On the planar Kirchhoff-type problem involving supercritical exponential growth

Limin Zhang; Xianhua Tang; Peng Chen

doi:10.1515/anona-2022-0250

Open Access Published by De Gruyter May 10, 2022

On the planar Kirchhoff-type problem involving supercritical exponential growth

Limin Zhang , Xianhua Tang and Peng Chen

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0250

Abstract

This article is concerned with the following nonlinear supercritical elliptic problem:

− M ( ‖ ∇ u ‖ 2 2 ) Δ u = f ( x , u ) , in B 1 ( 0 ) , u = 0 , on ∂ B 1 ( 0 ) ,

where B 1 ( 0 ) is the unit ball in R 2 , M : R + → R + is a Kirchhoff function, and f ( x , t ) has supercritical exponential growth on t , which behaves as exp [ ( β 0 + ∣ x ∣ α ) t 2 ] and exp ( β 0 t 2 + ∣ x ∣ α ) with β 0 , α > 0 . Based on a deep analysis and some detailed estimate, we obtain Nehari-type ground state solutions for the above problem by variational method. Moreover, we can determine a fine upper bound for the minimax level under weaker assumption on liminf t → ∞ t f ( x , t ) exp [ ( β 0 + ∣ x ∣ α ) t 2 ] and liminf t → ∞ t f ( x , t ) exp ( β 0 t 2 + ∣ x ∣ α ) , respectively. Our results generalize and improve the ones in G. M. Figueiredo and U. B. Severo (Ground state solution for a Kirchhoff problem with exponential critical growth, Milan J. Math. 84 (2016), no. 1, 23–39.) and Q. A. Ngó and V. H. Nguyen (Supercritical Moser-Trudinger inequalities and related elliptic problems, Calc. Var. Partial Differ. Equ. 59 (2020), no. 2, Paper No. 69, 30.) for M ( t ) = 1 . In particular, if the weighted term ∣ x ∣ α is vanishing, we can obtain the ones in S. T. Chen, X. H. Tang, and J. Y. Wei (2021) (Improved results on planar Kirchhoff-type elliptic problems with critical exponential growth, Z. Angew. Math. Phys. 72 (2021), no. 1, Paper No. 38, Theorem 1.3 and Theorem 1.4) immediately.

Keywords: Kirchhoff problem; Trudinger-Moser-type inequality; supercritical exponential growth

MSC 2010: 35J60; 35J20

1 Introduction

This article is dedicated to studying the following Kirchhoff-type problem with Trudinger-Moser-type nonlinearity:

(1.1) − M ( ‖ ∇ u ‖ 2 2 ) Δ u = f ( x , u ) , in B , u = 0 , on ∂ B ,

where B ≔ B 1 ( 0 ) ⊂ R 2 is the unit ball centered at the origin 0 ∈ R 2 , M : R + → R + , f : B × R → R are continuous functions and satisfy some mild conditions. More precisely, M satisfies the following basic assumptions:

M ∈ C ( R + , R + ) , ℳ ( t ) ≔ ∫ 0 t M ( s ) d s , and there exists m 0 > 0 such that M ( t ) ≥ m 0 for t ≥ 0 , and
ℳ ( t + s ) ≥ ℳ ( t ) + ℳ ( s ) , ∀ t , s ≥ 0 ;
M ∈ C ( R + , R + ) , and there exists m 0 > 0 such that M ( t ) ≥ m 0 for t ≥ 0 , and M ( t ) is nondecreasing on [ 0 , + ∞ ) ;
M ˜ ( t ) ≔ M ( t ) − m 0 , and there exists θ > 1 such that M ˜ ( t ) t θ − 1 is nonincreasing on ( 0 , ∞ ) .

Problem (1.1) is called nonlocal owning to the presence of the term M ( ‖ ∇ u ‖ 2 2 ) , which implies that problem (1.1) is no longer a pointwise identity. The main motivation to study problem (1.1) is due to the work of Kirchhoff [13], in which the author initially proposed the following hyperbolic equation:

ρ ∂ 2 u ∂ t 2 − ρ 0 h + E 2 L ∫ 0 L ∂ u ∂ x d x ∂ 2 u ∂ x 2 = 0 ,

which is a generalization of the classical D’Alembert’s wave equation. Furthermore, in the 1950s, Woinowsky-Krieger [30] considered the following evolution equation of Kirchhoff-type:

u t t + Δ 2 u − M ( ‖ ∇ u ‖ 2 ) Δ u = f ( x , u ) ,

which has some applications, for instance in physics [5] and mathematics [22,28,33]. Concerning other applications, see also [14,15,17,25,26,31,34,35,36].

Denote by H 0 , rad 1 ( B ) the subspace of the Sobolev space H 0 1 ( B ) , which consists of radially symmetric functions. Problem (1.1) has strictly critical exponent growth except in the origin about the well-known Moser-Trudinger inequality [6,18,21,27], which ensures

(1.2) MT β ≔ sup u ∈ H 0 , rad 1 ( B ) : ‖ ∇ u ‖ 2 ≤ 1 ∫ B exp ( β ∣ u ∣ 2 ) d x < + ∞

for any β ≤ 4 π . Moreover, inequality (1.2) is sharp (or critical): for any growth exp ( β ∣ u ∣ 2 ) with β > 4 π , the corresponding MT β is + ∞ . Generally, for any β > 0 ,

(1.3) MT ¯ β ≔ sup u ∈ H 0 , rad 1 ( B ) ∫ B exp ( β ∣ u ∣ 2 ) d x < + ∞ .

Since inequality (1.3) and its variants have many applications in many aspects of analysis, the generalization of (1.3) has already been a hot research topic and a huge set of works have already been written within the last two decades, see [1,8,11,23,24] and references therein. Apparently, the general critical growth used in these works differs from ours by the exponent ∣ x ∣ α .

It should be noted that the nonvanishing exponent ∣ x ∣ α is very crucial in our arguments. Based on this, in a very classical paper [19], Ni studied the existence of solutions for the following Hénon equation in B 1 ( 0 ) ⊂ R N ( N ≥ 3 ) with M ( ‖ ∇ u ‖ 2 2 ) = 1 and f ( x , u ) = ∣ x ∣ α ∣ u ∣ p − 2 u

− Δ u = ∣ x ∣ α ∣ u ∣ p − 2 u , in B 1 ( 0 ) , u = 0 , on ∂ B 1 ( 0 ) ,

where 1 < p < 2 ( N + α ) N − 2 . Under the compactness lemma of radial lemma for radial functions, that is, H 0 1 ( B 1 ( 0 ) ) ↪ ↪ L p ( B 1 ( 0 ) ) with 1 < p < 2 ( N + α ) N − 2 . When 2 ∗ < p < 2 ( N + α ) N − 2 , the above equation is a supercritical elliptic problem. In [7], Cao et al. studied nodal solutions for the following supercritical semilinear with M ( ‖ ∇ u ‖ 2 2 ) = 1 and f ( x , u ) = ∣ u ∣ 2 ∗ + ∣ x ∣ α − 2 u

− Δ u = ∣ u ∣ 2 ∗ + ∣ x ∣ α − 2 u , in B 1 ( 0 ) , u = 0 , on ∂ B 1 ( 0 ) ,

where 0 < α < min N 2 , N − 2 . It is worthwhile to mention that the weighted term ∣ x ∣ α plays similar role in the aforementioned two equations.

The case N = 2 , which is different from the case in which N ≥ 3 for the embedding H 0 1 ( B ) ↪ L 2 ∗ ( B ) 2 ∗ = 2 N N − 2 , is very special. The embedding H 0 1 ( B ) ↪ L ∞ ( B ) is no longer available. Although Trudinger’s inequality (1.3) provides us a perfect replacement at the origin, namely, there holds H 0 1 ( B ) ↪ e L ∞ ( B ) . As to the weighted term ∣ x ∣ α is nonvanishing, this is still an open question. But fortunately, Ngô and Nguyen [20] studied extensions for (1.2) in the presence of supercritical growth, which is a improvement for the classical Moser-Trudinger inequality. To be precise for β ≤ 4 π , they proved that the following two estimates hold:

(1.4) MT α 1 ( β ) ≔ sup u ∈ H 0 , rad 1 ( B ) : ‖ ∇ u ‖ 2 ≤ 1 ∫ B exp [ ( β + ∣ x ∣ α ) ∣ u ∣ 2 ] d x < + ∞

and

(1.5) MT α 2 ( β ) ≔ sup u ∈ H 0 , rad 1 ( B ) : ‖ ∇ u ‖ 2 ≤ 1 ∫ B exp ( β ∣ u ∣ 2 + ∣ x ∣ α ) d x < + ∞ ,

by using the crucial lemma as follows:

Lemma 1.1

(Concentration compactness principle of Lions type [20]) Assume { u n } ⊂ H 0 , rad 1 ( B ) satisfies that

‖ ∇ u n ‖ 2 = 1 , u n ⇀ u ¯ ≠ 0 in H 0 , rad 1 ( B ) .

Then for any 0 < p < 1 / ( 1 − ‖ ∇ u ¯ ‖ 2 2 ) , we have

sup n ∈ N ∫ B exp [ p ( 4 π + ∣ x ∣ α ) ∣ u n ∣ 2 ] d x < ∞

and

sup n ∈ N ∫ B exp ( 4 π p ∣ u n ∣ 2 + ∣ x ∣ α ) d x < ∞ .

Therefore, it is possible to solve (1.1) by variational method in H 0 , rad 1 ( B ) . In addition, both MT α 1 ( β ) and MT α 2 ( β ) are sharp in the sense that these become infinity for any β > 4 π .

On account of (1.4) and (1.5), we say f ( x , t ) has supercritical exponential growth at t = ± ∞ if it verifies the following assumption:

f ∈ C ( B × R , R ) , sup x ∈ B , ∣ s ∣ ≤ ∣ t ∣ ∣ f ( x , t ) ∣ < + ∞ for every t ∈ R , and there exists β 0 > 0 such that
lim ∣ t ∣ → ∞ ∣ f ( x , t ) ∣ exp [ ( β + ∣ x ∣ α ) t 2 ] = 0 , uniformly on x ∈ B for all β > β 0
and
lim ∣ t ∣ → ∞ ∣ f ( x , t ) ∣ exp [ ( β + ∣ x ∣ α ) t 2 ] = + ∞ , uniformly on x ∈ B for all β < β 0 .

Furthermore, in view of (F1), one can define a similar supercritical growth by replacing the exponent ( β + ∣ x ∣ α ) t 2 with the exponent β t 2 + ∣ x ∣ α , i.e.,

f ∈ C ( B × R , R ) , sup x ∈ B , ∣ s ∣ ≤ ∣ t ∣ ∣ f ( x , t ) ∣ < + ∞ for every t ∈ R , and there exists β 0 > 0 such that
lim ∣ t ∣ → ∞ ∣ f ( x , t ) ∣ exp ( β t 2 + ∣ x ∣ α ) = 0 , uniformly on x ∈ B for all β > β 0
and
lim ∣ t ∣ → ∞ ∣ f ( x , t ) ∣ exp ( β t 2 + ∣ x ∣ α ) = + ∞ , uniformly on x ∈ B for all β < β 0 .

As far as we know such a supercritical growth was only considered in the literature [20] under (F1) and (F1 ′ ). However, they just devoted to studying problem (1.1) with M ( ‖ ∇ u ‖ 2 2 ) = 1 , the purpose of this article is to fill this gap and provide a counterpart of the result of problem (1.1). It is natural to ask whether there exists any solution for problem (1.1).

In addition to (F1), (F1 ′ ), we make use of the following general conditions on the nonlinearity f :

f ( x , t ) = o ( ∣ t ∣ ) as t → 0 uniformly on x ∈ B ;
f ( x , t ) > 0 for all ( x , t ) ∈ B × ( 0 , + ∞ ) , and there exist M 0 > 0 and t 0 > 0 such that
F ( x , t ) ≤ M 0 f ( x , t ) , ∀ ( x , t ) ∈ B × [ t 0 , + ∞ ) ,
where F ( x , t ) = ∫ 0 t f ( x , s ) d s ;
liminf t → + ∞ t 2 F ( x , t ) exp [ ( β 0 + ∣ x ∣ α ) t 2 ] ≥ κ 1 > 2 e β 0 2 M 4 π β 0 uniformly on x ∈ B ;
liminf t → + ∞ t 2 F ( x , t ) exp ( β 0 t 2 + ∣ x ∣ α ) ≥ κ 2 > 2 e β 0 2 M 4 π β 0 uniformly on x ∈ B .

Obviously, if M 4 π β 0 = 1 , (F4) is a weak version of (F5) in [20]. It is worth noting that there is no specific relationship between κ 1 and κ 2 , and assumptions (F4), (F4 ′ ) are inspired by [9] in which, the authors proved the existence of nontrivial solutions and Nehari-type ground state solutions for problem (1.1) under the following assumption:

lim t → ∞ t 2 F ( x , t ) e β 0 t 2 ≥ κ 0 > 1 e β 0 2 M 4 π β 0 uniformly on x ∈ B ,

which implies (f3) is used in [12], since 1 β 0 2 M 4 π β 0 > 1 e β 0 2 M 4 π β 0 . Moreover, they obtained a more precise estimation than the ones in the existing literature about the minimax level. Due to the appearance of the exponent ( β + ∣ x ∣ α ) t 2 and β t 2 + ∣ x ∣ α , we cannot find a upper bound coefficient of 1 β 0 2 M 4 π β 0 like 1 / e from (F4 ″ ), which stems from the fact that

t 2 F ( x , t ) e β 0 t 2 ≥ max t 2 F ( x , t ) exp [ ( β 0 + ∣ x ∣ α ) t 2 ] , t 2 F ( x , t ) exp ( β 0 t 2 + ∣ x ∣ α ) .

But we can determine that a fine upper bound coefficient is exactly 2 / e , which shows that assumptions (F4) and (F4 ′ ) are weaker than [12, f3] at the origin. Simultaneously, (F4) and (F4 ′ ) play a crucial role to obtain the estimation of threshold c i ∗ < 1 2 ℳ 4 π β 0 ( i = 1 , 2 ) , see Section 3 for more details.

