Calculus of variations

1

Calculus of Variations

SOLO HERMELIN

"weak"neighbor

( )( )000 , txtA

( )( )fff txtB ,

( )txx

t

"strong"neighbor

( )ε,tx

http://www.solohermelin.com

http://www.solohermelin.com/

2

Table of Content

Calculus of VariationsSOLO

.Introduction

1. General Formulation of the Simplest Problem of Calculus of Variations

2. Solution Method

2.1 Neighborhoods and Variations

3. Variations of the Functional J

4. Necessary Conditions for Extremum

4.4 Special Cases

4.5 Examples

5. Boundary Conditions

6. Corner Conditions

7 Sufficient Conditions and Additional Necessary Conditions for a Weak Extremum

4.1 The First Fundamental Lemma of the Calculus of Variations

4.2 The Euler-Lagrange Equation4.3 The Second Fundamental Lemma of the Calculus of Variations

8 Legendre’s Necessary Conditions for a Weak Minimum (Maximum)

Table of Content (continue – 1)


9. Jacobi’s Differential Equation (1837) and Conjugate Points

9.1 Conjugate Points

9.2 Fields Definition

10. Hilbert’s Invariant Integral

11. The Weierstrass Necessary Condition for a Strong Minimum (Maximum)

Summary

12. Canonical Form of Euler-Lagrange Equations

12.1 Legendre’s Dual Transformation

12.2 Transversality Conditions (Canonical Variables )

12.3 Weierstrass-Erdmann Corner Conditions (Canonical Variables)

12.4 First Integrals of the Euler-Lagrange Equations

12.5 Equivalence Between Euler-Lagrange and Hamilton Functionals

12.6 Equivalent Functionals

12.7 Canonical Transformations

12.8 Caratheodory's Lemma

12.9 Hamilton-Jacobi EquationsJacobi’s Theorem

Table of Content (continue – 2)


References

Appendix 1: Implicit Functions Theorem

Appendix: Useful Mathematical Theorems

Appendix 2: Heine–Borel Theorem

Appendix 3: Ordinary Differential Equations

5

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)

Dido Maximum Area Problem

6

( ) ∫∫−

==a

axy

dxyAdJ maxmaxmax

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

The problem can be formulated as:

Dido Maximum Area Problem

HISTORY OF CALCULUS OF VARIATIONS

( ) ( ) ( ) ∫∫∫∫∫−−−

=+=

+=+==

a

a

a

a

a

a

dxdxdxxd

ydydxdsdP θθ sectan11 2

2

22

subject o the isoperimetric constraint:

where: θtan=

xd

yd

SOLO

Return to Table of Content

Rope of length P

( )xθ

x

y

a+a−

y

dx

7

1. General Formulation of the Simplest Problem of Calculus of Variations

Given:

(1) A Functional (function of functions) J [x (t)]


( )[ ] ( ) ( ) ( ) ( )( ) ( ) ( )∫∫

==

⋅ff t

t

t

t

nn dttxtxtFdttxtxtxtxtFtxJ00

,,,,,,,, 11

( ) ( ) ( )( ) Tn txtxtx ,,: 1 =

( ) ( ) ( )( ) ( ) ( )T

nT

n txdt

dtx

dt

dtxtxtx

==

⋅,,,,: 11

where:

( ) ( )

⋅

txtxtF ,,

( ) ( )txtxt⋅

,,

(2) shall be continuous and admit continuous partial derivatives of the first, second and third order in a domain which contains all points .

.

8

General Formulation of the Simplest Problem of Calculus of Variations


.

( ) ( ) ( ) ( ) ( )( ) Tn tftftftftx ,,, 21 ==(1) The vector of functions where t0 ≤ t ≤tf, fi (t)i=1,n being single

valued of t that minimizes (maximizes) the functional J in a weak neighborhood.

(2) fi (t)i=1,n are continuous and consist of a finite number of arcs of continuously turning tangent, not parallel to the x axis; i.e. fi (t) € D (1)

(3) passes through two points (constant vectors), defined or not.( )tx

( )tx xt, (4) lies in a given region of the space.

cornerpoints( )( )000 , txtA

( )( )fff txtB ,

( )txx

t

Find:

Figure: A Possible Solution for the Problem of Calculus of Variations

9



Examples of Calculus of Variations Problems

1. Brachistochrone Problem

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

10




1. Brachistochrone ProblemSince 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

11


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

12


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

13




2. 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.

14




3. 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).

15

SOLO

We have:

constS =constdSS =+

s

∫2

1

P

P

dsn

1P

2P

( ) ( ) ( )∫∫∫∫ =

+

+===

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

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 ++=

+

+=




3. Geometrical Optics and Fermat Principle

Paths of Rays Between Two Points

16

SOLO




4. Hamilton Principle for Conservative Systems

The motion of a conservative system, from time t0 to tf is such that the integral

( )∫=ft

t

dtqqLJ0

,

has a stationary value ( δJ = 0), where

( ) ( ) ( )qVqqTqqL −= ,:,

qq ,

( )qqT ,

( )qV

δ

is the Lagrangian of the system

are the generalized coordinate vector of the system and its derivatives

kinetic energy of the systempotential energy of the system

the variation that will be defined in the next section.

Since the system is conservative, the external forces acting on the system are given by

( ) ( )qVqQ ∇=

For a non-conservative system the Extended Hamilton Principle is

( ) ( ) 0,00

=+ ∫∫ff t

t

t

t

dtqqQdtqqT δδ The Hamilton Principle doesn’t require the minimization but only stationarity (vanishing of the first variation δJ = 0).

17

SOLO




5. Geodesics

Suppose we have a surface specified by two parameters u and 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

td

ud

u

r

td

vd

v

r

td

ud

u

rtd

td

rd

td

rdtd

td

rdds

2/12/1

∂∂+

∂∂⋅

∂∂+

∂∂=

⋅==

tdtd

vd

v

r

v

r

td

vd

td

ud

v

r

u

r

td

ud

u

r

u

r2/122

2

∂∂⋅

∂∂+

∂∂⋅

∂∂+

∂∂⋅

∂∂=

18

SOLO




5. Geodesics (continue)

A

B

( )vur ,

vdrv

udru

rd

The Shortest Path on a Surface

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:

∂∂⋅

∂∂=

u

r

u

rE

:

∂∂⋅

∂∂=

v

r

u

rF

:

∂∂⋅

∂∂=

v

r

v

rG

:

where


19

SOLO Calculus of Variations

2. Solution Method

( )tx( )ε,tx

To find a candidate for the minimizing (maximizing) trajectory , construct variations (neighbors) of this trajectory and find the conditions under which those variations increase (decrease) the value of the functional J [x (t)].

The results of this method are known as the Calculus of Variations.

"weak"neighbor

( )( )000 , txtA

( )( )fff txtB ,

( )txx

t

"strong"neighbor

( )ε,tx


20


2.1 Neighborhoods and Variations

"weak"neighbor

( )( )000 , txtA

( )( )fff txtB ,

( )txx

t

"strong"neighbor

( )ε,tx

( )tx( )ε,txLet define a function of the closeness of order k to

Weak Neighborhood

is a “weak” neighborhood of order k if:( )ε,tx

( ) ( )

( ) ( )

( ) ( )

∂∂=

∂∂

∂∂=

∂∂

=

→

→

→

txt

txt

txt

txt

txtx

k

k

k

k

ε

ε

ε

ε

ε

ε

,lim

,lim

,lim

0

0

0

( ) ( )( ) ( )( )00000 ,,, txtAtxt ∈εεε

( ) ( )( ) ( )( )fffff txtBtxt ,,, ∈εεε

Strong Neighborhood

If we only have (only for k = 0) then is called a “strong” neighborhood. If k > 0 then it is a “weak” neighborhood.

( ) ( )txtx =→

εε

,lim0

( )ε,tx

( ) ( ) ( ) ( ) ( ) ( )32

0

2

2

0

,

2

1,,:, εε

εεε

εεεε

εε

Ο/+∂

∂+∂

∂=−=∆==

dtx

dtx

txtxtxLet compute:

( )0lim

2

3

0→Ο/

→ εε

εwhere:

21


First and Second Variations

( )ε,tx

( )txFirst Variation of

( ) ( ) εε

εδε

dtx

tx0

,:

=∂∂=

i.e. the differential of as a function of ε

( )tx The First Variation of is defined as

( )txSecond Variation of

( )tx The Second Variation of is defined as

( ) ( ) 2

0

2

22 ,

: εε

εδε

dtx

tx=∂

∂=

( )( )fff txtB ,

x

t

( )2,εtx( )1,εtx

( )tx

ft

( )1εft

( )2εft

At the boundaries t0 and tf are functions of ε (see Figure)

22


First and Second Variations at the Boundary

Therefore at the boundaries we have ( )( ) fiitx ,0, =εε

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

( )xdxdxd

dtx

dtx

txtxtx

iitt

iiii

32

32

0

2

2

0

2

1

,

2

1,,:,

Ο/++=

=Ο/+∂

∂+

∂∂

=−=∆==

εεε

εεεε

εεεεεεεε

εε

ε dx

xdi

i

t

t0: =

∂∂=

202

22

02:&: ε

εε

εεε d

xxdd

xxd

i

i

i

i

t

t

t

t==

∂∂=

∂∂=

where:

( )( ) ( )( )

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

∂∂+

∂∂+

∂∂=

=

=

εεεεε

εε

εεεεεε

ε

εεεε

εεε

,,,

,,2

2

ii

ii

i

ii

txd

d

d

dt

d

dtx

td

dttx

td

d

txd

d

d

dtx

d

d

( )

∂∂+

∂∂∂+

∂∂+

∂∂

∂+∂∂=

2

22

2

22

2

2

εεεεε

εεεx

d

dt

t

x

d

td

t

x

d

dt

t

x

d

dt

t

x iiii

2

2

2

222

2

2

2εεεεε ∂

∂+∂∂+

∂∂∂+

∂∂= x

d

td

t

x

d

dt

t

x

d

dt

t

x iii

( )( ) ( )( ) ( ) ( )( )εεεε

εεεεεε

,,, ii

ii txd

dttx

ttx

d

d

∂∂+

∂∂=

We have:

( )( )fff txtB ,x

t

( )fxd 3Ο/

( )ε,tx

( )tx

ff dtt +( )0=εft

fxδ fxd

fx∆fxd 2

ff dtx•

fdt

Variations at the Boundary tf

23



( )( )fff txtB ,x

t

( )fxd 3Ο/

( )ε,tx

( )tx

ff dtt +( )0=εft

fxδ fxd

fx∆fxd 2

ff dtx•

fdt


( )( ) ( ) ( )( ) εεεε

εεεεε

εεε

dtxdd

dttx

txd i

iii

000

,,=== ∂

∂+

∂∂=

( )( ) ( )( )εεεεε

,,0

itt

i txxdt

xdtx

tii

••

=

===∂∂

( )i

i dtdd

dt ==

εεε

ε 0

( ) ( )( ) εεεε

δε

dtxtx ii0

,:=∂

∂=

( ) ( ) fitxdttxxd iii ,0=+=•

δ

But

Therefore we obtain:

( ) 2

0

2

22 : ε

εε

ε

dd

tdtd i

i

=

=and define:

24



( )( )fff txtB ,x

t

( )fxd 3Ο/

( )ε,tx

( )tx

ff dtt +( )0=εft

fxδ fxd

fx∆fxd 2

ff dtx•

fdt


( ) ( ) ( ) ( ) εεεεε

εε

εεε

εεεε

dd

dtd

t

txd

d

dt

t

txxd iiii

i00

2

00

2

22 ,

2,

====

∂∂

∂∂+

∂

∂=

( ) ( ) ( ) 2

0

2

22

0

2

2

0

,, εε

εεε

εε

εεε

dtx

dd

td

t

tx iii

=== ∂∂

+∂

∂+

( ) ( ) ( )iii txtx

dt

d

t

tx ••

=

==∂

∂2

2

0

2

2 ,

ε

ε

( ) ( )ii txdt

tx •

=

=

∂∂

∂∂ δεεε ε 0

,

( ) ( ) 2

0

2

22 ,

: εε

εδε

dtx

tx ii

=∂

∂=

( ) ( ) ( ) ( ) ( ) fitxtdtxdttxdttxxd iiiiiiiii ,02 2222 =+++=••••

δδ

Also we have:

But:

Therefore


25


3. Variations of the Functional J

The value of the functional J in the neighborhood of is ( )ε,tx ( )tx

( ) ( ) ( )( )

( )

∫

=

•ε

ε

εεεft

t

dttxtxtFJ0

,,,,

( ) ( )εε ,:, txt

tx∂∂=

•where

We can write:

( ) ( )

( ) ( )JJJdd

Jdd

d

dJ

JJJ

3232

02

2

0 2

1

2

1

0:

δδδεεε

εε

εε

εε

Ο/++=Ο/++=

=−=∆

==

where

εε

δε

dd

dJJ

0

:=

= the first variation of J

2

02

22 : ε

εδ

ε

dd

JdJ

=

= the second variation of J

( )0lim

2

3

0→Ο/

→ εε

ε

26


First Variation of the Functional J

( ) ( ) ( )( )

( )

= ∫

•ε

ε

εεεε

ε ft

t

dttxtxtFd

d

d

dJ

0

,,,,

( ) ( ) ( ) ( ) ( ) ( )εεεε

εε

εεd

dttxtxtF

d

dttxtxtF f

fff0

000 ,,,,,,,,

−

=

••

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

( )

∫

∂∂

+

∂∂

+

•••

•

ε

ε

εε

εεεε

εεft

tx

x dttxtxtxtFtxtxtxtF0

,,,,,,,,,,

T

n

nn

x x

F

x

F

x

F

x

F

x

F

x

F

F

x

x

x

xxtF

∂∂

∂∂

∂∂=

∂∂

∂∂∂∂

=

∂∂

∂∂

∂∂

=

•

,,,:,,21

2

1

2

1

T

nx x

F

x

F

x

FxxtF

∂∂

∂∂

∂∂=

•

•

,,,:,,21

and

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )∫

+

+

−

==

•••

••

=

•

ft

t

T

x

Tx

ffff

dttxtxtxtFtxtxtxtF

dttxtxtFdttxtxtFd

dJJ

0

,,,,

,,,, 00000

δδ

εεδ

ε

27


Second Variation of the Functional J

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

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

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

( )

∂∂

+

∂∂

+

+

−

+

=

=

∫•••

••

••

ε

ε

εε

εεεε

εεε

εε

εεε

εεεε

ε

εε

εεε

εε

εεεεεε

ft

t

Tx

Tx

ffff

ffff

dttxtxtxtFtxtxtxtFd

d

d

dt

d

dtxtxtF

d

dttxtxtF

d

d

d

dt

d

dtxtxtF

d

dttxtxtF

d

d

d

dJ

d

d

d

Jd

0

,,,,,,,,,,

,,,,,,,,

,,,,,,,,

0000

0000

2

2

In this equation we have:

( ) ( ) fix

d

dt

t

xF

x

d

dt

t

xF

d

dtFtxtxtF

d

d

it

iTx

iTx

itiii ,0,,,, =

∂∂+

∂∂+

∂∂+

∂∂+=

•••

εεεεεεε

ε

t

FFt ∂

∂=:( ) ( )

2

2

εε

εε

ε d

td

d

dt

d

d ii =

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

( )

∫

∂∂

+

∂∂

•••ε

εε

εεεε

εεε

ε

ft

t

Tx

Tx dttxtxtxtFtxtxtxtF

d

d

0

,,,,,,,,,,

( ) ( ) ( ) ( ) ( ) ( ) ( )εε

εε

εεεε

εεd

dttxtxtxtFtxtxtxtF f

ffffT

xffffT

x

∂∂

+

∂∂

=

•••,,,,,,,,,,

( ) ( ) ( ) ( ) ( ) ( ) ( )εεε

εεεε

εεε

d

dttxtxtxtFtxtxtxtF T

xT

x0

00000000 ,,,,,,,,,,

∂∂

+

∂∂

−

•••

( )

( )

tdx

Fxx

Fxx

Fx

Fxx

Fxx

Fxx

T

xx

T

T

x

t

txx

T

xx

TT

x

f

∂∂

∂∂+

∂∂

∂∂+

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂+

•••••

•••••∫ εεεεεεεεεε

ε

ε2

2

2

2

0

and

28


Second Variation of the Functional J (continue – 1)

[ ] [ ]

=

∂∂

∂∂

=∂∂=

∂∂

∂∂=

nnnn

n

n

n

xxxxxx

xxxxxx

xxxxxx

xxxTx

T

xx

FFF

FFF

FFF

FFF

x

x

Fxx

F

xF

21

21212

12111

21,,,:

1

1

[ ] [ ]

=

∂∂

∂∂

=∂

∂=

∂∂

∂

∂= •••

nnnn

n

n

n

xxxxxx

xxxxxx

xxxxxx

xxxTx

T

xx

FFF

FFF

FFF

FFF

x

x

Fxx

F

xF

21

21212

12111

21,,,:

1

1

[ ]

=

∂∂

∂∂

=

∂∂=

∂

∂∂∂= •• •

nnnn

n

n

n

xxxxxx

xxxxxx

xxxxxx

xxxT

x

T

xx

FFF

FFF

FFF

FFF

x

x

Fxx

F

xF

21

21212

12111

21,,,:

1

1

[ ]

=

∂∂

∂∂

=

∂

∂=

∂

∂

∂

∂= ••••

nnnn

n

n

n

xxxxxx

xxxxxx

xxxxxx

xxxT

x

T

xx

FFF

FFF

FFF

FFF

x

x

Fxx

F

xF

21

21212

12111

21,,,:

1

1

29



xxTxxxx

ijjixx FFF

x

F

xx

F

xF

jiij=→=

∂∂

∂∂=

∂∂

∂∂=:

xxxxTxxxx

ijjixx FFFF

x

F

xx

F

xF

jiij

≠=→=

∂∂

∂∂=

∂∂

∂∂=:

xxTxxxx

ijjixx FFF

x

F

xx

F

xF

jiij

=→=

∂∂

∂∂=

∂∂

∂∂=:

Let integrate by parts the term

( )

( )

( )

( )

( )

( )∫ ∂

∂

−

∂∂=∫ ∂

∂•••

•ε

ε

ε

ε

ε

ε εεε

ff

f t

t

T

x

t

t

T

x

t

t

T

xdt

xF

dt

dxFdt

xF

00

0

2

2

2

2

2

2

By using all those developments we get:

