Nontrivial solutions of discrete Kirchhoff-type problems via Morse theory

Yuhua Long

doi:10.1515/anona-2022-0251

Open Access Published by De Gruyter April 30, 2022

Nontrivial solutions of discrete Kirchhoff-type problems via Morse theory

Yuhua Long

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0251

Abstract

In this article, we study discrete Kirchhoff-type problems when the nonlinearity is resonant at both zero and infinity. We establish a series of results on the existence of nontrivial solutions by combining variational method with Morse theory. Several examples are provided to illustrate applications of our results.

Keywords: discrete Kirchhoff-type problem; nontrivial solution; variational method; Morse theory

MSC 2010: 39A10; 34B15; 35B38

1 Introduction

For a , b > 0 and positive integers m , n , this article is concerned with the existence of nontrivial solutions of the following discrete Kirchhoff-type problem:

(1.1) − a + b ∑ j = 1 n ∑ i = 1 m + 1 ∣ Δ 1 x ( i − 1 , j ) ∣ 2 + ∑ i = 1 m ∑ j = 1 n + 1 ∣ Δ 2 x ( i , j − 1 ) ∣ 2 ⋅ ( Δ 1 2 x ( i − 1 , j ) + Δ 2 2 x ( i , j − 1 ) ) = f ( ( i , j ) , x ( i , j ) ) , ∀ ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] ,

with the Dirichlet boundary conditions

(1.2) x ( i , 0 ) = x ( i , n + 1 ) = 0 i ∈ [ 0 , m + 1 ] , x ( 0 , j ) = x ( m + 1 , j ) = 0 j ∈ [ 0 , n + 1 ] ,

where [ s , t ] = { s , s + 1 , … , t } is a discrete segment with integers t ≥ s . Δ 1 x ( i , j ) = x ( i + 1 , j ) − x ( i , j ) , Δ 2 x ( i , j ) = x ( i , j + 1 ) − x ( i , j ) and Δ 2 x ( i , j ) = Δ ( Δ x ( i , j ) ) . f ( ( i , j ) , x ) is continuous in x for all ( i , j ) ∈ [ 0 , m + 1 ] × [ 0 , n + 1 ] .

(1.1) with (1.2) can be regarded as the discrete counterpart of the following Kirchhoff-type problem:

(1.3) − a + b ∫ Ω ∣ ∇ u ∣ 2 Δ u = f ( x , u ) in Ω , u = 0 , on Ω ,

which is the stationary case of a nonlinear wave equation

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

proposed by Kirchhoff [8] in 1883. (1.4) is an extension of the classical D’Alembert’s wave equation by considering the effects of the changes in the length of the string during the vibrations. Due to its strong physical background, (1.3) has captured much interest and been studied extensively by various methods. For example, [17] and [23] obtained nontrivial solutions via the Yang index and local linking theory, respectively. By minimax methods and invariant sets of descent flow, [16] gave sign-changing and multiple solutions. We refer the reader to [6,21] and references therein for more information.

On the other hand, with the rapid development of modern digital computing devices, more and more important information about the behavior of complex systems can be revealed by simulations by modern digital computing devices in a simple way, which contributes greatly to the increasing interest in discrete problems. As a result, numerous literature deal with difference equations, see [9,10,11, 12,24]. Partial difference equations with two or more discrete variables have been used in many fields such as image process, population models, digital control systems [4] and they are investigated in some literature [5,20,13]. However, as to (1.1), it contains Kirchhoff term b ∑ j = 1 n ∑ i = 1 m + 1 ∣ Δ 1 x ( i − 1 , j ) ∣ 2 + ∑ i = 1 m ∑ j = 1 n + 1 ∣ Δ 2 x ( i , j − 1 ) ∣ 2 ( Δ 1 2 x ( i − 1 , j ) + Δ 2 2 x ( i , j − 1 ) ) , which makes it much more difficult to deal with and there are rare works on it except for [14] deals with it by descending flow invariant set method and [15] studies multiple solutions of it.

Motivated by the aforementioned comments, we study (1.1) with (1.2) to obtain existence results of nontrivial solutions when the nonlinearity f ( ( i , j ) , x ) is resonant at both zero and infinity by combining variational method with Morse theory.

Before stating our main results, we give some notations. Let D be the discrete Laplacian defined as D x ( i , j ) = Δ 1 2 x ( i − 1 , j ) + Δ 2 2 x ( i , j − 1 ) . From [7], we know that − D is invertible and the distinct Dirichlet eigenvalues of − D on [ 1 , m ] × [ 1 , n ] can be denoted by 0 < λ 1 < λ 2 ≤ λ 3 ≤ ⋯ ≤ λ m n . Let ξ k ( 1 ≤ k ≤ m n ) denote the eigenvector corresponding to the eigenvalue λ k .

For any ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , consider

(1.5) − ∑ j = 1 n ∑ i = 1 m + 1 ∣ Δ 1 x ( i − 1 , j ) ∣ 2 + ∑ i = 1 m ∑ j = 1 n + 1 ∣ Δ 2 x ( i , j − 1 ) ∣ 2 ⋅ D = μ x 3 ( i , j ) .

Let μ 1 , μ 2 denote the minimum eigenvalue and the maximum eigenvalue of the nonlinear eigenvalue problem (1.5) with (1.2), respectively.

