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Infinitely many solutions for a class of Kirchhoff-type equations

  • Qin Zhou and Jing Zeng EMAIL logo
Published/Copyright: August 28, 2025
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Open Mathematics
From the journal Open Mathematics Volume 23 Issue 1

Abstract

In this article, we consider a class of Kirchhoff-type equations:

− a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u = f ( x , u ) , x ∈ Ω , u = 0 , x ∈ ∂ Ω .

It is a generalization of the classical wave equation and is often used to model wave propagation in various physical media. The nonlocal term b ( ∫ Ω ∣ ∇ u ∣ 2 d x ) Δ u in the equation causes the variational functional of the equation to have fundamentally different properties from the case of b = 0 . As far as we know, there are relatively few conclusions regarding infinitely many solutions of this equation. Under weaker assumptions, we obtain that the equation has infinitely many high-energy solutions. And our results extend the conclusions of Mao-Zhang (2009) and Zhang-Perera (2006).

Keywords: Kirchhoff-type equation; multiple solutions; high energy solution
MSC 2010: 35J20

1 Introduction and main results

This article is related to a class of Kirchhoff-type equations:

(1) − a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u = f ( x , u ) , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Ω is a smooth bounded domain in R N ( N = 1 , 2, or 3), a , b > 0 are constants, and f ( x , u ) : Ω ¯ × R → R is a continuous function. The equation is a generalization of the classical wave equation, capturing more complex wave phenomena such as dispersion and dissipation. It was initially formulated by Kirchhoff to address the limitations of the simple wave equation in describing wave propagation in acoustics [1]. It is connected with the stationary analogue of the equation:

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

where ρ denotes the mass density, u is the displacement, P 0 is the tension at the initial moment, h represents the cross-sectional area of the elastic string, and E and L denote Young’s modulus for the material and the length of the string, respectively [2]. Equation (1) is particularly important for understanding waves in media with varying properties, such as density and elasticity, which are common in seismic wave propagation and the vibration analysis of structures.

In physics, Kirchhoff-type equations find applications in various fields. In solid mechanics, they model the vibrations of elastic bodies, considering both the material’s inertia and its elastic properties. In fluid dynamics, they describe the behavior of surface waves under the influence of gravity and surface tension. In addition, these equations are integral to the theory of general relativity, where they predict the propagation of gravitational waves. By accounting for additional physical effects, Kirchhoff-type equations enhance our predictive capabilities for wave phenomena across different physical contexts [3].

It was not until Lions [4] proposed the following: Kirchhoff equation:

∂ 2 u ∂ t 2 − a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u = f ( x , u ) , x ∈ Ω u = 0 , x ∈ ∂ Ω

that equation (1) received increasing attention in mathematics, where a , b > 0 are constants and f ( x , u ) describes the external force. For nontrivial solutions, Pei and Zhang [5] obtained a nontrivial solution by combining a variant version of the mountain pass theorem with the moser-trudinger inequality for a = b = 1 . Song et al. [6] derived a nontrivial solution by applying the classical linking theorem and the G-linking theorem. By the morse theory and the minimax method, Sun et al. [7] computed the critical groups including the cases where zero is a mountain pass solution and the nonlinearity is resonant at zero. As an application, they demonstrated the multiplicity of nontrivial solutions for (1) when f ( x , u ) = g ( x , u ) + ∣ u ∣ 2 ∗ − 2 , where g ( x , u ) , an additional critical nonlinear term, satisfies certain conditions. Some results about nontrivial solutions of (1) can be found in [8–10] and the references therein.

Cheng and Wu [11] used the variational method to prove the existence of two positive solutions of (1), one dealing with the asymptotic behaviors of f ( x , u ) near zero and infinity, and the other addressing the 4-superlinear of f ( x , u ) at infinity. Yang and Zhang [12] used invariant sets of descent flow to obtain two solutions of (1), a positive one and a negative one. Zhang and Perera [13] assumed

(2) ν F ( x , t ) ≤ t f ( x , t ) , for x ∈ Ω , ∣ t ∣ ≥ L > 0 , ν > 4 ,

and obtained a positive, a negative and a sign-changing solution of (1) by invariant sets of descent flow. Later, Mao and Zhang [14] assumed F ˜ ( x , t ) = 1 4 f ( x , t ) t − F ( x , t ) → + ∞ , as ∣ t ∣ → + ∞ uniformly in x ∈ Ω , ∣ f ( x , t ) ∣ σ ≤ C F ˜ ( x , t ) ∣ t ∣ σ , for ∣ t ∣ large, σ > max { 1 , N 2 } , and obtained at least one positive, one negative, and one sign-changing solution of (1).

Moreover, Shuai [15] combined the constrained variational method with the deformation lemma to obtain a least energy sign-changing solution, whose energy is strictly larger than the ground state energy. By using analytical skills and a non-Nehari manifold method, Tang and Cheng [16] proved that (1) has a ground state sign-changing solution, whose energy is strictly larger than twice the ground state solutions. Liu et al. [17] discussed the existence of a positive and a negative solution by using an iterative technique and the mountain pass theorem, and a sign-changing solution by combining the iterative technique and the Nehari method.

In the present article, we assume:

( f 1 ) f ( x , t ) = o ( ∣ t ∣ ) , as ∣ t ∣ → 0 uniformly for any x ∈ Ω .

( f 2 ) There exist constants C > 0 and p ∈ ( 4 , 2 ∗ ) such that

∣ f ( x , t ) ∣ ≤ C ( 1 + ∣ t ∣ p − 1 ) ,

where 2 ∗ = + ∞ for N = 1 , 2 , and 2 ∗ = 2 N N − 2 for N ≥ 3 .

( f 3 ) f ( x , − t ) = − f ( x , t ) , for ( x , t ) ∈ Ω × R .

( f 4 ) F ( x , t ) t 4 → + ∞ , as ∣ t ∣ → + ∞ uniformly for x ∈ Ω , where F ( x , t ) = ∫ 0 t f ( x , s ) d s .

( f 5 ) F ˜ ( x , t ) ≔ 1 4 f ( x , t ) t − F ( x , t ) → + ∞ , as ∣ t ∣ → + ∞ uniformly in x ∈ Ω , and there exists C ˜ > 0 , σ > max { 1 , N 2 } , such that ∣ f ( x , t ) ∣ σ ≤ C ˜ F ˜ ( x , t ) ∣ t ∣ σ , for ∣ t ∣ large.

( f 6 ) t f ( x , t ) ≥ 4 F ( x , t ) , for ( x , t ) ∈ Ω × R .

Our main results are as follows.

Theorem 1.1

If the assumptions ( f 1 )–( f 5 ) hold, then equation (1) admits infinitely many high-energy solutions.

Remark 1.2

Assuming u f ( x , u ) ≥ 0 , conditions ( f 4 ) and ( f 5 ) are weaker than those in [13] (specifically compared to (2)). Examples of functions satisfying ( f 4 ) and ( f 5 ) but not (2) include

f ( x , u ) = u 3 ln ( 1 + ∣ u ∣ ) or F ( x , u ) = ∣ u ∣ μ + ( μ − 4 ) ∣ u ∣ μ − ε sin 2 ( ∣ u ∣ ε ε ) ,

where 4 < μ < 2 * , 0 < ε < 3 − μ 2 .

Corollary 1.3

Under the assumption of (2) and inf x ∈ Ω , ∣ u ∣ = L F ( x , u ) > 0 , we can certify that (1) has infinitely many high-energy solutions.

Remark 1.4

Our results generalize the results of [13] and [14]. Our assumptions are weaker than those of [13], but the number of solutions we obtained is greater than those in [13]. In [14], Mao-Zhang obtained three solutions of (1), and we have infinitely many high-energy solutions.

Theorem 1.2

If assumptions ( f 1 )–( f 4 ), and ( f 6 ) hold, then equation (1) has infinitely many high-energy solutions.

The article is organized as follows. Section 2 gives some lemmas to prove the theorems. Lemmas 2.1 and 2.2 are devoted to proving the functional corresponding to (1) satisfies the ( PS ) c condition. Section 3 proves Theorems 1.1 and 1.2.

2 Variational framework and lemmas

Let H ≔ H 0 1 ( Ω ) be a Hilbert space equipped with the inner product and norm

( u , v ) = ∫ Ω ∇ u ⋅ ∇ v d x , ‖ u ‖ = ∫ Ω ∣ ∇ u ∣ 2 d x 1 2 .

L p ( Ω ) ( 1 ≤ p < ∞ ) be the normal Banach space with the norm ∣ u ∣ p = ∫ Ω ∣ u ∣ p d x 1 p . Since Ω is a bounded domain, according to the Sobolev embedding theorem, H ↪ L s ( Ω ) continuously for s ∈ [ 1 , 2 ∗ ] , compactly for s ∈ [ 1 , 2 ∗ ) . Moreover, for any u ∈ H , there exists a constant η s > 0 such that

(3) ∣ u ∣ s ≤ η s ‖ u ‖ .

We choose a countable orthogonal basis { e i } of H with ‖ e i ‖ = 1 , i = 1 , 2 , … . Set

(4) X i = span { e i } , Y k = ⊕ i = 1 k X i , Z k = ⊕ i = k ∞ X i ¯ .

