Iteration with Periodic Point of Henon Map where
a>
− (1 + b ) 2
4
Iftichar Mudar Talb and Kawa Ahmad Hassan
Abstract
We study the dynamics of the two dimensional mapping the non
⎛ a − by − x 2 ⎞
⎟ .We
⎟
x
⎝
⎠
linear mapping H a,b ⎛⎜⎜ ⎞⎟⎟ = ⎜⎜
x
⎝ y⎠
study the iteration of all
points in the plane . We determine the regain which contain the set of
periodic points
1- Introducation
About 30 years ago the French astronomer –
mathematician Michel-Henon was searched for a simple twodimensional map possessing special properties of more complicated
systems .The result was a family of maps denoted by H a,b given by
⎛1 − ax 2 +
bx
⎝
H a,b ⎛⎜⎜ ⎞⎟⎟ = ⎜⎜
x
⎝ y⎠
y⎞
⎟
⎟
⎠
where a , b are real numbers .These maps defined in above are called
Henon maps [5].
we will prove one theorem about iteration of H a,b where
− (1 + b) 2
a>
4
b > 0 and
which is given in [3] . prove we prove that for a closed
region
71
S a,b ={ ⎛⎜⎜ ⎞⎟⎟ :
x
⎝ y⎠
x ≤ C a , b , y ≤ C a ,b }
if
a>
− (1 + b) 2
4
and
b>0
then for all
⎛ x⎞
⎜⎜ ⎟⎟ ∈ R 2 -S a,b
⎝ y⎠
either
a<
xn ⎯
⎯→ ∞ as n ⎯
⎯→ ∞ or
− (1 + b) 2
4
y −n ⎯
⎯→ ∞ as
n⎯
⎯→ ∞ ,
where
b > 0 and
, by finding some regions and prove some necessary
lemma for our proof..
2- The Set of Periodic Points
The main purpose of this section is to prove one theorem on
Henon map H a,b where
a>
− (1 + b ) 2
4
. To prove this theorem, we need
to prove some lemmas. To state and prove our lemmas, we fix
define two crucial
a − values
− (1 + b ) 2
1-
a 0 (b) =
2-
a1 (b) = 2(1 + b ) 2
4
and, for any particular
C a ,b =
1 + b + (1 + b ) 2 + 4a
2
a − value
,we define C=C a,b by
, we will go to prove necessary lemmas.
Lemma (2.1)[4]
72
b
and
(i) C is positive real and the larger root of C 2 − (1 + b ) C − a = 0 if and
only if
(ii)
a0 ≤ a .
a − bC > C
if and only if
proof: (i) Since
number, if
a0 ≤ a ,
a > a1
we have
(1 + b ) 2 + 4a ≥ 0 ,so
C 2 − (1 + b ) C − a = 0 ,since
distinct real roots which are
C is positive real
(1 + b ) 2 + 4a ≥ 0 .We
1 + b + (1 + b ) 2 + 4a
2
and
have two
1 + b − (1 + b ) 2 + 4a
2
,the first is positive and the second may be negative so C is larger
root .
(ii)If
(2.1)
a − b C > C ,this
We
have
(2.2) Now let
(1 + b ) = β ,
so
C 2 − β C = a .We
implies
that
a > (1 + b )C
.
2C − (1 + b ) = (1 + b ) 2 + 4a
then
2C − β = β 2 + 4a ,
put the value of
a
that is
(2C − β ) 2 = β 2 + 4a ,
in (2.1) ,we get that
C2 − β C > β C
(2.3)
so
C 2 > 2 β C ,since C
is positive we have
C > 2β
, that is
C > 2(1 + b ) .
(2.4)
Now by (2.4)
hence
(1 + b ) + (1 + b ) 2 + 4a > 4(1 + b )
a > 2(1 + b ) 2
,that is
4a + (1 + b ) 2 > 9(1 + b ) 2 ,
.
By the same way, we can prove that if
Lemma (2.2)[4]
73
a > a1
then
a − bC > C .
□
(i)
The image under H a ,b of the horizontal strip
region
bounded
by
a − b γ − y1 ≤ x1 ≤ a + b γ − y1
2
vertical strip
x0 ≤ γ
the
y0 ≤ γ
two
is the
parabolas
and the image under H a ,b of the
2
is the horizontal strip
y1 ≤ γ
, where
γ
is
positive real number .