Denote F ( ( i , j ) , x ) = ∫ 0 x f ( ( i , j ) , τ ) d τ . For all ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , let

(1.6) lim ∣ x ∣ → 0 f ( ( i , j ) , x ) a x = f 0 lim ∣ x ∣ → ∞ f ( ( i , j ) , x ) b x 3 = f ∞ ,

and

(1.7) ι = ♯ { λ k : λ k < f 0 , k ∈ [ 1 , m n ] } .

For all ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , we also need the following assumptions:

There exists δ 1 > 0 such that
F ( ( i , j ) , x ) ≥ a 2 k λ k x 2 + b 4 k λ k 2 x 4 , ∣ x ∣ ≤ δ 1 .
lim ∣ x ∣ → ∞ ( x f ( ( i , j ) , x ) − 4 F ( ( i , j ) , x ) = + ∞ .
There exist δ 2 > 0 and c ∈ a λ m n 2 , + ∞ such that
F ( ( i , j ) , x ) ≥ c x 2 + b f ∞ 4 x 4 , ∣ x ∣ ≥ δ 2 .

Our main results are the following two theorems.

Theorem 1.1

Let f 0 = λ k and ( F 0 ) hold, then (1.1) with (1.2) admits at least one nontrivial solution provided one of the following assumptions holds:

f ∞ < μ 1 ;
f ∞ > μ 2 and k ≠ m n ;
( F ∞ 1 ) and f ∞ = μ 1 ;
( F ∞ 2 ) and f ∞ = μ 2 and k ≠ m n .

Write Ξ , the set of the distinct Dirichlet eigenvalues of the discrete Laplacian − D on [ 1 , m ] × [ 1 , n ] .

Theorem 1.2

If f 0 ∉ ∑ , then (1.1) with (1.2) has at least one nontrivial solution in each of the following cases:

f ∞ < μ 1 and ι ≠ 0 ;
f ∞ > μ 2 and ι ≠ m n ;
( F ∞ 1 ), f ∞ = μ 1 and ι ≠ 0 ;
( F ∞ 2 ), f ∞ = μ 2 and ι ≠ m n .

Remark 1.1

According to [17] or [19], (1.1) with (1.2) is called resonant at zero if f 0 = λ k and resonant at infinity if f ∞ = μ , where μ is an eigenvalue of the nonlinearity eigenvalue problem (1.5) with (1.2). Therefore, cases (i) and (ii) in Theorem 1.1 consider (1.1) with (1.2) is resonant at zero, cases (iii) and (iv) in Theorem 1.1 concern about (1.1) with (1.2) is resonant at both zero and infinity. In Theorem 1.2, cases (i) and (ii) mean (1.1) with (1.2) is resonant at infinity, cases (iii) and (iv) mean (1.1) with (1.2) is non-resonant.

We organize the rest of this article as follows: we establish the variational functional of (1.1) with (1.2) and give preliminaries in Section 2. The detailed proofs of Theorems 1.1 and 1.2 are given in Section 3. Finally, we provide some examples to illustrate applications of our results as a conclusion in Section 4.

2 Preliminaries

Let E be an mn-dimensional Hilbert space with usual inner product ( ⋅ , ⋅ ) and norm ∣ ⋅ ∣ . Denote

X = { x : [ 0 , m + 1 ] × [ 0 , n + 1 ] → R such that x ( i , 0 ) = x ( i , n + 1 ) = 0 , i ∈ [ 0 , m + 1 ] and x ( 0 , j ) = x ( m + 1 , j ) = 0 , j ∈ [ 0 , n + 1 ] } .

For any x , y ∈ X , define the inner product ⟨ ⋅ , ⋅ ⟩ as

(2.1) ⟨ x , y ⟩ = ∑ j = 1 n ∑ i = 1 m + 1 ( Δ 1 x ( i − 1 , j ) ⋅ Δ 1 y ( i − 1 , j ) ) + ∑ i = 1 m ∑ j = 1 n + 1 ( Δ 2 x ( i , j − 1 ) ⋅ Δ 2 y ( i , j − 1 ) ) ,

then the induced norm ‖ ⋅ ‖ is

‖ x ‖ = ⟨ x , x ⟩ = ∑ j = 1 n ∑ i = 1 m + 1 ∣ Δ 1 x ( i − 1 , j ) ∣ 2 + ∑ i = 1 m ∑ j = 1 n + 1 ∣ Δ 2 x ( i , j − 1 ) ∣ 2 1 / 2 , ∀ x ∈ X .

Note that X is isomorphic to E , there and thereafter, we always deem x ∈ X as an extension of x ∈ E when it is needed.

Define

x i j = x ( i , j ) , 0 ≤ i ≤ m + 1 , 0 ≤ j ≤ n + 1 , x = ( x 11 , … , x m 1 , x 12 , … , x m 2 , … , x 1 n , … , x m n ) T , f ( x ) = ( f ( ( 1 , 1 ) , x 11 ) , … , f ( ( m , 1 ) , x m 1 ) , f ( ( 1 , 2 ) , x 12 ) , … , f ( ( m , 2 ) , x m 2 ) , … f ( ( 1 , n ) , x 1 n ) , … , f ( ( m , n ) , x m n ) ) T .

