Existence and concentration of solutions to Kirchhoff-type equations in ℝ2 with steep potential well vanishing at infinity and exponential critical nonlinearities

Jian Zhang; Xue Bao; Jianjun Zhang

doi:10.1515/anona-2022-0317

Article Open Access

Existence and concentration of solutions to Kirchhoff-type equations in ℝ² with steep potential well vanishing at infinity and exponential critical nonlinearities

Jian Zhang , Xue Bao and Jianjun Zhang

Published/Copyright: June 13, 2023

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From the journal Advances in Nonlinear Analysis Volume 12 Issue 1

Abstract

We are concerned with the following Kirchhoff-type equation with exponential critical nonlinearities

− a + b ∫ R 2 ∣ ∇ u ∣ 2 d x Δ u + ( h ( x ) + μ V ( x ) ) u = K ( x ) f ( u ) in R 2 ,

where a , b , μ > 0 , the potential V has a bounded set of zero points and decays at infinity as ∣ x ∣ − γ with γ ∈ ( 0 , 2 ) , the weight K has finite singular points and may have exponential growth at infinity. By using the truncation technique and working in some weighted Sobolev space, we obtain the existence of a mountain pass solution for μ > 0 large and the concentration behavior of solutions as μ → + ∞ .

Keywords: Kirchhoff-type equation; steep potential well; vanishing potential; Trudinger-Moser inequality; variational method

MSC 2010: 35A15; 35J60

1 Introduction and results

1.1 Background

In this article, we study the following Kirchhoff-type equation with steep potential well:

(1.1) − a + b ∫ R 2 ∣ ∇ u ∣ 2 d x Δ u + ( h ( x ) + μ V ( x ) ) u = K ( x ) f ( u ) in R 2 ,

where a , b > 0 are constants, μ > 0 is a parameter, the potential V decays at infinity as ∣ x ∣ − γ with γ ∈ ( 0 , 2 ) , the nonlinearity f has exponential growth at infinity, and the weight K is singular at finite points and may have exponential growth at infinity. Here, we say the nonlinearity f has exponential subcritical growth if for any α > 0 ,

(1.2) lim u → + ∞ f ( u ) e α u 2 = 0 ,

and the nonlinearity f has exponential critical growth if there exists α 0 > 0 such that

(1.3) lim u → + ∞ f ( u ) e α u 2 = 0 , ∀ α > α 0 , + ∞ , ∀ α < α 0 .

The Kirchhoff-type problem appears as a model of several physical phenomena. For example, it is related to the stationary analog of the equation:

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

where u is the lateral displacement at x and t , E is the Young’s modulus, ρ is the mass density, h is the cross-section area, L is the length, and P 0 is the initial axial tension. In the last decades, there have been many articles focusing on the existence, multiplicity, and concentration behavior of solutions to Kirchhoff-type equations. The readers may see [4,12,17,18,21–25,35,37–40] and the references therein.

In this article, we study Kirchhoff-type equations with steep potential well in R 2 . For the special case b = 0 , equation (1.1) is reduced to the Schrödinger equation. In [9], Bartsch and Wang studied the existence and concentration behavior of positive ground state solutions of the equation with steep potential well:

(1.5) − Δ u + ( 1 + μ V ( x ) ) u = u p − 1 in R N ,

where N ≥ 3 and 2 < p < 2 ∗ ≔ 2 N N − 2 . In [16], Ding and Tanaka constructed multi-bump positive solutions of Schrödinger equations with steep potential well. In [30], Sato and Tanaka considered sign-changing multi-bump solutions. For the critical case, Clapp and Ding [13] studied the existence and multiplicity of positive solutions of the equation with steep potential well:

(1.6) − Δ u + μ V ( x ) u = λ u + u 2 ∗ − 1 in R N ,

where N ≥ 4 , λ > 0 is small and μ > 0 is large. For other related results, we refer the readers to [7,8,15,19,20,31,33,34] and the references therein. For the general case b > 0 , Zhang and Du [38] studied the existence and concentration behavior of positive solutions of Kirchhoff-type equations with subcritical growth nonlinearity in dimension three. Zhang and Lou [39] obtained the multiplicity and concentration behavior of solutions of Kirchhoff-type equations with critical growth.

We remark that the potential V is always assumed to be bounded away from zero by a positive constant at infinity, i.e.,

int ( V − 1 ( 0 ) ) is nonempty with a smooth boundary, and there exists V 0 > 0 such that ∣ { x ∈ R N : V ( x ) ≤ V 0 } ∣ < ∞ .

The condition ( V ′ ) ensures studying equation (1.1) in the following complete subspace of H 1 ( R 2 ) :

H ≔ u ∈ H 1 ( R 2 ) : ∫ R 2 V ( x ) u 2 d x < + ∞ .

Thus, it is natural to ask what happens when the potential decays to 0 at infinity. To the best of our knowledge, the only result available in the literature is [11]. In [11], Chen and Wang studied the following quasilinear elliptic equation with steep potential well:

− Δ p u + λ V ( x ) ∣ u ∣ p − 2 u = K ( x ) ∣ u ∣ q − 2 u in R N ,

where N ≥ 2 , 1 < p < N , Δ p u = div ( ∣ ∇ u ∣ p − 2 ∇ u ) , V , and K satisfy the following conditions:

V and K are locally bounded nonnegative functions in R N .
There exist positive constants L , C 1 , C 2 , D 1 , and D 2 such that for ∣ x ∣ ≥ L ,
C 1 ≤ ∣ x ∣ b V ( x ) ≤ C 2 , D 1 ≤ ∣ x ∣ s K ( x ) ≤ D 2 .
There exists a closed subset Z ⊂ { x ∈ R N : ∣ x ∣ < L } with a non-empty interior Ω = int Z such that V ( x ) = 0 for x ∈ Z .
inf ∣ x ∣ ≤ L K ( x ) > 0 .

If b < p , b ≠ 0 , 0 < s b < 1 , and q = p ( N − p s b ) N − p , the authors obtained the existence of nontrivial solutions for λ > 0 large. For the special case of p = 2 and b > 0 , there holds N ≥ 3 , 0 < b < 2 , 0 < s < 2 , and q = 2 N N − 2 − 4 s b ( N − 2 ) , i.e., the potential decays to zero at infinity.

We recall that in [14], the authors studied elliptic equations in a bounded domain in R 2 with critical nonlinearities. See also [5,26,27,29] for Schrödinger equations with critical growth in R 2 and [17,18,37] for Kirchhoff-type equations. In particular, in [27], the authors studied Schrödinger equations with potentials vanishing at infinity. However, to the best of our knowledge, there have been no results studying the Kirchhoff-type equation with steep potential well vanishing at infinity and exponential critical nonlinearity in dimension two. On the other hand, we note that in [2,3], the authors studied nonlocal equations in R 2 involving vanishing potentials and exponential critical growth in a radial setting, where the potential and the coefficient of the nonlinearity may be singular at 0. By establishing a weighted Trudinger-Moser-type inequality, the authors obtained the existence of radial ground state solutions. Motivated by [2,3], in this article, we study (1.1) with the weight K being singular at finite points in Ω and exponential growth at infinity. More precisely, we assume the following conditions:

V ∈ C ( R 2 , R + ) .
Ω ≔ int V − 1 ( 0 ) is nonempty with smooth boundary and Ω ¯ = V − 1 ( 0 ) .
There exists γ ∈ ( 0 , 2 ) such that liminf ∣ x ∣ → ∞ V ( x ) ∣ x ∣ γ ≔ V 0 > 0 .
h ∈ C ( R 2 , R + ) and inf Ω ¯ h ≔ h 0 > 0 .
There exists ∪ i = 1 l { x i } ⊂ Ω with l ∈ N such that K ∈ C ( R 2 \ ∪ i = 1 l { x i } , R + ) .
There exist m , M > 0 such that K ( x ) ≤ M e m ∣ x ∣ 1 − γ 2 for all x ∈ R 2 \ Ω .
There exist r , d 1 , d 2 > 0 and m i ∈ ( − 2 , 0 ) with i = 1 , …, l such that for 0 < ∣ x − x i ∣ ≤ r ,
d 1 ∣ x − x i ∣ m i ≤ K ( x ) ≤ d 2 ∣ x − x i ∣ m i .
f ∈ C ( R + , R + ) .
There exists α 0 ∈ ( 0 , 4 π + 2 π m 0 ) such that
lim u → + ∞ f ( u ) e α u 2 = 0 , ∀ α > α 0 , + ∞ , ∀ α < α 0 ,
where m 0 ≔ min 1 ≤ i ≤ l m i .
There exists β > ( m 0 + 2 ) 2 [ a + 2 b π ( m 0 + 2 ) α 0 ] 2 e d α 0 r m 0 + 2 e r 2 ( m 0 + 2 ) max Ω ¯ h 4 [ a + 2 b π ( m 0 + 2 ) α 0 ] such that
lim u → + ∞ f ( u ) u e α 0 u 2 ≥ β ,
where d is the radius of an open ball contained in Ω .
The function f ( u ) u 3 is increasing for u ∈ R + \ { 0 } .
There exist M 0 , L 0 > 0 such that F ( u ) ≤ L 0 f ( u ) for u ≥ M 0 , where F ( u ) = ∫ 0 u f ( s ) d s .

Now we state the result. Define X the completion of C 0 ∞ ( R 2 ) with respect to the norm

‖ u ‖ ≔ ‖ ∇ u ‖ 2 2 + ∫ R 2 ( h ( x ) + V ( x ) ) u 2 d x 1 2 .

Theorem 1.1

Assume that ( V 1 )–( V 3 ), ( h 1 ) , ( K 1 )–( K 3 ) and ( f 1 )–( f 5 ) hold. Then, there exists μ 0 > 0 such that for μ > μ 0 , equation (1.1) has a nonnegative mountain pass solution u μ ∈ H 1 ( R 2 ) . Moreover, for any R > 0 with Ω ⊂ B R ( 0 ) , there exists μ ′ > 0 such that for μ > μ ′ , there exist c 1 ′ , c 2 ′ > 0 independent of μ such that

(1.7) u μ ( x ) ≤ c 2 ′ e − c 1 ′ μ ( ∣ x ∣ 1 − γ 2 − R 1 − γ 2 ) ∀ ∣ x ∣ ≥ R .