(ii)
The inverse image of the vertical strip
bounded by two parabolas
− γ + a − x −1 ≤ by −1 ≤ a + γ − x −1
2
the image of the horizontal strip
x −1 ≤ γ
y0 ≤ γ
and
2
is the vertical strip
.
Proof: (i) we have
and
is the region
x0 ≤ γ
y0 ≤ γ
b = −b ,
, so − γ
≤ y0 ≤ γ
, if
−1 < b < 0
then
hence
bγ ≤ −by 0 ≤ −bγ
− b γ ≤ −by0 ≤ b γ
.
(2.5)
If
0 < b <1
,then
bγ ≥ −by 0 ≥ −bγ , b = b
hence
− b γ ≤ −by0 ≤ b γ
.
(2.6)
Now in all case
− b γ ≤ −by0 ≤ b γ
a − b γ − y1 ≤ x1 ≤ a + b γ − y1
2
,so
a − b γ ≤ a − by0 ≤ a + b γ
,since
x0 = y1
.
2
(2.7)
Also if
x0 ≤ γ
, since
x0 = y1 ,we
under H a ,b of the vertical strip
have
x0 ≤ γ
74
y1 ≤ γ
that means the image
is the horizontal strip
y1 ≤ γ
.
(ii) We have
x0 ≤ γ
so
so
γ ≤ − x0 ≤ −γ
a − γ − x −1 ≤ by −1 ≤ a + γ − x −1
2
2
, y 0 = x −1
,
so
y 0 = x −1
,the image under H a ,b −1 of the horizontal strip
vertical strip
x −1 ≤ γ
y0 ≤ γ
is the
. □
Lemma (2.3)[2]
Let
⎛a⎞
⎜ ⎟
x
P ⎜ b ⎟ ={ ⎛⎜⎜ ⎞⎟⎟ : a − b γ − y 2 ≤ x ≤ a + b γ − y 2 }
⎝ y⎠
⎜γ ⎟
⎝ ⎠
S h ( α , β )=R × [ α , β ] and
S v ( α , β )=[ α , β ] × R, γ ∈ R and then for the Henon map H a ,b the
following are hold :
(i) H a ,b (S v ( − γ , γ ))= S h ( − γ , γ ).
(ii)
H a ,b (S h ( − γ , γ )) ⊂ P
(iii) H a,b (P
−1
⎛a⎞
⎜ ⎟
⎜b⎟ )⊂
⎜γ ⎟
⎝ ⎠
⎛a⎞
⎜ ⎟
⎜b⎟ .
⎜γ ⎟
⎝ ⎠
S h ( − γ ,γ ) .
Proof: (i) Let ⎛⎜⎜
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ S v ( − γ , γ ) such
⎝ y0 ⎠
⎛x ⎞
2
x = a − by0 − x0 , y = x0 ,since ⎜⎜ 0 ⎟⎟ ∈ [ − γ , γ ] × R
⎝ y0 ⎠
x⎞
⎟⎟ ∈ H a ,b (S v ( − γ , γ
⎝ y⎠
⎛ x0 ⎞ ⎛ x ⎞
⎟⎟ = ⎜⎜ ⎟⎟ ,
⎝ y0 ⎠ ⎝ y ⎠
that H a ,b ⎜⎜
we have
x0 ∈ [ − γ , γ
that is
] ,hence
)), so there exists
⎛ x⎞
⎜⎜ ⎟⎟ ∈ R × [ − γ , γ
⎝ y⎠
75
] ,that is
⎛ x⎞
⎜⎜ ⎟⎟ ∈
⎝ y⎠
S h ( − γ ,γ ) .
⎛ x⎞
⎜⎜ ⎟⎟ ∈ S h ( − γ , γ ),
then
⎝ y⎠
y
⎛
⎞
⎜
2 ⎟
a
−
x
−
y
⎜⎜
⎟⎟ ∈ [ − γ , γ ] × R ,
b
⎝
⎠
Conversely: Let
a − x − y2
∈R
b
,we have
,thus
⎛ x⎞
⎜⎜ ⎟⎟ ∈
⎝ y⎠
x ∈R
and
y ∈ [ − γ ,γ
⎛ x⎞
⎜⎜ ⎟⎟ =H a,b
⎝ y⎠
since
] ,so
y
⎛
⎜
a
−
x
− y2
⎜⎜
b
⎝
⎞
⎟
⎟⎟
⎠
H a,b (S v ( − γ , γ ) ).
