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Article

Existence of Sign-Changing Solutions for a Class of p(x)-Biharmonic Kirchhoff-Type Equations

School of Mathematics and Computer Science, Yunnan Minzu University, Kunming 650500, China
*
Author to whom correspondence should be addressed.
Axioms 2025, 14(7), 530; https://doi.org/10.3390/axioms14070530
Submission received: 5 June 2025 / Revised: 8 July 2025 / Accepted: 9 July 2025 / Published: 12 July 2025

Abstract

This paper mainly studies the existence of sign-changing solutions for the following p x -biharmonic Kirchhoff-type equations: a + b R N 1 p ( x ) | Δ u | p ( x ) d x Δ p ( x ) 2 u + V ( x ) | u | p ( x ) 2 u = K x f ( u ) , x R N , where Δ p ( x ) 2 u = Δ | Δ u | p ( x ) 2 Δ u is the p ( x ) biharmonic operator, a , b > 0 are constants, N 2 , V ( x ) , K ( x ) are positive continuous functions which vanish at infinity, and the nonlinearity f has subcritical growth. Using the Nehari manifold method, deformation lemma, and other techniques of analysis, it is demonstrated that there are precisely two nodal domains in the problem’s least energy sign-changing solution u b . In addition, the convergence property of u b as b 0 is also established.

