⎛ a − by − x 2 ⎞
⎟
⎟
x
⎝
⎠
The dynamics of Henon Map H a,b ⎛⎜⎜ ⎞⎟⎟ = ⎜⎜
x
⎝ y⎠
Where
a = 0 , −1 < b < 0
Iftichar Mudar Talb and Kawa Ahmad Hassan
Department of Mathematics, College of Education,
Babylon University.
Abstract
We prove the Henon map H 0,b , − 1 < b < 0 , has no periodic point for any period in the plane,.H 0,b has
⎛ 0⎞
⎛ 0⎞
attracting fixed point ⎜⎜ ⎟⎟ and a saddle fixed point..There exists an open set about ⎜⎜ ⎟⎟ in which all points
⎝ 0⎠
⎝ 0⎠
⎛ 0⎞
tend to ⎜⎜ ⎟⎟ under forward iteration of H 0,b .
⎝ 0⎠
1-Introduction
About 30 years ago the French astronomer –mathematician Michel-Henon was searched for a simple
two-dimensional map possessing special properties of more complicated systems .The result was a family of
⎛ x ⎞ ⎛1 − ax 2 + y ⎞
⎟ where a , b are real numbers .These maps defined in
maps denoted by H a,b given by H a,b ⎜⎜ ⎟⎟ = ⎜⎜
⎟
bx
⎝ y⎠ ⎝
⎠
above are called Henon maps [3]. Henon map H a,b :R 2 ⎯
⎯→ R 2 .The importance of Henon , s map for the
dynamics of all polynomial diffeomorphisms was recognized by S.Friedland and J.Milnor , who proved that
every quadratic polynomial diffeomorphism is conjugate to a Henon map or to an elementary maps have
trivial dynamics .Henon map is normal form for all quadratic polynomial diffeomorphism with non- trivial
dynamic.[4].This work we, care this form of Henon map H a,b
⎛ x ⎞ ⎛ a − by − x 2 ⎞
⎟ .
⎜⎜ ⎟⎟ = ⎜⎜
⎟
y
x
⎝ ⎠ ⎝
⎠
To determine the Henon map H 0,b , − 1 < b < 0 , has no periodic point for any period in the plane by dividing
the plane to some region which are shown in section four
⎛ 0⎞
we conclude that there exists an open set about ⎜⎜ ⎟⎟ in which all points tend to
⎝ 0⎠
⎛ 0⎞
⎜⎜ ⎟⎟ under forward iteration
⎝ 0⎠
of H .We found these regions Q1 , Q2,1 , Q2, 2 , Q3 , Q4 and R1 , R2 , R3 , R4 , R5 , R6 , R7 , we prove there are no
periodic point for Henon map in the plane .Also, we prove that the norm of iteration of Henon map H 0,b
tends to infinity for some regions shown in proposition (4-2) and (4-4).
In section two, we recall some necessary definitions and theorems. In section three,we introduce forward
and backward iteration of Henon map H a,b , where a = 0 ,-1< b < 0 .In section four ,we study the type of
fixed point of Henon map H 0,b with a basin of attraction of one of fixed points .In section five we study non
existence of periodic point of Henon map in R2 by finding some region in R2 .At the last ,we introduce
forward and backward iteration of Henon map H a,b , where a = 0 , b > 0 . we conclude that there exists an
⎛ 0⎞
open set about ⎜⎜ ⎟⎟ in which all points tend to
⎝ 0⎠
⎛ 0⎞
⎜⎜ ⎟⎟ under forward iteration of H ., we prove that the norm of
⎝ 0⎠
iteration of Henon map H 0,b tends to infinity for some regions
2-Preliminaries
The purpose of this section is to introduce some definitions and theorems necessary for this research, we
recall some fundamental definitions and necessary theorems. Let A be an n × n matrix .The real number λ
is called eigen value of A if there exists a non zero vector X in Rn such that AX= λ X
(2.1)
Every non zero vector X satisfying (2.1) is called an eigen vector of A associated with the eigen value
⎡ ⎛ x ⎞⎤
⎢ f ⎜⎜ ⎟⎟⎥
x
⎛
⎞
⎝ y ⎠⎥ ⎛ x ⎞
λ .In our work, we write the mapF: V ⎯
; ⎜⎜ ⎟⎟ ∈ V.Also,the
⎯→ R2 where V ⊆ R2, such that F ⎜⎜ ⎟⎟ = ⎢
⎝ y⎠ ⎢ ⎛ x⎞⎥ ⎝ y⎠
⎢ g ⎜⎜ ⎟⎟ ⎥
⎢⎣ ⎝ y ⎠ ⎥⎦
⎛x ⎞
⎛x ⎞ ⎛x ⎞
forward orbit of a vector ⎜⎜ 0 ⎟⎟ is the set of points ⎜⎜ 0 ⎟⎟ ,F ⎜⎜ 0 ⎟⎟ ,F 2
⎝ y0 ⎠
⎝ y0 ⎠ ⎝ y0 ⎠
⎛ x0 ⎞ 3 ⎛ x0 ⎞
⎜⎜ ⎟⎟ F ⎜⎜ ⎟⎟ ,…and denoted by
⎝ y0 ⎠
⎝ y0 ⎠
⎛x ⎞
O + ⎜⎜ 0 ⎟⎟ .If
⎝ y0 ⎠
⎛ x0 ⎞ ⎛ x0 ⎞
⎜⎜ ⎟⎟ ,O ⎜⎜ ⎟⎟ ,as the set of points
⎝ y0 ⎠ ⎝ y0 ⎠
F a homeomorphism ,we may define the full orbit of
⎛x ⎞
⎛x ⎞
⎛x ⎞
⎛x ⎞
⎛x ⎞
F n ⎜⎜ 0 ⎟⎟ for n ∈ Z ,and the backward orbit of ⎜⎜ 0 ⎟⎟ , O − ⎜⎜ 0 ⎟⎟ , as the set of points F −1 ⎜⎜ 0 ⎟⎟ ,F −2 ⎜⎜ 0 ⎟⎟ ,
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎛x ⎞
F −3 ⎜⎜ 0 ⎟⎟ ,…,where we have F n
⎝ y0 ⎠
⎛ x0 ⎞ n −1 ⎛ x1 ⎞
⎜⎜ ⎟⎟ = F ⎜⎜ ⎟⎟ =
⎝ y1 ⎠
⎝ y0 ⎠
⎛ xn ⎞
⎜⎜ ⎟⎟ .
