Home Existence and multiplicity of solutions for a new p(x)-Kirchhoff equation
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Existence and multiplicity of solutions for a new p(x)-Kirchhoff equation

  • Changmu Chu EMAIL logo and Jiaquan Liu
Published/Copyright: August 27, 2024

Abstract

This article is devoted to study a class of new p ( x ) -Kirchhoff equation. By means of perturbation technique, variational method, and the method invariant sets for the descending flow, the existence and multiplicity of solutions to this problem are obtained.

MSC 2010: 35J20; 35J60; 35J70; 47J10

1 Introduction and main results

Let Ω R N be a bounded domain with smooth boundary Ω . In this article, we consider the following p ( x ) -Kirchhoff problem:

(1.1) a λ Ω 1 p ( x ) u p ( x ) d x Δ p ( x ) u = u p + 2 u , in Ω , u = 0 , in Ω ,

where a > 0 , λ 0 is a parameter, and Δ p ( x ) u = div ( u p ( x ) 2 u ) represents the p ( x ) -Laplacian operator, where p ( x ) satisfies the following assumptions:

  1. p C ( Ω ¯ ) , p = min { p ( x ) x Ω ¯ } , p + = max { p ( x ) x Ω ¯ } ;

  2. 1 < p ( x ) < N , p + < p * = N p N p ;

  3. p ( 0 ) = p + , p ( x ) p + c x α for all x Ω ¯ , where c > 0 , 0 < α < p + .

Problem (1.1) is called a nonlocal problem because of the presence of the integral over the entire domain Ω , which implies that the equation in (1.1) is no longer pointwise identity. Nonlocal differential equations are also called Kirchhoff-type equations because Kirchhoff [13] has investigated an equation of the form

(1.2) ρ 2 u t 2 P 0 h + E 2 L 0 L u x 2 d x 2 u x 2 = 0 ,

which extends the classical D’Alembert’s wave equation, by considering the effect of change in the length of the string during the vibration. A distinguishing feature of (1.2) is that the equation contains a nonlocal coefficient P 0 h + E 2 L 0 L u x 2 d x , which depends on the average 1 2 L 0 L u x 2 d x , and hence, the equation is no longer a pointwise identity. The parameters in (1.2) have the following meanings: L is the length of the string, h is the area of the cross-section, E is the Young’s modulus of the material, ρ is the mass density, and P 0 is the initial tension.

In the past decade, the Kirchhoff-type problem is as follows:

(1.3) a + λ Ω u 2 d x Δ u = f ( x , u ) , in Ω , u = 0 , in Ω ,

which has been studied extensively, where a , λ > 0 (see [8,19,29] and references therein). In fact, if we replace a + λ Ω u 2 d x by a λ Ω u 2 d x , it turns to be a new nonlocal one. Due to the physical background of negative Young’s modulus [12], this class of new nonlocal problem has recently attracted some attention (see [6,1416,22,23,27,31,32]).

In recent years, p ( x ) -Laplacian operator has received considerable attention due to the fact that it can be applied to image processing, quantum mechanics, elastic mechanics, and electrorheological fluids (we refer the interested reader to [2,4,28]). In fact, the inhomogeneity of p ( x ) -Laplacian operator implies some difficulties. For example, some classical theories and methods, including the Lagrange multiplier theorem and the theory of Sobolev spaces, cannot be applied (see [21,25,26]). Moreover, there has been a great deal of work performed on the following p ( x ) -Kirchhoff-type equation:

(1.4) a λ Ω 1 p ( x ) u p ( x ) d x Δ p ( x ) u = f ( x , u ) , in Ω , u = 0 , in Ω ,

where Ω is a smooth bounded domain in R N , p ( x ) C ( Ω ¯ ) , a , b > 0 , and f ( x , u ) : Ω × R R satisfy certain condition. For the case f ( x , u ) = λ u p ( x ) 2 u + g ( x , u ) , the existence and multiplicity of solutions of Problem (1.4) are obtained in [10] and [33]. The authors in [10] require g ( x , u ) to satisfy the Ambrosetti-Rabinowitz-type growth condition (AR condition, for short), i.e.,

AR: there exist s A > 0 and θ > p + such that

0 g ( x , t ) t θ G ( x , t ) , for all t s A , x Ω , where G ( x , t ) = 0 t g ( x , s ) d s .

In fact, the AR condition guarantees the boundedness of the Palais-Smale sequence of the Euler-Lagrange functional. Moreover, some scholars, including the first author, studied Problem (1.4) under different assumptions of f ( x , u ) (see [30]). For more results of p ( x ) -Kirchhoff-type equation, we refer the interested readers to [3] and [9]. It is worth mentioning that many studies required p + < 2 p and the nonlinearity satisfied the superlinear growth condition

F ( x , u ) u p + + , as u + uniformly in x Ω ,

where F ( x , t ) = 0 t f ( x , s ) d s . However, it is easy to see that this condition will be violated with f ( x , u ) = u p + 2 u at some point of Ω . As described in [11], we need to overcome some difficulties to show the existence of nonnegative nontrivial solutions. Motivated by the aforementioned studies, our main purpose is to find some methods of dealing with such difficulties. To this end, we will study the special case Problem (1.1) of Problem (1.4).

The main results of this study are as follows.

Theorem 1.1

Suppose that ( P 1 ) ( P 3 ) hold. Then, there exists λ 0 > 0 such that Problem (1.1) has a nonnegative nontrivial solution for any λ [ 0 , λ 0 ] .

Theorem 1.2

Suppose that ( P 1 ) ( P 3 ) hold. Then, for any positive integer k, there exists λ k > 0 such that Problem (1.1) has at least k pairs of sign-changing solutions for any λ [ 0 , λ k ] .

Corollary 1.1

Suppose that ( P 1 ) ( P 3 ) hold and λ = 0 . Then, problem (1.1) has constant sign solutions ± u 0 and a series of sign-changing solutions ± u j , j = 1 , 2 , .

To discuss Problem (1.1), we need the functional space L p ( x ) ( Ω ) and W 1 , p ( x ) ( Ω ) . The variable exponent Lebesgue space L p ( x ) ( Ω ) is defined by

L p ( x ) ( Ω ) = u : u : Ω R is measurable , Ω u p ( x ) d x < ,

with the norm

u p ( x ) = inf λ > 0 : Ω u λ p ( x ) d x 1 .

The variable exponent Sobolev space W 1 , p ( x ) ( Ω ) is defined by

W 1 , p ( x ) ( Ω ) = { u L p ( x ) ( Ω ) : u L p ( x ) ( Ω ) } ,

with the norm

u 1 , p ( x ) = u p ( x ) + u p ( x ) .

