phys 705: classical mechanicscomplex.gmu.edu/ calculus of variatio… · variational calculus:...
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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
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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
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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
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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
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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
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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!
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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
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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
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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
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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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