Skip to main content

differential equations - How to access and change the state variable values in NDsolve`Iterate?


Essentially I am trying to calculate the Lyapunov exponent the naive way; i.e., using the method described here by Vaggelis_Z.


In each pass of the loop over time a new trajectory is spawned by adding a small displacement vector to the state and running it for a single time step. This is repeated at each time step after renormalisation of the displacement vector.


While invoking the complete NDSolve within the loop works OK, I can see that the costs for a 10+ dimensional system of ODEs will be prohibitive.


I want to speed up things by accessing and changing the state variables and resubmitting them in NDSolve`Iterate inside the loop. How can I do that?


MarcoB, here's the code I have got


 (* original ODE's for the Sprott system *)
f1 = (y1[t]*z1[t] + 0.01);
f2 = x1[t]*x1[t] - y1[t];

f3 = 1 - 4*x1[t] ;
(* cloned ODE's for shadowing the system *)
f4 = (y2[t]*z2[t] + 0.01);
f5 = x2[t]*x2[t] - y2[t];
f6 = 1 - 4*x2[t] ;
(* starting initial conditions for the original trajectory *)
x10 = 1; y10 = 1; z10 = 0;
dx0 = 10^-8;
(* starting initial conditions for the shadowing trajectory *)
x20 = x10 + dx0; y20 = y10; z20 = z10;

tin = 0; tfin = 1000;
dt = .01;
iterat = tfin/dt;
acc = 12;

lexp = {};
sum = 0;
d0 = Sqrt[(x10 - x20)^2 + (y10 - y20)^2 + (z10 - z20)^2];
Timing[For[i = 1, i < iterat, i++,
eqns = {x1'[t] == f1, y1'[t] == f2, z1'[t] == f3, x2'[t] == f4,

y2'[t] == f5, z2'[t] == f6, x1[0] == x10, y1[0] == y10,
z1[0] == z10, x2[0] == x20, y2[0] == y20, z2[0] == z20};
sol = NDSolve[
eqns, {x1[t], y1[t], z1[t], x2[t], y2[t], z2[t]}, {t, 0, dt},
MaxSteps -> Infinity, PrecisionGoal -> acc, AccuracyGoal -> acc];
(* solution functions *)
xx1[t_] = x1[t] /. sol[[1]];
yy1[t_] = y1[t] /. sol[[1]];
zz1[t_] = z1[t] /. sol[[1]];
xx2[t_] = x2[t] /. sol[[1]];

yy2[t_] = y2[t] /. sol[[1]];
zz2[t_] = z2[t] /. sol[[1]];
d1 = Sqrt[(xx1[dt] - xx2[dt])^2 + (yy1[dt] -
yy2[dt])^2 + (zz1[dt] - zz2[dt])^2];
sum += Log[d1/d0];
lyap = sum/(dt*i);
AppendTo[lexp, {dt*i, Log10[lyap]}];
(* components of displacement vector normalised to d0 *)
vc1 = (xx1[dt] - xx2[dt])*(d0/d1);
vc2 = (yy1[dt] - yy2[dt])*(d0/d1);

vc3 = (zz1[dt] - zz2[dt])*(d0/d1);
(*new initial conditions for the original trajectory *)
x10 = xx1[dt];
y10 = yy1[dt];
z10 = zz1[dt];
(* new initial conditions spawned for the shadowing trajectory *)
x20 = x10 + vc1;
y20 = y10 + vc2;
z20 = z10 + vc3;
i = i++;

If[Mod[dt*i, 10] == 0,
Print[" t = ", dt*i, " , ", "largest Lyapunov = ", lyap]]]]

Answer



The run time of the code in the question is 202 seconds on my PC. Even before employing the component structure of NDSolve, it is helpful to eliminate unnecessary code to both increase speed and increase readability. Doing so yields,


f1 = (y1[t]*z1[t] + 0.01);
f2 = x1[t]*x1[t] - y1[t];
f3 = 1 - 4*x1[t] ;
f4 = (y2[t]*z2[t] + 0.01);
f5 = x2[t]*x2[t] - y2[t];
f6 = 1 - 4*x2[t] ;

x10 = 1; y10 = 1; z10 = 0; dx0 = 10^-8;
x20 = x10 + dx0; y20 = y10; z20 = z10;
tin = 0; tfin = 1000; dt = .01; iterat = Round[tfin/dt]; acc = 12;
eqns = {x1'[t] == f1, y1'[t] == f2, z1'[t] == f3,
x2'[t] == f4, y2'[t] == f5, z2'[t] == f6};

