16—Calculus of Variations 38816.3 Brachistochrone
Now for a tougher example, again from the introduction. In Eq. (16.2), which of all the paths between
fixed initial and final points provides the path of least time for a particle sliding along it under gravity.
Such a path is called a brachistochrone. This problem was first proposed by Bernoulli (one of them), and
was solved by several people including Newton, though it’s unlikely that he used the methods developed
here, as the general structure we now call the calculus of variations was decades in the future.
Assume that the particle starts from rest so thatE= 0, then
T[y]=
∫x 00dx
√
1 +y′^2
√
2 gy
(16.13)
For the minimum time, compute the derivative and set it to zero.
√2 g
δT
δy
=−
√
1 +y′^2
2 y^3 /^2
−
d
dx
y′
2
√
y
√
1 +y′^2
= 0
This is starting to look intimidating, leading to an impossibly* messy differential equation. Is there
another way? Yes. Why mustxbe the independent variable? What about usingy? In the general
setup leading to Eq. (16.10) nothing changes except the symbols, and you have
I[x]=
∫
dyF(y,x,x′) −→
δI
δx
=
∂F
∂x
−
d
dy
(
∂F
∂x′
)
(16.14)
Equation (16.13) becomes
T[x]=
∫y 00dy
√
1 +x′^2
√
2 gy
(16.15)
The functionxdoes not appear explicitly in this integral, just its derivativex′=dx/dy. This simplifies
the functional derivative, and the minimum time now comes from the equation
δI
δx
= 0−
d
dy
(
∂F
∂x′
)
= 0 (16.16)
This is much easier.d()/dy= 0means that the object in parentheses is a constant.
∂F
∂x′
=C=
1
√
2 gy
x′
√
1 +x′^2
Solve this forx′and you get (letK=C
√
2 g)
x′=
dx
dy
=
√
K^2 y
1 −K^2 y
, so x(y) =
∫
dy
√
K^2 y
1 −K^2 y
This is an elementary integral. Let 2 a= 1/K^2
x(y) =
∫
dy
y
√
2 ay−y^2
=
∫
dy
y
√
a^2 −a^2 + 2ay−y^2
=
∫
dy
(y−a) +a
√
a^2 −(y−a)^2
* Only improbably. See problem16.12.