I have a polar differential equation (this is the reduced variant) as:

$$r' = 0, \theta' = 1$$

I figured out (from previous answers) that I can nicely convert this to Cartesian and use StreamPlot as:

  field = {0, 1};

  TransformedField["Polar" -> "Cartesian", field, {r, \[Theta]} -> {x, y}]


This generates the correct phase portrait and worked great. I was even able to animate it.

My question now is how can this field (polar DEQ) be solved as a polar DEQ? Can't this be solved as such and plotted using polar plots to arrive at the same phase portrait?

Am I missing something? I tried NDSolve without luck, so think I must be missing something basic here. Is something special required for polar type DEQs in Mathematica?


1 Answer 1


If your aim is just to solve this specific system:

{rd, an} = {r[t], s[t]} /. 
      First@DSolve[{r'[t] == 0, s'[t] == 1, r[0] == radius, 
         s[0] == 0}, {r[t], s[t]}, t]
f[u_, tm_] := 
 CoordinateTransform["Polar" -> "Cartesian", {rd, an}] /. {radius -> u,t->tm}
 ParametricPlot[Evaluate[Table[f[j, t], {j, Range[4]}]], {t, 0, 2 Pi}]

where $\theta$ is represented by s and I have assumed $\theta(0)=0$ for simplicity.

Animating (uniform circular motion):

enter image description here

  • $\begingroup$ Excellent! How did you do the animated uniform motion graphic? $\endgroup$
    – Amzoti
    Apr 4, 2014 at 6:21
  • $\begingroup$ @Amzoti I exported the following as an animated gif Table[ParametricPlot[Evaluate[Table[f[j, t], {j, Range[4]}]], {t, 0, 2 Pi},Epilog->(Point[{f[#,k]]&/@Range[4])],{k,Range[0,2Pi,0.1]}] This takes some processing time. You could randomise the starting position etc. $\endgroup$
    – ubpdqn
    Apr 4, 2014 at 7:23
  • $\begingroup$ Is there a syntax issue with the commented Table command? Regards $\endgroup$
    – Amzoti
    Apr 4, 2014 at 12:53
  • $\begingroup$ Yes. Sorry. correction: Table[ParametricPlot[Evaluate[Table[f[j, t], {j, Range[4]}]], {t, 0, 2 Pi},Epilog->(Point[{f[#,k]}]&/@Range[4])],{k,Range[0,2Pi,0.1]}] $\endgroup$
    – ubpdqn
    Apr 4, 2014 at 16:30
  • $\begingroup$ Is it possible to use something like PolarPlot so no conversion to Cartesian is necessary? $\endgroup$
    – Amzoti
    Apr 5, 2014 at 3:25

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