Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 10, pp. 1–14.

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

DOI: 10.58997/ejde.2024.10

GROUND STATE SOLUTIONS FOR FRACTIONAL KIRCHHOFF

TYPE EQUATIONS WITH CRITICAL GROWTH

KEXUE LI

Abstract. We study the nonlinear fractional Kirchhoff problem(
a + b

∫
R3
|(−∆)s/2u|2dx

)
(−∆)su + u = f(x, u) + |u|2

∗
s−2u in R3,

u ∈ Hs(R3),

where a, b > 0 are constants, s(3/4, 1), 2∗s = 6/(3−2s), (−∆)s is the fractional

Laplacian. Under some relaxed assumptions on f , we prove the existence of
ground state solutions.

1. Introduction

In this article, we consider the fractional Kirchhoff equation with critical growth(
a+ b

∫
R3

|(−∆)s/2u|2dx
)

(−∆)su+ u = f(x, u) + |u|2
∗
s−2u in R3,

u ∈ Hs(R3),

(1.1)

where a, b > 0, the fractional Laplacian (−∆)s is defined by (−∆)su = F−1(|ξ|2sFu),
F is the usual Fourier transform. The function f satisfies the following conditions:

(A1) f ∈ C(R3 × R,R), there exist constants C0 > 0 and q ∈ (2, 2∗s) such that
|f(x, t)| ≤ C0(1 + |t|q−1), where 2∗s = 6

3−2s is the fractional critical Sobolev
exponent.

(A2) f(x, t) ≥ 0 for t ≥ 0 and f(x, t) = o(|t|), |t| → 0, uniformly on R3;
(A3) For any r > 0 and τ ∈ R\{0}, f satisfies[f(x, τ)

τ3
− f(x, rτ)

(rτ)3

]
sign(1− r) +

|1− r2|
(rτ)2

≥ 0, ∀x ∈ R3;

(A4) There exists an open set Ω ⊂ R3 satisfying 0 ∈ Ω and

lim
|t|→∞

F (x, t)

|t|
4s

3−2s

= +∞, ∀x ∈ Ω,

where F (x, t) =
∫ t

0
f(x, s)ds.

2020 Mathematics Subject Classification. 35R11, 35B50, 34A08.

Key words and phrases. Ground state solution; fractional Kirchhoff equation; critical exponent.
©2024. This work is licensed under a CC BY 4.0 license.
Submitted February 13, 2023. Published January 29, 2024.

1



2 K. LI EJDE-2024/10

When s = 1, equation (1.1) reduces to the Kirchhoff equation

−
(
a+ b

∫
R3

|∇u|2
)

∆u+ u = f(x, u) + u5. (1.2)

Li and Ye [15] studied problem (1.2), under some assumptions on the sign-changing
function f(x, u), they proved the existence of positive solutions by variational meth-
ods.

In recent years, the Kirchhoff-type problem

−
(
a+ b

∫
Rn
|∇u|2dx

)
∆u+ V (x)u = f(x, u) in Rn,

u ∈ H1(Rn),

(1.3)

has attracted much attention, where V ∈ C(Rn,R), f ∈ C(Rn ×R,R) and a, b > 0
are constants. Many results of existence, multiplicity of solutions, ground states and
concentration phenomenon for problem (1.3) have been obtained when f satisfies
various conditions; we refer the reader to [7, 9, 13, 17, 20, 22, 21, 24, 25]. One
usually assumes that f(x, u) is subcritical or superlinear at u = 0 and satisfies the
Ambrosetti-Rabinowitz type condition:

(AR) there exists µ > 4 such that 0 < µF (x, t) ≤ tf(x, t) for all t ∈ R, where

F (x, t) =
∫ t

0
f(x, s)ds; or that 4-superlinear at t =∞

(F) lim|t|→∞
F (x,t)
t4 =∞ uniformly for x ∈ Rn; or the following variant convex

condition
(VC) f(x, t)/|t|3 is strictly increasing for t ∈ R\{0}.

Li and Ye [16] proved that (1.3) had a ground state solution in R3 when f(x, u) =
|u|p−1u and 2 < p ≤ 3. In this case, neither (AR) or (VC) was satisfied. Guo
[12] generalized the result of [16] to (1.3) with general nonlinearity. Guo considered
(1.3) with f(x, u) = f(u) and proved that (1.3) had a positive ground state solution
if f satisfies

• f ∈ C1(R+,R), f(t) = o(t) as t→ 0;

• limt→+∞
f ′(t)
t4 = 0;

• limt→+∞
f(t)
t = +∞;

• f(t)
t is strictly increasing in (0,+∞).

Anello[2] studied the existence and multiplicity of solutions for the nonlocal per-
turbed Kirchhoff problem

−
(
a+ b

∫
Rn
|∇u|2dx

)
∆u = λg(x, u) + f(x, u), in Ω,

u = 0, on ∂Ω,

where Ω is a bounded smooth domain in Rn, n > 4, a, b, λ > 0, and f, g : Ω×R→ R
are Carathéodory functions, with f subcritical, and g of arbitrary growth.

