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.
Suppose $\gamma$ is a geodedic on the paraboloid. Let's see how much information can we get from Clairaut's relation $r\cos\beta=c$, where $c$ is a constant:
$c\neq 0$. If this is not true, then $\cos\beta=0$. Since $\beta$ is the angle between $u'\frac{\partial}{\partial u}+v'\frac{\partial}{\partial v}$ and $\frac{\partial}{\partial u}$, we see $u'\equiv 0$, which is a contradiction. Then we know $\cos\beta\neq 0$, so $u'\neq 0$, we can assume $u'>0$.
Since $r=v$, from Clairaut's relation we see $v=c/\cos\beta$, so $v≥|c|$. We can also see that $v=|c|$ iff $v'=0$.
There is at most one $t\in \mathbb{R}$ such that $v(t)=|c|$. Otherewise, there would have $t_1<t_2$ satisfy this. Suppose $v$ attains its maximum on $[t_1,t_2]$ at $t_0$, then $v'(t_0)=0$ which implies $v(t_0)=|c|$ by previous analysis. So $\gamma$ has a segment coincident with the parallel $v=|c|$, which is impossible since $\gamma$ is geodesic but $v=|c|$ is not.
So either $v'\neq 0$ or $v'=0$ at exactly one point $t_0$. We now only consider the second situation, the first follows similarly (actually you can see that the first case is impossible ).
$v\to +\infty$ as $t\to +\infty$. Otherwise either $v'≤0$ on $[t_0,+\infty)$ or $v'≥0$ on $[t_0,+\infty)$ but $\limsup_{t\to +\infty} v<\infty$. In both case $\int_{t_0}^{\infty}|v'|$ would be finite. When $t$ is large enough we see $\epsilon<\cos\beta=|c|/v＜1-\epsilon$ for some fixed $\epsilon$, since $\cos\beta=\frac{u'v}{\sqrt{(u')^2v^2+(v')^2(1+4v^2)}}$, we conclude that $(u')^2≤k(v')^2$ for some constant $k$. So we have $\int_{t_0}^{\infty}|\gamma'|<+\infty$, which is a contradiction. Similarly we see $v\to +\infty$ as $t\to -\infty$.
$u$ is not bounded. From $\cos\beta=c/v$, we can compute that $(u')^2=(v')^2{\frac{c^2}{v^2-c^2}}(\frac{1}{v^2}+4)$, so $|u'|>l\frac{|v'|}{v}$ for some constant $l$ and integrate this inequality on both sides we see $u\to +\infty$ because $\ln v\to +\infty$.
Now we see on $(t_0,+\infty)$, $v$ is strictly increasing and on $(-\infty，t_0)$ is strictly decreasing. $u$ is always strictly increasing. So we can think of $u$ as a function of $v$ on each interval, say $u_1(v),u_2(v)$ and see that there would be infinite many $v$ makes $u_1(v)-u_2(v)$ an integral multiple of $2\pi$ since $u_1-u_2$ is strictly increasing to $+\infty$.