Define W 0 1 , p ( x ) ( Ω ) as the closure of C 0 ( Ω ) in W 1 , p ( x ) ( Ω ) . The spaces L p ( x ) ( Ω ) , W 1 , p ( x ) ( Ω ) , and W 0 1 , p ( x ) ( Ω ) are separable and reflexive Banach spaces if 1 < p p + < (see [7]). Moreover, there is a constant C > 0 such that

u p ( x ) C u p ( x ) , for any u W 0 1 , p ( x ) ( Ω ) .

Therefore, u = u p ( x ) and u 1 , p ( x ) are the equivalent norms on W 0 1 , p ( x ) ( Ω ) .

Lemma 1.3

[7] If q C ( Ω ¯ ) satisfies 1 q ( x ) < p * ( x ) ( p * ( x ) = N p ( x ) N p ( x ) , if N > p ( x ) ; p * ( x ) = + , if N p ( x ) ) for x Ω ¯ , then the embedding from W 1 , p ( x ) ( Ω ) to L q ( x ) ( Ω ) is compact and continuous.

Lemma 1.4

[7] Set ρ ( u ) = Ω u p ( x ) d x f o r u L p ( x ) ( Ω ) . If u L p ( x ) ( Ω ) and { u k } k N L p ( x ) ( Ω ) , then we have

  1. u p ( x ) < 1 ( = 1 ; > 1 ) ρ ( u ) < 1 ( = 1 ; > 1 ) ;

  2. u p ( x ) > 1 u p ( x ) p ρ ( u ) u p ( x ) p + ;

  3. u p ( x ) < 1 u p ( x ) p + ρ ( u ) u p ( x ) p ;

  4. lim k u k u p ( x ) = 0 lim k ρ ( u k u ) = 0 u k u in measure in Ω and lim k ρ ( u k ) = ρ ( u ) .

Similar to Lemma 1.5, it is easy to obtain the following lemma.

Lemma 1.5

Set L ( u ) = Ω u p ( x ) d x for u W 0 1 , p ( x ) ( Ω ) . If u W 0 1 , p ( x ) ( Ω ) and { u k } k N W 0 1 , p ( x ) ( Ω ) , we have

  1. u < 1 ( = 1 ; > 1 ) L ( u ) < 1 ( = 1 ; > 1 ) ;

  2. u > 1 u p L ( u ) u p + ;

  3. u < 1 u p + L ( u ) u p ;

  4. u k 0 L ( u k ) 0 ; u k L ( u k ) .

Equation (1.1) has a variational structure. The corresponding functional is

I λ ( u ) = a Ω u p ( x ) p ( x ) d x 1 2 λ Ω u p ( x ) p ( x ) d x 2 1 p + Ω u p + d x ,

for u W 0 1 , p ( x ) ( Ω ) . Noting that p ( 0 ) = p + , it is not easy to verify the boundedness of Palais-Smale sequence for the functional I λ . In order to overcome this difficulty, we use the perturbation technique proposed by the first author in [5].

Since p ( x ) is a continuous function, from ( P 2 ) , we see that there exists r > 0 ¯ such that

(1.5) 1 < p r < p + + r < p * .

Let ϕ ( t ) C 0 ( R , [ 0 , 1 ] ) be a smooth even function with the following properties: ϕ ( t ) = 1 for t 1 , ϕ ( t ) = 0 for t 2 , and ϕ ( t ) is monotonically decreasing in the interval ( 0 , + ) . Define

ψ ( t ) = 0 t ϕ ( τ ) d τ , b μ ( t ) = ϕ ( μ t ) , m μ ( t ) = 0 t b μ ( τ ) d τ ,

for μ ( 0 , 1 ] . Consider the perturbation functional

(1.6) I λ , μ ( u ) = a Ω u p ( x ) p ( x ) d x 1 2 ψ λ Ω u p ( x ) p ( x ) d x 2 Ω G μ ( u ) d x ,

for u W 0 1 , p ( x ) ( Ω ) , where g μ ( t ) = t m μ ( t ) r t p + 2 t , G μ ( t ) = 0 t g μ ( τ ) d τ . It is easy to know from the property of φ that if u is a critical point of I λ , μ and satisfies

(1.7) λ Ω 1 p ( x ) u p ( x ) d x 2 < 1 , u ( x ) < 1 μ , x Ω ¯ ,

then u is also a critical point of functional I λ . We first prove that the critical point of I λ , μ satisfies (1.7) and obtain the critical point of functional I λ , μ by means of variational method.

Throughout this article, let B δ = { x : x < δ } Ω and Ω δ = Ω \ B δ . We use to denote the usual norms of W 0 1 , p ( x ) ( Ω ) , and the letters C and C μ stand for positive constants, which may take different values at different places.

2 A priori estimate of the critical point of perturbation functional

In this section, we will show that any critical points of I λ , μ satisfy (1.7).

Proposition 2.1

Suppose that ( P 1 ) ( P 3 ) hold. If u W 0 1 , p ( x ) ( Ω ) is a critical point of the functional I λ , μ with I λ , μ ( u ) L , then there exists a positive constant C = C ( L ) depending only on L such that

Ω 1 p ( x ) u p ( x ) d x C , u ( x ) C , x Ω ¯ , f o r 0 λ λ a = 1 8 a 2 .

To prove Proposition 2.1, we need some preliminaries.

Lemma 2.2

[5] The aforementioned function G μ ( t ) satisfies the following inequality:

G μ ( t ) 1 p + t g μ ( t ) , G μ ( t ) 1 p + + r t g μ ( t ) + C μ ,

where C μ > 0 is a positive constant.

Let β = 0 and n = 0 in Corollary 2 on page 139 of [20], we obtain the following lemma.

Lemma 2.3

Let 1 p N , p q N p N p , and α 1 = 1 N ( q p ) p q 0 . Then,

u L q ( R N ) C x α 1 u L p ( R N ) ,

for all u D ( R N ) , where D ( R N ) is the space of functions in C ( R N ) with compact supports in R N .

Lemma 2.4

Let δ > 0 , then there exists C = C δ > 0 such that

Ω δ u p d x C Ω δ u p d x ,

for u W 0 1 , p ( x ) ( Ω ) .

Proof

If the conclusion does not hold, then there exists { u n } W 0 1 , p ( x ) ( Ω ) satisfying

Ω δ u n p d x = 1 and n Ω δ u n p d x Ω δ u n p d x .

Since { u n } is bounded, we can obtain a subsequence still denoted by { u n } such that

u n u , in W 1 , p ( Ω δ ) , u n u , in L q ( Ω δ ) , 1 q p * , u n u , in L q ( Ω δ ) , 1 q p + = ( N 1 ) p N p .

