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Advances in Difference Equations

, 2018:370 | Cite as

Nontrivial solutions for boundary value problems of a fourth order difference equation with sign-changing nonlinearity

  • Keyu Zhang
  • Donal O’Regan
  • Zhengqing Fu
Open Access
Research
  • 86 Downloads

Abstract

In this paper, using the topological degree theory, we establish two existence theorems for nontrivial solutions for boundary value problems of a fourth order difference equation with a sign-changing nonlinearity.

Keywords

Difference equations boundary value problems Sign-changing nonlinearity Nontrivial solutions Topological degree theory 

1 Introduction

For \(a,b\in\mathbb {Z}\), let \(\mathbb {T}_{a}^{b}=\{ a,a+1,a+2,\ldots,b\}\) with \(a< b\). In this paper we consider the existence of nontrivial solutions for boundary value problems of the following fourth order difference equation with a sign-changing nonlinearity
$$ \textstyle\begin{cases} \Delta^{4} u(t-2)=f(t,u(t)), \\ u(1)=u(T+1)=\Delta^{2} u(0)=\Delta^{2} u(T)=0, \end{cases} $$
(1.1)
where T is an integer with \(T\ge5\), and \(f:\mathbb {T}_{2}^{T}\times \mathbb {R}\to\mathbb {R}\) is a continuous function with \(\mathbb {T}_{2}^{T}=\{ 2,3,\ldots,T\}\) and \(\mathbb {R}=(-\infty,+\infty)\) (it is assumed to be continuous from the topological space \(\mathbb {T}_{2}^{T}\times\mathbb {R} \) into the topological space \(\mathbb {R}\), the topology on \(\mathbb {T}_{2}^{T}\) being the discrete topology).
Difference equations with discrete boundary value conditions have been widely studied in the literature; see, for example, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] and the references therein. However, as mentioned in [6], very few results are available with sign-changing nonlinearities; see [6, 7, 8, 9, 10, 11]. Other related work in this field can be found in [12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45] and the references therein. In [7], C.S. Goodrich used the Krasnosel’skiĭ fixed point theorem to obtain the existence of at least one positive solution to the following discrete fractional semipositone boundary value problem
$$ \textstyle\begin{cases} \Delta^{\nu}y(t)=\lambda f(t+\nu-1,y(t+\nu-1)), \quad t\in[0,T]\cap \mathbb {Z}, \\ y(\nu-1)=y(\nu+T)+\sum_{i=1}^{N} F(t_{i}, y(t_{i})), \end{cases} $$
(1.2)
where \(\Delta^{\nu}\) is the νth fractional difference with \(\nu \in(0,1)\), f is continuous, bounded below (i.e., \(f+M\ge0\) for some \(M>0\)), and
$$ \lim_{y\to+\infty}\frac {f(t,y)}{y}=0 \quad \text{uniformly for }t\in[\nu-1,\nu+T]_{\mathbb {Z}_{\nu-1}}. $$
(1.3)
In [10], J. Xu and D. O’Regan used the fixed point index to obtain the existence of nontrivial solutions for (1.2) with weaker conditions than that of (1.3), and also in [11], J. Xu et al. considered the existence of positive solutions for system (1.2), with adopted convex and concave functions to depict the coupling behavior of nonlinearities. In [40], Y. Cui used the \(u_{0}\)-positive operator to study the uniqueness of solutions for the following nonlinear fractional boundary value problems:
$$ \textstyle\begin{cases} D^{p}x(t)+p(t)f(t,x(t))+q(t)=0,\quad t\in(0,1), \\ x(0)=x'(0)=0,\qquad x(1)=0, \end{cases} $$
(1.4)
where \(D^{p}\) is the Riemann–Liouville fractional derivative, and f is a Lipschitz continuous function, with the Lipschitz constant associated with the first eigenvalue for the relevant operator. Using similar methods, the authors in [12, 39, 41] obtained some existence and nonexistence theorems for their problems.

Motivated by the works mentioned above, we consider the existence of nontrivial solutions for (1.1) involving sign-changing nonlinearities. Using the topological degree theory of a completely continuous field, and conditions concerning the first eigenvalue corresponding to the relevant linear problem, two existence theorems are obtained.

2 Preliminaries

For convenience, we let \(\mathbb {T}_{1}^{T+1}=\{1,2,3,\ldots,T,T+1\}\), \(\mathbb {T}_{0}^{T+2}=\{0,1,2,3,\ldots,T+1,T+2\}\), \(\mathbb {T}_{2}^{T}=\{ 2,3,\ldots,T\}\). Then we define our space E as the collection of all maps from \(\mathbb {T}_{0}^{T+2}\) to \(\mathbb {R}\) equipped with the norm \(\| u\|=\max_{j\in\mathbb {T}_{0}^{T+2}}|u(j)|\). Consequently, E is a Banach space, and we let \(P=\{u\in E: u(t)\ge0, t\in\mathbb {T}_{1}^{T+1}\}\). Then P is a cone on E. Throughout our paper, we let \(B_{\rho}=\{u\in E:\|u\|<\rho\}\) for \(\rho>0\). Now \(\partial B_{\rho}=\{ u\in E: \|u\|=\rho\}\) and \(\overline{B}_{\rho}=\{u\in E: \|u\|\le\rho \}\).

