PHYS 705: Classical MechanicsCalculus of Variations II

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Calculus of Variations: Generalization (no constraint yet)Suppose now that F depends on several dependent variables :

We need to find such that has a

stationary value

( ), ' ( ); 1, 2, ,i iF F y x y x x i N

( ), '( );B

i iA

I F y x y x x dx

We need to consider N variations in N dimensions.

( , ) (0, ) ( ) 1, 2, ,i i iy x y x x i N

2

as a parameter ( ) are -smooth variation at ( ) ( ) 0

i

i A i A

x C xx x

Again, we have,

1( ) N

iy x

1( ) N

iy x

2

Variational Calculus: Generalization (no constraint)Everything proceeds as before, and we get:

If all the variations are independent, this equation requires that each

coefficients in the integrant to vanish independently. Then, we have,

( ) 0'

B

A

x

ii i i

x

dI F d F x dxd y dx y

( )i x

0 1, 2, ,'i i

F d F i Ny dx y

This is the same result (Euler-Lagrange Eq) as we have before. However…

(note: sum over i comes from chain rule in taking the derivative:

)( , '; )i idF y y x d

iy

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Variational Calculus: With Constraint (2D)

When we have constraints, the coordinates and hence the variations

will NOT be independent!

e.g., the system is constraint to move on a lower dimensional surface (or

curve) and allowed variations are coupled in order to satisfy the constraint.

( )i x

To extend our analysis to systems with constraints, let first consider the 2D

case with as our dependent variables with one constraint.

( )iy x

1 2( , )y y

To specify the Holonomic constraint, we have

1 2( , ; ) 0g y y x linking the two dependent variables.

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Variational Calculus: With Constraint (2D)

With two dep variables, our variational equation can be written as,

Similarly with,

1 21 1 2 2

( ) ( ) 0 (*)' '

B

A

x

x

dI F d F F d Fx x dxd y dx y y dx y

1 1 1

2 2 2

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

y x y x xy x y x x

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Variational Calculus: With Constraint (2D)

The constraint links so that these two variations are

NOT independent! 1 2( ) to ( )x x

Thus, we can’t say that (Eq. *) requires the individual

coefficients within the integrant to go to zero independently as before.

0dI d

6

21

11 2 2

( ) ( ) 0 (*'

)'

B

A

x

x

dI x x dxF d F F d Fy dx y y dd x y

Variational Calculus: With Constraint (2D)

To get around this, we consider the total variation of g (total derivative) wrt :

The constraint links so that these two variations are

NOT independent! 1 2( ) to ( )x x

Thus, we can’t say that (Eq. *) requires the individual

coefficients within the integrant to go to zero independently as before.

0dI d

This expression needs to be identically zero since the NET variation must vanish and be consistent with the constraint (everything must stay on the surface defined by .1 2( , ; ) 0g y y x

1 2

1 2

0 (**)y ydg g gd y d y d

NOTE:

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Variational Calculus: With Constraint (2D)

1 21 2

0dg g gd y y

Putting these into Eq. (**),

Rearranging terms, this then gives,

12 1

2

g yg y

From the definition of the variations, we have

1 21 2( ) and ( )y yx x

(Note: the variations are linked through the equation of constraint g.)

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Variational Calculus: With Constraint (2D)

Substituting into Eq. (*), we have,

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1 1 2 2 2

0' '

B

A

x

x

g ydI F d F F d F dxd y dx y y dx y g y

2

Since is an independent variation, we can now argue that the coefficients { }

in front of it must vanish.

1

1

1 1 2 2 2

0' '

g yF d F F d Fy dx y y dx y g y

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Variational Calculus: With Constraint (2D)

Rewriting the expression with separately on the LHS and

RHS, we have the following equation:

1 1 1 2 2 2

1 1' '

F d F F d Fy dx y g y y dx y g y

And, we will call this as yet un-determined function (multiplier): ( )x

1 2( ) and ( )y x y x

And, since these two terms EQUAL to each other and depend on independently they can only be a function of x ONLY !

