Fixed-end point problems

Transcription

1 The Catenar Variational Methods & Optimal Control lecture 4 Matthew Roughan <matthew.roughan@adelaide.edu.au> The potential energ of the cable is L W p {}= mg(s)ds Where L is the length of the cable =() d g 1 Discipline of Applied Mathematics School of Mathematical Sciences Universit of Adelaide April 14, Catenar problem where we have pullies on top of each plon, and a large amount of cable. Under appropriate conditions it will reach an equilibrium shape. The critical features of this problem are that the end-points are fied but the length L of the cable is unconstrained. Variational Methods & Optimal Control: lecture 4 p.1/4 Variational Methods & Optimal Control: lecture 4 p.3/4 Fied end-point variational problem Fied-end point problems We ll start with the simplest functional maimization problem, and show how to solve b finding the first variation and deriving the Euler-Lagrange equations: ( ) d f d f = (, ) = () ( 1, 1 ) Variational Methods & Optimal Control: lecture 4 p.2/4 Variational Methods & Optimal Control: lecture 4 p.4/4

2 Formulation Define the functional F : C 2, 1 ] IR F{}= f(,, )d, where f is assumed to be function with (at least) continuous second-order partial derivatives, WRT,, and. Problem: determine C 2, 1 ] such that ( )= and ( 1 )= 1, such that F has a local etrema. The Catenar L W p {}= mg(s)ds Change of variables ds= 1+ d. So the functional of interest (the potential energ) is where W p {} = mg = mg f(,, ) = d, f(,, )d, The Catenar Variational Methods & Optimal Control: lecture 4 p.5/4 How do we tackle these problems Variational Methods & Optimal Control: lecture 4 p.7/4 L W p {}= mg(s)ds But I don t know how to evaluate this integral directl. Lets do a simple change of variables. The length of a line segment from (,) to (+δ,+δ) is δs δ 2 + δ 2 ( ) 2 δ = 1+ δ δ ds = 1+ d δ s δ δ look at small perturbations about the ma/min. =() ma + ε For a local maimum f(+ε) f() Conditions for etremals, i.e., f ()= Variational Methods & Optimal Control: lecture 4 p.6/4 Variational Methods & Optimal Control: lecture 4 p.8/4

3 Perturbations of functions Perturbations of functions = + ε η = + ε η 2 (, ) = () ( 1, 1 ) (, ) = () ( 1, 1 ) Perturbations of functions Variational Methods & Optimal Control: lecture 4 p.9/4 The Functional of interest. Variational Methods & Optimal Control: lecture 4 p.11/4 = + ε η 2 Define the functional F : C 2, 1 ] IR F{}= f(,, )d, (, ) = () ( 1, 1 ) where f is assumed to be function with continuous second-order partial derivatives, WRT,, and. Problem: determine C 2, 1 ] such that ( )= and ( 1 )= 1, such that F has a local etrema. The space of possible curves is S= { C 2, 1 ] ( )=,( 1 )= 1 } The vector space of allowable perturbations is H = { η C 2, 1 ] η( )=,η( 1 )= } Variational Methods & Optimal Control: lecture 4 p.1/4 Variational Methods & Optimal Control: lecture 4 p.12/4

4 Perturbation functions The vector space of allowable perturbations is H = { η C 2, 1 ] η( )=,η( 1 )= } η 1 The first variation For small ε the quantit F{+εη} F{} δf(η,)=lim = ε ε is called the First Variation. η f + f ] η d For F{} to be a minimum, for small ε, F{ŷ} F{}, so the sign of δf(η,) is determined b ε. As before, we can var the sign of ε, so for F{} to be a local minima it must be the case that δf(η,)=, η H Variational Methods & Optimal Control: lecture 4 p.13/4 Variational Methods & Optimal Control: lecture 4 p.15/4 What to do Regard f as a function of 3 independent variables:,, Take ŷ()=()+εη(), where S and η H. Talor s theorem (note is kept constant below) So f(,ŷ,ŷ )= f(,, )+ε η f + f ] η + O(ε 2 ) F{ŷ} F{} = = ε f(,ŷ,ŷ )d η f + η f f(,, )d ] d+o(ε 2 ) Analog to functions This condition on the first variation is analogous to all partial derivatives being zero! For a function of N variables to have a local etrema f i =, For a functional to be an etrema i=1,...,n δf(η,)= d dε F(+εη) ε= =, η H Note now that we have to minimize over an infinite dimensional spaceh, instead of IR n. Variational Methods & Optimal Control: lecture 4 p.14/4 Variational Methods & Optimal Control: lecture 4 p.16/4

