Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 53, pp. 1–12.

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

KIRCHHOFF-TYPE PROBLEMS WITH CRITICAL SOBOLEV

EXPONENT IN A HYPERBOLIC SPACE

PAULO CESAR CARRIÃO, AUGUSTO CÉSAR DOS REIS COSTA,

OLIMPIO HIROSHI MIYAGAKI, ANDRÉ VICENTE

Abstract. In this work we study a class of the critical Kirchhoff-type prob-

lems in a Hyperbolic space. Because of the Kirchhoff term, the nonlinearity

uq becomes “concave” for 2 < q < 4, This brings difficulties when proving the
boundedness of Palais Smale sequences. We overcome this difficulty by using

a scaled functional related with a Pohozaev manifold. In addition, we need to

overcome singularities on the unit sphere, so that we use variational methods
to obtain our results.

1. Introduction

In this article we study the Kirchhoff-type problem

−
(
a+ b

∫
B3

|∇B3u|2dVB3

)
∆B3u = λ|u|q−2u+ |u|4u in H1(B3), (1.1)

where a, b, λ are positive constants, 2 < q < 4, H1(B3) is the usual Sobolev space
on the disc of the Hyperbolic space B3, and ∆B3 denotes the Laplace Beltrami
operator on B3. Problem (1.1) defined in whole space RN , with N ≥ 3, and with
the non-linearity behaving as a polynomial function of degree 2∗ = 2N

N−2 was studied

by Brezis and Nirenberg [7]. Posteriorly, several authors have studied this class of
problems; see for instance Carrião, Costa, and Miyagaki [8].

In the Euclidean context, equation (1.1) is related to a stationary Kirchhoff
equation (see [25])

utt −M
(∫

Ω

|∇xu|2dx
)

∆xu = f(x, t), (x, t) ∈ Ω× (0,∞),

where Ω is a bounded domain of RN , M(s) = a+bs with a, b > 0, and f is a suitable
function, which is an extension of the classical D’Alembert’s wave equation. One
characteristic of this model is that it considers the effects of the changes in the
length of the strings during the vibrations. The main difficulty appears because
the equation does not satisfy a pointwise identity any longer. It is generated by the
presence of the term containing M in the equation, and it makes (1.1) a nonlocal
problem.

Ma and Rivera [27] were the pioneers to study this problem by employing min-
imizing methods. In [1], the mountain pass theorem was used, while in [30] the

2010 Mathematics Subject Classification. 58J05, 35R01, 35J60, 35B33.

Key words and phrases. Kirchhoff-type problem; variational methods; hyperbolic space.
c©2021. This work is licensed under a CC BY 4.0 license.

Submitted January 30, 2021. Published June 14, 2021.

1



2 P. C. CARRIÃO, A. C. R. COSTA, O. H. MIYAGAKI, A. VICENTE EJDE-2021/53

Yang index and critical groups was used. In [21] the equation was studied using
the minimization arguments and the Fountain theorem. Results can be seen in
[12, 17, 36]. Results involving the Kirchhoff equation and critical exponents can be
found in [2, 16, 19, 20, 26] and references therein. See also [11, 13, 14, 32] for some
related results.

We also would like to cite the recent works by Xiang, Zhang and Rǎdulescu
[38, 39]. In the first one, the authors studied the multiplicity of solutions for a class
of quasilinear Kirchhoff system involving the fractional p-Laplacian. In the second
paper, they proved the existence of local solution and a blow-up result for a class
of nonlocal Kirchhoff diffusion problems.

Our main result reads as follows.

Theorem 1.1. Under the assumptions that 2 < q < 4, for λ > 0 sufficiently large,
problem (1.1) has a nontrivial solution u ∈ H1(B3).

This result extends the result in [20] with respect to the existence in a hyperbolic
space. Also, in [9], when a = 1 and b = 0. It also extends [8], where the authors
studied (1.1) with 4 < q < 6 for λ > 0 arbitrary. We highlight that the case
2 < q < 4 is more delicate and it is necessary additional tools.

