The Calculus of Variations: An · PDF fileWhat does this have to do with the Calculus of Variations? What is the Calculus of Variations ... Challenging Integral to Solve

The Calculus of Variations: An Introduction

By Kolo Sunday Goshi

Some Greek Mythology

Queen Dido of Tyre– Fled Tyre after the death of her husband

– Arrived at what is present day Libya

Iarbas’ (King of Libya) offer– “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– Fundamental to the Calculus of Variations

Proving the Shortest Distance Between Two Points– In Euclidean Space

The Brachistochrone Problem– 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

– Example

b

a

z f x

1

12 2

00

cosx x dx

Functionals

The functionals dealt with in the calculus of variations are of the form

The goal is to find a y(x) that minimizes Г, or maximizes it.

Used in deriving the Euler-Lagrange equation

, ( ), ( )b

af x F x y x y x dx

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.

ba

y(b)

y(a)

y1

y0 = y

y2


Recalling

It can now be said that:

At the extremum yα = y0

= y and

The derivative of the

functional with respect

to α must be evaluated

and equated to zero

, ,b

ay F x y y dx

0

0d

d

, ,b

a

dF x y y dx

d

, ( ), ( )b

af x F x y x y x dx


The mathematics

involved

– Recalling

So, we can say

y x y x g x

, ,b

a

dF x y y dx

d

b

a

y yd F Fdx

d y y

b b b

a a a

d F F F F dgg g dx gdx dx

d y y y y dx


Integrate the first part by parts and get

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

b

a

d Fg dx

dx y

b

a

d F d Fg dx

d y dx y

b b

a a

d F F dggdx dx

d y y dx


So

We have said that y0 = y, y being the extremizing function, therefore

Since g(x) is an arbitrary function, the quantity in the brackets must equal zero

0 0

0b

a

F d Fg dx

y dx y

0b

a

F d Fg dx

y dx y

y1

y2

y0 = y

The Euler-Lagrange Equation

We now have the Euler-Lagrange Equation

When , where x is not included,

the modified equation is

0F d F

y dx y

,F F y y

FF 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

Therefore

s ds

2 2s dx dy

ds

adx

dy

b


Factoring a dx2 inside the square root and taking its square root we obtain

Now we can let

so

2

1b

a

dys dx

dx

dyy

dx

21b

as y dx

2 2s dx dy


Since

And we have said that

we see that

therefore

21b

ay dx

21F y

0F

y

21

F y

y y

, ( ), ( )b

af x F x y x y x dx


Recalling the Euler-Lagrange equation

Knowing that

A substitution can be made

Therefore the term in brackets must be a constant, since its derivative is 0.

20

1

d y

dx y

0F d F

y dx y

0F

y

21

F y

y y


More math to reach the solution

2

2 2 2

2 2 2

2

1

1

1

yC

y

y C y

y C C

y D

y M


Since

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 M

y Mx B

The Brachistochrone Problem

Brachistochrone

– Derived from two Greek words

brachistos meaning shortest

chronos meaning time

The problem

– Find the curve that will allow a particle to fall under the

action of gravity in minimum 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

Cycloid.nb

2 sin 22

1 cos22

Dx

Dy

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

2ˆ

kF r

r

0r

x

y

1

2

2r

kF

r


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

2

1

dst

v

21

2

kmv E

r


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

2

0

1

2

k kmv

r r

0

2 1 1kv

m r r

2

1

0

2 1 1

m dst

k

r r


r + dr

rdΘ

r

dr

ds

2 22 2ds dr r d

Determine an

expression for ds


We continue using a

polar coordinate system

An expression can be

determined for ds to put

into the time expression

2 22 2ds dr r d

2

22 2drds d r

d

2 2ds r r d


Here is the term for

time t

The function F is the

term in the integral

2 2

0

0

( )rr r rF

r r

2 22

0

10

( )

2

rr r rmt

k r r


Using the modified

Euler-Lagrange

equationF

F r Cr

2 220 0

2 2

0 0

( )

( )

rr r r rrr C

r r r r r r


More math involved in finding an integral to

be solved

2 22

2 2

0 0

( )

( )

r r r rr D

r r r r r r

2

2 2

0 ( )

r rD

r r r r

5

2 2

0 ( )

rG

r r r r


Reaching the integral

Solving the integral for r(Θ)

finds the equation for the

path that minimizes the time.

5 2

0

0

( )

( )

r r G r rdrr

d G r r

0

5 2

0

( )

( )

G r rdr d

r r G r r


Challenging Integral to Solve

– Brachistochrone.nb

Where to then?

– 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

Brachistochrone.nb

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 1 2 2sin sinin n

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.

Documents

The Calculus of Variations: An · PDF fileWhat does this have to do with the Calculus of Variations? What is the Calculus of Variations ... Challenging Integral to Solve