Obtaining a first-order ODE from a system of ODEs and then proving an asymptote exists

Question

I was debating whether to post this on the mathematics or physics StackExchange, and ultimately, I decided to post this here.

I have a system of differential equations which arose from a physics problem: $$\ddot{r}=r\dot{\theta}^2$$ $$r\ddot{\theta}=\dot{r}\dot{\theta}$$ These are functions of $t$. The problem first asks me to show that: $$\dot{r}=\sqrt{Ar^4+B}$$ And then show that $r=\infty$ in a finite amount of time.

To solve the first part of the problem, I was able to perform some manipulations to obtain: $$\ddot{r}\dot{r}=\frac{1}{3}r\dddot{r}$$ And I found that the equation $\dot{r}=\sqrt{Ar^4+B}$ does indeed satisfy this.

As for the second part, I realize that the problem amounts to proving that there exists a vertical asymptote for the function $r(t)$. Graphing the slope field, this seems to be the case, though I have no idea how to prove it.

Any thoughts are appreciated. I'm not too content with my "solution" to the first part either, as it is rather indirect.

Answer 1

You are correct that $\ddot{r} \dot{r} = r \dddot{r}/3$. But it's not quite enough to show that a solution of $\dot{r} = \sqrt{A r^4 + B}$ satisfies this equation, you'd have to show that every solution of $\ddot{r} \dot{r} = r \dddot{r}/3$ satisfies $\dot{r} = \sqrt{ A r^4 + B}$ for some constants $A$ and $B$. And if $A$ and $B$ are supposed to be real, that can't be true (unless there are constraints you're not telling us about), because you can't get any solution with $\dot{r}(0) < 0$ that way.

Also, there are solutions that don't have $r \to \infty$ in finite time, e.g. (in terms of the original equations) $\theta = \text{constant}$, $r = a t + b$.

Answer 2

From $\varphi=\dot\theta,$ we have \begin{align*} \ddot{r}&=r\varphi^2\\ r\dot\varphi&=\dot{r}\varphi \\ \varphi&=Cr \quad\text{solve second equation for } \varphi \\ \ddot{r}&=C^2r^3 \\ \ddot{r} \dot{r}&=C^2r^3\dot{r} \\ \frac{\dot{r}^2}{2}&=\frac{C^2r^4}{4}+B. \end{align*} The first result follows from solving for $\dot{r},$ throwing away the negative square root (presumably on physical grounds), and re-labeling the arbitrary constants.

I don't have anything yet on your second part.