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Existence of positive solutions for critical p-Laplacian equation with critical Neumann boundary condition

  • Yinbin Deng , Yulin Shi and Liangshun Xu EMAIL logo
Published/Copyright: September 18, 2025

Abstract

In this article, we consider the existence and nonexistence of positive solutions for the following critical Neumann problem:

Δ p u + λ u p 1 = u p * 1 + f ( x , u ) , x Ω ; u p 2 u ν = u p 1 , x Ω ,

where Ω R n is a bounded domain with C 1 boundary, n 3 , p * = n p n p , p = ( n 1 ) p n p , 1 < p < n , Δ p u = div ( u p 2 u ) is the p -Laplacian operator, ν denotes the unit outward normal to Ω , and f ( x , u ) is a subcritical perturbation. Employing the argument established by Brézis and Nirenberg, we establish the existence and nonexistence of positive solutions for aforementioned equation. Additionally, the extremal functions of the Sobolev constants involving p and p * are investigated.

MSC 2010: 35B09; 35B33; 35J66

1 Introduction and main results

In this article, we mainly investigate the following quasilinear elliptic equation with Neumann boundary conditions:

(1.1) Δ p u + λ u p 1 = u p * 1 + f ( x , u ) , x Ω ; u p 2 u ν = u p 1 , x Ω , u > 0 , x Ω ,

where Ω R n is a bounded domain with C 1 boundary, n 3 , p * = n p n p , p = ( n 1 ) p n p , 1 < p < n , Δ p u = div ( u p 2 u ) is the p -Laplacian operator, and ν denotes the unit outward normal to Ω . The function f ( x , u ) satisfies the following assumptions:

  1. f C ( R , R + ) , f ( x , 0 ) = 0 , and f ( x , t ) 0 for all x Ω if t < 0 ;

  2. lim u 0 f ( x , u ) u p 1 = 0 uniformly for x Ω ;

  3. there exists a constant γ ( p , p ) such that lim u f ( x , u ) u γ 1 = 0 uniformly for x Ω ;

  4. f ( x , t ) t p F ( x , t ) 0 for all t 0 and x Ω , where F ( x , t ) = 0 t f ( x , s ) d s .

The associated energy functional of (1.1) is defined as

(1.2) J ( u ) = Ω 1 p u p + λ p u p 1 p * u p * F ( x , u ) d x 1 p Ω u p d σ ,

which is a C 1 functional in W 1 , p ( Ω ) , where d σ is the measure on the boundary. Critical points of J are weak solutions to (1.1), namely, they satisfy

(1.3) Ω ( u p 2 u v + λ u p 2 u v u p * 2 u v f ( x , u ) v ) d x Ω u p 2 u v d σ = 0 ,

for all v W 1 , p ( Ω ) .

In the last few decades, elliptic problems involving critical Sobolev nonlinearities with Neumann boundary conditions have been extensively investigated (see [3,4,6,8,10,13,15,16,19,22,24] and references therein). The majority of these studies focus on nonlinearities with subcritical growth in the boundary conditions [3,6,8,13,15,16,22,24]. A natural question is whether there exists a positive solution when the Sobolev critical nonlinearities appear in both the equation and the boundary condition.

The early work addressing this issue was conducted by Deng et al. [10]. They considered the Neumann problem

(1.4) Δ u = u 2 * 1 + λ u , x Ω , u ν = u 2 1 , x Ω ,

and proved that (1.4) admits a positive solution when λ < 0 . Pierotti and Terracini [20] investigated another type of Neumann problem:

(1.5) Δ u = u 2 * 1 , x Ω , u ν + δ u 2 1 + g ( x , u ) = 0 , x Ω ,

in which the existence of a positive solution was also demonstrated when δ 0 and g ( x , u ) satisfies the growth condition g ( x , u ) C 1 + C 2 u θ 1 for some θ < 2 . For further exploration, we refer readers to [4] and [19].

Recently, the following critical Neumann problem

(1.6) Δ u α 2 ( x u ) = u 2 * 2 u + λ u + f ( u ) , in R + n , u ν = β u 2 2 u , on R + n ,

was investigated in [7,9,14], where α , β { 0 , 1 } , λ R , and f ( u ) is a nonlinear term. The main characteristic of equation (1.6) is that at least one critical term is involved. If α = β = 1 , λ R , and f ( u ) = 0 , Ferreira et al. [14] obtained that equation (1.6) has a positive solution for λ ( n 2 δ , n 2 ) and has no positive solutions for λ ( , n 4 ) [ n 2 , + ) , where δ > 0 is a small constant. Deng and Shi complemented and improved the existence result in [7,14]. More precisely, they found a lower bound Λ * [ n 4 , n 2 ) depending on n such that equation (1.6) when α = β = 1 and f ( u ) = 0 , has a positive solution if λ ( Λ * , n 2 ) . Particularly, they also estimated that Λ * ( n 4 , n 2 2 ) if n 5 and Λ * = n 4 if n = 4 , which gives the best lower bound of λ for the existence of a positive solution for (1.6) if n = 4 .

However, despite the fact that critical elliptic equations with critical Neumann boundary condition, for example (1.4) and (1.5) and (1.6), have been studied for 30 years, there are no existing results that address problems involving the p -Laplacian operator with critical exponents in both the equation and the boundary conditions. This gap in the literature is the main motivation for us to consider problem (1.1).

Let R + n = { x = ( x , x n ) : x R n 1 , x n 0 } denote the upper half n -dimensional Euclidean space. Note that both p * and p are the critical Sobolev exponents for the embedding W 1 , p ( R + n ) L p * ( R + n ) and W 1 , p ( R + n ) L p ( R + n ) , respectively. For any given positive constants a , b , one can define a sharp constant S a , b p as follows:

(1.7) S a , b p = inf u 0 , u W 1 , p ( R + n ) u p , R + n p a u p * , R + n p + b u p , R + n p ,

where

(1.8) p , R + n R + n p d x 1 p and p , R + n R + n p d x 1 p

for all p > 1 .

The sharp constant S a , b p has been extensively studied, including the search for its extremal functions. When b = 0 , a > 0 , and p > 1 , the extremal function of S a , 0 p can be deduced readily from the work of Aubin [1] and Talenti [21]. When a = 0 , b > 0 and p > 1 , the extremal function of S 0 , b p was obtained by Nazaret [18] and Escobar [12]. In the case a and b are positive, the situation for p = 2 was investigated by Nazaret [11] using conformal invariance argument. The general cases for a > 0 , b > 0 and p 2 remain open.

