Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 75, pp. 1–13.

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

POSITIVE SOLUTIONS FOR KIRCHHOFF-SCHRÖDINGER

EQUATIONS VIA POHOZAEV MANIFOLD

XIAN HU, YONG-YI LAN

Abstract. In this article we consider the Kirchhoff-Schrödinger equation

−
(

(a+ b

∫
R3
|∇u|2 dx

)
∆u+ λu = k(x)f(u), x ∈ R3,

where u ∈ H1(R3), λ > 0, a > 0, b ≥ 0 are real constants, k : R3 → R and
f ∈ C(R,R). To overcome the difficulties that k is non-symmetric and the

non-linear, and that f is non-homogeneous, we prove the existence a positive

solution using projections on a general Pohozaev type manifold, and the linking
theorem.

1. Introduction and main results

This article concerns the Kirchhoff-Schrödinger equation

−
(

(a+ b

∫
R3

|∇u|2 dx
)

∆u+ λu = k(x)f(u), x ∈ R3, (1.1)

where u ∈ H1(R3), λ > 0, a > 0, b ≥ 0 real constants, k : R3 → R and f : R→ R.
We use the following assumptions:

(H1) k ∈ C1(R3, [0,∞]), with k0 = infx∈R3 k(x) > 0;
(H2) k∞ = lim|y|→∞ k(y) <∞;

(H3) t 7→ k(tx) + 1
3∇k(tx) · (tx) is nondecreasing on (0,∞) for all x ∈ R3;

(H4) ∇k(x) · x ≥ 0 and k(x) + 1
3∇k(x) · x ≤ ( 6≡)k∞, for all x ∈ R3;

(H5) supR3 |k∞ − k(x)| ≤ β0(
∫
R3 F (w) dx)−1, where β0 is the unique positive

root of the equation

t2/3 + 2(m∞)1/3t = (m∞)2/3;

(H6) f ∈ C(R,R), tf(t) ≥ 0, and there exist q ∈ (2, 6) such that
lim|t|→∞ f(t)/|t|q−1 = 0;

(H7) limt→0 f(t)/t = 0;

(H8) f(t)t− 4F (t) ≥ 0 for all t ∈ R\{0}, where F (t) =
∫ t

0
f(s) ds.

We look for the weak solutions of (1.1) which are the same as the critical points
of the functional defined in H1(R3) by

I(u) =
1

2

∫
R3

(a|∇u|2 + λu2) dx+
b

4

(∫
R3

|∇u|2 dx
)2

−
∫
R3

k(x)F (u) dx. (1.2)

2020 Mathematics Subject Classification. 35J35, 35B38, 35J92.
Key words and phrases. Kirchhoff-Schrödinger equation; Pohozaev manifold;
Cerami sequence; linking theorem.
©2022. This work is licensed under a CC BY 4.0 license.
Submitted March 21, 2022. Published November 17, 2022.

1



2 X. HU, Y. LAN EJDE-2022/75

If k(x) ≡ k∞, then (1.1) reduces to the autonomous form

−
(

(a+ b

∫
R3

|∇u|2 dx
)

∆u+ λu = k∞f(u), x ∈ R3, (1.3)

with u ∈ H1(R3). Its energy functional is

I∞(u) =
1

2

∫
R3

(a|∇u|2 + λu2) dx+
b

4
(

∫
R3

|∇u|2 dx)2 − k∞
∫
R3

F (u) dx. (1.4)

Problem (1.1) is related to the stationary analogue of the equation

utt − (a+ b

∫
R3

|∇u|2 dx)4u = 0

which was proposed by Kirchhoff [8] as an extension of classical D’Alembert’s wave
equation. It has been applied widely to model various physics problems and ap-
pears in some biological systems. The nonlocal term (

∫
R3 |∇u|2 dx)4u, arises in

various models of physical and biological systems, and the research for related is-
sues gives rise to more mathematical difficulties and challenges; for more details
and backgrounds, we refer the reader to [1, 3, 6] and references therein. After the
pioneer work of Lions [12], Kirchhoff type problems began to attract the attention
of mathematicians, see for example [10, 11, 21].