Let λ 1 denote the first eigenvalue of − Δ with a Dirichlet boundary condition. To state our results, except (F1)–(F4 ′ ), we also introduce the following assumptions:

for every x ∈ B , t ↦ f ( x , t ) − m 0 λ 1 t t 2 θ − 1 is nondecreasing on ( 0 , ∞ ) ;
f ( x , t ) t − 2 θ F ( x , t ) + ( θ − 1 ) λ 1 t 2 ≥ 0 for all ( x , t ) ∈ B × [ 0 , + ∞ ) .

Here conditions (F5) and (F6) were introduced by Chen et al. in [9], and from its Remark 1.5 we know that they are weaker than the following assumptions:

for every x ∈ B , t ↦ f ( x , t ) t 3 is increasing on ( 0 , ∞ ) ;
f ( x , t ) t − 2 θ F ( x , t ) ≥ 0 for all ( x , t ) ∈ B × [ 0 , + ∞ ) ,

which were widely used in [3,4,10,23] and references therein.

Thanks to the work of [20], which inspired us to consider the existence of solutions for problem (1.1). Such a existence result, to the best of our knowledge, seems to be new. Motivated by the above works [9,12,20], we will study the nonlocal problem (1.1) under (F4) and (F4 ′ ), and aim to establish the existence of nontrivial and ground state solutions of problem (1.1) for the supercritical growth cases. It is worthwhile pointing out that the nonlocal problem together with the supercritical exponent bring some new difficulties in our analysis. Specifically, we list them as follows:

Certifying the energy functional associated with problem (1.1) has the mountain pass geometry, which possesses Cerami sequences { u n } ⊂ H 0 , rad 1 ( B ) at the mountain pass level c i ∗ > 0 ( i = 1 , 2 ) . This needs a careful analysis on the relationship between the sequence’s H 0 , rad 1 ( B ) norm and L 2 ( B ) norm in order to use the Trudinger-Moser-type inequality MT α 1 ( β ) and MT α 2 ( β ) , which require that the critical exponent β shall be less than 4 π .
Controlling the minimax levels by a fine threshold, which is very complicated and difficult to determine due to the appearance of both the term M ( ‖ ∇ u ‖ 2 2 ) and the supercritical growth nonlinearity f ( x , u ) . Actually, obtaining the estimation of c i ∗ < 1 2 ℳ 4 π β 0 ( i = 1 , 2 ) associated with energy functional I can help us to obtain β 0 ‖ u n − u ‖ 2 ≤ β < 4 π , which is essential to complete the proof of Theorems 1.1 and 1.3.
Proving weak limit u ¯ of { u n } is a solution of problem (1.1) by using deep analysis and some detailed estimate rather than using Lemma 1.1, which gives more precise information than [20, Lemma 1.2]. Indeed, if β is large enough, the integral is divergent. (Here we ignore the constraint ‖ u ¯ ‖ 2 ≤ 1 , so the set of functions consisting of the exponent β ∣ u ¯ ∣ 2 + ∣ x ∣ α is even larger than that of Lemma 1.2 in [20].) We cannot obtain a similar conclusion as MT ¯ β about MT α 1 ( β ) and MT α 2 ( β ) .
Showing that the supercritical growth nonlinearity f ( x , u ) ∈ L q ( B ) ( q ∈ ( 1 , 2 ) ) for the MT α 2 ( β ) , which aims at finding a ground state solution requires a careful analysis. Compared with the classical Trudinger-Moser inequality MT ¯ β with β > 0 , however, it does not hold for the supercritical exponential growth.

The main ingredient in our approach is the observation that a minimizing Cerami sequence for the energy functional can be found by employing some techniques and using an approaching argument and some detailed estimate, part of which comes from the author’s recent papers [9], who developed some new approaches to estimate precisely the minimax level of the energy functional and proved the existence of Nehari-type ground state solutions and nontrivial solutions for problem (1.1) when ∣ x ∣ α is vanishing, which is completely different from the one of Figueiredo and Severo [12]. In this article, based on [20], we further develop the approach in [9] to find ground state solution for problem (1.1) with supercritical exponential growth. It is worth noting that we will still use the classical Moser-type function w n (i.e., (3.0.1)) to estimate c i ∗ < 1 2 ℳ 4 π β 0 ( i = 1 , 2 ) instead of using the test function ω n , see [20, Lemma 4.3].

Our results for MT α 1 ( β ) are read as follows:

Theorem 1.1

Assume that (M1), (M2), and (F1)–(F5) hold, then problem (1.1) has a positive ground state solution.

Theorem 1.2

Assume that (M1 ′ ), (M2), (F1)–(F4), and (F6) hold, then problem (1.1) has a nontrivial solution.

When the nonlinearity has MT α 2 ( β ) -type supercritical exponential growth, we have the following theorems:

Theorem 1.3

Assume that (M1), (M2), ( F 1 ′ ) , and (F2)–(F5) hold, then problem (1.1) has a positive ground state solution.

Theorem 1.4

Assume that ( M 1 ′ ) , (M2), ( F 1 ′ ) , (F2)–(F4), and (F6) hold, then problem (1.1) has a nontrivial solution.

Remark 1.5

If the weighted term ∣ x ∣ α is vanishing about MT α 1 ( β ) and MT α 1 ( β ) , our Theorems 1.1, 1.2, and 1.3, 1.4 will reduce to [9, Theorems 1.3 and 1.4], respectively.

The article is organized as follows. Preliminaries and useful lemmas are given in Section 2. In Section 3, we establish the minimax estimates of the energy functional associated with MT α 1 ( β ) and MT α 2 ( β ) separately. Theorems 1.1–1.4 will be proved in Section 4.

Throughout the article, we make use of the following notations:

C 1 , C 2 , … denote positive constants possibly different in different places;
For any x ∈ B and r > 0 , B r ( x ) ≔ { y ∈ B : ∣ y − x ∣ < r } and B r = B r ( 0 ) ;
u ∈ H 0 , rad 1 ( B ) is radially symmetric about origin, we rewrite u ( x ) as u ( ∣ x ∣ ) .

2 Preliminaries

In the sequel of this article we let H ≔ H 0 , rad 1 ( B ) and

‖ u ‖ 2 ≔ ∫ B ∣ ∇ u ∣ 2 d x , ∀ u ∈ H .

Moreover, one has

(2.1) ‖ u ‖ 2 ≥ λ 1 ‖ u ‖ 2 2 , ∀ u ∈ H .

It is widely known that H is continuously embedded into L s ( B ) for s ≥ 2 , and by Rellich embedding theorem, for any s ∈ [ 2 , + ∞ ) , there exists γ s > 0 such that

(2.2) ‖ u ‖ s ≤ γ s ‖ u ‖ .

From (M1) (or (M1 ′ )), (F1) (or (F1 ′ )), (F2), we know that the weak solutions of problem (1.1) correspond to the critical points of the energy functional defined by

(2.3) I ( u ) = 1 2 ℳ ( ‖ u ‖ 2 ) − ∫ B F ( x , u ) d x .

Moreover, I belongs to C 1 ( H , R ) and

(2.4) ⟨ I ′ ( u ) , v ⟩ = M ( ‖ u ‖ 2 ) ∫ B ∇ u ∇ v d x − ∫ B f ( x , u ) v d x

for all u , v ∈ H .

Ground state solution associated with energy functional (2.3) refers to minimizers of the corresponding energy within the set of nontrivial solutions. We separately define the minimax energy

b 1 ≔ inf u ∈ N I ( u ) , b 2 ≔ inf u ∈ N I ( u )

under (F1) and (F1 ′ ), where

N ≔ { u ∈ H \ { 0 } : ⟨ I ′ ( u ) , u ⟩ = 0 } .

In order to prove our results, we first give the following preparatory propositions and lemmas.

Proposition 2.1

(Radial lemma [2]) The Radial inequality holds:

∣ u ( r ) ∣ ≤ − log r 2 π ‖ ∇ u ‖ 2 , ∀ u ∈ H .

Proposition 2.2

(Kirchhoff property [32]) Assume that (M1) and (M2) hold. Then for any t ≥ 0 , we have

ℳ ( t ) ≤ ℳ ( 1 ) ( 1 + t θ ) ;
θ ℳ ( t ) − M ( t ) t ≥ 0 ;
ℳ ( t 1 + t 2 ) ≤ ℳ ( t 1 ) + t 1 θ 1 + t 2 t 1 θ − 1 M ( t 1 ) for any t 1 , t 2 ≥ 0 .

Proposition 2.3

(Key inequality [9]) Assume that ( M 1 ) , ( M 2 ) , ( F 1 ) ( o r ( F1 ′ ) ) , ( F 2 ) , and ( F 5 ) hold. Then

(2.5) I ( u ) ≥ I ( s u ) + 1 − s 2 θ 2 θ ⟨ I ′ ( u ) , u ⟩ , ∀ u ∈ H , s ≥ 0 .

Lemma 2.4

Assume that ( M 1 ) ( or ( M1 ′ ) ) , ( M 2 ) , ( F 1 ) , ( F 2 ) , and (F3) hold. Then there exists a sequence { u n } ⊂ H satisfying

(2.6) I ( u n ) → c 1 ∗ , ∥ I ′ ( u n ) ∥ ( 1 + ∥ u n ∥ ) → 0 ,

where c 1 ∗ is given by

c 1 ∗ = inf γ ∈ Γ max t ∈ [ 0 , 1 ] I ( γ ( t ) ) , Γ = { γ ∈ C ( [ 0 , 1 ] , H ) : γ ( 0 ) = 0 , I ( γ ( 1 ) ) < 0 } .

Proof

Since (M1) implies (M1 ′ ), hence we only consider the case when (M1) holds. From (F1)–(F2), for β ∈ ( β 0 , 2 π ] , there exists C 1 > 0 such that

(2.7) ∣ F ( x , t ) ∣ ≤ m 0 4 γ 2 2 ∣ t ∣ 2 + C 1 ∣ t ∣ 3 exp [ ( β + ∣ x ∣ α ) ∣ t ∣ 2 ] , ∀ ( x , t ) ∈ B × R .

Claim. For any u ∈ H , there exists a constant C 0 > 0 such that

∫ B exp [ 2 ( β + ∣ x ∣ α ) u 2 ] d x ≤ C 0 , ∀ ‖ u ‖ ≤ 1 .

For any ‖ u ‖ ≤ 1 , from radial Lemma 2.1, one has

(2.8) r α ∣ u ( r ) ∣ 2 ≤ − r α log r 2 π ≔ g ( r ) .

Then, it follows from (1.2) and (2.8) that

∫ B exp [ 2 ( β + ∣ x ∣ α ) u 2 ] d x ≤ 2 π ∫ 0 1 exp [ 2 ( β + r α ) u 2 ] r d r = 2 π ∫ 0 δ exp [ 2 ( β + r α ) u 2 ] r d r + 2 π ∫ δ 1 exp [ 2 ( β + r α ) u 2 ] r d r ≤ exp ( sup r ∈ ( 0 , δ ) 2 g ( r ) ) ∫ B exp ( 2 β u 2 ) d x + 2 π ∫ δ 1 r − 1 ( 1 − r ) − 2 ρ d r ≤ exp ( sup r ∈ ( 0 , δ ) 2 g ( r ) ) MT 2 β + 2 π ( 1 − δ ) 1 − 2 ρ δ ( 1 − 2 ρ ) ≔ C 0 ,

where δ , ρ ∈ ( 0 , 1 ) such that r α ≤ 2 π ρ log r ( 1 − r ) . Thus, we complete the claim.

In view of (2.2), (2.7), the Hölder inequality, and the above Claim, we have

(2.9) ∫ B F ( x , u ) d x ≤ m 0 4 γ 2 2 ‖ u ‖ 2 2 + C 1 ∫ B ∣ u ∣ 3 exp [ ( β + ∣ x ∣ α ) u 2 ] d x ≤ m 0 4 ‖ u ‖ 2 + C 1 ‖ u ‖ 6 3 ∫ B exp [ 2 ( β + ∣ x ∣ α ) u 2 ] d x 1 / 2 ≤ m 0 4 ‖ u ‖ 2 + C 0 1 / 2 C 1 γ 6 3 ‖ u ‖ 3 , ∀ ‖ u ‖ ≤ 1 .

Hence, it follows from (M1), (2.3), and (2.9) that

I ( u ) = 1 2 ℳ ( ‖ u ‖ 2 ) − ∫ B F ( x , u ) d x ≥ m 0 4 ‖ u ‖ 2 − C 0 1 / 2 C 1 γ 6 3 ‖ u ‖ 3 , ∀ ‖ u ‖ ≤ 1 .

Therefore, there exist σ 1 > 0 and 0 < ϱ 1 < 1 such that

I ( u ) ≥ σ 1 , ∀ u ∈ S ≔ { u ∈ H : ‖ u ‖ = ϱ 1 } .

By (F1), it follows that there exists C 0 > 0 such that

(2.10) F ( x , t ) ≥ ∣ t ∣ 2 θ + 1 − C 0 , ∀ ( x , t ) ∈ B × R .

Choose u 1 ∈ H \ { 0 } , by (2.3), (2.10), and Proposition 2.2(i), we obtain

I ( t u 1 ) = 1 2 ℳ ( ‖ t u 1 ‖ 2 ) − ∫ B F ( x , t u 1 ) d x ≤ 1 2 ℳ ( 1 ) ( 1 + t 2 θ ‖ u 1 ‖ 2 θ ) − ∫ B ( ∣ t u 1 ∣ 2 θ + 1 − C 0 ) d x ≤ t 2 θ 1 2 ℳ ( 1 ) 1 t 2 θ + ‖ u 1 ‖ 2 θ − t ∫ B ∣ u 1 ∣ 2 θ + 1 d x + π C 0 .

It is easy to see that I ( t u 1 ) → − ∞ as t → ∞ . Thus, we can choose T 1 > 0 such that e 1 = T 1 u 1 ∈ { u ∈ H : ‖ u ‖ > ϱ 1 } and I ( e 1 ) < 0 , then in view of the mountain pass lemma [29], we deduce that there exist c 1 ∗ ∈ [ σ 1 , sup t ≥ 0 I ( t u 1 ) ] and a sequence { u n } ⊂ H satisfying (2.6).□

Lemma 2.5

Assume that ( M 1 ) ( o r ( M1 ′ ) ) , ( M 2 ) , ( F1 ′ ) , ( F 2 ) , and (F3) hold. Then there exists a sequence { u n } ⊂ H satisfying

(2.11) I ( u n ) → c 2 ∗ , ∥ I ′ ( u n ) ∥ ( 1 + ∥ u n ∥ ) → 0 ,

where c 2 ∗ is given by

c 2 ∗ = inf γ ∈ Γ max t ∈ [ 0 , 1 ] I ( γ ( t ) ) , Γ = { γ ∈ C ( [ 0 , 1 ] , H ) : γ ( 0 ) = 0 , I ( γ ( 1 ) ) < 0 } .