Recently, Maggi and Neumayer considered the following variational problem:

(1.9) Φ ( T ) = inf { u p , R + n : u p * , R + n = 1 , u p , R + n = T } , T 0 ,

in [17]. Employing the mass transportation argument developed by Nazaret et al. [5], they have the following result:

Theorem A

[17, Theorem 1.1] If n 2 and p ( 1 , n ) , then for any T ( 0 , T E ) , there exists a unique t 0 R such that

(1.10) U ε , x 0 , t 0 ( x , x n ) = ε n p p ( p 1 ) ε p p 1 + ( ( x n ε t 0 ) 2 + x x 0 2 ) p 2 ( p 1 ) n p p

uniquely minimizes Φ ( T ) up to dilations for ε and translations for x 0 R n 1 , where x = ( x , x n ) R + n and T E is a positive constant.

Taking α ( 1 a + b ( T E ) p , 1 a ) and multiplying both the numerator and the denominator of (1.7) by it, we obtain

(1.11) S a , b p = inf u 0 , u W 1 , p ( R + n ) α u p , R + n p α a u p * , R + n p + α b u p , R + n p = inf α v p , R + n p α a + α b v p , R + n p : v W 1 , p ( R + n ) , v p * , R + n = 1 , = α inf { v p , R + n : v W 1 , p ( R + n ) , v p * , R + n = 1 , α a + α b v p , R + n p = 1 } = α Φ ( T α ) ,

where T α = ( 1 a α b α ) 1 p . From Theorem A and (1.11), the following corollary follows.

Corollary 1.1

Let n 2 and p ( 1 , n ) . For any constants a > 0 and b > 0 , the Sobolev constant S a , b p defined in (1.7) is uniquely achieved by the function U ε , x 0 , t 0 ( x , x n ) up to dilations for ε > 0 and translations for x 0 R n 1 , respectively, where x = ( x , x n ) R + n and t 0 R is unique.

Let u ( x , x n ) = C U ε , x 0 , t 0 ( x , x n ) for any given ε > 0 , x 0 R n 1 and C > 0 . A standard variational argument demonstrates that u satisfies

(1.12) div ( u p 2 u ) = a S a , b p u p * , R + n p p * u p * 1 in R + n , u p 2 u ν = b S a , b p u p , R + n p p u p 1 on R + n ,

where ν = ( 0 , 1 ) R n 1 × R denotes the unit outward normal to R + n . Multiplying the boundary equation in (1.12) on both sides by u and integrating over R + n , we then obtain

(1.13) n p p 1 p 1 t 0 U ε , x 0 , t 0 p , R + n p = b S a , b p U ε , x 0 , t 0 p , R + n p .

Equation (1.13) indicates that the constant t 0 < 0 depends only on a , b , n , and p . Particularly, if b = 0 , it follows from (1.13) that t 0 = 0 . We then obtain the well-known result [1,21] that the Sobolev constant

(1.14) S a S a , 0 p = inf u 0 , u W 1 , p ( R + n ) u p , R + n p u p * , R + n p

is achieved by the function

(1.15) U ε , x 0 , 0 ( x , x n ) = ε n p p ( p 1 ) ε p p 1 + ( x n 2 + x x 0 2 ) p 2 ( p 1 ) n p p

up to dilations for ε > 0 and translations for x 0 R n 1 .

As an application, we shall employ the extremal function of S a , b p to establish the following result concerning the existence of weak solutions for equation (1.1).

Theorem 1.1

Let n 3 and 1 < p 1 2 ( n + 1 ) . Suppose that f ( x , u ) satisfies ( H 1 ) ( H 4 ) , then (1.1) admits at least one positive solution provided λ > 0 and has no positive solutions provided λ 0 .

To complete the proof of Theorem 1.1, we adopt the method established by Brézis and Nirenberg [2]. After obtaining the local compactness result for the functional J ( u ) , the difficulty is to show that the mountain pass level of the functional belongs to the compact range. This is performed by taking the minimizers of S a , b p as test functions and performing some delicate estimates on the asymptotic behavior of the test functions (see Lemmas 3.1 and 3.2 for further details).

For simplicity of notation, similar to (1.8), we will denote the integration over Ω and its boundary Ω by

p , Ω Ω p d x 1 p and p , Ω Ω p d σ 1 p ,

respectively, for all p > 1 . We consider our problem in the Sobolev space W 1 , p ( Ω ) equipped with the norm defined by

u = ( u p , Ω p + λ u p , Ω p ) 1 p , λ > 0 .

We will omit p and write S a , b instead of S a , b p throughout this article.

The structure of this article is as follows. In Section 2, we determine the energy threshold for λ > 0 . In Section 3, we first verify that the energy functional J ( u ) satisfies the (PS) c condition, and then use the Mountain Pass Lemma to establish the existence of a positive solution for (1.1), provided λ > 0 . Finally, we demonstrate the nonexistence of a positive solution when λ 0 by a contradiction argument.

2 Energy threshold

In this section, we shall analyze the critical energy of functional J ( u ) to satisfy the classical Palais-Smale (PS) condition when λ > 0 . Toward this purpose, we require some additional notations. We set

(2.1) c = inf ψ Γ max t [ 0 , 1 ] J ( ψ ( t ) ) ,

where

Γ = { γ C ( [ 0 , 1 ] , W 1 , p ( Ω ) ) : γ ( 0 ) = 0 , γ ( 1 ) = d } ,

and d being a constant large enough so that J ( t γ ( 1 ) ) < 0 for all t > 1 . It follows from ( H 1 ) ( H 3 ) that for any ε > 0 , there exists a constant C ε > 0 such that

Ω F ( x , u ) d x ε Ω u p d x + C ε Ω u p * d x .

Applying the Sobolev inequalities, we then have

(2.2) J ( u ) 1 p ε u p C 1 1 p * + C ε u p * C 2 p u p .

Taking ε > 0 small enough such that 1 p ε > 0 , we then conclude from (2.2) that c > 0 .

For θ ( 0 , 1 ] , in what follows, we define

(2.3) S θ = inf 0 u W 1 , p ( R + n ) u p , R + n p θ u p * , R + n p + ( 1 θ ) u p , R + n p ,

then S 1 = S . It follows from Corollary 1.1 that S θ is achieved by

(2.4) U ε , 0 , t 0 ( x , x n ) = ε n p p ( p 1 ) ε p p 1 + ( ( x n ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) n p p ,

for any ε > 0 . For some t 0 < 0 , a standard argument shows that U ε , 0 , t 0 ( x , x n ) solves the following system:

(2.5) div ( u p 2 u ) = θ S θ u p * , R + n p p * u p * 1 , in R + n , u p 2 u ν = ( 1 θ ) S θ u p , R + n p p u p 1 , on R + n .

Analysis similar to (1.13) shows that

(2.6) t 0 = ( 1 θ ) p 1 n p p 1 S θ U ε , 0 , t 0 p , R + n p p .