Recently, a lots of interesting results for problem (1.1) or similar problems have
been obtained, see for example [2, 12, 16, 17, 18, 20] for the radial symmetry case,
and [4, 5, 9, 13, 14, 19, 22, 23] for the non-radial symmetry case. As we known, the
radial symmetry plays a crucial role since which can restore the compactness of the
(PS)-sequence for the energy functional I. Salvatore [16] established the existence
of multiple radially symmetric solutions with the radially symmetric case where V
depends on |x|. Wang et al [20] obtained a least-energy sign-changing (or nodal)
solution by using constraint variational method and the quantitative deformation
lemma. When b = 0, the existence of solution was obtain by Strauss [17] and Lions
[12] if f is superlinear at infinity, also in [2, 18] if f is asymptotically linear at
infinity.

For non-radial symmetry case, problem (1.1) with k(x) > k∞ > 0 was also
solved in [13] by constrained minimization and concentration-compactness argu-
ments. There the role played by the inequality k(x) > k∞ in restoring compactness
in RN is used. However, in case k(x) ≤ ( 6≡)k∞ and f is superlinear at infinity,
nonsymmetric problem (1.1) cannot be solved by minimization [4]. Che and Chen
[5] considered existence and multiplicity of positive solutions by using the Nehari
manifold technique and the Ljusternik Schnirelmann category theory. Under proper
assumptions, Wang and Zhang [23] obtained a ground state solution for the above
problem with the help of Nehari manifold. In [14, 19], the authors studied the
existence of ground state solutions of Nehari-Pohozaev type. When b = 0, [9, 22]
studied a class of nonlinear Schrödinger equations by using concentration compact-
ness arguments and projections on a general Pohozaev type manifold.

Motivated by [9, 14, 19, 22], we investigate the existence of nontrivial solutions
of problem (1.1). In this article, the main obstacle is that the geometrical hy-
potheses on the potential k(x) does not allow us to use concentration compactness
arguments as in [4, 13]. In general, this difficulty is circumvented by assuming
symmetry properties of k(x). Our objective is to prove the existence of a positive
solution of (1.1) under k(x) ≤ (6≡)k∞ and k∞ = lim|x|→∞ k(x), but not requiring



EJDE-2022/75 KIRCHHOFF-SCHRÖDINGER EQUATIONS 3

any symmetry properties. Another obstacle is that the nonlinear term in (1.1) is
non-homogeneous and non-autonomous. Projections on Nehari manifold are not
possible in general, thus one is motivated to use the more suitable projections on
the set of points which satisfy the Pohozaev identity [15], the so-called the Pohozaev
manifold of (1.1).

Let a > 0 and b ≥ 0 be fixed. Throughout the paper we use the following
notation:
H1(R3) denotes the usual Sobolev space equipped with the norm

‖u‖2λ =

∫
R3

(a|∇u|2 + λu2) dx.

Ls(R3) (1 ≤ s <∞) denotes the Lebesgue space with the norm

‖u‖ss =

∫
R3

|u|s dx.

For u ∈ H1(R3)\{0}, ut(x) = u(x/t) for t > 0. For x ∈ R3 and r > 0, Br(x) =
{y ∈ R3 : |y − x| < r}. We denote various positive constants as c, ci, C, Ci (i =
0, 1, 2, 3, . . . ).

To state our results, we define two functionals on H1(R3) as follows:

P (u) =
a

2

∫
R3

|∇u|2 dx+
3λ

2

∫
R3

λu2 dx+
b

2

(∫
R3

|∇u|2 dx
)2

−
∫
R3

[3k(x) +∇k(x) · x]F (u) dx ,

(1.5)

P∞(u) =
a

2

∫
R3

|∇u|2 dx+
3λ

2

∫
R3

u2 dx+
b

2

(∫
R3

|∇u|2 dx
)2

− 3k∞

∫
R3

F (u) dx.

(1.6)

We define the Pohozaev manifold associated with (1.1) and (1.3) by

M = {u ∈ H1(R3)\{0} : P (u) = 0}, (1.7)

M∞ = {u ∈ H1(R3)\{0} : P∞(u) = 0}. (1.8)

We are now in position to state and prove our main result.

Theorem 1.1. Under assumptions (H1)–(H8), problem (1.1) has a positive solu-
tion u ∈ H1(R3)\{0}.

2. Proof of Theorem 1.1

Lemma 2.1. Suppose that
∫
R3 [λu

2

2 − k∞F (u)] dx < 0. Then there exists unique
tu > 0 and tu∗ > 0 such that utu ∈M and utu∗ ∈M∞.

Proof. First we define the function

ψ(t) = I(ut)

=
at

2

∫
R3

|∇u|2 dx+
λt3

2

∫
R3

u2 dx+
bt2

4

(∫
R3

|∇u|2 dx
)2

− t3
∫
R3

k(tx)F (u) dx.



4 X. HU, Y. LAN EJDE-2022/75

Taking the derivative of ψ(t), we obtain

ψ′(t) =
a

2

∫
R3

|∇u|2 dx+
3λt2

2

∫
R3

u2 dx+
bt

2

(∫
R3

|∇u|2 dx
)2

− 3t2
∫
R3

k(tx)F (u) dx− t3
∫
R3

∇k(tx) · xF (u) dx

=
a

2

∫
R3

|∇u|2 dx+
bt

2

(∫
R3

|∇u|2 dx
)2

+ 3t2
∫
R3

[
λu2

2
− k(tx)F (u)] dx

− t3
∫
R3

∇k(tx) · xF (u) dx.

By the Lebesgue Dominated Convergence Theorem,

lim
t→∞

∫
R3

[
λu2

2
− k(tx)F (u)] dx =

∫
R3

[
λu2

2
− k∞F (u)] dx < 0.

By (H2) and (H4), we have

∇k(x) · x→ 0, as |x| → ∞. (2.1)

Using again the Lebesgue Dominated Convergence Theorem,

lim
t→∞

∫
R3

∇k(tx) · (tx)F (u) dx = 0.

where we have used (H6) and (H7). Therefore, if t > 0 is sufficiently large, then
ψ′(t) < 0. On the other hand, taking t > 0 sufficiently small in the expression of
ψ′(t), we obtain ψ′(t) > 0. Since ψ′ is continuous, there exists at least one tu > 0
such that ψ′(tu) = 0. Then P (utu) = tψ′(tu) = 0 so that utu ∈M.

Moreover (H3) implies that

3t3[k(x)− k(tx)] + (t3 − 1)∇k(x) · x ≤ 0, ∀t ≥ 0, x ∈ R3. (2.2)

By this inequality, (H6) and (H7), for any u ∈ H1(R3), t > 0, one has

I(u)− I(ut)

=
a(1− t)

2
‖∇u‖22 +

λ(1− t3)

2
‖u‖22 +

b(1− t2)

4
‖∇u‖42

−
∫
R3

[k(x)− t3k(tx)]F (u) dx

=
1− t3

3
P (u) +

a(t3 − 3t+ 2)

6
‖∇u‖22 +

b(2t3 − 3t2 + 1)

12
‖∇u‖42

− 1

3

∫
R3

[3t3(k(x)− k(tx)) + (t3 − 1)∇k(x) · x]F (u) dx

≥ 1− t3

3
P (u) +

a(t3 − 3t+ 2)

6
‖∇u‖22 +

b(2t3 − 3t2 + 1)

12
‖∇u‖42.

(2.3)

Next we claim that tu is unique. In fact, for any given u satisfies
∫
R3 [λu

2

2 −
k∞F (u)] dx < 0. Let t1, t2 > 0 such that ut1 , ut2 ∈M. Then P (ut1) = P (ut2) = 0.
From this and (2.3), we have

I(ut1) ≥ I(ut2) +
t31 − t32

3t31
P (ut1) +

a(2t31 − 3t21t2 + t32)

6t31
‖∇ut1‖22

+
b(3t41 − 3t21t

2
2 − 2t31 + 2t32)

12t21
‖∇ut1‖42



EJDE-2022/75 KIRCHHOFF-SCHRÖDINGER EQUATIONS 5

= I(ut2) +
a(2t31 − 3t21t2 + t32)

6t31
‖∇ut1‖22 +

b(3t41 − 3t21t
2
2 − 2t31 + 2t32)

12t21
‖∇ut1‖42

and

I(ut2) ≥ I(ut1) +
t32 − t31

3t32
P (ut2) +

a(2t32 − 3t22t1 + t31)

6t32
‖∇ut2‖22

+
b(3t42 − 3t22t

2
1 − 2t32 + 2t31)

12t22
‖∇ut2‖42

= I(ut1) +
a(2t32 − 3t22t1 + t31)

6t32
‖∇ut2‖22 +

b(3t42 − 3t22t
2
1 − 2t32 + 2t31)

12t22
‖∇ut2‖42.

