Nontrivial solutions for a class of fractional difference boundary value problems and fixed-point problems

Abstract

In this work, we use the variant fountain theorem to study the existence of nontrivial solutions for the superquadratic fractional difference boundary value problem:

　　　　　　　　 \begin{array}{l}\{\begin{array}{l}{}_T{\mathrm{\Delta }}_{t-1}^{\nu }\left(t{\mathrm{\Delta }}_{\nu -1}^{\nu }x\left(t\right)\right)=f\left(x\right(t+\nu -1\left)\right),t\in [0,T{]}_{{\mathbb{N}}_0},\hfill \\ x(\nu -2)={\left[{}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x\left(t\right)\right]}_{t=T}=0.\hfill \end{array}\hfill \end{array}

The existence of nontrivial solutions is obtained in the case of super quadratic growth of the nonlinear term f by change of fountain theorem.

Keywords

Fractional difference / boundary value problem / fountain theorem / nontrivial solution

1 Introduction

Consider the following fractional difference equation boundary value problem:

(1.1)\begin{array}{l}\{\begin{array}{l}{}_T{\mathrm{\Delta }}_{t-1}^{\nu }\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)\right)=f(x(t+\nu -1)),\\ x(\nu -2)={\left[{}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)\right]}_{t=T}=0,\end{array}\end{array}

where \nu \in (0,1), {}_t{\mathrm{\Delta }}_{\nu -1}^{\nu } and {}_T{\mathrm{\Delta }}_t^{\nu } are the leftward and rightward fractional difference operators, respectively, t\in [0,T{]}_{{\mathbb{N}}_0}:=\{0,1,\dots ,T\},f\in C(\mathbb{R},\mathbb{R}).

In recent years, as the study of fractional problems has intensified, it is true that, as Zheng [10] said, “for fractional differential equations, discretization or discrete difference equations are inevitable when they are posed”. Cheng [5] subsequently summarizes some good results in this regard recently.

Guo-Krasnosel’ skii immobility point theorem on cones was used to obtain the existence and nonexistence of positive solutions to the following fractional difference equation side value problem [7]:

\{\begin{array}{l}-{\mathrm{\Delta }}^{\nu }y(t)=\lambda f(t+\nu -1,y(t+\nu -1),\\ y(\nu -2)=y(\nu +b+1)=0,\end{array}

where t\in [0,b+1{]}_{{\mathbb{N}}_0}, , f\in C(([\nu -1,\nu +b{]}_{{\mathbb{N}}_{\nu -1}}),{\mathbb{R}}^{\mathbb{+}}), b⩾2 and b is an is integer.

Chen and Tang considered the following fractional order difference equation boundary value problem [4]:

\{\begin{array}{l}-{\mathrm{\Delta }}_{\nu -2}^{\nu }y(t)=\lambda f(t,y(t+\nu -1)),t\in [0,b+1{]}_{{\mathbb{N}}_0},\\ \mathrm{\Delta }y(\nu -2)=\mathrm{\Delta }y(\nu +b)=0,\end{array}

and its corresponding integer order equation:

\{\begin{array}{l}-{\mathrm{\Delta }}^{2}y(t)=\lambda f(t,y(t+1)),t\in [0,b+1{]}_{{\mathrm{N}}_0},\\ \mathrm{\Delta }y(0)=\mathrm{\Delta }y(2+b)=0,\end{array}

where {\mathrm{\Delta }}_{\nu -2}^{\nu } is fractional difference operator, {\mathrm{\Delta }}^2 is integer difference operator, . They not only obtained the existence of solutions to the above two problems, but also compared the differences between them, see [4, Notes 3.1−3.3].

However, the application of variational methods and critical point theory to study fractional order difference equations is a relatively new topic, and few related studies are available [6, 8]. He and Hou [8] used the critical point theorem [3] to obtain the existence of multiple solutions of the following fractional order difference equation with p-Laplacian operator as the edge value problem:

\{\begin{array}{l}{}_{T-1}{\mathrm{\Delta }}_{t-1}^{\nu }\left({\phi }_p\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)\right)\right)=\lambda f(t+\nu -1,x(t+\nu -1)),t\in [0,T{]}_{{\mathbb{N}}_0},\\ x(\nu -2)={\displaystyle {\sum }_{s=\nu -1}^{T+\nu }(T-s{)}^{(-\nu )}x(s)=0,}\end{array}

where T⩾1, \nu \in (0,1), p>1, {\phi }_p(t)={\left|t\right|}^{p-2}t.