(ii) Let ⎛⎜⎜
x⎞
⎟⎟ ∈ H a ,b (S h ( − γ , γ
⎝ y⎠
⎛ x0 ⎞ ⎛ x ⎞
⎟⎟ = ⎜⎜ ⎟⎟ and
⎝ y0 ⎠ ⎝ y ⎠
)) ,so there exists
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ R × [ − γ , γ
⎝ y0 ⎠
H a,b ⎜⎜
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ S h ( − γ , γ
⎝ y0 ⎠
],so we get that
) such that
and by lemma
y0 ≤ γ
(2.2)
Part (i) a − b γ − y1
(iii)
x
Let ⎛⎜⎜ ⎞⎟⎟ ∈
⎝ y⎠
H a,b
−1
≤ x1 ≤ a + b γ − y1
2
H a ,b (P
−1
⎛ x0 ⎞ ⎛ x ⎞
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ,
⎝ y0 ⎠ ⎝ y ⎠
show that
⎛a⎞
⎜ ⎟
⎜b⎟ )
⎜γ ⎟
⎝ ⎠
hence
y ∈ [ − γ ,γ
2
so
⎛ x⎞
⎜⎜ ⎟⎟ ∈
⎝ y⎠
P
,so there exists
⎛a⎞
⎜ ⎟
⎜b⎟ .
⎜γ ⎟
⎝ ⎠
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈
⎝ y0 ⎠
P
a − b γ − y 0 ≤ x0 ≤ a + b γ − y 0
2
] , suppose that
y >γ
⎛a⎞
⎜ ⎟
⎜b⎟
⎜γ ⎟
⎝ ⎠
2
such that
. Clearly
then
x ∈ R.To
2
a − x0 − y 0
>γ
b
(2.8)
Now if b > 0 , then from (2.8) a − x0 − y 0 2 > b γ , so x0 < a − b γ − y0 2 which
is contradiction ,if b < 0 then from (2.8) a − x0 − y0 2 < bγ , so
2
x0 > a + b γ − y 0 which is contradiction .
To show that y ≥ −γ , if b > 0 then x0 ≤ a + b γ − y0 2 so − bγ ≤ a − x0 − y0 2 thus
2
a − x0 − y 0
−γ ≤
= y,
b
2
x0 ≥ a + bγ − y 0
so
− bγ ≥ a − x 0 − y 0
⎛ x⎞
⎜⎜ ⎟⎟ ∈ S h ( − γ , γ
⎝ y⎠
2
that is
thus
−γ ≤
y ≥ −γ
2
,if
a − x0 − y 0
= y,
b
). □
76
b < 0 then
that is
y ≥ −γ
b = −b ,
hence
hence y ∈ [ − γ , γ ]
Proposition (2.4)[2]
Let S a ,b = { ⎛⎜⎜ ⎞⎟⎟ :
x
⎝ y⎠
Henon map H a ,b , if
x ≤ C a , b , y ≤ C a ,b }
b≠0
be a closed region in R 2 ,for
then H a ,b (S a ,b )= P
⎛a⎞
⎜ ⎟
⎜b⎟ I
⎜C ⎟
⎝ ⎠
S h ( − C, C ).
Proof: From definition of S a ,b , we have S a ,b = S h ( − C, C ) I S v ( − C, C ).
So
(2.9)
Now
H a,b (S a,b )=
by
lemma
H a,b (S h ( − C, C )) I
H a,b (
H a ,b (S h ( − C, C )) ⊂
(2.3)(ii)
S v ( − C, C ))
.
⎛a⎞
⎜ ⎟
⎜b⎟
⎜C ⎟
⎝ ⎠
.
P
(2.10)
By lemma (2.3)(iii),since H a,b is diffeomorphism , we get that
P
⎛a⎞
⎜ ⎟
⎜ b ⎟ ⊂ H a,b (S h ( − C, C )
⎜C ⎟
⎝ ⎠
)
.