Fiscella and Valdinoci [10] considered the Kirchhoff type problem

M
(∫

R3

|(−∆)s/2u|2dx
)

(−∆)su = λf(x, u) + |u|2
∗
s−2u, in Ω,

u = 0 in Rn\Ω,
(1.4)

where Ω ⊂ Rn is an open bounded set, M and f are two continuous functions. This
equation models the nonlocal aspect of tension arising from nonlocal measurements
of the fractional length of the string. Under some assumptions on M and f , they



EJDE-2024/10 FRACTIONAL KIRCHHOFF TYPE EQUATIONS 3

showed the existence of non-negative solutions. Autuori, Fiscella and Pucci [3] con-
sidered the problem (1.4) and obtained the existence and the asymptotic behavior
of non-negative solutions when the Kirchhoff function M could be zero at zero.
Ambrosio and Isernia [1] dealt with the fractional Kirchhoff equation(

p+ q(1− s)
∫
Rn
|(−∆)s/2u|2dx

)
(−∆)su = g(u) in Rn,

where g : R → R is an odd function satisfying Berestycki-Lions type assumptions.
By using minimax arguments, they established a multiplicity result when q is suf-
ficiently small. Zhang, Tang, and Chen [26] studied(

a+ b

∫
R3

|(−∆)s/2u|2dx
)

(−∆)su+ V (x)u = f(x, u) + λ|u|p−2u in R3,

where a, b > 0, s ∈ (3/4, 1), p ≥ 2∗s = 6
3−2s , V, f satisfies some conditions, in

particular, f satisfies (VC). They showed the equation has a ground state solution
and a signed-changing solution. Gu, Tang and Yang [11] considered the fractional
Kirchhoff equation(

a+ b

∫
R3

|(−∆)s/2u|2dx
)

(−∆)su+ V (x)u = Q(x)f(u) + λ|u|p−2u in R3,

where a, b > 0, s ∈ ( 3
4 , 1), V vanishes at infinity, f satisfies some assumptions,

which contains the generalized (AR) condition

tf(t)− 4F (t) ≥ θtf(θt)− 4F (θt)

for all t ∈ R and θ ∈ [0, 1], where F (t) :=
∫ t

0
f(s)ds.

For any dimension n > 2s, Jin and Liu [14] studied the fractional Kirchhoff
equation (

a+ b

∫
Rn
|(−∆)s/2u|2dx

)
(−∆)su+ u = f(u) in Rn,

with a critical nonlinearity. They proved the existence of solutions without the
Ambrosetti-Rabinowitz condition when the parameter b is small and they obtained
the asymptotic behavior of solutions as b → 0. Liu, Squassina, Zhang [18] studied
the following nonlinear fractional Kirchhoff equation(

a+ b

∫
R3

|(−∆)s/2u|2dx
)

(−∆)su+ V (x)u = f(u) in Rn

u ∈ Hs(Rn), u > 0 in Rn,
(1.5)

where s ∈ (0, 1) and n > 2s. They proved the existence of ground states if f
satisfies the following assumptions:

(A5) f ∈ C1(R+,R), f(t) = 0 for all t ≤ 0 and limt→0
f(t)
t = 0;

(A6) limt→∞
f(t)

t2
∗
s−1 = 1;

(A7) there are D > 0 and 2 < q < 2∗s such that f(t) ≥ t2
∗
s−1 + Dtq−1 for any

t ≥ 0, and the potential V satisfies some conditions.

Remark 1.1. It should be pointed out that f ∈ C1, (A6) and (A7) are very
crucial in [18, 14]. Similar to [4], the smoothness condition f ∈ C1 is used to prove
some Pohozaev type identity. In this paper, we only need f ∈ C and f satisfying
(f1) − (f4). For example, if we take f(x, t) = f(t) = (|t|3 + α|t| 32 )t, 0 < α ≤ 8

√
2,

∀t ∈ R, then after some calculations, f satisfies (A1)–(A4). It is easy to verify that
f does not satisfy (AR), (VC) and the generalized (AR) condition in [11].



4 K. LI EJDE-2024/10

The main result of this paper is described as follows.

Theorem 1.2. Assume that (A1)–(A4) hold. Then for any a, b > 0, equation (1.1)
has a ground state solution.

Throughout this paper, C denotes generic positive constants, which may change
from line to line. The paper is organized as follows. In Section 2, we give some
preliminaries. In Section 3, we prove the main results.

