Consider the system $\frac{dr}{dt}=r(4-r^2 )$, $\frac{dθ}{dt}=1$. Given the initial condition $x(t)=0.1$, $y(0)=0$, sketch the approximate waveform of $x(t)$.

I'm not even sure if this problem can be done analytically. Assuming this system can be solved explicitly, I can convert the system from polar coordinates to rectangular coordinates. However, what would I do next?

And if the system cannot be solved, how would I go about trying to find an approximate trajectory?

  • $\begingroup$ Sketching can be done just by consideration of the sign of $r'$: here $r'>0$ initially and then it goes to zero as you move toward $r=2$. So the trajectory spirals out toward a circle, and eventually $x$ becomes sinusoidal. $\endgroup$ – Ian Nov 7 at 0:54

In your phase portrait you have the circle $r=2$ which is an attractor, so the orbit starting inside this circle will spiral out and gets closer and closer to the circle $r=2$

As a result the $x(t)$ starts jumping up and down the $t$ axis and its amplitude increases with time and approaches $2$.

So you have a wave of gaining amplitude but never passing the interval of $-2<x<2$

  • $\begingroup$ "Aptitude"? Don't we call it "amplitude"? $\endgroup$ – Jake Mirra Nov 7 at 1:33
  • 1
    $\begingroup$ @JakeMirra That is the word I was looking for. Thanks. $\endgroup$ – Mohammad Riazi-Kermani Nov 7 at 1:53

Note that $\theta(t) = t$ and $r(t)$ can be integrated analytically. Rewrite $\frac{dr}{dt}=r(4-r^2 )$ as,

$$\frac{dr}{r(4-r^2 )}=dt$$

Integrate with the initial condition $r(0)=0.1$,

$$\ln\frac{r^2}{0.1^2 } -\ln \frac{4-r^2}{4-0.1^2} =8t$$

Rearrange to obtain

$$ r = \frac{2}{\sqrt{399e^{-8t}+1} }$$

The expression for $x(t)$ is,

$$x(t)=\frac{2\cos t}{\sqrt{399e^{-8t}+1} }$$

which quickly degenerates to $2\cos t$, a waveform as shown below,

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.