In this section, we construct Green’s function associated with BVP (1.1), (1.2).

Let {r}_{1}, {r}_{2} be roots of the polynomial P(r)={r}^{2}-\alpha r+\beta, namely,

{r}_{1}=\frac{\alpha +\sqrt{{\alpha}^{2}-4\beta}}{2},\phantom{\rule{2em}{0ex}}{r}_{2}=\frac{\alpha -\sqrt{{\alpha}^{2}-4\beta}}{2}.

Then we have

\begin{array}{rl}{\mathrm{\Delta}}^{4}u(t-2)-\alpha {\mathrm{\Delta}}^{2}u(t-1)+\beta u(t)& =(-{\mathrm{\Delta}}^{2}L+{r}_{1})(-{\mathrm{\Delta}}^{2}L+{r}_{2})u(t)\\ =(-{\mathrm{\Delta}}^{2}L+{r}_{2})(-{\mathrm{\Delta}}^{2}L+{r}_{1})u(t),\end{array}

where u=(u(0),u(1),\dots ,u(T+2)), Lu=u(t-1), t\in {[2,T]}_{\mathbb{Z}}. Under the basic assumption on *α*, *β*, it is easy to see that {r}_{1}\ge {r}_{2}>-4{sin}^{2}\frac{\pi}{2(T-1)}.

Consider the two initial value problems:

\{\begin{array}{l}-{\mathrm{\Delta}}^{2}u(t-1)+{r}_{i}u(t)=0,\phantom{\rule{1em}{0ex}}t\in {[2,T]}_{\mathbb{Z}},\\ \mathrm{\Delta}u(1)=0,\phantom{\rule{2em}{0ex}}u(1)=1,\end{array}

(4.1)

\{\begin{array}{l}-{\mathrm{\Delta}}^{2}v(t-1)+{r}_{i}v(t)=0,\phantom{\rule{1em}{0ex}}t\in {[2,T]}_{\mathbb{Z}},\\ \mathrm{\Delta}v(T)=0,\phantom{\rule{2em}{0ex}}v(T)=1.\end{array}

(4.2)

By the direct computing, we get (4.1) has a unique solution

u(t)=\frac{{r}_{2}-1}{{r}_{1}({r}_{2}-{r}_{1})}{r}_{1}^{t}+\frac{1-{r}_{1}}{{r}_{2}({r}_{2}-{r}_{1})}{r}_{2}^{t},

and (4.2) has a unique solution

v(t)=\frac{{r}_{2}-1}{{r}_{1}^{T}({r}_{2}-{r}_{1})}{r}_{1}^{t}+\frac{1-{r}_{1}}{{r}_{2}^{T}({r}_{2}-{r}_{1})}{r}_{2}^{t}.

Let

\rho :=\frac{({r}_{1}-1)({r}_{2}-1)({r}_{2}^{T-1}-{r}_{1}^{T-1})}{{r}_{1}^{T-1}{r}_{2}^{T-1}}.

**Lemma 4.1** *Let* h:{[1,T]}_{\mathbb{Z}}\to \mathbb{R} *and* i\in \{1,2\} *be fixed*. *Then the problem*

\{\begin{array}{l}-{\mathrm{\Delta}}^{2}u(t-1)+{r}_{i}u(t)=h(t),\phantom{\rule{1em}{0ex}}t\in {[2,T]}_{\mathbb{Z}},\\ \mathrm{\Delta}u(1)=\mathrm{\Delta}u(T)=0\end{array}

(4.3)

*has a unique solution*

u(t)=\sum _{s=2}^{T}{G}_{i}(t,s)h(s),\phantom{\rule{1em}{0ex}}t\in {[1,T+1]}_{\mathbb{Z}},

*where*
{G}_{i}(t,s)
*is given by*

{G}_{i}(t,s)=\frac{1}{\rho}\{\begin{array}{ll}u(t)v(s),& 1\le s\le t\le T+1,\\ u(s)v(t),& 1\le t\le s\le T+1.\end{array}

**Remark 4.1** Green’s function {G}_{i}(t,s) defined by Lemma 4.1 is positive on {[1,T]}_{\mathbb{Z}}\times {[1,T]}_{\mathbb{Z}}.

