3
$\begingroup$

In an attempt to evaluate the point at which the cycle 2 becomes unstable for the given map:

\begin{equation} x_{n+1}=μ-x_n^4=f(x_n), \quad μ \in \mathbb{R} \end{equation}

I have so far managed to locate the two numerical values $\hat{x}_1, \hat{x}_2$ for which the period 2 trajectory appears:

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"}] 

But now I would like to locate the value of $μ$ for which this 2-cycle ceases to be stable and jumps onto the next one, a 4-cycle. The condition which has to hold is the following: \begin{equation} f'(\hat{x}_1)f'(\hat{x}_2)=-1 \Leftrightarrow 16\hat{x}_1^3 \hat{x_2}^3=-1 \end{equation} where $\hat{x}_i,i=1,2$ would be an expression of $μ$.

The problem is that Mathematica is not able to solve analytically the system: \begin{equation} \begin{cases} f(\hat{x}_1)=\hat{x}_2 \\ f(\hat{x}_2)=\hat{x}_1 \end{cases} \end{equation} in order to get the trajectory of the 2-cycle in a closed form, which will depend only on the parameter $μ$, which I can then plug it in the condition and find the value of that $μ$. What I get instead for

Reduce[{y == μ - z^4, z == μ - y^4}, {y, z}, Reals]

is the solution in Root form:

 ((μ == Root[27 + 256 #1^3 &, 1] && 
 y == Root[-μ + #1 + #1^4 &, 1]) || (Root[27 + 256 #1^3 &, 1] < 
 μ <= Root[-125 + 256 #1^3 &, 
 1] && (y == Root[-μ + #1 + #1^4 &, 1] || 
 y == Root[-μ + #1 + #1^4 &, 2])) || (μ > 
 Root[-125 + 256 #1^3 &, 1] && (y == Root[-μ + #1 + #1^4 &, 1] ||
 y == Root[
 1 - μ^3 - μ^2 #1 - μ #1^2 - #1^3 + 3 μ^2 #1^4 + 
 2 μ #1^5 + #1^6 - 3 μ #1^8 - #1^9 + #1^12 &, 1] || 
 y == Root[-μ + #1 + #1^4 &, 2] || 
 y == Root[
 1 - μ^3 - μ^2 #1 - μ #1^2 - #1^3 + 3 μ^2 #1^4 + 
 2 μ #1^5 + #1^6 - 3 μ #1^8 - #1^9 + #1^12 &, 2]))) && 
 z == -y^4 + μ

I understand that I am not able to solve this system analytically, but how can I then approximate this particular $μ$ value.

If I plug in the condition above the numerical values of $\hat{x}_1,\hat{x}_2$, I would have no $μ$ involved.

The point of this whole procedure is to calculate somewhow the universal $\delta$ Feigenbaum constant for this representative of the quartic equivalence class of discrete mappings.

Any help would be greatly appreciated. Thanks!

$\endgroup$
  • 1
    $\begingroup$ Replace the = with == in your Reduce and also check documentation for those respective symbols. $\endgroup$ – PlatoManiac Mar 10 '16 at 22:22
  • $\begingroup$ @PlatoManiac Thank you, you were right. But still, how can I then plug in those non-analytical roots into the condition mentioned above? Sorry for my perhaps naive questions I am still new to Mathematica, trying my best to make decent questions here. $\endgroup$ – Mitscaype Mar 10 '16 at 22:38
  • 1
    $\begingroup$ p = Reduce[{y == \[Mu] - z^4, z == \[Mu] - y^4}, {y, z, \[Mu]}, Reals, Backsubstitution -> True] // FullSimplify // N; First@p $\endgroup$ – Dr. belisarius Mar 10 '16 at 23:48
  • $\begingroup$ @Dr.belisarius Thank you for your help! But I cannot understand fully what did you do there. Mathematica returned to me the value $μ=6.29$. Is that the $μ$ which I would get if I knew the exact expression of $\hat{x}_1,\hat{x}_2$ and solved for $μ$ the condition: $ 16 {\hat{x}_1}^3{\hat{x}_2}^3=-1$ ? $\endgroup$ – Mitscaype Mar 10 '16 at 23:58
  • 1
    $\begingroup$ @Mitscaype p = Reduce[{y == \[Mu] - z^4, z == \[Mu] - y^4}, {y, z, \[Mu]}, Reals, Backsubstitution -> True] // FullSimplify // N; Solve[p[[1]]] $\endgroup$ – Dr. belisarius Mar 11 '16 at 0:13
6
$\begingroup$