These inequalities above imply t1 = t2. Therefore, tu > 0 is unique.
Similarly, we define the function

ϕ(t) = I∞(ut)

=
at

2

∫
R3

|∇u|2 dx+
λt3

2

∫
R3

u2 dx+
bt2

4

(∫
R3

|∇u|2 dx
)2

− k∞t3
∫
R3

F (u) dx.

Taking the derivative of ψ(t), we obtain

ϕ′(t) =
a

2

∫
R3

|∇u|2 dx+
3λt2

2

∫
R3

u2 dx+
bt

2

(∫
R3

|∇u|2 dx
)2

− 3t2k∞

∫
R3

F (u) dx

=
a

2

∫
R3

|∇u|2 dx+
bt

2

(∫
R3

|∇u|2 dx
)2

+ 3t2
∫
R3

[
λu2

2
− k∞F (u)] dx.

Therefore, if t > 0 is sufficiently large, then ϕ′(t) < 0. Taking t > 0 sufficiently
small, we obtain ϕ′(t) > 0. Since ϕ′ is continuous, there exists at least one tu∗ > 0
such that ϕ′(tu∗) = 0. Then P∞(utu∗ ) = tϕ′(tu∗) = 0 so that utu∗ ∈M∞. For any
u ∈ H1(R3), t > 0, one has

I∞(u)− I∞(ut)

=
a(1− t)

2
‖∇u‖22 +

λ(1− t3)

2
‖u‖22 +

b(1− t2)

4
‖∇u‖42

− k∞(1− t3)

∫
R3

F (u) dx

=
1− t3

3
P∞(u) +

a(t3 − 3t+ 2)

6
‖∇u‖22 +

b(2t3 − 3t2 + 1)

12
‖∇u‖42

=
1− t3

3
P∞(u) +

a(t3 − 3t+ 2)

6
‖∇u‖22 +

b(2t3 − 3t2 + 1)

12
‖∇u‖42.

(2.4)

Now we claim that tu∗ is unique. In fact, each u satisfies
∫
R3 [λu

2

2 −k∞F (u)] dx <
0. Let t3, t4 > 0 such that ut3 , ut4 ∈ M∞. Then P∞(ut3) = P∞(ut4) = 0. From
this and (2.4), we have

I∞(ut3)

= I∞(ut4) +
t33 − t34

3t33
P∞(ut3) +

a(2t33 − 3t23t4 + t34)

6t33
‖∇ut3‖22

+
b(3t43 − 3t23t

2
4 − 2t33 + 2t34)

12t23
‖∇ut3‖42

= I∞(ut4) +
a(2t33 − 3t23t4 + t34)

6t33
‖∇ut3‖22 +

b(3t43 − 3t23t
2
4 − 2t33 + 2t34)

12t23
‖∇ut3‖42



6 X. HU, Y. LAN EJDE-2022/75

and

I∞(ut4)

= I∞(ut3) +
t34 − t33

3t34
P∞(ut4) +

a(2t34 − 3t24t3 + t33)

6t34
‖∇ut4‖22

+
b(3t44 − 3t24t

2
3 − 2t34 + 2t33)

12t24
‖∇ut4‖42

= I∞(ut3) +
a(2t34 − 3t24t3 + t33)

6t34
‖∇ut4‖22 +

b(3t44 − 3t24t
2
3 − 2t34 + 2t33)

12t24
‖∇ut4‖42.

The two inequalities above imply t3 = t4. Therefore, tu∗ > 0 is unique. �

Lemma 2.2. If u ∈M∞, then there exists tu ≥ 1 such that utu ∈M.

