Group Abstract Group Abstract

Message Boards Message Boards

Precision in NDSolve

Francois Fayard, Inside Loop
Posted 11 years ago

Hi,

I would like to know if anyone has a clear understanding on the following parameters used to control the precision/accuracy of NDSolve.

  • PrecisionGoal / WorkingPrecision
  • WorkingPrecision

I have been working on the solution of the following differential equation

x'[s] = Cos[theta[s]]
y'[s] = Sin[theta[s]]
theta'[s] = 2 b - Sin[theta[s]] / x[s] (for s >0, and b for s = 0)

with the initial condition x[0] = 0, y[0] = 0, theta[0] = 0.

This differential equation has an exact solution x[s] = (1/b) Sin[b s], y[s] = (1/b) (1 - Cos[b s]), theta[s] = b s. Therefore, I can check the precision of the result. So I launch Mathematica and solve it with b = 1, PrecisionGoal -> 20, WorkingPrecision -> 30. But I can't seem to get something better than 9 digits of precision.

I have attached a Mathematica notebook to prove that.

I kind of understand that there is no way an algorithm can guarantee a given precision. But the documentation around PrecisionGoal looks pretty unclear to me. So here are my questions :

  • Is there a way to get a better precision with Mathematica ?
  • What does PrecisionGoal really control and can we get an idea of how it works ?
Attachments:
POSTED BY: Francois Fayard
5 Replies
Sort By:

From the Mathematica 10 documentation:

    InterpolationOrder   Automatic
POSTED BY: Frank Kampas

Hi Frank,

In this case, the internal rule is not really wise. As you could ask NDSolve[..., PrecisionGoal -> 30, ... ] without having any guarantee on the precision of the solution outside the computations points which are hidden in the "interpolation" object. I find this behaviour disturbing.
POSTED BY: Francois Fayard

According to the documentation, the default interpolating order is "Automatic". This probably means that Mathematica has some internal rule to chose the order.

POSTED BY: Frank Kampas

Hi,

I finally found the solution. NDSolve only guarantees the precision at the points where the solution is computed (which you don't really now expect if you inspect the interpolation object). A linear interpolation is made in between the points is made if no option is given. Therefore, the precision at some points could be much larger than the expected precision.

In order to have something better than a linear approximation, one should use the option InterpolationOrder->All . The precision seems to be the expected one everywhere on the interval. This is very misleading and it is strange that the option is not the default one.
POSTED BY: Francois Fayard

Try setting an AccuracyGoal as well.
POSTED BY: Frank Kampas
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard