Introduction to Calculus of Variations in Lagrangian Dynamics

phy 711 classical mechanics and mathematical n.w
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Explore the development of Lagrangian dynamics in Chapter 3 of Classical Mechanics, focusing on the calculus of variations for minimizing functions. Learn about extremizing functions, necessary conditions, and examples illustrating the concept. Dive into functional minimization functions and their applications in mathematical methods.

  • Calculus of Variations
  • Lagrangian Dynamics
  • Extremizing Functions
  • Classical Mechanics
  • Mathematical Methods
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  1. PHY 711 Classical Mechanics and Mathematical Methods 9-9:50 AM MWF Olin 107 Plan for Lecture 6: Start reading Chapter 3.17 1. Introduction to the calculus of variations 2. Example problems 3. Brachistochrone 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 1

  2. 2 9/08/2017 PHY 711 Fall 2017 -- Lecture 6

  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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 11

  12. + 2 + 3 = ( ) y x ln 2 1 x x 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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( )=- ? 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 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 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 18

  19. Parametric plot -- plot([theta-sin(theta), cos(theta)-1, theta = 0 .. Pi]) y x 9/08/2017 PHY 711 Fall 2017 -- Lecture 6 19

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