We have

Ω δ u p d x lim n Ω u n p d x = 0 , Ω δ u p d x = lim n Ω u n p d x = 1 ,

which implies that u is a nonzero constant function. Note that Ω Ω δ and u n W 0 1 , p ( x ) ( Ω ) , we have

Ω u q d s = lim n Ω u n q d s = 0 , 1 q < p + ,

which contradicts the fact that u is a nonzero constant function. The proof is complete.□

Lemma 2.5

Suppose that ( P 1 ) ( P 3 ) hold. If λ λ a = 1 8 a 2 , u is the critical point of the functional I λ , μ with I λ , μ ( u ) L , then there exists a constant C = C L > 0 depending only on L such that

Ω u p ( x ) d x C .

Proof

By Lemmas 2.2 and 2.4, we derive from ( P 3 ) that

L I λ , μ ( u ) 1 p + I λ , μ ( u ) , u = a Ω 1 p ( x ) 1 p + u p ( x ) d x 1 2 ψ λ Ω 1 p ( x ) u p ( x ) d x 2 + 1 p + ψ λ Ω 1 p ( x ) u p ( x ) d x 2 λ Ω 1 p ( x ) u p ( x ) d x Ω u p ( x ) d x + Ω 1 p + g μ ( u ) u G μ ( u ) d x C Ω x α u p ( x ) d x ,

which implies that Ω x α u p ( x ) d x C . For δ > 0 and 1 q p * , we have C = C δ > 0 by Lemma 2.4, such that

(2.1) Ω δ u q d x C Ω δ u p d x q p C 1 + Ω x α u p ( x ) d x q p C .

According to ( P 3 ) , we can choose r > 0 small enough such that

α < β = ( p + r ) N ( ( p + + r ) ( p + r ) ) p + + r .

Set δ 0 ( 0 , 1 2 ) such that B 2 δ 0 ¯ Ω and p ( x ) p + r for any x B 2 δ 0 ¯ . Let φ C 0 ( B 2 δ 0 ) satisfy φ 0 , φ ( x ) = 1 for x < δ 0 , φ ( x ) = 0 for x 2 δ 0 , and φ C for x Ω . It follows from Lemma 2.3 and (2.1) that

(2.2) B δ 0 u p + + r d x B 2 δ 0 φ u p + + r d x C B 2 δ 0 x β ( φ u ) p + r d x p + + r p + r C B 2 δ 0 x β φ u p + r d x + B 2 δ 0 \ B δ 0 x β u φ p + r d x p + + r p + r C 1 + B 2 δ 0 x α u p ( x ) d x + Ω δ 0 u p + r d x p + + r p + r C .

It implies from (2.1) and (2.2) that

Ω u p + + r d x C .

Since u satisfies the following equation:

(2.3) ( a M λ ( u ) ) Ω u p ( x ) 2 u v d x = Ω g μ ( u ) v d x ,

where M λ ( u ) = ψ λ Ω u p ( x ) p ( x ) d x 2 λ Ω u p ( x ) p ( x ) d x , φ W 0 1 , p ( x ) ( Ω ) . Note that

M λ ( u ) 2 λ 1 2 a , g μ ( t ) C ( t p + 1 + t p + + r 1 ) C ( 1 + t p + + r 1 ) .

Set v = u in (2.3), we deduce from (2.2) that

Ω u p ( x ) d x C Ω ( 1 + u p + + r ) d x C .

Lemma 2.6

If λ λ a = 1 8 a 2 , u W 0 1 , p ( x ) ( Ω ) is the critical point of the functional I λ , μ , and I λ , μ ( u ) L , then there exists a constant C = C L > 0 depending only on L such that

u ( x ) C , f o r a l l x Ω ¯ .

Proof

Let τ > 1 , s > 0 , and φ = u u τ s p W 0 1 , p ( x ) ( Ω ) as the test function in (2.3), where u τ is the truncation function of u ,

u τ = u , u τ , ± τ , ± u τ .

On the one hand, we have

(2.4) ( a M λ ( u ) ) Ω u p ( x ) 2 u φ d x C Ω u p ( x ) u τ s p d x C Ω u p u τ s p d x C Ω u τ s p d x C ( 1 + s ) p Ω ( u u τ s ) p d x C Ω u τ s p d x C ( 1 + s ) p Ω ( u u τ s ) p * d x p p * C Ω u τ s p d x .

On the other hand, we have

(2.5) Ω g μ ( u ) φ d x C Ω ( u p + 1 + u p + + r 1 ) u u τ s p d x C 1 + Ω u p + + r u τ s p d x .

It implies from (2.4) and (2.5) that

(2.6) Ω ( u u τ s ) p * d x p p * C ( 1 + s ) p 1 + Ω u p + + r u τ s p d x .

Set t = p + + r p < p * p , d = p * t p > 1 . According to Lemma 2.4, we know u W 0 1 , p ( x ) ( Ω ) W 0 1 , p ( Ω ) , Ω u p * d x C . Therefore, we obtain

(2.7) Ω u p + + r u τ s p d x Ω u t u ( 1 + s ) p d x Ω u p * d x t p * Ω u ( 1 + s ) p * d d x p * t p * C Ω u ( 1 + s ) p * d d x d p p * .

It implies from (2.6) and (2.7) that we have

Ω u ( 1 + s ) p * d x p p * C ( 1 + s ) p 1 + Ω u ( 1 + s ) p * d d x d p p * , as τ ,

which implies that

Ω u ( 1 + s ) p * d x 1 ( 1 + s ) p * C ( 1 + s ) 1 1 + s 1 + Ω u ( 1 + s ) p * d d x d ( 1 + s ) p * ( C ( 1 + s ) ) 1 1 + s max 1 , Ω v ( 1 + s ) p * d d x d ( 1 + s ) p * .

Now, we carry out an iteration process. Set s = s k and s k = d k 1 , k = 1 , 2 , . We have

max 1 , Ω v d k p * d x 1 d k p * ( C d k ) 1 d k max 1 , Ω v d k 1 p * d x 1 d k 1 p * Π j = 1 k ( C d j ) 1 d j max 1 , Ω v p * d x 1 p * C j = 1 k d j d j = 1 k j d j max 1 , Ω v p * d x 1 p * .

Since d > 1 , the series j = 1 d j and j = 1 j d j are convergent. Letting k , we conclude that u L ( Ω ) C . The proof is complete.□

3 Solutions of the perturbation equation

Lemma 3.1

Suppose that ( P 1 ) and ( P 2 ) hold. Then, for any μ ( 0 , 1 ] , I λ , μ satisfies the Palais-Smale condition for any 0 λ λ a = 1 8 a 2 .

