calculus of variations

1

Chapter 9 Calculus of variations

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 10 Euler equation

2

1. Introduction- Geodesic: a curve for a shortest distance between two points along a surface 1) On a plane, a straight line 2) On a sphere, a circle with a center identical to the sphere 3) On an arbitrary surface, ??? In this case, we can use the calculus of the variation.

cf. Because the geodesic is the shortest value, finding the geodesic is relevant to finding the max. or min. values.

3

- In the ordinary calculus with f(x), how can you find the max. or min? (or how can you make the quantity stationary) First, obtain the first derivative of f(x), and find the stationary points. : Stationary point to make f’(x)=0. f(x) becomes Max. and Min. points and more.

cf. But, we do not know if a given stationary point is a Max, Min, or a point of the inflection with a horizontal tangent.

4

- What is the quantity which we want to make stationary in this chapter?

dx

dyydxyyxFI

x

x' where )',,(

2

1

Ex. 1) shortest distance

22

1

2

222

'1)',,( ;'1

'1

yyyxFdxyI

dxydydxds

x

x

Ex. 2) brachistochrone problem: brachistos=shortest, chronos=time e.g., In what shape should you bend a wire joining two given points so that a bead will slide down from one point to the other in the shortest time? We must minimize “time”.ds : element of arc length, v=ds/dt : velocity

222 '11

)'.,( ;'11

,'111

yv

yyxFdxyv

dtdxyv

dsv

dt

5

Ex. 3) a soap film suspended between two rings What is the shape of the surface?The answer is the shape to minimize the surface area.

Other examples - chain suspended between two points hangs so that its center of gravity is as low as possible. - Fermat’s principle in optics. (light traveling between two given points follows the path requiring the least time.

6

2. Euler equation

1) Geodesic on a plane

2

1

22

1'1 minimize to)(

x

x

x

xdxydsIxyy We call this y(x) ‘extremal’.

We define a completely arbitrary function passing two points.

2 and 1at zero ishich function w a :)(

parameter, : extremal, desired :)(

)(ε)()(

xxx

xy

xxyxY

).()( 0,when ;

)parameter of(function '12

1

2

xyxY

dxYIx

x

7

Then, our problem is to make I() take its min. when = 0.

)'( 0'1

)(')('

)(''

),(')(')(' .'

'2'1

1

2

1

.0 ,0

0

2

1 20

2

1 2

yYdxy

xxy

d

dI

xd

dYxxyxYdx

d

dYY

Yd

dI

d

dI

x

x

x

x

8

)'( 0'1

)(')('0

2

1 20

yYdxy

xxy

d

dI x

x

- first term is zero because (x1)= (x2)=0.- (x) is an arbitrary function. In order for the second term to be zero,

const.or const.'1

' ,0

'1

'22

y'

y

y

y

y

dx

d

**From this, y(x) (geodesic on a plane) is a straight line.

0'1

')()(

'1

'

cf.

),( ,'1

' ,)(' ,'1/'

2

1 2

2

120

2

2

x

x

x

xdx

y

y

dx

dxx

y

y

d

dI

dxuvuvudvdx

xvdxy

y

dx

ddudxxdvyyu

9

2) Generalization

2

1

2

1

2

1

2

1

)(''

)('

'

).()()(for )',,()',,(

x

x

x

x

x

x

x

x

dxxY

Fx

Y

Fdx

d

dY

Y

F

d

dY

Y

F

d

dI

xxyxYdxYYxFdxyyxFI

For arbitrary , equation Lagrange)-Euler(or Euler .0'

y

F

dx

d

y

F

0)(''

)( ,0at and 0/ Using2

10

x

xdxx

y

Fx

y

F

d

dIεyYddI

0)('

.)('

)('

)(''

term,second For the

2

10

2

1

2

1

2

1

x

x

x

x

x

x

x

x

dxxy

F

dx

d

y

F

d

dI

dxxy

F

dx

dx

y

Fdxx

y

F

10

Example geodesic on a plane

2

1'1 minimize to)(

x

xdxyIxyy

.0'1

' ,0 ,

'1

'

' .'1)(

22

2

y

y

dx

d

y

F

y

y

y

FyxF

11

Chapter 9 Calculus of variations

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 11 Application of Euler equation

12

3. Using the Euler equation

1) Other variables

.0

/ where)',,(

θ

F

θ'

F

dr

d

dr,dθθ'drrF

.0

where),,(

x

F

x'

F

dt

d

dx/dt,x'dtx'xtF

Note. 1. The first derivative is with respect to the integration variable in the integral.2. The partial derivatives are with respect to the other variables and its derivatives.

