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Re: solve f(x)=0, where f:Rn+1 -> Rn

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26209] Re: solve f(x)=0, where f:Rn+1 -> Rn
  • From: Alois Steindl <Alois.Steindl+e325 at tuwien.ac.at>
  • Date: Sat, 2 Dec 2000 02:10:37 -0500 (EST)
  • Organization: Inst. f. Mechanics II, TU Vienna
  • References: <9074h5$b0j@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hello,
I don't know of a package for that matter in Mathematica, but there
exist programs in Fortran, eg. the package AUTO by Doedel.
Also Prof. Seydel (http://bifurcation.de) has a Fortran-package
BIFPACK, which he might send you.

Good luck
Alois
Pavel.Pokorny at vscht.cz writes:

>   Dear Mathematica friends
> 
> Is there a way in Mathematica 4.0 to solve (numerically) the problem 
>   f(x) = 0
> where f: R^{n+1} -> R^n,
> i.e. f has n+1 real arguments and n real results ?
> 
> The solution is (under certain conditions on f) 
> a curve in (n+1)-dim space.
> 
> Example 
>    x^2 + y^2 - 1 = 0 
> is a unit circle.
> 
> This problem is called "continuation" in nonlinear system analysis
> see 
> Seydel: Tutorial on Continuation
> Int.J.Bif.Chaos, Vol.1 No.1 (1991) pp 3-11.
> 
> -- 
> Pavel Pokorny
> Math Dept, Prague Institute of Chemical Technology
> http://staff.vscht.cz/mat/Pavel.Pokorny


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