Finally, we would like to emphasize that an extra difficulty of the present paper is
to prove that the Palais Smale sequence is bounded. To overcome this difficulty, we
use an appropriated modified functional (see Jθ(v) definition in next section). This
functional gives us an additional property of the Palais Smale sequence which is
fundamental to prove that the sequence is bounded (Lemmas 2.2 and 2.3). Precisely,
the scaled functional Jθ works coupled with another appropriated functional, G,
which has the property G(vk)→ 0, where (vk) is the Palais Smale sequence. Scaled
functional was used by Jeanjean [23] and Jeanjean and Le Coz [24]. See also [19]
and [22].

2. Proof of the main result

For the hyperbolic space Hn, we use the stereographic projection, where each
point P ′ ∈ Hn is projected to P ∈ Rn, where P is the intersection of the straight
line connecting P ′ and the point (0, . . . , 0,−1). Explicitly the projection operator
G : Rn → Hn and G−1 : Hn → Rn given by

G(x) = (x · p(x), (1 + |x|2)p/2) andquadG−1(y) =
1

yn+1
y, x, y ∈ Rn,

where p(x) = 2
1−|x|2 .

We consider the ball B1(0), and Bn endowed with the metric

ds = p(x)|dx|, where p(x) =
2

1− |x|2
.

With this notation, the gradient, the Dirichlet integral and the Laplace-Beltrami
operator corresponding to this metric are

∇Bnu =
∇u
p2
, Du =

∫
D′
|∇Bnu|2dVBn =

∫
D

|∇u|2pn−2dx,

∆Bnu = p−n div(pn−2∇u).

We denote byD ⊂ B1(0) the stereographic projection ofD′ ⊂ Hn. Details involving
the hyperbolic space can be found in [3, 18, 31, 33, 34].



EJDE-2021/53 KIRCHHOFF-TYPE PROBLEMS IN A HYPERBOLIC SPACE 3

Defining v := p1/2u, we have that u is solution of (1.1) if, and only if, v satisfies

(a+ b‖v‖2)(−∆v + (3/4)p2v) = λpα|v|q−2v + |v|4v, in B1(0)

v = 0, on ∂B1(0),
(2.1)

where α = (6− q)/2 and ‖v‖2 =
∫
B1(0)

(
|∇v|2 + (3/4)p2v2

)
.

We denote by H1
0,r(Ω), Ω := B1(0) the subspace of H1

0 (Ω) of the radial functions
which is endowed with the norm

‖v‖2 =

∫
Ω

(
|∇v|2 + (3/4)p2v2

)
.

Since the Euclidean sphere with center at the origin 0 ∈ RN is also a hyperbolic
sphere with center at the origin 0 ∈ Bn, H1

0,r(Ω) can also be seen as the subspace of

H1
0 (Ω) consisting of the hyperbolic radial functions. See this characterization as well

as others remarks in [3, Appendix], for instance, H1
0,r(Ω) is embedded compactly

in Lq(Ω) for 2 < q < 2∗, [3, Theorem 3.1]. Here, we use also [9, Lemma 3.1] and
recall that 2∗ = 6.

We consider the functional J : H1
0,r(Ω)→ R associated with problem (2.1),

J(v) =
a

2
‖v‖2 +

b

4
‖v‖4 − λ

q

∫
Ω

pα|v|q − 1

6

∫
Ω

|v|6, (2.2)

whose Gateaux derivative is

J ′(v)w = (a+ b‖v‖2)

∫
Ω

(
∇v ·∇w+

3

4
p2vw

)
−λ

∫
Ω

pα|v|q−2vw−
∫

Ω

|v|4vw. (2.3)

The proof uses variational methods, more exactly, the mountain pass theorem.
To this end, we have the following mountain pass geometry result.

Lemma 2.1 (Mountain pass geometry).

(a) There exist β > 0 and ρ > 0 such that J(v) ≥ β when ‖v‖ = ρ.
(b) There exists an element e ∈ H1

0,r(Ω) with ‖e‖ > ρ such that J(e) < 0.

