For $\frac{dr}{dt}=r(4-r^2 )$, $\frac{dθ}{dt}=1$, sketch the approximate waveform of $x(t)$

Question

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?

Answer 1

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 $-2<x<2$

Answer 2

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,