Besides, for any sequence μ n → + ∞ , there exists u 0 ∈ X such that u μ n → u 0 in X as n → ∞ , where u 0 is a nonnegative solution to the problem:

(1.8) − a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u + h ( x ) u = K ( x ) f ( u ) i n Ω .

When N = 2 , to deal with the exponential critical nonlinearity, it is crucial to estimate the upper bound on the energy. In [14], the authors considered a Dirichlet problem:

− Δ u = f ( x , u ) in Ω , u = 0 on ∂ Ω ,

where f satisfies the following growth condition:

There exists β > 4 3 α 0 d 2 such that
lim u → + ∞ f ( x , u ) u e α 0 u 2 ≥ β ,
where d is the radius of the largest open ball in Ω .

By using the Moser sequence of functions and the proof by contradiction, the authors deduced the upper bound on the energy. Related results can be found in [18,29,37]. In this article, since the weight K has singular points, we prove that the region of compactness of Palais-Smale sequences is related to the singularity of K , which is different from the aforementioned articles. By using the Moser sequence of functions, the estimate near singular points, and a direct argument, we obtain the desired upper bound on the energy.

To study the existence and concentration behavior of solutions, the main difficulty lies in the exponential critical growth of the nonlinearity. The Trudinger-Moser inequality plays an important role in dealing with critical nonlinearity. In this article, the weight K has finite singular points, and we do not require V and K to be radial. We establish a weighted singular Trudinger-Moser-type inequality in a bounded domain in Lemma 2.4, which is crucial to obtain the uniform upper bound on some integral sequences.

When using the Trudinger-Moser-type inequality, it is crucial to control the uniform X -norm of the sequence. Because of the presence of the nonlocal term, we cannot use the upper bound on the energy and the Ambrosetti-Rabinowitz-type condition to deduce the desired X -norm estimate, which is different from Schrödinger equations. We establish a compactness result restricted to a bounded domain in Lemma 2.6, which together with Lemma 2.4 helps to study the compactness in the bounded domain. Since the weight K may have exponential growth at infinity, we study the problem by penalizing the nonlinearity.

The outline of this article is as follows: in Section 2, we establish some important lemmas; in Section 3, we study the modified problem; and in Section 4, we prove that solutions to the modified problem are solutions to the original problem.

2 Preliminary lemmas

Without loss of generality, we assume that f ( u ) = 0 for u ≤ 0 . We provide some definitions. Denote C a universal positive constant (possibly different). Define ‖ u ‖ s ≔ ∫ R 2 ∣ u ( x ) ∣ s d x 1 s , where s ∈ [ 2 , ∞ ) . Define H 1 ( R 2 ) , the Hilbert space equipped with the norm

‖ u ‖ H 1 ≔ ( ‖ ∇ u ‖ 2 2 + ‖ u ‖ 2 2 ) 1 2 .

Let μ > 0 . Define X μ , the completion of C 0 ∞ ( R 2 ) with respect to the norm

‖ u ‖ μ ≔ ‖ ∇ u ‖ 2 2 + ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x 1 2 .

By ( V 1 ) , ( V 3 ) , and ( h 1 ) , we know that X μ is a uniformly convex Banach space. In particular, when μ = 1 , we denote X μ by X and ‖ . ‖ μ by ‖ . ‖ . We provide the following the Trudinger-Moser inequality in [28,32].

Lemma 2.1

Let Ω be a bounded domain in R 2 . If u ∈ H 0 1 ( Ω ) , then ∫ Ω e u 2 d x < ∞ .

Moreover, we provide a local version of the Trudinger-Moser inequality in [36].

Lemma 2.2

There exists C > 0 such that for all y ∈ R 2 , R > 0 , and u ∈ H 0 1 ( B R ( y ) ) with ‖ ∇ u ‖ 2 ≤ 1 ,

(2.1) ∫ B R ( y ) ( e 4 π u 2 − 1 ) d x ≤ C R 2 ∫ B R ( y ) ∣ ∇ u ∣ 2 d x .

We also provide a singular version of the Trudinger-Moser inequality in [1].

Lemma 2.3

If Ω is a bounded domain in R 2 containing the origin, u ∈ H 0 1 ( Ω ) and β ∈ [ 0 , 2 ) , then there exists C = C ( α , β , Ω ) such that

(2.2) sup ‖ ∇ u ‖ L 2 ( Ω ) ≤ 1 ∫ Ω ∣ x ∣ − β e α u 2 d x ≤ C

if and only if 0 < α ≤ 4 π 1 − β 2 .

By ( V 2 ) , ( K 1 ) , and ( K 3 ) , we can assume that B r ( x j ) ∩ B r ( x k ) = ∅ for all j ≠ k and ∪ i = 1 l B r ( x i ) ⊂ Ω . Since K is singular at x i , i = 1 , 2 , … , l , we establish the following Trudinger-Moser-type inequality in a bounded domain.

Lemma 2.4

Choose R > 0 such that Ω ⊂ B R 2 ( 0 ) . Assume that ( V 1 )–( V 3 ), ( h 1 ) , ( K 1 ) , and ( K 3 ) hold. If u ∈ X and α > 0 , then

(2.3) ∫ B R ( 0 ) K ( x ) ( e α u 2 − 1 ) d x < ∞ .

Moreover, if α ∈ ( 0 , 4 π + 2 π m 0 ) , then, for any τ > 0 , there exists R ′ > 0 such that for R > R ′ , there exists a constant C independent of μ ≥ 1 such that

(2.4) sup ‖ ∇ u ‖ 2 2 + τ ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x ≤ 1 ∫ B R ( 0 ) K ( x ) e α u 2 d x ≤ C .

Proof

By m 0 ∈ ( − 2 , 0 ) , we choose q > 1 such that q m 0 > − 2 . Then, for any i , we obtain

(2.5) ∫ B r ( x i ) ∣ x − x i ∣ m i q d x < ∞ .

Let q ′ = q q − 1 . If u ∈ X , by ( K 1 ) , ( K 3 ) , and the Hölder’s inequality,

(2.6) ∫ B R ( 0 ) K ( x ) ( e α u 2 − 1 ) d x = ∑ i = 1 l ∫ B r ( x i ) K ( x ) ( e α u 2 − 1 ) d x + ∫ B R ( 0 ) \ ∪ i = 1 l B r ( x i ) K ( x ) ( e α u 2 − 1 ) d x ≤ ∑ i = 1 l d 2 ∫ B r ( x i ) ∣ x − x i ∣ m i q d x 1 q ∫ B R ( 0 ) e q ′ α u 2 d x 1 q ′ + C ∫ B R ( 0 ) e α u 2 d x .

Define φ R ∈ C 0 ∞ ( R 2 ) such that φ R ( x ) = 1 for ∣ x ∣ ≤ R , φ R ( x ) = 0 for ∣ x ∣ ≥ 2 R , and 0 ≤ φ R ( x ) ≤ 1 and ∣ ∇ φ R ( x ) ∣ ≤ 2 R for x ∈ R 2 . Since u ∈ X , by ( V 1 ), ( V 2 ), and ( h 1 ) , we have that u ∈ H 1 ( B 2 R ( 0 ) ) . Moreover, u φ R ∈ H 0 1 ( B 2 R ( 0 ) ) . By Lemma 2.1, we obtain that for any α > 0 ,

(2.7) ∫ B R ( 0 ) e α u 2 d x ≤ ∫ B 2 R ( 0 ) e α ( φ R u ) 2 d x < + ∞ .

By (2.5)–(2.7), we obtain (2.3).

Let v i ≔ u ( . + x i ) , where i = 1 , …, k . Then, v i φ R ∈ H 0 1 ( B 2 R ( 0 ) ) . By ( V 3 ) , there exists R > 0 such that V ( x ) ≥ V 0 2 ∣ x ∣ γ for ∣ x ∣ ≥ R 2 . Thus, since ∣ x i ∣ ≤ R 2 for i = 1 ,… l , by the Young’s inequality, we derive that for any ε > 0 ,

(2.8) ∫ B 2 R ( 0 ) ∣ ∇ ( v i φ R ) ∣ 2 d x ≤ ( 1 + ε 2 ) ∫ B 2 R ( 0 ) ∣ ∇ v i ∣ 2 d x + ( 1 + ε − 2 ) ∫ B 2 R ( 0 ) ∣ ∇ φ R ∣ 2 v i 2 d x ≤ ( 1 + ε 2 ) ‖ ∇ u ‖ 2 2 + 4 ( 1 + ε − 2 ) R 2 ∫ R ≤ ∣ x ∣ ≤ 2 R v i 2 d x ≤ ( 1 + ε 2 ) ‖ ∇ u ‖ 2 2 + 8 ( 1 + ε − 2 ) ( 5 R 2 ) γ μ V 0 R 2 ∫ R 2 μ V ( x ) u 2 d x .

Let τ > 0 . Since α ∈ ( 0 , 4 π + 2 π m 0 ) , by (2.8), there exist ε > 0 small and R > 0 large such that for μ ≥ 1 ,

(2.9) sup ‖ ∇ u ‖ 2 2 + τ ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x ≤ 1 α ∫ B 2 R ( 0 ) ∣ ∇ ( v i φ R ) ∣ 2 d x ≤ 4 π + 2 π m 0 .

Similarly, there exists R > 0 large such that for μ ≥ 1 ,

(2.10) sup ‖ ∇ u ‖ 2 2 + τ ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x ≤ 1 α ∫ B 2 R ( 0 ) ∣ ∇ ( u φ R ) ∣ 2 d x ≤ 4 π .