2. Variational setting

For p ∈ [1,∞], we denote by ‖ · ‖p the usual norm of the space Lp(R3). Bσ(x)
denotes the open ball in R3 of radius σ centred at x. We recall some definitions of
fractional Sobolev spaces and the fractional Laplacian, for more details, we refer to
[8]. For any s ∈ (0, 1), the fractional Sobolev space Hs(R3) is defined as follows

Hs(R3) =
{
u ∈ L2(R3) :

∫
R3

(1 + |ξ|2s)|Fu(ξ)|2dξ <∞
}

with the norm

‖u‖Hs =
(∫

R3

(|Fu(ξ)|2 + |ξ|2s|Fu(ξ)|2)dξ
)1/2

, (2.1)

where Fu denotes the Fourier transform of u. By S(Rn), we denote the Schwartz
space of rapidly decaying C∞ functions in Rn. For u ∈ S(Rn) and s ∈ (0, 1), (−∆)s

is defined by

(−∆)su = F−1(|ξ|2s(Fu)), ∀ξ ∈ Rn.
By Plancherel’s theorem, we have |Fu|2 = |u|2, ||ξ|sFu|2 = |(−∆)s/2u|2. Then by
(2.1), we obtain the equivalent norm

‖u‖Hs =
(∫

R3

(|(−∆)s/2u(x)|2 + |u(x)|2)dx
)1/2

.

For any fixed constant a > 0, Hs(R3) can be equipped with the inner product

〈u, v〉 =

∫
R3

a(−∆)s/2u(−∆)s/2v dx+

∫
R3

uv dx

and the corresponding norm

‖u‖ =
(∫

R3

a|(−4)s/2u|2dx+

∫
R3

u2dx
)1/2

. (2.2)

By Theorem 6.5 in [8], we know that Hs(R3) is continuously embedded in Lq(R3)
for any q ∈ [2, 2∗s]. It is easy to see that the norm in (2.2) is equivalent to ‖ · ‖Hs .
For s ∈ (0, 1), the fractional Sobolev space Ds,2(R3) is defined as

Ds,2(R3) =
{
u ∈ L2∗s (R3) : |ξ|sFu(ξ) ∈ L2(R3)

}
,

which is the completion of C∞0 (R3) with respect to the norm

‖u‖Ds,2 =
(∫

R3

|(−∆)s/2u|2dx
)1/2

=
(∫

R3

|ξ|2s|Fu(ξ)|2dξ
)1/2

.

We define the best Sobolev constant

Ss := inf
u∈Ds,2(R3)\{0}

∫
R3 |(−∆)s/2u|2dx( ∫

R3 |u|2∗sdx
)2/2∗s . (2.3)



EJDE-2024/10 FRACTIONAL KIRCHHOFF TYPE EQUATIONS 5

It is known that [5] Ss is attained by the functions ũ(x) = κ
(
µ2 + |x− x0|2

)− 3−2s
2 ,

x ∈ R3, where κ ∈ R\{0}, µ > 0 and x0 ∈ R3 are fixed constants. We define

ū(x) = ũ(x)/‖ũ‖2∗s and let u∗(x) = ū(x/S
1
2s
s ), then by the [19, Claim 6], u∗ is a

solution of the problem

(−∆)su = |u|2
∗
s−2u in R3. (2.4)

and ‖u∗‖2
∗
s

2∗s
= S

3
2s
s . For any ε > 0, set Uε(x) = ε−

3−2s
2 u∗(xε ). Then Uε is a solution

of (2.4) and ‖Uε‖
2∗s
2∗s

= S
3
2s
s . Fix r > 0 and let ϕ ∈ C∞0 (R3, [0, 1]) be such that

suppϕ ⊂ B2r(0) and ϕ(x) = 1 for x ∈ Br(0). By Proposition 3.4 and Proposition
3.6 in [8], we have

‖(−∆)s/2u‖22 =

∫
R3

|ξ|2s|Fu(ξ)|2dξ =
C(s)

2

∫
R3

∫
R3

|u(x)− u(y)|2

|x− y|3+2s
dx dy, (2.5)

where

C(s) =
(∫

R3

1− cos(ξ1)

|ξ|3+2s
dξ
)−1

.

Without loss of generality, we assume that C(s) = 2.
Let vε(x) = ϕ(x)Uε(x). From (2.5), Proposition 21 and Proposition 22 in [19]

and (24) in [23], it follows that∫
R3

|(−∆)s/2vε(x)|2dx ≤ S
3
2s
s +O(ε3−2s), (2.6)∫

R3

|vε(x)|2
∗
sdx = S

3
2s
s +O(ε3), (2.7)

∫
R3

|vε(x)|pdx =


O(ε

(3−2s)p
2 ), p < 3

3−2s ,

O(ε
(2−p)3+2sp

2 | log ε|), p = 3
3−2s ,

O(ε
(2−p)3+2sp

2 ), 3
3−2s < p < 6

3−2s .

(2.8)

The functional associated with (1.1), I : Hs(R3)→ R is

I(u) =
1

2
‖u‖2 +

b

4

(∫
R3

|(−4)s/2u|2dx
)2

−
∫
R3

F (x, u)dx− 1

2∗s

∫
R3

|u|2
∗
sdx, (2.9)

where F (x, u) =
∫ u

0
f(x, t)dt.