⎝ yn ⎠
⎛ p⎞
⎛ p⎞
⎛ p⎞
Definition (2-1)[6]Any pair ⎜⎜ ⎟⎟ for which f ⎜⎜ ⎟⎟ = p , g ⎜⎜ ⎟⎟ = q
⎝q⎠
⎝q⎠
⎝q⎠
(2.2)
is called a fixed point of the two dimensional dynamical system
Definition (2-2)[1] Let F: R n ⎯
⎯→ R n be a map, x0 ∈ R n .The point x0 is a periodic point of period m if
F m (x0)= x0.The least positive integer m for which F m (x0)= x0 is called the prime period of x0 .
⎛ x ⎞ ⎛ y⎞
Example (2-3) Let F: R2 ⎯
⎯→ R2 is given by F ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ , then
⎝ y⎠ ⎝ x⎠
Definition (2-4)[2]
⎛ p⎞
Let ⎜⎜ ⎟⎟ be a fixed point of F, then
⎝q⎠
⎛ 1⎞
⎜⎜ ⎟⎟ is a periodic point of period 2 of F
⎝2⎠
⎛ p⎞
⎜⎜ ⎟⎟ is attracting fixed point if and only if there
⎝q⎠
⎛ p⎞
⎛ x⎞
⎛ p⎞
⎛ x⎞
is a disk centered at ⎜⎜ ⎟⎟ such that F(n) ⎜⎜ ⎟⎟ ⎯
⎯→ ⎜⎜ ⎟⎟ for every ⎜⎜ ⎟⎟ in the disk as n ⎯
⎯→ ∞ .By contrast
⎝q⎠
⎝ y⎠
⎝q⎠
⎝ y⎠
⎛u ⎞
⎛ p⎞
⎛ p⎞
⎛ p⎞
⎜⎜ ⎟⎟ is repelling fixed point if and only if there is a disk centered at ⎜⎜ ⎟⎟ such that F ⎜⎜ ⎟⎟ − F ⎜⎜ ⎟⎟ >
⎝v⎠
⎝q⎠
⎝q⎠
⎝q⎠
⎛u ⎞ ⎛ p⎞
⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ for every
⎝v⎠ ⎝ q ⎠
⎛u ⎞
⎜⎜ ⎟⎟ in the disk for which
⎝v⎠
⎛u ⎞ ⎛ p ⎞
⎜⎜ ⎟⎟ ≠ ⎜⎜ ⎟⎟ .
⎝v⎠ ⎝ q ⎠
⎛ p⎞
⎛ p⎞
Theorem (2-5)[2] Let ⎜⎜ ⎟⎟ be a fixed point of F. Assume that DF ⎜⎜ ⎟⎟ exists, with eigen values λ1 , λ 2 , then
⎝q⎠
⎝q⎠
:
⎛ p⎞
(1) ⎜⎜ ⎟⎟ is an attracting fixed point, if λ1 and λ 2 are less than one in absolute value .
⎝q⎠
⎛ p⎞
(2) ⎜⎜ ⎟⎟ is repelling fixed point, if λ1 and λ 2 are greater than one in absolute value .
⎝q⎠
⎛ p⎞
(3) ⎜⎜ ⎟⎟ is saddle point, if one of λ1 , λ 2 is larger and the other is less than one in absolute value.
⎝q⎠
⎛x⎞ ⎛ y
⎞
−1
⎛ 0⎞
⎟⎟ then ⎜ ⎟ is an attracting fixed point if
Example (2-6)[2] Let F ⎜⎜ ⎟⎟ = ⎜⎜
< a <0,
⎜ 0⎟
4
⎝ y ⎠ ⎝ a sin( x) − y ⎠
⎝ ⎠
⎛ 0⎞
⎜⎜ ⎟⎟ is a
⎝ 0⎠
⎛ 0⎞
repelling fixed point if a >2, ⎜⎜ ⎟⎟ is saddle fixed point if 0< a <2.
⎝ 0⎠
Definition (2-7)[1]
Let V,S ⊆ R
2
,F: V ⎯
⎯→ V and G:S ⎯
⎯→ S be two maps .Then F and G are said to be topologically
conjugate if there exist a homeomorphism H:V ⎯
⎯→ S such that H o F=G o H .The homeomorphism H is
called a topological conjugacy
Example (2-8)[6]
1
⎧
⎪ 2 x for 0 ≤ x ≤ 2
Let Q 4 ( x )=4 x (1- x ) for 0 ≤ x ≤ 1 and T( x )= ⎨
.
1
⎪2(1 − x) for
< x ≤1
2
⎩
Q 4 is topologically conjugate to T( x ). Where topological conjugacy is sin 2
π
2
x.
Definition (2-9)[3]
⎛ p⎞
⎛ p⎞
⎛ x⎞
Let ⎜⎜ ⎟⎟ be a fixed point of F .The basin of attraction of ⎜⎜ ⎟⎟ consists of all ⎜⎜ ⎟⎟ such that
⎝q⎠
⎝q⎠
⎝ y⎠
⎛ x⎞
⎛ p⎞
F n ⎜⎜ ⎟⎟ ⎯
⎯→ ∞ .
⎯→ ⎜⎜ ⎟⎟ as n ⎯
⎝ y⎠
⎝q⎠
Theorem (2-10)[1]
⎛ p⎞
⎛ p⎞
Suppose that F has an attracting fixed point at ⎜⎜ ⎟⎟ .Then there is an open set about ⎜⎜ ⎟⎟ , in which all
⎝q⎠
⎝q⎠
⎛ p⎞
points tend to ⎜⎜ ⎟⎟ under forward iteration of F.
⎝q⎠
Proposition (2.11)[1],[5]
Let H: R2 ⎯
⎯→ R2 be a Henon map and b be any fixed real number. Then
⎛ x⎞
(1) JH a,b ⎜⎜ ⎟⎟ =b ,
⎝ y⎠
∀ x, y in R.
⎛ x⎞
(2) If x2-b ≥ 0, then the eigen values of DH a,b ⎜⎜ ⎟⎟ are the real numbers - x ±
⎝ y⎠
one-to-one map
x 2 − b .(3) If b ≠ 0 , Ha,b is
(4) The Henon map Ha,b is C ∞ .
(5) If b ≠ 0 then Ha,b has an inverse.
(6) If b ≠ 0 Ha,b is diffeomorphism .
Proposition (2.12)[1]
For each value of b in R, there exists b 0 in R − U{0} such that, if a > b
points.
