Transcript Theoretical Mechanics

Theoretical Mechanics - PHY6200
Chapter 6
Introduction to the
calculus of variations
Prof. Claude A Pruneau,
Physics and Astronomy Department
Wayne State University
Introduction
Many mechanics problem more easily
analyzed/solved by means of calculus of
variation, Lagrange Eqs., etc.
 We will:

Consider some general principles
 Omit formal existence proofs


As an example, consider Fermat’s
principle.
Fermat’s Principle

Light travels between two points along the
path that takes the least amount of time.
L
2
air
L2
L1
water
d1
1
x1
x2
d2
Snell’s Law
nair sinair  nwater sinwater
2
2
By construction (geometry): L1  d1  x1
L2  d22  x22
2
2
2
n1 L1 n2 L2 n1 d1  x1 n2 d2  L  x1 



Total travel time: T (x1 ) 
c
c
c
c


n
L

x


dT
1
n
x


2
1
1 1
Extremum for :
0 


2
2
dx1
c  d12  x12
d2  L  x1  

2
Simplification/Substitution :
Snell’s Law :
0  n1 sin1  n2 sin2
n1 sin1  n2 sin 2
Statement of the problem



In classical mechanics, the problem amounts to finding a trajectory
between an initial and a final point (boundary conditions).
Express the problem to solve as a minimization problem.
Find a functional that once minimized will yield an optimal trajectory
which corresponds to the physical trajectory (according to Newton’s
laws of motion).
y
y2
2
y(x)
y1
1
x1
x2
x
Basic Formulation (1 dimension)
y
y2
2
y(x)
y1
1
x1

x2
x
Determine a function, y(x), such that the following integral is an extremum.
x2
J   f (y(x), y'(x) : x)dx
x1





f(y,y’;x) is a function considered as “given”
Limits of integration are fixed.
y(x), the trajectory is to be varied until an extremum is found for J.
y’(x)=dy(x)/dx
x : independent variable
Basic Formulation (cont’d)
y
y2
2
y(x)
y1
1
x1


ya (x)
x2
x
If the optimal trajectory y(x) produces a minimal
value for J, then any neighboring trajectory ya(x)
corresponds to a large value of J.
Hence by minimizing J, one finds the optimal
trajectory.
Parametric Representation

The function y(x) can be considered with a
parametric representation



ya(x) = y(a,x)
Where the variable a denotes a parameter that describes
functions close to y(x) but that differ by an “amount”
proportional to a.
By definition, a=0 corresponds to the optimal
trajectory y(x).

y0(x) = y(0,x)=y(x)
Parametric Representation (cont’d)

Consider the specific representation


y(a,x)=y(0,x)+ a(x)
where  (x) is some “arbitrary” function of
x, with continuous 1st derivative, and
vanishing at the boundaries

(x1) =  (x2) = 0
Parametric Representation (cont’d)

The functional J is written
x2
J(a )   f (y(a , x), y'(a , x); x)dx
x1

This integral has an extremum if
J(a )
0
a a  0


This must be true for all functions (x)
Note: This is a necessary but not sufficient condition.
Example 1





Consider the function f(y’;x)=(dy/dx)2.
Where y(x)=x.
Consider (x)=sin(x)
Find J(a) between x=0, and x=2.
Show that the stationary value (extremum) of J
occurs for a =0.
Example 1 - Solution

Let:


Note that by choice/construction, we have



y(a,x) = x + a sin(x)
(0) = 0
(2) = 0
Proceed to calculate…
Example 1 - Solution (cont’d)
y(a , x)  x  a sin x
dy(a , x)
 1  a cos x
dx
 dy(a , x) 
2
2
f 

1

2
a
cos
x

a
cos
x

 dx 
2
J(a )  
2
0


1  2a cos x  a 2 cos2 x dx
J(a )  2  a 2
J(a )  2  a   J(0) for a  0
2
Derivation of Euler’s Equation

Calculate the derivative of J
J


a a


x2
x1
f (y(a , x), y'(a , x); x)dx
For fixed integration limits, one can change the order
of operations


a

J

a

x2
x1

f (y(a , x), y'(a , x); x)dx
a
x2  f y
J
f y' 
 

dx

x
1
a
 y a y' a 

By construction:

Thus:
y
 (x)
a
dy(x)
d(x)
y' 
a
dx
dx
dy' d(x)

da
dx
x2  f
J
f (x) 
   (x) 
dx

x1
a
y' a 
 y
x2  f
J
f (x) 
   (x) 
 dx
x1
a
y
y'

a



Integrate 2nd term by parts  udv  uv   vdu

x2
x1
 f  
 y' x  dx 



x2
x1
x2
x2
f
f
d f
d 
(x)   (x)
dx
x
1
y'
y'
dx y'
x1
0
x2  f
x2
J
d f 
 f d f 
   (x)  (x)
dx   (x)  
dx


x1
x1
a
dx y' 
 y
 y dx y' 
The function (x) being totally arbitrary.
J
 We get
0

a
a 0
Provided the integrand itself vanishes.
 So…

Euler’s Equation
f
d f

0
y dx y'
This equation yields a solution for y(x) and y’(x)
which produces an extremum for J.
Example 2 - Brachistochrone
Consider a particle moving in a constant
force field, starting at rest from some point
(x1,y1) and ending at (x2,y2).
 Find the path (trajectory) that allows the
particle to accomplish the motion/transit in
the least amount of time.

Example 2 - Solution (1)
Constant force, neglect friction, implies a
conservative system.
 T+U=constant.
 Let U(x=0)=0, and T(0)+U(0)=0.
 T=0.5 mv2
y
 U=-Fx=-mgx
(x ,y )
 v  2gx
g (x ,y )

1
1
2
x
2
Example 2 - Solution (2)

Time required for transit:
x2
ds
t

v
x1
x2

dx
2
2gx
x1
 1 y' 
t  

2gx

x1
x2
2
 1 y' 
f 
 2gx 
2
 dy
1/2
1/ 2
dx

2 1/2
dx
Example 2 - Solution (3)

Use Euler’s Equation
 1  y '2 
f 
 x 
1/ 2
f
  1  y' 
 
y y  x 
2

Thus…
1/2
0
f
1
 constant 
y'
2a
f
d f

0
y dx y'
d f
0
dx y '
 1  y '2 
f 
 x 

f
1  1  y '2 
 
y ' 2  x 
1
x
x
x2
2
2a
y' 


x  2a  x  2ax  x 2

 1  
2a
1/ 2
1/ 2
2
y' 
x
1
2a
x
y' 
 1  y '2   y ' 
1

 x   x 
2a



2ax  x 
2
2
y
1 1  y'2 
 y' 
  
x
2a  x 
xdx

2ax  x 
2
2
2
2

 x
x
1

y'
y'2 

1  y'2


2a  x  2a

x
x
2
y'  1   

2a  2a

x  a(1  cos )
dx  a sin  d
a(1  cos )a sin  d
y
2a 2 (1  cos )  a 2 (1  cos )2

y



2a
a 2 (1  cos )sin  d
2
 2a 2 cos  a 2  2a 2 cos  a 2 cos 2 
a 2 (1  cos )sin  d
a
2

 a 2 cos 2 
  a(1  cos )d


Example 2 - Brachistochrone - Answer
y   a(1  cos )d
y  a   sin   k
x  a 1  cos 
Solution passes through (0,0)
(0,a)
(0,0)

(2a,0)
x
cycloid
(0,2a)
y
2nd Form of Euler’s Equation
f
0
 For a function “f” that does not explicitly depend on “x”. x
df
d
f y f y' f
1) dx  dx f (y, y'; x)  y x  y' x  x
df
f
f f
 y'  y''

dx
y
y' x
2)
d  f 
f
d f
y'   y''
 y'

dx  y' 
y'
dx y'
d  f  df f
f
d f
y'     y'  y'

dx  y'  dx x
y dx y'
 f d f 
y'  
0

 y dx y' 
df
f f
f
 y' 
 y''
dx
y x
y'
2nd Form of Euler’s Equation (cont’d)

We find:
d  f  df f
y'   

dx  y'  dx x
f d 
f 
  f  y'   0
x dx 
y' 
2nd Form of Euler’s Equation is thus
f
0
x
d
f 
f  y'   0

dx 
y' 
f
f  y'  constant
y'
If f does not depend on “x”, there is a conserved quantity
Function with Several Dependent
Variables

Consider a more general case where f is
a functional of several dependent
variables.

E.g. Motion of one particle in 3 dimensions,
or many particles…

f  f y (x), y (x),...; x,
f  f y1 (x), y1, (x), y2 (x), y2, (x),...; x
i
,
i

i  1,2,...,n
Several Dependent Variables…

Repeat our previous reasoning…
yi(a,x)=yi (0,x)+ ai(x)
J


a a


a

x2
x1
f (y(a , x), y'(a , x); x)dx

x2
 f
J
d f 
  

 (x)dx
, i
x
1
a
dx yi 
i  yi
Several Dependent Variables…

So we have
x2
 f
J
d f 
  

i (x)dx

,
x1
a
dx yi 
i  yi


Functions i are all independent.
Each term of the sum must be null for the above
derivative to vanish at a=0.
f
d f

 0,
,
yi dx yi
i  1,2,...,n
Euler’s Equations for many dependent variables
Euler Equations with Auxiliary Conditions




