PHYS 705: Classical Mechanics Calculus of Variations I 2

Calculus of Variations: The Problem

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

xB I  F y( x ), y '( x ); x  dx xA Comments:

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 called 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. 3

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.

4. The method does not explicitly give y(x). Instead, we will get a diff eq for y(x).

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

6. x is assumed to be unidirectional between xA and xB. If not, break it into different pieces. 4

Calculus of Variations

Consider a family of functions  y ( a , x )  parameterized by a.

y desired path yB yxyx(,)a (0,)  a () x y(0,x)  a is a parameter a(x )   (xC ) is a 2 -smooth variation at x

yA   (xA )   ( x A )  0 x xA xB

Note: y(,a xA )  y A  for all a, i.e., all sampled paths has the  same end points, x and x y(,a xB )  y B  A B

y(0, x ) is the desired path 5

Calculus of Variations

A necessary, but not sufficient, condition for the min(max)imization of I is: dI  0 da a 0 Taking the derivative of I wrt a,

dI d xB   F y( x ), y '( x ); x  dx  da d a  xA  xB d  F y( x ), y '( x ); x  dx (the end points xA and xB are fixed)  da xA xB Fy  Fy '    dx (x is the independent variable and  ya  y ' a does not vary with a) xA   6

Calculus of Variations

Recall, we have yxyx(,)a (0,)  a () x

y So, (x ) a

And, since we also have yxyx '(a , ) '(0, )  a '( x )

y' d So,  a dx

dIxB  F  F d  This gives,    dx da   yy  ' dx xA  

Now, to continue, we will integrate the second term by parts. 7

dI xB F F d    dx da  y y ' dx Calculus of Variations xA  

To integrate by parts, udv uv  vdu (with the fixed limits x and x )   A B F dF   Let, u  du    dx y' dx  y '  d dv  dx v  dx

xxB x So, BFd F B dF   dx() x () x dx       xA    x B   0 y'' dx y  dx  y ' xAxA x A   no variations @ end points udv uv  vdu 8

Calculus of Variations xBFd x B dF   So, we have, dx ( x )   dx y' dx  dx  y ' xA x A   dI Putting this back into our expression for , da dIxB  F dF    (x )   ( x )   dx  0 da   y dx  y ' xA    xB F d  F     (x ) dx  0  y dx  y ' xA    Since this has to be zero for any arbitrary C 2 variation  ( x ) ,

F d  F  Euler-Lagrange Equation (1744)    0 y dx  y '  (fundamental lamma of Calc. of Var.) 9

Calculus of Variations (Euler-Lagrange Equation) F d  F     0 Euler-Lagrange Equation y dx  y '  Comments: 1. This is the diff eq whose solution y(x) is the function we seek to min(max)imize with respect to F. dI 2. We only used a necessary condition for an extremum  0 da a 0 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.

3. It is important to keep track of which variable is the independent variable. Here, it is x. 10

Euler-Lagrange Equation (Special Case)

We will now derive an alternative form of the EL equation for the case when F does not explicitly depend on x, i.e., F( yy , ') (obviously, F depends on x implicitly thru y) Observe this: d F  d   F F dF yFy'  '   y '' dx y'  dx   y '' y dx

Fy  Fy'  F      yx  yx'  x 

d F  dF  F  F  F  y'  F   y'   y'' y '  y '' dx  y '  dx  y '  y ' y y ' 11

Euler-Lagrange Equation (Special Case)

Cancelling and grouping like-color terms,

dF  dFF    F F yFy'  '    y ''  y '' dx y'  dx   y ' y  y ' y '

Now, if y(x) satisfies the Euler-Lagrange Equation, the remaining term on the RHS will be zero as well. Thus, we have,

d F  F y' F   0 or y'  F  const dx y '  y '

This is a much easier equation to solve since it is 1st order only. We basically have done one integration already. 12

Few Classic Problems (#1: a Straight Line)

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

y Arc length is given by:

B I  ds A

where, 2 2 2 dy  x x x ds dx dy or ds 1   dx A B dx  Note: ds is written in So, we need to find y(x) that minimizes: Euclidian space. If it is B not (e.g. sphere), the result I1   y ' 2 dx (geodesic) will be different   A NOT a straight line. 13

Few Classic Problems (#1: a Straight Line)

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

F d  F  2    0 with F1   y '  y dx  y ' 

FF  12 1/2 y ' 0  1'y   2' y     2 y  y ' 2 1 y '  EL equation gives: d F  F y '    0   const dx y '  y ' 1 y ' 2 14

Few Classic Problems (#1: a Straight Line)

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

y'  const y mx  b (it is a line and m and b to be determined by the endpoints)

Alternatively, since F does not depend on x explicitly, we can use the second form for the EL equation,

d F  F y' F   0 y'  F  const dx y '  y '

y ' 2 y' 1' y   const 1 y ' 2 15

Few Classic Problems (#1: a Straight Line)

2 Multiplying 1   y '  to the whole equation, we have, y'1'2  ycy 2  1'    2 (c being a constant) 1c 1   y ' 2

y'  const and the same result, y mx  b 16