Proof

Let { u n } be a Palais-Smale sequence of I λ , μ in W 0 1 , p ( x ) ( Ω ) . This means that there exists L > 0 such that

(3.1) I λ , μ ( u n ) L , I λ , μ ( u n ) 0 , as n .

From (1.6), we have

L + 1 I λ , μ ( u n ) 1 p + + r I λ , μ ( u n ) , u n = a Ω 1 p ( x ) 1 p + + r u n p ( x ) d x 1 2 ψ λ Ω u n p ( x ) p ( x ) d x 2 + 1 p + + r ψ λ Ω u n p ( x ) p ( x ) d x 2 λ Ω u n p ( x ) p ( x ) d x Ω u n p ( x ) d x + Ω 1 p + + r g μ ( u n ) u n G μ ( u n ) d x C Ω u n p ( x ) d x C μ .

It implies from (3.1) and Lemma 1.6 that { u n } is bounded in W 0 1 , p ( x ) ( Ω ) . Set

ω n = M λ ( u n ) = ψ λ Ω u n p ( x ) p ( x ) d x 2 λ Ω 1 p ( x ) u n p ( x ) d x ,

then ω n 2 λ 1 2 a , ω = lim n ω n 1 2 a . For any integer pair ( i , j ) , we obtain

(3.2) o ( 1 ) = I λ , μ ( u n ) I λ , μ ( u m ) , u n u m = ( a ω ) Ω ( u n p ( x ) 2 u n u m p ( x ) 2 u m , u n u m ) d x Ω ( g μ ( u n ) g μ ( u m ) ) ( u n u m ) d x + o ( 1 ) .

Note that W 0 1 , p ( x ) ( Ω ) W 0 1 , p ( Ω ) , g μ ( t ) C ( t p + 1 + t p + + r 1 ) , and p + + r < p * . Up to a subsequence, we have u n u in L p + + r ( Ω ) . Using the Sobolev inequality and the Hölder inequality yields

(3.3) Ω ( g μ ( u n ) g μ ( u m ) ) ( u n u m ) d x C u n u m L p + + r ( Ω ) 0 ,

as n , m + . From (3.2) and (3.3), we have

Ω ( u n p ( x ) 2 u n u m p ( x ) 2 u m , u n u m ) d x 0 ,

as n , m + . By Young’s inequality, for any ε > 0 , we have

Ω ( u n p ( x ) 2 u n u m p ( x ) 2 u m , u n u m ) d x C p ( x ) 2 u n u m p ( x ) d x + C p ( x ) 2 u n u m 2 u n 2 p ( x ) u m 2 p ( x ) d x C p ( x ) 2 u n u m p ( x ) d x + 1 C ε p ( x ) 2 u n u m p ( x ) d x ε C ε p ( x ) 2 ( u n p ( x ) + u m p ( x ) ) .

Note that lim ε 0 C ε = + , according to the arbitrariness of ε > 0 , we obtain

Ω u n u m p ( x ) d x 0 , as n , m + .

It implies from Lemma 1.6 that { u n } contains a strongly convergent subsequence in u W 0 1 , p ( x ) ( Ω ) . Hence, I λ , μ satisfies the Palais-Smale condition. The proof is complete.□

In the following lemma, we will verify that I λ , μ possesses the mountain pass geometry.

Lemma 3.2

Suppose that ( P 1 ) ( P 3 ) hold. Then, for any 0 λ λ a = 1 8 a 2 such that the functional I λ , μ possesses the mountain pass geometry, namely,

  1. there exist m , ρ > 0 such that I λ , μ ( u ) m for any u W 0 1 , p ( x ) ( Ω ) with u = ρ ;

  2. there exists w W 0 1 , p ( x ) ( Ω ) such that w > ρ and I λ , μ ( w ) < 0 .

Proof

It follows from ψ ( t ) 2 t and G μ ( t ) C ( u p + + u p + + r ) that

(3.4) I λ , μ ( u ) = a Ω u p ( x ) p ( x ) d x 1 2 ψ λ Ω u p ( x ) p ( x ) d x 2 Ω G μ ( u ) d x ( a λ ) Ω u p ( x ) p ( x ) d x C Ω ( u p + + u p + + r ) d x .

By the Sobolev embedding theorem and Lemma 1.6, we obtain

Ω u p + + r d x C u W 0 1 , p ( x ) ( Ω ) p + + r C Ω u p ( x ) d x p + + r p + ,

for any u W 0 1 , p ( x ) ( Ω ) with Ω u p ( x ) d x < 1 . Let B ε Ω satisfy that there exists r ε > 0 such that p ( x ) p + r ε for any x Ω ε . By Lemma 1.6, the Hölder inequality, and the Sobolev embedding theorem, we have

Ω u p + d x = B ε u p + d x + Ω ε u p + d x B ε d x r p + + r B ε u p + + r d x p + p + + r + C u W 0 1 , p ( x ) ( Ω ) p + C ε N r p + + r Ω u p ( x ) d x + C Ω ε ( u p ( x ) + u p ( x ) ) d x p + p + r ε C ε N r p + + r Ω u p ( x ) d x + C Ω u p ( x ) d x p + p + r ε ,

for any u W 0 1 , p ( x ) ( Ω ) with Ω u p ( x ) d x < 1 . Set ρ 0 = Ω u p ( x ) d x . Fix μ ( 0 , 1 ] , and it implies from (3.4) that

I λ , μ ( u ) C 1 ρ 0 C 2 ρ 0 p + + r p + C 3 ρ 0 p + p + r ε C 4 ε N r p + + r ρ 0 ,

for ε , ρ 0 > 0 small enough. We have I λ , μ ( u ) 1 4 C 1 ρ 0 p + . By Lemma 1.6, we know that there exist m , ρ > 0 such that I λ , μ ( u ) > m for any u W 0 1 , p ( x ) ( Ω ) with u = ρ .

By the definition of function g μ , we know g μ ( t ) t p + 1 . Let φ C 0 ( Ω ε , [ 0 , 1 ] ) and φ 0 , such that

I λ , μ ( t φ ) a Ω 1 p ( x ) ( t φ ) p ( x ) d x Ω G μ ( t φ ) d x C 1 t p + r ε C 2 t p + < 0 ,

for t 1 . Choosing w = t φ with t > 0 sufficiently large, we have w > ρ and I λ , μ ( w ) < 0 . The proof is complete.□

Proof of Theorem 1.1

From Lemmas 3.1 and 3.2, we see that the functional I λ , μ satisfies the Palais-Smale condition and has the mountain pass geometry. Define

Γ = { γ γ C ( [ 0 , 1 ] , W 0 1 , p ( x ) ( Ω ) ) , γ ( 0 ) = 0 , γ ( 1 ) = w } , c 0 = inf γ Γ sup t [ 0 , 1 ] I λ , μ ( γ ( t ) ) .

