Bifurcation Patterns in Homogeneous Area-Preserving Piecewise-Linear Maps

Abstract

The dynamical behavior of a family of planar continuous piecewise linear maps with two zones is analyzed. Assuming homogeneity and preservation of areas we obtain a canonical form with only two parameters: the traces of the two matrices defining the map. It is shown the existence of sausage-like structures made by lobes linked at the nodes of a nonuniform grid in the parameter plane. In each one of these structures, called resonance regions, the rotation number of the associated circle map is a given rational number. The boundary of the lobes and a significant inner partition line are studied with the help of some Fibonacci polynomials.

Appendix A. Generalized Fibonacci Polynomials and Power of Matrices

The generalized Fibonacci polynomials were introduced in [7] as,

$$\begin{aligned} u_n(x,y)=xu_{n-1}(x,y)+ y u_{n-2}(x,y), \quad u_0(x,y)=0, u_1(x,y)=1. \end{aligned}$$

(A.1)

By using induction we can prove prove that

$$\begin{aligned} u_n(x,y)=\dfrac{\sigma ^n-(-y)^n \sigma ^{-n}}{\sigma +y\sigma ^{-1}}, \quad \text {where} \quad \sigma (x,y)=\dfrac{x-\sqrt{x^2+4y}}{2}. \end{aligned}$$

In the same spirit as that in Proposition 13 in [5], we obtain the derivative of $u_n$ with respect to the variable x as

$$\begin{aligned} u'_n(x,y)=\dfrac{(n-1) x u_{n}+2 n y u_{n-1}}{x^2+4y}=\dfrac{2n u_{n+1}- ( n+1) x u_{n}}{x^2+4y} \end{aligned}$$

(A.2)

Now, we are in position of give explicit expressions for the powers of a $2 \times 2$ matrix. First let us introduce the family of polynomials, closely related to the Fibonacci generalized polynomials, defined by the recursion

$$\begin{aligned} v_n(x,y)=x v_{n-1}(x,y)+y v_{n-2}(x,y), \quad v_0(x,y)=\beta , \quad v_1(x,y)=\alpha \end{aligned}$$

(A.3)

By induction it is direct to show that

$$\begin{aligned} v_{n}(x,y)=\alpha u_n(x,y)+\beta y u_{n-1}(x,y). \end{aligned}$$

(A.4)

Proposition A.1

If we denote by T and D the trace and the determinant of the $2 \times 2$ matrix $A= (a_{ij}),$ and $u_k=u_k(T,-D)$ the $k-$degree Fibonacci generalized polynomial, then

$$\begin{aligned}A^n= \left( \begin{array}{ll} a_{11}u_n-Du_{n-1} &{}\quad a_{12}u_n\\ a_{21}u_n &{}\quad a_{22}u_n-Du_{n-1} \end{array} \right) . \end{aligned}$$

Proof

From the Cayley Hamilton Theorem we have $A^n=T A^{n-1}- D A^{n-2}.$ for $n\geqslant 2.$ Then the entries of the matrix $A^n=(a_{ij}^{(n)}),$ satisfy the recursion

$$\begin{aligned} a_{ij}^{(n)}=T a_{ij}^{(n-1)}-D a_{ij}^{(n-2)} \quad \text {with} \quad \begin{array}{ll} a_{11}^{(0)}=1,&{}\quad a_{11}^{(1)}=a_{11},\\ a_{12}^{(0)}=0, &{}\quad a_{12}^{(1)}=a_{12},\\ a_{21}^{(0)}=0, &{}\quad a_{21}^{(1)}=a_{21},\\ a_{22}^{(0)}=1, &{}\quad a_{22}^{(1)}=a_{22}. \end{array} \end{aligned}$$

and from (A.3) and (A.4) the conclusion follows. $\square$

Next we consider some results about the powers of the matrix

$$\begin{aligned} \begin{array}{l} A(T)=\left( \begin{array}{ll} T &{}\quad -\,1\\ 1 &{}\quad 0 \end{array} \right) . \end{array} \end{aligned}$$

(A.5)

In order to facilitate the study we introduce the function

$$\begin{aligned} \Psi _n(T)=u_{n}(T,-1), \end{aligned}$$

(A.6)

where $u_n(x,y)$ is the generalized Fibonacci polynomial, that is

$$\begin{aligned} u_{n+1}(x,y)=x u_{n}(x,y)+y u_{n-1}(x,y), \quad \text {with}\quad u_0(x,y)=0, u_1(x,y)=1. \end{aligned}$$

Obviously, we have

$$\begin{aligned} \Psi _{n+1}(T)= T \Psi _n(T)- \Psi _{n-1}(T), \quad \text {with}\quad \Psi _0(T)=0,\quad \Psi _1(T)=1. \end{aligned}$$

Moreover, from Corollary 10 in [7], we easily deduce that

$$\begin{aligned} \Psi _{n}(T)=\prod _{k=1}^{n-1}\left( T-2\cos \left( \dfrac{k\pi }{n}\right) \right) \end{aligned}$$

(A.7)

Proposition A.2

The followings statements hold for the matrix (A.5).

