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Variational approach to Kirchhoff-type second-order impulsive differential systems

  • Wangjin Yao and Huiping Zhang EMAIL logo
Published/Copyright: August 13, 2024

Abstract

In this study, we consider a Kirchhoff-type second-order impulsive differential system with the Dirichlet boundary condition and obtain the existence and multiplicity of solutions to the impulsive problem via variational methods.

MSC 2010: 34B15; 34B37; 58E30

1 Introduction

In this study, we consider the impulsive nonlinear coupled differential system:

(1) M 1 ( u H 0 1 ( 0 , T ) 2 ) ( u ( t ) + λ u ( t ) ) = F u ( u ( t ) , v ( t ) ) , a.e. t [ 0 , T ] , M 2 ( v H 0 1 ( 0 , T ) 2 ) ( v ( t ) + λ v ( t ) ) = F v ( u ( t ) , v ( t ) ) , a.e. t [ 0 , T ] , Δ ( M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t k ) ) = I k ( u ( t k ) ) , k = 1 , 2 , , n , Δ ( M 2 ( v H 0 1 ( 0 , T ) 2 ) v ( t k ) ) = J k ( v ( t k ) ) , k = 1 , 2 , , n , u ( 0 ) = u ( T ) = v ( 0 ) = v ( T ) = 0 ,

where 0 = t 0 < t 1 < t 2 < < t n < t n + 1 = T , M 1 , M 2 : [ 0 , + ) R are the continuous functions such that there are two positive constants m 0 and m 1 with m 0 M 1 ( x ) , M 2 ( x ) m 1 for all x 0 , F ( u ( t ) , v ( t ) ) : R 2 R is a C 1 function, F u and F v denote the partial derivatives of F ( u ( t ) , v ( t ) ) with respect to u and v , respectively, norms u H 0 1 ( 0 , T ) and v H 0 1 ( 0 , T ) are specified later, I k , J k : R R , k = 1 , 2 , , n , are continuous, and

Δ ( M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t k ) ) = M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t k + ) M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t k ) , Δ ( M 2 ( v H 0 1 ( 0 , T ) 2 ) v ( t k ) ) = M 2 ( v H 0 1 ( 0 , T ) 2 ) v ( t k + ) M 2 ( v H 0 1 ( 0 , T ) 2 ) v ( t k ) ,

where

M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t k ± ) = lim t t k ± M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t ) , M 2 ( v H 0 1 ( 0 , T ) 2 ) v ( t k ± ) = lim t t k ± M 2 ( v H 0 1 ( 0 , T ) 2 ) v ( t ) .

The equation of Kirchhoff type is

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

where 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, which is proposed by Kirchhoff [1] as an extension of the classical D’Alembert’s wave equation for free vibrations of elastic strings. Some early classical studies dedicated to Kirchhoff equations can be seen in the studies of Bernstein [2] and Pohožaev [3]. However, after Lions [4] introduced an abstract framework to the following problem:

(2) u t t a + b Ω u 2 d x Δ u = f ( x , u ) ,

equation (2) began to arouse much attention. Some interesting results by variational methods can be found, for example, in [59].

On the other hand, impulsive differential equation is one of the main tools to study the dynamics of processes in which sudden, discontinuous jumps occur. In the last few years, variational methods are widely applied to study the existence and multiplicity of solutions for impulsive differential equations. The pioneering works of impulsive differential equations via variational methods were given in the studies of Tian and Ge [10] and Nieto and O’Regan [11]. Tian-Ge first studied the following impulsive differential equations by variational methods and obtained the existence of at least two positive solutions to the following problem:

( ρ ( t ) Φ p ( u ( t ) ) ) + s ( t ) Φ p ( u ( t ) ) = f ( t , u ( t ) ) , t t k , a.e. t [ 0 , T ] , Δ ( ρ ( t k ) Φ p ( u ( t k ) ) ) = I k ( u ( t k ) ) , k = 1 , 2 , , n , α u ( 0 ) β u ( 0 ) = A ¯ , γ u ( T ) + σ u ( T ) = B ¯ ,

where p > 1 , Φ p ( u ) = u p 2 u , 0 = t 0 < t 1 < t 2 < < t n < t n + 1 = T , Δ ( ρ ( t k ) Φ p ( u ( t k ) ) ) = ρ ( t k + ) Φ p ( u ( t k + ) ) ρ ( t k ) Φ p ( u ( t k ) ) , where u ( t k + ) and u ( t k ) denote the right and left limits, respectively, of u ( t ) at t = t k , I k C ( [ 0 , + ) , [ 0 , + ) ) , f C ( [ 0 , T ] × [ 0 , + ) , [ 0 , + ) ) , f ( t , 0 ) 0 , t [ 0 , T ] , ρ , s L [ 0 , T ] , ess inf [ 0 , T ] ρ > 0 , ess inf [ 0 , T ] s > 0 , 0 < ρ ( 0 ) , ρ ( T ) < , A ¯ 0 , B ¯ 0 , α , β , γ , σ > 0 , k = 1 , 2 , , n .

Then, Nieto-O’Regan considered two classes of impulsive problems below, one of which is the linear problem:

u ( t ) + λ u ( t ) = σ ( t ) , a.e. t [ 0 , T ] , Δ u ( t k ) = d k , k = 1 , 2 , , n , u ( 0 ) = u ( T ) = 0 ,

the other is the nonlinear problem:

u ( t ) + λ u ( t ) = f ( t , u ( t ) ) , a.e. t [ 0 , T ] , Δ u ( t k ) = I k ( u ( t k ) ) , k = 1 , 2 , , n , u ( 0 ) = u ( T ) = 0 ,

where 0 = t 0 < t 1 < t 2 < < t n < t n + 1 = T , I k : R R are continuous, f : [ 0 , T ] × R R is continuous, d k are the fixed constants, k = 1 , 2 , , n , σ L 2 ( 0 , T ) . They obtain the existence and multiplicity of solutions via variational methods.

Since then, there has been increasing research in this area. For some important and recent literature about such impulsive problems, we cite the works [1218] for the interested readers.

The Kirchhoff-type second-order impulsive differential problems have been considered only by a few authors up to now. More precisely, Heidarkhani et al. [19] considered the following Kirchhoff-type second-order impulsive differential problem:

(3) K 0 + ( u ( t ) 2 + y ( t ) u ( t ) 2 ) d t ( u ( t ) + y ( t ) u ( t ) ) = λ f ( t , u ( t ) ) , t [ 0 , + ) , t t k , Δ ( u ( t k ) ) = μ I k ( u ( t k ) ) , u ( 0 + ) = g ( u ( 0 ) ) , u ( + ) = 0 ,

where K : [ 0 , + ) R is a continuous function such that there are two positive constants m 0 and m 1 with m 0 K ( x ) m 1 for all x 0 , y L [ 0 , + ) , λ and μ are referred to as two control parameters, I k C ( R , R ) for 1 k p , 0 = t 0 < t 1 < t 2 < < t p < + , Δ ( u ( t k ) ) = u ( t k + ) u ( t k ) = lim t t k + u ( t ) lim t t k u ( t ) , f : [ 0 , + ) × R R is an L 2 -Carathéodory function, and g : R R is a Lipschitz continuous function. They obtained the existence of at least one weak solution and infinitely many weak solutions via variational methods when y ( t ) = q ( t ) and μ = λ , where q L [ 0 , + ) with 0 < q ̲ ess inf [ 0 , + ) q < q ¯ ess sup [ 0 , + ) q . When y ( t ) = q ( t ) , Caristi et al. [20] continued the study of problem (3) and obtained the existence of at least one weak solution, at least two weak solutions, and at least three weak solutions via variational approach. Based on previous works [19,20], Afrouzi et al. [21] also considered problem (3) when y ( t ) = q 2 , where q is a non-zero constant. The authors obtained the existence of at least three weak solutions using variational methods.

At present, to the best of our knowledge, there are few works on studying the existence and multiplicity of solutions for Kirchhoff-type second-order impulsive differential systems using variational methods. As a result, the goal of this study is to fill the gap in this area. Our main purpose is to study the existence and multiplicity of solutions of problem (1) under the nonlinearities F u and F v and impulsive functions I k , J k , k = 1 , 2 , , n satisfying different growth conditions.

Throughout this study, we always assume that λ > λ 1 and λ 1 is the first eigenvalue of the problem:

(4) u ( t ) = λ u ( t ) , t [ 0 , T ] , u ( 0 ) = u ( T ) = 0 .

In this study, we make the following assumptions:

  • (H1) F ( x , y ) is concave and there exist a , b > 0 and γ 1 , γ 2 [ 0 , 1 ) such that

    F x ( x , y ) a + b x γ 1 and F y ( x , y ) a + b y γ 2 , for ( x , y ) R 2 .

  • (H2) There exist a k , b k > 0 and β k [ 0 , 1 ) ( k = 1 , 2 , , n ) such that

    I k ( u ) a k + b k u β k and J k ( v ) a k + b k v β k , for u , v R .

  • (H3) There exist c 1 > 0 and μ ( 1 , + ) such that

    F s ( s , t ) , F t ( s , t ) c 1 ( 1 + s μ + t μ ) .

  • (H4) F s ( s , t ) , F t ( s , t ) = o ( s + t ) , as s + t 0 .

  • (H5) There exist c 2 > 0 , R 1 > 0 and 2 < θ < β such that

    F ( s , t ) c 2 ( s β + t β ) , s + t R 1 , and θ F ( s , t ) s F s ( s , t ) + t F t ( s , t ) ,

    where m 0 2 m 1 θ > 0 .