Proof. Since u ∈M∞, we have

P∞(u) =
a

2
‖∇u‖22 +

3λ

2
‖u‖22 +

b

2
‖∇u‖42 − 3k∞

∫
R3

F (u) dx = 0. (2.5)

In view of Lemma 2.1, there exists tu > 0 such that utu ∈ M. From (H4), (H6),
and (H7), one has

0 = P (utu)

=
atu
2
‖∇u‖22 +

3λt3u
2
‖u‖22 +

bt2u
2
‖∇u‖42 −

∫
R3

[3k(tux) +∇k(tux) · (tux)]F (u) dx

=
atu
2
‖∇u‖22 + t3u(−a

2
‖∇u‖22 −

b

2
‖∇u‖42 + 3k∞

∫
R3

F (u) dx)

+
bt2u
2
‖∇u‖42 −

∫
R3

[3k(tux) +∇k(tux) · (tux)]F (u) dx

=
a(tu − t3u)

2
‖∇u‖22 +

b(t2u − t3u)

2
‖∇u‖42

+ t3u

∫
R3

[3(k∞ − k(tux))−∇k(tux) · (tux)]F (u) dx

≥ a(tu − t3u)

2
‖∇u‖22 +

b(t2u − t3u)

2
‖∇u‖42,

which implies tu ≥ 1. �

Lemma 2.3. If u ∈M, then there exists tu ∈ (0, 1] such that utu ∈M∞.

Proof. Since u ∈M, we have

P (u) =
a

2
‖∇u‖22 +

3λ

2
‖u‖22 +

b

2
‖∇u‖42 −

∫
R3

[3k(x) +∇k(x) · x]F (u) dx = 0.

In view of Lemma 2.1, there exists tu > 0 such that utu ∈ M∞. From (H4), (H6)
and (H7), one has

0 = P∞(utu)

=
atu
2
‖∇u‖22 +

3λt3u
2
‖u‖22 +

bt2u
2
‖∇u‖42 − 3k∞

∫
R3

F (u) dx

=
atu
2
‖∇u‖22 + t3u(−a

2
‖∇u‖22 −

b

2
‖∇u‖42 +

∫
R3

[3k(x) +∇k(x) · x]F (u) dx)



EJDE-2022/75 KIRCHHOFF-SCHRÖDINGER EQUATIONS 7

+
bt2u
2
‖∇u‖42 − 3k∞

∫
R3

F (u) dx

=
a(tu − t3u)

2
‖∇u‖22 +

b(t2u − t3u)

2
‖∇u‖42

+ t3u

∫
R3

[3(k(x)− k∞) +∇k(x) · x]F (u) dx

≤ a(tu − t3u)

2
‖∇u‖22 +

b(t2u − t3u)

2
‖∇u‖42

which implies tu ≤ 1. Therefore tu ∈ (0, 1]. �

Lemma 2.4. If u ∈M∞, then u(·− y) ∈M∞ for all y ∈ R3. Moreover, for every
y ∈ R3, there exists ty ≥ 1 such that uty (· − y) ∈M and lim|y|→∞ ty = 1.

Proof. If u ∈ M∞, then from the translation invariance of I∞ it follows that
u(· − y) ∈ M∞ for all y ∈ R3. Furthermore, from Lemma 2.2 there exists ty ≥ 1
such that uty (· − y) ∈ M. By (2.1) and the Lebesgue Dominated Convergence
Theorem, we have

0 = lim inf
|y|→∞

t−3
y P (uty (· − y))

= lim inf
|y|→∞

[
at−2
y

2
‖∇u‖22 +

3λ

2
‖u‖22 +

bt−1
y

2
‖∇u‖42]

− lim inf
|y|→∞

∫
R3

[3k(tyx+ y) +∇k(tyx+ y) · (tyx+ y)]F (u) dx

= lim inf
|y|→∞

[
at−2
y

2
‖∇u‖22 +

bt−1
y

2
‖∇u‖42 −

a

2
‖∇u‖22 −

b

2
‖∇u‖42 + 3

∫
R3

k∞(x)F (u) dx]

− lim inf
|y|→∞

∫
R3

[3k(tyx+ y) +∇k(tyx+ y) · (tyx+ y)]F (u) dx

= lim inf
|y|→∞

[
a(t−2

y − 1)

2
‖∇u‖22 +

b(t−1
y − 1)

2
‖∇u‖42]

+ lim inf
|y|→∞

∫
R3

3[k∞ − k(tyx+ y)− 1

3
∇k(tyx+ y) · (tyx+ y)]F (u) dx

=
a

2
(lim inf
|y|→∞

t−2
y − 1)‖∇u‖22 +

b

2
(lim inf
|y|→∞

t−1
y − 1)‖∇u‖42

which implies lim sup|y|→∞ ty = 1, and so lim|y|→∞ ty = 1. �

From Jeanjean and Tanaka [7] have that

inf
u∈M∞

I∞(u) = m∞.