By ( K 1 ) and ( K 3 ) , we obtain

(2.11) ∫ B R ( 0 ) K ( x ) ( e α u 2 − 1 ) d x ≤ ∑ i = 1 l d 2 ∫ B r ( 0 ) ∣ x ∣ m i e α v i 2 d x + C ∫ B R ( 0 ) ( e α u 2 − 1 ) d x ≤ ∑ i = 1 l d 2 ∫ B 2 R ( 0 ) ∣ x ∣ m i e α ( v i φ R ) 2 d x + C ∫ B 2 R ( 0 ) ( e α ( u φ R ) 2 − 1 ) d x .

By (2.9)–(2.11), Lemmas 2.2 and 2.3, we obtain (2.4).□

Let

(2.12) c 0 ≔ a 2 4 π + 2 π m 0 α 0 + b 4 4 π + 2 π m 0 α 0 2 .

Corollary 2.1

Choose R > 0 such that Ω ⊂ B R 2 ( 0 ) . Assume that ( V 1 )–( V 3 ), ( h 1 ) , ( K 1 ) , and ( K 3 ) hold. Then, there exist q > 1 (close to 1), α > α 0 (close to α 0 ), and R 0 > 0 such that for R > R 0 , there exists a constant C independent of μ ≥ 1 such that

sup ‖ ∇ u ‖ 2 2 ≤ 4 π + 2 π m 0 α 0 − δ , ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x ≤ 4 c 0 ∫ B R ( 0 ) K ( x ) e q α u 2 d x ≤ C ,

where δ ∈ 0 , 4 π + 2 π m 0 α 0 . In particular, if

lim n → ∞ ‖ ∇ u n ‖ 2 2 < 4 π + 2 π m 0 α 0 , lim n → ∞ ∫ R 2 ( h ( x ) + μ V ( x ) ) u n 2 d x < 4 c 0 ,

then there exist q > 1 (close to 1), α > α 0 (close to α 0 ), and R 0 > 0 such that for R > R 0 , there exists a constant C independent of μ ≥ 1 such that for n large,

∫ B R ( 0 ) K ( x ) e q α u n 2 d x ≤ C .

Lemma 2.5

Choose R > 0 such that Ω ⊂ B R 2 ( 0 ) . Assume that ( V 1 )–( V 3 ), ( h 1 ) , ( K 1 ) , and ( K 3 ) hold. If p ≥ 2 , then there exists C = C ( R ) > 0 such that for all u ∈ X ,

(2.13) ∫ B R ( 0 ) K ( x ) ∣ u ∣ p d x ≤ C ‖ u ‖ p .

Proof

Choose q 0 > 1 such that m 0 q 0 > − 2 . Let q 0 ′ = q 0 q 0 − 1 . By ( K 1 ) and ( K 3 ) , we have that

(2.14) ∫ B R ( 0 ) K ( x ) ∣ u ∣ p d x = ∑ i = 1 l ∫ B r ( x i ) K ( x ) ∣ u ∣ p d x + ∫ B R ( 0 ) \ ∪ i = 1 l B r ( x i ) K ( x ) ∣ u ∣ p d x ≤ ∑ i = 1 l d 2 ∫ B r ( x i ) ∣ x − x i ∣ m i q 0 d x 1 q 0 ∫ B R ( 0 ) ∣ u ∣ p q 0 ′ d x 1 q 0 ′ + C ∫ B R ( 0 ) ∣ u ∣ p d x .

Define φ R ∈ C 0 ∞ ( R 2 ) such that φ R ( x ) = 1 for ∣ x ∣ ≤ R , φ R ( x ) = 0 for ∣ x ∣ ≥ 2 R , and 0 ≤ φ R ( x ) ≤ 1 and ∣ ∇ φ R ( x ) ∣ ≤ 2 R for x ∈ R 2 . By ( V 3 ) , there exists R > 0 such that V ( x ) ≥ V 0 2 ∣ x ∣ γ for ∣ x ∣ ≥ R 2 . Since φ R u ∈ H 0 1 ( B 2 R ( 0 ) ) , by (2.14) and the Sobolev embedding theorem, we derive that there exists c ′ > 0 such that for all u ∈ X ,

(2.15) ∫ B R ( 0 ) K ( x ) ∣ u ∣ p d x ≤ ∑ i = 1 l d 2 ∫ B r ( x i ) ∣ x − x i ∣ m i q 0 d x 1 q 0 ∫ B 2 R ( 0 ) ∣ φ R u ∣ p q 0 ′ d x 1 q 0 ′ + C ∫ B 2 R ( 0 ) ∣ φ R u ∣ p d x ≤ c ′ ∫ B 2 R ( 0 ) ∣ ∇ ( φ R u ) ∣ 2 d x p 2 ≤ 2 c ′ ‖ ∇ u ‖ 2 2 + 16 c ′ ( 2 R ) γ V 0 R 2 ∫ R 2 V ( x ) u 2 d x p 2 ,

which implies (2.13).□

Lemma 2.6

Let μ > 1 . Assume that ( V 1 )–( V 3 ), ( h 1 ) , ( K 1 ) , ( K 3 ) , ( f 1 ), ( f 2 ), and ( f 5 ) hold. If ‖ ∇ u n ‖ 2 2 , ∫ R 2 ( h ( x ) + V ( x ) ) u n 2 d x , and ∫ R 2 K ( x ) f ( u n ) u n d x are bounded and u n ( x ) → u ( x ) a.e. x ∈ R 2 , then, for any R > 0 ,

(2.16) lim n → ∞ ∫ B R ( 0 ) K ( x ) ∣ f ( u n ) − f ( u ) ∣ d x = 0 .

Moreover,

(2.17) lim n → ∞ ∫ B R ( 0 ) K ( x ) F ( u n ) d x = ∫ B R ( 0 ) K ( x ) F ( u ) d x .

Proof

By ( f 4 ) , we obtain lim u → 0 + f ( u ) u 3 < + ∞ . Moreover, by ( f 1 ) and ( f 2 ), for any ε > 0 , there exists C ε > 0 such that

(2.18) ∣ f ( u ) ∣ ≤ ε ∣ u ∣ + C ε ( e α u 2 − 1 ) , ∀ u ∈ R .

By Lemma 2.4, we obtain that for any u ∈ X ,

(2.19) ∫ B R ( 0 ) K ( x ) ( e 2 α u 2 − 1 ) d x < + ∞ .

By Lemma 2.5, we obtain ∫ B R ( 0 ) K ( x ) u n 2 d x is bounded. Together with (2.18) and (2.19), we have that ∫ B R ( 0 ) K ( x ) ∣ f ( u ) ∣ ∣ u n ∣ d x is bounded. Thus, for any ε > 0 , there exists M ε > 0 such that for M ≥ M ε ,

(2.20) ∫ B R ( 0 ) ∩ { ∣ u n ∣ ≥ M } K ( x ) ∣ f ( u n ) − f ( u ) ∣ d x ≤ 1 M ∫ B R ( 0 ) K ( x ) ∣ f ( u n ) u n − f ( u ) u n ∣ d x ≤ ε .

We note that for any ε > 0 , there exists C ε > 0 such that

(2.21) ∣ f ( u ) ∣ ≤ ε + C ε ( e α u 2 − 1 ) , ∀ u ∈ R .

By ( K 1 ) , ( K 3 ) , (2.21), and Lemma 2.4, we obtain ∫ B R ( 0 ) K ( x ) ∣ f ( u ) ∣ d x < ∞ for any u ∈ X . Thus, by K ∈ L 1 ( B R ( 0 ) ) and the Lebesgue dominated convergence theorem,

(2.22) lim n → ∞ ∫ B R ( 0 ) ∩ { ∣ u n ∣ ≤ M } K ( x ) ∣ f ( u n ) − f ( u ) ∣ d x = lim n → ∞ ∫ B R ( 0 ) K ( x ) ∣ f ( u n ) − f ( u ) ∣ χ { ∣ u n ∣ ≤ M } ( x ) d x = 0 .

By (2.20) and (2.22), we obtain (2.16).

By ( f 1 ) and ( f 5 ) , there exists C 0 > 0 such that

(2.23) F ( u ) ≤ C 0 + L 0 f ( u ) , ∀ u ∈ R .

By K ∈ L 1 ( B R ( 0 ) ) , (2.16), (2.23), and the generalized Lebesgue dominated convergence theorem, we obtain (2.17).□

We consider the Moser sequence of functions

(2.24) ω ¯ n ( x ) = 1 2 π ( log n ) 1 2 , 0 ≤ ∣ x ∣ ≤ 1 n , log 1 ∣ x ∣ ( log n ) 1 2 , 1 n ≤ ∣ x ∣ ≤ 1 , 0 , ∣ x ∣ ≥ 1 .

It is well known that ‖ ∇ ω ¯ n ‖ 2 2 = 1 and ‖ ω ¯ n ‖ 2 2 = 1 4 log n + o ( 1 log n ) . By ( K 3 ) , we may assume that B r ( 0 ) ⊂ Ω and K ( x ) ≥ d 1 ∣ x ∣ m 0 for 0 < ∣ x ∣ < r . Define the functions ω n ( x ) = ω ¯ n ( x r ) . Then, ‖ ∇ ω n ‖ 2 2 = 1 . Define the functional

I 0 ( u ) = a 2 ∫ Ω ∣ ∇ u ∣ 2 d x + 1 2 ∫ Ω h ( x ) u 2 d x + b 4 ∫ Ω ∣ ∇ u ∣ 2 d x 2 − ∫ Ω K ( x ) F ( u ) d x ,

where u ∈ H 0 1 ( Ω ) .

Lemma 2.7

For n large, there holds max t ≥ 0 I 0 ( t ω n ) < c 0 .

Proof

Let α > α 0 and q > 2 . By ( f 1 ), ( f 2 ), and ( f 4 ) , for any ε > 0 , there exists C ε > 0 such that for all u ∈ R ,

∣ f ( u ) ∣ ≤ ε ∣ u ∣ + C ε ∣ u ∣ q − 1 ( e α u 2 − 1 ) .