It is easy to see that I is well defined, I ∈ C1(Hs(R3,R)) and

〈I ′(u), v〉 =

∫
R3

(a(−∆)s/2u(−∆)s/2v + uv)dx

+ b

∫
R3

|(−∆)s/2u|2dx
∫
R3

(−∆)s/2u(−∆)s/2v dx

−
∫
R3

f(x, u)v dx−
∫
R3

|u|2
∗
s−2uv dx, ∀v ∈ Hs(R3).

(2.10)

A function u ∈ Hs(R3) is a weak solution of (1.1) if for any v ∈ Hs(R3),(
a+ b

∫
R3

|(−∆)s/2u|2dx
)∫

R3

(−∆)s/2u(−∆)s/2v

=

∫
R3

f(x, u)v dx+

∫
R3

|u|2
∗
s−2uv dx.

It is clear that the critical points of I are weak solutions of (1.1).



6 K. LI EJDE-2024/10

3. Ground state solutions

Lemma 3.1. Under the assumptions (A1), and (A2) we have

(i) there exist δ, ρ > 0 such that ‖u‖ = ρ implies that I(u) ≥ δ > 0;
(ii) there exists e ∈ Hs(R3) such that ‖e‖ > ρ implies that I(e) < 0.

Proof. (i) From (A1) and (A2), it follows that for any ε > 0, there exists Cε > 0
such that

|f(x, u)| ≤ ε|u|+ Cε|u|q−1. (3.1)

Then

I(u) ≥ 1

2
‖u‖2 − ε

2

∫
R3

|u|2dx− Cε
q

∫
R3

|u|qdx− 1

2∗s

∫
R3

|u|2
∗
sdx

≥
(1

2
− Cε

)
‖u‖2 − C‖u‖q − C‖u‖2

∗
s

For ε ∈ (0, 1
2C ), we can choose ρ and δ such that I(u) ≥ δ > 0 for ‖u‖ = ρ.

(ii) By (3.1), we have F (x, u) ≥ − ε2 |u|
2 − Cε

q |u|
q. Then for t > 0 and u0 ∈

C∞0 (R3),

I(tu0) ≤ t2

2
‖u0‖2 +

bt4

4

(∫
R3

|(−∆)s/2u0|2dx
)2

− εt2

2

∫
R3

|u0|2dx

− Cεt
q

q

∫
R3

|u0|qdx−
t2
∗
s

2∗s

∫
R3

|u0|2
∗
sdx,

since 2∗s > 4, it follows that limt→+∞ I(tu0) = −∞. We choose sufficiently large t0
and set e = t0u0 such that ‖e‖ > ρ and I(e) < 0. �

We denote the mountain pass value by

c = inf
γ∈Γ

max
t∈[0,1]

I(γ(t)),

where

Γ := {γ ∈ C([0, 1], Hs(R3)) : γ(0) = 0, I(γ(1)) < 0}.
By Lemma 3.1, there exists a Palais-Smale sequence {un} ⊂ Hs(R3) for I at the
level c (or (PS)c sequence, for short):

I(un)→ c and I ′(un)→ 0 as n→∞. (3.2)

Lemma 3.2. Assume that (A1)–(A3) hold. Then

0 < c < c∗ :=
aSs
2
T 3−2s +

bS2
s

4
T 6−4s − T 3

2∗s
,

where T > 0 is the unique maximum point of the function J(t) = aSs
2 t3−2s +

bS2
s

4 t6−4s − t3

2∗s
, t > 0.

Proof. By (i) of Lemma 3.1, c > 0. Let γε(t) = vε(·/t). From (2.9) and (2.5), it
follows that

I(γε(t)) =
at3−2s

2

∫
R3

|(−∆)s/2vε|2dx+
bt6−4s

4

(∫
R3

|(−∆)s/2vε|2dx
)2

+
t3

2

∫
R3

|vε(x)|2dx−
∫
R3

F (x, vε(x/t))dx−
t3

2∗s

∫
R3

|vε(x)|2
∗
sdx.

(3.3)



EJDE-2024/10 FRACTIONAL KIRCHHOFF TYPE EQUATIONS 7

By [19, (4.18)], Uε can be written as (with x0 = 0)

Uε(x) = k̃ε−
3−2s

2

(
µ2 +

∣∣ x

εS
1
2s
s

∣∣2)− 3−2s
2

(3.4)

where k̃ ∈ R\{0}. We choose k̃ > 0, from (3.4),

Uε(x) =
Cε

3−2s
2(

µ2S
1/s
s ε2 + |x|2

) 3−2s
2

, (3.5)

where C > 0. Since 0 ∈ Ω and Ω is an open set, then Bε(0) ⊂ Ω. Since vε(x/t)→
+∞ as t→ +∞, ε→ 0+. By (A4), (3.5),∫