Lemma 2.5. m = m∞.

Proof. Let u ∈ H1(R3) be the ground state solution (which is positive and radially
symmetric) of the problem at infinity, u ∈ M∞ and I∞(u) = m∞. From the
translation invariance of the integrals, given any y ∈ R3 such that u(· − y) ∈M∞,
I∞(u(· − y)) = m∞. From Lemma 2.4, for any y ∈ R3, there exists a ty ≥ 1 such
that uty (· − y) ∈M. Therefore,

|I(uty · (−y))−m∞|



8 X. HU, Y. LAN EJDE-2022/75

= |I(uty · (−y))− I∞(u · (−y))|

= |a(ty − 1)

2
‖∇u‖22 +

b(t2y − 1)

4
‖∇u‖42 +

λ(t3y − 1)

2

∫
R3

u2 dx

+

∫
R3

(k∞ − t3yk(tyx+ y))F (u) dx|

≤ |a(ty − 1)

2
‖∇u‖22|+ |

λ(t3y − 1)

2

∫
R3

u2 dx|+
∫
R3

|k∞ − t3yk(tyx+ y)||F (u)|dx.

Since ty → 1 as |y| → ∞, it follows that

|I(uty · (−y))−m∞| ≤ oy(1) + oy(1) +

∫
R3

|k∞ − k(x+ y)||F (u)|dx.

and since k(x+ y)→ k∞ as |y| → ∞, it follows that

lim
|y|→∞

I(uty · (−y)) = m∞.

Therefore, m = infu∈M I(u) ≤ m∞.
On the other hand, we consider u ∈ M and 0 < ty ≤ 1 such that uty ∈ M∞.

Since u ∈M, then P (u) = 0 and u satisfies

m = I(u) =
a

2
‖∇u‖22 +

λ

2
‖u‖22 +

b

4
‖∇u‖42 −

∫
R3

k(x)F (u) dx

=
1

3
P (u) +

a

3
‖∇u‖22 +

b

12
‖∇u‖42 +

1

3

∫
R3

∇k(x) · xF (u) dx

≥ aty
3
‖∇u‖22 +

bt2y
12
‖∇u‖42

≥ I∞(uty )− 1

3
P∞(uty )

= I∞(uty )

≥ m∞

where we have used (H4) and (H6). Thus, for any u ∈ M, I(u) ≥ m∞ and hence
infu∈M I(u) ≥ m∞. We conclude that m = m∞. �

Lemma 2.6. The functional I satisfies condition (Ce) at level d ∈ (m∞, 2m∞).

Proof. Since {un} ⊂ H1(R3) is a Cerami sequence (Ce)d, by (H8), we have

d+ o(1) = I(un)− 1

4
〈I ′(un), un〉

=
1

4

∫
R3

a|∇un|2 + λu2
n dx+

1

4

∫
R3

a(f(un)un − 4F (un)) dx

≥ 1

4
‖un‖2λ.

This shows {un} is bounded in H1(R3). Applying the splitting lemma cite[Lemma
4.6]l1, up to subsequences, we have

un −
k∑
j=1

uj(x− yjn)→ u in H1(R3),



EJDE-2022/75 KIRCHHOFF-SCHRÖDINGER EQUATIONS 9

where uj is a weak solution of the problem at infinity, |yjn| → ∞ and u is a weak
solution of (1.1). Moreover,

I(un) = I(u) +

k∑
j=1

I∞(uj) + on(1).

Since d < 2m∞, it follows that k < 2. If k = 1, we have two cases to distinguish:

(1) u 6= 0, which implies I(u) ≥ m∞ and hence I(un) ≥ 2m∞.
(2) u = 0, which yields I(un)→ I∞(u1).

In both cases we arrive at a contradiction with the fact that d ∈ (m∞, 2m∞).
Therefore, we must have k = 0 and the convergence un → u follows. �

Definition 2.7. Define the barycenter function of a given function u ∈ H1(R3)\{0}
as follows: let

µ(u)(x) =
1

|B1|

∫
B1(x)

|u(y)|dy,

with µ(u) ∈ L∞(R3)) and µ is a continuous function. Subsequently, take

µ̂(u)(x) = [µ(u)(x)− 1

2
maxµ(u)]+.

