Lorenz map for the Rössler system [duplicate]

Question

Possible Duplicate:
How to find all the local minima/maxima in a range

I have the solution of the following non-linear system:

sol1 = NDSolve[
  {x'[t] == -(y[t] + z[t]),
   y'[t] == x[t] + 0.2 y[t],
   z'[t] == 0.2 + x[t] z[t] - 5.7 z[t],
   x[0] == 1, y[0] == 1, z[0] == 1
  },
  {x, y, z},
  {t, 0, 100}
  ]

How can I find the $k^{th}$ local maximum of $z(t)$, i.e. $z(k)$, and then plot $z(k+1)$ vs. $z(k)$? There is an example in the "Mapping local maxima" section in Rössler attractor's wiki page. I am working with Wolfram Mathematica 8.0.

Answer 1

The maxima will occur at points where the derivative is zero and, except in special cases, they will alternate with minima. You can easily detect where the derivative zero using event detection. In V9, you do this like so.

Clear[x, y, z, sol, pts];
{{sol}, {pts}} = Reap[
   NDSolve[
    {x'[t] == -(y[t] + z[t]),
     y'[t] == x[t] + 0.2 y[t],
     z'[t] == 0.2 + x[t] z[t] - 5.7 z[t],
     x[0] == 1, y[0] == 1, z[0] == 1,
     WhenEvent[z'[t] == 0, Sow[t]]
     },
    {x, y, z}, {t, 0, 100}
    ]];
z = z /. sol;
maxPts = Last /@ Partition[pts, 2];
Plot[z[t], {t, 0, 100}, PlotRange -> All,
 Epilog -> Point[{#, z[#]} & /@ maxPts]]

Note that the extremes are found reliably during the solution of the differential equation and there's no need to numerically solve equations involving interpolating functions afterward.

Now that you've got the maxima in a list, you can do anything you want with them.

Answer 2

I do not understand the last part about plot z(k+1) vs. z(k) since these are just numbers. But this is a histogram of the result

eqs = {x'[t] == -(y[t] + z[t]), 
       y'[t] == x[t] + 0.2 y[t], 
       z'[t] == 0.2 + x[t] z[t] - 5.7 z[t]};

ic = {x[0] == 1, y[0] == 1, z[0] == 1};

sol1 = First@NDSolve[Flatten[{eqs, ic}], {x, y, z}, {t, 0, 100}];
f = z /. sol1;
Plot[f[t], {t, 0, 100}, PlotRange -> All]
data = FindMaximum[{f[t], 0 <= t <= 100}, {t, #}] & /@ Range[1, 100];
data[[All, 2]] = t /. # & /@ data[[All, 2]];

Histogram[data[[All, 2]], 50]

The peaks and their locations are all now in data.

data[[1 ;; 3]]    
(* {{1., 5.86338*10^-13}, {1., 5.86338*10^-13}, {0.0590077, 5.70096}} *)

You can do anything with it. You can for example sort them by value:

SortBy[data, #[[2]] &]

or plot them, etc...

list = Reverse[SortBy[data, #[[2]] &]];
ListPlot[list, Filling -> Axis]

again, I do not understand your main question of plotting z(k) vs. z(k+1) but may be this helps.