Given an m n × m n matrix A as

A = L − I m 0 0 ⋯ 0 0 0 0 − I m L − I m 0 ⋯ 0 0 0 0 0 − I m L − I m ⋯ 0 0 0 0 0 0 − I m L ⋯ 0 0 0 0 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ 0 0 0 0 ⋯ L − I m 0 0 0 0 0 0 ⋯ − I m L − I m 0 0 0 0 0 ⋯ 0 − I m L − I m 0 0 0 0 ⋯ 0 0 − I m L ,

where I m is an m × m identity matrix and

L = 4 − 1 0 ⋯ 0 0 0 − 1 4 − 1 ⋯ 0 0 0 0 − 1 4 ⋯ 0 0 0 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ 0 0 0 ⋯ 4 − 1 0 0 0 0 ⋯ − 1 4 − 1 0 0 0 ⋯ 0 − 1 4 m m ,

then (1.1) with the boundary conditions (1.2) can be rewritten in the form of

(2.2) ( a + b ‖ x ‖ 2 ) A x = f ( x ) .

Obviously, the eigenvalues of the matrix A are same as the Dirichlet eigenvalues of − D on [ 1 , m ] × [ 1 , n ] , namely, which can also be denoted by 0 < λ 1 < λ 2 ≤ λ 3 ≤ ⋯ ≤ λ m n [13]. At the same time, we still use ξ k = ( ξ k ( 1 ) , ξ k ( 2 ) , … , ξ k ( m n ) ) T , 1 ≤ k ≤ m n , to represent an eigenvector corresponding to the eigenvalue λ k . Therefore, the inner product ⟨ ⋅ , ⋅ ⟩ , defined by (2.1), can be expressed in an equivalent form

⟨ x , y ⟩ = x T A y .

For p ≥ 1 , we define another norm as ‖ x ‖ p = ∑ i = 1 m ∑ j = 1 n ∣ x ( i , j ) ∣ p 1 p . Then for all x ∈ E , we have

(2.3) λ 1 ‖ x ‖ 2 2 ≤ ‖ x ‖ 2 ≤ λ m n ‖ x ‖ 2 2

and

(2.4) ‖ x ‖ 4 2 ≤ ‖ x ‖ 2 2 ≤ m n ‖ x ‖ 4 2 .

Define a functional J ( x ) : E → R in the following form:

(2.5) J ( x ) = a 2 ∑ j = 1 n ∑ i = 1 m + 1 ∣ Δ 1 x ( i − 1 , j ) ∣ 2 + ∑ i = 1 m ∑ j = 1 n + 1 ∣ Δ 2 x ( i , j − 1 ) ∣ 2 + b 4 ∑ j = 1 n ∑ i = 1 m + 1 ∣ Δ 1 x ( i − 1 , j ) ∣ 2 + ∑ i = 1 m ∑ j = 1 n + 1 ∣ Δ 2 x ( i , j − 1 ) ∣ 2 2 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x ( i , j ) ) = a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x ( i , j ) ) .

Note that f ( ( i , j ) , x ) ∈ C 1 ( E , R ) , it is clear J ∈ C 2 ( E , R ) and solutions of problem (1.1) with (1.2) are precisely the critical points of J ( x ) . Moreover, for any x , z ∈ E , we have

(2.6) ( J ′ ( x ) , z ) = ( a + b ‖ x ‖ 2 ) ∑ j = 1 n ∑ i = 1 m + 1 ( Δ 1 x ( i − 1 , j ) ⋅ Δ 1 z ( i − 1 , j ) ) + ∑ i = 1 m ∑ j = 1 n + 1 ( Δ 2 x ( i , j − 1 ) ⋅ Δ 2 z ( i − 1 , j ) ) − ∑ i = 1 m ∑ j = 1 n ( f ( ( i , j ) , x ( i , j ) ) ⋅ z ( i , j ) ) = ( ( a + b ‖ x ‖ 2 ) x − A − 1 f ( x ) , z ) ,

and

(2.7) J ″ ( x ) = ( a + b ‖ x ‖ 2 ) I m n + 2 b x ( A x ) T − A − 1 f ′ ( x ) , x ∈ E .

Consider the eigenvalues of the nonlinear eigenvalue problem (1.5) with (1.2), in the same manner as [22], there has the following:

Proposition 2.1

(1.5) with (1.2) possesses m n eigenvalues which belong to [ λ 1 2 , m n λ m n 2 ] . Moreover,

(2.8) μ 1 ‖ x ‖ 4 4 ≤ ‖ x ‖ 4 ≤ μ 2 ‖ x ‖ 4 4 , ∀ x ∈ E .

Note that Morse theory is a very powerful tool to study the existence of multiple solutions of differential equations having variational structure. In Morse theory, the most important component is that the functional must possess deformation property. If J satisfies the Palais-Smale (PS) condition, then J satisfies the deformation condition [1,2].

Recall E is a Hilbert space and J ∈ C 2 ( E , R ) . In order to use Morse theory to prove our results, we state some basic facts.

Definition 2.1

J satisfies the PS condition if every sequence { x κ } ⊂ E such that { J ( x κ ) } is bounded and J ′ ( x κ ) → 0 as κ → ∞ contains a convergent subsequence.

Definition 2.2

[3] If x 0 is an isolated critical point of J with J ( x 0 ) = c and U is any neighborhood of x 0 , then the qth critical group of J at x 0 is defined by

C q ( J , x 0 ) ≔ H q ( J c ∩ U , ( J c ∩ U ) \ { x } ) , q ∈ N = { 0 , 1 , 2 , … } ,

where J c = { x ∈ E : J ( x ) ≤ c } and H q ( A , B ) is the qth singular relative homology group of the topological pair ( A , B ) with coefficients in a field F .

