Broken Extremals. Variational Methods & Optimal Control lecture 20. Broken extremals. Broken extremals

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1 Broken etremals Variational Methods & Optimal Control lecture 2 Matthew Roughan <matthew.roughan@adelaide.edu.au> Discipline of Applied Mathematics School of Mathematical Sciences Universit of Adelaide April 4, 26 Broken etremals are continuous etremals for which the gradient has a discontinuit at one of more points. If a variational problem has a smooth etremal (that therefore satisfies the E-L equations), this will be better than a broken one, e.g. Brachstochrone Variational Methods & Optimal Control: lecture 2 p./32 Variational Methods & Optimal Control: lecture 2 p.3/32 Broken etremals Broken Etremals Until now we have required that etremal curves have at least two well-defined derivatives. Obviousl this is not alwas true (see for instance Snell s law). In this lecture we consider the alternatives. But some problems don t admit smooth etremals Eample: Find () to minimize subject to ( )= and ()=. F{}= 2 ( ) 2 d Variational Methods & Optimal Control: lecture 2 p.2/32 Variational Methods & Optimal Control: lecture 2 p.4/32

2 Broken etremals eample Broken etremals eample There is no eplicit dependence inside the integral, so we can find H(, )= f f = const Find c and c 2 from ( c 2 ) 2 = 2 c 2 ( 2)( ) 2 ( ) 2 = c If c = we get the singular solutions 2 ( )( + 2 ) = c 2 ( )( ) = c 2 ( 2 ) = c = and =±+B Neither of these satisfies both end-points conditions ( )= and ()=, so c (we think) using the end-points. Combine the two equations ( ) = ( c 2 ) 2 = c () = ( c 2 ) 2 = c ( c 2 ) 2 = +(+c 2 ) 2 which has solutions c 2 = /4, and so c = 9/6 2 =(+/4) 2 9/6 Broken etremals eample Given c Variational Methods & Optimal Control: lecture 2 p.5/32 Broken etremals eample Variational Methods & Optimal Control: lecture 2 p.7/32 The end-points are on opposite branches of the hperbola! 2 ( 2 ) = c 2 = 2 c 2 d = ± 2 c d d = ± d 2 c (,) ( /4,) (,) = ± 2 c + c 2 ( c 2 ) 2 = 2 c The solution is a rectangular hperbola 2 There is NO smooth etremal curve that connects(,) and(,) Variational Methods & Optimal Control: lecture 2 p.6/32 Variational Methods & Optimal Control: lecture 2 p.8/32

3 Broken etremal sometimes there is no smooth etremal we must seek a broken etremal still want a continuous etremal what should we do? previous smoothness results suggest that we should use a smooth etremal when we can, and so we will tr to minimize the number of corners. We ll start b looking for curves with one corner But can we appl E-L equations? Smoothness theorem Theorem: If the smooth curve () gives an etremal of a functional F{} over the class of all admissible curves in some ε neighborhood of, then () also gives an etremal of a functional F{} over the class of all piecewise smooth curves in the same neighborhood. Meaning: we can etend our results to piecewise smooth curves (where a smooth result eists), not just curves with 2 continuous derivatives. (, ) = + ε η = () (, ) Variational Methods & Optimal Control: lecture 2 p.9/32 Variational Methods & Optimal Control: lecture 2 p./32 Broken etremal If we have an etremal like this, can we use E-L equations? Proof sketch The theorem assumes that there eists a smooth etremal (in this case a minimum for the purpose of illustration), then for an other smooth curve ŷ B ε () we know F{ŷ}>F{}. (, ) (, ) Assume for the moment that for a piecewise smooth function ỹ B ε () that F{ỹ}<F{}. We can approimate ỹ b a smooth curve ŷ δ B ε () b rounding off the edges of the discontinuit. Given that we can approimate the curve ỹ arbitraril closel b a smooth curve ŷ δ, for which we alread know F{ŷ δ }>F{}, we get a contradiction with F{ỹ}<F{}, and so no such alternative etremal can eist. * Variational Methods & Optimal Control: lecture 2 p./32 Variational Methods & Optimal Control: lecture 2 p.2/32

4 Proof sketch So what do we do? Break the functional into two parts: F{}=F {}+F 2 {}= f(,, )d+ f(, 2, 2)d where we require to have two continuous derivatives everwhere ecept at, and ( )= 2 ( ) (, ) * (, ) Variational Methods & Optimal Control: lecture 2 p.3/32 Variational Methods & Optimal Control: lecture 2 p.5/32 Proof sketch Possible perturbations = + ε η = () (, ) (, ) * * The location of the corner can also be perturbed. Variational Methods & Optimal Control: lecture 2 p.4/32 Variational Methods & Optimal Control: lecture 2 p.6/32

