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
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