I am reading a paper (https://arxiv.org/abs/math/9907210), and the author first gives the following differential equation:
$$\ddot{p}-\frac{\dot{p}^2}{p}+\frac{A}{p}-p^3=0$$
Later, they state that the first integral is given by
$$\dot{p}^2=p^4+Kp^2+A$$
where $K$ is a real constant. I can't get this; here is my attempt.
Multiply by $\dot{p}$ and integrate with respect to $s$,
$$\int \dot{p}\ddot{p} ds=\int (\frac{\dot{p}^2}{p}-\frac{A}{p}+p^3)\dot{p}ds$$
Left hand side is good, $\frac{1}{2}\dot{p}^2$, but the right hand side is not:
$$\int (\frac{\dot{p}^2}{p}-\frac{A}{p}+p^3)\dot{p}ds=\int\frac{\dot{p}^3}{p}ds-A\ln(p)+\frac{1}{4}p^4.$$
Maybe my naive understanding of "first integral" is incorrect. Can anyone spot my error?
Best Answer
I don't know if this is classified as an error. But the integral $\int\frac{\dot{p}^3}{p}ds$ depends on the path, not just the initial and the final states. To have an integral of motion, you should end up with an expression involving $p$ and $\dot{p}$ only, which is independent of the path you take. I am presenting a derivation of the integral of motion in the cited paper.
Multiply both sides of the DE by $\frac{2\dot{p}}{p^2}$, we get $$\frac{2\dot{p}\ddot{p}}{p^2}-\frac{2\dot{p}^3}{p^3}+\frac{2A\dot{p}}{p^3}-2p\dot{p}=0.$$ This is equivalent to $$\frac{d}{ds}\left(\frac{\dot{p}^2}{p^2}-\frac{A}{p^2}-p^2\right)=0.$$ Thus, there exists a constant $K$ such that $$\frac{\dot{p}^2}{p^2}-\frac{A}{p^2}-p^2=K.$$ That is, $$\dot{p}^2=p^4+Kp^2+A.$$