For any τ > 0 , we define

(2.7) ψ ε , τ ( x , x n ) = C ( n , p ) ε n p p ( p 1 ) ε p p 1 + ( ( x n τ ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) n p p ,

where

(2.8) C ( n , p ) = n n p p 1 p 1 n p p 2 and t 0 = n ( p 1 ) n p p 1 p .

Simple computations show that ψ ε , τ satisfies

(2.9) div ( u p 2 u ) = u p * 1 , in R + n ; u p 2 u ν = τ u p 1 , on R + n .

Define the constant

(2.10) θ = ψ ε , τ p * , R + n p * p ψ ε , τ p * , R + n p * p + τ ψ ε , τ p , R + n p p ,

which is independent of ε . Then, ψ ε , τ is a minimizer of S θ with

(2.11) S θ = ψ ε , τ p * , R + n p * p + τ ψ ε , τ p , R + n p p .

In the following discussion, we denote

(2.12) Ψ ( u ) = 1 p R + n u p d x 1 p * R + n u p * d x 1 p R + n u p d x

and

(2.13) A = inf u 0 sup t > 0 Ψ ( t u ) .

The following lemma shows that the constant A can be attained by some function.

Lemma 2.1

The infimum A is achieved by ψ ε ( x , x n ) ψ ε , 1 ( x , x n ) .

Proof

Define the constant

(2.14) A ˜ = inf τ 0 sup t > 0 Ψ ( t ψ ε , τ ) ,

which is independent of ε > 0 , then A A ˜ . Denote

(2.15) T ( τ ) = ψ ε , τ p * , R + n p ψ ε , τ p , R + n p .

Note that T ( τ ) is a continuous function and

(2.16) lim τ T ( τ ) = 0 .

We claim that A = A ˜ . On the contrary, assume that A < A ˜ , then there exists u W 1 , p ( R + n ) with u p , R + n = ψ ε , 0 p , R + n such that sup t > 0 Ψ ( t u ) < A ˜ . It follows from the well-known result of the Sobolev constant S defined in (1.14) that ψ ε , 0 reaches the infimum S 1 in (2.3). We then have

S ψ ε , 0 p * , R + n p = ψ ε , 0 p , R + n p = u p , R + n p S u p * , R + n p .

Therefore,

(2.17) ψ ε , 0 p * , R + n p u p * , R + n p .

We next discuss the following two cases to derive a contradiction. If

(2.18) u p * , R + n p u p , R + n p > ψ ε , 0 p * , R + n p ψ ε , 0 p , R + n p = T ( 0 )

holds, we then have from (2.17) that

(2.19) ψ ε , 0 p , R + n p > u p , R + n p .

Combining (2.17) and (2.19), we obtain that

sup t > 0 Ψ ( t u ) > sup t > 0 Ψ ( t ψ ε , 0 ) ,

which contracts that sup t > 0 Ψ ( t u ) < A ˜ .

If (2.18) does not hold, then there must be some τ 0 such that

(2.20) u p * , R + n p u p , R + n p = ψ ε , τ p * , R + n p ψ ε , τ p , R + n p = T ( τ ) .

In this case since ψ ε , τ reaches the infimum S θ for some θ , we have

(2.21) ψ ε , τ p , R + n p θ ψ ε , τ p * , R + n p + ( 1 θ ) ψ ε , τ p , R + n p u p , R + n p θ u p * , R + n p + ( 1 θ ) u p , R + n p .

Denote by k the constant defined by

(2.22) k u p * , R + n ψ ε , τ p * , R + n = u p , R + n ψ ε , τ p , R + n > 0 .

It follows from (2.21) and (2.22) that

(2.23) u p , R + n p ψ ε , τ p , R + n p k p .

Note that

(2.24) sup t > 0 Ψ ( t u ) = 1 p 1 p * t 1 p * u p * , R + n p * + 1 p 1 p t 1 p u p , R + n p ,

where t 1 > 0 satisfies

(2.25) t 1 p * p u p * , R + n p * + t 1 p p u p , R + n p u p , R + n p = 0 .

In the same way, we then obtain from (2.22) and (2.24) that

(2.26) sup t > 0 Ψ ( t ψ ε , τ ) = 1 p 1 p * t 2 p * ψ ε , τ p * , R + n p * + 1 p 1 p t 2 p ψ ε , τ p , R + n p = 1 p 1 p * t 2 k p * u p * , R + n p * + 1 p 1 p t 2 k p u p , R + n p ,

where t 2 > 0 satisfies

t 2 p * p ψ ε , τ p * , R + n p * + t 2 p p ψ ε , τ p , R + n p ψ ε , τ p , R + n p = 0 .

It follows from (2.22) and (2.23) that

(2.27) t 2 k p * p u p * , R + n p * + t 2 k p p u p , R + n p u p , R + n p 0 .

Noting that the function

f ( t ) = t p * p u p * , R + n p * + t p p u p , R + n p u p , R + n p

across t -axis at most one time in ( 0 , ) , we then obtain from (2.25) and (2.27) that

(2.28) t 2 k t 1 .

Substituting (2.28) into (2.26), we have

sup t > 0 Ψ ( t ψ ε , τ ) sup t > 0 Ψ ( t u ) ,

which contracts that sup t > 0 Ψ ( t u ) < A ˜ . Consequently, A = A ˜ .

Finally, we complete the proof by showing that the constant A ˜ is achieved by ψ ε , 1 . Note that ψ ε , 1 is the only function in { ψ ε , τ : τ 0 } that satisfies the Euler equation of Ψ ( u ) :

div ( ψ ε , 1 p 2 ψ ε , 1 ) = ψ ε , 1 p * 1 , in R + n ; ψ ε , 1 p 2 ψ ε , 1 ν = ψ ε , 1 p 1 , on R + n .

Therefore, ψ 1 , ε is a critical point of Ψ ( u ) , namely, Ψ ( ψ ε , 1 ) , φ = 0 for any φ W 1 , p ( R + n ) . We then conclude from A = A ˜ that

A = min { sup t > 0 Ψ ( t ψ ε , 0 ) , sup t > 0 Ψ ( t ψ ε , 1 ) , liminf τ sup t > 0 Ψ ( t ψ ε , τ ) } .

Since ψ ε , τ attains the infimum S θ with θ being defined in (2.10), it is easy to see that

ψ ε , 0 p * , R + n p ψ ε , 0 p , R + n p > ψ ε , 1 p * , R + n p ψ ε , 1 p , R + n p ,

with ψ ε , 0 p , R + n p = ( s ψ ε , 1 ) p , R + n p for some s > 0 . Then, the same argument as before shows that sup t > 0 Ψ ( t ψ ε , 0 ) > sup t > 0 Ψ ( t ψ ε , 1 ) . Therefore, it remains to show that liminf τ sup t > 0 Ψ ( t ψ ε , τ ) > A . Let us define

(2.29) ψ ˜ τ ( x ) = τ n p p 1 ψ ε , τ ( τ x ) .

