16—Calculus of Variations 389Make the substitution(y−a)^2 =zin the first half of the integral and(y−a) =asinθin the second
half.
x(y) =
1
2
∫
dz
1
√
a^2 −z
+
∫
a^2 cosθdθ
√
a^2 −a^2 sin^2 θ
=−
√
a^2 −z+aθ=−
√
a^2 −(y−a)^2 +asin−^1
(
y−a
a
)
+C′
The boundary condition thatx(0) = 0determinesC′=aπ/ 2 , and the other end of the curve determines
a:x(y 0 ) =x 0. You can rewrite this as
x(y) =−
√
2 ay−y^2 +acos−^1
(
a−y
a
)
(16.17)
This is a cycloid. What’s a cycloid and why does this equation describe one? See problem16.2.
x-independent
In Eqs. (16.15) and (16.16) there was a special case for which the dependent variable was missing from
F. That made the equation much simpler. What if the independent variable is missing? Does that
provide a comparable simplification? Yes, but it’s trickier to find.
I[y]=
∫
dxF(x,y,y′)−→
δI
δy
=
∂F
∂y
−
d
dx
(
∂F
∂y′
)
= 0 (16.18)
Use the chain rule to differentiateFwith respect tox.
dF
dx
=
∂F
∂x
+
dy
dx
∂F
∂y
+
dy′
dx
∂F
∂y′
(16.19)
Multiply the Lagrange equation (16.18) byy′to get
y′
∂F
∂y
−y′
d
dx
∂F
∂y′
= 0
Now substitute the termy′(∂F/∂y)from the preceding equation (16.19).
dF
dx
−
∂F
∂x
−
dy′
dx
∂F
∂y′
−y′
d
dx
∂F
∂y′
= 0
The last two terms are the derivative of a product.
dF
dx
−
∂F
∂x
−
d
dx
[
y′
∂F
∂y′
]
= 0 (16.20)
If the functionFhas no explicitxin it, the second term is zero, and the equation is now a derivative
d
dx
[
F−y′
∂F
∂y′
]
= 0 and y′
∂F
∂y′
−F=C (16.21)
This is already down to a first order differential equation. The combinationy′Fy′−Fthat appears on
the left of the second equation is important. It’s called the Hamiltonian.