Proof

From (F1 ′ )–(F2), for β ∈ ( β 0 , 2 π / ς 2 ] , there exists C 2 > 0 such that

(2.12) ∣ F ( x , u ) ∣ ≤ m 0 4 γ 2 2 ∣ u ∣ 2 + C 2 ∣ u ∣ 3 exp ( β ∣ u ∣ 2 + ∣ x ∣ α ) , ∀ ( x , u ) ∈ B × R ,

where ς ∈ [ ‖ u ‖ , 1 ) . Then, it follows from (1.5), (2.2), (2.12), and the Hölder inequality, one has

(2.13) ∫ B F ( x , u ) d x ≤ m 0 4 γ 2 2 ‖ u ‖ 2 2 + C 2 ∫ B ∣ u ∣ 3 exp ( β ∣ u ∣ 2 + ∣ x ∣ α ) d x

≤ m 0 4 ‖ u ‖ 2 + C 2 ‖ u ‖ 6 3 ∫ B exp ( 2 β ∣ u ∣ 2 + ∣ x ∣ α ) d x 1 / 2 ≤ m 0 4 ‖ u ‖ 2 + C 2 γ 6 3 ‖ u ‖ 3 ∫ B exp 2 β ‖ u ‖ 2 + ∣ x ∣ α u ‖ u ‖ 2 + ∣ x ∣ α d x 1 / 2 ≤ m 0 4 ‖ u ‖ 2 + MT α 2 ( 4 π ) C 2 γ 6 3 ‖ u ‖ 3 .

Hence, from (M1), (2.3), and (2.13) that

I ( u ) = 1 2 ℳ ( ‖ u ‖ 2 ) − ∫ B F ( x , u ) d x ≥ m 0 4 ‖ u ‖ 2 − MT α 2 ( 4 π ) C 2 γ 6 3 ‖ u ‖ 3 , ∀ ‖ u ‖ ≤ 1 .

Therefore, there exist σ 2 > 0 and 0 < ϱ 2 < 1 such that

I ( u ) ≥ σ 2 , ∀ u ∈ S ≔ { u ∈ H : ‖ u ‖ = ϱ 2 } .

Similar to the proof of Lemma 2.4, choose u 2 ∈ H \ { 0 } , one has I ( t u 2 ) → − ∞ as t → ∞ . Thus, we can choose T 2 > 0 such that e 2 = T 2 u 2 ∈ { u ∈ H : ‖ u ‖ > ϱ 2 } and I ( e 2 ) < 0 , then in view of the mountain pass lemma [29], we deduce that there exist c 2 ∗ ∈ [ σ 2 , sup t ≥ 0 I ( t u 2 ) ] and a sequence { u n } ⊂ H satisfying (2.11).□

Corollary 2.6

Assume that ( M 1 ) , ( M 2 ) , ( F 1 ) ( or ( F1 ′ ) ) , ( F 2 ) , and (F5) hold. Then

I ( u ) ≥ max t ≥ 0 I ( t u ) , ∀ u ∈ N .

Lemma 2.7

Assume that ( M 1 ) , ( M 2 ) , ( F 1 ) ( or ( F1 ′ ) ) , and (F2) hold. Then for any u ∈ H \ { 0 } , there exists t u > 0 such that t u u ∈ N .

From Corollary 2.6 and Lemma 2.7 we can easily prove the following lemmas.

Lemma 2.8

Assume that ( M 1 ) , ( M 2 ) , ( F 1 ) , ( F 2 ) , and ( F 5 ) hold. Then

c 1 ∗ = b 1 = inf u ∈ N I ( u ) = inf u ∈ H \ { 0 } max t ≥ 0 I ( t u ) .

Lemma 2.9

Assume that ( M 1 ) , ( M 2 ) , ( F 1 ′ ) , ( F 2 ) , and ( F 5 ) hold. Then

c 2 ∗ = b 2 = inf u ∈ N I ( u ) = inf u ∈ H \ { 0 } max t ≥ 0 I ( t u ) .

Lemma 2.10

Assume that (M1) (or ( M 1 ′ ) ), ( M 2 ) , ( F 1 ) , ( F 2 ) , and ( F 3 ) hold. Then any sequence { u n } satisfying (2.6) is bounded in H .

Lemma 2.11

Assume that (M1) (or ( M 1 ′ ) ), ( M 2 ) , ( F1 ′ ) , ( F 2 ) , and ( F 3 ) hold. Then any sequence { u n } satisfying (2.11) is bounded in H .

3 Minimax estimates

In this section, we depict the minimax estimate of the energy functional associated with problem (1.1) for MT α 1 and MT α 2 , which follows the approach of [9, Lemma 3.1] to give a more precise estimation. As in [16], we define Moser-type functions w n ( x ) supported in B as follows:

(3.0.1) w n ( x ) = 1 2 π ⋅ log n , 0 ≤ ∣ x ∣ ≤ 1 / n ; − log ∣ x ∣ log n , 1 / n ≤ ∣ x ∣ ≤ 1 .

Note that w n ∈ H , and by an elemental computation, we have

(3.0.2) ∥ w n ∥ 2 = ∫ B ∣ ∇ w n ∣ 2 d x = 1 .

3.1 Estimate with MT α 1

Lemma 3.1

Assume that ( M 1 ) , ( M 2 ) , and (F1)–(F4) hold. Then there exists n ¯ ∈ N such that

(3.1.1) c 1 ∗ ≤ max t ≥ 0 I ( t w n ¯ ) < 1 2 ℳ 4 π β 0 .

Proof

Note that κ 1 > 2 e β 0 2 M 4 π β 0 in (F4), we can choose ε > 0 such that

(3.1.2) ( κ 1 − ε ) 2 ( 1 + ε ) 2 > M 4 π β 0 e β 0 2

and

(3.1.3) log 2 ( 1 + ε ) 2 M 4 π β 0 ( κ 1 − ε ) β 0 2 < 1 − ε 1 + ε .

Using (F4), we know that there exists t ε > 0 such that

(3.1.4) t 2 F ( x , t ) ≥ ( κ 1 − ε ) exp [ ( β 0 + ∣ x ∣ α ) t 2 ] , ∀ x ∈ B , ∣ t ∣ ≥ t ε .

Let P ( n ) = 2 π ( β 0 + n − α ) − 1 / 2 . For convenience, we rewrite P ( n ) as P . There are four possible cases as follows. From now on, in the sequel, all inequalities hold for large n ∈ N .

Case (i) t ∈ [ 0 , P ] . It follows from (M1), (F3), (2.3), and (3.0.2) that

I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) ≤ 1 2 ℳ 2 π β 0 + n − α .

Clearly, there exists n ¯ ∈ N such that (3.1.1) holds.

Case (ii) t ∈ [ P , 2 P ] . Then t w n ( x ) ≥ t ε for x ∈ B 1 / n and for large n ∈ N . Let Q ( n ) = t 2 log n 2 π , which simplified to Q . Due to

(3.1.5) ∫ B 1 / n ( κ 1 − ε ) exp [ ( β 0 + ∣ x ∣ α ) t 2 w n 2 ] t 2 w n 2 d x = 2 π ( κ 1 − ε ) exp ( β 0 Q ) α Q 2 ∕ α + 1 ∫ 0 Q n α r 2 α − 1 e r d r ≥ 2 π ( κ 1 − ε ) α Q 2 ∕ α + 1 ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 Q n α 2 α − i exp [ ( β 0 + n − α ) Q ] , ∀ α ∈ ( 0 , 2 ] ; π ( κ 1 − ε ) Q n 2 exp ( β 0 Q ) , ∀ α ∈ ( 2 , + ∞ ] ,

where A 2 ∕ α − 1 i − 1 is permutation and combination. The following discussions are divided into two situations: (1) α ∈ ( 0 , 2 ] ; (2) α ∈ ( 2 , + ∞ ] .

(1) α ∈ ( 0 , 2 ] . From (F4), (3.1.4), and (3.1.5) it follows that

(3.1.6) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 1 − ε ) exp [ ( β 0 + ∣ x ∣ α ) t 2 w n 2 ] t 2 w n 2 d x = 2 π ( κ 1 − ε ) α Q 2 ∕ α + 1 ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 Q n α 2 α − i exp [ ( β 0 + n − α ) Q ] + ∫ 1 2 1 2 π ( κ 1 − ε ) exp [ ( β 0 + n − α r ) Q r 2 ] log n Q n 2 r r 2 d r ≥ 2 π ( κ 1 − ε ) α Q 2 ∕ α + 1 ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 Q n α 2 α − i exp [ ( β 0 + n − α ) Q ] − 2 π ( κ 1 − ε ) α Q 2 ∕ α + 1 ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 Q n α 2 α − i exp [ ( β 0 + n − α ) Q ] + 2 π 2 ( κ 1 − ε ) P − 2 ∫ 1 2 1 n [ ( 2 π ) − 1 ( β 0 + n − α r ) t 2 r − 2 ] r d r ≥ 2 π ( κ 1 − ε ) α ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 n α i − 2 ( P 2 log n / π ) i + 1 exp [ ( β 0 + n − α ) Q ] − 2 π ( κ 1 − ε ) α ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 n α i − 2 ( P 2 log n / ( 2 π ) ) i + 1 exp [ ( β 0 + n − α ) Q ] + 2 π 2 ( κ 1 − ε ) P − 2 ∫ 1 2 1 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 r − 2 d r = ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 2 π ( κ 1 − ε ) n α i − 2 α [ 2 ( β 0 + n − α ) − 1 log n ] i + 1 + 2 π 2 ( κ 1 − ε ) t 2 n 2 log n n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − 2 π ( κ 1 − ε ) α ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 n α i − 2 [ ( β 0 + n − α ) − 1 log n ] i + 1 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − 2 π 2 ( κ 1 − ε ) t 2 n 2 log n n ( 4 π ) − 1 ( β 0 + n − α ) t 2 ≥ π ( κ 1 − ε ) ( β 0 + n − α ) α n 2 log n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 ( β 0 n α + 1 ) i ( 2 log n ) i + α 2 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − 2 π ( κ 1 − ε ) ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 ( β 0 n α + 1 ) i + 1 α n 2 + α log i + 1 n n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − π ( κ 1 − ε ) ( β 0 + n − α ) n log n ≥ π ( κ 1 − ε ) ( β 0 + n − α ) α n 2 log n ∑ i = 2 k + 1 2 ∕ α 1 2 i − 2 ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 ( β 0 n α + 1 ) i log i n + α 2 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − O 1 n log n .

Let

R ( n ) ≔ ∑ i = 2 k + 1 2 ∕ α 1 2 i − 2 ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 ( β 0 n α + 1 ) i log i n + α 2 > α 2 .

It follows from (2.3) and (3.1.6) that

(3.1.7) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 1 − ε ) ( β 0 + n − α ) R ( n ) α n 2 log n exp β 0 + 1 n α t 2 log n 2 π + O 1 n log n ≔ φ n ( t ) + O 1 n log n .

Let t n > 0 such that φ n ′ ( t n ) = 0 . Then

M ( t n 2 ) = ( κ 1 − ε ) ( β 0 + n − α ) 2 R ( n ) α n 2 exp β 0 + 1 n α t n 2 log n 2 π ,

which yields

(3.1.8) lim n → ∞ t n 2 = lim n → ∞ 4 π β 0 + 1 n α − 1 = 4 π β 0 , ∀ α ∈ ( 0 , 2 ] .

Let

(3.1.9) D 1 = log 2 ( 1 + ε ) M 4 π β 0 ( κ 1 − ε ) β 0 2 ,

then (3.1.2) and (3.1.3) imply

(3.1.10) ( 1 + ε ) max { D 1 , 0 } − ( 1 − ε ) < 0 .

From (M1), (M2), (3.1.8), and (3.1.9), we have

t n 2 = 4 π β 0 + n − α 1 + log ( α M ( t n 2 ) ) − log ( ( κ 1 − ε ) ( β 0 + n − α ) 2 R ( n ) ) 2 log n ≤ 4 π β 0 + n − α + 2 π D 1 ( β 0 + n − α ) log n ,

which together with (3.1.7), (3.1.8), and Proposition 2.2(iii) that

(3.1.11) φ n ( t ) ≤ φ n ( t n ) = 1 2 ℳ ( t n 2 ) − π M ( t n 2 ) ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + 2 π D 1 ( β 0 + n − α ) log n − π ( 1 − ε ) M 4 π β 0 ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + ( 1 + ε ) max { D 1 , 0 } − ( 1 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 .

Hence, from (3.1.7), (3.1.10), and (3.1.11), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + n − α + ( 1 + ε ) max { D 1 , 0 } − ( 1 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 + O 1 n log n ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.1.1) holds.

(2) α ∈ ( 2 , + ∞ ] . It follows from (F4), (3.1.4), and (3.1.5) that

(3.1.12) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 1 − ε ) exp [ ( β 0 + ∣ x ∣ α ) t 2 w n 2 ] t 2 w n 2 d x ≥ π ( κ 1 − ε ) Q n 2 exp ( β 0 Q ) + ∫ 1 2 1 2 π ( κ 1 − ε ) exp [ ( β 0 + n − α r ) Q r 2 ] log n Q n 2 r r 2 d r ≥ 2 π 2 ( κ 1 − ε ) t 2 n 2 log n n ( 2 π ) − 1 β 0 t 2 + 2 π 2 ( κ 1 − ε ) P 2 ∫ 1 2 1 n [ ( 2 π ) − 1 ( β 0 + n − α r ) t 2 r − 2 ] r d r

≥ π ( κ 1 − ε ) ( β 0 + n − α ) n log n + 2 π 2 ( κ 1 − ε ) P 2 ∫ 1 2 1 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 r − 2 d r = π ( κ 1 − ε ) ( β 0 + n − α ) n log n + 2 π 2 ( κ 1 − ε ) t 2 n 2 log n [ n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − n ( 4 π ) − 1 ( β 0 + n − α ) t 2 ] ≥ π ( κ 1 − ε ) ( β 0 + n − α ) 2 n 2 log n n ( 2 π ) − 1 ( β 0 + n − α ) t 2 .

Then, it follows from (2.3) and (3.1.12) that

(3.1.13) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 1 − ε ) ( β 0 + n − α ) 2 n 2 log n exp β 0 + 1 n α t 2 log n 2 π ≔ φ ^ n ( t ) .

Let t ˆ n > 0 such that φ ^ n ′ ( t ˆ n ) = 0 . Hence,

(3.1.14) M ( t ˆ n 2 ) = ( κ 1 − ε ) ( β 0 + n − α ) 2 2 n 2 exp β 0 + 1 n α t ˆ n 2 log n 2 π ,

which yields

(3.1.15) lim n → ∞ t ˆ n 2 = lim n → ∞ 4 π β 0 + 1 n α − 1 = 4 π β 0 , ∀ α ∈ ( 2 , + ∞ ] .

From (M1), (M2), (3.1.9), and (3.1.15), we have

t ˆ n 2 = 4 π β 0 + n − α 1 + log ( 2 M ( t ˆ n 2 ) ) − log ( ( κ 1 − ε ) ( β 0 + n − α ) 2 ) 2 log n ≤ 4 π β 0 + n − α + 2 π D 1 ( β 0 + n − α ) log n ,

which together with (3.1.13), (3.1.14), and Proposition 2.2 (iii) that

(3.1.16) φ ^ n ( t ) ≤ φ ^ n ( t ˆ n ) = 1 2 ℳ ( t ˆ n 2 ) − π M ( t ˆ n 2 ) ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + 2 π D 1 ( β 0 + n − α ) log n − π ( 1 − ε ) M 4 π β 0 ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + ( 1 + ε ) max { D 1 , 0 } − ( 1 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 .

Hence, from (3.1.10), (3.1.13), and (3.1.16), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + n − α + ( 1 + ε ) max { D 1 , 0 } − ( 1 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.1.1) holds.

Case (iii) t ∈ [ 2 P , 2 ( 1 + ε ) P ] . Then t w n ( x ) ≥ t ε for x ∈ B 1 / n and for large n ∈ N . In this case, the following discussions are also divided into two situations: (1) α ∈ ( 0 , 2 ] ; (2) α ∈ ( 2 , + ∞ ] .

(1) α ∈ ( 0 , 2 ] . It follows from (F4), (3.1.4), and (3.1.5) that

(3.1.17) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 1 − ε ) exp [ ( β 0 + ∣ x ∣ α ) t 2 w n 2 ] t 2 w n 2 d x ≥ 2 π ( κ 1 − ε ) α Q 2 ∕ α + 1 ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 Q n α 2 α − i exp [ ( β 0 + n − α ) Q ] − 2 π ( κ 1 − ε ) α Q 2 ∕ α + 1 ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 Q n α 2 α − i exp [ ( β 0 + n − α ) Q ] + 2 π 2 ( κ 1 − ε ) ( 1 + ε ) n 2 P 2 ∫ 1 2 1 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 r 2 − 2 r + 2 d r ≥ 2 π ( κ 1 − ε ) α ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 n α i − 2 ( ( 1 + ε ) P 2 log n / π ) i + 1 exp [ ( β 0 + n − α ) Q ] − 2 π ( κ 1 − ε ) α ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 n α i − 2 ( P 2 log n / π ) i + 1 exp [ ( β 0 + n − α ) Q ] + 2 π 2 ( κ 1 − ε ) ( 1 + ε ) n 2 P 2 ∫ 1 − ε 1 n [ ( 1 − ε ) ( 2 π ) − 1 ( β 0 + n − α ) t 2 + 2 ε ] r d r ≥ ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 2 π ( κ 1 − ε ) α − 1 n α i − 2 [ 2 ( 1 + ε ) ( β 0 + n − α ) − 1 log n ] i + 1 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 + π 2 ( κ 1 − ε ) ( 1 + ε ) ( 1 − ε 2 + ε ) n 2 P 2 log n n ( 1 − ε ) ( 2 π ) − 1 ( β 0 + n − α ) t 2 + 2 ε − 2 π ( κ 1 − ε ) α ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 n α i − 2 [ 2 ( β 0 + n − α ) − 1 log n ] i + 1 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − π 2 ( κ 1 − ε ) ( 1 + ε ) ( 1 − ε 2 + ε ) n 2 P 2 log n n ( 1 − ε ) 2 ( 2 π ) − 1 ( β 0 + n − α ) t 2 + 2 ε ( 1 − ε ) ≥ π ( κ 1 − ε ) ( β 0 + n − α ) α ( 1 + ε ) n 2 log n ∑ i = 2 k + 1 2 ∕ α − ∑ i = 2 k 2 ∕ α ( 1 + ε ) i + 1 A 2 ∕ α − 1 i − 1 ( β 0 n α + 1 ) i [ 2 ( 1 + ε ) log n ] i + α 2 ( 1 − ε 2 + ε ) n 2 ε 2 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 − O 1 n 2 ε 2 log n .

Let

R ˜ ( n ) ≔ ∑ i = 2 k + 1 2 ∕ α − ∑ i = 2 k 2 ∕ α ( 1 + ε ) i + 1 A 2 ∕ α − 1 i − 1 ( β 0 n α + 1 ) i [ 2 ( 1 + ε ) log n ] i + α 2 ( 1 − ε 2 + ε ) n 2 ε 2 > α 2 .

It follows from (2.3) and (3.1.17) that

(3.1.18) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 1 − ε ) ( β 0 + n − α ) R ˜ ( n ) α ( 1 + ε ) n 2 log n exp β 0 + 1 n α t 2 log n 2 π + O 1 n 2 ε 2 log n ≔ φ ˜ n ( t ) + O 1 n 2 ε 2 log n .

Let t ˜ n > 0 such that φ ˜ n ′ ( t ˜ n ) = 0 . Then

(3.1.19) M ( t ˜ n 2 ) = ( κ 1 − ε ) ( β 0 + n − α ) 2 R ˜ ( n ) α ( 1 + ε ) n 2 exp β 0 + 1 n α t ˜ n 2 log n 2 π ,

which yields

(3.1.20) lim n → ∞ t ˜ n 2 = lim n → ∞ 4 π β 0 + 1 n α − 1 = 4 π β 0 , ∀ α ∈ ( 0 , 2 ] .

Let

(3.1.21) D ˜ 1 = log 2 ( 1 + ε ) 2 M 4 π β 0 ( κ 1 − ε ) β 0 2 ,

then (3.1.2) and (3.1.3) imply

(3.1.22) ( 1 + ε ) max { D ˜ 1 , 0 } − ( 1 − ε ) < 0 .

From (M1), (M2), (3.1.20), and (3.2.11), we have

t ˜ n 2 = 4 π β 0 + n − α 1 + log ( α ( 1 + ε ) M ( t ˜ n 2 ) ) − log ( ( κ 1 − ε ) ( β 0 + n − α ) 2 R ˜ ( n ) ) 2 log n ≤ 4 π β 0 + n − α + 2 π D ˜ 1 ( β 0 + n − α ) log n ,

which together with (3.1.18), (3.1.19), and Proposition 2.2(iii) give

(3.1.23) φ ˜ n ( t ) ≤ φ ˜ n ( t ˜ n ) = 1 2 ℳ ( t ˜ n 2 ) − π M ( t ˜ n 2 ) ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + 2 π D ˜ 1 ( β 0 + n − α ) log n − π ( 1 − ε ) M 4 π β 0 ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + ( 1 + ε ) max { D ˜ 1 , 0 } − ( 1 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 .

Hence, from (3.1.18), (3.1.22), and (3.1.23), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + n − α + ( 1 + ε ) max { D ˜ 1 , 0 } − ( 1 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 + O 1 n 2 ε 2 log n ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.1.1) holds.

(2) α ∈ ( 2 , + ∞ ] . It follows from (F4), (3.1.4), and (3.1.5) that

(3.1.24) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 1 − ε ) exp [ ( β 0 + ∣ x ∣ α ) t 2 w n 2 ] t 2 w n 2 d x ≥ π ( κ 1 − ε ) Q n 2 exp ( β 0 Q ) + 2 π 2 ( κ 1 − ε ) ( 1 + ε ) n 2 P 2 ∫ 1 2 1 n ( 2 π ) − 1 ( β 0 + n − α ) t 2 r 2 − 2 r + 2 d r

≥ 2 π 2 ( κ 1 − ε ) t 2 n 2 log n n ( 2 π ) − 1 β 0 t 2 + 2 π 2 ( κ 1 − ε ) ( 1 + ε ) n 2 P 2 ∫ 1 − ε 1 n [ ( 1 − ε ) ( 2 π ) − 1 ( β 0 + n − α ) t 2 + 2 ε ] r d r ≥ π 2 ( κ 1 − ε ) ( 1 + ε ) P 2 n 2 log n n ( 2 π ) − 1 ( β 0 + n − α ) t 2 + 1 ( 1 − ε 2 + ε ) n ( 1 − ε ) ( 2 π ) − 1 ( β 0 + n − α ) t 2 + 2 ε − O 1 n 2 ε 2 log n ≥ π ( κ 1 − ε ) ( β 0 + n − α ) ( 1 + ε ) 3 / 2 n 2 − ε log n n ( 2 − ε ) ( 4 π ) − 1 ( β 0 + n − α ) t 2 − O 1 n 2 ε 2 log n .

It follows from (2.3) and (3.1.24) that

(3.1.25) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 1 − ε ) ( β 0 + n − α ) ( 1 + ε ) 3 / 2 n 2 − ε log n exp β 0 + 1 n α ( 2 − ε ) t 2 log n 4 π + O 1 n 2 ε 2 log n ≔ φ ¯ n ( t ) + O 1 n 2 ε 2 log n .

Let t ¯ n > 0 such that φ ¯ n ′ ( t ¯ n ) = 0 . Then

(3.1.26) M ( t ¯ n 2 ) = ( 2 − ε ) ( κ 1 − ε ) ( β 0 + n − α ) 2 2 ( 1 + ε ) 3 / 2 n 2 − ε exp β 0 + 1 n α ( 2 − ε ) t ¯ n 2 log n 4 π ,

which yields

(3.1.27) lim n → ∞ t ¯ n 2 = lim n → ∞ 4 π β 0 + 1 n α − 1 = 4 π β 0 , ∀ α ∈ ( 2 , + ∞ ] .

Let

(3.1.28) D ¯ 1 = log 2 ( 1 + ε ) 2 M 4 π β 0 ( 2 − ε ) ( κ 1 − ε ) β 0 2 ,

then (3.1.2) and (3.1.3) imply

(3.1.29) ( 1 + ε ) max { D ¯ 1 , 0 } − ( 1 − ε ) < 0 .

From (M1), (M2), (3.1.27), and (3.1.28), we have

t ¯ n 2 = 4 π β 0 + n − α 1 + log ( 2 ( 1 + ε ) 3 / 2 M ( t ¯ n 2 ) ) − log ( ( 2 − ε ) ( κ 1 − ε ) ( β 0 + n − α ) 2 ) ( 2 − ε ) log n ≤ 4 π β 0 + n − α + 4 π D ¯ 1 ( 2 − ε ) ( β 0 + n − α ) log n ,

which together with (3.1.25), (3.1.26), and Proposition 2.2(iii) give

(3.1.30) φ ¯ n ( t ) ≤ φ ¯ n ( t ¯ n ) = 1 2 ℳ ( t ¯ n 2 ) − 2 π M ( t ¯ n 2 ) ( 2 − ε ) ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + 4 π D ¯ 1 ( 2 − ε ) ( β 0 + n − α ) log n − 2 π ( 1 − ε ) M 4 π β 0 ( 2 − ε ) ( β 0 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + n − α + 2 [ ( 1 + ε ) max { D ¯ 1 , 0 } − ( 1 − ε ) ] ( 2 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 .

Hence, from (3.1.25), (3.1.29), and (3.1.30), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + n − α + 2 [ ( 1 + ε ) max { D ¯ 1 , 0 } − ( 1 − ε ) ] ( 2 − ε ) ( β 0 + n − α ) log n π M 4 π β 0 + O 1 n 2 ε 2 log n ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.1.1) holds.

Case (iv) t ∈ ( 2 ( 1 + ε ) P , + ∞ ) . Then t w n ( x ) ≥ t ε for x ∈ B 1 / n and for large n ∈ N . In this case, the following discussions are also divided into two situations: (1) α ∈ ( 0 , 2 ] ; (2) α ∈ ( 2 , + ∞ ] .

(1) α ∈ ( 0 , 2 ] . It follows from (2.3), (3.0.2), and (3.1.5) that

I ( t w n ) = 1 2 ℳ ( t 2 w n 2 ) − ∫ B 1 / n F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − 2 π ( κ 1 − ε ) α Q 2 ∕ α + 1 ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 Q n α 2 α − i exp [ ( β 0 + n − α ) Q ] = 1 2 ℳ ( t 2 ) − 2 π ( κ 1 − ε ) α ∑ i = 1 2 ∕ α A 2 ∕ α − 1 i − 1 − 2 π t 2 log n i + 1 n α i − 2 exp ( β 0 + n − α ) t 2 log n 2 π ≤ 1 2 ℳ 4 π ( 1 + ε ) β 0 + n − α − 2 π ( κ 1 − ε ) α ∑ i = 1 2 ∕ α A 2 ∕ α − 1 i − 1 − β 0 + n − α 2 ( 1 + ε ) log n i + 1 n α i + 2 ε ,

which implies that there exists n ¯ ∈ N such that (3.1.1) holds. In the above derivation process, we use the fact that the function

1 2 ℳ ( t 2 ) − 2 π ( κ 1 − ε ) α ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 2 π t 2 log n i + 1 n α i − 2 exp ( β 0 + n − α ) t 2 log n 2 π

is decreasing on t ∈ ( 2 ( 1 + ε ) P , + ∞ ) , since its stagnation points tend to P as n → + ∞ .