5 Simplification Integrate the second term b parts δf(η,) = η f + f η = η f ] 1 η + ] d f d d ( )] f d A useful lemma Lemma 2.2.1: Let α,β IR, such that α<β. Then there is a function ν C 2 (IR), such that ν()> for all (α,β) and ν()= otherwise. Proof: b eample ν()= { ( α) 3 (β ) 3, if (α,β), otherwise. But note that b the problem definition η H, and so η( )=η( 1 )=, and so the first term is zero. The function inside the integral eists, and is continuous b our assumption that f has two continuous derivatives, so for E()= f d d ( f )] δf(η,)= η()e()d= η,e 2 = α β Variational Methods & Optimal Control: lecture 4 p.17/4 Variational Methods & Optimal Control: lecture 4 p.19/4 Euler-Lagrange equation Theorem 2.2.1: Let F : C 2, 1 ] IR be a functional of the form F{}= f(,, )d, where f has continuous partial derivatives of second order with respect to,, and, and < 1. Let S= { C 2, 1 ] ( )= and ( 1 )= 1 }, A second useful lemma Lemma 2.2.2: Suppose η,g = for all η H. If g :, 1 ] IR is a continuous function then g()= for all, 1 ]. Proof: Suppose g()> for α,β]. Choose ν as in Lemma ν(),g() 2 2 β = ν()g()d= ν()g()d > 1 α Hence a contradiction. Similar proof for g()<. where and 1 are real numbers. If S is an etremal for F, then for all, 1 ] ( ) d f d f = the Euler-Lagrange equation Variational Methods & Optimal Control: lecture 4 p.18/4 Variational Methods & Optimal Control: lecture 4 p.2/4

6 Proof of Euler-Lagrange equation As noted earlier, at an etremal the first variation δf(η,)= η(),e() 2 1 = η()e()d= for all η() H. From Lemma 2.2.2, we can therefore state that the Euler-Lagrange equation. f E()= d ( )] f d =, Eample: geodesics in a plane The arclength of a curve described b () will be Then So F{}= 1 1+ d ( ) ( ) d f d f = d = d 1+ is a constant, impling = const. Hence ()=c c 2, the equation of a straight line. Q: how do I know this is a minimum? Variational Methods & Optimal Control: lecture 4 p.21/4 Variational Methods & Optimal Control: lecture 4 p.23/4 Eample: geodesics in a plane Let(, )=(,) and( 1, 1 )=(1,1), find the shortest path between these two points. The length of a line segment from to +δ is δs = δ 2 + δ 2 ( ) 2 δ = 1+ δ δ ds = 1+ d So the total path length is F{}= =1 = ds= 1 1+ d δ s δ δ Special cases Now that we know the Euler-Lagrange (E-L) equations, we can use them directl, but there are some special cases for which the equations simplif, and make our life easier: f depends onl on f has no eplicit dependence on (autonomous case) f has no eplicit dependence on f = A(,) + B(,) (degenerate case) Variational Methods & Optimal Control: lecture 4 p.22/4 Variational Methods & Optimal Control: lecture 4 p.24/4

7 f depends onl on Special case 1 When f depends onl on the E-L equations simplif to f = const An eample of this is calculating geodesics in the plane (which we all know are straight lines). If f ( )=, then f( )=a + b. We will later see that problems in this form are degenerate, and solutions don t depend on the curve s shape. If =, then =c 1 +c 2. So for non-degenerate problems with onl dependence the etremals are straight lines e.g. geodesics in the plane f depends onl on Variational Methods & Optimal Control: lecture 4 p.25/4 Geodesics in the plane are a special case of f = f( ), with no eplicit dependence on. Appl the chain rule to the E-L equation and we get d f d = d 2 f( ) d d d = d 2 f( ) d = so one of the two following must be true f ( ) = = Eample f depends onl on Consider finding the etremals of such that ()= and (1)=b. The Euler-Lagrange equation is Variational Methods & Optimal Control: lecture 4 p.27/4 1 F{}= α 4 β,d d 4α 3 2β ] = d We could pla around with this for a while to solve, but we alread know the solutions are straight lines, so the etremal will be =b Variational Methods & Optimal Control: lecture 4 p.26/4 Variational Methods & Optimal Control: lecture 4 p.28/4