By Corollary 2.1 and Lemma 2.5, for t > 0 small,

∫ Ω K ( x ) ∣ t ω n ∣ q ( e α t 2 ω n 2 − 1 ) d x ≤ ∫ Ω K ( x ) ∣ t ω n ∣ 2 q d x 1 2 ∫ Ω K ( x ) ( e 2 α t 2 ω n 2 − 1 ) d x 1 2 ≤ C ∫ Ω [ ∣ ∇ ( t ω n ) ∣ 2 + h ( x ) ( t ω n ) 2 ] d x q

and

∫ Ω K ( x ) ∣ t ω n ∣ 2 d x ≤ C ∫ Ω [ ∣ ∇ ( t ω n ) ∣ 2 + h ( x ) ( t ω n ) 2 ] d x .

Then, I 0 ( t ω n ) > 0 for t > 0 small. Let p > 3 . By ( f 1 ), ( f 2 ), and ( f 4 ) , there exist c 1 , c 2 > 0 such that

(2.25) f ( u ) ≥ c 1 u p − c 2 u , ∀ u ∈ R + .

Then, I 0 ( t ω n ) < 0 for t > 0 large. Moreover, max t ≥ 0 I 0 ( t ω n ) > 0 is attained at a t n > 0 and ( I 0 ′ ( t n ω n ) , t n ω n ) = 0 .

By ( I 0 ′ ( t n ω n ) , t n ω n ) = 0 , we have that

(2.26) ( a + b t n 2 ) t n 2 + t n 2 ∫ Ω h ( x ) ω n 2 d x ≥ d 1 r m 0 + 2 ∫ B 1 ( 0 ) ∣ x ∣ m 0 f ( t n ω ¯ n ) t n ω ¯ n d x .

If lim n → ∞ t n = 0 , then lim n → ∞ I 0 ( t n ω n ) = 0 . So we may assume that lim n → ∞ t n = l ∈ ( 0 , + ∞ ] . By ( f 2 ) ,

lim t → + ∞ F ( t ) t − 2 e α 0 t 2 = lim t → + ∞ f ( t ) 2 α 0 t − 1 e α 0 t 2 .

Then, by ( f 3 ) , for any δ > 0 , there exists t δ > 0 such that for t ≥ t δ ,

(2.27) f ( t ) t ≥ ( β − δ ) e α 0 t 2 , F ( t ) t 2 ≥ β − δ 2 α 0 e α 0 t 2 .

By (2.26) and (2.27),

( a + b t n 2 ) t n 2 + max Ω ¯ h r 2 t n 2 ‖ ω ¯ n ‖ 2 2 ≥ d 1 ( β − δ ) r m 0 + 2 e α 0 t n 2 log n 2 π ∫ B 1 n ( 0 ) ∣ x ∣ m 0 d x = 2 d 1 π ( β − δ ) r m 0 + 2 m 0 + 2 e α 0 t n 2 2 π − ( m 0 + 2 ) log n ,

from which we derive that lim n → ∞ t n = l ∈ ( 0 , + ∞ ) . Moreover, l ∈ 0 , 2 π ( m 0 + 2 ) α 0 . If l ∈ 0 , 2 π ( m 0 + 2 ) α 0 , then lim n → ∞ I 0 ( t n ω n ) < c 0 .

Now we consider the case l = 2 π ( m 0 + 2 ) α 0 . By (2.27),

∫ Ω K ( x ) F ( t n ω n ) d x ≥ d 1 ( β − δ ) 2 α 0 ∫ A n t n − 2 ω n − 2 e α 0 t n 2 ω n 2 ∣ x ∣ m 0 d x ,

where A n ≔ { x ∈ B d ( 0 ) : t n ω n ( x ) ≥ t δ } . Let s ∈ ( 0 , 1 2 ) . Then, there exists N 1 ∈ N such that for N > N 1 , we have that t n ω n ( x ) ≥ t δ for all ∣ x ∣ ≤ r n s . So

(2.28) ∫ Ω K ( x ) F ( t n ω n ) d x ≥ d 1 ( β − δ ) r m 0 + 2 2 α 0 ∫ B 1 n s ( 0 ) t n − 2 ω ¯ n − 2 e α 0 t n 2 ω ¯ n 2 ∣ x ∣ m 0 d x .

We note that

(2.29) ∫ B 1 n s ( 0 ) t n − 2 ω ¯ n − 2 e α 0 t n 2 ω ¯ n 2 ∣ x ∣ m 0 d x = ∫ ∣ x ∣ ≤ 1 n 2 π n α 0 t n 2 2 π ∣ x ∣ m 0 t n 2 log n d x + ∫ 1 n ≤ ∣ x ∣ ≤ 1 n s 2 π log n e α 0 t n 2 2 π log n log 2 ∣ x ∣ ∣ x ∣ m 0 t n 2 log 2 ∣ x ∣ d x = 4 π 2 m 0 + 2 n α 0 t n 2 2 π − ( m 0 + 2 ) t n 2 log n + 4 π 2 log n t n 2 ∫ 1 n 1 n s x m 0 + 1 e α 0 t n 2 2 π log n log 2 x log 2 x d x .

Let C n = α 0 t n 2 2 π and t = C n log n log 1 x . Then,

(2.30) ∫ 1 n 1 n s x m 0 + 1 e α 0 t n 2 2 π log n log 2 x log 2 x d x = C n log n ∫ s C n C n n − ( m 0 + 2 ) x C n + x 2 C n x − 2 d x ≥ 1 log n ∫ s 1 n − ( m 0 + 2 ) x + C n x 2 d x .

Here,

(2.31) ∫ s 1 n − ( m 0 + 2 ) x + C n x 2 d x ≥ ∫ m 0 + 2 2 C n 1 n [ 2 C n − ( m 0 + 2 ) ] x − C n d x + ∫ s m 0 + 2 2 C n n − ( m 0 + 2 ) x d x = n − α 0 t n 2 2 π α 0 t n 2 π − ( m 0 + 2 ) log n n α 0 t n 2 π − ( m 0 + 2 ) − n ( m 0 + 2 ) − ( m 0 + 2 ) 2 π α 0 t n 2 + 1 ( m 0 + 2 ) log n n − ( m 0 + 2 ) s − n − ( m 0 + 2 ) 2 π α 0 t n 2 .

By (2.28)–(2.31), there exists C > 0 such that

(2.32) ∫ Ω K ( x ) F ( t n ω n ) d x ≥ 2 d 1 ( β − δ ) π 2 r m 0 + 2 α 0 ( m 0 + 2 ) n α 0 t n 2 2 π − ( m 0 + 2 ) t n 2 log n + 2 d 1 ( β − δ ) π 2 r m 0 + 2 α 0 ( m 0 + 2 ) n − ( m 0 + 2 ) s − n − ( m 0 + 2 ) 2 π α 0 t n 2 t n 2 log n + 2 d 1 ( β − δ ) π 2 r m 0 + 2 α 0 n − α 0 t n 2 2 π n α 0 t n 2 π − ( m 0 + 2 ) − n ( m 0 + 2 ) − ( m 0 + 2 ) 2 π α 0 t n 2 t n 2 α 0 t n 2 π − ( m 0 + 2 ) log n ≥ 2 d 1 ( β − δ ) π 2 r m 0 + 2 ( m 0 + 2 ) ( α 0 t n 2 − ( m 0 + 2 ) π ) n α 0 t n 2 2 π − ( m 0 + 2 ) log n + C n − ( m 0 + 2 ) s log n .

Then,

(2.33) I 0 ( t n ω n ) ≤ a 2 t n 2 + b 4 t n 4 + r 2 max Ω ¯ h 2 ‖ ω ¯ n ‖ 2 2 t n 2 − 2 d 1 ( β − δ ) π 2 r m 0 + 2 ( m 0 + 2 ) ( α 0 t n 2 − ( m 0 + 2 ) π ) n α 0 t n 2 2 π − ( m 0 + 2 ) log n .

Since lim n → ∞ t n = 2 π ( m 0 + 2 ) α 0 , we obtain that for any ε > 0 , there exists N 2 ∈ N such that for n > N 2 ,

1 α 0 t n 2 − ( m 0 + 2 ) π ≥ 1 π ( m 0 + 2 ) + ε .

Let

l n ( t ) = a 2 t 2 + b 4 t 4 + r 2 max Ω ¯ h 2 ‖ ω ¯ n ‖ 2 2 t 2 − 2 d 1 ( β − δ ) π 2 r m 0 + 2 ( m 0 + 2 ) ( π ( m 0 + 2 ) + ε ) n α 0 t 2 2 π − ( m 0 + 2 ) log n .

Then, I 0 ( t n ω n ) ≤ sup t ≥ 0 l n ( t ) . Obviously, sup t ≥ 0 l n ( t ) is attained at a t n ′ > 0 . By ( l n ′ ( t n ′ ) , t n ′ ) = 0 ,

(2.34) a + r 2 max Ω ¯ h ‖ ω ¯ n ‖ 2 2 + b ( t n ′ ) 2 = 2 d 1 α 0 π ( β − δ ) r m 0 + 2 ( m 0 + 2 ) ( π ( m 0 + 2 ) + ε ) n α 0 ( t n ′ ) 2 2 π − ( m 0 + 2 ) .

Then,

(2.35) I 0 ( t n ω n ) ≤ a 2 ( t n ′ ) 2 + b 4 ( t n ′ ) 4 + r 2 max Ω ¯ h 2 ‖ ω ¯ n ‖ 2 2 ( t n ′ ) 2 − π α 0 log n ( a + r 2 max Ω ¯ h ‖ ω ¯ n ‖ 2 2 + b ( t n ′ ) 2 ) .

By (2.34), we obtain lim n → ∞ α 0 ( t n ′ ) 2 = 2 π ( m 0 + 2 ) . Moreover,

(2.36) ( t n ′ ) 2 = 2 π ( m 0 + 2 ) α 0 + 2 π α 0 log ( m 0 + 2 ) ( π ( m 0 + 2 ) + ε ) ( a + r 2 max Ω ¯ h ‖ ω ¯ n ‖ 2 2 + b ( t n ′ ) 2 ) 2 d 1 α 0 π ( β − δ ) r m 0 + 2 log n ≔ 2 π ( m 0 + 2 ) α 0 + A n ,

where A n = O ( 1 log n ) . By (2.36), we have that

(2.37) a 2 ( t n ′ ) 2 + b 4 ( t n ′ ) 4 + r 2 max Ω ¯ h 2 ‖ ω ¯ n ‖ 2 2 ( t n ′ ) 2 ≤ c 0 + a 2 + b π ( m 0 + 2 ) α 0 A n + π r 2 ( m 0 + 2 ) max Ω ¯ h 4 α 0 log n + o 1 log n .