Meanwhile, a weak solution is the critical point of the functional

Φ ( u ) = a 2 ∫ Ω ∣ ∇ u ∣ 2 d x + b 4 ∫ Ω ∣ ∇ u ∣ 2 d x 2 − ∫ Ω F ( x , u ) d x ,

and for any u , v ∈ H ,

( Φ ′ ( u ) , v ) = ( a + b ‖ u ‖ 2 ) ∫ Ω ∇ u ∇ v d x − ∫ Ω f ( x , u ) v d x .

In this article, C and C i in different positions denote different positive constants.

Theorem 2.1

([18], Fountain theorem) Let X be a Banach space, and let { X i } be a sequence of subspaces of X with dim X i < ∞ for all i ∈ N . Moreover, let X = ⊕ i = 1 ∞ X i ¯ , Y k = ⊕ i = 1 k X i and Z k = ⊕ i = k ∞ X i ¯ . Consider an even functional I ∈ C 1 ( X , R ) , i.e., I ( u ) = I ( − u ) . Assume for every k ∈ N , there exist ρ k > γ k > 0 such that

  1. a k ≔ max u ∈ Y k , ‖ u ‖ = ρ k I ( u ) ≤ 0 ,

  2. b k ≔ inf u ∈ Z k , ‖ u ‖ = γ k I ( u ) → + ∞ , as k → ∞ ,

  3. I satisfies ( PS ) c condition for every c > 0 .

Then I has an unbounded sequence of critical values.

Lemma 2.1

Assume that ( f 1 ), ( f 2 ), ( f 4 ), and ( f 5 ) hold, then Φ ( u ) satisfies the ( PS ) c condition for c > 0 . That is, any sequence { u n } ⊂ H satisfying Φ ( u n ) → c > 0 and Φ ′ ( u n ) → 0 as n → + ∞ , has a convergent subsequence.

Proof

Following the standard approach, we need to prove that { u n } is bounded and that it contains a convergent subsequence.

If { u n } is unbounded, assume ‖ u n ‖ → + ∞ , as n → + ∞ . We have

c + 1 + ‖ u n ‖ ≥ Φ ( u n ) − 1 4 ( Φ ′ ( u n ) , u n ) = a 4 ∫ Ω ∣ ∇ u n ∣ 2 d x + ∫ Ω 1 4 f ( x , u n ) u n − F ( x , u n ) d x = a 4 ‖ u n ‖ 2 + ∫ Ω F ˜ ( x , u n ) d x .

Then

(5) c + 1 ≥ a 4 ‖ u n ‖ 2 − ‖ u n ‖ + ∫ Ω F ˜ ( x , u n ) d x ≥ ∫ Ω F ˜ ( x , u n ) d x .

Let v n = u n ‖ u n ‖ , then ‖ v n ‖ = 1 , meanwhile by (3), ∣ v n ∣ s ≤ η s ‖ v n ‖ = η s , for any s ∈ [ 1 , 2 ∗ ) . Note that

( Φ ′ ( u n ) , u n ) ‖ u n ‖ 4 = a ‖ u n ‖ 2 + b − ∫ Ω f ( x , u n ) u n d x ‖ u n ‖ 4 ,

then as n → + ∞ ,

(6) lim n → + ∞ ∫ Ω f ( x , u n ) u n ‖ u n ‖ 4 d x = b .

Next we certify that b = 0 , which will lead to a contradiction. For 0 ≤ α < β ≤ + ∞ , let

O n ( α , β ) ≔ { x ∈ Ω : α ≤ ∣ u n ( x ) ∣ < β } .

For large n and α < β , we have

(7) ∫ Ω f ( x , u n ) u n ‖ u n ‖ 4 d x = ∫ O n ( 0 , α ) f ( x , u n ) u n ‖ u n ‖ 4 d x + ∫ O n ( α , β ) f ( x , u n ) u n ‖ u n ‖ 4 d x + ∫ O n ( β , + ∞ ) f ( x , u n ) u n ‖ u n ‖ 4 d x .

Subsequently we will prove ∫ O n ( 0 , α ) f ( x , u n ) u n ‖ u n ‖ 4 d x → 0 , ∫ O n ( α , β ) f ( x , u n ) u n ‖ u n ‖ 4 d x → 0 , and ∫ O n ( β , + ∞ ) f ( x , u n ) u n ‖ u n ‖ 4 d x → 0 , as n → + ∞ , respectively.