(2.11)
Hence from (2.10) and (2.11), we get that H a ,b (S h ( − C, C ))= P
⎛a⎞
⎜ ⎟
⎜b⎟
⎜C ⎟
⎝ ⎠
(2.12)
From lemma (2.3)(i) we have H a,b (S v ( − C, C ))= S h ( − C, C )
(2.13)
Now by (2.9), (2.12) and (2.13) H a,b ( S a,b )= P
⎛a⎞
⎜ ⎟
⎜b⎟ I
⎜C ⎟
⎝ ⎠
S h ( − C, C ) . □
Remark (2.5)
We define some region and proof one theorem on Henon map
for a > a0 , the regions are M 1 , M 2 , M 3 and M 4 where :
77
.
− C − y −C − y
x
2
⎝ y⎠
x
M 2 = { ⎛⎜⎜ ⎞⎟⎟ : x ≥ y , y < −C }.
⎝ y⎠
x
M 3 = { ⎛⎜⎜ ⎞⎟⎟ : x ≥ − y , y > C }.
⎝ y⎠
x
M 4 = { ⎛⎜⎜ ⎞⎟⎟ : x > C , y < C }.
⎝ y⎠
M 1 = { ⎛⎜⎜ ⎞⎟⎟ : x <
}.
Theorem (2.6)[3]
Suppose S a ,b = { ⎛⎜⎜ ⎞⎟⎟ :
x
⎝ y⎠
a > a 0 (b)
or
and
y −n ⎯
⎯→ ∞
b > 0 ,then
as
x ≤ C a , b , y ≤ C a ,b }
for all
⎛ x⎞
⎜⎜ ⎟⎟ ∈ R 2 ⎝ y⎠
n⎯
⎯→ ∞ .
78
is a closed region in R 2 .If
S a ,b either
xn ⎯
⎯→ ∞
as
n⎯
⎯→ ∞
Lemma (2.7)
Let H a,b be a Henon map, ⎛⎜⎜
x⎞
⎟⎟ ∈
⎝ y⎠
M 1 and
is strictly decreasing sequence and
Proof: If
y0 ≠ C
Case 1:
y0 < C ,
b > 0 .Then
xn ⎯
⎯→ ∞
as
the sequence
n⎯
⎯→ ∞ .
then we have two cases:
hence
C − y0 = C − y0
thus
, so
x0 <
− (C − y 0 ) − C − y 0
2
=−C
.
x0 < −C < − y 0
(2.14)
⟨ xn ⟩
Now
x1 − x0 = a − by0 − x0 − x0 ,
2
so
x1 − x0 ≤ a + b y 0 − x0 − x0 .
2
(2.15)
Since
x0 < − y 0
,we have y0
< − x0 ,
so by (2.15)
2
x1 − x0 < a − b x0 − x0 − x0
(2.16)
a − b x0 − x0 − x0 = a − ( b + 1) x0 − x0
2
are
x0 =
than
m
− (1 + b ) m (1 + b ) 2 + 4a
2
2
but this equation has two roots which
, one of them is
−C,
for any value less
−C
a − ( b + 1) x0 − x0 < 0 .
2
hence by (2.16)
Case 2:
y0 > C ,
From (2.14), we have
x1 − x0 < 0 ,
hence
that is
x0 < −C ,
so
a − ( b + 1) x0 − x0 < 0 ,
2
x1 < x0 .
C − y0 = y0 − C ,
79
so
x0 <
− ( y0 − C ) − C − y0
2
=−
y0
and
y0 > C ,
thus
.
x0 < − y 0 < −C
(2.17)
Now
2
2
x1 − x0 = a − by0 − x0 − x0 ≤ a + b y 0 − x0 − x0
< a − b x0 − x0 2 − x0 = a − (1 + b ) x0 − x0 2
.
(2.18)
As case 1,
(2.17) ,
a − (1 + b ) x0 − x0
x0 < −C ,
Now since
x0 <
so
2
= 0 has two roots one of them is
a − (1 + b ) x0 − x0
− (C − y 0 ) − C − y 0
2
2
< 0 , hence by (2.18)
,we have
x0 < − C ,
− C .From
x1 < x0 .
that is
y1 < −C
so
y1 > C .
Hence
(2.19)
On the other hand
x1 < − x0 = − y1 .
