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

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?

• 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. – 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

• "Aptitude"? Don't we call it "amplitude"? – Jake Mirra Nov 7 at 1:33
• @JakeMirra That is the word I was looking for. Thanks. – 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,