Inspired by the above literature, in this paper, we apply the change fountain theorem to study the existence of non-trivial solutions to problem (1.1), for which we assume f satisfies the following conditions:

(H1) There exists d_1>0, \alpha >1 such that

\left|f(x)\right|⩽d_1(1+{\left|x\right|}^{\alpha }),\mathrm{\forall }x\in \mathbb{R},

(H2) F(x)⩾0,\mathrm{\forall }x\in \mathbb{R} and \underset{|x|\to \mathrm{\infty }}{lim}\frac{F(x)}{|x{|}^2}=\mathrm{\infty }, where F(x)={\int }_0^{x}f(s)\mathrm{d}s,

(H3) {\mathrm{l}\mathrm{i}\mathrm{m}}_{\left|x\right|\to \mathrm{\infty }}\frac{F(x)}{{\left|x\right|}^2}=0,

(H4) there exists d_2>0,L>0 such that

xf(x)⩾\left(2+\frac{d_2}{{\left|x\right|}^2}\right)F(x),\mathrm{\forall }|x\Vert ⩾L,

(H5) f(x)+f(-x)=0,\mathrm{\forall }x\in \mathbb{R}.

Note 1.1　The condition (H4) was first used in the literature [9], which is weaker than the classical Ambrosetti-Rabinowitz condition: the existence of \mu >2 makes

2 Preliminary knowledge

In this paper, we always assume that \sum _{i=j}^{m}x(i)=0, if . For any integer \beta, let {\mathbb{N}}_{\beta }:=\{\beta ,\beta +1,\dots \} and t^{(\nu )}=\frac{\mathrm{\Gamma }(t+1)}{\mathrm{\Gamma }(t+1-\nu )}, where t, \nu is defined by (1.1). For convenience, if t+1-\nu is the pole of the Gamma function, and t+1 is not, then t^{(\nu )}=0.

Definition 2.1 [1, 2]　The \nu (\nu >0) order fraction of f is defined as:

(2.1){\mathrm{\Delta }}_{\alpha }^{-\nu }f(t)=\frac{1}{\mathrm{\Gamma }(\nu )}\sum _{s=a}^{t-\nu }(t-s-1{)}^{\nu -1}f(s),t\in {\mathbb{N}}_{a-\nu }.

The fractional difference of \nu (\nu >0) is defined as: {\mathrm{\Delta }}^{\nu }f(t):={\mathrm{\Delta }}^N{\mathrm{\Delta }}^{\nu -N}f(t), where t\in {\mathbb{N}}_{a+N-\nu },N\in \mathbb{N} meets .

Definition 2.2 [1, 2]　If f is a real-valued function, \nu \in (0,1), then the leftward and rightward fractional order difference operators of f are defined as:

\begin{array}{rl}{}_t{\mathrm{\Delta }}_{\alpha }^{\nu }f(t)& ={\mathrm{\Delta }}_t{\mathrm{\Delta }}_{\alpha }^{-(1-\nu )}f(t)\\ & =\frac{1}{\mathrm{\Gamma }(1-\nu )}\mathrm{\Delta }\sum _{s=a}^{t+\nu -1}(t-s-1{)}^{-\nu }f(s),t\equiv a-\nu +1(\mathrm{m}\mathrm{o}\mathrm{d}1),\end{array}

\begin{array}{rl}{}_b{\mathrm{\Delta }}_t^{\nu }f(t)& =-{\mathrm{\Delta }}_b{\mathrm{\Delta }}_t^{-(1-\nu )}f(t)\\ & =\frac{1}{\mathrm{\Gamma }(1-\nu )}(-\mathrm{\Delta })\sum _{s=t+1-\nu }^b(s-t-1{)}^{-\nu }f(s),t\equiv b+\nu -1(\mathrm{m}\mathrm{o}\mathrm{d}1).\end{array}

Next, we first construct the variational framework of problem (1.1). Let

(2.2)X:=\{x=(x(\nu -1),x(\nu ),\dots ,x(\nu +T-1){)}^{\mathrm{T}}:x(\nu +i-1)\in \mathbb{R},i=0,1,\dots ,T\}.

Then X is T+1 dimensions Hilbert space. Assign the inner product and the corresponding parameters on it:

(2.3)(x,z)=\sum _{t=\nu -1}^{T+\nu -1}x(t)z(t),\Vert x\Vert ={\left(\sum _{t=\nu -1}^{T+\nu -1}|x(t){|}^2\right)}^{\frac{1}{2}},x,z\in X.

