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

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? • I removed my answer for the time being. It will have to be up to someone else to find period doubling bifurcations. Jun 30 '16 at 11:53 • @C.E. I can still the plot command you helped me with while giving time for someone to help with the period doubling. Could you please post again? Jun 30 '16 at 12:02 • ok, I've undeleted my answer. Jun 30 '16 at 12:15 • You have lumped together a simple mathemaitca question with one that is arguably out of scope of this site. At the very least you should show how you compute the "period" of the sequence. Jun 30 '16 at 14:51 • @george2079 Ok, I mistakenly took for granted that the period doubling was something widely known. I will edit the question. Jun 30 '16 at 14:53 ## 2 Answers 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[], {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}} • Thank you for taking time to help me with this one. On the first piece of code that you provided I tried to run it but it returns "aborted" error. I really dont know why this happens :/ The second part of your answer runs perfectly, but to be honest, I do not understand what exactly is it doing. Is it predefining in some way the parameters$α,β$? Because I would like to vary them. I will also edit the question a bit, have a look. Thanks again Jun 30 '16 at 18:45 • It is ok now, I managed to make it work. I used from your code just to print the$x\$ values and tried to see it as 1D. I find the correct results. Again, many many thanks! Jun 30 '16 at 20:53

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.

• Thank you for your answer. Does {1,1} inside your code stand for initial conditions? Also, what about the main part of the question? Do you know any possible way that this can be done? Again, thanks! Jun 30 '16 at 11:33
• Yes, {1,1} is the initial values. I don't know how to find the period doublings at this moment. Jun 30 '16 at 12:16
• Is it possible that you also put it in Manipulate? I know it sounds a bit silly, but I am doing something wrong with it. Jun 30 '16 at 12:39
• @Mitscaype Like this: Manipulate[ ListPlot[NestList[henon[alpha, 0.3], {1, 1}, 1000]], {alpha, 1.3, 1.4}, ContinuousAction -> False] but watch out, it seems like it will overflow if alpha is larger than 1.4, if you do enough iterations. Jun 30 '16 at 12:48