2

2

2

2

εεεεεεεε d

tdF

d

dtx

d

dt

t

xF

x

d

dt

t

xF

d

dtF

d

Jd f

t

f

t

fT

x

fTx

ft

f

f

+

∂∂+

∂∂+

∂∂+

∂∂+=

••

•

20

20000

0

0

εεεεεεε d

tdF

d

dtx

d

dt

t

xF

x

d

dt

t

xF

d

dtF

t

t

T

x

Txt +

∂∂+

∂∂+

∂∂+

∂∂+−

••

•

( ) ( )εε εεεεεεεεff

f

t

T

xt

T

x

t

T

x

Tx

f

t

T

x

Tx

xF

xF

d

dtxF

xF

d

dtxF

xF

00

2

2

2

20

∂∂−

∂∂+

∂∂+

∂∂−

∂∂+

∂∂+ ••••

••

( )

( )

dtx

Fxx

Fxx

Fx

Fxx

Fxx

Fdt

dF

xx

T

xx

T

T

x

t

txx

T

xx

TT

xx

f

∂∂

∂∂+

∂∂

∂∂+

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂

−+

•••••

••••••∫ εεεεεεεεεε

ε

ε2

2

2

2

0

30



ft

fffT

x

ffTx

ft d

tdF

x

d

dtx

d

dt

t

xF

d

dtx

d

dt

t

xF

d

dtF

d

Jd

+

∂∂+

∂∂+

∂∂+

∂∂+

∂∂+

=

••

• 2

2

2

222

2

2

22εεεεεεεεεε

0

20

2

2

20

2

000

2

0 22

t

T

x

Txt

d

tdF

x

d

dtx

d

dt

t

xF

d

dtx

d

dt

t

xF

d

dtF

+

∂∂+

∂∂+

∂∂+

∂∂+

∂∂+

−

••

•

εεεεεεεεε

( )

( )

dtx

Fxx

Fxx

Fxx

Fxx

Fdt

dF

xx

T

xx

Tt

txx

T

xx

TT

xx

f

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂

−+

••••

•••••∫ εεεεεεεεε

ε

ε0

2

2

( ) ( )ft

fffT

xff

Txft tdFxdtxdtxFdtxdtxFdtFd

d

JdJ

+

+++

++==

••••

=

•22222

02

22 22 δδδε

εδ

ε

( ) ( )ft

fffT

xff

Txft tdFxdtxdtxFdtxdtxFdtF

+

+++

++−

••••

•2222 22 δδδ

( ) ( ) ( ) ( )( )

( )∫

+

+

++

−+

••••

•••••

ε

εδδδδδδδδδ

ft

t xx

T

xx

T

xx

Txx

TT

xx dtxFxxFxxFxxFxxF

dt

dF

0

2

Therefore

31



( ) ( ) fitxdttxxd iii ,0=+=•

δ

But we found that:

( ) ( ) ( ) ( ) ( ) fitxtdtxdttxdttxxd iiiiiiiii ,02 2222 =+++=••••

δδ

( ) −

+

−+

−+=

••

•

ft

ffT

xff

Txft tdFtdxxdFdtdtxxdFdtFJ 22222 2δ

( ) −

+

−+

−+−

••

•

0

02

022

002

0 2t

T

x

Txt tdFtdxxdFdtdtxxdFdtF

( ) ( ) ( ) ( )( )

( )∫

+

+

++

−+

••••

•••••

ε

εδδδδδδδδδ

ft

t xx

T

xx

T

xx

Txx

TT

xx dtxFxxFxxFxxFxxF

dt

dF

0

2

Hence:

and the final result is:

( )

( )

( ) ( ) ( ) ( )( )

( )

∫

+

+

++

−+

+

−+++

−−

−+++

−=

••••

••

••

•••••

••

••

ε

ε

δδδδδδδδδ

δ

f

f

t

txx

T

xx

T

xx

Txx

TT

xx

t

T

x

T

x

Tx

Txt

t

fT

xf

T

xff

Txf

Txt

dtxFxxFxxFxxFxxFdt

dF

tdxFFxdFdtxdFdtxFF

tdxFFxdFdtxdFdtxFFJ

0

0

2

02

02

002

0

2222

2

2

32


4. Necessary Conditions for Extremum

We found that:

( ) ( ) ( ) ( )JJJdd

Jdd

d

dJJJJ 3232

0

2

2

0 2

1

2

10: δδδεε

εε

εεε

εε

Ο/++=Ο/++==−=∆==

For a Minimum Solution of the Functional J we must have:

ΔJ ≥ 0 for any small dε (see Figure)

( )( )ε,txJ

0=εε

Minimum of J as function of ε

For a Maximum Solution of the Functional J we must have:

ΔJ ≤ 0 for any small dε

To prevent that the sign of dε to change the same of ΔJ the Necessary Condition for Extremum is

000

===

Jordd

dJ δεε ε

This condition must be fulfilled for any admissible variation.

33


Necessary Conditions for Extremum (continue – 1)

Suppose that is an extremal solution with the fixed end points and . ( )tx * ( )0*

0* , xtA ( )0

*0

* , xtB

Let choose first all the variation that passes through those points (see Figure ).

( )( )000* , txtA

( )( )fff txtB ,*

( )tx*x

t

( )ε,1 tx

( )ε,2 tx

Variations Passing through Fixed End Points

0&0 022

0 ==== tdtddtdt ff

( ) ( ) ( ) ( ) 0&0 022

0 ==== txtxtxtx ff δδδδ

( ) ( ) ( ) ( ) 0&0 022

0 ==== txdtxdtxdtxd ff

Therefore:

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

=

+

= ∫

•••

•

ft

t

T

x

Tx dttxtxtxtFtxtxtxtFJ δδδ

where:

( ) ( ) εεε

δε

dtxtx0

,=∂

∂= ( ) ( ) εεε

δε

dtxtx0

,=

••

∂∂=

34



Transformation of the First Variation δ J by integration by parts

(a) First way:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫

−

=∫

••••

•••

ff

f t

t

T

x

t

t

T

x

t

t

T

xdttxtxtxtF

dt

dtxtxtxtFdttxtxtxtF

00

0

,,,,,, δδδ

Because we have( ) ( ) 00 == txtx f δδ

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

=

−

= ∫

••

•

ft

t

T

xx dttxtxtxtF

dt

dtxtxtFJ δδ

δ J must be zero for all admissible variations , where and dt0 = dtf = 0.

( )txδ ( ) ( ) 00 == txtx f δδ

Note:•••••••

+

+

=

• xxxtGxxxtGxxtGxxtGdt

d T

x

Txt ,,,,,,,,

therefore integration by parts assumes however, that not only , but also exists and is continuous in (t0, tf ).

•x

••x

End Note


35



4.1 The First Fundamental Lemma of the Calculus of Variations (Du Bois-Reymond-1879)

If M(t) is a continuous function of t in (t0, tf ) and if

( ) ( ) 00

=∫ft

t

tdtxtM δ

for all functions that vanish at t0 and tf and which admit a continuous derivative in (t0, tf ), then

( )txδ

( ) fttttM ≤≤= 00

Paul David Gustav Du Bois-Reymond

(1831-1889)

Proof:

Suppose M (t) ≠ 0, say greater than zero at a point t1 on the interval (t0, tf ).

Because M(t) is continuous exists a neighbor of t1 say (t1-ζ, t1+ζ) in which we chose

( )( ) ( ) ( )

+−∈−−+−

+−∉=

ζζζζ

ζζδ

112

12

1

11

,

,0

ttttttt

tttx

kk( )tM

tft0t 1t ζ+1tζ−1t

xδ

admits a continuous derivative in (t0, tf ) and vanishes at t0 and t1 and nevertheless makes

( ) ( ) 00

>∫ft

t

tdtxtM δcontrary to the hypothesis; therefore M (t) ≠ 0 is impossible for al t0≤ t ≤tf. q.e.d. Return to Table of Content

36



4.2 The Euler-Lagrange Equation

The Necessary Condition for an extremal is δ J = 0, where

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

=

−

= ∫

••

•

ft

t

T

xx dttxtxtxtF

dt

dtxtxtFJ δδ

For all variations satisfying and dt0 = dtf = 0.( )txδ ( ) ( ) 00 == txtx f δδ

( ) ( ) ( ) ( )( ) fT

n ttttxtxtxtx ≤≤= 021 ,,, δδδδ

By choosing for i=1,…,n δ xi(t) ≠ 0 and δ xj(t) = 0 for all j ≠ i and using the First Fundamental Lemma, we can see that δ J= 0 for all admissible variations if and only if ( )txδ

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

=

−

=

−

=

−

••

••

••

•

•

•

0,,,,

0,,,,

0,,,,

22

11

txtxtFdt

dtxtxtF

txtxtFdt

dtxtxtF

txtxtFdt

dtxtxtF

nn x

x

xx

xx

37



The Euler-Lagrange Equation (continue – 1)

As a matrix equation

( ) ( ) ( ) ( ) 0,,,, =

−

••

• txtxtFdt

dtxtxtF

xx

Euler-Lagrange Equation

It was discovered by Euler in 1744. Later in 1760 Lagrange discussed this equation and introduced the notation δ and the notion of Variation.

Leonhard Euler (1707-1783)

Joseph-Louis Lagrange (1736-1813)

By developing this equation we get:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0,,,,,,,, =

−

+

+

•••••••

•••• txtxtFtxtxtFtxtxtxtFtxtxtxtF xtxxxxx

This is a Nonhomogeneous, Second Order, Differential Equation.

( ) ( )

•

•• txtxtFxx

,,If is nonsingular on t0 ≤ t ≤tf, then

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

−

+

−=

••••−•••

•••• txtxtFtxtxtFtxtxtxtFtxtxtFtx xtxxxxx

,,,,,,,,1

The existence of is achieved if the matrix has an inverse for all t in (t0, tf ).If this condition is satisfied we have a Regular Problem. The problem is well defined if 2n boundary conditions are defined (see Appendix 3 ).

( )tx•• ( ) ( )

•

•• txtxtFxx

,,

38



The Euler-Lagrange Equation (continue – 2)

Leonhard Euler (1707-1783)


Therefore the general solutions of the Euler-Lagrange Equations are therefore two vector parameters solutions ( ) ( ) T

nT

n βββααα ,,,,, 11 ==

( ) ( )βαϕ ,,ttx =and those parameters are defined by the 2n boundary conditions.

39



(b) Second way:

Du Bois-Reymond and Hilbert integrated the first, instead of the second, term of δ J by parts

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

( ) ( )∫ ∫∫

∫

••••

•••

−

+

=

+

=

•

•

f ff

f

f

t

t

Tt

t

xx

t

t

t

t

Tx

t

t

T

x

Tx

dttxdtxxtFxxtFtxdtxxtF

dttxtxtxtFtxtxtxtFJ

0 00

0

0

,,,,,,

,,,,

δδ

δδδPaul David Gustav Du Bois-Reymond

(1831-1889)

David Hilbert (1862 – 1943)

Because , we have:( ) ( ) 00 == txtx f δδ

( ) 0,,,,0 0

=

−

= ∫ ∫

•••

•

f ft

t

Tt

t

xx

dttxdtxxtFxxtFJ δδ

δ J must be zero for all admissible variations , such that . ( )txδ ( ) ( ) 00 == txtx f δδ


40



4.3 The Second Fundamental Lemma of the Calculus of Variations (Du Bois-Reymond-1879)

Paul David Gustav Du Bois-Reymond

(1831-1889)

Proof:

q.e.d.

If N(t) is a continuous function of t in (t0, tf ) and if

for all functions of classes C(1) that vanish at t0 and tf then N (t) must be constant in (t0, tf ).

( )txδ

( ) ( ) 00

=∫•ft

t

dttxtN δ

Let subtract from the previous equation the identity: ( ) ( ) ( )[ ] 000

=−=∫•

txtxCdttxC f

t

t

f

δδδ

where C is a constant.

( )[ ] ( ) 00

=∫ −•ft

t

dttxCtN δ

From all the possible variations let choose the following particular variation:

( ) ( )[ ] 0>−=•

εεδ CtNtx

For this variation we must have: ( )[ ] ( ) ( )[ ] 000

2 =∫ −=∫ −• ff t

t

t

t

dtCtNdttxCtN δ

This is possible only if N (t) = C. Therefore N (t) = C is a necessary condition.The sufficiency condition is proven by substituting N (t) = C in the original equation.

41



The Second Fundamental Lemma of the Calculus of Variations (Du Bois-Reymond-1879) (continue – 1)

Let apply the Second Fundamental Lemma of the Calculus of Variations to the equation:

( ) 0,,,,0 0

=

−

= ∫ ∫

•••

•

f ft

t

Tt

t

xx

dttxdtxxtFxxtFJ δδ

( ) ( ) ( ) ( ) f

T

n ttttxtxtxtx ≤≤

=

••••

021 ,,, δδδδ where

We obtain the following form of the Euler-Lagrange Equation:

∫

+=

••

•

ft

t

xx

dtxxtFCxxtF0

,,,,

From this equation we can see that every solution of our problem with continuousfirst derivative – not only those admitting a second derivative – must satisfy the Euler-Lagrange Equation; i.e. the existence of is not necessary.( )tx

••


42


4.4 Special Cases

F doesn’t depend explicitly on the free variable t

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

( ) ( ) xxxFtd

dxxF

xxxFtd

dxxxFxxxFxxxFxxxFxxF

td

d

T

xx

T

xT

xT

xT

xT

x

−=

−−+=−

,,

,,,,,,

For an extremal the Euler-Lagrange equation applies, and we have

( ) ( ) ( ) ( ) 0,, =

−

••

• txtxFdt

dtxtxF

xx

( ) ( )[ ] 0,, =− xxxFxxFtd

d Tx

Therefore

( ) ( ) constCxxxFxxF Tx ==− ,,

Let perform the following:

43


Special Cases (continue – 1)

F is not an explicit function of x

In this case the Euler-Lagrange equation is:

( ) 0, =

•

• txtFdt

dx

that can be integrated to give

( ) constCtxtFx

==

•

• ,

F is not an explicit function of x

In this case the Euler-Lagrange equation is:

( )( ) 0, =txtFx

( )( ) ( ) ( ) 0,..,0,det =∀≠ xtFtsxttxtF xxIf we can find that satisfies this equation.

( )txx =

According to Implicit Function Theorem this solution is unique..

44


Special Cases (continue – 2)

F is an exact differential

( ) ( )( ) ( ) ( ) ( ) xxtVxtVxtVtd

dtxtxtF T

xt ,,,,, +=≡

If this is true than

( ) ( )( )( )

( )( )

( )

( )( ) ( )00

,

,

,

,

,,,,,000000

xtVxtVdtxtVtd

ddttxtxtF ff

xtP

xtP

xtP

xtP

ffffff

−=∫=∫

therefore the functional is independent on the integration path.

Let find what conditions F must satisfy in order to be an exact differential. Let compute

( ) ( )( ) ( ) ( ) xxtVxtVtxtxtF xxxtx ,,,, +=

( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) xxtVxtVtxtxtFtd

dxtVtxtxtF T

txtxxxx

,,,,,,, +=→=

From those relations we can see that the condition that F is an exact

differential if and only if the Euler-Lagrange equation is an identity.

( ) ( )( ) ( ) ( )( )txtxtFtxtxtFtd

dxx

,,,, ≡


45


4.5 Example 1: Brachistocrone

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

46


Example 1: Brachistocrone (continue – 1)

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 −=

47


Example 1: Brachistocrone (continue – 2)

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

48



( )

( )

−−=

−+=

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

49


Example 2: Minimum Surface of Revolution

x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y( )∫

+=

b

a

xdxd

ydxyJ

2

12π

For this problem we derived the cost function:

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.

We have

( ) ( ) ( ) ( )xd

ydxyxyxyyyxF =+= :12:,, 2 π

F doesn’t depend explicitly on the free variable x, therefore we can apply the results for this special case, with ( ) ( )xtyx ,, →

( ) ( ) Cy

yyyyyyFyyyF y ππ 2

112,,

2

22 =

+−+=−

21 yCy +=or

Separating variables, we obtainC

xd

C

y

C

yd

=

−

12

12

−

=

C

yy

50


Example 2: Minimum Surface of Revolution (continue – 1)

x

y

( )bybB ,

( )ayaA ,

( ) ( ) 22 ydxdsd +=

y

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 compute

2

expexp 11

−−+

−

= CCx

CCx

C

y ( )

−=

C

CxCxy 1coshor

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.

51

SOLO

Example 3: Geometrical Optics and Fermat Principle

We have:

constS =constdSS =+

s

∫2

1

P

P

dsn

1P

2P

( ) ( ) ( )∫∫∫∫ =

+

+===

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 ++=

+

+=


52

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


Example 3: Geometrical Optics and Fermat Principle (continue – 1)

53

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

∂∂=


Example 3: Geometrical Optics and Fermat Principle (continue –2)

54

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



55

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

∂∂=



56

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.




