Update: I made a number of mistakes in the original version of this post, although I think all the big ideas are right. I tried to fix everything, but I wouldn't be at all surprised if I've made additional errors.
I'm pretty sure that this is a very hard problem! I think I know how to get started, but I doubt I can finish.
I'll work in Schwarzschild coordinates, with $c=1$, Schwarzschild radius $R$, and $(+---)$ signature, so the metric is
$$
ds^2=(1-R/r)dt^2-(1-R/r)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta\,d\phi^2).
$$
All the action can be taken to lie in the equatorial plane $\theta=\pi/2$.
For a particle traveling on a geodesic in this geometry, the energy-like conserved quantity (i.e., the one arising from time-translation invariance of the metric) is $u_0$ where $u$ is the 4-velocity. I'll call this quantity $E$ (it's really energy per unit mass):
$$
E=\left(1-{R\over r}\right)\left(dt\over d\tau\right).
$$
I'm going to assume that this quantity is conserved even for our particle whizzing along in the tube. I'm pretty sure this is the correct generalization of the assumption that the constraint forces due to the tube do no work. I think you could prove this by looking at the physics in a local inertial reference frame in which the tube is at rest as the particle whizzes by. In that frame, the above energy conservation law is equivalent to the statement that the particle's speed is constant, which follows from a special-relativistic analysis in that frame, because the tube pushes in a direction perpendicular to the velocity.
Next we use the fact that the four-velocity has unit norm:
$$
1=u_\mu u^\mu=(1-R/r)\dot t^2-(1-R/r)^{-1}\dot r^2-r^2\dot\phi^2,
$$
where dots mean $d/d\tau$.
Divide through by $\dot t^2$:
$$
\dot t^{-2}=1-{R\over r}-\left(1-{R\over r}\right)^{-1}\left({dr\over dt}\right)^2-r^2\left({d\phi\over dt}\right)^2.
$$
Express $\dot t$ in terms of the energy, and rearrange:
$$
\left(1-{R\over r}\right)^{-1}\left(dr\over dt\right)^2+r^2\left(d\phi\over dt\right)^2
=1-{R
\over r}-\left(1-R/r\over E\right)^2.
$$
Say the particle starts from rest at position $r_0$. Then $E=(1-R/r_0)^{1/2}$. So
$$
dt=\sqrt{(1-R/r)^{-1}(dr/d\phi)^2+r^2\over 1-R/r-(1-R/r)^2/(1-R/r_0)}\,d\phi.
$$
If our initial points are $(r_0,0)$ and $(r_0,\alpha$), then the quantity we want to minimize is
$$
t=\int_0^\alpha \sqrt{(1-R/r)^{-1}(dr/d\phi)^2+r^2\over 1-R/r-(1-R/r)^2/(1-R/r_0)}\,d\phi.
$$
You can in principle use standard calculus of variations techniques from here to get $r(\phi)$.
That's enough for me! You said in the comments that you'd be happy with just the functional. Are you happy?
A good question! This is best answered by making use of intrinsic coordinates. The reason for this is twofold: firstly, as you have very nearly found, the equation of a cycloid is particularly simple when written in intrinsic coordinates; and secondly, the variational equation for the curve also takes on a a particularly simple form in these coordinates. I will explain both of these points now.
Intrinsic coordinates
In any curvilinear coordinate system , a curve is described by specifying the values of the coordinates as functions of each other e.g. $y=f(x)$, or as a functions of some parameter, e.g. $y=y(t), x=x(t)$. Here, each coordinate is understood to be defined relative to some 'grid'. An alternative, and somewhat esoteric, way to describe a curve is to specify its arc length, $s$, and its angle of inclination (relative to some chosen direction in space), $\psi$. This does not require reference to any fixed grid of coordinates, and instead uses the 'intrinsic' properties of the curve itself.
(N.B. this image shows the y axis pointing up - we will take it to be pointing down instead.)
Cartesian coordinates and intrinsic coordinates are related by
$$
\frac{\text{d}y}{\text{d}s} = \sin\psi, \qquad \frac{\text{d}x}{\text{d}s} = \cos\psi
$$
These awkward equations are part of the reason that intrinsic coordinates are so rarely used.
(I haven't quite figured out Geogebra yet!)
As you have seen, for a cycloid described by parametric equations
$$
x(\phi) = r(\phi-\sin\phi), \qquad y(\phi) = r(1-\cos\phi)
$$
we get
$$
x' = r(1-\cos\phi), \qquad y' = r\sin\phi
$$
so that
$$
\frac{\text{d}y}{\text{d}\phi} \frac{\text{d}\phi}{\text{d}x}=\frac{r\sin\phi}{r(1-\cos\phi)} = \cot\frac{\phi}{2}= \frac{\text{d}y}{\text{d}s}\frac{\text{d}s}{\text{d}x} = \tan\psi \implies \psi = \frac{\pi}{2}-\frac{\phi}{2}.
$$
We also have
$$
s(\phi) = \int_0^\phi \text{d}\phi'\sqrt{x'^2+y'^2} = r^2\int_0^\phi \text{d}\phi' \sqrt{2-2\cos\phi'}=4r-2r\sqrt{2}\sqrt{1-\cos\phi}\cot\frac{\phi}{2}.
$$
so that, combining the two, we get the very simple
$$
s(\psi) = 4r(1-\sin\psi).
$$
This is the first part done - we have used intrinsic coordinates to show how such a simple expression arises. I will now demonstrate why it is so simple in these coordinates.
Equations of motion in intrinsic coordinates
We now consider the velocity of a particle with position $\mathbf{r}(t)$, and rewrite it in terms of the arc length:
$$
\mathbf{v}(t) = \frac{\text{d}\mathbf{r}}{\text{d}t} = \frac{\text{d}\mathbf{r}}{\text{d}s}\frac{\text{d}s}{\text{d}t} = \hat{\boldsymbol{\tau}}\dot{s},
$$
where $\hat{\boldsymbol{\tau}} = \text{d}\mathbf{r}/\text{d}s$ is a unit vector tangent to the path of the particle, and $\text{d}s/\text{d}t=v$ is its speed. Now, using the Frenet-Serret formulas, the time derivative of $\hat{\boldsymbol{\tau}}$ is
$$
\frac{\text{d}\hat{\boldsymbol{\tau}}}{\text{d}t} =\frac{\text{d}\hat{\boldsymbol{\tau}}}{\text{d}s}\frac{\text{d}s}{\text{d}t}= \frac{1}{\rho} \hat{\mathbf{n}} \dot{s}
$$
where $\hat{\mathbf{n}}$ is a unit vector orthogonal to $\hat{\boldsymbol{\tau}}$. This means that the acceleration of the particle is
$$
\mathbf{a} =\frac{\text{d}\mathbf{v}}{\text{d}t} = \frac{\dot{s}^2}{\rho}\hat{\mathbf{n}}+\ddot{s}\hat{\boldsymbol{\tau}}.
$$
If the particle moves under gravity, then the equation of motion is therefore
$$
m\frac{\dot{s}^2}{\rho}\hat{\mathbf{n}}+m\ddot{s}\hat{\boldsymbol{\tau}} = mg\hat{\mathbf{y}},
$$
where $\hat{\mathbf{y}}$ is a unit vector in the $y$ direction. Taking the dot product with $\hat{\boldsymbol{\tau}}$, and using $\hat{\boldsymbol{\tau}}\cdot\hat{\mathbf{y}} = \sin\phi$, we get the equation of motion for the speed of the particle:
$$
\ddot{s} = g\sin\psi.\tag{2}
$$
This equation is very important. Since for the Brachistochrone we have the initial conditions $s(0)=\dot{s}(0)=0$, the above equation means that the form of $s(t)$ is entirely determined by $\psi(t)$. In other words, the function $\psi(t)$ entirely determines the shape of the curve.
Functional minimisation
Finally, consider the functional to be minimised for the Brachistochrone problem: the total time $T$ to traverse a curve $C$ under the influence of gravity is
$$
T = \int_C \text{d}t
$$
As it stands, the integrand is not in a form that allows the functional to be varied: the path dependence of $T$ is not explicit. In order to apply the Euler-Lagrange equations, it is necessary to rewrite the integral to be over a fixed domain. Two common ways to do this are
$$
T = \int_C \frac{\text{d}s}{\dot{s}} = \int \sqrt{\frac{1+y'(x)^2}{2gy(x)}}\text{d}x
$$
and
$$
T = \int \sqrt{\frac{\dot{x}(t)^2+\dot{y}(t)^2}{2gy(t)}}\text{d}t.
$$
This works because $y(x)$, or $x(t)$ and $y(t)$ together are sufficient to specify the shape of the curve. This is true more generally: any form of the integrand (over a fixed domain) that will allow the Euler Lagrange equations to return a complete description of the curve is sufficient. But we have just found a single function that does just that, namely $\psi(t)$! Assuming $\psi(t)$ is invertible (which it is over the domain we are interested in) we are therefore allowed to rewrite the integrand as
$$
T = \int_C \text{d}t = \int\frac{\text{d}t}{\text{d}\psi} \text{d}\psi. \tag{3}
$$
Applying the E-L equations
$$
\frac{\text{d}}{\text{d}\psi}\frac{\partial\mathcal{L}}{\partial t'} = \frac{\partial\mathcal{L}}{\partial t}
$$
with $\mathcal{L} = \frac{\text{d}t}{\text{d}\psi} = t'$, we find
$$
\frac{\text{d}}{\text{d}\psi} t' =0 \implies t' = \text{const} \implies t=A\psi+B \text{ i.e. } \psi(t) = at+b
$$
This is exactly what you found before. As I will show below, the parameter $a$ is your $-\omega/2$. (I am stuck on one final detail here - clearly we need $b=\pi/2$, but I can't see a simple reason for this).
To reiterate, it is only possible to write equation (3) because $s(t)$ is entirely determined by $\psi(t)$ by equation (2). This is the reason we are not allowed to write e.g. $T= \int(\text{d}t/\text{d}x) \text{d}x$ - because the function $x(t)$ alone is not sufficient to describe the curve - $y(t)$ is also required.
Finally, setting $b=\pi/2$ and $a=-\omega/2$ integrating the equation of motion (2), we find
$$
\frac{\text{d}^2 s}{\text{d}t^2} = g\sin\left(\frac{\pi}{2}-\frac{\omega}{2} t\right) \implies s(t) = -\frac{4g}{\omega^2}\sin\left(\frac{\pi}{2}-\frac{\omega}{2} t\right) +ct+d.
$$
With $s(0) = \dot{s}(0) = 0$, we find $c=0, d = 4g/\omega^2$ so that
$$
s(\psi) = \frac{4g}{\omega^2}(1- \sin\psi)
$$
as before.
References:
- https://en.wikipedia.org/wiki/Curvilinear_coordinates
- https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas
- https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec06.pdf
Best Answer
Yes there is, the curve is a a hypocycloid.
See for instance:
http://en.wikipedia.org/wiki/Hypocycloid
http://demonstrations.wolfram.com/SphereWithTunnelBrachistochrone/
http://www.physics.unlv.edu/~maxham/gravitytrain.pdf