It follows that û ∈ C0(R3). Now define the barycenter of u by

β(u)(x) =
1

‖û‖

∫
R3

xû(x) dx ∈ R3.

Since û has compact support, by definition, β(u) is well defined.

Now we define

b = inf{I(u) : u ∈M, β(u) = 0}.
It is clear that b ≥ m∞.

Lemma 2.8. b > m∞.

Proof. By contradiction, suppose that b = m∞. By the definition of b, there exists
a (minimizing) sequence {un} ∈ {u ∈ M, β(u) = 0} such that I(u

n
) → b. By

Lemma 2.8, the sequence {un} is bounded. Since m = m∞ by Lemma 2.6, then
{un} is also a minimizing sequence of I on M. By Ekeland Variational Principle
[24, Theorem 8.5] there exists another sequence {ũn} ∈ M such that:

(i) I(ũn)→ m;
(ii) I ′(ũn)→ 0;

(iii) ‖ũn − un‖ → 0.

Moreover, {un} is bounded, β(un) = 0 and ‖ũn−un‖ → 0 imply that the sequence
{ũn} is bounded and |β(ũn)− β(un)| → 0, since β is a continuous function. So we
have that β(ũn) is bounded.

Therefore, the sequence {ũn} satisfies the assumptions of [9, Corollary 4.8] and
since m = m∞ and is not attained, then the splitting lemma holds with k = 1.
This yields

ũn(x)→ u1(x− yn),

where yn ∈ R3, |y| → ∞, and u1 is a solution of the problem at infinity. By making
a translation, we obtain

ũn(x+ yn) = u1(x) + on(1).



10 X. HU, Y. LAN EJDE-2022/75

Calculating the barycenter function on both sides, we have

β(ũn(x+ yn)) = β(ũn)− yn,
where β(ũn) is bounded and

β(u1(x) + on(1))→ β(u1(x)),

since β is a continuous function. On one side, β(u1(x)) is a fixed real value and,
on the other, |yn| → ∞ so we arrive at a contradiction. Therefore, we must have
b > m∞. �

Inspired by [9], let w ∈ H1(R3) be the positive, radially symmetric, ground state
solution of (1.3). We define the operator Π : R3 →M by

Π[y](x) = w(
x− y
ty

) = wty (x− y).

Proof of Theorem 1.1. By Lemma 2.2, for any w ∈ M∞, then there exists ty ≥ 1
such that wty = Π[y] ∈M. Therefore P (Π[y]) = 0 for any y ∈ R3, and we have

I(Π[y])

=
a

2
‖∇Π[y]‖22 +

λ

2
‖Π[y]‖22 +

b

4
‖∇Π[y]‖42 −

∫
R3

k(x)F (Π[y]) dx

=
1

3
P (Π[y]) +

a

3
‖∇Π[y]‖22 +

b

12
‖∇Π[y]‖42 +

1

3

∫
R3

∇k(x) · xF (Π[y]) dx

=
a

3
‖∇Π[y]‖22 +

b

12
‖∇Π[y]‖42 +

1

3

∫
R3

∇k(x) · xF (Π[y]) dx

=
aty
3
‖∇w‖22 +

bt2y
12
‖∇w‖42 +

t3y
3

∫
R3

∇k(tyx+ y) · (tyx+ y)F (w) dx.

(2.6)

Moreover, since w ∈M∞, we have

I∞ =
1

3
P∞(w) +

aty
3
‖∇w‖22 +

bt2y
12
‖∇w‖42

=
aty
3
‖∇w‖22 +

bt2y
12
‖∇w‖42.

Combing (2.6) and the above equality yields

I(Π[y]) = I∞ +
t3y
3

∫
R3

∇k(tyx+ y) · (tyx+ y)F (w) dx.

By (2.2), it follows that I(Π[y]) → m∞, as |y| → ∞. In view of Lemma 2.8, we
have b > m∞. Then there exists ρ̄ > 0 such that for every ρ ≥ ρ̄,

m∞ < max
|y|=ρ

I(Π[y]) < b.

To apply the Linking Theorem, we take Q = Π(Bρ̄(0)) and S = {u ∈ M : β(u) =
0}. From [9, Lemma 4.13], we have

β(Π[y](x)) = y, ∀y ∈ R3.