(2) α ∈ ( 2 , + ∞ ] . It follows from (2.3), (3.0.2), and (3.1.5) that

I ( t w n ) = 1 2 ℳ ( t 2 w n 2 ) − ∫ B 1 / n F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 1 − ε ) Q n 2 exp ( β 0 Q ) = 1 2 ℳ ( t 2 ) − 2 π 2 ( κ 1 − ε ) t 2 n 2 log n exp β 0 t 2 log n 2 π ≤ 1 2 ℳ 4 π ( 1 + ε ) β 0 + n − α − π ( κ 1 − ε ) ( β 0 + n − α ) 2 ( 1 + ε ) n 2 log n exp 2 ( 1 + ε ) β 0 log n β 0 + n − α ,

which implies that there exists n ¯ ∈ N such that (3.1.1) holds. In the above derivation process, we use the fact that the function

1 2 ℳ ( t 2 ) − 2 π 2 ( κ 1 − ε ) t 2 n 2 log n exp β 0 t 2 log n 2 π

is decreasing on t ∈ ( 2 ( 1 + ε ) P , + ∞ ) , since its stagnation points tend to P as n → + ∞ .□

3.2 Estimate with MT α 2

Different from the test function ω n defined in [20, Lemma 4.3] by

(3.2.1) ω n ( x ) = c + 1 4 π c [ 1 − log ( 1 + π n 2 ∣ x ∣ 2 ) ] , 0 ≤ ∣ x ∣ ≤ log n n ; − 1 2 π c log ∣ x ∣ , log n n ≤ ∣ x ∣ ≤ 1 ,

where c ≔ C 1 n > 0 and

C 2 1 n = 1 2 π log n + 1 4 π log π − 1 4 π + O 1 log 2 n .

We still use Moser-type functions w n ( x ) , which are given in (3.0.1). It is easier to estimate energy functional I in our following lemma instead of using (3.2.1).

Lemma 3.2

Assume that (M1), (M2), ( F 1 ′ ) , (F2), (F3), and ( F 4 ′ ) hold. Then there exists n ̲ > 0 such that

(3.2.2) c 2 ∗ ≤ max t ≥ 0 I ( t ω n ̲ ) < 1 2 ℳ 4 π β 0 .

Proof

Note that κ 2 > 2 e β 0 2 M 4 π β 0 in (F4 ′ ), we can choose ε > 0 such that

(3.2.3) κ 2 − ε 2 ( 1 + ε ) > M 4 π β 0 e β 0 2

and

(3.2.4) log 2 ( 1 + ε ) M 4 π β 0 ( κ 2 − ε ) β 0 2 < 1 − ε 1 + ε .

Using (F4 ′ ), we know that there exists t ε > 0 such that

(3.2.5) t 2 F ( x , t ) ≥ ( κ 2 − ε ) exp ( β 0 t 2 + ∣ x ∣ α ) , ∀ x ∈ B , ∣ t ∣ ≥ t ε .

Let P ˜ ≔ P ˜ ( n ) = 2 π ( 2 / β 0 ) ( 2 + n − α ) − 1 , there are four possible cases as follows. From now on, in the sequel, all inequalities hold for large n ∈ N .

Case (i) t ∈ [ 0 , P ˜ / 2 ] . It follows from (M1), (F3), (2.3), and (3.0.2) that

I ( t ω n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) ≤ 1 2 ℳ 2 π β 0 .

Clearly, there exists n ̲ ∈ N such that (3.2.2) holds.

Case (ii) t ∈ [ P ˜ / 2 , P ˜ ] . Then t w n ( x ) ≥ t ε for x ∈ B 1 / n and for large n ∈ N . Recall Q = t 2 log n 2 π . Due to

(3.2.6) ∫ B 1 / n ( κ 2 − ε ) exp [ β 0 ( t w n ) 2 + ∣ x ∣ α ] t 2 w n 2 d x = ∑ j = 1 n 2 π ( κ 2 − ε ) β 0 j Q j − 1 α j ! j 2 ∕ α ∫ 0 j n α r 2 α − 1 Q r / 2 d r ≥ 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q , ∀ α ∈ ( 0 , 2 ] ; π ( κ 2 − ε ) ∑ j = 1 n β 0 j Q j − 1 j ! n 2 , ∀ α ∈ ( 2 , + ∞ ] ,

where A 2 ∕ α − 1 i − 1 is permutation and combination. So the following discussions are divided into two situations: (1) α ∈ ( 0 , 2 ] ; (2) α ∈ ( 2 , + ∞ ] .

(1) α ∈ ( 0 , 2 ] . From (F4 ′ ), (3.2.5), and (3.2.6) that

(3.2.7) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 2 − ε ) exp [ β 0 ( t w n ) 2 + ∣ x ∣ α ] t 2 w n 2 d x ≥ 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q + 2 π ( κ 2 − ε ) log n Q ∫ 1 2 1 exp [ β 0 ( Q 1 / 2 r ) 2 + n − α r ] n 2 r r 2 d r ≥ 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q + 2 π ( κ 2 − ε ) log n Q ∫ 1 2 1 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α r 2 − 2 r d r ≥ 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i ( P ˜ 2 log n / ( 2 π ) ) − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i ( P ˜ 2 log n / ( 4 π ) ) + 4 π 2 ( κ 2 − ε ) P ˜ 2 ∫ 1 2 1 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α r − 2 d r ≥ ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 2 π ( κ 2 − ε ) β 0 j Q j − 1 α j ! ( j / 2 ) i n 4 − α i log i ( 2 log n / β 0 ) + 2 π 2 ( κ 2 − ε ) t 2 n 2 log n × n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − 2 π 2 ( κ 2 − ε ) t 2 n 2 log n n 2 − 1 β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i ( log n / β 0 ) ≥ π ( κ 2 − ε ) β 0 α n 2 ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 2 log j − 1 n j ! ( j / 2 ) i n 2 − α i log i ( 2 log n / β 0 ) + α 2 log n × n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − π ( κ 2 − ε ) β 0 2 n log n − 2 π ( κ 2 − ε ) ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 ( 2 log n ) j − 1 α j ! ( j / 2 ) i n 2 − α i log i ( log n / β 0 ) ≥ ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α 1 2 j − 1 − ∑ i = 2 k 2 ∕ α 2 j A 2 ∕ α − 1 i − 1 log j − 1 n j ! ( j / 2 ) i n 2 − α i log i − 1 ( log n / β 0 ) + α 2 π ( κ 2 − ε ) β 0 α n 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − O 1 n log n .

Let

S ( n ) ≔ ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α 1 2 j − 1 − ∑ i = 2 k 2 ∕ α 2 j A 2 ∕ α − 1 i − 1 log j − 1 n j ! ( j / 2 ) i n 2 − α i log i − 1 ( log n / β 0 ) + α 2 > α 2 .

It follows from (2.3) and (3.2.7) that

(3.2.8) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 2 − ε ) β 0 S ( n ) α n 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + O 1 n log n ≔ ϕ n ( t ) + O 1 n log n .

Let t n > 0 such that ϕ n ′ ( t n ) = 0 . Then

(3.2.9) M ( t n 2 ) = π ( κ 2 − ε ) β 0 2 ( 2 + n − α ) S ( n ) t n n − α α ( 2 π ) 1 + 2 − 1 n − α n 2 n β 0 ( ( 2 π ) − 1 / 2 t n ) 2 + n − α ,

which yields

(3.2.10) lim n → ∞ t n 2 = lim n → ∞ t n 2 + n − α = lim n → ∞ 2 ( 2 π ) 1 + 2 − 1 n − α β 0 = 4 π β 0 , ∀ α ∈ ( 0 , 2 ] .

Let

(3.2.11) D 2 = log 2 ( 1 + ε ) M 4 π β 0 ( κ 2 − ε ) β 0 2 ,

then (3.2.3) and (3.2.4) imply

(3.2.12) ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) < 0 .

Then from (M1), (M2), (3.2.10), and (3.2.11), we have

t n 2 = 4 π β 0 1 + log ( α ( 2 π ) 1 + 2 − 1 n − α M ( t n 2 ) ) − log ( π ( κ 2 − ε ) ( 2 + n − α ) β 0 2 S ( n ) ) 2 log n ≤ 4 π β 0 + 2 π D 2 β 0 log n ,

which together with (3.2.8), (3.2.9), and Proposition 2.2(iii) have

(3.2.13) ϕ n ( t ) ≤ ϕ n ( t n ) = 1 2 ℳ ( t n 2 ) − ( 2 π ) 1 + 2 − 1 n − α M ( t n 2 ) β 0 t n n − α ( 2 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + 2 π D 2 β 0 log n − π ( 1 − ε ) M 4 π β 0 β 0 log n ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 .

Hence, from (3.2.12), (3.1.10), (3.2.8), and (3.2.13), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 + O 1 n log n ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.2.2) holds.

(2) α ∈ ( 2 , + ∞ ] . It follows from (F4 ′ ), (3.2.5), and (3.2.6) that

(3.2.14) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 2 − ε ) exp [ β 0 ( t w n ) 2 + ∣ x ∣ α ] t 2 w n 2 d x ≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 j Q j − 1 j ! n 2 + 2 π ( κ 2 − ε ) log n Q ∫ 1 2 1 exp [ β 0 ( Q 1 / 2 r ) 2 + n − α r ] n 2 r r 2 d r

≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 j ( t 2 log n / ( 2 π ) ) j − 1 j ! n 2 + 2 π ( κ 2 − ε ) log n Q ∫ 1 2 1 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α r 2 − 2 r d r ≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 j ( P ˜ 2 log n / ( 4 π ) ) j − 1 j ! n 2 + 4 π 2 ( κ 2 − ε ) P ˜ 2 ∫ 1 2 1 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α r − 2 d r ≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 log j − 1 n j ! n 2 + 2 π 2 ( κ 2 − ε ) t 2 n 2 log n [ n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − n 2 − 1 β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α ] ≥ π ( κ 2 − ε ) β 0 n 2 log n ∑ j = 1 n log j n j ! + 1 2 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − O 1 n log n .

Let

S ^ ( n ) ≔ ∑ j = 1 n log j n j ! + 1 2 > 1 2 .

It follows from (2.3) and (3.2.14) that

(3.2.15) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 2 − ε ) β 0 S ^ ( n ) n 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + O 1 n log n ≔ ϕ ^ n ( t ) + O 1 n log n .

Let t ˆ n > 0 such that ϕ ^ n ′ ( t ˆ n ) = 0 . Then

(3.2.16) M ( t ˆ n 2 ) = π ( κ 2 − ε ) β 0 2 ( 2 + n − α ) S ^ ( n ) t ˆ n n − α ( 2 π ) 1 + 2 − 1 n − α n 2 n β 0 ( ( 2 π ) − 1 / 2 t ˆ n ) 2 + n − α ,

which yields

(3.2.17) lim n → ∞ t ˆ n 2 = lim n → ∞ t ˆ n 2 + n − α = lim n → ∞ 2 ( 2 π ) 1 + 2 − 1 n − α β 0 = 4 π β 0 , ∀ α ∈ ( 2 , + ∞ ] .

Recall (3.2.11), then it follows from (M1), (M2), and (3.2.17) that

t ˆ n 2 = 4 π β 0 1 + log ( ( 2 π ) 1 + 2 − 1 n − α M ( t ˆ n 2 ) ) − log ( π ( κ 2 − ε ) β 0 2 t ˆ n n − α ( 2 + n − α ) S ^ ( n ) ) 2 log n ≤ 4 π β 0 + 2 π D 2 β 0 log n ,

which together with (3.2.15), (3.2.16), and Proposition 2.2(iii) that

(3.2.18) ϕ ^ n ( t ) ≤ ϕ ^ n ( t ˆ n ) = 1 2 ℳ ( t ˆ n 2 ) − ( 2 π ) 1 + 2 − 1 n − α M ( t ˆ n 2 ) β 0 t ˆ n n − α ( 2 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + 2 π D 2 β 0 log n − π ( 1 − ε ) M 4 π β 0 β 0 log n ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 .

Hence, in virtue of (3.1.10), (3.2.15), and (3.2.18), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 + O 1 n log n ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.2.2) holds.

Case (iii) t ∈ [ P ˜ , 1 + ε P ˜ ] . Then t w n ( x ) ≥ t ε for x ∈ B 1 / n and for large n ∈ N . In this case, the following discussions are also divided into two situations: (1) α ∈ ( 0 , 2 ] ; (2) α ∈ ( 2 , + ∞ ] .

(1) α ∈ ( 0 , 2 ] . It follows from (F4 ′ ), (3.2.5), and (3.2.6) that

(3.2.19) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 2 − ε ) exp [ β 0 ( t w n ) 2 + ∣ x ∣ α ] t 2 w n 2 d x ≥ 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q + 2 π ( κ 2 − ε ) log n Q ∫ 1 2 1 exp [ β 0 ( Q 1 / 2 r ) 2 + n − α r ] n 2 r r 2 d r ≥ 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q + 2 π ( κ 2 − ε ) log n Q n 2 ∫ 1 2 1 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α r 2 − 2 r + 2 d r ≥ 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i ( ( 1 + ε ) P ˜ 2 log n / ( 2 π ) ) − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 × β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i ( P ˜ 2 log n / ( 2 π ) ) + 4 π 2 ( κ 2 − ε ) ( 1 + ε ) P ˜ 2 n 2 ∫ 1 − ε 1 n [ ( 1 − ε ) β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + 2 ε ] r d r ≥ ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 2 π ( κ 2 − ε ) β 0 j Q j − 1 α j ! ( j / 2 ) i n 4 − α i log i ( 2 ( 1 + ε ) log n / β 0 ) + π ( κ 2 − ε ) β 0 2 ( 1 − ε 3 + 2 ε ) n 2 + 2 ε 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − π ( κ 2 − ε ) β 0 2 ( 1 − ε 3 + 2 ε ) n 2 log n n ( 1 − ε ) 2 β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + 2 ε ( 1 − ε ) − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i ( 2 log n / β 0 )

≥ π ( κ 2 − ε ) β 0 α n 2 ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α A 2 ∕ α − 1 i − 1 2 j log j − 1 n j ! ( j / 2 ) i n 2 − α i log i ( 2 ( 1 + ε ) log n / β 0 ) + α 2 ( 1 − ε 3 + 2 ε ) n 2 ε 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − π ( κ 2 − ε ) β 0 2 ( 1 − ε 3 + 2 ε ) n 4 ε 2 log n − 2 π ( κ 2 − ε ) ∑ j = 1 n ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 β 0 ( 2 ( 1 + ε ) log n ) j − 1 α j ! ( j / 2 ) i n 2 − α i log i ( 2 log n / β 0 ) ≥ π ( κ 2 − ε ) β 0 α n 2 log n ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α − ( 1 + ε ) j − 1 ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 2 j log j − 1 n j ! ( j / 2 ) i n 2 − α i log i − 1 ( 2 log n / β 0 ) + α 2 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − O 1 n 4 ε 2 log n .