By the mountain pass lemma (see [1] or [24]), we see that c 0 m is the critical point of I λ , μ , and there are u W 0 1 , p ( x ) ( Ω ) , I λ , μ ( u ) = c 0 , I λ , μ ( u ) = 0 . In order to obtain the nonnegative critical point of I λ , μ , we consider the functional

I λ , μ + ( u ) = a Ω u p ( x ) p ( x ) d x 1 2 ψ λ Ω u p ( x ) p ( x ) d x 2 Ω G μ ( u + ) d x ,

for u W 0 1 , p ( x ) ( Ω ) . Define

c 0 + = inf γ Γ sup t [ 0 , 1 ] I λ , μ + ( u ) ,

then c 0 + m is the critical point of I λ , μ + , and there are u W 0 1 , p ( x ) ( Ω ) , I λ , μ + ( u ) = c 0 + , I λ , μ + ( u ) = 0 . It can be seen from the equation satisfied by u that u is nonnegative, so u is also the critical point of I λ , μ , we have

I λ , μ ( u ) = I λ , μ + ( u ) = c 0 + m and I λ , μ ( u ) = I λ , μ + ( u ) = 0 .

It implies that

I λ , μ ( u ) = c 0 + L = sup t > 0 J ( t φ ) ,

where J ( u ) = a Ω 1 p ( x ) u p ( x ) d x 1 p + Ω u p + d x . By Proposition 2.1, there exist λ 0 ( λ a ) and μ 0 > 0 . It is easy to see that u is, indeed, a nonnegative nontrivial solution of I λ for λ λ 0 with μ μ 0 , and I λ ( u ) = I λ , μ ( u ) > 0 , I λ ( u ) = I λ , μ ( u ) = 0 .□

4 Existence of sign-changing solutions for perturbation problems

In this section, we prove the existence of multiple sign-changing solutions of the perturbation problem by the following abstract critical point theorem (see Theorem A of [18]). Let X be a Banach space and f be an even C 1 -functional on X . Let P j , Q j , j = 1 , , k be a family of open convex sets of X , Q j = P j . Set

W = j = 1 k ( P j Q j ) and Σ = j = 1 k ( P j Q j ) .

Define

Γ j = { E E X , E compact , E = E , γ ( E η 1 ( Σ ) j ) , for η Λ } ,

Λ = { η η C ( X , X ) , η odd , η ( P j ) P j , η ( Q j ) Q j , η ( x ) = x , if f ( x ) < 0 } ,

c j = inf E Γ j sup x E \ W f ( x ) , K c = { x D f ( x ) = 0 , f ( x ) = c } , K c * = K c \ W ,

where γ is the genus of symmetric sets, γ ( E ) = inf { n , there exists an odd map φ : E R N \ { 0 } } .

Assume there exists an odd continuous map A : X X satisfying

( A 1 ) Given c 0 , b 0 > 0 , there exists B = B ( c 0 , b 0 ) such that if D f ( x ) b 0 , f ( x ) c 0 , then

D f ( x ) , x A x B x A x > 0 .

( A 2 ) A ( P j ) P j , A ( Q j ) Q j , j = 1 , , k .

Moreover, assume

( I 1 ) f satisfies the Palais-Smale condition.

( I 2 ) c * = inf x f ( x ) > 0 .

( Γ ) Γ j is nonempty.

The following abstract critical point theorem is from [18].

Theorem 4.1

Assume ( I 1 ) , ( I 2 ) , ( A 1 ) , ( A 2 ) , and ( Γ ) hold. Then,

  1. c j c * , K c j * .

  2. c j , as j .

  3. If c j = c j + 1 = c j + k 1 = c , then γ ( K c * ) k .

Let u W 0 1 , p ( x ) ( Ω ) , and define v = A u W 0 1 , p ( x ) ( Ω ) as the unique solution of the following equation:

(4.1) 1 2 ( a M λ ( u ) ) Ω v p ( x ) 2 v φ d x + Ω ( v u ) p ( x ) 2 ( v u ) φ d x + Ω u p ( x ) 2 u φ d x = Ω g μ ( u ) φ d x ,

for φ W 0 1 , p ( x ) ( Ω ) . Equation (4.1) has a unique solution v that depends continuously on u .

Lemma 4.2

Let u W 0 1 , p ( x ) ( Ω ) , v = A u , and we have I λ , μ ( u ) , u v C Ω ( u v ) p ( x ) d x and

  1. If p + 2 , then

    I λ , μ ( u ) C Ω ( u v ) p ( x ) d x 1 1 p + Ω ( u v ) p ( x ) d x 1 1 p + ;

  2. If p 2 < p + , then

    I λ , μ ( u ) C Ω ( u v ) p ( x ) d x 1 1 p + Ω ( u v ) p ( x ) d x 1 1 p + + C μ I λ , μ ( u ) 1 2 p + Ω ( u v ) p ( x ) d x 1 p + ;

  3. If p > 2 , then

    I λ , μ ( u ) C Ω ( u v ) p ( x ) d x 1 1 p + Ω ( u v ) p ( x ) d x 1 1 p + + ( C μ + C I λ , μ ( u ) 1 2 p + ) Ω ( u v ) p ( x ) d x 1 p + + ( C μ + C I λ , μ ( u ) 1 2 p ) Ω ( u v ) p ( x ) d x 1 p .

Proof

By the definition of operator A , we have

(4.2) I λ , μ ( u ) , φ = ( a M λ ( u ) ) Ω u p ( x ) 2 u φ d x Ω g μ ( u ) φ d x = 1 2 ( a M λ ( u ) ) Ω ( u v ) p ( x ) 2 ( u v ) φ d x + Ω ( u p ( x ) 2 u v p ( x ) 2 v ) φ d x ,

for φ W 0 1 , p ( x ) ( Ω ) . Thus, taking φ = u v in (4.2), we obtain

I λ , μ ( u ) , u v 1 2 ( a M λ ( u ) ) Ω ( u v ) p ( x ) d x C Ω ( u v ) p ( x ) d x .