If u ∈ S, then β(u) = 0, and if u ∈ ∂Q, then β(u) = y 6= 0, because of equality
|y| = ρ̄; therefore ∂Q ∩ S = ∅.

For any h ∈ H = {h ∈ C(Q,M) : h|∂Q = id}, we define T : Bρ̄(0) → R3 as
T [y] = β ◦ h ◦ Π[y]. The function T is continuous. Moreover, for any |y| = ρ̄, we



EJDE-2022/75 KIRCHHOFF-SCHRÖDINGER EQUATIONS 11

have Π[y] ∈ ∂Q, thus h ◦ Π[y] = Π[y], T (y) = β(Π[y]) = y. By Brouwer’s Fixed
Point Theorem we conclude that there exists ỹ ∈ Bρ̄(0) such that T (ỹ) = 0, which
implies h(Π[ỹ]) ∈ S. Therefore h(Q) ∩ S 6= ∅ and S and ∂Q link.

If h is fixed, then there exists z ∈ S such that z also belongs to h(Q), which
means that z = h(v) form some v ∈ Π(Bρ̄(0)). Therefore,

I(z) ≥ inf
u∈S

I(u) and max
u∈Q

I(h(u)) ≥ I(h(v)).

This gives

max
u∈Q

I(h(u)) ≥ I(h(v)) = I(z) ≥ inf
u∈S

I(u) = b,

and hence

d = inf
h∈H

max
u∈Q

I(h(u)) ≥ b > m∞.

Since w ∈M∞ and m∞ = I∞(w), it follows that m∞ = a
3‖∇w‖

2
2 + b

12‖∇w‖
4
2, and

P∞(w) =
a

2
‖∇w‖22 +

3λ

2
‖w‖22 +

b

2
‖∇w‖42 − 3k∞

∫
R3

F (w) dx = 0.

We set

t∗ = [
m∞

m∞ − 2β0
]1/2.

Since β0 is the unique positive root of (H5), then 1 < t∗ <∞. Hence

I(Π[y]) =
aty
2
‖∇w‖22 +

λt3y
2

+
bt2y
4
‖∇w‖42 − t3y

∫
R3

k(tyx+ y)F (w) dx

= t3yP
∞(w) +

a(3ty − t3y)

6
‖∇w‖22 +

b(3t2y − 2t3y)

12
‖∇w‖42

+ t3y

∫
R3

[k∞ − k(tyx+ y)]F (w) dx

≤
a(3ty − t3y)

6
‖∇w‖22 +

b(3t2y − 2t3y)

12
‖∇w‖42 + β0t

3
y

≤ a(3t∗ − t3∗)
6

‖∇w‖22 +
b(3t2∗ − 2t3∗)

12
‖∇w‖42 + β0t

3
∗

<
a

3
‖∇w‖22 +

b

12
‖∇w‖42 + β0[

m∞

m∞ − 2β0
]3/2

= 2m∞.

Furthermore, if we take h = id, then

d = inf
h∈H

max
u∈Q

I(h(u)) < max
u∈Q

I(u) < 2m∞.

Then we have d ∈ (m∞, 2m∞), thus from Lemma 2.6, (Ce) condition is satisfied
at level d. Therefore, we can apply the Linking Theorem and conclude that d is
a critical level for the functional I . This guarantees the existence of a nontrivial
solution u ∈ H1(R3) of (1.1). Reasoning as usual, because of the hypotheses on f ,
and using the maximum principle we may conclude that u is positive, which implies
the proof. �

Acknowledgments. This work was supported by the Natural Science Founda-
tion of Fujian Province (No. 2022J01339; No. 2020J01708), and by the National
Foundation Training Program of Jimei University (ZP2020057).



12 X. HU, Y. LAN EJDE-2022/75

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Xian Hu

School of Sciences, Jimei University, Xiamen 361021, China
Email address: huxian19972021@163.com



EJDE-2022/75 KIRCHHOFF-SCHRÖDINGER EQUATIONS 13

Yong-Yi Lan (corresponding author)

School of Sciences, Jimei University, Xiamen 361021, China

Email address: lanyongyi@jmu.edu.cn


	1. Introduction and main results
	2. Proof of Theorem ??
	Acknowledgments

	References