Define operators K,f,A:E\to E, respectively, by

(Ku)(t)=\sum _{s=2}^{T}\sum _{k=2}^{T}{G}_{2}(t,k){G}_{1}(k,s)u(s),\phantom{\rule{1em}{0ex}}u\in E,t\in {[1,T]}_{\mathbb{Z}};

(4.4)

\begin{array}{r}(fu)(t)=f(t,u(t)),\phantom{\rule{1em}{0ex}}u\in E,t\in {[1,T]}_{\mathbb{Z}};\\ A=Kf.\end{array}

(4.5)

Now, from Lemma 4.1, it is easy to see that BVP (1.1), (1.2) has a solution u=u(t) if and only if *u* is a fixed point of the operator *A*. It follows from the continuity of *f* that A:E\to E is completely continuous.

**Lemma 4.2** *Let* h:{[2,T]}_{\mathbb{Z}}\to R. *Then the linear discrete fourth*-*order boundary value problem*

\{\begin{array}{l}{\mathrm{\Delta}}^{4}u(t-2)-\alpha {\mathrm{\Delta}}^{2}u(t-1)+\beta u(t)=h(t),\phantom{\rule{1em}{0ex}}t\in {[2,T]}_{\mathbb{Z}},\\ \mathrm{\Delta}u(1)=\mathrm{\Delta}u(T)={\mathrm{\Delta}}^{3}u(0)={\mathrm{\Delta}}^{3}u(T-1)=0\end{array}

(4.6)

*has a unique solution*

\begin{array}{rl}u(t)& =\sum _{s=2}^{T}(\sum _{k=2}^{T}{G}_{1}(t,k){G}_{2}(k,s)h(s))\\ =\sum _{s=2}^{T}(\sum _{k=2}^{T}{G}_{2}(t,k){G}_{1}(k,s)h(s)),\phantom{\rule{1em}{0ex}}t\in {[2,T]}_{\mathbb{Z}},\end{array}

*and*

u(0)=u(3),\phantom{\rule{2em}{0ex}}u(1)=u(2),\phantom{\rule{2em}{0ex}}u(T)=u(T+1),\phantom{\rule{2em}{0ex}}u(T-1)=u(T+2).

*Proof* The conclusion is obvious, so we omit it. □

We will use the following assumptions.

(H1) *α*, *β* are real parameters and satisfy

{\alpha}^{2}\ge 4\beta ,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\alpha -\sqrt{{\alpha}^{2}-4\beta}>-8{sin}^{2}\frac{\pi}{2(T-1)}.

(H2) f:{[2,T]}_{\mathbb{Z}}\times \mathbb{R}\to \mathbb{R} is continuous, for any t\in {[2,T]}_{\mathbb{Z}}, f(t,0)=0; for any t\in {[2,T]}_{\mathbb{Z}} and x\in \mathbb{R}, xf(t,x)\ge 0.

(H3) There exists an even number {k}_{0}\in {[0,T-2]}_{\mathbb{Z}} such that

\frac{1}{{\lambda}_{{k}_{0}}^{2}}<{\beta}_{0}<\frac{1}{{\lambda}_{{k}_{0}+1}^{2}},

(4.7)

where {lim}_{x\to 0}\frac{f(t,x)}{x}={\beta}_{0} uniformly for t\in {[2,T]}_{\mathbb{Z}}, {\lambda}_{k} is defined in Lemma 3.1 and {\lambda}_{T-1}\triangleq \mathrm{\infty}.

(H4) There exists an even number {k}_{1}\in {[0,T-2]}_{\mathbb{Z}} such that

\frac{1}{{\lambda}_{{k}_{1}}^{2}}<{\beta}_{\mathrm{\infty}}<\frac{1}{{\lambda}_{{k}_{1}+1}^{2}},

(4.8)

where {lim}_{x\to \mathrm{\infty}}\frac{f(t,x)}{x}={\beta}_{\mathrm{\infty}} uniformly for t\in {[2,T]}_{\mathbb{Z}}, and {\lambda}_{0},{\lambda}_{1},\dots ,{\lambda}_{T-1} are given in the condition (H3).

(H5) There exists a constant M>0 such that for any (t,x)\in {[2,T]}_{\mathbb{Z}}\times [-M,M],

|f(t,x)|<{\omega}^{-1}M,

(4.9)

where \omega ={max}_{t\in {[2,T]}_{\mathbb{Z}}}{\sum}_{s=2}^{T}{\sum}_{k=2}^{T}{G}_{2}(t,k){G}_{1}(k,s).

**Lemma 4.3** *Suppose that* (H2) *holds and* u\in P\mathrm{\setminus}\{\theta \} *is a solution of BVP* (1.1), (1.2). *Then* u\in {P}^{\circ}.