• (a) By using the function $\Psi _n(T)$ defined in (A.6) we have

  $$\begin{aligned} A^n(T)=\left( \begin{array}{ll} \Psi _{n+1}(T) &{}\quad -\,\Psi _{n}(T)\\ \Psi _{n}(T) &{}\quad -\,\Psi _{n-1}(T) \end{array} \right) , \end{aligned}$$

• (b) If $\vert T \vert <2,$ we can put $T=2 \cos \alpha$ with $0< \alpha < \pi ,$ and then

  $$\begin{aligned} A^n(2 \cos \alpha )=\dfrac{1}{\sin \alpha }\left( \begin{array}{cc} \sin (n+1)\alpha &{}\quad -\,\sin n \alpha \\ \sin n\alpha &{}\quad -\,\sin (n-1) \alpha \\ \end{array} \right) , \end{aligned}$$

• (c) If $\vert T \vert >2,$ we can put $T=2 \gamma \cosh \alpha$ with $\alpha \in \mathbb {R}, \gamma =\pm 1$ and then

  $$\begin{aligned} A^n(2 \gamma \cosh \alpha )=\dfrac{\gamma ^n}{\sinh \alpha }\left( \begin{array}{ll} \sinh (n+1)\alpha &{}\quad -\,\gamma \sinh n \alpha \\ \gamma \sinh n\alpha &{}\quad -\,\sinh (n-1) \alpha \\ \end{array} \right) . \end{aligned}$$

• (d) If $\vert T \vert =2,$ we can put $T=2 \gamma$ with $\gamma =\pm 1$ and then

  $$\begin{aligned} A^n(2 \gamma )=\gamma ^n\left( \begin{array}{ll} n+1&{}\quad -\,\gamma n \\ \gamma n &{}\quad -\,(n-1) \\ \end{array} \right) . \end{aligned}$$

Proof

(a) The statement follows from Proposition A.1 by taking

$$\begin{aligned} a_{11}=T, \quad a_{12}=-1, \quad a_{21}=1, \quad a_{22}=0, \end{aligned}$$

and using the function $\Psi _n(T),$ see (A.6).

(b) If $T=2 \cos \alpha =e^{i \alpha }+ e^{-i \alpha },$ then $A^n=M D^n M^{-1}$ where

$$\begin{aligned} M=\left( \begin{array}{ll} 1 &{}\quad 1\\ e^{i \alpha } &{}\quad e^{-i \alpha } \end{array} \right) \quad \text {and}\quad D=\left( \begin{array}{ll} e^{-i \alpha } &{}\quad 0\\ 0 &{}\quad e^{i \alpha } \end{array} \right) , \end{aligned}$$

and the statement is straightforward.

(c) If $T=2 \gamma \cosh \alpha ,$ then $A^n=M D^n M^{-1}$ where

$$\begin{aligned} M=\left( \begin{array}{ll} 1 &{}\quad 1\\ \gamma e^{- \alpha } &{}\quad \gamma e^{ \alpha } \end{array} \right) \quad \text {and}\quad D=\left( \begin{array}{ll} \gamma e^{\alpha } &{}\quad 0\\ 0 &{}\quad \gamma e^{- \alpha } \end{array} \right) , \end{aligned}$$

and the statement follows.

Statement (d) can be easily proved by induction. $\square$

Remark A.3

As a by-product of Proposition A.2 we get for $\gamma =\pm 1,$

$$\begin{aligned} \Psi _n(2 \cos \alpha ) = \dfrac{ \sin n \alpha }{\sin \alpha },\quad \Psi _n(2 \gamma \cosh \alpha )= \dfrac{\gamma ^{n+1} \sinh n \alpha }{\sinh \alpha }, \quad \Psi _n(2 \gamma )= \gamma ^{n+1}n. \end{aligned}$$

In addition, by using the notations

$$\begin{aligned} T_{n}=2 \cos \left( \dfrac{\pi }{n}\right) , \quad \widehat{T}_n=T_{n-\frac{1}{2}}=2 \cos \left( \frac{\pi }{n-\frac{1}{2}}\right) \end{aligned}$$

(A.8)

we have

$$\begin{aligned}\begin{array}{lll} \Psi _{n-1}\left( T_n \right) = 1,&{}\quad \Psi _{n}\left( T_n \right) =0, &{}\quad \Psi _{n+1}\left( T_n \right) =-1\\ \\ \Psi _{2n-2}\left( \widehat{T}_{n} \right) = -1, &{}\quad \Psi _{2n-1}\left( \widehat{T}_{n} \right) =0, &{}\quad \Psi _{2n}\left( \widehat{T}_{n} \right) =1. \end{array} \end{aligned}$$

and so, $\quad A^n\left( T_n\right) =-I, \quad A^{2n-1}\left( \widehat{T}_{n}\right) =I.$

The proof of the following lemmata is direct from expression  A.2 and Remark A.3.