1. Introduction and Main Results

In this paper, we mainly study the existence of the following p ( x ) -biharmonic Kirchhoff-type equations with sign-changing solutions:
a + b R N 1 p ( x ) | Δ u | p ( x ) d x Δ p ( x ) 2 u + V ( x ) | u | p ( x ) 2 u = K x f ( u ) , x R N
where a , b > 0 are constants, 1 < p p + < N , the potentials V ( x ) and K ( x ) : R N R are positive continuous functions that vanish at infinity, and function f: R R has subcritical growth. The variable exponent p ( x ) is Lipschitz continuous and satisfies
1 < p : = inf x R N p ( x ) p + : = sup x R N p ( x ) < N .
For example, when N 3 , p x = 2 + 0.3 sin x , then p x 1.7 , 2.3 , which satisfies condition (2).
The problem (1) simplifies to the following problem if b 0
a Δ p ( x ) 2 u + V ( x ) | u | p ( x ) 2 u = K x f ( u ) .
And functions V ( x ) and K ( x ) satisfy the following assumptions:
( v k 1 )
V x , K x > 0 , x R N , and K x L R N ;
( v k 2 )
If B n n R N is a sequence of Borel sets such that the Lebesgue measure m e a s B n R , for some R > 0 and n N , then lim r + B n B r c 0 K x d x = 0 uniformly in n N , where B r c 0 : = { x R N : | x | > r } ;
( v k 3 )
K / V L R N ; or
( v k 4 )
There exists p 0 x p x , p * x such that
lim x K x / V x p * x p 0 x / p * x p x = 0 ,
where
p * ( x ) = N p ( x ) / ( N 2 p ( x ) ) , p ( x ) < N / 2 , p ( x ) N / 2 .
In reference [1], if V and K satisfy the above conditions, we say that ( V , K ) K and apply it to references [2,3].
We need the following constraints on the C 1 function f : R R and its primitive F ( u ) : = 0 u f ( s ) d s in order to explain our key results:
( f 1 )
f u = o ( | u | 2 p x 1 ) , as | u | 0 if ( v k 3 ) hold;
( f 1 )
f u = o ( | u | p 0 x 1 ) , as | u | 0 if ( v k 4 ) hold;
( f 2 )
lim | u | + f u / u p * ( x ) 1 = 0 ;
( f 3 )
There is μ > 2 p + such that 0 < μ F u u f u , u > 0 ;
( f 4 )
For every s > 0 and τ R 0 ;
f ( s τ ) τ s 2 p + 1 f ( τ ) τ 0 , for s 1 ; p s 2 p 1 p + f ( τ ) τ f ( s τ ) τ 0 , for 0 < s 1 .
For example, f x = u μ 2 u , where μ = 5 , and the functional f satisfies the above conditions.
The Kirchhoff-type problem, given by the system
( a + b Ω | u | 2 d x ) Δ u + V ( x ) u = f ( x , u ) , x Ω , u H 0 1 ( Ω ) ,
has been extensively studied over recent decades. Here, Ω R N is an open set, which may be unbounded with either an empty or smooth boundary, V : Ω R , f C Ω × R , R , and a , b > 0 are constants. The problem (4) is referred to as a nonlocal problem due to the nonlocal term Ω | u | 2 d x Δ u . This term introduces significant mathematical challenges, making the analysis of (4) particularly intriguing. Assuming V x = 0 , and given a limited domain Ω R N in (4), we obtain
a + b Ω | u | 2 d x Δ u = f ( u ) , x Ω , u = 0 , x Ω .
This issue is linked to the Kirchhoff equation’s stationary analogue
u t t a + b Ω | u | 2 d x Δ u = f x , t ,
which was first proposed by Kirchhoff [4]. By taking into account the change in the string’s length during vibration, this model expands on the traditional D’Alembert wave equation. Problem (6), as mentioned in [5], represents a variety of biological and physical systems, where u denotes a process that depends on its own average, such as population density. For further mathematical and physical context on Kirchhoff-type problems, see [6,7,8,9,10]. Variational approaches have recently yielded important conclusions about the existence of ground state solutions, multiple solutions, and positive solutions to (5), where the function f meets certain constraints. For detailed references, see [11,12,13,14,15,16,17,18,19,20,21]. The study of sign-changing solutions to Kirchhoff-type problems (5) has yielded important results on their existence, using techniques such as invariant sets of descent flow, the deformation lemma, and other analytical methods (see [22,23,24,25] and the references therein).
Recently, many researchers have shifted their focus to the Kirchhoff problem defined on the entire space R N . In [26], Wang, Zhang, and Cheng extended the results of [24,25] to the entire space. Using variational methods, including the deformation lemma and Miranda’s theorem, they proved the existence of a least energy sign-changing solution to problem (5) with precisely two nodal domains. In [27], Han, Ma, and He studied a class of p-Laplacian Kirchhoff problems:
( a + b R N | u | p d x ) Δ p u + V ( x ) u p 2 u = K x f ( u ) , x R N , u D 1 , p ( R N ) .
The least energy sign-changing solution to the problem was found by applying the Nehari manifold approach and the minimization argument (7). The existence of a least energy sign-changing solution u 0 with two nodal domains whose energy is strictly greater than the ground state energy was established by Xu and Chen [28] using the deformation lemma.
More recently, there has been growing interest in Kirchhoff-type problems with variable exponent p x . In [29], Shen and Shang studied a class of p x -Laplacian Kirchhoff-type equations:
a + b R N 1 p ( x ) | u | p ( x ) d x Δ p ( x ) u + V ( x ) | u | p ( x ) 2 u = f ( x , u ) , x R N
and using methods such as the Nehari manifold, deformation lemma, and topological degree theory, they obtained the least energy sign-changing solution to the problem. Zhang and Hai investigated a class of Choquard-type problems with p ( x ) -biharmonic operators in [30]:
M Ω 1 p x | Δ u | p x d x Δ p x 2 u Δ p x u = λ Ω | u ( y ) | q y | x y | α x , y d y | u | q x 2 u + | u | p * x 2 u , u = Δ u = 0 on Ω ,
using the calculus of variations and the concentration-compactness principle to obtain multiple results for the solution of problem (8). For the p x -biharmonic Kirchhoff-type Equation (1), there are, as far as we are aware, no results regarding the presence of sign-changing solutions.
Motivated by the aforementioned studies, the main goal of this study is to examine if sign-changing solutions exist for the issue (1) and what characteristics they have. The inclusion of the p x -biharmonic Kirchhoff term introduces additional complexity compared to the standard p-Laplacian Kirchhoff-type problem. We also need to overcome the lack of compactness of the space R N . To overcome these difficulties, first, the lack of compactness issue is addressed by using the compact embedding theorem obtained by Alves and Liu in [31] and the term V x . The problem’s least energy sign-changing solution with precisely two nodal domains is then demonstrated by using the deformation lemma, degree theory, the Nehari manifold approach, and other analytical techniques. Finally, the convergence property of u b as the parameter b 0 is also established.
To solve problem (1), we first define and explain the basic properties of the variable exponent Lebesgue spaces L p x R N and the variable exponent Sobolev spaces W k , p x R N .
For any p ( x ) C ( R N ) , L p ( x ) ( R N ) is defined by
L p ( x ) ( R N ) = u : u is a measurable real-valued function R N | u | p ( x ) d x < ,
with the norm
u L p x ( R N ) = inf λ > 0 : R N u ( x ) λ p ( x ) d x 1 .
Its dual space is L p ( · ) ( R N ) , where 1 p ( x ) + 1 p ( x ) = 1 . Then,
R N u v d x 1 p + 1 p u L p ( · ) ( R N ) v L p ( · ) ( R N ) 2 u L p ( · ) ( R N ) v L p ( · ) ( R N ) ,              
for any u L p ( · ) ( R N ) and v L p ( · ) ( R N ) ; see reference [32].
We define the weighted Lebesgue spaces L K q R N as follows:
L K q R N = u : R N R u is a measurable and R N K x | u | q d x < + ,
with the norm
u q , K q : = R N K ( x ) | u | q d x .
W k , p x R N is defined by
W k , p ( x ) R N = u L p ( x ) R N D α u L p ( x ) R N , α k ,
with the norm
u W k , p ( x ) = a k D α u p ( x ) ,
where α = α 1 , , α N , α = i = 1 N α i , D α u = | α | u α 1 x 1 α 2 x 2 α N x N . W 0 k , p ( x ) R N is the closure of C 0 ( R N ) in W k , p x R N , W 0 k , p ( x ) R N and W k , p x R N are separable reflexive Banach Spaces.
In this paper, we work in the subspace defined as follows:
E = u W 2 , p ( x ) R N : R N | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x < + ,
with the norm
u = inf λ > 0 : R N Δ u λ p ( x ) + V ( x ) u λ p ( x ) d x 1 .
The potential V ( x ) is assumed to meet the following requirements:
( V )
V C ( R N , R ) , there exists α > 0 , such that V ( x ) α > 0 for all x R , and
meas x R N : < V ( x ) v 0 < + , for all v 0 R .
Definition 1. 
If φ E satisfies the following integral identity, then u E is a weak solution of Equation (1).
a R N | Δ u | p x 2 Δ u Δ φ d x + b R N 1 p x | Δ u | p x d x R N | Δ u | p x 2 Δ u Δ φ d x + R N V x | u | p x 2 u φ d x = R N K ( x ) f ( u ) φ d x .
Hence, the energy functional Φ b u : E R given by
Φ b u = a R N 1 p ( x ) | Δ u | p ( x ) d x + b 2 R N 1 p ( x ) | Δ u | p ( x ) d x 2 + R N V ( x ) p ( x ) | u | p ( x ) d x R N K x F ( u ) d x .
For any u , φ E , we can calculate
Φ b ( u ) , φ = a R N | Δ u | p ( x ) 2 Δ u Δ φ d x + R N V ( x ) | u | p ( x ) 2 u φ d x + b R N 1 p ( x ) | Δ u | p ( x ) d x R N | Δ u | p ( x ) 2 Δ u Δ φ d x R N K x f ( u ) φ d x
The weak solutions of (1) are undoubtedly the important locations of the above functional. Additionally, if u E is a solution of (1) and u ± 0 , then u is a solution of (1) that changes sign, where
u + x = max { u ( x ) , 0 }
u x = min { u ( x ) , 0 }
We say that u E is the least energy nodal solution of problem (1) if u is a sign-changing solution of (1) and
Φ b u = inf Φ b v : v ± 0 , Φ b v = 0
For u = u + + u , by (11), one has
Φ b u = Φ b u + + u = a R N 1 p ( x ) | Δ u + + u | p ( x ) d x + b 2 R N 1 p ( x ) | Δ u + + u | p ( x ) d x 2 + R N V ( x ) p ( x ) | u + + u | p ( x ) d x R N K x F ( u + + u ) d x = a R N 1 p ( x ) | Δ u + | p ( x ) d x + a R N 1 p ( x ) | Δ u | p ( x ) d x + b 2 R N 1 p ( x ) | Δ u + | p ( x ) d x 2 + b 2 R N 1 p ( x ) | Δ u | p ( x ) d x 2 + b R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x R N K x F ( u + ) d x R N K x F ( u ) d x = Φ b ( u + ) + Φ b ( u ) + b R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x .
Similarly, by calculation we have
Φ b ( u ) , u +   = Φ b ( u + ) , u + + b R N 1 p ( x ) | Δ u | p ( x ) d x R N | Δ u + | p ( x ) d x
and
Φ b ( u ) , u   = Φ b ( u ) , u + b R N 1 p ( x ) | Δ u + | p ( x ) d x R N | Δ u | p ( x ) d x .
Problem (1) does not contain the nonlocal term R N ( 1 / p ( x ) ) | Δ u | p ( x ) d x Δ 2 p ( x ) u , when b = 0 . The energy functional Φ 0 u : E R for problem (3) is given by
Φ 0 u = a R N 1 p ( x ) | Δ u | p ( x ) d x + R N V ( x ) p ( x ) | u | p ( x ) d x R N K x F ( u ) d x .
And, for any u , φ E , we have
Φ 0 ( u ) , φ = a R N | Δ u | p ( x ) 2 Δ u Δ φ d x + R N V ( x ) | u | p ( x ) 2 u φ d x R N K x f ( u ) φ d x .
To obtain the least energy sign-changing solutions, we define the following constrained sets:
M b : = u E : u ± 0 , Φ b u , u + = Φ b u , u = 0 ,
and
M 0 : = u E : u ± 0 , Φ 0 u , u + = Φ 0 u , u = 0 .
The minimizers correspond to the sign-changing solutions for Equations (1) and (3). We also define the Nehari manifolds for these equations as follows:
N b : = { u E : u 0 , Φ b ( u ) , u = 0 } ,
and
N 0 : = { u E : u 0 , Φ 0 ( u ) , u = 0 } .
Clearly, M b N b contains all sign-changing solutions of (1). Thus, we define the critical level as
m b = inf u M b Φ b u ,
and
m 0 = inf u M 0 Φ 0 u .
The following are the primary results of this paper.
Theorem 1. 
Assume that ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) are satisfied. Then, problems (1) and (3) each have a least energy sign-changing solution, denoted by u b and u 0 , respectively, with exactly two nodal domains. Furthermore, for any sequence b n with b n 0 as n , there exists subsequence (still denoted by b n ) such that u b n u 0 in E, where u 0 M 0 is the least energy sign-changing solution of (3).
The following is how this document is structured. Section 2 establishes a number of important results and offers some tentative lemmas. The purpose of Section 3 is to show that the convergence problem has a least energy sign-changing solution and to prove the existence results for problems (1) and (3).