By (2.35)–(2.37), we derive that

I 0 ( t n ω n ) ≤ c 0 + a 2 + b π ( m 0 + 2 ) α 0 A n + π r 2 ( m 0 + 2 ) max Ω ¯ h 4 α 0 log n + o 1 log n − π α 0 log n a + 2 b π ( m 0 + 2 ) α 0 .

Since β > ( m 0 + 2 ) 2 [ a + 2 b π ( m 0 + 2 ) α 0 ] 2 e d 1 α 0 r m 0 + 2 e r 2 ( m 0 + 2 ) max Ω ¯ h 4 [ a + 2 b π ( m 0 + 2 ) α 0 ] , by choosing δ , ε small, and n large, we obtain that I 0 ( t n ω n ) < c 0 .□

3 The modified problem

Let d > 0 . Define

χ ( x ) = 1 , x ∈ Ω d , 0 , x ∈ R 2 \ Ω d ,

where Ω d ≔ { x ∈ R 2 : dist ( x , Ω ) < d } . For x ∈ R 2 \ Ω d , define

f ˆ ( x , u ) = min { K ( x ) f ( u ) , κ V ( x ) u + } ,

where u + = max { u , 0 } and κ ∈ ( 0 , 1 ) . Define

g ( x , u ) = χ ( x ) K ( x ) f ( u ) + ( 1 − χ ( x ) ) f ˆ ( x , u ) .

Then,

G ( x , u ) = ∫ 0 u g ( x , s ) d s = χ ( x ) K ( x ) F ( u ) + ( 1 − χ ( x ) ) F ˆ ( x , u ) ,

where F ˆ ( x , u ) = ∫ 0 u f ˆ ( x , s ) d s . By ( f 4 ) and the structure of f ˆ , we derive that for all ( x , u ) ∈ R 2 × R ,

(3.1) f ( u ) u − 4 F ( u ) ≥ 0 , f ˆ ( x , u ) u − 2 F ˆ ( x , u ) ≥ 0 .

Instead of studying (1.1), we consider the following modified problem:

(3.2) − ( a + b ‖ ∇ u ‖ 2 2 ) Δ u + ( h ( x ) + μ V ( x ) ) u = g ( x , u ) in R 2 .

The functional associated with (3.2) is

I ˆ μ ( u ) = a 2 ‖ ∇ u ‖ 2 2 + b 4 ‖ ∇ u ‖ 2 4 + 1 2 ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x − ∫ R 2 G ( x , u ) d x , u ∈ X μ .

Obviously, I ˆ μ ∈ C 1 ( X μ , R ) and critical points of I ˆ μ are weak solutions of (3.2).

Lemma 3.1

Let μ > 1 . Define l ( t ) = I ˆ μ ( t u ) , where t ≥ 0 and u ∈ X μ with ∣ supp u ∩ Ω d ∣ > 0 . Then, there exists a unique t 0 > 0 such that l ′ ( t 0 ) = 0 , l ′ ( t ) > 0 for t ∈ ( 0 , t 0 ) , and l ′ ( t ) < 0 for t > t 0 .

Proof

Obviously, l ( 0 ) = 0 . Let α > α 0 and q > 2 . By ( f 1 ), ( f 2 ), and ( f 4 ) , we derive that for any ε > 0 , there exists C ε > 0 such that for all ( x , u ) ∈ R 2 × R ,

(3.3) ∣ g ( x , u ) ∣ ≤ ε χ ( x ) K ( x ) ∣ u ∣ + C ε χ ( x ) K ( x ) ∣ u ∣ q − 1 ( e α u 2 − 1 ) + κ V ( x ) ∣ u ∣ .

Then, for all ( x , u ) ∈ R 2 × R ,

(3.4) ∣ G ( x , u ) ∣ ≤ ε 2 χ ( x ) K ( x ) ∣ u ∣ 2 + C ε q χ ( x ) K ( x ) ∣ u ∣ q ( e α u 2 − 1 ) + κ 2 V ( x ) ∣ u ∣ 2 .

By Corollary 2.1 and Lemma 2.5, we obtain that there exists ρ > 0 small such that for ‖ u ‖ μ ≤ ρ ,

(3.5) ∫ R 2 χ ( x ) K ( x ) ∣ u ∣ q ( e α u 2 − 1 ) d x ≤ ∫ ∣ x ∣ ≤ R K ( x ) ∣ u ∣ 2 q d x 1 2 ∫ ∣ x ∣ ≤ R K ( x ) ( e 2 α u 2 − 1 ) d x 1 2 ≤ C ‖ u ‖ μ q ,

and there exists C > 0 such that for all u ∈ X μ ,

(3.6) ∫ R 2 χ ( x ) K ( x ) ∣ u ∣ 2 d x ≤ C ‖ u ‖ μ 2 .

By (3.4)–(3.6), we obtain l ( t ) > 0 for t > 0 small. Let p > 3 . By (2.25), we obtain l ( t ) < 0 for t > 0 large. Moreover, max t ≥ 0 l ( t ) is attained at a t 0 > 0 and l ′ ( t 0 ) = 0 .

Let

y ( t ) = ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x − ∫ R 2 \ Ω d f ˆ ( x , t u ) u t d x + t 2 a t 2 + b ‖ ∇ u ‖ 2 2 ‖ ∇ u ‖ 2 2 − ∫ Ω d K ( x ) f ( t u ) u t 3 d x , t > 0 .

Then, y ( t 0 ) = 0 . By the structure of g , we derive that

(3.7) ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x − ∫ R 2 \ Ω d f ˆ ( x , t 0 u ) u t 0 d x > 0 , a t 0 2 + b ‖ ∇ u ‖ 2 2 ‖ ∇ u ‖ 2 2 − ∫ Ω d K ( x ) f ( t 0 u ) u t 0 3 d x < 0 .

By (3.7) and ( f 4 ) , we obtain t 0 is unique. Moreover, l ′ ( t ) > 0 for t ∈ ( 0 , t 0 ) and l ′ ( t ) < 0 for t > t 0 .□

Define

(3.8) c ˆ μ ≔ inf γ ∈ Γ max t ∈ [ 0 , 1 ] I ˆ μ ( γ ( t ) ) ,

where Γ ≔ { γ ∈ C ( [ 0 , 1 ] , X μ ) : γ ( 0 ) = 0 , I ˆ μ ( γ ( 1 ) ) < 0 } .

Lemma 3.2

Let μ > 1 . Then, there exists a sequence { u n } ⊂ X μ such that lim n → ∞ I ˆ μ ( u n ) = c ˆ μ ∈ ( 0 , max t ≥ 0 I 0 ( t ω n ) ] and lim n → ∞ I ˆ μ ′ ( u n ) = 0 .

Proof

By (3.4)–(3.6), we obtain that there exist ρ , η > 0 such that I μ ( u ) ≥ η for ‖ u ‖ μ = ρ . Let v ∈ C 0 ∞ ( Ω ) \ { 0 } be a nonnegative function. By (2.25), we have that lim t → + ∞ I μ ( t v ) = − ∞ . Thus, by the mountain pass lemma in [6], there exists a sequence { u n } ⊂ X μ such that lim n → ∞ I ˆ μ ( u n ) = c ˆ μ > 0 and lim n → ∞ I ˆ μ ′ ( u n ) = 0 . By the definition of c ˆ μ , we obtain c ˆ μ ≤ max t ≥ 0 I ˆ μ ( t ω n ) = max t ≥ 0 I 0 ( t ω n ) .□

Lemma 3.3

Let μ > max { 1 , 2 κ } . If { u n } ⊂ X μ is a sequence such that I ˆ μ ( u n ) → c ˆ μ ∈ ( 0 , max t ≥ 0 I 0 ( t ω n ) ] and I ˆ μ ′ ( u n ) → 0 , then there exists u μ ∈ X μ \ { 0 } such that u n → u μ in X μ .

Proof

By ( f 4 ) and the structure of g , we obtain that

(3.9) c ˆ μ + o n ( 1 ) + o n ( 1 ) ‖ u n ‖ μ = I ˆ μ ( u n ) − 1 4 ( I ˆ μ ′ ( u n ) , u n ) ≥ a 4 ‖ ∇ u n ‖ 2 2 + 1 4 ∫ R 2 [ h ( x ) + ( μ − κ ) V ( x ) ] u n 2 d x .

Then, ‖ u n ‖ μ is bounded. Moreover, ∫ R 2 K ( x ) f ( u n ) u n d x is bounded.

We claim that u n ⇀ u μ ≠ 0 weakly in X μ . Otherwise, u n ⇀ 0 weakly in X μ . By Lemma 2.6, we obtain lim n → ∞ ∫ Ω d K ( x ) F ( u n ) d x = 0 . So

(3.10) c ˆ μ ≥ a 2 lim n → ∞ ‖ ∇ u n ‖ 2 2 + b 4 lim n → ∞ ‖ ∇ u n ‖ 2 4 + 1 2 lim n → ∞ ∫ R 2 [ h ( x ) + ( μ − κ ) V ( x ) ] u n 2 d x .

Moreover, by Lemma 2.7,

(3.11) lim n → ∞ ‖ ∇ u n ‖ 2 2 < 4 π + 2 π m 0 α 0 , lim n → ∞ ∫ R 2 [ h ( x ) + μ V ( x ) ] u n 2 d x < 4 c 0 .