First, we prove ∫ O n ( β , + ∞ ) f ( x , u n ) u n ‖ u n ‖ 4 d x → 0 , as n → + ∞ . ‖ u n ‖ → + ∞ as n → + ∞ shows that there exists an integer N 0 > 0 such that ‖ u n ‖ ≥ 1 if n > N 0 . For σ > max { 1 , N 2 } , let τ = 2 σ σ − 1 , then τ ∈ ( 2 , 2 ∗ ) and σ = τ τ − 2 . By ( f 5 ) and β → + ∞ ,

(8) ∫ O n ( β , + ∞ ) f ( x , u n ) u n ‖ u n ‖ 4 d x ≤ ∫ O n ( β , + ∞ ) ∣ f ( x , u n ) ∣ ∣ u n ∣ ‖ u n ‖ v n 2 ‖ u n ‖ d x ≤ 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ∣ f ( x , u n ) ∣ σ ∣ u n ∣ σ d x 1 σ 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ∣ v n ∣ τ d x 2 τ ≤ C ˜ ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) F ˜ ( x , u n ) d x 1 σ 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ∣ v n ∣ τ d x 2 τ .

For (8), we will certify C ˜ ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) F ˜ ( x , u n ) d x 1 σ is bounded and 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ∣ v n ∣ τ d x 2 τ → 0 , as β → + ∞ . By (5), then

(9) c + 1 ‖ u n ‖ 2 ≥ 1 ‖ u n ‖ 2 ∫ Ω F ˜ ( x , u n ) d x ≥ 1 ‖ u n ‖ 2 ∫ O n ( 0 , α ) F ˜ ( x , u n ) d x + 1 ‖ u n ‖ 2 ∫ O n ( α , β ) F ˜ ( x , u n ) d x + 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) F ˜ ( x , u n ) d x .

By (9), 1 ‖ u n ‖ 2 ∫ Ω F ˜ ( x , u n ) d x is bounded. To establish that 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) F ˜ ( x , u n ) d x is bounded, we must prove 1 ‖ u n ‖ 2 ∫ O n ( 0 , α ) F ˜ ( x , u n ) d x and 1 ‖ u n ‖ 2 ∫ O n ( α , β ) F ˜ ( x , u n ) d x are bounded.

By ( f 1 ), for all ε > 0 , there exists δ > 0 such that

(10) ∣ f ( x , u n ) ∣ ≤ ε ∣ u n ∣ ,

as ∣ u n ∣ < δ . By ( f 2 ),

(11) ∣ f ( x , u n ) ∣ ≤ C + C ∣ u n ∣ p − 1 ≤ C + C 1 δ p − 1 ∣ u n ∣ p − 1 ,

as ∣ u n ∣ ≥ δ . By combining (10) with (11), we obtain

(12) ∣ f ( x , u n ) ∣ ≤ ε ∣ u n ∣ + C δ p − 1 + C ∣ u n ∣ p − 1 = ε ∣ u n ∣ + C 1 ∣ u n ∣ p − 1 ,

where C 1 = C δ p − 1 + C . Then

(13) ∣ F ( x , u n ) ∣ ≤ ε 2 ∣ u n ∣ 2 + C 1 p ∣ u n ∣ p .

Therefore, for ∣ u n ∣ ∈ [ 0 , α ) ,

(14) ∣ F ˜ ( x , u n ) ∣ ≤ 1 4 ∣ f ( x , u n ) ∣ ∣ u n ∣ + ∣ F ( x , u n ) ∣ ≤ 1 4 ε ∣ u n ∣ 2 + C 1 4 ∣ u n ∣ p + ε 2 ∣ u n ∣ 2 + C 1 p ∣ u n ∣ p ≤ C 2 ∣ u n ∣ 2 + C 2 ∣ u n ∣ p ≤ C 3 ∣ u n ∣ 2 ,

where C 2 = max 3 4 ε , 4 + p 4 p C 1 , C 3 = C 2 ( 1 + α p − 2 ) .

By (14) and (3),

1 ‖ u n ‖ 2 ∫ O n ( 0 , α ) F ˜ ( x , u n ) d x ≤ C 3 ‖ u n ‖ 2 ∫ O n ( 0 , α ) ∣ u n ∣ 2 d x = C 3 ∫ O n ( 0 , α ) ∣ v n ∣ 2 d x ≤ C 3 ∫ Ω ∣ v n ∣ 2 d x ≤ C 3 η 2 2 ‖ v n ‖ 2 = C 3 η 2 2 ,

then 1 ‖ u n ‖ 2 ∫ O n ( 0 , α ) F ˜ ( x , u n ) d x is bounded. The boundedness of 1 ‖ u n ‖ 2 ∫ O n ( α , β ) F ˜ ( x , u n ) d x can be proved in the same way. Therefore, 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) F ˜ ( x , u n ) d x is bounded.

In the following, we will prove ( 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ∣ v n ∣ τ d x ) 2 τ → 0 , as β → + ∞ .

According to the Hölder inequality, for any τ ∈ ( 2 , 2 ∗ ) , β → + ∞ , let q > 0 be a constant such that τ q ∈ ( 0 , 1 ) , by (3), then

(15) 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ∣ v n ∣ τ d x ≤ 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ( ∣ v n ∣ τ ) q τ d x τ q ∫ O n ( β , + ∞ ) 1 q q − τ d x q − τ q ≤ ‖ u n ‖ − 2 η q τ ‖ v n ‖ τ ∣ O n ( β , + ∞ ) ∣ q − τ q = η q τ 1 ‖ u n ‖ 2 τ q O n ( β , + ∞ ) ‖ u n ‖ 2 q − τ q .

Let r ≥ 0 ,

g ( r ) ≔ inf { F ˜ ( x , u ) : x ∈ Ω , u ∈ R , ∣ u ∣ ≥ r } , S α β ≔ inf F ˜ ( x , u ) ∣ u ∣ 2 : x ∈ Ω , u ∈ R , α ≤ ∣ u ∣ < β .

( f 5 ) implies that g ( r ) → + ∞ as r → + ∞ , and g ( α ) > 0 , S α β > 0 for large α > 0 . Moreover, for any x ∈ O n ( α , β ) ,

(16) F ˜ ( x , u n ) ≥ S α β ∣ u n ∣ 2 .

Meantime, 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) F ˜ ( x , u n ) d x ≥ g ( β ) ∣ O n ( β , + ∞ ) ∣ ‖ u n ‖ 2 . g ( β ) → + ∞ as β → + ∞ and the proven boundedness of 1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) F ˜ ( x , u n ) d x , imply ∣ O n ( β , + ∞ ) ∣ ‖ u n ‖ 2 → 0 as β → + ∞ . By (15),

1 ‖ u n ‖ 2 ∫ O n ( β , + ∞ ) ∣ v n ∣ τ d x → 0 , as β → + ∞ ,

then ∫ O n ( β , + ∞ ) f ( x , u n ) u n ‖ u n ‖ 4 d x in (8) converges to 0, as n → + ∞ .

Second, we will prove ∫ O n ( α , β ) f ( x , u n ) u n ‖ u n ‖ 4 d x → 0 , as n → + ∞ . By (12), for ∣ u n ∣ ∈ [ α , β ) , ∣ f ( x , u n ) ∣ ∣ u n ∣ ≤ ε ∣ u n ∣ 2 + C 1 ( ε ) ∣ u n ∣ p ≤ C 4 ∣ u n ∣ 2 . By (16),

∫ O n ( α , β ) f ( x , u n ) u n ‖ u n ‖ 4 d x ≤ ∫ O n ( α , β ) C 4 ∣ u n ∣ 2 ‖ u n ‖ 4 d x ≤ 1 ‖ u n ‖ 4 S α β ∫ O n ( α , β ) F ˜ ( x , u n ) d x .

As the estimation of (14), for ∣ u n ∣ ∈ [ α , β ) , F ˜ ( x , u ) ≤ C 5 ∣ u n ∣ 2 , then

∫ O n ( α , β ) F ˜ ( x , u n ) d x ≤ ∫ O n ( α , β ) C 5 ∣ u n ∣ 2 d x ≤ ∫ Ω C 5 ∣ u n ∣ 2 d x ≤ C 6 ,

that is, ∫ O n ( α , β ) F ˜ ( x , u n ) d x is bounded. Then 1 ‖ u n ‖ 4 S α β ∫ O n ( α , β ) F ˜ ( x , u n ) d x → 0 , as n → + ∞ . So as n → + ∞ , ∫ O n ( α , β ) f ( x , u n ) u n ‖ u n ‖ 4 d x → 0 .

Finally, we will demonstrate ∫ O n ( 0 , α ) f ( x , u n ) u n ‖ u n ‖ 4 d x → 0 , as n → + ∞ . By ( f 1 ), ∣ f ( x , u ) ∣ ≤ C 7 ∣ u ∣ , and by (3),

∫ O n ( 0 , α ) f ( x , u n ) u n ‖ u n ‖ 4 d x ≤ ∫ O n ( 0 , α ) ∣ f ( x , u n ) ∣ ∣ u n ∣ ‖ u n ‖ 4 d x ≤ ∫ O n ( 0 , α ) C 7 ∣ u n ∣ 2 ‖ u n ‖ 4 C 7 η 2 2 ‖ u n ‖ 2 d x → 0 ,

as n → + ∞ . Hence, by (6) and (7), we have b = 0 . This is a contradiction with b > 0 . Therefore, the assumption ‖ u n ‖ → + ∞ is invalid, { u n } is bounded.

