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.
2 Answers
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}]
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}}]
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]
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]
The above shows the A
matrix in $x'=Ax$. So first row reads $x_1'(t)=x_2$ and second row reads $x'_2(t)=-16 x_1$
Update to answer comment
The following can be done to automate plotting StreamPlot
directly from the state space ss
result
A = First@Normal[ss];
vars = {x1, x2}; (*state space variables*)
eqs = A . vars;
StreamPlot[eqs, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
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$\begingroup$ Can you method plot
y''[x]+2 y'[x]+3 y[x]==2 x
? $\endgroup$– yodeMar 27, 2022 at 8:59 -
$\begingroup$ @yode Phase portrait are used for homogeneous ode's. Systems of the form $x'=A x$ and not $x'=A x + u$. Since it shows the behaviour of the system itself, independent of any forcing functions (the stuff on the RHS). This behavior is given by phase portrait diagram. The reason is, it is only the $A$ matrix eigenvalues and eigenvectors that determines this behaviour, and $A$ depends only on the system itself, without any external input being there. $\endgroup$– NasserMar 27, 2022 at 14:08
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$\begingroup$ @yoda Yes. I've updated the above with what I think you are asking for. Hope this helps. $\endgroup$– NasserMar 29, 2022 at 15:23
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]
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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 $\endgroup$ Mar 15, 2020 at 6:04 -
1$\begingroup$ This plot is from
ParametricPlot
, notEquationTrekker
, but in v12.0EquationTrekker
gives me plots, although I do get PropertyValue errors. $\endgroup$ Mar 15, 2020 at 7:40