Let

S ˜ ( n ) ≔ ∑ j = 1 n ∑ i = 2 k + 1 2 ∕ α − ( 1 + ε ) j − 1 ∑ i = 2 k 2 ∕ α A 2 ∕ α − 1 i − 1 2 j log j − 1 n j ! ( j / 2 ) i n 2 − α i log i − 1 ( 2 log n / β 0 ) + α 2 > α 2 .

It follows from (2.3) and (3.2.19) that

(3.2.20) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 2 − ε ) β 0 S ˜ ( n ) α n 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + O 1 n 4 ε 2 log n ≔ ϕ ˜ n ( t ) + O 1 n 4 ε 2 log n .

Let t ˜ n > 0 such that ϕ ˜ n ′ ( t ˜ n ) = 0 . Then,

(3.2.21) M ( t ˜ n 2 ) = π ( κ 2 − ε ) β 0 2 ( 2 + n − α ) S ˜ ( n ) t ˜ n n − α α ( 2 π ) 1 + 2 − 1 n − α n 2 n β 0 ( ( 2 π ) − 1 / 2 t ˜ n ) 2 + n − α ,

which yields

(3.2.22) lim n → ∞ t ˜ n 2 = lim n → ∞ t ˜ n 2 + n − α = lim n → ∞ 2 ( 2 π ) 1 + 2 − 1 n − α β 0 = 4 π β 0 , ∀ α ∈ ( 0 , 2 ] .

Combining with (M1), (M2), (3.1.9), and (3.2.22), we have

t ˜ n 2 = 4 π β 0 1 + log ( α ( 2 π ) 1 + 2 − 1 n − α M ( t ˜ n 2 ) ) − log ( π ( κ 2 − ε ) β 0 2 ( 2 + n − α ) S ˜ ( n ) ) 2 log n ≤ 4 π β 0 + 2 π D 2 β 0 log n ,

which together with (3.2.20), (3.2.21), and Proposition 2.2(iii) that

(3.2.23) ϕ ˜ n ( t ) ≤ ϕ ˜ n ( t ˜ n ) = 1 2 ℳ ( t ˜ n 2 ) − ( 2 π ) 1 + 2 − 1 n − α M ( t ˜ n 2 ) β 0 t ˜ n n − α ( 2 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + 2 π D 2 β 0 log n − π ( 1 − ε ) M 4 π β 0 β 0 log n ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 .

Hence, from (3.2.12), (3.2.20), and (3.2.23), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 + O 1 n 4 ε 2 log n ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.2.2) holds.

(2) α ∈ ( 2 , + ∞ ] . It follows from (F4 ′ ), (3.2.5), and (3.2.6) that

(3.2.24) ∫ B F ( x , t w n ) d x ≥ ∫ B 1 / n F ( x , t w n ) d x ≥ ∫ B 1 / n ( κ 2 − ε ) exp [ β 0 ( t w n ) 2 + ∣ x ∣ α ] t 2 w n 2 d x ≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 j Q j − 1 j ! n 2 + 2 π ( κ 2 − ε ) log n Q ∫ 1 2 1 exp [ β 0 ( Q 1 / 2 r ) 2 + n − α r ] n 2 r r 2 d r ≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 j ( t 2 log n / ( 2 π ) ) j − 1 j ! n 2 + 2 π ( κ 2 − ε ) log n Q n 2 ∫ 1 2 1 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α r 2 − 2 r + 2 d r ≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 j ( P ˜ 2 log n / ( 2 π ) ) j − 1 j ! n 2 + 4 π 2 ( κ 2 − ε ) ( 1 + ε ) P ˜ 2 n 2 ∫ 1 − ε 1 n [ ( 1 − ε ) β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + 2 ε ] r d r ≥ π ( κ 2 − ε ) ∑ j = 1 n β 0 ( 2 log n ) j − 1 j ! n 2 + π ( κ 2 − ε ) β 0 2 ( 1 − ε 3 + 2 ε ) n 2 + 2 ε 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − π ( κ 2 − ε ) β 0 2 ( 1 − ε 3 + 2 ε ) n 2 log n n ( 1 − ε ) 2 β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + 2 ε ( 1 − ε ) ≥ π ( κ 2 − ε ) β 0 n 2 log n ∑ j = 1 n 2 j − 1 log j n j ! n 2 + 2 ε + 1 2 ( 1 − ε 3 + 2 ε ) n 2 ε 2 n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α − O 1 n 4 ε 2 log n .

Let

S ¯ ( n ) : = ∑ j = 1 n 2 j − 1 log j n j ! n 2 + 2 ε + 1 2 ( 1 − ε 3 + 2 ε ) n 2 ε 2 > 1 2 .

It follows from (2.3) and (3.2.24) that

(3.2.25) I ( t w n ) = 1 2 ℳ ( t 2 ∥ w n ∥ 2 ) − ∫ B F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 2 − ε ) β 0 S ¯ ( n ) n 2 log n n β 0 ( ( 2 π ) − 1 / 2 t ) 2 + n − α + O 1 n 4 ε 2 log n ≔ ϕ ¯ n ( t ) + O 1 n 4 ε 2 log n .

Let t ¯ n > 0 such that ϕ ¯ n ′ ( t ¯ n ) = 0 . Then

(3.2.26) M ( t ¯ n 2 ) = π ( κ 2 − ε ) β 0 2 ( 2 + n − α ) S ¯ ( n ) t ¯ n n − α ( 2 π ) 1 + 2 − 1 n − α n 2 n β 0 ( ( 2 π ) − 1 / 2 t ¯ n ) 2 + n − α ,

which yields

(3.2.27) lim n → ∞ t ¯ n 2 = lim n → ∞ t ¯ n 2 + n − α = lim n → ∞ 2 ( 2 π ) 1 + 2 − 1 n − α β 0 = 4 π β 0 , ∀ α ∈ ( 2 , + ∞ ] .

Combining with (M1), (M2), (3.1.9), and (3.2.27), we have

t ¯ n 2 = 4 π β 0 1 + log ( ( 2 π ) 1 + 2 − 1 n − α M ( t ¯ n 2 ) ) − log ( π ( κ 2 − ε ) β 0 2 ( 2 + n − α ) S ¯ ( n ) ) 2 log n ≤ 4 π β 0 + 2 π D 2 β 0 log n ,

which together with (3.2.25), (3.2.26), and Proposition 2.2(iii) that

(3.2.28) ϕ ¯ n ( t ) ≤ ϕ ¯ n ( t ¯ n ) = 1 2 ℳ ( t ¯ n 2 ) − ( 2 π ) 1 + 2 − 1 n − α M ( t ¯ n 2 ) β 0 t ¯ n n − α ( 2 + n − α ) log n ≤ 1 2 ℳ 4 π β 0 + 2 π D 2 β 0 log n − π ( 1 − ε ) M 4 π β 0 β 0 log n ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 .

Hence, from (3.2.12) and (3.2.28), one has

I ( t w n ) ≤ 1 2 ℳ 4 π β 0 + ( 1 + ε ) max { D 2 , 0 } − ( 1 − ε ) β 0 log n π M 4 π β 0 + O 1 n 4 ε 2 log n ≤ 1 2 ℳ 4 π β 0 ,

which shows (3.2.2) holds.

Case (iv) t ∈ ( 1 + ε P ˜ , + ∞ ) . Then t w n ( x ) ≥ t ε for x ∈ B 1 / n and for large n ∈ N . In this case, the following discussions are also divided into two situations: (1) α ∈ ( 0 , 2 ] ; (2) α ∈ ( 2 , + ∞ ] .

(1) α ∈ ( 0 , 2 ] . It follows from (2.3) and (3.2.6) that

I ( t w n ) = 1 2 ℳ ( t 2 w n 2 ) − ∫ B 1 / n F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 β 0 j Q j − 1 j ! ( j / 2 ) i n 2 − α i log i Q = 1 2 ℳ ( t 2 ) − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 β 0 j ( t 2 log n / 2 π ) j − 1 j ! ( j / 2 ) i n 2 − α i log i ( t 2 log n / 2 π ) ≤ 1 2 ℳ 4 π ( 1 + ε ) β 0 − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 × β 0 j ( ( 1 + ε ) P ˜ 2 log n / 2 π ) j ( 1 + 2 − 1 n − α ) − 1 j ! ( j / 2 ) i n 2 − α i log i ( ( 1 + ε ) P ˜ 2 log n / 2 π ) ,

which implies that there exists n ¯ ∈ N such that (3.2.2) holds. Note that in the above derivation process, we use the fact that the function

1 2 ℳ ( t 2 ) − 2 π ( κ 2 − ε ) α ∑ j = 1 n ∑ i = 1 2 ∕ α ( − 1 ) i + 1 A 2 ∕ α − 1 i − 1 β 0 j ( t 2 log n / 2 π ) j − 1 j ! ( j / 2 ) i n 2 − α i log i ( t 2 log n / 2 π )

is decreasing on t ∈ ( 1 + ε P ˜ , + ∞ ) , since its stagnation points tend to P ˜ as n → + ∞ .

(2) α ∈ ( 2 , + ∞ ] . It follows from (2.3) and (3.2.6) that

I ( t w n ) = 1 2 ℳ ( t 2 w n 2 ) − ∫ B 1 / n F ( x , t w n ) d x ≤ 1 2 ℳ ( t 2 ) − π ( κ 2 − ε ) ∑ j = 1 n β 0 j Q j − 1 j ! n 2 = 1 2 ℳ ( t 2 ) − π ( κ 2 − ε ) ∑ j = 1 n β 0 j ( t 2 log n / ( 2 π ) ) j − 1 j ! n 2 ≤ 1 2 ℳ 4 π ( 1 + ε ) β 0 − π ( κ 2 − ε ) ∑ j = 1 n β 0 ( 2 ( 1 + ε ) log n ) j − 1 j ! n 2 ,

which implies that there exists n ¯ ∈ N such that (3.2.2) holds. Note that in the above derivation process, we use the fact that the function

1 2 ℳ ( t 2 ) − π ( κ 2 − ε ) ∑ j = 1 n β 0 j ( t 2 log n / ( 2 π ) ) j − 1 j ! n 2

is decreasing on t ∈ ( 1 + ε P ˜ , + ∞ ) , since its stagnation points tend to P ˜ as n → + ∞ .□

4 The proofs of main results

Proof of Theorem 1.1

Applying Lemmas 2.4 and 2.10, we deduce that there exists a sequence { u n } ⊂ H satisfying (2.6) and ∥ u n ∥ ≤ C 3 for some constant C 3 > 0 . Up to a subsequence if necessary, we may assume that

(4.1) u n ⇀ u ¯ , in H , u n → u ¯ , in L s ( B ) , for s ∈ [ 2 , ∞ ) , u n → u ¯ , a.e. on B .

It follows from (M1), (M2), (2.3), and (2.4) that

(4.2) ∫ B f ( x , u n ) u n d x ≤ C 4 .

By (4.2) and [11, Lemma 2.1], we have

lim n → ∞ ∫ B f ( x , u n ) d x = ∫ B f ( x , u ¯ ) d x

and therefore using (F1), (F3), and generalized Lebesgue dominated convergence theorem, we can see that

(4.3) lim n → ∞ ∫ B f ( x , u n ) φ d x = ∫ B f ( x , u ¯ ) φ d x , ∀ φ ∈ C 0 ∞ ( B )

and

(4.4) lim n → ∞ ∫ B F ( x , u n ) d x = ∫ B F ( x , u ¯ ) d x .

Now, we prove that u ¯ ≠ 0 . Arguing by the contradiction, suppose that u ¯ ≡ 0 , which implies that F ( x , u ¯ ) ≡ 0 . Then, it follows from (4.4), one has

(4.5) lim n → ∞ ∫ B F ( x , u n ) d x = 0 .

Without loss of generality, we may assume ‖ u n ‖ ≤ 1 . Let β > β 0 , by (F1), (F2), (4.1), and Hölder inequality with p > 2 , we obtain

(4.6) ∫ B f ( x , u n ) u n d x ≤ C 5 ∫ B ∣ u n ∣ 2 d x + C 6 ∫ B ∣ u n ∣ exp [ ( β + ∣ x ∣ α ) u n 2 ] d x ≤ C 5 ∥ u n ∥ 2 2 + C 6 ∥ u n ∥ p p − 1 ∫ B exp [ p ( β + ∣ x ∣ α ) u n 2 ] d x 1 p ≤ C 5 ∥ u n ∥ 2 2 + C 6 ∥ u n ∥ p p − 1 exp ( sup r ∈ ( 0 , δ ) p g ( r ) ) MT p β + 2 π ( 1 − δ ) 1 − p ρ δ p β / ( 2 π ) ( 1 − p ρ ) 1 p = o ( 1 ) ,

where δ , ρ ∈ ( 0 , 1 ) such that r α ≤ 2 π ρ log r ( 1 − r ) , and g is given in (2.8). Hence, from (2.4), (2.6), and (4.6), we can deduce that M ( ∥ u n ∥ 2 ) ∥ u n ∥ 2 = o ( 1 ) . Consequently, by (M1), (4.5), and therefore I ( u n ) = o ( 1 ) , which is a contradiction and thus we must have u ¯ ≠ 0 .