For any ε > 0 , we then have

ψ ˜ τ ( x ) = τ n p p 1 C ( n , p ) ε n p p ( p 1 ) ε p p 1 + ( ( τ x n τ ε t 0 ) 2 + τ x 2 ) p 2 ( p 1 ) n p p = C ( n , p ) ε n p p ( p 1 ) ( τ 1 ε ) p p 1 + ( ( x n ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) n p p ψ ˜ 0 ( x ) C ( n , p ) ε n p p ( p 1 ) ( ( x n ε t 0 ) 2 + x 2 ) n p 2 ( p 1 ) , uniformly as τ .

Moreover,

A ( ψ ε , τ ) = A ( ψ ˜ τ ) A ( ψ ˜ 0 ) , as τ ,

where A ( ψ ε , τ ) sup t > 0 Ψ ( t ψ ε , τ ) . Since ψ ˜ 0 does not satisfy the Euler equation of Ψ ( u ) , it follows that sup t > 0 Ψ ( t ψ ˜ 0 ) > A ˜ . Therefore,

A ˜ = sup t > 0 Ψ ( t ψ ε , 1 ) .

This completes the proof.□

Similarly, we next consider the infimum defined on the bounded domain Ω . In what follows, we denote

(2.30) Ψ Ω ( u ) = 1 p Ω u p d x 1 p * Ω u p * d x 1 p Ω u p d σ .

Let B 1 = { x R n , x < 1 } be a ball and denote B 1 + = B 1 { x n > h ( x ) } , where h ( x ) is a given function defined on { x R n 1 : x < 1 } with h ( 0 ) = h ( 0 ) = 0 . Define

(2.31) S θ ( B 1 + ) = inf u W 0 1 , p ( B 1 + ) u p , B 1 + p θ u p * , B 1 + p + ( 1 θ ) u p , B 1 + p .

We now establish the following lemma:

Lemma 2.2

Let { v n } W 1 , p ( Ω ) be a sequence with v n 0 weakly in W 1 , p ( Ω ) , then v n 0 strongly in W 1 , p ( Ω ) or

(2.32) liminf n sup t > 0 Ψ Ω ( t v n ) A .

Proof

For any u W 0 1 , p ( B 1 + ) W 1 , p ( B 1 + ) , we define u ˜ = u for x B 1 + and u ˜ = 0 for x R n \ B 1 + . Making a translation such that B 1 + R + n , we then have

(2.33) u p , B 1 + p θ u p * , B 1 + p + ( 1 θ ) u p , B 1 + p = u ˜ p , R + n p θ u ˜ p * , R + n p + ( 1 θ ) u ˜ p , R + n p S θ .

Therefore, S θ ( B 1 + ) S θ .

Choose a nonnegative cut-off function η δ ( x , x n ) such that η δ ( x , x n ) = 1 for x δ 2 and η δ ( x , x n ) = 0 for x δ , where 0 < δ < 1 . Let us define

(2.34) u ( x , x n ) = U ε , 0 , t 0 ( x , x n ) η δ ( x , x n ) .

Making the transformation y = x and y n = x n h ( x ) and denoting y = ( y , y n ) , we then have

(2.35) B 1 + u ( x , x n ) p * d x = B 1 + ε U 1,0 , t 0 ( x , x n ) p * η δ ε ( x , x n ) p * d x = R + n U 1,0 , t 0 ( y , y n ) p * η δ ε ( y , y n ) p * d y + O ( ε n p 1 ) = R + n U 1,0 , t 0 ( y , y n ) p * d y + o ( 1 ) , as ε 0 .

The same argument shows that as ε 0 ,

(2.36) B 1 + u ( x , x n ) p d σ = R + n U 1,0 , t 0 ( y , y n ) p d y + o ( 1 )

and

(2.37) B 1 + u ( x , x n ) p d x = R + n U 1,0 , t 0 ( y , y n ) p d y + o ( 1 ) .

We then obtain from (2.35)–(2.37) that

(2.38) u ( x , x n ) p , B 1 + p θ u ( x , x n ) p * , B 1 + p + ( 1 θ ) u ( x , x n ) p , B 1 + p = S θ + o ( 1 ) , as ε 0 .

Therefore, S θ ( B 1 + ) = S θ .

Assume that liminf n v n p , Ω > 0 ; otherwise, we can conclude that there exists a subsequence of { v n } such that v n 0 strongly in W 1 , p ( Ω ) . Let ( φ j ) j = 1 N be a unit partition of Ω ¯ such that the diameter of the support of each φ j , denoted diam ( supp φ j ) , is less than δ . Since Ω is of class C 1 , it follows from equation (2.38) that

(2.39) ( v n φ j 1 p ) p , Ω p S θ ( θ v n φ j 1 p p * , Ω p + ( 1 θ ) v n φ j 1 p p , Ω p ) , 1 j N ,

for any u W 1 , p ( Ω ) . We then have

(2.40) θ v n p * , Ω p + ( 1 θ ) v n p , Ω p = θ j = 1 N φ j v n p p * p , Ω + ( 1 θ ) j = 1 N φ j v n p p p , Ω = θ j = 1 N φ j 1 p v n p * , Ω p + ( 1 θ ) j = 1 N φ j 1 p v n p , Ω p .

Employing inequalities (2.39), we then obtain that

(2.41) θ v n p * , Ω p + ( 1 θ ) v n p , Ω p S θ 1 j = 1 N ( v n φ j 1 p ) p , Ω p S θ 1 j = 1 N ( 1 + ε ) Ω v n p φ j d x + C ε Ω v n p φ j 1 p p d x S θ 1 ( 1 + ε ) v n p , Ω p + o ( 1 ) , as n ,

where ε > 0 is an arbitrary constant. We now utilize (2.41) to check that

(2.42) sup t > 0 Ψ Ω ( t v n ) A + o ( 1 ) , for all n N + .

The proof of (2.42) is the same as in Lemma 2.1. For the reader’s convenience, we provide some details here. On the contrary, assume that sup t > 0 Ψ Ω ( t v n ) < A for some n N + . Then, there exists s R + such that s v n p , Ω = ψ ε , 0 p , R + n . Taking θ = 1 in (2.41), we then have

(2.43) S ψ ε , 0 p * , R + n p = ψ ε , 0 p , R + n p = s v n p , Ω p S s v n p * , Ω p .