Define P ( x , t ) = g ( x , t ) t and Q ( x , t ) = K ( x ) t ( e α t 2 − 1 ) . By (3.11), Corollary 2.1, and Lemma 2.5, there exist α > α 0 (close to α 0 ), q > 1 (close to 1), and R 0 > 0 such that for R > R 0 , there exists C > 0 independent of n such that

(3.12) ∫ ∣ x ∣ ≤ R Q ( x , u n ) d x ≤ ∫ ∣ x ∣ ≤ R K ( x ) ∣ u n ∣ q ′ d x 1 q ′ ∫ ∣ x ∣ ≤ R K ( x ) ( e q α u n 2 − 1 ) d x 1 q ≤ C ,

where q ′ = q q − 1 . By ( f 2 ) , we have that

(3.13) lim t → ∞ P ( x , t ) Q ( x , t ) = 0 uniformly for ∣ x ∣ ≤ R .

Furthermore,

(3.14) lim m → ∞ P ( x , u n ( x ) ) = 0 a.e. ∣ x ∣ ≤ R .

By (3.13) and ( f 1 ) , for any ε > 0 , there exists C ε > 0 independent of n such that

(3.15) P ( x , u n ) ≤ ε Q ( x , u n ) + C ε K ( x ) , ∀ ∣ x ∣ ≤ R .

By (3.12) and (3.15),

(3.16) lim T → ∞ ∣ { P ( x , u n ) ≥ T } ∩ { ∣ x ∣ ≤ R } ∣ ≤ lim T → ∞ ∫ { P ( x , u n ) ≥ T } ∩ { ∣ x ∣ ≤ R } 1 T P ( x , u n ) d x ≤ lim T → ∞ ∫ ∣ x ∣ ≤ R 1 T ( ε Q ( x , u n ) + C ε K ( x ) ) d x = 0 .

Moreover,

(3.17) lim T → ∞ ∫ { P ( x , u n ) ≥ T } ∩ { ∣ x ∣ ≤ R } P ( x , u n ) d x ≤ C ε + C ε lim T → ∞ ∫ { P ( x , u n ) ≥ T } ∩ { ∣ x ∣ ≤ R } K ( x ) d x = C ε .

By the Lebesgue dominated convergence theorem, we have that

(3.18) lim n → ∞ ∫ { P ( x , u n ) ≤ T } ∩ { ∣ x ∣ ≤ R } P ( x , u n ) d x = 0 .

By (3.17)–(3.18), we derive that

(3.19) lim n → ∞ ∫ B R ( 0 ) g ( x , u n ) u n d x = 0 .

Since g ( x , u n ) u n ≤ κ V ( x ) u n 2 for ∣ x ∣ ≥ R , by ( I ˆ μ ′ ( u n ) , u n ) → 0 and (3.19), we obtain that u n → 0 in X μ , a contradiction with c ˆ μ > 0 .

Let A = lim n → ∞ ‖ ∇ u n ‖ 2 2 . Define the functional

J μ ( u ) = a + b A 2 ‖ ∇ u ‖ 2 2 + 1 2 ∫ R 2 ( h ( x ) + μ V ( x ) ) u 2 d x − ∫ R 2 G ( x , u ) d x , u ∈ X μ .

By I ˆ μ ′ ( u n ) = o n ( 1 ) , we obtain J μ ′ ( u n ) = o n ( 1 ) . Then, J μ ′ ( u μ ) = 0 . We claim that lim n → ∞ ‖ u n ‖ μ 2 = ‖ u μ ‖ μ 2 . Otherwise, ‖ u μ ‖ μ 2 < lim n → ∞ ‖ u n ‖ μ 2 . Then, ( I ˆ μ ′ ( u μ ) , u μ ) < 0 . Since u μ ≠ 0 , we obtain ∣ supp u μ ∩ Ω d ∣ > 0 . By Lemma 3.1, there exists a unique t μ with t μ ∈ ( 0 , 1 ) such that ( I ˆ μ ′ ( t μ u μ ) , t μ u μ ) = 0 . By the structure of g , we have that for x ∈ R 2 \ Ω d ,

(3.20) μ 4 V ( x ) u n 2 + 1 4 f ˆ ( x , u n ) u n − F ˆ ( x , u n ) ≥ 0 .

By (3.20), ( f 4 ) , and Fatou’s lemma, we derive that

(3.21) c ˆ μ = I ˆ μ ( u n ) − 1 4 ( I ˆ μ ′ ( u n ) , u n ) + o n ( 1 ) ≥ a 4 ‖ ∇ u μ ‖ 2 2 + 1 4 ∫ R 2 ( h ( x ) + μ V ( x ) ) u μ 2 d x + ∫ R 2 \ Ω d 1 4 f ˆ ( x , u μ ) u μ − F ˆ ( x , u μ ) d x + ∫ Ω d 1 4 K ( x ) f ( u μ ) u μ − K ( x ) F ( u μ ) d x + o n ( 1 ) .

Since t μ ∈ ( 0 , 1 ) , by ( f 4 ) , we obtain

(3.22) ∫ Ω d 1 4 K ( x ) f ( u μ ) u μ − K ( x ) F ( u μ ) d x ≥ ∫ Ω d 1 4 K ( x ) f ( t μ u μ ) t μ u μ − K ( x ) F ( t μ u μ ) d x .

By ( K 1 ) , ( V 1 ) , and ( f 4 ) , we derive that for any x ∈ R 2 \ Ω d , if K ( x ) > 0 , then there exists a unique u x > 0 such that K ( x ) f ( u ) = κ V ( x ) u for u = u x , K ( x ) f ( u ) < κ V ( x ) u for u < u x , and K ( x ) f ( u ) > κ V ( x ) u for u > u x . Thus,

(3.23) μ 4 ∫ R 2 V ( x ) u μ 2 d x + ∫ R 2 \ Ω d 1 4 f ˆ ( x , u μ ) u μ − F ˆ ( x , u μ ) d x ≥ μ 4 ∫ R 2 V ( x ) ( t μ u μ ) 2 d x + ∫ R 2 \ Ω d 1 4 f ˆ ( x , t μ u μ ) t μ u μ − F ˆ ( x , t μ u μ ) d x .

By (3.21)–(3.23) and the definition of c ˆ μ , we have that

(3.24) c ˆ μ > I ˆ μ ( t μ u μ ) = max t ≥ 0 I ˆ μ ( t u μ ) ≥ c ˆ μ ,

a contradiction. So lim n → ∞ ‖ u n ‖ μ 2 = ‖ u μ ‖ μ 2 . Moreover, u n → u μ in X μ as n → ∞ and I ˆ μ ′ ( u μ ) = 0 .□

4 The original problem

Lemma 4.1

If I ˆ μ n ( u μ ) = c ˆ μ and I ˆ μ ′ ( u μ ) = 0 , then u μ → u 0 in X as μ → + ∞ , where u 0 ≠ 0 is a nonnegative solution of the equation

(4.1) − a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u + h ( x ) u = K ( x ) f ( u ) in O m e g a ,

and u 0 ( x ) = 0 a.e. x ∈ R 2 \ Ω ¯ .

Proof

We only need to prove that for any sequence { μ n } with μ n → ∞ as n → ∞ , if I ˆ μ n ( u μ n ) = c ˆ μ n and I ˆ μ n ′ ( u μ n ) = 0 , then u μ n → u 0 in X as n → ∞ .

Let A k = { x ∈ R 2 : V ( x ) ≥ 1 k } , where k ∈ N . Then, ∪ k = 1 ∞ A k = R 2 \ Ω ¯ . By I ˆ μ n ( u μ n ) = c ˆ μ n , I ˆ μ n ′ ( u μ n ) = 0 and ( f 4 ) , we have that ‖ ∇ u μ n ‖ 2 2 , ∫ R 2 ( h ( x ) + μ n V ( x ) ) u μ n 2 d x , and ∫ R 2 K ( x ) f ( u μ n ) u μ n d x are bounded. Assume that u μ n ⇀ u 0 weakly in X . By Fatou’s Lemma, we obtain ∫ R 2 V ( x ) u 0 2 d x = 0 . Moreover, ∫ A k u 0 2 d x = 0 . Then, u 0 ( x ) = 0 a.e. x ∈ R 2 \ Ω ¯ .

We claim that u μ n ⇀ u 0 ≠ 0 weakly in X . Otherwise, u μ n ⇀ 0 weakly in X . By Lemmas 2.7 and 3.2, we have that c ˆ μ n ≤ max t ≥ 0 I 0 ( t ω n ) < c 0 . Since ∫ R 2 K ( x ) f ( u μ n ) u μ n d x is bounded, by Lemma 2.6, we obtain that

c 0 > a 2 lim n → ∞ ‖ ∇ u μ n ‖ 2 2 + b 4 lim n → ∞ ‖ ∇ u μ n ‖ 2 4 + 1 2 lim n → ∞ ∫ R 2 [ h ( x ) + ( μ n − κ ) V ( x ) ] u μ n 2 d x .

Moreover,

(4.2) lim n → ∞ ‖ ∇ u μ n ‖ 2 2 < 4 π + 2 π m 0 α 0 , lim n → ∞ ∫ R 2 ( h ( x ) + μ n V ( x ) ) u μ n 2 d x < 4 c 0 .

Let R > 0 be such that B R ( 0 ) ⊃ Ω d . Similar to the argument in Lemma 3.3, we can prove that lim n → ∞ ∫ B R ( 0 ) g ( x , u μ n ) u μ n d x = 0 . Moreover, by ( I ˆ μ n ′ ( u μ n ) , u μ n ) = 0 , we have that lim n → ∞ ‖ u μ n ‖ μ n = 0 . Let α > α 0 and t > 2 . By ( I ˆ μ n ′ ( u μ n ) , u μ n ) = 0 and (3.3), we obtain that for any ε > 0 , there exists C ε > 0 such that

(4.3) a ‖ ∇ u μ n ‖ 2 2 + ∫ R 2 [ h ( x ) + ( μ n − κ ) V ( x ) ] u μ n 2 d x ≤ ε ∫ B R ( 0 ) K ( x ) u μ n 2 d x + C ε ∫ B R ( 0 ) K ( x ) ∣ u μ n ∣ t ( e α u μ n 2 − 1 ) d x .

