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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 CARRI ˜AO, AUGUSTO C ´ESAR DOS REIS COSTA, OLIMPIO HIROSHI MIYAGAKI, ANDR ´E 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 uqbecomes “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

Z

B3

|∇B3u|2dVB3

B3u=λ|u|q−2u+|u|4u in H1(B3), (1.1) where a, b, λare 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 spaceRN, withN ≥3, and with the non-linearity behaving as a polynomial function of degree 2= N−22N was studied by Brezis and Nirenberg [7]. Posteriorly, several authors have studied this class of problems; see for instance Carri˜ao, Costa, and Miyagaki [8].

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

utt−MZ

|∇xu|2dx

xu=f(x, t), (x, t)∈Ω×(0,∞),

where Ω is a bounded domain ofRN,M(s) =a+bswitha, b >0, andfis 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 containingM 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

2010Mathematics 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

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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 Rˇadulescu [38, 39]. In the first one, the authors studied the multiplicity of solutions for a class of quasilinear Kirchhoff system involving the fractionalp-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 that2< q <4, forλ >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 andb = 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 (seeJθ(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 propertyG(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 pointP0 ∈ Hn is projected to P ∈Rn, whereP is the intersection of the straight line connecting P0 and the point (0, . . . ,0,−1). Explicitly the projection operator G:Rn→ Hn andG−1:Hn →Rn given by

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

yn+1y, x, y∈Rn, wherep(x) = 1−|x|2 2.

We consider the ballB1(0), andBn 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= Z

D0

|∇Bnu|2dVBn= Z

D

|∇u|2pn−2dx,

Bnu=p−ndiv(pn−2∇u).

We denote byD⊂B1(0) the stereographic projection ofD0⊂ Hn. Details involving the hyperbolic space can be found in [3, 18, 31, 33, 34].

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Definingv:=p1/2u, we have thatuis solution of (1.1) if, and only if,v satisfies (a+bkvk2)(−∆v+ (3/4)p2v) =λpα|v|q−2v+|v|4v, inB1(0)

v= 0, on∂B1(0), (2.1)

whereα= (6−q)/2 andkvk2=R

B1(0) |∇v|2+ (3/4)p2v2 .

We denote byH0,r1 (Ω), Ω :=B1(0) the subspace ofH01(Ω) of the radial functions which is endowed with the norm

kvk2= Z

|∇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,H0,r1 (Ω) can also be seen as the subspace of H01(Ω) consisting of the hyperbolic radial functions. See this characterization as well as others remarks in [3, Appendix], for instance, H0,r1 (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 functionalJ :H0,r1 (Ω)→Rassociated with problem (2.1), J(v) =a

2kvk2+b

4kvk4−λ q Z

pα|v|q−1 6

Z

|v|6, (2.2) whose Gateaux derivative is

J0(v)w= (a+bkvk2) Z

∇v· ∇w+3 4p2vw

−λ Z

pα|v|q−2vw− Z

|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 andρ >0 such that J(v)≥β when kvk=ρ.

(b) There exists an element e∈H0,r1 (Ω) with kek> ρsuch thatJ(e)<0.

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

Z

pαvq ≤CZ

|∇v|2q/2

≤ChZ

|∇v|2+ (3/4)p2v2iq/2 .

Therefore, J(u)≥a

2kvk2+b

4kvk4−Cλ q

hZ

|∇v|2+ (3/4)p2v2iq/2

−1 6

Z

|v|6,

and by the Sobolev continuous embedding, there exists a constantC >e 0, satisfying J(u)≥ a

2kvk2+ b

4kvk4−Cλ

q kvkq−Ce

6kvk6≥β, where the conclusion follows by makingkvk=ρsufficiently small.

Now, we prove the item (b). We take 0< v∈H0,r1 (Ω) and 0< t. Therefore, J(tv) = at2

2 kvk2+bt4

4 kvk4−λtq q

Z

pα|v|q−t6 6

Z

|v|6.

ThereforeJ(tv)→ −∞, ast→+∞. Consequently,J satisfies the Mountain Pass

Theorem geometry.

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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], H0,r1 (Ω)) : γ(0) = 0, J(γ(1)) <0}. For each θ > 0, we define the functional

Jθ(v) =a 2 Z

|∇v|2+3 4

1 ep2 x

e

v2 +b

4 hZ

|∇v|2+3 4

1 ep2 x

e

v2i2

−λ q Z

pα x e

vq−1 6

Z

|v|6.