57


5. Boundary Conditions

Until now we considered only the variations passing through the end points and . But those are not all the admissible variations. If or are not specified then if we consider all admissible variations (see Figure), then δ J will be given by:

( )*0

*0

* , xtA

( )*** , ff xtB ( )00 , xtA ( )ff xtB ,

( )txδ

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫

+

+

−

=

•••••

•

ft

t

T

xxffff dttxtxtxtFtxtxtxtFdttxtxtFdttxtxtFJ

0

,,,,,,,, 0000 δδδ

( )( )000 , txtA

( )( )fff txtB ,

( )tx*x

t

( )ε,1 tx

( )ε,2 tx

Variations that Satisfy the Boundary Conditions

Integrating by parts the second term of the integral as before we have:

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

( ) ( ) ( ) ( ) ( )∫

−

+

+

−

+

=

••

••••

•

••

f

f

t

t

T

xx

t

T

xt

T

x

dttxtxtxtFdt

dtxtxtF

xtxtxtFdttxtxtFxtxtxtFdttxtxtFJ

0

0

,,,,

,,,,,,,,

δ

δδδ

58


Boundary Conditions (continue – 1)

( )( )000 , txtA

( )( )fff txtB ,

( )tx*x

t

( )ε,1 tx

( )ε,2 tx

Variations that Satisfy the Boundary Conditions

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

( ) ( ) ( ) ( ) ( )∫

−

+

+

−

+

=

••

••••

•

••

f

f

t

t

T

xx

t

T

xt

T

x

dttxtxtxtFdt

dtxtxtF

xtxtxtFdttxtxtFxtxtxtFdttxtxtFJ

0

0

,,,,

,,,,,,,,

δ

δδδ

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

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

( ) ( ) ( ) ( ) ( )∫

−

+

+

−

−

+

−

=

••

••••

••••

•

••

••

f

f

t

t

T

xx

t

T

x

T

x

t

T

x

T

x

dttxtxtxtFdt

dtxtxtF

xdtxtxtFdtxtxtxtFtxtxtF

xdtxtxtFdtxtxtxtFtxtxtFJ

0

0

,,,,

,,,,,,

,,,,,,

δ

δ

But ( ) ( ) ( ) iiiii dttxtxdtx•

−= δ

Therefore:

59



We found before that the necessary conditions such that δ J is zero for those admissible solutions passing through the points and are the Euler-Lagrange Equation:( )*

0*0

* , xtA ( )*** , ff xtB

( ) ( ) ( ) ( ) 0,,,, =

−

••

• txtxtFdt

dtxtxtF

xx

For other admissible variations we shall need to add the additional necessary conditions, such that δ J is zero, called Transversality Conditions Equations:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) fitxdtxtxtFdttxtxtxtFtxtxtF iiiix

iiiiiT

xiii ,00,,,,,, ==

+

−

••••

••

(a) Suppose that the following relation defines the boundary:

( ) ( ) ( ) ( ) ( ) fidttdttdt

dtxdttx iitiiiii ,0=Ψ=Ψ=→Ψ=

then the Transversality Conditions Equations are:

( ) ( ) ( ) ( ) ( ) ( ) fitxttxtxtFtxtxtF iitiiiT

xiii ,00,,,, ==

−Ψ

+

•••

•

60



Geometric Interpretation of the Transversality Conditions

Let plot as a function of . The hyper-plane tangent at ( ) ( )

=

•txtxtF ,,η

•= xξ

( ) ( ) ( )

==

••

iiiiii txtxtFtx ,,, ηξ is given by

( ) ( ) ( ) ( ) ( )

+

−

=

•••

iiiiiiiT

x txtxtFtxtxtxtF ,,,, ξη

( ) ( ) ( ) ( ) ( ) ( ) fitxttxtxtFtxtxtF iitiiiT

xiii ,00,,,, ==

−Ψ

+

•••

•

We can see that for η = 0 the last equation is identical to the Transversality Conditions Equation.

Geometric Representation of the Transversality Conditions

( ) ( ) ( ) ( ) ( ) fidttdttdt

dtxdttx iitiiiii ,0=Ψ=Ψ=→Ψ=

61



Transversality Conditions

Suppose that ti and are not defined and is not a function of ti, then dti and are independent differentials and therefore both coefficients of dti and must be zero.

ixix ixd

ixd

( ) ( ) ( ) ( ) ( ) 0,,,, =

−

•••

• iiiiT

xiii txtxtxtFtxtxtF

( ) ( ) 0,, =

•

• iiix

txtxtF

Or, by using the second equation to simplify the first we get:

( ) ( )

( ) ( )fi

txtxtF

txtxtF

iiix

iii

,0

0,,

0,,

=

=

=

•

•

•

Those Equations are called Natural Boundary Conditions because they arise naturally when the original problem doesn’t specify boundary conditions.

62



Example: 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 Return to Table of Content

63


6. Corner Conditions

In the development of the Euler-Lagrange Equation we assumed that not only is continuous, but also . However, there are a number of problems, for which this assumption is not true, for example problems of reflection or refraction.

( )tx

( )tx•

We define such problems as follows:

Find the curve that passes through the boundary points (given or not) and and extremizes the functional . This curve should reach the point after having been reflected by a given function (see Figure).

( )tx ( )00 , xtA

( )ff xtB , ( )[ ] ( ) ( )∫

=

•ft

t

dttxtxtFtxJ0

,,

( )ff xtB , ( )tx Ψ=

cornerpoint

( )( )000 , txtA( )( )fff txtB ,

( )tx*x

tct

( )tΨ

The Corner Point of the Trajectories

64


Corner Conditions (continue – 1)

cornerpoint

( )( )000 , txtA( )( )fff txtB ,

( )tx*x

tct

( )tΨ


Solution:

Let define by tc the unknown time when the extremal is reflected. Then we can express the functional J in the form:

( )tx•

( )[ ] ( ) ( ) ( ) ( )∫∫

+

=

•• f

c

c t

t

t

t

dttxtxtFdttxtxtFtxJ ,,,,0

We suppose that is continuous in each of the intervals (t0, tc-), (tc+, tf). Then for both intervals we have:

( )tx•

(1) The Euler-Lagrange Equation is:

( ) ( ) ( ) ( ) cfx

x ttttttxtxtFdt

dtxtxtF ≠≤≤=

−

••

• 00,,,, ( )tx••

if exists,

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )∫

∫

+

•

−

•

+=

+=

••

••

f

c

c

t

t

xx

t

t

xx

dttxtxtFCtxtxtF

dttxtxtFCtxtxtF

0

0

0

,,,,

,,,,

( )tx••

if doesn’t exist

65



cornerpoint

( )( )000 , txtA( )( )fff txtB ,

( )tx*x

tct

( )tΨ


Solution (continue – 1):

(2) The Transversality Conditions at the initial (0) and final (f) points are:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) fitxdtxtxtFdttxtxtxtFtxtxtF iiiiT

xiiiii

T

xiii ,00,,,,,, ==

+

−

••••

••

Then: ( ) ( ) 000000000

=

+

−−

+

−= ++

•

−−

•

+

•

+

•

−

•

−

• ct

T

xc

t

T

xc

t

T

xc

t

T

xtxdFdtxFFtxdFdtxFFJ

cccc

δ

But tc- = tc+ = tc and , therefore ccc xdxdxd == +− 00

( ) 00000

=

−+

−−

−=

+

•

−

•

+

•

−

•

••

ct

T

xt

T

xc

t

T

xt

T

xtxdFFdtxFFxFFJ

cccc

δ

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

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

,,,,

,,,,

00

000

000

=

−

+

+

−

−

+

•

−

•

+

•

+

•

+

•

−

•

−

•

−

•

••

•

•

ccccx

cccx

cccccx

ccc

ccccx

ccc

txdtxtxtFtxtxtF

dttxtxtxtFtxtxtF

txtxtxtFtxtxtF

The necessary conditions for extremal at the corners are:

Those are the Weierstrass-Erdmann Corner Conditions. Those equations were developed independently by Weierstrass and Erdmann in 1877.

Karl Theodor Wilhelm Weierstrass1815-1897

66



cornerpoint

( )( )000 , txtA( )( )fff txtB ,

( )tx*x

tct

( )tΨ



(a) If they are a priori conditions at the corner like:

( ) ( ) ( ) ( ) ( ) cctccccc dttdttdt

dtxdttx Ψ=Ψ=→Ψ=

then the necessary conditions at the corner are:

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

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

−Ψ

+

−Ψ

+

+

•

+

•

+

•

−

•

−

•

−

•

•

•

000

000

,,,,

,,,,

cctcccT

xccc

cctcccT

xccc

txttxtxtFtxtxtF

txttxtxtFtxtxtF

67




(b) If they are not a priory conditions at the corner; i.e. the function is not a priori defined then dtc and are independent variables and

( ) ( )cc ttx Ψ=

cxd

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

=

−

=

−

+

•

−

•

+

•

+

•

+

•

−

•

−

•

−

•

••

•

•

00

000

000

,,,,

,,,,

,,,,

cccx

cccx

ccccT

xccc

ccccT

xccc

txtxtFtxtxtF

txtxtxtFtxtxtF

txtxtxtFtxtxtF

cornerpoint

( )( )000 , txtA( )( )fff txtB ,

( )tx*x

tct

( )tΨ


68



Geometric Interpretation of the Corner Conditions

Since the Corner Conditions where derived from the Transversality Conditions, we have a similar geometrical interpretation.

Let plot as a function of .( ) ( )

=

•txtxtF ,,η

•= xξ

Since the hyper-plane tangent at is given by ( ) ( )+− = cc txtx ( ) ( ) ( )

== −

•

−−

•

− cccccc txtxtFtx ,,, ηξ

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )−

•

−

•

−

•

−

•

−

•

−

•

−

•

−

+

=

+

−

=

ccccT

xccccccT

x

cccccccT

x

txtxtxtFtxtxtFtxtxtF

txtxtFtxtxtxtF

,,,,,,

,,,,

ξ

ξη

The hyper-plane tangent at is given by ( ) ( ) ( )

== +

•

++

•

+ cccccc txtxtFtx ,,, ηξ

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )+

•

+

•

+

•

+

•

+

•

+

•

+

•

−

+

=

+

−

=

ccccT

x

ccccccT

x

cccccccT

x

txtxtxtF

txtxtFtxtxtF

txtxtFtxtxtxtF

,,

,,,,

,,,,

ξ

ξη

But according to the Corner Conditions the two tangent hyper-planes are the same (see Figure )



Example: 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.



Example: 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.




( ) 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




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



7 Sufficient Conditions and Additional Necessary Conditions for a Weak Extremum

We found that:

( ) ( ) ( ) ( )JJJdd

Jdd

d

dJJJJ 3232

02

2

0 2

1

2

10: δδδεε

εε

εεε

εε

Ο/++=Ο/++==−=∆==

The Necessary Condition that Δ J ≥ 0 or ≤ 0 for all small dε is

000

===

Jord

dJ δε ε

For Sufficient Conditions for a “Weak” Local Extremum we must add the following:

( ) ( )0000 2

02

2

≤≥≤≥=

Jord

Jd δε ε

for a minimum (maximum) solution.

The expression for δ2J is

( )

( )

( ) ( ) ( ) ( )( )

( )

∫

+

+

++

−+

−+++

−−

−+++

−=

••••

••

••

•••••

••

••

ε

ε

δδδδδδδδδ

δ

f

f

t

txx

T

xx

T

xx

Txx

TT

xx

t

T

x

T

x

Tx

Txt

t

fT

xf

T

xff

Txf

Txt

dtxFxxFxxFxxFxxFdt

dF

tdxFFxdFdtxdFdtxFF

tdxFFxdFdtxdFdtxFFJ

0

0

2

02

02

002

0

2222

2

2


Sufficient Conditions and Additional Necessary Conditions for a Weak Extremum(continue -1)

Suppose first that the end points are fixed; i.e.:

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 00

00

00

20

20

20

20

20

20

====

====

====

ff

ff

ff

txdtxdtxdtxd

txtxtxtx

tdtddtdt

δδδδ

In this case:

( ) ( ) ( ) ( )( )

( )

∫

+

+

++

−=

••••

•••••

ε

ε

δδδδδδδδδδft

txx

T

xx

T

xx

Txx

TT

xx dtxFxxFxxFxxFxxF

dt

dFJ

0

22

For an extremal solution the Euler-Lagrange Equation holds, therefore 0=− •x

x Fdt

dF

( ) ( ) ( ) ( )( )

( )

( )( )

( )∫

=

∫

+

+

+=

•

•

••••

•••

•

••••

ε

ε

ε

ε

δ

δδδ

δδδδδδδδδ

f

f

t

txxxx

xxxxT

T

t

t xx

T

xx

T

xx

Txx

T

dtx

x

FF

FFxx

dtxFxxFxxFxxFxJ

0

0

2

We have the following properties of the derivatives

Txxxx

Txxxx

Txxxx FFFFFF ===


Sufficient Conditions and Additional Necessary Conditions for a Weak Extremum(continue -2)

Let define:TT

xxxxT

xxxx

TTxxxx RFFRFFQPFFP ======== •• :,:,:

( )( )

( )

( )( )

( )

( )( )

( )

∫

∫

∫

++=

+++=

= •

•==

==

ε

ε

ε

ε

ε

ε

δδδδδδ

δδδδδδδδ

δ

δδδδ

f

f

fii

ii

t

t

TTT

t

t

TTTTT

t

tT

TT

xdxd

tddt

dtxPxxQxxPx

dtxPxxQxxQxxPx

dtx

x

RQ

QPxxJ

0

0

0

2

2

2

00

00

2

Therefore:



8 Legendre’s Necessary Conditions for a Weak Minimum (Maximum) – 1786

Adrien-Marie Legendre1752-1833

Proof:

Let start from the necessary condition of an Minimal Optimal Trajectory, that

( )( )

( )

020

2

2

00

00

2 ≥++== ∫==

==

ε

ε

δδδδδδδf

ii

ii

t

t

TTTxdxd

tddtdtxPxxQxxPxJ

(The same reasoning applies for a Maximal Optimal trajectory where it is required that δ2J ≤ 0)

Suppose that R is not positive definite and we have a constant vector such that at some point t = τ, on our curve. Since R (t) was assumed continuous, this inequality will hold over some sufficiently small interval [τ-h, τ+h] . We now define the function so that it vanishes outside and at the end points of the interval, it has all the necessary derivatives; is sufficiently small in absolute value in the interval, but performs fairly rapid oscillations.

v 0<vRv T

( )txδ

( )

+<<

−−

≤<−

−+

=

elsewhere

hth

th

thh

th

tx

0

1

1

ττντε

ττντε

δ ( )

+<<−

≤<−

=⇒

elsewhere

hth

thh

tx

0

ττνε

ττνε

δ

The matrix must be Positive (Negative) Definite along a Minimal (Maximal) Optimal Trajectory

xxFR =


Legendre’s Necessary Conditions for a Weak Minimum (Maximum) – 1786(continue – 1)


Proof (continue – 1):

ixδ

t1h−τ 2h−τ 2h+τ1h+τ

hε

+<<

−−

<<−

−+

=

elsewhere

hth

th

thh

th

x

0

1

1

ττντε

ττντε

δ

+<<−

<<−

=

elsewhere

hth

thh

x

0

ττνε

ττνε

δ t

1h−τ 2h−τ2h+τ1h+τ

h

ε

ixδ

Since P (t) and Q (t) are continuous matrix functions in the interval t € [0, tf] we can find two positive numbers M1 and M2 such that:

( ) ( ) [ ]fTT ttMvtQvMvtPv ,0, 21 ∈∀<<



Proof (continue – 2):ixδ


hε

+<<

−−

<<−

−+

=

elsewhere

hth

th

thh

th

x

0

1

1

ττντε

ττντε

δ

+<<−

<<−

=

elsewhere

hth

thh

x

0

ττνε

ττνε

δ t


h

ε

ixδ

( ) ( ) [ ]fTT ttMvtQvMvtPv ,0, 21 ∈∀<<

22

0

1

0

1

22

0

0

22

211

1100

Mhdth

tdt

h

tM

dtQvh

tdtQv

h

tdtxQxdtxQx

h

h

h

hTT

t

t

Tt

t

Tff

εττε

ντεντεδδδδ

τ

τ

τ

τ

≤

∫

−−+∫

−+≤

∫ ∫

−−+

−+=∫≤∫

+

<

−

<

−

+

122

0

1

20

1

2

12

0

0

22

22

211

1100

Mhdth

tdt

h

tMh

dtPvh

thdtPv

h

thdtxPxdtxPx

h

h

h

hTT

t

t

Tt

t

Tff

εττε

ντεντεδδδδ

τ

τ

τ

τ

≤

∫

−−+∫

−+≤

∫ ∫

−−+

−+=∫≤∫

+

<

−

<

−

+

We have

( ) ( ) ( ) ( )[ ]

( )[ ] 10212

21

2

22

0

≤≤−+=

+−+−=∫=∫+

−

λλτε

ντλτλενεδδτ

τ

someforhRvh

hhhRvh

dttRvh

dtxRx

T

Th

h

Tt

t

Tf



ixδ


hε

+<<

−−

<<−

−+

=

elsewhere

hth

th

thh

th

x

0

1

1

ττντε

ττντε

δ

+<<−

<<−

=

elsewhere

hth

thh

x

0

ττνε

ττνε

δ t


h

ε

ixδ

( ) ( ) ( ) ( )[ ]

( )[ ] 10212

21

2

22

0

≤≤−+=

+−+−=∫=∫+

−

λλτε

ντλτλενεδδτ

τ

someforhRvh

hhhRvh

dttRvh

dtxRx

T

Th

h

Tt

t

Tf

Since by assumption and R (t) is a continuous matrix functions in the interval t € [0, tf] we can find a small h1 such that for all h ≤ h1 we have

( ) 0<vRv T τ

( )[ ] 021 2 <−≤−+ µνλτ hR

Therefore 1222

0

hhdtxRxft

t

T ≤∀−≤∫ µεδδ

Using the previous results we can write

( )( )

( )

( )

( )

( )

( )

( )( )

( )

( ) 12

2122

00

00

2

22

220000

2

2

hhMhMh

dtxPxdtxQxdtxPxdtxPxxQxxPxJffff

ii

ii

t

t

T

t

t

T

t

t

T

t

t

TTTxdxd

tddt

≤∀−+≤

++≤++= ∫∫∫∫==

==

µε

δδδδδδδδδδδδδε

ε

ε

ε

ε

ε

ε

ε

Since we can find a small h2 ≤ h1 such that for all h ≤ h1 ( ) 02 02

212 =−+ =hMhMh µ

( ) 22

21222 022 hhMhMhJ ≤∀<−+≤ µεδ

δ2 J turn out negative, which contradicts the before mentioned necessary condition for minimum; i.e. δ2 J ≥ 0.

Therefore must be Positive Definite along the trajectory to have a minimum.xxFR =q.e.d.



