⎛ a − by − x 2 ⎞
⎟
⎟
x
⎝
⎠
The Non Wandering Points of Henon map H a,b ⎛⎜⎜ ⎞⎟⎟ = ⎜⎜
x
⎝ y⎠
Iftichar Mudar Talb and Kawa Ahmad Hassan
Department of Mathematics, College of Education,
Babylon University.
Abstract
In this work we study the non wandering point of Henon map.The main goal
of this work is to prove theorems on the Henon map H a,b , b > 0 .Thus we divide
the plane into three regions such that the union of this three regions covers the
plane. We prove there is no non wandering point for Henon map H a,b , where
b > 0 and a <
− (1 + b) 2
4
.
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 maps denoted by H a,b
given
⎛ x⎞
⎝ y⎠
⎛1 − ax 2 +
bx
⎝
by H a,b ⎜⎜ ⎟⎟ = ⎜⎜
y⎞
⎟ where a , b are real numbers .These maps
⎟
⎠
defined in above are called Henon maps [3]. In this work we, care this form of
⎛ x ⎞ ⎛ a − by − x 2 ⎞
⎟ [1] .Also we study the non wandering point
⎟
x
⎝ ⎠ ⎝
⎠
Henon map H a,b ⎜⎜ ⎟⎟ = ⎜⎜
y
for Henon map H a,b where a <
Definition (1-1) [3]
− (1 + b) 2
by using trapping region.
4
Let V be a subset of R2, and v0 be any element in
2
R2.Consider F:V ⎯
⎯→ R be a map. Further more assume that the first partials
1
of the coordinate maps f and g of F exist at v0 .The differential of F at v0
⎛ ∂f
⎜ (v 0 )
⎜ ∂x
2
is the linear map DF(v0 ) defined on R by DF(v0 ) = ⎜
⎜
⎜ ∂g (v )
⎜ ∂x 0
⎝
∂f
⎞
(v 0 ) ⎟
∂y
⎟
⎟,
⎟
∂g
(v0 ) ⎟⎟
∂y
⎠
for all v in R2 .The determinant of DF(v0 ) is called the Jacobian of F at v 0
and is denoted by J=det DF(v 0 ) .
Example (1-2)[3]
⎛x⎞
2
Let F: R2 ⎯
⎯→ R be given by F ⎜⎜ ⎟⎟ =
⎝ y⎠
we have for f = y and g = a sin( x) + by that
1⎞
⎛ x0 ⎞ ⎛ 0
⎟⎟ = ⎜⎜
⎟⎟ for all
⎝ y 0 ⎠ ⎝ a cos( x0 ) b ⎠
DF ⎜⎜
⎛ x0 ⎞
⎛y
⎞
⎜⎜
⎟⎟ .To find DF ⎜⎜ ⎟⎟
⎝ a sin( x) + by ⎠
⎝ y0 ⎠
∂g
∂f
∂f
∂g
=0,
=1, = a cos( x) , =b, so
∂y
∂y
∂x
∂x
⎛ x0 ⎞
⎜⎜ ⎟⎟ in R2, J= − a cos( x0 )
⎝ y0 ⎠
2- The Non Wandering Point of Henon Map Where
a<−
(1 + b) 2
4
b>0
and
.
The main goal of this section is to prove two theorems on the Henon map
H a,b , b > 0 . To prove these theorems , we divide the plane into three regions
such that the union of this three regions covers the plane. we suppose
− (1 + b) 2
, prove some lemmas with respect to parameters a, b and regions
a0 =
4
till we get the main purpose ,the regions are the following :
⎛ x⎞
⎝ y⎠
S 1 = { ⎜⎜ ⎟⎟ : x, y ∈ R, x ≤ − y }
2
⎛ x⎞
⎝ y⎠
S 2 = { ⎜⎜ ⎟⎟ : x, y ∈ R, x ≥ − y , y ≤ 0 }
⎛ x⎞
⎝ y⎠
S 3 = { ⎜⎜ ⎟⎟ : x, y ∈ R, x ≥ − y , y > 0 }
Definition (2-1)[1]
Let F: R n ⎯
⎯→ R n be a map .A closed region Q ⊂ R n is trapping region for
F if F(Q) is contained in the interior of Q.
Lemma (2-2)
Let a < a 0 , b > 0 .Then S 1 is trapping region for H a,b .
Proof: clearly S 1 is closed region .To show that H a,b ( S 1 ) ⊂ S 1 0 .