From (2.19), we get
x2 < x1 <
.
y1 − C = C − y1
x1 < x0
x1 <
− (C − y1 ) − C − y1
2
and
x0
is a negative real number so
− (C − y1 ) − C − y1
2
. As above we get that
.
(2.20)
Now y1 > C , from (2.20) ,we have x1 = y 2 < −C so y 2 − C = C −
(2.21)
Also x2 < x1 and x1 is a negative real number so x2 < − x1 = − y2 .
From (2.21) we get
x2 <
− (C − y 2 ) − C − y 2
2
.As (2.18), we get
y2
.
x3 < x 2 .
continuing in this procedure, we get that for a positive integer n
x n < x n −1 < … < x 2 < x1 < x0 , that is ⟨ x n ⟩ is strictly decreasing sequence .
For the second part ,if possible ⟨ xn ⟩ is bounded, since it is monotone,
we get
⟨ xn ⟩
convergent , since
x n = y n +1 ,
80
that is
⟨ xn ⟩ ,
also convergent ,
H a ,b is continuous. So there exits a fixed point for H a ,b in M 1 which is
contradiction, hence
is not bounded ,that is
⟨ xn ⟩
xn ⎯
⎯→ ∞
as
n⎯
⎯→ ∞ .□
Remark (2.8)[6]
In theorem (2.7) the equality holds only for x0 = −C , y0 = mC .
Proof: If b ≥ 0 , then x1 − x0 = a − b y 0 − x0 2 − x0 if x0 = −C , y 0 = −C , then
x1 − x0 = a + (1 + b )C − C 2 ,
If
b < 0,
then
by lemma (2.1)
2
x1 − x0 = a + b y 0 − x0 − x0
x1 − x0 = a + (1 + b )C − C 2 ,
x1 − x0 = 0 ,
that is
x1 = x0 .
if x0 = −C , y0 = C , then
by lemma (2.1)
x1 − x0 = 0 ,
that is
x1 = x0 .
□
Lemma (2.9)
Let H a,b
b > 0 .Then
y −n ⎯
⎯→ ∞
Proof: Let
−1
be the inverse of Henon map , suppose
the sequence
as
is strictly decreasing sequence and
⟨ y −n ⟩
n⎯
⎯→ ∞ .
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ M 2 ,
⎝ y0 ⎠
from remark (2.5)
2
b( y −1 − y 0 ) = by −1 − by 0 = a − x0 − y 0 − by0
x0 ≥ y 0 , y 0 < −C
.
(2.22)
Since
x0 ≥ y 0 ,
⎛ x⎞
⎜⎜ ⎟⎟ ∈ M 2 and
⎝ y⎠
we have
2
2
a − x0 − y 0 − by 0 ≤ a − y 0 − y 0 − by0
= a − (1 + b) y0 − y0 2 .
(2.23)
81
, consider
From (2.22) and (2.23), we get
b( y −1 − y 0 ) ≤ a − (1 + b) y 0 − y 0
2
.
(2.24)
Now the quadratic equation
a − (1 + b) y 0 − y 0
2
= 0 has a negative root
y 0− = − C
if we take another value
y = y 0*
such that
y 0* < −C ,
becomes a − (1 + b) y − y 2 < 0 , and we have y 0 < −C , so a − (1 + b) y0 − y0 2 < 0 .
From (2.24), we get that b( y −1 − y 0 ) < 0 ,since b > 0 , we get y −1 < y0 < −C
and since
x −1 = y 0 ,we
have
y −1 < x −1
that is
⎛ x −1 ⎞
⎜⎜ ⎟⎟ ∈ M 2
⎝ y −1 ⎠
.
Now if possible y −k < y −( k −1) < … < y −1 < y0 , since
y − k < −C .
Also
we
have
x − k = y −( k −1) ,
(2.25)
2
y 0 < −C ,
we have
.
y −k < x−k
2
b( y − ( k +1) − y − k ) ≤ a − x − k − y − k − by − k < a − y − k − y − k − by −k
= a − (1 + b) y −k − y −k 2 .