We also define Φ : R×H0,r1 (Ω) → H0,r1 (Ω) by Φ(θ, v) = eθv ex

and I : R× H0,r1 (Ω)→RbyI(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

˜ c= inf

˜γ∈˜Γ

sup

t∈[0,1]

I(˜γ(t)), where ˜Γ =

˜

γ∈C([0,1],R×H0,r1 (Ω)); ˜γ(0) = (0,0), I(˜γ(1))<0 , we have c = ˜c because Γ ={Φ◦˜γ; ˜γ∈Γ}.˜

Now, we defineG:H0,r1 (Ω)→Rby G(v) = 2a

Z

|∇v|2+9a 8

Z

p2v2+ 2bZ

|∇v|22

+21b 8

Z

|∇v|2 Z

p2v2 +27b

32 Z

p2v22

−λ q(q+ 6)

Z

pαvq−2 Z

|v|6.

As it was mentioned in the introduction, the functional Gworks coupled with the scaled functionalJθ. The functionalG 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 ofG.

Lemma 2.2. There exists a sequence (vk)⊂H0,r1 (Ω) such that J(vk)→c J0(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, I0k, vk)→0, θk →0.

We note that I(θ, v) = a

2 Z

∇ eθv x

e

2+3 4

1 ep2 x

ev2 x e

e

+ b 4

Z

∇ eθv x

e

2+3 4

1 ep2 x

ev2 x e

e2

−λ q Z

pα x e

eθv x

e q

−1 6 Z

eθv x e

6

= a 2

e Z

|∇v|2+3 4e

Z

p2v2 + b

4 h

eZ

|∇v|22 +3

4e Z

|∇v|2 Z

p2v2

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+ 9 16eZ

p2v22i

−λ qeθ(q+6)

Z

pαvq−1 6212θ

Z

|v|6. Thus

∂I

∂θ = 2ae Z

|∇v|2+9ae 8

Z

p2v2+21 b be

Z

|∇v|2 Z

p2v2 +3

2 9 16eZ

p2v22

−λ

q(q+ 6)eθ(q+6) Z

pαvq−2e12θ Z

|v|6.

(2.4)

Consideringθk →0, by (2.4) and the definition ofGfor all >0 there existsk0∈N such thatk≥k0

∂I

∂θ(θk, vk)−G(vk)

< . (2.5)

SinceI0k, vk)→0, by (2.5) we conclude that G(vk)→0.

On the other hand, sinceI(θk, vk)→candI0k, vk)→0 we obtain respectively

|I(θk, vk)−J(vk)|< , (2.6)

|I0k, vk)(ξ, w)−J0(vk)(w)|< , (2.7) for all k≥k0. Using the facts thatI(θk, vk)→c andI0k, vk)→0 by (2.6) and (2.7) we haveJ(vk)→c andJ0(vk)→0 respectively.

Next Lemma gives us the boundness for Palais Smale sequence.

Lemma 2.3. The sequence(vk)⊂H0,r1 (Ω) obtained in Lemma2.2 is bounded.

Proof. We note that J(vk)−G(vk) =a1

2− 2 q+ 6

Z

|∇vk|2+3a 8

1− 3

q+ 6 Z

p2v2k +b1

4− 2 q+ 6

Z

|∇vk|2+3b 8

1− 7

q+ 6 Z

|∇vk|2 Z

p2v2k + 9b

64

1− 6 q+ 6

Z

p2vk22

+ 2 q+ 6 −1

6 Z

|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 numberSis 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 < 14abS3+241b3S6+241(b2S4+ 4aS)3/2, where S:= inf

u∈H10,r(Ω)

R

|∇u|2 R

u61/3.

Proof. First, we observe that it is sufficient to show that there exists a v0 ∈ H0,r1 (Ω), v06= 0, such that

sup

t≥0

J(tv0)< 1

4abS3+ 1

24b3S6+ 1

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

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Indeed, observing that J(tv0) → −∞ as t → ∞, there exists R > 0 such that J(Rv0)<0. Now, we writeu1:=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

4abS3+1

24b3S6+1

24(b2S4+4aS)3/2. Therefore, we are going to prove the existence of a functionv0such that (2.8) holds.

We consider 0< R < 12 a fixed number and letϕ∈C0(Ω) be a cut-off function with support at B2R, such that ϕis identically 1 onBR and 0 ≤ϕ≤1 on B2R. Here,Brdenotes the ball in R3 with center at the origin and radiusr.

