I'm trying to make a phase portrait for the ODE x'' + 16x = 0, with initial conditions x[0]=-1 & x'[0]=0. I know how to solve the ODE and find the integration constants; the solution comes out to be x(t) = -cos(4t) and x'(t) = 4sin(4t). But I don't know how to make a phase portrait out of it. I've looked at this link Plotting a Phase Portrait but I couldn't replicate mine based off of it.


Phase portrait for any second order autonomous ODE can be found as follows.

Convert the ODE to state space. This results in 2 first order ODE's. Then call StreamPlot with these 2 equations.

Let the state variables be $x_1=x,x_2=x'(t)$, then taking derivatives w.r.t time gives $x'{_1}=x_2,x'{_2}=x''(t)=-16 x_1$. Now, using StreamPlot gives

    StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -2, 2}]

Mathematica graphics

To see the line that passes through the initial conditions $x_1(0)=1,x_2(0)=0.1$, add the option StreamPoints

StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -5, 5}, 
 StreamPoints -> {{{{1, .1}, Red}, Automatic}}]

Mathematica graphics

To verify the above is the correct phase plot, you can do

ClearAll[x, t]
ode = x''[t] + 16 x[t] == 0;
ic = {x[0] == 1, x'[0] == 1/10};
sol = x[t] /. First@(DSolve[{ode, ic}, x[t], t]);
ParametricPlot[Evaluate[{sol, D[sol, t]}], {t, 0, 3}, PlotStyle -> Red]

Mathematica graphics

The advatage of phase plot, is that one does not have to solve the ODE first (so it works for nonlinear hard to solve ODE's).

All what you have to do is convert the ODE to state space and use function like StreamPlot

If you want to automate the part of converting the ODE to state space, you can also use Mathematica for that. Simply use StateSpaceModel and just read of the equations.

eq = x''[t] + 16 x[t] == 0;
ss = StateSpaceModel[{eq}, {{x[t], 0}, {x'[t], 0}}, {}, {x[t]}, t]

Mathematica graphics

The above shows the A matrix in $x'=Ax$. So first row reads $x_1'(t)=x_2$ and second row reads $x'_2(t)=-16 x_1$

| improve this answer | |

EquationTrekker works for me, but if you are not interested in looking at a range of solutions, it might be easier to just do it with ParametricPlot

x[t_] := -Cos[4 t]

ParametricPlot[{x[t], x'[t]} // Evaluate, {t, 0, 2 π}, 
 Axes -> False, PlotLabel -> PhaseTrajectory, Frame -> True, 
 FrameLabel -> {x[t], x'[t]}, GridLines -> Automatic]

enter image description here

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  • $\begingroup$ What version is this on, Bill? Someone in the QA that OP links to says EquationTrekker is broken for them on v11.0 $\endgroup$ – CA Trevillian Mar 15 at 6:04
  • 1
    $\begingroup$ This plot is from ParametricPlot, not EquationTrekker, but in v12.0 EquationTrekker gives me plots, although I do get PropertyValue errors. $\endgroup$ – Bill Watts Mar 15 at 7:40

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