Lemma A.4

The two first derivatives of the function $\Psi _n(x),$$n\geqslant 2,$ are

$$\begin{aligned} \Psi '_n(x)= & {} \dfrac{2n \Psi _{n+1}(x)-(n+1) x\Psi _{n}(x)}{x^2-4}, \nonumber \\ \Psi ''_n(x)= & {} \dfrac{2n \Psi '_{n+1}(x)-(n+1) \Psi _{n}(x)-(n+3) x \Psi '_{n}(x)}{x^2-4}. \end{aligned}$$

(A.9)

In particular, we have

$$\begin{aligned} \Psi '_{n-1}(T_n)=\Psi '_{n+1}(T_n)=\dfrac{-n T_n}{T_n^2-4},\quad \Psi '_{n}(T_n)=\dfrac{-2 n}{T_n^2-4}, \end{aligned}$$

and

$$\begin{aligned} \Psi ''_{n}(T_n)=\dfrac{6 n T_n}{(T_n^2-4)^2},\quad \Psi ''_{n \pm 1}(T_n)=\dfrac{ n(T_n^2+8) \mp n^2(T_n^2-4)}{(T_n^2-4)^2}. \end{aligned}$$

The next lemmata deals with the values of some functions $\Psi _k$ and its derivatives at the point $\widehat{T}_n.$

Lemma A.5

The values of the functions $\Psi _{n-2}, \Psi _{n-1}, \Psi _{n}, \Psi _{n+1},$ at the point $\widehat{T}_n$ are,

$$\begin{aligned} \Psi _{n-1}(\widehat{T}_n)=-\Psi _{n}(\widehat{T}_n)=\dfrac{1}{\sqrt{2+\widehat{T}_n}}, \; \Psi _{n-2}\left( \widehat{T}_n\right) =-\Psi _{n+1}(\widehat{T}_n)=\dfrac{1+\widehat{T}_n}{\sqrt{2+\widehat{T}_n}}. \end{aligned}$$

(A.10)

and its derivatives at the point $\widehat{T}_n$ are

$$\begin{aligned} \Psi '_{n-2}\left( \widehat{T}_n\right)= & {} \dfrac{(n-1)(\widehat{T}_n^2+\widehat{T}_n-2)+2}{\widehat{T}_n^2-4}\Psi _n\left( \widehat{T}_n\right) , \nonumber \\ \Psi '_{n-1}\left( \widehat{T}_n\right)= & {} \dfrac{2n-2+n \widehat{T}_n}{\widehat{T}_n^2-4}\Psi _n\left( \widehat{T}_n\right) ,\nonumber \\ \Psi '_{n}\left( \widehat{T}_n\right)= & {} \dfrac{2n+(n-1)\widehat{T}_n}{\widehat{T}_n^2-4}\Psi _n\left( \widehat{T}_n\right) ,\nonumber \\ \Psi '_{n+1}\left( \widehat{T}_n\right)= & {} \dfrac{n(\widehat{T}_n^2+\widehat{T}_n-2)-2}{\widehat{T}_n^2-4}\Psi _n\left( \widehat{T}_p\right) . \end{aligned}$$

(A.11)

Proof

We start by considering the equality,

$$\begin{aligned} A^{2n-1}(\hat{T}_{n})=A^{n-1}(\widehat{T}_{n})A^{n}(\widehat{T}_{n})=I. \end{aligned}$$

Then from (3.2), by dropping the functional dependence, we get

$$\begin{aligned}\begin{array}{ll} \Psi _{n} \Psi _{n+1}-\Psi _{n-1} \Psi _{n}=1, &{}\quad \Psi ^2_{n-1}-\Psi ^2_{n}=0, \\ \Psi _{n-1} \Psi _{n+1}-\Psi _{n-2} \Psi _{n}=0, &{}\quad \Psi _{n-2} \Psi _{n-1}-\Psi _{n-1} \Psi _{n}=1, \end{array} \end{aligned}$$

and by resolving the above equations we obtain A.10.

Finally, from Lemma A.4, we compute the derivatives (A.11). $\square$