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13.8.3  Solving the Brachistochrone Problem

To solve the brachistochrone problem (see Section 13.8.1), you can first find the Euler-Lagrange equations for the Lagrangian

  L(x,y(x),y′(x))=
1+y′(x)2
g y(x)
.

You can simplify this somewhat by assuming that you are using units where 2g=1.

assume(y>=0):; euler_lagrange(sqrt((1+y'^2)/y),x,y)
     











1









d
dx
y
x



2



 
+1





y
x
=K2,
d2
dx2
y
x
=



d
dx
y
x



2



 
−1
2 y
x











          

It is easier to solve the first equation for y which can be rewritten as

  
dy
dx
=−
C
x
−1

for appropriate C, which can be solved by separation of variables, getting you the parametric equations

  








    x=
1
2
C
2θ −sin(2θ)
    y=
1
2
C
1 −cos(2θ)

which parameterize a cycloid. This implicitly defines a function y=y(x) as the only stationary function for L. The problem is to prove that it minimizes T, which would be easy if the integrand L was convex. However, it’s not the case here:

assume(y>=0):; convex(sqrt((1+y'^2)/y),y(x))
     









d
dx
y
x



2



 
+3≥ 0





          

This is equivalent to |y′(t)|≤√3, which is certainly not satisfied by the cycloid y near the point x=0.

Using the substitution y(x)=z(x)2/2, you get y′(x)=z′(xz(x) and

  L(x,y(x),y′(x))=P(x,z(x),z′(x))=
2(z(x)−2+z′(x)2)
.

The function P is convex:

assume(z>=0):; convex(sqrt(2*(z^(-2)+z'^2)),z(x))
     
true           

Hence the function z(t)=√y(t), stationary for P (verified directly), minimizes the objective functional

  U(z)=
x1


0
P(x,z(x),z′(x)) dx.

From here and U(z)=T(y) it easily follows that y minimizes T and is therefore the brachistochrone.


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