Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 102, pp. 1–11.

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

p-LAPLACIAN EQUATION WITH FINITELY MANY CRITICAL

NONLINEARITIES

PENGCHENG XIA, YU SU

Abstract. This article concerns the ground state solution of the p-Laplacian
equation with finitely many critical nonlinearities. By using the refined Sobolev

inequality with Morrey norm and variational methods, we establish the exis-

tence of nonnegative ground state solution.

1. Introduction

We consider the p-Laplacian equation

−∆pu− ζ
|u|p−2u

|x|p
=

k∑
i=1

(
Iαi ∗ |u|

p∗αi
)
|u|p

∗
αi
−2u+ |u|p

∗−2u, x ∈ RN , (1.1)

where N > 3, p ∈ (1, N), ζ ∈ (0,Λ), Λ = (N−pp )p, ∆p := div(|∇u|p−2∇u) is

the p-Laplacian, p∗αi = p
2 (N+αi

N−p ) are the Hardy-Littlewood-Sobolev critical upper

exponents, and the parameters αi satisfy the following assumption:

(H1) 0 < α1 < · · · < αk < N , for k ∈ N, 2 6 k <∞.

Problem (1.1) is related to the nonlinear Choquard equation

−∆u+ V (x)u = (Iα ∗ |u|q) |u|q−2u, x ∈ RN , (1.2)

where N+α
N 6 q 6 N+α

N−2 and α ∈ (0, N). For q = 2 and α = 2, problem (1.2)
goes back to the description of the quantum theory of a polaron at rest by Pekar
in 1954 [10] and the modeling of an electron trapped in its own hole in 1976 in
the work of Choquard, as a certain approximation to Hartree-Fock theory of one-
component plasma [11]. The existence and qualitative properties of solutions of
Choquard equations or other related equations have been widely studied in the
previous decades, see [2, 3, 4, 5, 7, 8, 12, 15].

Recall that the Sobolev space D1,p(RN ) is the completion of C∞0 (RN ) with
respect to the semi-norm

‖u‖p
D1,p(RN )

=

∫
RN
|∇u|pdx.

2010 Mathematics Subject Classification. 35A15, 35J20.
Key words and phrases. p-Laplacian equation; critical exponent; ground state solution;

mountain pass theorem.
©2021. This work is licensed under a CC BY 4.0 license.

Submitted June 22, 2021. Published December 24, 2021.

1



2 P. XIA, Y. SU EJDE-2021/102

It is well known that Λ =
(
N−p
p

)p
is the best constant in the Hardy inequality

Λ

∫
RN

|u|p

|x|p
dx 6 ‖u‖p

D1,p(RN )
for all u ∈ D1,p(RN ).

By the Hardy inequality for ζ ∈ [0,Λ), we derive that ‖u‖pζ = ‖u‖p
D1,p(RN )

−
ζ
∫
RN
|u|p
|x|p dx is an equivalent norm in D1,p(RN ), and

(
1− ζ

Λ

)
‖u‖p

D1,p(RN )
6 ‖u‖pζ 6 ‖u‖

p
D1,p(RN )

.

For α ∈ [0, N), ζ ∈ [0,Λ) and p ∈ (1, N), we define the best constant:

Sζ,α := inf
u∈D1,p(RN )\{0}

‖u‖p
D1,p(RN )

− ζ
∫
RN
|u|p
|x|p dx( ∫

RN
∫
RN
|u(x)|p∗α |u(y)|p∗α
|x−y|N−α dxdy

) p
2·p∗α

. (1.3)

Lemma 1.1 ([6]). Let t, r > 1 and 0 < α < N with 1
t + 1

r + N−α
N = 2, f ∈ Lt(RN )

and h ∈ Lr(RN ). There exists a sharp constant C(N,α, t, r) > 0, independent of
f, g such that ∫

RN

∫
RN

|f(x)||h(y)|
|x− y|N−α

dxdy 6 C(N,α, t, r)‖f‖t‖h‖r .

If t = r = 2N
N+α , then

C(N,α, t, r) = C(N,α) = π
N−α

2
Γ(α2 )

Γ(N + α
2 )

(Γ(N2 )

Γ(N)

)α/N
.

