I'm trying to get a Mathematica program to plot a phase portrait of this differential equation:

$$\frac{dv}{dt} = g - \frac{k}{m}v^2$$

where g, k, m are physical constants. I want to make it such that the x axis is time t and y axis is velocity v(t). How do I use VectorPlot or StreamPlot to accomplish this?

  • 1
    $\begingroup$ Related: this and this. Also check this. $\endgroup$
    – codebpr
    Commented Mar 4 at 5:49
  • $\begingroup$ I've checked all of those; I am not quite sure if any of them have the time as the x-axis. I wanted to have a phase portrait of time vs velocity to show the critical points that are asymptotically stable with some solution curves.(So as time increases, any solution curve will approach x*...). $\endgroup$
    – Boltzmann
    Commented Mar 4 at 9:05
  • $\begingroup$ What have you tried? $\endgroup$
    – bbgodfrey
    Commented Mar 4 at 12:19

1 Answer 1


I've checked all of those; I am not quite sure if any of them have the time as the x-axis

The syntax to generate phase plot for first order ode of form $$ y'(x) = \frac{A(x,y)}{B(x,y)} $$



So using the above on your ode, where now y is your v and x is your t gives

\begin{align*} \frac{dv}{dt} &= g - \frac{k}{m}v^2\\ &= \frac{g - \frac{k}{m}v^2}{1} \end{align*}

Hence setting some values for g and m and k gives

g = 9.81; m = 10; k = 0.1;
StreamPlot[{1, g - k/m*v^2}, {t, 0, 10}, {v, -10, 10}, 
 FrameLabel -> {"t", "v(t)"}, BaseStyle -> 20]

Mathematica graphics

The above shows solutions curves. If you have a specific IC, then one of these curves will be the solution.

Here is a Manipulate to make it easier to analyze the system

 Module[{g = 9.81},
  StreamPlot[{1, g - k/m*v^2}, {t, 0, maxtime}, {v, -maxV, maxV}, 
   FrameLabel -> {"t", "v(t)"}, BaseStyle -> 20]
 {{m, 10, "mass"}, 0.1, 100, .1, Appearance -> "Labeled"},
 {{k, 1, "stiffness k"}, 0.1, 10, .1, Appearance -> "Labeled"},
 {{maxtime, 1, "time range"}, 0.01, 10, .01, 
  Appearance -> "Labeled"},
 {{maxV, 1, "velocity range"}, 0.01, 30, .01, Appearance -> "Labeled"},
 TrackedSymbols :> {m, k, maxtime, maxV}

enter image description here

I thought I update with a method I have been using to make slope field without using StreamPlot. This can be useful sometimes where StreamPlot does not work as well.

The idea is simple. Solve the ode and generate curves for different values of constant of integrations.

So for the above problem the code becomes

ClearAll[v, t];
g = 981/100; m = 10; k = 1/10;
ode = v'[t] == g - k/m*v[t]^2;
sol = Flatten@Values[DSolve[ode , v[t], t]];
data = Table[sol /. C[1] -> n, {n, -3, 3, .1}];
Plot[data, {t, -30, 30}]

enter image description here

The above shows different integral curves for different C constants and the asymptotically lines. The only tricky part of this, is that one needs to make few tries first to pick the correct range of the C's to use.


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