For \alpha ⩾1, define the \alpha -norm on X: \Vert x{\Vert }_{\alpha }={\left(\sum _{t=\nu -1}^{T+\nu -1}|x(t){|}^{\alpha }\right)}^{\frac{1}{\alpha }}, if \alpha =\mathrm{\infty }, then \Vert x{\Vert }_{\alpha }=\underset{t\in [0,T{]}_{N_0}}{max}|u(t)|. Since dim, for any x\in X, there exists c_{1\alpha },c_{2\alpha },c_{1\mathrm{\infty }},c_{2\mathrm{\infty }}>0 such that

(2.4)c_{1\alpha }\Vert x\Vert ⩽\Vert x{\Vert }_{\alpha }⩽c_{2\alpha }\Vert x\Vert ,

(2.5)c_{1\mathrm{\infty }}\Vert x\Vert ⩽\Vert x{\Vert }_{\mathrm{\infty }}⩽c_{2\mathrm{\infty }}\Vert x\Vert .

According to [6, (2.9)], the energy generalization function on X can be defined as:

(2.6)I(x)=\frac{1}{2}\sum _{t=-1}^T{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)\right)}^2-\sum _{t=-1}^{T}F(x(t+\nu -1)),x\in X,

where F(x(t+\nu -1))={\int }_0^{x(t+\nu -1)}f(s)\mathrm{d}s,x(\nu -2)=0,

{\left[t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)\right]}_{t=T}=\frac{-\nu }{\mathrm{\Gamma }(1-\nu )}\sum _{s=\nu -1}^{T+\nu }(T+s-1{)}^{(-\nu -1)}x(s)=0.

Obviously I(\theta )=0. Let

(2.7)\begin{array}{rl}E:=& \{\chi =(x(\nu -2),x(\nu -1),\dots ,x(\nu +T){)}^{\mathrm{T}}\in {\mathbb{R}}^{T+3}:\\ & x(\nu -2)=0,{\left[{}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)\right]}_{t=T}=0\}.\end{array}

Then, according to the boundary condition (1.1), E is isomorphic to X. When referring to x\in X, we always consider that x can be extended to \chi \in E, if necessary. If x=(x(\nu -1),x(\nu ),\dots ,x(\nu +T-1){)}^{\mathrm{T}}\in X is a critical point of I, then \chi =(x(\nu -2),x(\nu -1),\dots ,x(\nu +T){)}^{\mathrm{T}}\in E is a solution of (1.1). In fact, since I can be viewed as a continuous differentiable general function defined on a finite-dimensional Hilbert space X, its derivatives I^{\prime }(x)=0 if and only if \frac{\mathrm{\partial }I(x)}{\mathrm{\partial }x(i)}=0, \mathrm{\forall }i=\nu -1,\nu ,\dots ,\nu +T-1. By the definition of the Gâteaux derivative, we have:

(2.8)\begin{array}{rl}\left(I^{\mathrm{\prime }}(x),y\right)& =\underset{\epsilon \to 0}{lim}\frac{I(x+\epsilon y)-I(x)}{\epsilon }\\ & =\frac{1}{\epsilon }[\frac{1}{2}\sum _{t=-1}^T\left[{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }(x(t)+\epsilon y(t))\right)}^2-{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)\right)}^2\right]\\ & -\sum _{t=-1}^T[F((x+\epsilon y)(t+\nu -1))-F(x(t+\nu -1))]]\\ & =\sum _{t=-1}^T{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t{)}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }y(t)-\sum _{t=-1}^{T}f(x(t+\nu -1))y(t+\nu -1).\end{array}

Therefore, in order to obtain the existence of the solution of (1.1), we only need to study the existence of the critical point of the energy generalization I on X. Noting Definition 1.2, for t\in [-1,T{]}_{N-1}, let

(2.9)\begin{array}{rl}{}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }x(t)& =\mathrm{\Delta }\frac{1}{\mathrm{\Gamma }(1-\nu )}\sum _{s=\nu -1}^{t-(1-\nu )}(t-s-1{)}^{(-\nu )}x(s)\\ & :=\mathrm{\Delta }z(t+\nu -1).\end{array}

Then

\begin{array}{rl}& z(\nu -2)=0,\\ & z(\nu -1)=\frac{1}{\mathrm{\Gamma }(1-\nu )}\sum _{s=\nu -1}^{0-(1-\nu )}(-s-1{)}^{(-\nu )}x(s)=x(\nu -1),\\ & z(\nu )=\frac{1}{\mathrm{\Gamma }(1-\nu )}\sum _{s=\nu -1}^{1-(1-\nu )}(1-s-1{)}^{(-\nu )}x(s)=(1-\nu )x(\nu -1)+x(\nu ),\\ & z(\nu +1)=\frac{1}{\mathrm{\Gamma }(1-\nu )}\sum _{s=\nu -1}^{2-(1-\nu )}(2-s-1{)}^{(-\nu )}x(s),\\ & =\frac{(2-\nu )(1-\nu )}{2!}x(\nu -1)+(1-\nu )x(\nu )+x(\nu +1),\\ & ⋮\\ & z(\nu +T-1)=\frac{1}{\mathrm{\Gamma }(1-\nu )}\sum _{s=\nu -1}^{T-(1-\nu )}(T-s-1{)}^{(-\nu )}x(s)\\ & =\frac{(T-\nu )(T-1-\nu )\cdots (1-\nu )}{T!}x(\nu -1)\\ & +\frac{(T-1-\nu )(T-2-\nu )\cdots (1-\nu )}{(T-1)!}x(\nu )\\ & +\cdots +(1-\nu )x(\nu +T-2)+x(\nu +T-1).\end{array}

That is, z=Bx, where z=(z(\nu -1),z(\nu ),\dots ,z(\nu +T-1){)}^{\mathrm{T}},x=(x(\nu -1),x(\nu ),\dots ,x(\nu +T-1){)}^{\mathrm{T}},

\begin{array}{rl}& B=\\ & {\left(\begin{array}{ccccc}1& 0& 0& \cdots & 0\\ 1-nu& 1& 0& \cdots & 0\\ {\displaystyle \frac{(2-\nu )(1-\nu )}{2!}}& 1-\nu & 1& \cdots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ {\displaystyle \frac{(T-\nu )(T-1-\nu )\cdots (1-\nu )}{T!}}& {\displaystyle \frac{(T-1-\nu )(T-2-\nu )\cdots (1-\nu )}{(T-1)!}}& \cdots & \cdots & 1\end{array}\right)}_{(T+1)\times (T+1)}.\end{array}

Obviously (B^{-1}{)}^{\mathrm{T}}{B}^{-1} is the positive definite matrix and all its eigenvalues are positive. Let {\lambda }_{min} and {\lambda }_{max} be the minimum and maximum eigenvalues of (B^{-1}{)}^{\mathrm{T}}{B}^{-1}. Since x=B^{-1}z, we have:

(2.10)\begin{array}{l}{\lambda }_{min}\Vert z{\Vert }^2⩽\Vert x{\Vert }^2=\left(z^{\mathrm{T}}{\left(B^{-1}\right)}^{\mathrm{T}},B^{-1}z\right)⩽{\lambda }_{max}\Vert z{\Vert }^2.\end{array}

Let

\begin{array}{l}A={\left(\begin{array}{cccccc}2& -1& 0& \cdots & 0& 0\\ -1& 2& -1& \cdots & 0& 0\\ 0& -1& 2& \cdots & 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 2& -1\\ 0& 0& 0& \cdots & -1& 1\end{array}\right)}_{(T+1)\times (T+1)}.\end{array}

Obviously A is also a positive definite matrix. Let {\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{T+1} be the eigenvalues of A and {\eta }_1,{\eta }_2,\dots ,{\eta }_{T+1} be the corresponding orthogonal canonical vectors, where . Obviously, X:=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{{\eta }_1,{\eta }_2,\dots ,{\eta }_{\mathrm{T}+1}\}. Let Y_k:=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{{\eta }_1,{\eta }_2,\dots ,{\eta }_{\mathrm{k}}\}, Z_k:=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{{\eta }_{\mathrm{k}},{\eta }_{\mathrm{k}}+1,\dots ,{\eta }_{\mathrm{T}}+1\}, k\in [1,T+1{]}_{{\mathbb{N}}_0}. Let (X,\Vert \cdot \Vert ) be Banach space, X=\overline{\underset{j\in \mathbb{N}}{⨁}{X}_j} and . Denote Y_k=\underset{j=1}{\overset{k}{⨁}}{X}_j,Z_k=\overline{\underset{j=k}{\overset{\mathrm{\infty }}{⨁}}{X}_j}, consider C^1- generalized function {\mathrm{\Phi }}_{\lambda }:X\to \mathbb{R} as follows:

(2.11)\begin{array}{l}{\mathrm{\Phi }}_{\lambda }(u)=A(u)-\lambda B(u),\lambda \in [1,2].\end{array}

Lemma 2.1 [11, Theorem 2.1]　 The generalized function {\mathrm{\Phi }}_{\lambda } defined by (2.11) satisfies:

(C1) For any \lambda \in [1,2], {\mathrm{\Phi }}_{\lambda } reflecting the bounded set as the bounded set, and {\mathrm{\Phi }}_{\lambda }(u)={\mathrm{\Phi }}_{\lambda }(-u),\mathrm{\forall }(\lambda ,u)\in [1,2]\times X;

(C2) B(u)⩾0,\mathrm{\forall }u\in X, and A(u)\to \mathrm{\infty } or B(u)\leftarrow \mathrm{\infty },\Vert u\Vert \to \mathrm{\infty };

(C3) There exists r_k>{\rho }_k>0 such that

{\alpha }_k(\lambda )=\underset{u\in Z_k,\Vert u\Vert ={\rho }_k}{inf}{\mathrm{\Phi }}_{\lambda }(u)>{\beta }_k(\lambda )=\underset{u\in Y_k,\Vert u\Vert =r_k}{max}{\mathrm{\Phi }}_{\lambda }(u),\mathrm{\forall }\lambda \in [1,2].

