Quasilinear Schrödinger equations with singular and vanishing potentials involving exponential critical growth in R2

Nonlinear Analysis 196 (2020) 111800

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Quasilinear Schrödinger equations with singular and vanishing potentials involving exponential critical growth in R2 ✩ Uberlandio B. Severo a ,∗, Gilson M. de Carvalho b a

Departamento de Matemática, Universidade Federal da Paraíba, 58051-900, João Pessoa – PB, Brazil Departamento de Matemática, Universidade Federal Rural de Pernambuco, 52171-900, Recife – PE, Brazil b

article

info

Article history: Received 28 August 2019 Accepted 30 January 2020 Communicated by Vicentiu D Radulescu MSC: 35J20 35J25 35J60 35Q60 Keywords: Schrödinger equations Singular potentials Trudinger–Moser inequality Ground state solution

abstract We study the existence and nonexistence of solution for the following class of quasilinear Schrödinger equations:

{

− ∆u + V (|x|)u − [∆(u2 )]u = Q(|x|)h(u), u(x) → 0

as

x ∈ R2 ,

|x| → ∞,

where V and Q are potentials that can be singular at the origin, unbounded or vanishing at infinity and the nonlinearity h(s) is allowed to satisfy the exponential critical growth with respect to the Trudinger–Moser inequality. By combining variational methods in a suitable weighted Orlicz space with a version of the Trudinger–Moser inequality for this space, we obtain the existence of a nonnegative ground state solution. For this, we have used some regularity results and we need to explore a symmetric criticality type argument. Moreover, under some conditions on V , Q and h(s), we prove that this problem does not have positive radial solution. Schrödinger equations of this type have been studied as models of several physical phenomena. © 2020 Elsevier Ltd. All rights reserved.

1. Introduction and main results In this paper, we deal with the existence and nonexistence of solution for quasilinear Schr¨odinger equations of the form − ∆u + V (x)u − κ[∆(u2 )]u = g(x, u), x ∈ R2 , (1.1) where κ is a nonnegative parameter and V, g are continuous functions satisfying appropriate conditions which will be introduced later. ✩ Research partially supported CAPES, Brazil - Finance Code 001, CNPq, Brazil grant 308735/2016-1 and Grant 2019/0014 Para´ıba State Research Foundation (Fapesq), Brazil. ∗ Corresponding author. E-mail addresses: [email protected] (U.B. Severo), [email protected] (G.M. de Carvalho).

https://doi.org/10.1016/j.na.2020.111800 0362-546X/© 2020 Elsevier Ltd. All rights reserved.

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U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

Such equations arise in various branches of mathematical physics and they have been the subject of extensive study in recent years. Part of the interest is due to the fact that solutions of equations of the type (1.1) are related to the existence of solitary wave solutions for quasilinear Schr¨odinger equations of the form [ ] 2 2 2 i∂t ψ = −∆x ψ + W (x)ψ − ϕ(x, |ψ| )ψ − κ ∆ρ(|ψ| ) ρ′ (|ψ| )ψ, x ∈ RN , t ∈ R (1.2) where N ≥ 1, ψ : RN ×R → C, W : RN → R is a given potential, ϕ : RN ×R+ → R, κ ≥ 0 is a parameter and ρ : R+ → R is an appropriate function. Considering ρ(s) = s and putting ψ(x, t) = exp(−iωt)u(x), we obtain the elliptic equation (1.1), where W (x) = V (x) − ω and g(x, u) = ϕ(x, u2 )u. For example, the case ρ(s) = s was used for the superfluid film equation in plasma physics by Kurihara in [16] and for ρ(s) = (1 + s)1/2 , equation (1.2) models the self-channeling of a high-power ultra short laser in matter, see [8,9]. For more details and physical applications see the papers [17,31] and references therein. Motivated by these physical aspects, Eq. (1.1) has attracted the attention of many researchers and some existence and multiplicity results have been obtained. The semilinear case (κ = 0) has been studied extensively in recent years with a large variety of hypotheses on the potential V (x) and on the nonlinearity g(x, s), see for example [4,10,23,27,30] and references therein. Compared to the semilinear case, the quasilinear one (κ > 0) becomes much more complicated due to the effects of the quasilinear and nonconvex term ∆(u2 )u. One of the main difficulties to study (1.1) is that there is no suitable space on which the associated energy functional is well defined and belongs to C 1 -class, except for the one-dimensional case (see [31]). Another feature of the quasilinear equation (1.1) is that the exponential critical growth changes 4 for eαs , α > 0, as observed in [20,24]. As we know, in the semilinear case, the standard critical growth 2 2 4 in dimension two is eαs , α > 0. The change of growth of eαs for eαs is because of the presence of the nonlinear term ∆(u2 )u. To the best of our knowledge, the first existence result for quasilinear problems of the type (1.1) involving variational methods was due to [31] for the one-dimensional case or V being radially symmetric for high dimensions, where the authors have used a constrained minimization argument (see also [17] for the more general case). After them, some ideas and approaches were developed to overcome the difficulties, see [13,17,32] for Nehari manifold arguments, gauge transforms and derivation process. By using a change of variables (dual approach), the author in [19] reduced the quasilinear equation (1.1) to a semilinear one, and an Orlicz space framework was used to obtain solution (for more details about this change of variables, see the pioneering works [12,18]). The same method was also applied in [24], but the usual Sobolev space framework was used as the working space. Quasilinear equations of type (1.1) involving exponential subcritical or critical growth in R2 were studied by various authors, see [1,19,20,24–26,29,34,35,40]. In all these papers, it was assumed that the potential V (x) is bounded from below by a positive constant. Here, we want to consider singular potentials or vanishing at infinity. From the physical point of view, singular potential corresponds to collision of particles with the center of force, see for instance [22] for more details. The main purpose of this paper is to show that, using a variational framework based on a nonstandard Orlicz space, it is possible to find sufficient conditions for the existence of nonnegative and nonzero solution for the problem { − ∆u + V (|x|)u − [∆(u2 )]u = Q(|x|)h(u), x ∈ R2 , (P ) u(x) → 0 as |x| → ∞, where V, Q : (0, ∞) → R are potentials that can be singular at the origin, unbounded or vanishing at infinity and the nonlinear term h(s) is allowed to satisfy the exponential critical growth with respect to the Trudinger–Moser inequality. Problems involving exponential critical growth in the plane have been extensively studied after the works [21,39] (see also [2,10,14] and references therein). There exists a vast literature about versions of the famous Trudinger–Moser inequality. In this paper, it also was necessary for us to obtain a version of it for our variational setting (see Theorem 4.4). As far as we know, there are not any reference and results about quasilinear problems of the type (P ) involving critical growth in R2 and

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

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these features on the potentials V and Q. Even in the case Q ≡ 1, our results are new and complement the previous works, because we prove the existence of ground state solution. Moreover, we also obtain a nonexistence result. A version of quasilinear problem (P ) in dimensions N ≥ 3 was obtained in [33]. Next, for ease of reference we state our assumptions in a more precise way. We assume that the potentials V and Q satisfy the following hypotheses: (V ) V : (0, ∞) → R is continuous, V (r) > 0 for all r > 0 and there exist constants a0 > −2 and a > −2 such that V (r) V (r) lim sup a < ∞ and 0 < lim inf a ; 0 r→+∞ r r + r→0 (Q) Q : (0, ∞) → R is continuous, Q(r) > 0 for all r > 0 and there exist −2 < b0 ≤ 0 and −2 < b < a satisfying Q(r) Q(r) lim sup b < ∞ and lim sup b < ∞. 0 r r + r→+∞ r→0 We observe that conditions (V ) and (Q) allow that V (r) and Q(r) are singular at the origin and vanishing at infinity (see example below). Example 1.1. Let V, Q : (0, ∞) → R be given by V (r) = r−1 (cos r + 2)

and

3

2

Q(r) = 3r− 2 (1 + e−r sin r).

Clearly V (r) and Q(r) are continuous and positive functions. Moreover, lim inf r→+∞

V (r) V (r) = 1, lim sup −1 = 3 r−1 r + r→0

and

lim sup r→+∞

Q(r) Q(r) = lim sup −3/2 = 3, r−3/2 r + r→0

that is, V and Q satisfy conditions (V ) and (Q) respectively, with a = a0 = −1 and b = b0 = −3/2. Note that they are singular at the origin and vanishing at infinity. With respect to the function h(s), we suppose that h ∈ C(R, R+ ) and the following assumptions hold: (h1 ) (Exponential critical growth) There exists λ0 > 0 such that { h(s) 0, ∀ λ > λ0 lim = +∞, ∀ λ < λ0 ; s→+∞ eλs4 (h2 ) lim sups→0+ h(s)/s < +∞; ∫s (h3 ) there exists µ > 2 such that 2µH(s) ≤ sh(s) for all s ≥ 0, where H(s) = 0 h(t)dt; (h4 ) there exists ξ > 0 and p > 2 such that H(s) ≥

ξ p s , p

∀ s ∈ [0, 1].

As already mentioned, condition (h1 ) establishes the definition of exponential critical growth for this class of quasilinear problems. Note that we are not requiring that h(s)/s → 0 as s → 0+ as was assumed in the works [20,26,29,34,35]. We also emphasize that in [4,20,25–27,34,35,38], the authors consider condition (h4 ) for all s ≥ 0. Here, we ask only for s ∈ [0, 1]. Deriving H(s)/s2µ for s > 0, it is easily seen that (h3 ) is equivalent to condition (ˆ h3 ) H(s)/s2µ is nondecreasing for s > 0. In particular, if we denote C = H(1) then H(s) ≤ Cs2µ , ∀ s ∈ [0, 1] and therefore we must have p ≥ 2µ in (h4 ).

(1.3)

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

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Example 1.2. A simple example of nonlinearity satisfying the hypotheses (h1 ) − (h4 ) is given by ′ h(s) = H (s), with { 4 ξsτ es , if s ≥ 0, H(s) = 0, if s < 0, where ξ > 0 and τ > 4. Here, µ = τ /2 and p = τ . In order to state our main results, we introduce some spaces. For 1 ≤ q < ∞, we define the Lebesgue space with weight Q(|x|) by { } ∫ q Lq (R2 ; Q) = u : R2 → R; u is measurable and Q(|x|)|u| dx < ∞ , R2

endowed with the norm (∫ ∥u∥q,Q =

)1/q q Q(|x|)|u| dx .