In what follows, we establish the Green’s function for (1.1). As in [3, 4], we transform (1.1) into its equivalent sum equation
$$ u(t)= \sum_{s=2}^{T} H(t,s) \sum_{j=2}^{T} H(s,j) f\bigl(j,u(j)\bigr), \quad t\in\mathbb {T}_{1}^{T+1}, $$
(2.1)
where
$$ H(t,s)=\frac{1}{T} \textstyle\begin{cases} (t-1)(T+1-s), & 1\le t\le s\le T, \\ (s-1)(T+1-t), & 2\le s\le t\le T+1. \end{cases} $$
(2.2)

Lemma 2.1

Green’s functionHhas the following properties:
  1. (i)

    \(H(t,s)>0\)for\((t,s)\in\mathbb {T}_{2}^{T}\times\mathbb {T}_{2}^{T}\),

     
  2. (ii)

    \(\frac{1}{T}H(t,t)H(s,s)\le H(t,s)\le H(s,s)\)for\((t,s)\in \mathbb {T}_{2}^{T}\times\mathbb {T}_{1}^{T+1}\).

     

Proof

We only need to prove the first inequality of (ii). Indeed, for all \((t,s)\in\mathbb {T}_{2}^{T}\times\mathbb {T}_{1}^{T+1}\), from the definitions of \(H(t,s)\) and \(H(s,s)\) we have
$$\frac{H(t,s)}{H(s,s)}= \textstyle\begin{cases} \frac{t-1}{s-1}\ge\frac{t-1}{T}\ge\frac{t-1}{T} \frac{T+1-t}{T}= \frac{1}{T} H(t,t) , & 1\le t\le s\le T, \\ \frac{T+1-t}{T+1-s}\ge\frac{T+1-t}{T} \ge \frac{T+1-t}{T} \frac {t-1}{T}= \frac{1}{T} H(t,t), & 2\le s\le t\le T+1. \end{cases} $$
Then we have \(H(t,s)\ge \frac{1}{T} H(t,t)H(s,s) \) for \((t,s)\in \mathbb {T}_{2}^{T}\times\mathbb {T}_{1}^{T+1}\). This completes the proof. □
We define an operator \(A: E\to E\) as follows:
$$ (Au) (t)= \sum_{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) f\bigl(j,u(j) \bigr),\quad t\in\mathbb {T}_{1}^{T+1}. $$
(2.3)
The existence of solutions for (1.1) is equivalent to that of fixed points of A.
From [4], we know that \(\sin\frac{\pi(t-1)}{T}:=\varphi _{0}(t)\), \(t\in\mathbb {T}_{2}^{T}\) is the eigenfunction related to the eigenvalue \(\frac{1}{16} \sin^{-4} \frac{\pi}{2T}\) of the eigenproblem
$$\textstyle\begin{cases} \Delta^{4} u(t-2)=\lambda u(t), \quad t\in\mathbb {T}_{2}^{T}, \\ u(1)=u(T+1)=\Delta^{2} u(0)=\Delta^{2} u(T)=0, \end{cases} $$
i.e., the following two equations hold:
$$\begin{aligned}& \sum_{s=2}^{T} \sum _{j=2}^{T} H(t,s) H(s,j)\sin\frac{\pi(j-1)}{T}= \frac{1}{16} \sin^{-4} \frac{\pi}{2T} \sin\frac{\pi(t-1)}{T} , \quad t\in\mathbb {T}_{2}^{T}, \end{aligned}$$
(2.4)
$$\begin{aligned}& \sum_{s=2}^{T} \sum _{t=2}^{T} H(t,s) H(s,j)\sin\frac{\pi(t-1)}{T}= \frac{1}{16} \sin^{-4} \frac{\pi}{2T} \sin\frac{\pi(j-1)}{T} , \quad t\in\mathbb {T}_{2}^{T}. \end{aligned}$$
(2.5)

Lemma 2.2

Let\(e(t)= \frac{1}{T} H(t,t) \)and\(P_{0}=\{u\in P: u(t)\ge e(t)\|u\|, t\in\mathbb {T}_{1}^{T+1}\}\). Then\(L(P)\subset P_{0}\), where
$$ (Lu) (t)=\sum_{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) u(j), \quad t\in \mathbb {T}_{1}^{T+1}. $$
(2.6)

This is a direct result from Lemma 2.1(ii), so we omit its proof.

Now, we offer two basic theorems from the topological degree theory; for details we refer the reader to [46].

Lemma 2.3

LetEbe a Banach space and Ω a bounded open set inE. Suppose that\(A: \Omega\to E\)is a continuous compact operator. If there exists\(u_{0}\in E\setminus \{0\}\)such that
$$u-Au\neq \mu u_{0}, \quad \forall u\in\partial\Omega, \mu\ge0, $$
then the topological degree\(\deg(I-A,\Omega,0)=0\).

Lemma 2.4

LetEbe a Banach space and Ω a bounded open set inEwith\(0\in\Omega\). Suppose that\(A: \Omega\to E\)is a continuous compact operator. If
$$Au\neq \mu u, \quad \forall u\in\partial\Omega, \mu\ge1, $$
then the topological degree\(\deg(I-A,\Omega,0)=1\).

3 Nontrivial solutions for (1.1)

Now we present some assumptions for our nonlinearity f.
  1. (H1)
    There exist two constants \(a>0\), \(b>0\) and a function \(k\in C(\mathbb {R}, \mathbb {R}^{+})\) such that
    $$f(t,u)\ge-a-bk(u),\quad \forall u\in\mathbb {R}, t\in\mathbb {T}_{2}^{T}. $$
     
  2. (H2)

    \(\lim_{|u|\to+\infty} \frac{k(u)}{|u|}=0\).

     
  3. (H3)

    \(\liminf_{|u|\to+\infty}\frac{f(t,u)}{|u|}>16 \sin^{4} \frac {\pi}{2T}\) uniformly on \(t\in\mathbb {T}_{2}^{T}\),

     
  4. (H4)

    \(\limsup_{|u|\to0}\frac{|f(t,u)|}{|u|}<16 \sin^{4} \frac{\pi }{2T}\) uniformly on \(t\in\mathbb {T}_{2}^{T}\),

     
  5. (H5)

    \(\liminf_{u\to0^{+}}\frac{f(t,u)}{u}>16 \sin^{4} \frac{\pi}{2T}\), \(\limsup_{u\to0^{-}}\frac{f(t,u)}{u}<16 \sin^{4} \frac{\pi}{2T}\), uniformly on \(t\in\mathbb {T}_{2}^{T}\),

     
  6. (H6)

    \(\limsup_{|u|\to+\infty}\frac{|f(t,u)|}{|u|}<16 \sin^{4} \frac{\pi}{2T}\) uniformly on \(t\in\mathbb {T}_{2}^{T}\).

     

Theorem 3.1

Suppose that (H1)(H4) hold. Then (1.1) has at least one nontrivial solution.

Proof

From (H3) there exist \(\varepsilon_{0}>0\) and \(X_{0}>0\) such that
$$ f(t,u)\ge \biggl(16 \sin^{4} \frac{\pi }{2T}+\varepsilon_{0} \biggr)|u|, \quad \forall t\in\mathbb {T}_{2}^{T}, |u|>X_{0}. $$
(3.1)
For any given ε with \(\varepsilon_{0} - b\varepsilon>0\), and from (H2), there exists \(X_{1}>X_{0}\) such that
$$ k(u)\le\varepsilon|u|,\quad \forall|u|>X_{1}. $$
(3.2)
Now since \(a>0\), \(b>0\) and k is a nonnegative function, we have
$$\begin{aligned} f(t,u)&\ge \biggl(16 \sin^{4} \frac{\pi }{2T}+ \varepsilon_{0} \biggr)|u|-a-bk(u) \\ & \ge \biggl(16 \sin^{4} \frac{\pi}{2T}+\varepsilon_{0} \biggr)|u|-a-b \varepsilon|u|, \quad \forall|u|>X_{1}. \end{aligned}$$
(3.3)
Now we choose \(c_{1}= (16 \sin^{4} \frac{\pi}{2T}+\varepsilon _{0}-b \varepsilon )X_{1}+\max_{t\in\mathbb {T}_{2}^{T}, |u|\le X_{1}}|f(t,u)|\) and \(k^{*}=\max_{|u|\le X_{1}}k(u)\). Then we have
$$\begin{aligned} f(t,u)&\ge \biggl(16 \sin^{4} \frac{\pi }{2T}+\varepsilon_{0}-b \varepsilon \biggr)|u|-a-c_{1} \\ &= \biggl(16 \sin ^{4} \frac{\pi}{2T}+\varepsilon_{0}-b \varepsilon \biggr)|u|-c_{2}, \quad \forall t\in\mathbb {T}_{2}^{T}, u\in\mathbb {R}, \end{aligned}$$
(3.4)
where \(c_{2}=c_{1}+a\). Note that ε can be chosen arbitrarily small, and we let
$$\begin{aligned} R >& \max \biggl\{ \frac{(c_{2}+bk^{*}) [ (\varepsilon _{0}-b \varepsilon )\sum_{s=2}^{T} H(s,s)\sum_{j=2}^{T} H(s,j)+ (16 \sin^{4} \frac{\pi}{2T}+\varepsilon_{0}-b \varepsilon ) \sum_{s=2}^{T} \sum_{j=2}^{T} H(s,j) ]}{\varepsilon_{0}-b \varepsilon-b\varepsilon [ (\varepsilon_{0}-b \varepsilon )\sum_{s=2}^{T} H(s,s)\sum_{j=2}^{T} H(s,j)+ (16 \sin^{4} \frac{\pi}{2T}+\varepsilon_{0}-b \varepsilon ) \sum_{s=2}^{T} \sum_{j=2}^{T} H(s,j) ]}, \\ & \frac{\sum_{s=2}^{T} H(s,s)\sum_{j=2}^{T} H(s,j)(c_{2}+bk^{*})}{1-b\varepsilon\sum_{s=2}^{T} H(s,s)\sum_{j=2}^{T} H(s,j) }, 0 \biggr\} . \end{aligned}$$
Now we prove that
$$ u-Au\neq \mu\varphi_{0},\quad \forall u\in \partial B_{R}, \mu\ge0. $$
(3.5)
From (2.4) and Lemma 2.2, we have \(\varphi_{0}=16 \sin^{4} \frac{\pi}{2T}L\varphi_{0}\in P_{0} \). Indeed, if (3.5) isn’t true, then there exist \(u_{0}\in\partial B_{R}\) and \(\mu_{0}>0\) such that
$$ u_{0}-Au_{0}=\mu_{0} \varphi_{0}. $$
(3.6)
Let \(\tilde{u}(t)=\sum_{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j)(a+bk(u_{0})+c_{1})\). Then
$$\begin{aligned} \tilde{u}(t)&\le\sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \bigl(c_{2}+b \varepsilon|u_{0}|+bk^{*}\bigr) \\ &\le\sum_{s=2}^{T} H(s,s)\sum_{j=2}^{T} H(s,j) \bigl(c_{2}+b\varepsilon\|u_{0}\|+bk^{*}\bigr). \end{aligned}$$
Therefore,
$$ \|\tilde{u}\|\le\sum_{s=2}^{T} H(s,s)\sum _{j=2}^{T} H(s,j) \bigl(c_{2}+b \varepsilon R+bk^{*}\bigr). $$
(3.7)
Then from \(L(P)\subset P_{0}\), \(\varphi_{0}\in P_{0}\), and
$$\begin{aligned} u_{0}(t)+\tilde{u}(t)&=\tilde{u}(t)+(Au_{0}) (t)+\mu _{0}\varphi_{0}(t) \\ & =\sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \bigl( f\bigl(j,u_{0}(j) \bigr)+bk\bigl(u_{0}(j)\bigr)+a+c_{1}\bigr)+ \mu_{0}\varphi_{0}(t), \end{aligned}$$
we have
$$u_{0}+\tilde{u}\in P_{0}. $$
As a result, we obtain
$$\begin{aligned} &(Au_{0}) (t)+\tilde{u}(t) \\ &\quad =\sum_{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) \bigl( f \bigl(j,u_{0}(j)\bigr)+bk\bigl(u_{0}(j)\bigr)+c_{2} \bigr) \\ &\quad \ge\sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \biggl( \biggl(16 \sin ^{4} \frac{\pi}{2T}+\varepsilon_{0}-b \varepsilon \biggr) \bigl\vert u_{0}(j) \bigr\vert -c_{2}+bk \bigl(u_{0}(j)\bigr)+c_{2} \biggr) \\ &\quad \ge \biggl(16 \sin^{4} \frac{\pi}{2T}+\varepsilon_{0}-b \varepsilon \biggr) \sum_{s=2}^{T} H(t,s) \sum_{j=2}^{T} H(s,j) \bigl\vert u_{0}(j) \bigr\vert \\ &\quad \ge \biggl(16 \sin^{4} \frac{\pi}{2T}+\varepsilon_{0}-b \varepsilon \biggr) \sum_{s=2}^{T} H(t,s) \sum_{j=2}^{T} H(s,j) u_{0}(j). \end{aligned}$$
(3.8)
On the other hand, from the definition of L, we get
$$\begin{aligned}& \biggl(16 \sin^{4} \frac{\pi }{2T} +\varepsilon_{0}-b \varepsilon \biggr) \sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) u_{0}(j) \\& \quad = 16 \sin^{4} \frac{\pi}{2T} \sum _{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) \bigl(u_{0}(j)+\tilde{u}(j)\bigr) \\& \qquad {}-16 \sin^{4} \frac{\pi}{2T} \sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \tilde{u}(j) \\& \qquad {}+ (\varepsilon_{0}-b \varepsilon ) \sum _{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) u_{0}(j) \\& \quad \ge16 \sin^{4} \frac{\pi}{2T} \sum _{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) \bigl(u_{0}(j)+\tilde{u}(j)\bigr); \end{aligned}$$
(3.9)
in order to obtain the above inequality, we prove that