Example: Geometrical Optics and Fermat’s Principle

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=


( )[ ] [ ] [ ]

( )[ ] 2/322

2

2/322

2

2/1222/1222

2

1

1

111

,,

zy

zn

zy

yn

zy

n

zy

yzyxn

yy

F

++

+=

++−

++=

++∂∂

=∂∂ ( )

[ ] [ ] 2/3222/122

2

11

,,

zy

zyn

zy

zzyxn

yzy

F

++−=

++∂∂=

∂∂∂

( )[ ]

( )[ ] 2/322

2

2/1222

2

1

1

1

,,

zy

yn

zy

zzyxn

zz

F

++

+=

++∂∂

=∂∂

From those equations we obtain:( )

[ ]( )

( )

+−

−+

++=

2

2

2/322''1''

1

1

,,

yyx

zyz

zy

zyxnF XX

Let use Sylvester Theorem to check the positiveness of ''XXF

[ ]( )

( ) [ ] ( )( )[ ] [ ] 01

1111''

1det

1det 2/122

22222/3222

2

2/322'' >++

=−++++

=

+−

−+

++=

zy

nzyyz

zy

n

yyx

zyz

zy

nF XX

1

( ) 01 2 >+ z2

We can see that according to Sylvester Theorem is Positive Definite. ''XXF

James Joseph Sylvester1814-1897



9. Jacobi’s Differential Equation (1837) and Conjugate Points

Let start from the necessary condition of a Minimal Optimal Trajectory, that

( )( )

( )02

0

2

2

00

00

2 ≥∫ ++===

==

ε

εδδδδδδδ

fii

ii

t

t

TTTxdxd

tddtdtxRxxQxxPxJ

Define ( ) xRxxQxxPxxx TTT δδδδδδδδ ++=Ω 2:,

Therefore ( ) ( )( )

( )

∫Ω===

==

ε

ε

δδδδf

ii

ii

t

t

xdxd

tddtdtxxxJ

0

2

2,

00

00

2

We have( )

( )xRxQ

x

xx

xQxPx

xx T

δδδ

δδ

δδδ

δδ

22,

22,

+=∂

Ω∂

+=∂

Ω∂

We can see that ( ) ( ) ( )x

x

xxx

x

xxxx

TT

δ

δδδδ

δδδδδ

∂

Ω∂+

∂

Ω∂=Ω ,,,2

Since ( )( )

( ) ( ) ( )∫

∂

Ω∂−

∂

Ω∂=∫

∂

Ω∂ =

=

f

f

f

f t

t

Tt

t

Ttx

tx

t

t

T

T

dtxx

xx

td

dx

x

xxdtx

x

xx

0

0

0

0

,,,

0

0

0δ

δδδδ

δδδδ

δδδ δ

δ

we have

( ) ( )( )

( ) ( ) ( )

( ) ( )

( ) ( )∫

+−+=

∫

∂

Ω∂−∂

Ω∂=

∫

∂

Ω∂+

∂

Ω∂=∫ Ω===

==

f

f

ffii

ii

t

t

TTTTT

t

t

T

t

t

TTt

t

xdxd

tddt

dtxRxQxdt

dQxPx

dtxx

xx

dt

d

x

xx

dtxx

xxx

x

xxdtxxxJ

0

0

00

2

2

,,

2

1

,,

2

1,

00

00

2

δδδδδ

δδ

δδδ

δδ

δδ

δδδδ

δδδδδδε

ε


Jacobi’s Differential Equation (1837) and Conjugate Points(continue – 1)

we have ( ) ( ) ( )∫

+−+=

ft

t

TTTTT dtxRxQxdt

dQxPxxJ

0

2 δδδδδδδ

Gilbert Ames Bliss(1876 –1951)

Gilbert A. Bliss (1876-1951) suggested to show that the minimum of δ2Jfor all possible is non-negative. If this is true thanxδ

( ) ( ) 0min 22 ≥> xJxJx

δδδδδ

We obtain the following

Secondary (Accessory) Variational Problem:

( ) ( )( )

( )∫ Ω=

==

==

ε

εδδδδδδ

fii

ii

t

tx

xdxd

tddtxdtxxxJ

0

2

2,minmin

00

00

2

The necessary conditions for a minimum are satisfaction of 1.Euler-Lagrange Equations and 2.Transversality 3.Weierstrass-Erdmann Corner Conditions

Euler-Lagrange Equation for the Secondary Variational Problem:

( ) ( )0

,, =∂

Ω∂−∂

Ω∂x

xx

dt

d

x

xx

δδδ

δδδ ( ) ( ) 0**** =+−+ xQxPxRxQ

dt

d T δδδδor



( ) ( ) 0**** =+−+ xQxPxRxQdt

d T δδδδ

Euler-Lagrange Equation for the Secondary Variational Problem:

We can see that the extremal makes .*xδ ( ) 0*00

00

2

2

2

==

===

ii

ii

xdxd

tddtxJ δδ

Assume that det R ≠ 0 in t € [0,tf], i.e. R is non-singular and has an inverse, in this interval, then

0*** 11 =

−+

−++ −− xPQ

dt

dRxQQR

dt

dRx TT δδδ

Jacobi’s Differential Equation

Carl Gustav Jacob Jacobi

1804-1851

This is a Second Order Vectorial Homogeneous Linear Differential Equation with continuous coefficients.



0*** 11 =

−+

−++ −− xPQ

dt

dRxQQR

dt

dRx TT δδδ


1804-1851

We apply the general existence and uniqueness theorems for linear differential equations and we obtain n solutions, for the initial conditions:

U (t) is a nxn matrix and contains the n independent solutions of the Vectorial Homogeneous Linear Differential Equation:

0112

2

=

−+

−++ −− uP

td

QdR

td

udQQ

td

RdR

td

ud TT

Where is a vector( )

( )( )

( )

=

tu

tu

tu

tu

n

2

1

If are the n solutions of the Jacobi’s Vectorial Differential Equation,with initial conditions:

( ) ( ) ( )tututu n,,, 21

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

( ) ( ) ( )1,,0,0:&0

0,,1,0:&0

0,,0,1:&0

0

2022

10101

===

======

nnn etutu

etutu

etutu

then define: ( ) ( ) ( ) ( )[ ]tutututU n,,,: 21 =




1804-1851

Theorem

For a weak minimum (maximum) it is necessary that:

is Positive (Negative) Definite for t ϵ [0, tf ] ( ) xxFtR =( ) ( ) ( ) ( )[ ]tutututU n,,,: 21 =If is the n solution matrix of the

Jacobi’s Homogeneous Linear Differential Equation:

0112

2

=

−+

−++ −− uP

td

QdR

td

udQQ

td

RdR

td

ud TT

then U (t) must be nonsingular in t ϵ [0, tf ] .

Proof

Assume that for , then there exists n constants(c1, c2,…,cn) ≠ (0, 0,…,0), such that

( ) ( ) ( ) 0det,0 =→∈ cjcjfcj tUsingularistUtt

( ) ( ) ( ) 02211 =+++ cjnncjcj tuctuctuc

Define ( )( ) ( ) ( )

≤<

≤≤+++=

fcj

cjnn

ttt

ttttuctuctuctx

0:*

02211 δ

We see that ( ) ( ) 0** 0 == cjtxtx δδ




1804-1851

Proof (continue – 1)

Let check the Weierstrass-Erdmann corner conditions at t - tcj

( ) ( )+− == ∂

Ω∂=∂

Ω∂

cjcj ttttx

xx

x

xx

δδδ

δδδ ,,

We have( ) ( ) ( ) ( ) ( ) ( ) ( ) 0222

,

0

≠=+=∂

Ω∂−−−−−−

−

cjcjcjcjcjcj

t

txtRtxtRtxtQx

xx

cj

δδδ

δδδ

The expression is nonzero since R (tcj) is positive definite and (otherwise is uniquely defined by the terminal conditions ).

( ) 0≠−cjtxδ( ) ( ) [ ]fttttxtx ,0 0∈∀== δδ

( ) ( ) [ ]fttttxtx ,0 0∈== δδ

( ) ( ) ( ) ( ) ( ) 022,

00

=+=∂

Ω∂+−++

= +

cjcjcjcj

tt

txtRtxtQx

xx

cj

δδδ

δδ

Therefore( ) ( )

0,, =

∂Ω∂≠

∂Ω∂

+− == cjcj ttttx

xx

x

xx

δδδ

δδδ

The Weierstrass-Erdmann corner conditions at t = tcj are not satisfied, hence

is not the minimum of the second variation, therefore exists a variation such that .

( ) 0*00

00

2

2

2

==

===

ii

ii

xdxd

tddtxJ δδ

xδ ( ) 02 <xJ δδq.e.d.




9.1 Conjugate Points

If U (t) is singular in t ϵ [0, tf ] we say that we have a conjugate point. In this case an optimal solution doesn’t exist.

The geometric meaning of the conjugate points is as follows:

The Second Order Euler-Lagrange Equation has a two-parameter family of solutions.Through any point here passes in general, a one-parameter family of extremals.Let denote this parameter by α and the solutions by .

( )0tx

( )α,tx

The solution must satisfy the Euler-Lagrange Equation:

( ) ( ) ( ) ( ) 0,,,,,,,, =

−

••

• αααα txtxtFdt

dtxtxtF

xx

Let take the partial derivative with respect to t of previous equation

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

+

−

+

=

••••αααααααααααα αααα ,,,,,,,,,,,,,,,,,,,,0 txtxtxtFtxtxtxtF

dt

dtxtxtxtFtxtxtxtF xxxxxxxx

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

( ) ( ) ( ) ( ) ( ) ( )αααααα

αααααααααααα

αα

αααα

,,,,,,,,,,

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

txtxtxtFtxtxtxtFdt

d

txtxtxtFtxtxtxtFdt

dtxtxtxtFtxtxtxtF

xxxx

xxxxxxxx

−

−

−

−

+

=

••

••••



Conjugate Points (continue – 1)

Rearrange the previous equation

( ) ( ) ( )

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

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

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

,,,,,

=

−

+

−

+

+

••

•••

•

ααααα

ααααααα

ααα

α

α

α

txtxtxtFtxtxtFdt

d

txtxtxtFtxtxtFtxtxtFdt

d

txtxtxtF

xxxx

xxxxxx

xx

Using TTxxxx

T

xxxx

TTxxxx RFFRFFQPFFP ======== •• :&:&:

we can write ( ) ( ) ( ) 0,,, =

−+

−++ ααα ααα txPQ

td

dtxQQR

td

dtxR TT

which is identical to the Jacobi Equation.

Since is a solution of the Jacobi Equation if we have for any tcj ϵ [0, tf ] than we have

( )α,tx ( ) 0, =αα tx

( ) ( ) ( ) ( ) 0, 2211 =+++= cjnncjcjcj tuctuctuctx αα

where were defined as the independent solutions of the Jacobi Equation. Therefore U (t) is singular if and according to Theorem the problem doesn’t have a minimum (maximum).

( ) ( ) ( ) ( )[ ]tutututU n,,,: 21 =( ) [ ]fcjcj tttfortx ,0, 0∈=αα



Conjugate Points (continue – 2)

Let tray to understand the meaning of .( ) [ ]fcjcj tttfortx

,0, 0∈=∂∂ α

αSuppose that two close solutions of the family intersect at tcj ϵ [0, tf ]. ( )α,tx

( ) ( )ααα ,, cjcj txdtx =+In this case the family of solutions has an envelope (see Figure )

ExtremalTrajectories Envelope of

ExtremalTrajectories

G( )0tx

ftconjt

321

0t

( )ftx

( )conjtx

Description of Conjugate Points

We have ( ) ( ) ( ) [ ]fcjcjcj

dcj tttfor

d

txdtxt

x,0

,,lim, 0

0∈=

−+=

∂∂

→ αααα

αα α

If such a family has an envelope G, then a point of contact of an extremal with the envelope is called a conjugate point to on that extremal.In the Figure point is conjugate to between 0 and tf.

( )0tx

( )conjtx ( )0tx

On a minimizing (maximizing) extremal curve connecting point and with nonsingular at each point of it, there can be no point conjugate to , between t0 and tf .

( )00 =tx ( )ftx

xxFR = ( )conjtx

( )0tx



Examples of Conjugate Points:

1. The shortest path between two points A and B on the surface of a sphere is on that one great circle passing trough those two points. If the points are on an opposite diameter there are an infinity of great circles passing through, we don’t have one extremal and those two points are conjugate to each other.

A'

B'

B

A

Poles as Conjugate Points on a Sphere

2. Rays from a point source refracted by a lens. The refracted rays forms an envelope called caustic. The point P’2 where the reflected ray touches the envelope 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.

Field of rays passing through a lens and generating a caustic


91



9.2 Fields Definition

Let represent a one parameter family of solutions of the Euler-Lagrange equation in a simply connected region of . This family of solution defines a field if they are not conjugate points in this region. This means that through any point of this regionpasses one and only one curve of the family.

( )α,tx

( )xt,

( )xt,

Field around a solution of Euler-Lagrange equation

In Figure we can see a solution of the Euler-Lagrange equation passing trough and . A field of solutions are shown in a simply connected region that contains the solution. The conjugate point is shown outside this region. We say that the solutionis Embedded in the Field.

( )0, =αtx

( )00 , xt ( )ff xt ,

( )0, =αtx


92


10. Hilbert’s Invariant Integral

Suppose that defines a field of solutions of Euler-Lagrange equation; i.e. ( )α,tx

( ) ( ) ( ) ( ) 0,,,,,,,, =

−

••

• αααα txtxtFdt

dtxtxtF

xx

Let define any curve C in the field region, that starts at and ends at , and passes through a point with a slope (instead of Field slope ).Trough and passes also the unique extremal solution .

( )00 , xt ( )ff xt ,

( )xt, ( )xtX , ( )xtx ,

( )00 , xt ( )ff xt , ( )0, =αtx

Hilbert’s Integral

( )( ) ( )( ) ( ) ( )( )[ ]∫ −−ft

t

TxC tdxtXxtxxtxxtFxtxxtF

0

,,,,,,,,

is invariant on the path C as long as this curve remains in the field of the unique extremal solution.

is the field slope and is the path C slope at the point of C. ( )xtx , ( )xtX , ( )xt,


93


Hilbert’s Invariant Integral (continue – 1)

Hilbert’s Invariant Integral


Proof

Since on C we can writeC

td

xdX =

( )( ) ( )( ) ( )[ ] ( )( )∫ +−ft

t

Tx

TxC xdxtxxtFtdxtxxtxxtFxtxxtF

0

,,,,,,,,,,

( ) ( )( ) ( )( ) ( )( ) ( )( )xtxxtFxtN

xtxxtxxtFxtxxtFxtM

x

Tx

,,,:,

,,,,,,,:,

=−=Define

Rewrite ( ) ( )∫ +ft

t

TC xdxtNtdxtM

0

,,

This integral is path independent if there exists a function

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

( ) ( )xtVxtN

xtVxtM

xdxtNdtxtMxdxtVdtxtVxtVd

x

t

TTxt

,,

,,

,,,,,

==

+≡+=

The following condition must be satisfied

( ) ( ) ( ) ( )xtNt

xtMx

xtVt

xtVx xt ,,,,

∂∂≡

∂∂→

∂∂≡

∂∂

94



Hilbert’s Invariant Integral


Proof (continue – 1)

The following condition must be satisfied ( ) ( )xtNt

xtMx

,,∂∂≡

∂∂

Let check that this is satisfied

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

( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )xtxxtFx

xtxxtx

x

xtxxtxxtFxtxxtxxtF

xtxxtFx

xtxxtxxtFxtM

x

x

TT

xxxx

x

T

x

,,,,

,,

,,,,,,,

,,,,

,,,,

∂∂−

∂∂−−

∂∂+=

∂∂

( ) ( )( ) ( )( ) ( )t

xtxxtxxtFxtxxtFxtN

t xxtxt ∂∂+=

∂∂ ,

,,,,,,,

Let compute

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

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

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

,,,

,,

,,,,,,,,,,

,,,,,,,,,

=−++

∂

∂+∂

∂=

∂∂++−

∂∂+=

∂∂−

∂∂

xtxxtFxtxxtFxtxxtxxtFxtxx

xtx

t

xtxxtxxtF

xtxx

xtxxtxxtFxtxxtxxtFxtxxtF

t

xtxxtxxtFxtxxtFxtM

xxtN

t

xtxxx

T

xx

T

xxxxx

xxtxt

But ( ) ( ) ( ) ( ) ( )xtxtd

xtxdxtx

x

xtx

t

xtx T

,,

,,,

==

∂∂+

∂∂

Since satisfies the Euler-Lagrange equation, that is given by( )xtx ,

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0,,,,,,,, =

−

+

+

•••••••

•••• txtxtFtxtxtFtxtxtxtFtxtxtxtF xtxxxxx

we can see that along C we have ( ) ( ) 0,, =∂∂−

∂∂

xtMx

xtNt t

q.e.d.

95




10.1 Example: 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 ∂

∂==∂∂== ,,,,,

96




Example: Geometrical Optics and Fermat’s Principle (continue – 1)

( ) ( ) ( )[ ]( )

( )( ) ( ) 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.


Integration Path through a Ray Bundle



11. The Weierstrass Necessary Condition for a Strong Minimum (Maximum) –1879


11.1 Derivation from Hilbert’s Invariant Integral

Along the unique extremal path (denoted as C* - see Figure ), that passes through andwe have and the Hilbert Integral becomes

( )00 , xt ( )ff xt ,

( ) ( )xtxxtX ,, =

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

( )( ) [ ]xtJtdxtxxtF

tdxtXxtxxtxxtFxtxxtF

f

f

t

t

C

t

t

TxC

,*,,,

,,,,,,,,

0

0

* ==

−−

∫

∫

where is the minimum of the functional[ ]xtJ ,*

( )[ ] ( ) ( )( )∫=ft

t

C dttXtxtFtxJ0

,,

Suppose that the extremal is a strong minimum and•C* represents the strong minimum curve•C represents a strong neighbor of C*We can compute

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

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

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

∫∫

∫∫

−−−=

−−−=

−=∆

f

ff

ff

t

t

TxC

t

t

TxC

t

t

C

t

t

C

t

t

C

tdxtxxtXxtxxtFxtxxtFtXtxtF

tdxtXxtxxtxxtFxtxxtFdttXtxtF

dttxtxtFdttXtxtFtxJ

0

00

00

,,,,,,,,,,

,,,,,,,,,,

,,,, *


The Weierstrass Necessary Condition for a Strong Minimum (Maximum) (continue – 1)


Derivation from Hilbert’s Invariant Integral (continue – 1)

Let define the Weierstrass E-function:

( ) ( ) ( ) ( ) ( )xXxxtFxxtFXxtFXxxtE Tx

−−−= ,,,,,,:,,,

Therefore the strong minimum condition is

( )[ ] ( ) ( )( ) ( )( ) ( )( ) ( ) ( )( )[ ]∫ −−−=∆ft

t

TxC tdxtxxtXxtxxtFxtxxtFtXtxtFtxJ

0

,,,,,,,,,,

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

≥∫=∆ft

t

C dttXtxtxtEtxJ

Weierstrass Necessary Conditions for a Strong Minimum (Maximum) is:

( ) ( )00,,, ≤≥XxxtE for every admissible set ( )Xxt ,,




11.2 Weierstrass Derivation

Weierstrass, that first derived this necessary condition for a Strong Minimum (Maximum)used the following derivation:

1t δ+1tδε+1t

( )txx =

( ) ( ) ( )[ ]111 txtXtx −+ δε

12

3

0t ft

( )( ) [ ] [ ]

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

++∈−−

−++

+∈−−+

+∈

=

δδεεδε

δε

δ

ε

11111

11111

110

,1

,

,,

,

ttttxtXtt

tx

ttttxtXtttx

ttttttx

tx

f

Weierstrass Strong Variation

Suppose is a candidate trajectory passing trough points 1 and 2, such that it contains no points of discontinuity of and no conjugate points between those points. Let take an arbitrary curve through point 1 such that . Let point 3 a movable point on at t = t1 + εδ . Let connect points 3 with point 2 on The arc 1, 3, 2 constitutes a strong variation (by δ tacking as small as we want). This variation , has a discontinuous derivative at point 1.