Now, we estimate the second integral at the right-hand side of (4.2), we have

(4.3) Ω ( u p ( x ) 2 u v p ( x ) 2 v ) φ d x = p ( x ) 2 ( u p ( x ) 2 u v p ( x ) 2 v ) φ d x + p ( x ) > 2 ( u p ( x ) 2 u v p ( x ) 2 v ) φ d x C p ( x ) 2 ( u v ) p ( x ) 1 φ d x + C p ( x ) > 2 ( u p ( x ) 2 + v p ( x ) 2 ) ( u v ) φ d x C p ( x ) 2 ( u v ) p ( x ) 1 φ d x + C p ( x ) > 2 ( u p ( x ) 2 + ( u v ) p ( x ) 2 ) ( u v ) φ d x C Ω ( u v ) p ( x ) 1 φ d x + C p ( x ) > 2 u p ( x ) 2 ( u v ) φ d x .

It follows from (4.2) and (4.3) that

(4.4) I λ , μ ( u ) , φ C Ω ( u v ) p ( x ) 1 φ d x + C p ( x ) > 2 u p ( x ) 2 ( u v ) φ d x .

Now, we first estimate the first integral of (4.4). Let φ W 0 1 , p ( x ) ( Ω ) satisfy φ = 1 , i.e., Ω φ p ( x ) d x = 1 , we have

(4.5) Ω ( u v ) p ( x ) 1 φ d x Ω ( ε 1 ( u v ) p ( x ) + ε p ( x ) 1 φ p ( x ) ) d x ,

for ε > 0 . If Ω ( u v ) p ( x ) d x < 1 , take ε = Ω ( u v ) p ( x ) d x 1 p < 1 from (4.5), we have

(4.6) Ω ( u v ) p ( x ) 1 φ d x Ω ( ε 1 ( u v ) p ( x ) + ε p ( x ) 1 φ p ( x ) ) d x ε 1 Ω ( u v ) p ( x ) d x + ε p 1 Ω φ p ( x ) d x = 2 Ω ( u v ) p ( x ) d x 1 1 p .

If Ω ( u v ) p ( x ) d x 1 , take ε = Ω ( u v ) p ( x ) d x 1 p + 1 from (4.5), we obtain

(4.7) Ω ( u v ) p ( x ) 1 φ d x Ω ( ε 1 ( u v ) p ( x ) + ε p ( x ) 1 φ p ( x ) ) d x ε 1 Ω ( u v ) p ( x ) d x + ε p + 1 Ω φ p ( x ) d x = 2 Ω ( u v ) p ( x ) d x 1 1 p + .

From (4.6) and (4.7), we have

(4.8) Ω ( u v ) p ( x ) 1 φ d x 2 C Ω ( u v ) p ( x ) d x 1 1 p + C Ω ( u v ) p ( x ) d x 1 1 p + ,

for φ W 0 1 , p ( x ) ( Ω ) , φ = 1 . In order to estimate the second integral of (4.4), we need to discuss three cases.

In the case p + 2 , the set { x x Ω , p ( x ) > 2 } is empty,

p ( x ) > 2 u p ( x ) 2 ( u v ) φ d x = 0 .

It implies from (4.4) and (4.8) that

I λ , μ ( u ) = sup φ = 1 I λ , μ ( u ) , φ C Ω ( u v ) p ( x ) d x 1 1 p + Ω ( u v ) p ( x ) d x 1 1 p + ,

for φ W 0 1 , p ( x ) ( Ω ) .

In the case p > 2 , let Ω = { x x Ω , p ( x ) > 2 } , ε > 0 , we have

(4.9) p ( x ) > 2 u p ( x ) 2 ( u v ) φ d x = Ω u p ( x ) 2 ( u v ) φ d x Ω ε 1 u p ( x ) + ε p ( x ) 2 2 ( u v ) p ( x ) 2 φ p ( x ) 2 d x ,

for φ W 0 1 , p ( x ) ( Ω ) , φ = 1 . If Ω u p ( x ) d x 2 < Ω ( u v ) p ( x ) d x , take ε = Ω u p ( x ) d x 2 Ω ( u v ) p ( x ) d x 1 p < 1 , and it follows from (4.9) that

(4.10) Ω u p ( x ) 2 ( u v ) φ d x Ω ε 1 u p ( x ) + ε p ( x ) 2 2 ( u v ) p ( x ) 2 φ p ( x ) 2 d x ε 1 Ω u p ( x ) d x + ε p 2 2 Ω ( u v ) p ( x ) d x 1 2 Ω φ p ( x ) d x 1 2 = 2 Ω u p ( x ) d x 1 2 p Ω ( u v ) p ( x ) d x 1 p ,

for φ W 0 1 , p ( x ) ( Ω ) , φ = 1 . If Ω u p ( x ) d x 2 Ω ( u v ) p ( x ) d x , take ε = Ω u p ( x ) d x 2 Ω ( u v ) p ( x ) d x 1 p + 1 , and we deduce from (4.9) that

(4.11) Ω u p ( x ) 2 ( u v ) φ d x Ω ε 1 u p ( x ) + ε p ( x ) 2 2 ( u v ) p ( x ) 2 φ p ( x ) 2 d x ε 1 Ω u p ( x ) d x + ε p + 2 2 Ω ( u v ) p ( x ) d x 1 2 Ω φ p ( x ) d x 1 2 = 2 Ω u p ( x ) d x 1 2 p + Ω ( u v ) p ( x ) d x 1 p + ,

for φ W 0 1 , p ( x ) ( Ω ) , φ = 1 . Combining (4.10) with (4.11), we obtain

I λ , μ ( u ) = sup φ = 1 I λ , μ ( u ) , φ C Ω ( u v ) p ( x ) d x 1 1 p + Ω ( u v ) p ( x ) d x 1 1 p + + C Ω u p ( x ) d x 1 2 p × Ω ( u v ) p ( x ) d x 1 p + Ω u p ( x ) d x 1 2 p + Ω ( u v ) p ( x ) d x 1 p + ,

for φ W 0 1 , p ( x ) ( Ω ) .