Therefore,

(2.44) ψ ε , 0 p * , R + n p s v n p * , Ω p .

As mentioned previously, we split into two cases to drive a contradiction. If

(2.45) v n p * , Ω p v n p , Ω p > ψ ε , 0 p * , R + n p ψ ε , 0 p , R + n p = T ( 0 )

holds, we obtain from (2.44) that

(2.46) ψ ε , 0 p , R + n p > s v n p , Ω p .

Combining (2.44) and (2.46), we have

(2.47) sup t > 0 Ψ Ω ( t v n ) = sup t > 0 Ψ Ω ( t s v n ) > sup t > 0 Ψ ( t ψ ε , 0 ) > A ,

which is a contradiction to our assumption. If (2.45) does not hold, then there must be some τ 0 such that

(2.48) v n p * , Ω p v n p , Ω p = ψ ε , τ p * , R + n p ψ ε , τ p , R + n p = T ( τ ) .

A contradiction can be derived from (2.48) using the same argument as in Lemma 2.1. We omit the details here.

Therefore, for any sequence { v n } as in the statement of the lemma, if liminf n v n p , Ω > 0 , we obtain from (2.42) that

(2.49) liminf n sup t > 0 Ψ Ω ( t v n ) A .

This completes the proof.□

Lemma 2.3

Assume that f ( x , u ) satisfies ( H 1 ) ( H 4 ) . Let { u n } W 1 , p ( Ω ) be a sequence such that

(2.50) J ( u n ) 0 and J ( u n ) c < A ,

then there exists a strongly convergent subsequence of { u n } in W 1 , p ( Ω ) .

Proof

Let { u n } W 1 , p ( Ω ) be a sequence that satisfies (2.50). We then have

(2.51) 1 p Ω u n p d x + λ p Ω u n p d x 1 p * Ω u n p * d x 1 p Ω u n p d σ Ω F ( x , u n ) d x = c + o ( 1 )

and

J ( u n ) , ϕ = Ω u n p 2 u n ϕ d x + λ Ω u n p 2 u n ϕ d x Ω u n p * 2 u n ϕ d x Ω u n p 2 u n ϕ d σ Ω f ( x , u n ) ϕ d x = o ( 1 ) ,

for all ϕ W 1 , p ( Ω ) . Taking ϕ = u n , we then have

(2.52) J ( u n ) , u n = Ω u n p d x + λ Ω u n p d x Ω u n p * d x Ω u n p d σ Ω f ( x , u n ) u n d x = o ( 1 ) .

We then obtain from (2.51) and (2.52) that

(2.53) 1 p 1 p * Ω u n p * d x + 1 p 1 p Ω u n p d σ = Ω F ( x , u n ) 1 p f ( x , u n ) u n d x + c + o ( 1 ) .

It follows from assumptions ( H 1 ) ( H 3 ) that for any ε > 0 , there exists a constant C ε > 0 such that

(2.54) Ω F ( x , u n ) 1 p f ( x , u n ) u n d x ε Ω u n p * d x + C ε Ω u n p d x ε Ω u n p * d x + C ε ˜ Ω u n p * d x p p * .

Taking suitable ε > 0 , we then have

(2.55) 1 p 1 p * ε Ω u n p * d x + 1 p 1 p Ω u n p d σ c + C ε ˜ Ω u n p * d x p p * .

We then obtain from (2.55) that u n p * , Ω and u n p , Ω are bounded for all n . Combining with (2.51), we conclude from (2.55) that { u n } is bounded in W 1 , p ( Ω ) . Extract a subsequence, still denoted by { u n } , so that

u n u , weakly in W 1 , p ( Ω ) and L p * ( Ω ) ; u n u , weakly in L p ( Ω ) ; u n u , strongly in L s ( Ω ) , for p s < p * ,

for some u W 1 , p ( Ω ) . Since f ( x , u ) is of lower order than u p * 1 at infinity, we have

(2.56) lim n Ω f ( x , u n ) u n d x = Ω f ( x , u ) u d x and lim n Ω F ( x , u n ) d x = Ω F ( x , u ) d x .

We now write u n = u + v n with v n 0 weakly in W 1 , p ( Ω ) as n . Employing Brézis-Lieb lemma, we obtain that

(2.57) J ( u n ) = J ( u ) + Ψ Ω ( v n ) + o ( 1 )

and

(2.58) o ( 1 ) = J ( u n ) , u n = Ψ ( v n ) , v n ,

where Ψ Ω is defined in (2.30). It follows from (2.58) that

(2.59) Ψ Ω ( v n ) = 1 p 1 p * Ω v n p * d x + 1 p 1 p Ω v n p d σ + o ( 1 )

and

(2.60) J ( u n ) = 1 p 1 p * Ω u n p * d x + 1 p 1 p Ω u n p d σ + Ω 1 p f ( x , u n ) u n F ( x , u n ) d x + o ( 1 ) .

Passing to the limit in (2.60), it follows from assumption ( H 4 ) that J ( u ) 0 . We then obtain from (2.57) and (2.50) that Ψ Ω ( v n ) c < A .

On the other hand, note that

(2.61) d d t Ψ Ω ( t v n ) = t p 1 Ω v n p d x t p * 1 Ω v n p * d x t p 1 Ω v n p d σ = 0

admits a unique zero point and denote it by t n . It follows from (2.58) and (2.61) that t n = 1 + o ( 1 ) as n . We then have

liminf n sup t > 0 Ψ Ω ( t v n ) liminf n Ψ Ω ( t n v n ) = Ψ Ω ( t n v n ) + o ( 1 ) = Ψ Ω ( v n ) + o ( 1 ) c + o ( 1 ) < A .

This implies that v n 0 strongly in W 1 , p ( Ω ) by Lemma 2.2. The lemma is thus proved.□

3 Proof of Theorem 1.1

In this section, we shall complete the proof of Theorem 1.1. We start with using the test function ψ ε ( x , x n ) = ψ ε , 1 ( x , x n ) to validate that the energy functional J ( u ) of problem (1.1) satisfies ( PS ) c condition with c < A . To this end, we utilize the technique introduced in [22,23]. Let B ( x ¯ , R ) be a ball containing Ω so that B ( x ¯ , R ) Ω and x 0 B ( x ¯ , R ) Ω . Take a suitable translation and rotation so that x 0 is the origin and Ω R + n . Then, the set B ( x ¯ , R ) Ω can be described by

B ( x ¯ , R ) Ω = { x B ( x ¯ , R ) ; x n h ( x ) } ,

where h ( x ) : R n 1 R + is a bounded function. Denote the principle curvatures of Ω at x 0 (relative to the inner normal) by α i , 1 i n 1 , then α i > R 1 . For convenience, we denote

g ( x ) = 1 2 i = 1 n 1 α i x i 2 .

Near the origin, we then have

(3.1) x n = h ( x ) = g ( x ) + o ( x 2 ) , ( x 1 , , x n 1 ) D ( 0 , δ ) ,

for some δ > 0 , where D ( 0 , δ ) = { x R n 1 ; x < δ } . In what follows, we denote

(3.2) K 1 ( ε ) = Ω ψ ε ( x ) p d x , K 2 ( ε ) = Ω ψ ε ( x ) p * d x , K 3 ( ε ) = Ω ψ ε ( x ) p d σ , K 4 ( ε , γ ) = Ω ψ ε ( x ) γ d x .

Without special notations, throughout the whole section, we always assume that 1 < p 1 2 ( n + 1 ) . We next derive the estimates of K i ( i = 1 , 2, 3, 4) as ε 0 in the following lemma.

Lemma 3.1

As ε 0 , the following estimates hold:

(3.3) K 1 ( ε ) = R + n ψ ε p d x I ε + o ( ε ) + O ε n p p 1 , 2 p < n + 1 , R + n ψ ε p d x C ε ln ε + O ( ε ) , 2 p = n + 1 ,

(3.4) K 2 ( ε ) = R + n ψ ε p * d x Λ ε + o ( ε ) + O ε n p 1 ,

(3.5) K 3 ( ε ) = R + n ψ ε p d x + M ε + o ( ε ) + O ε n 1 p 1 ,

(3.6) K 4 ( ε , γ ) = O ε ( n p ) γ p ( p 1 ) , p < γ < n ( p 1 ) n p , O ε n p ln ε , γ = n ( p 1 ) n p , O ε n ( n p ) γ p , γ > n ( p 1 ) n p ,

and

(3.7) K 4 ( ε , p ) = O ε n p p 1 , p < n < p 2 , O ( ε p ln ε ) , n = p 2 , O ( ε p ) , n > p 2 ,

where

(3.8) I n p p 1 p C ( n , p ) p R n 1 ( t 0 2 + x 2 ) p 2 ( p 1 ) g ( x ) 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n d x ,

(3.9) Λ C ( n , p ) p * R n 1 g ( x ) 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n d x ,

(3.10) M p ( n 1 ) p 1 C ( n , p ) p t 0 R n 1 ( x 2 + t 0 2 ) 2 p 2 ( p 1 ) g ( x ) 1 + ( x 2 + t 0 2 ) p 2 ( p 1 ) n d x

and C > 0 is a positive constant.

Proof

We first estimate K 1 ( ε ) . If 2 p < n + 1 ,

(3.11) K 1 ( ε ) = R + n ψ ε p d x D ( 0 , δ ) d x 0 h ( x ) ψ ε p d x n + O ε n p p 1 = R + n ψ ε p d x R n 1 d x 0 g ( x ) ψ ε p d x n D ( 0 , δ ) d x g ( x ) h ( x ) ψ ε p d x n + O ε n p p 1 R + n ψ ε p d x I 1 ( ε ) I 2 ( ε ) + O ε n p p 1 .

We next estimate I 1 ( ε ) and I 2 ( ε ) . Simple computations show

(3.12) I 1 ( ε ) = n p p 1 p C ( n , p ) p ε n p p 1 R n 1 d x 0 g ( x ) ( ( x n ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) ε p p 1 + ( ( x n ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) n d x n = n p p 1 p C ( n , p ) p R n 1 d x 0 ε g ( x ) ( ( x n t 0 ) 2 + x 2 ) p 2 ( p 1 ) 1 + ( ( x n t 0 ) 2 + x 2 ) p 2 ( p 1 ) n d x n .

It follows from (3.12) that lim ε 0 I ( ε ) = I . Using L’Hôpital’s rule, we have

(3.13) lim ε 0 ε 1 I 1 ( ε ) = I .

As for I 2 ( ε ) , we obtain that

I 2 ( ε ) C ε n p p 1 D ( 0 , δ ) d x g ( x ) h ( x ) ( x 2 + x n ε t 0 2 ) p 2 ( p 1 ) ε p p 1 + ( x 2 + x n ε t 0 2 ) p 2 ( p 1 ) n d x n C ε n p p 1 D ( 0 , δ ) h ( x ) g ( x ) ε p p 1 + ( x 2 + x n ε t 0 2 ) p 2 ( p 1 ) n 1 d x .

Noting that h ( x ) g ( x ) = o ( x 2 ) and 2 p < n + 1 , it follows that

(3.14) I 2 ( ε ) o ( ε ) R n 1 x 2 1 + x p ( p 1 ) n 1 d x = o ( ε ) .

Combining of (3.11), (3.13), and (3.14) gives that

K 1 ( ε ) = R + n ψ ε p d x I ε + o ( ε ) + O ε n p p 1 .

If 2 p = n + 1 , we have

(3.15) K 1 ( ε ) = R + n ψ ε p d x D ( 0 , δ ) d x 0 h ( x ) ψ ε p d x n + O ( ε ) .

Let c 1 , c 2 > 0 be two positive constants such that c 1 x 2 h ( x ) c 2 x 2 for x D ( 0 , δ ) . We then have

(3.16) D ( 0 , δ ) d x 0 h ( x ) ψ ε p d x n D ( 0 , δ ) d x 0 c 1 x 2 ψ ε p d x n n p p 1 p C ( n , p ) p D ( 0 , δ ε ) d x 0 ε c 1 x 2 ( ( x n t 0 ) 2 + x 2 ) p 2 ( p 1 ) 1 + ( ( x n t 0 ) 2 + x 2 ) p 2 ( p 1 ) n d x n C 1 ε D ( 0 , δ ε ) ( t 0 2 + x 2 ) p 2 ( p 1 ) x 2 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n d x C 1 ε ln ε .

By the same argument,

(3.17) D ( 0 , δ ) d x 0 h ( x ) ψ ε p d x n D ( 0 , δ ) d x 0 c 2 x 2 ψ ε p d x n n p p 1 p C ( n , p ) p D ( 0 , δ ε ) d x 0 ε c 2 x 2 ( ( x n t 0 ) 2 + x 2 ) p 2 ( p 1 ) 1 + ( ( x n t 0 ) 2 + x 2 ) p 2 ( p 1 ) n d x n C 2 ε D ( 0 , δ ε ) ( t 0 2 + x 2 ) p 2 ( p 1 ) x 2 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n d x C 2 ε ln ε .

Combining (3.15), (3.16), and (3.17), we obtain

K 1 ( ε ) = R + n ψ ε p d x C ε ln ε + O ( ε ) , as ε 0

for some positive constant C > 0 . This completes the estimate of K 1 ( ε ) in (3.3).