( )txx =x

( )tXx = ( ) ( )11 tXtx ≠( )tXx = ( )δ+= 1txx

( )ε,txx =



Weierstrass Derivation (continue – 1)

1t δ+1tδε+1t

( )txx =

( ) ( ) ( )[ ]111 txtXtx −+ δε

12

3

0t ft

( )( ) [ ] [ ]

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

++∈−−

−++

+∈−−+

+∈

=

δδεεδε

δε

δ

ε

11111

11111

110

,1

,

,,

,

ttttxtXtt

tx

ttttxtXtttx

ttttttx

tx

f

Weierstrass Strong Variation

( )( ) [ ] [ ]

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

++∈−−

−++

+∈−−+

+∈

=

δδεεδε

δε

δ

ε

11111

11111

110

,1

,

,,

,

ttttxtXtt

tx

ttttxtXtttx

ttttttx

tx

f

This variation fails to lie within any weak neighborhood of , no matter how small is δ.

( )txx =

Since is a strong minimum, we have:( )[ ]txJ

( )[ ] ( )[ ]

( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ] ( )∫∫+

+

+

−

−

−−+−−+=

−≤δ

δε

δε

εεεε

ε1

1

1

1

,,1

,,,,,,,,

,0

111111

t

t

t

t

dtxxtFtxtXtxtxtFdtxxtFtxtXtxtxtF

txJtxJ

Let assume δ→0 ( )[ ] ( )[ ]

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

εε

εεε

ε

δεε

δ

−

−

−

−−

+−=

−≤→

1

,,1

,,,

,,,,,

,lim0

111

1

0

xxtFtxtXtxtxtF

xxtFtXtxtF

txJtxJ



Weierstrass Derivation (continue – 2)

1t δ+1tδε+1t

( )txx =

( ) ( ) ( )[ ]111 txtXtx −+ δε

12

3

0t ft

( )( ) [ ] [ ]

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

++∈−−

−++

+∈−−+

+∈

=

δδεεδε

δε

δ

ε

11111

11111

110

,1

,

,,

,

ttttxtXtt

tx

ttttxtXtttx

ttttttx

tx

f

Now let take ε→0

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

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

XxxtE

x

Tx

txtXxxtFxxtFtXtxtF

xxtFtxtXxxtFxxtF

xxtFtXtxtF

,,,

111

211

0

1

,,,,,,

1

,,,,1

,,

lim

,,,,0

−−−=

−

−

Ο/+−

−−

+

−≤

→

εε

εε

ε

ε

This Inequality is the Weierstrass Necessary Conditions for a Strong Minimum (Maximum)

( ) 0,,, ≥XxxtE or

Since the Weierstrass condition directly concerns minimality, rather than stationarity as did Euler-Lagrange condition, it entails no further supporting statements analogous to the Legendre and Jacobi conditions that support the Euler-Lagrange stationary condition. A weak variation is included in the strong variations, therefore a condition that is necessary for a weak local minimum (maximum) is also necessary for a strong local minimum (maximum).



11.3 Geometric Interpretation of Weierstrass Conditions

Let plot as a function of (see Figure). The hyper-plane tangent at is given by

( ) ( )

=

•txtxtF ,,η •

= xξ( ) ( ) ( )

==

••

iiiiii txtxtFtx ,,,ηξ

( ) ( ) ( ) ( ) ( )

+

−

=

•••

iiiiiiiT

x txtxtFtxtxtxtF ,,,, ξη

The E function will be given by the difference between and the tangenthyper-plane. We can see that the condition for minimality is that the tangent hyper-plane remains bellow the surface .

( ) ( )

==

•XtxtxtF ,,η

( ) ( )

=

•txtxtF ,,η

Geometric Representation of Weierstrass Condition



11.4 Example: Geometrical Optics and Fermat’s Principle

( ) ( ) ( ) 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



Summary

Necessary Conditions for a Weak Relative Minimum (Maximum)

Satisfies Boundary Conditions


xiiiii

xiii ,00,,,,,, ==

+

−

••••

••

Satisfies Weierstrass-Erdmann Corner Conditions

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

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

,,,,,,,,

00

000000

=

−

+

+

−

−

+

•

−

•

+

•

+

•

+

•

−

•

−

•

−

•

••

••

ccccT

xccc

T

x

cccccT

xccccccc

T

xccc

txdtxtxtFtxtxtF

dttxtxtxtFtxtxtFtxtxtxtFtxtxtF

Satisfies the Euler-Lagrange Equation [ ]fx

x tttforFdt

dF ,0 0∈=− •

1

Satisfies Legendre (Clebsh) Condition 2

is Positive (Negative) Definite for ( ) xxFtR = [ ]fttt ,0∈

It contains no Conjugate Point for 3 [ ]fttt ,0∈

Necessary Conditions for a Strong Relative Minimum (Maximum).

Satisfies (1), (2) and (3) and additionally:

Weirestrass Necessary Conditions for a Strong Minimum (Maximum) is that:

( ) ( ) ( ) ( ) ( ) ( )00,,,,,,:,,, ≤≥−−−= xXxxtFxxtFXxtFXxxtE Tx

4

for every admissible set ( )Xxt ,,



12. Canonical Form of Euler-Lagrange Equations

We found that the first variation of the cost function J is given by:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫

−

+

+

−

=

••••••

•••

ff t

t

T

xx

t

t

T

x

T

xdttxtxtxtF

dt

dtxtxtFxdtxtxtFdtxtxtxtFtxtxtFJ

00

,,,,,,,,,, δδ

Let define

( ) ( )T

xxx n

FFtxtxtFp

=

= •••

•,,,,:

1

and suppose that

[ ]

=

∂∂

∂∂

=

∂

∂=

∂

∂

∂

∂≡ ••••

nnnn

n

n

n

xxxxxx

xxxxxx

xxxxxx

xxx

n

T

x

T

xx

FFF

FFF

FFF

FFF

x

x

Fxx

F

xF

21

21212

12111

21,,,

1

is nonsingular for t ϵ [t0, tf ] (regular problem), then we can solve ) (locally, because of the Implicit Function Theorem) as a function of , by usingLegendre’s Dual Transformation.

( )tx•

( ) ( )tptxt ,,



Canonical Form of Euler-Lagrange Equations (continuous – 1)

12.1 Legendre’s Dual Transformation


Let consider a function of n variables xi, m variables, and time t: ii xy ≡

( )mn yyxxtF ,,,,,, 11

and introduce a new set of n variables pi defined by the transformation:

niy

Fp

ii ,,2,1: =

∂∂=

We can see that for t ϵ (t0, tf )

∂∂∂

∂∂∂

∂∂∂

∂∂∂

+

∂∂

∂∂∂

∂∂∂

∂∂

=

mmnn

m

nnn

n

n dx

dx

dx

xy

F

xy

F

xy

F

xy

F

dy

dy

dy

y

F

yy

F

yy

F

y

F

dp

dp

dp

2

1

2

1

2

1

2

11

2

2

1

2

2

1

2

1

2

21

2

2

1

We want to replace the variables dyi (i=1,2,…,n) by the new variables dpi (i=1,2,…,n).We can see that the new n variables are independent if the Hessian Matrix

ni

njji

ni

njji xx

F

yy

F,,1

,,1

2,,1

,,1

2

=

=

=

=

∂∂∂=

∂∂∂

is nonsingular in the interval t ϵ (t0, tf ).

According to the Implicit Function Theorem (Appendix 1) we can obtain a unique function in the interval t ϵ (t0, tf ).( )pxtxy ii ,,: =


Canonical Form of Euler-Lagrange Equations (continuous –2)

Let define a new function H (Hamiltonian) of the variables t, xi, pi.

( )nm

n

iix

n

iii ppxxtHxFFypFH

i,,,,,,: 11

11

=+−=+−= ∑∑==

Then:

( ) ∑∑∑∑∑=====

∂∂−++

∂∂−

∂∂−=++

∂∂−

∂∂−

∂∂−=

n

ii

iiii

n

jj

j

n

iiiii

n

ii

i

n

jj

j

dyy

Fpdpydx

x

Fdt

t

Fdpydypdy

y

Fdx

x

Fdt

t

FdH

11111

But because ( )nm ppxxtHH ,,,,,, 11 =

∑ ∑∂∂+

∂∂+

∂∂=

= =

n

j

n

ii

ij

j

dpp

Hdx

x

Hdt

t

HdH

1 1

and because all the variations are independent we have:

;,,1;,,1&, mjx

F

x

Hni

y

Fp

p

Hy

t

F

t

H

jjii

ii =

∂∂−=

∂∂=

∂∂=

∂∂=

∂∂−=

∂∂

Now we can define the Dual Legendre’s Transformation from

( ) ( ) ( )yxtFyppxtHtoyxtFn

iii ,,,,,,

1

−= ∑=

by using

nip

Hxy

nix

F

y

Fp

iii

iii

,,2,1

,,2,1

=∂∂==

=∂∂=

∂∂=

The variables t, , and H are called Canonical Variables corresponding to the functional J.

x p



If we apply now the Legendre Transformation to and H we obtainp

( ) ( )yxtFpxtHypn

iii ,,,,

1

=−∑=

The Legendre Transformation is an involution, i.e. a transformation which is its own inverse.

( ) ( ) ( ) ( )∫

−

++−=

•f

f

t

t

T

x

t

tT dttxp

dt

dtxtxtFxdpdtHJ

0

0,, δδ

Let write δJ in terms of the Canonical Variables:

From this expression the necessary conditions such that δJ is zero are

( ) 00

=+− tT xdpdtH

( ) ( )

=

•txtxtFp

dt

dx ,,

( ) 0=+−ft

T xdpdtH

We found before that the necessary conditions such that δJ is zero for those admissible solutions passing through the points and are the Euler-Lagrange Equations:

( )*0*0

* , xtA ( )*** , ff xtB

( ) ( ) ( ) ( ) 0,,,, =

−

••

• txtxtFdt

dtxtxtF

xx

niFy

Fp x ,,2,1: ==

∂∂=Since the two equations are identical.




∂∂=⇒=

∂∂==

p

Hxni

p

Hxy

iii

,,2,1and

x

H

x

FFmj

x

F

x

Hx

jj ∂∂−=

∂∂=⇒=

∂∂−=

∂∂

:;,,1also

( ) ( )

=

•txtxtFp

dt

dx ,,we obtain

Canonical Euler-Lagrange Equation or Hamilton’s Equations

x

H

td

pd

p

H

td

xd

∂∂−=

∂∂=

we can write the Euler-Lagrange Equations in the form

x

H

∂∂−=

( ) ( )

=

•

• txtxtFpx

,,:using

William Rowan Hamilton (1805-1855)



Canonical Euler-Lagrange Equation or Hamilton’s Equations

x

H

td

pd

p

H

td

xd

∂∂−=

∂∂=

Therefore we transformed the n Second Order Euler-Lagrange Differential Equations in 2 n First Order Hamilton’s Equations. Let find out when this problem is well-posed, i.e.:• a solution exists.• the solution is unique.• the solution depends continuously on the initial values.

First we must remember that the Hamilton’s Equation where derived only for regular problems: is nonsingular for t ϵ (t0, tf ). xxF

From the theory of First Order Differential Equations (see Appendix 3) the solution exists if

∂∂

∂∂

x

H

p

H, exists, are continuous in t ϵ (t0, tf ) (except a finite number of points).

This implies that is continuous and has continuous partial derivatives. ( ) ( )yxtFyppxtHn

iii ,,,,

1

−= ∑=

The solution is unique and depends continuously on the initial values if in addition the problem has 2 n defined boundary conditions.

Therefore the general solutions of the Hamilton’s Equations are therefore two vector parameters solutions . Those parameters are defined by 2n boundary conditions.

( ) ( ) Tn

Tn βββααα ,,,,, 11 == ( ) ( )βαϕ ,,ttx =




12.2 Transversality Conditions (Canonical Variables )

For other admissible variations we shall need to add the additional necessary conditions, such that δJ is zero, called Transversality Conditions Equations:


xiiiii

T

xiii ,00,,,,,, ==

+

−

••••

••

( ) ( )( ) ( ) fitxdpdttptxtH iT

iiii ,00,, ==+−or

(a) Suppose that the following relation defines the boundary:

( ) ( ) ( ) ( ) ( ) iitiiiii dttdttdt


then the Transversality Conditions Equations are:

( ) ( ) ( ) ( ) ( ) ( ) fitxttxtxtFtxtxtF iiiiiT

xiii ,00,,,, ==

−Ψ

+

•••

•

( ) ( )( ) ( ) ( ) fittptptxtH iT

iiii ,00,, ==Ψ+−or

(b) Suppose that ti and are not defined, and is not a function of ti , therefore d ti and are independent differentials and the Transversality Conditions will be:

ix ix

ixd

( ) ( ) ( ) ( ) ( )

( ) ( ) 0,,

0,,,,

=

=

−

•

•••

•

•

iiix

iiiiT

xiii

txtxtF

txtxtxtFtxtxtF ( ) ( )( )( ) 0

0,,

==

i

iii

tp

tptxtHor




12.3 Weierstrass-Erdmann Corner Conditions (Canonical Variables)

At the corners c we found:

( ) 00000

=

−+

−−

−=

+

•

−

•

+

•

−

•

••

ct

T

xt

T

xc

t

T

xt

T

xtxdFFdtxFFxFFJ

cccc

δ

Using the Canonical Variables we obtain:

( ) ( )[ ] ( ) ( )[ ] ( ) 00000 =−++−= +−+− ccT

cT

ccc txdtptpdttHtHJδ

(a) If they are apriori conditions at the corner like:

( ) ( ) ( ) ( ) ( ) cctccccc dttdttdt


then the necessary conditions at the corner are:

( ) ( ) ( ) ( ) ( ) ( )0000 ++−− −Ψ=−Ψ cctcT

cctcT tHttptHttp

(b) If they are not apriori conditions at the corner; i.e. the function is not apriori defined then dti and independent variables and

( ) ( )cc ttx Ψ=

cxd

( ) ( ) ( ) ( )0000 & +−+− == cccc tptptHtH




12.4 First Integrals of the Euler-Lagrange Equations

A First Integral of a system of differential equations is a function which has a constant value along each integral curve of the system.

We defined ( ) ( ) ( )yxtFypyxtFyppxtH Tn

iii ,,,,,,

1

−=−= ∑=

and( ) ( ) ( ) ( )

td

pd

p

pxtH

td

xd

x

pxtH

t

pxtH

td

pxtHdTT

∂

∂+

∂

∂+∂

∂= ,,,,,,,,

Using the Canonical Euler-Lagrange Equations

x

H

td

pd

p

H

td

xd

∂∂−=

∂∂=

we obtain from ( ) ( ) ( )yxtFypyxtFyppxtH Tn

iii ,,,,,,

1

−=−= ∑=

t

H

x

H

p

H

p

H

x

H

t

H

td

HdTT

∂∂=

∂∂

∂∂−

∂∂

∂

∂+∂

∂=

If does not depend on t explicitly, H doesn’t depend on t explicitly and isis constant over the optimal path, i.e. is a First Integral of the Euler-Lagrange Equations.

( )yxtF ,, ( )pxH ,

0=∂

∂=t

H

td

Hd



First Integrals of the Euler-Lagrange Equations (continue – 1)

Consider an arbitrary function ( )pxt ,,Φ=Φ

and compute

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )x

pxtH

p

pxt

p

pxtH

x

pxt

t

pxt

td

pd

p

pxt

td

xd

x

pxt

t

pxt

td

pxtd

TT

TT

∂∂

∂

Φ∂−∂

∂

∂

Φ∂+∂

Φ∂=

∂

Φ∂+

∂

Φ∂+∂

Φ∂=Φ

,,,,,,,,,,

,,,,,,,,

Define

Poisson Bracket

[ ] ( ) ( ) ( ) ( )x

pxtH

p

pxt

p

pxtH

x

pxtH

TT

∂∂

∂

Φ∂−∂

∂

∂

Φ∂=Φ ,,,,,,,,:,

Siméon Denis Poisson1781-1840

From which [ ]Httd

d,Φ+

∂Φ∂=Φ

is constant over the optimal path, i.e. is a First Integral of the Euler-Lagrange Equations iff1.F and Φ do not depend on t explicitly.2.[Φ,H] = 0.