⎛x ⎞
⎛ x⎞
Let ⎜⎜ ⎟⎟ ∈ H a,b ( S 1 ), then there exists ⎜⎜ 0 ⎟⎟ in S 1 , such that H a,b
⎝ y⎠
⎝ y0 ⎠
2
x = a − by0 − x0 , y = x0 , since x 0 ∈ R we have two cases :
3
⎛ x0 ⎞
⎜⎜ ⎟⎟ =
⎝ y0 ⎠
⎛ x⎞
⎜⎜ ⎟⎟ so
⎝ y⎠
case1: y 0 ≥ 0 then y0 = y0 , x0 ≤ − y 0 , since b > 0 , we have by 0 ≤ −bx0 .
(2.1)
Thus a + by0 − x0 2 ≤ a − bx0 − x0 2 .
(2.2)
Since − by 0 is a negative real number we have − by 0 < by0 hence
2
2
a − by0 − x0 < a + by0 − x0 .
(2.3)
Now from (2.2) and (2.3) we get x < a − bx0 − x0 2 .
Since a < a 0 from (2.4), we get x < −
(1 + b ) 2
4
(1 + b ) 2
=- [
4
1+ b
=−[
2
− bx 0 − x 0
(2.4)
2
2
+ (1 + b ) x 0 + x0 ] + x0
+ x0 ] 2 + x0 .
So x < x 0 , that is x < y . On the other hand x0 < 0 , so y = − y , x < − y , hence
⎛ x⎞
⎜⎜ ⎟⎟ ∈ S 1 0 ,that is H a ,b ( S 1 ) ⊂ S 1 0 .
⎝ y⎠
Case2: y 0 < 0 then y 0 = − y 0 ,so x0 ≤ y 0 , since b > 0 , we have − by 0 ≤ −bx0
(2.5)
thus a − by0 − x0 2 ≤ a − bx0 − x0 2 .
(2.6)
Now as the same as above case , we get x < x 0 , x0 ≤ y 0 < 0 so y = − y thus
⎛ x⎞
x < − y , hence ⎜⎜ ⎟⎟ ∈ S 1 0 , that is H a ,b ( S 1 ) ⊂ S 1 0 . □
⎝ y⎠
Lemma (2-3)
Let a < a 0 , b > 0 .Then S 2 is trapping region for H a,b
Proof: clearly, S 2 is closed region. To show that H a ,b
⎛ x⎞
⎜⎜ ⎟⎟ ∈ H a,b
⎝ y⎠
−1
−1
−1
.
(S 2 ) ⊂ S 2 0 , let
⎛ x0 ⎞
⎟⎟ in S 2 , such that H a,b
y
⎝ 0⎠
(S 2 ) ,then there exists ⎜⎜
4
−1
⎛ x0 ⎞
⎜⎜ ⎟⎟ =
⎝ y0 ⎠
⎛ x⎞
⎜⎜ ⎟⎟ , so
⎝ y⎠
x = y0 , y =
a − x0 − y 0
b
2
⎛x ⎞
, since ⎜⎜ 0 ⎟⎟ ∈ S 2 we have x 0 ≥ − y 0 and y 0 ≤ 0 that is
⎝ y0 ⎠
x0 ≥ y0 .
(2.7)
Now since b > 0 , a < a 0 from (2.7) we have :
by = a − x0 − y 0 < −
(1 + b) 2
2
− x0 − y 0
4
<−
(1 + b) 2
2
− y0 − y0
4
2
=−[
(1 + b) 2
2
+ y 0 + by 0 − by 0 + y 0 ]
4
(1 + b) 2
2
=−[
+ (1 + b) y 0 + y 0 ] + by 0
4
=−[
(1 + b)
+ y 0 ] 2 + by 0 .
2
Hence by < by 0 ,since b > 0 , we get y < y 0 ,that is y < x , since x = y 0 ≤ 0 , we get
⎛ x⎞
y = − y , y < x . Hence − y < x , so ⎜⎜ ⎟⎟ ∈ S 2 0 , hence H a ,b
⎝ y⎠
−1
(S 2 ) ⊂ S 2 0 . □
Proposition (2-4)[2]
Let a < a 0 , b > 0 .
(i)
H a ,b (S 1 U S 3 ) ⊂ S 1 0
(ii)
⟨ xn ⟩
is strictly decreasing sequence along H
n⎛ x⎞
H a,b ⎜⎜ ⎟⎟ ⎯
⎯→ ∞ as n ⎯
⎯→ ∞ , for all
⎝ y⎠
(iii)
H a ,b
−1
(S 2 U S 3 ) ⊂ S 2 0 .
