# Geodesics on paraboloid self-interesect in an infinite number of points

I'm trying to solve an exercise of Do Carmo's Riemannian geometry. Specifically, I have to prove that on a paraboloid(that is the revolution surface of a paraboloid $$\{(v\cos u,v\sin u,v^2):v\in(0,\infty),u\in(0,2\pi)\}$$) the geodesics which are not meridians (that is $$u\neq$$constant)self-intersects in an infinite amount of points.

Using Clairaut's relation it is possible to show that geodesics are characterized (at least locally) by the following ODE's system:

$$\begin{cases} (1+4v(t)^2)v'(t)^2+u'(t)^2=c_0\\ u'(t)v(t)^2=c_1 \end{cases}$$ where $$c_0,c_1$$ are unkown constants. This system reduces to the equation $$(1+4v(t)^2)v'(t)^4-c_0v(t)^2+c_1=0$$

Which I have absolutely no idea on how to solve. Any hint is very much appreciated.

P.S: Do Carmo suggest using Clairaut's relation, which I've already used, but it is possible there is a more tricky application.

• What about a geodesic which is just an intersection of the paraboloid with a plane through the $z$-axis? Apr 18 '20 at 20:05
• @EthanDlugie I forgot to mention meridians are the one exception. Thank you. Apr 18 '20 at 21:02
• You don't care about $t$. You want $dv/du$ or $du/dv$ so that you can study the angle as a function of the distance from the axis. Think about what happens at the bottom-most point of the geodesic. Apr 18 '20 at 21:14
• @TedShifrin I don't see how you can consider $dv/du$ since $v$ is not a function of $u$. Apr 18 '20 at 22:00
• Standard chain rule, of course. $dv/du = \dfrac{v'(t)}{u'(t)}$. If you want to see concrete examples, look at pp. 72 ff. of my differential geometry text. Also see exercises 23 ff. at the end of that section. Apr 18 '20 at 22:05