We introduce the energy functional associated with problem (1.1) by

Iζ(u) =
1

p
‖u‖pζ −

k∑
i=1

1

2 · p∗αi

∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy − 1

p∗

∫
RN
|ūn|p

∗
dx.

We define the Nehari manifold

Nζ = {u ∈ D1,p(RN ) : 〈I ′ζ(u), u〉 = 0, u 6= 0},

and

c̄ζ = inf
u∈Nζ

Iζ(u), ¯̄cζ = inf
u∈D1,p(RN )

max
t>0

Iζ(tu), cζ = inf
Υ∈Γ

max
t∈[0,1]

Iζ(Υ(t)),

where

Γ = {Υ ∈ C([0, 1], D1,p(RN )) : Υ(0) = 0, Iζ(Υ(1)) < 0}.

Because of lack of compactness of the Sobolev embedding D1,p(RN ) ↪→ Lp
∗
(RN )

and that the functional Iζ is invariant under the weighted dilation, it is hard to
show that the Palais-Smale sequence of Iζ has a convergent subsequence.

Recently, Su et al. [14] studied the existence of ground state solution for (1.1)
with the additional condition

(H2)
S0,0

C(N,αi)

p
2·p∗αi

> (k + 1)
− p
N+αk for i = 1, . . . , k, where C(N,αi) and S0,0 are

defined in Lemma 2.1 and (1.3) respectively.



EJDE-2021/102 p-LAPLACIAN EQUATION 3

Applying this condition, they showed that

c̄0 > c̄ζ . (1.4)

As an application of inequality (1.4), they proved that the dilated subsequence of
Palais-Smale sequence weak converges to nonzero function. And then they estab-
lished the existence of ground state solution to (1.1). Hence, condition (H2) plays
a key role in [14].

It is natural to ask

Can we find a nontrivial solution to (1.1) without assuming (H2)?

To the best of our knowledge, there is no affirmative answer in the literature. An
answer to this question is given in the main result of this article:

Theorem 1.2. Assume that N > 3, p ∈ (1, N), ζ ∈ (0,Λ) and (H1) holds. Then
equation (1.1) has a nonnegative ground state solution.

This article is organized as follows: In Section 2, we study the ground state
solution of limit equation. In Section 3, we prove Theorem 1.2.

2. Ground state solution of the limit equation

To study (1.1), we consider the problem

−∆pu =

k∑
i=1

(
Iαi ∗ |u|

p∗αi
)
|u|p

∗
αi
−2u+ |u|p

∗−2u, x ∈ RN . (2.1)

We introduce the energy functional associated with problem (2.1) by

I0(u) =
1

p
‖u‖p

D1,p(RN )
−

k∑
i=1

1

2 · p∗αi

∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy− 1

p∗

∫
RN
|u|p

∗
dx.

We define

c0 = inf
Υ∈Γ

max
t∈[0,1]

I0(Υ(t)),

where

Γ = {Υ ∈ C([0, 1], D1,p(RN )) : Υ(0) = 0, I0(Υ(1)) < 0}.

Lemma 2.1. Assuming the conditions of Theorem 1.2. The following conclusions
hold:

(i) there exists {vn} ⊂ D1,p(RN ) such that

I0(vn)→ c0, ‖I ′0(vn)‖D−1,p(RN ) → 0 as n→∞

and {vn} is uniformly bounded in D1,p(RN ), and limn→∞
∫
RN |vn|

p∗dx > 0;

(ii) for each u ∈ D1,p(RN ) \ {0}, there exists a unique tu > 0 such that tuu ∈
N 0;

(iii) c0 = c̄0 = ¯̄c0 = infu∈N 0 I0(u) > 0.

Proof. (i) Clearly, I0 satisfies the mountain pass geometry. Then there exists a
(PS)c sequence {vn} ⊂ D1,p(RN ) of I0 at level c0 > 0. It is not hard to prove that
{vn} is uniformly bounded in D1,p(RN ).

