$\dot{r}=ar^3, \dot{\theta}=1$ If $a>0$, then $r(t)$ goes to infinity monotonically and the origin is an unstable spiral.

Question

In the book by Strogatz, Nonlinear Dynamics and Chaos (1994), following problem is discussed

$\dot{r}=ar^3, \dot{\theta}=1$, in polar coordinates. If $a>0$, then $r(t)$ goes to infinity monotonically and the origin is an unstable spiral.

I tried to interpret the differential equation by solving it. I understood other cases (for $a<0$ and $a=0$). For $r(0)=r_0$, we obtain

$$r^2=\frac{1}{-2at+\frac{1}{r_0^2}} \text{ (in polar coordinates)}$$

When $a<0$, $-2at>0$. Then as $t$ goes to infinity, $r$ goes to $0$ the behavior is stable spiral and when $a=0$ it is center.

But, when $a>0$ I can't see that $r(t)$ goes to infinity as $t$ goes to infinity? (I don't want to interpret with derivative test, I want to understand the behavior by solving it explicitly)

edit: Since $r^2$ is greater or equal to $0$, $t$ is less or equal to $1/2ar_0^2$ and the limit of $r^2$ goes to infinity as $t$ approaches to $1/2ar_0^2$?

I would be very appreciate if someone could help.

Answer 1

You can interpret that by looking at the cartesian coordinates given by

$$(x(t),y(t))=(r(t)\cos(t-t_0+\theta_0),r(t)\sin(t-t_0+\theta_0))$$ where $\theta_0$ is the initial condition for $\theta$ at $t=t_0$. So, you can see that depending on the behavior of $r(t)$, we have different trajectories. For instance,

• If $r(t)=r_0$ is a constant, then the trajectory is just a circle.
• If $r(t)$ is monotonically increasing without bound, then the trajectory is a spiral spiraling outward.
• If $r(t)$ is monotonically decreasing to zero, then the trajectory is a spiral spiraling inwards towards zero.

In your case, one can see that starting from $r(0)>0$, we have that $\dot{r}(0)>0$ since $a>0$. Since $r^3$ is a monotonically increasing function of $r$, then this implies that $r(t)$ is an increasing function of times and that $r(t)$ will grow without bound. So, the trajectory will be spiraling out.

More quantitatively, the solution to $\dot{r}=ar^3$ is given by

$$r(t)=\dfrac{r_0}{\sqrt{1-2ar_0^2(t-t_0)}}$$ where $r(t_0)=r_0$. This expression can be established by exploiting the fact that the differential equation is separable; i.e. one can reformulate it as

$$\int_{r_0}^{r(t)}\dfrac{dr}{r^3}=a\int_{t_0}^tdt.$$

From this explicit solution one can see that $r(t)$ as a finite escape time at $t=t_0+1/(2ar_0^2)$.