Integrals and Invariants of EulerLagrange Equations


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1 Lecture 16 Integrals and Invarants of EulerLagrange Equatons ME 256 at the Indan Insttute of Scence, Bengaluru Varatonal Methods and Structural Optmzaton G. K. Ananthasuresh Professor, Mechancal Engneerng, Indan Insttute of Scence, Banagalore 1
2 Outlne of the lecture Frst ntegrals of EulerLagrange equatons Noether s ntegral Parametrc form of EL equatons Invarance of EL equatons What we wll learn: How to smplfy the EL equatons to easytosolve dfferental equatons n some cases How to take advantage of parametrc forms and change of varables 2
3 More than formulatng equatons So far, we have learnt how to get dfferental equatons and boundary condtons usng the technques of calculus of varatons. Indeed t s powerful. We have learnt varous generalzatons: Multple dervatves Multple functons Two and three ndependent varables Equalty and nequalty constrants Varable end condtons Broken extremals and corner condtons There are a few concepts that become useful when we also want to solve them usng analytcal (rather than numercal) technques. We wll stll not get a soluton rght away but we get a smpler or easly solvable form of dfferental equatons. In some cases, we get some nsght nto the problem. Ths s the am of the content of ths lecture. 3
4 Consder the brachstochrone problem From Slde 14 n Lecture 11 A Mnmze H g L B And we have Drchlet (essental) boundary condtons at both the ends. 4
5 Looks formdable to solve At frst sght, ths dfferental equaton looks to be too complcated to solve analytcally And we are far from showng that the soluton of ths s a cyclod. Frst ntegral of EulerLagrange equatons provdes a way out of ths. 5
6 Frst ntegrals of specal forms Solvng dfferental equatons means that we are ntegratng them. Ths s what we do whether we do t analytcally or numercally. So, frst ntegrals mply that we are ntegratng the dfferental equaton to some extent. For EulerLagrange equatons, some specal forms, are amenable for wrtng the frst ntegrals and thereby reduce ther degree and hence ther complexty. 6
7 Integrand of the form EulerLagrange equaton has only one term, n ths case. It s smply an algebrac equaton; not a dfferental equaton. So, there s nothng to ntegrate here. Notce also that t does not have a boundary condton too. Recall that the smplest boundary condton term nvolves y. 7
8 Integrand of the form EulerLagrange equaton has only one term, n ths case too. We can express y n ths form and now t can be drectly ntegrated ether analytcally (when t s possble to do) or numercally. See Slde 13 n Lecture 11 for an example. 8
9 Integrand of the form EulerLagrange equaton has two terms. Expanded. A smple contracton of the terms. An elegant frst ntegral. Multply by y through out. 9
10 Brachstochrone problem has the form Mnmze Now, nstead of that, we get ths. 10
11 Smplfcaton of the Brachstochrone dfferental equaton A much smpler form to solve. 11
12 An nsght wth the frst ntegral Consder the acton ntegral for the dynamcs of a sprngmass system: Sde 33 n Lecture 3 f Opt xt () Opt xt () Opt xt () T 2 1 dx 1 2 A = m kx + fx dt 0 T 0 T 0 { } A = KE PE dt A = 2 dt 2 Ldt Ths s of the form and hence s amenable for the elegant frst ntegral. Hamlton s prncple for dynamcs. 12
13 An nsght wth the frst ntegral: conservaton of energy Opt xt () T 2 1 dx 1 2 A = m kx + fx dt 0 2 dt 2 F yf = C= y F  xf = C x constant ( ) = mx + kx fx = c 2 2 KE + PE = constant 2 2 mx kx fx x mx C Thus, the frst ntegral gave rse to the prncple of conservaton of energy. 13
14 An ntegrand of the form F y ( Fy ) = 0 2 fy fy 1+ y = y y y y fy 1+ y fx f 0 2 y f = 2 2 3/2 1+ y 1+ y 1 y fy f + y fy fy = 1+ y y(1 ) x y 0 2 fy 1+ y 2 fy fy x = 2 ( + ) 0 Not ntegrated, but s a smpler form to deal wth. 14
15 Now, try to solve ths functonal It s of the form: Therefore, 2 2 y Thus, y + y y = C 2 2 y + y y C y y = + No sght of soluton yet! (despte usng the frst ntegral) Let us try change of varables: 15
16 Change of varables Now, ths s a new functonal n u and v where we need to fnd v(u). What would be the EulerLagrange equatons for ths? 16
17 New functonal satsfes the old equaton! s satsfed by v(u) just as y(x) satsfes F d F = y dx y 0 So, we need to get the new functonal n the form shown above, when we change varables. 17
18 An example Wth And notng that becomes Check the algebra by workng t out n detal. 18
19 Example (contd.) Thus, the soluton of the dfferental equaton n slde 15 s Thus, Wth or 19