Subsequently, we will prove { u n } has a convergent subsequence. Taking a subsequence of { u n } , as well called { u n } , such that u n ⇀ u in H , u n → u in L s ( Ω ) for s ∈ [ 1 , 2 ∗ ) , u n ( x ) → u ( x ) a.e. x ∈ R N , as n → + ∞ . According to ( Φ ′ ( u n ) − Φ ′ ( u ) , u n − u ) → 0 and

( Φ ′ ( u n ) − Φ ′ ( u ) , u n − u ) = a + b ∫ Ω ∣ ∇ u n ∣ 2 d x ∫ Ω ∇ u n ∇ ( u n − u ) d x − a + b ∫ Ω ∣ ∇ u ∣ 2 d x ∫ Ω ∇ u ∇ ( u n − u ) d x − ∫ Ω ( f ( x , u n ) − f ( x , u ) ) ( u n − u ) d x = ( a + b ‖ u n ‖ 2 ) ‖ u n − u ‖ 2 − b ∫ Ω ∣ ∇ u ∣ 2 d x − ∫ Ω ∣ ∇ u n ∣ 2 d x ∫ Ω ∇ u ∇ ( u n − u ) d x − ∫ Ω ( f ( x , u n ) − f ( x , u ) ) ( u n − u ) d x ,

then

(17) ( a + b ‖ u n ‖ 2 ) ‖ u n − u ‖ 2 = ( Φ ′ ( u n ) − Φ ′ ( u ) , u n − u ) + b ∫ Ω ∣ ∇ u ∣ 2 d x − ∫ Ω ∣ ∇ u n ∣ 2 d x ∫ Ω ∇ u ∇ ( u n − u ) d x + ∫ Ω ( f ( x , u n ) − f ( x , u ) ) ( u n − u ) d x .

Since { u n } is bounded and u n ⇀ u , then

(18) b ∫ Ω ∣ ∇ u ∣ 2 d x − ∫ Ω ∣ ∇ u n ∣ 2 d x ∫ Ω ∇ u ∇ ( u n − u ) d x = b ( ‖ u ‖ 2 − ‖ u n ‖ 2 ) ( u , u n − u ) → 0 ,

for n → + ∞ . In addition, by (12),

(19) ∫ Ω ( f ( x , u n ) − f ( x , u ) ) ( u n − u ) d x ≤ ∫ Ω ( ∣ f ( x , u n ) ∣ + ∣ f ( x , u ) ∣ ) ∣ u n − u ∣ d x ≤ ∫ Ω ( ε ( ∣ u n ∣ + ∣ u ∣ ) + C 1 ( ε ) ( ∣ u n ∣ p − 1 + ∣ u ∣ p − 1 ) ) ∣ u n − u ∣ d x ≤ ε ∫ Ω ∣ u n ∣ 2 d x 1 2 ∫ Ω ∣ u n − u ∣ 2 d x 1 2 + ε ∫ Ω ∣ u ∣ 2 d x 1 2 ∫ Ω ∣ u n − u ∣ 2 d x 1 2 + C 1 ( ε ) ∫ Ω ∣ u n ∣ p d x p − 1 p ∫ Ω ∣ u n − u ∣ p d x 1 p + C 1 ( ε ) ∫ Ω ∣ u ∣ p d x p − 1 p ∫ Ω ∣ u n − u ∣ p d x 1 p = ε ( ∣ u n ∣ 2 + ∣ u ∣ 2 ) ∣ u n − u ∣ 2 + C 1 ( ε ) ( ∣ u n ∣ p p − 1 + ∣ u ∣ p p − 1 ) ∣ u n − u ∣ p .

Since H ↪ L s ( Ω ) is compact for s ∈ [ 1 , 2 ∗ ) , then ∣ u n − u ∣ 2 → 0 , ∣ u n − u ∣ p → 0 in L s ( Ω ) , as n → + ∞ . Therefore, ∫ Ω ( f ( x , u n ) − f ( x , u ) ) ( u n − u ) d x → 0 as n → + ∞ . By combining (17), (18), and (19), we conclude that ‖ u n − u ‖ → 0 as n → ∞ . This implies that { u n } contains a convergent subsequence.□

Lemma 2.2

Assume that ( f 1 ), ( f 2 ), ( f 4 ), and ( f 6 ) hold, then Φ ( u ) satisfies the ( PS ) c condition for c > 0 .