We now show that limn→∞
∫
RN |vn|

p∗dx > 0. Suppose on the contrary that

lim
n→∞

∫
RN
|vn|p

∗
dx = 0. (2.2)



4 P. XIA, Y. SU EJDE-2021/102

It follows from (2.2) and Lemma 1.1 that

lim
n→∞

∫
RN

∫
RN

|vn(x)|p
∗
αi |vn(y)|p

∗
αi

|x− y|N−αi
dxdy = 0, for i = 1, . . . , k. (2.3)

By using (2.3) and the definition of (PS)c0 sequence, we obtain

c0 + o(1) =
1

p
‖un‖pD1,p(RN )

,

o(1) = ‖un‖pD1,p(RN )
.

These equalities yield c0 = 0 which contradicts c0 > 0.

(ii) For u ∈ D1,p(RN ) \ {0} and t ∈ (0,∞), we set

g1(t) = I0(tu)

=
tp

p
‖u‖p

D1,p(RN )
−

k∑
i=1

t2·p
∗
αi

2 · p∗αi

∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy

− tp
∗

p∗

∫
RN
|u|p

∗
dx,

and

g′1(t) = tp−1‖u‖p
D1,p(RN )

−
k∑
i=1

t2·p
∗
αi
−1

∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy

− tp
∗−1

∫
RN
|u|p

∗
dx.

We know that g′1(·) = 0 if and only if

‖u‖p
D1,p(RN )

=

k∑
i=1

t2·p
∗
αi
−p
∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy + tp

∗−p
∫
RN
|u|p

∗
dx.

We set

g2(t) =

k∑
i=1

t2·p
∗
αi
−p
∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy + tp

∗−p
∫
RN
|u|p

∗
dx.

Then we obtain that limt→0 g2(t) → 0, limt→∞ g2(t) → ∞, and g2(·) is strictly
increasing on (0,∞). Then there exists a unique 0 < tu <∞ such that

g2(t)


> ‖u‖p

D1,p(RN )
, tu < t <∞,

= ‖u‖p
D1,p(RN )

, t = tu,

< ‖u‖p
D1,p(RN )

, 0 < t < tu.

This shows that tuu ∈ N 0. Moreover, we have

g′1(t)


< 0, tu < t <∞,
= 0, t = tu,

> 0, 0 < t < tu.

This implies that g1(·) admits a unique critical point tu on (0,∞) such that g1(·)
takes the maximum at tu.

(iii) Clearly, I0 is bounded from below on N 0, and c̄0 > 0. Indeed, it follows
form Lemma 3.1 (ii) that c̄0 = ¯̄c0. Notice that for any u ∈ D1,p(RN ) \ {0}, there



EJDE-2021/102 p-LAPLACIAN EQUATION 5

exists t̃ > 0 large, such that I0(t̃u) < 0. We define a path γ : [0, 1]→ D1,p(RN ) by
γ(t) = tt̃u. Clearly, γ ∈ Γ and consequently, c0 6 ¯̄c0. On the other hand, for every
path γ ∈ Γ, we let g3(t) := 〈I ′0(γ(t)), γ(t)〉. Then g3(0) = 0 and g3(t) > 0 for t > 0
small. We have

I0(γ(1))− 1

p
〈I ′0(γ(1)), γ(1)〉

>
k∑
i=1

(1

p
− 1

2 · p∗αi

)∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy −

(1

p
− 1

p∗

)∫
RN
|u|p

∗
dx > 0

which implies

〈I ′0(γ(1)), γ(1)〉 6 pI0(γ(1)) = pI0(t̃u) < 0.

Then there exists ˜̃t ∈ (0, 1) such that g3(˜̃t) = 0, i.e., γ(˜̃t) ∈ N 0. So c0 > c̄0. �

We recall that a measurable function u : RN → R belongs to the Morrey space
‖u‖Lq,$(RN ) with q ∈ [1,∞) and $ ∈ (0, N ], if and only if

‖u‖qLq,$(RN )
= sup
R>0,x∈RN

R$−N
∫
B(x,R)

|u(y)|qdy <∞.