In the case p 2 < p + , if p ( x ) > 2 u p ( x ) d x 2 p ( x ) > 2 ( u v ) p ( x ) d x , take ε = p ( x ) > 2 u p ( x ) d x 2 p ( x ) > 2 ( u v ) p ( x ) d x 1 p + 1 ; from (4.9), we have

(4.12) p ( x ) > 2 u p ( x ) 2 ( u v ) φ d x ε 1 p ( x ) > 2 u p ( x ) d x + ε p + 2 2 p ( x ) > 2 ( u v ) p ( x ) d x 1 2 p ( x ) > 2 φ p ( x ) d x 1 2 2 p ( x ) > 2 u p ( x ) d x 1 2 p + p ( x ) > 2 ( u v ) p ( x ) d x 1 p + ,

for φ W 0 1 , p ( x ) ( Ω ) , φ = 1 . If p ( x ) > 2 u p ( x ) d x 2 < p ( x ) > 2 ( u v ) p ( x ) d x , take ε = 1 , it implies from (4.9) that

(4.13) p ( x ) > 2 u p ( x ) 2 ( u v ) φ d x p ( x ) > 2 u p ( x ) d x + p ( x ) > 2 ( u v ) p ( x ) d x 1 2 p ( x ) > 2 φ p ( x ) d x 1 2 2 p ( x ) > 2 ( u v ) p ( x ) d x 1 2 2 p ( x ) > 2 ( u v ) p ( x ) d x 1 1 p + 2 p ( x ) > 2 ( u v ) p ( x ) d x 1 1 p + ,

for φ W 0 1 , p ( x ) ( Ω ) , φ = 1 . It follows from (4.12) and (4.13) that

I λ , μ ( u ) = sup φ = 1 I λ , μ ( u ) , φ C Ω ( u v ) p ( x ) d x 1 1 p + Ω ( u v ) p ( x ) d x 1 1 p + + Ω u p ( x ) d x 1 2 p + Ω ( u v ) p ( x ) d x 1 p + ,

for φ W 0 1 , p ( x ) ( Ω ) . In order to estimate I λ , μ ( u ) , we need to estimate the integral term Ω u p ( x ) d x . From (4.2), we have

(4.14) I λ , μ ( u ) , u C Ω ( u v ) p ( x ) 1 u d x + C p ( x ) 2 ( u v ) p ( x ) 1 u d x + C p ( x ) > 2 ( u p ( x ) 2 + v p ( x ) 2 ) ( u v ) u d x C Ω ( u v ) p ( x ) 1 u d x + C p ( x ) > 2 u p ( x ) 1 ( u v ) d x ε Ω u p ( x ) d x + C ε Ω ( u v ) p ( x ) d x ,

for u W 0 1 , p ( x ) ( Ω ) . On the other hand,

(4.15) I λ , μ ( u ) 1 p + + r I λ , μ ( u ) , u = a Ω 1 p ( x ) 1 p + + r u p ( x ) d x 1 2 ψ λ Ω u p ( x ) p ( x ) d x 2 + 1 p + + r ψ λ Ω u p ( x ) p ( x ) d x 2 λ Ω u p ( x ) p ( x ) d x Ω u p ( x ) d x + Ω 1 p + + r g μ ( u ) u G μ ( u ) d x C Ω u p ( x ) d x C μ .

It implies from (4.14) and (4.15) that

Ω u p ( x ) d x C I λ , μ ( u ) + Ω ( u v ) p ( x ) d x + C μ ,

for ε > 0 small enough. The proof is complete.□

Next, we verify that functional I λ , μ satisfies all assumptions of Theorem 4.1.

Lemma 4.3

Given c 0 , σ > 0 , there exists C = C ( c 0 , σ ) such that

I λ , μ ( u ) , u v C u v .

Proof

Let u W 0 1 , p ( x ) ( Ω ) , I λ , μ ( u ) c 0 , I λ , μ ( u ) > σ , and v = A u . From Lemma 4.2, then there exists σ > 0 such that Ω ( u v ) p ( x ) d x C σ .

If Ω ( u v ) p ( x ) d x 1 , from Lemma 4.2,

I λ , μ ( u ) , u v C 1 Ω ( u v ) p ( x ) d x C 1 Ω ( u v ) p ( x ) d x 1 p C 1 u v .

If Ω ( u v ) p ( x ) d x < 1 , from Lemma 4.1,

I λ , μ ( u ) , u v C 1 Ω ( u v ) p ( x ) d x C 1 Ω ( u v ) p ( x ) d x 1 1 p + Ω ( u v ) p ( x ) d x 1 p + C 1 σ 1 1 p + u v .

Define

P = { u u W 0 1 , p ( x ) ( Ω ) , u L p + + r ( Ω ) < σ } , Q = { u u W 0 1 , p ( x ) ( Ω ) , u + L p + + r ( Ω ) < σ } ,

for σ > 0 .

Lemma 4.4

There exists σ 0 = σ 0 ( u ) > 0 such that A ( P ) P and A ( Q ) Q for 0 < σ < σ 0 .

Proof

Set u Q , u + L p + + r ( Ω ) = σ , and v = A u . We take the test function φ = v + in the definition of operator A , and we have

(4.16) Ω v + p ( x ) d x C Ω u m μ ( u ) r u p + 2 u v + d x C Ω u + m μ ( u + ) r u + p + 1 v + d x C Ω u + p + + r 1 v + d x + Ω u + p + 1 v + d x .

It implies from (4.16) that

(4.17) Ω v + p ( x ) d x < 1 ,

for σ 0 small enough. We obtain the left-hand side of (4.16) as

(4.18) Ω v + p ( x ) d x v + p + C v + L p + + r ( Ω ) p + .

The first integral on the right-hand side of (4.16) is

(4.19) Ω u + p + + r 1 v + d x u + L p + + r ( Ω ) p + + r v + L p + + r ( Ω ) = σ p + + r v + L p + + r ( Ω ) .

Let ε > 0 , Ω ε = { x x Ω , x > ε } . Then, there exists r ε > 0 , p ( x ) p + r ε . We estimate the second integral on the right-hand side of (4.16),

(4.20) Ω u + p + 1 v + d x = B ε u + p + 1 v + d x + Ω ε u + p + 1 v + d x C B ε d x r p + + r u + L p + + r ( B ε ) p + 1 v + L p + + r ( B ε ) + C Ω ε u + p + + r d x p + 1 p + + r Ω ε v + p + + r d x 1 p + + r C ε N r p + + r σ p + 1 v + L p + + r ( Ω ) + C σ p + 1 Ω ε v + p + + r d x 1 p + + r .

Note that there is p ( x ) p + r ε in Ω ε , according to Lemma 2.3,

(4.21) Ω ε v + p + + r d x 1 p + + r C v + L p ( Ω ε ) C v + W 1 , p ( x ) ( Ω ε ) C Ω ε v + p ( x ) d x 1 p + r ε C Ω v + p ( x ) d x 1 p + r ε C Ω ( u + p + + r 1 + u + p + 1 ) v + d x 1 p + r ε C ( C u + L p + + r ( Ω ) p + + r 1 + u + L p + + r ( Ω ) p + 1 ) ( v + L p + + r ( Ω ) ) 1 p + r ε C u + L p + + r ( Ω ) p + 1 p + r ε v + L p + + r ( Ω ) 1 p + r ε = C σ p + 1 p + r ε v + L p + + r ( Ω ) 1 p + r ε .

