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What we need to find out the periods of a periodic difference equation. (English) Zbl 1180.39021
The author studies a set of periods of a $$p$$-periodic difference equation satisfying a generic condition.
Consider a set $$X$$, a sequence $$(f_n)_{n\in\mathbb{Z}^+}$$ of functions $$f_n: X\to X$$, $$n\in\mathbb{Z}^+$$, a sequence $$(x_n)_{n\in\mathbb{Z}^+}$$ of elements of $$X$$ and a difference equation
$(\forall n\in\mathbb{Z}^+)\;(x_{n+1}= f_n(x_n)).\tag{1}$ A sequence $$(s_n)_{n\in\mathbb{Z}^+}$$ is called a solution of (1) if
$(\forall n\in\mathbb{Z}^+)\;(s_{n+1}= f_n(s_n)).$ If $$q\in\mathbb{Z}^+$$, a solution $$(s_n)_{n\in\mathbb{Z}^+}$$ of (1) is called $$q$$-periodic if $$q$$ is the smallest positive integer satisfying
$(\forall i\in\mathbb{Z}^+)\;(s_{i+q}= s_i).$ A positive integer $$q$$ is called a period of (1) if there exists a $$q$$-periodic solution of (1). Denote by $${\mathcal P}$$ the set of periods of equation (1). If $$p\in{\mathcal P}$$, the equation (1) is called $$p$$-periodic if $$p$$ is the smallest positive integer satisfying
$(\forall n\in\mathbb{Z}^+)\;(f_{n+p}= f_n).$ Denote by $$Q= \{n\in{\mathcal P}; p\nmid n\}$$ the set of periods of $$p$$-periodic equation (1) which are not divisible by $$p$$.
If (1) is a $$p$$-periodic difference equation suppose
$(\forall(i,j)\in\mathbb{Z}^+\times \mathbb{Z}^+)(i\neq j\pmod p)(\text{card}\{x\i X; f_i(x)= f_j(x)\}<+\infty).\tag{2}$ The author studies the set $$Q$$ of the $$p$$-periodic difference equation (1) satisfying the generic condition (2).
Define $$g_0= \text{id}_X$$ and
$$(\forall n\in \mathbb{Z}^+)$$ $$(g_n= f_{n-1}\circ g_{n-1})$$,
$$(\forall m\in\mathbb{Z}^+)$$ $$(\forall(i,j)\in \mathbb{Z}^+\times \mathbb{Z}^+)(j= i\pmod m)(X_m(i)=\{x\in X; f_i(x)= f_j(x)\})$$,
$$(\forall m\in\mathbb{Z}^+)$$ $$(X_m= \bigcap^{m-1}_{i= 0} g^{-1}_i(X_m(i)))$$
and the functions $$h_m: X_m\to X$$, $$m\in\mathbb{Z}^+$$, the restrictions of $$g_m$$ to $$X_m$$, $$m\in\mathbb{Z}^+$$. Let $$d$$ be a divisor of $$p$$ satisfying $$1\leq d< p$$.
Define the matrix $$M_d$$ putting $$M_d= 0$$ if $$X_d=\emptyset$$ and
$M_d(i,j)= \begin{cases} 0, &h_d(x_j)\neq x_i,\\ 1, &h_d(x_j)= x_i\end{cases}$ for $$(i,j)\in \mathbb{Z}^+\times \mathbb{Z}^+$$, if $$X_d= \{x_1,\dots, x_k\}$$.
If $$I$$ denotes the identity matrix define the polynomial $$P_d(z)= \text{det}(I- zM_d)$$. Define
${\mathcal R}_d= \{q\in \mathbb{Z}^+;\;\text{gcd}(p, q)= d\wedge(1- z^{qd^{-1}})\mid P_d(z)\},$
${\mathcal R}= \bigcup{\mathcal R}_d,$
$(1\leq i\leq k)\,(d_i= \text{gcd}(p, r_i)\wedge b_i= \text{tr\,}M_{d_i}^{r_i d^{-1}_i})$ if $$\{r_1,\dots, r_k\}\subset{\mathcal R}$$ and consider the linear system
$A\begin{pmatrix} \alpha_1\\ \vdots\\ \alpha_k\end{pmatrix}= \begin{pmatrix} b_1\\ \vdots\\ b_k\end{pmatrix},\tag{3}$ where $$A$$ is a $$k\times k$$ matrix defined by
$A(i,j)= \begin{cases} 1,\quad & r_j|r_i,\\ 0,\quad & \text{otherwise},\end{cases}\qquad (i,j)\in \mathbb{Z}^+\times \mathbb{Z}^+.$ The author proves the following results:
(i)
the set $$Q$$ is finite,
(ii)
if $$q\in Q$$ and $$d= \text{gcd}(p,q)< p$$ then the polynomial $$1- z^{qd^{-1}}$$ divides $$P_d(z)$$,
(iii)
if $$(\alpha_1,\dots, \alpha_k)$$ is a unique solution of (3) then the number of $$r_i$$-periodic solutions of (1) is $$\alpha_i$$ and $$Q= \{r_i\in{\mathcal R}$$; $$(1\leq i\leq k)$$ $$(\alpha_i\neq 0)\}$$.

##### MSC:
 39A23 Periodic solutions of difference equations
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##### References:
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