Based on Proposition 2.3 and Lemma 2.7, from [9, Assertion 2], we obtain ⟨ I ′ ( u ¯ ) , u ¯ ⟩ ≥ 0 . Using this information and (F5), (2.1), (2.3), (2.4), Proposition 2.2(ii), one has

(4.7) I ( u ¯ ) ≥ I ( u ¯ ) − 1 2 θ ⟨ I ′ ( u ¯ ) , u ¯ ⟩ = 1 2 ℳ ( ‖ u ¯ ‖ 2 ) − 1 2 θ M ( ‖ u ¯ ‖ 2 ) ‖ u ¯ ‖ 2 + ∫ B 1 2 θ f ( x , u ¯ ) u ¯ − F ( x , u ¯ ) d x = 1 2 ℳ ˜ ( ‖ u ¯ ‖ 2 ) − 1 2 θ M ˜ ( ‖ u ¯ ‖ 2 ) ‖ u ¯ ‖ 2 + ( θ − 1 ) m 0 2 θ ( ‖ u ¯ ‖ 2 − λ 1 ‖ u ¯ ‖ 2 2 ) + ∫ B 1 2 θ f ( x , u ¯ ) u ¯ − F ( x , u ¯ ) + ( θ − 1 ) m 0 λ 1 2 θ u ¯ 2 d x ≥ 0 .

Then, it follows from (M1), (2.3), (2.6), (4.1), (4.4), and (4.7) that

(4.8) c 1 ∗ + o ( 1 ) = I ( u n ) = 1 2 ℳ ( ∥ u n ∥ 2 ) − ∫ B F ( x , u n ) d x = 1 2 ℳ ( ‖ u ¯ ‖ 2 + ∥ u n − u ¯ ∥ 2 ) − ∫ B F ( x , u ¯ ) d x + o ( 1 ) ≥ I ( u ¯ ) + 1 2 ℳ ( ∥ u n − u ¯ ∥ 2 ) + o ( 1 ) ≥ 1 2 ℳ ( ∥ u n − u ¯ ∥ 2 ) + o ( 1 ) .

Hence, by (4.8) and Lemma 3.1 that

lim n → ∞ 1 2 ℳ ( ‖ u n − u ¯ ‖ 2 ) ≤ c 1 ∗ < 1 2 ℳ 4 π β 0 ,

which, together with (M1), implies that there exists β > 0 such that

(4.9) β 0 ‖ u n − u ¯ ‖ 2 ≤ β < 4 π .

Let β = ( 1 − 3 ε ¯ ) 4 π . Then (4.9) implies that ε ¯ > 0 . Now we choose q ∈ ( 1 , 2 ) such that

(4.10) ( 1 + ε ¯ ) 2 ( 1 − 3 ε ¯ ) q 3 1 − ε ¯ < 1 .

Using the growth of f and supercritical Truding-Moser inequality (1.4), similar to [12, Assertion 1], we can obtain u ¯ > 0 in B . Then, from this and (4.10), Proposition 2.1, we obtain

(4.11) ∫ B exp [ q ˆ ( β 0 + ∣ x ∣ α ) u ¯ 2 ] d x ≤ exp ( q ˆ sup r ∈ ( 0 , δ ) g ( r ) ) ∫ B exp ( q ˆ β 0 u ¯ 2 ) d x + 2 π ∫ δ 1 r − q ˆ β 0 / ( 2 π ) ( 1 − r ) − q ˆ ρ d r ≤ MT ¯ q ˆ β 0 exp ( q ˆ sup r ∈ ( 0 , δ ) g ( r ) ) + 2 π ( 1 − δ ) 1 − q ˆ ρ δ q ˆ β 0 / ( 2 π ) ( 1 − q ˆ ρ )

and

(4.12) ∫ B exp [ ( 1 + ε ¯ ) 2 q 2 ( β 0 + ∣ x ∣ α ) ( u n − u ¯ ) 2 ] d x ≤ ∫ B exp [ ( 1 + ε ¯ ) 2 q 3 β 0 ( u n − u ¯ ) 2 ] d x 1 / q ∫ B exp [ ( 1 + ε ¯ ) 2 q 2 q ′ ∣ x ∣ α ( u n − u ¯ ) 2 ] d x 1 / q ′ ≤ ∫ B exp ( 1 + ε ¯ ) 2 q 3 β 0 ‖ u n − u ¯ ‖ 2 u n − u ¯ ‖ u n − u ¯ ‖ 2 d x 1 / q ∫ B exp [ ( 1 + ε ¯ ) 2 q 2 q ′ ∣ x ∣ α ( u n − u ¯ ) 2 ] d x 1 / q ′ ≤ ( MT q ˇ ) 1 / q 2 π ∫ 0 δ exp [ ( 1 + ε ¯ ) 2 q 2 q ′ r α ( u n − u ¯ ) 2 ] r d r + 2 π ∫ δ 1 r ( 1 − r ) − ε ¯ ρ q q ˆ d r 1 / q ′ ≤ ( MT q ˇ ) 1 / q δ exp ( ε ¯ q q ˆ sup r ∈ ( 0 , δ ) g ( r ) ) + 2 π ( 1 − δ ) 1 − ε ¯ ρ q q ˆ 1 − ε ¯ ρ q q ˆ 1 / q ′ ,

where q ′ = q / ( q − 1 ) and

(4.13) q ˆ = ( 1 + ε ¯ ) 2 q 2 ε ¯ ( q − 1 ) , q ˇ ≔ ( 1 + ε ¯ ) 2 q 3 β 0 ‖ u n − u ¯ ‖ 2 .

By (F1), there exists C 7 > 0 such that

(4.14) ∣ f ( x , t ) ∣ q ≤ C 7 exp [ ( 1 + ε ¯ ) q ( β 0 + ∣ x ∣ α ) t 2 ] , ∀ ( x , t ) ∈ B × [ 0 , + ∞ ) .

Hence, from (1.2), (1.3), (4.9), (4.11), (4.12), (4.14), and the Young’s inequality, we have

(4.15) ∫ B ∣ f ( x , u n ) ∣ q d x ≤ C 7 ∫ B exp [ ( 1 + ε ¯ ) q ( β 0 + ∣ x ∣ α ) u n 2 ] d x ≤ C 7 ∫ B exp [ ( 1 + ε ¯ ) 2 ε ¯ − 1 q ( β 0 + ∣ x ∣ α ) u ¯ 2 ] exp [ ( 1 + ε ¯ ) 2 q ( β 0 + ∣ x ∣ α ) ( u n − u ¯ ) 2 ] d x ≤ C 7 q ′ ∫ B exp [ q ˆ ( β 0 + ∣ x ∣ α ) u ¯ 2 ] d x + C 7 q ∫ B exp [ ( 1 + ε ¯ ) 2 q 2 ( β 0 + ∣ x ∣ α ) ( u n − u ¯ ) 2 ] d x ≤ C 7 q ′ MT ¯ q ˆ β 0 exp ( q ˆ sup r ∈ ( 0 , δ ) g ( r ) ) + 2 π ( 1 − δ ) 1 − q ˆ ρ δ q ˆ β 0 / ( 2 π ) ( 1 − q ˆ ρ ) + C 7 q ( MT q ˇ ) 1 / q δ exp ( ε ¯ q q ˆ sup r ∈ ( 0 , δ ) g ( r ) ) + 2 π ( 1 − δ ) 1 − ε ¯ ρ q q ˆ 1 − ε ¯ ρ q q ˆ 1 / q ′ .

Thus, by (4.1), (4.15), and the Hölder inequality, we obtain

(4.16) ∫ B f ( x , u n ) ( u n − u ¯ ) d x ≤ ∫ B ∣ f ( x , u n ) ∣ q d x 1 / q ∥ u n − u ¯ ∥ q ′ ≤ C 7 q ′ MT ¯ q ˆ β 0 exp ( q ˆ sup r ∈ ( 0 , δ ) g ( r ) ) + 2 π ( 1 − δ ) 1 − q ˆ ρ δ q ˆ β 0 / ( 2 π ) ( 1 − q ˆ ρ ) + C 7 q ( MT q ˇ ) 1 / q δ exp ( ε ¯ q q ˆ sup r ∈ ( 0 , δ ) g ( r ) ) + 2 π ( 1 − δ ) 1 − ε ¯ ρ q q ˆ 1 − ε ¯ ρ q q ˆ 1 / q ′ 1 / q ∥ u n − u ¯ ∥ q ′ ≤ C 8 ∥ u n − u ¯ ∥ q ′ = o ( 1 ) .

Therefore, it follows from (2.4), (2.6), and (4.16) that

o ( 1 ) = ⟨ I ′ ( u n ) , u n − u ¯ ⟩ = M ( ∥ u n ∥ 2 ) ∫ B ∇ u n ⋅ ( ∇ u n − ∇ u ¯ ) d x − ∫ B f ( x , u n ) ( u n − u ¯ ) d x = M ( ∥ u n ∥ 2 ) ∥ u n ∥ 2 − M ( ∥ u n ∥ 2 ) ∫ B ∇ u n ⋅ ∇ u ¯ d x + o ( 1 ) = M ( A ) A − M ( A ) ‖ u ¯ ‖ 2 + o ( 1 ) ,

which implies that A ≔ lim n → ∞ ∥ u n ∥ 2 = ‖ u ¯ ‖ 2 . Thus, u n → u ¯ . In view of the continuity of I and I ′ , we must have I ( u ¯ ) = c 1 ∗ and I ′ ( u ¯ ) = 0 .□

Proof of Theorem 1.2

In view of the proof of Theorem 1.1, there exists a bounded sequence { u n } ⊂ H satisfying

u n ⇀ u ¯ , in H , u n → u ¯ , in L s ( B ) , for s ∈ [ 2 , ∞ ) , u n → u ¯ , a.e. on B .

Moreover, there hold (4.2), (4.3), and (4.4).

Now, we only need to prove that (4.9) still holds if (M1) and (F5) are replaced by (M1 ′ ) and (F6), respectively. Without loss of generality, we may assume ‖ u n ‖ ≤ 1 . From (1.3) and Proposition 2.1, we have

(4.17) ∫ B exp [ 2 ( β 0 + ε + ∣ x ∣ α ) u ¯ 2 ] d x ≤ exp ( sup r ∈ ( 0 , δ ) 2 g ( r ) ) ∫ B exp [ 2 ( β 0 + ε ) u ¯ 2 ] d x + 2 π ∫ δ 1 r 1 − ( β 0 + ε ) / π ( 1 − r ) − 2 ρ d r ≤ exp ( sup r ∈ ( 0 , δ ) 2 g ( r ) ) MT ¯ 2 ( β 0 + ε ) + 2 π ( 1 − δ ) 1 − 2 ρ δ ( β 0 + ε ) / π ( 1 − 2 ρ ) ,

where ρ ∈ ( 0 , 1 ) such that r α ≤ 2 π ρ log r ( 1 − r ) . By (F1) and (F2), for any ε > 0 there exists C 9 > 0 such that

(4.18) ∣ f ( x , t ) ∣ ≤ ∣ t ∣ + C 9 exp [ ( β 0 + ε + ∣ x ∣ α ) t 2 ] , ∀ ( x , t ) ∈ B × [ 0 , + ∞ ) .

Note that u ¯ ∈ H , there exists { φ n } ⊂ C 0 ∞ ( B ) such that ∥ φ n − u ¯ ∥ = o ( 1 ) . Then, by (2.2), (4.17), (4.18), and the Hölder inequality, one has

(4.19) ∫ B f ( x , u ¯ ) ( φ n − u ¯ ) d x ≤ ∫ B { ∣ u ¯ ∣ + C 9 exp [ ( β 0 + ε + ∣ x ∣ α ) u ¯ 2 ] } ∣ φ n − u ¯ ∣ d x ≤ ‖ u ¯ ‖ 2 ∥ φ n − u ¯ ∥ 2 + C 9 ∫ B exp [ 2 ( β 0 + ε + ∣ x ∣ α ) u ¯ 2 ] d x 1 / 2 ∥ φ n − u ¯ ∥ 2

≤ γ 2 ‖ u ¯ ‖ 2 ∥ φ n − u ¯ ∥ + C 9 exp ( sup r ∈ ( 0 , δ ) 2 g ( r ) ) MT ¯ 2 ( β 0 + ε ) + 2 π ( 1 − δ ) 1 − 2 ρ δ ( β 0 + ε ) / π ( 1 − 2 ρ ) 1 / 2 ∥ φ n − u ¯ ∥ 2 ≤ γ 2 ‖ u ¯ ‖ 2 ∥ φ n − u ¯ ∥ + C 10 ∥ φ n − u ¯ ∥ = o ( 1 ) .

From (2.4), (2.6), and (4.3), arguing as in the proof of Assertion 5 in [9, Theorem 1.4], we obtain

(4.20) M ( A ) ∫ B ∇ u ¯ ⋅ ∇ φ d x = ∫ B f ( x , u ¯ ) φ d x , ∀ φ ∈ C 0 ∞ ( B ) .

Hence, it follows from (4.19) and (4.20) that

(4.21) M ( A ) ‖ u ¯ ‖ 2 = M ( A ) ∫ B ∇ u ¯ ⋅ ∇ u ¯ d x = M ( A ) lim n → ∞ ∫ B ∇ u ¯ ⋅ ∇ φ n d x = lim n → ∞ ∫ B f ( x , u ¯ ) φ n d x = ∫ B f ( x , u ¯ ) u ¯ d x .

Recall that A ≔ lim n → ∞ ∥ u n ∥ 2 . Using this information and combining with (2.4) and (2.6), we have

0 = lim n → ∞ ⟨ I ′ ( u n ) , u n ⟩ = lim n → ∞ M ( ∥ u n ∥ 2 ) ∥ u n ∥ 2 − ∫ B f ( x , u n ) u n d x = M ( A ) A − lim n → ∞ ∫ B f ( x , u n ) u n d x ,

which, together with (4.21), implies that

(4.22) lim n → ∞ ∫ B [ f ( x , u n ) u n − f ( x , u ¯ ) u ¯ ] d x = M ( A ) ( A − ‖ u ¯ ‖ 2 ) .