8 Fermat s principle Fermat s principle of geometrical optics: Light travels along a path between an two points such that the time taken is minimized Take the speed of light to be dependent on the media, e.g. c=c(,), the time taken b light along a path () is Eample Consider c(, ) = 1/g() f has no eplicit dependence on so T{}= g() 1+ d f(,, )=g() T{}= c(,) d Fermat s principle sas the actual path of light will be a minima of this functional. f g() 1+ = const = const Variational Methods & Optimal Control: lecture 4 p.29/4 Variational Methods & Optimal Control: lecture 4 p.31/4 Speed of light The speed of light (EM radiation) is onl constant in a vacuum medium speed (km/s) refractive inde vacuum 3, 1. water 231, 1.3 glass 2, 1.5 diamond 125, 2.4 silicon 75, 4. Refractive inde = c/v Eample (ii) g() = ( ) 1 c2 1 g() 2 = = c 1 = = c 2 1 g() 2 implies c 2 1 g() 2 c 2 1 g() 2(1+ ) c 2 1 g() 2 c 2 1 g() 2 c 2 1 Variational Methods & Optimal Control: lecture 4 p.3/4 Variational Methods & Optimal Control: lecture 4 p.32/4

9 Eample (iii) c 2 1 = g() 2 c = c 1 g()2 /c 2 1 1d+c 2 The constants, c 1 and c 2 are determined b the fied end points. so not all etremals are straight lines we had to include an term here to make it more interesting How we might solve Break into two problems, with a boundar point(, ), which has a fied value of (the location of the boundar), but a movable value for. (*, *) φ φ 1 ( 1, 1 ) (, ) Variational Methods & Optimal Control: lecture 4 p.33/4 Variational Methods & Optimal Control: lecture 4 p.35/4 What we can t do (et) Remember, f must have at least two continuous derivatives. If the speed of light c(,) has discontinuities, then we are in trouble. ( 1, 1 ) The functional F{}= d+ d c c 1 Separate into two problems, as if we knew(, ). Each is a geodesic in the plane problem. So the solutions are straight lines ()= { ( ) + ( ) Now we can eplicitl compute F{} as a function of, b differentiating, and then we can treat it as a minimization problem in one variable. (, ) Variational Methods & Optimal Control: lecture 4 p.34/4 Variational Methods & Optimal Control: lecture 4 p.36/4

10 The total time taken We can simplif the integrals b noting from Pthagoras that the lengths of the two lines are ( ) 2 +( ) 2 and ( 1 ) 2 +( 1 ) 2 and that the time take to traverse the pair of line segments will be More than one boundar Snell s law applies at each boundar ( 1, 1 ) T{}= ( ) 2 +( ) 2 ( 1 ) + 2 +( 1 ) 2 c c 1 (*, *) φ φ 1 dt d = ( ) c ( ) 2 +( ) 2 ] 1/2 ( 1 ) c 1 ( 1 ) 2 +( 1 ) 2 ] 1/2 (, ) Variational Methods & Optimal Control: lecture 4 p.37/4 Variational Methods & Optimal Control: lecture 4 p.39/4 The result dt d = ( ) c ( ) 2 +( ) 2 ] 1/2 ( 1 ) c 1 ( 1 ) 2 +( 1 ) 2 ] 1/2 = sinφ c sinφ 1 c 1 which we require to be zero to find the minimum. Hence Dealing with kinks We ll spend a fair bit of time later on dealing with kinks in curves Underling point The integral can still be well defined even if etremal isn t smooth But the Euler-Lagrange equations don t work at the kinks Use the Euler-Lagrange equations everwhere ecept the kinks Do something else at the kinks sinφ c = sinφ 1 c 1 Snell s law for refraction Hence there are often was around discontinuities, though it ma involve some pain (e.g. what about internal reflection) Variational Methods & Optimal Control: lecture 4 p.38/4 Variational Methods & Optimal Control: lecture 4 p.4/4