk} ⊂ RN such that

lim
k→+∞

∫
yk+BR

ρnk
dx ≥ α > 0.

Lemma 5.2 ([28, Lemma 2.3]). Let 1 < p < ∞, 1 ≤ q < ∞, with q ̸= Np
N−p if

p < N . Assume that {un} is bounded in Lq(RN ), {|∇un|} is bounded in Lp(RN )
and

sup
y∈RN

∫
y+BR

|un|q dx→ 0 for some R > 0 as , n→ ∞.

Then un → 0 in Lα(RN ), for α between q and Np
N−p .

Lemma 5.3. Assume that a(x) satisfies (A1). It follows that

C1∥u∥pW 1,p(RN )
≤

∫
RN

(
|∇u|p + a(x)|u|p − µ

|u|p

|x|p
)
dx ≤ C2∥u∥pW 1,p(RN )

, (5.3)

where C1 and C2 are positive constants.

The proof of the above Lemma is obtained based on condition (A1).

Lemma 5.4. Let {un} be a Palais-Smale sequence of I at level d ∈ R. Then d ≥ 0
and {un} ⊂ W 1,p(RN ) is bounded. Moreover, every Palais-Smale sequence for I at
a level zero converges strongly to zero.

Proof. From (A4), it follows that

1

p+ θ
unf(un) ≥ F (x, un),

1

p+ θ
>

1

p∗
. (5.4)

Thus from (5.3) and (5.4), we have

d+ 1 + o(∥un∥) ≥ I(un)−
1

p+ θ
⟨I ′(un), un⟩

=
(1
p
− 1

p+ θ

)∫
RN

(∣∣∇un|p − µ
|un|p

|x|p
+ a(x)|un|p

)
dx

+
1

p+ θ

∫
RN

unf(x, un) dx−
∫
RN

F (x, un) dx

≥ C
(1
p
− 1

p+ θ

)∫
RN

(∣∣∇un|p − µ
|un|p

|x|p
+ a(x)|un|p

)
dx

≥ C∥un∥pW 1,p(RN )
.

(5.5)

It follows from (5.5) that {un} is bounded in W 1,p(RN ). Since

d = lim
n→∞

I(un)−
1

p+ θ
⟨I ′(un), un⟩ ≥ C lim sup

n→∞
∥un∥pW 1,p(RN )

,

then we have d ≥ 0. Suppose now that d = 0, we obtain from the above inequality
that

lim
n→∞

∥un∥W 1,p(RN ) = 0.

□



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 19

Lemma 5.5. Let {un} be a Palais-Smale sequence of I at level d ∈ R. Then {u+n }
is also a Palais-Smale sequence of I at level d when u+n = max{un, 0}.

Proof. By the definition of I, we have that as n→ ∞

I(un) =
1

p

∫
RN

(
|∇un|p + a(x)|un|p

)
dx− µ

p

∫
RN

|un|p

|x|p
dx

− 1

p∗

∫
RN

(
u+n )

p∗
dx−

∫
RN

F (x, un)un dx→ d,

and

⟨I ′(un), ϕ⟩ =
∫
RN

|∇un|p−2∇un∇ϕdx+

∫
RN

a(x)|un|p−2unϕdx

− µ

∫
RN

|un|p−2unϕ

|x|p
dx−

∫
RN

f(x, un)ϕdx

−
∫
RN

(
u+n

)p∗−1
ϕdx→ 0, for all ϕ ∈W 1,p(RN ).

Taking ϕ = −u−n = min{un, 0}, from
un = u+n − u−n , u+nu

−
n = 0, (5.6)

we have

⟨I ′(un),−u−n ⟩

= −
∫
RN

|∇un|p−2∇un∇u−n dx−
∫
RN

a(x)|un|p−2unu
−
n dx

+ µ

∫
RN

|un|p−2unu
−
n

|x|p
dx−

∫
RN

f(x, un)u
−
n dx+

∫
RN

(
u+n

)p∗−1
u−n dx

=

∫
RN

(
|∇u−n |p + a(x)|u−n |p

)
dx− µ

∫
RN

|u−n |p

|x|p
dx→ 0.

(5.7)

From (A1), (5.7), u+n ≥ 0, and u−n ≥ 0, it follows that

∥u−n ∥W 1,p(RN ) → 0. (5.8)

Thus

lim
n→∞

I(u+n ) = lim
n→∞

I(un) = d,

I ′(u+n , ϕ) = I ′(un, ϕ) → 0,

as n→ ∞. This complete the proof. □

Lemma 5.6. All nontrivial critical points of Iµ are the positive solutions.