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Note that these two terms are identical with 1 2y y

( )x

1 2( ) and ( )y x y x

Variational Calculus: With Constraint (2D)

With this multiplier , we can rewrite the LHS & RHS as two decoupled

equations:

( )x

( ) 0 1,2 (1)'i i i

F d F gx iy dx y y

Together with , these 3 equations can be solved for the

three unknowns: 1 2( ), ( ), ( )y x y x and x

is call the “Lagrange undetermined multiplier”

( )x

1 2( , ; ) 0g y y x

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is a functional which takes in two functions and gives back out

a real number and the first two terms on the LHS of EL equations

Variational Calculus: “Geometric” View Point

, 1,2'i i i

I F d F iy y dx y

1 2( ) and ( )y x y x

is in fact a functional derivative of I [yi] with respect to the function yi.

Similar to regular derivatives in vector calculus, it gives the rate of change

of I with respect to a change in the function yi.

In our previous 2D problem, the integral

[ ] ( ), ' ( );B

i i iA

I y F y x y x x dx

'( ) 0

i i i

gxF d Fy dx y y

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as a functional “gradient” similar to a gradient vector in multivariable calculus. This

“gradient” will point along the “direction” of the largest increase of I in the space of

functions corresponding to and normal to the level curves of I.

Variational Calculus: “Geometric” View Point

Similarly, we can make a same association with the functional derivative of

g with respect to

So, one can think of the following combined “vector” object

1 2

[ ] ,iI II yy y

1 2( ) and ( )y x y x

1 2( ) and ( )y x y x

1 2 1 2

[ ] , ,ig g g gg yy y y y

(Previously, we didn’t make this distinction but g is in fact a functional.)

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points in the direction

“normal” to the constraint curve

defined by

Variational Calculus: “Geometric” View Point

In the space of ,

[ ]ig y

1 2( ), ( )y x y x

1 2[ , ] 0g y y

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1 2, 0g y y

y1

y2

[ ]ig y

[ ]iI y

1 2[ , ] 0g y y - defines a curve in the

space 1 2( ), ( )y x y x

points in the direction

“normal” to the level curves defined by

Variational Calculus: “Geometric” View Point

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1 2,I y y c

[ ]iI y

y1

y2

[ ]iI y

1 2,I y y c define sets of level curves

in the space of 1 2( ), ( )y x y x

In the space of , 1 2( ), ( )y x y x

Then, the Euler-Lagrange Equation has a specific “geometric” meaning:

Variational Calculus: “Geometric” View Point

( )'i i i

F d F gxy dx y y

[ ] ( ) [ ]i iI y x g y

*[ ]iI y

The solution at are

chosen such that

* *1 2,y y

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1 2, 0g y y

y1

y2

*[ ]ig y

*[ ]iI y

*& [ ]ig y

are collinear!

1 2, 0g y y

- AND, now we are also requiring that

the solution to be consistent with the

constraint .

Variational Calculus: “Geometric” View Point

- The solution at to the E-L

is also a stationary point to .

* *1 2,y y

1 2[ , ] 0g y y

y1

y2

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1 2,I y y

*[ ]ig y

*[ ]iI y

Recall that…

It is a problem in constrained optimization.

1 2,I y y min, max, or inflection pt

Only at the location

where the two curves touch

tangentially, the I value will not

change as we move along the

constrained solution.

This geometric condition can be

compactly written as:

As we vary along the constraint

curve (red) , we will

typically traverse different level

curves of I (blue curves).

1 2, 0g y y

Variational Calculus: “Geometric” View Point

y1

y2

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[ ] [ ]i iI y g y

*[ ]ig y

*[ ]iI y1 2[ , ] 0g y y

* *1 2,y y

Each of these M equations describes a (N-1) dimensional surface and the

intersection of them will be a (N-M) dimensional surface on which the

dynamics must live.

