Re: ODEs and phase portraits
- To: mathgroup at smc.vnet.net
- Subject: [mg14743] Re: ODEs and phase portraits
- From: "Kevin J. McCann" <kevinmccann at Home.com>
- Date: Thu, 12 Nov 1998 02:17:42 -0500
- Organization: @Home Network
- References: <728kp1$ehc@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Try this one:
Remove[de]
de[epsilon_, x0_, v0_, T_] :=
Module[{eqn, x, y, t},
eqn = {Derivative[1][Derivative[1][x]][t] +
epsilon*(x[t]^2 - 1)*
Derivative[1][x][t] + x[t] == 0,
x[0] == x0, Derivative[1][x][0] == v0};
y = x /. NDSolve[eqn, x, {t, 0, T}][[1]];
ParametricPlot[{y[t], Derivative[1][y][t]},
{t, 0, T}, PlotStyle -> RGBColor[1, 0, 0],
Epilog ->
{RGBColor[0, 0, 1], AbsolutePointSize[5],
Point[{y[0], Derivative[1][y][0]}]}]; ]
Sample usage:
de[-.01,1,0,100]
If you copy the above into a NB, highlight the bracket, and type
Ctl-Shif-N, it will look a lot better. Anyway, try it.
Kevin
phantomlord at my-dejanews.com wrote in message <728kp1$ehc at smc.vnet.net>...
>I am trying to write a function such that I have the following ODE:
>x''(t)+epsilon*(x(t)^2-1)*x'(t)+x(t)==0 [1] where epsilon is to be one
>of the parameters in the function. I want to beable to draw out the
>phase portraits for the equation for different values of epsilon.
>
>To deduce the phase portraits in a mathematical procedure I multiply
>equation[1] by dx(t)/dt and integrate w.r.t. t. To do this in
>Mathematica is trivial, so I'll skip past this - it is the next step
>that I would like assistance with:
>
>Q:Is there a way that I can decompose the result of the above (I'll call
>it [2]) into the corresponding pair (below) of ODEs to deduce the
>trajectory of the phase portrait?
>
>x'(t) = y(t)
>y'(t) = f(x,y)
>some f(x,y) function of x and y.
>
>Perhaps there is a built in function that will allow me to do this? If
>not do, can anybody suggest another way to do this?
>
>Also after plotting the trajectory is there any way to determine the
>direction of it in Mathematica?
>
>thanks for you time.
>Paul
>
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