Then

{\alpha }_k(\lambda )⩽{\zeta }_k(\lambda )=\underset{\gamma \in {\mathrm{\Gamma }}_k}{inf}\underset{u\in B_k}{max}{\mathrm{\Phi }}_{\lambda }(\gamma (u)),\mathrm{\forall }\lambda \in [1,2],

where B_k=\left\{u\in Y_k:\Vert u\Vert ⩽r_k\right\},{\mathrm{\Gamma }}_k=\{\gamma \in C\left(B_k,X\right):\gamma is odd, \gamma {|}_{\mathrm{\partial }{B}_k}=\mathrm{i}\mathrm{d}\}. Moreover, for \lambda \in [1,2] almost everywhere, there exists a sequence \{u_m^k(\lambda ){\}}_{m=1}^{\mathrm{\infty }} such that

3 Main conclusions and proof

To apply Lemma 2.1 to prove our conclusions, define the generalized functions A,B and {\mathrm{\Phi }}_{\lambda } as follows:

A(u)=\frac{1}{2}\sum _{t=-1}^T{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }u(t)\right)}^2,B(u)=\sum _{t=-1}^{T}F(u(t+\nu -1)),

{\mathrm{\Phi }}_{\lambda }(u)=\frac{1}{2}\sum _{t=-1}^T{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }u(t)\right)}^2-\lambda \sum _{t=-1}^{T}F(u(t+\nu -1)),u\in X.

Lemma 3.1　 If (H1) and (H2) hold, then B(u)⩾0,\mathrm{\forall }u\in X. Further, when \Vert u\Vert \to \mathrm{\infty }, A(u)\to \mathrm{\infty } or B(u)\to \mathrm{\infty }.

Proof　Since F(u)⩾0, it is obvious that B(u)⩾0,\mathrm{\forall }u\in X. The following proof exists \epsilon >0 such that

(3.1)\mathrm{♯}\left(\left\{t\in [0,T{\}}_{{\mathbb{N}}_0}:|u|⩾\epsilon \Vert u\Vert \right\}\right)=c_{\epsilon }>0,\mathrm{\forall }u\in X\mathrm{\setminus }\{0\},

where \mathrm{♯}(\cdot ) denotes the base of the set. Otherwise, for any n\in \mathbb{N}, there exists a sequence \{u_n\}\in X\mathrm{\setminus }\{0\} such that

\mathrm{♯}\left(\left\{t\in [0,T{]}_{{\mathbf{N}}_0}:\left|u_n(t)\right|⩾\frac{\Vert u\Vert }{n}\right\}\right)=0.

Let v_n=\frac{u_n}{\Vert u_n\Vert }\in X, then \Vert v_n\Vert =1 and

\mathrm{♯}\left(\left\{t\in [0,T{]}_{{\mathbf{N}}_0}:\left|v_n(t)\right|⩾\frac{1}{n}\right\}\right)=0.

Since dim, in the sense of sub-sequence, there exists v_0\in X such that v_n\to v_0, and obviously \Vert v_0\Vert =1. According to (2.4) and (2.5), we have

(3.2)\sum _{t=-1}^T|v_n-v_0|\to 0,n\to \mathrm{\infty }

and \Vert v_0{\Vert }_{\mathrm{\infty }}>0. Then there exists {\delta }_0>0 such that

\mathrm{♯}\left(\left\{t\in [0,T{]}_{{\mathbf{N}}_0}:\left|v_0(t)\right|⩾{\delta }_0\right\}\right)=c_{{\delta }_0}>0.

For \mathrm{\forall }n\in \mathbb{N}, let

{\mathrm{\Lambda }}_n^c=\left\{t\in [0,T{]}_{{\mathbb{N}}_0}:\left|v_n(t)\right|⩾\frac{1}{n}\right\}.