R3

F (x, vε(x/t))dx ≥
∫
Bε(0)

F (x, vε(x/t))dx

≥M
∫
Bε(0)

[ (t2ε)
3−2s

2

(µ2t2S
1/s
s ε2 + |x|2)

3−2s
2

] 4s
3−2s dx

≥M
∫
Bε(0)

( t2ε

µ2t2S
1/s
s ε2 + ε2

)2s

dx

=
Mt4s

(µ2t2S
1/s
s + 1)2s

ε3−2s,

(3.6)

where M is a large positive number.
It follows from (2.6)-(2.8), (3.6) that for any ε > 0 small enough, I(γε(t))→ −∞

as t→ +∞. There exists t0 > 0 such that I(γε(t0)) < 0. Since

‖γε(t)‖2Hs =

∫
R3

(
|(−∆)s/2γε(t)|2 + |γε(t)|2

)
dx

= t3−2s

∫
R3

|(−∆)s/2vε(x)|2dx+ t3
∫
R3

|vε(x)|2dx,

by (2.6) and (2.7), we see that limt→0+ ‖γε(t)‖2Hs = 0 for sufficiently small ε > 0.
We set γε(0) = 0, thus γε(t0·) ∈ Γ. Let γ0(·) = γε(t0·). Then

c ≤ max
t∈[0,1]

I(γ0(t)) = max
t∈[0,1]

I(γε(t0t)) = max
t∈[0,t0]

I(γε(t)) ≤ sup
t≥0

I(γε(t)). (3.7)

By (3.3), supt≥0 I(γε(t)) is attainable at tε = t(γε) > 0. By (2.6)-(2.8), (3.6)

and (3.3), we have that I(γε(t)) → 0+ as t → 0+ and I(γε(t)) → −∞ as t → +∞
uniformly for ε > 0 small enough. Then there exist constants t1, t2 > 0, independent
of ε, such that 0 < t1 ≤ tε ≤ t2 <∞.

Next we need to prove supt≥0 I(γε(t)) < c∗. We define

gε(t) :=
at3−2s

2

∫
R3

|(−∆)s/2vε(x)|2dx+
bt6−4s

4

(∫
R3

|(−∆)s/2vε(x)|2dx
)2

− t3

2∗s

∫
R3

|vε(x)|2
∗
sdx.

Then by (2.8), we have

sup
t≥0

I(γε(t)) ≤ sup
t≥0

gε(t) +O(ε3−2s)−
∫
R3

F (x, vε(x/tε))dx (3.8)



8 K. LI EJDE-2024/10

By (3.6),

sup
t≥0

I(γε(t)) ≤ sup
t≥0

gε(t) +O(ε3−2s)−MCε3−2s, (3.9)

where C > 0 is a constant. By a similar argument as above, there exist constants
t3, t4 > 0, independent of ε, such that supt≥0 gε(t) = supt∈[t3,t4] gε(t). Thus by

(2.6), (2.7), we obtain

sup
t≥0

I(γε(t)) ≤ sup
t≥0

J(S
1
2s
s t) +O(ε3−2s)−MCε3−2s, (3.10)

where

J(t) =
aSs
2
t3−2s +

bS2
s

4
t6−4s − t3

2∗s
. (3.11)

Since M can be arbitrarily large, from (3.10), we see that

sup
t≥0

I(γε(t)) < sup
t≥0

J(S
1
2s
s t). (3.12)

By (3.11),

J ′(t) =
(3− 2s)t2−2s

2
J̃(t),

where

J̃(t) = aSs + bS2
s t

3−2s − t2s,

J̃ ′(t) = t2−2s(bS2
s (3− 2s)− 2st4s−3).

Since s > 3/4, there exists a unique T > 0 such that J̃(t) > 0 for t ∈ (0, T ) and

J̃(t) < 0 for t > T . Then T is the unique maximum point of J . Therefore, by (3.7)
and (3.12), we have c < c∗. �

Lemma 3.3. For any u ∈ Hs(R3) and t > 0, the following inequality holds

I(u) ≥ I(tu) +
1− t4

4
〈I ′(u), u〉+

a(1− t2)2

4

∫
R3

|(−∆)s/2u|2dx. (3.13)

Proof. By (A3), for any r ≥ 0 and τ ∈ R\{0}, we have

(1− r4)τf(x, τ)

4
+ F (x, rτ)− F (x, τ) +

1

4
(1− r2)2τ2

=

∫ 1

r

(f(x, τ)

τ3
− f(x, ντ)

(ντ)3
+

1− ν2

(ντ)2

)
ν3τ4dν ≥ 0.