2. Some Preliminary Lemmas

This section presents key preliminary lemmas essential for proving our results.
Proposition 1 
([33]). Let
ρ ( u ) = R N | Δ u | p ( x ) + V x | u | p ( x ) d x , u E .
Then,
(1) 
ρ u < 1 ( = 1 ; > 1 ) u < 1 ( = 1 ; > 1 ) .
(2) 
If u > 1 , then u p ρ u u p + .
(3) 
If u < 1 , then u p + ρ u u p .
The above result can be obtained from Proposition 2.4 of the literature [33].
Proposition 2 
([34]). Let p x : R N R be Lipschitz continuous and (2) be satisfied. Also, let q x : R N R be a measurable function.
(1) 
If p x q x p * x , then W 2 , p x R N is continuously embedded into L q x R N .
(2) 
If p x q x < p * x and inf x R N p * ( x ) q ( x ) > 0 , then W 2 , p x R N is compactly embedded into L l o c q x R N .
Lemma 1 
([31]). If assumption (V) holds for the weight function V, then we have the following:
(1) 
There is a compact embedding E L p ( x ) R N ;
(2) 
For any measurable function q : R N R with p ( x ) < q ( x ) for all x R N , there is a compact embedding E L q ( x ) R N , if inf x R N p * ( x ) q ( x ) > 0 .
Lemma 2. 
Assume that ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, then,
Φ b u Φ b s u + + t u + 1 μ 2 2 p + Φ b ( u ) , u + + 1 η 2 2 p + Φ b ( u ) , u + min 1 , a 1 μ 2 2 p + R N | Δ u + | p ( x ) + V ( x ) | u + | p ( x ) d x + min 1 , a 1 η 2 2 p + R N | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x + b μ η 2 2 R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x .
where
μ = s p 0 < s 1 ; s p + s 1 , η = t p 0 < t 1 ; t p + t 1 .
Proof. 
Based on the different cases of s and t mentioned above, we divide the proof into four cases. Here, we provide the detailed proof for the case where 0 < s 1 , 0 < t 1 , and omit the similar proof for the other cases.
From ( f 4 ), when 0 < s 1 , it can be obtained that
1 t 2 p 2 p + f ( τ ) τ + F ( t τ ) F ( τ ) = t 1 p s 2 p 1 p + f ( τ ) τ f ( s τ ) τ d s 0 ,
and for s 1
1 t 2 p + 2 p + f ( τ ) τ + F ( t τ ) F ( τ ) = 1 t f ( s τ ) τ s 2 p + 1 f ( τ ) τ d s 0 .
From (11) and (13), we have
Ψ : = Φ b ( u ) Φ b ( s u + + t u ) = Φ b ( u + ) + Φ b ( u ) + b R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x Φ b ( s u + ) Φ b ( t u ) b R N 1 p ( x ) | Δ s u + | p ( x ) d x R N 1 p ( x ) | Δ t u | p ( x ) d x .
Since 0 < s 1 , 0 < t 1 , then from (2), one has
Ψ a ( 1 s p ) p + R N | Δ u + | p ( x ) d x + b ( 1 s 2 p ) 2 p + R N 1 p ( x ) | Δ u + | p ( x ) d x R N | Δ u + | p ( x ) d x + 1 s p p + R N V ( x ) | u + | p ( x ) d x 1 s 2 p 2 p + R N f ( u + ) u + d x + a ( 1 t p ) p + R N | Δ u | p ( x ) d x + b 1 t 2 p 2 p + R N 1 p ( x ) | Δ u | p ( x ) d x R N | Δ u | p ( x ) d x + 1 t p p + R N V ( x ) | u | p ( x ) d x 1 t 2 p 2 p + R N f ( u ) u d x + b 1 s p t p R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x + R N K x 1 s 2 p 2 p + f ( u + ) u + + F ( s u + ) F ( u + ) d x + R N K x 1 t 2 p 2 p + f ( u ) u + F ( t u ) F ( u ) d x .
By (20) and K ( x ) > 0 , we obtain
R N K x 1 s 2 p 2 p + f ( u + ) u + + F ( s u + ) F ( u + ) d x 0 ,
and
R N K x 1 t 2 p 2 p + f ( u ) u + F ( t u ) F ( u ) d x 0 .
Then, from (12), (14) and (15), we have
Ψ 1 s 2 p 2 p + Φ b ( u ) , u + b 1 s 2 p 2 p + R N 1 p ( x ) | Δ u | p ( x ) d x R N | Δ u + | p ( x ) d x + 1 t 2 p 2 p + Φ b ( u ) , u b 1 t 2 p 2 p + R N 1 p ( x ) | Δ u + | p ( x ) d x R N | Δ u | p ( x ) d x + a 1 s p 2 2 p + R N | Δ u + | p ( x ) d x + 1 s p 2 2 p + R N V ( x ) | u + | p ( x ) d x + a 1 t p 2 2 p + R N | Δ u | p ( x ) d x + 1 t p 2 2 p + R N V ( x ) | u | p ( x ) d x + b 1 s p t p R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x .
From (2), we have p ( x ) < p + , then
b 1 s 2 p 2 p + R N 1 p ( x ) | Δ u | p ( x ) d x R N | Δ u + | p ( x ) d x b 1 s 2 p 2 R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x
and
b 1 t 2 p 2 p + R N 1 p ( x ) | Δ u + | p ( x ) d x R N | Δ u | p ( x ) d x b 1 t 2 p 2 R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x .
Also since
b 1 s 2 p 2 b 1 t 2 p 2 + b 1 s p t p = b s p t p 2 2 ,
and
R N a | Δ u + | p ( x ) + V ( x ) | u + | p ( x ) d x min 1 , a R N | Δ u + | p ( x ) + V ( x ) | u + | p ( x ) d x ,
R N a | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x min 1 , a R N | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x ,
then we obtain
Ψ 1 s 2 p 2 p + Φ b ( u ) , u + + 1 t 2 p 2 p + Φ b ( u ) , u + min 1 , a 1 s p 2 2 p + R N | Δ u + | p ( x ) + V ( x ) | u + | p ( x ) d x + min 1 , a 1 t p 2 2 p + R N | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x + b s p t p 2 2 R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x .
The proof is complete. □
Corollary 1. 
Assume that ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, let u = u + + u M b , then
Φ b ( u ) Φ b ( s u + + t u ) + min 1 , a 1 μ 2 2 p + R N Δ u + | p ( x ) + V ( x ) | u + | p ( x ) d x + min 1 , a 1 η 2 2 p + R N | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x + b ( μ η ) 2 2 R N 1 p ( x ) | Δ u + | p ( x ) d x R N 1 p ( x ) | Δ u | p ( x ) d x
Corollary 2. 
If ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, let u = u + + u M b , then
Φ b ( u + + u ) = max s , t 0 Φ b ( s u + + t u ) .
Lemma 3. 
If hypothesis ( f 4 ) holds, then
1 2 p + f ( τ ) τ F ( τ ) 0 , τ R .
Proof. 
The above result can be obtained by taking t = 0 in Equations (20) and (21). □
Lemma 4. 
If ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, then there exists a constant β > 0 such that
(1) 
For any u M b , min u ± p + , u ± p β ;
(2) 
For any u N b , Φ b ( u ) min a , 1 2 p + β and min u p + , u p β .
Proof. 
(1) From ( f 2 ) we can deduce that
f ( u ) u C 3 | u | p * ( x ) ,
where C 3 is a positive constant.
For any u M b , we have Φ b ( u ± ) , u ± < 0 . It implies that
R N a | Δ u ± | p ( x ) + V ( x ) | u ± | p ( x ) d x R N a | Δ u ± | p ( x ) + V ( x ) | u ± | p ( x ) d x + b R N | Δ u ± | p ( x ) d x R N 1 p ( x ) | Δ u ± | p ( x ) d x < R N K x f ( u ± ) u ± d x .
From the assumption of (V), ( v k 1 ), and (25), one has
