16—Calculus of Variations 397The potential energy is a functionUofrandφ. With the of the Lagrangian defined asT−U, the
variational derivative determines the equations of motion to be
S[r,φ]=
∫t 2t 1L
(
r(t),φ(t)
)
dt→
δS
δr
=
∂L
∂r
−
d
dt
∂L
∂r ̇
=mrφ ̇^2 −
∂U
∂r
−m ̈r= 0
δS
δφ
=
∂L
∂φ
−
d
dt
∂L
∂φ ̇
=−
∂U
∂φ
−m
d
dt
(
r^2 φ ̇
)
= 0
These are the components ofF~=m~ain polar coordinates. If the potential energy is independent of
φ, the second equation says that angular momentum is conserved:mr^2 φ ̇.
What do the discrete approximate equations (16.34) or (16.35) look like in this context? Look
at the case of one-dimensional motion to understand an example. The Lagrangian is
L=
m
2
x ̇^2 −U(x)
Take the expression in Eq. (16.34) and set it to zero.
−
dU
dx
(x`) +
1
2∆
[
m(x−x− 2 )/2∆−m(x+2−x)/2∆
]
= 0
or m
x+2− 2 x+x`− 2
(2∆)^2
=−
dU
dx
(x`) (16.39)
This is the discrete approximation to the second derivative, as in Eq. (11.12).
16.8 Endpoint Variation
Following Eq. (16.6) I restricted the variation of the path so that the endpoints are kept fixed. What
if you don’t? As before, keep terms to the first order, so that for example∆tbδyis out. Because the
most common application of this method involves integrals with respect to time, I’ll use that integration
variable
∆S=
∫tb+∆tbta+∆tadtL
(
t,y(t) +δy(t),y ̇(t) +δy ̇(t)
)
−
∫tbtadtL
(
t,y(t),y ̇(t)
)
=
∫tb+∆tbta+∆tadt
[
L(t,y,y ̇) +
∂L
∂y
δy+
∂L
∂y ̇
δy ̇
]
−
∫tbtadtL
(
t,y(t),y ̇(x)
)
=
[∫
tb+∆tbtb−
∫ta+∆tata]
dtL(t,y,y ̇) +
∫tbtadt
[
∂L
∂y
δy+
∂L
∂y ̇
δy ̇
]
Drop quadratic terms in the second line: anything involving(δy)^2 orδyδy ̇or(δy ̇)^2. Similarly, drop
terms such as∆taδyin going to the third line. Do a partial integration on the last term
∫tbtadt
∂L
∂y ̇
dδy
dt
=
∂L
∂y ̇
δy
∣
∣
∣∣
tbta−
∫tbtadt
d
dt
(
∂L
∂y ̇
)
δy(t) (16.40)
The first two terms, with the∆taand∆tb, are to first order
[∫
tb+∆tbtb−
∫ta+∆tata