Definition 2.3

[3] Let J ∈ C 2 ( E , R ) and J ′ ( x 0 ) = 0 . If J ″ ( x 0 ) has a bounded inverse, then x 0 is a non-degenerate critical point of J . Let m 0 and n 0 be the Morse index and nullity of J at x 0 , respectively. Then

m 0 = dim E − , n 0 = dim ker J ″ ( x 0 ) ,

where E − is the supremum of the vector subspaces of E such that J ″ ( x 0 ) is negative definite.

Lemma 2.1

([18] Proposition 2.2) Let J ∈ C 2 ( E , R ) and 0 be an isolated critical point. Assume J has a local linking at 0 with respect to E = E + ⊕ E − and l = dim E − , i.e., there exists a small ρ > 0 such that J ( x ) ≤ J ( 0 ) , for x ∈ E − , ‖ x ‖ ≤ ρ ; J ( x ) > J ( 0 ) , for x ∈ E + , 0 < ‖ x ‖ ≤ ρ . Then

C q ( J , 0 ) ≅ δ q , l F , for l = m 0 or l = m 0 + n 0 .

Lemma 2.2

[3] Let J ∈ C 2 ( E , R ) and x 0 be an isolated critical point. Then the following conclusions hold:

If x 0 is a minimum point, then C q ( J , x 0 ) ≅ δ q , 0 F ;
If x 0 is a minimum point and dim E = l < ∞ , then C q ( J , x 0 ) ≅ δ q , l F ;
If x 0 is a non-degenerate critical point with the Morse index m 0 , then C q ( J , x 0 ) ≅ δ q , m 0 F .

In applications we always need to verify that J satisfies the following compactness condition. Denoted by Ξ is the set of all eigenvalues of (1.5) with (1.2).

Lemma 2.3

Suppose f ∞ ∉ Ξ , then the functional J ( x ) satisfies the PS condition.

Proof

For some constant M , suppose { x κ } κ ∈ N ⊂ E be such that

(2.9) J ( x κ ) → M , J ′ ( x κ ) → 0 , κ → ∞ .

In the following, we are to show that { x κ } possesses a convergent subsequence. Since E is finite dimensional, it suffices to prove that { x κ } is bounded. Arguing indirectly, we assume ‖ x κ ‖ → + ∞ as κ → + ∞ and set y κ = x κ ‖ x κ ‖ , then ‖ y κ ‖ = 1 and there is a subsequence, still denoted by { y κ } , such that y κ ( i , j ) → y ( i , j ) as κ → ∞ for all ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] . From (2.6) and (2.9), for all z ∈ E , we have

(2.10) ( J ′ ( x κ ) , z ) ‖ x κ ‖ 3 = a ⟨ y κ , z ⟩ ‖ x κ ‖ 2 + b ‖ y κ ‖ 2 ⟨ y κ , z ⟩ − ∑ i = 1 m ∑ j = 1 n f ( ( i , j ) , x κ ( i , j ) ) x κ 3 ( i , j ) y κ 3 ( i , j ) , z ( i , j ) → 0 , as κ → ∞ .

Choose z = y κ − y in (2.10), then ‖ y ‖ = 1 . Together with lim ∣ x ∣ → ∞ f ( ( i , j ) , x ) b x 3 = f ∞ , it yields that

(2.11) ‖ y ‖ 2 ⟨ y , z ⟩ = ∑ i = 1 m ∑ j = 1 n f ∞ ( y 3 ( i , j ) , z ( i , j ) ) , ∀ z ∈ E .

Denote y + = max { y , 0 } . By (2.1) and (2.11), we obtain

(2.12) ‖ y ‖ 2 ∑ j = 1 n ∑ i = 1 m + 1 ( Δ 1 y ( i − 1 , j ) ⋅ Δ 1 z ( i − 1 , j ) ) + ∑ i = 1 m ∑ j = 1 n + 1 ( Δ 2 y ( i , j − 1 ) ⋅ Δ 2 z ( i , j − 1 ) ) = ∑ i = 1 m ∑ j = 1 n f ∞ ( y + 3 ( i , j ) , z ( i , j ) ) , ∀ z ∈ E .

Therefore, combining (2.12) with the boundary condition (1.2), we have

− ‖ y ‖ 2 ( Δ 1 2 y ( i − 1 , j ) + Δ 2 2 y ( i , j − 1 ) ) = f ∞ y 3 + ( i , j ) , ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] y ( i , 0 ) = y ( i , n + 1 ) = 0 y ( 0 , j ) = y ( m + 1 , j ) = 0 , i ∈ [ 0 , m + 1 ] j ∈ [ 0 , n + 1 ]

By the maximum principle, y = y + ≥ 0 . Since f ∞ ∉ Ξ , then y ≡ 0 , which contradicts ‖ y ‖ = 1 . Subsequently, { x κ } is bounded and J satisfies the PS condition.□

Lemma 2.4

Let f ∞ = μ 1 and ( F ∞ 1 ) hold, then the functional J ( x ) satisfies the PS condition.

Proof

Assume f ∞ = μ 1 , that is, lim ∣ x ∣ → ∞ F ( ( i , j ) , x ) x 4 = b μ 1 4 . Write

(2.13) f ( ( i , j ) , x ) = b μ 1 x 3 + g ( ( i , j ) , x ) and F ( ( i , j ) , x ) = b μ 1 4 x 4 + G ( ( i , j ) , x ) ,

where G ( ( i , j ) , x ) is the primitive function of g ( ( i , j ) , x ) . If ( F ∞ 1 ) holds, then

lim ∣ x ∣ → ∞ G ( ( i , j ) , x ) x 4 = lim ∣ x ∣ → ∞ F ( ( i , j ) , x ) − b μ 1 4 x 4 x 4 = 0

and

(2.14) lim ∣ x ∣ → ∞ ( g ( ( i , j ) , x ) x − 4 G ( ( i , j ) , x ) ) = lim ∣ x ∣ → ∞ ( x f ( ( i , j ) , x ) − 4 F ( ( i , j ) , x ) ) = + ∞ .

