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MathGroup Archive 2005

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Re: NDSolve/InterpolatingFunction and vectors

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53722] Re: NDSolve/InterpolatingFunction and vectors
  • From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
  • Date: Tue, 25 Jan 2005 05:03:19 -0500 (EST)
  • Organization: Uni Leipzig
  • References: <csqrsc$1sl$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

soln = NDSolve[{
xx'[t] == {{0, 1}, {-1, 0}}.xx[t], xx[0] == {{0}, {1}}
}, xx, {t, 0, 10}]



Needs["DifferentialEquations`NDSolveUtilities`"]

and

time = DifferentialEquations`NDSolveUtilities`Private`GetTimeData[soln];
grid = First[
DifferentialEquations`NDSolveUtilities`Private`GetGridData[soln]];
ip1 = Interpolation[Transpose[{time, #[[1, 1, 1]] & /@ Transpose[grid]}]];
ip2 = Interpolation[Transpose[{time, #[[1, 2, 1]] & /@ Transpose[grid]}]]

helps not ?

at least

Plot[{ip1[t], ip2[t]}, {t, 0, 10}]

work fine.

Regards

  Jens



"D Herring" <dherring at at.uiuc.dot.edu> schrieb im Newsbeitrag 
news:csqrsc$1sl$1 at smc.vnet.net...
> Hi all,
>
> For numerous reasons (such as dot products), I would like to use NDSolve
> with vector-valued functions.
>
> For example, the Sine and Cosine could be defined as
> soln=NDSolve[{
>       xx'[t]\[Equal]{{0,1},{-1,0}}.xx[t],xx[0]\[Equal]{{0},{1}}
>       },xx,{t,0,10}]
>
> Then I have the multi-valued function
> f[t_]=(xx/.soln[[1]])[t]
>
> such that f[Pi] is roughly {{0},{-1}} as expected.
>
> The trouble comes when trying to extract scalar values from f[t].
> f[3.14][[1,1]] ~= 0 but f[t][[1,1]] throws an error.
> Likewise, Plot[f[t],{t,0,10}]  pukes because f[t] doesn't return a
> scalar when evaluated.
>
> My current solution uses dot products to extract values.
> {1,0}.f[t] doesn't error since it holds until f is evaluated.
> Plot[{{1,0},{0,1}}.f[t],{t,0,10}] still bombs, but
> Plot[{{1,0}.f[t],{0,1}.f[t]},{t,0,10}] works fine.
>
> Can anyone suggest a better overall method for embedding dot products in
> NDSolve?  The system I have looks something like
> {p'[t]=dp[p]+(p[t]-a[s[t]]).dads[s[t]], s'[t]=p'[t].dads[s[t]]/...)
> where a[s] is given.  t and s[t] are scalar; the other variables are of
> dimension 4 (or more).
>
> Thanks,
> Daniel
> 



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