min { a , 1 } u ± p + R N a | Δ u ± | p ( x ) + V ( x ) | u ± | p ( x ) d x R N K x f ( u ± ) u ± d x C 3 K x R N | u ± | p * ( x ) d x C 3 K x α R N V x | u ± | p * ( x ) d x C 3 K x α u ± p * , for 0 < u ± 1 ,
and
min { a , 1 } u ± p C 3 K x α u ± p * + , for u ± 1 .
Thus, there is β such that β min u ± p + , u ± p .
(2) For any u N b , by (11) and (12), one has
Φ b ( u ) = Φ b ( u ) 1 2 p + Φ b ( u ) , u = a R N 1 p ( x ) | Δ u | p ( x ) d x + b 2 R N 1 p ( x ) | Δ u | p ( x ) d x 2 + R N V ( x ) p ( x ) | u | p ( x ) d x R N K x F ( u ) d x a 2 p + R N | Δ u | p ( x ) d x + b 2 p + R N 1 p ( x ) | Δ u | p ( x ) d x R N | Δ u | p ( x ) d x + 1 2 p + R N V ( x ) | u | p ( x ) d x 1 2 p + R N K x f ( u ) u d x .
From (2), we have p ( x ) < p + , then
a 2 p + R N | Δ u | p ( x ) d x a 2 R N 1 p ( x ) | Δ u | p ( x ) d x ,
b 2 p + R N 1 p ( x ) | Δ u | p ( x ) d x R N | Δ u | p ( x ) d x b 2 R N 1 p ( x ) | Δ u | p ( x ) d x 2
and
1 2 p + R N V ( x ) | u | p ( x ) d x 1 2 R N V ( x ) p ( x ) | u | p ( x ) d x .
Then, from K ( x ) > 0 , Lemma 3 and (1) of Lemma 4, one has
Φ b ( u ) a 2 R N 1 p ( x ) | Δ u | p ( x ) d x + 1 2 R N V ( x ) p ( x ) | u | p ( x ) d x + R N K x 1 2 p + f u u F u d x a 2 R N 1 p ( x ) | Δ u | p ( x ) d x + 1 2 R N V ( x ) p ( x ) | u | p ( x ) d x 1 2 p + R N a | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x min a , 1 2 p + min u p + , u p min a , 1 2 p + β .
This proof is complete. □
Corollary 3. 
If ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, for any u N b , one has
Φ b u min 1 , a 2 p + min u p + , u p .
In particular, from Proposition 1, we obtain
Φ b u + and Φ b u C 1 u p as u ,
where C 1 = min 1 , a / 2 p + .
Lemma 5. 
If ( v k 1 )–( v k 4 ) hold, then E is a continuous embedding in L K q x R N . Moreover, if q x p ( x ) , p * ( x ) and inf x R N p * ( x ) q ( x ) > 0 , the embedding E L K q x R N is compact.
Proof. 
To demonstrate embedded continuity, we must first demonstrate that E L K q x R N . From Lemma 1, the E L q x R N is continuous as q x p ( x ) , p * ( x ) . Thus, we have u L q x R N for every u E . And then there is
R N | u | q x d x <
From Lemma 4, we know that m b min a , 1 2 p + β > 0 . Let u n M b be such that Φ b u n m b . First, prove that u n is bounded in E. In fact, when n N is sufficiently large, one has
m b + 1 Φ b ( u n ) 1 2 p + Φ b ( u n ) , u n min a , 1 2 p + min u n p + , u n p ,
thus u n is bounded. From (V) and ( v k 2 ), one has
R N K x | u | q x d x K R N | u | q x d x K α R N V x | u | q x d x K α max u q , u q + < .
hence, u L K x q x R N . That is, the embedding E L K q x R N is continuous.
Next, we prove that the embedding is compact. Since u n is bounded in E, then there exists u b E such that u n ± u b ± in E. From Lemma 1, the compactness of the embedding E L q x R N since inf x R N p * ( x ) q ( x ) > 0 , so u n u b in L q x R N . Then
R N K x | u n u b | q x d x K R N | u n u b | q x d x 0 , n .
Therefore, we have u n u b strongly in L K q x R N , that is, the embedding E L K q x R N is compact. □
Lemma 6. 
Assume that f satisfies ( f 1 ) (or ( f 1 ))–( f 2 ) and V,K satisfy ( v k 1 )–( v k 4 ). If u n u in E, then
R N K x f u n u n d x = R N K x f u u d x + o 1 ,
and
R N K ( x ) F ( u n ) d x = R N K ( x ) F ( u ) d x + o ( 1 ) .
Proof. 
The proof follows that of Proposition 2.1 in [1], details omitted. □
Lemma 7. 
Under the assumptions that (V), ( v k 1 )–( v k 4 ), and ( f 1 ) (or ( f 1 ))–( f 4 ) are satisfied, the functional Φ b meets the (C)-condition.
Proof. 
Let { u n } be a (C)-sequence in E for Φ b . This means that
{ Φ b ( u n ) } is bounded and Φ b u n E * 1 + u n E 0 as n .
Consequently, s u p Φ b u n M and Φ b u n , u n = o 1 as n , where M is a positive constant. In Lemma 5, we have already shown that { u n } is bounded in E. A subsequence of { u n } is still denoted by { u n } and u 0 E such that u n u 0 . By Lemma 1, we have
u n u 0 a . e . i n R N , u n u 0 i n L p · R N a n d L q · R N as n .
Next, we demonstrate that { u n } converges strongly to u 0 in E. Let φ E be fixed, and define the linear functional J φ v on E by
J φ v = R N | Δ φ | p ( x ) 2 Δ φ · Δ v d x , f o r a l l v E
Clearly, J φ is continuous. Hence, from (29), one has
lim n R N 1 p ( x ) | Δ u n | p ( x ) d x R N 1 p ( x ) | Δ u 0 | p ( x ) d x J u 0 u n u 0 = 0 ,
as R N 1 p ( x ) | Δ u n | p ( x ) d x R N 1 p ( x ) | Δ u 0 | p ( x ) d x is bounded in R . By Lemma 6, we have
lim n R N K x f ( u n ) f ( u 0 ) u n u 0 d x = 0
Since u n u 0 in E and Φ b ( u n ) 0 in E * as n , it follows that
Φ b u n Φ b u 0 , u n u 0 0 as n .
Therefore, from (29)–(31), one has
o 1 = Φ b u n Φ b u 0 , u n u 0 = a + b R N 1 p ( x ) | Δ u n | p ( x ) d x J u n u n u 0 J u 0 u n u 0 + b R N 1 p ( x ) | Δ u n | p ( x ) d x b R N 1 p ( x ) | Δ u 0 | p ( x ) d x J u 0 u n u 0 + R N V ( x ) | u n | p ( x ) 2 u n V ( x ) | u 0 | p ( x ) 2 u n u n u 0 d x + o 1 R N K x f ( u n ) f ( u 0 ) u n u 0 d x ,
that is,
lim n a + b R N 1 p ( x ) | Δ u n | p ( x ) d x J u n u n u 0 J u 0 u n u 0 + R N V ( x ) | u n | p ( x ) 2 u n V ( x ) | u 0 | p ( x ) 2 u n u n u 0 d x = 0
Obviously,
a + b R N 1 p ( x ) | Δ u n | p ( x ) d x J u n u n u 0 J u 0 u n u 0 0
and
R N V ( x ) | u n | p ( x ) 2 u n V ( x ) | u 0 | p ( x ) 2 u n u n u 0 d x 0 .
Thus,
J u n u n u 0 J u 0 u n u 0 = 0
and
R N V ( x ) | u n | p ( x ) 2 u n V ( x ) | u 0 | p ( x ) 2 u n u n u 0 d x = 0 .
The well-known vector inequalities
ρ σ p ( x ) C ( ρ p ( x ) 2 ρ σ p ( x ) 2 σ ) · ( ρ σ ) for x γ 1 C [ ( ρ p ( x ) 2 ρ σ p ( x ) 2 σ ) · ( ρ σ ) ] p ( x ) 2 × ( ρ p ( x ) + σ p ( x ) ) 2 p ( x ) 2 for x γ 2 and ( ρ , σ ) ( 0 , 0 ) ,
hold for all ρ , σ R N , where C is a positive constant depending only on p ( · ) , γ 1 = x R N : p x 2 , γ 2 = x R N : 1 < p x < 2 ; see proposition 3.3 of reference [35].
If x γ 1 , by (32) and (34) as n ,
γ 1 Δ u n Δ u 0 p ( x ) d x C γ 1 ( Δ u n p ( x ) 2 Δ u n Δ u 0 p ( x ) 2 Δ u 0 ) · ( Δ u n Δ u 0 ) d x C J u n ( u n u 0 ) J u 0 ( u n u 0 ) = o ( 1 ) .
Likewise, using (V), (33) and (34) as n ,