Proof. (a) We observe that by [9, Lemma 2.1] (see also to [5, 6]) there exists a
constant C > 0, such that∫

Ω

pαvq ≤ C
(∫

Ω

|∇v|2
)q/2

≤ C
[ ∫

Ω

(
|∇v|2 + (3/4)p2v2

) ]q/2
.

Therefore,

J(u) ≥ a

2
‖v‖2 +

b

4
‖v‖4 − Cλ

q

[ ∫
Ω

(
|∇v|2 + (3/4)p2v2

) ]q/2
− 1

6

∫
Ω

|v|6,

and by the Sobolev continuous embedding, there exists a constant C̃ > 0, satisfying

J(u) ≥ a

2
‖v‖2 +

b

4
‖v‖4 − Cλ

q
‖v‖q − C̃

6
‖v‖6 ≥ β,

where the conclusion follows by making ‖v‖ = ρ sufficiently small.
Now, we prove the item (b). We take 0 < v ∈ H1

0,r(Ω) and 0 < t. Therefore,

J(tv) =
at2

2
‖v‖2 +

bt4

4
‖v‖4 − λtq

q

∫
Ω

pα|v|q − t6

6

∫
Ω

|v|6.

Therefore J(tv) → −∞, as t → +∞. Consequently, J satisfies the Mountain Pass
Theorem geometry. �



4 P. C. CARRIÃO, A. C. R. COSTA, O. H. MIYAGAKI, A. VICENTE EJDE-2021/53

We recall that the pass mountain level is defined by

c = inf
γ∈Γ

sup
t∈[0,1]

J(γ(t)),

where Γ = {γ ∈ C([0, 1], H1
0,r(Ω)) : γ(0) = 0, J(γ(1)) < 0}. For each θ > 0, we

define the functional

Jθ(v) =
a

2

∫
Ω

(
|∇v|2 +

3

4

1

e5θ
p2
( x

e2θ

)
v2
)

+
b

4

[ ∫
Ω

|∇v|2 +
3

4

1

e5θ
p2
( x

e2θ

)
v2
]2

− λ

q

∫
Ω

pα
( x

e2θ

)
vq − 1

6

∫
Ω

|v|6.

We also define Φ : R × H1
0,r(Ω) → H1

0,r(Ω) by Φ(θ, v) = eθv
(
x
e2θ

)
and I : R ×

H1
0,r(Ω)→ R by I(θ, v) = Jθ(Φ(θ, v)).
Using Lemma 2.1, we have that the functional I satisfies the geometry of the

Mountain Pass Theorem. Taking

c̃ = inf
γ̃∈Γ̃

sup
t∈[0,1]

I(γ̃(t)),

where Γ̃ =
{
γ̃ ∈ C([0, 1],R×H1

0,r(Ω)); γ̃(0) = (0, 0), I(γ̃(1)) < 0
}

, we have c = c̃

because Γ = {Φ ◦ γ̃; γ̃ ∈ Γ̃}.
Now, we define G : H1

0,r(Ω)→ R by

G(v) = 2a

∫
Ω

|∇v|2 +
9a

8

∫
Ω

p2v2 + 2b
(∫

Ω

|∇v|2
)2

+
21b

8

∫
Ω

|∇v|2
∫

Ω

p2v2

+
27b

32

(∫
Ω

p2v2
)2

− λ

q
(q + 6)

∫
Ω

pαvq − 2

∫
Ω

|v|6.

As it was mentioned in the introduction, the functional G works coupled with the
scaled functional Jθ. The functional G is a class of Pohozaev functional and it is
defined to prove the boundedness of the Palais Smale sequence. The lemma below
gives us the main property of G.

Lemma 2.2. There exists a sequence (vk) ⊂ H1
0,r(Ω) such that

J(vk)→ c J ′(vk)→ 0, G(vk)→ 0.