Givenε >0 we setψε(x) :=ϕ(x)ωε(x), where ωε(x) = (3ε)1/4 1

(ε+|x|2)1/2, andωε satisfies (see[35])

Z

R3

|∇ωε|2= Z

R3

ε|6=S3/2. (2.9)

From the definition ofωε, it can be shown that Z

BR

|∇ωε|2≤ Z

BR

ε|6, (2.10)

Z

B1−BR

|∇ψε|2=O(ε1/2) as ε→0. (2.11) Now, we define

vε:= ψε

R

B2Rψε61/6 andXε:=R

B1|∇vε|2. Then, we have Xε=

Z

BR

|∇ψε| B2 +

Z

B2R−BR

|∇ψε| B2 , whereB:= R

B2Rψε61/6

. Thus, sinceϕ≡1, and consequently∇ϕ≡0 onBR, we have

Xε= 1 B2

Z

BR

|∇ωε|2+ Z

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

t→+∞lim J(tvε) =−∞, ∀ε >0.

This implies that there existstε>0 such that sup

t≥0

J(tvε) =J(tεvε). (2.13) Now, we are going to prove an estimate fortε. From (2.13), we have

d

dtJ(tvε)|t=tε = 0, thus,

atεkvεk2+bt3εkvεk4−λtq−1ε Z

pα|vε|q−t5ε Z

|vε|6= 0,

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which implies

akvεk2+bt2εkvεk4−λtq−2ε Z

pα|vε|q−t4ε Z

|vε|6= 0.

SinceR

|vε|6= 1, we have

−akvεk2−bt2εkvεk4+t4ε≤0.

Hence

0≤t2ε≤bkvεk4+

(bkvεk4)2+ 4akvεk21/2

2 :=t0.

Since the functiont 7→ a2t2kvεk2+4bt4kvεk4t66 is increasing on [0, t0), denoting C1=akvεk2and C2=bkvεk4, we have

J(tεvε)≤ C1C2 4 +C23

24 + 1

24(C22+ 4C1)3/2−λtqε q

Z

pαvqε. Considering A= 3/4R

p2v2ε, 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

Z

pαvqε

≤ab

4(S+O(ε1/2) +A)3+ b3

24(S+O(ε1/2) +A)6 + 1

24 h

b2(S+O(ε1/2) +A)4+ 4a(S+O(ε1/2) +A)i3/2

−λtqε q

Z

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) +

Z

B2R

3C

4 p2v2ε−λCεpαvεq ,

(2.14)

for some constantC >0, whereCε=tqε/q.

At this point, we can assume that there exists a positive constantC0 such that Cε≥C0>0 for all ε >0. If it is not true, then we can find a sequenceεk →0 as k → ∞, such that tεk →0 as k→ ∞, sinceCε ≥0. Now, up to a subsequence, that we still denote byεk, we havetεkvεk→0, ask→ ∞. Therefore,

0< c≤sup

t≥0

J(tvεk) =J(tεkvεk) =J(0) = 0, which is a contradiction.

Observing thatR

B2Rp2v2ε<∞, we claim that

ε→0lim 1 ε1/2

Z

B2R

3C

4 p2v2ε−Cελpαvqε

=−∞.

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Assuming the Claim is proved, from (2.14) we have J(tεvε)< abS3

4 +b3S6 24 + 1

24(b2S4+ 4aS)3/2, for someε >0 sufficiently small, and the proof is complete.

Now, we prove the Claim. For this, it is sufficient to show that

ε→0lim 1 ε1/2

Z

BR

3C

4 p2ωε2−Cελpαωεq

=−∞ (2.15)

Z

B2R−BR

3C

4 p2vε2−Cελpαvqε

=O(ε1/2). (2.16) First, we consider

Jε= 1 ε1/2

Z

BR

3C

4 p2ωε2−Cελpαωεq

= 3C 4ε1/2

Z

BR

2 1− |x|2

2 (3ε)1/2

(ε+|x|2)−λCε ε1/2 Z

BR

2 1− |x|2

α (3ε)q/4 (ε+|x|2)q/2

= ˜C Z

BR

2 1− |x|2

2 1

(ε+|x|2)−λC˜εε(q−2)4 Z

BR

2 1− |x|2

α 1 (ε+|x|2)q/2

=J1−J2,

(2.17) for some constantC >e 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≤ 4 ˜C

(1−R2)2ωε1/2

Z −1/2

0

r2

(1 +r2)dr, (2.19) for some constantC >e 0. Similarly, forJ2, we have

J2≥λC˜ε2αq4+1

Z −1/2

0

r2

(1 +r2)q/2dr, (2.20) whereCeεis a positive constant. Thus, combining (2.17), (2.19) and (2.20) we obtain

Jε≤ 4 ˜C

(1−R2)2ωε1/2

Z −1/2

0

r2 (1 +r2)dr

−λC˜ε2αq4+1

Z −1/2

0

r2 (1 +r2)q/2dr.

