The Periodic Orbits of a Dynamical System Associated with a Family of QRT-Maps

Abstract

We study the QRT-maps associated with the family of biquadratic curves $C_d(K)$ with equations $x^2y^2 - dxy - 1 + K(x^2 + y^2) = 0$. With the Prime Number Theorem and the geometry of elliptic cubics we determine the periods of periodic orbits of the dynamical systems defined by these QRT-maps, and prove sensitivity to its initial conditions.

Appendices

Appendix 1: 4 and 6 are Minimal Periods

(a) We will find a necessary and sufficient condition for 4 being a minimal period.

Let ${\mathcal {V}}$ and ${\mathcal {H}}$ be the involutions defined by vertical [resp. horizontal] alignment of two points on the same curve ${\mathcal {C}}_d(K)$, so that we have $T_d={\mathcal {V}}\circ {\mathcal {H}}$ and $T_d^{-1}={\mathcal {H}}\circ {\mathcal {V}}$ (excepted for the points $M_0$, $M_1$, $N_0$, $N_1$).

From Proposition 11 or Corollary 12, we see that $T_d$ is 4-periodic on ${\mathcal {C}}_d(K)$ if and only if (with an exception that we will see later) $A_0:= (m,m)$ is 4-periodic for $T_d$ (m is given by relation (7)), or the same condition for $B_0:=(-m,-m)$. The condition $T_d^4(A_0)=A_0$ can be written

$$\begin{aligned} {\mathcal {V}}\circ {\mathcal {H}}\circ {\mathcal {V}}\circ {\mathcal {H}}(A_0)={\mathcal {H}}\circ {\mathcal {V}}\circ {\mathcal {H}}\circ {\mathcal {V}}(A_0). \end{aligned}$$

(61)

But these two points are symmetric with respect to the diagonal (because the curve is symmetric). So they can be equal if and only if they are both equal to $A_0$ or to $B_0$. If they are equal to $A_0$, we would have $T_d^2(A_0)=A_0$, and $T_d$ would be 2-periodic, which is false. So we must have the equality

$$\begin{aligned} T_d(A_0)=T_d^{-1}(B_0). \end{aligned}$$

(62)

If we put $T_d(A_0)=(X,Y)$ and $T_d^{-1}(B_0)=(X_1,Y_1)$, the condition (61) is $X-X_1=0$ and $Y-Y_1=0$, and the easy (with Maple) computation gives

$$\begin{aligned} d=\frac{\sqrt{2}(K^2+1)}{\sqrt{K^2-1}}. \end{aligned}$$

(63)

As an example, for $d=5$ and $K=3$, $T_5$ is 4-periodic on the curve ${\mathcal {C}}_5(3)$.

An exception happens when the two denominators (which are the same ones) of $X-X_1$ and $Y-Y_1$ are 0, that is when

$$\begin{aligned} m^{12}-5dm^{10}+3(d^2+1)m^8-(d^3-2d)m^6+3(d^2+1)m^4+3dm^2+1=0.\nonumber \\ \end{aligned}$$

(64)

But it is easy to see that relations (64) and (62) are incompatible. So this exception is not a curve on which $T_d$ is 4-periodic.

It is also interesting to note that the condition that ${\mathcal {H}}(A_0)=N_0$, so that $Y=\infty$, id est $T_d(A_0)=V$, is exactly the relation (61), and by symmetry it is also the relation for having $T_d^{-1}(B_0)=H$. So we find again the previous exception.

(b) Now we will find a condition for 6 being a minimal period.

With the same reasoning we find that the condition is

$$\begin{aligned} ({\mathcal {V}}\circ {\mathcal {H}})^{3\circ }(A_0)=({\mathcal {H}}\circ {\mathcal {V}})^{3\circ }(A_0), \end{aligned}$$

(65)

so that, by symmetry, these to points must be equal to $A_0$ or to $B_0$. If it would $A_0$, $T_d$ would be 3-periodic, and 6 would not be minimal. So the good condition is

$$\begin{aligned} T_d^2(A_0)=T_d^{-1}(B_0). \end{aligned}$$

(66)

It is now easy, with the use of Maple, and by computing $T_d^2(A_0)=(X_2,Y_2)$, to find the condition for 6 being a minimal period (at least, it is a sufficient condition)

$$\begin{aligned} K^2d^4-(K^4-1)(K^2+1)d^2+3(K^2+1)^4=0, \end{aligned}$$

(67)

which gives easily d as a function of K

$$\begin{aligned} d=\frac{(K^2+1)\sqrt{K^2-1\pm \sqrt{K^4-14K^2+1}}}{K\sqrt{2}}. \end{aligned}$$

(68)

As an example, for $d=4\sqrt{3+2\sqrt{3}}$ and $K=2+\sqrt{3}$, we obtain a curve on which 6 is a minimal period.

