[sjastro]

Originally Posted by mithrandir

However, the function [image that won't paste] is particularly nice since x does not appear explicitly. Therefore, [another image] and we can immediately use the Beltrami identity ...
http://mathworld.wolfram.com/Brachis...neProblem.html

The Beltrami identity doesn't apply.
If you look at equation (4) if the endpoints are vertical, dx=0 hence ds=dy.
The integrand in equation (6) becomes dy/(2gy)^.5

The functional (http://en.wikipedia.org/wiki/Functio...tics)#Integral ) in this case is of the format f(y, x, x')dy not
f(x, y, y')dx as in the mathworld link where there is both a vertical and horizontal component of the endpoints.

In the f(y,x, x')dy format the partial derivatives of the Euler Lagrange equation are with respect to x and x' (for f(x,y,y')dx are these are with respect to y and y').

Since x and x' do not appear explicitly in the integrand, all the partial derivatives in the Euler LaGrange equation are zero.
Hence we have the trivial case of 0=0.

Are you suggesting there is no limit as delta x approaches zero, and the the vertical line is a discontinuity? Physics and discontinuities don't get on well together.

I'm not sugesting anything of the kind.

The Euler Lagrange equations applies to functionals not functions. A functional that satisfies the EL equation has a stationary or extremal value. Vertical endpoints are simply a particular case. Here the functional does not have a stationary solution.

Doesn't that assume a non-rotating frame of reference? A falling object in a rotating frame follows a parabola. (Ignoring any effects of air resistance.)

The Brachistrome problem assumes there are no fictitious forces.

Regards

Steven