Proof. Let u ̸≡ 0 and u ∈ W 1,p(RN ) be a nontrivial critical point of Iµ. First,
arguing as in the proof of Lemma 5.5 (similar to (5.7) and (5.8)), we can obtain
that ∥u−∥W 1,p(RN ) = 0 which gives that u ≥ 0 a.e. in RN . By the maximum

principle we can obtain u > 0 in RN . □

Let {un} be a Palais-Smale sequence. Up to a subsequence, we assume that

un ⇀ u in W 1,p(RN ) as n→ ∞.

Obviously, we have I ′(u) = 0. Let vn = un − u, then as n→ ∞,

vn ⇀ 0 in W 1,p(RN ), (5.9)



20 L. JIN, S. WEI EJDE-2024/79

vn → 0 in Lq
loc(R

N ) for all 1 < q < p∗. (5.10)

As a consequence, we have the following Lemma.

Lemma 5.7. {vn} is a Palais-Smale sequence for I at level d0 = d− I(u).

Proof. By the Brézis-Lieb Lemma in [3] and vn ⇀ 0 in W 1,p(RN ), as n → ∞, we
have ∫

RN

F (x, vn)dx =

∫
RN

F (x, un)dx−
∫
RN

F (x, u)dx+ o(1), (5.11)∫
RN

∣∣vn|p
|x|p

dx =

∫
RN

|un|p

|x|p
dx−

∫
RN

|u|p

|x|p
dx+ o(1), (5.12)∫

RN

|∇vn|p dx =

∫
RN

|∇un|p dx−
∫
RN

|∇u|p dx+ o(1). (5.13)

Hence I(vn) = I(un)− I(u) + o(1) = d− I(u) + o(1).
For ϕ ∈ C∞

0 (RN ), there exists a B(0, r) such that suppϕ ⊂ B(0, r). Then as
n→ ∞,∣∣ ∫

RN

f(x, vn)ϕdx
∣∣ ≤ c

∣∣ ∫
B(0,r)

(|vn|q−1 + |vn|p−1)ϕdx
∣∣ = o(1), (5.14)

and from the Lebesgue convergence theorem∣∣ ∫
RN

|vn|p
∗−2vnϕ

|x|p
dx

∣∣ ≤ ∣∣ ∫
|x|≤r

|vn|p
∗−2vnϕ

|x|p
dx

∣∣ = o(1). (5.15)

By (5.9), (5.14) and (5.15), we have ⟨ϕ, I ′(vn)⟩ = o(1) as n→ ∞. □

Lemma 5.8. The assumption (A1) holds naturally if a(x) ∈ C(RN ) satisfies

(1) a(x) → ā > 0 as x→ ∞;
(2) −m ≤ a(x) and the set {x ∈ RN : −m ≤ a(x) ≤ 0} is nonempty and

bounded, where m ∈ (0,m∗) and m∗ is a small positive constant.

Proof. By assumptions (1) and (2), we can find ρ > 0 such that

{x ∈ RN : a(x) ⩽ 0} ⊂ B(0, ρ),

inf
RN\B(0,ρ)

a(x) >
ā

2
.

We claim that

(|a|2 + |b|2 + 2a · b)p/2 − (|a|2)p/2 ⩾ p|a|p−2a · b.

From Cauchy’s mean value theorem, we have(
|a|2 + |b|2 + 2a · b

)p/2 − (
|a|2

)p/2
=
p

2
ξ

p−2
2

(
|b|2 + 2a · b

)
.

If |a+ b|2 ⩾ |a|2, i.e., |b|2 + 2a · b ⩾ 0, |a|2 ⩽ ξ ⩽ |a+ b|2, thus

ξ
p−2
2

(
|b|2 + 2a · b

)
⩾ |a|p−2(|b|2 + 2a · b) ⩾ 2|a|p−2a · b.

If |a+ b|2 ⩽ |a|2, i.e., |b|2 + 2a · b ⩽ 0, |a+ b|2 ⩽ ξ ⩽ |a|2, thus

|ξ|
p−2
2 (|b|2 + 2a · b) ⩾ |a|p−2(|b|2 + 2a · b) ⩾ 2|a|p−2a · b.



EJDE-2024/79 A GLOBAL COMPACTNESS RESULT 21

For R > ρ, we can choose φ(x) ∈ C∞
0 (B(0, R)) satisfying 0 ⩽ φ(x) ⩽ 1, φ(x) = 1

in B(0, ρ), φ(x) = 0 in RN\B(0, R) and |∇φ| ⩽ 1
2(R−ρ)p . For each u ∈ W 1,p(RN ),

let

a = ∇(φu), b = ∇((1− φ)u).

