Consider the catenoid $\{(x,y,z) \in \mathbb{R}^3: \cosh z = \sqrt{x^2+y^2}\}$, given with the parameterization $$ r(u,v) = (\cosh u \cos v, \cosh u \sin v, u) \ \ . $$
I'm trying to show that if $\gamma(t) = r(u(t),v(t))$ is a geodisic curve, such that $\gamma$ is bounded (for every $t\in\mathbb{R}$), then the image $\gamma$ is the circle $\{x^2 + y^2 =1, z=0\}$.
By Clairaut's relation, it seems that that $\gamma$ is geodesic if and on if the following are true: $$ \cosh(u)^2 \dot v \equiv c\\ \cosh(u)^2 (\dot u^2 + \dot v^2) \equiv 1 \ \ , $$ where $c$ is constant, and $u,v$ are understood as functions of $t$. The first equation comes from Clairaut's relation, and the second comes from the first fundamental form.
However, trying all sorts of algebric manipulations, I couldn't find a way to show that if $\gamma$ is not the said circle, then it is unbounded. I'd be glad for any hint.
Your second equation isn't right. If you assume an arclength parametrization by $t$, then the second equation should be $\dot u^2 + \cosh^2u\,\dot v^2 = 1$, right? Then we have $$\frac{dv}{du} = \frac{\dot v}{\dot u} = \frac c{\cosh u\sqrt{\cosh^2u-c^2}},$$ and so $$v = c \int \frac{du}{\cosh u\sqrt{\cosh^2 u-c^2}} + c'$$ for some constants $c$ and $c'$. This formulation is good for understanding how the geodesic winds around.
First thing to understand: What value of $c$ gives the belt circle?
Next thing to understand: Is there any way such an integral will diverge as $u$ stays bounded? How can a geodesic meet a parallel circle tangentially? It really is better to think about Clairaut geometrically: The expression $\cosh^2 u\,\dot v = c$ tells you that the radius of the parallel circle times the cosine of the angle between the geodesic and the parallel must stay constant.