PHYS 705: Classical Mechanicscomplex.gmu.edu/ Calculus of... · 2017-09-11 · Few Classic Problems...

PHYS 705: Classical MechanicsCalculus of Variations I

1

Calculus of Variations: The Problem

Determine the function y(x) such that the following integral is min(max)imized.

Comments:

( ), '( );B

A

x

x

I F y x y x x dx

1. Notation: • x is the independent variable (x =t in our mech prob)

• y(x) is a function of x and y’(x) = dy/dx

•F{ } is a functional on y(x)

2. Intuition: Let say y(x) denote a route in the x-y plane and I is the

amount of gas needed for the trip. The problem is to

find the route which uses the least gas.

2

Calculus of Variations: The Problem

Comments (cont):

3. To keep it simple, we will fix the end points (xA and xB). We

assume that they are known. Method can be extended to

situation where (xA and xB) can be varied as well.

The method does not explicitly give y(x). Instead, we will get

a diff eq for y(x).

4.

F is assumed to be at least twice differentiable, i.e., C 2 in all

its arguments.

5.

x is assumed to be unidirectional between xA and xB. If not,

break it into different pieces.

6.

3

Calculus of Variations

Consider a family of functions parameterized by .

Note:

( , )y x

( , ) (0, ) ( )y x y x x

2

is a parameter ( ) is a -smooth variation at ( ) ( ) 0A A

x C xx x

x

y

xBxA

yA

yB

desired path

for all , i.e., all sampled paths has the same end points, xA and xB

( , )( , )

A A

B B

y x yy x y

(0, )y x is the desired path

y(0,x)( )x

4


A necessary, but not sufficient, condition for the min(max)imization of I is:

0

0dId

(the end points xA and xB are fixed)

Taking the derivative of I wrt , we have,

( ), '( );B

A

x

x

dI d F y x y x x dxd d

(x is the independent variable and does not vary with )

5

( ), '( );

''

B

A

B

A

x

x

x

x

d F y x y x x dxd

F y F y dxy y


Recall, we have

And,

'

B

A

x

x

ddx

dI F F dxd y y

( , ) (0, ) ( )y x y x x

So, ( )y x

'( , ) '(0, ) '( )y x y x x

So, 'y ddx

This gives,

Now, to continue, we will integrate the second term by parts.

6


So, ( ) ( )

' ' '

BB B

A AA

xx x

x xx

F d F d Fdx x x dxy dx y dx y

To integrate by parts, udv uv vdu

Let, ' '

F d Fu du dxy dx yddvdx

dx v

(with the fixed limits xA and xB)

vduuvudv

no variations

@ end points

0A Bx x

'

B

A

x

x

F ddIy

F dxd dxy

7


Putting this back into our expression for ,

( )' '

B B

A A

x x

x x

F d d Fdx x dxy dx dx y

So, we have,

dId

( ) ( ) 0'

( ) 0'

B

A

B

A

x

x

x

x

dI F d Fx x dxd y dx y

F d F x dxy dx y

Since this has to be zero for any arbitrary C 2 variation ,( )x

0'

F d Fy dx y

Euler-Lagrange Equation (1744)(fundamental lamma of Calc. of Var.)

8

Calculus of Variations (Euler-Lagrange Equation)

Comments:

0'

F d Fy dx y

Euler-Lagrange Equation

This is the diff eq whose solution y(x) is the function we seek

to min(max)imize with respect to F.

1.

We only used a necessary condition for an extremum 2. 0

0dId

I might end up to be an inflection point. We can always check

afterward. In Hamilton’s principle (later), the condition is only

for a stationary value.

It is important to keep track of which variable is the independent

variable. Here, it is x.

3.

9

Euler-Lagrange Equation (Special Case)

We will now derive a particular useful form of the EL equation for the case

when F does not explicitly depend on x, i.e.,

' ' ''' ' '

d F d F F dFy F y ydx y dx y y dx

Observe this:

( , ')F y y (obviously, F depends

on x implicitly thru y)

''

F y F y Fy x y x x

'' ''''' '

''

'd F Fy ydx y

d Fy Fd

F Fy yyx yy y

10

Euler-Lagrange Equation (Special Case)

Cancelling and grouping like-color terms,

' ' ''' ' '

d F d F F Fy F y ydx y dx y y y

'''

F yy

Now, if y(x) satisfies the Euler-Lagrange Equation, the remaining term on the

RHS will be zero as well. Thus, we have,

' 0'

d Fy Fdx y

or ''

Fy F consty

This is a much easier equation to solve since it is 1st order

only. We basically have done one integration already.