Taking {\mathrm{\Lambda }}_0=\{t\in [0,T{]}_{{\mathbb{N}}_0}:|v_0(t)⩾{\delta }_0|\}, then for sufficiently large n, we have

\begin{array}{l}\mathrm{♯}\left({\mathrm{\Lambda }}_n\cap {\mathrm{\Lambda }}_0\right)⩾\mathrm{♯}\left({\mathrm{\Lambda }}_0\right)-\mathrm{♯}\left({\mathrm{\Lambda }}_n^c\right)⩾c_{{\delta }_0}>0.\end{array}

As a result, when n is large enough, we can introduce

\begin{array}{rl}\sum _{t=-1}^T\left|v_n-v_0\right|& ⩾\sum _{{\mathrm{\Lambda }}_n\cap {\mathrm{\Lambda }}_0}\left|v_n-v_0\right|⩾\sum _{{\mathrm{\Lambda }}_n\cap {\mathrm{\Lambda }}_0}\left(\left|v_0\right|-\left|v_n\right|\right)\\ & ⩾\left({\delta }_0-\frac{1}{n}\right)\mathrm{♯}\left({\mathrm{\Lambda }}_n\cap {\mathrm{\Lambda }}_0\right)⩾\frac{{\delta }_0}{2}{c}_{{\delta }_0}>0.\end{array}

This contradicts (3.2), so (3.1) holds. Let {\mathrm{\Lambda }}_u=\left\{t\in [0,T{]}_{{\mathbb{N}}_0}:|u|⩾\epsilon \Vert u\Vert \right\}, then \mathrm{♯}({\mathrm{\Lambda }}_u)=c_{\epsilon }>0.

According to (H2), there exists d_3>0 and sufficiently large R>0 such that

F(x)⩾d_3|x{|}^2,\mathrm{\forall }|x|⩾R.

Notice that if the above equation is restricted to {\mathrm{\Lambda }}_u, there are

B(u)=\sum _{t=-1}^{T}F(u(t+\nu -1))⩾\sum _{{\mathrm{\Lambda }}_u}F(u(t+\nu -1))⩾c_{\epsilon }{d}_3{\epsilon }^2\Vert u{\Vert }^2.

This indicates that B(u)\to \mathrm{\infty },\Vert u\Vert \to \mathrm{\infty }. □

Lemma 2.2　 If (H1) and (H2) hold, then there exists r_k>{\rho }_k>0 such that

(3.3){\alpha }_k(\lambda )=\underset{u\in Z_k,\Vert u\Vert ={\rho }_k}{inf}{\mathrm{\Phi }}_{\lambda }(u)>0,Z_k:=\mathrm{span}\left\{{\eta }_k,{\eta }_{k+1},\dots ,{\eta }_{T+1}\right\},

and

(3.4)

Proof　From (H1), it follows that there exists d_4>0 such that

(3.5)|F(x)|⩽d_1(|x|+|x{|}^{\alpha +1})+d_4,\mathrm{\forall }x\in \mathbb{R}.

According to (2.4), for any u\in Z_k, we have

\begin{array}{rl}{\mathrm{\Phi }}_{\lambda }(u)& ⩾\frac{1}{2}\sum _{t=-1}^T{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }u(t)\right)}^2-2\sum _{t=-1}^{T}F(u(t+\nu -1))\\ & ⩾\frac{{\lambda }_k}{2{\lambda }_{max}}\Vert u{\Vert }^2-2d_1\sum _{t=0}^T|u(t+\nu -1)|-2d_1\sum _{t=0}^T|u(t+\nu -1){|}^{\alpha +1}-2d_4(T+1)\\ & ⩾\frac{{\lambda }_k}{2{\lambda }_{max}}\Vert u{\Vert }^2-2d_1{c}_{21}\Vert u\Vert -2d_1{c}_{2,\alpha +1}^{\alpha +1}\Vert u{\Vert }^{\alpha +1}-2d_4(T+1)\text{.}\end{array}

Let {\rho }_k={\left(\frac{16{\lambda }_{max}}{{\lambda }_k}{d}_1{c}_{2,\alpha +1}^{\alpha +1}\right)}^{\frac{1}{1-\alpha }}>max\left\{\frac{16{\lambda }_{max}}{{\lambda }_k}{d}_1{c}_{21}+1,\frac{16{\lambda }_{max}}{{\lambda }_k}{d}_4(T+1)\right\}. Then we can obtain

{\alpha }_k(\lambda )=\underset{u\in Z_k,\Vert u\Vert ={\rho }_k}{inf}{\mathrm{\Phi }}_{\lambda }(u)⩾\frac{{\lambda }_k}{4{\lambda }_{max}}{\rho }_k^2>0.

On the other hand, it is known from (H2) that there exists M_1>\frac{{\lambda }_k}{{\lambda }_{\mathrm{m}\mathrm{i}\mathrm{n}}},M_2>0 such that

F(x)⩾M_1|x{|}^2-M_2,\mathrm{\forall }x\in \mathbb{R}.

