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Re: sketch a phase portrait for the nonlinear system

  • To: mathgroup at smc.vnet.net
  • Subject: [mg106551] Re: sketch a phase portrait for the nonlinear system
  • From: "Norbert P." <bertapozar at gmail.com>
  • Date: Sat, 16 Jan 2010 06:10:38 -0500 (EST)
  • References: <hip89t$ss1$1@smc.vnet.net>

Hi Hongtao,

this is more a math question than a Mathematica question ;)

The system has 3 stationary points: saddle at (0,0) and 2 spirals at
(-1,0) and (1,0). The initial conditions (+-0.1, +- 0.16) are chosen
such that they lie very close to the stable and unstable manifolds of
the saddle point at (0,0). That's why you see the nice cross at
(0,0).

I don't know how much you tried to change them, but as long as you
keep them inside the plot range -3<=x0<=3 and -3<=y0<=3, you should
see the trajectory in the plot.

Best,
Norbert

On Jan 15, 12:17=C2 am, =E6=B6=9B =E6=B4=AA <hongtao... at yahoo.com.cn> wro=
te:
> Hi,everyone,
> I am new in Mathematica.When I try to sketch a phase portrait for the non=
linear system, I don't know how to plan the input values for the ODEs to ge=
t the right FIGURE.
> My example is from a book as follows:
> ode1[x0_,y0_]:=NDSolve[{x=E2=80=99[t]==y[t],y=E2=80=99[=
t]==x[t](1-(x[t])^2)+y[t],x[0]==x0,y[0]==y0},{x[t],y[t]},{t,-10=
,10}];
> sol[1]=ode1[0.1,0.16];sol[2]=ode1[0.1,-0.06];
> sol[3]=ode1[-0.1,-0.16];sol[4]=ode1[-0.1,0.06];
> sol[5]=ode1[1.3,0];sol[6]=ode1[-1.3,0];
> p=ParametricPlot[Evaluate[Table[{x[t],y[t]}/.sol[i],{i,6}]],{t,-10,10},=
PlotRange->{{-3,3},{-3,3}}];
> Show[p,PlotRange->{{-3,3},{-3,3}},AxesLabel->{"x","y"}]
> Would you like to tell me why we choose the input values for the ODEs suc=
h as ode1[0.1,0.16]; ode1[0.1,-0.06];...? You see, when I tried to change t=
he values ,I can't see the figure any more.
> Best Regards,
> hongtao



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