0
the Henon map has two fixed
3-Type of fixed points with basin of attraction
Our goal of this section is to determine type of fixed points of H 0,b with the basin of attraction of a fixed
⎛ 0⎞
point ⎜⎜ ⎟⎟ . We will use maximum norm, where the maximum norm of
⎝ 0⎠
⎛ x⎞
⎜⎜ ⎟⎟ =
⎝ y⎠
⎛ x⎞
⎜⎜ ⎟⎟ =max{ x , y }.
⎝ y⎠
Proposition (3-1)
⎛ 0⎞
If − 1 < b < 0 , then H 0,b has attracting fixed point ⎜⎜ ⎟⎟ and a saddle fixed point
⎝ 0⎠
⎛ − (b + 1) ⎞
⎟⎟ .
⎜⎜
⎝ − (b + 1) ⎠
⎛ 0⎞
⎛ − (b + 1) ⎞
⎟⎟ are fixed points for
Proof: Since a = 0 and − 1 < b < 0 . By proposition (2.12) P+= ⎜⎜ ⎟⎟ , P-= ⎜⎜
⎝ 0⎠
⎝ − (b + 1) ⎠
⎛ x⎞
H 0,b .By proposition (2.11) λ1, 2 = ± − b are eigen values of DH 0,b ⎜⎜ ⎟⎟ at P+ .. Since − 1 < b < 0 then
⎝ y⎠
⎛ 0⎞
0 < − b < 1 and λ1 = λ2 < 1, so ⎜⎜ ⎟⎟ is attracting fixed point. To show that P- is a saddle fixed point . From
⎝ 0⎠
proposition (2.11) , we have two eigen values, λ1 = −(b + 1) + (b + 1) 2 − b , λ 2 = −(b + 1) − (b + 1) 2 − b
(3.1)
Since b 2 + b + 1 < (b + 2) 2 thus
On the other hand, since
b 2 + b + 1 < (b + 2) 2 = b + 2 hence − (b + 1) + b 2 + b + 1 < 1
(3.2)
b 2 + b + 1 = (b + 1) 2 − b > (b + 1) 2 = b + 1 = b + 1
so − (b + 1) + b 2 + b + 1 > 0 , hence by (3.2) − (b + 1) + (b + 1) 2 − b = λ 2 < 1 .
(3.3)
For λ1 since − 1 < b < 0 we have
(3.4)
b2 + b + 1 > b2 = b = − b .
By adding (b + 1) for both sides of (3.4), we get (b + 1) +
then (b + 1) + (b + 1) 2 − b = λ1 > 1 .
theorem (2-5), P- is a saddle fixed point. □
b 2 + b + 1 >1
(3.5)
Hence from (3.4), (3.5) and
⎛ 0⎞
Remark (3-2) In view of theorem (2-10), there exists an open set about ⎜⎜ ⎟⎟ in which all points tend to
⎝ 0⎠
⎛ 0⎞
⎜⎜ ⎟⎟ under forward iteration of H 0,b .The next theorem shows this open set.
⎝ 0⎠
Theorem(3-3)Suppose
⎛ x⎞
{ ⎜⎜ ⎟⎟ : x, y ∈ R, x ≤ 1 − b , y ≤ 1 − b , b < 1}
⎝ y⎠
S=
then
for
all
⎛ x⎞
⎛ x⎞
⎛ 0⎞
⎜⎜ ⎟⎟ ∈ S o ,H 0n,b ⎜⎜ ⎟⎟ ⎯
⎯→ ⎜⎜ ⎟⎟ as n ⎯
⎯→ ∞ .
⎝ y⎠
⎝ y⎠
⎝ 0⎠
⎛ x1 ⎞
Proof: We claim that H 0,b (S o ) ⊂ S o , to show this , let ⎜⎜ ⎟⎟ be any element in H 0,b (S o ) then there is
⎝ y1 ⎠
⎛ x ⎞ ⎛ x1 ⎞ ⎛ − by − x 2
S o such that H 0,b ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ = ⎜⎜
x
⎝ y ⎠ ⎝ y1 ⎠ ⎝
⎞
⎟ .From this y1 = x < 1 − b
⎟
⎠
2
(3.6)
2
2
and x1 = by + x 2 ≤ b y + x < b ( 1 − b ) + (1 − b ) 2 = b − b + 1 − 2 b + b = 1 − b
⎛ x1 ⎞
Now from (3.6), (3.7) ⎜⎜ ⎟⎟ ∈ S o . Let r =
⎝ y1 ⎠
r <1 − b .
We
⎛ x⎞
⎜⎜ ⎟⎟ in
⎝ y⎠
(3.7)
⎛ x⎞
⎜⎜ ⎟⎟ , so x ≤ r , y ≤ r and since x ≤ 1 − b , y ≤ 1 − b , then
⎝ y⎠
⎛ x⎞
H 0,b n ⎜⎜ ⎟⎟ for
⎝ y⎠
denote
⎛ x⎞
⎜⎜ ⎟⎟ in
⎝ y⎠
S o ,by
⎛ xn ⎞
2
⎜⎜ ⎟⎟ ,then y 2 = x1 = by + x 2 ≤ b y + x ≤ b r + r 2 = r ( b + r )
⎝ yn ⎠
(3.8)
2
x2 = by1 + x1 ≤ b r + r 2 ( b + r ) 2
(3.9)
Now since 0 < b + r < 1, we have 0 < r 2 ( b + r ) < r 2 , then b r + r 2 ( b + r ) 2 < b r + r 2 = r ( b + r ) , hence
by (3.8), (3.9) x2 < r ( b + r )
x2n < r ( b + r )
Now we claim that
(3.10)
n
, y 2n < r ( b + r )
n
.
(3.11)
We prove it by using mathematical induction from (3.8), (3.10),it is true for n = 1 , suppose it is true for
k then x 2 k < r ( b + r ) k , y 2 k < r ( b + r ) k to show that it is true for k + 1 ,
y 2 ( k +1) = − by 2( k +1)− 2 − x 2 2( k +1)− 2 = − by 2 k − x 2 2 k .
y 2 ( k +1) = by 2 k + x 2 k
2
2
≤ b y 2k + x2k ≤ b r ( b + r ) k + r 2 ( b + r )
2 k
(3.12)
Now 0 < ( b + r ) k < 1, so r 2 ( b + r )
k
Hence r 2 ( b + r )
< r( b + r)
2 k
+b r ( b +r )
k
+ b r < b r + r 2 = r( b + r)
Hence from (3.12),we get y 2 ( k +1) < r ( b + r )
k +1
(3.13)
k +1
(3.14)
Also we have x 2 ( k +1) = − by 2 ( k +1) −1 − x 22( k +1) −1 = − by 2 k +1 − x 2 2 k +1 = − bx 2 k − y 2 2 ( k +1) , since x n = y n +1
2
so x 2 ( k +1) = bx 2 k + y 2 2 ( k +1) < b x 2 k + y 2( k +1) < b r ( b + r ) k + r 2 ( b + r )
x 2 ( k +1) < r ( b + r )
= x 2 n then
k +1
x
lim H 0,b 2 n ⎛⎜⎜ ⎞⎟⎟ = lim x 2 n = lim r ( b + r ) n = 0 , so by definition of maximum norm
n⎯
⎯→∞
⎝ y⎠
n⎯
⎯→∞
2n
n⎯
⎯→∞
⎛ x⎞
⎛ 0⎞
⎜⎜ ⎟⎟ ⎯
⎯→ ⎜⎜ ⎟⎟ as n ⎯
⎯→ ∞ . (3.15)
⎝ y⎠
⎝ 0⎠
2n ⎛ x ⎞
If H 0,b ⎜⎜ ⎟⎟ = y 2 n , as the same as (3.13), H 0,b
⎝ y⎠
2n
⎛ x⎞
⎛ 0⎞
⎜⎜ ⎟⎟ ⎯
⎯→ ⎜⎜ ⎟⎟ as n ⎯
⎯→ ∞ and since H 0,b (S o ) ⊂ S o
y
0
⎝ ⎠
⎝ ⎠
⎛ x⎞
⎛ x⎞
then ∀ ⎜⎜ ⎟⎟ ∈ S o , we have H 0,b ⎜⎜ ⎟⎟ ∈ S o by (3.15), so H 0,b
⎝ y⎠
⎝ y⎠
2 n +1
hence from (3.13) we get that
⎛x ⎞
2n ⎛ x ⎞
2n ⎛ x ⎞
, hence it is true for k + 1 .Now H 0,b ⎜⎜ ⎟⎟ = ⎜⎜ 2 n ⎟⎟ =max{ x 2 n , y 2 n },if H 0,b ⎜⎜ ⎟⎟
⎝ y ⎠ ⎝ y2n ⎠
⎝ y⎠
2n ⎛ x ⎞
of H 0,b ⎜⎜ ⎟⎟ . lim y 2 n = 0 so H 0,b
⎝ y ⎠ n ⎯⎯→∞
H 0 ,b
2 k
⎛ x⎞
⎛ 0⎞
⎜⎜ ⎟⎟ ⎯
⎯→ ⎜⎜ ⎟⎟ as n ⎯
⎯→ ∞ , hence H 0,b
⎝ y⎠
⎝ 0⎠
n
2n
⎛ x⎞
⎛ 0⎞
( H 0,b ⎜⎜ ⎟⎟ ) ⎯
⎯→ ⎜⎜ ⎟⎟ as n ⎯
⎯→ ∞ , so
⎝ y⎠
⎝ 0⎠
⎛ x⎞
⎛ x⎞
⎛ 0⎞
⎜⎜ ⎟⎟ ⎯
⎯→ ⎜⎜ ⎟⎟ as n ⎯
⎯→ ∞ , ∀⎜⎜ ⎟⎟ ∈ S o . □
⎝ y⎠
⎝ y⎠
⎝ 0⎠
4- The Periodic Points for Henon Map Where
a = 0 , −1 < b < 0
In this section, we will show that Henon map H 0,b , − 1 < b < 0 has no periodic points other than fixed
points in the plane. To prove this, we divide the proof to fifteen lemmas, thus we find some regions such that
the union of all regions covers the plane. We prove that there is no periodic point in each of them until we
get the main purpose. The regions are the following Q1 , Q2,1 , Q2, 2 , Q3 , Q4 and in Q3 define the following
regions R1 , R2 , R3 , R4 , R5 , R6 , R7 .
⎛ x⎞
Q1 ={ ⎜⎜ ⎟⎟ : x ≥ 0 , y ≥ 0 }.
⎝ y⎠
⎛ x⎞
2
Q2,1 ={ ⎜⎜ ⎟⎟ : x < 0 , y ≥ 0 , x ≥ − y }.
⎝ y⎠
⎛ x⎞
2
Q2, 2 ={ ⎜⎜ ⎟⎟ : x < 0 , y > 0 , x < − y }.
⎝ y⎠
⎛ x⎞
Q4 = { ⎜⎜ ⎟⎟ : x ≥ 0 , y < 0 }.
⎝ y⎠
⎛ x⎞
2
R1 = { ⎜⎜ ⎟⎟ : x ≥ − y }.
⎝ y⎠
⎛ x⎞
2
R2 = { ⎜⎜ ⎟⎟ : x ≤ b − y − y }
y
⎝ ⎠
⎛ x⎞
− x − x2
R3 ={ ⎜⎜ ⎟⎟ : y <
,− y
b
⎝ y⎠
2
2
− by < x < − y , x < y }.
⎛ x⎞
− x − x2
R4 ={ ⎜⎜ ⎟⎟ : y <
,b − y − y
b
⎝ y⎠
⎛ x⎞
− x − x2
R5 ={ ⎜⎜ ⎟⎟ : y <
,− y
b
⎝ y⎠
2
2
2
< x ≤ −by − y }
2
− by < x < − y , x ≥ y , by + x 2 > 1 }.
⎛ x⎞
2
R6 ={ ⎜⎜ ⎟⎟ : y < −1 − b , x < − y , x ≥ y , by + x 2 ≤ 1 , x > − y 2 − by , x > b − y − y 2 }.
y
⎝ ⎠
⎛ x⎞
− x − x2
R7 ={ ⎜⎜ ⎟⎟ : x < −1 − b , y > −1 − b , y > x , y ≥
, − y 2 − by > x > b − y − y 2 }.
b
⎝ y⎠
Lemma (4-1)
H 0,b is topologically conjugate to H 0−.1b−1
⎛ x ⎞ ⎛ by ⎞
Proof: Let G: R2 ⎯
⎯→ R2 be a map defined by G ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ . Clearly G is continuous and bijective , G −1 is
⎝ y ⎠ ⎝ bx ⎠
continuous, so G is homeomorphism and
GoH
−1
0.b −1
y
⎛
⎛ x⎞
⎜⎜ ⎟⎟ =G ⎜⎜ − x − y 2
⎜
⎝ y⎠
⎝ b −1
⎞ ⎛ − x − y2
⎟ ⎜b
b −1
⎟⎟ = ⎜⎜
by
⎠ ⎝
⎞ ⎛ − x − y2
⎟ ⎜
−2
⎟⎟ = ⎜⎜ b
⎠ ⎝ by
⎞
⎟
⎟⎟
⎠
(4.1)
2
⎛ x⎞
⎛ by ⎞ ⎛ − b(bx) − (by ) 2 ⎞ ⎛ b 2 (− x − y 2 ) ⎞ ⎛⎜ − x − y
−2
⎟=⎜
⎟=
H 0,b o G ⎜⎜ ⎟⎟ = H 0,b ⎜⎜ ⎟⎟ = ⎜⎜
⎟ ⎜
⎟ ⎜⎜ b
by
by
⎝ y⎠
⎝ bx ⎠ ⎝
⎠ ⎝
⎠ ⎝ by
⎞
⎟
⎟⎟ . (4.2)
⎠
From (4.1) and (4.2), G o H 0−.1b −1 = H 0,b o G. Hence by definition (2-7), H 0,b is topologically conjugate to
H 0−.1b−1 . □
Lemma (4-2)
There are no periodic points in Q1 other than fixed points.