R2

We also define the spaces { Y :=

u ∈ L2loc (R2 ) : |∇u| ∈ L2 (R2 ) and

∫

V (|x|)u2 dx < ∞

}

R2

and Yrad = {u ∈ Y : u(x) = u(gx), ∀g ∈ O(2)} , endowed with the inner product ∫ ⟨u, v⟩ =

(∇u∇v + V (|x|)uv)dx

(1.4)

R2

and the correspondent norm (∫ ∥u∥Y =

2

2

) 21

[|∇u| + V (|x|)u ]dx

.

R2

In fact, due to condition (V ), the equality in (1.4) defines an inner product on Y . Moreover, we can show that Y and Yrad are Hilbert spaces. We point out that these spaces are suitable for our purposes. 2 In this work, we say that a function u : R2 → R is a weak solution of (P ) if u ∈ Y ∩ L∞ loc (R ) and it holds the equality ∫ ∫ ∫ ∫ 2 2 (1 + 2u )∇u∇ϕ dx + 2u|∇u| ϕdx + V (|x|)uϕ dx = Q(|x|)h(u)ϕ dx, (1.5) R2

R2

R2

R2

for all ϕ ∈ C0∞ (R2 ). We observe from (h2 ) that u ≡ 0 is the trivial solution for (P ). Thus, our purpose is to show the existence of nonzero solution for (P ). Next, we state our first main result. Theorem 1.3.

Suppose that (V ), (Q) and (h1 ) − (h4 ) are satisfied and, in condition (h4 ), ⎧ ⎫ p ] p−2 ⎨ [ µλ0 (p − 2)∥Q∥ 1 2 2 ⎬ 2ξ L (B1 ) 1 , ξ ≥ max ξ1 , ⎩ pp−1 ⎭ 2p/2 π(1 + b0 /2)

where ξ1 :=

2(p−2)/2 p(4π + ∥V ∥L1 (B2 ) ) . ∥Q∥L1 (B1 )

Then, problem (P ) has a nonnegative and nonzero weak solution in Yrad .

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

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The main difficulties to prove Theorem 1.3 is the lack of compactness since we are working in all R2 , the features of the potentials V and Q, the presence of the quasilinear term ∆(u2 )u and the exponential critical growth of h(s). We cannot apply directly variational methods, because the associated energy functional to (P ), given by ∫ ∫ ∫ 1 1 2 J(u) = (1 + 2u2 )|∇u| dx + V (|x|)u2 dx − Q(|x|)H(u)dx 2 R2 2 R2 R2 ∫ 2 is not well defined in linear subspaces of H 1 (R2 ), because of the term R2 u2 |∇u| dx. We observe that weak solutions of (P ) can be seen as critical points of the functional J on an appropriate functions space, since the formal Gateaux derivative of J is given by ∫ ∫ ∫ 2 ′ 2 J (u).ϕ = (1 + 2u )∇u∇ϕ dx + 2 u|∇u| ϕ dx + V (|x|)uϕ dx 2 R∫ R2 R2 − Q(|x|)h(u)ϕ dx. R2

In order to overcoming these problems, as in [12,18] we make a change of variables to transform the quasilinear problem in a semilinear one, which has an associated functional I well defined and Gateauxdifferentiable on a weight Orlicz space E (see Section 2), thanks to a version of the Trudinger–Moser inequality proved for functions in E (see Theorem 4.4). Thus, using an argument of symmetric criticality principle type, we show that critical points of I are weak solutions of (P ). We guarantee that I satisfies the geometric conditions of a version of the Mountain Pass Theorem and due to a compactness result, we are able to show that I satisfies the Palais–Smale condition in a convenient interval. From this, we obtain a nonnegative and nonzero critical point for I and hence a nonnegative and nonzero weak solution for (P ). 2 ′ ∞ 2 We observe that u ∈ Y ∩ L∞ loc (R ) is a solution of (P ) if J (u).ϕ = 0 for every direction ϕ ∈ C0 (R ). As defined in [13], we say that a weak solution u of (P ) is a ground state if J(u) = inf{J(w) : w is a nontrivial weak solution of (P )}. As a consequence of Theorem 1.3, we have the following result: Corollary 1.4. Under the assumptions of Theorem 1.3, if we suppose in addition that h(s)/s3 is increasing for s > 0, then the solution obtained in Theorem 1.3 is a ground state. Remark 1.5. If a > 0 and b0 = b = 0, the potential Q(r) ≡ 1 satisfies condition (Q). Thus, our result assures the existence of a nonnegative and nonzero solution at the mountain pass level for the problem { − ∆u + V (|x|)u − [∆(u2 )]u = h(u), x ∈ R2 , u(x) → 0,

as |x| → ∞.

In this case, V can also be singular at the origin. We emphasize that we do not assume the standard condition ∃ M > 0, s0 > 0 such that H(s) ≤ M h(s), ∀ s ≥ s0 , which is often used in semilinear and quasilinear problems involving exponential critical growth. Furthermore, we can obtain a ground state solution. In this direction, our paper improves and complements some results in the literature, see for instance [4,23,27,30,35]. In this work, we also establish a nonexistence result of positive solution for problem (P ). For this, we consider the following hypotheses on the potentials V and Q:

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

6

(Vˆ ) V : (0, ∞) → R is continuous, V (r) ≥ 0 and there exists a < −2 such that lim sup r→+∞

V (r) < ∞. ra

ˆ Q : (0, ∞) → R is continuous, Q(r) > 0 and there exists b > −2 such that (Q) lim inf r→+∞

Q(r) > 0. rb

About the function h(s), we require that h : R → R is continuous and satisfies (ˆ h) There exist ξ > 0 and p > 1 such that h(s) ≥ ξsp ,

∀ s ≥ 0.

ˆ and (ˆ Theorem 1.6. Suppose that (Vˆ ), (Q) h) are satisfied. Then, problem (P ) does not have radial positive classical solution. We observe that in condition (Vˆ ) we require that the number a ∈ (−∞, −2), while in condition (V ) we ˆ we do not impose a behavior at the origin. In order to prove assume that a > −2. Moreover, in (Vˆ ) and (Q), this result, we adapt some arguments used in [6]. However, the computations in our case are more delicate due to the presence of the quasilinear term −∆(u2 )u. The outline of the paper is as follows: in the forthcoming section, we prove a compactness result which is essential in the sequel of the work. In Section 3, we introduce a weight Orlicz space E, its convergence properties and some embedding results involving this space and the weight Lebesgue spaces, as well as the reformulation of the problem including the appropriate variational setting to study the quasilinear equation. Section 4 is devoted to prove a version of the Trudinger–Moser inequality for the space E. In Section 5, we prove some properties of the energy functional associated to the dual problem and we relate critical points of this functional with weak solutions of (P ). Sections 6 and 7 are devoted to the proofs of the main results. Throughout this paper, we use ∥ · ∥p to denote the norm of the Lebesgue space Lp (R2 ), 1 ≤ p ≤ ∞ and the symbols C, Ci , i = 0, 1, 2, . . . will denote various constants. 2. Preliminaries In this section, we obtain some results which will be used in the sequel of the paper. We shall prove a compact embedding involving Yrad and the spaces Lq (R2 , Q). For this, we will need of a version of the Radial Lemma for functions in Yrad , whose proof can be found in [37, Lemma 4]. Lemma 2.1.

Suppose that a > −2 and lim inf V (r)/ra > 0. r→+∞

Then, there exist C > 0 and R > 1 such that, for all u ∈ Yrad , we have − 2+a 4

|u(x)| ≤ C|x|

∥u∥Y ,

for |x| ≥ R.

Proposition 2.2. Suppose that (V ) and (Q) hold. Then, Yrad is continuously and compactly immersed in Lq (R2 , Q) for each 2 ≤ q < ∞.

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

7

Proof . For q ≥ 2, consider the number ∫ Sq :=

inf

2

(|∇u| + V (|x|)u2 )dx (∫ )2 . q Q(|x|)|u| dx q R2

R2

u∈Yrad , u̸=0

To prove that the embedding Yrad ↪→ Lq (RN , Q) is continuous, it suffices to show that Sq is positive. Suppose, by contradiction, that Sq = 0. Then, there exists a sequence (un ) ⊂ Yrad \ {0} such that ∫ ∫ 2 q 2 (|∇un | + V (|x|)un )dx → 0 and Q(|x|)|un | dx = 1, ∀ n ∈ N. (2.1) R2

R2

From conditions (V ) and (Q), for R > 1 given by Lemma 2.1, there exist C1 , C2 > 0 satisfying Q(|x|) ≤ C1 |x|

b

a

∀ |x| ≥ R.

2

q−2

V (|x|) ≥ C2 |x| ,

and

Thus, since b < a and using Lemma 2.1 we get ∫ ∫ q b q Q(|x|)|un | dx ≤ C1 |x| |un | dx c BR

c BR

C1 C2 ∫ ≤C

∫

b−a

|x|

=

c BR

a

C2 |x| |un | |un | b−a

c BR

∥un ∥q−2 Y |x| a+2 4

≤ CRb−a−(q−2)

a+2 4

≤ CRb−a−(q−2)

dx

− a+2 4 (q−2)

|x|

∥un ∥Yq−2

2

V (|x|)|un | dx

∫

(2.2)

2

c BR

V (|x|)|un | dx

∥un ∥qY .

Moreover, in view of (Q) given 0 < r0 < 1/2 there exists C3 > 0 such that b

Q(|x|) ≤ C3 |x| 0 ,

∀ 0 < |x| < r0 .

(2.3)

∞ Now, consider a function φ ∈ C0,rad (B1 ) satisfying 0 ≤ φ(x) ≤ 1 for all x ∈ B1 , φ(x) = 1 if |x| < 1/2, φ(x) = 0 if 3/4 < |x| < 1 and |∇φ(x)| ≤ C for all x ∈ B1 for some C > 0. We have that φun ∈ H01 (B1 ) and we can choose σ > 1 so that b0 σ > −2. Thus, by inequality (2.3) and H¨older’s inequality, we obtain ∫ ∫ q b q Q(|x|)|un | dx ≤ C3 |x| 0 |φun | dx Br0

Br0

(∫

b0 σ

≤ C3

|x|

) σ1 (∫ dx

Br0

≤

Cr0b0 σ+2

qσ σ−1

|φun |

) σ−1 σ dx

Br0

(∫ |φun |

qσ σ−1

) σ−1 σ dx .