Then we can derive that∫
RN

|∇u|p dx

⩾
∫
RN

|∇(φu)|p dx+ p

∫
RN

|∇(φu)|p−2(∇u · u · ∇φ(1− 2φ)− |∇φ|2u2) dx

⩾
1

2

∫
RN

|∇(φu)|p dx−
∫
RN

(|∇u||u||∇φ|)p/2pp/22p−2 dx−
∫
RN

|∇φ|p|u|ppp/22p−2 dx

⩾
1

2

∫
B(0,R)

|∇(φu)|p dx− 2p−3pp/2
∫
RN

|∇u|p dx− 3pp/22p−3

∫
RN

|∇φ|p|u|p dx

⩾
λ1,p(B(0, R))

2

∫
B(0,ρ)

|u|p dx− 3 · 2p−3 · pp/2

(2(R− ρ)p)p

∫
RN

|u|p dx− 2p−3pp/2
∫
RN

|∇u|p dx

⩾
1

2(p− 1)Rp

∫
B(0,ρ)

|u|p dx− 3

8

1

(R− ρ)p
· 1

pp/2

∫
RN

|u|p dx− 2p−3pp/2
∫
RN

|∇u|p dx,

where λ1,p(B(0, R)) ≥ (2π)p

(p−1)(p sin(π/p)2R)p is the first eigenvalue of the operator

(−∆)p in W 1,p
0 (B(0, R)) (refer to [16]).

We set d = 1− ( p
N−p )

pµ, choose R large enough such that

1

2(p− 1)Rp
− 3

8
· 1

pp/2
1

(R− ρ)p
⩾

1

8(p− 1)Rp
, and

3d

8(R− ρ)ppp/2
<
ā

4
.

Then∫
RN

(
d|∇u|p + a(x)|u|p

)
dx

⩾
d

2(p− 1)Rp

∫
B(0,ρ)

|u|p dx+

∫
RN

a(x)|u|p dx

− 3d

8(R− ρ)ppp/2

∫
RN

|u|p dx− 2p−3pp/2d

∫
RN

|∇u|p dx

=

∫
B(0,ρ)

[ d

2(p− 1)Rp
− 3d

8(R− ρ)ppp/2
+ a(x)

]
|u|p dx

+

∫
RN\B(0,ρ)

[
a(x)− 3d

8(R− ρ)ppp/2
]
|u|p dx− 2p−3pp/2d

∫
RN

|∇u|p dx.

Therefore

(1 + 2p−3pp/2)

∫
RN

(d|∇u|p + a(x)|u|p) dx

⩾
∫
Bp(0)

[
d

8(p− 1)Rp
+ (1 + 2p−3p

p
2 )a(x)

]
|u|p dx+

ā

4

∫
RN\B(0,ρ)

|u|p dx.

Let m∗ = 1

(1+2p−3p
p
2 )

· d
16(p−1)Rp . For 0 ⩽ m ⩽ m∗, it follows that

(1 + 2p−3pp/2)

∫
RN

(d|∇u|p + a(x)|u|p) dx



22 L. JIN, S. WEI EJDE-2024/79

⩾
d

16(p− 1)Rp

∫
B(0,ρ)

|u|p dx+
ā

4

∫
RN\Bρ(0)

|u|p dx

⩾ min

{
d

16(p− 1)Rp
,
ā

4

}∫
RN

|u|p dx.

If we set λ∗ = 1

1+2p−3p
p
2
min

{
d

16(p−1)Rp ,
ā
4

}
, then we have∫

RN

[
(1−

( p

N − p

)p
µ)|∇u|p + a(x)|u|p

]
dx =

∫
RN

(d|∇u|p + a(x)|u|p) dx

⩾ λ∗
∫
RN

|u|p dx

=
λ∗

ā+m∗

∫
RN

(ā+m∗)|u|p dx

⩾
λ∗

ā+m∗

∫
RN

(ā− a(x))|u|p dx.

We can complete the proof by taking λ1 = λ∗

ā+m∗ . □

Acknowledgements: S. Wei was supported by the NSFC grants No. 12001203 and
No. 12171109, and by the Guangzhou Basic and Applied Basic Research Foundation
grant No. 2024A04J4907.

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Lingyu Jin
Department of Mathematics, South China Agricultural University, Guangzhou 510642,

China

Email address: jinlingyu300@126.com

Suting Wei (corresponding author)

Department of Mathematics, South China Agricultural University, Guangzhou 510642,
China

Email address: stwei@scau.edu.cn


	1. Introduction
	2. Main results
	3. Non-compactness analysis
	4. Proof of Theorem ??
	5. Appendix
	Acknowledgements:

	References