⎛ x0 ⎞
Proof: Let ⎜⎜ ⎟⎟ ∈ Q1 , we define norm
⎝ y0 ⎠
as H and H
−1
0.b −1
as H
−1
H
−n
⎛ x0 ⎞
⎛x ⎞
⎜⎜ ⎟⎟ by ⎜⎜ 0 ⎟⎟ = x0 2 + y 0 2 From now for simply we refer to H 0,b
⎝ y0 ⎠
⎝ y0 ⎠
y −n
⎞
⎛ x0 ⎞ ⎛ x −( n +1) ⎞ ⎛⎜
⎟
⎟ = −1
⎜⎜ ⎟⎟ = ⎜⎜
2
⎟ ⎜ ( x−n + y −n ) ⎟
y
y
⎝ 0 ⎠ ⎝ −( n +1) ⎠ ⎝ b
⎠
We have − 1 < b < 0 , x −1 = y 0 , y −1 =
(4.3)
−1
−1
−1
( x0 + y 0 ) then y −1 =
( x0 + y 0 ) >
x0 > x0
b
b
b
2
2
2
2
2
y −1 > x0 and since y −1 , x0 are non negative y −1 > x0 .That is y −1 + y 0 > x0 + y 0
2
2
2
2
2
2
2
2
2
thus y −1 + x −1 > x0 + y 0 so y −1 + x −1 > x0 + y 0 ,
⎛x ⎞ ⎛x ⎞
⎛x ⎞ ⎛x ⎞
hence ⎜⎜ −1 ⎟⎟ > ⎜⎜ 0 ⎟⎟ , that is H −1 ⎜⎜ 0 ⎟⎟ > ⎜⎜ 0 ⎟⎟ .
⎝ y −1 ⎠ ⎝ y 0 ⎠
⎝ y0 ⎠ ⎝ y0 ⎠
Since x −1 = y 0 ≥ 0 , y −1 =
−1
( x 0 + y 0 ) ≥ 0 ,we have H
b
⎛x ⎞
⎛x ⎞
H − n ⎜⎜ 0 ⎟⎟ > H − n +1 ⎜⎜ 0 ⎟⎟ ∀n ∈ Z + ,
⎝ y0 ⎠
⎝ y0 ⎠
(4.4)
−1
(Q1 ) ⊂ Q1 , then (4.4) is true for all n ∈ Z + ,hence
(4.5)
so the norms of the points in the orbit are strictly increasing. Hence there are no periodic point in the region
Q1 . □
Lemma (4-3)
There are no periodic points in Q2,1 .
Proof:
x−1 > 0 ,
⎛ x0 ⎞
Let ⎜⎜ ⎟⎟ ∈ Q2,1 ,
⎝ y0 ⎠
since
H
−1
y0
⎞
⎛ x0 ⎞ ⎛⎜
⎟
⎜⎜ ⎟⎟ = − 1
2
⎜ ( x0 + y 0 ) ⎟ , y 0 > 0 and
y
⎝ 0⎠ ⎝ b
⎠
2
x0 ≥ − y 0 ,
hence
⎛ x −1 ⎞
−1
2
( x 0 + y 0 ) ≥ 0 , so x−1 > 0 , y −1 ≥ 0 , ⎜⎜ ⎟⎟ ∈ Q1 , that means Q2,1 maps into Q1 then by lemma (4b
⎝ y −1 ⎠
.2) there are no periodic point for H in Q2,1 . □
Lemma (4-4)
There are no periodic points of even period in Q2, 2 .
⎛ x0 ⎞
−1
2
Proof: Let ⎜⎜ ⎟⎟ ∈ Q2, 2 .From (4.3),since y − n = ( x −( n −1) + y −( n −1) ) , x − n = y −( n −1) . Since x0 < 0, y 0 ≥ 0 ,so
b
⎝ y0 ⎠
x −1 = y 0 > 0 , y −1 =
−1
−1
−1
2
2
( x 0 + y 0 ) < 0 , hence x− 2 = y −1 < 0 , y −2 =
x −1 . Since − 1 < b < 0 ,
( x −1 + y −1 ) >
b
b
b
−1
−1
x −( n −1) > x −( n −1) = y −( n − 2 ) , so ⟨ y − 2 n ⟩ is strictly
x −1 > x −1 = y 0 > 0 ,hence y −2 > y 0 > 0 .In general y − n >
b
b
⎛ x0 ⎞
increasing sequence ,H −2 n ⎜⎜ ⎟⎟ in Q2,1 U Q2, 2 then we have two cases.
⎝ y0 ⎠
⎛ x0 ⎞
Case 1: If H −2 n ⎜⎜ ⎟⎟ ∈ Q2,1 , then by lemma (4-3) there are no periodic points of even period for H.
⎝ y0 ⎠
⎛ x0 ⎞
Case 2: If H −2 n ⎜⎜ ⎟⎟ ∈ Q2, 2 and since ⟨ y − 2 n ⟩ is strictly increasing sequence so there are no periodic points
⎝ y0 ⎠
of even period for H. □
Lemma (4-5)
There are no periodic points of odd period in Q4 .