B1

Since min1/2≤r≤3/4 V (r) > 0, invoking the continuous embedding H01 (B1 ) ↪→ Ls (B1 ) for s ≥ 2 and the definition of φ, we reach (∫ ) 2q ∫ q 2 b0 σ+2 Q(|x|)|un | dx ≤ Cr0 |∇(φun )| dx Br0

B1

≤

Cr0b0 σ+2

(∫

2

2

|un ∇φ| dx + B1

B1

⎞ 2q

⎛ ∫ b0 σ+2 ⎝ ≤ C4 r0

2

∫

V (|x|)|un | dx +

B 3 \B 1 4

≤

) 2q |φ∇un | dx

∫

C4 r0b0 σ+2 ∥un ∥qY

2

.

B1

2 |∇un | dx⎠

(2.4)

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

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1 Next, by virtue of the continuous embedding Hrad (BR \Br0 ) ↪→ Lq (BR \Br0 ) and since minr0 ≤r≤R V (r) > 0 one has ∫ ∫ q q Q(|x|)|un | dx ≤ C5 |un | dx ≤ C∥un ∥qY . (2.5) BR \Br0

BR \Br0

Hence, from estimates (2.2), (2.4) and (2.5) it follows that ∫ a+2 q Q(|x|)|un | dx ≤ C(Rb−a−(q−2) 4 + r0b0 σ+2 + 1)∥un ∥qY → 0, R2

which is a contradiction with (2.1). Therefore, Sq > 0 and the first part is proved. To prove the compactness of the embedding, let (un ) be a bounded sequence in Yrad . Hence, there exists u ∈ Yrad and a subsequence of (un ), which will be denoted still by (un ), such that un ⇀ u in Yrad . Without loss of generality, we can suppose that u = 0. By (2.2) and (2.4), we obtain ∫ a+2 a+2 q Q(|x|)|un | dx ≤ CRb−a−(q−2) 4 ∥un ∥qY ≤ C1 Rb−a−(q−2) 4 and c BR

∫ Br0

q

Q(|x|)|un | dx ≤ Cr0b0 σ+2 ∥un ∥qY ≤ C2 r0b0 σ+2 .

Hence, given ε > 0, we can take R > 0 sufficiently large and r0 > 0 sufficiently ∫ ∫ ε q q Q(|x|)|un | dx ≤ and Q(|x|)|un | dx ≤ c 3 Br B 0

R

small such that ε . 3

(2.6)

1 Furthermore, since the embedding Hrad (BR \ Br0 ) ↪→ Lq (BR \ Br0 ) is compact, one has that there exists n0 ∈ N such that ∫ ∫ ε q q Q(|x|)|un | dx ≤ C5 |un | dx ≤ , ∀ n ≥ n0 . (2.7) 3 BR \Br BR \Br 0

0

Thus, from estimates (2.6) and (2.7), we have ∫ q Q(|x|)|un | dx ≤ ε,

∀ n ≥ n0

R2

and therefore un → 0 in Lq (R2 , Q). This finalizes the proof.

□

1,2 Remark 2.3. We observe that we are not working in a subspace of Drad (R2 ), as it was used by Su, Wang 1,2 and Willem in [36] when N = 2, because in this dimension we do not know embeddings involving Drad (R2 ) and we believe that the space Yrad is more appropriate. We also do not consider the space { } ∫ X := u ∈ H 1 (R2 ) : V (|x|)u2 dx < ∞ , R2

because we cannot prove that it is a Banach space. Remark 2.4. Let u be in Y . For each R > 0, let φ be in C0∞ (R2 ) satisfying supp(φ) ⊂ BR+1 and φ ≡ 1 in BR . Thus, by the Poincar´e inequality we reach ∫ ∫ ∫ 2 2 2 |u| dx ≤ |φu| dx ≤ C1 |∇(φu)| dx BR BR+1 BR+1 ∫ ∫ 2 2 2 2 ≤ C2 |∇φ| |u| dx + |φ| |∇u| dx BR+1 \BR

C2 ∥∇φ∥2∞ ≤ MR ≤ CR ∥u∥2 ,

∫ BR+1 \BR

BR+1 2

V (|x|)|u| dx + ∥φ∥2∞

∫

2

|∇u| dx BR+1

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

9

where MR = minx∈BR+1 \BR V (|x|) > 0. Therefore, u ∈ L2 (BR ) and, for each open ball BR ⊂ R2 , the space Y is continuously immersed in H 1 (BR ). We emphasize that the previous estimate is crucial to show that Y is complete. 3. The Orlicz space E As observed in the Introduction, we cannot apply directly variational methods to deal with problem (P ) ∫ 2 due the presence of the term R2 u2 |∇u| dx in the functional J. To overcome this difficulty, as in [12,28] we −1 make the change of variable v = f (u) where f ′ (t)

=

f (t)

=

1 (1 + 2f 2 (t))1/2 −f (−t)

in [0, +∞), in (−∞, 0].

Hence, we obtain the new functional I(v) := J(f (v)) =

1 2

∫

2

[|∇v| + V (|x|)f 2 (v)]dx −

∫

R2

Q(|x|)H(f (v))dx. R2

For an easy reference, we recall here some properties of f (t), see [12,15,28]. Lemma 3.1. (1) (2) (3) (4) (5) (6) (7) (8) (9)

The function f (t) satisfies the following properties:

f is uniquely defined, C ∞ and invertible; |f ′ (t)| ≤ 1 for all t ∈ R; |f (t)| ≤ |t| for all t ∈ R; f (t)/t → 1 as t → 0; √ f (t)/ t → 21/4 as t → +∞; f (t)/2 ≤ tf ′ (t) ≤ f (t) for all t ≥ 0; 1/2 |f (t)| ≤ 21/4 |t| for all t ∈ R; 2 the function f (t) is strictly convex; there exists a positive constant C such that { |f (t)| ≥

C|t|, 1/2 C|t| ,

|t| ≤ 1 |t| ≥ 1;

(10) there exist positive constants C1 and C2 such that |t| ≤ C1 |f (t)| + C2 |f (t)|

2

f or all t ∈ R;

√

(11) |f (t)f ′ (t)| ≤ 1/ 2 for all t ∈ R; (12) f 2 (ρt) ≤ ρ2 f 2 (t) for all 0 ≤ ρ ≤ 1 and t ∈ R; (13) f 2 (ρt) ≤ ρf 2 (t) for all ρ ≥ 1 and t ∈ R. Note that as a consequence immediate of (12) and (13) we have f 2 (ρt) ≤ max{ρ, ρ2 }f 2 (t) ≤ (ρ + ρ2 )f 2 (t), ∀ ρ ≥ 0 and t ∈ R.

(3.1)

Moreover, taking t = 1 in item (6) above, we get f 2 (1) ≥ [f ′ (1)]2 = 1/(1+2f 2 (1)), that is, 2f 4 (1)+f 2 (1) ≥ 1. √ Since f (1) > 0, a simple computation shows that f (1) ≥ 1/ 2 and it is immediate to verify by virtue of item (6) that f (t)/t is decreasing. Therefore, 1 f (t) ≥ √ t, 2

∀ t ∈ [0, 1].

(3.2)

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10

Now, we introduce the following defined by { ∫ E = v ∈ L2loc (R2 ) : v is radial, |∇v| ∈ L2 (R2 ) and

} V (|x|)f 2 (v)dx < ∞

R2

endowed with the norm given by [ ] ∫ 1 1+ V (|x|)f 2 (λv)dx . λ>0 λ R2

∥v∥ = ∥∇v∥2 + inf

(3.3)

We can show that (E, ∥ · ∥) is a Banach space (it is similar to Proposition 2.7 in [28]). The main result of this section consists in proving that the map v ↦→ f (v) from E into Lq (R2 , Q) is continuous and compact for all q ≥ 2. For this, we shall need of some auxiliary lemmas. Lemma 3.2. (1) There exists C > 0 such that ∫

V (|x|)f 2 (v) dx R2 ]1/2 ≤ C∥v∥, ∀ v ∈ E; [ ∫ 1 + R2 V (|x|)f 2 (v) dx (2) If vn → v in E then ∫ V (|x|)|f 2 (vn ) − f 2 (v)| dx → 0

∫

2

V (|x|)|f (vn ) − f (v)| dx → 0;

and

R2

R2

(3) If vn → v almost everywhere in R2 and ∫ ∫ V (|x|)f 2 (vn ) dx → R2

V (|x|)f 2 (v) dx,

R2

then [ ] ∫ 1 2 inf 1+ V (|x|)f (ξ(vn − v)) dx → 0. ξ>0 ξ R2 Proof . See Proposition 2.1 in [28]. □ Corollary 3.3. Suppose that conditions (V ) and (Q) holds. Then, the embedding Yrad ↪→ E is continuous. Proof . By item (3) of Lemma 3.1, one has Yrad ⊂ E. Let (vn ) ⊂ Yrad be such that vn → 0 in Yrad . Then, ∫ ∫ 2 |∇vn | dx → 0 and V (|x|)vn2 dx → 0. R2

R2

As V (r) > 0 for all r > 0, we have vn → 0 almost everywhere in R2 and by virtue of (3) in Lemma 3.1, it holds ∫ V (|x|)f 2 (vn )dx → 0. R2

Hence, invoking item (3) of Lemma 3.2, we conclude that vn → 0 in E. □ Proposition 3.4. If conditions (V ) and (Q) hold, then the map v ↦→ f (v) from E into Lq (R2 , Q) is continuous and compact for all q ∈ [2, ∞).

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11

Proof . By item (3) of Lemma 3.1 and Proposition 2.2, the application v ↦→ f (v) is well defined. Let (vn ) ⊂ E be a sequence such that vn → v in E. In particular, ∂vn ∂v → ∂xi ∂xi

in

L2 (R2 )

for i = 1, 2. Hence, there exists hi ∈ L2 (R2 ) such that, up to a subsequence, |∂vn /∂xi | ≤ hi almost everywhere in R2 . Moreover, ∂f (vn ) ∂f (v) → a.e. in R2 . ∂xi ∂xi Invoking the Lebesgue Dominated Convergence Theorem, it follows that ∂f (v) ∂f (vn ) → ∂xi ∂xi

L2 (R2 )

in

(3.4)

for i = 1, 2. On the other hand, by using (2) of Lemma 3.2, one has ∫

2

V (|x|)|f (vn ) − f (v)| dx → 0.

