The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Numerical ODEs
Replies: 10   Last Post: May 16, 2013 4:22 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Jose Carlos Santos

Posts: 4,896
Registered: 12/4/04
Re: Numerical ODEs
Posted: May 13, 2013 11:30 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 13/05/2013 14:53, wrote:

>> This question is perhaps too vague to have a meaningful answer, but
>> here it goes.
>> In what follows, I am only interested in functions defined in some
>> interval of the type [0,a], with a > 0.
>> Suppose that I want to solve numerically the ODE f'(x) = 2*sqrt(f(x)),
>> under the condition f(0) = 0. Of course, the null function is a
>> solution of this ODE. The problem is that I am not interested in that
>> solution; the solution that I am after is f(x) = x^2.
>> For my purposes, numerical solutions are enough, but if I try to solve
>> numerically an ODE of the type f'(x) = g(f(x)) (with g(0) = 0) and
>> f(0) = 0, what I get is the null function. So far, my way of dealing
>> with this has been to solve numerically the ODE f'(x) = g(f(x)) and
>> f(0) = k, where _k_ is positive but very small and to hope that the
>> solution that I get is very close to the solution of the ODE that I am
>> interested in (that is, the one with k = 0). Do you know a better way
>> of dealing with this problem?

> I suspect you may never get the sort of answer you want here.
> There are those numerical methods out there. The proof that
> they work typically depends on certain hypotheses, which
> hypotheses typical entail the existence and uniqueness of
> the solution. You have a DE that does not satisfy those
> hypotheses, so you shouldn't expect those methods to
> give the solution you want.

You are probably right. :(

Best regards,

Jose Carlos Santos

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.