(3.14)



EJDE-2024/10 FRACTIONAL KIRCHHOFF TYPE EQUATIONS 9

Then, for any u ∈ Hs(R3) and t ≥ 0,

I(u)− I(tu)− 1− t4

4
〈I ′(u), u〉

=
(1− t2)2

4
‖u‖2 +

1− t4

4

∫
R3

f(x, u)u dx+

∫
R3

F (x, tu)dx−
∫
R3

F (x, u)dx

+
(
− 1

2∗s
+
t2
∗
s

2∗s
+

1− t4

4

)∫
R3

|u|2
∗
sdx

≥ a(1− t2)2

4

∫
R3

|(−∆)s/2u|2dx

+

∫
R3

[1− t4
4

f(x, u)u+ F (x, tu)− F (x, u) +
(1− t2)2

4
|u|2
]
dx

+
(
− 1

2∗s
+
t2
∗
s

2∗s
+

1− t4

4

)∫
R3

|u|2
∗
sdx.

(3.15)

Next we prove ξ(t) = − 1
2∗s

+ t2
∗
s

2∗s
+ 1−t4

4 ≥ 0 for all t ≥ 0. In fact, since 2∗s =
6

3−2s > 4, it is easy to see that ξ(t) decreases on (0, 1) and increases on (1,∞), then

mint≥0 ξ(t) = ξ(1) = 0 and ξ(t) ≥ 0. This together with (3.14) and (3.15) yield
that

I(u)− I(tu)− 1− t4

4
〈I ′(u), u〉 ≥ a(1− t2)2

4

∫
R3

|(−∆)s/2u|2dx. �

Lemma 3.4. Any sequence satisfying (3.2) is bounded in Hs(R3). There exists u ∈
Hs(R3), such that, up to a subsequence, un ⇀ u weakly in Hs(R3) and 〈I ′(u), u〉 ≤
0.

Proof. Let {un} ⊂ Hs(R3) be a sequence satisfying (3.2). By (A3), for any τ ∈
R\{0},

1

4
τf(x, τ)− F (x, τ) +

1

4
τ2 =

∫ 1

0

(
f(x, τ)

τ3
− f(x, ντ)

(ντ)3
+

1− ν2

(ντ)2

)
ν3τ4dν

≥ 0.

(3.16)

For τ = 0, it is obvious that 1
4τf(x, τ) − F (x, τ) + 1

4τ
2 = 0. Since I(un) → c and

I ′(un)→ 0 as n→∞, it follows that

c+ 1 + o(1)

≥ I(un)− 1

4
〈I ′(un), un〉

=
1

4
‖un‖2 +

∫
R3

[1
4
unf(x, un)− F (x, un)

]
dx+ (

1

4
− 1

2∗s
)

∫
R3

|un|2
∗
sdx

≥ a

4

∫
R3

|(−∆)s/2un|2dx.

(3.17)



10 K. LI EJDE-2024/10

By (3.17) and (2.3), there exists a constant Mc = M(a, c, s) > 0 such that
|(−∆)s/2un|2 ≤Mc and |un|2∗s ≤Mc. It follows from I(un) = c+ on(1) that

1

2

∫
R3

|un|2dx = c+ on(1) +

∫
R3

F (x, un)dx+
1

2∗s

∫
R3

|un|2
∗
sdx

− a

2

∫
R3

|(−∆)s/2un|2dx−
b

4

(∫
R3

|(−∆)s/2un|2dx
)2

≤ c+ on(1) + ε|un|22 + Cε|un|qq +
Mc

2∗s
.

(3.18)

By an Lp interpolation inequality and (2.3),

|un|qq ≤ |un|
qθ
2 |un|

q(1−θ)
2∗s

≤ C|un|qθ2 |(−∆)s/2un|q(1−θ)2 , (3.19)

where θ ∈ (0, 1), 1
q = θ

2 + 1−θ
2∗s

. Since qθ ∈ (0, 2), by Young inequality and (3.19),

we have

Cε|un|qq ≤ ε|un|22 + C̃ε|(−∆)s/2un|
2q(1−θ)
2−qθ

2 . (3.20)

If we take ε = 1/5, from (3.18) and (3.20), it follows that |un|2 is bounded. There-
fore {un} is bounded in Hs(R3).

Let A2 := limn→∞
∫
R3 |(−∆)s/2un|2dx. By (2.5) and Fatou’s lemma,∫

R3

|(−∆)s/2u|2dx ≤ A2. (3.21)

Since un ⇀ u, then u is a weak solution of

(a+ bA2)(−∆)s/2u+ u = f(x, u) + |u|2
∗
s−2u. (3.22)

Thus,

〈I ′(u), u〉 = ‖u‖2 + b
(∫

R3

|(−∆)s/2|2dx
)2

−
∫
R3

f(x, u)u dx−
∫
R3

|u|2
∗
sdx

≤ (a+ bA2)

∫
R3

|(−∆)s/2u|2dx+

∫
R3

|u|2dx−
∫
R3

f(x, u)u dx

−
∫
R3

|u|2
∗
sdx = 0.

(3.23)

�

Lemma 3.5. For any sequence {un} satisfying (3.2), there exists δ0 > 0 such that

δ0 = lim
n→∞

sup
y∈R3

∫
B1(y)

u2
ndx > 0.