This leads to

\begin{array}{rl}{\mathrm{\Phi }}_{\lambda }(u)& ⩽\frac{1}{2}\sum _{t=-1}^T{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }u(t)\right)}^2-\sum _{t=-1}^{T}F(u(t+\nu -1))\\ & ⩽\frac{{\lambda }_k}{2{\lambda }_{min}}\Vert u{\Vert }^2-\sum _{t=0}^T\left(M_1|u(t+\nu -1){|}^2-M_2\right)\\ & =\left(\frac{{\lambda }_k}{2{\lambda }_{min}}-M_1\right)\Vert u{\Vert }^2+M_2(T+1)\\ & ⩽-\frac{{\lambda }_k}{2{\lambda }_{min}}\Vert u{\Vert }^2+M_2(T+1).\end{array}

If taken

r_k>max\left\{{\rho }_k,\sqrt{\frac{2{\lambda }_{min}{M}_2(T+1)}{{\lambda }_k}}\right\},

then

□

According to Lemmas 3.1 and 3.2, we can obtain the main theorem of this paper.

Theorem 3.1　 If (H1)−(H5) holds, then there exists at least one nontrivial solution to (1.1).

Proof　By (2.4) and (3.5), {\mathrm{\Phi }}_{\lambda } maps bounded sets to bounded sets, which holds consistently for \lambda \in [1,2]. Then we know from (H5) that {\mathrm{\Phi }}_{\lambda } is an even generalized function, and thus Lemma 2.1(C1) holds. Lemmas 3.1 and 3.2 show that Lemma 2.1(C2),(C3) also holds. Therefore, according to Lemma 2.1, for any k⩾1 and for almost all \lambda \in [1,2], there exists a sequence \{u_m^k(\lambda ){\}}_{m=1}^{\mathrm{\infty }} such that

(3.6)

where {\zeta }_k(\lambda )=\underset{\gamma \in {\mathrm{\Gamma }}_k}{inf}\underset{u\in B_k}{max}{\mathrm{\Phi }}_{\lambda }(\gamma (u)),\mathrm{\forall }\lambda \in [1,2], B_k=\{u\in Y_k:\Vert u\Vert ⩽r_k\},{\mathrm{\Gamma }}_k=\{\gamma \in C(B_k,X):\gamma is odd, \gamma {|}_{\mathrm{\partial }{B}_k}=\mathrm{i}\mathrm{d}\}. And from the proof of Lemma 3.2, we know that {\zeta }_k(\lambda )\in [{\overline{\alpha }}_k,{\overline{\zeta }}_k],\mathrm{\forall }k⩾1, where {\overline{\zeta }}_k=\underset{u\in B_k}{max}{\mathrm{\Phi }}_1(u),{\overline{\alpha }}_k=\frac{{\lambda }_k}{4{\lambda }_{max}}{\rho }_k^2.

Since the sequence \{u_m^k(\lambda ){\}}_{m=1}^{\mathrm{\infty }} is bounded, and thus for all k⩾1, {\lambda }_n\to 1 can be chosen such that the sequence \{u_m^k(\lambda ){\}}_{m=1}^{\mathrm{\infty }} has a strongly convergent sub-sequence. Without loss of generality, it can be assumed that

\begin{array}{l}\underset{m\to \mathrm{\infty }}{lim}{u}_m^k\left({\lambda }_n\right)=u_n^k,\mathrm{\forall }n\in \mathbb{N},k⩾1.\end{array}

Combined with (3.6), we can see that

(3.7)\begin{array}{l}{\mathrm{\Phi }}_{{\lambda }_n}^{\mathrm{\prime }}\left(u_n^k\right)=0,{\mathrm{\Phi }}_{{\lambda }_n}\left(u_n^k\right)\in \left[{\overline{\alpha }}_k,{\overline{\zeta }}_k\right],\mathrm{\forall }n\in \mathbb{N},k⩾1.\end{array}

The following proof \{u_n^k{\}}_{\mathrm{n}=1}^{\mathrm{\infty }} is bounded in X and there exists a strongly convergent sub-sequence, and its limit is u^k\in X, for simplicity, the following notation u_n=u_n^k.

Disproof method. If \{u_n\} is unbounded, in the sense of a sub-sequence, then \Vert u_n\Vert \to \mathrm{\infty },n\to \mathrm{\infty }. Let v_n=\frac{u_n}{\Vert u_n\Vert }. Then \{v_n\} is bounded in X such that there exist sub-sequences (still recorded as \{v_n\}) and v_0\in X such that