( )px,Φ




12.5 Equivalence Between Euler-Lagrange and Hamilton Functionals

The Euler-Lagrange Functional [ ] ( )∫=ft

t

dtxxtFxJ0

,,

is optimized by the solution of the Euler-Lagrange Equations:

( ) ( ) ( ) ( ) 0,,,, =

−

••

• txtxtFdt

dtxtxtF

xx

We set ( ) ( )

=

•

• txtxtFpx

,,:

and the Hamiltonian ( ) ( ) ( ) ( ) xppxtHxxtFxxtFxppxtH TT +−=⇒−= ,,,,,,:,,

We define the Hamilton Functional

[ ] ( )[ ]∫ +−=ft

t

T dtxppxtHpxJ0

,,:,

Since: ( ) ( )xxtFxppxtH T ,,,, =+−

( )[ ] ( )∫∫ =+−ff t

t

t

t

T dtxxtFdtxppxtH00

,,,, William Rowan Hamilton

(1805-1855)



Equivalence Between Euler-Lagrange and Hamilton Functionals (continue – 1)

Hamilton Functional [ ] ( )[ ]∫ +−=ft

t

T dtxppxtHpxJ0

,,:,


Let find the Euler-Lagrange Equations for the Hamilton Functional

( )[ ] ( )[ ]

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

0,,,,

=

+−∂∂−+−

∂∂

=

+−∂∂−+−

∂∂

xppxtHptd

dxppxtH

p

xppxtHxtd

dxppxtH

x

TT

TT

or0

0

=+∂∂−

=−∂∂−

td

xd

p

H

td

pd

x

H

We recovered the Canonical Euler-Lagrange (Hamilton) Equations(William R. Hamilton 1835)




12.6 Equivalent Functionals

Two functionals are said to be equivalent if they have the same extremal trajectories.Suppose we have an arbitrary continuous and differentiable function: ( )xtS ,

Define ( ) ( ) xx

S

t

SxtS

td

dxxt

T

∂∂+

∂∂==Ψ ,:,,

Let compute ( ) xx

S

xt

Sxxt

x

2

22

,,∂∂+

∂∂∂=Ψ

∂∂

( ) xx

S

tx

S

xtd

d

x

Sxxt

x

2

22

,,∂∂+

∂∂∂=

∂Ψ∂⇒

∂∂=Ψ

∂∂

Since tx

S

xt

S

∂∂∂=

∂∂∂ 22

0=

∂Ψ∂−

∂Ψ∂

xtd

d

x Similar to Euler-Lagrange (E.-L.) Equations

The functionals[ ] ( )∫=

ft

t

dtxxtFxJ0

,, 0..

=

∂∂−

∂∂⇒

−

x

F

td

d

x

FLE

[ ] ( ) ( )[ ]∫ Ψ−=ft

t

dtxxtxxtFxJ0

,,,,~ 0

..

=

∂∂−

∂∂=

∂Ψ∂+

∂Ψ∂−

∂∂−

∂∂⇒

−

x

F

td

d

x

F

xtd

d

xx

F

td

d

x

FLE

and

have the same Euler-Lagrange Equations.



Equivalent Functionals (continue – 1)

Two functionals are said to be equivalent if they have the same extremal trajectories.

We can see that

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

( ) ( )[ ] ( )[ ]

[ ] ( )[ ] ( )[ ]00

00

,,

,,,,

,,,,,,,

~

0

00

txtStxtSxJ

txtStxtSdtxxtF

dttd

xtdSxxtFdtxxtxxtFxJ

ff

ff

t

t

t

t

t

t

f

ff

+−=

+−=

−=Ψ−=

∫

∫∫

The functionals and are called equivalent functionals.( )xJ ( )xJ~




12.7 Canonical Transformations

The functional [ ] ( )[ ]∫ +−=ft

t

T dtxppxtHpxJ0

,,:,

is optimized by the trajectories derived from the Canonical Euler-Lagrange (Hamilton) Equations

x

H

td

pd

p

H

td

xd

∂∂−=

∂∂=

Let perform a change of variables, from to according to px, px ~,~

( )( )pxpp

pxxx

,~~,~~

==

Since

∂∂

∂∂

∂∂

∂∂

=

pd

xd

p

p

x

p

p

x

x

x

pd

xd~~

~~

~

~

this is possible iff

( )( ) 0,

~,~:~~

~~

det ≠∂∂=

∂∂

∂∂

∂∂

∂∂

px

px

p

p

x

p

p

x

x

x



Canonical Transformations (continue – 1)

We look for transformations under which the Canonical Euler-Lagrange (Hamilton) Equations preserve their form. They are called Canonical Transformations.

x

H

td

pd

p

H

td

xd

~

~~

~

~~

∂∂−=

∂∂=

and optimize the functional [ ] ( )[ ]∫ +−=ft

t

T dtxppxtHpxJ0

~~~,~,~

:~,~

The two functional are equivalent if

( ) ( ) ( )xtStd

d

td

xdppxtH

td

xdppxtH TT ,

~~~,~,

~,, −+−=+−

From which ( ) ( ) ( )( )pxxtSdxdptdpxtHxdptdpxtH TT ~,~,~~~,~,~

,, −+−=+−

or ( )( ) ( ) ( )[ ] tdpxtHpxtHxdpxdppxxtSd TT ,,~~,~,~~~,~, −+−=



Canonical Transformations (continue – 2)

