Download The Calculus of Variations: An Introduction

The Calculus of Variations: An
Introduction
By Kolo Sunday Goshi
Some Greek Mythology

Queen Dido of Tyre
–
–

Iarbas’ (King of Libya) offer
–

Fled Tyre after the death of her husband
Arrived at what is present day Libya
“Tell them, that this their Queen of theirs may
have as much land as she can cover with the hide
of an ox.”
What does this have to do with the Calculus
of Variations?
What is the Calculus of Variations



“Calculus of variations seeks to find the path,
curve, surface, etc., for which a given
function has a stationary value (which, in
physical problems, is usually a minimum or
maximum).” (MathWorld Website)
Variational calculus had its beginnings in
1696 with John Bernoulli
Applicable in Physics
Calculus of Variations


Understanding of a Functional
Euler-Lagrange Equation
–

Proving the Shortest Distance Between Two Points
–


In Euclidean Space
The Brachistochrone Problem
–

Fundamental to the Calculus of Variations
In an Inverse Square Field
Some Other Applications
Conclusion of Queen Dido’s Story
What is a Functional?


The quantity z is called a functional of f(x) in
the interval [a,b] if it depends on all the
values of f(x) in [a,b].
Notation
b
z    f  x  
a
–
Example
1
2
  x    cos  x 2  dx
0
0
1
Functionals

The functionals dealt with in the calculus of
variations are of the form
  f  x     F  x, y( x), y( x)  dx
a
b


The goal is to find a y(x) that minimizes Г, or
maximizes it.
Used in deriving the Euler-Lagrange
equation
Deriving the Euler-Lagrange Equation

I set forth the following equation:
y  x   y  x    g  x 
Where yα(x) is all the possibilities of y(x) that
extremize a functional, y(x) is the answer, α is a
constant, and g(x) is a random function.
y1
y(b)
y0 = y
y(a)
y2
a
b
Deriving the Euler-Lagrange Equation
  f  x    F  x, y( x), y( x)  dx
a
b

Recalling

It can now be said that:   y    F  x, y , y  dx
b
a

At the extremum yα = y0
= y and

The derivative of the
d
functional with respect

to α must be evaluated d
and equated to zero
d
0
d  0
 

F
x
,
y
,
y
a       dx
b
Deriving the Euler-Lagrange Equation

The mathematics
b 
d
involved

 
F  x, y , y   dx
a
d
 

b  F y
d
F y 

 

dx
a
d
 y  y  
–

Recalling
y  x   y  x    g  x 
So, we can say
b  F
b F
b F dg
d
F 
 
g
g dx  
gdx  
dx
a
a
a
d
y 
y
y dx
 y
Deriving the Euler-Lagrange Equation
b F
b F dg
d

gdx  
dx
a
a
d
y
y dx

Integrate the first part by parts and get

b
a
d  F
g

dx  y

 dx


So

Since we stated earlier that the derivative of Г with respect to α equals
zero at α=0, the extremum, we can equate the integral to zero
 F
b
d
d  F
 g


a
d
 y dx  y

  dx

Deriving the Euler-Lagrange Equation

So
0
b
a

 F d  F  
g
 
  dx
 y0 dx  y0  
We have said that y0 = y, y
being the extremizing
function, therefore
y1
y0 = y
y2

Since g(x) is an arbitrary
function, the quantity in the
brackets must equal zero
0
b
a
 F d  F  
g
 
  dx
 y dx  y  
The Euler-Lagrange Equation

We now have the Euler-Lagrange Equation
F d  F 


0
y dx  y 
 When F  F  y, y , where x is not included,
the modified equation is
F
Fy
C
y
The Shortest Distance Between Two
Points on a Euclidean Plane

What function describes the shortest
distance between two points?
–
Most would say it is a straight line



Logically, this is true
Mathematically, can it be proven?
The Euler-Lagrange equation can be used to
prove this
Proving The Shortest Distance
Between Two Points

Define the distance to be s, so
s   ds
b
ds
dy
dx
a

Therefore
s   dx 2  dy 2
Proving The Shortest Distance
Between Two Points

Factoring a dx2 inside the square root and
taking its square root we obtain
s   dx  dy
2
2
s
b
a
 dy 
1 
 dx
 dx 
dy
 Now we can let y 
dx

so
s
b
a
2
1  y 2 dx  
Proving The Shortest Distance
Between Two Points

Since

b
a
1  y 2 dx
  f  x     F  x, y( x), y( x)  dx
a
b

And we have said that

we see that
F  1 y2
therefore
F
0
y

F
y

y
1 y2
Proving The Shortest Distance
Between Two Points

Recalling the Euler-Lagrange
equation
F d  F 
 
0
y dx  y 

Knowing that
F
0
y

A substitution can be made

Therefore the term in brackets
must be a constant, since its
derivative is 0.
F
y

y
1 y2
d  y 
 
0
2
dx  1  y 
Proving The Shortest Distance
Between Two Points

More math to reach the solution
y
1 y2
C
y 2  C 2 1  y 2 
y 2 1  C 2   C 2
y2  D
yM
Proving The Shortest Distance
Between Two Points

Since
yM
We see that the derivative or slope of the
minimum path between two points is a
constant, M in this case.
The solution therefore is:
y  Mx  B
The Brachistochrone Problem

Brachistochrone
–
Derived from two Greek words



The problem
–
Find the curve that will allow a particle to fall under the
action of gravity in minimum time.


brachistos meaning shortest
chronos meaning time
Led to the field of variational calculus
First posed by John Bernoulli in 1696
–
Solved by him and others
The Brachistochrone Problem

The Problem restated
–

Find the curve that will allow a particle to fall under the
action of gravity in minimum time.
The Solution
–
–
A cycloid
Represented by the parametric equations
D
x    2  sin 2 
2
D
y  1  cos 2 
2

Cycloid.nb
The Brachistochrone Problem In an
Inverse Square Force Field

The Problem
–
–
Find the curve that will
allow a particle to fall
under the action of an
inverse square force field
defined by k/r2 in
minimum time.
Mathematically, the force
is defined as
k
Fr   2
r
y
1
2
r0
F 
k
rˆ
r2
x
The Brachistochrone Problem In an
Inverse Square Force Field


Since the minimum time is
being considered, an
expression for time must be
determined
An expression for the
velocity v must found and
this can be done using the
fact that KE + PE = E
t
2
1
ds
v
1
k
2
mv   E
2
r
The Brachistochrone Problem In an
Inverse Square Force Field

The initial position r0 is
known, so the total energy E
is given to be –k/r0, so
An expression can be found
for velocity and the desired
expression for time is found
1
k
k
2
mv   
2
r
r0
v
2k  1 1 
  
m  r r0 
m 2
ds
t
2k 1  1 1 
  
 r r0 
The Brachistochrone Problem In an
Inverse Square Force Field
Determine an
expression for ds
rdΘ
r
ds
dr
r + dr
ds   dr   r  d 
2
2
2
2
The Brachistochrone Problem In an
Inverse Square Force Field


We continue using a
polar coordinate system
An expression can be
determined for ds to put
into the time expression
ds   dr   r  d 
2
2
2
2
2


2  dr 
2
2
ds   d  
 r 
 d 

ds  r 2  r 2 d
The Brachistochrone Problem In an
Inverse Square Force Field


Here is the term for
time t
The function F is the
term in the integral
rr0 (r  r )
r0  r
2
m
t
2k 1
2
rr0 (r  r )
F
r0  r
2
2
2
The Brachistochrone Problem In an
Inverse Square Force Field

Using the modified
Euler-Lagrange
equation
F
F r
C
r
rr0 (r 2  r 2 )
 r2
r0  r
rr0
C
2
2
 r0  r  (r  r )
The Brachistochrone Problem In an
Inverse Square Force Field

More math involved in finding an integral to
be solved
r (r 2  r 2 )
 r2
r0  r
r2 r
2
2
r

r
(
r

r
)
0 
D
r
D
2
2
 r0  r  (r  r )
r5
G
2
2
 r0  r  (r  r )
The Brachistochrone Problem In an
Inverse Square Force Field


Reaching the integral
Solving the integral for r(Θ)
finds the equation for the
path that minimizes the time.
r 5  r 2G(r0  r )
dr
r

d
G (r0  r )

G(r0  r )
dr    d
5
2
r  r G (r0  r )
The Brachistochrone Problem In an
Inverse Square Force Field

Challenging Integral to Solve
–

Where to then?
–
–

Brachistochrone.nb
Use numerical methods to solve the integral
Consider using elliptical coordinates
Why Solve this?
–
Might apply to a cable stretched out into space to
transport supplies
Some Other Applications

The Catenary Problem
–
–
–

Derived from Greek for
“chain”
A chain or cable
supported at its end to
hang freely in a uniform
gravitational field
Turns out to be a
hyperbolic cosine curve
Derivation of Snell’s
Law
n1 sin i  n2 sin 2
Conclusion of Queen Dido’s Story


Her problem was to find the figure bounded by a line
which has the maximum area for a given perimeter
Cut ox hide into infinitesimally small strips
–
–
–

Used to enclose an area
Shape unknown
City of Carthage
Isoperimetric Problem
–
–
Find a closed plane curve of a given perimeter which
encloses the greatest area
Solution turns out to be a semicircle or circle
References





Atherton, G., Dido: Queen of Hearts, Atherton
Company, New York, 1929.
Boas, M. L., Mathematical Methods in the Physical
Sciences, Second Edition, Courier Companies, Inc.,
United States of America, 1983.
Lanczos, C, The Variational Principles of Mechanics,
Dover Publications, Inc., New York, 1970.
Ward, D., Notes on Calculus of Variations
Weinstock, R., Calculus of Variations, Dover
Publications, Inc., New York, 1974.