Lemma 2.2 ([9, Theorem 2]). For p ∈ (1, N), there exists C > 0 such that for ι
and ϑ satisfying p

p∗ 6 ι < 1, 1 6 ϑ < p∗, we have(∫
RN
|u|p

∗
dx
)1/p∗

6 C‖u‖ιD1,p(RN )‖u‖
1−ι

Lϑ,
ϑ(N−p)

p (RN )

,

for all u ∈ D1,p(RN ).

Proof of ground state solution for (2.1). We divided our proof into two steps.

Step 1. Note that {vn} is a bounded sequence in D1,p(RN ). Up to a subsequence,
we assume that

vn ⇀ v in D1,p(RN ), vn → v a.e. in RN , vn → v in Lrloc(RN )

for all r ∈ [p, p∗). From Lemmas 2.1 (i) and 2.2, there exists C > 0 such that

‖vn‖Lp,N−p(RN ) > C > 0.

On the other hand, since the sequence is bounded in D1,p(RN ), and (see [13]),

D1,p(RN ) ↪→ Lp
∗
(RN ) ↪→ Lp,N−p(RN ),

we have

‖vn‖Lp,N−p(RN ) 6 C.

Hence, there exists a positive constant which we denote again by C such that for
any n we obtain

C 6 ‖vn‖Lp,N−p(RN ) 6 C
−1.

So we may find σn > 0 and xn ∈ RN such that

1

σpn

∫
B(xn,σn)

|un(y)|pdy > ‖un‖pLp,N−p(RN )
− C

2n
> C1 > 0.

Let v̄n(x) = σ
N−p
p

n vn(xn + σnx). We need to verify that

I0(v̄n) = I0(un)→ c0, ‖I ′0(v̄n)‖D−1,p(RN ) → 0 as n→∞,



6 P. XIA, Y. SU EJDE-2021/102

For all ϕ ∈ D1,p(RN ), we have

|〈I ′0(v̄n), ϕ〉| = |〈I ′0(vn), ϕ̄n〉|
6 ‖I ′0(vn)‖D−1,p(RN )‖ϕ̄n‖D1,p(RN )

= o(1)‖ϕ̄n‖D1,p(RN ) ,

where ϕ̄n = σ
−N−p

p
n ϕ(x−xnσn

). From ‖ϕ̄n‖D1,p(RN ) = ‖ϕ‖D1,p(RN ), we obtain

‖I ′0(v̄n)‖D−1,p(RN ) → 0 as n→∞.

Thus there exists v̄ such that

v̄n ⇀ v̄ in D1,p(RN ), v̄n → v̄a.e. in RN , v̄n → v̄ in Lrloc(RN )

for all r ∈ [p, p∗). Then∫
B(0,1)

|v̄n(y)|pdy =
1

σpn

∫
B(xn,σn)

|vn(y)|pdy > C1 > 0.

This implies v̄ 6≡ 0.

Step 2. For any ϕ ∈ D1,p(RN ), applying 〈I ′0(v̄n), ϕ〉 → 0 and v̄n ⇀ v̄ weakly in
D1,p(RN ), we obtain

〈I ′0(v̄), ϕ〉 = 0. (2.4)

Moreover, by (2.4) and v̄ 6≡ 0, we obtain ū ∈ N 0. By the Brézis-Lieb Lemma [1],
we have ∫

RN
|ūn|p

∗
dx >

∫
RN
|ū|p

∗
dx+ o(1), (2.5)∫

RN

∫
RN

|ūn(x)|p∗α |ūn(y)|p∗α
|x− y|N−α

dxdy >
∫
RN

∫
RN

|ū(x)|p∗α |ū(y)|p∗α
|x− y|N−α

dxdy + o(1). (2.6)

We set

K(u) =

k∑
i=1

(1

p
− 1

2 · p∗αi

)∫
RN

∫
RN

|u(x)|p
∗
αi |u(y)|p

∗
αi

|x− y|N−αi
dxdy+

(1

p
− 1

p∗

)∫
RN
|u|p

∗
dx.

(2.7)
Applying Lemma 2.1, (2.5), (2.6), (2.7) and ū ∈ N 0, we obtain

c̄0 = c0 = I0(v̄n)− 1

p
〈I ′0(v̄n), v̄n〉 = lim

n→∞
K(v̄n)

> K(v̄) = I0(v̄)− 1

p
〈I ′0(v̄), v̄〉

= I0(v̄) > c̄0.

Therefore, the inequalities above have to be equalities. We obtain I0(v̄) = c0, which
means that ṽ is a ground state solution of problem (2.1) at the energy level c0. We
know that |v̄| ∈ D1,p(RN ) and |∇|v̄|| = |∇v̄| a.e. in RN . Hence, we can choose
v̄ > 0. �



EJDE-2021/102 p-LAPLACIAN EQUATION 7

3. Proof of Theorem 1.2

As in Lemma 2.1, we have the following results without proof.

Lemma 3.1. Under the conditions of Theorem 1.2, the following conclusions hold:

(i) there exists {un} ⊂ D1,p(RN ) such that

Iζ(un)→ cζ , ‖I ′ζ(un)‖D−1,p(RN ) → 0 as n→∞,

and {un} is uniformly bounded in D1,p(RN ), and limn→∞
∫
RN |un|

p∗dx > 0;

(ii) for each u ∈ D1,p(RN ) \ {0}, there exists a unique tu > 0 such that tuu ∈
N ζ ;

(iii) cζ = c̄ζ = ¯̄cζ = infu∈N ζ Iζ(u) > 0.

We now prove inequality (1.4).

Lemma 3.2. Assume that the conditions of Theorem 1.2 hold. Then c̄0 > c̄ζ for
all ζ ∈ (0,Λ).

Proof. Since v̄ is a nonnegative ground state solution of equation (2.1), so we have
I0(v̄) = c0 and

‖v̄‖p
D1,p(RN )

=

k∑
i=1

∫
RN

∫
RN

|v̄(x)|p
∗
αi |v̄(y)|p

∗
αi

|x− y|N−αi
dxdy +

∫
RN
|v̄|p

∗
dx. (3.1)

By Lemma 3.1, there exists tv̄ > 0 such that tv̄ v̄ ∈ N ζ . We now claim tv̄ < 1. It
follows from tv̄ v̄ ∈ N ζ that

‖v̄‖p
D1,p(RN )

− ζ
∫
RN

|v̄|p

|x|p
dx

=

k∑
i=1

t
2·p∗αi−p
v̄

∫
RN

∫
RN

|v̄(x)|p
∗
αi |v̄(y)|p

∗
αi

|x− y|N−αi
dxdy + tp

∗−p
v̄

∫
RN
|v̄|p

∗
dx.

(3.2)

Putting (3.1) into (3.2), we have

− ζ
∫
RN

|v̄|p

|x|p
dx

=

k∑
i=1

(
t
2·p∗αi−p
v̄ − 1

) ∫
RN

∫
RN

|v̄(x)|p
∗
αi |v̄(y)|p

∗
αi

|x− y|N−αi
dxdy + (tp

∗−p
v̄ − 1)

∫
RN
|v̄|p

∗
dx,

which implies

0 >

k∑
i=1

(
t
2·p∗αi−p
v̄ − 1

) ∫
RN

∫
RN

|v̄(x)|p
∗
αi |v̄(y)|p

∗
αi

|x− y|N−αi
dxdy + (tp

∗−p
v̄ − 1)

∫
RN
|v̄|p

∗
dx.



8 P. XIA, Y. SU EJDE-2021/102

Note that 2p∗αi > p and p∗ > p. Then tv̄ < 1. It follows from (3.1) that

I0(v̄)− I0(tv̄ v̄)

=
1− tpv̄
p
‖v̄‖p

D1,p(RN )
− 1− tp

∗

v̄

p∗

∫
RN
|v̄|p

∗
dx

−
k∑
i=1

1− t
2·p∗αi
v̄

2 · p∗αi

∫
RN

∫
RN

|v̄(x)|p
∗
αi |v̄(y)|p

∗
αi

|x− y|N−αi
dxdy

=

k∑
i=1

[1

p
− tpv̄
p
− 1

2 · p∗αi
+
t
2·p∗αi
v̄

2 · p∗αi

] ∫
RN

∫
RN

|v̄(x)|p
∗
αi |v̄(y)|p

∗
αi

|x− y|N−αi
dxdy

+
[1

p
− tpv̄
p
− 1

p∗
+
tp

∗

v̄

p∗

] ∫
RN
|v̄|p

∗
dx.