Proof

By ( f 6 ),

c + 1 + ‖ u n ‖ ≥ Φ ( u n ) − 1 4 ( Φ ′ ( u n ) , u n ) = a 4 ‖ u n ‖ 2 + 1 4 ∫ Ω ( f ( x , u n ) u n − 4 F ( x , u n ) ) d x ≥ a 4 ‖ u n ‖ 2 .

Since a > 0 , then { u n } is bounded in H .

Next we will prove { u n } has a convergent subsequence. Taking a subsequence of { u n } , as well called { u n } , such that u n ⇀ u in H , u n → u in L s ( Ω ) for s ∈ [ 1 , 2 ∗ ) , u n → u a.e. x ∈ R N , as n → + ∞ . By computation,

(20) ( a + b ‖ u n ‖ 2 ) ( ( u n , u ) − ( u n , u n ) ) = ( a + b ‖ u n ‖ 2 ) ∫ Ω ∇ u n ∇ ( u − u n ) d x − ∫ Ω f ( x , u n ) ( u − u n ) d x + ∫ Ω f ( x , u n ) ( u − u n ) d x = ( Φ ′ ( u n ) , ( u − u n ) ) + ∫ Ω f ( x , u n ) ( u − u n ) d x .

For { u n } is bounded in H and Φ ′ ( u n ) → 0 , then

(21) ( Φ ′ ( u n ) , ( u − u n ) ) ≤ ‖ Φ ′ ( u n ) ‖ ‖ u − u n ‖ ≤ ‖ Φ ′ ( u n ) ‖ ( ‖ u ‖ + ‖ u n ‖ ) → 0 ,

as n → + ∞ . Due to (12) and u n → u in L s ( Ω ) for s ∈ [ 1 , 2 ∗ ) , we know that

(22) ∫ Ω ∣ f ( x , u n ) ( u − u n ) ∣ d x ≤ ∫ Ω ( ε ∣ u n ∣ + C 1 ( ε ) ∣ u n ∣ p − 1 ) ∣ u − u n ∣ d x ≤ ε ∫ Ω ∣ u n ∣ 2 d x 1 2 ∫ Ω ∣ u − u n ∣ 2 d x 1 2 + C 1 ( ε ) ∫ Ω ∣ u n ∣ p d x p − 1 p ∫ Ω ∣ u − u n ∣ p d x 1 p = ε ∣ u n ∣ 2 ∣ u − u n ∣ 2 + C 1 ( ε ) ∣ u n ∣ p p − 1 ∣ u − u n ∣ p → 0 ,

as n → + ∞ . By combining (20), (21), and (22), we obtain ( u n , u ) − ( u n , u n ) → 0 . Meanwhile by u n ⇀ u in H and ( u n , u ) − ( u n , u n ) = ( u n − u , u ) + ( u , u ) − ( u n , u n ) , then ( u n , u n ) → ( u , u ) , as n → + ∞ . In summary, u n → u in H , n → + ∞ . Therefore, Φ ( u ) satisfies the ( PS ) c condition.□

Remark 2.2

The method of proving that { u n } has a convergent subsequence in Lemma 2.2 can be applied to prove Lemma 2.1.

3 Proof of Theorems 1.1 and 1.2

Proof of Theorem 1.1

Firstly, we prove a k ≔ max u ∈ Y k , ‖ u ‖ = ρ k Φ ( u ) ≤ 0 , where Y k is defined in (4). According to ( f 4 ), for any M > 0 , there exist δ > 0 , such that, for ∣ u n ∣ ≥ δ ,

(23) F ( x , u n ) ≥ M ∣ u n ∣ 4 .

( f 1 ) implies that there exists δ ¯ > 0 such that f ( x , u n ) u n ≤ 1 for 0 < ∣ u n ∣ < δ ¯ . By ( f 2 ), for δ ¯ ≤ ∣ u n ∣ ≤ δ , there exists M 1 > 0 such that

f ( x , u n ) u n ≤ C ( 1 + ∣ u n ∣ p − 1 ) ∣ u n ∣ ≤ M 1 .