(3.5)

R2

Thus, convergences (3.4) and (3.5) imply that f (vn ) → f (v) in Yrad and invoking Proposition 2.2 we conclude that f (vn ) → f (v) in Lq (R2 , Q) for each 2 ≤ q < ∞ and this proves the continuity of the application v ↦→ f (v). To prove the compactness, let (vn ) be a bounded sequence in E. From (1) of Lemma 3.2, the sequence ∫ ( R2 V (|x|)f 2 (vn )dx) is bounded. Moreover, by item (2) of Lemma 3.1 we have ∫

∫

2

|∇f (vn )| dx ≤ R2

2

2

|f ′ (vn )| |∇vn | dx ≤

∫

R2

2

|∇vn | dx ≤ C. R2

Thus, (f (vn )) is bounded in Yrad and therefore there exists u ∈ Yrad such that, up to a subsequence, f (vn ) ⇀ u in Yrad . According to Proposition 2.2, f (vn ) → u in Lq (R2 , Q) for each 2 ≤ q < ∞ and the compactness is proved. □ As a consequence of Proposition 3.4, we have the following result: Corollary 3.5. If (V ) and (Q) hold then the embedding E ↪→ Lq (R2 , Q) is continuous and compact for each 2 ≤ q < ∞. Proof . By using (10) of Lemma 3.1 and Proposition 3.4, it follows that E ⊂ Lq (R2 , Q) for each q ∈ [2, ∞). Now, consider (vn ) ⊂ E such that vn → 0 in E. Again by (10) of Lemma 3.1 and Proposition 3.4, we obtain ∫

q

∫

Q(|x|)|vn | dx ≤ C R2

q

∫

R2

2q

Q(|x|)|f (vn )| dx → 0,

Q(|x|)|f (vn )| dx + C R2

which shows the continuity of the embedding. Next, let (vn ) be a bounded sequence in E. From Proposition 3.4 there exists u ∈ Lq (R2 , Q) satisfying un = f (vn ) → u in Lq (R2 , Q). In particular, un → u almost everywhere in R2 and there exists hq ∈ Lq (R2 , Q) such that |un | ≤ hq almost everywhere in R2 . Item (10) of Lemma 3.1 implies that |vn | ≤ C(hq + h22q ) almost everywhere in R2 and by using Lebesgue’s Dominated Convergence Theorem, we conclude vn → v = f −1 (u) in Lq (R2 , Q) and the compactness is proved. □

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12

4. A Trudinger–Moser inequality In this section, we are going to prove a version of the Trudinger–Moser inequality involving the space E. This was necessary because the Trudinger–Moser inequalities, obtained in [4,10,30], are not applicable for this class of problems due to the Orlicz setting and the fact that functions in E do not necessarily belong to L2 (R2 ). For this, we need some preliminary results. Lemma 4.1.

Suppose that (V ) is satisfied. Then there exist C > 0 and R > 0 such that − a+2 4

|f (v(x))| ≤ CΥ (v)|x| where (∫ Υ (v) :=

∀ |x| ≥ R and v ∈ E,

,

) 21 . [|∇v| + V (|x|)f (v)]dx 2

2

R2

Proof . If v ∈ E then f (v) ∈ Yrad . Thus, from Lemma 2.1 there exist C > 0 and R > 0 such that |f (v(x))| ≤ C∥f (v)∥Y |x|

− a+2 4

,

for |x| ≥ R.

By item (2) of Lemma 3.1, it follows that ∫ 2 2 ∥f (v)∥2Y = [|f ′ (v)| |∇v| + V (|x|)f 2 (v)]dx ≤ Υ (v)2 R2

and the lemma is proved. Corollary 4.2.

□

If D > 0 is a constant then there exist C > 0 and R > 0 such that − a+2 4

|v(x)| ≤ CD|x|

, for |x| ≥ R

and for all v ∈ E satisfying Υ (v) ≤ D. Proof . From Lemma 4.1, there exist C > 0 and R > 0 such that |f (v(x))| ≤ CDR−

a+2 4

∀ |x| ≥ R

(4.1)

and for all v ∈ E satisfying Υ (v) ≤ D. By item (4) of Lemma 3.1, f −1 (t)/t → 1 as t → 0. Thus, there exists r0 > 0 such that |f −1 (t)| ≤ 2|t| for all |t| ≤ r0 . Moreover, there exists r1 > 0 such that |t| ≤ 2|f (t)| for all |t| ≤ r1 . Without loss of generality, we can suppose R > 0 large enough so that 2CDR−(a+2)/4 < min{r0 , r1 }. Hence, if v ∈ E satisfies Υ (v) ≤ D then by (4.1) one has − a+2 4

|v(x)| = |f −1 (f (v(x)))| ≤ 2|f (v(x))| ≤ 2CDR0

< r1 ,

∀ |x| ≥ R

and by using again Lemma 4.1, we obtain − a+2 4

|v(x)| ≤ 2|f (v(x))| ≤ CD|x| which concludes the proof.

∀ |x| ≥ R,

□

In order to prove a version of the Trudinger–Moser inequality, which will be usual for our problem, we shall use the following singular Trudinger–Moser inequality due to Adimurthi–Sandeep (see [3]):

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13

Lemma 4.3. Let Ω be a bounded domain in R2 and u ∈ H01 (Ω ). Then, for all γ > 0 and d ∈ [0, 2) we have 2 −d |x| eγ|u| ∈ L1 (Ω ). Moreover, ∫ sup ∥∇u∥2 ≤1

Proof . See Theorem 2.1 in [3].

Ω

2

eγ|u|

γ d + ≤ 1. 4π 2

dx < ∞ ⇐⇒

d

|x|

□

Now, we are ready to prove the main result of this section. 2

Theorem 4.4. Suppose that (V ) and (Q) are satisfied. Then Q(|x|)[eλv − 1] ∈ L1 (R2 ) for all v ∈ E and λ > 0. Furthermore, if λ < 4π(1 + b0 /2) and D > 0 then there exists C = C(a, b, b0 , λ, D) > 0 such that ∫ 2 Q(|x|)(eλv − 1)dx ≤ C, sup R2

v∈E; ∥∇v∥2 ≤1

whenever Υ (v) ≤ D. Proof . Let R1 > 1 be such that the conclusion of Corollary 4.2 is valid for |x| = R1 . By condition (Q), there exists C1 > 0 such that Q(r) ≤ C1 rb0 , ∀ 0 < r < R1

Q(r) ≤ C1 rb , ∀ r > R1 .

and

(4.2)

Fixing R > R1 , we have ∫

λv 2

Q(|x|)(e

∫ − 1)dx =

Q(|x|) c BR

c BR

∫ Q(|x|)

= c BR

≤ C1

∞ ∑ λj v 2j

j!

j=1 ∞ ∑

λj v 2j dx + j!

j=2

∫ ∞ ∑ λj j=2

j!

dx ∫

Q(|x|)λv 2 dx

c BR

b

|x| v 2j dx + λ

c BR

∫

(4.3)

Q(|x|)v 2 dx.

R2

From Corollary 3.5 and by virtue of the definition of norm in E, there exists C2 > 0 satisfying ∥v∥22,Q ≤ C2 ∥v∥2 ≤ C2 [1 + Υ (v) + Υ (v)2 ]2 , On the other hand, by Corollary 4.2 ∫ ∫ b |x| v 2j dx ≤ (CD)2j c BR

b

∀ v ∈ E.

−2j a+2 4

|x| |x|

(4.4)

dx

c BR

and since a > −2 and b − j(a + 2)/2 + 2 < b − a < 0 for all j ≥ 2, we obtain ∫ ∫ ∞ j(a+2) b |x| v 2j dx ≤ (CD)2j 2π sb− 2 +1 ds c BR

R

2π(CD) Rb− 2j

=−

j(a+2) +2 2

b − j(a+2) +2 2 2j b−a 2π(CD) R ≤ . a−b

(4.5)

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14

From (4.3), (4.4) and (4.5), it follows that ∫ ∞ ∑ 2 λj 2π(CD)2j + C2 [1 + Υ (v) + Υ (v)2 ]2 Q(|x|)(eλv − 1)dx ≤ C1 a−b c j! (a − b)R B j=1

(4.6)

R

≤ C3 e

λ(CD)2

2 2

+ λC2 (1 + D + D ) ,

1 where C3 = C3 (a, b, R). Since v ∈ Hrad (Bρ ) for each ρ > 0 (see Remark 2.4), without loss of generality we 2 can suppose that v is continuous in R \ {0} (cf. Corollary 1.3 - page 251 in [7]) and we denote v(R) = v(x) 1 for |x| = R. Defining u : BR → R by u(x) = v(x) − v(R), we have u ∈ H0,rad (BR ) and ∥∇u∥2 = ∥∇v∥2 . Now, we take ε > 0 so that λ(1 + ε) < 4π(1 + b0 /2). By using Young’s inequality and Corollary 4.2, we obtain a+2 v 2 (x) ≤ (1 + ε)u2 (x) + (1 + Cε )v(R)2 ≤ (1 + ε)u2 (x) + (1 + 1/ε)(CD)2 R− 2 (4.7)

for all x ∈ BR . Moreover, we can choose R so that 2

eλ(1+1/ε)(CD)

R

− a+2 2

2

≤ e4π(1+1/ε)(CD)

R

− a+2 2

≤ 1.

Hence, by using (4.2), Lemma 4.3 and (4.7) we reach ∫ ∫ ∫ 2 2 2 eλv − 1 eλ(1+ε)u − 1 Q(|x|)(eλv − 1)dx ≤ C1 dx ≤ C dx < ∞. 1 −b0 −b |x| 0 BR BR |x| BR

(4.8)

2

From the estimates (4.6) and (4.8) we conclude that Q(|x|)(eλv − 1) ∈ L1 (R2 ). Moreover, again by (4.6), (4.8) and Lemma 4.3, we conclude that there exists C = C(a, b, b0 , λ, D) > 0 such that ∫ 2 Q(|x|)(eλv − 1)dx ≤ C, R2

for all v ∈ E with ∥∇v∥2 ≤ 1 and Υ (v) ≤ D. □ 5. Critical points of I and weak solutions of (P ) We start this section presenting some properties of the functional I, namely: Proposition 5.1.