5
⎛ x⎞
⎜⎜ ⎟⎟ in S 1 .
⎝ y⎠
a ,b
-orbits and
(iv)
⟨ y −n ⟩ is strictly increasing sequence along H
−n ⎛ x ⎞
H a,b ⎜⎜ ⎟⎟ ⎯
⎯→ ∞ as n ⎯
⎯→ ∞ for all
⎝ y⎠
−1
a,b
-orbits
and
⎛ x⎞
⎜⎜ ⎟⎟ in S 2 .
⎝ y⎠
Proof: See [2]
Remark: From Proposition (2-4) we note that S 1 U S 3 is trapping region for
H a,b and S 2 U S 3 is trapping region for H a,b
−1
.
Theorem (2-5)
Let a < a 0 (b) , b > 0 ,the Henon map H a,b has no periodic point for any
period, that is Per n ( H a,b ) = φ .
⎛ x⎞
Proof: Suppose that there exists a periodic point ⎜⎜ ⎟⎟ for H a,b of period n , where
y
⎝ ⎠
2
n ∈ N then must be in R
and from the partition with respect to S 1 , S 2 and
⎛ x⎞
⎛ x⎞
⎛ x⎞
⎝ ⎠
⎝ ⎠
⎝ ⎠
S 3 .Since ⎜⎜ ⎟⎟ ∈ R 2 then either ⎜⎜ ⎟⎟ in S 1 U S 2 or S 3 . If ⎜⎜ ⎟⎟ ∈ S 1 U S 2
y
y
y
, then by lemma(2-4)(ii) ⟨ x n ⟩ is strictly decreasing sequence along H a ,b -orbits
⎛ x⎞
⎛ x⎞
⎝ ⎠
⎝ ⎠
in S 1 and H a,b n ⎜⎜ ⎟⎟ ⎯
⎯→ ∞ as n ⎯
⎯→ ∞ for all ⎜⎜ ⎟⎟ in S 1 and from lemma(2-4)(iv)
y
y
⟨ y −n ⟩ is strictly increasing sequence along H
−n ⎛ x ⎞
H a,b ⎜⎜ ⎟⎟ ⎯
⎯→ ∞ as n ⎯
⎯→ ∞ for all
⎝ y⎠
−1
a,b
orbits in S
2
and
⎛ x⎞
⎜⎜ ⎟⎟ in S 2 that means there is no finite orbit
⎝ y⎠
⎛ x⎞
for any point in S 1 U S 2 , so ⎜⎜ ⎟⎟ has no finite orbit which is contradiction .
y
⎝ ⎠
6
⎛ x⎞
If ⎜⎜ ⎟⎟ ∈ S 3 , by lemma (2-4)(i) H a,b (S 3 ) ⊂ S 1 0 and S 1 0 ⊂ S 1 , thus S 3 maps into
y
⎝ ⎠
⎛ x⎞
⎛ x⎞
⎝ ⎠
⎝ ⎠
S 1 and by our supposition ⎜⎜ ⎟⎟ is periodic, so H a,b ⎜⎜ ⎟⎟ is a periodic point in S 1
y
y
which is contradiction . So there is no periodic point for H a ,b in S 1 U S 2 U S 3
and since S 1 U S 2 U S 3 = R2. We get H a ,b has no periodic point for any
period .□
Definition (2-6)[5]
⎛ p⎞
2
Let F: R2 ⎯
⎯→ R be a map. A point ⎜⎜ ⎟⎟ is called non wandering provided
q
⎝ ⎠
⎛ p⎞
that for every neighborhood U of ⎜⎜ ⎟⎟ , there is an integer n > 0 such that
⎝q⎠
⎛r⎞
⎛r⎞
⎝ ⎠
⎝ ⎠
F n (U) I U ≠ φ .Thus ,there is a point ⎜⎜ ⎟⎟ ∈ U with F n ⎜⎜ ⎟⎟ ∈ U. The set of all non
s
s
wandering points for F is called the non wandering set and is denoted by
Ω (F).
Definition (2-7)[4]
2
2
Let F: R2 ⎯
⎯→ R be a map. Let K + (F) denotes the set of points in R with
bounded forward orbits .Let K − (F) denotes the set of points in R2 with
bounded backward orbits then the set K(F)= K + (F) I K − (F) is called filled
Julia set .
Definition (2-8)[4]
2
Let F: R2 ⎯
⎯→ R be a map. Let J ± (F)= ∂ K ± (F),and J(F)= J + (F) I J − (F)
then J ± (F) is called the forward /backward Julia set and J(F) is the Julia set of F.