By (4.2) and Corollary 2.1, there exist q > 1 (close to 1) and α > α 0 (close to α 0 ) such that

(4.4) ∫ B R ( 0 ) K ( x ) e q α u μ n 2 d x ≤ C ,

where C > 0 is independent of n . By (4.3), (4.4), and Lemma 2.5, we derive that ‖ u μ n ‖ μ n 2 ≤ C ′ ‖ u μ n ‖ μ n t , a contradiction to lim n → ∞ ‖ u μ n ‖ μ n = 0 .

Let A = lim n → ∞ ‖ ∇ u μ n ‖ 2 2 . Define the functionals I ¯ 0 and I ˜ 0 on X by

I ¯ 0 ( u ) = a 2 + b A 4 ∫ R 2 ∣ ∇ u ∣ 2 d x + 1 2 ∫ R 2 ( h ( x ) + V ( x ) ) u 2 d x − ∫ R 2 G ( x , u ) d x , I ˜ 0 ( u ) = a 2 + b A 2 ∫ R 2 ∣ ∇ u ∣ 2 d x + 1 2 ∫ R 2 ( h ( x ) + V ( x ) ) u 2 d x − ∫ R 2 G ( x , u ) d x .

Since u μ n ⇀ u 0 weakly in X with u 0 ( x ) = 0 a.e. x ∈ R 2 \ Ω ¯ , we have that ( I ˜ 0 ′ ( u 0 ) , u 0 ) = 0 . By ( f 4 ) , we obtain I ¯ 0 ( u 0 ) ≥ 0 . Let w μ n = u μ n − u 0 . Then, w μ n ⇀ 0 weakly in X . Let R > 0 be such that B R ( 0 ) ⊃ Ω d . Since u 0 ( x ) = 0 a.e. x ∈ R 2 \ Ω ¯ , we have that

(4.5) c ˆ μ n − I ¯ 0 ( u 0 ) = a 2 + b A 4 ∫ R 2 ∣ ∇ w μ n ∣ 2 d x + 1 2 ∫ R 2 ( h ( x ) + μ n V ( x ) ) ∣ w μ n ∣ 2 d x − ∫ R 2 \ B R ( 0 ) K ( x ) F ( w μ n ) d x − ∫ B R ( 0 ) K ( x ) [ F ( u μ n ) − F ( u 0 ) ] d x + o n ( 1 ) ,

and

(4.6) o n ( 1 ) = ( a + b A ) ∫ R 2 ∣ ∇ w μ n ∣ 2 d x + ∫ R 2 ( h ( x ) + μ n V ( x ) ) ∣ w μ n ∣ 2 d x − ∫ R 2 \ B R ( 0 ) K ( x ) f ( w μ n ) w μ n d x − ∫ B R ( 0 ) K ( x ) [ f ( u μ n ) u μ n − f ( u 0 ) u 0 ] d x .

Since ‖ ∇ u μ n ‖ 2 2 , ∫ R 2 ( h ( x ) + μ n V ( x ) ) u μ n 2 d x , and ∫ R 2 K ( x ) f ( u μ n ) u μ n d x are bounded, by Lemma 2.6, we can derive that

(4.7) lim n → ∞ ∫ B R ( 0 ) K ( x ) ( F ( u μ n ) − F ( u 0 ) ) d x = 0 .

Furthermore,

(4.8) ∫ R 2 \ B R ( 0 ) K ( x ) F ( w μ n ) d x ≤ κ 2 ∫ R 2 \ B R ( 0 ) V ( x ) w μ n 2 d x .

By (4.5), (4.7), and (4.8), we obtain that

(4.9) lim n → ∞ ‖ ∇ w μ n ‖ 2 2 < 2 π ( m 0 + 2 ) α 0 .

We claim that

(4.10) ∫ B R ( 0 ) K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ ∣ φ ∣ d x = o n ( 1 ) ‖ φ ‖

uniformly for φ ∈ X . Let α > α 0 . By ( f 1 ), ( f 2 ), and ( f 4 ) , for any ε > 0 , there exists C ε > 0 such that

(4.11) ∣ f ( u ) ∣ ≤ ε ( e α u 2 − 1 ) + C ε ∣ u ∣ , ∀ u ∈ R .

Then,

(4.12) ∫ B R ( 0 ) K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ ∣ φ ∣ d x ≤ ε ∫ B R ( 0 ) K ( x ) [ e α u μ n 2 + ( e α u 0 2 − 1 ) ] ∣ φ ∣ d x + C ε ∫ ∪ i = 1 l B r ( x i ) K ( x ) ( ∣ u μ n ∣ + ∣ u 0 ∣ ) ∣ φ ∣ d x + C ε ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≥ 1 } K ( x ) ( ∣ u μ n ∣ + ∣ u 0 ∣ ) ∣ φ ∣ d x + ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ ∣ φ ∣ d x .

By Lemmas 2.4 and 2.5,

(4.13) ∫ B R ( 0 ) K ( x ) ( e α u 0 2 − 1 ) ∣ φ ∣ d x ≤ ∫ B R ( 0 ) K ( x ) ( e 2 α u 0 2 − 1 ) d x 1 2 ∫ B R ( 0 ) K ( x ) φ 2 d x 1 2 ≤ C ‖ φ ‖ .

For any δ > 0 , we have that

(4.14) u μ n 2 ≤ ( 1 + δ 2 ) w μ n 2 + 1 + 1 δ 2 u 0 2 .

Since ∫ R 2 ( h ( x ) + μ n V ( x ) ) u μ n 2 d x is bounded, by (4.9) and Corollary 2.1, there exist p > 1 (close to 1), α > α 0 (close to α 0 ), and δ > 0 (close to 0) such that

(4.15) ∫ B R ( 0 ) K ( x ) e α p ( 1 + δ 2 ) w μ n 2 d x ≤ C ,

where C > 0 is independent of n . Let t > 1 , t ′ = t t − 1 , and p ′ = p p − 1 . By (4.14), (4.15), and Lemmas 2.4 and 2.5, we derive that

(4.16) ∫ B R ( 0 ) K ( x ) e α u μ n 2 ∣ φ ∣ d x ≤ ∫ B R ( 0 ) K ( x ) e α ( 1 + δ 2 ) w μ n 2 + 1 + 1 δ 2 u 0 2 ∣ φ ∣ d x ≤ ∫ B R ( 0 ) K ( x ) e α p ( 1 + δ 2 ) w μ n 2 d x 1 p ∫ B R ( 0 ) K ( x ) e α p ′ 1 + 1 δ 2 u 0 2 ∣ φ ∣ p ′ d x 1 p ′ ≤ C ∫ B R ( 0 ) K ( x ) e α p ′ t 1 + 1 δ 2 u 0 2 d x 1 p ′ t ∫ B R ( 0 ) K ( x ) ∣ φ ∣ p ′ t ′ d x 1 p ′ t ′ ≤ C ‖ φ ‖ .

Noting that

(4.17) ∫ ∪ i = 1 l B r ( x i ) K ( x ) ( ∣ u μ n ∣ + ∣ u 0 ∣ ) ∣ φ ∣ d x ≤ ∫ ∪ i = 1 l B r ( x i ) 2 K ( x ) ( ∣ u μ n ∣ 2 + ∣ u 0 ∣ 2 ) d x 1 2 ∫ R 2 K ( x ) φ 2 d x 1 2 .

Define φ R ∈ C 0 ∞ ( R 2 ) such that φ R ( x ) = 1 for ∣ x ∣ ≤ R , φ R ( x ) = 0 for ∣ x ∣ ≥ 2 R , and 0 ≤ φ R ( x ) ≤ 1 and ∣ ∇ φ R ( x ) ∣ ≤ 2 R for x ∈ R 2 . By u μ n ⇀ u weakly in X , we obtain { φ R u μ n } is bounded in H 0 1 ( B 2 R ( 0 ) ) . Choose q 0 > 1 such that m 0 q 0 > − 2 and let q 0 ′ = q 0 q 0 − 1 . Then, by ( K 1 ) , ( K 3 ) , and the Sobolev embedding theorem, we can choose r > 0 small such that

(4.18) ∫ ∪ i = 1 l B r ( x i ) K ( x ) ( ∣ u μ n ∣ 2 + ∣ u 0 ∣ 2 ) d x ≤ ∑ i = 1 l d 2 ∫ B r ( x i ) ∣ x − x i ∣ m i q 0 d x 1 q 0 ∫ B R ( 0 ) ∣ u μ n ∣ 2 q 0 ′ d x 1 q 0 ′ + ∫ B R ( 0 ) ∣ u 0 ∣ 2 q 0 ′ d x 1 q 0 ′ ≤ ∑ i = 1 l d 2 ∫ B r ( x i ) ∣ x − x i ∣ m i q 0 d x 1 q 0 ∫ B 2 R ( 0 ) ∣ φ R u μ n ∣ 2 q 0 ′ d x 1 q 0 ′ + ∫ B R ( 0 ) ∣ u 0 ∣ 2 q 0 ′ d x 1 q 0 ′ ≤ C ∑ i = 1 l ∫ B r ( x i ) ∣ x − x i ∣ m i q 0 d x 1 q 0 ≤ C ε .

Noting that

(4.19) ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≥ 1 } K ( x ) ( ∣ u μ n ∣ + ∣ u 0 ∣ ) ∣ φ ∣ d x ≤ ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≥ 1 } 2 K ( x ) ( ∣ u μ n ∣ 2 + ∣ u 0 ∣ 2 ) d x 1 2 ∫ R 2 K ( x ) φ 2 d x 1 2 .

By ( K 1 ) and ( K 3 ) , we have that lim n → ∞ ∫ B R ( 0 ) K ( x ) ∣ u μ n − u 0 ∣ 2 d x = 0 . Moreover,

(4.20) lim n → ∞ B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≥ 1 } = 0 .