  • (H6)

    • (i) I k ( k = 1 , 2 , , n ) satisfy I k ( x ) x 0 , for all x R ;

    • (ii) There exist θ > 2 , δ k > 0 , k = 1 , 2 , , n , such that 0 x I k ( s ) d s δ k x θ , for x R \ { 0 } ;

    • (iii) I k ( x ) x θ 0 x I k ( s ) d s for x R \ { 0 } ;

      and

    • (i)′ J k ( k = 1 , 2 , , n ) satisfy J k ( x ) x 0 , for all x R ;

    • (ii)′ There exist θ > 2 , δ k > 0 , k = 1 , 2 , , n , such that 0 x J k ( s ) d s δ k x θ , for x R \ { 0 } ;

    • (iii)′ J k ( x ) x θ 0 x J k ( s ) d s , for x R \ { 0 } .

  • (H7) There exist θ > 2 , δ k > 0 , k = 1 , 2 , , n such that

    • (i) I k ( x ) x θ 0 x I k ( s ) d s < 0 , for x R \ { 0 } ;

    • (ii) 0 x I k ( s ) d s δ k x θ , for x R \ { 0 } ;

      and

    • (i)′ J k ( x ) x θ 0 x J k ( s ) d s < 0 , for x R \ { 0 } ;

    • (ii)′ 0 x J k ( s ) d s δ k x θ , for x R \ { 0 } .

The rest of this article is organized as follows. In Section 2, we present some preliminaries. In Section 3, we prove the main results via variational methods.

2 Variational structure

The following definition, lemma, and theorems will be needed in our argument.

Definition 1

([22], (PS) condition) Let X be a real reflexive Banach space. If any sequence { u k } X such that { Φ ( u k ) } is bounded and Φ ( u k ) 0 as k possesses a convergent subsequence, then we say that Φ satisfies the Palais-Smale condition.

Lemma 1

[22] Let X be a reflexive Banach space and Φ : X R be continuously Fréchet differentiable. If Φ is weakly lower semi-continuous and has a bounded minimizing sequence, then Φ has a minimum on X.

Theorem 1

([23], Mountain pass theorem) Let X be a real Banach space and Φ C 1 ( X , R ) satisfy the (PS) condition with Φ ( 0 ) = 0 . If Φ satisfies the following conditions:

  1. there exist constants ρ , α > 0 such that Φ B ρ α ,

  2. there exists an e X \ B ρ , such that Φ ( e ) 0 ,

then Φ possesses a critical value c α . Moreover, c is given by c = inf g Γ max s [ 0 , 1 ] Φ ( g ( s ) ) , where

Γ = { g C ( [ 0 , 1 ] , X ) g ( 0 ) = 0 , g ( 1 ) = e } .

Theorem 2

([23], Symmetric mountain pass theorem) Let X be an infinite dimensional real Banach space. Let Φ C 1 ( X , R ) be an even functional that satisfies the (PS) condition, and Φ ( 0 ) = 0 . Suppose that X = V E , where V is finite dimensional, and Φ satisfies that

  1. there exist α > 0 and ρ > 0 such that Φ B ρ E α ;

  2. for any finite dimensional subspace W X , there is R = R ( W ) such that Φ ( u ) 0 on W \ B R ( W ) .

Then, Φ possesses an unbounded sequence of critical values.

Let C ( [ 0 , T ] ) be the space of all continuous functions on [ 0 , T ] with the norm

u = max t [ 0 , T ] u ( t ) .

In the Sobolev space H 0 1 ( 0 , T ) , we consider the inner product

( u , v ) = 0 T u ( t ) v ( t ) d t ,

which induces the corresponding norm

u = 0 T u ( t ) 2 d t 1 2 .

It is a consequence of Poincaré’s inequality that

0 T u 2 ( t ) d t 1 2 1 λ 1 0 T ( u ( t ) ) 2 d t 1 2 ,

where λ 1 is the first eigenvalue of the Dirichlet problem (4). We can also define the inner product

( u , v ) H 0 1 ( 0 , T ) = 0 T ( u ( t ) v ( t ) + λ u ( t ) v ( t ) ) d t ,

which induces the corresponding norm

(5) u H 0 1 ( 0 , T ) = 0 T ( u ( t ) 2 + λ u ( t ) 2 ) d t 1 2 .

Then, the norm is equivalent to the norm H 0 1 ( 0 , T ) .

Set X = H 0 1 ( 0 , T ) × H 0 1 ( 0 , T ) , in the Sobolev space X , for any ( u , v ) X , we set the norm

( u , v ) 2 = u 2 + v 2 ,

and we also introduce the norm

( u , v ) X 2 = u H 0 1 ( 0 , T ) 2 + v H 0 1 ( 0 , T ) 2 = 0 T ( u ( t ) 2 + λ u ( t ) 2 ) d t + 0 T ( v ( t ) 2 + λ v ( t ) 2 ) d t .

