Henon map

Invertible 2D Henon map is
    x' = a + x2 + by,
    y' = x
.
Unfortunately they use different parametrizations for the map. E.g. we will get -x2 term after substitution (a,b,x,y) → (-a,b,-x,-y).
Inverse Henon map is
    x = y',
    y = (x' - y' 2 - a)/b
.
The map is decomposed in three simple operations:
uniform squeezing b times in the y direction (to the left the squeezing is omitted, i.e. b = 1)
    x' = x,
    y' = by
,
bending in the same direction
    x' = x,
    y' = a + x2 + y

reflection in the diagonal y = x
    x' = y,
    y' = x
.
To the left you see mapping of the (± 3.9, ± 3.9) square. Next iterations (press "+" button) will draw horseshoe ends together and mix red and blue regions.

Click mouse to zoom the image.

The map has two fixed points
    x1 = y1 = (1 - b)/2 + [(1 - b)2/4 - a]1/2     x2 = y2 = (1 - b)/2 - [(1 - b)2/4 - a]1/2.
For a = -8, b = 1 we get x = y = ± 2.83. These two points are marked by white crosses (in the blue and yellow regions where colors of the square and horseshoe coincide).
For small perturbation (δx, δy) of the point (x, y) corresponding final deviation is
    (
δx'
δy'
)   =   (
∂ x'/∂ x   ∂ x'/∂ y
∂ y'/∂ x ∂ y'/∂ y
)(
δx
δy
)   =   (
2x   b
1 0
)(
δx
δy
)   =   J (
δx
δy
).
Points of a small circle around (x, y) are mapped into an ellipse around (x', y'). E.g. for real eigenvalues of the matrix J
    λ1,2 = x ± (x2 + b)1/2
principal axis of this ellipse coinside with eigenvectors of the matrix and deformation of the initial circle is determined by the λ1,2 values. Thus for a = 1.4 and b = 0.3 the fixed point x2 = y2 = -0.884 is unstable with λ1 = 0.156 and λ2 = -1.92 . For a = -8, b = 1 correspondingly x = y = ± 2.83 and λ1 = ± 5.83 and 2 = ± 0.17 . To the left you see direct and inverse mappings of small vicinities (with radiuses R) of the fixed points.
Jacobian of the Henon map
    Det(J) =   |
2x   b
1 0
|   =   λ1 λ2   =   -b.
The map is contracting for |b| < 1. All bounded attracting orbits are located in this region and attractors have zero measure. Under iterations of the map ellipses become narrower and elongated. For n-th iterations
    (
δxn
δyn
)   =   (
2xn-1   b
1 0
)(
δxn-1
δyn-1
)   =   [ i=0,n-1 (
2xi   b
1 0
)](
δxo
δyo
).
As since one eigenvalue of the matrix product grows and the other decreases therefore it is rather difficult to compute both values accurately (due to roundoff errors). But one can replace the big eigenvalue by Sp(J) = λ1 + λ2 ≈ λmax with good accuracy. Then the biggest Lyapunov exponent is approximately
    Lmax = 1/n ln|λmax| ≈ 1/n ln|Sp(J)|.
Contents   Previous: Smale horseshoe and 2D strange repellers   Next: Homoclinic strucrures and horseshoe
updated 3 July 07