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How does one show the uniqueness of the solution to the brachistochrone problem? Doesn't the fact that the solution is of the form $x=a-c(2t+\sin2t)$ and $y=c(1+\cos2t)$ naturally guarantee uniqueness given the 2 endpoints of the path -- 2 unknowns $(a,c)$ and 2 restraints (the 2 endpoints)?

Thanks!

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Sure, you have those conditions. Now, how sure are you an algebraic curve doesn't suit the bill? That's where solving the associated differential equations comes in... –  J. M. Jul 27 '11 at 9:26
    
Hi, J.M., I'm afraid I don't quite understand... Would you mind elaborating a bit? Thanks! –  Robin H Jul 27 '11 at 9:37
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I can for instance construct a parametric cubic which matches the boundary conditions, but does not satisfy the brachistochrone condition... –  J. M. Jul 27 '11 at 9:41

1 Answer 1

I'm not sure I understand your question clearly. But generally speaking, Picard's theorem asserts that a differential equation whose coefficients are continuous has only one solution through any given point. The brachistochrone is a second order linear equation, and thus, a "point" in the equations definition area consists of values for t, y(t) and y'(t). This can also be separated two two different boundary conditions, e.g. two tuples (t_0,y(t_0)) and (t_1,y'(t_1)). The representation of the brachistochrone as a linear equation asserts a certain location at a given time, and a maximal speed (which is the derivative of location, roughly speaking) at another time, which give initial conditions on both y and y'.

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