(3.8)\begin{array}{l}{v}_n\to v_0,n\to \mathrm{\infty }.\end{array}

From the definition of the generic function {\mathrm{\Phi }}_{\lambda }, we have

\begin{array}{rl}{\lambda }_n\sum _{t=-1}^{T}F\left(u_n(t+\nu -1)\right)& =\frac{1}{2}\sum _{t=-1}^T{\left({}_t{\mathrm{\Delta }}_{\nu -1}^{\nu }{u}_n(t)\right)}^2-{\mathrm{\Phi }}_{{\lambda }_n}\left(u_n\right)\\ & ⩽\frac{{\lambda }_{T+1}}{2{\lambda }_{min}}{\Vert u_n\Vert }^2-{\mathrm{\Phi }}_{{\lambda }_n}\left(u_n\right).\end{array}

Also according to (3.7), there exists M>0 such that |{\mathrm{\Phi }}_{{\lambda }_n}(u_n)|⩽M. Therefore

(3.9)\left|{\lambda }_n\sum _{t=-1}^T\frac{F\left(u_n(t+\nu -1)\right)}{{\Vert u_n\Vert }^2}-\frac{{\lambda }_{T+1}}{2{\lambda }_{min}}\right|⩽\frac{M}{{\Vert u_n\Vert }^2}\to 0.

The following two cases are discussed.

Case 1: v_0 is not constant equal to 0.

Let {\mathrm{\Lambda }}_{v_0}=\{t\in [0,T{]}_{{\mathbb{N}}_0}:|v_0(t)|>0\}, then \mathrm{♯}{\mathrm{\Lambda }}_{v_0}>0. Since \Vert u_n\Vert \to 0, so in {\mathrm{\Lambda }}_{v_0}, |u_n|\to \mathrm{\infty },n\to \mathrm{\infty }. Thus by (H2) and (3.8), we have

\underset{\left|u_n\right|\to \mathrm{\infty }}{lim}\sum _{t=-1}^T\frac{F\left(u_n(t+\nu -1)\right)}{{\Vert u_n\Vert }^2}⩾\underset{\left|u_n\right|\to \mathrm{\infty }}{lim}\sum _{{\mathrm{\Lambda }}_{v_0}}\frac{F\left(u_n(t+\nu -1)\right)}{{\left|u_n\right|}^2}{\left|v_n\right|}^2=\mathrm{\infty }.

This contradicts (3.9).

Case 2: v_0 is constantly equal to 0.

At this point, according to (3.9), noting {\lambda }_n\to 1, we have

(3.10)\underset{n\to \mathrm{\infty }}{lim}\sum _{t=-1}^T\frac{F\left(u_n(t+\nu -1)\right)}{{\Vert u_n\Vert }^2}=\frac{{\lambda }_{T+1}}{2{\lambda }_{min}}.

From (H3) and (H4), it follows that there exists d_5>0 such that

\frac{d_2}{|x{|}^2}F(x)⩽xf(x)-2F(x)+d_5,\mathrm{\forall }x\in \mathbb{R}.

Then

\begin{array}{rl}{\lambda }_n\sum _{t=-1}^T\frac{F\left(u_n(t+\nu -1)\right)}{{\Vert u_n\Vert }^2}& ={\lambda }_n\sum _{t=-1}^T\frac{F\left(u_n(t+\nu -1)\right)}{{\left|u_n\right|}^2}{\left|v_n\right|}^2\\ & ⩽{\lambda }_n\sum _{t=-1}^T\frac{F\left(u_n(t+\nu -1)\right)}{{\left|u_n\right|}^2}{\Vert v_n\Vert }_{\mathrm{\infty }}^2\\ & ⩽\frac{{\lambda }_n}{d_2}\sum _{t=-1}^T\left(u_{n}f\left(u_n\right)-2F\left(u_n\right)+d_5\right){\Vert v_n\Vert }_{\mathrm{\infty }}^2\\ & =\frac{1}{d_2}\left(2{\mathrm{\Phi }}_{{\lambda }_n}\left(u_n\right)-\left({\mathrm{\Phi }}_{{\lambda }_n}^{\mathrm{\prime }}\left(u_n\right),u_n\right)\right){\Vert v_n\Vert }_{\mathrm{\infty }}^2+\frac{{\lambda }_n}{d_2}\sum _{t=-1}^T{d}_5{\Vert v_n\Vert }_{\mathrm{\infty }}^2\\ & \to 0.\end{array}

This contradicts (3.10).

In summary, \{u_n\} is bounded in X. And since X is a finite dimensional space, \{u_n\} has strongly convergent sub-sequences. Notice that (3.7), for k⩾1, the limit u^k is a critical point of {\mathrm{\Phi }}_1 and {\mathrm{\Phi }}_1(u_k)\in [{\overline{\alpha }}_k,{\overline{\zeta }}_k]. Since k\in [1,T+1{]}_{{\mathbb{N}}_0}. k takes different values and {\overline{\zeta }}_k also takes different values. Thus (1.1) has at least one nontrivial solution.□

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