In many problems, additional constrains come
into play.
Consider e.g. motion constrained to a spherical
shell, or some other type of curved surface.
Then obviously the path (motion) must be on
the surface, and thereby satisfy the equation of
the surface, which can be generally be written:
g{yi;x}=0
One therefore introduce constraint equations.
Euler Equations with Auxiliary Conditions

x2
 f
J
d f 
  

 (x)dx
, i
x1
a
y
dx
y
i 
i
i 
J

a
Also include constraints:
The variations
y
a

f  f yi , yi, ; x  f y, y', z, z'; x
Consider a case:
and
z
a

x2
x1
 f
d f  y  f
d f  z 
 
 a   z  dx z '  a  dx

y
dx
y'


g yi ; x g y, z; x
are thus no longer independent.
 g y g z 
dg  

da  0

 y a z a 
No terms in x appear since
x
0
a
y
 1 (x)
a
y(a , x)  y(x)  a1 (x)
z(a , x)  z(x)  a2 (x)
We write:
z
 2 (x)
a
The constraints equation becomes:
or:
J

a
J

a
J

a



x2
x1
x2
x1
x2
x1
g
g
1 (x)   2 (x)
y
z
2 (x)
g y

1 (x)
g z
 f
d f  y  f
d f  z 




 a  z dx z '  a  dx

y
dx
y'


 f

d f 
d f 
 f


(x)



(x)


 2  dx
 1

y
dx
y'
z
dx
z
'


 f
d f   f
d f   g y  
 
   z  dx z '   g z   1 (x)dx

y
dx
y'


J

a

x2
x1
 f
d f   f
d f   g y  
 
   z  dx z '   g z   1 (x)dx

y
dx
y'


 f d f   f d f   g y 
 y  dx y'    z  dx z '   g z   0
 f
d f   g 

 y dx y'   y 
1
d f   g 
 f
 
 z dx z '   z 
1
The left hand side involves only derivatives of f and g with respect to y and y’,
whereas the right hand side involves only derivatives of f and g with respect to z
and z’. Because y and z are both functions of x, the two sides may be set equal to
some function of x, which we note: -l(x).
f d f
g

 l (x)  0
y dx y'
y
f d f
g

 l (x)  0
z dx z '
z
The complete solution requires one finds three functions: y(x), z(x), and l(x).
Fortunately, three relations may be used: the above two equations and the
constraints Eq.
The function l(x) is known as Lagrange underdetermined multiplier.
In general, if there are many external constraints, one has:
g j
f
d f

  l j (x)
0
yi dx yi '
y
j
i
g j yi ; x  0
Note that the constraints can also be written:
g j
 y
j
i
dyi  0,
i  1,2,..., m

 j  1,2,...,n
Often more “useful” in problem solutions….
m equations
m+n unknowns
n equations
The d Notation

To simplify writing in calculus of variations, one uses a
shorthand notation. Consider:
x2  f
J
d f  y
da    
da dx

x
1
a
 y dx y'  a
x2
 f
d f 
dJ    
d ydx

x
 y dx y' 
1
J
da
a
y
dy 
da
a
dJ 
The d Notation
x2

Condition of extremum
d J  d  f {y, y'; x}dx  0

Swap d and integral:
d J   d fdx
x1
x2
x1


Note:

So…

 dy  d
 d y 
 dx  dx
d y'  d 
dJ 



x2
x1
x2
x1
 f
f d 
 y d y  y' dx d y  dx
 f
d f 

 y dx y'  d ydx
x2
x1
 f

f
d
y

d
y'
dx
 y

y' 
The d Notation


Do keep in mind that the d notation is only a
shorthand expression for a more detailed and
precise quantity.
You can visualize the varied path dy as a virtual
displacement from the actual path consistent
with all forces and constraints. It is to be
distinguished from dy (differential displacement)
by the condition that dt=0 (fixed time)
Geodesic on a sphere

ds   d  sin  d
2
2

 '  sin  
2 1/2

2
s   ds    d 2  sin 2  d
  d  2

s        sin2  
 d 

1
2

f   '  sin 
2
2

 '  sin  
2
2
 '2
 '  sin  
2
d
sin 4
2


  a  '  sin  
sin 2   a  '2  sin 2 
1/2
2
2
1/2
2
sin 4   a 2 sin2   a 2 '2

1/2
d
1
  '
d
d
a csc 2 

d
1  a 2 csc 2 

f
 constant
y'
1/2

2 1/2
f
0

f  y'
1/2
2

'
 '2  sin 2 
 '


1/2
 constant=a

1/2
 cot  
c
 b 
  sin 1 
cot   bsin   c
1/2
a