*Proof* According to (H2) and the positivity of Green’s function defined in Lemma 4.1, we can easily get the desired conclusion. □

**Remark 4.2** Similarly to Lemma 4.3, we also know that if (H2) holds and u\in (-P)\mathrm{\setminus}\{\theta \} is a solution of BVP (1.1), (1.2), then u\in {(-P)}^{\circ}.

**Lemma 4.4** *Suppose that* (H2)-(H4) *hold*. *Then the operator* *A* *is Fréchet differentiable at* *θ* *and* ∞, *where the operator* *A* *is defined by* (4.5). *Moreover*, {A}^{\prime}(\theta )={\beta}_{0}K *and* {A}^{\prime}(\mathrm{\infty})={\beta}_{\mathrm{\infty}}K.

*Proof* By (H3), for any \u03f5>0, there exists \delta >0 such that |f(t,x)-{\beta}_{0}x|<\u03f5|x| for any 0<|x|<\delta, t\in {[2,T]}_{\mathbb{Z}}. Hence, noticing that f(t,0)=0 for any t\in {[2,T]}_{\mathbb{Z}}, we have

\begin{array}{rl}\parallel Au-A\theta -{\beta}_{0}Ku\parallel & =\parallel K(fu-{\beta}_{0}u)\parallel \\ \le \parallel K\parallel \underset{t\in {[2,T]}_{\mathbb{Z}}}{max}|f(t,u(t))-{\beta}_{0}u(t)|<\u03f5\parallel K\parallel \parallel u\parallel \end{array}

(4.10)

for any u\in E with 0<|u|<\delta, where \parallel K\parallel ={max}_{t\in {[2,T]}_{\mathbb{Z}}}{\sum}_{s=2}^{T}{\sum}_{k=2}^{T}|{G}_{2}(t,k)||{G}_{1}(k,s)|. Consequently,

\underset{\parallel u\parallel \to 0}{lim}\frac{\parallel Au-A\theta -{\beta}_{0}Ku\parallel}{\parallel u\parallel}=0.

(4.11)

This means that the nonlinear operator *A* is Fréchet differentiable at *θ* and {A}^{\prime}(\theta )={\beta}_{0}K.

By (H4), for any \u03f5>0, there exists W>0 such that |f(t,x)-{\beta}_{\mathrm{\infty}}x|<\u03f5|x| for any |x|>W, t\in {[2,T]}_{\mathbb{Z}}. Let c={max}_{(t,x)\in {[2,T]}_{\mathbb{Z}}\times [-W,W]}|f(t,x)-{\beta}_{\mathrm{\infty}}x|. By the continuity of f(t,x) with respect to *x*, we have c<+\mathrm{\infty}. Then, for any (t,x)\in {[2,T]}_{\mathbb{Z}}\times \mathbb{R}, |f(t,x)-{\beta}_{\mathrm{\infty}}x|<\u03f5|x|+c. Thus

\parallel Au-{\beta}_{\mathrm{\infty}}Ku\parallel \le \parallel K\parallel \cdot \underset{t\in {[2,T]}_{\mathbb{Z}}}{max}|f(t,u)-{\beta}_{\mathrm{\infty}}u(t)|<\parallel K\parallel (\u03f5\parallel u\parallel +c)

(4.12)

for any u\in E. Consequently,

\underset{\parallel u\parallel \to \mathrm{\infty}}{lim}\frac{\parallel Au-A\theta -{\beta}_{\mathrm{\infty}}Ku\parallel}{\parallel u\parallel}=0,

(4.13)

which implies that operator *A* is Fréchet differentiable at ∞ and {A}^{\prime}(\mathrm{\infty})={\beta}_{\mathrm{\infty}}K. The proof is completed. □

**Lemma 4.5** *Let* *M* *be given in the condition* (H5). *Suppose that* (H1)-(H4) *hold*. *Then* A(P)\subset P, A(-P)\subset (-P). *Moreover*, *one has the following*.

(i) *There exists an* {r}_{0}\in (0,M) *such that for any* 0<r\le {r}_{0},

i(A,P\cap B(\theta ,r),P)=0,\phantom{\rule{2em}{0ex}}i(A,(-P)\cap B(\theta ,r),-P)=0.

(4.14)

(ii) *There exists an* {R}_{0}>M *such that for any* R\ge {R}_{0},

i(A,P\cap B(\theta ,R),P)=0,\phantom{\rule{2em}{0ex}}i(A,(-P)\cap B(\theta ,R),-P)=0.