Proof. By contradiction, we assume that limn→∞ supy∈R3

∫
B1(y)

u2
ndx = 0. By [6,

lemma 2.3], for any 2 < r < 2∗s,

un → 0 in Lr(R3) for r ∈ (2, 2∗s).

By (A1) and (A2),

lim
n→∞

∫
R3

unf(x, un)dx→ 0 andquad lim
n→∞

∫
R3

F (x, un)dx→ 0.

This and (3.2) yield

1

2
‖un‖2 +

b

4

(∫
R3

|(−∆)s/2un|2dx
)2

− 1

2∗s

∫
R3

|un|2
∗
sdx = c+ o(1), (3.24)



EJDE-2024/10 FRACTIONAL KIRCHHOFF TYPE EQUATIONS 11

and

‖un‖2 + b
(∫

R3

|(−∆)s/2un|2dx
)2

−
∫
R3

|un|2
∗
sdx = o(1). (3.25)

Up to a subsequence, we assume that there exist li ≥ 0 (i = 1, 2, 3) such that

‖un‖2 → l1, b
(∫

R3

|(−∆)s/2un|2dx
)2

→ l2,

∫
R3

|un|2
∗
sdx→ l33. (3.26)

by (3.24) and (3.25), we have

l1 + l2 = l33, (3.27)(1

2
− 1

2∗s

)
l1 +

(1

4
− 1

2∗s

)
l2 = c. (3.28)

From (2.3), it follows that

‖un‖2 ≥ aSs
(∫

R3

|un|2
∗
sdx
) 2

2∗s , (3.29)

b
(∫

R3

|(−∆)s/2un|2dx
)2

≥ bS2
s

(∫
R3

|un|2
∗
sdx
) 4

2∗s . (3.30)

Then by (3.25),

aSs

(∫
R3

|un|2
∗
sdx
)2/2∗s

+ bS2
s

(∫
R3

|un|2
∗
sdx
)4/2∗s

≤
∫
R3

|un|2
∗
sdx+ on(1). (3.31)

By (3.31),

J ′(l3) =
(3− 2s)l−1

3

2
(aSsl

3−2s
3 + bS2

s l
6−4s
3 − l33) ≤ 0, (3.32)

where J is defined in (3.11). Since T is the unique maximum point of J , then
l3 ≥ T .

From (3.26), (3.28), (3.29), (3.30) and note that J ′(T ) = 0, we obtain

c ≥
(1

2
− 1

2∗s

)
aSsl

3−2s
3 +

(1

4
− 1

2∗s

)
bS2
s l

6−4s
3

≥
(1

2
− 1

2∗s

)
aSsT

3−2s +
(1

4
− 1

2∗s

)
bS2
sT

6−4s

=
1

2
aSsT

3−2s +
1

4
bS2
sT

6−4s − 1

2∗s
T 3 = c∗.

This contradicts c < c∗. The proof is complete. �

Lemma 3.6. Any Palais-Smale sequence for I at the level c ∈ (0, c∗) possesses a
strongly convergent subsequence.

Proof. From [8, Corollary 7.2 ], it is known that the embedding Hs(R3) ↪→ Lq(R3)
locally compact for q ∈ [1, 2∗s). Suppose {un} is the (PS)c sequence for I, then
I(un)→ c and I ′(un)→ 0 as n→∞. By Lemma 3.4, {un} is bounded in Hs(R3).
Up to a subsequence, there exists u ∈ Hs(R3) such that

un ⇀ u in Hs(R3),

un → u in Lqloc(R
3), q ∈ [1, 2∗s),

un → u a.e. in R3.

(3.33)



12 K. LI EJDE-2024/10

By Lemma 3.5 and (3.33), ∫
B1(y)

u2dx ≥ δ0
2
> 0.

Then ũ(x) := u(x+ y) 6= 0. Without loss of generality, we assume that u 6= 0. By
(3.15) and Fatou’s lemma, for n large, we have

c+ on(1) = I(un)− 1

4
〈I ′(un), un〉

=
1

4
‖un‖2 +

∫
R3

[1
4
unf(x, un)− F (x, un)

]
dx+

(1

4
− 1

2∗s

) ∫
R3

|un|2
∗
sdx

=
a

4

∫
R3

|(−∆)s/2un|2dx+

∫
R3

[1
4
unf(x, un)− F (x, un) +

1

4
u2
n

]
dx

+
(1

4
− 1

2∗s

) ∫
R3

|un|2
∗
sdx

≥ a

4

∫
R3

|(−∆)s/2u|2dx+

∫
R3

[1
4
uf(x, u)− F (x, u) +

1

4
u2
]
dx

+
(1

4
− 1

2∗s

) ∫
R3

|u|2
∗
sdx

= I(u)− 1

4
〈I ′(u), u〉.