Lemma 2

For the Sobolev space X, the norm and the norm X are equivalent, i.e., there exist C 1 , C 2 > 0 , such that

C 1 ( u , v ) 2 ( u , v ) X 2 C 2 ( u , v ) 2 , ( u , v ) X .

Lemma 3

For any ( u , v ) X , there exists C > 0 such that u , v C ( u , v ) X .

Remark 1

Lemmas 2 and 3 are deduced from Lemmas 2.1 and 2.2 of [24], so the proving processes are omitted here.

Lemma 4

We say that ( u , v ) X is a weak solution of problem (1) if ( u , v ) satisfies

(6) M 1 ( u H 0 1 ( 0 , T ) 2 ) 0 T ( u ( t ) φ ( t ) + λ u ( t ) φ ( t ) ) d t + M 2 ( v H 0 1 ( 0 , T ) 2 ) 0 T ( v ( t ) ϕ ( t ) + λ v ( t ) ϕ ( t ) ) d t + k = 1 n I k ( u ( t k ) ) φ ( t k ) + k = 1 n J k ( v ( t k ) ) ϕ ( t k ) 0 T F u ( u ( t ) , v ( t ) ) φ ( t ) d t 0 T F v ( u ( t ) , v ( t ) ) ϕ ( t ) d t = 0 ,

for any ( φ , ϕ ) X .

Proof

Taking φ ( t ) H 0 1 ( 0 , T ) , multiplying the first equation of problem (1), and integrating from 0 to T , we obtain

0 T M 1 ( u H 0 1 ( 0 , T ) 2 ) ( u ( t ) + λ u ( t ) ) φ ( t ) d t = 0 T F u ( u ( t ) , v ( t ) ) φ ( t ) d t .

In view of the impulse conditions and Dirichlet boundary condition, we have

0 T M 1 ( u H 0 1 ( 0 , T ) 2 ) ( u ( t ) + λ u ( t ) ) φ ( t ) d t = k = 0 n t k t k + 1 M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t ) φ ( t ) d t + 0 T λ M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t ) φ ( t ) d t = k = 1 n Δ ( M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t k ) ) φ ( t k ) + M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( 0 ) φ ( 0 ) M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( T ) φ ( T ) + 0 T M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t ) φ ( t ) d t + 0 T λ M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t ) φ ( t ) d t = k = 1 n I k ( u ( t k ) ) φ ( t k ) + 0 T M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t ) φ ( t ) d t + 0 T λ M 1 ( u H 0 1 ( 0 , T ) 2 ) u ( t ) φ ( t ) d t .

Hence,

(7) k = 1 n I k ( u ( t k ) ) φ ( t k ) + M 1 ( u H 0 1 ( 0 , T ) 2 ) 0 T ( u ( t ) φ ( t ) + λ u ( t ) φ ( t ) ) d t = 0 T F u ( u ( t ) , v ( t ) ) φ ( t ) d t .

Similarly,

(8) k = 1 n J k ( v ( t k ) ) ϕ ( t k ) + M 2 ( v H 0 1 ( 0 , T ) 2 ) 0 T ( v ( t ) ϕ ( t ) + λ v ( t ) ϕ ( t ) ) d t = 0 T F v ( u ( t ) , v ( t ) ) ϕ ( t ) d t .

Combining (7) with (8), we obtain that (6) holds.□

Next, we will use variational methods to find the critical points of the functional Φ : X R , defined by

(9) Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u ( t ) , v ( t ) ) d t ,

for each ( u , v ) X , where M 1 ˜ ( t ) = 0 t M 1 ( ξ ) d ξ , M 2 ˜ ( t ) = 0 t M 2 ( ξ ) d ξ . It is clear that Φ is Fréchet differentiable at any ( u , v ) X and Φ is continuous, with

(10) Φ ( u , v ) , ( φ , ϕ ) = M 1 ( u H 0 1 ( 0 , T ) 2 ) 0 T ( u ( t ) φ ( t ) + λ u ( t ) φ ( t ) ) d t + M 2 ( v H 0 1 ( 0 , T ) 2 ) 0 T ( v ( t ) ϕ ( t ) + λ v ( t ) ϕ ( t ) ) d t + k = 1 n I k ( u ( t k ) ) φ ( t k ) + k = 1 n J k ( v ( t k ) ) ϕ ( t k ) 0 T F u ( u ( t ) , v ( t ) ) φ ( t ) d t 0 T F v ( u ( t ) , v ( t ) ) ϕ ( t ) d t .

Hence, the weak solutions of problem (1) are the corresponding critical points of Φ .

3 Proof of main results

Lemma 5

Assume that (H1) and (H2) are satisfied. Then, the functional Φ defined by (9) is continuously Fréchet differentiable and weakly lower semi-continuous.

