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Re: directionfields from StreamPlot looks different from solution
*To*: mathgroup at smc.vnet.net
*Subject*: [mg100483] Re: [mg100444] directionfields from StreamPlot looks different from solution
*From*: sean k <sean_incali at yahoo.com>
*Date*: Fri, 5 Jun 2009 03:03:22 -0400 (EDT)
*Reply-to*: sean_incali at yahoo.com
Hi Murray,
Thanks for the suggestion regarding 1 dimensional ode.
How would you use StreamPlot for the following system?
a'[t] == - k1 (t ^-h) a[t] - k2 (t ^-h) b[t],
b'[t] == k1 (t ^-h) a[t]
--- On Thu, 6/4/09, Murray Eisenberg <murray at math.umass.edu> wrote:
> From: Murray Eisenberg <murray at math.umass.edu>
> Subject: Re: [mg100444] directionfields from StreamPlot looks different f=
rom solution
> To: mathgroup at smc.vnet.net
> Date: Thursday, June 4, 2009, 8:25 AM
> You're not. For the "direction
> field" of a 1-dimensional ordinary differential equation y'
> == f[t, y], the vector field you want to plot is {1,
> f[t,y]}. So...
>
> streams =
> StreamPlot[{1, t^2 - y}, {t, -4, 4}, {y, -2, 10},
> StreamStyle ->
> Directive[Orange]];
> sol = Table[NDSolve[{y'[t] == t^2 - y[t], y[0] ==
> y0}, y[t],
> {t, -4, 4}],
> {y0, 0, 6, 0.5}];
> solutionCurves =
> Plot[y[t] /. sol, {t, -4, 4},
> PlotRange -> {{-4, 4}, {-2, 10}}];
> Show[{solutionCurves, streams}
>
> Note that you do not need any parentheses around t^2. And
> in the current version of Mathematica, you no longer need to
> wrap y[t]/.sol with Evaluate -- unless you would like the
> curves automatically to be given different colors.
>
> sean_incali at yahoo.com
> wrote:
> > I don't think I'm using StreamPlot properly.
> >
> > Consider the following non-autonomous ODE
> >
> > y'[t] == (t^2) - y[t]
> >
> > Solutions for various ICs can be viewed by the
> following.
> >
> > sol= Table[NDSolve[{y'[t] == (t^2) - y[t], y[0] ==
> y0}, y[t], {t, -4,
> > 4}], {y0, 0, 6, 0.5}];
> >
> > Plot[Evaluate[y[t] /. sol], {t, -4, 4}, PlotRange
> -> {{-4, 4}, {-2,
> > 10}}]
> >
> > Shouldn't Vector fields be similar to the solutions
> above? If I plot t
> > on x-axis vs. t^2-y on y axis...
> >
> > VectorPlot[{t, (t^2) - y}, {t, -4, 4}, {y, -2, 10}]
> >
> > StreamPlot[{t, (t^2) - y}, {t, -4, 4}, {y, -2, 10}]
> >
> > I don't get similar results...
> >
> > What is the reason for this?
> >
> > Thanks much in advance
> >
> > Sean
> >
> >
> >
> >
>
> -- Murray Eisenberg
> murray at math.umass.edu
> Mathematics & Statistics Dept.
> Lederle Graduate Research Tower phone
> 413 549-1020 (H)
> University of Massachusetts
> 413 545-2859 (W)
> 710 North Pleasant Street
> fax 413 545-1801
> Amherst, MA 01003-9305
> =0A=0A=0A
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