From (2.14), we have that for every M 1 > 0 , there exists R M 1 > 0 such that

g ( ( i , j ) , x ) x − 4 G ( ( i , j ) , x ) ≥ M 1 , ∀ ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , ∣ x ∣ ≥ R M 1 .

Integrating the equality d d x G ( ( i , j ) , x ) x 4 = g ( ( i , j ) , x ) x − 4 G ( i , j ) x 5 over the interval [ x , X ] ⊂ [ R M 1 , + ∞ ) , it yields that

(2.15) G ( ( i , j ) , X ) X 4 − G ( ( i , j ) , x ) x 4 ≥ − M 1 4 1 X 4 − 1 x 4 .

Let ∣ X ∣ → + ∞ , (2.15) implies that

(2.16) G ( ( i , j ) , x ) ≤ − M 1 4 , ∀ ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , ∣ x ∣ ≥ R M 1 .

For the arbitrariness of M 1 , (2.16) ensures that

(2.17) lim ∣ x ∣ → ∞ G ( ( i , j ) , x ) = − ∞ , ∀ ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] .

Let { x κ } ⊂ E be such that ‖ x κ ‖ → + ∞ as κ → + ∞ , and J ( x κ ) ≤ M ∗ for some constant M ∗ . Owing to (2.8), (2.13) and (2.17), there has been

M ∗ ≥ J ( x κ ) = a 2 ‖ x κ ‖ 2 + b 4 ‖ x κ ‖ 4 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x κ ( i , j ) ) = a 2 ‖ x κ ‖ 2 + b 4 ‖ x κ ‖ 4 − ∑ i = 1 m ∑ j = 1 n b μ 1 4 x κ 4 ( i , j ) + G ( ( i , j ) , x κ ( i , j ) ) = a 2 ‖ x κ ‖ 2 + b 4 ‖ x κ ‖ 4 − b μ 1 4 ‖ x κ ‖ 4 4 − ∑ i = 1 m ∑ j = 1 n G ( ( i , j ) , x κ ( i , j ) ) = a 2 ‖ x κ ‖ 2 − ∑ i = 1 m ∑ j = 1 n G ( ( i , j ) , x κ ( i , j ) ) → + ∞ , as κ → + ∞ ,

which is impossible. Therefore, { x κ } is bounded and J satisfies the PS condition.□

Lemma 2.5

Let f ∞ = μ 2 and ( F ∞ 2 ) hold, then the functional J ( x ) satisfies the PS condition.

Proof

For some constant M ∗ ∗ , suppose { x κ } κ ∈ N ⊂ E be such that

(2.18) J ( x κ ) → M ∗ ∗ , J ′ ( x κ ) → 0 , as κ → ∞ .

Due to the finite dimension of the Hilbert space E , we only need to show that { x κ } is bounded. Suppose, by the way of contradiction, that

(2.19) ‖ x κ ‖ → + ∞ , as κ → ∞ .

Since f ∞ = μ 2 , ( F ∞ 2 ) and the continuity of F , it follows that there exists M 2 > 0 such that

(2.20) F ( ( i , j ) , x ) ≥ c x 2 + b μ 2 4 x 4 − M 2 , ∀ x ∈ R , ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] .

Hence, together (2.20) with (2.3), (2.8), one concludes that

(2.21) J ( x κ ) = a 2 ‖ x κ ‖ 2 + b 4 ‖ x κ ‖ 4 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x κ ( i , j ) ) ≤ a 2 ‖ x κ ‖ 2 + b 4 ‖ x κ ‖ 4 − ∑ i = 1 m ∑ j = 1 n b μ 2 4 x κ 4 ( i , j ) + c x κ ( i , j ) − M 2 = a 2 ‖ x κ ‖ 2 + b 4 ‖ x κ ‖ 4 − b μ 2 4 ‖ x κ ‖ 4 4 − c ‖ x κ ‖ 2 2 + m n M 2 ≤ a 2 − c λ m n ‖ x κ ‖ 2 + m n M 2 .

Note that c ∈ a λ m n 2 , + ∞ , i.e., a 2 − c λ m n < 0 . Then (2.19) and (2.21) lead to

J ( x κ ) → − ∞ , as κ → ∞ ,

which contradicts (2.18). And this completes the proof.□

3 Proofs of main results

With the help of Lemmas 2.3, 2.4 and 2.5, we have verified that J ( x ) satisfies the PS condition under the given conditions in Theorems 1.1 and 1.2. Subsequently, we will present detailed proofs of Theorems 1.1 and 1.2 by Lemmas 2.1 and 2.2.

Proof of Theorem 1.1

We begin the proof with verifying that J has a local linking at 0 with conditions f 0 = λ k and ( F 0 ).