Then f ( x , u n ) ≥ − ( M 1 + 1 ) ∣ u n ∣ , for 0 ≤ ∣ u n ∣ ≤ δ . By the definition of F ( x , u n ) = ∫ 0 u n f ( x , s ) d s ,

(24) F ( x , u n ) ≥ − C 8 ∣ u n ∣ 2 ,

for 0 ≤ ∣ u n ∣ ≤ δ , where C 8 = 1 2 ( M 1 + 1 ) . By combining (23) with (24), we obtain F ( x , u ) ≥ M ∣ u ∣ 4 − C 8 ∣ u ∣ 2 . Since all norms are equivalent in Y k , a finite-dimensional space, there exist constants d 1 , d 2 > 0 such that in Y k :

Φ ( u ) ≤ a 2 ‖ u ‖ 2 + b 4 ‖ u ‖ 4 − M ∣ u ∣ 4 4 + C 8 ∣ u ∣ 2 2 = a 2 ‖ u ‖ 2 + b 4 ‖ u ‖ 4 − M d 1 ‖ u ‖ 4 + C 8 d 2 ‖ u ‖ 2 = b 4 − M d 1 ‖ u ‖ 4 + a 2 + C 8 d 2 ‖ u ‖ 2 .

As b 4 − M d 1 < 0 , that is, M > b 4 d 1 , by the aforementioned inequality, let ρ k > 0 is big enough such that a k ≤ 0 . In fact, ρ k > γ k > 0 , where γ k will be defined in (25).

Next, we will prove b k ≔ inf u ∈ Z k , ‖ u ‖ = γ k Φ ( u ) → + ∞ , as k → ∞ , where Z k is defined in (4).

By the conclusion of Lemma 3.8 in [18], we have β k ≔ sup u ∈ Z k , ‖ u ‖ = 1 ∣ u ∣ s → 0 , as k → ∞ , for s ∈ [ 2 , 2 ∗ ) . Let ε ∈ 0 , a β k 2 , for u ∈ Z k , ‖ u ‖ = γ k , by β k → 0 as k → ∞ , we can see that β k = sup u ∈ Z k , ‖ u ‖ = γ k u ‖ u ‖ s . Therefore, ∣ u ∣ s ≤ β k ‖ u ‖ . By (3) and (13),

Φ ( u ) ≥ a 2 ‖ u ‖ 2 + b 4 ‖ u ‖ 4 − ε 2 ∫ Ω ∣ u ∣ 2 d x − C 1 p ∫ Ω ∣ u ∣ p d x ≥ a 2 ‖ u ‖ 2 + b 4 ‖ u ‖ 4 − ε 2 β k 2 ‖ u ‖ 2 − C 1 p β k p ‖ u ‖ p ≥ a − ε β k 2 2 − C 1 β k p p ‖ u ‖ p − 2 ‖ u ‖ 2 .

Let

a − ε β k 2 2 − C 1 β k p p ‖ u ‖ p − 2 = a − ε β k 2 4 ,

then

(25) γ k = ‖ u ‖ = ( a − ε β k 2 ) p 4 C 1 β k p 1 p − 2 .

Therefore,

b k ≥ a − ε β k 2 4 ( a − ε β k 2 ) p 4 C 1 β k p 2 p − 2 .

According to p ∈ ( 4 , 2 ∗ ) and β k → 0 as k → ∞ , then b k → + ∞ as k → ∞ . Lemma 2.1 establishes that Φ ( u ) satisfies the ( PS ) c condition. Furthermore, from ( f 3 ), it is immediate that Φ ( u ) = Φ ( − u ) .

Hence, by Theorem 2.1, Φ ( u ) has an unbounded sequence of critical values. Therefore, (1) has infinitely many high-energy solutions.□

Proof of Theorem 1.2

The proofs that a k ≤ 0 and b k → + ∞ as k → ∞ follow the same method as in Theorem 1.1. Lemma 2.2 proves that Φ ( u ) satisfies the ( PS ) c condition; moreover, Φ ( u ) = Φ ( − u ) , and hence, Φ ( u ) has an unbounded sequence of critical values. Therefore, (1) has infinitely many high-energy solutions.□

Acknowledgments

Jing Zeng was supported by the National Science Foundation of China (Grant No. 11501110) and Fujian Natural Science Foundation (Grant No. 2018J01656). The authors sincerely thank the anonymous reviewers for their constructive feedback and valuable insights, which have significantly improved the rigor and clarity of this manuscript. We also appreciate the editorial team’s professional handling of the submission process.

  1. Funding information: Jing Zeng was supported by the National Science Foundation of China (Grant No. 11501110) and Fujian Natural Science Foundation (Grant No. 2018J01656).

  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. QZ: methodology (equal); writing – original draft (equal). JZ: methodology (equal); writing – original draft (equal); writing – review and editing.

  3. Conflict of interest: The authors have no conflicts to disclose.

  4. Data availability statement: Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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Received: 2024-10-16
Revised: 2025-04-29
Accepted: 2025-06-11
Published Online: 2025-08-28

© 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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https://doi.org/10.1515/math-2025-0179
Keywords for this article
Kirchhoff-type equation; multiple solutions; high energy solution
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