(2.21)

Observing that

Z −1/2

0

r2

1 +r2dr=Rε−1/2−tan−1(Rε−1/2) we obtain

Jε≤C−Cε1/2tan−1(Rε−1/2)−λCεq4+1

Z −1/2

0

r2

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

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Now, as

Z −1/2

0

r2

(1 +r2)q/2dr≥

Z −1/2

0

1

1 +r2dr≥C >0,

for allε < ε0, withε0 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 involvingλ. Since, by assumption,λ is positive and sufficiently large, we can takeλ=ε12 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 inH0,r1 (Ω). Thus, we can assume, passing to a subsequence, thatvn* v, weakly inH0,r1 (Ω) asn→ ∞. Arguing as in [9], we have

J0(vn)w=o(1), ∀w∈H0,r1 (Ω). (3.1) Now, we observe that

|J0(vn)w−J0(v)w| →0, (3.2) as n → ∞, for all w ∈ Cc,rad (Ω). From this, it follows that J0(v)w = 0, for all w∈Cc,rad (Ω). By denseness, we conclude that

J0(v)w= 0, ∀w∈H0,r1 (Ω), (3.3) andv is a critical point of the functionalJ restricted to the spaceH0,r1 (Ω).

Now, we follow the ideas in [4, 10, 15] (see also [29]). Since H0,r1 (Ω) is a closed subspace ofH01(Ω), we can write

H01(Ω) =H0,r1 (Ω)⊕H0,r1 (Ω),

where · denotes the orthogonal complement of the space. Therefore, for each w∈H01(Ω), there existϑ∈H0,r1 (Ω) andϑ∈H0,r1 (Ω) such that

w=ϑ+ϑ. (3.4)

AsH0,r1 (Ω) is a Hilbert space and J0(v)∈H0,r1 (Ω), from the Riesz Representa- tion Theorem there existsz∈H0,r1 (Ω) such that

J0(v)w= Z

∇z· ∇w, ∀w∈H0,r1 (Ω).

Thus, asz∈H0,r1 (Ω) andϑ∈H0,r1 (Ω), we have

J0(v)ϑ= 0. (3.5)

From (3.3), (3.4) and (3.5), for eachw∈H01(Ω), we obtain J0(v)w=J0(v)ϑ+I0(v)ϑ= 0.

This allows us to conclude that v is a critical point of the functional J in H01(Ω) and consequentlyvis 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

J0(vn)vn=akvnk2+bkvnk4−λ Z

pα|vn|q− Z

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

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By [9, Lemma 3.1], we obtain λ

Z

pα|vn|q →0, asn→ ∞, (3.7) LetL1>0,L2>0 be such that

akvnk2→L1 and bkvnk4→L2, as n→ ∞. (3.8) By (3.6), (3.7), and (3.8),

Z

|vn|6→L1+L2, asn→ ∞. (3.9) But

SZ

vn61/3

≤ Z

|∇vn|2, (3.10)

which implies aSZ

v6n1/3

≤a Z

|∇vn|2≤a Z

|∇vn|2+ (3/4)p2v2n

=akvnk2, (3.11) bS2Z

v6n2/3

≤bhZ

|∇vn|2i2

≤bhZ

|∇vn|2+ (3/4)p2vn2i2

=bkvnk4. (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

3L1+ 1

12L2≥1

3aS(L1+L2)1/3+ 1

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

≥ 1

4abS3+ 1

24b3S6+ 1

24(b2S4+ 4aS)3/2,

which is a contradiction to Lemma 2.4. Therefore, we conclude thatv6= 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.

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Paulo Cesar Carri˜ao

Departamento de Matem´atica, Universidade Federal de Minas Gerais, Belo Horizonte, MG 31270-901, Brazil

Email address:[email protected]

Augusto C´esar dos Reis Costa

Faculdade de Matem´atica, Instituto de Ciˆencias Exatas e Naturais, Universidade Fed- eral do Par´a, Bel´em, PA 66075-110, Brazil

Email address:[email protected]

Olimpio Hiroshi Miyagaki

Departamento de Matem´atica, Universidade Federal de Juiz de Fora, Juiz de Fora, MG 36036-330, Brazil

Email address:[email protected]

Andr´e Vicente

Centro de Ciˆencias Exatas e Tecnol´ogicas, Universidade Estadual do Oeste do Paran´a, Cascavel, PR 85819-110, Brazil

Email address:[email protected]

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