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MathGroup Archive 2004

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Re: populating a list from system of odes

  • To: mathgroup at smc.vnet.net
  • Subject: [mg45933] Re: populating a list from system of odes
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Fri, 30 Jan 2004 04:16:00 -0500 (EST)
  • Organization: Universitaet Leipzig
  • References: <bvaq98$ah0$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de
  • Sender: owner-wri-mathgroup at wolfram.com

Hi, you mean


eqn = {a'[t] == k7 - k5 a[t] + k2 ab[t] - k1 a[t] b[t], 
    ab'[t] == -k2 ab[t] + k1 a[t] b[t], 
    b'[t] == k8 + k2 ab[t] - k6 b[t] - k1 a[t] b[t], 
    c'[t] == k3 ab[t] - k4 c[t]};

Union[Cases[eqn, _[t], Infinity]] /. a_Derivative[_][_] :> Sequence[]


Regards
  Jens

sean kim wrote:
> 
> hello group.
> 
> thanks all for the replies. they have been tremendiously helpful.
> 
> (about rossler vz lorenz. you are right, jens. what i did was to copy
> the first system that had attractor from the help menu.  lol)
> 
> let's say you have a system of ode's like..
> 
> {
> a'[t] == k7 - k5 a[t] + k2 ab[t] - k1 a[t] b[t],
> ab'[t] == -k2 ab[t] + k1 a[t] b[t],
> b'[t] == k8 + k2 ab[t] - k6 b[t] - k1 a[t] b[t],
> c'[t] == k3 ab[t] - k4 c[t]
> }
> 
> now what I want to do is get a list of the  variables in the system
> without having to type them out.
> 
> so that I would get {a[t],ab[t], b[t], c[t]}
> 
> as you can see from my previous posts, the problems i'm having are
> very similar.  I'm sur eit has to do with no tunderstanding the
> fundamentals of mathematica... but meanwhile, how would you guys
> approach that problem above?
> 
> once i have that, i can change the system to steady state easily with
> 
> % /. _'[t] -> 0
> 
> which gives
> 
> {
> 0 == k7 - k5 a[t] + k2 ab[t] - k1 a[t] b[t],
> 0 == -k2 ab[t] + k1 a[t] b[t],
> 0 == k8 + k2 ab[t] - k6 b[t] - k1 a[t] b[t],
> 0 == k3 ab[t] - k4 c[t]
> }
> 
> which i can use with the list i'm tryign to populate in Solve[]
> routine which then can be used in NDSolve[] with the codes you guys
> explained.
> 
> thank you all very much in advance.
> 
> sean


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