The functional I has the following properties:

(1) I : E → R is well defined and I is continuous; (2) I : E → R is Gateaux differentiable and its Gateaux derivative is given by ∫ ∫ I ′ (u).ϕ = (∇u∇ϕ + V (|x|)f (u)f ′ (u)ϕ) dx − Q(|x|)h(f (u))f ′ (u)ϕ dx, RN

u, ϕ ∈ E;

RN

(3) I ′ (u) ∈ E ′ for all u ∈ E and if vn → v in E then I ′ (vn ) → I ′ (v) in the weak star topology of E ′ . Proof . It is similar to the proof of Proposition 3.4 in [29] and we omit it.

□

Next, we relate critical points of I with weak solutions of (P ). First, we obtain some technical results. Proposition 5.2. Suppose that (V ), (Q) and (h1 ) − (h2 ) are satisfied. Then, for each v ∈ E, there exists C = C(b0 , ∥v∥) > 0 such that ⏐∫ ⏐ ⏐ ⏐ ′ ⏐ Q(|x|)h(f (v))f (v)wdx⏐⏐ ≤ C∥w∥Y , ∀ w ∈ Y. ⏐ R2

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15

Proof . By (h1 ) and (h2 ), for λ > λ0 and p ≥ 0 we obtain p

|h(s)| ≤ C1 |s| + C2 |s| (eλf

4

(s)

− 1),

which together items (2) and (11) of Lemma 3.1 implies ⏐∫ ⏐ ⏐ ⏐ ′ ⏐ ⏐ ⏐ 2 Q(|x|)h(f (v))f (v)wdx⏐ ≤ R ∫ ∫ C1 Q(|x|)|f (v)||w|dx + C2 R2

∀ s ∈ R,

(5.1)

(5.2) Q(|x|)(e

λf 4 (v)

− 1)|w|dx,

R2

for v ∈ E and w ∈ Y . Considering R > 1 and according to conditions (V ), (Q) and H¨older’s inequality, we reach ∫ ∫ b Q(|x|)|f (v)||w|dx ≤ C |x| |f (v)||w|dx c BR

c BR

∫

a

|x| |f (v)||w|dx

≤C c BR

∫ ≤C

(5.3)

V (|x|)|f (v)||w|dx c BR

(∫

) 21 (∫

2

≤C

V (|x|)f (v)dx

) 12 V (|x|)w dx 2

R2

R2

≤ C∥w∥Y , where C = C(∥v∥). On the other hand, H¨ older’s inequality and Proposition 3.4 show that (∫ ) 12 (∫ ) 12 ∫ 2 2 Q(|x|)|f (v)||w|dx ≤ Q(|x|)f (v)dx Q(|x|)w dx BR

BR

BR

(5.4)

) 12 2 , Q(|x|)w dx

(∫ ≤ C3 BR

b

where C3 = C3 (∥v∥). By (Q), there exist C, r0 > 0 such that Q(|x|) ≤ C|x| 0 for all x ∈ R2 with |x| < r0 . ∞ Now, we consider 1 < q < −2/b0 (1 < q < ∞ if b0 = 0) and φ ∈ C0,rad (R2 ) satisfying 0 ≤ φ(x) ≤ 1 for 2 c x ∈ R , φ(x) = 1 for x ∈ BR , φ(x) = 0 for x ∈ B2R and |∇φ(x)| ≤ C for x ∈ R2 , for some C > 0. Note that φw ∈ H01 (B2R ). Thus, by the H¨ older inequality and the continuous embedding of H01 (B2R ) into Lt (B2R ) for t ≥ 2, we have ∫ ∫ b

Q(|x|)w2 dx ≤ C

|x| 0 w2 dx

BR

BR

(∫ ≤C

|x|

b0 q

) 1q (∫ dx

w

BR

BR

(∫ ≤ C0

|φw|

2q q−1

) q−1 q

2q q−1

) q−1 q dx (5.5)

dx

B2R

∫ ≤ C0

2

|∇(φw)| dx, B2R

where C0 = C(b0 , R). If mR := minr∈B2R \BR V (r) then by (5.5) and the continuous embedding of H 1 (B2R \BR ) into L2 (B2R \BR ) we get ∫ ∫ ∫ 2 2 Q(|x|)w2 dx ≤ C0 |w∇φ| dx + C0 |∇w| dx BR B2R \BR B2R ∫ ∫ C0 2 2 (5.6) V (|x|)|w| dx + C0 |∇w| dx ≤ mR B2R \BR B2R ≤ C1 ∥w∥2Y ,

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16

where C1 = C1 (b0 , R). Therefore, by (5.3), (5.4) and (5.6) we conclude ∫ Q(|x|)|f (v)||w|dx ≤ C∥w∥Y ,

(5.7)

R2

where C = C(∥v∥, b0 , R). Next, according to the H¨older inequality, item (7) of Lemma 3.1, (V ), (Q) and Theorem 4.4 we have (∫ ) 12 (∫ ) 21 ∫ 2 λf 4 (v) 2λf 4 (v) Q(|x|)(e Q(|x|)|w| dx Q(|x|)(e − 1)|w|dx ≤ − 1)dx c BR

c BR

c BR

(∫ ≤

Q(|x|)(e

4λv 2

) 21

) 21 ( ∫ − 1)dx C

b

2

|x| |w| dx

c BR

R2

) 12

(∫

a

≤C

(5.8)

2

|x| |w| dx c BR

) 12

(∫

2

≤C

V (|x|)|w| dx c BR

≤ C∥w∥Y , where C = C(∥v∥). On the other hand, Theorem 4.4 and estimate (5.6) imply (∫ ) 21 (∫ ∫ 4 4 Q(|x|)(eλf (v) − 1)|w|dx ≤ Q(|x|)(e2λf (v) − 1)dx BR

) 12 Q(|x|)w2 dx

BR

BR

(∫

Q(|x|)w2 dx

≤C

) 12

(5.9)

BR

≤ C∥w∥Y . From (5.8) and (5.9) we obtain ∫

Q(|x|)(eλf

4

(v)

− 1)|w|dx ≤ C∥w∥Y ,

(5.10)

R2

where C = C(∥v∥, b0 ). Thus, from (5.2), (5.7) and (5.10), the proof of the proposition follows.

□

If v ∈ E then the linear functional Tv : Y → R defined by ∫ ∫ ∫ ′ Tv .w = ∇v∇wdx + V (|x|)f (v)f (v)wdx − Q(|x|)h(f (v))f ′ (v)wdx

Corollary 5.3.

R2

R2

R2

is continuous. Proof . It suffices to show that there exists C > 0 such that |Tv .w| ≤ C∥w∥Y , for all w ∈ Y . By the H¨older inequality, ⏐∫ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ 2 ∇v∇wdx⏐ ≤ ∥∇v∥2 ∥w∥Y and R ⏐∫ ⏐ (∫ ) 21 ⏐ ⏐ ′ 2 ⏐ ⏐≤ ∥w∥Y . V (|x|)f (v)f (v)wdx V (|x|)f (v)dx ⏐ ⏐ R2

R2

From these estimates together with Proposition 5.2, the proof follows.

□

The main result of this section is as follows: Proposition 5.4. Suppose that (V ), (Q) and (h1 )−(h2 ) hold. If v ∈ E is a critical point of I, then u = f (v) is a weak solution of (P ).

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17

Proof . Let v ∈ E be a critical point of I. First, we are going to show that v ∈ L∞ (R2 ). It suffices to prove 2 that v ∈ L∞ loc (R ) because v(x) → 0 as |x| → ∞. If R > 0 then v satisfies −∆v = f ′ (v)[Q(|x|)h(f (v)) − V (|x|)f (v)] =: w

in BR

in the weak sense. Using (5.1) there exists C1 > 0 such that [ ] 4 |w(x)| ≤ C1 |f ′ (v(x))f (v(x))| Q(|x|) + Q(|x|)(eλf (v(x)) − 1) + V (|x|) , ∀ x ∈ BR and from this inequality, (V ), (Q) and item (12) of Lemma 3.1, we obtain [ ] 4 b a |w(x)| ≤ C2 |x| 0 + Q(|x|)(eλf (v(x)) − 1) + |x| 0 .

(5.11)

Now, we analyze four cases: Case 1: a0 ≥ 0 and b0 ≥ 0. In this case, for each t > 1 we have [ ] 2 t |w(x)| ≤ C3 Rtb0 + Q(|x|)t−1 Q(|x|)(e2tλv(x) − 1) + Rta0 [ ] 2 ≤ C4 1 + Q(|x|)(e2tλv(x) − 1) . Hence, by Theorem 4.4 we conclude that w ∈ Lt (BR ) for all t > 1. Case 2: −2 < a0 < 0 and −2 < b0 < 0. Using condition (Q) and estimate (5.11), there exists C3 > 0 such that [ ] 4 b b a |w(x)| ≤ C3 |x| 0 + |x| 0 (eλf (v(x)) − 1) + |x| 0 .

(5.12)

Let t0 > 1 be such that −2 < a0 t0 < 0 and −2 < b0 t0 < 0. Thus, by (5.12) and Lemma 4.3, we have ∫ ∫ [ ] 2 t b t b t a t |w(x)| 0 dx ≤ C4 |x| 0 0 + |x| 0 0 (e2tλv(x) − 1) + |x| 0 0 dx < ∞, BR

BR

that is, w ∈ Lt0 (BR ). The cases (a0 , b0 ) ∈ [0, +∞) × (−2, 0) and (a0 , b0 ) ∈ (−2, 0) × [0, +∞) are treated in a similar way. Therefore, in any case, we assure the existence of t > 1 such that w ∈ Lt (BR ). By a regularity result due 0,γ to H. J. Choe (see Section 2 of [11]) we conclude that v ∈ Cloc (R2 ) for some γ ∈ (0, 1) and, in particular, ∞ 2 v ∈ Lloc (R ). 2 2 2 2 For the second part, we observe that |∇u| = |f ′ (v)| |∇v| ≤ |∇v| . Consequently, u ∈ Yrad (R2 )∩L∞ (R2 ). From the notation of Corollary 5.3, we have Tv .w = 0 for all w ∈ Yrad ⊂ E. Since Tv is linear and continuous on Y , by the Riesz Representation Theorem there exists a unique vˆ ∈ Y such that Tv .ˆ v = ∥ˆ v ∥2Y = ∥Tv ∥2Y ′ . Thus, by using change of variables, one has for each w ∈ Y Tv .(gw) = Tv .w

and

∥gw∥Y = ∥w∥Y

∀ g ∈ O(2).

Taking w = vˆ, by uniqueness it follows that gˆ v = vˆ for all g ∈ O(2), that is, vˆ ∈ Yrad . Hence, Tv .ˆ v = 0 and therefore ∥Tv ∥Y ′ = 0 which is equivalent to ∫ ∫ ∫ ∇v∇wdx + V (|x|)f (v)f ′ (v)wdx = Q(|x|)h(f (v))f ′ (v)wdx (5.13) R2

R2

R2

for each w ∈ Y . Since (f −1 )′ (t) = 1/f ′ (f −1 (t)), it follows that (f −1 )′ (t) = [1 + 2(f (f −1 (t)))2 ]1/2 = (1 + 2t2 )1/2

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18

which implies ∇v = (f −1 )′ (u)∇u = (1 + 2u2 )1/2 ∇u.

(5.14)

For all φ ∈ C0∞ (R2 ), one has f ′ (v)−1 φ = (1 + 2u2 )1/2 φ ∈ Y and ∇(f ′ (v)−1 φ) = 2(1 + 2u2 )−1/2 uφ∇u + (1 + 2u2 )1/2 ∇φ.

(5.15)

Taking w = f ′ (v)−1 φ in (5.13) and using (5.14)–(5.15), we obtain (1.5) which shows that u = f (v) is a weak solution of (P ) and this concludes the proof. □ Remark 5.5. Using the same arguments as in [12,28], we have that if v ∈ C 2 (R2 ) ∩ E is a critical point of the functional I then u = f (v) is a classical solution of (P ). Moreover, supposing V , Q and h are locally 2,γ H¨ older continuous functions, by using Schauder’s regularity we can conclude that v ∈ Cloc (R2 ) for some γ ∈ (0, 1). According to Proposition 5.4, in order to find a nonzero solution for (P ), it suffices to show that I has a nonzero critical point. 6. Proof of Theorem 1.3 and Corollary 1.4 To prove our existence result, we shall use the following version of the Mountain-Pass Theorem (see for example [5]): Theorem 6.1. Let X be a Banach space and Φ ∈ C(X, R) Gateaux differentiable in X with Gateaux derivative Φ ′ (v) ∈ X ′ continuous from the norm topology of X to the weak ∗ topology of X ′ and Φ(0) = 0. Let S be a closed subset of X which disconnects (archwise) X. Let v0 = 0 and v1 ∈ X be points belonging to distinct connected components of X \ S. Suppose that inf Φ ≥ ρ > 0 S

and

Φ(v1 ) ≤ 0

(6.1)

and let Γ = {γ ∈ C([0, 1]; X); γ(0) = 0

and

γ(1) = v1 } .

Then c = inf max Φ(γ(t)) ≥ ρ γ∈Γ t∈[0,1]

and there exists a (P S)c 1 sequence for Φ. The number c is called the Mountain-Pass Level of Φ. Since h(0) = 0 and as we are interested in nonnegative solutions, we set h(s) = 0 for all s < 0. At first, we show that I satisfies the geometric conditions (6.1). Moreover, we are going to verify that I satisfies the Palais–Smale condition in certain levels of energy. For each ρ > 0, let us define the set Sρ = {v ∈ E : Υ (v) = ρ}. We have that Sρ ⊂ E is closed and disconnects E in two components. Thus, we are ready to prove (6.1). Proposition 6.2.

The functional I satisfies the following conditions:

(i) there exist ρ, σ0 > 0 with ρ < 1 such that I(v) ≥ σ0 for all v ∈ Sρ ; (ii) there exists v ∈ E satisfying Υ (v) > ρ and I(v) < 0. 1

We recall that (un ) in E is a (P S)c sequence for Φ if Φ (un ) → c and Φ ′ (un ) → 0.

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

19

Proof . (i) According to (1.3), H(s)/s2 → 0 as s → 0+ . From this and (h1 ), given ε > 0 and λ > λ0 , there exists C > 0 such that 4 ε 2 3 H(s) ≤ |s| + C|s| (eλs − 1), ∀ s ∈ R. 2 Considering ρ > 0 and v ∈ Sρ , one has ρ2 ε − 2 2

I(v) ≥

∫

∫

2

3

Q(|x|)|f (v)| (eλf

Q(|x|)|f (v)| dx − C RN

4

(v)

− 1) dx.

(6.2)

R2

Choosing r1 , r2 > 1 so that 1/r1 + 1/r2 = 1 and by using H¨older’s inequality, we obtain ∫ 4 3 Q(|x|)|f (v)| (eλf (v) − 1)dx R2

(∫

3r1

≤

Q(|x|)|f (v)|

) r1 (∫ 1 dx

λr2 f 4 (v)

Q(|x|)(e

) r1 2 − 1)dx .

R2

R2

Thus, by item (7) of Lemma 3.1 and since f (v) ∈ Yrad whenever v ∈ E, by Proposition 2.2, we reach ∫

3

Q(|x|)|f (v)| (e

λf 4 (v)

− 1)dx ≤

R2

C∥f (v)∥3Y 3

(∫ Q(|x|)(e

) r1 2 − 1)dx

R2

(∫

2λr2 v 2

≤ Cρ

2λr2 v 2

Q(|x|)(e

) r1 2 − 1)dx .

(6.3)

R2

Now, we observe that ∫

2

Q(|x|)(e2λr2 v − 1)dx =

R2

2λr2 Υ (v)2

∫

Q(|x|)(e ∫R2

− 1)dx (6.4)

Q(|x|)(e

=

v2 Υ (v)2

2 2λr2 ρ2 ( v ρ)

− 1)dx

R2

and take 0 < ρ < 1 small enough so that 2λr2 ρ2 < 4π(1 + b0 /2). As 1/ρ > 1 by item (13) of Lemma 3.1 we have that if v ∈ Sρ and vˆ := v/ρ then Υ (ˆ v ) ≤ 1 and consequently ∥∇ˆ v ∥2 ≤ 1. Thus, Theorem 4.4 in combination with estimates (6.3) and (6.4) implies that there exists C > 0 such that ∫ 4 3 Q(|x|)|f (v)| (eλf (v) − 1)dx ≤ Cρ3 . R2

From this inequality and (6.2), we have I(v) ≥

) 1 C1 ε − ρ2 − Cρ3 2 2 ( ) 1 C1 ε = ρ2 − − Cρ =: σ0 . 2 2

1 2 ε ρ − ∥f (v)∥22,Q − C∥f (v)∥3r1 ,Q ≥ 2 2

(

Thus, if ε < 1/C1 and ρ < 2(1 − C1 ε)/C then I(v) ≥ σ0 > 0 for all v ∈ E with Υ (v) = ρ. (ii) By condition (h3 ), there exist constants C1 , C2 > 0 such that H(s) ≥ C1 s2µ − C2 for all s ≥ 0. Let ∞ φ ∈ C0,rad (R2 ) be such that 0 ≤ φ ≤ 1 and 0 ̸∈ supp(φ) =: K. Hence, we have ∫ ∫ ∫ t2 1 2 2 I(tφ) = |∇φ| dx + V (|x|)f (tφ)dx − Q(|x|)H(f (tφ))dx 2 K 2 K K ∫ ∫ ∫ t2 2 ≤ (|∇φ| + V (|x|)φ2 )dx − C1 Q(|x|)f 2µ (tφ)dx + C2 Q(|x|)dx. 2 K K K

(6.5)

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20

Since f (s)/s is decreasing in (0, +∞) and 0 ≤ tφ ≤ t for t > 0, it follows that f (tφ) ≥ f (t)φ and returning to (6.5), we get ∫ ∫ t2 2 I(tφ) ≤ (|∇φ| + V (|x|)φ2 )dx − C1 Q(|x|)f 2µ (t)φ2µ dx + C 2 K K ] [∫ f 2µ (t) t2 2 2 (|∇φ| + V (|x|)φ )dx − C2 2 + C → −∞, as t → +∞, ≤ 2 t K where we have used item (9) of Lemma 3.1. Therefore, for t sufficiently large and taking v = tφ we conclude that I(v) < 0 and the proof is finished. □ Lemma 6.3.

If (vn ) ⊂ E is a Palais–Smale sequence for I at any level c ∈ R, then (vn ) is bounded in E.

Proof . Let c ∈ R and (vn ) ⊂ E be such that I(vn ) → c and I ′ (vn ) → 0. Thus, |I ′ (vn )vn | ≤ ∥I ′ (vn )∥∥vn ∥ = on (1)∥vn ∥. By item (6) of Lemma 3.1 and using (h3 ), we obtain ( )∫ ( )∫ 1 ′ 1 1 1 1 2 I(vn ) − I (vn )vn ≥ − |∇vn | dx + − V (|x|)f 2 (vn )dx µ 2 µ 2 µ 2 2 R R ∫ 1 + Q(|x|)[h(f (vn ))f (vn ) − 2µH(f (vn ))]dx 2µ R2 )∫ ( )∫ ( 1 1 1 1 2 − |∇vn | dx + − V (|x|)f 2 (vn )dx. ≥ 2 µ 2 µ R2 R2

(6.6)

(6.7)

Hence, from (6.6) and (6.7) we reach ( )∫ ( )∫ 1 1 1 1 2 − |∇vn | dx + − V (|x|)f 2 (vn )dx ≤ c + on (1) + on (1)∥vn ∥ 2 µ 2 µ R2 R2 (∫ ) 21 ∫ 2 ≤ c + on (1) + on (1) |∇vn | dx + on (1) V (|x|)f 2 (vn )dx, R2

R2

which implies )∫ 1 1 − − on (1) V (|x|)f 2 (vn )dx ≤ |∇vn | dx + 2 µ R2 R2 (∫ ) 12 2 . c + on (1) + on (1) |∇vn | dx R2 (∫ ) 2 Thus, the sequence R2 |∇vn | dx is bounded. Again by using (6.8), it follows that (

1 1 − 2 µ

)∫

2

(

(6.8)

)∫ (∫ ) 21 1 1 2 2 − − on (1) V (|x|)f (vn )dx ≤ C + |∇vn | dx 2 µ R2 R2 (∫ ) and consequently R2 V (|x|)f 2 (vn )dx is also bounded. Now, since (

(∫ ∥vn ∥ ≤

2

) 21

|∇vn | dx R2

∫ +1+

V (|x|)f 2 (vn )dx,

R2

we conclude that (vn ) is bounded in E. □ Corollary 6.4.

