Understanding Calculus of Variations in Lagrangian Dynamics

phy 711 classical mechanics and mathematical n.w
1 / 19
Embed
Share

Explore the development of Lagrangian dynamics in Chapter 3 of Classical Mechanics, focusing on the calculus of variations. Learn about functional minimization, extremizing functions, and necessary conditions for optimization.

  • Lagrangian dynamics
  • Calculus of variations
  • Classical Mechanics
  • Optimization
  • Extremizing functions
rayaanacharya
ship_ris Follow

Uploaded on | 0 Views

Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.

Download Presentation

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.

E N D

Presentation Transcript

  1. PHY 711 Classical Mechanics and Mathematical Methods 11-11:50 AM MWF Olin 107 Plan for Lecture 5: Start reading Chapter 3.17 1. Introduction to the calculus of variations 2. Example problems 3. Brachistochrone 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 1

  2. 2 9/12/2016 PHY 711 Fall 2016 -- Lecture 5

  3. In Chapter 3, the notion of Lagrangian dynamics is developed; reformulating Newton s laws in terms of minimization of related functions. In preparation, we need to develop a mathematical tool known as the calculus of variation . Minimization of a simple function local minimum dV = 0 dx global minimum 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 3

  4. Minimization of a simple function , ) ( function a Given x V value(s) the find of V x x for which ( minimized is ) maximized) (or . dV = Necessary condition : 0 dx local minimum dV = 0 dx global minimum 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 4

  5. Functional minimization functions of family a Consider ( with the , ) x end points y dy = = ( ) and ( ) function a and ( ), , . y x y y x y L y x x i i f f dx dy function the Find ( which ) extremizes ( ), , . y x L y x x dx = Necessary condition : 0 L Example : 1 , 1 ( ) dx ( ) dy ) 2 2 = + L ( 0 , 0 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 5

  6. Example : 1 , 1 ( ) dx ( ) dy 2 2 = + L ( 0 , 0 ) 2 1 dy 0 = + 1 dx dx Sample functions : 1 1 = = + 4789 . 1 = ( ) 1 y x x L dx 1 4 x 0 1 = = + = 4142 . 1 = ( ) x 1 1 2 y x L dx 2 0 1 = = + . 1 = 2 2 ( ) 1 4 4789 y x x L x dx 2 0 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 6

  7. Calculus of variation example for a pure integral functions dy function the Find ( which ) extremizes ( ), , y x L y x x dx x dy dy f x where ( ), , ( ), , . L y x x f y x x dx dx dx i = Necessary condition : 0 L + At any let , ( ) ( ) x ( ) x y x y x y x ( ) ( ) ( ) dy x dy dy x + dx dx dx Formally : x f f dy f = + . L y dx ( ) / y dy dx dx dy x , , x x y i dx 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 7

  8. After some derivations, we find , x f f dy f = + L y dx ( ) / y dy dx dx dy x , x x y i dx x f d f f = dx y = 0 for all x x x ( ) i f / y dx dy dx dy x , , x x y i dx f d f = 0 for all x x x ( ) i f / y dx dy dx dy , , x x y dx 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 8

  9. = = Example : End points - - ) 0 ( y ; 0 ) 1 ( y 1 2 2 1 dy dy dy = + = + 1 ( ), , 1 L dx f y x x dx dx dx 0 f d f = 0 ( ) / y dx dy dx dy , , x x y dx / d dy dx = 0 ( ) dx 2 + 1 / dy dx Solution: + = / dy dx dy dx K = ' K K ( ) 2 2 1 K 1 / dy dx = ( ) y x x = + ( ) y x ' K x C 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 9

  10. Example : Lamp shade shape y(x) x 2 2 dy dy dy f x = + = + 2 1 ( ), , 1 A x dx f y x x x dx dx dx i f d f y = 0 ( ) / y dx dy dx dy , , x x y dx xi yi / d xdy dx = 0 ( ) dx 2 + 1 / dy dx x xf yf 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 10

  11. / d dx xdy dx = 0 ( ) 2 + 1 / dy dx / xdy dx = K 1 ( ) 2 + 1 / dy dx 1 dy dx = 2 x 1 K 1 2 x x K = + ( ) y x ln 1 K K 2 1 2 K 1 1 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 11

  12. + 2 + 3 = ( ) y x ln 2 1 x x 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 12

  13. Another example: (Courtesy of F. B. Hildebrand, Methods of Applied Mathematics) ( ) x ( ) 0 ( ) 1 = = Consider curves all with and 0 1 y y y minimize that integral the : 2 1 dy = 2 constant for 0 I ay dx a dx 0 Euler - Lagrange equation : 2 d y 2 + = 0 ay dx ( ) sin a x = ( ) ( ) y x sin a 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 13

  14. dy Review for : ( ), , , f y x x dx x dy f x necessary a condition extremize to : f y(x), ,x dx dx i f d f dy = 0 Euler-Lagrange equation ( ) / y dx dy dx dy , , x x y dx dy Note that for ( ), , , f y x x dx df f dy f d f = + + ( ) / dx y dx dy dx dx dx x d f dy f d dy f dy = + + ( ) ( ) / / dx dy dx dx dy dx dx dx x d f f Alternate Euler-Lagrange equation = f ( ) / dx dy dx dx x 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 14

  15. Brachistochrone problem: (solved by Newton in 1696) http://mathworld.wolfram.com/BrachistochroneProblem.html A particle of weight mg travels frictionlessly down a path of shape y(x). What is the shape of the path y(x) that minimizes the travel time from y(0)=0 to y( )=- ? 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 15

  16. 2 dy dx gy + 1 x y x f f f ds v = = = 2 because 1 2 T dx mv mgy 2 i i x y x i 2 dy dx y + 1 dy dx Note that for the original form of Euler-Lagrange equation: = ( ), y x , f x d dx f dy f y d dx f = = 0 f 0, ( ) ( ) / / dy dx dx dy dx dy dx , x , x y differential equation is more complicated: 1 d dx 2 dy dx y dy dx = 0 + 1 1 2 d dx 2 dy dx = 0 + 1 y 3 2 dy dx + 1 y 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 16

  17. 2 dy + 1 dy dx = ( ), , f y x x dx y d f dy f = f ( ) / dx dy dx dx x 2 1 d dy = + = 0 1 2 y K a dx dx 2 dy + 1 y dx 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 17

  18. 2 dy ( ) d + = = = a cos a 2 2 1 2 y K a Let 2 dy sin cos 1 y a a dx 2 2 2 2 sin = = dx 2 dy a 2 a = 1 1 1 dx y 2 2 2 sin y a dy 2 = dx ( ) ( ) 0 = d = 1 cos ' ' sin x a a a 1 y Parametric equations for Brachistochrone: ( ( cos = a y ) = sin x a ) 1 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 18

  19. Parametric plot -- plot([theta-sin(theta), cos(theta)-1, theta = 0 .. Pi y x 9/12/2016 PHY 711 Fall 2016 -- Lecture 5 19

Related


No related presentations.

More Related Content