Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 69, pp. 1–19.

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

DOI: 10.58997/ejde.2025.69

EXISTENCE OF NONTRIVIAL SOLUTIONS FOR BIHARMONIC

EQUATIONS WITH CRITICAL GROWTH

JUHUA HE, KE WU, FEN ZHOU

Abstract. We consider the biharmonic equation with critical Sobolev exponent,

∆2u−∆u−∆(u2)u+ V (x)u = |u|2
∗∗−2u+ α|u|p−2u, in RN ,

where N > 4, α > 0, V (x) is a given potential, 2∗∗ = 2N
N−4

is the Sobolev critical exponent

and 2 < p < 2∗∗. Under the combined influence of the biharmonic, quasilinear terms, and

critical nonlinearities, looking for solutions with N ∈ {5, 6} is totally different from the case
when N ≥ 7. For the case N ∈ {5, 6}, we show that this equation has a nontrivial solution,

using a variational method.

1. Introduction

We consider the existence of nontrivial solutions for the critical biharmonic equation

∆2u−∆u−∆(u2)u+ V (x)u = |u|2
∗∗−2u+ α|u|p−2u, in RN , (1.1)

where N > 4, α > 0, V (x) is a given potential, 2∗∗ = 2N
N−4 is the Sobolev critical exponent and

2 < p < 2∗∗.
Equation (1.1) without the quasilinear term is the biharmonic problem

∆2u−∆u+ V (x)u = |u|2
∗∗−2u+ α|u|p−2u, in RN .

The biharmonic operator ∆2 is used to study the impact of higher-order dispersion terms in
the nonlinear Schrödinger equation with a fourth-order dispersion term [7, 8]. In physics, the
biharmonic equation can be simulating the static deflection of an elastic plate in a fluid [18]. In
[11], this type of equation also furnishes a model for studying the traveling waves in suspension
bridges.

Recently, Liang, Zhang, and Luo [12] proved the existence and multiplicity of solutions for the
perturbed biharmonic equation

ε4∆2u+ V (x)u = |u|2
∗∗−2u+ h(x, u), x ∈ RN ,

u(x) → 0, as |x| → ∞,
(1.2)

by a variational method. In [2], the authors apply the concentration compactness principle to
obtain a nontrivial solution for the equation

∆2u+ a(x)u = h(x)|u|q−1u+ k(x)|u|p−1u, in RN ,

u ∈ H2(RN ), N ≥ 5,
(1.3)

2020 Mathematics Subject Classification. 35J35, 35J60, 35J62.

Key words and phrases. Critical exponent; nontrivial solutions; biharmonic operator; quasilinear problem.
©2025. This work is licensed under a CC BY 4.0 license.
Submitted January 6, 2025. Published July 7, 2025.

1



2 J. HE, K. WU, F. ZHOU EJDE-2025/69

where 1 < q < p ≤ 2∗∗ − 1 = N+4
N−4 and a, h, k are bounded, nonnegative and continuous functions.

Alves, do Ó, and Miyagaki [1] showed the existence of nontrivial solutions for the equation

∆2u+ V (x)|u|q−1u = |u|2
∗∗−2u, in Ω ⊂ RN ,

u ∈ D2,2
0 (Ω), N ≥ 5,

(1.4)

using the mountain pass theorem and the Hardy inequality, where 1 ≤ q < 2∗∗ − 1, 2∗∗ = 2N
N−4 , Ω

is an open domain and V is a potential that changes sign in Ω with some points of singularities in
Ω. Additionally, we refer readers to [17, 28, 6, 21, 23] for more studies of biharmonic equations.

Equation (1.1) is also related to the known quasilinear problem with critical growth,

−∆u−∆(u2)u+ V (x)u = |u|2
∗∗−2u+ α|u|p−2u, in RN . (1.5)

Solutions of (1.5) are standing waves for the quasilinear equation

iψt +∆ψ −W (x)ψ +∆(h(|ψ|2))h′(ψ2)ψ + (|ψ|2
∗∗−2 + α|ψ|p−2)ψ = 0, in [0,∞)× RN , (1.6)

i.e. solutions of the form

ψ(t, x) = exp{−iβt}u(x),
where W (x) = V (x) + β, h is a real function and β ∈ R. (1.6) plays an important role in various
fields in physics. For example, it can be used for the superfluid film equation in plasma physics
[9]. It also appears in fluid mechanics [10] and condensed matter theory [16]. We refer readers to
[20, 13] for more details of (1.6).

We mention some work relating to the quasilinear problems. Wang [24] used the method of
change of variables to establish the existence of nontrivial solutions for the equation

−∆u+ V (x)u+
κ

2
∆(u2)u = l(u), x ∈ RN , (1.7)

where l(u) = λ|u|p−2u + |u|q−2u, p ≥ 22∗, 4 < q < 22∗, 2∗ = 2N
N−2 . Recall that 22∗ = 4N

N−2 ,

N ≥ 3, is the corresponding critical exponent for the quasiliner term ∆(u2)u. The problem
studied in [24] is critical or supercritical. The same change of variables was also used in [27, 5, 22]
to deal with quasilinear equations involving critical exponent. Wu and Wu [26] obtained the
existence of standing wave solutions for generalized quasilinear equations with critical growth by
the perturbation method. The existence of solutions for quasilinear equations can also be obtained
by Nehari method [14] and minimization process [15, 19].

In this article, we investigate the existence of nontrivial solutions for (1.1) with critical nonlin-
earities. Our main motivation in mathematics comes from the following fact. Compared to the
pure critical biharmonic problems and quasilinear ones, three different cases occur as far as (1.1)
is concerned:

(i) if N = 5, then 22∗ < 2∗∗. The critical exponent 22∗ for the quasilinear term is actually
a subcritical one for the whole equation (1.1). In this case, it seems that the quasilinear
term has barely effect on the existence of the solution of (1.1).

(ii) if N = 6, then 22∗ = 2∗∗. In this case the exponent 2∗∗ is the critical exponent for both
the biharmonic operator and the quasilinear term whose combined effects make our study
of (1.1) more difficult.

(iii) if N ≥ 7, then 22∗ > 2∗∗. In this case (1.1) is not a critical problem anymore, and 2∗∗

is nothing but a common subcritical exponent. However, it is worth noting that, for the
case N ≥ 7, the domain of the functional corresponding to (1.1) is not a vector space.

We will use the variational method to find solutions of (1.1). To do this, we need to estimate the
energy of the functional carefully. Some special techniques are also applied.

We make the following assumption of the potential V (x).

(A1) V ∈ C(RN ,R) satisfies 0 < V0 ≤ V (x) ≤ lim|x|→∞ V (x) := V∞ ≤ +∞.

Our main result is the following theorem.

Theorem 1.1. Let N ∈ {5, 6} and assume that (A1) holds.

(i) If 8
N−4 < p < 2∗∗, then for any α > 0, (1.1) has a nontrivial solution.



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 3

(ii) If one of the following two conditions hold:
(a) 2 < p ≤ 8

N−4 and V∞ = +∞;

(b) 4 ≤ p ≤ 8
N−4 and V∞ < +∞;

then there exists a constant α∗ > 0 such that, for all α ∈ (α∗,∞), (1.1) has a nontrivial
solution.

Remark 1.2. From the above theorem, we see that the existence of solutions for (1.1) might
depend on α, p, and the limit V∞. However, if 8

N−4 < p < 2∗∗, then the result is independent of
α and V∞.

In this article, we use the following notation: For p ∈ [1,+∞], we denote the usual Lp(RN )
norm by ∥ · ∥p. For y ∈ RN and r > 0, we denote Br(y) := {x ∈ RN : |x− y| < r}, Br := B(0, r).
C and Ci denote positive constants.

2. Preliminaries

Throughout this article, we assume that N ∈ {5, 6}. Set

E = {u ∈ H2(RN ) :

∫
RN

V (x)u2dx <∞},

where
H2(RN ) := {u ∈ L2(RN ) : Dαu ∈ L2(RN ),∀α ∈ ZN

+ , |α| ≤ 2}
is the usual Hilbert space with the scalar product

⟨u, v⟩H2 =
∑
|α|≤2

∫
RN

DαuDαv dx

and the norm ∥u∥H2 = ⟨u, u⟩1/2H2 . We define the inner product

⟨u, v⟩ =
∫
RN

[∆u∆v +∇u · ∇v + V (x)uv] dx

and the norm ∥u∥ = ⟨u, u⟩1/2 on E. Then E is a Hilbert space. Moreover, if V∞ = +∞ in the
assumption (A1), then the continuous embedding E ↪→ Ls(RN ), 2 ≤ s < 2∗∗, is compact [3].

Consider the functional defined on E by

I(u) =
1

2

∫
RN

[(∆u)2 + |∇u|2 + V (x)u2] dx+

∫
RN

u2|∇u|2 dx− 1

2∗∗

∫
RN

|u|2
∗∗
dx− α

p

∫
RN

|u|p dx.

In view of the proof of [4, Proposition 2.1], for any u ∈ C∞
c (RN ), we conclude from Sobolev

inequality that there exists a constant C > 0 such that

∥u∥2∗ ≤ C∥∇u∥2 , ∥∂iu∥2∗ ≤ C∥∇∂iu∥2 , for i = 1, 2, . . . , N.

Notice that ∑
i,j

∫
RN

|∂iju|2dx =

∫
RN

|∆u|2dx.

Then we have
∥∇u∥2∗ ≤ C∥∆u∥2.

Recall that N ∈ {5, 6}. It follows from Hölder inequality that∫
RN

u2|∇u|2dx ≤
(∫

RN

|u|Ndx
)2/N(∫

RN

|∇u|2
∗
dx

)N−2
N

= ∥u∥2N∥∇u∥22∗
≤ C∥u∥2N∥∆u∥22 <∞.

(2.1)

On the other hand,∫
RN

u2|∇u|2dx =
1

4

∫
RN

|∇(u2)|2dx ≥ C
(∫

RN

(u2)2
∗
dx

)2/2∗

.



4 J. HE, K. WU, F. ZHOU EJDE-2025/69

Therefore, the functional I is well defined in E. Moreover, it is easy to check that I ∈ C1(E,R)
and

⟨I ′(u), φ⟩ =
∫
RN

[∆u∆φ+∇u∇φ+ V (x)uφ]dx+ 2

∫
RN

(u2∇u∇φ+ uφ|∇u|2) dx

−
∫
RN

|u|2
∗∗−2uφdx− α

∫
RN

|u|p−2uφdx

for all u, φ ∈ E (see [3]). Clearly, solutions of (1.1) are critical points of the functional I.
The following lemma shows that the functional I has the mountain pass geometric structure.

Lemma 2.1. (i) There exist constants ρ, β > 0 such that inf∥u∥=ρ I(u) ≥ β;
(ii) There exists an e ∈ E such that ∥e∥ > ρ and I(e) < 0.

Proof. (i) By the Sobolev inequality, for each u ∈ E with ∥u∥ = ρ, we have

I(u) ≥ 1

2

∫
RN

[(∆u)2 + |∇u|2 + V (x)u2]dx− 1

2∗∗

∫
RN

|u|2
∗∗
dx− α

p

∫
RN

|u|pdx

≥ 1

2
∥u∥2 − C1∥u∥2

∗∗
− C2∥u∥p

=
1

2
ρ2 − C1ρ

2∗∗ − C2ρ
p,

Choose ρ > 0 with 1
2ρ

2 − C1ρ
2∗∗ − C2ρ

p = 1
4ρ

2 := β > 0, then inf∥u∥=ρ I(u) ≥ β.
(ii) Let u ∈ E \ {0} be fixed. Remark that N ∈ {5, 6}. For t ≥ 0, according to Hölder inequality
and (2.1), we have

I(tu) ≤ t2

2

∫
RN

[(∆u)2 + |∇u|2 + V (x)u2]dx+ t4
∫
RN

u2|∇u|2dx− t2
∗∗

2∗∗

∫
RN

|u|2
∗∗
dx

≤ t2

2
∥u∥2 + Ct4∥u∥2N∥∇u∥22∗ − C1t

2∗∗∥u∥2
∗∗

2∗∗

≤ t2

2
∥u∥2 + C2t

4∥u∥2N∥∆u∥22 − C1t
2∗∗∥u∥2

∗∗

2∗∗ → −∞

as t → +∞, which implies that there exists a large t > 0 such that I(tu) < 0. Let e = tu. Then
I(e) < 0. The proof is complete. □

We define the mountain pass level c of I by

c = inf
γ∈Γ

max
t∈[0,1]

I(γ(t)), (2.2)

where Γ = {γ ∈ C([0, 1], E) : γ(0) = 0 and I(γ(1)) < 0}.
To obtain nontrivial solutions of (1.1), we first estimate the mountain pass level value c. We

define the best constant S∗∗ for the Sobolev embedding D2,2(RN ) ↪→ L2∗∗(RN ) by

S∗∗ := inf
{∫

RN

(∆u)2dx : u ∈ D2,2(RN ), ∥u∥2∗∗ = 1
}
, (2.3)

where D2,2(RN ) is the completion of the space C∞
c (RN ) with respect to the norm ∥u∥2,2 =( ∫

RN (∆u)2dx
)1/2

. Clearly,D2,2(RN ) is a Hilbert space with the scalar product (u, v) =
∫
RN ∆u∆v dx,

(see [6]). It is known that the best constant S∗∗ is attained by the function

uε = (N(N − 4)(N2 − 4))(N−4)/8 ε(N−4)/2

(ε2 + |x|2)(N−4)/2
, ∀ε > 0,

see [17]. Moreover, ∥∆uε∥22 = ∥uε∥2
∗∗

2∗∗ = S
N
4
∗∗ and uε satisfies the equation ∆2u = u2

∗∗−1 in RN ,
N ≥ 5.

Recalling that the best constant S∗ for the Sobolev embedding D1,2(RN ) ↪→ L2∗(RN ) is given
by

S∗ := inf
u∈D1,2(RN )∥u∥2∗=1

∥∇u∥22, (2.4)



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 5

where D1,2(RN ) := {u ∈ L2∗(RN ) : ∇u ∈ L2(RN )} is the Hilbert space with the scalar product
(u, v) =

∫
RN ∇u∇v dx. Let

vε = (N(N − 2))(N−2)/8 ε(N−2)/4

(ε2 + |x|2)(N−2)/4
, ∀ε > 0.

We know that the best constant S∗ is attained by the function v2ε for any ε > 0, v2ε satisfies the
equation −∆u = u2

∗−1 in RN , where N ≥ 3. Moreover, by direct computation, we have

(
3

2
)3/2S

3/2
∗∗ = S3

∗ (2.5)

if N = 6.
Let 0 < R < 1 and wε = ϕuε, where ϕ is a smooth cut-off function satisfying ϕ(x) = 1 for

|x| ≤ R and ϕ(x) = 0 for |x| ≥ 2R. Moreover, by (2.5) and arguments as in the proofs of (5.1)-(5.6)
in Appendix, we have the following estimates∫

RN

|wε|2
∗∗
dx = S

N
4
∗∗ +O(εN ) (2.6)∫

RN

|∆wε|2dx = S
N
4
∗∗ +O(εN−4) (2.7)∫

RN

|∇wε|2dx = O(εN−4) (2.8)∫
RN

|∇(w2
ε)|2dx =

{
O(ε2), N = 5
√
6
2 S

3/2
∗∗ +O(ε4), N = 6

(2.9)∫
RN

|wε|qdx = O(εN− q
2 (N−4)),

N

N − 4
< q < 2∗∗ (2.10)∫

RN

|wε|2dx = O(εN−4). (2.11)

For the mountain pass level value c given in (2.2), we have the following estimates.

Lemma 2.2. Let

c∗ =

{
2
5S

5/4
∗∗ , N = 5

( 5
32

√
6 + 11

96

√
22)S

3/2
∗∗ , N = 6.

(i) If 8
N−4 < p < 2∗∗, then c < c∗ for any α > 0.

(ii) If 2 < p ≤ 8
N−4 , then there exists a constant α∗ > 0 such that c < c∗ for all α > α∗.

Proof. Case 1: N = 5. (i) We first consider the case where 8 < p < 2∗∗. Arguing in a similar
way to [27], we define tε > 0 satisfying I(tεwε) = supt≥0 I(twε). We claim that there exist ε0 > 0
and positive constants t1 and t2 such that t1 ≤ tε ≤ t2 for all ε ∈ (0, ε0). From (2.6)-(2.11), there
exists a small ε2 > 0 such that

I(twε) ≤
t2

2

∫
R5

[(∆wε)
2 + |∇wε|2 + V (x)w2

ε ]dx+
t4

4

∫
R5

|∇(w2
ε)|2dx− t2

∗∗

2∗∗

∫
R5

|wε|2
∗∗
dx

≤ t2

2
S
5/4
∗∗ +

t4

4
− t2

∗∗

2∗∗
S
5/4
∗∗

(2.12)

for all ε ∈ (0, ε2). Since I(tεwε) = supt≥0 I(twε) and I(0) = 0, we have I(tεwε) ≥ 0. Hence
t2

∗∗
ε

2∗∗ S
5/4
∗∗ ≤ t2ε

2 S
5/4
∗∗ +

t4ε
4 , which implies that there exists a constant t2 > 0 such that tε ≤ t2 for all

ε ∈ (0, ε2).
Note that 5 < 8 < p < 2∗∗. Again by (2.6)-(2.11), there exists a small ε1 ∈ (0, ε2) such that

I(twε) ≥
t2

2

∫
R5

(∆wε)
2dx− t2

∗∗

2∗∗

∫
R5

|wε|2
∗∗
dx− α

tp

p

∫
R5

|wε|pdx

≥ t2

4
S
5/4
∗∗ − t2

∗∗

2∗∗
S
5/4
∗∗ − αCε5−

p
2 tp



6 J. HE, K. WU, F. ZHOU EJDE-2025/69

for all ε ∈ (0, ε1). Let η = max0≤t≤1(
t2

4 − t2
∗∗

2∗∗ )S
5/4
∗∗ , it is clear that η > 0. Since 5 − p

2 > 0, we

can find a small ε0 < ε1 such that αCε5−
p
2 ≤ η

2 for all ε ∈ (0, ε0). Hence,

I(tεwε) ≥ max
0≤t≤1

{ t
2

4
S
5/4
∗∗ − t2

∗∗

2∗∗
S
5/4
∗∗ − αCε5−

p
2 tp} ≥ η

2
.

It follows from (2.12) that η
2 ≤ I(tεwε) ≤ t2ε

2 S
5/4
∗∗ +

t4ε
4 − t2

∗∗
ε

2∗∗ S
5/4
∗∗ , which implies that there exists

a constant t1 > 0 such that tε ≥ t1 for all ε ∈ (0, ε0). Hence, the claim is true.
For ε ∈ (0, ε0), by (2.6)-(2.11), we have

I(tεwε) ≤
t2ε
2

∫
R5

(∆wε)
2dx− t2

∗∗

ε

2∗∗

∫
R5

|wε|2
∗∗
dx+

t22
2

∫
R5

|∇wε|2dx

+
t22
2

∫
R5

V (x)w2
εdx+

t42
4

∫
R5

|∇(w2
ε)|2dx− α

tp1
p

∫
R5

|wε|pdx

≤ (
t2ε
2
− t2

∗∗

ε

2∗∗
)S

5/4
∗∗ +O(ε) +O(ε2)− αCε5−

p
2

≤ 2

5
S
5/4
∗∗ +O(ε)− αCε5−

p
2 .

Noticing that 5 − p
2 < 1, we see that I(tεwε) <

2
5S

5/4
∗∗ for small ε > 0. Then we can find a small

ε̃ > 0 such that

sup
t≥0

I(twε̃) = I(tε̃wε̃) <
2

5
S
5/4
∗∗ .

Moreover, from (2.12), we conclude that I(twε̃) → −∞ as t→ ∞. Hence, there exists a t̃ > 0 such

that I(t̃wε̃) < 0. Let γ̃(t) = tt̃wε̃. Then γ̃ ∈ Γ and c ≤ maxt∈[0,1] I(γ̃(t)) <
2
5S

5/4
∗∗ for all α > 0.

(ii) We consider the case where 2 < p ≤ 8. For simplicity of notation, we rewrite the functional
I as Iα. Let w0 ∈ C∞

c (R5)\{0}. We define tα > 0 such that Iα(tαw0) = supt≥0 Iα(tw0). We claim
that tα → 0 as α→ +∞. Indeed, if the claim is not true. Then there exist a constant t0 > 0 and
a sequence {αn} such that αn → +∞ and tαn

≥ t0 for all n. Assume that αn ≥ 1 for all n. Set
tn = tαn

and I1 = Iα|α=1, then 0 ≤ Iαn
(tnw0) ≤ I1(tnw0), which implies that tn is bounded from

above. Moreover, we have

Iαn
(tnw0) =

t2n
2

∫
R5

[(∆w0)
2 + |∇w0|2 + V (x)w2

0]dx+
t4n
4

∫
R5

|∇(w2
0)|2dx− t2

∗∗

n

2∗∗

∫
R5

|w0|2
∗∗
dx

− αn
tpn
p

∫
R5

|w0|pdx

≤ t2n
2

∫
R5

[(∆w0)
2 + |∇w0|2 + V (x)w2

0]dx+
t4n
4

∫
R5

|∇(w2
0)|2dx− αn

tpn
p

∫
R5

|w0|pdx

≤ C − αn
tp0
p

∫
R5

|w0|pdx→ −∞

as n→ ∞. This contradicts Iαn
(tnw0) ≥ 0. Hence the claim holds and tα → 0 as α→ +∞.

Clearly,

Iα(tαw0) ≤
t2α
2

∫
R5

[(∆w0)
2 + |∇w0|2 + V (x)w2

0]dx+
t4α
4

∫
R5

|∇(w2
0)|2dx.

This implies that Iα(tαw0) → 0 as α → +∞. Hence, there exists a α∗ > 0 such that Iα(tαw0) =

supt≥0 Iα(tw0) <
2
5S

5/4
∗∗ for all α > α∗. Consequently, c < 2

5S
5/4
∗∗ for all α > α∗.

Case 2: N = 6. (i) 4 < p < 2∗∗. Arguing in a similar way to the proof of (i) with N = 5,
there exist ε0 > 0 and positive constants t1 and t2 such that t1 ≤ tε ≤ t2. For ε ∈ (0, ε0), by



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 7

(2.6)-(2.11), we have

I(tεwε) ≤
t2ε
2

∫
R6

(∆wε)
2dx− t2

∗∗

ε

2∗∗

∫
R6

|wε|2
∗∗
dx+

t4ε
4

∫
R6

|∇(w2
ε)|2dx

+
t22
2

∫
R6

V (x)w2
εdx+

t22
2

∫
R6

|∇wε|2dx− α
tp1
p

∫
R6

|wε|pdx

≤ (
t2ε
2
− t2

∗∗

ε

2∗∗
+

√
6

8
t4ε)S

3/2
∗∗ +O(ε2) +O(ε4)− αCε6−p

≤ (
5

32

√
6 +

11

96

√
22)S

3/2
∗∗ +O(ε2)− αCε6−p.

(2.13)

Note that 6− p < 2. We conclude that c < ( 5
32

√
6 + 11

96

√
22)S

3/2
∗∗ for any α > 0.

(ii) The desired result can be deduced by the same arguments we used in the proof of (ii) in
the case N = 5. □

Remark 2.3. If N = 6, then 2∗∗ = 22∗. For this case, with the aid of (2.5), we can also take
ṽε = ϕvε as a test function to obtain the same estimates in Lemma 2.2. We will show this
statement in the Appendix.

3. (PS)c sequence

Recall that, for any c ∈ R, {un} is a (PS)c sequence of I if I(un) → c and I ′(un) → 0 as
n→ ∞. We have the following results about (PS)c sequence of I.

Lemma 3.1. Assume that the condition (A1) holds and: 2 < p < 2∗∗ if V∞ = +∞, and 4 ≤ p <
2∗∗ if V∞ < +∞. Then any (PS)c sequence of the functional I is bounded in E.

Proof. Let {un} be a (PS)c sequence of the functional I. We deal with two cases separately.

Case 1: 4 ≤ p < 2∗∗ and V∞ < +∞. We have

c+ o(1) = I(un)−
1

p
⟨I ′(un), un⟩

= (
1

2
− 1

p
)

∫
RN

[(∆un)
2 + |∇un|2 + V (x)u2n]dx+ (

1

4
− 1

p
)

∫
RN

|∇(u2n)|2dx

+ (
1

p
− 1

2∗∗
)

∫
RN

|un|2
∗∗
dx

≥ (
1

2
− 1

p
)

∫
RN

[(∆un)
2 + |∇un|2 + V (x)u2n]dx,

which implies that {un} is bounded in E.

Case 2: 2 < p < 2∗∗ and V∞ = +∞. In this case, we have lim|x|→∞ V (x) = +∞. Hence,
for each M > 0, there exists an R > 0 such that V (x) > M as |x| > R. This implies that
meas{x ∈ RN : V (x) ≤M} ≤ meas{x ∈ BR : V (x) ≤M} <∞, where BR := {x ∈ RN : |x| ≤ R}.

We define two real functions f(t) = |t|2∗∗−2t+α|t|p−2t and F (t) =
∫ t

0
f(s)ds = 1

2∗∗ |t|
2∗∗ + α

p |t|
p.

Also we choose a fixed constant q ∈ (4, 2∗∗). Then limt→0
tf(t)−qF (t)

t2 = 0, limt→∞
tf(t)−qF (t)

tq =

+∞, and limt→∞
tf(t)−qF (t)

t2∗∗
= d > 0. Hence, there exists r > 0 such that

tf(t)− qF (t) ≥ 0, ∀|t| ≥ r. (3.1)

Furthermore, for any ε > 0, there exists a positive constant C(ε) such that

|tf(t)− qF (t)| ≤ ε|t|2 + C(ε)|t|2
∗∗
, ∀|t| ∈ R. (3.2)

It follows from (3.1) that

c+ o(1) = I(un)−
1

q
⟨I ′(un), un⟩

= (
1

2
− 1

q
)

∫
RN

[(∆un)
2 + |∇un|2]dx+ (

1

2
− 1

q
)

∫
RN

V (x)u2ndx



8 J. HE, K. WU, F. ZHOU EJDE-2025/69

+ (
1

4
− 1

q
)

∫
RN

|∇(u2n)|2dx+

∫
RN

[
1

q
f(un)un − F (un)]dx

≥ (
1

2
− 1

q
)

∫
RN

[(∆un)
2 + |∇un|2]dx+ (

1

2
− 1

q
)

∫
RN

V (x)u2ndx

+

∫
|un|≤r

[
1

q
f(un)un − F (un)]dx.

By (3.2), there exists a constant M > V0 such that

|1
q
tf(t)− F (t)| ≤ (

1

4
− 1

2q
)Mt2, ∀|t| ≤ r, (3.3)

where V0 is the constant given in the assumption (A1).
By (3.3) and assumption (A1), we have

(
1

4
− 1

2q
)

∫
RN

V (x)u2ndx+

∫
|un|≤r

[
1

q
f(un)un − F (un)]dx

≥ (
1

4
− 1

2q
)

∫
RN

V (x)u2ndx−
∫
|un|≤r

(
1

4
− 1

2q
)Mu2ndx

≥ (
1

4
− 1

2q
)

∫
|un|≤r

(V (x)−M)u2ndx

≥ (
1

4
− 1

2q
)

∫
|un|≤r,V (x)≤M

(V (x)−M)u2ndx

≥ (
1

4
− 1

2q
)(V0 −M)r2(meas({x ∈ RN : V (x) ≤M} ∩ {x ∈ RN : |un| ≤ r}))

≥ (
1

4
− 1

2q
)(V0 −M)r2(meas{x ∈ RN : V (x) ≤M}),

which implies that

(
1

2
− 1

q
)

∫
RN

[(∆un)
2 + |∇un|2]dx+ (

1

4
− 1

2q
)

∫
RN

V (x)u2ndx

≤ (
1

4
− 1

2q
)(M − V0)r

2(meas{x ∈ RN : V (x) ≤M}) + c+ o(1).

Hence {un} is bounded in E. □

Lemma 3.2. Let ρ > 0 and {un} ⊂ E be a bounded (PS)c sequence of I. If 0 < c < c∗, then
there exist a sequence {yn} ⊂ RN and a constant ξ > 0 such that

lim sup
n→∞

∫
Bρ(yn)

|un|2dx ≥ ξ.

Proof. Suppose that the conclusion does not hold, it follows from [25, Lemma 1.21] that∫
RN

|un|sdx→ 0, ∀s ∈ (2, 2∗∗). (3.4)

Case 1: N = 5 and c < c∗ = 2
5S

5/4
∗∗ . From (3.4) and (2.1), we have

o(1) = ⟨I ′(un), un⟩ =
∫
R5

[(∆un)
2 + |∇un|2 + V (x)u2n]dx−

∫
R5

|un|2
∗∗
dx+ o(1).

This yields ∥un∥2 −
∫
R5 |un|2

∗∗
dx = o(1). We may assume that

∥un∥2 → b,

∫
R5

|un|2
∗∗
dx→ b.

Since c > 0, it is easy to check that b > 0. From the definition of S∗∗, we have

∥un∥2 ≥ ∥∆un∥22 ≥ S∗∗∥un∥22∗∗ , (3.5)



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 9

which implies that b ≥ S∗∗b
1
5 . Thus b ≥ S

5/4
∗∗ and

c = lim
n→∞

I(un)

= lim
n→∞

[
1

2

∫
R5

((∆un)
2 + |∇un|2 + V (x)u2n)dx+

1

4

∫
R5

|∇(u2n)|2dx− 1

2∗∗

∫
R5

|un|2
∗∗
dx]

≥ lim
n→∞

[
1

2

∫
R5

((∆un)
2 + |∇un|2 + V (x)u2n)dx− 1

2∗∗

∫
R5

|un|2
∗∗
dx]

= (
1

2
− 1

2∗∗
)b

≥ 2

5
S
5/4
∗∗ ,

which contradicts c < 2
5S

5/4
∗∗ .

Case 2: N = 6 and c < c∗ = ( 5
32

√
6 + 11

96

√
22)S

3/2
∗∗ . Applying (3.4) again, we have

o(1) = ⟨I ′(un), un⟩ = ∥un∥2 +
∫
R6

|∇(u2n)|2dx−
∫
R6

|un|2
∗∗
dx+ o(1).

We assume that

∥un∥2 +
∫
R6

|∇(u2n)|2dx→ b > 0,∫
R6

|un|2
∗∗
dx→ b.

Recall that N = 6, we have 22∗ = 2∗∗. It follows from the definition of S∗ that∫
R6

|∇(u2n)|2dx ≥ S∗

(∫
R6

|un|2
∗∗
dx

)2/2∗

. (3.6)

Combining this with (3.5) and (3.6), we obtain

∥un∥2 +
∫
R6

|∇(u2n)|2dx ≥ S∗∗∥un∥22∗∗ +

∫
R6

|∇(u2n)|2dx ≥ S∗∗∥un∥22∗∗ + S∗

(∫
R6

|un|2
∗∗
dx

)2/2∗

.

Hence, b ≥ S∗∗b
2/2∗∗ + S∗b

2/2∗ > 0. Then we have

b ≥
(−S∗ +

√
S2
∗ + 4S∗∗

2S∗∗

)−3

(3.7)

because b > 0. It then follows from (2.5), (3.4), (3.5), and (3.7) that

c = lim
n→∞

I(un)

= lim
n→∞

[
1

4
∥un∥2 +

1

4
∥un∥2 +

1

4

∫
R6

|∇(u2n)|2dx− 1

2∗∗

∫
R6

|un|2
∗∗
dx]

≥ lim
n→∞

[
1

4
S∗∗∥un∥22∗∗ +

1

4
(∥un∥2 +

∫
R6

|∇(u2n)|2dx)−
1

2∗∗

∫
R6

|un|2
∗∗
dx]

=
1

4
S∗∗b

2/2∗∗ + (
1

4
− 1

2∗∗
)b

≥ 1

4
S∗∗

(−S∗ +
√
S2
∗ + 4S∗∗

2S∗∗

)− 6
2∗∗

+
(1
4
− 1

2∗∗
)(−S∗ +

√
S2
∗ + 4S∗∗

2S∗∗

)−3

=
(1
2

1√
11
2 −

√
3
2

+
2

3

1

(
√

11
2 −

√
3
2 )

3

)
S
3/2
∗∗

=
[
(
2

32

√
6 +

6

96

√
22) + (

3

32

√
6 +

5

96

√
22)

]
S
3/2
∗∗

=
( 5

32

√
6 +

11

96

√
22
)
S
3/2
∗∗

which contradicts c < ( 5
32

√
6 + 11

96

√
22)S

3/2
∗∗ . The proof is complete. □



10 J. HE, K. WU, F. ZHOU EJDE-2025/69

4. Proof of main results

To prove Theorem 1.1, we need some lemmas. We define a C1 functional I∞ : H2(RN ) → R by

I∞(u) =
1

2

∫
RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+

∫
RN

u2|∇u|2dx− 1

2∗∗

∫
RN

|u|2
∗∗
dx− α

p

∫
RN

|u|pdx,

and define

c∞ = inf
γ∈Γ

max
t∈[0,1]

I∞(γ(t)),

where Γ = {γ ∈ C([0, 1], H2(RN )) : γ(0) = 0 and I∞(γ(1)) < 0}.

Lemma 4.1. Assume that (A1) holds and V∞ < +∞. Then c ≤ c∞, where c is the mountain
pass level given by (2.2).

Proof. By condition (A1), we have V (x) ≤ V∞ for any x ∈ RN , then
∫
RN V (x)u2dx ≤

∫
RN V∞u

2dx

for all u ∈ E. Hence, I(u) ≤ I∞(u) for any u ∈ E. By the definition of Γ and Γ, we have Γ ⊂ Γ.
Therefore,

inf
γ∈Γ

max
t∈[0,1]

I∞(γ(t)) ≥ inf
γ∈Γ

max
t∈[0,1]

I∞(γ(t)) ≥ inf
γ∈Γ

max
t∈[0,1]

I(γ(t)).

The proof is complete. □

We define the Nehari manifold

M := {u ∈ E\{0} : ⟨I ′∞(u), u⟩ = 0},
and m = infu∈M I∞(u).

Lemma 4.2. Assume that (A1) holds and V∞ < +∞. Then for any u ∈ E\{0}, there exists
t(u) > 0 such that t(u)u ∈M .

Proof. Let u ∈ E\{0} and f(t) = I∞(tu), t ∈ [0,∞). Then

f(t) = I∞(tu)

=
t2

2

∫
RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+ t4

∫
RN

u2|∇u|2dx

− t2
∗∗

2∗∗

∫
RN

|u|2
∗∗
dx− αtp

p

∫
RN

|u|pdx.

(4.1)

Obviously, we have f ′(t) = 0 ⇔ tu ∈M which is also equivalent to∫
RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+ t2

∫
RN

|∇(u2)|2dx = t2
∗∗−2∥u∥2

∗∗

2∗∗ + αtp−2∥u∥pp.

It is clear that f(0) = 0, f(t) > 0 for small t > 0 and f(t) < 0 for large t > 0. Hence,
maxt∈[0,∞) I∞(tu) is achieved at some t = t(u). So f ′(t(u)) = 0 and t(u)u ∈ M . The proof is
complete. □

Lemma 4.3. Assume that (A1) holds. If V∞ < +∞ and 4 ≤ p < 2∗∗, then for all u ∈ M , it
holds I∞(u) ≥ I∞(tu) for all t ≥ 0.

Proof. Our proof depends on the following inequality. For all 1 < r ≤ s, it holds

tr − 1

r
≤ ts − 1

s
, ∀t ≥ 0. (4.2)

Indeed, it is easy to check that the maximum of function h(t) = tr

r − ts

s is h(1).
For u ∈M , we have ⟨I ′∞(u), u⟩ = 0, hence∫

RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+

∫
RN

|∇(u2)|2dx =

∫
RN

|u|2
∗∗
dx+ α

∫
RN

|u|pdx.

Combining this with (4.2) we have

I∞(u)− I∞(tu)



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 11

= (
1

2
− t2

2
)

∫
RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+ (

1

4
− t4

4
)

∫
RN

|∇(u2)|2dx

+ (
t2

∗∗

2∗∗
− 1

2∗∗
)

∫
RN

|u|2
∗∗
dx+ (

tp

p
− 1

p
)α

∫
RN

|u|pdx

≥ (
1

2
− t2

2
)

∫
RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+ (

1

4
− t4

4
)

∫
RN

|∇(u2)|2dx

+ (
tp

p
− 1

p
)[

∫
RN

|u|2
∗∗
dx+ α

∫
RN

|u|pdx]

= (
1− t2

2
+
tp − 1

p
)

∫
RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+ (

1− t4

4
+
tp − 1

p
)

∫
RN

|∇(u2)|2dx.

The proof is complete. □

Lemma 4.4. Assume that (A1) holds and V∞ < +∞. Then m = c∞.

Proof. We define

c1 = inf
u∈E\{0}

max
t≥0

I∞(tu).

Note that p ≥ 4. We see that, for each u ∈ E

I∞(u)− 1

4
⟨I ′∞(u), u⟩

=
1

4

∫
RN

[(∆u)2 + |∇u|2 + V∞u
2]dx+ (

1

4
− 1

2∗∗
)

∫
RN

|u|2
∗∗
dx+ (

α

4
− α

p
)

∫
RN

|u|pdx ≥ 0.

For each γ ∈ Γ, let h(t) = ⟨I ′∞(γ(t)), γ(t)⟩. By the Sobolev inequality, it is easy to check that
there exists a constant ρ > 0 such that

inf
∥u∥=ρ

⟨I ′∞(u), u⟩ > 0

and ∥γ(1)∥ > ρ. This implies that there exists a t1 ∈ (0, 1) such that h(t1) > 0. By the definition
of Γ, we have h(1) = ⟨I ′∞(γ(1)), γ(1)⟩ ≤ 4I∞(γ(1)) < 0 for all γ ∈ Γ. Hence, there exists a
tγ0 ∈ (t1, 1) such that h(tγ0) = 0, which implies that γ(tγ0) ∈M . Therefore,

max
t∈[0,1]

I∞(γ(t)) ≥ I∞(γ(tγ0)) ≥ inf
u∈M

I∞(u) = m.

Then

c∞ = inf
γ∈Γ

max
t∈[0,1]

I∞(γ(t)) ≥ m.

On the other hand, for any u ∈ E\{0}, we have I∞(0) = 0, I∞(tu) > 0 for small t > 0 and
I∞(tu) < 0 for large t > 0. Hence, there exists a t∗ > 0 such that I∞(t∗u) < 0 for all t ≥ t∗. Let
γu(t) = tt∗u, t ∈ [0, 1], then γu ∈ Γ. We conclude that

c∞ ≤ max
t∈[0,1]

I∞(γu(t)) = max
t∈[0,1]

I∞(tt∗u) ≤ max
t≥0

I∞(tu),

which implies that

c∞ ≤ inf
u∈E\{0}

max
t≥0

I∞(tu) = c1.

For u ∈M , by Lemma 4.3, we obtain I∞(u) ≥ maxt≥0 I∞(tu). Therefore,

m = inf
u∈M

I∞(u) ≥ inf
u∈M

max
t≥0

I∞(tu) ≥ inf
u∈E\{0}

max
t≥0

I∞(tu) = c1,

and so m = c∞. The proof is complete. □

Proof of Theorem 1.1. (i) Let c be the mountain pass level given in (2.2). By the mountain pass
theorem [25, Theorem 1.15] and Lemma 2.1, there exists a sequence {un} ⊂ E such that I(un) → c
and I ′(un) → 0 as n → ∞. From Lemma 3.1, {un} is bounded in E under the assumptions of
Theorem 1.1. Up to a subsequence, we may assume that un ⇀ u in E and un → u in Ls

loc(RN ),
2 ≤ s < 2∗∗. Hence ⟨I ′(un), φ⟩ → ⟨I ′(u), φ⟩ for any φ ∈ C∞

c (RN ), that is, u is a weak solution of



12 J. HE, K. WU, F. ZHOU EJDE-2025/69

problem (1.1). We have to show that u ̸= 0 or find a nontrivial solution if u = 0. We distinguish
two cases:

Case 1: V∞ < +∞ and 8
N−4 < p < 2∗∗. If V∞ ̸= V (x), we verify that u ̸= 0. Indeed, suppose

by contradiction that u = 0. Since lim|x|→∞ V (x) = V∞, for all ε > 0, there exists an R > 0 such
that |V (x)− V∞| < ε as |x| > R. Hence,

|I(un)− I∞(un)| = |1
2

∫
RN

(V (x)− V∞)u2ndx|

≤ 1

2

∫
RN\BR(0)

|V (x)− V∞|u2ndx+
1

2

∫
BR(0)

|V (x)− V∞|u2ndx

≤ 1

2
ε

∫
RN\BR(0)

u2ndx+ C

∫
BR(0)

u2ndx

≤ Cε+ o(1)

as n→ ∞. Combining this with I(un) → c, we have I∞(un) → c. Moreover, for each φ ∈ C∞
c (RN ),

|⟨I ′(un), φ⟩ − ⟨I ′∞(un), φ⟩|

= |
∫
RN

(V (x)− V∞)unφdx|

≤
∫
RN\BR(0)

|V (x)− V∞||un||φ| dx+

∫
BR(0)

|V (x)− V∞||un||φ| dx

≤ ε

∫
RN\BR(0)

|un||φ| dx+ C

∫
BR(0)

|un||φ| dx

≤ ε
(∫

RN\BR(0)

u2n dx
)1/2(∫

RN\BR(0)

φ2 dx
)1/2

+ C
(∫

BR(0)

u2n dx
)1/2(∫

BR(0)

φ2 dx
)1/2

≤ Cε+ o(1)

as n → ∞. Combining with I ′(un) → 0, we have I ′∞(un) → 0. Therefore, {un} is a (PS)c
sequence of the functional I∞.

By Lemmas 2.2 and 3.2, for a fixed ρ > 0, for all α > 0, there exist {yn} ⊂ RN and ξ > 0 such
that

lim sup
n→∞

∫
Bρ(yn)

|un|2dx ≥ ξ. (4.3)

It is easy to verify that {yn} is unbounded in RN . Indeed, if {yn} is bounded, then there exists an
r > 0 such that Bρ(yn) ⊂ Br(0). According to (4.3), we have ξ ≤

∫
Bρ(yn)

|un|2dx ≤
∫
Br(0)

|un|2dx.
Recall that un → u in L2

loc(RN ). We obtain ξ ≤ limn→∞
∫
Br(0)

|un|2dx =
∫
Br(0)

|u|2dx = 0. This

is a contradiction. Thus, {yn} is unbounded. Up to a subsequence, we may assume that |yn| → ∞
as n→ ∞.

Recall that V (x) ̸= V∞. It follows from condition (A1) that there exist a constant ρ̃ > 0,
x0 ∈ RN and a neighborhood Bρ̃(x0) of x0 such that σ := V∞ − V (x0) > 0 and V∞ − V (x) > 1

2σ
for all x ∈ Bρ̃(x0). Let vn(x) = un(x + yn − x0). Then {vn} is bounded in E. We may assume
that vn ⇀ v in E and vn → v in L2

loc(RN ). By (4.3), we have∫
Bρ(x0)

|v(x)|2dx = lim
n→∞

∫
Bρ(x0)

|vn(x)|2dx

= lim
n→∞

∫
Bρ(yn)

|vn(x− yn + x0)|2dx

= lim
n→∞

∫
Bρ(yn)

|un(x)|2dx ≥ ξ.

(4.4)

This implies that v ̸= 0 and∫
RN

(V∞ − V (x))v2dx ≥
∫
Bρ(x0)

(V∞ − V (x))v2dx ≥ 1

2
σξ > 0.



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 13

Moreover, since {un} is a (PS)c sequence of I∞, then {vn} is also a (PS)c sequence of I∞. Thus
v is a critical point of I∞ and v ∈ M . Applying Lemma 4.4, Fatou’s lemma and the weak lower
semi-continuity, we have

c∞ = m = inf
u∈M

I∞(u) ≤ I∞(v)

= I∞(v)− 1

4
⟨I ′∞(v), v⟩

=
1

4

∫
RN

[(∆v)2 + |∇v|2 + V∞v
2]dx+ (

1

4
− 1

2∗∗
)

∫
RN

|v|2
∗∗
dx+ (

α

4
− α

p
)

∫
RN

|v|pdx

≤ 1

4

∫
RN

[(∆v)2 + |∇v|2]dx+
1

4
lim inf
n→∞

∫
RN

V (x+ yn − x0)v
2
ndx+ (

1

4
− 1

2∗∗
)

∫
RN

|v|2
∗∗
dx

+ (
α

4
− α

p
)

∫
RN

|v|pdx

≤ 1

4
lim inf
n→∞

∫
RN

[(∆vn)
2 + |∇vn|2]dx+

1

4
lim inf
n→∞

∫
RN

V (x+ yn − x0)v
2
ndx

+ (
1

4
− 1

2∗∗
) lim inf

n→∞

∫
RN

|vn|2
∗∗
dx+ (

α

4
− α

p
) lim inf

n→∞

∫
RN

|vn|pdx

≤ 1

4
lim inf
n→∞

∫
RN

[(∆un(x+ yn − x0))
2 + |∇un(x+ yn − x0)|2

+ V (x+ yn − x0)|un(x+ yn − x0)|2]dx+ (
1

4
− 1

2∗∗
) lim inf

n→∞

∫
RN

|un(x+ yn − x0)|2
∗∗
dx

+ (
α

4
− α

p
) lim inf

n→∞

∫
RN

|un(x+ yn − x0)|pdx

≤ lim
n→∞

(I(un)−
1

4
⟨I ′(un), un⟩) = c.

It then follows from Lemma 4.1 that

c = I∞(v). (4.5)

We define m′ = infu∈M ′ I(u), where M ′ = {u ∈ E\{0} : ⟨I ′(u), u⟩ = 0} Arguing in a similar
way to lemmas 4.2 and 4.4, we conclude that c = m′ and there exists a t0(v) > 0 such that
t0(v)v ∈M ′. Note that 2∗∗ > p > 8

N−4 ≥ 4. In view of (4.5) and Lemma 4.3, we have

c = I∞(v) ≥ max
t≥0

I∞(tv)

≥ I∞(t0(v)v)

= I(t0(v)v) +
1

2

∫
RN

(V∞ − V (x))(t0(v)v)
2dx

≥ inf
u∈M ′

I(u) +
1

2

∫
RN

(V∞ − V (x))(t0(v)v)
2dx

= c+
1

2

∫
RN

(V∞ − V (x))(t0(v)v)
2dx > c,

This is a contradiction. Hence, u ̸= 0 is a nontrivial solution of (1.1) if V∞ ̸= V (x)
Now we turn to prove that (1.1) has a nontrivial solution for each α > 0 if V∞ < +∞,

V (x) ≡ V∞ and 8
N−4 < p < 2∗∗. In this case, the conclusion follows if u ̸= 0. If u = 0, by the

same argument as used above, {un} is a (PS)c sequence of the functional I∞. Moreover, we can
find a sequence {yn} ⊂ RN and a constant ρ > 0 such that |yn| → ∞ and

lim sup
n→∞

∫
Bρ(yn)

|un|2dx ≥ ξ.

We define vn(x) = un(x + yn). Then {vn} is a bounded (PS)c sequence of I∞. Assume that
vn ⇀ v in E. Then v ̸= 0 is a critical point of I∞. Notice that I = I∞. v is a nontrivial solution
of (1.1).



14 J. HE, K. WU, F. ZHOU EJDE-2025/69

Case 2: V∞ = +∞ and 8
N−4 < p < 2∗∗. It is known that the embedding E ↪→ L2(RN ) is

compact if V∞ = +∞. Thus un → u in L2(RN ). Again by Lemma 2.2 and Lemma 3.2, for a fixed
ρ > 0, for any α > 0, there exist {yn} ⊂ RN and ξ > 0 such that

lim sup
n→∞

∫
Bρ(yn)

|un|2dx ≥ ξ.

This implies that ∫
RN

u2dx = lim
n→∞

∫
RN

|un|2dx ≥ ξ

and u is a nontrivial solution of (1.1).
Conclusion (ii) can be shown in the same way as in the proof of (i) in Theorem 1.1. The proof

is complete. □

5. Appendix

The aim of this section is to give some critical estimates for the test function ṽε we mentioned
in the Section 2. We mainly focus on the case N = 6. As a result, the same estimates as ones of
Lemma 2.2 are obtained if N = 6. Recall that

vε = (N(N − 2))(N−2)/8 ε(N−2)/4

(ε2 + |x|2)(N−2)/4
, ∀ε > 0.

It is known that ∥∇v2ε∥22 = ∥v2ε∥2
∗

2∗ = S
N
2
∗ and v2ε satisfies the equation −∆u = u2

∗−1 in RN , N ≥ 3.
We are in the position to verify the following estimates∫

RN

|ṽε|2
∗∗
dx = S

N
2
∗ +O(εN ) (5.1)∫

RN

|∇(ṽ2ε)|2dx = S
N
2
∗ +O(εN−2) (5.2)∫

RN

|∆ṽε|2dx =
2

3
S3
∗ +O(ε2), N = 6, (5.3)∫

RN

|∇ṽε|2dx = O(ε
N−2

2 |lnε|) (5.4)∫
RN

|ṽε|qdx = O(εN− q
4 (N−2)), 2∗ < q < 2∗∗ (5.5)∫

RN

|ṽε|2dx = O(ε
N−2

2 ). (5.6)

We see that

∂vε
∂xi

= (N(N − 2))(N−2)/8ε(N−2)/4(−N − 2

2
)

xi
(ε2 + |x|2)(N+2)/4

,

|∇vε|2 = (N(N − 2))(N−2)/4ε(N−2)/2(−N − 2

2
)2

|x|2

(ε2 + |x|2)(N+2)/2
,

∆vε = (N(N − 2))(N−2)/8(−N − 2

2
)ε(N−2)/4 Nε2 + N−2

2 |x|2

(ε2 + |x|2)(N+6)/4
.

Then, ∫
RN

|ṽε|2
∗∗
dx =

∫
RN

|ϕvε|22
∗
dx =

∫
|x|<R

|vε|22
∗
dx+

∫
R≤|x|≤2R

|ϕvε|22
∗
dx,

and ∫
|x|<R

|vε|22
∗
dx ≤

∫
RN

|vε|22
∗
dx+

∫
RN\|x|<R

|vε|22
∗
dx

= S
N
2
∗ + (N(N − 2)

N
2 εN

∫
RN\|x|<R

1

(ε2 + |x|2)N
dx



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 15

= S
N
2
∗ + (N(N − 2)

N
2 ωN−1

∫ +∞

R/ε

rN−1

(1 + r2)N
dr

≤ S
N
2
∗ + (N(N − 2)

N
2 ωN−1

∫ +∞

R/ε

rN−1

r2N
dr

≤ S
N
2
∗ + CεN ,

where ωN−1 is the surface area of the unit sphere in RN . Similarly,
∫
R≤|x|≤2R

|ϕvε|22
∗
dx ≤ CεN .

Hence,
∫
RN |ṽε|2

∗∗
dx = S

N
2
∗ +O(εN ) and (5.1) holds. Moreover, we have∫

RN

|∇(ṽ2ε)|2dx =

∫
RN

[ϕ4|∇(v2ε)|2 + 2ϕ2∇(v2ε)v
2
ε∇ϕ2 + v4ε |∇ϕ2|2]dx

=

∫
|x|<R

|∇(v2ε)|2dx+

∫
R≤|x|≤2R

ϕ4|∇(v2ε)|2dx

+

∫
R≤|x|≤2R

2ϕ2∇(v2ε)v
2
ε∇ϕ2dx+

∫
R≤|x|≤2R

v4ε |∇ϕ2|2dx,

∫
|x|<R

|∇(v2ε)|2dx =

∫
RN

|∇(v2ε)|2dx+

∫
RN\|x|<R

|∇(v2ε)|2dx

= S
N
2
∗ + (N(N − 2)

N−2
2 (N − 2)2εN−2

∫
RN\|x|<R

|x|2

(ε2 + |x|2)N
dx

= S
N
2
∗ + (N(N − 2)

N−2
2 (N − 2)2ωN−1

∫ +∞

R/ε

rN+1

(1 + r2)N
dr

≤ S
N
2
∗ + (N(N − 2)

N−2
2 (N − 2)2ωN−1

∫ +∞

R/ε

rN+1

r2N
dr

≤ S
N
2
∗ + CεN−2

and∫
R≤|x|≤2R

2ϕ2∇(v2ε)v
2
ε∇ϕ2dx ≤ C

(∫
R≤|x|≤2R

|∇(v2ε)|2dx
)1/2(∫

R≤|x|≤2R

v4εdx
)1/2

≤ (CεN−2)1/2
(
(N(N − 2))

N−2
2 εN−2

∫
R≤|x|≤2R

1

(ε2 + |x|2)N−2
dx

)1/2

= (CεN−2)1/2
(
(N(N − 2))

N−2
2 ε2ωN−1

∫ 2R/ε

R/ε

rN−1

(1 + r2)N−2
dr
)1/2

≤ (CεN−2)1/2(CεN−2)1/2

= CεN−2.

Similarly,
∫
R≤|x|≤2R

ϕ4|∇(v2ε)|2dx ≤ CεN−2 and
∫
R≤|x|≤2R

v4ε |∇ϕ2|2dx ≤ CεN−2. Therefore,∫
RN |∇(ṽ2ε)|2dx = S

N
2
∗ +O(εN−2) and (5.2) is true.

Since ∆(ϕvε) = div(∇(ϕvε)) = vε∆ϕ+ 2∇vε∇ϕ+ ϕ∆vε, we obtain∫
RN

|∆ṽε|2dx =

∫
RN

[(∆ϕ)2v2ε + 4(∇ϕ∇vε)2 + (∆vε)
2ϕ2 + 4vε∆ϕ∇vε∇ϕ]dx

+

∫
RN

[4∇vε∇ϕ∆vεϕ+ 2vε∆ϕ∆vεϕ]dx,



16 J. HE, K. WU, F. ZHOU EJDE-2025/69∫
RN

(∆ϕ)2v2εdx ≤ C

∫
R≤|x|≤2R

v2εdx

= C((N(N − 2))
N−2

4 ε
N−2

2

∫
R≤|x|≤2R

1

(ε2 + |x|2)N−2
2

dx

= C((N(N − 2))
N−2

4 ε
N+2

2 ωN−1

∫ 2R/ε

R/ε

rN−1

(1 + r2)
N−2

2

dr

≤ Cε
N−2

2 ,∫
RN

(∇ϕ∇vε)2dx ≤ C

∫
R≤|x|≤2R

|∇vε|2dx

= C((N(N − 2))
N−2

4 (
N − 2

2
)2ε

N−2
2

∫
R≤|x|≤2R

|x|2

(ε2 + |x|2)N+2
2

dx

= C((N(N − 2))
N−2

4 (
N − 2

2
)2ε

N−2
2 ωN−1

∫ 2R/ε

R/ε

rN+1

(1 + r2)
N+2

2

dr

≤ Cε
N−2

2 |lnε|.

It is easy to check that

(∆vε)
2 = (N(N − 2))

N−2
4 (

N − 2

2
)2ε

N−2
2
N2ε4 + (N−2)2

4 |x|4 + (N − 2)Nε2|x|2

(ε2 + |x|2)N+6
2

,

(∆uε)
2 = (N(N − 4)(N2 − 4))

N−4
4 (N − 4)2εN−4N

2ε4 + 4|x|4 + 4Nε2|x|2

(ε2 + |x|2)N
.

If N = 6, it follows from (2.5) that∫
RN

(∆vε)
2dx =

1

4

N1/2(N − 2)
5
2

(N − 4)
N+4

4 (N + 2)
N−4

4

∫
RN

(∆uε)
2dx

=
1

4

N1/2(N − 2)
5
2

(N − 4)
N+4

4 (N + 2)
N−4

4

S
N
4
∗∗

=
2

3
S

N
2
∗

which implies ∫
RN

(∆vε)
2ϕ2dx =

∫
|x|≤R

(∆vε)
2dx+

∫
R≤|x|≤2R

(∆vε)
2ϕ2dx

and∫
|x|≤R

(∆vε)
2dx

≤
∫
RN

(∆vε)
2 dx+

∫
RN\|x|≤R

(∆vε)
2 dx

=
2

3
S

N
2
∗ +

∫
RN\|x|≤R

(N(N − 2))
N−2

4 (
N − 2

2
)2ε

N−2
2
N2ε4 + (N−2)2

4 |x|4 + (N − 2)Nε2|x|2

(ε2 + |x|2)N+6
2

dx

≤ 2

3
S

N
2
∗ + C1ε

N
2 −3

∫ +∞

R/ε

rN−1

rN+6
dr + C2ε

N
2 −3

∫ +∞

R/ε

rN+3

rN+6
dr + C3ε

N
2 −3

∫ +∞

R/ε

rN+1

rN+6
dr

≤ 2

3
S

N
2
∗ + C4ε

N
2 +3 + C5ε

N−2
2 + C6ε

N+2
2

≤ 2

3
S

N
2
∗ + Cε

N−2
2 .

Similarly,
∫
R≤|x|≤2R

(∆vε)
2ϕ2dx ≤ Cε

N−2
2 . Therefore,

∫
RN (∆vε)

2ϕ2dx ≤ 2
3S

2
N
∗ + Cε

N−2
2 .



EJDE-2025/69 BIHARMONIC EQUATIONS WITH CRITICAL GROWTH 17

It is easy to verify that∫
RN

vε∆ϕ∇vε∇ϕdx =

∫
R≤|x|≤2R

vε∆ϕ∇vε∇ϕdx

≤
(∫

R≤|x|≤2R

v2εdx
)1/2(∫

R≤|x|≤2R

|∇vε|2dx
)1/2

≤ Cε
N−2

2 .

Similarly, we have
∫
RN ∇vε∇ϕ∆vεϕdx ≤ Cε

N−2
2 and

∫
RN vε∆ϕ∆vεϕdx ≤ Cε

N−2
2 . Therefore,∫

RN |∆ṽε|2dx = 2
3S

N
2
∗ +O(ε

N−2
2 ) and (5.3) follows.

By the same arguments as in the proof of (5.1)-(5.3), we have
∫
RN |∇ṽε|2dx ≤ Cε

N−2
2 |lnε|,∫

RN |ṽε|qdx ≤ CεN− q
4 (N−2), 2∗ < q < 2∗∗ and

∫
RN |ṽε|2dx ≤ Cε

N−2
2 . We then conclude (5.4)-

(5.6).
According to (5.1)-(5.6), we have the following conclusion.

Lemma 5.1. Let N = 6.

(i) If 2(N+2)
N−2 < p < 2∗∗, then c < ( 5

32

√
6 + 11

96

√
22)S

N
4
∗∗ for any α > 0.

(ii) If 2 < p ≤ 2(N+2)
N−2 , then there exists a constant α∗ > 0 such that c < ( 5

32

√
6 + 11

96

√
22)S

N
4
∗∗

for all α > α∗.

Note that 2(N+2)
N−2 = 8

N−4 if N = 6. Hence, if N = 6, then the results in Lemma 5.1 are the
same as ones in Lemma 2.2.

Proof. Case 1: 4 < p < 2∗∗. Arguing as in [27], we define tε > 0 satisfying I(tεṽε) = supt≥0 I(tṽε).
We claim that there exist ε0 > 0 and positive constants t1 and t2 such that t1 ≤ tε ≤ t2 for each
ε ∈ (0, ε0). From (5.1)-(5.6), there exists a small ε2 > 0 such that

I(tṽε) ≤
t2

2

∫
R6

[(∆ṽε)
2 + |∇ṽε|2 + V (x)ṽ2ε ]dx+

t4

4

∫
R6

|∇(ṽ2ε)|2dx− t2
∗∗

2∗∗

∫
R6

|ṽε|2
∗∗
dx

≤ t2

3
S3
∗ +

t4

4
S3
∗ − t2

∗∗

2∗∗
S3
∗

(5.7)

for all ε ∈ (0, ε2). Since I(tεṽε) = supt≥0 I(tṽε) and I(0) = 0, we have I(tεṽε) ≥ 0. Hence
t2

∗∗
ε

2∗∗ S
3
∗ ≤ t2ε

3 S
3
∗ +

t4ε
4 S

3
∗ , which implies that there exists a constant t2 > 0 such that tε ≤ t2 for all

ε ∈ (0, ε2).
Note that 2∗ < 4 < p < 2∗∗. Again by (5.1)-(5.6), there exists a small ε1 ∈ (0, ε2) such that

I(tṽε) ≥
t2

2

∫
R6

(∆ṽε)
2dx+

t4

4

∫
R6

|∇(ṽ2ε)|2dx− t2
∗∗

2∗∗

∫
R6

|ṽε|2
∗∗
dx− α

tp

p

∫
R6

|ṽε|pdx

≥ t2

4
× 2

3
S3
∗ +

t4

4
S3
∗ − t2

∗∗

2∗∗
S3
∗ − αCε6−ptp

for all ε ∈ (0, ε1). Let η = max0≤t≤1(
t2

6 + t4

4 − t2
∗∗

2∗∗ )S
3
∗ , it is clear that η > 0. Since 6− p > 0, we

can find a small ε0 < ε1 such that αCε6−p ≤ η
2 for all ε ∈ (0, ε0). Therefore,

I(tεṽε) ≥ max
0≤t≤1

{ t
2

6
S3
∗ +

t4

4
S3
∗ − t2

∗∗

2∗∗
S3
∗ − αCε6−ptp} ≥ η

2
.

It follows from (5.7) that η
2 ≤ I(tεṽε) ≤ t2ε

3 S
3
∗ +

t4ε
4 S

3
∗ − t2

∗∗
ε

2∗∗ S
3
∗ , which implies that there exists a

constant t1 > 0 such that tε ≥ t1 for all ε ∈ (0, ε0). The claim is true.
For ε ∈ (0, ε0), by (5.1)-(5.6), we have

I(tεṽε) ≤
t2ε
2

∫
R6

(∆ṽε)
2dx− t2

∗∗

ε

2∗∗

∫
R6

|ṽε|2
∗∗
dx+

t4ε
4

∫
R6

|∇(ṽ2ε)|2dx

+
t22
2

∫
R6

V (x)ṽ2εdx+
t22
2

∫
R6

|∇ṽε|2dx− α
tp1
p

∫
R6

|ṽε|pdx



18 J. HE, K. WU, F. ZHOU EJDE-2025/69

≤ (
t2ε
3
+
t4ε
4
− t2

∗∗

ε

2∗∗
)S3

∗ +O(ε2|lnε|) +O(ε2)− αCε6−p

≤ (
11

72

√
11

3
+

5

24
)S3

∗ +O(ε2|lnε|)− αCε6−p.

Noticing that 6 − p < 2, we see that I(tεṽε) < ( 1172

√
11
3 + 5

24 )S
3
∗ for small ε > 0. Combining this

with (2.5), we have I(tεṽε) < ( 1196
√
22 + 5

32

√
6)S

3/2
∗∗ . Hence we can find a small ε̃ > 0 such that

sup
t≥0

I(tṽε̃) = I(tε̃ṽε̃) < (
11

96

√
22 +

5

32

√
6)S

3/2
∗∗ .

Moreover, from (5.7), we conclude that I(tṽε̃) → −∞ as t→ ∞. Hence, there exists a t̃ > 0 such

that I(t̃ṽε̃) < 0. Let γ̃(t) = tt̃ṽε̃, then γ̃ ∈ Γ and c ≤ maxt∈[0,1] I(γ̃(t)) < ( 1196
√
22 + 5

32

√
6)S

3/2
∗∗ for

any α > 0.

Case 2: 2 < p ≤ 4. We first rewrite the functional I as Iα. Let ṽ0 ∈ C∞
c (R6)\{0}. We define

tα > 0 such that Iα(tαṽ0) = supt≥0 Iα(tṽ0). We claim that tα → 0 as α → +∞. Indeed, if the
claim is not true, then there exists a constant t0 > 0 and a sequence {αn} such that αn → +∞
and tαn

≥ t0 for all n. Assume that αn ≥ 1 for all n. Set tn = tαn
and I1 = Iα|α=1, then

0 ≤ Iαn
(tnṽ0) ≤ I1(tnṽ0), which implies that tn is bounded from above. Moreover, we have

Iαn(tnṽ0) =
t2n
2

∫
R6

[(∆ṽ0)
2 + |∇ṽ0|2 + V (x)ṽ20 ]dx+

t4n
4

∫
R6

|∇(ṽ20)|2dx− t2
∗∗

n

2∗∗

∫
R6

|ṽ0|2
∗∗
dx

− αn
tpn
p

∫
R6

|ṽ0|pdx

≤ t2n
2

∫
R6

[(∆ṽ0)
2 + |∇ṽ0|2 + V (x)ṽ20 ]dx+

t4n
4

∫
R6

|∇(ṽ20)|2dx− αn
tpn
p

∫
R6

|ṽ0|pdx

≤ C − αn
tp0
p

∫
R6

|ṽ0|pdx→ −∞

as n→ ∞. This contradicts Iαn
(tnṽ0) ≥ 0. Hence the claim holds and tα → 0 as α→ +∞.

Clearly,

Iα(tαṽ0) ≤
t2α
2

∫
R6

[(∆ṽ0)
2 + |∇ṽ0|2 + V (x)ṽ20 ]dx+

t4α
4

∫
R6

|∇(ṽ20)|2dx.

This implies that Iα(tαṽ0) → 0 as α → +∞. Hence, there exists a constant α∗ > 0 such that

Iα(tαṽ0) = supt≥0 Iα(tṽ0) < ( 1196
√
22 + 5

32

√
6)S

3/2
∗∗ for all α > α∗. Consequently, c < ( 1196

√
22 +

5
32

√
6)S

3/2
∗∗ for all α > α∗. □

Acknowledgments. This work is supported by the NSFC (12061090) and the Natural Science
Foundation of Yunnan (20241AT070105).

References

[1] C. O. Alves, J. M. do Ó, O. H. Miyagaki; On a class of singular biharmonic problems involving critical
exponents. J. Math. Anal. Appl., 277 (2003), no. 1, 12–26.

[2] C. O. Alves, J. M. do Ó, O. H. Miyagaki; Nontrivial solutions for a class of semilinear biharmonic problems
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Juhua He (corresponding author)

Department of Mathematics, Yunnan Normal University, Kunming 650500, China
Email address: 1745190963@qq.com

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

Email address: wuke2002@126.com

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

Email address: zhoufen 85@163.com