Proof

First, using the continuity of F u , F v , M 1 , M 2 , I k , and J k , k = 1 , 2 , , n , we can obtain Φ C 1 ( X , R ) . Next, we prove that Φ is weakly lower semi-continuous. Let { ( u m , v m ) } X with ( u m , v m ) ( u , v ) . Since the embedding H 0 1 ( 0 , T ) C ( [ 0 , T ] ) is compact ([25, Theorem 8.8]), we see that { u m } , { v m } converge uniformly to u , v on [ 0 , T ] , respectively, and liminf m u m H 0 1 ( 0 , T ) u H 0 1 ( 0 , T ) , liminf m v m H 0 1 ( 0 , T ) v H 0 1 ( 0 , T ) . Thus,

liminf m Φ ( u m , v m ) = liminf m 1 2 M 1 ˜ ( u m H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v m H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u m ( t k ) I k ( t ) d t + k = 1 n 0 v m ( t k ) J k ( t ) d t 0 T F ( u m ( t ) , v m ( t ) ) d t 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u ( t ) , v ( t ) ) d t = Φ ( u , v ) .

This implies that the functional Φ is weakly lower semi-continuous.□

Theorem 3

Suppose that (H1) and (H2) hold, then problem (1) has at least one nontrivial solution.

Proof

For any ( u , v ) X , by (H1), (H2), and Lemma 3, we have

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 0 u H 0 1 ( 0 , T ) 2 + 1 2 m 0 v H 0 1 ( 0 , T ) 2 + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 0 ( u , v ) X 2 k = 1 n 0 u ( t k ) ( a k + b k t β k ) d t k = 1 n 0 v ( t k ) ( a k + b k t β k ) d t 0 T ( a u + a v + b u γ 1 + 1 + b v γ 2 + 1 ) d t 1 2 m 0 ( u , v ) X 2 n A u B k = 1 n u β k + 1 n A v B k = 1 n v β k + 1 a T ( u + v ) b T ( u γ 1 + 1 + v γ 2 + 1 ) 1 2 m 0 ( u , v ) X 2 2 n A C ( u , v ) X 2 B k = 1 n C β k + 1 ( u , v ) X β k + 1 2 a T C ( u , v ) X b T C γ 1 + 1 ( u , v ) X γ 1 + 1 b T C γ 2 + 1 ( u , v ) X γ 2 + 1 ,

where A max { a 1 , a 2 , , a n } , B max { b 1 , b 2 , , b n } . Since γ 1 , γ 2 , β k [ 0 , 1 ) , k = 1 , 2 , , n , this implies that the functional lim ( u , v ) X Φ ( u , v ) = + , i.e., Φ is coercive on X . By Lemmas 1 and 5, we obtain that Φ has a minimum point on X . Hence, problem (1) has at least one nontrivial solution.□

Lemma 6

Assume that (H2), (H3), and (H5) hold, then Φ satisfies the (PS) condition.

Proof

If a sequence { ( u m , v m ) } X such that { Φ ( u m , v m ) } is bounded and Φ ( u m , v m ) 0 , then there is positive constant R 2 such that

R 2 + ( u m , v m ) X Φ ( u m , v m ) 1 θ Φ ( u m , v m ) , ( u m , v m ) = 1 2 M 1 ˜ ( u m H 0 1 ( 0 , T ) 2 ) 1 θ M 1 ( u m H 0 1 ( 0 , T ) 2 ) u m H 0 1 ( 0 , T ) 2 + 1 2 M 2 ˜ ( v m H 0 1 ( 0 , T ) 2 ) 1 θ M 2 ( v m H 0 1 ( 0 , T ) 2 ) v m H 0 1 ( 0 , T ) 2 + k = 1 n 0 u m ( t k ) I k ( t ) d t 1 θ k = 1 n I k ( u m ( t k ) ) u m ( t k ) + k = 1 n 0 v m ( t k ) J k ( t ) d t 1 θ k = 1 n J k ( v m ( t k ) ) v m ( t k ) + 1 θ 0 T F u ( u m , v m ) u m d t + 1 θ 0 T F v ( u m , v m ) v m d t 0 T F ( u m , v m ) d t m 0 2 m 1 θ ( u m , v m ) X 2 k = 1 n ( a k u m + b k u m β k + 1 ) 1 θ k = 1 n ( a k u m + b k u m β k + 1 ) k = 1 n ( a k v m + b k v m β k + 1 ) 1 θ k = 1 n ( a k v m + b k v m β k + 1 ) m 0 2 m 1 θ ( u m , v m ) X 2 2 ( θ + 1 ) θ n A C ( u m , v m ) X + B k = 1 n C β k + 1 ( u m , v m ) X β k + 1 ,

for large m . Since m 0 2 m 1 θ > 0 , we obtain { ( u m , v m ) } is bounded in X . Going if necessary to a subsequence, we can assume that there exist { ( u m , v m ) } X such that