Since { φ R u μ n } is bounded in H 0 1 ( B 2 R ( 0 ) ) , we obtain φ R u μ n → v in L p ( B R ( 0 ) ) for any p > 2 . By Theorem 4.9 in [10], we may assume that there exists h 1 ∈ L p ( B R ( 0 ) ) such that for any n , there holds ∣ φ R u μ n ( x ) ∣ ≤ h 1 ( x ) a.e. on B R ( 0 ) . So by (4.20),

(4.21) ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≥ 1 } K ( x ) ( ∣ u μ n ∣ 2 + ∣ u 0 ∣ 2 ) d x ≤ ( max B R ( 0 ) \ ∪ i = 1 l B r ( x i ) K ) ∫ B R ( 0 ) ∣ h 1 ∣ p d x 2 p B R ( 0 ) \ ∪ i = 1 l B d ( x i ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≥ 1 } p − 2 p + ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≥ 1 } K ( x ) ∣ u 0 ∣ 2 d x → 0 , n → ∞ .

By (4.11), for any L > 0 ,

(4.22) ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ ∣ φ ∣ d x ≤ ε ∫ B R ( 0 ) K ( x ) [ e α u μ n 2 + ( e α u 0 2 − 1 ) ] ∣ φ ∣ d x + C ε ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } ∩ { ∣ u 0 ( x ) ∣ ≥ L } K ( x ) ( ∣ u μ n ∣ + ∣ u 0 ∣ ) ∣ φ ∣ d x + C ε ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } ∩ { ∣ u 0 ( x ) ∣ ≤ L } K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ ∣ φ ∣ d x ,

where

(4.23) ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } ∩ { ∣ u 0 ( x ) ∣ ≥ L } K ( x ) ( ∣ u μ n ∣ + ∣ u 0 ∣ ) ∣ φ ∣ d x ≤ ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u 0 ( x ) ∣ ≥ L } 2 K ( x ) ( ∣ u μ n ∣ 2 + ∣ u 0 ∣ 2 ) d x 1 2 ∫ R 2 K ( x ) φ 2 d x 1 2 ,

and

(4.24) ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } ∩ { ∣ u 0 ( x ) ∣ ≤ L } K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ ∣ φ ∣ d x ≤ ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } ∩ { ∣ u 0 ( x ) ∣ ≤ L } K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ 2 d x 1 2 × ∫ R 2 K ( x ) φ 2 d x 1 2 .

By Lemma 2.5, we obtain ∫ B R ( 0 ) \ ∪ i = 1 l B r ( x i ) K ( x ) u 0 2 d x < ∞ . Moreover, by ( K 1 ) , we have that

(4.25) lim L → + ∞ B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ∩ { ∣ u 0 ( x ) ∣ ≥ L } = 0 .

Then,

(4.26) lim L → + ∞ ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u 0 ( x ) ∣ ≥ L } K ( x ) ( ∣ u μ n ∣ 2 + ∣ u 0 ∣ 2 ) d x ≤ lim L → + ∞ ∫ ( B R ( 0 ) \ ∪ i = 1 l B r ( x i ) ) ∩ { ∣ u 0 ( x ) ∣ ≥ L } K ( x ) ( ∣ h 1 ∣ 2 + ∣ u 0 ∣ 2 ) d x = 0 .

By the Lebesgue dominated convergence theorem,

(4.27) lim n → ∞ ∫ ( B R ( 0 ) \ ∪ i = 1 k B r ( x i ) ) ∩ { ∣ u μ n ( x ) − u 0 ( x ) ∣ ≤ 1 } ∩ { ∣ u 0 ( x ) ∣ ≤ L } K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ 2 d x = 0 .

Combining (4.12), (4.13), (4.16)–(4.19), (4.21)–(4.24), (4.26), and (4.27), we obtain (4.10).

By (4.10), we have that

(4.28) lim n → ∞ ∫ B R ( 0 ) K ( x ) ∣ f ( u μ n ) − f ( u 0 ) ∣ ∣ u μ n ∣ d x = 0 .

By ( K 1 ) , ( K 3 ) , ( f 1 ), and ( f 2 ), we obtain that for any u ∈ X ,

∫ B R ( 0 ) K ( x ) f ( u 0 ) ∣ u ∣ d x ≤ ∫ B R ( 0 ) K ( x ) f 2 ( u 0 ) d x 1 2 ∫ B R ( 0 ) K ( x ) ∣ u ∣ 2 d x 1 2 < + ∞ .

Then, by u μ n ⇀ u 0 weakly in X , we have that

(4.29) lim n → ∞ ∫ B R ( 0 ) K ( x ) f ( u 0 ) u μ n d x = ∫ B R ( 0 ) K ( x ) f ( u 0 ) u 0 d x .

By (4.28)–(4.29), we obtain

(4.30) lim n → ∞ ∫ B R ( 0 ) K ( x ) f ( u μ n ) u μ n d x = ∫ B R ( 0 ) K ( x ) f ( u 0 ) u 0 d x .

By (4.6) and (4.30), we obtain that lim n → ∞ ‖ w μ n ‖ μ n = 0 . Moreover, lim n → ∞ ‖ u μ n − u 0 ‖ = 0 .□

Let R > 0 be such that Ω 2 d ⊂ B R ( 0 ) . Since inf x ∈ B 2 R + d 4 ( 0 ) ¯ \ Ω d 4 V ( x ) > 0 , by Lemma 4.1 and a standard argument, we can derive the following result.

Lemma 4.2

Let R > 0 be such that Ω 2 d ⊂ B R ( 0 ) . Then, there exists C R > 0 independent of μ > max { 1 , 2 κ } such that

(4.31) ‖ u μ ‖ L ∞ ( B 2 R ( 0 ) ¯ \ Ω d 2 ) ≤ C R ‖ u μ ‖ H 1 ( B 3 R ( 0 ) ¯ \ Ω d 4 ) .

Lemma 4.3

There exists μ ′ > 0 such that for μ > μ ′ , there exist c 1 ′ , c 2 ′ > 0 independent of μ such that

(4.32) u μ ( x ) ≤ c 2 ′ e − c 1 ′ μ ( ∣ x ∣ 1 − γ 2 − R 1 − γ 2 ) ∀ ∣ x ∣ ≥ R .

Proof

By the structure of g ,

− ( a + b ‖ ∇ u μ ‖ 2 2 ) Δ u μ + ( μ − κ ) V ( x ) u μ ≤ 0 ∀ ∣ x ∣ ≥ R .

By I ˆ μ ( u μ ) = c ˆ μ , ( I ˆ μ ′ ( u μ ) , u μ ) = 0 , Lemmas 2.7 and 3.2, we obtain that ‖ ∇ u μ ‖ 2 2 is bounded. So by ( V 1 )–( V 2 ), there exists c ′ > 0 such that

(4.33) − Δ u μ + c ′ μ V ( x ) u μ ≤ 0 ∀ ∣ x ∣ ≥ R .

Define w μ ( x ) = e − c 1 ′ μ ( ∣ x ∣ 1 − γ 2 − R 1 − γ 2 ) . By a direct calculation,

− Δ w μ = − ( c 1 ′ ) 2 μ 1 − γ 2 2 ∣ x ∣ − γ w μ − c 1 ′ γ μ 2 1 − γ 2 ∣ x ∣ − γ 2 − 1 w μ + c 1 ′ 1 − γ 2 μ ∣ x ∣ − γ 2 − 1 w μ .

By ( V 3 ) , there exists μ ′ > 1 and c 1 ′ ∈ ( 0 , 1 ) such that for μ > μ ′ ,

(4.34) − Δ w μ + c 1 ′ μ V ( x ) w μ ≥ 0 ∀ ∣ x ∣ ≥ R .

By Lemma 4.2, we obtain u μ ( x ) ≤ ‖ u μ ‖ L ∞ ( R ≤ ∣ x ∣ ≤ 2 R ) w μ ( x ) for ∣ x ∣ = R . Thus, by (4.33), (4.34), and the maximum principle, we obtain (4.32).□

Lemma 4.4

u μ ∈ H 1 ( R 2 ) is the nonnegative solution of (1.1) for μ > 0 large.

Proof

By Lemma 4.3, we have that

(4.35) u μ ( x ) ≤ c 2 ′ e − c 1 ′ μ 1 − 2 − 1 + γ 2 ∣ x ∣ 1 − γ 2 ∀ ∣ x ∣ ≥ 2 R .

By ( K 2 ) , ( f 1 ), ( f 2 ), and ( f 4 ) , there exists C > 0 such that

(4.36) K ( x ) f ( u μ ) u μ ≤ C e m ∣ x ∣ 1 − γ 2 [ u μ 2 + u μ 2 ( e α u μ 2 − 1 ) ] .

Recall that V ( x ) ≥ V 0 2 ∣ x ∣ γ for ∣ x ∣ ≥ 2 R . Then, by (4.35) and (4.36), we derive that for μ > 0 large,

(4.37) K ( x ) f ( u μ ) u μ ≤ κ V ( x ) ∀ ∣ x ∣ ≥ 2 R .

Since inf x ∈ B 2 R ( 0 ) \ Ω d V ( x ) > 0 , by (4.36), and Lemmas 4.1 and 4.2, we derive that for μ > 0 large,

(4.38) K ( x ) f ( u μ ) u μ ≤ κ V ( x ) ∀ x ∈ B 2 R ( 0 ) \ Ω d .

By (4.37) and (4.38), we know that u μ is the nonnegative solution of (1.1). Moreover, by (4.35), we obtain ∫ ∣ x ∣ ≥ 2 R ∣ u μ ∣ 2 d x < ∞ , which implies that u μ ∈ H 1 ( R 2 ) .□

Proof of Theorem 1.1

By Lemmas 4.1, 4.3, and 4.4, we obtain the result.□

Acknowledgements

The authors would like to express their sincere gratitude to the anonymous referee for his/her valuable suggestions and comments.

Funding information: Jian Zhang was supported by the Fundamental Research Funds for the Central Universities (19CX02054A). Jianjun Zhang was supported by National Natural Science Foundation of China (No. 11871123).
Conflict of interest: There are no conflicts of interest.

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Received: 2023-01-20

Revised: 2023-03-22

Accepted: 2023-03-24

Published Online: 2023-06-13

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

Articles in the same Issue

https://doi.org/10.1515/anona-2022-0317

Keywords for this article

Kirchhoff-type equation; steep potential well; vanishing potential; Trudinger-Moser inequality; variational method

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