Variational Calculus: Generalization

[ ] 1,2, ,k ig y k M

If the system has more than one constraint, let say M:

( ; ) 0 1, 2, , 1,2,k ig y x i N k M (# dep vars) (# constraints)

There will also be M “normal” (vectors) to each of these M (N-1)-dimensional

surfaces given by:

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Variational Calculus: Generalization

In a similar geometric fashion, points in the direction away from (normal

to) the level set of and in the direction of its greatest rate of change.

[ ]iI y[ ]iI y

Each of these “gradients” will point in a direction AWAY from

(normal to) its prospective constraint surface .

[ ]k ig y( ; ) 0k ig y x

0k ig y [ ]k ig y

level set of [ ]iI y

1[ ]iI y c

2[ ]iI y c

[ ]iI y

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Variational Calculus: Generalization

1

[ ] [ ]M

i k k ik

I y g y

The “geometric” condition for to have a stationary value with

respect to the constraints will be similar to our 2D previous case, i.e., the

optimal set of is the one with,

This means that , can be written as a linear combination of the M

“normal vectors” from the M constraint equations.

[ ]iI y

[ ] [ ]i k iI y span g y

[ ]iI y

Instead of one, we have M

independent Lagrange Multipliers

1, , Ny y

[ ] [ ] 0i k i kI y g y

Or, equivalently

(points normal to ALL the

constrain surfaces.)

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Variational Calculus: Generalization

Putting this back into the original differential equation format, we can write

this condition as,

1( ) 0 1, 2, ,

'1, 2, ,( ; ) 0

Mk

kki i i

k i

gF d F x i Ny dx y y

k Mg y x

Here, we have N+M unknowns: and ( )iy x ( )k xAnd, we have N+M equations: top (N) and bottom (M)

This problem in general can be solved!

(for stationarity)

(to be on constraints)

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Hamilton’s Principle

Now, back to mechanics and we will define the Action as,

where L = T- U is the Lagrangian of the system

2

1

, ,i iI L q q t dt

Configuration Space: The space of the qi’s. Each point gives the full

configuration of the system. The motion of the system is a path through

this configuration space. (This is not necessary the real space.)

Hamilton’s Principle: The motion of a system from t1 to t2 is such that

the action evaluated along the actual path is stationary.

(Note: Here we don’t require all the generalized coordinates to be necessarily

independent. The can be linked through constraints.)iq

'iq s

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We also further assume that the system is monogenic, i.e., all forces

except forces of constraint are derivable from a potential function

which can be a function of

Lagrange Equation of Motion

So, we apply our variational calculus results to the action integral.

1

( ) 0 1,2, ,1, 2, ,

( ; ) 0

Mk

kki i i

k i

gL d L t i Nq dt q q

k Mg q t

The resultant equation is the Lagrange Equation of Motion with N generalized

coordinates (not necessary proper) and M Holonomic constraints:

, , and i iq q t

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, ,i iU q q t

1. The Lagrange EOM can be formally written as:

Forces of Constraint

Comments:

1

( ) 1,2, ,M

kk i

ki i i

gL d L t Q i Nq dt q q

where the Qi are the generalized forces which give the magnitudes of the

forces needed to produce the individual constraints.

- since the choice of the sign for k is arbitrary, the direction of the

forces of constraint forces cannot be determined.

- the generalized coordinates are NOT necessary independent and they are

linked through constraints.iq

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2. If one chooses a set of “proper” (independent) generalized

coordinates in which the (N-M) are no longer linked through the

constraints, then the Lagrange EOM reduces to:

Proper Generalized Coordinates

Comments:

0 1, 2, ,j j

L d L jq dt q

N M

- In practice, one typically will explicitly use the constraint equations

to reduce the number of variables to the (N-M) proper set of

generalized coordinates.

- However, one CAN’T solve for the forces of constraint here

'jq s

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