11

Few Classic Problems (#1: a Straight Line)

1. What plane curve connecting two given points has the shortest length?

where,

Arc length is given by:

x

y

xBxA

B

A

I ds

22 2 1 dyds dx dy or ds dx

dx

So, we need to find y(x) that minimizes:

21 'B

A

I y dx

Note: ds is written in Euclidian space. If it is not (e.g. sphere), the result (geodesic) will be different NOT a straight line.

12


To solve this problem, we apply the Euler-Lagrange equation,

with

EL equation gives:

21 'F y 0'

F d Fy dx y

1/22

2

1 '0 1 ' 2 '' 2 1 '

F F yy yy y y

0'

d Fdx y

2

'' 1 '

F y consty y

13


' 0'

d Fy Fdx y

(it is a line and m and b to be

determined by the endpoints)

2

2

''

'' 1 '1 '

Fy F consty

yy y consty

'y const

Solving for y’ gives y’ being equals to another constant,

y mx b

Alternatively, since F does not depend on x explicitly, we can use the second

form for the EL equation,

14


'y const

Multiplying to the whole equation, we have,

y mx b

2 2 2

2

' 1 ' 1 '

1 1 '

y y c y

c y

21 'y

and the same result,

(c being a constant)

15

Few Classic Problems (#2: Catenoid)

Two points in an xy-plane are given. A curve in the plane connecting the

two points is revolved about an axis. Find the curve that results in the

minimum surface area of revolution.

(note: x is our indep var and it is unidirectional)

2. Minimum Surface of Revolution Problem (Soap Film bet 2 Wire Loops)

y

ds

element of surf area da

22 2 1 'da y ds y y dx 1

x

y2

2 2ds dx dy

Area to be minimize is:

2

1

22

1

2 1 'x

x

I da y y dx

(Goldstein is not correct on this)

16


For the given integrant, the EL equation is:

with

With F not explicitly depends on x (the independent variable), we can use the

alternative form for the EL equation instead:

2( , '; ) 1 'F y y x y y 0'

F d Fy dx y

1/22

2

1 '1 ' 2 '' 2 1 '

F yyy y yy y

' where c is an integration constant'

Fy F cy

17


Putting it into the alternative form of the EL equation:

2

2

'' 1 '1 '

yyy y y cy

2'yy 21 'y y

2

2

1 '

1 '

cy

y cy

222 21 ' ' 1yc y y or y

c

18


So the diff eq that we need to solve is now:

2 21dy y cdx c

1

2 2coshdy yx dx c c b

cy c

This curve is called a

Catenary and the surf.

of revolution is called a

Catenoid.

(both c and b are integration

constants which will depend on

the end points x1 and x2)

cosh x by cc

x

y21

19


A rendering of the catenoid using Mathematica

y

Soap film between two loops

20


There are more interesting subtleties to this problem:

1. When one considers a family of catenaries that go through one fixed

point (let say the left one : pt 1) …

x

- All the caternaries will tangent on

an envelope curve (blue) which is a

parabola with its focus at pt #1

- Interesting observations:

If we choose pt 2 at this location (other

similar intersection pts), there will be 2 solns.

There will be no soln

here.

21

y

2

1fixed

2’

Few Classic Problems (#2: Catenoid)2. What we have considered so far involve only twice-differentiable solutions.

There is also a nondifferentiable solution that has physical relevance !

x

y21

x

y21

Goldschmidt Solution

Soap film bet

two rings

Smooth Caternary Solution

Keeps moving the two loops apart until the red surf area > blue surf area

22

Few Classic Problems (#3: Brachistochrone problem)

Find the path that goes between pt 1 and pt 2

with the least time under gravity.

x

+y

2

1

v g

2

1 2

1

minimize dstv

With (Goldstein, p.43), the solution is a cycloid given by:0 0T

sin , 1 cosx a y a

a

+y

x

(HW is to derive the cycloid with .)20 0 2T mv

2a

23

Few Classic Problems (#3: Brachistochrone problem)

Animation for the cycloid:

http://www.youtube.com/watch?v=li-an5VUrIA

Movie on YouTube:

24

PHYS 705: Classical Mechanicscomplex.gmu.edu/ Calculus of... · 2017-09-11 · Few Classic Problems...

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Transcript of PHYS 705: Classical Mechanicscomplex.gmu.edu/ Calculus of... · 2017-09-11 · Few Classic Problems...