(2.26)
Now the quadratic equation
−
=−C
y −k
a − (1 + b) y −k − y −k
if we take another value
a − (1 + b) y − y 2 < 0
and we have
(2.26) we get that
we get that
y = y −* k
y − k < −C ,
2
= 0 has a negative root
such that
so
is
becomes
a − (1 + b) y −k − y −k < 0 ,
2
b( y −( k +1) − y − k ) < 0 ,since b > 0 ,
y − n < y − (n −1) < … < y −1 < y 0 ,that
y −* k < −C
we get
⟨ y −n ⟩
so from
y −( k +1) < y − k ,
so
is strictly decreasing
sequence .
For the second part ,if possible
we get
⟨ y −n ⟩ convergent
,since
⟨ y −n ⟩
is bounded, since it is monotone
x −( n +1) = y − n ,that
is
⟨ x −n ⟩
also convergent ,
H a,b is continuous so there exits a fixed point for H a,b in M 2 which is
82
contradiction hence
⟨ y −n ⟩
not bounded that is
y −n ⎯
⎯→ ∞
as
n⎯
⎯→ ∞ .□
Lemma(2.10)
The region M 3 maps into M 2 under backward iteration of Henon
map H a ,b , provided a > a0 , b > 0 .
Proof: Let
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ M 3 ,
⎝ y0 ⎠
hence by remark (2.5) we have
x0 ≥ − y 0
and
y 0 > C a ,b
so
.
2
b( y −1 − y 0 ) = by −1 − by 0 = a − x0 − y 0 − by0
(2.27)
Since
x0 ≥ − y 0
and
y 0 > 0 we
have
− by 0 < by 0
b( y −1 − y 0 ) ≤ a + y 0 − y 0 + by0 = a + (1 + b) y 0 − y 0
2
so by (2.27), we have
.
2
(2.28)
Now by lemma (2.1)(i) the quadratic equation
positive real root
y 0+ = C ,for
negative and we have
b > 0 ,we
get
y −1 < y 0 and
any value
y0 > C
since
Now we have to show that
we have
y 0* > C
y0 − (1 + b) y 0 − a = 0
2
this quadratic equation is
so from (2.28)
x −1 = y 0 ,we
have
2
y −1 < x −1
b( y −1 − y 0 ) < 0 ,since
.
2
since
y −1 < −C
has a
y −1 + C =
a − x0 − y 0
+ C , y0 > 0 ,
b
2
b( y −1 + C ) = a − x0 − y 0 + bC < a + y 0 − y 0 + by 0
= a + (1 + b) y − y 2 .
(2.29)
As above and since
we get that
y0 > C ,
we have
b( y −1 + C ) < 0 ,since b > 0 ,
a + (1 + b) y − y 2 < 0
we get
83
, hence by (2.29)
⎛x ⎞
y −1 < −C ,hence ⎜⎜ −1 ⎟⎟ ∈ M 2 .□
⎝ y −1 ⎠
Lemma (2.11)
The region M 4 maps into M 1 under forward iteration of the
Henon map H a,b , provided a > a0 , b > 0 .
Proof: Let
⎛ x0 ⎞
⎜⎜ ⎟⎟ ∈ M 4
⎝ y0 ⎠
hence from remark (2.5), we have
y0 < C
and
x0 > C
from
definition
of
Henon
map
2
x1 + C = a − by 0 − x0 + C
.
(2.30)
Since x0 > C , we have
x0 > C 2 .
2
Furthermore
y0 < C
so
− by 0 < bC .
Now from (2.30) and (2.31) we get that
x1 + C < a + bC − C 2 + C = a + (1 + b)C − C 2
(2.32)
From lemma (2.1)(i) , C is a positive real root
so
a + (1 + b)C − C 2 = 0 ,
y1 = x0 , x0 > C
from (2.32), we get
and
we
have
y 2 − (1 + b) y − a = 0
x1 + C < 0 ,
x1 < −C ,
that is
we
get
x1 < −C ,since
y1 > x1
.
(2.33)
Now
x0 + x1 = x0 + a − by0 − x0
.
2
(2.34)
Since
x0 > C > y 0
If y0 < 0 then
From
,so if
y0 ≥ 0
x0 > − y 0 that
(2.34),
we
is
then
x0 > y 0 ,hence bx0 > by 0 > −by 0 .
bx0 > −by 0 .
get
x0 + x1 < x0 + a + bx0 − x0
(2.35)
84
2
= a + (1 + b) x0 − x0 2
.
