Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

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


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

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 ]
Niels Diepeveen

Posts: 47
Registered: 5/28/10
Re: Numerical ODEs
Posted: May 14, 2013 7:07 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

José Carlos Santos wrote:

> Hi all,
>
> 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?


There is a standard trick for dealing with this type of problem for
ODEs of the form dy/dx = g(y).
Note that y = f(x) is a solution of this equation, then so is
y = f(C + x) with an arbitrary constant C.
Also, y = f(C - x) is a solution of dy/dx = -g(y).
You can use this to work backward and forward from an arbitrary point
where the equation is not singular and then translate the solution
along the x-axis to satisfy the boundary condition.

For the case you mention, you might start by solving the IVP
dy/dx = -2 sqrt y
y(0) = 1 (instead of 1 any positive number will do)
This should give you a solution that looks like f_1(x) = (1-x)^2
on [0, 1]. (What happens past 1 probably depends on the algorithm used).
Because f_1(1) = 0, you know that there is a solution to your original
problem that looks like (1 - (1 - x))^2 on [(1-1), (1-0)].
You can then complete the solution by solving the IVP
dy/dx = 2 sqrt y
y(1) = 1

--
Niels Diepeveen



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

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.