As for K 2 ( ε ) , by direct computations, we have

(3.18) K 2 ( ε ) = R + n ψ ε p * d x D ( 0 , δ ) d x 0 h ( x ) ψ ε p * d x n + O ( ε n p 1 ) = R + n ψ ε p * d x R n 1 d x 0 g ( x ) ψ ε p * d x n D ( 0 , δ ) d x g ( x ) h ( x ) ψ ε p * d x n + O ( ε n p 1 ) R + n ψ ε p * d x Λ 1 ( ε ) Λ 2 ( ε ) + O ( ε n p 1 ) .

We next estimate Λ 1 ( ε ) and Λ 2 ( ε ) as follows:

Λ 1 ( ε ) = C ( n , p ) p * ε n p 1 R n 1 d x 0 g ( x ) 1 ε p p 1 + ( ( x n ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) n d x n = C ( n , p ) p * R n 1 d x 0 ε g ( x ) 1 1 + ( ( x n t 0 ) 2 + x 2 ) p 2 ( p 1 ) n d x n .

In the same way, we have

(3.19) lim ε 0 ε 1 Λ 1 ( ε ) = C ( n , p ) p * R n 1 g ( x ) 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n d x Λ .

Similarly, using h ( x ) g ( x ) = o ( x 2 ) , we have

(3.20) Λ 2 ( ε ) D ( 0 , δ ) d x g ( x ) h ( x ) ψ ε p * d x n = o ( ε ) , as ε 0 .

We then obtain from (3.18)–(3.20) that

K 2 ( ε ) = R + n ψ ε p * d x Λ ε + o ( ε ) + O ( ε n p 1 ) .

We now estimate K 3 ( ε ) . By the definition of h ( x ) , we have h ( x ) + h ( x ) 2 = O ( x 2 ) as x . Recalling (3.1), we have the relation between the surface elements d σ = 1 + h 2 d x . It follows that

(3.21) K 3 ( ε ) = Ω ψ ε ( x ) p d σ = C ( n , p ) p ε n 1 p 1 D ( 0 , δ ) 1 + h 2 ε p p 1 + ( ( h ( x ) ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) n 1 d x + O ε n 1 p 1 = C ( n , p ) p R n 1 1 + h ( ε x ) 2 1 + x 2 + 1 ε h ( ε x ) t 0 2 p 2 ( p 1 ) n 1 d x + O ε n 1 p 1 .

Taking a limit as ε 0 , we obtain

(3.22) lim ε 0 K 3 ( ε ) = R n 1 ψ ε ( x , 0 ) p d x = R + n ψ ε ( x ) p d x .

Since h ( x ) = g ( x ) + o ( x 2 ) as x 0 , we have

(3.23) d d ε K 3 ( ε ) ε = 0 = p ( n 1 ) p 1 C ( n , p ) p t 0 R n 1 ( x 2 + t 0 2 ) 2 p 2 ( p 1 ) g ( x ) 1 + ( x 2 + t 0 2 ) p 2 ( p 1 ) n d x M

We then obtain from (3.21)–(3.23) that

K 3 ( ε ) = R + n ψ ε ( x ) p d x + M ε + o ( ε ) + O ε n 1 p 1 .

As for K 4 ( ε , γ ) , when p < γ < p * , simple computations show

Ω ψ ε γ d x = Ω C ( n , p ) γ ε ( n p ) γ p ( p 1 ) ε p p 1 + ( ( x n ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) ( n p ) γ p d x = ε n ( n p ) γ p Ω ε C ( n , p ) γ 1 + ( ( x n ε t 0 ) 2 + x 2 ) p 2 ( p 1 ) ( n p ) γ p d x = O ( ε ( n p ) γ p ( p 1 ) ) , p < γ < n ( p 1 ) n p , O ε n p ln ε , γ = n ( p 1 ) n p , O ( ε n ( n p ) γ p ) , n ( p 1 ) n p < γ < p * .

We can treat K 4 ( ε , p ) in the same way and omit it here. This completes the proof.□

Applying the estimates of K i ( i = 1 , 2, 3, 4) in Lemma 3.1, we next prove that the energy functional J ( u ) satisfies ( PS ) c condition with c < A .

Lemma 3.2

Assume that f ( x , u ) satisfies ( H 1 ) ( H 4 ) . For λ > 0 and ε small enough, if 2 p n + 1 , we have

(3.24) sup t > 0 J ( t ψ ε ) < A .

Proof

We shall complete the proof by discussing the case 2 p < n + 1 and the case 2 p = n + 1 .

Note that, for any ε > 0 , there exists a constant C ε > 0 such that

(3.25) Ω F ( x , ψ ε ) d x ε K 4 ( ε , p ) + C ε K 4 ( ε , γ ) ,

for some γ ( p , p ) . It follows from (3.6), (3.7), and (3.25) that

(3.26) Ω F ( x , ψ ε ) d x = o ( ε ) , if n > 2 p 1 , O ( ε ) , if n = 2 p 1 .

Case 1: n > 2 p 1 . From (3.26), we have

(3.27) J ( t ψ ε ) = t p p K 1 ( ε ) t p * p * K 2 ( ε ) t p p K 3 ( ε ) + λ t p p K 4 ( ε , p ) Ω F ( x , t ψ ε ) d x = t p p K 1 ( ε ) t p * p * K 2 ( ε ) t p p K 3 ( ε ) + o ( ε ) t p + o ( ε ) t γ .

As ε > 0 is small enough, it follows from (3.27) that J ( 0 ) = 0 and lim t J ( t ψ ε ) = , which implies that J ( t ψ ε ) admits a maximum value in ( 0 , ) . Let t ε be a constant such that

J ( t ε ψ ε ) = sup t > 0 J ( t ψ ε ) .

We conclude from (3.27) that t ε must be a zero point of equation

K 1 ( ε ) t p * p K 2 ( ε ) t p p K 3 ( ε ) + o ( ε ) = 0 .

Combining with estimates of K i ( i = 1,2,3 ) , it follows that t ε satisfies

(3.28) R + n ψ ε p d x t ε p * p R + n ψ ε p * d x t ε p p R + n ψ ε p d x + P ( t ε ) ε + o ( ε ) = 0 ,

where

P ( t ε ) ( I + o ( 1 ) ) + ( Λ + o ( 1 ) ) t ε p * p ( M + o ( 1 ) ) t ε p p = O ( 1 ) , as ε 0 .

Observing that equation (3.28) has a unique zero point and

R + n ψ ε p d x R + n ψ ε p * d x R + n ψ ε p d x = 0 ,

we then obtain that

(3.29) t ε = 1 + O ( ε ) .

By the definition of ψ ε and Lemma 2.1, we have

(3.30) A = 1 p R + n ψ ε p d x 1 p * R + n ψ ε p * d x 1 p R + n ψ ε p d x .

We then have from Lemma 3.1, (3.29), and (3.30) that

(3.31) J ( t ε ψ ε ) = t ε p p R + n ψ ε p d x I ε + o ( ε ) t ε p * p * R + n ψ ε p * d x Λ ε + o ( ε ) t ε p p R + n ψ ε p d x + M ε + o ( ε ) + λ t ε p p K 4 ( ε , p ) Ω F ( x , t ψ ε ) d x = A 1 p I 1 p * Λ + 1 p M ε + o ( ε ) .

For convenience, we denote

(3.32) 1 p I 1 p * Λ + 1 p M

and set

α 0 = 1 2 ( n 1 ) i = 1 n 1 α i .

In the following, we shall check > 0 to obtain (3.24). Note that

(3.33) 1 p I + 1 p M = R n 1 ( t 0 2 + x 2 ) 2 p 2 ( p 1 ) g ( x ) 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n G ( x ) d x = α 0 R n 1 ( t 0 2 + x 2 ) 2 p 2 ( p 1 ) x 2 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n G ( x ) d x ,

where

G ( x ) = 1 p n p p 1 p C ( n , p ) p ( t 0 2 + x 2 ) + 1 p p ( n 1 ) p 1 C ( n , p ) p t 0 .

Simple computations give that

(3.34) G ( x ) = 1 p 1 n p p 1 n p p n + 2 p n n + p 2 p + 1 p n p p 1 n p n + p p n n p p x 2 .

We then obtain from (3.33) and (3.34) that

(3.35) 1 p I + 1 p M = 1 p 1 n p p 1 n p p n + 2 p n n + p 2 p α 0 R n 1 ( t 0 2 + x 2 ) 2 p 2 ( p 1 ) x 2 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n + 1 p n p p 1 n p n + p p n n p p α 0 R n 1 ( t 0 2 + x 2 ) 2 p 2 ( p 1 ) x 4 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n 1 p 1 n p p 1 n p p n + 2 p n n + p 2 p A + 1 p n p p 1 n p n + p p n n p p ,

where

A = α 0 ω n 1 0 ( t 0 2 + r 2 ) 2 p 2 ( p 1 ) r n 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n d r

and

= α 0 ω n 1 0 ( t 0 2 + r 2 ) 2 p 2 ( p 1 ) r n + 2 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n d r .

Furthermore,

(3.36) 1 p * Λ = 1 p * C ( n , p ) p * α 0 R n 1 x 2 1 + ( t 0 2 + x 2 ) p 2 ( p 1 ) n d x = n p p n n p p n p p 1 n ( p 1 ) p α 0 ω n 1 0 r n 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n d r .

Since 2 p < n + 1 , using the integration by parts, we obtain

(3.37) p ( n 1 ) p 1 0 ( t 0 2 + r 2 ) 2 p 2 ( p 1 ) r n + 2 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n d r = 0 r n + 1 d 1 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n 1 = ( n + 1 ) 0 r n 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n d r + ( n + 1 ) 0 r n + 2 ( t 0 2 + r 2 ) 2 p 2 ( p 1 ) 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n d r + ( n + 1 ) t 0 2 0 r n ( t 0 2 + r 2 ) 2 p 2 ( p 1 ) 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n d r .

By the definition of A and , it follows from (3.37) that

(3.38) ω n 1 α 0 0 r n 1 + ( t 0 2 + r 2 ) p 2 ( p 1 ) n = n + 1 2 p ( p 1 ) ( n + 1 ) n 2 ( p 1 ) p n p p 1 2 ( p 1 ) p A .

Combining (3.35), (3.36), and (3.38) gives that

(3.39) = 1 p I + 1 p M 1 p * Λ = 1 p 1 n p p 1 n p p n + 2 p n n + p 2 p A + 1 p n p p 1 n p n + p p n n p p n p p n n p p n p p 1 n ( p 1 ) p n + 1 2 p ( p 1 ) ( n + 1 ) n 2 ( p 1 ) p n p p 1 2 ( p 1 ) p A = 2 n + 1 n p p 1 n p n + p p n n p p > 0 .

Hence, for n > 2 p 1 and ε > 0 small enough, it follows from (3.31) and (3.39) that

(3.40) J ( t ε ψ ε ) = A ε + o ( ε ) < A .

Case 2: n = 2 p 1 . Let t ε > 0 , so that

(3.41) J ( t ε ψ ε ) = sup t > 0 J ( t ψ ε ) .

By the similar argument, we infer that

(3.42) t ε = 1 + O ( ε ) .

Hence, for ε > 0 small enough,

(3.43) J ( t ε ψ ε ) = A 1 p C ε ln ε + O ( ε ) < A .

This completes the proof.□

Finally, let us complete the proof of Theorem 1.1.

Proof of Theorem 1.1

First, we check that J ( u ) satisfies the mountain pass geometry. Clearly, J ( 0 ) = 0 holds. For any u W 1 , p , u 0 , and t > 0 , we have

J ( t u ) , as t .

Hence, we can set v t u with t > 0 large enough to obtain J ( v ) < 0 . Taking suitable ε > 0 such that 1 p ε > 0 in (2.2) , we then conclude that there exists r > 0 such that

(3.44) inf u = r J ( u ) > 0 = φ ( 0 ) .

This shows that J ( u ) satisfies the mountain pass geometry.

On the other hand, it follows from Lemmas 3.2 and 2.3 that J ( u ) satisfies ( PS ) c condition, where c is defined as in (2.1). Applying the mountain pass lemma and the maximum principle, we then obtain a positive solution of (1.1) when λ > 0 .

We now show that there is no positive solutions for (1.1) when λ 0 . On the contrary, suppose that ϕ 0 satisfies (1.1). For any ψ W 1 , p ( Ω ) , we have

(3.45) Ω ϕ p 2 ϕ ψ d x + λ Ω ϕ p 1 ψ d x Ω ϕ p * 1 ψ d x Ω ϕ p 1 ψ d σ Ω f ( x , ϕ ) ψ d x = 0 .

However, taking ψ = 1 for x Ω , since f ( x , ϕ ) 0 , it is easy to see that the equality in (3.45) is invalid. This completes the proof of Theorem 1.1.□

  1. Funding information: The research was supported by the Natural Science Foundation of China (No. 12301243, 12271196) and Guangxi Young Talent Project (Grant No. 202401965).

  2. Author contributions: All authors contributed equally to this article. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors state no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article.

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Received: 2024-11-16
Revised: 2025-06-13
Accepted: 2025-07-29
Published Online: 2025-09-18

© 2025 the author(s), published by De Gruyter

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

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