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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.

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  • $\begingroup$ What about a geodesic which is just an intersection of the paraboloid with a plane through the $z$-axis? $\endgroup$ Apr 18 '20 at 20:05
  • $\begingroup$ @EthanDlugie I forgot to mention meridians are the one exception. Thank you. $\endgroup$ Apr 18 '20 at 21:02
  • $\begingroup$ 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. $\endgroup$ Apr 18 '20 at 21:14
  • $\begingroup$ @TedShifrin I don't see how you can consider $dv/du$ since $v$ is not a function of $u$. $\endgroup$ Apr 18 '20 at 22:00
  • $\begingroup$ 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. $\endgroup$ Apr 18 '20 at 22:05

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