Henon Map Fixed Points Plot versus Iterations and Plot of the Map

Question

For the Henon Map: $$ x_{n+1}=y_n+1-αx_n^2, \quad y_{n+1}=β x_n$$ I would like to discover period doubling bifurcations by varying the parameter $α$ and plotting $x_n,y_n$ versus the iterations $n$ of the map and then print below the values of the points in order to distinguish period cycles.

For one dimensional systems of the form $x_{n+1}=f(x_n)$ the period-$2$ cycles happen when the system: $$ x_1=f(x_2) \\ x_2=f(x_1) $$ has a unique solution. In other words, the trajetory "jumps" between $x_1$ and $x_2$ so while iterating the map they keep appearing (until a certain value of the parameter of the system). I have managed to do something similar for the map

$$ x_{n+1}=μ-x_n^4$$ as shown below:

Manipulate[Module[{list = NestList[μ - #^4 &, x0, 100]}, list2 = list;
Column[{ListLinePlot[list, PlotRange -> {-1, 1.5}, 
ImageSize -> {450, 375}], 
TableForm[Transpose@{Range[86, 101], list[[-16 ;;]]}, 
TableHeadings -> {None, {"point", "x"}}]}]], {{μ, 0.2, 
"parameter μ"}, 0, 4, 
Appearance -> "Labeled"}, {{x0, 0.4, 
"Initial \!\(\*SubscriptBox[\(x\), \(0\)]\)"}, 0, 1, 
Appearance -> "Labeled"}] 

and one can see the period two cycle (red and green are the points that repeat themselves) for a certain value of $μ$.

For a 2D system, in our case the Henon map, period-$2$ cycle means that the system: $$ 1)x_1=y_2+1-αx_2^2,\quad y_1=β x_2 \\ 2)x_2=y_1+1-αx_1^2, \quad y_2=β x_1 $$ has a unique solution and that this solution consists of two pairs of points $(x_1,y_1)$ and $(x_2,y_2)$. In this case the trajectory jumps between these two points on the plane (while at the 1D case it would be on a line).

I am actually curious how one would do it. I mean now, it has to return a pair of points each time right?

Finally, I would like to plot the Henon map after lets say $n=100$ iterations for $β=0.3$ fixed and $α$ varying until $. How can I do that?

Thank you all in advance!

EDIT Since the Henon Map is defined as above $$ x_{n+1}=y_n+1-αx_n^2, \quad y_{n+1}=β x_n$$ all I actually need is to do the same but just for the $x_{n+1}$ of the system because (and I just realized this) $y_{n+1}$ is just $β$ times the $x_n$ part! Therefore, I just want to plot $x_{n+1}$ vs $n$ and then I can deduce the value of $y_n$ but definition of the map!

I guess that is correct, right?

Answer 1

Here is the equivalent of your 1-d approach, using @ ce's henon

henon[alpha_, beta_][{x_, y_}] := {y + 1 - alpha x^2, beta x}

Manipulate[
  list = NestWhileList[henon[a, b], {1, 1}, Max[Abs[#]] < 200 &, 1, 2000];
  ListPlot[list[[-Min[20, Length@list] ;;]], 
      PlotRange -> All], {{a, -.3}, -1, 1}, {{b, -.4}, -1, 1}]

the trick here is to use NestWhile set up to abort when the sequence diverges.

I didn't have any luck finding solutions, but it should get you started.

Edit: another approach:

(this take a few minutes)

err[a_?NumericQ, b_] :=
 StandardDeviation[#[[-Min[20, Length@#] ;;, 1]] &@
   NestWhileList[henon[a, b], {1, 1}, Max[Abs[#]] < 200 &, 1, 2000]]
s = NMinimize[ err[a, b] , {a, b}]

{0., {a -> -0.329763, b -> -0.485251}}

 NestWhileList[henon[a, b] /. s[[2]], {1, 1}, 
    Max[Abs[#]] < 200 &, 1, 20]

{{1, 1}, {2.32976, -0.485251}, {2.30463, -1.13052}, {1.62096, -1.11832}, {0.748128, -0.786571}, {0.397995, -0.36303}, {0.689205, -0.193128}, {0.963511, -0.334437}, {0.971699, -0.467544}, {0.843817, -0.471518}, {0.763282, -0.409463}, {0.782657, -0.370383}, {0.831613, -0.379785}, {0.848273, -0.403541}, {0.833745, -0.411625}, {0.817603, -0.404575}, {0.815862, -0.396743}, {0.822758, -0.395898}, {0.827328, -0.399244}, {0.826469, -0.401462}, {0.823783, -0.401045}}

Answer 2

First define the map:

henon[alpha_, beta_][{x_, y_}] := {y + 1 - alpha x^2, beta x}

And then you can do something like

list = NestList[henon[1.4, 0.3], {1, 1}, 10000];
ListPlot[list]

It is straightforward to wrap this in Manipulate.