⎛ x0 ⎞
−1
−1
−1
2
2
Proof: Let ⎜⎜ ⎟⎟ ∈ Q4 since x −1 = y 0 < 0 , y −1 =
( x 0 + y 0 ) , x0 > 0 so y −1 =
( x0 + y 0 ) >
x0 > x0 ,
b
b
b
⎝ y0 ⎠
y −1 > 0 ,from
hence
x − ( n +1) = y − n
(4.3)
,
we
have x−2 > 0 ,then
y −3 =
−1
−1
2
( x−2 + y −2 ) >
x − 2 > x − 2 = y −1 > 0 ,hence
b
b
y −5 =
−1
−1
2
( x−4 + y −4 ) >
x − 4 > x − 4 = y −3 > 0 , hence y −( 2 n +1) > y −( 2 n −1) > ... > y −5 > y −3 > y −1 > 0
b
b
x−4 > 0
by
the
same
way
(4.6)
⎛ x0 ⎞
⎛ x0 ⎞
Either x − 2 n −1 > 0 or x − 2 n −1 < 0 so H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q1 or H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q2
⎝ y0 ⎠
⎝ y0 ⎠
⎛ x0 ⎞
Case 1: If H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q1 , by lemma (4-2), there are no periodic point of odd period
⎝ y0 ⎠
⎛ x0 ⎞
Case 2: If H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q2 since ⟨ y − 2n −1 ⟩ is strictly increasing sequence there are no periodic points of odd
⎝ y0 ⎠
period . □
Lemma (4-6)
There are no periodic points of even period in Q4 .
⎛ x0 ⎞
Proof: Suppose there exists ⎜⎜ ⎟⎟ in Q4 such that
⎝ y0 ⎠
⎛ x0 ⎞
⎜⎜ ⎟⎟ is a periodic point
⎝ y0 ⎠
⎛ x0 ⎞ ⎛ x0 ⎞
of even period. That is there is 2n in Z + , H −2 n ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ .
⎝ y0 ⎠ ⎝ y0 ⎠
(4.7)
Since x − n = y − n −1 , y − n =
all n ∈ N .If
−1
( x −( n −1) + y −2( n −1) ) ,then
b
⎛ x0 ⎞
H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q2,1 ,from
⎝ y0 ⎠
(4.7)
either
⎛ x0 ⎞
H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q2,1
⎝ y0 ⎠
⎛ x0 ⎞ ⎛ x0 ⎞
H −2 n ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ so
⎝ y0 ⎠ ⎝ y0 ⎠
or
⎛ x0 ⎞
H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q1
⎝ y0 ⎠
⎛ x0 ⎞
H −2 n −1 ⎜⎜ ⎟⎟ =
⎝ y0 ⎠
⎛ x0 ⎞
H −1 ⎜⎜ ⎟⎟ ,
⎝ y0 ⎠
for
hence
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x0 ⎞
H −2 n (H −1 ⎜⎜ ⎟⎟ )= H −1 ⎜⎜ ⎟⎟ , H −1 ⎜⎜ ⎟⎟ ∈ Q2,1 so Q2,1 has a periodic point which is contradiction .If
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x0 ⎞
H −2 n −1 ⎜⎜ ⎟⎟ ∈ Q2, 2 , in the same way, H −2 n (H −1 ⎜⎜ ⎟⎟ )= H −1 ⎜⎜ ⎟⎟ , H −1 ⎜⎜ ⎟⎟ ∈ Q1 ,that means there is a
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
periodic point of even period in Q1 which is contradiction ,so there are no periodic points of even period
in Q4 . □
Lemma (4-7)
There are no periodic points of odd period in Q2, 2 .
⎛ x0 ⎞
Proof: Suppose that there is a periodic point ⎜⎜ ⎟⎟ of odd period m .That is there exists at least positive
⎝ y0 ⎠
integer
n
such
2
that
m = 2n + 1 ,
x0 < − y 0 , x −1 = y 0 ≥ 0 , also y −1 =
and
⎛ x0 ⎞ ⎛ x0 ⎞
H −2 n −1 ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ,
⎝ y0 ⎠ ⎝ y0 ⎠
⎛ x0 ⎞
since ⎜⎜ ⎟⎟ ∈ Q2, 2
⎝ y0 ⎠
,
we
have
⎛ x0 ⎞
⎛ x0 ⎞
−1
2
( x 0 + y 0 ) < 0 ,hence H −1 ⎜⎜ ⎟⎟ ∈ Q4 , that is H −1 (H −2 n −1 ⎜⎜ ⎟⎟ )=
b
⎝ y0 ⎠
⎝ y0 ⎠
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x −1 ⎞
H −1 ⎜⎜ ⎟⎟ ,also H −2 n −1 (H −1 ⎜⎜ ⎟⎟ )= H −1 ⎜⎜ ⎟⎟ hence ⎜⎜ ⎟⎟ is a periodic point of odd period in Q 4 which is
⎝ y −1 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
contradiction by lemma (4-5),hence there are no periodic points of odd period in Q2, 2 .
Lemma (4-8)
There are no periodic points in R1 .
⎛ x0 ⎞
⎛ x0 ⎞
2
Proof: Let ⎜⎜ ⎟⎟ be any element in R1 , then ⎜⎜ ⎟⎟ ∈ Q3 , x0 ≥ − y 0 ,that is y 0 < 0
⎝ y0 ⎠
⎝ y0 ⎠
2
x0 < 0 , x0 ≥ − y 0 , so x −1 = y 0 < 0 , y −1 =
⎛ x0 ⎞
⎛ x −1 ⎞
−1
2
( x 0 + y 0 ) ≥ 0 , hence ⎜⎜ ⎟⎟ ∈ Q2 .That is H −1 ⎜⎜ ⎟⎟ ∈ Q2 that
b
⎝ y −1 ⎠
⎝ y0 ⎠
means R1 maps into Q2 , so by lemma (4-4) and lemma (4-5) there are no periodic points in R1 . □
Lemma (4-9)
There are no periodic points in R2 .
⎛ x0 ⎞
Proof : Let ⎜⎜ ⎟⎟ ∈ R2 then
⎝ y0 ⎠
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ Q3 , x0 ≤ b − y 0 − y 0 2 ,that is y 0 < 0 , x0 < 0
⎝ y0 ⎠
2
2
x0 ≤ b − y 0 − y 0 .Thus x0 + y 0 ≤ b − y 0 so
−1
2
( x 0 + y 0 ) ≤ − − y 0 , that is
b
⎛ x0 ⎞
2
2
2
y −1 ≤ − − y 0 .Hence y −1 ≥ − y 0 , that is y −1 ≥ − x1 so − y −1 ≤ x1 , hence H −1 ⎜⎜ ⎟⎟ ∈ R1 .Hence R2 maps into
⎝ y0 ⎠
R1 , by lemma (4-8) ,there are no periodic point in R2 . □
Lemma (4-10)
There are no periodic point in R3 .