( u m , v m ) ( u , v ) , in X , u m u , v m v , uniformly in C ( [ 0 , T ] ) ,

as m + . Hence, by (H3) and the boundedness of { u m } in X , we obtain

( M 1 ( u m H 0 1 ( 0 , T ) 2 ) M 1 ( u H 0 1 ( 0 , T ) 2 ) ) 0 T ( u ( u m u ) + λ u ( u m u ) ) d t 0 , k = 1 n I k ( u m ( t k ) ) k = 1 n I k ( u ( t k ) ) ( u m ( t k ) u ( t k ) ) 0 , 0 T ( F u ( u , v ) F u ( u m , v m ) ) ( u m u ) d t 0 ,

as m + . Moreover, a computation shows that

Φ ( u m , v m ) Φ ( u , v ) , ( u m u , 0 ) = M 1 ( u m H 0 1 ( 0 , T ) 2 ) 0 T ( u ( u m u ) + λ u m ( u m u ) ) d t + k = 1 n I k ( u m ( t k ) ) ( u m ( t k ) u ( t k ) ) 0 T F u ( u m , v m ) ( u m u ) d t M 1 ( u H 0 1 ( 0 , T ) 2 ) 0 T ( u ( u m u ) + λ u ( u m u ) ) d t k = 1 n I k ( u ( t k ) ) ( u m ( t k ) u ( t k ) ) + 0 T F u ( u , v ) ( u m u ) d t = M 1 ( u m H 0 1 ( 0 , T ) 2 ) u m u H 0 1 ( 0 , T ) 2 + ( M 1 ( u m H 0 1 ( 0 , T ) 2 ) M 1 ( u H 0 1 ( 0 , T ) 2 ) ) 0 T ( u ( u m u ) + λ u ( u m u ) ) d t + k = 1 n I k ( u m ( t k ) ) k = 1 n I k ( u ( t k ) ) ( u m ( t k ) u ( t k ) ) + 0 T ( F u ( u , v ) F u ( u m , v m ) ) ( u m u ) d t .

So u m u H 0 1 ( 0 , T ) 2 0 as m + . Using the same methods, we can prove v m v H 0 1 ( 0 , T ) 2 0 . Hence, ( u m , v m ) ( u , v ) in X . Therefore, Φ satisfies the (PS) condition.□

Remark 2

Assume M 1 , M 2 : [ 0 , + ) R are non-increasing. By the Lagrange mean value theorem, we obtain

M 1 ˜ ( t ) M 1 ( t ) t , M 2 ˜ ( t ) M 2 ( t ) t , for t 0 ,

which can be directly applied to demonstrate the boundness of (PS) sequence { ( u m , v m ) } in X , without the need of the condition m 0 2 m 1 θ > 0 in (H5). However, in the Kirchhoff case, to the original meaning of the M 1 (or M 2 ) that takes the form M 1 ( t ) = a + b t with a , b > 0 (or M 2 ( t ) = c + d t with c , d > 0 ), it should be an increasing function. Then,

M 1 ˜ ( t ) < M 1 ( t ) t , and M 2 ˜ ( t ) < M 2 ( t ) t , for t 0 ,

and therefore, we have to introduce the condition m 0 2 m 1 θ > 0 to prove the (PS) condition, especially in the case that M 1 and M 2 are non-decreasing continuous functions.

Theorem 4

Suppose that (H2)–(H5) and the following conditions hold:

  1. I k ( x ) , J k ( x ) , k = 1 , 2 , , n , are odd about x;

  2. F s ( s , t ) , F t ( s , t ) are odd about s, t, respectively.

If I k ( x ) , J k ( x ) are non-decreasing, then problem (1) has infinitely many weak solutions.

Proof

Since F u , F v , M 1 , M 2 , I k and J k , k = 1 , 2 , , n , are continuous, we obtain that Φ C 1 ( X , R ) . In view of (9), it is obvious that Φ is even and Φ ( 0 , 0 ) = 0 . By Lemma 6, Φ satisfies the (PS) condition. According to eigenvalue problem (4), we know that λ n = ( n π T ) 2 ( n = 1 , 2 , ) denote the sequence of eigenvalues. Let Y n denote the eigenspace corresponding to λ n , then we obtain the Sobolev space H 0 1 ( 0 , T ) = i N Y i ¯ and X = H 0 1 ( 0 , T ) × H 0 1 ( 0 , T ) = i N Y i × Y i ¯ . Assume that V = i = 1 2 Y i × Y i and E = i = 3 + Y i × Y i ¯ , then X = V E , where V is finite dimensional. By the Sobolev embedding theorem, for any ( u , v ) X , there exists c 3 > 0 such that

(11) ( u , v ) 2 2 c 3 ( u , v ) X 2 .