( )( ) ( ) ( )[ ] tdpxtHpxtHxdpxdppxxtSd TT ,,~~,~,~~~,~, −+−=

Fromxd

x

x

p

xxd

p

xpdpd

p

xxd

x

xxd ~

~~~~~

~~

~

11

∂∂

∂∂−

∂∂=⇒

∂∂+

∂∂=

−−

we can use instead of to obtain xx ~, px ~,~

( )( ) ( )

( ) ( )[ ] tdpxtHpxtHxdpxdp

xdx

Sxd

x

Std

t

SxxtSdpxxtSd

TT

TT

,,~~,~,~~

~~

~,,~,~,

−+−=

∂∂+

∂∂+

∂∂==

Finally we obtain:

( ) ( )t

SpxtHpxtH

x

Sp

x

Sp

∂∂=−

∂∂−=

∂∂= ,,

~~,~,~~




12.8 Caratheodory's Lemma

Consider the problem of minimizing the functional [ ] ( ) ( )∫

=

⋅fxt

xt

dttxtxtFxtJ,

, 00

,,,

defined in a simple connected domain Ω in the plane. Given an initial point suppose that one and only one extremal of exists between the initial point and everypoint .

( )xt, ( ) Ω∈00 , xt

[ ]xtJ ,

( ) Ω∈xt,

Assume that is continuous for all and all admissible . Define

⋅

xxtF ,, ( ) Ω∈xt, x

( )( )

( ) ( )∫

=

⋅

Ω∈

fxt

xtxt

dttxtxtFxtS,

,,

00

,,min:,

called the geodesic distance (Hamilton called it optical distance) or the Hamilton's characteristic function. Since we have one and only one extremal connecting with along a curve Γ Ω. is a single-valued function for all .ϵ

( ) Ω∈00 , xt ( ) Ω∈xt,( )xtS , ( ) Ω∈xt,

Field of Extremals Starting from

( ) Ω∈00 , xt



Caratheodory's Lemma (continue – 1)


( ) Ω∈00 , xt

Suppose that we calculate the functional along a neighbor curve Γε Ω that connects with . ϵ

[ ]xtJ ,

( )00 , xt ( )xt,

By definition of ( )xtS ,

[ ] ( ) ( ) ( ) ( ) ( )∫Γ

⋅

Γ Ω∈∀≥

−

=−

fxt

xt

xtdtxtStd

dtxtxtFxtSxtJ

,

, 00

,0,,,,,

ε

ε

where we use the fact that ( ) ( ) ( ) ( )xtwithxtconnectingdtxtStd

dxtS

fxt

xt

,,,, 00

,

, 00

ε

ε

Γ∀

= ∫

Γ

Along , if is continuous and differentiable relative to t and we haveΩ∈Γε ( )xtS , x

( )( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )εεεεεεε ,,,,,,,,,,,, txtxtStxtStxt

txtSx

txtSt

txtStd

d Txt

T

+=∂∂

∂∂+

∂∂=

we obtain:

[ ] ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )∫Γ

⋅

Γ Ω∈∀≥

−−

=−

fxt

xt

Txt xtdttxtxtStxtStxtxtFxtSxtJ

,

, 00

,0,,,,,,,,,,,

ε

εεεεεε

Since this is true for all and curve is the only optimal curve, the last equation is equivalent to:

( ) Ω∈xt, Ω∈Γ

( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )εεεεεεε ε ,&,,0,,,,,,,,, txtxttxtxtStxtStxtxtF Txt

Γ≠Γ∈∀>−−

⋅

( ) ( ) ( )( ) ( )( ) ( ) ( )( ) Γ∈=−−

⋅

txtfortxtxtStxtStxtxtF Txt ,0,,,, and



Caratheodory's Lemma (continue – 2)


( ) Ω∈00 , xt

Since at any point the change in the curve direction is defined by its slope we can write

( ) Ω∈xt,

( )xtX ,

Carathéodory's Lemma

If the Hamiltonian's characteristic function is defined on an admissible set of simple connected region of terminations Ω and if S is continuous and differentiable on it's arguments, then every element of an optimal trajectory that lies entirely in Ω is characterized by

( )xtS ,

( )xxt ,,

( ) ( ) ( ) ( ) ( )[ ] 0,,,,min,,,, =−−=−−

⋅

XxtSxtSXxtFxxtSxtSxxtF Txt

X

Txt

Constantin Carathéodory

(1873-1950)

R.E Bellman arrived to the Carathéodory's Lemma in a different way which will be described elsewhere. Return to Table of Content



12.9 Hamilton-Jacobi Equations

From the Carathéodory's Lemma we can derive the following

Theorem

(a) If is continuous for all , is not constraint and if is continuous and differentiable on it's arguments, then for every element of an optimal trajectory that lies entirely in Ω, except for the corners of , the following two conditions have to hold

⋅

xxtF ,, ( ) Ω∈xt, x ( )xtS ,

( )xxt ,,

x

( ) xxxtFxxtFxtS T

xt

−

=

⋅⋅,,,,,

( )

=

⋅xxtFxtS

xx ,,,

(b) Here is the uniquely determined slope of the optimal trajectory at the point and is viewed as a function of t and .

x ( ) Ω∈xt,

x

First Proof of (a)

Using the Carathéodory's Lemma, let define:

( ) ( ) ( ) 0,,,,:,, ≥−−=

⋅

XxtSxtSXxtFxxtE Txt

we see that if is not constraint, from the ordinary differential calculus, the necessary conditions for the minimum are, that on the optimal trajectory .

x

( )xxt ,,



Hamilton-Jacobi Equations (continue – 1)

First Proof of (a) (continue – 1)

( ) ( ) ( )[ ] ( ) ( ) 0,,,,,,, =−=−−∂∂=

∂∂

xtSxxtFxxtSxtSxxtFxx

Exx

Txt

and this gives ( )

=

⋅xxtFxtS

xx ,,,

( ) ( ) ( )[ ] ( ) 0,,,,,,22

2

≥=−−∂

∂=∂∂

xxtFxxtSxtSxxtFxx

Exx

Txt

This is the Legendre's Necessary Condition.

( )

=

⋅xxtFxtS

xx ,,, If we substitute in Carathéodory's Lemma

( ) ( ) ( ) ( ) ( )[ ] 0,,,,min,,,, =−−=−−

⋅


X

Txt

we obtain for the optimal trajectory ( )xxt ,,

( ) ( ) ( ) 0,,,,,,,,, =

−−

=−−

⋅⋅⋅

xxxtFxtSxxtFxxtSxtSxxtF Txt

Txt

If we substitute and in we obtain

( ) xxxtFxxtFxtS T

xt

−

=

⋅⋅,,,,, ( )

=

⋅xxtFxtS

xx ,,,

( ) ( ) ( ) 0,,,,:,, ≥−−=

⋅

XxtSxtSXxtFxxtE Txt

( ) ( ) ( ) ( ) 0,,,,,,,, ≥−−−=

⋅

xXxxtFxxtFXxtFxxtE Tx

Weierstrass' ConditionE is called Weierstrass' Excess Function.

End of Proof (a)




Second Proof of (a) (Geometrical)

Suppose that is a point on the optimal trajectory . ( )( ) Ω∈ii txt ,

Let plot( ) ( )

=

•txtxtF ,,η

( ) ( ) xxtSxtS TxtS

,, +=η

•= xξas a function of and the hyper-plane

From Carathéodory's Lemma those two functions intersect at and thehyper-plane must stay on one side of , therefore it is tangent to it.

( )( ) Ω∈ii txt ,

( ) ( )

=

•txtxtF ,,η




Second Proof of (a) (Geometrical) (continue – 1)

On the other side the hyper-plane tangent at is given by ( ) ( ) ( )

==

••

iiiiii txtxtFtx ,,,ηξ

( ) ( ) ( ) ( ) ( )

+

−

=

•••

iiiiiiiT

xT txtxtFtxtxtxtF ,,,, ξη

Since ηT ≡ ηS we must have

( ) xxxtFxxtFxtS T

xt

−

=

⋅⋅,,,,, ( )

=

⋅xxtFxtS

xx ,,,

We can see from Figure that E is given by the difference betweenand the tangent to hyper-plane. We can see that the condition for minimality is that the tangent hyper-plane remains bellow the surface .

( ) ( )

==

•XtxtxtF ,,η

( ) ( )

=

•txtxtF ,,η

End of Geometrical Proof of (a)

Geometric Representation of , and Weierstrass Condition ( ) ( )

=

•txtxtF ,,η ( ) ( ) xxtSxtS xtS

,, +=η




Proof of (b)

Let use the Canonical Forms. Start with definition

( )xxtFx

Fp x

,,: =

∂∂=

If the problem is regular, i.e. is nonsingular, we proved, that according to the Implicit Function Theorem we obtain a unique function in the interval t ϵ (t0, tf ).

( )xxtF xx

,,

( )pxtxx ,,: =

End of Proof of (b)

(b) Here is the uniquely determined slope of the optimal trajectory at the point and is viewed as a function of t and .

x ( ) Ω∈xt,

x

Theorem (continue)

Note

( )

=

⋅xxtFxtS

xx ,,, ( )xxtFx

Fp x

,,: =

∂∂=Using and

( ) ( ) ( )xtppxtSxxtFx

Fp xx ,,,,: =⇒==

∂∂=

End Note




We defined, also, the function H (Hamiltonian) of the variables t, px,

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

( )( )xtHpxtHxxxtFxxtFxxxtpxxtFH

xtpppxtxxT

xT ,

~,,,,,,,,,,:

,,,: ==

==+−=+−=

If we compare this expression with , we obtain:( ) xxxtFxxtFxtS T

xt

−

=

⋅⋅,,,,,

( ) ( )pxtHxtS t ,,, −=

If we compare with , we obtain:( )xxtFx

Fp x

,,: =

∂∂= ( )

=

⋅xxtFxtS

xx ,,,

( ) pxtS x =,

Therefore the Carathéodory's Lemma equation

( ) ( ) ( ) ( ) ( )[ ] 0,,,,min,,,, =−−=−−

⋅


Xxt

can be rewritten as

( ) ( ) 0,,, =+ xt SxtHxtS Hamilton-Jacobi Equation



1804-1851

The Hamilton-Jacobi Equation is a Partial Differential Equation in which is in general nonlinear.( )xtS ,




Jacobi’s Theorem


1804-1851

Let be a general solution of the Hamilton-Jacobi equation:( )α,, xtS

( ) ( ) 0,,,,,, =

∂∂+

∂∂ αα xt

x

SxtHxt

t

S

depending on the parameters ( )nT ααα ,,1 =

Assume also that ( ) 0,,detdet2

≠

∂∂

∂= ααα xt

x

SS x

Let n arbitrary constants.( )nT βββ ,,1 =

The two-parameter family of solutions of the Hamilton Equations

( ) ( )βαβα ,,,,, tpptxx ==

x

H

td

pd

p

H

td

xd

∂∂−=

∂∂=

are obtained from

( ) βαα

=∂∂

,, xtS

together with

( )α,, xtx

Sp

∂∂=



Proof of Jacobi’s Theorem


1804-1851

Since det Sαx ≠ 0 using the Implicit Function Theorem (see Appendix 1) we can use to uniquely find as a function of ( ) βα

α=

∂∂

,, xtS x ( )βα ,,t

( ) ( )βαβαα α

,,,,

Im

0dettxxxt

S TheoremFunction

plicit

S x

=⇒=∂∂

≠

Substitute this back in and differentiate with respect to t ( ) βαα =,, xtS

( )( ) 0,,,,22

=∂∂

∂+∂∂

∂=

∂∂

td

xd

x

S

t

Stxt

S

td

d

αααβα

α

Now, take the partial differential of the Hamilton-Jacobi equation with respect to α

( ) ( ) 0,,,,,,22

=∂∂

∂∂∂+

∂∂∂=

∂∂

∂∂+

∂∂

∂∂

xS

H

x

S

t

Sxt

x

SxtHxt

t

S

ααα

αα

α

If we use in this equation the fact that Sαt = Stα and , we obtainxSp =

( ) ( ) 0,,,,,,22

=∂∂

∂∂∂+

∂∂∂=

∂∂

∂∂+

∂∂

∂∂

p

H

x

S

t

Sxt

x

SxtHxt

t

S

ααα

αα

αtherefore

02

=

∂∂−

∂∂∂

p

H

td

xd

x

S

α

+

-

Since det Sαx ≠ 0 the previous equation is satisfied only if

0=∂∂−

p

H

td

xd



Proof of Jacobi’s Theorem (continue – 1)


1804-1851

( )α,, xtx

Sp

∂∂=Let differentiate with respect to t

( )p

H

xx

S

xt

S

td

xd

xx

S

xt

Sxt

x

S

td

d

td

pd

∂∂

∂∂∂+

∂∂∂=

∂∂∂+

∂∂∂=

∂∂=

2222

,, α

Now, take the partial differential of the Hamilton-Jacobi equation with respect to x

( ) ( ) 0,,,,,,22

=∂∂

∂∂∂+

∂∂+

∂∂∂=

∂∂

∂∂+

∂∂

∂∂

xS

H

xx

S

x

H

tx

Sxt

x

SxtH

xxt

t

S

xαα

If we use in this equation the fact that Sαt = Stα and , we obtainxSp =

x

H

p

H

xx

S

xt

S

∂∂−=

∂∂

∂∂∂+

∂∂∂ 22

x

H

td

pd

∂∂−=

We obtain

q.e.d.




Example: Geometrical Optics and Fermat’s Principle

( ) ( ) ( ) 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF ++=

+

+=


Define ( )[ ]

( )[ ] 2/122

2/122

1

,,:

1

,,:

zy

zzyxn

z

Fp

zy

yzyxn

y

Fp

z

y

++=

∂∂=

++=

∂∂=

Adding the square of those two equations gives

( ) ( ) ( )2222222 1 zynzypp zy +=+++

from which ( ) ( )222

2221

zy ppn

nzy

+−=++

Substitute this equation in that of F

( ) ( )222

2

,,,,zy

zy

ppn

nppzyxF

+−=

( )

( )222

222

zy

z

zy

y

ppn

pz

ppn

py

+−=

+−=

solve for zy ,



Example: Geometrical Optics and Fermat’s Principle (continue – 1)

Define the Hamiltonian

( ) ( ) ( ) ( ) ( )( ) ( )222

222

2

222

2

222

2

,,

,,,,:,,,,

zy

zy

z

zy

y

zy

zyzyzy

ppzyxn

ppn

p

ppn

p

ppn

nzpypppzyxFppzyxH

+−−=

+−+

+−+

+−−=++−=

The canonical equations are

( )

( )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

zn

n

z

H

xd

pd

ppn

yn

n

y

H

xd

pd

+−

∂∂

−=∂∂−=

+−

∂∂

−=∂∂−=

We recover the previous equations.

We can also see that if n is constant, than H is not an explicit function of x, y, z and is also constant since from previous equations both py and pz are constant.


136

[1] C. Carathéodory, “Calculus of Variations and Partial Differential Equations of the First Order”, Part I and Part II, Holden-Day Inc, 1965, English translation from German 1935

References


[2] O. Bolza, “Lectures on the Calculus of Variations”, Dover Publications, New York, 1961, Republication of a work published by Univ. of Chicago 1904

[3] G.A. Bliss, “Lectures on the Calculus of Variations”, Univ. of Chicago Press, Chicago, 1946

[4] W.S. Kimball, “Calculus of Variations, by Parallel Displacement”, Butterworths Scientific Publications, 1952

[5] L.E. Elsgolc, , “Calculus of Variations”, Pergamon Press, Addison-Wesley, 1962

[6] I.M. Gelfand, S.V. Fomin, “Calculus of Variations”, Prentice-Hall, 1963

[7] G. Leitmann, “Calculus of Variations and Optimal Control, An Introduction”, Plenum Press, 1981

[8] H. Sagan, “Introduction to Calculus of Variations”, Dover Publication, New York, 1969

[9] D. Lovelock, H. Rund, “Tensors Differential Forms, and Variational Principles”, Dover Publication, New York, 1975, 1989

[10] J.L. Troutman, “Variational Calculus with Elementary Convexity”, Springer-Verlag, 1983

[11] R. Weinstock, “Calculus of Variations with Applications to Physics and Engineering”, Dover Publication, New York, 1952, 1974

137

SOLOReferences Calculus of Variations

138

SOLO

References (continue – 1)



February 20, 2015 139

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

SOLO

Appendix: Useful Mathematical Theorems


The following mathematical theorems are useful in the Calculus of Variations:

•Implicit Function Theorem

•Heine-Borel Theorem

•Ordinary Differential Equations Theorems

•Euler-Lagrange Ordinary Differential Equations Theorems

•Partial Differential Equations of the First Order Theorems

SOLO

Appendix 1: Implicit Functions Theorem


Let continuous functions on a domain D of the parameters:( ) ( ) 0,,:, 1 == Tnffuxf

( ) nT

n Rxxx ∈= ,,: 1

( ) mT

m Ruuu ∈= ,,: 1

having continuous first partial derivatives

∂∂

∂∂

∂∂

∂∂

=∂∂=

∂∂

∂∂

∂∂

∂∂

=∂∂=

m

nn

m

u

n

nn

n

x

u

f

u

f

u

f

u

f

u

ff

x

f

x

f

x

f

x

f

x

ff

,,

,,

:

,,

,,

:

1

1

1

1

1

1

1

1

Consider an interior point of the domain of definition of for which ( )00 ,uxP ( )uxf ,

( ) ( )( ) 0, 00 =uxf

and the following Jacobian is nonzero: ( ) ( )( )( ) ( )( )

( ) ( )( )

0

,,

,,

00

00

00

,1

1

1

1

,,

≠

∂∂

∂∂

∂∂

∂∂

=∂∂=

uxn

nn

n

uxuxx

x

f

x

f

x

f

x

f

x

ff

SOLO

Appendix 1: Implicit Functions Theorem (continue – 1)


Then there exists a certain neighborhood of this point a unique system of continuous functions that satisfies the conditions:

( )( ) ( ) δδ ≤−= 00 : uuuuN

( )ux ϕ=

( ) ( )( )00 ux ϕ=a

( )( ) ( )( )00:, uNuuuf δϕ ∈∀=b

u∂∂ϕc exists in the same neighborhood, are continuous and are found by solving:

( )( ) ( ) ( )0

,,:

, =∂

∂+∂∂

∂∂=

u

uxf

ux

uxf

ud

uufd ϕϕ

If are of class C(p) in than is also of class C(p). u( )uxf , ( )uϕ

Proof

Existence ( ) ( ) 0,:,

1

2 ≥= ∑=

n

ii uxfuxF Define the scalar

and the neighborhoods:( )( ) ( ) ( )( ) ( )

≤−=

≤−=

δ

ρ

δ

ρ

00

00

:

:

uuuuN

xxxxN

( )0u

( )0x

( ) δ+0u( ) δ−0u

( ) ρ+0x

( ) ρ−0x

( ) σ−0x

( ) σ+0x

u

( ) 0, =uxfD

SOLO



Proof of Existence (continue – 1)

( )0u

( )0x

( ) δ+0u( ) δ−0u

( ) ρ+0x

( ) ρ−0x

( ) σ−0x

( ) σ+0x

u

( ) 0, =uxfD

Let be the set of boundary points of defined as ( )( )0xN ρ∂ ( )( )0xN ρ

( )( ) ( ) ρρ =−=∂ 00 : xxxxN

( ) ( )( ) 0, 00 =uxfAccording to we have

( ) ( )( ) ( ) ( )( ) 0,:,1

00200 == ∑=

n

ii uxfuxF

Let choose such thatu ( )( )0uNu δ∈

Since is continuous and compact (bounded and closed) in and for and chosen .

( )uxf ,( )( )0xN ρ

( )( )0uNδ( )( )0xNx ρ∂∈ ( )( )0uNu δ∈

( ) ( ) 0,,1

2 >= ∑=

n

ii uxfuxF

and attains its minimum value m on ( )( )0xN ρ∂ ( )( )( )( )

( )uxFm

uNu

xNx,inf

0

0

δ

ρ∈

∂∈=

Since we can find a σ < ρ such that and( ) ( )( ) 0, 00 =uxF ( )( ) ( )( )00 xNxN ρσ ⊂

( )( )( )( )

( ) ( ) ( )( )0

2,

2,inf

0

0xNxfor

muxF

muxF

uNuxNx

σ

δ

σ

∂∈>→=∈

∂∈

SOLO



Proof of Existence (continue – 2)

( )0u

( )0x

( ) δ+0u( ) δ−0u

( ) ρ+0x

( ) ρ−0x

( ) σ−0x

( ) σ+0x

u

( ) 0, =uxfD

Also since we can diminish σ < ρ such that( ) 0, 00 =uxF

( ) ( )( )0

2, xNinsidexfor

muxF σ<

Since on the boundary of and , at some point inside , the scalar attains its minimum inside and

( )2

,m

uxF > ( )( )0xNσ

( )2

,m

uxF < ( )( )0xNσ

( )uxF , ( )( )0xNσ

( ) ( ) ( ) ( ) ( )0

,,

,,,

1 11 1

=

∂

∂⋅=∂

∂⋅= ∑ ∑∑∑= == =

n

jj

n

i j

ii

n

i

n

jj

j

ii dx

x

uxfuxfdx

x

uxfuxfuxFd

This must hold for each , therefore ( )( )0xNxd j σ∈ ( ) ( )nj

x

uxfuxf

n

i j

ii ,,10

,,

1

=∀=∂

∂⋅∑=

( ) ( )

( ) ( )

( )

( )( ) ( ) 0,

,

,

,

,,

,,

1

1

1

1

1

=

∂

∂=

∂∂

∂∂

∂∂

∂∂

uxfx

uxf

uxf

uxf

x

uxf

x

uxf

x

uxf

x

uxf

n

n

nn

n

or

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

( ) ( )( )

0

,,

,,

00

00

00

,1

1

1

1

,,

≠

∂∂

∂∂

∂∂

∂∂

=∂∂=

uxn

nn

n

uxuxx

x

f

x

f

x

f

x

f

x

ff

Since it follows that the previous equality is possible only if: ( ) 0, =uxf

We proved that for every , exist at least one such that . ( )( )0uNu δ∈ ( )( )0xNx σ∈ ( ) 0, =uxf

SOLO



Proof of Uniqueness

( )0u

( )0x

( ) δ+0u( ) δ−0u

( ) ρ+0x

( ) ρ−0x

( ) σ−0x

( ) σ+0x

u

( ) 0, =uxfD

Suppose that for a given we have two values such that .

( )( )0uNu δ∈( ) ( ) ( )( )021 , xNxx σ∈ ( )( ) ( )( ) 0,, 21 == uxfuxf

We can write this equation as( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )uxxxfuxxxfuxfuxf niniii ,,,,,,,,,, 11

21

122

22

112 −=−

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )uxxxxfuxxxxf

uxxxxfuxxxxf

uxxxxfuxxxxf

uxxxxfuxxxxf

nini

nini

nini

nini

,,,,,,,,,,

,,,,,,,,,,

,,,,,,,,,,

,,,,,,,,,,

113

12

11

213

12

11

113

12

11

223

12

11

223

12

11

223

22

11

223

22

11

223

22

21

−+

−+

−+

−=

( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )( )12

1

0

12212111

2111

2211

,,,,,

,,,,,,,,,,

jjijjjnjjjj

i

njinji

xxBxxduxxxxxx

f

uxxxfuxxxf

−=−−+∂∂=

−

∫ θθ

We can write

where ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )∫ −+∂∂

= +−

1

0

221

12111

11

21 ,,,,,,,:,, θθ duxxxxxxxx

fuxxB njjjjj

j

iij

We can see that ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( )( )uxx

fduxxxxx

x

fuxxB

j

injjj

j

iij ,,,,,,,,,, 1

1

0

221

111

11

11

∂∂

=∫ ∂∂

= +− θ

SOLO



Proof of Uniqueness (continue – 1)

( )0u

( )0x

( ) δ+0u( ) δ−0u

( ) ρ+0x

( ) ρ−0x

( ) σ−0x

( ) σ+0x

u

( ) 0, =uxfD

Therefore we can write( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )

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

,,,,,,,,,,

1

1221

112

11

222

21

12

=−=

−=−

∑=

n

jjjij

niniii

xxuxxB

uxxxfuxxxfuxfuxf

Those equations hold for i= 1,2,…,j, therefore we obtain ( ) ( )

( ) ( )

( ) ( )

( ) ( )( )[ ] ( ) ( )( ) 0,, 1221

12

12

22

11

21

21

22221

11211

=−=

−

−

−

xxuxxB

xx

xx

xx

BBB

BBB

BBB

ij

nnnnnn

n

n

Since we have( )

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

( ) ( )( )

0

,,

,,

,,

00

0000

,1

1

1

1

,

000

,≠

∂∂

∂∂

∂∂

∂∂

=∂∂==

uxn

nn

n

ux

ijuxx

x

f

x

f

x

f

x

f

x

fuxxBf

and is continuous in , if we choose σ small enough we can assure that and the equation is satisfied only if .

( )uxf x , ( )ux,( ) ( )( )[ ] 0,det 21 ≠xxBij

( ) ( )( )[ ] ( ) ( )( ) 0, 1221 =− xxxxBij( ) ( )12 xx =

This proves that for every , exist at least one (σ small enough) such that we can write .

( )( )0uNu δ∈ ( )( )0xNx σ∈

( ) 0, =uxf ( )ux ϕ=

SOLO



Proof of Continuity

( )0u

( )0x

( ) δ+0u( ) δ−0u

( ) ρ+0x

( ) ρ−0x

( ) σ−0x

( ) σ+0x

u

( ) 0, =uxfD

By taking the derivative of with respect to , we obtain ( )( )uuf ,ϕ u

( )( ) ( ) ( )0

,,, =∂

∂+∂∂

∂∂=

u

uxf

ux

uxf

ud

uufd ϕϕ

from which we can see that exists in the same neighborhood, and are continuous because and exist and are continuous.