(4.15)

*Proof* By (H2) and the fact that {G}_{i}(t,s) is positive on {[2,T]}_{\mathbb{Z}}\times {[2,T]}_{\mathbb{Z}}, we get that for any t\in {[2,T]}_{\mathbb{Z}}, f(t,P)\subset P, f(t,-P)\subset -P, and K(P)\subset P, K(-P)\subset -P. Then A(P)\subset P and A(-P)\subset -P.

We only need to prove conclusion (i). The proof of conclusion (ii) is similar and will be omitted here. Let {\gamma}_{0}={inf}_{\parallel u\parallel =1}\parallel u-{\beta}_{0}Ku\parallel. The condition (H3) yields {\gamma}_{0}>0. It follows from (4.12) that there exists {r}_{0}\in [0,M] such that

\parallel Au-{\beta}_{\mathrm{\infty}}Ku\parallel <\frac{1}{2}{\gamma}_{0}\parallel u\parallel ,

(4.16)

where 0<u\le {r}_{0}. Setting H(s,u)=sAu+(1-s){\beta}_{0}Ku, then H:[0,1]\times E\to E is completely continuous. For any s\in [0,1] and 0<u\le {r}_{0}, we obtain that

\parallel u-H(s,u)\parallel \ge \parallel u-{\beta}_{0}Ku\parallel -s\parallel Au-{\beta}_{0}Ku\parallel \ge {\gamma}_{0}\parallel u\parallel -\frac{1}{2}{\gamma}_{0}\parallel u\parallel >0.

(4.17)

According to the homotopy invariance of the fixed point index, for any 0<r\le {r}_{0}, we have

i(A,P\cap B(\theta ,r),P)=i({\beta}_{0}K,P\cap B(\theta ,r),P),

(4.18)

i(A,-P\cap B(\theta ,r),-P)=i({\beta}_{0}K,-P\cap B(\theta ,r),-P).

(4.19)

Let {\phi}_{0}(t)=1. Then K{\phi}_{0}={\lambda}_{0}^{2}{\phi}_{0} and {\phi}_{0}\in P (see Lemma 3.1 and the proof of Lemma 4.2). We claim

u-{\beta}_{0}Ku\ne \sigma {\phi}_{0},\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in P\cap \partial B(\theta ,r),\sigma \ge 0.

(4.20)

Indeed, we assume that there exist {u}_{1}\in P\cap \partial B(\theta ,r) and {\sigma}_{1}\ge 0 such that {u}_{1}-{\beta}_{0}K{u}_{1}={\sigma}_{1}{\phi}_{0}. Obviously, {u}_{1}={\beta}_{0}K{u}_{1}+{\sigma}_{1}{\phi}_{0}\ge {\sigma}_{1}{\phi}_{0}. Since {\beta}_{0}\ne {\lambda}_{k}^{-2}, k=1,2,\dots ,T-2, then {\sigma}_{1}>0. Set {\sigma}_{\mathrm{max}}=sup\{\sigma :{u}_{1}\ge \sigma {\phi}_{0}\}. It is clear that {\sigma}_{1}\le {\sigma}_{\mathrm{max}}<\mathrm{\infty} and {u}_{1}\ge {\sigma}_{\mathrm{max}}{\phi}_{0}. Then

{u}_{1}={\beta}_{0}K{u}_{1}+{\sigma}_{1}{\phi}_{0}\ge {\beta}_{0}K{\sigma}_{\mathrm{max}}{\phi}_{0}+{\phi}_{0}{\sigma}_{1}=({\beta}_{0}{\lambda}_{0}^{2}{\sigma}_{\mathrm{max}}+{\sigma}_{1}){\phi}_{0}.

(4.21)

Since {\beta}_{0}{\lambda}_{0}^{2}>1, then {\beta}_{0}{\lambda}_{0}^{2}{\sigma}_{\mathrm{max}}+{\sigma}_{1}>{\sigma}_{\mathrm{max}}, which contradicts the definition of {\sigma}_{\mathrm{max}}. This proves (4.21).

It follows from Lemma 2.1 and (4.20) that

i({\beta}_{0}K,P\cap B(\theta ,r),P)=0.

(4.22)

Similarly to (4.22), we know also that

i({\beta}_{0}K,-P\cap B(\theta ,r),-P)=0.

(4.23)

By (4.18), (4.19), (4.22), and (4.23), we conclude

i(A,P\cap B(\theta ,r),P)=0,\phantom{\rule{2em}{0ex}}i(A,-P\cap B(\theta ,r),-P)=0.

(4.24)

□