phys 705: classical mechanicscomplex.gmu.edu/ calculus of variatio… · variational calculus:...
TRANSCRIPT
PHYS 705: Classical MechanicsCalculus of Variations II
1
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
3
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.
4
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
5
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:
7
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.)
8
Variational Calculus: With Constraint (2D)
Substituting into Eq. (*), we have,
11
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
9
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 !
10
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
11
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
12
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.)
13
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
14
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
15
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
16
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
17
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
18
[ ] [ ]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:
19
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
20
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.)
21
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)
22
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
23
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
24
, ,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
25
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
26