γ 1 V ( x ) | u n u 0 | p ( x ) d x C γ 1 V ( x ) ( | u n | p ( x ) 2 u n | u 0 | p ( x ) 2 u 0 ) ( u n u 0 ) d x = o ( 1 ) .
Next, the case x γ 2 is considered. Since { u n } is bounded in E, we have R N | Δ u n | p ( x ) d x M for all n N , where M is a positive constant. From (9), (32), and (34), one has
γ 2 Δ u n Δ u 0 p x d x C γ 2 [ ( Δ u n p x 2 Δ u n Δ u 0 p x 2 Δ u 0 ) · ( Δ u n Δ u 0 ) ] p x 2 ( Δ u n p x + Δ u 0 p ( x ) ) 2 p x 2 d x 2 C ( γ 2 ( Δ u n p x 2 Δ u n Δ u 0 p x 2 Δ u 0 ) · ( Δ u n Δ u 0 ) d x ) α γ 2 Δ u n p ( x ) + Δ u 0 p x d x β 4 C 2 M β J u n ( u n u 0 ) J u 0 ( u n u 0 ) α = o ( 1 ) ,
where α = p / 2 o r p + / 2 and β = 2 p + / 2 o r 2 p / 2 . Likewise, using (29), R N V ( x ) | u n | p ( x ) d x K for all n N , K is a positive constant. By (9), (33), and (34), we have
γ 2 V ( x ) | u n u 0 | p x d x 2 C R N V ( x ) ( | u n | p x 2 u n | u 0 | p x 2 u 0 ) ( u n u 0 ) d x α ·   γ 2 V ( x ) u n p ( x ) + V ( x ) u 0 p x d x β 4 C 2 K β R N V ( x ) ( | u n | p x 2 u n | u 0 | p x 2 u 0 ) ( u n u 0 ) d x α = o ( 1 ) .
From (35)–(38), one has
R N Δ u n Δ u 0 p x d x + R N V ( x ) u n u 0 p x d x 0 as n .
Therefore, by Proposition 1, we have u n u 0 E 0 as n . This implies that Φ b satisfies the (C)-condition. □
Lemma 8. 
Let u E with u ± 0 , ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, and then there exists a unique pair ( s u , t u ) with s u > 0 , t u > 0 such that s u u + + t u u M b .
Proof. 
Firstly, for any u E with u ± 0 , prove the existence of ( s u , t u ) . To prove that s u u + + t u u M b , just prove that Φ b ( s u u + + t u u ) , s u u + = Φ b ( s u u + + t u u ) , t u u = 0 .
Let
g 1 ( s , t ) = Φ b ( s u + + t u ) , s u + = R N a | Δ ( s u + ) | p ( x ) + V ( x ) | s u + | p ( x ) d x +   b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( s u + ) | p ( x ) d x +   b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( t u ) | p ( x ) d x R N K x f s u + s u +
and
g 2 ( s , t ) = Φ b ( s u + + t u ) , t u = R N a | Δ ( t u ) | p ( x ) + V ( x ) | t u | p ( x ) d x +   b R N | Δ ( t u ) | p ( x ) d x R N 1 p ( x ) | Δ ( t u ) | p ( x ) d x +   b R N | Δ ( t u ) | p ( x ) d x R N 1 p ( x ) | Δ ( s u + ) | p ( x ) d x R N K x f ( t u ) t u
By assumptions ( f 1 )–( f 2 ) or ( f 1 ) –( f 2 ), respectively, one has that, for any ε > 0 , a positive constant C ε > 0 exists such that
| f ( t ) t | ε | t | p x + C ε | t | p * x , t R ,
| f ( t ) t | ε | t | p 0 x + C ε | t | p * x , t R .
From (41), ( v k 1 ), ( v k 3 ), and Proposition 2, one has
R N K x f s u + s u + R N K x ε | s u + | p x + C ε | s u + | p * x d x ε R N K x V x V x | s u + | p x d x + C ε R N K x | s u + | p * x d x ε K x V x R N a | Δ ( s u + ) | p ( x ) + V ( x ) | s u + | p ( x ) d x +   C ε K x R N | s u + | p * x d x ε K x V x R N a | Δ ( s u + ) | p ( x ) + V ( x ) | s u + | p ( x ) d x +   C ˜ ε K x s u + p * x
Therefore, taking ε = 1 / ( 2 K / V ) , based on (43) and Proposition 2, we have
g 1 ( s , t ) 1 2 R N a | Δ ( s u + ) | p ( x ) + V ( x ) | s u + | p ( x ) d x +   b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( s u + ) | p ( x ) d x +   b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( t u ) | p ( x ) d x   C ˜ ε K x s u + p * x .
Similarly, from (42), ( v k 1 ), ( v k 4 ) and Lemma 5, we derive
R N K x f ( t u ) t u R N K x ε | t u | p 0 x + C ε | t u | p * x d x ε R N K x | t u | p 0 x d x + C ε K x R N | t u | p * x d x ε C 1 t u p 0 x + C ε K x t u p * x
which means that
g 2 ( s , t ) R N a | Δ ( t u ) | p ( x ) + V ( x ) | t u | p ( x ) d x +   b R N | Δ ( t u ) | p ( x ) d x R N 1 p ( x ) | Δ ( t u ) | p ( x ) d x +   b R N | Δ ( t u ) | p ( x ) d x R N 1 p ( x ) | Δ ( s u + ) | p ( x ) d x   ε C 1 t u p 0 x C ε K x t u p * x
So, for s > 0 that is sufficiently small, it holds that
g 1 ( s , t ) > 0 , for all t > 0
Similarly, for t > 0 that is sufficiently small, it also holds that
g 2 ( s , t ) > 0 , for all s > 0
Moreover, from ( f 3 ), there is C 2 > 0 such that
F ( u ) C 2 u μ , for every u sufficiently large ,
which implies
g 1 ( s , t ) R N a | Δ ( s u + ) | p ( x ) + V ( x ) | s u + | p ( x ) d x +   b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( s u + ) | p ( x ) d x +   b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( t u ) | p ( x ) d x   C 2 s μ R N K x | u + | μ d x
Thus, since μ > 2 p + , for s > 0 large enough, one has
g 1 ( s , t ) < 0 , for all t > 0
Similarly, for t > 0 that is sufficiently small, it also holds that
g 2 ( s , t ) < 0 , for all s > 0
Thus there exists 0 < r < R such that
g 1 ( r , r ) > 0 , g 2 ( r , r ) > 0 ; g 1 ( R , R ) < 0 , g 2 ( R , R ) < 0 .
Then, from (39), (40), and (47), we have
g 1 ( r , t ) > 0 , g 1 ( R , t ) < 0 , t r , R
and
g 2 ( s , r ) > 0 , g 2 ( s , R ) < 0 , s r , R
After that, we demonstrate that g 1 ( s , t ) and g 2 ( s , t ) are continuous with regard to ( s , t ) on [ r , R ] 2 . We just demonstrate the continuity of g 1 here; the continuity of g 2 can be attained in a similar manner. For convenience, we note
A s = R N a | Δ ( s u + ) | p ( x ) + V ( x ) | s u + | p ( x ) d x ,
B s = b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( s u + ) | p ( x ) d x ,
C s , t = b R N | Δ ( s u + ) | p ( x ) d x R N 1 p ( x ) | Δ ( t u ) | p ( x ) d x ,
D s = R N K x f s u + s u + .
Then, g 1 s , t = A s + B s + C s , t D s . Since s p ( x ) is continuous, p ( x ) < N and u + E , then
a | Δ ( s u + ) | p ( x ) + V ( x ) | s u + | p ( x ) max s p + , s p a | Δ u + | p ( x ) + V ( x ) | u + | p ( x ) C | Δ u + | p ( x ) + V ( x ) | u + | p ( x ) L 1
From the dominated convergence theorem, we have
A s n A s 0 a s s n s 0 .
Likewise, because a continuous function’s integral stays continuous, we obtain
B s n B s 0 a s s n s 0 .
Since K ( x ) > 0 and f ( u ) C 1 , then D ( s ) C 1 and D s n D s 0 a s s n s 0 .
  • Let P s = b R N | Δ ( s u + ) | p ( x ) d x , Q t = R N 1 p ( x ) | Δ ( t u ) | p ( x ) d x , that is, C ( s , t ) = P ( s ) Q ( t ) . Then, similar to the continuity of A ( s ) , P ( s ) and Q ( t ) are also continuous. Moreover, because u + is fixed, P ( s ) and Q ( t ) are bounded, that is