Since (H8) and I k ( x ) , J k ( x ) , k = 1 , 2 , , n , are non-decreasing, we have

(12) k = 1 n 0 u ( t k ) I k ( t ) d t 0 , k = 1 n 0 v ( t k ) J k ( t ) d t 0 .

By (H4), we have

F ( u , v ) = o ( u 2 + v 2 ) , as u + v 0 .

There exists δ > 0 such that u , v < δ , which implies

(13) F ( u , v ) m 0 4 c 3 ( u 2 + v 2 ) .

Assume that ( u , v ) X δ C , by (9), (12), (13), and Lemma 3, we have

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) 0 T F ( u , v ) d t 1 2 m 0 ( u , v ) X 2 m 0 4 c 3 0 T ( u 2 + v 2 ) d t 1 2 m 0 ( u , v ) X 2 1 4 m 0 ( u , v ) X 2 1 4 m 0 ( u , v ) X 2 .

Choose α = m 0 δ 2 4 C 2 , ρ = δ C , then Φ ( u , v ) α , for any ( u , v ) B ρ E . Thus, Φ satisfies condition (i) of Theorem 2.

Next, we verify condition (ii) of Theorem 2, i.e., for any finite dimensional subspace W X , there is R = R ( W ) such that Φ ( u , v ) 0 on W \ B R ( W ) . By (H4) and (H5), there exists c 4 > 0 such that

(14) F ( u , v ) c 2 ( u β + v β ) c 4 , ( u , v ) R 2 .

Since all norms are equivalent in the finite dimensional space, there exists c 5 > 0 such that

(15) ( u , v ) X β c 5 ( u 2 β + v 2 β ) , ( u , v ) W .

By (9), (H2), and Lemma 3, we have

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 1 ( u , v ) X 2 + n A u + B k = 1 n u β k + 1 + n A v + B k = 1 n v β k + 1 0 T ( c 2 ( u β + v β ) c 4 ) d t 1 2 m 1 ( u , v ) X 2 + 2 n A C ( u , v ) X + 2 B k = 1 n C β k + 1 ( u , v ) X β k + 1 c 2 ( u β β + v β β ) + c 4 T 1 2 m 1 ( u , v ) X 2 + 2 n A C ( u , v ) X + 2 B k = 1 n C β k + 1 ( u , v ) X β k + 1 c 2 c 5 ( u , v ) X β + c 4 T

for any ( u , v ) W . Since β > 2 , β k + 1 [ 1 , 2 ) , the aforementioned inequality implies that Φ ( u , v ) as ( u , v ) X . So there exists R = R ( W ) such that Φ ( u , v ) 0 on W \ B R ( W ) . According to Theorem 2, the functional Φ ( u , v ) possesses infinitely many critical points, i.e., problem (1) has infinitely many weak solutions.□

Theorem 5

Suppose that (H4)–(H6) hold, then problem (1) has at least one weak solution.

Proof

Obviously, Φ C 1 ( X , R ) and Φ ( 0 , 0 ) = 0 . By (H5) and (H6), we can obtain Φ that satisfies the (PS) condition. The proof of (PS) condition is similar to Lemma 6, and we omit it. We use Theorem 1 to prove Theorem 5. The proof of (i) of Theorem 1 is also similar to Theorem 4, and we omit it. By (9), (14), and (H6), we obtain

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 1 ( u , v ) X 2 + k = 1 n δ k u ( t k ) θ + k = 1 n δ k v ( t k ) θ 0 T ( c 2 ( u β + v β ) c 4 ) d t 1 2 m 1 ( u , v ) X 2 + k = 1 n δ k u θ + k = 1 n δ k v θ c 2 ( u β β + v β β ) + c 4 T 1 2 m 1 ( u , v ) X 2 + 2 C θ ( u , v ) X θ k = 1 n δ k c 2 u β β c 2 v β β + c 4 T .

Let u H 0 1 ( 0 , T ) = v H 0 1 ( 0 , T ) = 1 , and we have

Φ ( N u , N v ) 1 2 m 1 ( N u , N v ) X 2 + 2 C θ ( N u , N v ) X θ k = 1 n δ k c 2 N u β β c 2 N v β β + c 4 T m 1 N 2 + 2 2 + θ 2 C θ N θ k = 1 n δ k c 2 N β u β β c 2 N β v β β + c 4 T .

Since 2 < θ < β , u β β > 0 , v β β > 0 , Φ ( N u , N v ) as N + . Therefore, there exists N 0 > ρ such that Φ ( N 0 u , N 0 v ) 0 , which satisfy (ii) of Theorem 1. Φ has at least a critical point by Theorem 1. Consequently, we obtain at least one weak solution of problem (1).□

Theorem 6

Suppose that (H4)–(H6), (H8), and (H9) hold, then problem (1) has infinitely many weak solutions.