5 The First Variation: part We get first component of the first variation b considering a problem with onl one fied end-point, and allowing to var, so that [ ˆ ] δf (η,) = lim f(,ŷ,ŷ ε ε )d f(,, )d And as with transversals, we get an integral term which results in the E-L equation, plus the additional term p δ H δ where δ( ) = X and δ( ) = Y H = f f and p = f Conditions We rearrange to give δf(η,)= (p p 2 )δ (H H 2 )δ Note that the point of discontinuit ma var freel, so we ma independentl var δ and δ or set one or both to zero. Hence, we can separate the condition to get two conditions p p 2 H H 2 = = The First Variation: part 2 Variational Methods & Optimal Control: lecture 2 p.7/32 Weierstrass-Erdman Variational Methods & Optimal Control: lecture 2 p.9/32 Note that, for the second component of the First Variation we get a similar etra term, e.g. δf 2 (η,) introduces the term p 2 δ+h 2 δ We can write the conditions as p = p 2 the sign is reversed because it corresponds to the term in the transversal problem (as opposed to the term for δf. The combined second variation (minus the terms that result from the E-L equation which must be zero) is δf(η,)=δf (η,)+δf 2 (η,)= p δ H δ p 2 δ+h 2 δ H = H 2 Called the Weierstrass-Erdman Corner Conditions Rather than separating into and 2 we ma write the corner conditions in terms of limits from the left and right, e.g. p = p + H = H + Variational Methods & Optimal Control: lecture 2 p.8/32 Variational Methods & Optimal Control: lecture 2 p.2/32

6 Solution So the broken etremal solution must satisf Eample The actual etremal (in red) the E-L Equations the Weierstrass-Erdman Corner Conditions p = p + H = H + (,) * (,) must hold at an corner 2 Obviousl, this is onl valid if we allow non-smooth solutions. Variational Methods & Optimal Control: lecture 2 p.2/32 Variational Methods & Optimal Control: lecture 2 p.23/32 Eample In the eample considered, p = f = 2 2 ( ) H = f f = 2 ( 2 ) Remember that = and =+A are valid solutions to the E-L equations, and that for both of these solutions p=h =, so we can put a corner where needed. The solution must also satisf the end-point conditions, so ( )= and ()=, and therefore, as valid solution has = and = for [, ] 2 = for [,] More insight sometimes we have a constraint on where the corner can appear: sometimes the discontinuit arise from the problem itself, e.g., a discontinuous boundar such as in refraction (see Fermat s principle, and Snell s law in earlier lectures) in these cases, we need to go back to the condition δf(η,)= (p p 2 )δ (H H 2 )δ = and look at whether δ or δ are forced to be zero, or if there is a relationship between them, and use that to form a constraint such as we had for transversals. Variational Methods & Optimal Control: lecture 2 p.22/32 Variational Methods & Optimal Control: lecture 2 p.24/32

7 General strateg solve E-L equations look for solutions for each end condition match up the solutions at a corner so that ( )= 2 ( ) the Weierstrass-Erdman Corner Conditions are satisfied in theor can allow more than one corner, but this would get ver painful! Newton s aerodnamical problem Newton s aerodnamical problem Find etremal of air resistance F{}= Variational Methods & Optimal Control: lecture 2 p.25/32 R d, + 2 subject to ()=L and (R)= with solutions. =const for [, ] 2. u [u,u 2 ] (u)= c ( ) u (+u2 ) 2 = c u + 2u+u3. (u)=l c ( lnu A+u ) u4 Newton s aerodnamical problem Variational Methods & Optimal Control: lecture 2 p.27/32 we could find u b tring to minimize F as a function of u, but this is hard because we onl have a numerical solution to get u 2. alternative is to use corner conditions. at the corner (a) = (u ) is free (b) =L is fied 2. corner condition of interest is H = H + Trick bit is working out u which sets the location of the corner, and fies A, c and u 2. Variational Methods & Optimal Control: lecture 2 p.26/32 Variational Methods & Optimal Control: lecture 2 p.28/32

8 Newton s aerodnamical problem Newton s aerodnamical problem Calculating H H = f f = 2 2 (+ 2 ) 2 = = (+ 2 ) [ 2 2 (+ 2 ) 2 +(+ 2 ) ] [ 3 2 (+ 2 ) 2 + ] H = = H + (+u 2 ) 2 = 3u 2 + u 4 u 2 = u 2 (u 2 ) = [ 3u 2 (+u 2 ) 2 + ] u = or ± but = u> so u= is the onl valid solution, hence u = and the rest of the solution follows from there. Variational Methods & Optimal Control: lecture 2 p.29/32 Variational Methods & Optimal Control: lecture 2 p.3/32 Newton s aerodnamical problem Corner condition H = [ 2 2 (+ 2 ) 2 + ] Now on the LHS of = we have =, so H = On the RHS, remember = u (from Lecture 6) H = [ + 3u 2 (+u 2 ) 2 + ] Newton s aerodnamical problem real rockets don t look like this. resistance functional is onl approimate (a) ignores friction (b) ignores shock waves 2. rockets must pass through multiple laers of atmosphere, at varing speeds additional constraints:. nose cone is tangent to rocket at joint 2. nose is eas to build reall, we need to do CFD++ (R)= Variational Methods & Optimal Control: lecture 2 p.3/32 Variational Methods & Optimal Control: lecture 2 p.32/32