s u p s r , R P s M 1 , s u p t r , R Q t M 2 .
Then, by the continuity of P ( s ) and Q ( t ) , we have
P s n Q t n P s 0 Q t 0 P s n Q t n Q t 0 + P s n P s 0 Q t n 0 , a s s n , t n s 0 , t 0 ,
That is, C s n , t n C s 0 , t 0 a s s n , t n s 0 , t 0 .
  • Therefore,
lim s n , t n s 0 , t 0 g 1 s n , t n = g 1 s 0 , t 0
That is, g 1 is continuous with regard to ( s , t ) . Similarly, g 2 is also continuous with respect to ( s , t ) .
Finally, by the Miranda Theorem [36] (see also [25] Lemma 2.4), there exists some point ( s u , t u ) with s u , t u r , R such that g 1 ( s u , t u ) = g 2 ( s u , t u ) = 0 . Therefore, s u u + + t u u M b .
Next, prove the uniqueness of ( s u , t u ) . Suppose there exist ( s i , t i ) ( i = 1 , 2 ) such that s i u + + t i u M b . We assume s 1 < s 2 and t 1 < t 2 . From Corollary 1, we have
Φ b ( s 1 u + + t 1 u ) Φ b ( s 2 u + + t 2 u ) + min 1 , a s 1 p + s 2 p + 2 2 p + s 1 p + R N | Δ u + | p ( x ) + V ( x ) | u + | p ( x ) d x + min 1 , a t 1 p + t 2 p + 2 2 p + t 1 p + R N | Δ u | p ( x ) + V ( x ) | u | p ( x ) d x
and
Φ b ( s 2 u + + t 2 u ) Φ b ( s 1 u + + t 1 u ) + min 1 , a s 2 p s 1 p 2 2 p + s 2 p R N | Δ u + | p ( x ) + V ( x ) | | u + | p ( x ) d x + min 1 , a t 2 p t 1 p 2 2 p + t 2 p R N | Δ u | p ( x ) + V ( x ) | | u | p ( x ) d x .
Combining (51) and (52) implies s 1 = s 2 , t 1 = t 2 . □
Lemma 9. 
If ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, then
m b = inf u M b Φ b ( u ) = inf u E , u ± 0 max s , t 0 Φ b ( s u + + t u ) .
Proof. 
The proof follows that of Lemma 3.5 in [29], details omitted. □
Lemma 10. 
If ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, then m b > 0 can be achieved.
Proof. 
From Lemma 5, we know that u n is bounded in E, then there exists u b E such that u n ± u b ± in E. Since u n M b , we have Φ b ( u n ) , u n ± = 0 , that is
R N a | Δ u n ± | p x + V x | u n ± | p x d x +   b R N 1 p x | Δ u n ± | p x d x R N | Δ u n ± | p x d x = R N K x f ( u n ± ) u n ± .
Through Lemma 4 we know that min u n ± p + , u n ± p β for n N . By (26) and the boundedness of u n , we have
min a , 1 β min { a , 1 } min u n ± p + , u n ± p R N a | Δ u n ± | p x + V x | u n ± | p x d x < R N K x f ( u n ± ) u n ± d x = R N K x f u b ± u b ± d x + o 1 ,
which means that u b ± 0 .
By the Fatou lemma and Lemma 6, we have
R N [ a | Δ u b ± | p ( x ) + V ( x ) | u b ± | p ( x ) ] d x + b R n 1 p ( x ) | Δ u b | p ( x ) d x R n | Δ u b ± | p ( x ) d x liminf R n [ a | Δ u n ± | p ( x ) + V ( x ) | u n ± | p ( x ) ] d x + liminf b R n 1 p ( x ) | Δ u n | p ( x ) d x R n | Δ u n ± | p ( x ) d x = liminf R N K x f ( u n ± ) u n ± d x = R N K x f ( u b ± ) u b ± d x
which means that Φ b ( u b ) , u b ± 0 . Using the Fatou lemma, we have
m b = lim n Φ b u n 1 2 p + Φ b ( u n ) , u n lim inf n a R N 1 p x 1 2 p + | Δ u n | p x d x + lim inf n b 2 R N 1 p x | Δ u n | p x d x R N 1 p x 1 p + | Δ u n | p x d x + lim inf n R N 1 p x 1 2 p + V x | u n | p x d x + lim inf n R N K x 1 2 p + f ( u n ) u n F ( u n ) d x a R N 1 p x 1 2 p + | Δ u b | p x d x + R N 1 p x 1 2 p + V x | u b | p x d x + R N K x 1 2 p + f ( u b ) u b F ( u b ) d x + b 2 R N 1 p x | Δ u b | p x d x R N 1 p x 1 p + | Δ u b | p x d x = Φ b u b 1 2 p + Φ b ( u b ) , u b .
From Lemma 2, one has
Φ b u b Φ b ( s u b + + t u b ) + 1 s 2 p 2 p + Φ b u b , u b + + 1 t 2 p 2 p + Φ b u b , u b ,
then
m b sup s , t 0 Φ b ( s u b + + t u b ) + 1 s 2 p 2 p + Φ b u b , u b + + 1 t 2 p 2 p + Φ b u b , u b 1 2 p + Φ b ( u b ) , u b = sup s , t 0 Φ b ( s u b + + t u b ) s 2 p 2 p + Φ b u b , u b + t 2 p 2 p + Φ b u b , u b max s , t 0 Φ b ( s u b + + t u b ) m b .
Thus, Φ b u b = m b and u b M b . □
Lemma 11. 
If ( v k 1 )–( v k 4 ) and ( f 1 ) (or ( f 1 ))–( f 4 ) hold, u 0 M b and Φ b u 0 = m b , then u 0 is the critical point of Φ b .
Proof. 
It is demonstrated that Φ b u b = 0 using the deformation lemma. If Φ b u b 0 is a contradiction, then ω > 0 and λ > 0 exist such that
u E , Φ b u E λ for all u u 0 3 ω .
For any s , t 0 , we have
Φ b ( s u 0 + + t u 0 ) Φ b u 0 min 1 , a ( 1 μ ) 2 2 p + R N | Δ u 0 + | p ( x ) + V ( x ) | u 0 + | p ( x ) d x min 1 , a ( 1 η ) 2 2 p + R N | Δ u 0 | p ( x ) + V ( x ) | u 0 | p ( x ) d x , s , t 0 .
Let D = 1 / 2 , 3 / 2 × 1 / 2 , 3 / 2 . From (54), one has
κ : = max ( s , t ) D Φ b ( s u 0 + + t u 0 ) < m b
For ε : = min m b κ / 3 , 1 , ω λ / 8 . By the deformation lemma [37], one has
(i)
η ( 1 , u ) = u if u Φ b 1 m b 2 ε , m b + 2 ε S 2 ω ;
(ii)
η 1 , Φ b m b + ε B ( u 0 , ω ) Φ b m b ε ;
(iii)
Φ b η 1 , u Φ b u , u E .
From Corollary 2, one has Φ b ( s u 0 + + t u 0 ) < Φ b ( u 0 ) = m b for s , t > 0 . By (ii), we have
Φ b η ( 1 , s u 0 + + t u 0 ) m b ε , s , t 0 , | s 1 | 2 + | t 1 | 2 < ω 2 / u 0 2
Also, with (iii) and (54), we have
Φ b η ( 1 , s u 0 + + t u 0 ) Φ b ( s u 0 + + t u 0 ) m b min 1 , a ( 1 μ ) 2 2 p + R N | Δ u 0 + | p ( x ) + V ( x ) | u 0 + | p ( x ) d x min 1 , a ( 1 η ) 2 2 p + R N | Δ u 0 | p ( x ) + V ( x ) | u 0 | p ( x ) d x m b min { 1 , a } ω 2 2 p + u 0 2 × R N | Δ u 0 + | p ( x ) + V ( x ) | u 0 + | p ( x ) d x , R N | Δ u 0 | p x + V ( x ) | u 0 | p x d x s , t 0 , | s 1 | 2 + | t 1 | 2 < ω 2 / u 0 2 .
With (56) and (57) we can get
max ( s , t ) D ¯ Φ b η ( 1 , s u 0 + + t u 0 ) < m b
Next, we show that η ( 1 , s u 0 + + t u 0 ) M b , which contradicts the definition of m b .
Define
Ψ 0 ( s , t ) = Φ b ( s u 0 + + t u 0 ) , u 0 + , Φ b ( s u 0 + + t u 0 ) , u 0 ,
and
Ψ 1 ( s , t ) = Φ b η ( 1 , s u 0 + + t u 0 ) , η ( 1 , s u 0 + + t u 0 ) + s , Φ b η ( 1 , s u 0 + + t u 0 ) , η ( 1 , s u 0 + + t u 0 ) t
From Lemma 2 and the degree theory, we can infer that deg ( Ψ 0 , D , 0 ) = 1 . The combination of formula (55) and (i) can be seen
η ( 1 , s u 0 + + t u 0 ) = s u 0 + + t u 0
Therefore Ψ 1 ( s 0 , t 0 ) = 0 for some ( s 0 , t 0 ) D , meaning η ( 1 , s u 0 + + t u 0 ) M b , which contradicts (58). This leads to the conclusion that u 0 is critical for Φ b . □