Proof

Obviously, Φ ( 0 , 0 ) = 0 , Φ C 1 ( X , R ) and satisfies the (PS) condition. By (H8) and (H9), we obtain that Φ is even. Due to the proof of condition (i) of Theorem 2 is similar to Theorem 4, we omit it here. By (9), (14), (15), and (H6), we obtain

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 1 ( u , v ) X 2 + k = 1 n δ k u ( t k ) θ + k = 1 n δ k v ( t k ) θ 0 T ( c 2 ( u β + v β ) c 4 ) d t 1 2 m 1 ( u , v ) X 2 + k = 1 n δ k u θ + k = 1 n δ k v θ c 2 ( u β β + v β β ) + c 4 T 1 2 m 1 ( u , v ) X 2 + 2 C θ k = 1 n δ k ( u , v ) X θ c 2 c 5 ( u , v ) X β + c 4 T , for all ( u , v ) W .

Since 2 < θ < β , the aforementioned inequality implies that Φ ( u , v ) as ( u , v ) X . So there exists R = R ( W ) such that Φ ( u , v ) 0 on W \ B R ( W ) . Hence, condition (ii) of Theorem 2 is proved. We obtain that problem (1) has infinitely many weak solutions by Theorem 2.□

Theorem 7

Suppose that (H4), (H5), and (H7) hold, then problem (1) has at least one weak solution.

Proof

Obviously, Φ ( 0 , 0 ) = 0 , Φ C 1 ( X , R ) . By (H5) and (H7), the (PS) condition is fulfilled, reasoning as in Lemma 6. By (9), (11), (13), and (H7), we have

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 0 ( u , v ) X 2 k = 1 n δ k u ( t k ) θ k = 1 n δ k v ( t k ) θ 1 4 m 0 ( u , v ) X 2 1 4 m 0 ( u , v ) X 2 2 k = 1 n δ k C θ ( u , v ) X θ .

Since θ > 2 , one can choose a ρ > 0 small enough such that Φ ( u , v ) > 0 . Hence, condition (i) of Theorem 1 is proved.

Next, by (14) and (H7), we have

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 1 ( u , v ) X 2 c 2 u β β c 2 v β β + c 4 T .

Let u H 0 1 ( 0 , T ) = v H 0 1 ( 0 , T ) = 1 , and we have

Φ ( N u , N v ) 1 2 m 1 ( N u , N v ) X 2 c 2 N u β β c 2 N v β β + c 4 T m 1 N 2 c 2 N β u β β c 2 N β v β β + c 4 T .

Since β > 2 , u β β > 0 , v β β > 0 , Φ ( N u , N v ) as N + . Therefore, there exists N 0 > ρ such that Φ ( N 0 u , N 0 v ) 0 . By Theorem 1, we obtain at least one weak solution of problem (1).□

Theorem 8

Suppose that (H4), (H5), and (H7)–(H9) hold, then problem (1) has infinitely many weak solutions.

Proof

We apply Theorem 2 to complete the proof. Clearly, Φ ( 0 , 0 ) = 0 , Φ C 1 ( X , R ) is even, and Φ satisfies the (PS) condition. In the same way as in Theorem 7, we can easily verify that condition (i) of Theorem 2 is satisfied. Next, we prove condition (ii) of Theorem 2. By (9), (14), (15), and (H7), we have

Φ ( u , v ) = 1 2 M 1 ˜ ( u H 0 1 ( 0 , T ) 2 ) + 1 2 M 2 ˜ ( v H 0 1 ( 0 , T ) 2 ) + k = 1 n 0 u ( t k ) I k ( t ) d t + k = 1 n 0 v ( t k ) J k ( t ) d t 0 T F ( u , v ) d t 1 2 m 1 ( u , v ) X 2 c 2 c 5 ( u , v ) X β + c 4 T , for all ( u , v ) W .

Since β > 2 , the aforementioned inequality implies that Φ ( u , v ) . So there exists R = R ( W ) such that Φ ( u , v ) 0 on W \ B R ( W ) . According to Theorem 2, Φ possesses infinitely many critical points, i.e., problem (1) has infinitely many weak solutions.□

Acknowledgment

The authors are very grateful to the referees for their very helpful comments and suggestions, which help to enhance the quality of this article.

  1. Funding information: The work was supported by the Natural Science Foundation of Fujian Province (Grant Nos. 2023J01994, 2023J01995, 2021J05237), the Program for Innovative Research Team in Science and Technology in Fujian Province University (Grant No. 2018-39), the Education and Research Project for Middle and Young Teachers in Fujian Province (Grant No. JAT231093), and the Mathematics Discipline Alliance Project in Fujian Province University (Grant No. 2024SXLMMS05).

  2. Author contributions: The 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. The authors declare that they have contributed equally to this article.

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

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2024-01-24
Revised: 2024-04-29
Accepted: 2024-05-26
Published Online: 2024-08-13

© 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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