(3.3)

We set

f1(tv̄) =
1

p
− tpv̄
p
− 1

2 · p∗αi
+
t
2·p∗αi
v̄

2 · p∗αi
,

f2(tv̄) =
1

p
− tpv̄
p
− 1

p∗
+
tp

∗

v̄

p∗
.

It is easy to see that

f1(0) > 0, f1(1) = 0, f ′1(tv̄) = tp−1
v̄ (t

2·p∗αi−p
v̄ − 1),

f2(0) > 0, f2(1) = 0, f ′2(tv̄) = tp−1
v̄ (tp

∗−p
v̄ − 1).

Note that 0 < tv̄ < 1, 2p∗αi > p, and p∗ > p. Then

f ′1(tv̄) < 0 for tv̄ ∈ (0, 1),

f ′2(tv̄) < 0 for tv̄ ∈ (0, 1).

Hence,

f1(tv̄) > 0 for tv̄ ∈ (0, 1),

f2(tv̄) > 0 for tv̄ ∈ (0, 1).

Putting two results above into (3.3), we have

I0(v̄) > I0(tv̄ v̄).

On the other hand,

Iζ(tv̄ v̄) = I0(tv̄ v̄)− tpv̄ζ
∫
RN

|v̄|p

|x|p
dx < I0(tv̄ v̄).

So in general, we can obtain

c̄ζ 6 Iζ(tv̄ v̄) < I0(tv̄ v̄) < I0(v̄) = c̄0.

The proof is complete. �

Proof of Theorem 1.2. We divided our proof into four steps.

Step 1. Note that {un} is a bounded sequence in D1,p(RN ). Up to a subsequence,
we assume that

un ⇀ u in D1,p(RN ), un → ua.e. in RN , un → u in Lrloc(RN )



EJDE-2021/102 p-LAPLACIAN EQUATION 9

for all r ∈ [p, p∗). By Lemmas 2.2 and 3.1 (i), there exists C > 0 such that

‖un‖Lp,N−p(RN ) > C > 0.

On the other hand, since the sequence is bounded in D1,p(RN ), and (see [13])

D1,p(RN ) ↪→ Lp
∗
(RN ) ↪→ Lp,N−p(RN ),

we have

‖un‖Lp,N−p(RN ) 6 C,

for some C > 0 independent of n. Hence, there exists a positive constant which we
denote again by C such that for any n we obtain

C 6 ‖un‖Lp,N−p(RN ) 6 C
−1.

So we may find σn > 0 and xn ∈ RN such that

1

σpn

∫
B(xn,σn)

|un(y)|pdy > ‖un‖pLp,N−p(RN )
− C

2n
> C6 > 0.

Let ūn(x) = σ
N−p
p

n un(xn + σnx). We need to verify that

Ĩζ(ūn) = Iζ(un)→ cζ , Ĩζ
′
(ūn)→ 0 as n→∞,

where

Ĩζ(ūn) =
1

p
‖ūn‖pD1,p(RN )

− ζ

p

∫
RN

|ūn|p

|x+ xn
σn
|p

dx

−
k∑
i=1

1

2 · p∗αi

∫
RN

∫
RN

|ūn(x)|p
∗
αi |ūn(y)|p

∗
αi

|x− y|αi
dxdy − 1

p∗

∫
RN
|ūn|p

∗
dx.

Now, for ϕ ∈ D1,p(RN ), we obtain

|〈Ĩζ
′
(ūn), ϕ〉| = |〈I ′ζ(un), ϕ̄n〉|

6 ‖I ′ζ(un)‖D−1,p(RN )‖ϕ̄n‖D1,p(RN )

= o(1)‖ϕ̄n‖D1,p(RN ),

where ϕ̄n = σ
−N−p

p
n ϕ(x−xnσn

). Since ‖ϕ̄n‖D1,p(RN ) = ‖ϕ‖D1,p(RN ), we obtain

Ĩζ
′
(ūn)→ 0 as n→∞.