By substituting (4.21) into (4.20), it follows from (4.16)–(4.21) that

v + L p + + r ( Ω ) p + C 1 σ p + + r v + L p + + r ( Ω ) + C 2 ε N r p + + r σ p + 1 v + L p + + r ( Ω ) + C 3 σ p + 1 + p + 1 p + r ε v + L p + + r ( Ω ) 1 p + r ε .

It implies that

v + L p + + r ( Ω ) ( 3 C 1 σ r ) 1 p + 1 σ ( 3 C 1 σ 0 r ) 1 p + 1 σ ,

or

v + L p + + r ( Ω ) ( 3 C 2 ε N r p + + r ) 1 p + 1 σ ,

or

v + L p + + r ( Ω ) ( 3 C 3 σ r ε p + r ε ) p + r ε p + ( p + r ε ) 1 σ ( 3 C 3 σ 0 r ε p + r ε ) p + r ε p + ( p + r ε ) 1 σ .

We can choose ε and σ 0 such that

max ( 3 C 1 σ 0 r ) 1 p + 1 , ( 3 C 2 ε N r p + + r ) 1 p + 1 , ( 3 C 3 σ 0 r ε p + r ε ) p + r ε p + ( p + r ε ) 1 1 2 ,

which implies that

v + 1 2 σ .

Thus, for u Q , there is v = A v Q , namely, A ( Q ) Q . It also has A ( P ) P .□

Lemma 4.5

For sufficiently small σ 0 , when σ < σ 0 there exists m > 0 satisfying

c * = inf u Ξ I λ , μ ( u ) m .

Proof

Set Σ = P Q = { u u W 0 1 , p ( x ) ( Ω ) , u + L p + + r ( Ω ) = u L p + + r ( Ω ) = σ } , and we have

(4.22) I λ , μ ( u ) = a Ω u p ( x ) p ( x ) d x 1 2 ψ λ Ω u p ( x ) p ( x ) d x 2 Ω G μ ( u ) d x ( a 2 λ ) Ω u p ( x ) p ( x ) d x C Ω ( u p + + u p + + r ) d x C Ω u p ( x ) d x C Ω ( u p + + u p + + r ) d x ,

for u Σ . Now, we estimate Ω u p + d x . Let ε > 0 , there is p ( x ) p + r ε in Ω ε = { x x Ω , x > ε } , we have

(4.23) Ω u p + d x = B ε u p + d x + Ω ε u p + d x B ε d x r p + + r B ε u p + + r d x p + p + + r + C Ω ε u p + + r d x p + p + + r C ε N r p + + r Ω u p + + r d x p + p + + r + C Ω ε u p + + r d x r ε p + + r Ω ε u p ( x ) d x C 2 ε N r p + + r σ p + + C 3 σ r ε Ω u p ( x ) d x .

It follows from (4.22) and (4.23) that

I λ , μ ( u ) ( C 1 C 3 σ r ε ) Ω u p ( x ) d x C 2 ε N r p + + r σ p + C 4 σ p + + r .

Note that

(4.24) Ω u p ( x ) d x u p + C u L p + + r ( Ω ) p + = C σ p + .

It implies from (4.23) and (4.24) that

I λ , μ ( u ) c 0 σ p + = m ,

for ε , σ 0 small enough with σ σ 0 .□

Denote W = P Q , Σ = P Q . Define c j ( λ , μ ) = inf E Γ j sup u E \ W I λ , μ , where

Γ j = { E E H 0 1 ( Ω ) , E compact , E = E , γ ( E σ 1 ( η ) ) j , for η Λ } ,

Λ = { η η C ( H 0 1 ( Ω ) , H 0 1 ( Ω ) ) , η odd , η ( P ) P , η ( Q ) Q , η ( u ) = u , if I λ , μ 0 } .

Proof of Theorem 1.2

Set ε > 0 , there is p ( x ) p + r ε in Ω ε , and take C 0 ( Ω ε ) a column of linear independent function φ 1 , φ 2 , . Denote

E j = span { φ 1 , φ 2 , , φ j + 1 } , B j ( R ) = { u E j u R } .

Let u B j ( R ) , one has

I λ , μ ( u ) = a Ω u p ( x ) p ( x ) d x 1 2 ψ λ Ω u p ( x ) p ( x ) d x 2 Ω G μ ( u ) d x a Ω u p ( x ) d x 1 p + Ω u p + d x C 1 R p + r ε C 2 R + < 0 ,

for R sufficiently large. By Lemmas 4.1 and 4.2 of [17], we have B j ( R ) Γ j for R large enough. Therefore, for j = 1 , 2 , , c j ( λ , μ ) c * > 0 are the critical values of I λ , μ , K c j ( λ , μ ) * = K c j ( λ , μ ) \ W , and c j ( λ , μ ) + , as j . Moreover, if c j ( λ , μ ) = = c j + k 1 ( λ , μ ) = c , then γ ( K c * ) k . Hence, given an integer k , for μ ( 0 , 1 ] , the functional I μ ( u ) has k pairs of sign-changing critical points ± u j ( μ ) , and the corresponding critical values are c j ( μ ) . In addition, there exist L j > 0 independent of μ such that

c j ( λ , μ ) sup u E j I λ , μ ( u ) sup u S j J ( u ) = L j ,

where S j = span { φ 1 , φ 2 , , φ j + 1 } , J ( u ) = a Ω u p ( x ) d x 1 p + Ω u p + d x . Note that L j has nothing to do with λ λ a = 1 8 a 2 and μ > 0 . According to Proposition 2.1, for any positive integer k , there exist λ k ( λ a ) and μ k > 0 . I λ ( u j ( λ , μ ) ) = I λ , μ ( u j ( λ , μ ) ) = c j ( λ , μ ) , I λ ( u j ( λ , μ ) ) = I λ , μ ( u j ( λ , μ ) ) = 0 , j = 1 , 2 , , k . c j ( λ , μ ) , j = 1 , 2 , , k is also the critical point of I λ , and the relative sign-changing critical point u j ( λ , μ ) W 0 1 , p ( x ) ( Ω ) \ M is also the sign-changing critical point of I λ .□

  1. Funding information: The first author was supported by National Natural Science Foundation of China (No. 11861021). The second author was supported by National Natural Science Foundation of China (No. 12071438).

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript. J. Liu came up with the main idea. C. Chu prepared the manuscript with contributions from all co-authors.

  3. Conflict of interest: The authors declare that they have no competing interests.

  4. Data availability statement: The authors declare that our manuscript has no associated date.

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Received: 2023-06-22
Revised: 2024-01-04
Accepted: 2024-05-06
Published Online: 2024-08-27

© 2024 the author(s), published by De Gruyter

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

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