Theorem (2-9)[4]
7
Let F be a hyperbolic regular polynomial automorphism of C n with
det DF ≤ 1 .Then Ω (F) = J(F) U { α 1 , α 2 , … , α m }, where the α i are the attracting
periodic points of F.
Theorem (2-10)
Let a < a0 (b) , b > 0 ,the Henon map H a ,b has no non wandering point , that is
Ω ( H a ,b )= φ .
⎛ − 2x − b ⎞
⎟ = b < 1,
0 ⎟⎠
⎝ 1
Proof: since b < 1 and from definition (1-1) det DF = det⎜⎜
⎛ x⎞
⎝ y⎠
for ⎜⎜ ⎟⎟ ∈ R2 we have three cases :
⎛ x⎞
case 1 : If ⎜⎜ ⎟⎟ ∈ S 1 ,from lemma (2-4) part (ii) , ⟨ x n ⟩ is strictly decreasing
⎝ y⎠
sequence along H a,b -orbits and ⟨ xn ⟩ is not bounded below that is ⟨ x n ⟩ ⎯
⎯→ −∞
⎛ x⎞
as n ⎯
⎯→ ∞ for all ⎜⎜ ⎟⎟ in S 1 ,that is there is no element in S 1 such that has
⎝ y⎠
bounded forward orbits ,so by definition (2-7) K + ( H a ,b )= φ and by definition
⎛ x⎞
(2-8) J + ( H a,b )= ∂ ( φ ) for all ⎜⎜ ⎟⎟ in S 1 ,since ∂ ( φ )= φ ,we have J + ( H a,b )= φ
⎝ y⎠
hence by definition (2-8) J(H a,b )= φ I J − ( H a,b ) = φ .
⎛ x⎞
Case 2:If ⎜⎜ ⎟⎟ ∈ S 2 , from lemma (2-4) part (iv) the sequence ⟨ y −n ⟩ is strictly
⎝ y⎠
decreasing sequence along H a ,b
−1
-orbit and it is not bounded below , that is
⎛ x⎞
⟨ y −n ⟩ ⎯
⎯→ −∞ as n ⎯
⎯→ ∞ for all ⎜⎜ ⎟⎟ in S 2 ,that is there is no element in S 2
⎝ y⎠
such that has bounded backward orbit so by definition (2-7) K − ( H a ,b )= φ and
8
⎛ x⎞
by definition (2-8) J − ( H a,b )= ∂ ( φ ) for all ⎜⎜ ⎟⎟ in S 2 ,we have ∂ ( φ )= φ so
⎝ y⎠
J − ( H a ,b )= φ , hence by definition (2-8) we get J(H a ,b )= J + ( H a ,b ) I φ = φ .
⎛ x⎞
Case 3: If ⎜⎜ ⎟⎟ ∈ S 3 ,by lemma (2-4) part (i)
⎝ y⎠
H
a ,b
(S 3 ) ⊂ S
0
1
,hence
H a,b (S 3 ) I S 3 ⊂ S 3 I S 1 0 , since S 3 I S 1 0 = φ , we have H a,b (S 3 ) I S 3 = φ
H a ,b (S 3 ) ⊂ S 1 0 , so S 3 maps into S 1 , hence by case 1 ,we have J + ( H a ,b )= φ . So
⎛ x⎞
from definition (2-8), J(H a,b )= φ I J + ( H a,b ) = φ .Now for all ⎜⎜ ⎟⎟ in R2, we get
⎝ y⎠
J(H a,b )= φ ,by theorem (2-5), H a,b has no periodic point for any period ,hence
by theorem (2-9) Ω ( H a,b )= φ . □
Referencess
[1] Devaney ,R.L.An Introduction to Chaotic Dynamical Systems ,Second
education ,Addison -Wesley ,1989.
[2] Devaney ,R.L .And Nitecki,Z.Shift Automorphisms in The Henon Mapping
,commun.Math,phys 67, (1979)137-146
[3] Gulick ,D.Encounters with chaos .McGraw –Hill,Inc.U.S.A,1992.
[4] Shafikov .R,wolf.C,Filtrations , Hyperbolicity and Dimension for
Polynomial Automorphism of Cn
http://www.math.uwo.ca/~shafikov/research/papers/michigan-4.pdf .
[5] Xlaoling.L.,A. Charactrization of Henon map from R 2 to R 2 .
http://lagrange.math.trinity.edu/tumath/research/studpapers/s33.pdf .
9