PHYS 705: Classical Mechanics. Calculus of Variations I

Transcription

1 1 PHYS 705: Classical Mechanics Calculus of Variations I

2 Calculus of Variations: The Problem Determine the function y() such that the following integral is min(ma)imized. Comments: I F y( ), y'( ); d 1. Notation: is the independent variable ( =t in our mech prob) y() is a function of and y () = dy/d F{ } is a functional on y(). Intuition: Let say y() denote a route in the -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 3 Calculus of Variations: The Problem Comments (cont): 3. To keep it simple, we will fi the end points ( and ). We assume that they are known. Method can be etended to situation where ( and ) can be varied as well The method does not eplicitly give y(). Instead, we will get a diff eq for y(). F is assumed to be at least twice differentiable, i.e., C in all its arguments. is assumed to be unidirectional between and. If not, break it into different pieces.

4 4 Calculus of Variations Consider a family of functions parameterized by. y(, ) y y desired path y(, ) y(0, ) ( ) y(0,) y ( ) is a parameter ( ) is a C -smooth variation at ( ) ( ) 0 Note: y(, ) y y(, ) y for all, i.e., all sampled paths has the same end points, and y(0, ) is the desired path

5 5 Calculus of Variations necessary, but not sufficient, condition for the min(ma)imization of I is: di d 0 0 Taking the derivative of I wrt, we have, di d Fy( ), y'( ); d d d d F y ( ), y '( ); d d F y F y' d y y' (the end points and are fied) ( is the independent variable and does not vary with )

6 6 Calculus of Variations Recall, we have y(, ) y(0, ) ( ) So, nd, y ( ) y'(, ) y'(0, ) '( ) So, y' d d This gives, di F F d y y' d d d Now, to continue, we will integrate the second term by parts.

7 7 Calculus of Variations di d F Fd d y y ' d To integrate by parts, udv uv vdu Let, F d F u du d y' dy' d dv d v d (with the fied limits and ) So, Fd F df d ( ) ( ) d y' d y' d y' 0 no end points udv uv vdu

8 8 Calculus of Variations So, we have, Fd df d ( ) d y' d d y' di Putting this back into our epression for, d di F d F ( ) ( ) d 0 d y d y' F d F ( d ) 0 y d y' Since this has to be zero for any arbitrary C variation, ( ) F d F y dy' 0 Euler-Lagrange Equation (1744) (fundamental lamma of Calc. of Var.)

9 9 Calculus of Variations (Euler-Lagrange Equation) F d F y dy' 0 Euler-Lagrange Equation Comments: 1. This is the diff eq whose solution y() is the function we seek to min(ma)imize with respect to F.. We only used a necessary condition for an etremum di 0 d 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 It is important to keep track of which variable is the independent variable. Here, it is.

10 10 Euler-Lagrange Equation (Special Case) We will now derive a particular useful form of the EL equation for the case when F does not eplicitly depend on, i.e., Observe this: F( y, y ') d ' F ' d F y F y y'' F df d y ' d y ' y ' d (obviously, F depends on implicitlythru y) F y F y' F y y' d d F y' F y ' d F F F y' y'' d y ' y ' y y' F y ' y''

11 11 Euler-Lagrange Equation (Special Case) Cancelling and grouping like-color terms, d ' F ' d y F y F F y'' F d y ' d y ' y y ' F y '' y ' Now, if y() satisfies the Euler-Lagrange Equation, the remaining term on the RHS will be zero as well. Thus, we have, d d F y' F 0 y ' or F y ' F const y ' This is a much easier equation to solve since it is 1 st order only. We basically have done one integration already.

12 1 Few Classic Problems (#1: a Straight Line) 1. What plane curve connecting two given points has the shortest length? rc length is given by: y I ds where, dy ds d dy or ds 1 d d So, we need to find y() that minimizes: I 1 y ' d Note: ds is written in Euclidian space. If it is not (e.g. sphere), the result (geodesic) will be different NOT a straight line.

13 13 Few Classic Problems (#1: a Straight Line) To solve this problem, we apply the Euler-Lagrange equation, F d F y dy' 0 with F y 1 ' EL equation gives: F 1 1/ ' 0 F 1 y' y' y y y' 1 y ' d d F y ' 0 F y' y ' 1 y ' const

14 14 Few Classic Problems (#1: a Straight Line) Solving for y gives y being equals to another constant, y' const y mb (it is a line and m and b to be determined by the endpoints) lternatively, since F does not depend on eplicitly, we can use the second form for the EL equation, d d F y' F 0 y ' F y' F const y ' y ' y ' 1 y ' const 1 y '

15 15 Few Classic Problems (#1: a Straight Line) y 1 ' Multiplying to the whole equation, we have, y' 1 y' c 1 y' (c being a constant) 1c 1 y' y' const and the same result, y mb

16 16 Few Classic Problems (#: Catenoid). Minimum Surface of Revolution Problem (Soap Film bet Wire Loops) Two points in an y-plane are given. curve in the plane connecting the two points is revolved about an ais. Find the curve that results in the minimum surface area of revolution. ds y element of surf area da 1 y (note: is our indep var and it is unidirectional) (Goldstein is not correct on this) ds d dy da y ds y 1 y ' d rea to be minimize is: 1 I da y 1 y' d 1

17 17 Few Classic Problems (#: Catenoid) For the given integrant, the EL equation is: F d F y dy' 0 with F( yy, '; ) y 1 y' With F not eplicitly depends on (the independent variable), we can use the alternative form for the EL equation instead: F y' F c y ' where c is an integration constant F 1 1/ yy' y 1 y' y' y ' 1 y '

18 18 Few Classic Problems (#: Catenoid) Putting it into the alternative form of the EL equation: yy ' y' y 1 y' c 1 y ' yy ' y 1 y' y 1 y ' 1 y ' c c y c 1 y' y or y' 1 c

19 19 Few Classic Problems (#: Catenoid) So the diff eq that we need to solve is now: dy d 1 c y c dy y c 1 dc ccosh b y c (both c and b are integration constants which will depend on the end points 1 and ) y ccosh b c 1 y This curve is called a Catenary and the surf. of revolution is called a Catenoid.

20 0 Few Classic Problems (#: Catenoid) y rendering of the catenoid using Mathematica Soap film between two loops

21 1 Few Classic Problems (#: Catenoid) There are more interesting subtleties to this problem: 1. When one considers a family of catenaries that go through one fied point (let say the left one : pt 1) 1 fied y - ll the caternaries will tangent on an envelope curve (blue) which is a parabola with its focus at pt #1 - Interesting observations: If we choose pt at this location (other similar intersection pts), there will be solns. There will be no soln here.

22 Few Classic Problems (#: Catenoid). What we have considered so far involve only twice-differentiable solutions. There is also a nondifferentiable solution that has physical relevance! Smooth Caternary Solution y 1 y 1 Goldschmidt Solution Soap film bet two rings Keeps moving the two loops apart until the red surf area > blue surf area

23 3 Few Classic Problems (#3: rachistochrone problem) +y 1 v g Find the path that goes between pt 1 and pt with the least time under gravity. minimize t 1 ds v With T (Goldstein, p.43), the solution is a cycloid given by: 0 0 +y a a sin, 1 cos a y a 0 mv0 (HW is to derive the cycloid with.) T 1

24 4 Few Classic Problems (#3: rachistochrone problem) nimation for the cycloid: Movie on YouTube: