calculus of variation problems

1

Calculus of VariationsProblems

SOLO HERMELIN

2

SOLO Calculus of Variations - Problems

Table of Content

Introduction

1. Brachistochrone Problems

Brachistochrone Problem 1


2. Isoperimetric Problems

Dido Maximum Area Problem

Maximum volume with given surface area

Shape of a liquid drop on a horizontal surface

3. Problem of Minimum Surface of Revolution

4. Geodesics Problems

Geodesic in 3 Dimensional Spaces

Geodesics in Riemannian Space

Geodesics for an Implicit Equation of the Surface, G (x,y,z) = 0

5. Geometrical Optics and Fermat Principle

References

3

SOLO

Introduction

The Mathematical Theory of Calculus of Variation was given in another presentation:“Calculus of Variations”, by the same author.

Calculus of Variations tackles problems of finding Functions that Extremize a given Cost Functional . The Solutions must satisfy Boundary Conditions and some giving constraints. Solving those problems require finding solution to Differential Equations.

Calculus of Variations - Problems

In general few problems have analytic solutions and therefore the results can be obtained only numerically.

In this presentation we give example f problems that could be solved analytical, in order to facilitate the understanding of the subject.

Return to Table of Content

4

SOLO

A particle slides on a frictionless wire between two fixed points A(0,0) and B (xfc, yfc) in a constant gravity field g. The curve such that the particle takes the least time to go from A to B is called brachistochrone (βραχιστόσ Greek for

“shortest“, χρόνοσ greek for “time). The brachistochrone problem was posed by John Bernoulli in 1696, and played an important part in the development of calculus of variations. The problem was solved by Johann Bernoulli, Jacob Bernoulli, Isaac Newton, Gottfried Leibniz and Guillaume de L’Hôpital.

Let choose a system of coordinates with the origin at point A (0,0) and the y axis in the constant g direction

x

y

V

( )tγ

( )fcfc yxB ,fcx

fcy

N

( )0,0A




5

SOLO

Since the motion of the particle is in a frictionless fixed gravitational field the total energy is conserved

( ) ygVyVygVV 22

1

2

1 20

220 +=→−=

x

y

V

( )tγ

( )fcfc yxB ,fcx

fcy

N

( )0,0A

Second way to get this relation is:

( ) ygVVdygdVVsd

ydggV

sd

Vd

td

sd

sd

Vd

td

Vd =−→=→==== 20

2

2

1sin γ

where V0 is the velocity of the particle at point A and ( ) ( ) 22 ydxdsd +=

td

xd

xd

yd

td

yd

td

xd

td

sdV

222

1

+=

+

==

We have xdygV

xd

yd

xdV

xd

yd

td2

11

20

22

+

+

=

+

=

The cost function is ∫∫

=

+

+

=cfcf

xx

xdxd

ydyxFxd

ygV

xdyd

J00

20

2

,,2

1




6

HISTORY OF CALCULUS OF VARIATIONS

The brachistochrone problem

In 1696 proposed the Brachistochrone (“shortest time”) Problem:Given two points A and B in the vertical plane, what is the curve traced by a point acted only by gravity, which starts at A and reaches B in the shortest time.

Johann Bernoulli 1667 - 1748

SOLO

7


Jacob Bernoulli(1654-1705)

Gottfried Wilhelmvon Leibniz(1646-1716)

Isaac Newton(1643-1727)

The solutions of Leibniz, Johann Bernoulli, Jacob Bernoulliand Newton were published on May 1697 publication ofActa Eruditorum. L’Hôpital solution was published only in 1988.

Guillaume FrançoisAntoine de L’Hôpital

(1661-1704)

SOLOHISTORY OF CALCULUS OF VARIATIONS

8

SOLO

A particle slides on a frictionless wire between two fixed points A(0,0) and B (xfc, yfc) in a constant gravity field g. The curve such that the particle takes the least time to go from A to B is called brachistochrone.

x

y

V

( )tγ

( )fcfc yxB ,fcx

fcy

N

( )0,0A

∫∫

=

+

+

=cfcf xx

xdxd

ydyxFxd

ygV

xd

yd

J00

20

2

,,2

1

We derived the cost function:

xd

ydy

ygV

y

xd

ydyxF =

+

+=

:

2

1:,,

20

2

where

F doesn’t depend explicitly on the free variable x, therefore if we replace and we use the result obtained for F not depending explicitly on x, we obtain

( ) ( )xtyx ,, →

( ) ( ) constygVy

y

ygV

yyyFyyyF y ==

++−

+

+=− α

212

1,,

20

2

2

20

2

constygVy

==++

α21

12

02

or




9

SOLO

x

y

V

( )tγ

( )fcfc yxB ,fcx

fcy

N

( )0,0A

constygVy

==++

α21

12

02

Let define a parameter τ such that

τcos

1

12

=

+

xd

yd

and constygV

ygVxd

yd==

+=

+

+

ατ

2

cos

21

12

020

2

From which ( ) ( )τταα

τ2cos12cos1

4

1

2

cos

2 22

220 +=+==+ r

ggg

Vy

Tacking the derivative of this equation with respect to τ we obtain ττ

2sin2 rd

yd −=




10

SOLO

τ

ττ

τ

ττ

222

2

2cos

/1

1 =

+

=

+

d

yd

d

xd

d

xd

d

xd

d

yd

( )

( )ττ

τττ

ττττ

τττττ

2sin2

2cos12cos4

cossin16sin

cos2sin4cos

0

2

4222

2

2222

22

+±=→

+±=±=→

=

→

+

=

→

rxx

rrd

xd

rd

xd

rd

xd

d

xd

Let change variables to 2τ = θ – π, to get

( )

( )θ

θθ

cos12

sin2

0

0

−=+

−+=

rg

Vy

rxx

θsinr

θcosr

θr

x

y

0x

0V

g

V

2

20

r

rA

B

),( yxθ

We obtain the equation of a cycloid generated by a circle of radius r rolling upon the horizontal line

and starting at the point

g

Vy

2

20−=

−−

g

Vx

2,

20

0




11

HISTORY OF CALCULUS OF VARIATIONS


( )

( )

−−=

−+=

g

Vry

rxx

2cos1

sin2

0

0

θ

θθ Cycloid Equation

∫∫∫∫

=

+

+

===cfcfcf xxxt

xdxd

ydyxFxd

ygV

xd

yd

V

sdtdJ

002

0

2

00

,,2

1

Minimization Problem

Solution of the Brachistochrone Problem:

SOLO

Johann Bernoulli 1667 - 1748

12

SOLO

( )

( )θ

θθ

cos12

sin2

0

0

−=+

−+=

rg

Vy

rxxθsinr

θcosr

θr

x

y

0x

0V

g

V

2

20

r

rA

B

),( yxθ


We have

( )

( )td

dr

td

dr

td

d

d

xd

d

xdV

rd

yd

rd

xd

θθθθθθθ

θθ

θθ

=−=

+

=

=

−=

2sin2cos12

sin

cos1

22

( ) rgrgrgygVV

=

=−=+=

2sin2

2sin4cos122 22

0

θθθ

constr

g

td

d ==θ

( )0*

0

θθθθ

θ

−=== ∫∫ fABg

rd

g

rtdt

f

From those two equations we obtain

and the minimum time to travel between A and B will be




13


Brachistochrone Problem 2 (Bryson Sec 2.7, Problem 6, pg. 81, Sec. 3.11, Example 1, pg. 119, Sec. 4.3, Example 1, pg. 142)

A particle slides on a frictionless wire between two points A (0,0) and B (xfc, yfc) in a constant gravity field. The particle has an initial velocity V0 at point A. What is the shape of the wire:

• that will produce a minimum time path between the two points with the constraint that y ≤ x tanθ + h, where θ and h are constant.

• That provides a maximum xf for given tf and yf.

hxy +< θtan

x

y

V

( )tγ

( )fcfc yxB ,fcx

fcy

N


14



Solution to Brachistochrone Problem 2

Since the motion of the particle is in a frictionless fixed gravitational field the total energy is conserved

( ) ygVyVygVV 22

1

2

1 20

220 +=→−=

The equations of motions are

( ) ( ) ( )( ) ( ) ( ) fcf

fcf

ytytyyVy

xtxtxyVx

===

===

0sin

0cos

0

0

γγ

We have 2 First Order Differential Equations in x (t) and y (t) and 1 control γ (t) with 4 Boundary Conditions

To find the wire path we must define the angle γ (t) such that

ftJγγ

minmin =


15


Solution to Brachistochrone Problem 2 (continue – 1)

The unconstrained problem: y ≤ x tanθc + h

( )

=−

=−=

0

0:,

fcf

fcf

ff yy

xxtxψ

Let adjoin the constraints to obtain the augmented cost function:

( ) ( ) ( )( ) ( )( )[ ]∫ −+−+−+−+=ft

t

yxfcffcff dtyyVxyVxxxxtJ0

sincos21 γλγλνν

1. The Hamiltonian ( ) ( )γλγλ sincos: yxyVH +=

( ) ( ) ( )fcffcffff xxxxttx −+−+=Φ 21:, ννand

2. Euler-Lagrange Equations

( )fxfxT tH Φ=−= λλ

( ) ( ) ( ) 10 νλλλλ ==→=∂∂−= fxfxxx ttt

x

H

( ) ( ) 2sincos νλγλγλλ =+−=∂∂−= fyyxy t

yd

Vd

y

H


16



( ) ( )x

yyxyVH

λλ

γγλγλγ =→=+−= tan0cossin:

Since H is not an explicit function of time we have: ( ) ( )ftHconsttH ** ==

Using the Boundary Conditions: ( ) ( ) 1* −=Φ−= ftf ttH

Therefore instead of solving the differential equation, let use: ( )tyλ

( ) ( ) ( ) ( ) ( ) ( ) 1sin

sincoscotsincossincos:* −==+=

+=+=

γλ

γγγλγγλλλγλγλ y

yy

xyyx yVanyVyVyVH

From which ( )yVy

γλ sin−=

Note: If for the end point B (xfc, yfc) only xfc, is defined and yfc is free, then

( ) ( )( ) ( ) 0sin0

sin=→=−= f

f

ffy t

yV

tt

c

γγ

λ

End Note


17



Also

( ) ( ) ( ) ( ) ( ) ( ) 1cos

sintancossincossincos:* −==+=

+=+=

γλγγγλγ

λλ

γλγλγλ xx

x

yxyx yVyVyVyVH

Therefore( )

0

0

cos

1

cos γλγV

constyV

x

==−=

or ( )x

yVλ

γcos−=

Note:

( )0

*

cos01

γλ xVH =−=

20

πγ =If V0 = V (0)=0, to keep we must have

End Note

If we take the time derivative of the last equation we obtain:( ) ( ) ( )

td

dyV

yd

yVd

td

yd

yd

yVd

x

γλ

γγ sinsin ==

or using ( ) ygVyV 220 +=

( ) ( ) ( ) ( ) constgyVyV

gyV

yd

yVd

td

dxxx ===== ωλλλγ


18



Let find now the shape of the wire

Let use and( ) ( ) 0cos 0 == txyVtd

xd γ ( )x

yVλ

γcos−=

( )x

x yVgd

xd

td

d

d

xd

td

xd

λγγλ

γγ

γ

2coscos −====

or( )

−+−−=

+−=−=→−=

∫∫

002

22

2

2

2

2sin2sin224

1

2

2cos11cos

1

cos00

γγγγλ

γγλ

γγλ

λγ

γ

γ

γ

γ

γ

x

xx

x

g

dg

dg

x

gd

xd

From the equations: and , we obtain: ( ) ygVyV 220 += ( )

x

yVλ

γcos−=

( )g

V

gg

V

gyygV

xxx2

2cos14

1

2

cos

2

1cos2

20

2

20

2

2

2

22

0 −+=−

=→=+ γ

λλγ

λγ


19



The wire shape is given by:

( )

( )g

V

gy

gx

x

x

22cos1

4

1

2sin2sin224

1

20

2

002

−+=

−+−−=

γλ

γγγγλ

For V0 ≠ 0 from which ( ) 0

00 cos

0

cos

VyVx

γγλ −==

−= 00

cosγλωγV

gg

td

dx −===

( )

( )g

V

g

Vy

g

Vx

22cos1

cos4

2sin2sin22cos4

20

02

20

000

2

20

−+=

−+−−=

γγ

γγγγγ

The wire shape is given by:


20



By using

unknowng

Vr 0

02

20

cos4: γ

γ= πγθ += 2: ( ) unknownrx 0000 sin: θθθ −−=

we obtain:( )

( ) EquationCycloid

g

Vry

rxx

−−=

−+=

2cos1

sin2

0

0

θ

θθ

To find r and x0 we must solve the following 4 equations with additional 2 unknowns θ0 and θ f

( )

( ) ( )

( )( )ff

fff

rg

Vy

rxx

g

V

g

Vrr

g

V

rx

c

c

θ

θθ

θθθ

θθ

cos12

sin

2sin4

cos12cos1

2

sin0

20

0

02

20

0

20

0

20

000

−=+

−+=

=−

=→−=

−+=

For V0 ≠ 0


21



For V0 = 0 we found 0,02 000 ==→= xθπγ

( )( )

−=−=

θθθ

cos1

sin

ry

rx

where 24

1

xgr

λ=

and it can be calculated using the following 2 equations with 2 unknowns r and θf

( )( )

−=

−=

ff

fff

ry

rx

c

c

θθθ

cos1

sin

θf is given by the following transcedental equation

( )( )f

ff

f

f

c

c

y

x

θθθ

cos1

sin

−−

=

and ( )f

fcy

rθcos1−

=

Now we can compute( )

cf

fx ygrg

θλ

cos1

2

1

2

1 −==

( )const

y

g

y

g

r

gg

td

d

td

d

cc f

f

f

fx =

=

−=====

22sin

cos1

2

1

2

1

2

1 θθλωθγand

∫∫

===f

c

g

ydtdt f

f

f

AB

θ

θ

θ

θω0

*2

2sin

2

2

1


22

SOLO




If yfc is free, we found that

( ) ( )( ) ( ) ( ) ( ) πππγθπγγγ

λ 2,2,00sin0sin

=+=→=→=→=−= fffff

ffy ttt

yV

tt

c

Since for V0 = 0 we have θ0 = 0, we must have θf = π (γf = 0)

Using this result we obtain:

( )π

πθθ c

c

f

fff

xrrrx =→=−= sin

( )π

θ c

c

f

ff

xrry 22cos1 ==−=

cfx xgrg

πλ2

1

2

1 ==

constx

g

r

gg

td

d

td

d

cfx ====== πλωθγ

2

1

2

1

2

1

∫∫ ===π π

θω0

*

2

1

g

xdtdt cf

AB

For V0 = 0


23

HISTORY OF CALCULUS OF VARIATIONSSOLO

“When the Tyrian princess Dido landed on the Mediterranean sea she was welcomed by a local chieftain. He offered her all the land that she could enclose between the shoreline and a rope of knotted cowhide. While the legend does not tell us, we may assume that Princess Dido arrived at the correct solution by stretching the rope into the shape of a circular arc and thereby maximized the area of the land upon which she was to found Carthage. This story of the founding of Carthage is apocryphal. Nonetheless it is probably the first account of a problem of the kind that inspired an entire mathematical discipline, the calculus of variations and its extensions such as the theory of optimal control.” (George Leitmann “The Calculus of Variations and Optimal Control – An Introduction” Plenum Press, 1981)

1. Dido Maximum Area Problem1. Isoperimetric Problems

24

Given a rope of length P connected to each end of straight line of length 2 a < P find the shape of the rope necessary to enclose the maximum area between the rope and the straight line.

( ) ( ) ( ) ∫∫∫∫∫−−−

=+=

+=+==

a

a

a

a

a

a

dxdxdxxd

ydydxdsdP θθ sectan11 2

2

22

subject o the isoperimetric constraint:

where: θtan=xd

yd

SOLO

Rope of length P

( )xθ

x

y

a+a−

y

dx


( ) ∫∫−

==a

axy

dxyAdJ maxmaxmax The problem can be formulated as:

Solution

1. Dido Maximum Area ProblemIsoperimetric Problems

25


Therefore we have the following differential equations as constraint

( ) ( ) 0&0tan ==−= ayayxd

yd θ

( ) ( ) Papapxd

pd ==−= &0cos

1

θ

1. Define the Hamiltonian:θ

λθλcos

1tan: pyyH ++=

and the end constraint function ( ) ( )Ppypyx py −+=Φ νν:,,

2. Euler-Lagrange and Boundary Conditions:

( ) ( ) ( )xaxy

ay

H

xd

dyyyy

y −+=→=∂

Φ∂=−=∂∂−= νλνλ

λ1

( ) ( ) pypyp x

pa

p

H

xd

dνλνλ

λ=→=

∂Φ∂==

∂∂−= 0

p

y

p

ypy

xaH

νν

λλ

θθ

θλθ

λθ

−+−=−=→+=

∂∂= *

22sin

cos

sin

cos

10

3. The extremal of the Hamiltonian is given for

**

***

** coscos

1tansin

cos

1tan θλ

θθθλ

θθ

λλ

λ ppp

yp yyyH +=

+−+=

++=

Rope of length P

( )xθ

x

y

a+a−

y

dx

Solution (continue – 1)


26


Since H is not an explicit function of x and

( ) constpyxHx

H

xd

Hdpy =→=

∂∂= λλθ ,,,,,0

Therefore ( ) ( )**

0**

0

**0

coscos

cos0cos0

ff

fpp axHaxH

θθθθ

θλθλ

±=→=→

+===+=−=

Fromp

y xa

νν

θ−+

−=*sin

we have ( ) ( )p

y

p

y axa

axνν

θν

νθ

±−==±=

+−=−= ** sin

2sin

This equation is defined only when ± → -, and in this case p

y

p

y a

νν

νν

=+

−2

( ) ( ) 0** θπθπθ −==−=−= axax

and ay −=νp

x

νθ =→ *sin

We have and2

* 1cos

−=

p

x

νθ 2

*

1

tan

−

==

p

p

x

x

xd

yd

ν

νθ

Rope of length P

( )xθ

x

y

a+a−

y

dx



27


We obtained 2

*

1

tan

−

==

p

p

x

x

xd

yd

ν

νθ

Let integrate this equation ( )22

2

2

0

11

12

1

−+

−−=

−

−

−=−− ∫− p

pp

p

x

a

p

pp

ax

x

xd

ayyν

νν

ν

ν

νν

or 2222 xay pp −−=−− νν

Rope of length P

( )ax +=θx

y

a+a−

( )ax −=θ

pR ν=22 ap −ν

0θ

and finally

222

22pp xay νν =+

−−

This the equation of a circular arc with radius R = vp and center at

− 22,0 apν



28



Rope of length P

( )ax +=θx

y

a+a−

( )ax −=θ

pR ν=22 ap −ν

0θ

The radius R of the circular arc is found from the perimeter constraint

=

=

−

== −

=

−=

−

−−∫∫

pp

ax

axpp

a

a

p

a

a

ax

x

xdxdP

νν

νν

ν

θ11

2*sin2sin

1cos

or

=

pp

Pa

νν

2sin

From the equations*sinθν px =

*0

*

22

coscos11 θνθνν

νν

ν ppp

pp

p

axy −=

−−

−=

The maximum Area enclosed by the Rope is

( )

−= 00

2 2sin2

1* θθν pA

( ) ( ) ( )∫∫∫∫−−−−

−

+=−==0

0

0

0

0

0

**0

2**

2***0

** sincos2

2cos1coscoscos

θ

θ

θ

θπ

θ

θπ

θθνθθνθθνθνθν dddxdyA ppppp

a

a

( )( ) ( ) ( )

−=

−+=

−+

=

−

002

0002

02 2sin

2

12sin2sin

2

1sincos

2

2sin21 0

θθνθθθνθθθθ

ν

θ

θπ

ppp

The maximum Area



29


Given an area of canvas, A, to built a tent, find the shape of the canvas necessary to cover a circular floor area of radius a ( π a2 < A ) with maximum volume inside the tent.

x

y

z

rdr

Solution

( ) ( )∫∫ ==aa

drrrzdrrrzV00

max22maxmax ππ

where z (r) is the height of the canvas surface at radius r.

Define ( ) ( ) ( ) rdrdrd

zdzdrd

rd

rzd

θθ

cos

11tan

2

22 =

+=+→=

and ( ) ( ) ∫∫∫ =

+=+=

aa

rdr

rdrd

zdzdrdrA

00

2

22

cos2122

θπππ

Isoperimetric Problems

2. Maximum volume with given surface area

30


x

y

z

rdr


Summarize

∫=a

drrzJ0

2π

( ) ( )

( ) ( ) 0&0tan

&00cos

2

==

===

azdefinednotzrd

zd

AaAAr

rd

Ad

θ

θπ

where

1. Define Hamiltonian θλθ

πλπ tancos

22: zA

rrzH ++=

2. Euler-Lagrange Equations and Boundary Conditions

constA

H

rd

dA

A ==∂∂−= λλ

0

( ) ( )( ) 20sin002 rdefinednotzcerz

H

rd

dzz

z πλλπλ −=→=−=∂∂−=

22

22 21cos

222sin

cos

1

cos

sin20

−=→==−=→++=

∂∂=

AAAA

zzA

rr

r

r

rr

H

λθ

λλππ

λπλθ

θλ

θθπλ

θ




31


x

y

z

rdr


The following relations are developed

2

21

2

cos

sintan

−

===

A

A

r

r

rd

zd

λ

λθθθ

from which ∫∫

−

−=

−

=

−

=−a

AA

A

AA

a

A

A r

r

rd

rdr

r

tzz0

2

2

2

020 1

212

21

22

21

2

λλ

λ

λλ

λ

λ

that can be written as ( ) ( ) 2220 22 AA rzz λλ =++−

Since 222 yxr +=

we obtain ( ) ( ) 22220 22 AA yxzz λλ =+++−



32



we obtained ( ) ( ) 22220 22 AA yxzz λλ =+++−

This is the equation of a spherical surface with radius 2 λ A and center on the vertical axis passing trough the center of the circular floor and located at . Azz λ20 −=

We want to find the two unknowns z0 and λA

(1) by using the given A, the area of the canvas

( )

( ) ( )

−−=

−−=

−−=

−

==

=

=

∫∫

22

2

2

0

2

2

02

2

2

0

22222

118

218

21

222

cos2

aa

r

r

rd

rdr

A

AAAA

A

ar

rA

A

a

A

AA

a

λλλπλ

πλ

λπλ

λ

λλπ

θπ

This is a transcedental equation in 2 λ A .



33





(2) by using the condition z (r=a) = 0

( ) ( ) ( ) 220

2220 2222 azaz AAAA −±=→=++− λλλλ

We can see that for 2 λ A = a22 aA π=

020 =+− Az λ

the spherical surface has the radius a and its center on the center floor.

For A > 2 π a2 we have

( ) 220 22 az AA −=− λλ the center of the spherical canvas surface is above the floor

For A< 2 π a2 we have

( ) 220 22 az AA −−=− λλ the center of the spherical canvas surface is below the floor


34


A liquid drop on a horizontal surface assume an axially symmetric shape that minimize the sum of the liquid’s potential energy in the earth’s gravitational field plus the liquid’s surface energy.

Solution:

Because of the axially symmetry let consider

a) a disc, at a height z ,that touches the external surface and has a radius r(z) , and width dz that has the potential energy

( )( )zdzrzgmdzgUd vol2πρ−=−=

from which we obtain

( ) rdrd

zdrzgzdzrzgU

a

vol ∫∫

−=−=

0

22 ρπρπ

b) the change in potential energy is given by the change in the axially symmetric surface due to surface tension

( ) ( ) ( )( ) rdrd

zdrrzdrdrsdrUd surface

2

22 1222

+=+== πσπσπσ

from which we obtain ∫

+=

a

surface rdrd

zdrU

0

2

12 σπ

x

yr

ds σσ

dr

σ σσ

dz

a

)(zrz


3. Shape of a liquid drop on a horizontal surface

35


Solution (continue – 1):

x

yr

ds σσ

dr

σ σσ

dz

a

)(zrz

The sum of the liquid’s potential energy in the earth’s gravitational field plus the liquid’s surface energy that is minimized is given by:

∫∫

++−=+=

aa

surfacevol rdrd

zdrrd

rd

zdrzgUUU

0

2

0

2 12 σπρπ

The volume of the liquid drop is fixed rdrd

zdrV

a

∫

=

0

2π

Let define urd

zd == θtan

Summarizing ∫∫ ++−=aa

rdurrdurzgJ0

2

0

2 12 σπρπ

a is not defined and we have the constraints

( ) ( )

( ) ( ) VaVVurrd

Vd

azdefinednotzurd

zd

===

==

&00

0&0

2π



36



x

yr

ds σσ

dr

σ σ σdz

a

)(zrz1. The Hamiltonian is given by

uruururgzH Vz222 12 πλλσπρπ ++++−=

2. Euler-Lagrange Equations and Boundary Conditions

constV

H

rd

dV

V ==∂∂−= λλ

0

( ) ( )( )definednotzceugrz

H

rd

dz

z 0sin002 ==∂∂−= λρπλ

2

2

2

120 r

u

urrgz

u

HVz πλλσπρπ ++

++−=

∂∂=


Let take the derivative of the last equation relative to r:

( )rrgu

rd

ud

u

u

ur

u

urgzrgu Vπλρπσπσπρπρπ 2

11

12

1220 2

22

2

22

2 ++

+−

++

++−−=

( ) σλ

σρ Vz

g

u

u

rrd

ud

u−=−

++

+232 1

1

1

1or



37





x

yr

ds σσ

dr

σ σ σdz

a

)(zrz( ) σ

λσ

ρ Vzg

u

u

rrd

ud

u−=−

++

+232 1

1

1

1

For u << 1 and using we obtain urd

zd =

σλ

σρ Vz

g

rd

zd

rrd

zd −=−+ 12

2

which has the solution

( )

+

+= r

gKBr

gIA

grz V

σρ

σρ

ρλ

00

where I0 and K0 are Bessel functions of order zero with imaginary argument, and A, B and λV to be defined.


38

SOLO

Problem of Minimum Surface of Revolution

Given two points A (a,ya) and B (b, yb) a≠b in the plane. Find the curve that joints thesetwo points with a continuous derivative, in such a way that the surface generated by therotation of this curve about the x axis has the smallest possible area.

x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y Minimum Surface of Revolution

The surface generated by the rotation of y (x) curve about the x – axis can be calculated using

( ) ( ) xdxd

ydyydxdysdydS

2

22 1222

+=+== πππ

Therefore

( )∫

+==

b

a

xdxd

ydxySJ

2

12: π2

1,,

+=

xd

ydy

xd

ydyxF

We can see that Fis not an explicit function of x.


39

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y

Minimum Surface of Revolution

Solution:


1. Euler-Lagrange Equation 0=− yy Fxd

dF

2

1,,

+=

xd

ydy

xd

ydyxF

212 yFy += π

212

y

yyFy

+= π

( ) ( )

++

+=

+−

+

+=3

22

2

32

2

2

2

112

112

y

yy

y

y

y

yyy

y

yyyF

xd

dy

ππ

with

Substituting in the Euler-Lagrange Equation gives

( ) ( )( )

0

1

2

1

12

11

12

1112

2

32

2

322322

22 =

+

+−+

=

+−

+=

+−

+−+=−

y

y

xd

d

y

or

y

yyy

y

yy

yy

yy

y

yyF

xd

dF yy

π

π

ππ

40

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y

Minimum Surface of Revolution



( )0

1

2

1

12

2

32

2

=

+

+−+

=−

y

y

xd

d

y

or

y

yyy

Fxd

dF yy

π

π

E-L. Equation

01 2 =−+ yyy (1) nonlinear second order differential equation

01 2

=

+ y

y

xd

d

)2(

Second Way to solve Euler-Lagrange Equation

Since F is not an explicit function of x, we can write:

( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( )

−=−−+=− yyF

xd

dyyFyyyF

xd

dyyyFyyyFyyyFyyyFyyyF

xd

dyyyyyyy ,,,,,,,,

In our case ( ) ( ) Cy

yyyyyyFyyyF y ππ 2

112,,

2

22 =

+−+=−

21 yCy +=or We obtained the same result

We obtain twoequivalent equations

constCy

y ==+ 21

41

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

ySolution (continue – 2):


Third Way to obtain Euler-Lagrange Equation

Start from the expression of d S and change the free variable from x to y

( ) ( ) ydyd

xdyydxdysdydS

2

22 1222

+=+== πππ

Define ( )yd

xdxxyxxyF =←+= :12:,,

~ 2 π

Euler-Lagrange Equation ( ) ( ) 0,,~

,,~ =− xxyF

yd

dxxyF xx

Since is not a function of x( ) 212:,,~

xyxxyF += π ( ) 0,,~ =xxyFx

Therefore ( ) ( )1

2

11

2

1

2,,

~0,,

~2

2

2

+

=

+=

+==→=

xd

yd

y

x

y

x

xyxxyFconstxxyF

yd

dxx

πππ

We recovered equation 21 yCy +=

42

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y



Solving the Euler-Lagrange Equation

12

−

=

C

yyRewrite the E-L Equation as

Separating variables, we obtain C

xd

C

y

C

yd

=

−

12

Integration of this equation, gives

−

+=− 1ln

2

1 C

y

C

yCCx

from which 1exp2

1 −

+=

−

C

y

C

y

C

Cx

take the square

1exp211212122exp 1

222

1 −

−=−

−

+=−

+−

=

−

C

Cx

C

y

C

y

C

y

C

y

C

y

C

y

C

y

C

Cx

From this equation we can compute2

expexp 11

−−+

−

= C

Cx

C

Cx

C

y

43

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y



Solving the Euler-Lagrange Equation (continue – 1)

( )

−=

C

CxCxy 1coshWe obtained

The solution is a curve called a catenary (catena = chain in Latin) and the Surface of Revolution which is generated is called a catenoid of revolution.

The two parameters C and C1 are defined by the two End Conditions:

−=

−=

C

CbCy

C

CaCy

b

a

1

1

cosh

cosh

Since we have only two parameters C and C1 and three independent variables ya, yb, b-a , that define the end conditions, a solution not always exists.

44

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y




To see when a solution exists , without loss of generality choose

aybBCCa /&/&0 1 === λ

−=

−=

C

CbCy

C

CaCy

b

a

1

1

cosh

cosh

We have λcoshay

C =

( )λ

λλcosh

coshcosh −= B

y

y

a

b

Since( ) ( )

tttt

t ∀≥−+=2

expexpcosh

1cosh

≤λ

λ

therefore 0cosh

sinhcosh

cosh1

coshmax

2=−=

→≤

λλλλ

λλ

λλλ

λ d

d

The minimum is obtained for a positive λ0 that is a solution of . 0sinhcosh 000 =− λλλ

( )λ

λλ

λλ

λλλ

λλcoshcoshcosh

cosh

cosh

coshcosh −≥−=−≥−= BBBB

y

y

a

bwe have

We obtained

45

SOLO





We can see that if we don't have a solution to the minimum problem.0

0

cosh λλ

−<aa

b

y

b

y

y

λλ

λλ

coshcosh−=−≥

aa

b

y

bB

y

y

A solution to the minimum surface problem exists if:

0sinhcosh 000 =− λλλwhere λ > λ0 and λ0 is given by

Let define ( ) aa

b yy

by

−=

0

00 cosh

:λ

λλ

( ) existscatenaryayyIf bb 0λ≥

( ) existstdoesncatenaryayyIf bb '0λ<

( )ayaA ,

x

y

( )bybB ,( ) ( ) 22 ydxdsd +=

y x

y

( )bybB ,

( )ayaA ,

Catenary existsCatenary does

not exists

46

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Legendre's Condition for a Weak Local MinimumLet compute

( ) ( ) 01

2

11

12

12

32

32

2

22≥

+=

+−

+=

+∂∂=

∂∂=

y

y

y

y

yy

y

yy

yF

yF yyy

πππ

We can see that the Legendre's Necessary Condition for a minimum is satisfied.

47

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Weierstrass' Necessary Condition for a Strong Local Minimum

Let compute the Weierstrass' Excess Function E

( ) ( ) ( ) ( ) ( )

( )YyyYy

y

y

yYyyYy

y

yyYyYyyyFyYyyFYyFE y

−+++

=

+−++−+=

+−−+−+=−−−=

22

22

222

2

22

111

2

1

112

1112,,,

ππ

π

By adding to both sides of the inequality , we obtain 22 Yy 01 22 >++ yY

( ) ( ) 22222222 111 YyyYYyyY >++=+++

By taking the positive square root of this inequality, we obtain YyyY >++ 22 11

Therefore ( ) 0111

2 22

2>−++

+= YyyY

y

yE

π

We can see that the Weierstrass' Necessary Condition for a strong minimum is satisfied.

48

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Conjugate Points and Jacobi’s Equation

To have an extremal for x є [a,b] we must show that there are no conjugate points toA (a, ya) in this interval. This can be shown by finding the non-trivial solution (u(x)≠0)of the Jacobi’s Equation:

02

2

=

−+

−++ uP

xd

Qd

xd

udQQ

xd

Rd

xd

udR

TT

where yyyyyy FRFQFP === :::

212 yFy += πWe found

we can see that 0:,0: ==== yyyy FQFP

( ) 01

23

2≥

+==

y

yFR yy

πWe also found

constCy

y ==+ 21 Using in the previous equation we obtain 0

2

1

22

3

2≥=

+==

y

C

y

CFR yy

ππ

−+=

−==

C

CxCC

C

CxCCyC

xd

ud 12

1222 2cosh12

~cosh

~~

2

2

3

2

2 ~20 yC

xd

udconst

xd

ud

y

C

xd

udR

xd

udR

xd

d

xd

ud

xd

Rd

xd

udR =→==→=

=+ π

Substitute those results in the Jacobi’s Equation

49

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Conjugate Points and Jacobi’s Equation

−+=

C

CxCC

xd

ud 12

2cosh12

~

( ) ( ) ∫

−+=−

x

axd

C

CxCCauxu 1

2

2cosh12

~

( ) ( )

−−

−+−=

−++=

= C

Ca

C

CxCax

CC

C

CxCx

CCauxu

x

ax

112

12

0

2sinh2sinh22

~2sinh

22

~

( ) bxaC

Cax

C

axCax

CCxu ≤<>

−+

−+−= 0

2coshsinh

2

~1

2

Therefore there are non-trivial solution (u(x)≠0) for a < x ≤ a and there are no Conjugate Points in this interval.

50

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Canonical Equations

( ) 212:,, yyyyxF += π

Define21

2:

y

yyFp y

+== π

from which

( )222

222

222

2222222

4

41

441

py

yy

py

pyyypy

−=+→

−=→=+

ππ

ππ

Substituting this in equation gives( )yyxF ,, ( ) ( )222

22

4

4,,:,,

~

py

yyyxFpyxF

yFp −==

= ππ

1. The Hamiltonian is

( ) ( ) 222

222

2

222

22

444

4,,:,, py

py

p

py

yypyyxFpyxH −−=

−+

−−=+−= π

πππ

2. The canonical equations are

( )

( )222

2

222

4

4,,

4

,,

py

y

y

pyxH

xd

pd

py

p

p

pyxH

xd

yd

−=

∂∂−=

−=

∂∂=

ππ

π

51

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Canonical Equations

The canonical equations are

( )

( )222

2

222

4

4,,

4

,,

py

y

y

pyxH

xd

pd

py

p

p

pyxH

xd

yd

−=

∂∂−=

−=

∂∂=

ππ

π

If we divide the first equation by the second, we obtain y

p

pd

yd24 π

=

or pdpydy =24 πIntegrating this equation gives

22222 4.4 Cconstpy ππ ==−

The reason that we choose the constant as const .= 4 π2C2 is to recover previous results , as we shall see. If we substitute this expression in the Hamiltonian expression we obtain:

( ) CpypyxH ππ 24,, 222 −=−−=

The fact that the Hamiltonian is constant on the extremal trajectory is a consequence of the fact that H is not an explicit function of x.

The canonical equations become

yCxd

pd

pCxd

yd

ππ

2

2

1

=

=

52

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Canonical Equations

The canonical equations are

yCxd

pd

pCxd

yd

ππ

2

2

1

=

=

Let Differentiate first equation with respect to x and use the second, to obtain

yCxd

pd

Cxd

yd22

2 1

2

1 ==π

The general solution of this Ordinary Differential Equation is:

( )

−=

C

CxCxy 1cosh

We recovered the “catenary” equation.

53

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Hamilton-Jacoby Equation

( ) ( ) ( )22

22

1

2

112,,,,,

y

y

y

yyyyyyxFyyyxFyxS yx

+=

+−+=−= ππ

( ) ( ) py

yyyyxFyxS yy =

+==

21

2,,,

π

Using in the first equation, we obtainCy

y =+ 21

( ) Cy

yyxSx ππ

21

2,

2=

+=

from which we obtain 12

2 −

=

C

yy

Substituting this equation in the Sy (x,y) equation, we obtain:

( ) 12

12

1

2,

2

2

2−

=

−

=+

=C

yC

CyCy

y

y

yyyxS y π

ππ

( ) ( ) 022,,, =−=+ CCSyxHyxS yx ππWe have Hamilton-Jacoby Equation

We find that the Hamiltonian is ( ) CpypyxH ππ 24,, 222 −=−−=

54

SOLO


x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y


Hamilton-Jacoby Equation

−

+=+= yd

C

yxdCydSxdSSd yx 12

2

π

Integrating this, to obtain:

0

22

0

2

1ln2

12

1212 S

C

y

C

yC

C

yyxCSyd

C

yxdCS +

−

+−−

+=+

−

+= ∫∫ ππ

Using the expressions for Sx (x,y) and Sy (x,y), we obtain


55

SOLO

4. Geodesics Problems


1. Geodesic in 3 Dimensional Spaces

Suppose we have a surface specified by two parameters u, v and the vector .( )vur ,

The shortest path lying on the surface and connecting to points of the surface is called a Geodesic.

A

B

( )vur ,

vdrv

udru

rd

The Shortest Path on a Surface

The arc length differential is

tdtd

vd

v

r

v

r

td

vd

td

ud

v

r

u

r

td

ud

u

r

u

r

tdtd

vd

v

r

td

ud

u

r

td

vd

v

r

td

ud

u

rtd

td

rd

td

rdtd

td

rdds

2/122

2/12/1

2

∂∂⋅

∂∂+

∂∂⋅

∂∂+

∂∂⋅

∂∂=

∂∂+

∂∂⋅

∂∂+

∂∂=

⋅==

The length of the path between the two points A and B is:

∫

+

+

==

B

A

r

r

tdtd

vdG

td

vd

td

udF

td

udESJ

2/122

2:

where ( ) ( ) ( )

∂∂⋅

∂∂=

∂∂⋅

∂∂=

∂∂⋅

∂∂=

v

r

v

rvuG

v

r

u

rvuF

u

r

u

rvuE

:,,:,,:,

56

SOLO

Geodesics Problems



A

B

( )vur ,

vdrv

udru

rd

The Shortest Path on a Surface

We have

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]dt

dsvvuGvuvuFuvuEvuvutF =++=

2/122 ','',2',',',,,

where td

vdv

td

udu == :',:'

1. Euler-Lagrange Equations

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 0

','',2',

',',

','',2',2

','',2',

',',,,',',,,

2/1222/122

22

'

=++

+−++

++=

−

vvuGvuvuFuvuE

vvuFuvuE

td

d

vvuGvuvuFuvuE

vvuGvuvuFuvuE

vuvutFtd

dvuvutF

uuu

uu

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 0

','',2',

',',

','',2',2

','',2',

',',,,',',,,

2/1222/122

22

'

=++

+−++

++=

−

vvuGvuvuFuvuE

vvuGuvuF

td

d

vvuGvuvuFuvuE

vvuGvuvuFuvuE

vuvutFtd

dvuvutF

vvv

vv

57

SOLO

Geodesics Problems



If instead of t we use the variable s, we obtain

( ) ( )

( ) ( ) ( ) ( ) ( )0

,,

2

,,2,

',',,,',',,,

22

'

=+

−

+

+

=

−

dt

dsdtds

dsdv

vuFdtds

dsdu

vuE

dt

ds

sd

d

dt

dsdt

ds

ds

dvvuG

dt

ds

ds

dv

dt

ds

ds

duvuF

dt

ds

ds

duvuE

vuvutFtd

dvuvutF

uuu

uu

A

B

( )vur ,

vdrv

udru

rd

( ) ( )

( ) ( ) ( ) ( ) ( )0

,,

2

,,2,

',',,,',',,,

22

'

=+

−

+

+

=

−

dt

dsdtds

dsdv

vuGdtds

dsdu

vuF

dt

ds

sd

d

dt

dsdt

ds

ds

dvvuG

dt

ds

ds

dv

dt

ds

ds

duvuF

dt

ds

ds

duvuE

vuvutFtd

dvuvutF

vvv

vv

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) 0,,,,2,2

1

0,,,,2,2

1

22

22

=

+−

+

+

=

+−

+

+

ds

dvvuG

ds

duvuF

sd

d

ds

dvvuG

ds

dv

ds

duvuF

ds

duvuE

ds

dvvuF

ds

duvuE

sd

d

ds

dvvuG

ds

dv

ds

duvuF

ds

duvuE

vvv

uuu

or

58

SOLO

Geodesics Problems



Example: Spherical Surface

A spherical surface is defined by

or ( ) ( )kjiRr ˆcosˆsinsinˆcossin, θφθφθθφ ++=

θφφ

φθ

θθ

22

2

sin

0

Rrr

G

rrF

Rrr

E

=∂∂⋅

∂∂=

=∂∂⋅

∂∂=

=∂∂⋅

∂∂=

( ) ( ) ( ) ( ) ( ) ( ) θθφθφθθφφθθ d

d

dRddRdGddFdEsd

2222222 sin1sin2

+=+=++=

θθθφθφθφ cossin20 2RGGFFEE ======

Let write the Euler-Lagrange Equations for this particular problem

022

1

022

1

22

22

=

+−

+

+

=

+−

+

+

ds

dvG

ds

duF

sd

d

ds

dvG

ds

dv

ds

duF

ds

duE

ds

dvF

ds

duE

sd

d

ds

dvG

ds

dv

ds

duF

ds

duE

vvv

uuu

φθθ dRdvdRdu sin, ==

0sin

0cossin

22

2

23

2

3

=

−

=−

ds

dR

sd

d

sd

dR

sd

dR

φθ

θφθθ

φθφθφ cos,sinsin,cossin RzRyRx ===

59

SOLO

Geodesics Problems



Example: Spherical Surface (continue – 1)

0sin

0cossin

22

2

23

2

3

=

−

=−

ds

dR

sd

d

sd

dR

sd

dR

φθ

θφθθ

From last equation and we obtainθθφθ d

d

dRsd

22sin1

+=

constAR

dd

dd

R

ds

d

d

dR

ds

dR ==

+

==2

2

2

2222

sin1

sinsinsin

θφθ

θφθθ

θφθφθ

2222

24

22

2

sinsin

sin1

sin

+=

⇒=

+

θφθ

θφθ

θφθ

θφθ

d

dAA

d

dA

dd

dd

This equation can be rewritten as

θθθθθ

φ

2

22

22

sin1sin

sinsin A

A

A

A

d

d

−

±=−

±=

60

SOLO

Geodesics Problems



Example: Spherical Surface (continue – 2)

We obtain

θθ

θφ

2

22

sin1sin

A

A

d

d

−

±=

To solve this equation let write the constant A as the sinus of the constant A = sin αand define a new independent variable w as w = ctg θ.

θθθ

22

2 sin

11,

sin=+−= w

dwdTherefore

( ) ( ) αα

ααθ

θαθ

αθθφφ

2222

2

2

22 tan1

tan

sin11

sinsin

sinsin

1sin

sin

wwdw

d

d

d

dw

d

−=

+−=−

−

±==−⇒±

and

ffffffff RzRyRxB

RzRyRxA

φθφθφφθφθφcos,sinsin,cossin:

cos,sinsin,cossin: 00000000

======

The Geodesic must pass through the points A and B given by:

.tan1

tan22

constdww

d =−

= αα

αφThe Geodesic O,D.E. is

61

SOLO

Geodesics Problems


1. Geodesic in 3 Dimensional SpacesExample: Spherical Surface (continue – 2)

We obtained the O. D. E.

The solution is defined by 4 parameters ϕ0, θ0, ϕ f. θf that define the initial and final points A and B.

This satisfies the point A (ϕ 0, θ0 ) constraints.

.tan1

tan22

constdww

d =−

= αα

αφ

( ) ( )

−

+=−+= −−

=−−

0

1100

110 tan

tancos

tan

tancostancostancos

θα

θαφααφφ

θctgw

ww

The O.D.E. can be easily integrated

from which we obtain ( )θαφφ

tan

tancos 0 =−

To simplify the expressions, without affect the solution generality, let choose the direction of x and y axes, relative to points A and B such that

0tan

tancos

0

1 =

−

θα

( ) ff θφφα tancostan 0−=and

62

SOLO

Geodesics Problems



Example: Spherical Surface (continue –3)

Great Circle

( )θαφφ

tan

tancos 0 =−

Multiply this equation by R sinϕ to obtain

αθφφθφφθ tancossinsinsincoscossin 00 RRR =+

Using , in the previous equation, we obtain

φθφθφ cos,sinsin,cossin RzRyRx ===

αφφ tansincos 00 zyx =+

This is the equation of a plane passing through x=y=z=0 (center of the sphere).

Thus the geodesic of a spherical surface lies in a plane passing through the center of the sphere, therefore a Great Circle.


63

SOLO

Geodesics Problems


2. Geodesics in Riemannian Space

Riemann Spaces

Let assume a n dimensional space Vn in which we have a m (m , n) dimensional subspace Vm (Vm submersed in Vn). nm VV ⊂

Let take a point M defined by the vector as function of m independent parameters x1, x2,…,xm:

r

( ) ( ) ( ) nkexxxexxxxxxr kmkk

mkm ,,2,1,,,,,,,,, 212121

=== ξξ

→

1N

M

C→

2r

→

1r

2V3V

→r

Τ

dt

dr→

→

1r

→

2r

→

1N

constx =1Τ

and define the tangent hyper-surface T, defined by the m vectors tangents to C at the point M:

mix

rr

ii ,,2,1,:

=

∂∂=

We assume that those vectors are linear independent since the parameters x1, x2,…,xm are independent.

64

SOLO

Geodesics Problems



Riemann Spaces →

1N

M

C→

2r

→

1r

2V3V

→r

Τ

dt

dr→

→

1r

→

2r

→

1N

constx =1Τ

mxxx ,,, 21 mxxx ,,, 21 Change of Coordinates from to

( ) ( )mm xxxrxxxr ,,,,,, 2121

=

At each point since the parameters are independent the Jacobian is nonzero:

( )( ) 0det

,,,

,,,

121

1

2

1

1

1

21

21

≠

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=∂∂

m

mm

m

m

m

x

x

x

x

x

x

x

x

x

x

x

x

xxx

xxx

We have: jj

jj

i

ii

ixd

x

rxd

x

x

x

rxd

x

rrd

∂∂=

∂∂

∂∂=

∂∂=

ji

j

iij

i

ij

i

jj

jmm

i

i

jm

m

i

i

ji

i

jjj

j

ii

rx

xrr

x

x

x

r

x

x

x

rr

x

x

x

xxd

x

x

x

xxd

x

xxdxd

x

xxd

∂∂=

∂∂=

∂∂

∂∂=

∂∂=

=∂∂

∂∂⇒

∂∂

∂∂=

∂∂=

∂∂=

&:

& δ

65

SOLO

Geodesics Problems



Riemann Spaces →

1N

M

C→

2r

→

1r

2V3V

→r

Τ

dt

dr→

→

1r

→

2r

→

1N

constx =1Τ

mjx

rr

jj ,...,2,1: =

∂∂=

mix

rr

ii ,...,2,1: =

∂∂=

The vectors are also in T since are

given by linear combinations of

Metric in T over the base . mi

x

rr

ii ,...,2,1: =

∂∂=

Let assume that a Scalar Product is defined in T and define the Metric in T over the base

asmix

rr

ii ,...,2,1: =

∂∂=

( ) ( ) jiij

jijiij grr

x

r

x

rrrg ==

∂∂

∂∂==

,,,:

Dual Base to . mix

rr

ii ,...,2,1: =

∂∂=

Since are linear independent; i.e.:

miri ,...,2,1=

{ }( ) ( ){ } { } 0det,...,10,,0

,...,100

≠=⇔=∀=⇔===→

→=∀=⇔=

iji

iji

jii

jii

ii

i

ggmigrrrr

mir

αααααα

66

SOLO

Geodesics Problems



Riemann Spaces →

1N

M

C→

2r

→

1r

2V3V

→r

Τ

dt

dr→

→

1r

→

2r

→

1N

constx =1Τ

the mxm matrix is nonsingular and we can define the matrix:

{ }ijgG =

{ } { } { }{ } mxmijij

ijij IgggGg =→== −−

∆11

ijkj

ik gg δ=or

Let define now: mix

rgrgr

k

ikk

iki ,...,2,1: ==∂∂==

We can see that ( ) ( ) ( ) ijkj

ikjk

ikjk

ikj

i ggrrgrrgrr δ==== ,,,

mir i ,...,2,1=miri ,...,2,1=

Therefore are dual to .

Moreover

( ) ( ) ( ) ijjk

ikjk

ikjk

ikji ggrrgrrgrr ==== δ,,,

gik is the Metric of the dual base . mir i ,...,2,1=

67

SOLO

Geodesics Problems



In a Riemann space the length of a space curve, defined by it’s coordinates xi (t), as a function of a parameter t, is given by

∫

=

1

0

2/1t

t

dttd

xd

td

xdgs βα

βα

where we used the tensor index summation rule

∑ ∑= =

⇔m m

td

xd

td

xdg

td

xd

td

xdg

1 1α β

βαβα

βαβα

mix

rr

ii ,...,2,1: =

∂∂=

( ) ( ) jiijji

jiij grrx

r

x

rrrg ==

∂∂

∂∂==

,,,:

Define mtd

sd

td

xd

td

xdg

td

xdxtF ,,1,:,,

2/1

==

=

βαβα

βα

To find the geodesics (the curve of the minimum length) let apply the Euler-Lagrange Equations

mix

F

x

F

td

d

ii

,,10

==∂∂−

∂∂

→

1N

M

C→

2r

→

1r

2V3V

→r

Τ

dt

dr→

→

1r

→

2r

→

1N

constx =1Τ

68

SOLO

Geodesics Problems



mtd

sd

td

xd

td

xdg

td

xdxtF ,,1,:,,

2/1

==

=

βαβα

βα

1. To find the geodesics (the curve of the minimum length) let apply the Euler-Lagrange Equations

mix

F

x

F

td

d

ii

,,10

==∂∂−

∂∂

→

1N

M

C→

2r

→

1r

2V3V

→r

Τ

dt

dr→

→

1r

→

2r

→

1N

constx =1Τ

( ) mitd

xdg

td

xdg

td

sd

dtdstd

xd

td

xd

x

g

td

xd

td

xd

x

g

td

xdg

td

xdg

dtds

dtds

td

xdg

tdxd

g

dt

d

x

F

dt

d

iiii

ii

ii

i

,,1/2

1

/2

1

/2

2

2

22

2

2

2

=

+−

∂∂

+∂∂

++=

+=

∂∂

ββ

αα

βα

α

ββα

β

αββ

αα

ββ

αα

mitd

xd

td

xd

x

g

dtdsx

F

ii

,,1/2

1=

∂

∂=

∂∂ βαβα

We have

69

SOLO

Geodesics Problems


2. Geodesics in Riemannian Space→

1N

M

C→

2r

→

1r

2V3V

→r

Τ

dt

dr→

→

1r

→

2r

→

1N

constx =1Τ

If we choose 0,12

2

==→=td

sd

dt

dsst

misd

xd

sd

xd

x

g

x

F

ii

,,12

1=

∂

∂=

∂∂ βαβα

misd

xd

sd

xd

x

g

x

g

sd

xdg

sd

xd

sd

xd

x

g

sd

xd

sd

xd

x

g

sd

xdg

sd

xdg

x

F

ds

d

iii

iiii

i

,,12

1

2

1

2

2

2

2

2

2

=

∂

∂+

∂∂

+=

∂∂

+∂∂

++=

∂∂

βα

α

β

β

ααα

βα

α

ββα

β

αββ

αα

Therefore we obtain: misd

xd

sd

xd

x

g

x

g

x

g

sd

xdg

i

iii ,,1

2

12

2

=

∂

∂−

∂∂

+∂∂

+ βαβα

α

β

β

ααα

Multiplying by gri and summing on i, we obtain:

misd

xd

sd

xd

x

g

x

g

x

gg

sd

xd

i

iiir

r ,,122

2

=

∂

∂−

∂∂

+∂∂

+ βαβα

α

β

β

α

We define

∂

∂−

∂∂

+∂∂

=Γi

iiir

r

x

g

x

g

x

gg βα

α

β

β

αβα 2: Christoffel Symbol of Second Kind

misd

xd

sd

xd

sd

xd rr ,,12

2

=Γ+ βαβα Differential Equations of the Geodesics


70

SOLO

Geodesics Problems


3. Geodesics for an Implicit Equation of the Surface

Suppose that we want to find the geodesics on a surface defined by an implicit function:

( ) 0,, =zyxG

We want to find the extremal of ∫

+

+

=

ft

t

dttd

zd

td

yd

td

xds

0

222

Let adjoin the first equation to the second using the Lagrange multiplier μ (t).

( ) ( )∫

+

+

+

=

ft

t

dtzyxGttd

zd

td

yd

td

xds

0

,,222

µ

Define ( ) ( )zyxGttd

zd

td

yd

td

xd

td

zd

td

yd

td

xdzyxtF ,,:,,,,,,

222

µ+

+

+

=

dt

ds

td

zd

td

yd

td

xdf =

+

+

=

222

:and

71

SOLO

Geodesics Problems



The Euler-Lagrange Equations are:

( )

( )

( ) 01

01

01

=∂∂−

=∂∂−

=∂∂−

z

Gt

td

zd

ftd

d

y

Gt

td

yd

ftd

d

x

Gt

td

xd

ftd

d

µ

µ

µ

( ) ( )zyxGttd

zd

td

yd

td

xdf

td

zd

td

yd

td

xdzyxtF ,,,,:,,,,,, µ+

=

dt

ds

td

zd

td

yd

td

xdf =

+

+

=

222

:

μ (t) can be eliminated to obtain ( )t

zG

tdzd

ftdd

yG

tdyd

ftdd

xG

tdxd

ftdd

µ=

∂∂

=

∂∂

=

∂∂

111

Using f = ds/dt we can write the last expression as

( )t

z

G

td

zd

sd

td

sd

d

td

sd

y

G

td

yd

sd

td

sd

d

td

sd

x

G

td

xd

sd

td

sd

d

td

sd

µ=

∂∂

=

∂∂

=

∂∂

( )

td

sdt

z

G

sd

zd

sd

d

y

G

sd

yd

sd

d

x

G

sd

xd

sd

d

µ=

∂∂

=

∂∂

=

∂∂

or

72

SOLO

Geodesics Problems



( )

td

sdt

z

G

sd

zd

sd

d

y

G

sd

yd

sd

d

x

G

sd

xd

sd

d

µ=

∂∂

=

∂∂

=

∂∂

We found

From Differential Geometry we know that for a three dimensional curve

( ) ( ) ( ) kszjsyisxr ˆˆˆ ++=

we have ( ) ( ) ( )k

sd

szdj

sd

sydi

sd

sxd

sd

rdt ˆˆˆ: ++==

the unit vector tangent to the curve

( ) ( ) ( )nk

sd

szd

sd

dj

sd

syd

sd

di

sd

sxd

sd

d

sd

rd

sd

d

sd

tdˆˆˆˆ κ=

+

+

=

=

is the unit vector that defines the principal normal to the curve and κ is the magnitude of the curvature. Since is a unit vector

n

t

0ˆ

ˆˆˆ =⋅=⋅sd

tdtnt

We also have kz

Gj

y

Gi

x

GG ˆˆˆ

∂∂+

∂∂+

∂∂=∇

We can see that the principal normal to any point of the geodesic curve is parallel to the normal to the surface.

73

SOLO

Geodesics Problems



Example: Geodesics on a Spherical Surface

( ) 0,, 2222 =−++= RzyxzyxGSpherical Surface:

kzjyixkz

Gj

y

Gi

x

GG ˆ2ˆ2ˆ2ˆˆˆ ++=

∂∂+

∂∂+

∂∂=∇

Let differentiate twice G (x,y,z) = 0 as a function of s

0=++sd

zdz

sd

ydy

sd

xdx ( ) 0,, =zyxG

sd

d

( ) 0,,2

2

=zyxGsd

d02

2

2

2

2

2222

=+++

+

+

sd

zdz

sd

ydy

sd

xdx

sd

zd

sd

yd

sd

xd

1222

=

+

+

sd

zd

sd

yd

sd

xd

dt

ds

td

zd

td

yd

td

xdf =

+

+

=

222

:

dsdt =

( )

td

sdt

z

G

sd

zd

sd

d

y

G

sd

yd

sd

d

x

G

sd

xd

sd

d

µ=

∂∂

=

∂∂

=

∂∂

( )tkz

sd

zd

y

sd

yd

x

sd

xd

===222

2

2

2

2

2

2

( ) ( ) ( ) ( )2

2222

2

102121

RtkRtkzyxtk −=→=+=+++

74

SOLO

Geodesics Problems




( )2

2

2

2

2

2

2

2

1

222 Rtk

z

sdzd

y

sdyd

x

sdxd

−====We found

+

−

+

−

+

−

R

sC

R

sCz

R

sC

R

sCy

R

sC

R

sCx

cossin

cossin

cossin

65

43

21

The solutions of those equation are

0

cos

sin

1

65

43

21

=

−

R

s

R

s

CCz

CCy

CCx

0

0

0

22

2

22

2

22

2

=+

=+

=+

R

z

sd

zd

R

y

sd

yd

R

x

sd

xd

75

SOLO

Geodesics Problems




0

cos

sin

1

65

43

21

=

−

R

s

R

s

CCz

CCy

CCx

We found

The solution exists only if

0det

65

43

21

=

CCz

CCy

CCx

0=++ CzByAx

Thus the Geodesic on a Spherical Surface lies in a plane passing through the center of the sphere, therefore a Great Circle.

Great Circle


76

SOLO

Geometrical Optics and Fermat Principle

The Principle of Fermat (principle of the shortest optical path) asserts that the optical length

of an actual ray between any two points is shorter than the optical ray of any other curve that joints these two points and which is in a certain neighborhood of it. An other formulation of the Fermat’s Principle requires only Stationarity (instead of minimal length).

∫2

1

P

P

dsn

An other form of the Fermat’s Principle is:

Principle of Least Time The path following by a ray in going from one point in space to another is the path that makes the time of transit of the associated wave stationary (usually a minimum).


77

SOLO

We have:

( ) ( ) ( )∫∫∫∫ =

+

+===

2

1

2

1

2

1

,,,,1

1,,1

,,1

0

22

00

P

P

P

P

P

P

xdzyzyxFc

xdxd

zd

xd

ydzyxn

cdszyxn

ctdJ

Let find the stationarity conditions of the Optical Path using the Calculus of Variations

( ) ( ) ( ) xdxd

zd

xd

ydzdydxdds

22

222 1

+

+=++=

Define:

xd

zdz

xd

ydy == &:

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=

constS =constdSS =+

s

∫2

1

P

P

dsn

1P

2P

Paths of Rays Between Two Points



78

SOLO

Necessary Conditions for Stationarity (Euler-Lagrange Equations)

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=

0=∂∂−

∂∂

y

F

y

F

dx

d

( )[ ] 2/1221

,,

zy

yzyxn

y

F

++=

∂∂ [ ] ( )

y

zyxnzy

y

F

∂∂++=

∂∂ ,,

1 2/122

( )[ ] [ ] 011

,, 2/122

2/122=

∂∂++−

++ y

nzy

zy

yzyxn

xd

d

0=∂∂−

∂∂

z

F

z

F

dx

d

[ ] [ ] 011

2/1222/122=

∂∂

−

++++ y

n

zy

yn

xdzy

d



79

SOLO

Necessary Conditions for Stationarity (continue - 1)

We have

[ ] 01

2/122=

∂∂−

++ y

n

zy

yn

sd

d

y

n

sd

ydn

sd

d

∂∂=

In the same way

[ ] 01

2/122=

∂∂−

++ z

n

zy

zn

sd

d

z

n

sd

zdn

sd

d

∂∂=



80

SOLO


Using ( ) ( ) ( ) xdxd

zd

xd

ydzdydxdds

22

222 1

+

+=++=

we obtain 1222

=

+

+

sd

zd

sd

yd

sd

xd

Differentiate this equation with respect to s and multiply by n

sd

d

0=

+

+

sd

zd

sd

dn

sd

zd

sd

yd

sd

dn

sd

yd

sd

xd

sd

dn

sd

xd

sd

nd

sd

zd

sd

nd

sd

yd

sd

nd

sd

xd

sd

nd =

+

+

222

sd

nd

and

sd

nd

sd

zdn

sd

d

sd

zd

sd

ydn

sd

d

sd

yd

sd

xdn

sd

d

sd

xd =

+

+

add those two equations



81

SOLO


sd

nd

sd

zdn

sd

d

sd

zd

sd

ydn

sd

d

sd

yd

sd

xdn

sd

d

sd

xd =

+

+

Multiply this by and use the fact that to obtainxd

sd

cd

ad

cd

bd

bd

ad =

xd

nd

sd

zdn

sd

d

xd

zd

sd

ydn

sd

d

xd

yd

sd

xdn

sd

d =

+

+

Substitute and in this equation to obtainy

n

sd

ydn

sd

d

∂∂=

z

n

sd

zdn

sd

d

∂∂=

xd

zd

z

n

xd

yd

y

n

xd

nd

sd

xdn

sd

d

∂∂−

∂∂−=

Since n is a function of x, y, zx

n

xd

zd

z

n

xd

yd

y

n

xd

ndzd

z

nyd

y

nxd

x

nnd

∂∂=

∂∂−

∂∂−→

∂∂+

∂∂+

∂∂=

and the previous equation becomes

x

n

sd

xdn

sd

d

∂∂=



82

SOLO


We obtained the Euler-Lagrange Equations:

x

n

sd

xdn

sd

d

∂∂=

y

n

sd

ydn

sd

d

∂∂=

z

n

sd

zdn

sd

d

∂∂=

ksd

zdj

sd

ydi

sd

xd

sd

rd

kzjyixr

ˆˆˆ

ˆˆˆ

++=

++=

Define the unit vectors in the x, y, z directionskji ˆ,ˆ,ˆ

The Euler-Lagrange Equations can be written as:

nsd

rdn

sd

d ∇=

This is the Eikonal Equation from Geometrical Optics.



83

SOLO

Transversality Conditions for Geometrical Optics and Fermat’s Principle

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=

For the Geometrical Optics we obtained:

Assume that the initial and final boundaries are defined by the surfaces A (x0, y0, z0) and B (xf, yf, zf) respectively. The transversality conditions at the boundaries i=0,f are defined by

( ) ( ) ( )[ ]( ) ( ) 0,,,,,,,,

,,,,,,,,,,,,

=++

−−

iziy

izy

dzzyzyxFdyzyzyxF

dxzyzyxFzzyzyxFyzyzyxF

( ) [ ] [ ] [ ]

[ ] ( )sd

xdzyxn

zy

n

zy

znz

zy

ynyzynFzFyzyzyxF zy

,,1

111,,,,

2/122

2/1222/122

2/122

=++

=

++−

++−++=−−

( )[ ] ( )

( )[ ] ( )

sd

zdzyxn

zy

zzyxn

z

FF

sd

ydzyxn

zy

yzyxn

y

FF

z

y

,,1

,,

,,1

,,

2/122

2/122

=++

=∂∂=

=++

=∂∂=

For are tangent to the boundary surfaces A (x0, y0, z0) and B (xf, yf, zf). fird i ,0=

From Transversality Conditions we can see that the rays are normal (transversal) to the boundary surfaces (see Figure).

Transversality Conditions



SOLO

Corner Conditions for Geometrical Optics and Fermat’s Principle

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=


Let examine the following two cases:

1. The optical path passes between two regions with different refractive indexes n1 to n2

(see Figure)In region (1) we have:In region (2) we have:

( ) ( ) 2211 1,,,,,, zyzyxnzyzyxF ++=

( ) ( ) 2222 1,,,,,, zyzyxnzyzyxF ++=

( ) ( ) ( )[ ]{( ) ( ) ( )[ ]}

( ) ( )[ ]( ) ( )[ ] 0,,,,,,,,

,,,,,,,,

,,,,,,,,,,,,

,,,,,,,,,,,,

222111

222111

22222222222

11111111111

=−+

−+

−−−

−−

dzzyzyxFzyzyxF

dyzyzyxFzyzyxF

dxzyzyxFzzyzyxFyzyzyxF

zyzyxFzzyzyxFyzyzyxF

zz

yy

zy

zy

The Weierstrass-Erdmann necessary condition at the boundary between the two regions is

where dx, dy, dz are on the boundary between the two regions.



SOLO

Corner Conditions for Geometrical Optics and Fermat’s Principle (continue – 1)

( ) [ ] [ ] [ ]

[ ] ( )sd

xdzyxn

zy

n

zy

znz

zy

ynyzynFzFyzyzyxF zy

,,1

111,,,,

2/122

2/1222/122

2/122

=++

=

++−

++−++=−−

( )[ ] ( )

( )[ ] ( )

sd

zdzyxn

zy

zzyxn

z

FF

sd

ydzyxn

zy

yzyxn

y

FF

z

y

,,1

,,

,,1

,,

2/122

2/122

=++

=∂∂=

=++

=∂∂=

( ) ( )0

2121 =⋅

− rd

sd

rdn

sd

rdn rayray

where is on the boundary between the two regions andrd

( ) ( )sd

rds

sd

rds rayray 2

:ˆ,1

:ˆ 21

==

are the unit vectors in the direction of propagation of the rays.



SOLO


( ) 0ˆˆ 2211 =⋅− rdsnsn

2211 ˆˆ snsn −Therefore is normal to . rd

Since can be in any direction on the boundary between the two regions (see Figure ) is parallel to the unit vector normal to the boundary surface, and we have

rd

2211 ˆˆ snsn − 21ˆ −n

( ) 0ˆˆˆ 221121 =−×− snsnn

This the Snell’s Law of Geometrical Optics



SOLO


2. The optical path is reflected at the boundary.

( ) ( ) ( ) 0ˆˆ21

21 =⋅−=⋅

− rdssrd

sd

rd

sd

rd rayray

n1 = n2 , we obtain

i.e. is normal to , i.e. to the boundary where the reflection occurs.Also we can write

21 ˆˆ ss − rd

( ) 0ˆˆˆ 2121 =−×− ssn

( ) ( ) ( ) 0ˆˆ21

221121 =⋅−=⋅

− rdsnsnrd

sd

rdn

sd

rdn rayray

In this case, if we substitute in the equation



88

SOLO

Hilbert’s Invariant Integral

David Hilbert (1862 – 1943)

Geometrical Optics and Fermat’s Principle

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=


The Hilbert’s Invariant Integral is

( ) ( )( ) ( ) ( )[ ] ( ) ( )( ){( )

( )

( ) ( )[ ] ( ) ( )( )} xdzyxzzyxyzyxFzyxZzyxz

zyxzzyxyzyxFzyxYzyxyzyxzzyxyzyxF

z

zyxP

zyxPyC

ffff

,,,,,,,,,,,,

,,,,,,,,,,,,,,,,,,,,,,

,, 0000

−−

∫ −−

This is known as Hilbert’s Invariant Integral because it is invariant on the path C as long as this curve remains in the field of the unique extremal solution.

( ) ( ) ( ) ( )zyxx

zzyxzzyx

x

yzyxy ,,,,,,,,,

∂∂=

∂∂= is the field slope and

( ) ( )CC

x

zzyxZ

x

yzyxY

∂∂=

∂∂= :,,,:,, is the path C slope at the point (x,y,z) of C

we have on path C ( ) ( ) dxx

zdxzyxZzddx

x

ydxzyxYyd

CC

CC ∂

∂==∂∂== ,,,,,


89

SOLO

Hilbert’s Invariant Integral (continue – 1)

David Hilbert (1862 – 1943)

( ) ( ) ( )[ ]{( )

( )( ) ( ) }zdzyzyxFydzyzyxFxdzyzyxFzzyzyxFyzyzyxF zy

zyxP

zyxP

zyC

ffff

,,,,,,,,,,,,,,,,,,,,,,

,, 0000

−−−−∫

The Hilbert’s Invariant Integral is

We can write

( ) [ ] [ ] [ ] [ ] ( )sd

xdzyxn

zy

n

zy

znz

zy

ynyzynFzFyzyzyxF zy ,,

1111,,,, 2/1222/1222/122

2/122 =++

=++

−++

−++=−−

( )[ ] ( )

( )[ ] ( )

sd

zdzyxn

zy

zzyxn

z

FF

sd

ydzyxn

zy

yzyxn

y

FF

z

y

,,1

,,

,,1

,,

2/122

2/122

=++

=∂∂=

=++

=∂∂=

Now we can write the Hilbert’s Invariant Integral as

( )

( )

( )

( )

∫∫ ⋅=⋅ffffffff zyxP

zyxP

zyxP

zyxP

ray rdsnrdsd

rdn

,,

,,

,,

,, 1000010000

ˆ

This is the Lagrange’s Invariant Integral from Geometrical Optics.

Joseph-Louis Lagrange (1736-1813)

Integration Path through a Ray Bundle



90

SOLO

Second Order Conditions: Legendre’s Condition for a Weak Minimum

Adrien-Marie Legendre1752-1833

From( )

[ ] ( )sd

ydzyxn

zy

yzyxn

y

FFy ,,

1

,,2/122

=++

=∂∂=

we obtain

( )[ ] [ ] [ ]

( )[ ] 2/322

2

2/322

2

2/1222/1222

2

1

1

111

,,

zy

zn

zy

yn

zy

n

zy

yzyxn

yy

F

++

+=

++−

++=

++∂∂

=∂∂

From

we obtain

( )[ ] [ ] 2/3222/122

2

11

,,

zy

zyn

zy

zzyxn

yzy

F

++−=

++∂∂=

∂∂∂

( )[ ] ( )

sd

zdzyxn

zy

zzyxn

z

FFz ,,

1

,,2/122

=++

=∂∂=

( )[ ]

( )[ ] 2/322

2

2/1222

2

1

1

1

,,

zy

yn

zy

zzyxn

zz

F

++

+=

++∂∂=

∂∂

From those equations we obtain

( )[ ]

( )( )

+−

−+

++=

∂∂

∂∂∂

∂∂∂

∂∂

=2

2

2/322

2

22

2

2

2

''1''

1

1

,,

yyx

zyz

zy

zyxn

z

F

yz

F

zy

F

y

F

F XX



91

SOLO

Second Order Conditions: Legendre’s Condition for a Weak Minimum (continue – 1)

( )[ ]

( )( )

+−−+

++=

∂∂

∂∂∂

∂∂∂

∂∂

=2

2

2/322

2

22

2

2

2

''1''

1

1

,,

yyx

zyz

zy

zyxn

z

F

yz

F

zy

F

y

F

F XX

Let use Sylvester’s Theorem to check if/when is positive definite (i.e. check that the determinants of all principal minors are positive)

''XXF

James Joseph Sylvester

1814-1897[ ]

( )( )

+−−+

++=

2

2

2/322''1''

1det

1det

yyx

zyz

zy

nF XX

[ ] ( )( )[ ] [ ] 01

111

2/122

2222

2/322>

++=−++

++=

zy

nzyyz

zy

n

( ) 01 2 >+ z

According to Sylvester’s Theorem is positive definite''XXF ( )0'' >XXF

According to Legendre’s Condition ,if the Jacobi’s Condition is satisfied (no conjugate points between P1 and P2), every extremal is a weak minimum.

( )0'' >XXF



SOLO

The Weierstrass Necessary Condition for a Strong Minimum (Maximum)

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=


Weierstrass E Function is defined as

( ) ( ) ( ) ( ) ( ) ( ) ( )zyzyxFzZzyzyxFyYzyzyxFZYzyzyxFZYzyzyxE zy ,,,,,,,,,,,,,,,,,,:,,,,,, −−−−−=

[ ] [ ] ( ) [ ] ( ) [ ][ ] [ ] ( ) ( )[ ]

[ ] [ ] ( )

[ ] [ ] [ ] ( )InequalitySchwarzzyZY

zZyYZYn

zZyYzy

nZYn

zzZyyYzyzy

nZYn

zy

znzZ

zy

ynyYzynZYn

011

111

11

1

11

1

1''

111

2/1222/122

2/122

2/122

2/122

222/122

2/122

2/1222/122

2/1222/122

≥

++++++−++=

++++

−++=

−+−+++++

−++=

++−−

++−−++−++=

According to Weierstrass Condition if the Jacobi Condition (no conjugate points between and ) is satisfied every extremal is a strong minimum.

( )( )0',',',',',',,, ≥ZYXzyxzyxE



93

SOLO

Hamilton’s Canonical Equations

Define ( )[ ] ( )

( )[ ] ( )

sd

zdzyxn

zy

zzyxn

z

Fp

sd

ydzyxn

zy

yzyxn

y

Fp

z

y

,,1

,,:

,,1

,,:

2/122

2/122

=++

=∂∂=

=++

=∂∂=

( ) ( ) ( )2222222 1 zynzypp zy +=+++Adding the square of twose two equations gives

( ) ( )2

222

2221

=

+−=++

xd

sd

ppn

nzy

zy

from which

Substituting in ( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=

gives( ) ( )222

2

,,,,zy

zy

ppn

nppzyxF

+−=



94

SOLO

Hamilton’s Canonical Equations (continue – 1)

From ( )[ ] ( )

( )[ ] ( )

sd

zdzyxn

zy

zzyxn

z

Fp

sd

ydzyxn

zy

yzyxn

y

Fp

z

y

,,1

,,:

,,1

,,:

2/122

2/122

=++

=∂∂=

=++

=∂∂=

solve for

( ) ( )222

2

,,,,zy

zy

ppn

nppzyxF

+−=and

( )

( )222

222

zy

z

zy

y

ppn

pz

ppn

py

+−=

+−=

Define the Hamiltonian( ) ( )

( ) ( ) ( )( ) ( ) ( )

sd

xdzyxnppzyxn

ppn

p

ppn

p

ppn

n

zpypppzyxFppzyxH

zy

zy

z

zy

y

zy

zyzyzy

,,,,

,,,,:,,,,

222

222

2

222

2

222

2

−=+−−=

+−+

+−+

+−−=

++−=



95

SOLO


From

We obtain the Hamilton’s Canonical Equations

( ) ( ) ( ) ( )sd

xdzyxnppzyxnppzyxH zyzy ,,,,,,,, 222 −=+−−=

( )

( )222

222

zy

z

z

zy

y

y

ppn

p

p

H

xd

zdz

ppn

p

p

H

xd

ydy

+−=

∂∂==

+−=

∂∂==

( )

( )222

222

zy

z

zy

y

ppn

z

nn

z

H

xd

pd

ppn

y

nn

y

H

xd

pd

+−

∂∂

−=∂∂−=

+−

∂∂

−=∂∂−=



96

SOLO


From( ) ( ) ( ) ( )sd

xdzyxnppzyxnppzyxH zyzy ,,,,,,,, 222 −=+−−= ( )222

zy ppn

n

sd

xd

+−=

By similarity with ( )sd

ydzyxnpy ,,=

define ( ) ( ) ( ) ( )222 ,,,,,,,,: zyzyx ppzyxnppzyxHsd

xdzyxnp +−=−==

Let differentiate px with respect to x ( ) x

H

xd

Hd

ppn

x

nn

xd

pd

zy

x

∂∂−=−=

+−

∂∂

=222

Let compute

( )( )

x

n

n

ppn

ppn

x

nn

sd

xd

xd

pd

sd

pd zy

zy

xx

∂∂=

+−

+−

∂∂

==222

222



97

SOLO


and

( )( )

x

n

n

ppn

ppn

xn

n

sd

xd

xd

pd

sd

pd zy

zy

xx

∂∂=

+−

+−

∂∂

==222

222

( )( )

y

n

n

ppn

ppn

yn

n

sd

xd

xd

pd

sd

pd zy

zy

yy

∂∂=

+−

+−

∂∂

==222

222

( )( )

z

n

n

ppn

ppn

zn

n

sd

xd

xd

pd

sd

pd zy

zy

zz

∂∂=

+−

+−

∂∂

==222

222

nsd

pd ∇=

xpnsd

xd 1=

( )( )

y

zy

zy

y pnn

ppn

ppn

p

sd

xd

xd

yd

sd

yd 1222

222=

+−

+−==

( )( )

z

zy

zy

z pnn

ppn

ppn

p

sd

xd

xd

zd

sd

zd 1222

222=

+−

+−==

pnsd

rd ray

1=

We recover the result from Geometrical Optics



98

William Rowan Hamilton (1805-1855) Canonical Equations of Motion 1835

where H is the Hamiltonian defined as:

Hamilton-Jacobi Theory

SOLO CALCULUS OF VARIATIONS

William Rowan Hamilton1805-1855

Carl Gustav Jacob Jacobi

1804-1851

Hamilton-Jacobi Equation

99

SOLO

Hamilton-Jacobi Equation (continue – 1)

( ) ( ) ( ) [ ]( )

[ ]( )

[ ]( )

[ ] ( )sd

xdzyxn

zy

zyxn

zy

zzyxnz

zy

yzyxny

zyzyxnFzFyzyzyxFzyxS zyx

,,1

,,

1

,,

1

,,

1,,,,,,,,

2/1222/1222/122

2/122

=++

=++

−++

−

++=−−=

( ) ( )[ ] ( )

sd

ydzyxnp

zy

yzyxn

y

FzyxS yy ,,:

1

,,,,

2/122==

++=

∂∂=

( ) ( )[ ] ( )

sd

zdzyxnp

zy

zzyxn

z

FzyxS zz ,,:

1

,,,,

2/122==

++=

∂∂=

We obtain

snsd

rdnS ray ˆ==∇

From this

22 ˆˆ nssnSS =⋅=∇⋅∇

We recovered again the Eikonal Equation



100

SOLO

Hamilton-Jacobi Equation (continue – 2)

( ) ( ) sdzyxnsdsd

zd

sd

yd

sd

xdzyxn

sdsd

zdSsd

sd

ydSsd

sd

xdSzdSydSxdSSd zyxzyx

,,,,

1

222

=

+

+

=

++=++=

( ) ( ) ( )∫=− sdzyxnzyxSzyxS ,,,,,, 000Along an extremal trajectory we obtain

( ) ( )sd

ydzyxnzyxS y ,,,, =

( ) ( )sd

zdzyxnzyxS z ,,,, =

( ) ( )2

222

2221

=

+−=++

xd

sd

ppn

nzy

zy

( ) ( ) ( )[ ] ( )

sd

xdzyxn

zy

zyxnFzFyzyzyxFzyxS zyx ,,

1

,,,,,,,, 2/122

=++

=−−=

( ) ( )[ ] ( )[ ]

( )

zy ppzyxH

zyx ppnzy

zyxnzyxS

,,,,

2/12222/1221

,,,,

−

+−=++

=

( ) ( ) 0,,,,,, ≡+ zyx ppzyxHzyxS

Hamilton-Jacobi Equation is an identity.



101

SOLO

If the regularity condition (optical ray not intersecting) doesn’t hold, the optical ray may not be a minimum, as we can see from the Figure, where the optical ray reflected from the planar mirror and reaches the point P2 (P1MP2) is longer than the direct ray from P1 to P2.



102

SOLO

On other example is given in Figure bellow on the rays from a point source refracted by a lens. The refracted rays form an envelope called caustic. The point P’2 where the refracted ray touches the caustic is called a conjugate point. From the Figure we can see that this point is reached by, at least, two rays with different optical paths.



103

SOLO

Example of the stationarity of the Fermat’s Principle

Suppose that we have a elliptical mirror and a point source locate at one of it’s foci P1.

The elliptical mirror has the following properties:

1. The sum of the distances from the two foci to any point R on the ellipse is constant.

2121 PRRPPRRP EE +=+ 2. The normal at any point R on the ellipse bisects the angle P1RP2.

2P1P

Point Source

Elliptic Mirror

RER

RnERs

Rs

According to Snell’s Law, all the rays originated at the focus P1 will be reflected by the elliptical mirror and intersect at the second foci P2.

Since the rays travel in the same media and the geometrical paths are equal, the optical paths will be equal also.

( ) ( )2121 PRRPnPRRPn EE +=+ Since all the optical paths reflected by the mirror reach the point P2, we call P2 the conjugate point to P1.



104

SOLO

Example of the stationarity of the Fermat’s Principle (continue – 1)

Now replace the elliptical mirror with a planar one normal to at the point R.Rn

2P1P

Point Source

Planar Mirror

Elliptic Mirror

RER

PR

RnERs

PRs

Rs

For this reason the ray will be reflected at R and reach the point P2, in the same way as for the elliptical mirror.

RP1

From the Figure we can see that:

( ) ( ) 222121 PRRRPRPRRRPnPRRPn PPEEPPE +<←+<+

In this case the Fermat’s Principle will give a minimum for the optical path

( )21 PRRPn +



105

SOLO

Example of the stationarity of the Fermat’s Principle (continue – 3)

Now replace the elliptical mirror with a circular one normal to at the point R(The mirror diameter is smaller than the maximum axis of the ellipse).

Rn

For this reason the ray will be reflected at R and reach the point P2, in the same way as for the elliptical mirror.

RP1

From the Figure we can see that:

In this case the Fermat’s Principle will give a maximum for the optical path

( )21 PRRPn +

2P1P

Point Source

Elliptic Mirror

CircularMirror

R

CR

ER

RnCRs

ERsPRs

Rs

( ) ( ) 222121 PRRRPRPRRPnPRRPn EECCCC +<←+>+




106


References

Bryson & Ho “Applied Optimal Control”, Ginn and Company, 1969, Sec 3.1, pg. 91-95

George Leitmann “The Calculus of Variations and Optimal Control – An Introduction”, Plenum Press, 1981

O. Bolza, “Lectures on the Calculus of Variations”, Dover Publications, New York, 1961, Republication of a work published by Univ. of Chicago 1904, pp.1,27-28,41,48-49,64, 77-78, 153

W.S. Kimball, “Calculus of Variations, by Parallel Displacement”, Butterworths Scientific Publications, 1952, § 13, pp.432-475

L.E. Elsgolc, , “Calculus of Variations”, Pergamon Press, Addison-Wesley, 1962, pp.37-38

I.M. Gelfand, S.V. Fomin, “Calculus of Variations”, Prentice-Hall, 1963, pp.20-21

H. Sagan, “Introduction to Calculus of Variations”, Dover Publication, New York, 1969, pp.62-65 H. Tolle, “Optimization Methods”, Springer Verlag, Berlin Heidelberg New York, 1975, pp.15-18

107


References

S. Hermelin, “Calculus of Variations”, http://www.solohermelin.com Math Folder

S. Hermelin, “Foundation of Geometrical Optics”, http://www.solohermelin.com Optics Folder


February 23, 2015 108

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

calculus of variation problems

Science

brachistochrone greek

solocalculus of variations

solohistory of calculus

development of calculus

johann bernoulli

minimum time

soloa particle slides

john bernoulli