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I'm trying to solve an ODE system with the NDSolve method. This my ODE system, with the BC and the functions V and dV defined:

ODE system, with BC in (4) and the function V(B) and dV(B) defined in (5)

    s = NDSolve[{F'[
  t] == -(2 /Sqrt[3]) (F[t]^2/2 + a[t]^2*  V[B[ t]])^(1/2)*F[t] - 
  a[t]^2*dV[B[t]], 
a'[t] == a[t]/Sqrt[3] (F[t]^2/2 + a[t]^2*  V[B[ t]])^(1/2), 
B'[t] == F[t], F[tmin] == Subscript[v, 0], a[tmin] == a0 , 
B[tmin] == Subscript[B, 0] }, {F, a, B}, {t, tmin, tmax}, Method->"ExplicitRungeKutta"]

Where the solution functions are:

F(t), a(t), B(t)

I need the solutions for a big interval, like:

t = [-10^(10), -10^(-10)]

So:

tmin = -10^(10)
tmax = -10^(-10)

But the system seems to be stiff. I'm trying to replace the "ExplicitRungeKutta" method for "StiffnessSwitching method", replacing the method with:

Method -> {"StiffnessSwitching", Method -> {"ExplicitRungeKutta", Automatic}}

But the system seems to remain stiff.

The complete code is:

Needs["DifferentialEquations`NDSolveProblems`"];
Needs["DifferentialEquations`NDSolveUtilities`"];
Needs["NumericalCalculus`"];

V[B_] := 3;
dV[B_] :=  V'[B];

B0 = 1;
F0 = 0;
a0 = 10^(-10);

tmin = -10^(10);
tmax = -10^(-10);

  s = NDSolve[{F'[
  t] == -(2 /Sqrt[3]) (F[t]^2/2 + a[t]^2*  V[B[ t]])^(1/2)*F[t] - 
  a[t]^2*dV[B[t]], 
a'[t] == a[t]/Sqrt[3] (F[t]^2/2 + a[t]^2*  V[B[ t]])^(1/2), 
B'[t] == F[t], F[tmin] == F0, a[tmin] == a0 , B[tmin] == B0 }, {F,
 a, B}, {t, tmin, tmax},  Method->"ExplicitRungeKutta"]

Question: Is there any way to obtain the solutions of these equations without stiffness? What method should I use?

I'd appreciate much all the answers.

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  • $\begingroup$ What if you just use Method -> "StiffnessSwitching" without invoking RK at all? $\endgroup$
    – J. M.'s lack of A.I. ♦
    Jun 21, 2016 at 23:03
  • 2
    $\begingroup$ (1) What's v0? (2) What if it's a singularity and not stiffness? $\endgroup$
    – Michael E2
    Jun 22, 2016 at 1:02
  • $\begingroup$ Are tmin & tmax switched, or is tmin supposed to be greater than tmax? $\endgroup$
    – Michael E2
    Jun 22, 2016 at 3:09
  • $\begingroup$ V0 was F0, sorry for the mistake. And both variables tmin and tmax were negative (another mistake writting the code here). $\endgroup$
    – Guillermo Martínez Somonte
    Jun 22, 2016 at 23:57
  • $\begingroup$ And the stiffnessSwitching method alone does not solve the (aparent) stiffness problem. $\endgroup$
    – Guillermo Martínez Somonte
    Jun 22, 2016 at 23:59

1 Answer 1

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Let's read the error message:

NDSolve`Iterate::ndsz: At t == -1.*10^10, step size is effectively zero; singularity or stiff system suspected.

The "step size is effectively zero" means that in floating-point arithmetic t + dt is equal to t for the computed time step dt. If this is the problem, increasing working precision might help.

de = {F'[t] == -(2/Sqrt[3]) (F[t]^2/2 + a[t]^2*V[B[t]])^(1/2)*F[t] - 
     a[t]^2*dV[B[t]], 
   a'[t] == a[t]/Sqrt[3] (F[t]^2/2 + a[t]^2*V[B[t]])^(1/2), 
   B'[t] == F[t]};
ics = {F[tmin] == F0, a[tmin] == a0, B[tmin] == B0};
s = 
 NDSolve[{de, ics}, {F, a, B}, {t, tmin, tmax},
  PrecisionGoal -> 8, AccuracyGoal -> 8, WorkingPrecision -> 20,
  Method -> "StiffnessSwitching"]

Mathematica graphics

Note: The first of the OP's differential equations has

F'[t] == <..stuff..> * F[t]

because dV[B] is zero and the second terms goes away. Since F[t] starts out at F[tmin] == 0, that means F[t] is a constant 0 (zero). Thus the third differential equation B'[t] == F[t] and its initial condition imply B[t] is a constant 1 (one).

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