Proof. Applying [37, Theorem 2.8] as in [19] and [22, Proposition 4.2], we obtain a
sequence (θk, vk) such that

I(θk, vk)→ c, I ′(θk, vk)→ 0, θk → 0.

We note that

I(θ, v) =
a

2

(∫
Ω

∣∣∇(eθv( x
e2θ

))∣∣2 +
3

4

1

e5θ
p2 x

e2θ
v2
( x
e2θ

)
e2θ
)

+
b

4

(∫
Ω

∣∣∇(eθv( x
e2θ

))∣∣2 +
3

4

1

e5θ
p2 x

e2θ
v2
( x
e2θ

)
e2θ
)2

− λ

q

∫
Ω

pα
x

e2θ

(
eθv
( x
e2θ

))q
− 1

6

∫
Ω

∣∣eθv( x
e2θ

)∣∣6
=
a

2

(
e4θ

∫
Ω

|∇v|2 +
3

4
e3θ

∫
Ω

p2v2
)

+
b

4

[
e8θ
(∫

Ω

|∇v|2
)2

+
3

4
e7θ

∫
Ω

|∇v|2
∫

Ω

p2v2



EJDE-2021/53 KIRCHHOFF-TYPE PROBLEMS IN A HYPERBOLIC SPACE 5

+
9

16
e6θ
(∫

Ω

p2v2
)2]
− λ

q
eθ(q+6)

∫
Ω

pαvq − 1

6
212θ

∫
Ω

|v|6.

Thus

∂I

∂θ
= 2ae4θ

∫
Ω

|∇v|2 +
9ae3θ

8

∫
Ω

p2v2 +
21

b
be7θ

∫
Ω

|∇v|2
∫

Ω

p2v2

+
3

2

9

16
e6θ
(∫

Ω

p2v2
)2

− λ

q
(q + 6)eθ(q+6)

∫
Ω

pαvq − 2e12θ

∫
Ω

|v|6.
(2.4)

Considering θk → 0, by (2.4) and the definition of G for all ε > 0 there exists k0 ∈ N
such that k ≥ k0 ∣∣∂I

∂θ
(θk, vk)−G(vk)

∣∣ < ε. (2.5)

Since I ′(θk, vk)→ 0, by (2.5) we conclude that G(vk)→ 0.
On the other hand, since I(θk, vk)→ c and I ′(θk, vk)→ 0 we obtain respectively

|I(θk, vk)− J(vk)| < ε, (2.6)

|I ′(θk, vk)(ξ, w)− J ′(vk)(w)| < ε, (2.7)

for all k ≥ k0. Using the facts that I(θk, vk) → c and I ′(θk, vk) → 0 by (2.6) and
(2.7) we have J(vk)→ c and J ′(vk)→ 0 respectively. �

Next Lemma gives us the boundness for Palais Smale sequence.

Lemma 2.3. The sequence (vk) ⊂ H1
0,r(Ω) obtained in Lemma 2.2 is bounded.

Proof. We note that

J(vk)−G(vk) = a
(1

2
− 2

q + 6

)∫
Ω

|∇vk|2 +
3a

8

(
1− 3

q + 6

)∫
Ω

p2v2
k

+ b
(1

4
− 2

q + 6

)∫
Ω

|∇vk|2 +
3b

8

(
1− 7

q + 6

)∫
Ω

|∇vk|2
∫

Ω

p2v2
k

+
9b

64

(
1− 6

q + 6

)(∫
Ω

p2v2
k

)2

+
( 2

q + 6
− 1

6

)∫
Ω

|vk|6.

Since all the coefficients of the terms involving the integrals, on the right side of
the equality are positive, J(vk) → c and G(vk) → 0 by Lemma 2.2, we have (vk)
bounded. �

In next lemma, the number S is the best constant of Sobolev (see [35]). We follow
the arguments of [7]. See also [9, 20, 19, 28]. We are going to omit some calculus,
the reader can found the details in [8] where was studied the case 4 < q < 6.

Lemma 2.4. We have c < 1
4abS

3 + 1
24b

3S6 + 1
24 (b2S4 + 4aS)3/2, where

S := inf
u∈H1

0,r(Ω)

∫
Ω
|∇u|2( ∫

Ω
u6
)1/3 .

Proof. First, we observe that it is sufficient to show that there exists a v0 ∈
H1

0,r(Ω), v0 6= 0, such that

sup
t≥0

J(tv0) <
1

4
abS3 +

1

24
b3S6 +

1

24
(b2S4 + 4aS)3/2. (2.8)



6 P. C. CARRIÃO, A. C. R. COSTA, O. H. MIYAGAKI, A. VICENTE EJDE-2021/53

Indeed, observing that J(tv0) → −∞ as t → ∞, there exists R > 0 such that
J(Rv0) < 0. Now, we write u1 := Rv0, and from Lemma 2.1, we have

0 < β ≤ c = inf
γ∈Γ

max
τ∈[0,1]

J(γ(τ)) ≤ sup
t≥0

J(tv0) <
1

4
abS3+

1

24
b3S6+

1

24
(b2S4+4aS)3/2.

Therefore, we are going to prove the existence of a function v0 such that (2.8) holds.
We consider 0 < R < 1

2 a fixed number and let ϕ ∈ C∞0 (Ω) be a cut-off function
with support at B2R, such that ϕ is identically 1 on BR and 0 ≤ ϕ ≤ 1 on B2R.
Here, Br denotes the ball in R3 with center at the origin and radius r.

Given ε > 0 we set ψε(x) := ϕ(x)ωε(x), where

ωε(x) = (3ε)1/4 1

(ε+ |x|2)1/2
,

and ωε satisfies (see[35]) ∫
R3

|∇ωε|2 =

∫
R3

|ωε|6 = S3/2. (2.9)

From the definition of ωε, it can be shown that∫
BR

|∇ωε|2 ≤
∫
BR

|ωε|6, (2.10)∫
B1−BR

|∇ψε|2 = O(ε1/2) as ε→ 0. (2.11)

Now, we define

vε :=
ψε( ∫

B2R
ψ6
ε

)1/6
and Xε :=

∫
B1
|∇vε|2. Then, we have

Xε =

∫
BR

|∇ψε|
B2

+

∫
B2R−BR

|∇ψε|
B2

,

where B :=
( ∫

B2R
ψ6
ε

)1/6
. Thus, since ϕ ≡ 1, and consequently ∇ϕ ≡ 0 on BR, we

have

Xε =
1

B2

∫
BR

|∇ωε|2 +

∫
B2R−BR

|∇ψε|2.

By (2.10) and (2.11) we obtain

Xε ≤ S +O(ε1/2). (2.12)

On the other hand, we have

lim
t→+∞

J(tvε) = −∞, ∀ε > 0.

This implies that there exists tε > 0 such that

sup
t≥0

J(tvε) = J(tεvε). (2.13)

Now, we are going to prove an estimate for tε. From (2.13), we have

d

dt
J(tvε)|t=tε = 0,

thus,

atε‖vε‖2 + bt3ε‖vε‖4 − λtq−1
ε

∫
Ω

pα|vε|q − t5ε
∫

Ω

|vε|6 = 0,



EJDE-2021/53 KIRCHHOFF-TYPE PROBLEMS IN A HYPERBOLIC SPACE 7

which implies

a‖vε‖2 + bt2ε‖vε‖4 − λtq−2
ε

∫
Ω

pα|vε|q − t4ε
∫

Ω

|vε|6 = 0.

Since
∫

Ω
|vε|6 = 1, we have

−a‖vε‖2 − bt2ε‖vε‖4 + t4ε ≤ 0.

Hence

0 ≤ t2ε ≤
b‖vε‖4 +

[
(b‖vε‖4)2 + 4a‖vε‖2

]1/2
2

:= t0.

Since the function t 7→ a
2 t

2‖vε‖2 + b
4 t

4‖vε‖4 − t6

6 is increasing on [0, t0), denoting

C1 = a‖vε‖2 and C2 = b‖vε‖4, we have

J(tεvε) ≤
C1C2

4
+
C3

2

24
+

1

24
(C2

2 + 4C1)3/2 − λtqε
q

∫
Ω

pαvqε .

Considering A = 3/4
∫

Ω
p2v2

ε , by definition of the norm, and the inequality (2.12),
we obtain

J(tεvε) ≤
ab

4
(Xε +A)3 +

b3

24
(Xε +A)6 +

1

24

[
b2(Xε + 4)4 + 4a(Xε +A)

]3/2
− λtqε

q

∫
Ω

pαvqε

≤ ab

4
(S +O(ε1/2) +A)3 +

b3

24
(S +O(ε1/2) +A)6

+
1

24

[
b2(S +O(ε1/2) +A)4 + 4a(S +O(ε1/2) +A)

]3/2
− λtqε

q

∫
Ω

pαvqε .

Using several times the standard inequality (see e.g. [28, Page 778])

(a+ b)β ≤ aβ + β(a+ b)β−1b, ∀ β ≥ 1, ∀a, b > 0,

we infer that

J(tεvε) ≤
abS3

4
+
b3S6

24
+

1

24
(b2S4 + 4aS)3/2 +O(ε1/2)

+

∫
B2R

(3C

4
p2v2

ε − λCεpαvqε
)
,

(2.14)

for some constant C > 0, where Cε = tqε/q.
At this point, we can assume that there exists a positive constant C0 such that

Cε ≥ C0 > 0 for all ε > 0. If it is not true, then we can find a sequence εk → 0 as
k → ∞, such that tεk → 0 as k → ∞, since Cε ≥ 0. Now, up to a subsequence,
that we still denote by εk, we have tεkvεk → 0, as k →∞. Therefore,

0 < c ≤ sup
t≥0

J(tvεk) = J(tεkvεk) = J(0) = 0,

which is a contradiction.
Observing that

∫
B2R

p2v2
ε <∞, we claim that

lim
ε→0

1

ε1/2

∫
B2R

(3C

4
p2v2

ε − Cελpαvqε
)

= −∞.



8 P. C. CARRIÃO, A. C. R. COSTA, O. H. MIYAGAKI, A. VICENTE EJDE-2021/53

Assuming the Claim is proved, from (2.14) we have

J(tεvε) <
abS3

4
+
b3S6

24
+

1

24
(b2S4 + 4aS)3/2,

for some ε > 0 sufficiently small, and the proof is complete.
Now, we prove the Claim. For this, it is sufficient to show that

lim
ε→0

1

ε1/2

(∫
BR

(3C

4
p2ω2

ε − Cελpαωqε
))

= −∞ (2.15)∫
B2R−BR

(3C

4
p2v2

ε − Cελpαvqε
)

= O(ε1/2). (2.16)

First, we consider

Jε =
1

ε1/2

∫
BR

(3C

4
p2ω2

ε − Cελpαωqε
)

=
3C

4ε1/2

∫
BR

( 2

1− |x|2
)2 (3ε)1/2

(ε+ |x|2)
− λCε
ε1/2

∫
BR

( 2

1− |x|2
)α (3ε)q/4

(ε+ |x|2)q/2

= C̃

∫
BR

( 2

1− |x|2
)2 1

(ε+ |x|2)
− λC̃εε

(q−2)
4

∫
BR

( 2

1− |x|2
)α 1

(ε+ |x|2)q/2

= J1 − J2,

(2.17)

for some constant C̃ > 0. We observe that on BR,

2 <
2

1− |x|2
≤ 2

1−R2
. (2.18)

Therefore, making the change of variables x = ε1/2y and using the polar coordi-
nates, we obtain

J1 ≤
4C̃

(1−R2)2
ωε1/2

∫ Rε−1/2

0

r2

(1 + r2)
dr, (2.19)

for some constant C̃ > 0. Similarly, for J2, we have

J2 ≥ λC̃ε2αwε−
q
4 +1

∫ Rε−1/2

0

r2

(1 + r2)q/2
dr, (2.20)

where C̃ε is a positive constant. Thus, combining (2.17), (2.19) and (2.20) we obtain

Jε ≤
4C̃

(1−R2)2
ωε1/2

∫ Rε−1/2

0

r2

(1 + r2)
dr

− λC̃ε2αwε−
q
4 +1

∫ Rε−1/2

0

r2

(1 + r2)q/2
dr.

(2.21)

Observing that ∫ Rε−1/2

0

r2

1 + r2
dr = Rε−1/2 − tan−1(Rε−1/2)

we obtain

Jε ≤ C − Cε1/2 tan−1(Rε−1/2)− λCε−
q
4 +1

∫ Rε−1/2

0

r2

(1 + r2)q/2
dr. (2.22)



EJDE-2021/53 KIRCHHOFF-TYPE PROBLEMS IN A HYPERBOLIC SPACE 9

Now, as ∫ Rε−1/2

0

r2

(1 + r2)q/2
dr ≥

∫ Rε−1/2

0

1

1 + r2
dr ≥ C > 0,

for all ε < ε0, with ε0 small enough. At this moment, it is possible to see the main
difference with the proof of [8, Lemma 2.3]. To control the sign of the expression
of (2.15) it is necessary to use the assumption involving λ. Since, by assumption, λ

is positive and sufficiently large, we can take λ = ε−
1
2 and we conclude that (2.15)

holds.
The proof of (2.16) is the same of [8, (2.13)], This completes the proof. �

3. Proof of Theorem 1.1

Let {vn} be the sequence given by Lemma 2.2. Lemma 2.3 implies that {vn} is
bounded in H1

0,r(Ω). Thus, we can assume, passing to a subsequence, that vn ⇀ v,

weakly in H1
0,r(Ω) as n→∞. Arguing as in [9], we have

J ′(vn)w = o(1), ∀w ∈ H1
0,r(Ω). (3.1)

Now, we observe that
|J ′(vn)w − J ′(v)w| → 0, (3.2)

as n → ∞, for all w ∈ C∞c,rad(Ω). From this, it follows that J ′(v)w = 0, for all

w ∈ C∞c,rad(Ω). By denseness, we conclude that

J ′(v)w = 0, ∀w ∈ H1
0,r(Ω), (3.3)

and v is a critical point of the functional J restricted to the space H1
0,r(Ω).

Now, we follow the ideas in [4, 10, 15] (see also [29]). Since H1
0,r(Ω) is a closed

subspace of H1
0 (Ω), we can write

H1
0 (Ω) = H1

0,r(Ω)⊕H1
0,r(Ω)⊥,

where ·⊥ denotes the orthogonal complement of the space. Therefore, for each
w ∈ H1

0 (Ω), there exist ϑ ∈ H1
0,r(Ω) and ϑ⊥ ∈ H1

0,r(Ω)⊥ such that

w = ϑ+ ϑ⊥. (3.4)

As H1
0,r(Ω) is a Hilbert space and J ′(v) ∈ H1

0,r(Ω)∗, from the Riesz Representa-

tion Theorem there exists z ∈ H1
0,r(Ω) such that

J ′(v)w =

∫
Ω

∇z · ∇w, ∀w ∈ H1
0,r(Ω).

Thus, as z ∈ H1
0,r(Ω) and ϑ⊥ ∈ H1

0,r(Ω)⊥, we have

J ′(v)ϑ⊥ = 0. (3.5)

From (3.3), (3.4) and (3.5), for each w ∈ H1
0 (Ω), we obtain

J ′(v)w = J ′(v)ϑ+ I ′(v)ϑ⊥ = 0.

