Brachistochrone on the surface of a sphere

[TimJ]

1. The problem statement

I'm trying to solve the brachistochrone problem between two points on the surface of a sphere.

2. The attempt at a solution

The Lagrangian for this problem in spherical coordinates is
<br /> L=\frac{1}{2}m \left(r^2 \left (\frac{d \theta}{dt}\right)^2+r^2 \sin^2(\theta) \left(\frac{d \phi}{dt}\right)^2\right)-mgr\cos(\theta)<br />

After applying the Euler-Lagrange equation we get:

For \theta:

\frac{\partial L}{\partial (\frac{d \theta}{dt})} = m r^2\frac{d \theta}{dt}

\frac{\partial L}{\partial \theta} = m r^2 \left(\frac{d \phi}{dt}\right)^2 \sin(\theta) \cos(\theta) + mg r \sin(\theta)

<br /> \frac{\partial L}{\partial \theta} - \frac{d}{dt} \left ( \frac{\partial L}{\partial (\frac{d \theta}{dt})} \right ) = 0 \;\; \Rightarrow \;\; <br /> m r^2 \left(\frac{d \phi}{dt}\right)^2 \sin(\theta) \cos(\theta) + mg r \sin(\theta) -<br /> m r^2 \frac{d^2 \theta}{dt^2}=0<br />

For \phi:

\frac{\partial L}{\partial \phi} = 0 \;\; \Rightarrow \;\; \frac{\partial L}{\partial (\frac{d \phi}{dt})}=m r^2 \sin^2(\theta) \frac{d \phi}{dt}= const.

Now we have two equations:

<br /> \frac{d^2 \theta}{dt^2}-\frac{g}{r}\sin(\theta)-\sin(\theta) \cos(\theta) \left(\frac{d \phi}{dt}\right)^2=0<br />

and

<br /> m r^2 \sin^2(\theta) \frac{d \phi}{dt}= const. = A<br />

And here is where it stopped. I don't know how to solve this two equations.

 

[Hurkyl]

The two things I first observe about that pair of equations is:
(1) It's easy to eliminate one of the variables
(2) A change of variable could help simplify things

 

[TimJ]

Then I get this equation:

<br /> \frac{d^2 \theta}{dt^2}-\frac{g}{r}\sin(\theta)- <br /> \frac{A^2 \cos(\theta)}{(m r^2)^2 \sin^3(\theta) }=0<br />

I tried to solve it with Mathematica, but i didn't get a result.