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I am given the following system of differential equations in polar coordiantes

$$\begin{cases} \dot{r}=r(1-r)\\ \dot{\phi}=\sin^2(\phi)+r(1-r)^3 \end{cases}$$

I am being asked to show that if $r(0)=1$, i.e. the solution starts on the unit circle for $t=0$, that it stays on the unit circle for all $t$. I was thinking that since $\dot{r}(0)=0$, which means that $r$ does not change and therefore $r(t)$ would be 1 for all other $t$. My question is whether this kind of reasoning would be correct, and if not, how could I show what I am asked?

Thanks in Advance!

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2 Answers 2

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You're absolutely right and your reasoning is also correct. Just a side note: Since $\dot{r} < 0$ ($r$ decreases) for $r > 1$ and $\dot{r} > 0$ ($r$ increases) for $r > 1$, the solution (also called limit cycle) you've found is stable.

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Note that $r=\dfrac{Ce^t}{1+Ce^t}$ is an answer to the differential equations for $C\ge 0$. Therefore if $r(0)=0$ we have $C=0$ and $r=0$ for every t.

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