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Physics 129a
Calculus of Variations
071113 Frank Porter
Revision 081120
1
Introduction
Many problems in physics have to do with extrema.
When the problem
involves fnding a Function that satisfes some extremum criterion, we may
attack it with various methods under the rubric oF “calculus oF variations”.
The basic approach is analogous with that oF fnding the extremum oF a
Function in ordinary calculus.
2
The Brachistochrone Problem
Historically and pedagogically, the prototype problem introducing the cal
culus oF variations is the “brachistochrone”, From the Greek For “shortest
time”. We suppose that a particle oF mass
m
moves along some curve under
the inﬂuence oF gravity. We’ll assume motion in two dimensions here, and
that the particle moves, starting at rest, From fxed point
a
to fxed point
b
.
We could imagine that the particle is a bead that moves along a rigid wire
without Friction [±ig. 1(a)]. The question is: what is the shape oF the wire
Forwh
ichthet
imetogetFrom
a
to
b
is minimized?
±irst, it seems that such a path must exist – the two outer paths in
±ig. 2(b) presumably bracket the correct path, or at least can be made to
bracket the path. ±or example, the upper path can be adjusted to take an
arbitrarily long time by making the frst part more and more horizontal. The
lower path can also be adjusted to take an arbitrarily long time by making
the dip deeper and deeper. The straightline path From
a
to
b
must take
a shorter time than both oF these alternatives, though it may not be the
shortest.
It is also readily observed that the optimal path must be singlevalued in
x
, see ±ig. 1(c). A path that wiggles back and Forth in
x
can be shortened in
time simply by dropping a vertical path through the wiggles. Thus, we can
describe path
C
as a Function
y
(
x
).
1
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a
b
a
b
(a)
(b)
y
x
(c)
a
b
.
.
.
.
.
.
Figure 1: The Brachistochrone Problem: (a) Illustration of the problem; (b)
Schematic to argue that a shortesttime path must exist; (c) Schematic to
argue that we needn’t worry about paths folding back on themselves.
We’ll choose a coordinate system with the origin at point
a
and the
y
axis
directed downward (Fig. 1). We choose the zero of potential energy so that
it is given by:
V
(
y
)=
−
mgy.
The kinetic energy is
T
(
y
−
V
(
y
1
2
mv
2
,
for zero total energy. Thus, the speed of the particle is
v
(
y
q
2
gy.
An element of distance traversed is:
ds
=
q
(
dx
)
2
+(
dy
)
2
=
v
u
u
t
1+
dy
dx
!
2
dx.
Thus, the element of time to traverse
ds
is:
dt
=
ds
v
=
r
⎥
dy
dx
±
2
√
2
gy
dx,
and the total time of descent is:
T
=
Z
x
b
0
r
⎥
dy
dx
±
2
√
2
dx.
2
Diferent Functions
y
(
x
) will typically yield diferent values For
T
;weca
l
l
T
a “Functional” oF
y
. Our problem is to ±nd the minimum oF this Functional
with respect to possible Functions
y
.N
o
t
et
h
a
t
y
must be continuous – it
would require an in±nite speed to generate a discontinuity. Also, the accel
eration must exist and hence the second derivative
d
2
y/dx
2
. We’ll proceed
to Formulate this problem as an example oF a more general class oF problems
in “variational calculus”.
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This note was uploaded on 03/21/2012 for the course PHYSICS 129 taught by Professor Johnh.schwarz during the Winter '12 term at Caltech.
 Winter '12
 JohnH.Schwarz
 Physics

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