$$\begin{aligned}& -16 \sin^{4} \frac{\pi}{2T} \sum _{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) \tilde{u}(j) \\& \quad {}+ (\varepsilon_{0}-b \varepsilon ) \sum _{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) u_{0}(j)\ge0. \end{aligned}$$
(3.10)
Indeed, since \(u_{0}+\tilde{u}\in P_{0}\), we have \(u_{0}(t)+\tilde {u}(t)\ge e(t)\|u_{0}+\tilde{u}\|\ge e(t) (\|u_{0}\|-\|\tilde{u}\| )\). Note that \(H(t,s)\) vanishes at \(t=1\) and \(t=T+1\), \(H(t,s)\) is symmetric on \(\mathbb {T}_{2}^{T}\), i.e., \(H(t,s)=H(s,t)\). Then
$$\begin{aligned}& (\varepsilon_{0} -b \varepsilon ) \sum_{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) \bigl( \tilde{u}(j)+ u_{0}(j)\bigr) \\& \qquad {}- \biggl(16 \sin ^{4} \frac{\pi}{2T}+\varepsilon_{0}-b \varepsilon \biggr) \sum _{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) \tilde{u}(j) \\& \quad \ge (\varepsilon_{0}-b \varepsilon ) \bigl(R-\|\tilde{u}\|\bigr) \sum _{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) e(j) \\& \qquad {}- \biggl(16 \sin^{4} \frac{\pi}{2T}+ \varepsilon_{0}-b \varepsilon \biggr) \sum _{s=2}^{T} H(t,s) \\& \qquad {}\times\sum_{j=2}^{T} H(s,j) e(j) \Biggl(\sum_{s=2}^{T} \sum _{j=2}^{T} H(s,j) \bigl(c_{2}+b \varepsilon R+bk^{*}\bigr) \Biggr) \\& \quad \ge0. \end{aligned}$$
Combining (3.8), (3.9) and (3.10), we have
$$\begin{aligned} (Au_{0}) (t)+\tilde{u}(t)&\ge16 \sin^{4} \frac{\pi }{2T} \sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \bigl(u_{0}(j)+\tilde {u}(j)\bigr) \\ &=16 \sin^{4} \frac{\pi}{2T} \bigl(L(u_{0}+ \tilde{u})\bigr) (t). \end{aligned}$$
(3.11)
Using (3.6) we obtain
$$ u_{0}+\tilde{u}=Au_{0}+\tilde{u}+\mu_{0} \varphi _{0}\ge16 \sin^{4} \frac{\pi}{2T} L(u_{0}+\tilde{u})+\mu_{0} \varphi_{0}\ge \mu_{0} \varphi_{0}. $$
(3.12)
Define
$$\mu^{*}=\sup\{\mu>0:u_{0}+\tilde{u}\ge\mu\varphi_{0}\}. $$
Note that \(\mu_{0}\in\{\mu>0:u_{0}+\tilde{u}\ge\mu\varphi_{0}\} \), and then \(\mu^{*}\ge\mu_{0}\), \(u_{0}+\tilde{u}\ge\mu^{*} \varphi_{0}\). From (2.4) we have
$$16 \sin^{4} \frac{\pi}{2T} L(u_{0}+\tilde{u})\ge\mu^{*} 16 \sin ^{4} \frac{\pi}{2T} L \varphi_{0}=\mu^{*} \varphi_{0} , $$
and hence
$$u_{0}+\tilde{u}\ge16 \sin^{4} \frac{\pi}{2T} L(u_{0}+\tilde {u})+\mu_{0} \varphi_{0}\ge\bigl( \mu_{0}+\mu^{*}\bigr)\varphi_{0}, $$
which contradicts the definition of \(\mu^{*}\). Therefore, (3.5) holds, and from Lemma 2.3 we obtain
$$ \deg(I-A,B_{R},0)=0. $$
(3.13)
On the other hand, from (H4), there exist \(\varepsilon_{1}\in(0,16 \sin ^{4} \frac{\pi}{2T})\) and \(r\in(0,R)\) such that
$$ \bigl\vert f(t,u) \bigr\vert \le{ \biggl(16 \sin^{4} \frac {\pi}{2T}-\varepsilon_{1} \biggr)|u|}, \quad \forall t\in\mathbb {T}_{2}^{T}, |u|< r. $$
(3.14)
Now for this r, we show that
$$ Au\neq \mu u,\quad u\in\partial B_{r}, \mu\ge1. $$
(3.15)
Otherwise, there would exist \(u_{1}\in\partial B_{r}\), \(\mu_{1}\ge1\) such that
$$\begin{aligned} \bigl\vert u_{1}(t) \bigr\vert &=\frac{1}{\mu_{1}} \bigl\vert (Au_{1}) (t) \bigr\vert \le \bigl\vert (Au_{1}) (t) \bigr\vert \\ &= \Biggl\vert \sum_{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) f \bigl(j,u_{1}(j)\bigr) \Biggr\vert \\ & \le\sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \bigl\vert f\bigl(j,u_{1}(j) \bigr) \bigr\vert \\ & \le{ \biggl(16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{1} \biggr) } \sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \bigl\vert u_{1}(j) \bigr\vert . \end{aligned}$$
Multiplying both sides of the above inequality by \(\sin\frac{\pi (t-1)}{T}\), then summing from 2 to T, and using (2.5), we obtain
$$\begin{aligned} &\sum_{t=2}^{T} \bigl\vert u_{1}(t) \bigr\vert \sin\frac{\pi(t-1)}{T} \\ &\quad \le{ \biggl(16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{1} \biggr)} \sum _{t=2}^{T} \Biggl[\sum _{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) \bigl\vert u_{1}(j) \bigr\vert \Biggr]\sin \frac {\pi(t-1)}{T} \\ &\quad =\frac{16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{1}}{16 \sin^{4} \frac{\pi}{2T}} \sum_{t=2}^{T} \bigl\vert u_{1}(t) \bigr\vert \sin\frac{\pi (t-1)}{T}. \end{aligned}$$
This implies that \(\sum_{t=2}^{T}|u_{1}(t)|\sin\frac{\pi(t-1)}{T}=0\), and whence \(u_{1}(t)\equiv0\), which contradicts \(u_{1}\in\partial B_{r}\). Hence, (3.15) holds, and from Lemma 2.4 we obtain
$$ \deg(I-A,B_{r},0)=1. $$
(3.16)
This, together with (3.13), implies that
$$\deg(I-A,B_{R}\setminus \overline{B}_{r},0)= \deg(I-A,B_{R},0)-\deg(I-A,B_{r},0)=-1. $$
Therefore, the operator A has at least one fixed point in \(B_{R}\setminus \overline{B}_{r}\), and (1.1) has at least one nontrivial solution. This completes the proof. □

Theorem 3.2

Suppose that (H5)(H6) hold. Then (1.1) has at least one nontrivial solution.

Proof

From (H5), there are \(\varepsilon_{2}\in(0, 16 \sin^{4} \frac{\pi}{2T})\) and \(r>0\) such that
$$f(t,u)\ge \biggl(16 \sin^{4} \frac{\pi}{2T}+\varepsilon_{2} \biggr)u,\quad \forall u\in[0,r], t\in\mathbb {T}_{2}^{T}, $$
and
$$f(t,u)\ge \biggl(16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{2} \biggr)u,\quad \forall u\in[-r,0], t\in\mathbb {T}_{2}^{T}. $$
The above two inequalities enable us to obtain
$$\begin{aligned}& f(t,u)\ge \biggl(16 \sin^{4} \frac{\pi }{2T}+ \varepsilon_{2} \biggr)u, \quad \forall u\in[-r,r], t\in\mathbb {T}_{2}^{T}, \end{aligned}$$
(3.17)
$$\begin{aligned}& f(t,u)\ge \biggl(16 \sin^{4} \frac{\pi }{2T}- \varepsilon_{2} \biggr)u,\quad \forall u\in[-r,r], t\in\mathbb {T}_{2}^{T}. \end{aligned}$$
(3.18)
Define a cone \(P_{1}\) as follows:
$$P_{1}= \Biggl\{ u\in P: \sum_{t=2}^{T} u(t)\sin\frac{\pi(t-1)}{T}\ge \delta\|u\| \Biggr\} , $$
where \(\delta=\sum_{t=2}^{T} e(t) \sin\frac{\pi(t-1)}{T} \). Then we claim
$$ L(P)\subset P_{1}. $$
(3.19)
Indeed, for \(u\in P\), from Lemma 2.1 we have
$$\begin{aligned} \sum_{t=2}^{T} (Lu) (t)\sin \frac{\pi(t-1)}{T}&= \sum_{t=2}^{T} \sum _{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) u(j) \sin\frac{\pi (t-1)}{T} \\ & \ge\sum_{t=2}^{T} \sum _{s=2}^{T} e(t) H(\tau,s)\sum _{j=2}^{T} H(s,j) u(j) \sin\frac{\pi(t-1)}{T} \\ & = \delta(Lu) (\tau),\quad \forall\tau\in\mathbb {T}_{2}^{T}, \end{aligned}$$
and thus
$$\sum_{t=2}^{T} (Lu) (t)\sin \frac{\pi(t-1)}{T}\ge\delta\|Lu\|. $$
Moreover, \(\varphi_{0}\in P_{1}\) since \(\varphi_{0}=16 \sin^{4} \frac{\pi }{2T}L\varphi_{0}\in P_{1} \). Now we claim that
$$ u-Au\neq \mu\varphi_{0}, \quad \forall u\in \partial B_{r}, \mu\ge0. $$
(3.20)
If the claim is false, then there exist \(u_{2}\in\partial B_{r}\) and \(\mu _{2}\ge0\) such that
$$ u_{2}-Au_{2} = \mu_{2} \varphi_{0}. $$
(3.21)
From (3.17) we have \(Au_{2}\ge(16 \sin^{4} \frac{\pi }{2T}+\varepsilon_{2}) Lu_{2}\) and so \(u_{2}\ge(16 \sin^{4} \frac{\pi }{2T}+\varepsilon_{2}) Lu_{2}\), i.e.,
$$u_{2}(t)\ge \biggl(16 \sin^{4} \frac{\pi}{2T}+ \varepsilon_{2} \biggr)\sum_{s=2}^{T} H(t,s)\sum_{j=2}^{T} H(s,j) u_{2}(j). $$
Multiplying both sides of the above inequality by \(\sin\frac{\pi (t-1)}{T}\), then summing from 2 to T, and using (2.5), we obtain
$$\begin{aligned} &\sum_{t=2}^{T} u_{2}(t)\sin \frac{\pi(t-1)}{T} \\ &\quad \ge{ \biggl(16 \sin^{4} \frac{\pi}{2T}+ \varepsilon_{2} \biggr)} \sum_{t=2}^{T} \Biggl[\sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) u_{2}(j) \Biggr]\sin \frac{\pi (t-1)}{T} \\ &\quad = \frac{16 \sin^{4} \frac{\pi}{2T}+\varepsilon_{2}}{16 \sin^{4} \frac{\pi}{2T}} \sum_{t=2}^{T} u_{2}(t)\sin\frac{\pi (t-1)}{T}, \end{aligned}$$
which implies that
$$ \sum_{t=2}^{T} u_{2}(t)\sin\frac{\pi (t-1)}{T}\le0. $$
(3.22)
On the other hand, from (3.21) we have
$$\begin{aligned} &u_{2}(t)-{ \biggl(16 \sin^{4} \frac{\pi}{2T}- \varepsilon _{2} \biggr)}(Lu_{2}) (t) \\ &\quad =(Au_{2}) (t)-{ \biggl(16 \sin^{4} \frac{\pi }{2T}-\varepsilon_{2} \biggr)}(Lu_{2}) (t)+\mu_{2} \varphi_{0}(t) \\ &\quad =\sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \biggl[f\bigl(j,u_{2}(j) \bigr)- { \biggl(16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{2} \biggr)} u_{2}(j) \biggr]+\mu_{2} \varphi_{0}(t). \end{aligned}$$
Then (3.18), (3.19) and \(\varphi_{0}\in P_{1}\) enable us to find \(u_{2}-(16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{2})Lu_{2}\in P_{1}\), and thus
$$\begin{aligned} &\biggl\Vert u_{2}-{ \biggl(16 \sin^{4} \frac{\pi}{2T}-\varepsilon _{2} \biggr)}Lu_{2} \biggr\Vert \\ &\quad \le\frac{1}{\delta} \sum_{t=2}^{T} \biggl[u_{2}(t)- \biggl(16 \sin^{4} \frac{\pi}{2T}- \varepsilon_{2} \biggr) (Lu_{2}) (t) \biggr] \sin \frac{\pi(t-1)}{T} \\ &\quad =\frac{\varepsilon_{2}}{\delta16 \sin^{4} \frac{\pi}{2T}}\sum_{t=2}^{T} u_{2}(t)\sin\frac{\pi(t-1)}{T}\le0. \end{aligned}$$
Note that \((16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{2})r(L)<1\), where \(r(L)\) is the spectral radius of L. Hence, we have \(u_{2}=0\), contradicting \(u_{2}\in\partial B_{r}\). This implies that (3.20) holds, and from Lemma 2.3 we have
$$ \deg(I-A,B_{r},0)=0. $$
(3.23)
On the other hand, from (H6) there exist \(\varepsilon_{3}\in(0,16 \sin ^{4} \frac{\pi}{2T})\) and \(c_{3}>0\) such that
$$ \bigl\vert f(t,u) \bigr\vert \le \biggl(16 \sin^{4} \frac{\pi }{2T}-\varepsilon_{3} \biggr) \vert u \vert +c_{3},\quad \forall t\in\mathbb {T}_{2}^{T}, u\in \mathbb {R}. $$
(3.24)
Let \(\mathcal {M}=\{u\in E: u=\lambda Au, \lambda\in[0,1]\}\). Then we prove that \(\mathcal {M}\) is bounded in E. If \(u\in\mathcal {M}\), then from (3.24) we have
$$\begin{aligned} \bigl\vert u(t) \bigr\vert &=\lambda \bigl\vert (Au) (t) \bigr\vert \le \sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \bigl\vert f\bigl(j,u(j)\bigr) \bigr\vert \\ &\le\sum_{s=2}^{T} H(t,s)\sum _{j=2}^{T} H(s,j) \biggl[ \biggl(16 \sin^{4} \frac{\pi}{2T}-\varepsilon_{3} \biggr) \bigl\vert u(j) \bigr\vert +c_{3} \biggr]. \end{aligned}$$
Multiplying both sides of the above inequality by \(\sin\frac{\pi (t-1)}{T}\), then summing from 2 to T, and using (2.5), we obtain
$$\sum_{t=2}^{T} \bigl\vert u(t) \bigr\vert \sin\frac{\pi(t-1)}{T}\le\frac{1}{16 \sin^{4} \frac{\pi}{2T}} \sum _{t=2}^{T} \biggl[ \biggl(16 \sin^{4} \frac{\pi }{2T}-\varepsilon_{3} \biggr) \bigl\vert u(t) \bigr\vert +c_{3} \biggr] \sin\frac{\pi(t-1)}{T}, $$
and then
$$\sum_{t=2}^{T} \bigl|u(t)\bigr|\sin\frac{\pi(t-1)}{T} \le c_{3}\varepsilon _{3}^{-1}\sum _{t=2}^{T} \sin\frac{\pi(t-1)}{T}. $$
We know that there is a \(t_{0}\in\mathbb {T}_{2}^{T}\) such that \(\|u\| =|u(t_{0})|\), and thus
$$\bigl\vert u(t_{0}) \bigr\vert \sin\frac{\pi(t_{0}-1)}{T}\le\sum _{t=2}^{T} \bigl\vert u(t) \bigr\vert \sin\frac {\pi(t-1)}{T}. $$
This implies that
$$\|u\|\le c_{3}\varepsilon_{3}^{-1} \sin^{-1}\frac{\pi(t_{0}-1)}{T}\sum_{t=2}^{T} \sin\frac{\pi(t-1)}{T}, $$
proving the boundedness of \(\mathcal {M}\). Choose \(R>\max\{\sup_{u\in \mathcal {M}} \|u\|, r\} \) (r is defined by (3.17)), then
$$ \lambda Au\neq u,\quad u\in\partial B_{R}, \lambda \in[0, 1]. $$
(3.25)
Lemma 2.4 implies that
$$ \deg(I-A,B_{R},0)=1. $$
(3.26)
This, together with (3.23), implies that
$$\deg(I-A,B_{R}\setminus \overline{B}_{r},0)= \deg(I-A,B_{R},0)-\deg(I-A,B_{r},0)=1. $$
Therefore, the operator A has at least one fixed point in \(B_{R}\setminus \overline{B}_{r}\), and (1.1) has at least one nontrivial solution. This completes the proof. □

Example 3.3

Let \(f(t,x)= a|x|-bk(x)\), \(k(x)=\ln(|x|+1)\), \(x\in\mathbb {R}\), where \(a\in(16 \sin^{4} \frac{\pi}{2T}, +\infty)\) and \(b\in(0, a+16 \sin^{4} \frac{\pi }{2T})\). Then \(\lim_{|x|\to+\infty}\frac{k(x)}{|x|}=0\), and \(\lim_{|x|\to+\infty} \frac{a|x|-b\ln(|x|+1)}{|x|}=a>16 \sin^{4} \frac {\pi}{2T}\), \(\lim_{|x|\to0} \frac{|a|x|-b\ln(|x|+1)|}{|x|} =|a-b|<16 \sin^{4} \frac{\pi}{2T} \). Therefore, (H1)–(H4) hold.

Example 3.4

Let \(f(t,x)=\scriptsize{ \bigl \{ \begin{array}{l@{\quad}l} ax+b \sin x,& x\ge0, \\ ax-be^{x}+b, &x\le0, \end{array} \bigr .} \) where \(a,b>0\) with \(a<16 \sin^{4} \frac{\pi}{2T}\), \(a+b> 16 \sin^{4} \frac{\pi}{2T} \) and \(a-b<16 \sin^{4} \frac{\pi}{2T}\). Then \(\lim_{x\to0^{+}} \frac{ax+b \sin x}{x}=a+b\), \(\lim_{x\to 0^{-}}\frac{ax-be^{x}+b}{x}=a-b\), \(\lim_{x\to+\infty} \vert \frac{ax+b \sin x}{x} \vert =a\), and \(\lim_{x\to-\infty} \vert \frac{ax-be^{x}+b}{x} \vert =a\). Therefore, (H5)–(H6) hold.

4 Conclusions

In this paper, we established the existence of nontrivial solutions for the boundary value problems of the fourth order difference equation (1.1) with sign-changing nonlinearity using the topological degree theory. Under some conditions concerning the first eigenvalue corresponding to the relevant linear problem, the results here improve and generalize those obtained in [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11].

Notes

Acknowledgements

The authors are grateful to the referees for their valuable suggestions and comments.

Availability of data and materials

Not applicable.

Authors’ contributions

This paper is the result of joint work of all authors who contributed equally to the final version of this paper. All authors read and approved the final manuscript.

Funding

This work is supported by Natural Science Foundation of Shandong Province (ZR2018MA009, ZR2015AM014).

Competing interests

The authors declare that they have no competing interests.

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Authors and Affiliations

  1. 1.School of MathematicsQilu Normal UniversityJinanChina
  2. 2.School of Mathematics, Statistics and Applied MathematicsNational University of IrelandGalwayIreland
  3. 3.College of Mathematics and System SciencesShandong University of Science and TechnologyQingdaoChina

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