2
⎛ x0 ⎞
− x0 − x0
2
2
2
, − y 0 − by 0 < x0 < − y 0 , x0 < y 0 ,we have x1 = −by0 − x0 , y1 = x0 ,
Proof: Let ⎜⎜ ⎟⎟ ∈ R3 , so y 0 <
y
b
⎝ 0⎠
2
2
2
since − x0 < y 0 + by 0 , x0 > y 0 2 , we get that
2
2
y 0 + by 0 < x0 + by 0 , so − x0 < x0 + by0 , that is
, x0 > x1 ,since x0 = y1 , we have y1 > x1 .
(4.8)
2
2
Suppose that − x1 ≥ y1 + by1 , that is , − x1 ≥ x0 + bx0 , bx0 > by 0 , so − x1 > x0 + by0 = − x1 which is
2
2
contradiction, hence x1 > − y1 − by1 .
(4.9)
2
2
2
Clearly, since − by 0 < 0 , then − by0 − x0 < − x0 ,hence x1 < − y1 .
(4.10)
2
To
show
that
2
− x1 − x1
,
y1 <
b
since
x0 < y 0 , b < 0 ,
we
have
2
2
x1 > − x0 − bx0 = − x0 − by1 .
(4.11)
2
− x0 − x0
,
y0 <
b
From
2
2
we
2
have
by 0 > − x0 − x0
2
2
2
− x1 − x1
b
x0 > x1 ,since
,hence
x 0 < 0 , x0 < x1 ,thus − x0 − by1 > − x1 − by1 then from (4.11),we get that
by1 > − x1 − x1 ,hence y1 <
2
− x0 − by0 > − x0 − bx0 ,so
2
x1 > − x1 − by1 ,that is
2
.
(4.12)
Now from (4.9),(4.10), (4.11)and (4.12), we can say that H ( R3 ) ⊂ R3 ,so H 2 ( R3 ) ⊂ R3 and so on
2
H ( R3 ) ⊂ H
n
n −1
( R3 ) ⊂ R3 ,where
2
n is
positive
integer
hence
,
2
− xn − xn
xn < y n , y n <
,that
b
is
by n > − x n − x n ,so x n > − x n − by n = xn +1 ,hence ⟨ x n ⟩ is strictly decreasing sequence in R3 , so there are no
periodic points in R3 .□
Lemma (4-11)
There are no periodic point in R4 .
2
⎛ x0 ⎞
− x0 − x0
Proof: Let ⎜⎜ ⎟⎟ be any element in R4 , then y 0 <
b
⎝ y0 ⎠
2
2
and b − y 0 − y 0 < x0 ≤ −by0 − y 0 .
(4.13)
(4.14)
2
2
By (3.28) by 0 > − x0 − x0 or x0 > − by0 − x0 = x1 , from (3.18) y1 = x0 ,so y1 > x1 (4.15)
2
2
Since x1 = − by0 − x0 , − by 0 < 0 , we have x1 < − y1 .
Now
,to
show
2
x1 > − y1 − by1 ,
that
2
2
(4.16)
by0 − x0 > − x0 − x0 + y 0 + by0 ,
2
since
2
− x0 + y 0 < 0 ,
so
that
2
2
by 0 > − x0 − x0 , − x0 ≥ by 0 + y 0 ,we
is
2
, x0 >
2
y 0 hence
get
x0 > y 0 ,since
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ Q3 , x0 = − x0 , y 0 = − y 0 ,thus x0 < y 0 , so y1 < y 0 ,since b < 0 ,we have x0 2 + by 0 < x0 2 + by1 , hence,
⎝ y0 ⎠
2
− x1 < x0 + by1 . Now since x0 = y1 , we get that − x1 < y1 2 + by1 that is , x1 > − y1 2 − by1 .
2
2
(4.17)
2
On the other hand , x0 < y 0 ,thus , − x0 − by0 > − x0 − bx0 ,hence we get that x1 > − x0 − by1 .
2
From by 0 > − x0 − x0 ,then
− x0
2
2
x0 > − x0 − by 0 , so
x0 > x1 ,
2
2
2
x 0 , x1 are negative ,so x0 < x1 , thus
− x1 − x1
− by1 > − x1 − by1 ,hence from (4.18) x1 > − x1 − by1 , that is y1 <
b
2
(4.18)
2
(4.19)
⎛ x1 ⎞
Now from (4.15), (4.16), (4.17) and (4.19), we get that ⎜⎜ ⎟⎟ ∈ R3 ,so R4 maps into R3 ,then by lemma (4-10)
⎝ y1 ⎠
there are no periodic points in R4 . □
Lemma (4-12)
There are no periodic points in R5 .
⎛ x0 ⎞
Let ⎜⎜ ⎟⎟ ∈ R5 ,
⎝ y0 ⎠
Proof:
2
so y 0 <
− x0 − x0
2
2
, − y 0 − by 0 < x0 < − y 0 , x0 > y 0
b
2
by0 + x0 > 1 ,we
2
2
have
2
x1 = −by0 − x0 , y1 = x0 .We claim that R5 maps into R4 , since x0 ≥ y 0 ,we have − x0 − bx0 ≥ − x0 − by0
2
2
also since − x0 − by 0 = x1 x0 = y1 , we have x1 ≤ − y1 − by1 .
2
On the other hand, since we have x0 < − y 0 ,then
2
− x0 >
(4.20)
2
y 0 = y 0 = − y 0 so, by 0 < −b − x0 , hence
2
2
by 0 + x0 < x0 − b − x0 ,thus − x1 < y1 − b − y1 .
(4.21)
2
− x1 − x1
2
2
To show that y1 <
,since by0 + x0 > 1 , we have x1 < −1 , so x1 > − x1 .Since bx0 > 0 ,we get
b
2
− x1 − x1
− x1 − x1
bx0 > − x1 − x1 ,that is x0 <
but y1 = x0 , so y1 <
b
b
2
2
.
(4.22)
Now from (4.20), (4.21)and (4.22) ,our claim is true, so by lemma (4-11) there are no periodic points
in R5 .□
Lemma (4-13)
The region R6 maps into R2 U R7 under backward iteration .
⎛ x0 ⎞
⎛ x0 ⎞
Proof: Let ⎜⎜ ⎟⎟ be any element in R6 .We must show that H −1 ⎜⎜ ⎟⎟ ∈ R2 U R7 .
⎝ y0 ⎠
⎝ y0 ⎠
− x0 − y 0
From (4.3) x −1 = y 0 , y −1 =
b
2
, since y 0 < −1 − b , we have x − 1 < −1 − b .
(4.23)
Since x0 > − y 0
2
− x0 − y 0
− by 0 , we have y 0 <
b
2
,that is x−1 < y −1 .