u∂∂ϕ

( )x

uxf

∂∂ ,

( )u

uxf

∂∂ ,

If are of class C(p) in than is also of class C(p). u( )uxf , ( )uϕ

q.e.d.

SOLO



Extension of the Implicit Functions Theorem to all Domain of Definition of ( ) ( ) 0,,:, 1 == Tnffuxf

We found the unique function that satisfies the conditions:( )ux ϕ=

( ) ( )( )00 ux ϕ=( )( ) ( )( )00, uNuuuf δϕ ∈∀=

( )uxf , We want to extend this result to the domain D where the functions are defined provided that

( ) ( ) 0,&,0

,,

,,

1

1

1

1

=∈∀≠

∂∂

∂∂

∂∂

∂∂

=∂∂= uxfDux

x

f

x

f

x

f

x

f

x

ff

n

nn

n

x

a)

b) D is compact (closed and bounded)

For this purpose we use the Heine-Borel Theorem ( see Appendix 2 for proof):

A compact domain S can be covered by a given finite number of open covering sub-domains.

SOLO



Extension of the Implicit Functions Theorem to all Domain of Definition of (continue – 1)

( ) ( ) 0,,:, 1 == Tnffuxf

We are using the following procedure:

Choose ( ) ( )( )12 uNu δ∈

Find the unique ( ) ( )( ) ( )( )( )01

11

1 xNux σϕ ∈=

2) Define a neighborhood ( )( )1uNδ

1) Choose

Find the unique ( ) ( )( ) ( )( )( )12

22

2 xNux σϕ ∈=

Choose ( ) ( )( )1−∈ NN uNu δ

N) Define a neighborhood ( )( )NuNδ

Find the unique ( ) ( )( ) ( )( )( )1

2−∈= NNN xNux Nσϕ

( )0u

( )0x

( )1u

( ) 0, =uxfD

( )1x

According to Heine-Borel Theorem the compact domain D for whichis covered by a finite number N of sets.

( ) 0, =uxf


150

Heinrich Eduard Heine

( 1821 - 1881)

Félix Édouard Justin Émile Borel

(1871 –1956)

Appendix 2: Heine–Borel Theorem



Proof of Heine–Borel Theorem

Both S and the sub-domains Ti are given beforehand, since it is no hard to pick out a single open interval which completely covers a bounded set S.

Let S be contained in the interval -N ≤ x ≤ N (S is bounded). Now divide this closed interval into two equal intervals(1) -N ≤ x ≤ 0(2) 0 ≤ x ≤ N

Any element x of S will belong to either (1) or (2). If the theorem is false, it will not be possible to cover the points of S in both (1) and (2), by a finite number of sub-domains of T, so the points of S in either (1) or (2) require an infinite covering. Assume that the elements of S in (1) still require an infinite covering, We subdivide this interval into two equal parts and repeat the above argument. In this way we construct a sequence of sets such that each Si is closed and bounded and such that the diameters of the

⊃⊃⊃⊃⊃ iSSSS 321

0lim →∞→ iiS

151

Heinrich Eduard Heine

( 1821 - 1881)

Félix Édouard Justin Émile Borel

(1871 –1956)

Appendix 2: Heine–Borel Theorem (continue – 1)



Proof of Heine–Borel Theorem (continue – 1)

Because the sub-domains are nested there exists a unique point Pwhich is contained in each Si. Since P is in Si, one of the open intervals of T, say Tp, will cover P. This Tp has a finite nonzero diameter so that eventually one of the Si will be contained in Tp, since . But by assumption all the elements of this Si require an infinite number of the sub-domains in T to cover them. This is a direct contradiction to the fact that a single Tp covers them. Hence our original assumption is wrong, and the theorem is proved.

0lim →∞→ iiS


152

Appendix 3: Ordinary Differential Equations (ODE)


is a normal system of ordinary differential equations.

Well-Posed Problems

A differential equation problem is well-posed if:

•A solution exists.

•The solution is unique.

•The solution depends continuously on the initial values. The well-posedness requires proving theorems of existence (there is a solution),

uniqueness (there is only one solution), and continuity (the solution depends continuously on the initial value).

( )

givenConditionsInitialn

nitxxxftd

xdnii

i ,,1,,,,,1 ==

153

Appendix 3: Ordinary Differential Equations (ODE) (continue – 1)

SOLOCalculus of Variations

( )


nitxxxftd

xdnii

i ,,1,,,,,1 ==


Definitions:

• Solution of an ODE means an explicit solution xi = φi (t) defined in a region R.

• A neighborhood of a point is defined as a sphere contained this point as is center, satisfying (r some positive constant)

( )00 , tx

( ) ( ) 220

20 rttxx <−+−

• A point is an interior point of R if it contains a neighborhood that is wholly contained in R. It is an exterior point of R if it contains a neighborhood

that doesn’t contain any point of R.

( )00 , tx

• A point is a boundary point of R if every neighborhood has both interior and exterior points in R.

( )00 , tx

( ) ( )txxxtx ni ,,,,,:, 1 =•

Neighborhood

InteriorPoint

BoundaryPoint

154

Appendix 3: Ordinary Differential Equations (ODE) (continue – 2


( )


txftd

xd,=


Definitions continue – 1):

InteriorPoint

BoundaryPoint

• Limit of a Sequence is equivalent to for each ε >0 exist an integer N such that .

( ) ( )00 ,, txtxi

ii

∞→⇒

( ) ( ) Nittxx ii >∀<−+− 220

20 ε

• Limit of a Function in a region R. We say that if for every sequence in R such that converges to the same limit A.

( )txf ,

( ) ( )( ) Atxf

txtx=

→,lim

00 ,,( ) ( )00 ,, txtx

i

ii

∞→⇒

( )txf ,

• A Function is Continuous at a point if ( )txf , ( )00 , tx( ) ( )

( ) ( )00,,

,,lim00

txftxftxtx

=→

• Uniform Convergence

A sequence Uniform Converge to a limit if for each positive ε there is an integer N, independent on , such that

( )ii tx , ( )tx,

x( ) ( ) Nitxtxi >∀<− ε

155



Derivation of a Solution of the Ordinary Differential Equations.

( ) givenConditionsInitialntxftd

xd,=

Theorem I (Existence) (Cauchy-Peano) If the functions are continuous in a closed and bounded region R of

n+1 dimensional space , then through each interior point of the region there exists at least one continuously derivable curve which is defined in an interval |t-t0| < a

( )txf ,( )tx, ( )00 , tx

( )txx =

Giuseppe Peano1858 - 1932

Augustin-Louis Cauchy

1789 –1857

ProofSince R is closed and bounded, the functions are uniformly bounded in R and there exists a positive number M such that

( )txf ,

( ) ( ) RtxniMtxf i ∈∀=< ,,,1,

( ) ( )( ) ( )

( ) ( )1111

121112

010001

,

,

,

−−−− −+=

−+=

−+=

NNNNNN tttxfxx

tttxfxx

tttxfxx

A solution can be build by dividing the interval |t-t0| < a in small intervals t0 < t1 <…<tj<…< tN < t0+a & |tj+1-tj| <δ. Start from and perform ( )00 , tx

InteriorPoint

156



Derivation of a Solution of the Ordinary Differential Equations (continue – 1).

Proof (continue -1)This construction defines a polygon, that for small enough, approximate a solution ( )txx =

( ) ( )∫+=t

dxfxtx0

0 , ττ

We can see that ( ) ( ) ( )

( ) ( ) ( ) ( )001211

01211012110

ttMttMttMttM

xxxxxxxxxxxxxx

jjjjj

jjjjjjjjj

−=−++−+−≤

−++−+−≤−++−+−=−

−−−

−−−−−−

Therefore the solution is bounded in the region ( )00 ttMxx −≤−

We can obtain also this result from

( ) ( ) ( ) ( )0

000

0 ,, ttMdMdxfdxfxtxttt

−=≤≤=− ∫∫∫ τττττInterior

Point

Slope +M

Slope -MThe Theorem that proves only “Existence”,

not “Uniqueness”, was first discovered by Giuseppe Peano in 1890.

This solution is due to Augustin Cauchy and the polygon is called “Cauchy Polygon”.

157



Uniqueness of a Solution of the Ordinary Differential Equations.

Rudolf Lipschitz(1832 – 1903)

( )


txftd

xd,=

To obtain Uniqueness of a Solution of the Ordinary Differential Equations we need in addition to the condition

the condition

( ) ( ) RtxniMtxf i ∈∀=< ,,,1,

( ) ( ) ( ) ( ) Rtxtxconstknixxktxftxf iiii ∈∀==−<− ,~&,.,,,1~,~, Lipschitz Conditio

nUnder suitable hypothesis the Mean Value Theorem, gives ( ) ( ) ( ) ( ) xxxx

x

tftxftxf i

ii~~,

,~, ≥≥−∂

∂=− ηη

This means that we can replace the Lipschitz Condition with the more restrictive condition

( ) ( ) Rtxconstknikx

txfii

i ∈∀==≤∂

∂,.,,,1

,

158




( )txftd

xd,=

Theorem (Existence and Uniqueness) (Picard–Lindelöf Theorem)

In a Bounded region R , let be continuous and satisfying ( )txf ,

( ) ( ) RtxniMtxf ii ∈∀=< ,,,1,

as well as ( ) ( ) ( ) ( ) Rtxtxconstknixxktxftxf iiii ∈∀==−<− ,~&,.,,,1~,~,

or . ( ) ( ) Rtxconstknikx

txfii

i ∈∀==≤∂

∂,.,,,1

,

The ODE has one, and only one, solution containing the internal point .The solution lies in the shadow region (defined by ) and can be extended to the right and the left of until it meets the boundary of R.

( )txx = ( )00 , tx

InteriorPoint

Slope +M

Slope -M( ) ii Mtxf <,

0x

Lipschitz Conditio

n

Those conditions are “sufficient” but not “necessary” for existence and uniqueness of solutions. There are cases when those conditions are not satisfied and a unique solution exists.

159




Proof of Theorem (Existence and Uniqueness) (Picard–Lindelöf Theorem)

We introduce the Picard Successive Approximations, called also the Picard Iterations, after the French Mathematician Charles Picard:

Slope +M

Slope -M

ShadedRegion

Charles Émile Picard 1856 - 1941

( )

( )

( )txftd

xdx

txftd

xdx

txftd

xdx

nn

n ,:'

,:'

,:'

1

12

2

01

1

−==

==

==

Ernst Leonard Lindelöf1870 - 1946

Let show first that all those curves are defined fora ≤ t ≤ b and lie in the Shaded Region defined by .

( )txfx ii ,' 1−=

00 ttxx −≤−

Assume that the graph is defined for a ≤ t ≤ b and lies in the Shaded Region, then for a ≤ t ≤ b ;hence , and

( )txx n=( ) mtxf n ≤,

( ) mtxfx nn ≤=+ ,' 1

( ) ( ) 011000

'' ttmdttxdttxxxt

t n

t

t n −≤≤=− ∫∫ ++

Therefore lies in the Shadow Region.( )txx n=

160




Proof of Theorem (Existence and Uniqueness) (continue – 1)

Slope +M

Slope -M

ShadedRegion

Let estimate now the difference between two successive approximations.Define: ( ) ( ) ( ) ( ) ( ) ( ) 0&:

0

01

0

001 =−=−= −− txtxtwtxtxtw nnnnnn

( ) ( )txfxandtxfx nnnn ,',' 11 −+ ==We have

Subtract and take the norm

( ) ( ) 111 ,,'' −−+ −≤−=− nn

Lipschitz

nnnn xxktxftxfxx

or ( ) ( ) ( )twktwtwtd

dn

Lipschitz

nn ≤= ++ 11 '

( ) ( ) ( ) ( ) 0000001100

, ttmtdmtdtxftxtxtwt

t

t

t−=≤=−= ∫∫Let compute:

( ) ( ) ( )twktwtd

dtwk nnn ≤≤− +1 ( ) ( ) ( ) 00121

1

ttmktwktwtd

dtwk

n

−=≤≤−⇒=

Integrating from t0 to t (since the integrand doesn’t change sign the inequalities are preserved after integration):

( ) ( ) ( )22

2

00000221

2

00

1

0

ttmkttmktwtwtwk

ttmk

t

t

n −=−≤−≤−=

−−⇒ ∫

=

161





Slope +M

Slope -M

ShadedRegion

( ) ( ) ( )twktwtd

dtwk nnn ≤≤− +1Start from:

( ) ( )22

2

00

0

022

2

00

1 ttkmtwtw

ttkm

n −≤−≤

−−⇒

=

( ) ( )!3!3

3

020

0

033

3

020

2 ttkmtwtw

ttkm

n −≤−≤

−−⇒

=

( ) ( )!!01

0

0

001

0

1

n

ttkmtwtw

n

ttkm

n

nnn

n

nn −

≤−≤−

−⇒ −−−

( ) ( ) ( )22

2

020232

2

020

2 ttkmtwktw

td

dtwk

ttkm

n −=≤≤−=

−−⇒

=

Integration

Induction

162




Proof of Theorem (Existence and Uniqueness) (continue – 2) Slope +M

Slope -M

ShadedRegion

We obtained: ( ) ( )!!01

0

0

001

0

1

n

ttkmtwtw

n

ttkm

n

nnn

n

nn −

≤−≤−

−⇒ −−−

Therefore: ( ) ( )!

00

n

ttk

k

mtw

n

n

−≤

Let compute: ( ) ( ) ( ) ( )[ ] ( ) ( )[ ] ( ) ( )[ ] 11012110 wwwtxtxtxtxtxtxtxtx nnnnnnn +++=−++−+−=− −−−−

( ) ( ) ( ) ( )1!

00

1

00110 −≤

−≤+++≤− −

=− ∑ ttk

n

i

i

nnn ek

m

i

ttk

k

mwwwtxtx

( ) ( ) ( )1lim 000 −≤− −

∞→

ttkn

ne

k

mtxtxLet take the limit n→∞ of the previous expression:

Therefore the limit : ( ) ( )txtxnn

=∞→

lim

exists as an Uniform Limit on the interval a ≤ t ≤ b.

Using again Lipschitz Condition: ( ) ( ) ( ) ( ) btatxtxktxftxf nn ≤≤−≤− ,,

we obtain: ( ) ( ) btatxftxf nn

≤≤=∞→

,,limq.e.d. Existence

163




Proof of Theorem (Existence and Uniqueness) (continue – 3)Slope +M

Slope -M

ShadedRegionUniqueness:

Assume the existence of another solution, such that:

( ) ( ) ( )00~,,~

~txtxtxf

td

xd ==

( ) ( ) ( )001 ,, txtxtxftd

xdnn

n == −Use the sequence:

( ) ( ) ( ) 00000

,~~ ttmtdmtdtxftxtxt

t

t

t−=≤=− ∫∫Compute:

Subtracting those equations and taking the norm, we obtain:

( ) ( ) 11~,,~

~−− −≤−=− n

Lipschitz

nn xxktxftxftd

xd

td

xd

001

1 ~~

ttkmxxktd

xd

td

xdn

−≤−≤−⇒=

164





Slope +M

Slope -M

ShadedRegion

Uniqueness (continue – 1):

01

0

1 ~ttkm

td

xd

td

xdttkm

n

−≤−≤−−⇒=

( ) ( )2

~2

2

01

2

01 tt

kmtxtxtt

kmn −

≤−≤−

−⇒=

11~

~~

−− −≤−≤−− nn

n xxktd

xd

td

xdxxk

Integration

We obtained:

2~

~~

2

2

021

21

2

022 tt

kmxxktd

xd

td

xdxxk

ttkm

n −≤−≤−≤−−≤

−−⇒

=

Integration

!3~~~

!3

3

02121

3

022 tt

kmxxkxxxxktt

kmn −

≤−≤−≤−−≤−

−⇒=

( )( ) ( ) ( ) ( )

( )!1~

!1

1

0

1

0

+−

≤−≤+

−−⇒

++

n

ttk

k

mtxtx

n

ttk

k

mn

n

nn

Induction

165





Slope +M

Slope -M

ShadedRegion

Uniqueness (continue – 1):

We obtained: ( ) ( ) ( )( )!1

~1

0

+−

≤−+

n

ttk

k

mtxtx

n

n

We see that: ( ) ( ) 0~ ∞→⇒−n

n txtx

Therefore the limit : ( ) ( )txtxnn

~lim =∞→

is Unique and exists as an Uniform Limit on the interval a ≤ t ≤ b.

q.e.d.

166



Continuous Dependence of Solution of the Ordinary Differential Equations.

Thomas Hakon Grönwall(1877 – 1932)

We want to show that the Solution of the Ordinary Differential Equations depends continuously on the Initial Values. For this let state the:

Grönwall Inequality

Let u and v be continuous function satisfying u (x) > 0 and v (x) ≥ 0 on [a,b]. Let c ≥ 0 be a constant. If

( ) ( ) ( ) bxatdtvtucxvx

a≤≤+≤ ∫ ,

then

( ) ( ) bxatdtucxvx

a≤≤≤ ∫ ,exp

Prof of Grönwall Inequality :

First assume c > 0 and define ( ) ( ) ( ) bxatdtvtucxVx

a≤≤+= ∫ ,:

Then V (x) ≥ v (x), and since u and v are nonnegative, V (x) ≥ V (a) = c on [a,b].Moreover, V’(x) = u (x) v (x0 ≤ u (x) V (x). Dividing by V (x), we get

167




Thomas Hakon Grönwall(1877 – 1932)

Prof of Grönwall Inequality (continue – 1):

If we take c → 0+, we get v (x) ≤ 0, which is the same result obtained fromGrönwall Inequality with c = 0.

( ) ( )( ) bxaxV

xVxu ≤≤≥ ,

'

Integrate both sides of this equation, from a to x

( ) ( )( ) ( ) ( )( )cxVsVsdsV

sVsdsu

x

a

x

a

x

a/lnln

' ==≥ ∫∫

Since logarithmic and exponential function are increasing function with their argument, we can resolve and preserve the inequality

( ) ( ) bxasdsucxVx

a≤≤≤ ∫ ,exp

Since V (x) ≥ v (x) we proved that .( ) ( ) 0&,exp >≤≤≤ ∫ cbxatdtucxvx

a

q.e.d. Grönwall Inequality

168




Let use Grönwall Inequality to prove the following

Continuous Dependence of ODE on Initial Value

( ) ( ) hkexxtxtx 0000~,,~ −≤−φφ

( )tx ,~0φ

where k is any positive constant such that for all .Moreover as approaches , the solution approaches uniformly in .

kxf ≤∂∂ / ( ) 0, Rtx ⊂

0x0~x ( )tx ,0φ

htt ≤− 0

Let continuous functions on an open rectangle containing the point . Assume that for all sufficiently close to , the solution of the ODE exists on the interval and the graph lies within a closed region . Then, for

( ) xfandtxf ∂∂ /, ( ) dxcbtatxR <<<<= ,:,

( )00 , tx0

~x 0x( )tx ,~

0φhtt ≤− 0RR ⊂0

htt ≤− 0

169




Proof of Continuous Dependence of ODE on Initial Value:

The Solution of ODE satisfies the Integral Equation: ( )tx ,0φ

( ) ( )( ) httsdssxfxtxt

t≤−+= ∫ 0000 ,,,,

0

φφ

Similarly the Solution of ODE satisfies the Integral Equation: ( )tx ,~0φ


t≤−+= ∫ 0000 ,,,~~,~

0

φφ

Subtracting the second equation from the first gives:

( ) ( ) ( )( ) ( )( )[ ]∫ −+−=−t

tsdssxfssxfxxtxtx

0

,,~,,~,~, 00000 φφφφ

Assume that t > t0 and tacking the norm of both sides

( ) ( ) ( )( ) ( )( )∫ −+−≤−t

tsdssxfssxfxxtxtx

0

,,~,,~,~, 00000 φφφφ

kxf ≤∂∂ /Since or satisfies the Lipschitz Condition

( )txf ,

( )( ) ( )( ) ( ) ( )sxsxkssxfssxf ,~,,,~,, 0000 φφφφ −≤−

( ) ( ) ( ) ( )∫ −+−≤−t

tsdsxsxkxxtxtx

0

,~,~,~, 00000 φφφφWe obtain:

170




Proof of Continuous Dependence of ODE on Initial Value (continue – 1):

( ) ( ) ( ) ( )∫ −+−≤−t

tsdsxsxkxxtxtx

0

,~,~,~, 00000 φφφφWe obtained:

Use Grönwall Inequality:

( ) ( ) ( )( ) ( ) 0&0,0

,

≥≤≤≥>

≤≤+≤ ∫cbxaonxvxu

bxatdtvtucxvx

a ( ) ( ) bxatdtucxvx

a≤≤≤ ∫ ,exp

( ) ( ) ( ) ( ) 0~:&0,~,:,0: 000 ≥−=≥−=>= xxctxtxtvktu φφwith:

( ) ( ) ( ) hkhtt

ttkt

texxexxsdkxxtxtx ~~exp~,~, 00000

00

0

−≤−=−≤−≤−

−∫φφ

For t < t0, we can use t0 – t instead of t.

( )tx ,~0φWe see that as approaches , the solution approaches

uniformly in . 0

~x 0x ( )tx ,0φhtt ≤− 0

q.e.d.

171



Let use Grönwall Inequality to prove the following

Continuous Dependence on of Solutions of ODE

( ) ( ) ( ) RtxtxftxF ⊂∀≤− ,,,, ε

Continuous Dependence on of Solution of the Ordinary Differential Equations.( )txf ,

( )txf ,

Let continuous functions on an open rectangle containing the point . Let be continuous in R and assume that

( ) xfandtxf ∂∂ /, ( ) dxcbtatxR <<<<= ,:,

( )00 , tx ( )txF ,

Let be the solution to the initial value problem: ( )tx,φ ( ) ( ) 00,, xtxtxftd

xd ==

Let be the solution to the initial value problem: ( )tx,ψ ( ) ( ) 00,, xtxtxFtd

xd ==

RR ⊂0

Assume both solutions exist on [t0 – h, t0 +h] and their graphs lie in a closed region . Then for |t-t0| ≤ h,

( ) ( )txandtx ,, ψφ

( ) ( ) hkehtxtx εψφ ≤− ,,

where k is any positive constant such that for all .Moreover as approaches uniformly on R, that is, as ε→0+, the solution approaches uniformly in .

kxf ≤∂∂ / ( ) 0, Rtx ⊂

fF

htt ≤− 0( )tx,ψ ( )tx,φ

172




Proof of Continuous Dependence on of Solutions of ODE( )txf ,


t≤−+= ∫ 00 ,,,,

0

φφ

The integral representation of , is ( ) ( )txandtx ,, ψφ

( ) ( )( ) httsdssxFxtxt

t≤−+= ∫ 00 ,,,,

0

ψψ

Subtracting those equations gives

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

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

sdssxFsdssxftxtx

t

t

t

t

t

t

t

t

≤−−+−=

−=−

∫∫∫∫

000

00

,,,,,,,,

,,,,,,

ψψψφ

ψφψφ

Applying the norm and using the Triangle Inequality, we obtain

( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) httsdssxFssxfsdssxfssxftxtxt

t

t

t≤−−+−≤− ∫∫ 0

00

,,,,,,,,,, ψψψφψφ

( )( ) ( )( ) ( ) ( )sxsxkssxfssxfkxf

Lipschitzor

,,,,,,/

ψφψφ −≤−≤∂∂

use

and ( )( ) ( )( ) εψψ ≤− ssxFssxf ,,,,

( ) ( ) ( ) ( ) ( ) ( ) htthsdsxsxksdsdsxsxktxtxt

t

t

t

t

t≤−+−≤+−≤− ∫∫∫ 0

000

,,,,,, εψφεψφψφ

173




We obtained:

Use Grönwall Inequality:

( ) ( ) ( )( ) ( ) 0&0,0

,

≥≤≤≥>

≤≤+≤ ∫cbxaonxvxu

bxatdtvtucxvx

a ( ) ( ) bxatdtucxvx

a≤≤≤ ∫ ,exp

( ) ( ) ( ) ( ) hctxtxtvktu εψφ =≥−=>= :&,0,,:,0:with:

( ) ( ) ( ) hkhtt

ttkt

tehehsdkhtxtx εεεψφ

≤−− ≤=≤− ∫

00

0

exp,,


Proof of Continuous Dependence on of Solutions of ODE (continue – 1)( )txf ,

( ) ( ) ( ) ( ) httsdsxsxkhtxtxt

t≤−−+≤− ∫ 0

0

,,,, ψφεψφ

In particular as approaches uniformly on R, that is, as ε→0+, the solution approaches uniformly in .

fFhtt ≤− 0( )tx,ψ ( )tx,φ

q.e.d.

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Calculus of variations