3. Proof of Main Results

To show the main result of Theorem 1, this section is separated into two subsections.

3.1. Sign-Changing Solutions for Problem (1)

Proof of Theorem 1. 
First, we demonstrate the presence of a least energy sign-changing solution u b for Equation (1), which changes sign only once. Using Lemmas 10 and 11, we find u b M b such that Φ b u b = m b and Φ b ( u b ) = 0 . By definition, u b is the least energy sign-changing solution to Equation (1).
Next, we demonstrate that u b has precisely two nodal domains. Suppose, for contradiction, that u b = u 1 + u 2 + u 3 , where
u i 0 , u 1 0 , u 2 0 and suppt ( u i ) suppt ( u j ) = , for i j , i , j = 1 , 2 , 3 .
Setting υ = u 1 + u 2 , where υ + = u 1 and υ = u 2 , i . e . , υ ± 0 . Since Φ b u b = 0 , we have
Φ b ( v ) , v + = a R N | Δ v + | p ( x ) d x + b R N 1 p ( x ) | Δ v + | p ( x ) d x R N | Δ v + | p ( x ) d x +   b R N 1 p ( x ) | Δ v | p ( x ) d x R N | Δ v + | p ( x ) d x + R N V ( x ) | v + | p ( x ) d x R N K x f ( v + ) v + d x .
and
Φ b ( u b ) , v + = a R N | Δ v + | p ( x ) d x + b R N 1 p ( x ) | Δ v + | p ( x ) d x R N | Δ v + | p ( x ) d x +   b R N 1 p ( x ) | Δ v | p ( x ) d x R N | Δ v + | p ( x ) d x +   b R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v + | p ( x ) d x + R N V ( x ) | v + | p ( x ) d x R N K x f ( v + ) v + d x .
From (59) and (60), we can deduce that
Φ b ( v ) , v + = Φ b ( u b ) , v + b R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v + | p ( x ) d x = b R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v + | p ( x ) d x .
Using the same method, we have
Φ b ( v ) , v = Φ b ( u b ) , v b R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v | p ( x ) d x = b R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v | p ( x ) d x
Using (61), (62), Lemma 2 and Lemma 9, we have
m b = Φ b ( u b ) = Φ b ( u b ) 1 2 p + Φ b ( u b ) , u b = Φ b ( v ) + Φ b ( u 3 ) + b R N 1 p ( x ) | Δ v | p ( x ) d x R N 1 p ( x ) | Δ u 3 | p ( x ) d x 1 2 p + Φ b ( v ) , v + Φ b ( u 3 ) , u 3 + b R N 1 p ( x ) | Δ v | p ( x ) d x R N | Δ u 3 | p ( x ) d x +   b R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v | p ( x ) d x sup s , t 0 Φ b ( s v + + t v ) + 1 μ 2 2 p + Φ b ( v ) , v + + 1 η 2 2 p + Φ b ( v ) , v 1 2 p + Φ b ( v ) , v + Φ b ( u 3 ) 1 2 p + Φ b ( u 3 ) , u 3 sup s , t 0 Φ b ( s v + + t v ) + b μ 2 2 p + R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v + | p ( x ) d x + b η 2 2 p + R N 1 p ( x ) | Δ u 3 | p ( x ) d x R N | Δ v | p ( x ) d x + a 2 R N 1 p ( x ) | Δ u 3 | p ( x ) d x + 1 2 R N V ( x ) p ( x ) | u 3 | p ( x ) d x max s , t 0 Φ b ( s v + + t v ) + 1 2 p + R N a | Δ u 3 | p ( x ) + V ( x ) | u 3 | p ( x ) d x m b + min a , 1 2 p + min u 3 p + , u 3 p .
Therefore, by Lemma 8, we conclude that u 3 = 0 , that is u b = v . Consequently, Φ b v = 0 and Φ b ( v ) , v + = Φ b ( v ) , v = 0 , then, v M b . Hence, it follows that u b has exactly two nodal domains. □