From (1.6), lim ∣ x ∣ → 0 f ( ( i , j ) , x ) a x = f 0 = λ k = lim ∣ x ∣ → 0 2 F ( ( i , j ) , x ) a x 2 , there exists ρ > 0 such that

(3.1) ρ λ 1 = δ 1 ≤ δ

and

(3.2) a 2 ( λ k − ε ) x 2 ≤ F ( ( i , j ) , x ) ≤ a 2 ( λ k + ε ) x 2 , ∣ x ∣ ≤ δ ,

together with ( F 0 ), it follows that

(3.3) a 2 λ k x 2 + b 4 k λ k 2 x 4 ≤ F ( ( i , j ) , x ) ≤ a 2 ( λ k + ε ) x 2 , ∣ x ∣ ≤ δ .

Let E = E 0 + ⊕ E 0 − with E 0 − = span { ξ 1 , ξ 2 , … , ξ k } and E 0 + = ( E 0 − ) ⊥ , then dim E 0 − = k < + ∞ . By (2.3) and (2.4), we have

(3.4) λ 1 ‖ x ‖ 2 2 ≤ ‖ x ‖ 2 ≤ λ k ‖ x ‖ 2 2 ≤ k λ k ‖ x ‖ 4 2 , x ∈ E 0 − ,

and

(3.5) λ k + 1 ‖ x ‖ 4 2 ≤ λ k + 1 ‖ x ‖ 2 2 ≤ ‖ x ‖ 2 ≤ λ m n ‖ x ‖ 2 2 , x ∈ E 0 + .

From F ( ( i , j ) , x ) = ∫ 0 x f ( ( i , j ) , τ ) d τ , for all ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] and x ∈ R , it yields that F ( ( i , j ) , 0 ) = 0 , that is, J ( 0 ) = 0 . For ρ and δ given in (3.1), if x ∈ E 0 − with ‖ x ‖ ≤ ρ , then ∣ x ( i , j ) ∣ ≤ δ . Making use of (3.3), we obtain

J ( x ) = a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x ( i , j ) ) ≤ a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n a 2 λ k x 2 ( i , j ) + b 4 k λ k 2 x 4 ( i , j ) ≤ a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − a 2 ‖ x ‖ 2 − b 4 k λ k 2 ‖ x ‖ 4 4 ≤ b 4 k λ k 2 ‖ x ‖ 4 4 − b 4 k λ k 2 ‖ x ‖ 4 4 = 0 .

Therefore,

(3.6) J ( x ) ≤ J ( 0 ) = 0 , for x ∈ E 0 − , ‖ x ‖ ≤ ρ .

Similarly, for x ∈ E 0 + with 0 < ‖ x ‖ ≤ ρ , there holds

J ( x ) = a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x ( i , j ) ) ≥ a 2 ‖ x ‖ 2 − ∑ i = 1 m ∑ j = 1 n a 2 ( λ k + ε ) x 2 ( i , j ) ≥ a 2 ‖ x ‖ 2 − a 2 ( λ k + ε ) ‖ x ‖ 2 2 ≥ a 2 1 − λ k + ε λ k + 1 ‖ x ‖ 2 .

Note that λ k < λ k + 1 , which leads to that there exists a sufficiently small ε such that

(3.7) J ( x ) ≥ J ( 0 ) = 0 , for x ∈ E 0 + , 0 < ‖ x ‖ ≤ ρ .

Thus, (3.6) and (3.7) indicate that J has a local linking at 0. By Lemma 2.1, we achieve

(3.8) C q ( J , 0 ) ≅ δ q , k F .

(i) f ∞ < μ 1 . Since lim ∣ x ∣ → 0 f ( ( i , j ) , x ) b x 3 = f ∞ and F is continuous, for any ε > 0 , there exists M ε > 0 such that

(3.9) F ( ( i , j ) , x ) ≤ b 4 ( μ 1 − ε ) x 4 + M ε , ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , x ∈ R .

Combining (3.9) with (2.8), it yields that

J ( x ) = a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x ( i , j ) ) ≥ a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n b 4 ( μ 1 − ε ) x 4 ( i , j ) + M ε = a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − b 4 ( μ 1 − ε ) ‖ x ‖ 4 4 − m n M ε ≥ a 2 ‖ x ‖ 2 − m n M ε ,

which implies that

(3.10) J ( x ) → + ∞ , as ‖ x ‖ → + ∞ .

Hence, (3.10) and J ( x ) satisfies the PS condition, which ensures that J attains a minimum point x 1 ∈ E . Because of Lemma 2.2 (i), there is

(3.11) C q ( J , x 1 ) ≅ δ q , 0 F .

Therefore, according to (3.8), (3.11) and k ≠ 0 , we draw a conclusion that x 1 ≠ 0 . And J admits at least one nontrivial critical point.

(ii) f ∞ > μ 2 and k ≠ m n . In the same manner as (3.9), we obtain

(3.12) F ( ( i , j ) , x ) ≥ b 4 ( μ 2 + ε ) x 4 − M ε , ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , x ∈ R ,

in combination with (2.3), (2.4) and (2.8), we also obtain

J ( x ) ≤ a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n b 4 ( μ 2 + ε ) x 4 ( i , j ) − M ε ≤ a 2 ‖ x ‖ 2 − b ε 4 m n λ m n 2 ‖ x ‖ 4 + m n M ε .

Hence,

(3.13) J ( x ) → − ∞ , as ‖ x ‖ → + ∞ ,

which means that J has a maximum when the PS condition is satisfied. According to Lemma 2.2 (ii), we have

(3.14) C q ( J , x 2 ) ≅ δ q , m n F .

Therefore, (3.11), (3.14) and the condition k ≠ m n guarantee x 2 ≠ 0 . Consequently, J possesses at least one nontrivial solution.