13

.const.1

011'

1

' ,0

,0'

equation Euler theUsing

22

2222

22222

Kr

rdr

d

r

rr

dr

drr

dr

dF

dr

dF

FF

dr

d

Example 1. Find the path followed by a light ray if n (refractive index) is prop. to r-2 (polar coord.).

.0 without1

.1 1set

222

22222222

FrrF

drrrdrdrrdsrcndsdt

14

const.arcsin n table,integratio theUsing

1

,)1('or 1

22

22222222

Kr

rK

K

dr

d

KrKrK

15

2) First integrals of the Euler equation

const.'

0'

,0 If

y

F

y

F

dx

d

y

F

In this case, we can integrate the Euler equation once.Such a equation (F/ y’) is called a first integral of the Euler equation.

0'

FF

dr

d

16

Example 2. Find the curve so that the surface area of revolution is minimized.

p1p2curve, y(x)

revolving the curve about x-axis

a surface of revolution

nintegratio of variable:,'1 ,'12

'1 form usual of instead

,'11

2

22

2

22

yxyFdyxyI

dxyds

dyxdyds

ydsI

dydx

17

,'

.'1 ,'1

'

.0'1

'0

'

21

2

1

21

2212

2

0

cy

c

dy

dxx

xcxycx

yx

x

yx

dy

d

x

F

x

F

dy

d x

F

‘catenary line ( 현수선 )’

.cosh

,cosh n table,integratio theUsing

1

21

21

11

c

cxcy

cc

ycx

18

- Like the above, if F(y,y’) does not have the independent variable x, we had better change to y as integration variable.

dyxdydy

dxdx

xy

dx

dy

dy

dxx ' ,

'

1' ,'

1

Example 3. . no where,'1 2

xdxy

yI

const.1'

'

01'

'

'0

'

.)',(1'

,1'''1'1

2

2

/

2222

xy

x

xy

x

dy

d

x

F

dy

d

x

F

x

F

dy

d

dyxyFy

dyxIdyxdyxydxy

xF

19

Example 4. Find the geodesics on the cone z^2=8(x^2+y^2) using the cylindrical coordinate.

,'99

98

,8,8,8

22222

222222222222

22

drrdrdrdsI

drdrdrdrdrdzdrdrds

drdzrzrz

zzx

yyxr

zzryrx

zrzyx

,tan,

,sin,cos

),,(),,( :coordinate lCylindrica

122

,3

9)('

),'9('4 ,const.'9

'

' ,0

'

22

24242

22224

22

2

Krr

KdrdKrKr

rKrKr

rFF

dr

d

20

.cosor 3

cos

n)integratio of const.( arccos1

3 table, theFrom

Krr

K

r

K

KK

21

4. Brachistochrone problem: cycloids

A bead slides along a curve from (x1,y1)=(0.0) to (x2,y2). Find the curve to minimize the during time.

‘gravity’

reference

mgy

dt

dsmmv

energy potential

,energy kinetic2

212

21

The sum of two energies is zero initially and therefore zero at any time and position.

.2or 0221 gyvmgymv

22

The integral we want to minimize is

3)section 3, (example '1

2

1

2

2

1

2x

xdx

y

y

ggy

ds

v

dsdt

.1'

'

01'

'

'0

'

.)',(1'

,1'''1'1

equation)Euler theof integral(First

2

2

0/

2222

cxy

x

xy

x

dy

d

x

F

dy

d

x

F

x

F

dy

d

dyxyFy

dyxIdyxdyxydxy

xF

23

')21arccos(2

1 table, theFrom

,or 1

'

1'

'

2

2

2

ccyc

yc

yx

yc

y

ydydx

cy

cy

dy

dxx

cxy

x

c’=0 for (x1, y1) = (0,0)

“This is the equation of a cycloid.”

24

- Cycloid A circle (radius a) rolls along x-axis. It start at origin O. Place a mark on the circle at O. As the circle rolls, the mark traces out a cycloid.

“trace of a mark on the circle when the circle rolls.”