Let's start with a brute-force bifurcation diagram to see what's happening, modifying some code from @bbgodfrey's answer to this question:

f[x_, μ_] := μ - x^4;

res[μ_] := 
  DeleteDuplicatesBy[
   NestList[f[#, μ] &, Nest[f[#, μ] &, μ, 1000], 100], 
   Round[#, 10^-6] &];

ListPlot[Catenate[Table[{μ, #} & /@ res[μ], {μ, 0, 1.2, 0.001}]], 
 PlotStyle -> PointSize[Tiny], PlotRange -> All]

enter image description here

Looks like the μ you're looking for is near 1.1. To find a 2-cycle for a given μ numerically, we can try

μ = 1.05;
FindRoot[f[f[x, μ], μ] == x, {x, -0.5}]

(* {x -> -0.162297} *)

The bifurcation occurs when the two-cycle goes through a period-doubling bifurcation (eigenvalue=-1). We can numerically find the the bifurcation point by simultaneously solving for the two-cycle and when it loses stability:

Clear[μ];
FindRoot[{f[f[x, μ], μ] == x, D[f[f[x,μ], μ], x] == -1}, {x, -0.5}, {μ, 1.1}]

(* {x -> -0.359832, μ -> 1.11964} *)

Edit:

To generalize to higher order cycles (e.g. here's where the 4-cycle gives way to an 8-cycle):

FindRoot[{Nest[f[#, μ] &, x, 4] == x, D[Nest[f[#, μ] &, x, 4], x] == -1}, 
    {x, -0.64}, {μ, 1.16}]

(* {x -> -0.646515, μ -> 1.16166} *)

Of course it gets more delicate, so you'll probably want better initial guesses (which I got from looking at a zoomed-in version of the bifurcation diagram).

I don't know of any way to directly calculate the Feigenbaum constant here.

Edit 2:

I applied this approach to the first 7 period-doublings. The ratio of bifurcation points as in this wikipedia page looks like {7.90629, 7.25658, 7.3238, 7.28859, 7.28785}.

$\endgroup$
  • $\begingroup$ Actually that is pretty correct. I was guessing the bifurcation values of $μ$ from a $x_{n+1}$ vs $n$ plot and I was around 1.13 for a 4-cycle. But to locate the 8-cycle is pretty hard this way. Is there also a code for the Feigenbaum $\delta$ in this case? $\endgroup$ – Mitscaype Mar 11 '16 at 0:43
  • 1
    $\begingroup$ I edited my answer to give code to generalize to finding the bifurcation of an n-cycle. I also tried calculating that δ for you: looks like around 7.288 if I didn't mess that up. $\endgroup$ – Chris K Mar 11 '16 at 2:49
  • $\begingroup$ Found this article which says it's 7.28468..., which makes me feel OK. $\endgroup$ – Chris K Mar 11 '16 at 3:04
  • $\begingroup$ Well, I guess I will have to thank like, forever! Yeah, I was able to approximate that $\delta$ too, but only by the first 2 iterations. I was around 8.28, which means wayyyyy out when it comes to this kind of systems. Again, many thanks. $\endgroup$ – Mitscaype Mar 12 '16 at 20:23
  • $\begingroup$ Not a bad paper either. I guess I had missed that somehow. Again, thanks $\endgroup$ – Mitscaype Mar 12 '16 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.