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

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...