TITLE: Geodesics on a hyperbolic paraboloid
QUESTION [8 upvotes]: Given any two points on a hyperbolic paraboloid ($xy = z$ or $z = (x^2 - y^2)/2$) how does one find the geodesic between them?
I know that since the hyperbolic paraboloid is doubly ruled, some of the geodesics are lines. However, I have very little idea of how to find the geodesics between arbitrary points. 
If an exact answer cannot be given, then I would be interested in a technique of approximating the geodesics. 
Finally, I'd like to add that this is outside my area of expertise (I'm an algebraist!) so you'll have to explain it to me.
Thanks!

REPLY [7 votes]: I was working on answering another question involving integrating the geodesic equations on a surface, and the links there lead me back to this question, which I hadn't noticed before.  In case anyone is interested, there is a more direct way to get to the geodesics of the induced metric on the hyperbolic paraboloid, so I thought that would go ahead and input this approach.
As Greg pointed out, the geodesic equations in the graphing coordinates are
$$
x'' = -\frac{2yx'y'}{(1+x^2+y^2)}\qquad\text{and}\qquad y'' = -\frac{2xx'y'}{(1+x^2+y^2)}.
$$
These equations have two first integrals that are quadratic in velocity:
$$
Q_1 = (1+y^2)\,x'^2 + 2xy\,x'\,y' + (1+x^2)\,y'^2,
$$
i.e., the induced metric itself, which is, of course, the statement that geodesics have constant speed, and
$$
Q_2 = 2(1+x^2+y^2)\,x'\,y',
$$
which is easily verified.  (It's a classical fact that the geodesic equations on any (non-flat) surface of degree $2$ in Euclidean $3$-space have a second first integral that is quadratic in velocities, namely $Q_2 = |K|^{-3/4}I\!I$ in addition to the obvious first integral $Q_1 = I$.)
Since there are two independent first integrals quadratic in velocity, this is a Liouville metric and hence can be put in Liouville form.  This is an algorithmic procedure; following it leads to the result that, if one establishes new coordinates $z$ and $w$ on the surface
by 
$$
 x = \sinh \left(\frac{z+w}2\right)\quad\text{and}\quad 
 y = \sinh\left(\frac{z-w}2\right),
$$ 
then we have the first integrals
$$
\begin{aligned}
Q_1 &= \tfrac14\bigl(\cosh(z)+\cosh(w)\bigr)\bigl(\,\cosh(z)\,z'^2 +\cosh(w)\,w'^2\,\bigr)\\
Q_2 &= \tfrac14\bigl(\cosh(z)+\cosh(w)\bigr)\bigl(\,\cosh(w)\cosh(z)\,z'^2 -\cosh(z)\cosh(w)\,w'^2\,\bigr)
\end{aligned} 
$$
(The actual Liouville coordinates would be $(u,v)$ where $\mathrm{d}u = \sqrt{\cosh z}\,\mathrm{d}z$ and $\mathrm{d}v = \sqrt{\cosh w}\,\mathrm{d}w$, but these are elliptic integrals, and it seems pointless to change to these coordinates.) 
In particular, for a unit speed geodesic, i.e., one for which $Q_1 = 1$ and $Q_2 = c$ for some constant $c$, we have $Q_2-c\,Q_1 =0$, so
$$
(\cosh(w)-c)\cosh(z)\,z'^2 -(\cosh(z)+c)\cosh(w)\,w'^2,
$$ 
and one can thus separate variables, yielding
$$
\frac{\cosh(z)\,\mathrm{d}z^2}{(\cosh(z)+c)} = \frac{\cosh(w)\,\mathrm{d}w^2}{(\cosh(w)-c)}.
$$
Now, $c=0$ corresponds to $z\pm w$ being constant, i.e., $x$ or $y$ is constant, which are straight lines in the surface.  When $c\not=0$, the geodesic has to stay in the region where $\cosh z + c$ and $\cosh w - c$ are non-negative, and we have an equation that can be integrated 'by quadratures' to yield two foliations by geodesics of the region where $\cosh z + c$ and $\cosh w - c$ are positive.
$$
\sqrt{\frac{\cosh z}{\cosh z + c}}\,\mathrm{d}z = \pm\, \sqrt{\frac{\cosh w}{\cosh w - c}}\,\mathrm{d}w
$$
This will include a family that envelopes either $\cosh z + c  = 0$ (if $c<-1$) or $\cosh w - c = 0$ (if $c>1$).  (The curves $z = 0$ and $w=0$ are geodesics.)
In order explicitly to compute the distance between two points using these formulae, one would need to compute the corredponding $z$ and $w$ coordinates of the two points (easy), find the $c$ belonging to the (unique) geodesic joining those two points (nontrivial), and then compute the elliptic integrals as above.  
I expect that the likelihood that one could actually carry this out and find a 'closed form' expression for the distance function on the surface is rather low.