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