(3.34)
By (3.13) and (3.23), we have

I(u)− 1

4
〈I ′(u), u〉 ≥ max

t≥0

[
I(tu) +

1− t4

4
〈I ′(u), u〉

]
− 1

4
〈I ′(u), u〉

= max
t≥0

[
I(tu)− t4

4
〈I ′(u), u〉

]
≥ max

t≥0
I(tu).

(3.35)

Since u 6= 0, for any u ∈ Hs(R3)\{0}, by (ii) of Lemma 3.1, for sufficiently large
t̃ > 0, I(t̃u) < 0. Let γ1(t) = tt̃u. Then γ1 ∈ C([0, 1], Hs(R3)) and γ1 ∈ Γ. We
have

c ≤ max
t∈[0,1]

I(γ1(t)) = max
t∈[0,1]

I(tt̃u) = max
t∈[0,t̃]

I(tu) ≤ max
t≥0

I(tu). (3.36)

From (3.35) and (3.36), it follows that

I(u)− 1

4
〈I ′(u), u〉 ≥ c. (3.37)

By (3.34) and (3.36), it is easy to see that

lim
n→∞

∫
R3

|(−∆)s/2un|2dx =

∫
R3

|(−∆)s/2u|2dx, (3.38)

lim
n→∞

∫
R3

u2
ndx =

∫
R3

u2dx. (3.39)

By (3.38) and (3.39), we have

‖un‖ → ‖u‖ as n→∞.

This and un ⇀ u yield un → u in Hs(R3). �



EJDE-2024/10 FRACTIONAL KIRCHHOFF TYPE EQUATIONS 13

4. Proof of Theorem 1.2

We define c1 := infu∈U I(u), where U = {u ∈ Hs(R3)\{0} : I ′(u) = 0}.

Lemma 4.1. 0 < c1 < c∗.

Proof. For I(γε(t)), there exists tε = t(γε) > 0 such that I(γε(tε)) = supt>0 I(γε(t)),
where I(γε(t)) is the same as in Lemma 3.2. By Lemma 3.2, c1 ≤ supt≥0 I(γε(t)) <
c∗. Next we show c1 > 0. For any u ∈ U,

‖u‖2 ≤ ‖u‖2 + b
(∫

R3

|(−∆)s/2u|2dx
)2

=

∫
R3

uf(x, u)dx+

∫
R3

|u|2
∗
sdx

=
ε

2
‖u‖2 + C‖u‖q + C‖u‖2

∗
s .

We take ε = 1/2, then there exists α > 0 such that

‖u‖ ≥ α > 0. (4.1)

For any u ∈ U, it follows that 〈I ′(u), u〉 = 0, then by (3.15), we have

I(u) = I(u)− 1

4
〈I ′(u), u〉

=
1

4
‖u‖2 +

∫
R3

[
1

4
uf(x, u)− F (x, u)

]
dx+

(1

4
− 1

2∗s

) ∫
R3

|u|2
∗
sdx

≥ a

4

∫
R3

|(−∆)s/2u|2dx.

(4.2)

Thus c1 ≥ 0. We need to prove c1 6= 0. Assume that c1 = 0 and {un} is the corre-
sponding minimizing sequence, that is, {un} ⊂ U, and I(un) → 0 as n → ∞. By
(4.2), limn→∞

∫
R3 |(−∆)s/2un|2dx = 0. This and (2.3) yield limn→∞

∫
R3 |un|2

∗
sdx =

0. Then by (3.18)-(3.20), we obtain limn→∞
∫
R3 |un|2dx = 0. Therefore, we have

limn→∞ ‖un‖2 = 0. Since (4.1) holds for any u ∈ U, we obtain a contradiction. The
proof is complete. �

Proof of Theorem 1.2. Suppose {un} ⊂ U is the minimizing sequence for c1 =
infu∈U I(u). Then {un} is the (PS)c1 sequence for I. By Lemma 4.1 and Lemma
3.6, there exists u ∈ Hs(R3) such that I ′(u) = 0 and I(u) = c1. Then u is a ground
state solution of (1.1). �

5. Conclusion

In this article, we have considered a nonlinear fractional Kirchhoff problem with
critical growth. We have proved the existence of ground state solutions by varia-
tional methods. The main feature of the result is that we only need the nonlinear
term f ∈ C and f satisfying (f1) − (f4). In fact, such f may not satisfy the con-
dition f ∈ C1, which is used to prove some Pohozaev type identity, see [18, 14, 4].
And such f may not satisfiy the classical (AR) condition or (VC) condition in [12]
or the generalized (AR) condition in [11], see Remark 1.1.

Acknowledgments. This work was supported by grant 2024JC-YBMS-065.



14 K. LI EJDE-2024/10

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Kexue Li

School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an,710049, China

Email address: kxli@mail.xjtu.edu.cn


	1. Introduction
	2. Variational setting
	3. Ground state solutions
	4. Proof of Theorem ??
	5. Conclusion
	Acknowledgments

	References