Thus there exists ū such that

ūn ⇀ ū in D1,p(RN ), ūn → ū a.e. in RN , ūn → ū in Lrloc(RN )

for all r ∈ [p, p∗). Then∫
B(0,1)

|ūn(y)|pdy =
1

σpn

∫
B(xn,σn)

|un(y)|pdy > C6 > 0.

This implies ū 6≡ 0.

Step 2. We now show that {xn/σn} is bounded. If xn/σn → ∞, then for any
ϕ ∈ D1,p(RN ), we obtain

lim
n→∞

∫
RN

|ūn|p−2ūnϕ

|x+ xn
σn
|p

dx = 0. (3.4)



10 P. XIA, Y. SU EJDE-2021/102

Using that 〈Ĩζ
′
(ūn), ϕ〉 → 0 and (3.4), we obtain

〈I ′0(ū), ϕ〉 = 0. (3.5)

Moreover, from (3.5) and ū 6≡ 0, we obtain ū ∈ N 0. Applying Lemma 3.1, Lemma
3.2, (3.3), (3.4), (3.5) and ū ∈ N 0, we obtain

c̄0 > c̄ζ = cζ

= Iζ(ūn)− 1

p
〈I ′ζ(ūn), ūn〉 = lim

n→∞
K(ūn)

> K(ū) = I0(ū)− 1

p
〈I ′0(ū), ū〉

= I0(ū) > c̄0,

which yields a contradiction. Hence, {xn/σn} is bounded.

Step 3. Let ũn(x) = σ
N−p
p

n un(σnx). Then we can verify that

Iζ(ũn) = Iζ(un)→ cζ , I ′ζ(ũn)→ 0 as n→∞.

Arguing as before, we have

ũn ⇀ ũ in D1,p(RN ), ũn → ũ a.e. in RN , ũn → ũ in Lrloc(RN )

for all r ∈ [p, p∗). Note that {xnσn} is bounded. Then there exists R̃ > 0 such that∫
B(0,R̃)

|ũn(y)|pdy >
∫
B( xnσn ,1)

|ũn(y)|pdy =
1

σpn

∫
B(xn,σn)

|un(y)|pdy > C5 > 0.

As a result, ũ 6≡ 0. An argument similar to one in Step 2 yields

〈I ′ζ(ũ), ϕ〉 = 0. (3.6)

From this equality and ũ 6≡ 0, we obtain ũ ∈ N ζ .

Step 4. It follows from ũ ∈ N ζ , (3.3), (3.4) and (3.5) that

c̄ζ = cζ = Iζ(ũn)− 1

p
〈I ′ζ(ũn), ũn〉

= lim
n→∞

K(ũn)

> K(ũ) = Iζ(ũ)− 1

p
〈I ′ζ(ũ), ũ〉

= Iζ(ũ) > c̄ζ .

Therefore, the inequalities above have to be equalities. We obtain Iζ(ũ) = cζ , which
means that ũ is a ground state solution of problem (1.1) at the energy level cζ . We
know that |ũ| ∈ D1,p(RN ) and |∇|ũ|| = |∇ũ| a.e. in RN . Hence, we can choose
ũ > 0. �

Acknowledgments. This research was supported by the Key Program of Univer-
sity Natural Science Research Fund of Anhui Province (Grant No. KJ2020A0294).



EJDE-2021/102 p-LAPLACIAN EQUATION 11

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Pengcheng Xia
School of Mathematics and Big Data, Anhui University of Science and Technology,

Huainan, Anhui 232001, China

Email address: pcxia@aust.edu.cn

Yu Su (corresponding author)

School of Mathematics and Big Data, Anhui University of Science and Technology,

Huainan, Anhui 232001, China
Email address: yusumath@aust.edu.cn


	1. Introduction
	2. Ground state solution of the limit equation
	3. Proof of Theorem ??
	Acknowledgments

	References