If (vn ) ⊂ E is a Palais–Smale sequence for I at level c, then Υ (vn )2 ≤

2µ c + on (1). µ−2

(6.9)

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

21

Proof . Since (vn ) is bounded in E, from (6.8) we obtain (

1 1 − 2 µ

)∫

2

(

|∇vn | dx + R2

1 1 − 2 µ

)∫

V (|x|)f 2 (vn )dx ≤ c + on (1)

R2

and therefore Υ (vn )2 =

∫

2

∫

|∇vn | dx + R2

V (|x|)f 2 (vn )dx ≤

R2

2µ c + on (1). □ µ−2

Estimate (6.9) will play an important role in proving that I satisfies the Palais–Smale condition in certain energy levels. Proposition 6.5. If c < π(1 + b0 /2)(µ − 2)/(λ0 µ) then the functional I satisfies the Palais–Smale condition at level c. Proof . Let c < π(1 + b0 /2)(µ − 2)/(λ0 µ) and (vn ) ⊂ E be a Palais–Smale sequence for I at level c. By ) (∫ ) (∫ 2 Lemma 6.3, (vn ) is bounded in E and, in particular, the sequences R2 |∇vn | dx and R2 V (|x|)f 2 (vn )dx are bounded. For n ∈ N, consider un := f (vn ). By item (2) of Lemma 3.1, we have |∇un | = |f ′ (vn )∇vn | ≤ |∇vn |, which implies that (un ) is also bounded in Yrad . Thus, there exists u ∈ Yrad such that un ⇀ u in Yrad . By Proposition 2.2, un → u in Lq (R2 , Q) for all q ≥ 2 and since Q > 0 in R2 , up to a subsequence, it holds un → u almost everywhere in R2 . On the other hand, by using Corollary 3.5, for each q ≥ 2 there exists v ∈ Lq (R2 , Q) such that vn → v in Lq (R2 , Q) and vn → v almost everywhere in R2 . The continuity of f −1 implies that f −1 (u) = v. Observe that v ∈ E. Since (Υ (vn )) is bounded, up to a subsequence, there exists τ ≥ 0 satisfying Υ (vn ) → τ . At this point, we are going to consider two possibilities: (1) τ = 0; In this case, we have ∫ ∫ 2 0 ≤ ∥un ∥2Y ≤ |f ′ (vn )∇vn | dx + V (|x|)f 2 (vn )dx ≤ Υ (vn )2 → 0. R2

R2

Thus, un → 0 in Yrad and therefore v = f −1 (u) = 0. By item (3) of Lemma 3.2, we obtain [ ] ∫ 1 1+ V (|x|)f 2 (ξ(vn )) dx → 0, ξ>0 ξ R2 inf

which shows that vn → 0 in E and the Palais–Smale condition at c is proved. (2) τ > 0; For this case, we define vˆn = vn /Υ (vn ). There exists n0 ∈ N such that if n ≥ n0 then Υ (vn ) ≤ τ + 1. By using (3.1), one has ( ) ( ) vn 1 1 ≤ + 2 f 2 (vn ). f 2 (ˆ vn ) = f 2 Υ (vn ) Υ (vn ) Υ (vn ) Thus, ( )∫ ∫ 1 1 1 2 Υ (ˆ vn ) ≤ |∇vn | dx + + V (|x|)f 2 (vn )dx Υ (vn )2 R2 Υ (vn ) Υ (vn )2 R2 ( ) (∫ ) 1 1 2 2 ≤ + [|∇v | + V (|x|)f (v )]dx n n Υ (vn ) Υ (vn )2 R2 = 1 + Υ (vn ) 2

≤ 2 + τ.

(6.10)

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

22

Moreover, ∥∇ˆ vn ∥2 ≤ 1. By (5.1), (6.10), items (2) and (11) of Lemma 3.1 and H¨older’s inequality, we get ⏐ ⏐∫ ∫ ⏐ ⏐ ′ ⏐ ⏐ ⏐ 2 Q(|x|)h(f (vn ))f (vn )(vn − v)dx⏐ ≤ C1 2 Q(|x|)|f (vn )||vn − v|dx R R ∫ ′ λf 4 (vn ) + C1 Q(|x|)|f (vn )f (vn )|(e − 1)|vn − v|dx R2

) 21 Q(|x|)|vn − v| dx

) 21 (∫ Q(|x|)f (vn )dx

(∫

2

2

≤ C1

(6.11)

R2

R2

(∫

2

Q(|x|)(e2r1 λvn

+ C1

) r1 (∫ 1 − 1)dx

R2

) r1 2 r2 Q(|x|)|vn − v| dx

R2

(∫ ≤ C∥vn − v∥2,Q + C

2

Q(|x|)(e2r1 λvn − 1)dx

R2

) r1

1

∥vn − v∥r2 ,Q ,

where ε > 0, λ > λ0 and r1 , r2 > 1 with 1/r1 + 1/r2 = 1. If the sequence ( bounded, then from (6.11), we conclude ⏐ ⏐∫ ⏐ ⏐ ′ ⏐ Q(|x|)h(f (vn ))f (vn )(vn − v)dx⏐⏐ → 0, ⏐

∫

2

R2

Q(|x|)(e2r1 λvn − 1)dx) is

(6.12)

R2

since vn → v in Lq (R2 , Q) for all q ≥ 2. To prove this, first we observe that ∫ ∫ 2 2 2 2r1 λvn vn Q(|x|)(e − 1)dx = Q(|x|)(e2r1 λΥ (vn )ˆ − 1)dx. R2

(6.13)

R2

Moreover, from Corollary 6.4 we get 2r1 λΥ 2 (vn ) ≤ 4r1

λ λ0 µ c + on (1). λ0 µ − 2

By the hypothesis on c, we can choose r1 > 1 sufficiently close de 1 and λ > λ0 sufficiently close to λ0 satisfying ( ) b0 2 2r1 λΥ (vn ) ≤ ρ0 < 4π 1 + , ∀ n ≥ n0 2 for some ρ0 > 0 and n0 ∈ N. Thus, since ∥∇ˆ vn ∥2 ≤ 1 and Υ (ˆ vn )2 ≤ 2 + τ , by using (6.13) and Theorem 4.4, we reach ∫ 2 sup Q(|x|)(e2r1 λvn − 1)dx < ∞ n∈N

R2

and, consequently, the convergence (6.12) holds true. Now, since f 2 (s) is convex we have that Υ 2 (v) is also convex and therefore 1 2 1 1 Υ (v) − Υ 2 (vn ) ≥ (Υ 2 )′ (vn )(v − vn ), 2 2 2 that is, ∫ ∫ ∫ ∫ 1 1 1 1 2 2 |∇v| dx + V (|x|)f 2 (v)dx − |∇vn | dx − V (|x|)f 2 (vn )dx 2 R2 2 R2 2 R2 2 R2 ∫ ′ ≥ I (vn )(v − vn ) + Q(|x|)h(f (vn ))f ′ (vn )(vn − v)dx. R2

Hence, the convergences (6.12) and ∥I ′ (vn )∥ → 0 show that [∫ ∫ ∫ ∫ 2 2 |∇v| dx + V (|x|)f 2 (v)dx ≥ lim inf |∇vn | dx + R2

R2

n→∞

R2

On the other hand, the Fatou Lemma implies ∫ ∫ V (|x|)f 2 (v)dx ≤ lim inf R2

n→∞

R2

] V (|x|)f 2 (vn )dx .

(6.14)

R2

V (|x|)f 2 (vn )dx.

(6.15)

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

We claim that

∫

∫

2

2

|∇v| dx ≤ lim inf n→∞

R2

|∇vn | dx.

23

(6.16)

R2

In order to prove this claim, we define the Hilbert space { ∫ YQ = u ∈ L2loc (R2 ) : u is radial, |∇u| ∈ L2 (R2 ) and

} Q(|x|)u2 dx < ∞ ,

R2

endowed with the inner product ⟨u, v⟩Q =

∫

R2

[∇u∇v + Q(|x|)uv]dx and its correspondent norm

(∫ ∥u∥Q =

) 12 . [|∇u| + Q(|x|)u ]dx 2

2

R2

Since the embedding E ↪→ L2 (R2 , Q) is continuous, we have E ⊂ YQ and this embedding is also continuous. Thus, (vn ) is also bounded in YQ and there exists w ∈ YQ such that, up to a subsequence, vn ⇀ w in YQ . Moreover, the embedding YQ ↪→ L2 (R2 , Q) is also continuous, which implies that vn ⇀ w in L2 (R2 , Q). As vn → v in L2 (R2 , Q), we have vn ⇀ v in L2 (R2 , Q) and therefore by the uniqueness of weak limit w = v. ∫ 2 The functional Φ(v) := R2 |∇v| dx is convex and continuous on YQ and consequently it is weakly lower semi-continuous. Thus, (6.16) is proved. From (6.14), (6.15) and (6.16), up to a subsequence, we obtain ∫ ∫ ∫ ∫ 2 2 lim |∇vn | dx = |∇v| dx and lim V (|x|)f 2 (vn )dx = V (|x|)f 2 (v)dx. n→∞

R2

n→∞

R2

R2

R2

Hence, (3) of Lemma 3.2 implies [ ] ∫ 1 2 inf 1+ V (|x|)f (ξ(vn − v)) dx → 0, ξ>0 ξ R2 which shows that vn → v in E and the proof is complete.