As above
is a positive real root ,
C
2
a + (1 + b) x0 − x0 < 0
x 2 − (1 + b) x − a = 0 ,for x0 > C
, from (2.35) ,we get that, x0 + x1 < 0 that is
x1 < − x0 = − y1 .
Now we have x1 < y1 < − x1
so x1 < − y1
, x0 =
y1 > C ,which
becomes
⎛ x1 ⎞
⎜⎜ ⎟⎟ ∈ M 1 .□
⎝ y1 ⎠
proof of theorem (2.6)
Let
⎛ x⎞
⎜⎜ ⎟⎟ ∈
⎝ y⎠
R 2 - S a ,b , by remark (2.5)
⎛ x⎞
⎜⎜ ⎟⎟ ∈ U 4i =1 M i
⎝ y⎠
so we have the
following cases :
Case I: If
⎛ x⎞
⎜⎜ ⎟⎟ ∈ M 1 ,
⎝ y⎠
by lemma (2.7) the sequence
decreasing sequence and
Case II: If
⎛ x⎞
⎜⎜ ⎟⎟ ∈ M 2 ,
⎝ y⎠
⎛ x⎞
⎜⎜ ⎟⎟ ∈ M 2
⎝ y⎠
as
y −n ⎯
⎯→ ∞
as
is strictly
n⎯
⎯→ ∞ .
by lemma (2.9) the sequence
decreasing sequence and
Case III: If
xn ⎯
⎯→ ∞
⟨ xn ⟩
⟨ y −n ⟩ is
strictly
n⎯
⎯→ ∞ .
by lemma (2.10)
⎛ x⎞
⎜⎜ ⎟⎟ maps
⎝ y⎠
into M 3 under
backward iteration of Henon map H a ,b .So from case II the sequence
⟨ y −n ⟩ is
strictly decreasing sequence and
85
y −n ⎯
⎯→ ∞
as
n⎯
⎯→ ∞ .
Case IV: If
⎛ x⎞
⎜⎜ ⎟⎟ ∈ M 4 by
⎝ y⎠
lemma (2.11)
⎛ x⎞
⎜⎜ ⎟⎟ maps
⎝ y⎠
into M 1 under forward
iteration of the Henon map H a ,b so from case I the sequence
strictly decreasing sequence and
xn ⎯
⎯→ ∞
as
n⎯
⎯→ ∞ .
⟨ x n ⟩ is
□
Corollary (2.12)[2]
If
Proof: Let
⎛ x⎞
⎜⎜ ⎟⎟ ∈
⎝ y⎠
⎛ x⎞
⎜⎜ ⎟⎟ ∈ R 2 -S a ,b
⎝ y⎠
y −n ⎯
⎯→ ∞
as
then Per n (H a ,b ) ⊂ S a ,b .
a > a0 (b)
Per(H a ,b ),if
⎛ x⎞
⎜⎜ ⎟⎟ ∉
⎝ y⎠
S a ,b ,then
, by theorem (2.5) either
n⎯
⎯→ ∞ , x n = y n +1 , x − ( n +1) = y − n ,
which is contradiction .So
⎛ x⎞
⎜⎜ ⎟⎟ ∈
⎝ y⎠
xn ⎯
⎯→ ∞
(S a ,b ) C ,that is
as
n⎯
⎯→ ∞ ,
or
so there is no finite orbit for H a ,b of
⎛ x⎞
⎜⎜ ⎟⎟ ∈
⎝ y⎠
⎛ x⎞
⎜⎜ ⎟⎟
⎝ y⎠
S a ,b ,that is Per n (H a ,b ) ⊂ S a ,b .□
Definition (2.13)[2]
For
a > a0 (b) we
⎛a⎞
b by Λ ⎜⎜ ⎟⎟ =
⎝b⎠
define non-escape set of H a ,b with respect to
a
and
x⎞
n⎛ x⎞
⎟⎟ ∈ R 2 : Lim H a,b ⎜⎜ ⎟⎟ ⎯
⎯→ ∞ } C .
n ⎯⎯→ ±∞
⎝ y⎠
⎝ y⎠
{ ⎛⎜⎜
Corollary (2.14)[2]
For the Henon map H a ,b , Λ ⎛⎜⎜ ⎞⎟⎟ =
a
⎝b⎠
Proof: Let
⎛ x⎞
⎛a⎞
⎜⎜ ⎟⎟ ∈ Λ ⎜⎜ ⎟⎟ .To
⎝ y⎠
⎝b⎠
I H a ,b ( S a ,b ) .
m
m∈Z
show that
that
86
⎛ x⎞
⎜⎜ ⎟⎟ ∈ I H a ,b m ( S a ,b ) ,
⎝ y ⎠ m∈Z
if we suppose
⎛ x⎞
⎜⎜ ⎟⎟ ∉ I H a ,b m ( S a ,b ) ,then
⎝ y ⎠ m∈Z
means there exists
r∈Z
H a ,b ( S a ,b ) ∈ (S a,b ) C ,so
r
(2.5)
there exists
such that
k ∈ Z ,such
that
⎛ x⎞
k
⎜⎜ ⎟⎟ ∉ H a ,b ( S a ,b ) ,that
⎝ y⎠
H a ,b ( S a ,b ) ∉ S a ,b ,that
r
by theorem
⎛ x⎞
⎛a⎞
p⎛ x⎞
H a,b ⎜⎜ ⎟⎟ ⎯
⎯→ ∞ as p ⎯
⎯→ ∞ so ⎜⎜ ⎟⎟ ∈ ( Λ ⎜⎜ ⎟⎟ ) C ,that
⎝ y⎠
⎝b⎠
⎝ y⎠
which is contradiction hence
To show that
for all
m in
is
is
⎛ x⎞
⎛a⎞
⎜⎜ ⎟⎟ ∉ Λ ⎜⎜ ⎟⎟
⎝ y⎠
⎝b⎠
⎛ x⎞
⎜⎜ ⎟⎟ ∈ I H a ,b m ( S a ,b ) .
⎝ y ⎠ m∈Z
⎛a⎞
m
I H a ,b ( S a ,b ) ⊂ Λ⎜⎜ ⎟⎟ ,
m∈Z
⎝b⎠
Z, hence H a ,b n ⎛⎜⎜
x⎞
⎟⎟ ∈ S a ,b
⎝ y⎠
let
⎛ x⎞
⎜⎜ ⎟⎟ ∈ I H a ,b m ( S a ,b )
⎝ y ⎠ m∈Z
for all
n∈Z ,
so
⎛ x⎞
m
⎜⎜ ⎟⎟ ∈ H a ,b ( S a ,b )
⎝ y⎠
that means if n is very
large
n⎛ x⎞
H a ,b ⎜⎜ ⎟⎟ ∈ S a ,b ,
⎝ y⎠
Lim
n ⎯⎯→ ±∞
that is
n⎛ x ⎞
H a, b ⎜⎜ ⎟⎟
⎝ y⎠
where
n⎯
⎯→ ∞
or
n⎯
⎯→ −∞
is real number that is
then
⎛ x⎞
Lim H a ,b n ⎜⎜ ⎟⎟ ∈ S a ,b
n ⎯⎯→ ±∞
⎝ y⎠
⎛ x⎞
⎛a⎞
⎛a⎞
⎜⎜ ⎟⎟ ∈ Λ⎜⎜ ⎟⎟ so I H a ,b m ( S a ,b ) ⊂ Λ⎜⎜ ⎟⎟ ,
m∈Z
⎝ y⎠
⎝b⎠
⎝b⎠
⎛a⎞
m
Λ ⎜⎜ ⎟⎟ = I H a ,b ( S a ,b ) .□
m∈Z
b
⎝ ⎠
Referencess
[1] Adrian ,S. An Application of The Least –Square Method to System
Parameters Extraction from Experimental Data, 24 November 2001.Preprint.
[2] Alaaddin Al-Dhahir,the Henon Map. http://doc.utwente.nl/fid/1502
87
so
[3] Devaney ,R.L.An Introduction to Chaotic Dynamical Systems ,Second
education ,Addison -Wesley ,1989.
[4] Devaney ,R.L .And Nitecki,Z.Shift Automorphisms in The Henon Mapping
,commun.Math,phys 67, (1979)137-146
[5] ] Gulick ,D.Encounters with chaos .McGraw –Hill,Inc.U.S.A,1992.
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sydniy, Applied Mathematics ,Math 2006 . Preprint.
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