(iii) f ∞ = μ 1 and ( F ∞ 1 ). Write G ( ( i , j ) , x ) = F ( ( i , j ) , x ) − b μ 1 4 x 4 , recall (2.17), there has been the following:

lim ∣ x ∣ → + ∞ ∑ i = 1 m ∑ j = 1 n G ( ( i , j ) , x ( i , j ) ) = ∑ i = 1 m ∑ j = 1 n lim ∣ x ∣ → + ∞ G ( ( i , j ) , x ( i , j ) ) = − ∞ .

It follows that

J ( x ) = a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n F ( ( i , j ) , x ( i , j ) ) = a 2 ‖ x ‖ 2 + b 4 ‖ x ‖ 4 − ∑ i = 1 m ∑ j = 1 n G ( ( i , j ) , x ( i , j ) ) + b μ 1 4 x 4 ( i , j ) ≥ a 2 ‖ x ‖ 2 − ∑ i = 1 m ∑ j = 1 n G ( ( i , j ) , x ( i , j ) ) → + ∞ , as ‖ x ‖ → + ∞ .

Since J satisfies the PS condition, J attains a minimum at x 3 . Thus,

(3.15) C q ( J , x 3 ) ≅ δ q , 0 F .

Together with (3.8) and the fact k ≠ 0 , it yields that x 3 ≠ 0 , which ensures that J has at least one nontrivial solution.

(iv) f ∞ = μ 2 and ( F ∞ 2 ) . Similar to (2.21), there has been the following:

J ( x ) ≤ a 2 − c λ m n ‖ x ‖ 2 + m n M ˜ → − ∞ , as ‖ x ‖ → + ∞

for c > a 2 λ m n . Consider the PS condition is met, then J achieves a maximum point at x 4 . Therefore,

(3.16) C q ( J , x 4 ) ≅ δ q , m n F .

Owing to (3.8), (3.16) and k ≠ m n , we obtain x 4 ≠ 0 . So J has at least one nontrivial solution and this completes the proof.□

Proof of Theorem 1.2

Remind f 0 ∉ ∑ , that is, lim ∣ x ∣ → 0 f ( ( i , j ) , x ) a x = f 0 ≠ λ s , s ∈ [ 1 , m n ] . Using (2.7), we obtain

(3.17) J ″ ( 0 ) = a I m n − A − 1 ∂ f ( ( i , j ) , 0 ) ∂ x , ∀ ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] .

Then x = 0 is a non-degenerate critical point of J with m 0 = ι . From Lemma 2.2 (iii), it is obtained that

(3.18) C q ( J , 0 ) ≅ δ q , ι F .

f ∞ < μ 1 . Since lim ∣ x ∣ → + ∞ f ( ( i , j ) , x ) b x 3 = f ∞ < μ 1 , (3.11) is true. Then (3.11), (3.18) and ι ≠ 0 guarantee that x 1 ≠ 0 , which implies that J has at least one nontrivial solution.
f ∞ > μ 2 . For lim ∣ x ∣ → + ∞ f ( ( i , j ) , x ) b x 3 = f ∞ > μ 2 , we have proved (3.14). Combining (3.14), (3.18) with ι ≠ m n , it yields that J has a nontrivial critical point.
( F ∞ 1 ) and f ∞ = μ 1 . lim ∣ x ∣ → + ∞ f ( ( i , j ) , x ) b x 3 = f ∞ = μ 1 and ( F ∞ 1 ) means that (3.15) holds. Note that ι ≠ 0 , then (3.15) and (3.18) lead to x 2 ≠ 0 , which is a nontrivial critical point of J .
( F ∞ 2 ) and f ∞ = μ 2 . By the proof of (iv) of Theorem 1.1, there holds (3.16). For ι ≠ m n , (3.16) and (3.18) bring a nontrivial critical point for J . Therefore, J has at least one nontrivial solution and the proof is complete.□

4 Examples

In this section, we provide three examples to help understand our main results.

Example 4.1

For ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , let

(4.1) f ( ( i , j ) , x ) = a λ 1 x + b λ 1 2 x 3 , ∣ x ∣ ≤ 1 f 1 ( x ) , 1 < ∣ x ∣ < 10 b μ 1 3 x 3 , ∣ x ∣ ≥ 10 ,

where f 1 is a smooth connection such that f ( ( i , j ) , x ) is a C 1 function in x for all x ∈ R and ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] .

Direct computation gives that ( F 0 ) is true and lim ∣ x ∣ → 0 f ( ( i , j ) , x ) a x = f 0 = λ 1 , lim ∣ x ∣ → + ∞ f ( ( i , j ) , x ) b x 3 = f ∞ = μ 1 3 < μ 1 . Then Theorem 1.1 (i) guarantees that problem (1.1) with (1.2) admits at least one nontrivial solution.

Example 4.2

For ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] , let

(4.2) f ( ( i , j ) , x ) = a λ 1 x + b λ 1 2 x 3 , ∣ x ∣ ≤ 1 f 2 ( x ) , 1 < ∣ x ∣ < 10 b μ 1 x 3 − x , ∣ x ∣ ≥ 10 ,

where f 2 is a smooth connection such that f ( ( i , j ) , x ) is a C 1 function in x for all x ∈ R and ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] .