□

Now, we are going to estimate the mountain pass level c0 := inf γ∈Γ maxt∈[0,1] I(γ(t)) > 0. Proposition 6.6 (Minimax Estimate). Suppose that (h4 ) is satisfied with ⎧ ⎫ p ] p−2 ⎨ [ µλ0 (p − 2)∥Q∥ 1 2 2 ⎬ 2ξ1 L (B1 ) ξ ≥ max ξ1 , , ⎩ pp−1 ⎭ 2p/2 π(1 + b0 /2) where ξ1 :=

p2(p−2)/2 (4π + ∥V ∥L1 (B2 ) ) . ∥Q∥L1 (B1 )

Then, c0 < π[1 + (b0 /2)](µ − 2)/µλ0 . ∞ Proof . First, we are going to consider a function φ0 ∈ C0,rad (R2 ) given by φ0 (x) = 1 if |x| ≤ 1, φ0 (x) = 0 if |x| ≥ 2, 0 ≤ φ0 (x) ≤ 1 for all x ∈ R2 and |∇φ0 (x)| ≤ 1 for all x ∈ R2 . By (h4 ) and (3.2) we infer that if ξ ≥ ξ1 then ∫ ∫ 1 ξ1 2 I(φ0 ) ≤ [|∇φ0 | + V (|x|)φ20 ]dx − Q(|x|)f p (φ0 )dx 2 B2 p B2 1 ξ1 f p (1) < 2π + ∥V ∥L1 (B2 ) − ∥Q∥L1 (B1 ) 2 p 1 ξ1 ≤ [4π + ∥V ∥L1 (B2 ) ] − p/2 ∥Q∥L1 (B1 ) = 0. 2 p2

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24

In particular, 1 2

∫

2

[|∇φ0 | + V (|x|)φ20 ]dx <

B2

ξ1 ∥Q∥L1 (B1 ) . p2p/2

(6.17)

According to the definition of φ0 , (6.17) and the hypothesis on ξ, a simple calculation shows that [ 2 (∫ ) ] ∫ t ξ 2 2 p c0 ≤ max I(tφ0 ) ≤ max [|∇φ0 | + V (|x|)φ0 ]dx − Q(|x|)f (tφ0 )dx p B2 t∈[0,1] t∈[0,1] 2 B2 ] [ ξ ξ1 ∥Q∥L1 (B1 ) t2 − f p (t)∥Q∥L1 (B1 ) < max p t∈[0,1] p2p/2 [ ] ∥Q∥L1 (B1 ) ≤ max ξ1 t2 − ξtp t≥0 p2p/2 Calculating the maximum, we obtain c0 <

∥Q∥L1 (B1 )

1

p2p/2

ξ 2/(p−2)

2/(p−2)

(p − 2)2

Thus, if [ ξ≥

µλ0 (p − 2)∥Q∥L1 (B1 )

] p−2 2

2p/2 π(1 + b0 /2)

(

ξ1 p

)p/(p−2) .

p

2ξ12 pp−1

then c0 < π[1 + (b0 /2)](µ − 2)/µλ0 and proof of the proposition is done.

□

Finalizing proof of Theorem 1.3: According to Propositions 5.1 and 6.2 and Theorem 6.1, there exists a Palais–Smale sequence (vn ) ⊂ E for I at minimax level c0 . Now, by Propositions 6.5 and 6.6, I satisfies the Palais–Smale condition at level c0 and therefore we obtain v ∈ E such that I(v) = c0 and I ′ (v) = 0. Taking as test function −v − ∈ E and since V (|x|)f (v)f ′ (v)(−v − ) ≥ 0 it follows that v = v + ≥ 0 and thus, by Proposition 5.4, u = f (v) is nonnegative and nonzero weak solution for (P ). Now, we proceed to prove Corollary 1.4. Setting m = inf I(v) v∈S

and

S := {v ∈ E : v ̸= 0 and I ′ (v) = 0},

2 by similar arguments as in [34, Proposition 1], we can show that u ∈ Yrad ∩ L∞ loc (R ) is a weak solution of −1 (P ) if, and only if, v = f (u) ∈ E is a critical point of I and J(u) = I(v). Thus, it is enough to prove that the mountain pass level c0 of I satisfies c0 ≤ m. Let v ∈ S and define ζ : (0, +∞) → R by ζ(t) = I(tv). We have that ζ is differentiable and ∫ ∫ ∫ 2 ζ ′ (t) = I ′ (tv)v = t |∇v| dx + V (|x|)f (tv)f ′ (tv)v dx − Q(|x|)h(f (tv))f ′ (tv)v dx. R2

R2

R2

′

Since I (v) · v = 0, we get ∫ ∫ 2 |∇v| dx = − R2

R2

V (|x|)f (v)f ′ (v)v dx +

∫

Q(|x|)h(f (v))f ′ (v)vdx

R2

and by using the two last identities we reach [ ] ∫ f (tv)f ′ (tv) f (v)f ′ (v) 2 ′ ζ (t) =t − v dx V (|x|) tv v {v>0} ] [ ∫ h(f (v)) f 3 (v)f ′ (v) h(f (tv)) f 3 (tv)f ′ (tv) 2 +t Q(|x|) − v dx. f 3 (v) v f 3 (tv) tv {v>0} By Corollary 2.3 of [29], the function f (s)f ′ (s)s−1 is decreasing for s > 0 and the f 3 (s)f ′ (s)s−1 is increasing for s > 0. Thus, it follows that ζ ′ (t) > 0 for 0 < t < 1, ζ ′ (t) < 0 for t > 1 and ζ ′ (1) = 0, which implies that

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

25

I(v) = maxt≥0 I(tv). Now, defining γ : [0, 1] → E, γ(t) = tt0 v, where t0 is such that I(t0 v) = e, we have γ ∈ Γ and therefore c0 ≤ max I(γ(t)) ≤ max I(tv) = I(v). t≥0

t∈[0,1]

Since v ∈ S is arbitrary, c0 ≤ m and we finish the proof. 7. Proof of Theorem 1.6 In this section, we prove our nonexistence result of radial positive classical solution for problem (P ). For this we shall use radial coordinates. If u ∈ C 2 (R2 , R) is radial, r = |x| and v(r) = u(x) then [ ] 1 v(r) ∆u = v ′′ (r) + v ′ (r) and ∆(u2 ) = 2v(r)v ′′ (r) + 2 + v ′ (r) v ′ (r). (7.1) r r ˆ imply that there exist positive constants C1 , C2 and R such that Conditions (Vˆ ) and (Q) V (r) ≤ C1 ra

and

Q(r) ≥ C2 rb ,

∀ r ≥ R.

(7.2)

Moreover, from (ˆ h) there exist ξ > 0 and p > 2 such that if v(r) ≥ 0 then Q(r)h(v(r)) ≥ ξQ(r)v(r)p .

(7.3)

Suppose, by contradiction, that there exists u ∈ C 2 (R2 , R), which is radial and positive classical solution for (P ). If we set v(r) = u(x) where r = |x|, then by (7.1), (7.2) and (7.3) we obtain 1 [1 + 2v(r)2 ]v ′′ (r) + [1 + 2v(r)2 ]v ′ (r) + 2v(r)[v ′ (r)]2 − V (r)v(r) + ξC2 rb v(r)p ≤ 0, r

(7.4)

for all r ≥ R. For r > R, we set t = log(r) and we define w(t) = rm v(r) where m = (b + 2)/(p − 1) > 0. Differentiating w(t) we have w′ (t) = mrm v(r) + rm+1 v ′ (r) ′′

2 m

w (t) = m r v(r) + mr

and

m+1 ′

v (r) + (m + 1)rm+1 v ′ (r) + rm+2 v ′′ (r).

Setting l1 = −2m and l2 = m2 , we get w′′ (t) + l1 [1 + 2v(r)2 ]w′ (t) + [(1 + 2v(r)2 )l2 − V (r)r2 ]w(t) + C2 ξw(t)p ] [ 1 m+2 2 ′′ 2 ′ b p =r (1 + 2v(r) )v (r) + (1 + 2v(r) )v (r) − V (r)v(r) + C2 ξr v(r) r

(7.5)

and from (7.4) and (7.5) it follows that [ ] [1 + 2v(r)2 ]w′′ (t) + l1 [1 + 2v(r)2 ]w′ (t) + (1 + 2v(r)2 )l2 − V (r)r2 w(t) + C1 ξw(t)p ≤ 0.

(7.6)

At this point we affirm that there exists t0 > 0 such that w(t) is nondecreasing for all t > t0 . Indeed, suppose by contradiction that for each t0 > 0 there exists T > t0 verifying w′ (T ) < 0. By (7.2) we have { ( 1 } )− a+2 C1 2 2 2 a+2 (1 + 2v(r) )l2 − V (r)r ≥ m − C1 r > 0, ∀ r > R0 := max R, . (7.7) m2 and according to (7.6) we get w′′ (t) + l1 w′ (t) ≤ 0,

∀ t > log R0 .

(7.8)

U.B. Severo and G.M. de Carvalho / Nonlinear Analysis 196 (2020) 111800

26

We can suppose without loss of generality that T > log R0 . Now, for t > T , multiplying inequality (7.8) by el1 t and integrating in the interval [T, t] we reach ∫

t

0≥

[w′′ (s) + l1 w′ (s)]el1 s ds =

T

∫

t

T

) d ( ′ w (s)el1 s ds = w′ (t)el1 t − w′ (T )el1 T ds

and consequently w′ (t) ≤ w′ (T )el1 (T −t) . Integrating again in [T, t], we obtain w(t) ≤ w(T ) +

w′ (T ) −l1 (t−T ) [e − 1] −l1

and this implies that for t sufficiently large we must have w(t) < 0, which is a contradiction. Therefore, our claim is proved. Next, we shall analyze two cases: Case 1: w is bounded. In this case, there exists w∞ > 0 such that w(t) → w∞ as t → +∞. Thus, by (7.6) and (7.7) we get [ ] p > 0, 0 ≥ lim inf (1 + 2v(r)2 )l2 − V (r)r2 + C2 ξw(t)p−1 w(t) ≥ lim inf [C2 ξw(t)p ] ≥ C2 ξw∞ t→∞

t→∞

which is an absurd and therefore this case cannot happen. Case 2: w is unbounded. For this situation, we define the functions l1

σ(t) = e 2 t w(t)

and

D(t) = l2 −

V (r)r2 l2 C2 ξw(t)p−1 − 1 + . 2 1 + 2v(r) 4 1 + 2v(r)2

Thus, by (7.6) we have [1 + 2v(r)2 ][σ ′′ (t) + D(t)σ(t)] l1 [ ] = e 2 t (1 + 2v(r)2 )w′′ (t) + l1 (1 + 2v(r)2 )w′ (t) + l2 (1 + 2v(r)2 )w(t) − V (r)r2 w(t) + Cw(t)p < 0. Therefore, if k ∈ N and t ∈ (2kπ, (2k + 1)π) then σ ′′ (t) sin(t) + D(t)σ(t) sin(t) < 0. Integrating by parts twice this inequality, we get ∫

(2k+1)π

[D(t) − 1]σ(t) sin(t)dt < −σ((2k + 1)π) − σ(2kπ) < 0. 2kπ

Since w(t) is unbounded, for k ∈ N sufficiently large we have D(t) > 1 and hence ∫

(2k+1)π

[D(t) − 1]σ(t) sin(t)dt > 0, 2kπ

which is also a contradiction. Hence, this case also does not occur and the proof of Theorem 1.6 is finalized. Acknowledgment We would like to thank the referee by the carefully reading of the paper with many useful comments and important suggestions which substantially helped improving the quality of the paper.

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