This allows us to conclude that v is a critical point of the functional J in H1
0 (Ω)

and consequently v is a weak solution for problem (2.1).
If v 6= 0 we are done. Now, we suppose that v ≡ 0. Considering vn ⇀ 0, as

n→∞, we have

J ′(vn)vn = a‖vn‖2 + b‖vn‖4 − λ
∫

Ω

pα|vn|q −
∫

Ω

|vn|6 = on(1). (3.6)



10 P. C. CARRIÃO, A. C. R. COSTA, O. H. MIYAGAKI, A. VICENTE EJDE-2021/53

By [9, Lemma 3.1], we obtain

λ

∫
Ω

pα|vn|q → 0, as n→∞, (3.7)

Let L1 > 0, L2 > 0 be such that

a‖vn‖2 → L1 and b‖vn‖4 → L2, as n→∞. (3.8)

By (3.6), (3.7), and (3.8),∫
Ω

|vn|6 → L1 + L2, as n→∞. (3.9)

But

S
(∫

Ω

v6
n

)1/3

≤
∫

Ω

|∇vn|2, (3.10)

which implies

aS
(∫

Ω

v6
n

)1/3

≤ a
∫

Ω

|∇vn|2 ≤ a
∫

Ω

(
|∇vn|2 + (3/4)p2v2

n

)
= a‖vn‖2, (3.11)

bS2
(∫

Ω

v6
n

)2/3

≤ b
[ ∫

Ω

|∇vn|2
]2
≤ b
[ ∫

Ω

(
|∇vn|2 + (3/4)p2v2

n

) ]2
= b‖vn‖4.

(3.12)

Thus, by (3.8), (3.9), (3.11) and (3.12),

L1 ≥ aS(L1 + L2)1/3 and L2 ≥ bS2(L1 + L2)2/3. (3.13)

On the other hand, J(vn) = c+ o(1). So

c =
L1

2
+
L2

4
− 1

6
(L1 + L2) =

L1

3
+
L2

12
. (3.14)

By (3.13) we have

(L1 + L2)1/3 ≥ bs2 + (b2s4 + 4as)1/2

2
. (3.15)

Hence by (3.13), (3.14) and (3.15),

c ≥ 1

3
L1 +

1

12
L2 ≥

1

3
aS(L1 + L2)1/3 +

1

12
bS2[(L1 + L2)1/3]2

≥ 1

4
abS3 +

1

24
b3S6 +

1

24
(b2S4 + 4aS)3/2,

which is a contradiction to Lemma 2.4. Therefore, we conclude that v 6= 0.

Acknowledgments. This work was done while P. C. C. was visiting the Depart-
ment of Mathematics of UFJF, under financial support by FAPEMIG CEX APQ
00063 15. A. C. R. C. was supported in part by PNPD CAPES 2017 PGM/UFJF.
O. H. M. received research grants from CNPq/Brazil 307061/2018-3, FAPEMIG
CEX APQ 00063/15 and INCTMAT/CNPQ/Brazil. The authors would like to
thank the anonymous referees for all insightful comments, which allow us to im-
prove our original version.



EJDE-2021/53 KIRCHHOFF-TYPE PROBLEMS IN A HYPERBOLIC SPACE 11

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Paulo Cesar Carrião

Departamento de Matemática, Universidade Federal de Minas Gerais, Belo Horizonte,

MG 31270-901, Brazil
Email address: pauloceca@gmail.com

Augusto César dos Reis Costa
Faculdade de Matemática, Instituto de Ciências Exatas e Naturais, Universidade Fed-

eral do Pará, Belém, PA 66075-110, Brazil

Email address: aug@ufpa.br

Olimpio Hiroshi Miyagaki

Departamento de Matemática, Universidade Federal de Juiz de Fora, Juiz de Fora, MG
36036-330, Brazil

Email address: ohmiyagaki@gmail.com

André Vicente
Centro de Ciências Exatas e Tecnológicas, Universidade Estadual do Oeste do Paraná,
Cascavel, PR 85819-110, Brazil

Email address: andre.vicente@unioeste.br


	1. Introduction
	2. Proof of the main result
	3. Proof of Theorem ??
	Acknowledgments

	References