To prove (4.9) still holds, arguing by contradiction, suppose that A ≥ ‖ u ¯ ‖ 2 + 4 π β 0 . Then, from (M1 ′ ), (M2), (F6), (2.3), (2.4), (4.4), and (4.22), we obtain

c 1 ∗ = lim n → ∞ I ( u n ) − 1 2 θ ⟨ I ′ ( u n ) , u n ⟩ = 1 2 θ lim n → ∞ θ ℳ ( ∥ u n ∥ 2 ) − M ( ∥ u n ∥ 2 ) ∥ u n ∥ 2 + ∫ B [ f ( x , u n ) u n − 2 θ F ( x , u n ) ] d x = 1 2 θ θ ℳ ( A ) − M ( A ) A + lim n → ∞ ∫ B f ( x , u n ) u n d x − ∫ B F ( x , u ¯ ) d x = 1 2 θ [ θ ℳ ˜ ( A ) − M ˜ ( A ) A + ( θ − 1 ) m 0 ( A − λ 1 ‖ u ¯ ‖ 2 2 ) ] + 1 2 θ lim n → ∞ ∫ B [ f ( x , u n ) u n − f ( x , u ¯ ) u ¯ ] d x + ∫ B 1 2 θ f ( x , u ¯ ) u ¯ − F ( x , u ¯ ) + ( θ − 1 ) m 0 λ 1 2 θ u ¯ 2 d x

≥ 1 2 θ [ θ ℳ ˜ ( A ) − M ˜ ( A ) A + ( θ − 1 ) m 0 ( A − λ 1 ‖ u ¯ ‖ 2 2 ) ] + 1 2 θ lim n → ∞ ∫ B [ f ( x , u n ) u n − f ( x , u ¯ ) u ¯ ] d x ≥ 1 2 θ θ ℳ ˜ 4 π β 0 − M ˜ 4 π β 0 4 π β 0 + ( θ − 1 ) m 0 4 π α 0 + 1 2 θ lim n → ∞ ∫ B [ f ( x , u n ) u n − f ( x , u ¯ ) u ¯ ] d x = θ ℳ 4 π β 0 − M 4 π β 0 4 π β 0 + 1 2 θ lim n → ∞ ∫ B [ f ( x , u n ) u n − f ( x , u ¯ ) u ¯ ] d x = 1 2 ℳ 4 π β 0 + 1 2 θ − M 4 π α 0 4 π β 0 + lim n → ∞ ∫ B [ f ( x , u n ) u n − f ( x , u ¯ ) u ¯ ] d x = 1 2 M 4 π β 0 + 1 2 θ M ( A ) ( A − ‖ u ¯ ‖ 2 ) − M 4 π β 0 4 π β 0 ≥ 1 2 M 4 π β 0 ,

which is a contradiction.□

Proof of Theorem 1.3

Applying Lemmas 2.5 and 2.11, we deduce that there exists a sequence { u n } ⊂ H satisfying (2.11) and ∥ u n ∥ ≤ C 11 for some constant C 11 > 0 . Up to a subsequence if necessary, we may assume that

(4.23) u n ⇀ u ¯ , in H , u n → u ¯ , in L s ( B ) , for s ∈ [ 2 , ∞ ) , u n → u ¯ , a.e. on B .

From (M1), (M2), (2.3), and (2.4), there exists C 12 > 0 such that

(4.24) ∫ B f ( x , u n ) u n d x ≤ C 12 .

By (4.24) and [11, Lemma 2.1], we have

lim n → ∞ ∫ B f ( x , u n ) d x = ∫ B f ( x , u ¯ ) d x

and therefore using (F1 ′ ), (F3), and generalized Lebesgue dominated convergence theorem, we can see that

(4.25) lim n → ∞ ∫ B f ( x , u n ) φ d x = ∫ B f ( x , u ¯ ) φ d x , ∀ φ ∈ C 0 ∞ ( B )

and

(4.26) lim n → ∞ ∫ B F ( x , u n ) d x = ∫ B F ( x , u ¯ ) d x .

From (4.23), (4.26), and Lemma 3.2, similar to the proof of Theorem 1.1, one has

lim n → ∞ 1 2 ℳ ( ‖ u n − u ¯ ‖ 2 ) ≤ c 2 ∗ < 1 2 ℳ 4 π β 0 ,

which, together with (M1), implies that there exists β > 0 such that

(4.27) β 0 ‖ u n − u ¯ ‖ 2 ≤ β < 4 π .

Let β = ( 1 − 3 ε ¯ ) 4 π . Then (4.27) implies that ε ¯ > 0 . Now we choose q ∈ ( 1 , 2 ) such that

(4.28) ( 1 + ε ¯ ) 2 ( 1 − 3 ε ¯ ) q 2 1 − ε ¯ < 1 .

Hence, for ( ε ¯ , q ) close to ( 0 + , 1 + ) , it follows from (4.27) that

(4.29) ∥ u n − u ¯ ∥ ∣ x ∣ α < 4 π ( 1 + ε ¯ ) 2 q 2 β 0 ∣ x ∣ α / 2 .

Combining with (4.28) and (4.29), we obtain

(4.30) ∫ B σ exp [ ( 1 + ε ¯ ) 2 q 2 β 0 ∣ u n − u ¯ ∣ 2 + ∣ x ∣ α ] d x = ∫ B σ exp ( 1 + ε ¯ ) 2 q 2 β 0 ∥ u n − u ¯ ∥ 2 + ∣ x ∣ α ∣ u n − u ¯ ∣ ∥ u n − u ¯ ∥ 2 + ∣ x ∣ α d x ≤ ∫ B σ exp ( 1 + ε ¯ ) 2 q 2 β 0 ∥ u n − u ¯ ∥ 2 ∣ u n − u ¯ ∣ ∥ u n − u ¯ ∥ 2 + ∣ x ∣ α d x ≤ ∫ B σ exp ( 1 − ε ¯ ) 4 π ∣ u n − u ¯ ∣ ∥ u n − u ¯ ∥ 2 + ∣ x ∣ α d x ≤ MT α 2 ( ( 1 − ε ¯ ) 4 π ) ,

where σ close to 0 + and q ˆ is given in (4.13). By (F1 ′ ), there exists C 13 > 0 such that

(4.31) ∣ f ( x , t ) ∣ q ≤ C 13 exp [ ( 1 + ε ¯ ) q β 0 t 2 + ∣ x ∣ α ] , ∀ ( x , t ) ∈ B × [ 0 , + ∞ ) .

Thus, it follows from (4.30), (4.31), and the Young’s inequality that

(4.32) ∫ B σ ∣ f ( x , u n ) ∣ q d x ≤ C 13 ∫ B σ exp [ ( 1 + ε ¯ ) q β 0 u n 2 + ∣ x ∣ α ] d x ≤ C 13 ∫ B σ exp [ ( 1 + ε ¯ ) 2 ε ¯ − 1 q β 0 u ¯ 2 + ∣ x ∣ α ] exp [ ( 1 + ε ¯ ) 2 q β 0 ∣ u n − u ¯ ∣ 2 + ∣ x ∣ α ] d x ≤ C 13 q ′ ∫ B σ exp [ ( 1 + ε ¯ ) 2 ε ¯ − 1 q q ′ β 0 u ¯ 2 + ∣ x ∣ α ] d x + C 13 q ∫ B σ exp [ ( 1 + ε ¯ ) 2 q 2 β 0 ∣ u n − u ¯ ∣ 2 + ∣ x ∣ α ] d x ≤ C 13 q ′ MT ¯ q ˆ + C 13 q MT α 2 ( ( 1 − ε ¯ ) 4 π ) ,

where q ′ = q / ( q − 1 ) .

On the other hand, without loss of generality, we may assume ‖ u n ‖ ≤ 1 , then it follows from Proposition 2.1 that there exists L 2 ≔ − log σ 2 π such that ∣ u n ∣ ≤ L a.e. B \ B σ . Hence,

(4.33) ∫ B ⧹ B σ ∣ f ( x , u n ) ∣ q d x ≤ π sup B × [ 0 , L ] ∣ f ( x , t ) ∣ q < + ∞ .

Thus, by (1.3), (1.5), (4.23), (4.32), (4.33), and the Hölder inequality, we obtain

(4.34) ∫ B f ( x , u n ) ( u n − u ¯ ) d x ≤ ∫ B ∣ f ( x , u n ) ∣ q d x 1 / q ∥ u n − u ¯ ∥ q ′ ≤ ∫ B σ ∣ f ( x , u n ) ∣ q d x + ∫ B ⧹ B σ ∣ f ( x , u n ) ∣ q d x 1 / q ∥ u n − u ¯ ∥ q ′

≤ C 13 q ′ MT ¯ q ˆ + C 13 q MT α 2 ( ( 1 − ε ¯ ) 4 π ) + π sup B × [ 0 , L ) ∣ f ( x , t ) ∣ q 1 / q ∥ u n − u ¯ ∥ q ′ ≤ C 14 ∥ u n − u ¯ ∥ q ′ = o ( 1 ) .

Therefore, it follows from (2.4), (2.6), and (4.34) that

which implies that A ≔ lim n → ∞ ∥ u n ∥ 2 = ‖ u ¯ ‖ 2 . Thus u n → u ¯ . In view of the continuity of I and I ′ , we must have I ( u ¯ ) = c 2 ∗ and I ′ ( u ¯ ) = 0 .□

Proof of Theorem 1.4

In view of the proof of Theorem 1.3, there exists a bounded sequence { u n } ⊂ H satisfying

(4.35) u n ⇀ u ¯ , in H , u n → u ¯ , in L s ( B ) , for s ∈ [ 2 , ∞ ) , u n → u ¯ , a.e. on B .

Moreover, there holds (4.25).

Now, we only need to prove that (4.27) still holds if (M1) and (F5) are replaced by (M1 ′ ) and (F6), respectively. Without loss of generality, we may assume ‖ u n ‖ ≤ 1 . Let σ be close to 0 + , by Proposition 2.1, we have

(4.36) ∫ B exp [ 2 ( β 0 + ε ) u ¯ 2 + ∣ x ∣ α ] d x = ∫ B σ exp [ 2 ( β 0 + ε ) u ¯ 2 + ∣ x ∣ α ] d x + ∫ B \ B σ exp [ 2 ( β 0 + ε ) u ¯ 2 + ∣ x ∣ α ] d x ≤ MT ¯ 2 ( β 0 + ε ) + π sup B × [ 0 , L ] exp [ 2 ( β 0 + ε ) u 2 + ∣ x ∣ α ] ,

where L is given in (4.33). By (F1 ′ ), for any ε > 0 , there exists C 15 > 0 such that

(4.37) ∣ f ( x , t ) ∣ ≤ ∣ t ∣ + C 15 exp [ ( β 0 + ε ¯ ) t 2 + ∣ x ∣ α ] , ∀ ( x , t ) ∈ B × [ 0 , + ∞ ) .

Note that u ¯ ∈ H , there exists { φ n } ⊂ C 0 ∞ ( B ) such that ∥ φ n − u ¯ ∥ = o ( 1 ) . Hence, from (1.3), (2.2), (4.35), (4.36), (4.37), and the Hölder inequality, one has

(4.38) ∫ B f ( x , u ¯ ) ( φ n − u ¯ ) d x ≤ ∫ B ∣ f ( x , u ¯ ) ∣ ∣ φ n − u ¯ ∣ d x ≤ ∫ B { ∣ u ¯ ∣ + C 13 exp [ ( β 0 + ε ) u ¯ 2 + ∣ x ∣ α ] } ∣ φ n − u ¯ ∣ d x ≤ ‖ u ¯ ‖ 2 ∥ φ n − u ¯ ∥ 2 + C 15 ∫ B exp [ 2 ( β 0 + ε ) u ¯ 2 + ∣ x ∣ α ] d x 1 / 2 ∥ φ n − u ¯ ∥ 2 ≤ γ 2 ‖ u ¯ ‖ 2 ∥ φ n − u ¯ ∥ + C 15 { MT ¯ 2 ( β 0 + ε ) + π sup B × [ 0 , L ] exp [ 2 ( β 0 + ε ) u 2 + ∣ x ∣ α ] } 1 / 2 ∥ φ n − u ¯ ∥ 2 ≤ γ 2 ‖ u ¯ ‖ 2 ∥ φ n − u ¯ ∥ + C 16 ∥ φ n − u ¯ ∥ = o ( 1 ) .

From (4.25), arguing as in the proof of Assertion 5 in [9, Theorem 1.4], we obtain

(4.39) M ( A ) ∫ B ∇ u ¯ ⋅ ∇ φ d x = ∫ B f ( x , u ¯ ) φ d x , ∀ φ ∈ C 0 ∞ ( B ) .

Hence, it follows from (4.38) and (4.39) that

M ( A ) ‖ u ¯ ‖ 2 = M ( A ) ∫ B ∇ u ¯ ⋅ ∇ u ¯ d x = M ( A ) lim n → ∞ ∫ B ∇ u ¯ ⋅ ∇ φ n d x = lim n → ∞ ∫ B f ( x , u ¯ ) φ n d x = ∫ B f ( x , u ¯ ) u ¯ d x ,

which, together with (2.4) and (2.6), imply that

(4.40) lim n → ∞ ∫ B [ f ( x , u n ) u n − f ( x , u ¯ ) u ¯ ] d x = M ( A ) ( A − ‖ u ¯ ‖ 2 ) .

Actually, after a careful look at the proof of Theorem 1.2, using (4.40), we can deduce that A < ‖ u ¯ ‖ 2 + 4 π β 0 , which implies (4.27) also holds.□

Acknowledgements

This work was partially supported by the National Natural Science Foundation of China (Nos. 11801574, 11971485, 12161091), Natural Science Foundation of Hubei Province of China (No. 2021CFB473), and Central Universities of Central South University (Grant Nos. 2021zzts0041, 2021zzts0042).

Conflict of interest: Authors state no conflict of interest.

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Received: 2022-01-05

Revised: 2022-03-06

Accepted: 2022-04-03

Published Online: 2022-05-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

On the planar Kirchhoff-type problem involving supercritical exponential growth

Abstract

1 Introduction

Lemma 1.1

Theorem 1.1

Theorem 1.2

Theorem 1.3

Theorem 1.4

Remark 1.5

2 Preliminaries

Proposition 2.1

Proposition 2.2

Proposition 2.3

Lemma 2.4

Proof

Lemma 2.5

Proof

Corollary 2.6

Lemma 2.7

Lemma 2.8

Lemma 2.9

Lemma 2.10

Lemma 2.11

3 Minimax estimates

3.1 Estimate with MT α 1

Lemma 3.1

Proof

3.2 Estimate with MT α 2

Lemma 3.2

Proof

4 The proofs of main results

Proof of Theorem 1.1

Proof of Theorem 1.2

Proof of Theorem 1.3

Proof of Theorem 1.4

Acknowledgements

References

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