Small coeff ODE

[Muzialis-disabl]

Hi All,

Iam having fun with the interseting topic of ODe with arbitrary small coefficients, which I try to solve using two - scale methods.

In the attached sheet you will find an approximate solution compared to the one calculated by Odesolve. the accuracy of the MathCad solver though degenrates very quickly once eps approaches 0.01.

anybody has a suggestion to overcome this?

Thank you very much

Best Regards

Muzialis

[ptc-1368288]

>anybody has a suggestion to overcome this ? <<br> _____________________________

Your setup is incorrect and your function as well. You must respect the rules about DE's, especially the 2nd order that you MUST feed as normalised. What you are doing does not respect linearity. For that kind of DE, go by the Laplace solver (added), that will surely fail above 11.2a.

jmG

[TomGutman]

Your y1 is wrong. Your discrepencies are mostly due to that fact. It's actually fairly good at ε=.01, big discrepencies occur at larger values of ε.

For even smaller ε (around .001) the derivatives become quite large and the fixed step size inadequate. You need to use the adaptive algorithm (Rkadapt) rather than the fixed interval algorithm (rkfixed).
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� � � � Tom Gutman

[Muzialis-disabl]

jMG,

thank you for your comment.

Tom,

I am not sure what you mean by "y1 is wrong".

Certianly it is an approximation ,which I said, and the fact I have found it using the two scale method means that the approximation is "valid" for "small" eps.

Also the y2 you found is quite close to y1.

Could you please clarify?

Thank you

Best Regards

Marco

[TomGutman]

y1 is just as you say, an approximation. Valid only for small ε. Comparing the numerical solution to y1 is meaningless until you have determined the accuracy of y1. Indeed, the discrepencies between the numerical solution and y1 are due not to problems with the numerical solution but to the errors in y1 itself. y2 is an exact solution (subject to the usual caveats about machine precision and rounding errors) and is what you need to use to evaluate the adequacy of the numerical solution.
__________________
� � � � Tom Gutman

[ptc-1368288]

Yes, y1 is a"pure function" representing, maybe ? the pure DE. So by falsifying the pure DE, it will no longer match the still be pure function.

jmG

[Muzialis-disabl]

Tom,

I see what you mean, thanks for ading some words.

JmG,

thanks for your post.

Best Regards

Muzialis