3.2. Sign-Changing Solutions for Problem (3)

In the argument above, it is clear that the case b = 0 is valid. Thus, under the conditions of Theorem 1, Equation (3) has the least energy sign-changing solution that changes sign only once. Next, we discuss the convergence issue. Let b > 0 , and u b M b be the least energy sign-changing solution of Equation (1) with only two nodal domains. We first prove that the sequence { u b n } is bounded. Without loss of generality, let ω 0 C 0 R N be such that ω 0 ± 0 .
By ( f 3 ), it can be deduced that there exist C 2 > 0 such that
F u C 2 u μ , for every u sufficiently large .
Hence, for any b [ 0 , 1 ] , by Lemma 8 and (63), we have
Φ b ( u b ) = m b max s , t 0 Φ b ( s ω 0 + + t ω 0 ) max s , t 0 a R N 1 p ( x ) | Δ ( s ω 0 + ) | p ( x ) d x + b 2 R N 1 p ( x ) | Δ ( s ω 0 + ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | s ω 0 + | p ( x ) d x C 2 R N K x | s ω 0 + | μ d x + a R N 1 p ( x ) | Δ ( t ω 0 ) | p ( x ) d x + b 2 R N 1 p ( x ) | Δ ( t ω 0 ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | t ω 0 | p ( x ) d x C 2 R N K x | t ω 0 | μ d x + b 2 R N 1 p ( x ) | Δ ( s ω 0 + ) | p ( x ) d x 2 + b 2 R N 1 p ( x ) | Δ ( t ω 0 ) | p ( x ) d x 2 max s , t 0 a R N 1 p ( x ) | Δ ( s ω 0 + ) | p ( x ) d x + R N 1 p ( x ) | Δ ( s ω 0 + ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | s ω 0 + | p ( x ) d x + a R N 1 p ( x ) | Δ ( t ω 0 ) | p ( x ) d x + R N 1 p ( x ) | Δ ( t ω 0 ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | t ω 0 | p ( x ) d x : = Λ 0 ( 0 , + ) .
Thus, for any sequence b n with b n 0 as n , using Lemma 9 and (64), we have for large n N
Λ 0 + 1 Φ b n ( u b n ) 1 2 p + Φ b n ( u b n ) , u b n min a , 1 2 p + min u b n p + , u b n p .
Thus, the sequence { u b n } is bounded in E. Next, we prove that Φ 0 ( u 0 ) = 0 . A subsequence of { u b n } , still denoted by { u b n } , and u 0 E such that u b n u 0 in E. Then u 0 is a weak solution of (1). By Lemma 7, we have u b n u 0 0 , n in E.
In fact,
Φ 0 ( u 0 ) , φ = a R N | Δ u 0 | p ( x ) 2 Δ u 0 Δ φ d x + R N V ( x ) | u 0 | p ( x ) 2 u 0 φ d x R N K x f ( u 0 ) φ d x = lim n a + b n R N 1 p ( x ) | Δ u b n | p ( x ) d x R N | Δ u b n | p ( x ) 2 Δ u b n Δ φ d x + R N V ( x ) | u b n | p ( x ) 2 u b n φ d x R N K x f ( u b n ) φ d x = lim n Φ b n u b n , φ = 0 , for all φ C 0 ( R N ) .
Therefore, we can get Φ 0 ( u 0 ) = 0 . Then u 0 M 0 and Φ 0 u 0 m 0 . Next, we prove Φ 0 u 0 = m 0 . Let b n 0 , 1 . By condition ( f 3 ), there exists K 0 > 0 such that
Φ b n s v 0 + + t v 0 = a R N 1 p ( x ) | Δ ( s v 0 + ) | p ( x ) d x + b n 2 R N 1 p ( x ) | Δ ( s v 0 + ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | s v 0 + | p ( x ) d x R N K x F ( s v 0 + ) d x + a R N 1 p ( x ) | Δ ( t v 0 ) | p ( x ) d x + b n 2 R N 1 p ( x ) | Δ ( t v 0 ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | t v 0 | p ( x ) d x R N K x F ( t v 0 ) d x + b n R N 1 p ( x ) | Δ ( s v 0 + ) | p ( x ) d x R N 1 p ( x ) | Δ ( t v 0 ) | p ( x ) d x a R N 1 p ( x ) | Δ ( s v 0 + ) | p ( x ) d x + R N 1 p ( x ) | Δ ( s v 0 + ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | s v 0 + | p ( x ) d x C 2 R N | s v 0 + | μ d x + a R N 1 p ( x ) | Δ ( t v 0 ) | p ( x ) d x + R N 1 p ( x ) | Δ ( t v 0 ) | p ( x ) d x 2 + R N V ( x ) p ( x ) | t v 0 | p ( x ) d x C 2 R N | t v 0 | μ d x < 0 , s , t K 0
By Lemma 8, there exists s n , t n such that s n v 0 + + t n v 0 M b n . From Lemma 10 and (65), we have 0 < s n , t n < K 0 . Because Φ 0 ( v 0 ) = 0 , then from (16) and Lemma 2, we have
m 0 = Φ 0 ( v 0 ) = Φ b n ( v 0 ) b n 2 R N 1 p ( x ) | Δ v 0 | p ( x ) d x 2 Φ b n s n v 0 + + t n v 0 + 1 s n 2 p 2 p + Φ b n ( v 0 ) , v 0 + + 1 t n 2 p 2 p + Φ b n ( v 0 ) , v 0 b n 2 R N 1 p ( x ) | Δ v 0 | p ( x ) d x 2 m b n 1 + K 0 2 p 2 p + | Φ b n ( v 0 ) , v 0 + | 1 + K 0 2 p 2 p + | Φ b n ( v 0 ) , v 0 | b n 2 R N 1 p ( x ) | Δ v 0 | p ( x ) d x 2 = m b n 1 + K 0 2 p b n 2 p + R N 1 p ( x ) | Δ v 0 | p ( x ) d x R N | Δ v 0 + | p ( x ) d x 1 + K 0 2 p b n 2 p + R N 1 p ( x ) | Δ v 0 | p ( x ) d x R N | Δ v 0 | p ( x ) d x b n 2 R N 1 p ( x ) | Δ v 0 | p ( x ) d x 2 ,
which implies
limsup n m b n m 0
From (11), (16), and (67), one has
m 0 Φ 0 ( u 0 ) = limsup n Φ b n u b n = limsup n m b n m 0 .
Thus, Φ 0 u 0 = m 0 .

Author Contributions

Methodology Q.M.; writing—original draft preparation, R.D.; writing—review and editing, Q.M. and R.D.; supervision, Q.M. All authors have read and agreed to the published version of the manuscript.

Funding

This project was supported by the National Natural Science Foundation of China (No. 11861078).

Data Availability Statement

The original contributions presented in the study are included in the article; further inquiries can be directed to the corresponding author.

Acknowledgments

The authors express their profound gratitude to the anonymous reviewer for their intelligent recommendations and perceptive comments, which have significantly raised the manuscript’s overall caliber.

Conflicts of Interest

The authors declare no conflicts of interest.

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Deng, R.; Miao, Q. Existence of Sign-Changing Solutions for a Class of p(x)-Biharmonic Kirchhoff-Type Equations. Axioms 2025, 14, 530. https://doi.org/10.3390/axioms14070530

AMA Style

Deng R, Miao Q. Existence of Sign-Changing Solutions for a Class of p(x)-Biharmonic Kirchhoff-Type Equations. Axioms. 2025; 14(7):530. https://doi.org/10.3390/axioms14070530

Chicago/Turabian Style

Deng, Rui, and Qing Miao. 2025. "Existence of Sign-Changing Solutions for a Class of p(x)-Biharmonic Kirchhoff-Type Equations" Axioms 14, no. 7: 530. https://doi.org/10.3390/axioms14070530

APA Style

Deng, R., & Miao, Q. (2025). Existence of Sign-Changing Solutions for a Class of p(x)-Biharmonic Kirchhoff-Type Equations. Axioms, 14(7), 530. https://doi.org/10.3390/axioms14070530

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