20 A note about change of varables Change of varables s a great way to solve an otherwse dffcult problem. But nobody can tell us whch change of varables wll work for a gven problem. You just have to know or guess. But note that calculus of varatons lets you use change of varables. 20
21 Parametrc form and EulerLagrange equatons Parametrc form dx = Where y ' = x dt y x F xt ( ), yt ( ), y ψ = x x Then, we have 2 2 y Mn J = F( x, y, y ) dx = F x( t), y( t), x dt y( x) x xt (), yt () x x 1 1 t 2 Mn J= ψ ( xyxy,,, ) dt t 1 and t satsfes the followng EL equatons. and t t should not depend on t explctly. 21
22 A comment We saw that the change of varables and the parametrc form do not alter the form of EulerLagrange equatons. It s very useful n a number of stuatons. Parametrc form s especally useful when y(x) s to denote a closed curve. It s also useful n dealng wth dynamcs problems too. There s a more general theorem related to nvarance of EulerLagrange theorem. It s called Noether s theorem. Noether s theorem s related to the frst ntegrals we dscussed earler n ths lecture. It leads to conserved quanttes. Proved by a German mathematcan Emmy Noether, ths theorem was prased by Ensten for ts penetratng thnkng. It s used wdely n mathematcal physcs. Noether s theorem next 22
23 Invarance under transformatons Consder we say that the functonal s nvarant under the transformaton shown above. 23
24 Noether s theorem Consder A oneparameter transformaton. If we say that the functonal s nvarant under the transformaton shown above. Then, ( F ) ( F yf ) y ψ φ y = α α= 0 α α= 0 constant 24
25 Noether s theorem (case of many functons) If Consder xˆ = φ( x, y, y,, y, y, y,, y, α) x2 2 dy ˆ dyˆ J = F x, y, dx = J = F xˆ, yˆ, dxˆ dx dxˆ x 1 2 n 1 2 yˆ = ψ ( x, y, y,, y, y, y,, y, α), = 1,2,,n 1 2 n 1 2 n xˆ 1 1 we say that the functonal s nvarant under the transformaton shown above. Then, xˆ n n ( ) F ( ) y F yf y ψ φ = α α = 1 α = 0 α = 0 constant 25
26 An applcaton of Noether s theorem Consder a system of n partcles wth poston coordnates: x(t), y(t), z( t) ( = 1, 2,, n) n 1 KE = m x y z = 1 Let the potental energy be = PE = U ( x1, y1, z1,, xn, yn, zn) The knetc energy of such a system = ( 2 2 2) Consder the acton ntegral = ( ) Consder x = x cosθ + y * y = x snθ + y z * * = z snθ cosθ t t 1 A = KE PE dt 0 A oneparameter famly of transformatons for the rotaton of the system of partcles about the zaxs. Suppose that A s nvarant under the above transformaton. 26
27 Compare wth the generc transformaton. xˆ = φ( xyy,,, α) yˆ = ψ ( xyy,,, α) = φ(t,, y,,, y,, θ) x (t, x,y, z, x,y, z, θ) y (t, x,y, z, x,y, z, θ) z (t, x,y, z, x,y, z, θ) * t x z x z * = ψ1 * = ψ2 * = ψ3 t = t φ No transformaton n the x = x cosθ + y * j j j y = x snθ + y z * j j j * j = z j ndependent varable. snθ cosθ j = 1, 2, N Now, as per Noether s theorem, we have n ( ) ψ ( ) φ Fy F yf constant y = = 1 α α = 0 α α = 0 ψ = 1,2, 3 N Note that N KE ψ 1 KE ψ 2 KE ψ 3 φ + + = constant = 0 = 1 x θ θ= 0 y θ θ= 0 z θ θ θ θ = 0 = 0 N = # partcles (contd.) 27
28 Noether s theorem applcaton (contd.) Note that (contd.) N x y z * θ * θ * θ θ = 0 θ = 0 θ = 0 = y 0 = x = {( mx )( y) ( my )( x) } constant { } r p = = constant = 1 = 1 and N φ θ θ = 0 = n KE ψ KE ψ KE ψ = 1 x θ θ= 0 y θ θ= 0 z θ θ = 0 where p = ( mx,my,m z ) ad n r = ( x, y, z ) = Lnear momentum vector Poston vector 0 28 constant Conservaton of angular momentum!
29 Why s Noether s theorem mportant? Because t lets us fnd conserved quanttes for any calculus of varatons problems leadng to frst ntegrals. It can be extended to multple functons. It can be extended to multple dervatves. In mechancs, conservaton of energy, conservaton of lnear momentum, and conservaton of angular momentum, etc., follow from Noether s theorem. The prevous example llustrated t for the conservaton of angular momentum. 29
30 The end note Frst ntegrals for varous forms of functonals Frst ntegrals and nvarance of Euler Lagrange equatons Ways to smplfy EulerLagrange equatons and thereby solve them analytcally. Change of varables does not alter the form of EulerLagrange equatons. Parametrc form too does not alter the form of the El equatons. Invarant transformatons and conserved quanttes usng Noether s theorem Thanks 30
Integrals and Invariants of
Lecture 16 Integrals and Invarants of Euler Lagrange Equatons NPTEL Course Varatonal Methods and Structural Optmzaton G. K. Ananthasuresh Professor, Mechancal Engneerng, Indan Insttute of Scence, Banagalore
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