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

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.

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, [111] and the references therein. However, as mentioned in [6], very few results are available with sign-changing nonlinearities; see [611]. Other related work in this field can be found in [1245] 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.

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 function H has 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

Let E be a Banach space and Ω a bounded open set in E. 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

Let E be a Banach space and Ω a bounded open set in E with \(0\in\Omega\). 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 \(\deg(I-A,\Omega,0)=1\).

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.

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 [111].

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Acknowledgements

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

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

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Zhang, K., O’Regan, D. & Fu, Z. Nontrivial solutions for boundary value problems of a fourth order difference equation with sign-changing nonlinearity. Adv Differ Equ 2018, 370 (2018). https://doi.org/10.1186/s13662-018-1840-3

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Keywords

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