Obviously, ( F 0 ) holds and f 0 = λ 1 , f ∞ = μ 1 . Furthermore, lim ∣ x ∣ → + ∞ ( x f ( ( i , j ) , x ) − 4 F ( ( i , j ) , x ) ) = lim ∣ x ∣ → + ∞ x 2 = + ∞ , which means ( F ∞ 1 ) is satisfied. Then all conditions of Theorem 1.1 (iii) are met and problem (1.1) with (1.2) has at least one nontrivial solution.

Example 4.3

Let

(4.3) f ( ( i , j ) , x ) = a 2 λ 1 x + b λ 1 2 x 3 + b μ 2 x 3 , ∀ x ∈ R , ( i , j ) ∈ [ 1 , m ] × [ 1 , n ] .

By (4.3), we have lim ∣ x ∣ → 0 f ( ( i , j ) , x ) a x = f 0 = λ 1 2 ∉ Σ and lim ∣ x ∣ → + ∞ f ( ( i , j ) , x ) b x 3 = f ∞ = λ 1 2 + μ 2 > μ 2 . Moreover, the definition of ι , see (1.7), and f 0 = λ 1 2 < λ 1 imply that ι = 0 ≠ m n . Hence, by Theorem 1.2 (ii), we can draw a conclusion that problem (1.1) with (1.2) possesses at least one nontrivial solution.

In particular, let m = n = 2 , write

S = { ( i , j ) : 1 ≤ i ≤ m , 1 ≤ j ≤ n } = { ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , ( 2 , 2 ) } ,

then

∂ S = { ( 1 , 0 ) , ( 2 , 0 ) , ( 0 , 1 ) , ( 0 , 2 ) , ( 3 , 1 ) , ( 3 , 2 ) , ( 1 , 3 ) , ( 2 , 3 ) } .

Now the matrix A can be expressed by

A = 4 − 1 − 1 0 − 1 4 0 − 1 − 1 0 4 − 1 0 − 1 − 1 4 ,

then the eigenvalues of A are

λ 1 = 2 , λ 2 = λ 3 = 4 , λ 4 = 6 .

From Proposition 2.1, we obtain μ 2 = 144 .

Take a = b = 1 , consider problem (1.1) with (1.2) in the following form:

(4.4) 1 + ∑ j = 1 2 ∑ i = 1 3 ∣ Δ 1 x ( i − 1 , j ) ∣ 2 + ∑ i = 1 2 ∑ j = 1 3 ∣ Δ 2 x ( i , j − 1 ) ∣ 2 ( Δ 1 2 x ( i − 1 , j ) + Δ 2 2 x ( i , j − 1 ) ) = − ( x ( i , j ) + 148 x 3 ( i , j ) ) , ( i , j ) ∈ S ; x ( i , j ) = 0 , ( i , j ) ∈ ∂ S .

It is easy to see that f 0 = 1 < λ 1 , f ∞ = 148 > μ 2 and ι = 0 , then conditions of Theorem 1.2(ii) are satisfied and (4.4) has at least one nontrivial solution.

For simplicity, we denote

x ( 1 , 1 ) = x 1 x ( 1 , 2 ) = x 2 x ( 2 , 1 ) = x 3 x ( 2 , 2 ) = x 4 ,

according to the definition of ‖ ⋅ ‖ , it follows that

‖ x ‖ 2 = ⟨ x , x ⟩ = ( x 1 , x 2 , x 3 , x 4 ) 4 − 1 − 1 0 − 1 4 0 − 1 − 1 0 4 − 1 0 − 1 − 1 4 x 1 x 2 x 3 x 4 .

Therefore, (4.4) can be rewritten as the system

1 + x 1 x 2 x 3 x 4 T 4 − 1 − 1 0 − 1 4 0 − 1 − 1 0 4 − 1 0 − 1 − 1 4 x 1 x 2 x 3 x 4 4 − 1 − 1 0 − 1 4 0 − 1 − 1 0 4 − 1 0 − 1 − 1 4 x 1 x 2 x 3 x 4 = x 1 + 148 x 1 3 x 2 + 148 x 2 3 x 3 + 148 x 3 3 x 4 + 148 x 4 3 .

Using MATLAB, we find that problem (4.4) has 53 solutions, including 1 trivial solution and 52 nontrivial solutions. Among those 52 nontrivial solutions, 9 are positive, 9 are negative and the other 34 are sign-changing. Here we list a few: (0.09525372, 0.09525372, 0.0751692, 0.0751692), ( − 0.11293459 , − 0.071431939 , − 0.07143194 , − 0.05404505 ), ( 1.11803399 , − 1.11803399 , − 1.11803399 , 1.11803399 ).

Acknowledgments

The author wishes to thank the handling editor and the anonymous referees for their valuable comments and suggestions.

Funding information: This work was supported by the National Natural Science Foundation of China (Grant No. 11971126).
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no data sets were generated or analyzed during the current study.

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Received: 2021-11-18

Revised: 2022-03-09

Accepted: 2022-03-10

Published Online: 2022-04-30

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

Nontrivial solutions of discrete Kirchhoff-type problems via Morse theory

Abstract

1 Introduction

Theorem 1.1

Theorem 1.2

Remark 1.1

2 Preliminaries

Proposition 2.1

Definition 2.1

Definition 2.2

Definition 2.3

Lemma 2.1

Lemma 2.2

Lemma 2.3

Proof

Lemma 2.4

Proof

Lemma 2.5

Proof

3 Proofs of main results

Proof of Theorem 1.1

Proof of Theorem 1.2

4 Examples

Example 4.1

Example 4.2

Example 4.3

Acknowledgments

References

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