lexp = ConstantArray[0, iterat - 1]; sum = 0;
d0 = Norm[{x10 , y10, z10} - {x20 , y20, z20}];
Timing[Do [
ic = {x1[0] == x10, y1[0] == y10, z1[0] == z10,

x2[0] == x20, y2[0] == y20, z2[0] == z20};
sol = Last /@ Flatten@NDSolve[{eqns, ic}, {x1[dt], y1[dt], z1[dt],
x2[dt], y2[dt], z2[dt]}, {t, 0, dt},
MaxSteps -> Infinity, PrecisionGoal -> acc, AccuracyGoal -> acc];
d1 = Norm[sol[[1 ;; 3]] - sol[[4 ;; 6]]];
sum += Log[d1/d0];
lyap = sum/(dt*i);
lexp[[i]] = {dt*i, Log10[lyap]};
{x10 , y10, z10} = sol[[1 ;; 3]];
{x20 , y20, z20} = {x10 , y10, z10} + (sol[[1 ;; 3]] - sol[[4 ;; 6]]) (d0/d1);

If[Mod[dt*i, 10] == 0,
Print[" t = ", dt*i, " , ", "largest Lyapunov = ", lyap]], {i, iterat - 1}]]

The run time now is only 114 seconds. Although many changes were made, most of the savings were achieved by replacing AppendTo[lexp, {dt*i, Log10[lyap]}] by lexp[[i]] = {dt*i, Log10[lyap]}. AppendTo is notoriously slow for large arrays, here 100 000 elements. Asking NDSolve to return only the final values, {x1[dt], y1[dt], z1[dt], x2[dt], y2[dt], z2[dt]}, instead of {x1[t], y1[t], z1[t], x2[t], y2[t], z2[t]} also helps.


Now, employ the NDSolve Component Structure, which requires only minor modifications to the preceding code.


f1 = (y1[t]*z1[t] + 0.01);
f2 = x1[t]*x1[t] - y1[t];
f3 = 1 - 4*x1[t] ;
f4 = (y2[t]*z2[t] + 0.01);
f5 = x2[t]*x2[t] - y2[t];

f6 = 1 - 4*x2[t] ;
x10 = 1; y10 = 1; z10 = 0; dx0 = 10^-8;
x20 = x10 + dx0; y20 = y10; z20 = z10;
tin = 0; tfin = 1000; dt = .01; iterat = Round[tfin/dt]; acc = 12;
eqns = {x1'[t] == f1, y1'[t] == f2, z1'[t] == f3,
x2'[t] == f4, y2'[t] == f5, z2'[t] == f6};
ic = {x1[0] == x10, y1[0] == y10, z1[0] == z10, x2[0] == x20, y2[0] == y20, z2[0] == z20};
state = First@NDSolve`ProcessEquations[{eqns, ic},
{x1[dt], y1[dt], z1[dt], x2[dt], y2[dt], z2[dt]}, t ,
MaxSteps -> Infinity, PrecisionGoal -> acc, AccuracyGoal -> acc];


lexp = ConstantArray[0, iterat - 1]; sum = 0;
d0 = Norm[{x10 , y10, z10} - {x20 , y20, z20}];
Timing[Do [
ic = {x1[0] == x10, y1[0] == y10, z1[0] == z10,
x2[0] == x20, y2[0] == y20, z2[0] == z20};
newstate = First@NDSolve`Reinitialize[state, ic];
NDSolve`Iterate[newstate, dt];
sol = Last /@ NDSolve`ProcessSolutions[newstate];
d1 = Norm[sol[[1 ;; 3]] - sol[[4 ;; 6]]];

sum += Log[d1/d0];
lyap = sum/(dt*i);
lexp[[i]] = {dt*i, Log10[lyap]};
{x10 , y10, z10} = sol[[1 ;; 3]];
{x20 , y20, z20} = {x10 , y10, z10} + (sol[[1 ;; 3]] - sol[[4 ;; 6]]) (d0/d1);
If[Mod[dt*i, 10] == 0, Print[" t = ", dt*i, " , ", "largest Lyapunov = ", lyap]],
{i, iterat - 1}]]

Run time now is only 34 seconds. This large additional savings results from moving NDSolve`ProcessEquations out of the loop


Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....