(4.24)
2
2
Also, since x0 > b − y 0 − y 0 ,we have
x0 + y 0
< − y0
b
,so − y −1 <
− y0
2
thus, y 0 < − y −1 ,hence
2
x −1 < − y −1 .
(4.25)
2
2
2
− x0 − y 0
− y0 − y0
− x −1 − x −1
From x0 ≥ y 0 ,we have
,hence y −1 ≥
.
≥
b
b
b
(4.26)
⎛ x0 ⎞
⎛ x −1 ⎞
Now from (3.38), (3.39),(3.40)and (3.41) ⎜⎜ ⎟⎟ ∈ R2 U R7 ,that is H −1 ⎜⎜ ⎟⎟ in R2 U R7 . □
⎝ y −1 ⎠
⎝ y0 ⎠
Lemma (4-14)
There are no periodic points in R7 .
2
⎛ x0 ⎞
− x0 − x0
− x − x2
, the graph of the map y =
Proof: Let ⎜⎜ ⎟⎟ be any element in R7 , since y 0 ≥
intersects xb
b
⎝ y0 ⎠
2
− x − y0
2
2
axis at point (-1,0),thus x0 > −1 ,so b 0
+ y 0 < 1 , that is by −1 + x −1 < 1 .
b
− x0 > 1 + b ,
Since
y 0 > −1 − b ,
we
have
(4.27)
2
− y 0 − x0 > (1 + b) − (1 + b) 2 = b( −1 − b) ,hence
2
− x0 − y 0
< −1 − b , y −1 < −1 − b .
b
(4.28)
On the other hand , x0 < − y 0
2
2
− x0 − y 0
− by 0 , thus y 0 >
, so x−1 > y −1 .
b
(4.29)
2
Also , since x0 > b − y 0 − y 0
2
,we get
x0 + y 0
< − y 0 ,thus − y −1 <
b
2
− y 0 so y −1 < − y 0 ,that is
x−1 < − y −1 2 .
(4.30)
Hence, from (4.27), (4.28), (4.29) and (4.30) R7 maps into R6 , by lemma (4-13) R6 maps into R2 U R7 , so
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x0 ⎞
there is no point ⎜⎜ ⎟⎟ in R7 such that H − 2 n −1 ⎜⎜ ⎟⎟ ∈ R7 .That is, there is no point ⎜⎜ ⎟⎟ in R7 such that
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎛ x0 ⎞ ⎛ x0 ⎞
H − 2 n −1 ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ hence there are no periodic points of odd period in R7 .To show that there are no periodic
⎝ y0 ⎠ ⎝ y0 ⎠
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x0 ⎞
⎛ x0 ⎞
points of even period in R7 .Let P={ ⎜⎜ ⎟⎟ ∈ R7 : H − 2 n −1 ⎜⎜ ⎟⎟ ∈ R6 , H − 2 n ⎜⎜ ⎟⎟ ∈ R7 } ∀n ∈ N ,if ⎜⎜ ⎟⎟ ∈ P,we
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
⎝ y0 ⎠
claim that the sequence ⟨ y − 2 n ⟩ is strictly increasing ,since y 0 > −1 − b ,we have y 0 ( −1 − b) < (1 + b) 2 . (4.31)
Also
2
2
2
y 0 < (1 + b) 2 , − x0 > 1 + b ,so
− x0 − y 0 > − b(1 + b) ,
hence
− x0 − y 0
< −1 − b that
b
is
2
,(
− x0 − y 0 2
) > (1 + b) 2 .
b
(4.32)
2
− x − y0 2
2
Now from (4.31),(4.32), we get that ( 0
) > y 0 (−1 − b) = − y 0 − by 0 so , by0 > − y 0 − y −1 , that is
b
2
⎛ x−2n ⎞
− x −1 − y −1
⎟⎟ ∈ R7 ,we have,
y0 <
= y − 2 .In general, since ⎜⎜
b
⎝ y −2n ⎠
2
y − 2 n ( −1 − b) < (1 + b) 2 , y − 2 n < (1 + b) 2 .Hence
2
y − 2 n > −1 − b , x − 2 n < −1 − b , so
− x −2 n − y −2 n > − b(1 + b) ,
that
2
is
− x−2n − y −2n 2
2
(
) > (1 + b) 2 > ( −1 − b) y − 2 n .So y −2 n −1 > (−1 − b) y −2 n = − y − 2 n − by − 2 n = − x − 2 n −1 − by − 2 n , that
b
2
is , we get that by −2 n > − x −2 n −1 − y −2 n −1 , hence y − 2 n
2
− x − 2 n −1 − y − 2 n −1
<
= y − 2 n − 2 , for all n ≥ 1 .
b
So, ⟨ y − 2 n ⟩ is strictly increasing sequence in R7 , so there are no periodic points of even period in P. □
Lemma (4-15)
There are no periodic points in R6 .
Proof: by lemma (4-13), R6 maps into R2 U R7 , there are no periodic points in R2 U R7 , so there are no
periodic points in R6 . □
Theorem (4-16)
There are no periodic points of H 0,b in R2 other than fixed points.
Proof:
If x0 > −1 − b, y 0 = −1 − b ,we
have
2
− x0 > −(1 + b) 2
,
− by 0 = b + b 2 ,
hence
⎛ x⎞
2
− x0 − by 0 > b + b 2 − 1 − b 2 − 2b = −1 − b , so x1 > −1 − b , y1 = x0 > −1 − b ,all points ⎜⎜ ⎟⎟ in R 3 satisfiy the
⎝ y⎠
equation
y<
− x2
,if
b
y 0 > −1 − b, x0 = −1 − b ,we
get
that
− by 0 > b + b 2 ,
x 2 = (1 + b) 2
so
− x 2 − by > −1 − b , that is x1 > −1 − b , y1 = x0 = −1 − b
Hence from theorem (3-3) ,and previous lemmas (4-1),(4-2),…,(4-14),,(4-15) , since a periodic point under
forwa rd iteration implies one under backwards iteration, and viceversa . □
Referencess
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,1989.
[2] Galloway .Introduction to Non Linear Systems and Chaos. The niversity of sydniy, Applied Mathematics
,Math 2006 . Preprint.
[3] Gulick ,D.Encounters with Chaos .McGraw –Hill,Inc.U.S.A,1992.
[4] Hayes .S and wolf .C. Dynamics of a One –Parameter Family of Henon Maps .Department of
Mathematics ,university Wichita ,ks 67260,E-mail address:
c wolf @t math.wichita.edu .
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69-77
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