Appendix 2: On the Forms of the Curves ${\mathcal {C}}_d(K)$ (Proof of Lemma 5)

The plan of proof of Lemma 5 is simple :

(1) The curves are starlike with respect the point (0, 0) : we put $x=\rho u$ and $y=\rho v$, with a unit vector (u, v). If $uv\not =0$ we have with equation of ${\mathcal {C}}_d(K)$ a quadratic equation in $\rho ^2$ with a unique positive solution, so $\rho =\pm \alpha$: the curve is starlike. If $uv = 0$, $\rho =\pm \frac{1}{\sqrt{K}}$ if $K>0$, and if $K\le 0$ and $uv=0$, there is no solution: the curves ${\mathcal {C}}_d(K)$ does not cut the axes.

(2) There is no double point in finite distance (see Lemma 6).

(3) If $K>0$, one see the inclusion ${\mathcal {C}}_d(K)\subset B\Big ((0,0),\sqrt{\frac{1+d^2/4}{K}}\Big )$: one has $\displaystyle x^2+y^2=\frac{1+dxy-x^2y^2}{K}\le \frac{1+d^2/4}{K}$. So, from the three previous points, one see that the curves are homeomorphic to circles for $K>0$.

(4) If $K<0$, it is easy to find the asymptotes of the curve ${\mathcal {C}}_d(K)$. $\square$

Appendix 3: Previous Particular Results of the Authors About QRT-Maps

The invention of QRT-maps was originally in [18], for physical reasons, and the essential exemple was on [22], but these papers were not easy to find. So the first papers ([3, 11] and [4]) were slightly different. They studied the now called “symmetric special QRT-maps”, where the family of biquadratic curves ${\mathcal {C}}(K)$ with equations $Q_1(x,y)-KQ_2(x,y)=0$ were symmetric with respect the diagonal, and with $Q_2(x,y)=xy$; and the QRT-map was defined by the following way: if ${\mathcal {C}}(K)$ is the curve passing through the point M, we cut it in $M_1$ by the horizontal line which contains M, and then T(M) is the symmetric of $M_1$ with respect the diagonal (it is on ${\mathcal {C}}(K)$).

These cases correspond to the study of difference equations of the form

$$\begin{aligned} u_{n+2}u_n=\frac{au_{n+1}^2+bu_{n+1}+c}{du_{n+1}^2+eu_{n+1}+f}. \end{aligned}$$

(69)

In [8] we studied the case of difference equations $u_{n+2}+u_n=\psi (u_{n+1})$ which is associated with “symmetric QRT-maps” defined by families of the forms $Q_1(x,y)+K=0$.

In each of these papers, we determine the possible periods of periodic orbits, prove the density of periodic points and not periodic points, and a form of sensitivity to initial conditions.

In the other works, we studied classical QRT-maps, associated with a couple of difference equations of the form $u_{n+1}u_n=f(v_n), v_{n+1}v_n=g(u_{n+1})$. We give some examples :

$$\begin{aligned}&[7] \quad u_{n+1}u_n=c+\frac{d}{v_n},\,\,v_{n+1}v_n \\&\quad =c+\frac{d}{v_{n+1}} \quad \mathrm{with\,\,curves} \quad xy(x+y)+xy+d-Kxy=0. \\&[5]\,\, u_{n+1}u_n=v_n^2-bv_n+c,\,v_{n+1}v_n\\&\quad =u_{n+1}^2-au_{n+1}+c\,\, \mathrm{with\,\,curves}\,\, x^2+y^2-ax-by+c-Kxy=0. \\&[6]\,\, u_{n+1}u_n=av_n+b,\,v_{n+1}v_n\\&\quad =u_{n+1}+\frac{b}{u_{n+1}}\,\, \mathrm{with\,\,curves}\,\, xy^2+x^2+ay+b-Kxy=0. \end{aligned}$$

In [2] we studied the 2-periodic Lyness’ equation $u_{n+2}u_n=u_{n+1}+a_n$, with $a_n$ 2-periodic.

In [12] we study the particular case of the dynamical systems $(x,y)\mapsto T_d(x,y)= (X,Y)$ given by

$$\begin{aligned} (X,Y)=\Big (\frac{1}{x}\frac{dy^2-20y+16}{y^2-5y+d}, \frac{1}{y}\frac{dX^2-20X+16}{X^2-5X+d}\Big ), \end{aligned}$$

with curves ${\mathcal {C}}_d(K)$ equations of them are

$$\begin{aligned} x^2y^2-5xy(x+y)+d(x^2+y^2)-20(x+y)+16-Kxy=0. \end{aligned}$$

In the present paper we begin study of non-special QRT-dynamical systems, associated with QRT-families of curves with equations $Q_1(x,y)-KQ_2(x,y)=0$ with $Q_2(x,y)$ not of the form xy.

In fine, in [9] and [10] we present examples of QRT-families of degree four, but such that each of the curves of the family has genus zero, and studied the correspondent dynamical systems.