25

When a circle rolled a little,

)cos1(cos

),sin(sin

aaaBCACABy

aaaPBOAx

aPAOA

‘parametric equation of a cycloid’

- Parametric equation of a cycloid

26

.cos21 )21arccos(let we

),21arccos(2

1 chronebranchisto for theresult previous theFrom 2

cycy

cyc

yc

yx

cos12

1 ,sin

2

1

2

1sin

2

1

sin4

1cos1

4

1cos1cos12

4

1

),cos1(2

1 ,cos21

22

22

2

22

cy

cccx

cccy

c

yc

ycy

“parametric equation of brachistochrone or parametric equation of cycloid.”

27

- Cycloids differ from each other only in size, not in shape.

22

2

y

x21

1

y

x23

3

y

x

- Rather surprisingly, when a bead slides from origin to P3 in the least time, it goes down to P2 and backs up to P3 !!

At P2, x/y = /2. For P1, x/y < /2. For P3, x/y > /2.

28

Chapter 9 Calculus of variations

Mathematical methods in the physical sciences 2nd edition Mary L. Boas

Lecture 12 Lagrange’s equations

29

5. Several dependent variables; Lagrange’s equation- Necessary condition for a minimum point in ordinary calculus,

for an one-variable function z=z(x), dz/dx=0,for a two-variable function z=z(x, y), z/x=0 and

z/y=0.

.0'

,0'

z

F

z

F

dx

d

y

F

y

F

dx

d

- The similar idea is applied to Euler equation. When for F=F(x, y, z, dy/dx, dz/dx) we find two curves y(x) and z(x) to minimize I = F dx, we need two Euler equations.

30

It is a very important application to mechanics; Lagrangian based on Hamilton’s principle

- Lagrangian: L = T – V where T : kinetic energy, V : potential energy- Hamilton’s principle: any particle or system of particles always moves

in such a way that I = L dt is stationary.

In this case, Euler equation is called Lagrange’s equation.

31

Example 1. Equation of motion of a single particle moving (near the earth) under gravity. (three dimensional motion)

.

,222

21

222212

21

mgzzyxmVTL

mgzVzyxmmvT

gz

y

x

mgzmdt

d

ymdt

d

xmdt

d

.

const.,

const.,

or

,0

,0

,0

equation sLagrange' .0 ,0 ,0

z

L

z

L

dt

d

y

L

y

L

dt

d

x

L

x

L

dt

d

32

- In some cases, it would be simpler to use elementary method from Newton’s equation. However, in some cases with many variables it would be much simpler to use Lagrange’ equation, because we treat one scalar function, Lagrangian L = T – V.

33

Example 2. Equation of motion in terms of polar coordinate variable r, .

. , 2222

222

22222 rrdt

dr

dt

dr

dt

dsvdrdrds

centripetal v^2/r Coriolis acceleration

).,(

).,( ,

22221

22221

rVrrmVTL

rVVrrmT

.0,0

.0 ,0

22

V

mrdt

d

r

Vmrrm

dt

d

LL

dt

d

r

L

r

L

dt

d

.2,

122 :motion ofequation

:motion ofequation

2

2

2

rrarra

V

rmarrm

Vrrrm

r

Vmarrmr

r

r

34

Example 3. m1 moves on the cone. (spherical coord. , , ) m2 is joined to m1 and move vertically up and down. (z-component)

,: ,2/3cos: :

.: ,sin: :

21

2222

22412222222

22

1

lzmzmV

zvmdt

dsvmT

Here, the cone ( =30) is a constraint for motion.

.const.

,0,330cos,30sin

2221

21

21

lzzlz mmmm

Using the above,

.3 21212

22122

412

121 lgmgmmmVTL

35

const.or 04/0

,034

0

.3

221

2121

21

21

21212

22122

412

121

mdt

dLL

dt

d

gmgmm

mmdt

dLL

dt

d

lgmgmmmVTL

36

cf. Rolling disk

sin

4

3

4

1

2

1

4

1

2

1

2

1

2

1 22

222222

MgyV

RyyMR

yMRyMMRyMIyMT

.sin3

2

,0sin2

30

sin4

3 2

gy

MgyMdt

d

y

L

y

L

dt

d

MgyyMVTL

37

cf. Atwood’s machine I

.,,, 2121 xvvvwhere

a

vx

dt

xldvx

dt

dxv

xlgmgxmV

a

xIxmxmIvmvmT

21

22

22

122

222

11 2

1

2

1

2

1

2

1

2

1

2

1

xlgmgxma

xIxmxmVTL

21

22

22

1 2

1

2

1

2

1

38

221

2121221

21

221221

21

22

22

1

0

2

1

2

1

2

1

a

Imm

mmgxmmgx

a

Imm

mmgx

L

xa

Imm

x

L

dt

dx

a

Imm

x

L

x

L

x

L

dt

d

xlgmgxma

xIxmxmVTL

39

cf. Atwood’s machine II

112222

1233

11222

1222

112

111

,2

1:

,2

1:

,2

1:

xlxlgmVxxmTm

xlxgmVxxmTm

gxmVxmTm

40

cos,2

1

2

1 2222 grmgrMVrrmrMT

cf. Swing atwood’s machine

41

cf. Double pendulum

,coscos,sinsin

,cos,sin

22

11

llyllx

lylx

cos22

1

2

1

sinsincoscos22

1

2

1

sinsincoscos2

1sincos

2

1

2

1

2

1

222222

222222

2222

22

22

21

21

lllmml

lllmml

llllmllm

yxmyxmT

coscos221 mglmglmgymgyV

42

2222

2

22

22

2

1

2

1

2

1

2

1cos1cos1

2

1

sin & sin

ions,approximat some Using

2

1

2

1

klmglmgl

yxkmglmglV

ml

lyllylxllx

ymxmT

cf. Prob. 19

43

6. Isoperimetric problems

Ex. To find a curve to make largest area ( y dx = Max.) with a given length ( ds = l)

cf. Lagrange multiplier (Max. or min. (stationary point) problem with a constraint)

.0 ,0 ,0

),,( ),,,(

z

F

y

F

x

F

gfFczyxgzyxf

alue.constant vgiven :)',,(

stationary make tointegral :)',,(

2

1

2

1

x

x

x

x

dxyyxGJ

dxyyxFI

By using the Lagrange multiplier method,

equation.Euler esatisfy th should

.stationary be should 2

1

GF

dxGFx

x

- Good news: Sometimes we do not need to find .

44

.'1 ,'1 ,

.'1 ,

22

2

1

22

1

2

1

yyGFyGyF

ldxydsJydxIx

x

x

x

x

x

Example 1. Find the shape of the curve of constant length joining two points x_1 and x_2 on the x-axis which, with the x axis, encloses the largest area.

x1 x2

The curve length is fixed.(l > x2-x1)

01'1

'1 and

'1

'

' 22

y

y

dx

dGF

yy

yGF

y

45

,

)(

,'

,'1''1

'

22

2222

2222

2

cx

dxcxdy

cxcxy

ycxycxy

y

.)'()('

,'

222222

22

cycxcxcy

cxcy

46

7. Variation notation

- : differentiation with respect to . just like the symbol d in a differential except that , not x, is a differential variable.

0.at ),('),,(, of / aldifferenti totalajust

;''

' ,''

'

variable.a is if aldifferenti a likejust ;

)(')('),(' ),()(),(

,

0

0

0

xYxYxFdFdF

yy

Fy

y

FF

dxdxdx

dy

dx

ddxd

Yy

dYdxdY

y

xxyxYxxyxY

dd

dII

47

.''

''

2

1

2

1

2

1

2

1

dxdxy

Fdx

y

F

dxyy

Fy

y

F

FdxFdxI

x

x

x

x

x

x

x

x

0)(''

)( cf. 2

10

x

xdxx

y

Fx

y

F

d

dI

- The meanings of two statement are the same. (a) I is stationary; that is, dI/d=0 at =0. (b) The variation of I is zero; that is, I=0

48

H. W (Due 13th of Nov.)Chap. 9

2-13-4,55-4 (G1), 5(G2), 9(G3)6-1, 2(G4)

49

Problem

5-4 Use Lagrange’s equations to find the equation of motion of a simple pendulum.5-5. Find the equation of motion of a particle moving along the x axis if the potential energy is V=(1/2)kx^2.

5-9 A mass m moves without friction on the surface of the corner r = z under gravity acting in the negative z direction. Find the Lagrangian and Lagrange’s equation in terms of r, .

6-2 The plane area between the curve and a straight line joining the points is a maximum.