Analyse Master Cours de Francis Clarke
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Chapter 10 The Lipschitz theory Analyse Master 1 : Cours de Francis Clarke (2011) Can we be satisfied with considering only smooth solutions of the basic problem, as we have done in the previous section? By the middle of the 19th century, it was becoming apparent that we cannot. The following example illustrates the need to go beyond differentiable arcs. 10.1 Example. Consider the basic problem minimize J(x) = 1 ?1 x(t)2 [x (t)?1 ]2 dt subject to x(?1) = 0, x(1) = 1. Note that J(x) 0 ?x. If x lies in C2[?1,1] and satisfies the given boundary con- ditions, there exists ? ? (?1,1) such that x (?) = 1/2. Then, in a neighborhood of ? , x vanishes at most once and x = 1; it follows that J(x) > 0. Consider now the continuous, piecewise-smooth function x?(t) = 0 if ?1 t 0 t if 0 t 1, which has a corner at t = 0. Then J(x?) = 0. Furthermore, it is easy to show that the infimum of J over the admissible functions in C2[?1,1] is 0, an infimum that is therefore not attained.
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Chapter 10 The Lipschitz theory Analyse Master 1 : Cours de Francis Clarke (2011) Can we be satisfied with considering only smooth solutions of the basic problem, as we have done in the previous section? By the middle of the 19th century, it was becoming apparent that we cannot. The following example illustrates the need to go beyond differentiable arcs. 10.1 Example. Consider the basic problem minimize J(x) = 1 ?1 x(t)2 [x (t)?1 ]2 dt subject to x(?1) = 0, x(1) = 1. Note that J(x) 0 ?x. If x lies in C2[?1,1] and satisfies the given boundary con- ditions, there exists ? ? (?1,1) such that x (?) = 1/2. Then, in a neighborhood of ? , x vanishes at most once and x = 1; it follows that J(x) > 0. Consider now the continuous, piecewise-smooth function x?(t) = 0 if ?1 t 0 t if 0 t 1, which has a corner at t = 0. Then J(x?) = 0. Furthermore, it is easy to show that the infimum of J over the admissible functions in C2[?1,1] is 0, an infimum that is therefore not attained.
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Chapter 10 The Lipschitz theory
Analyse Master 1 : Cours de Francis Clarke (2011)
Can we be satisfied with considering only smooth solutions of the basic problem, as we have done in the previous section? By the middle of the 19th century, it was becoming apparent that we cannot. The following example illustrates the need to go beyond differentiable arcs.
10.1 Example.Consider the basic problem 1 22 minimizeJ(x) =x(t) [x(t)−1]dtsubject tox(−1) =0,x(1) =1. −1
2 Note thatJ(x)0∀x. Ifxlies inC[−1,1]and satisfies the given boundary con- ditions, there existsτ∈(−1,1)such thatx(τ) =1/2. Then, in a neighborhood of τ,xvanishes at most once andx=1; it follows thatJ(x)>0. Consider now the continuous, piecewise-smooth function 0 if−1t0 x∗(t) = tif 0t1,
which has acorneratt=0. ThenJ(x∗) =0. Furthermore, it is easy to show that 2 the infimum ofJover the admissible functions inC[−1,1]is 0, an infimum that is therefore not attained. To summarize, the problem has a natural solution in the class of piecewise-smooth functions, but has no solution in that of smooth functions. Thus, the very existence of solutions is an impetus for admitting nonsmooth arcs.
The need to admit nonsmooth functionsxin the basic problem became apparent in physical applications as well (soap bubbles, for example, generally have corners
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and creases). Spurred by these considerations, the theory of the basic problem was extended to the context of piecewise-smooth functions. In this section, we develop this theory, but within the more general class of Lipschitz functionsx, absolutely continuous functions whose derivativesxare bounded and exist almost everywhere. Of course, we are only able to deal with this class since we have the benefit of Lebesgue’s legacy in measure and integration (which was not available until roughly 1900). We extend the setting of the basic problem in one more way, by allowingxto be vector-valued. The admissible class now becomes the Lipschitz functions mapping n [a,b]toR, for which we usually write simply Lip[a,b]. Thus each component ofx is an element of the space AC[a,b]introduced in Example 1.12. The hypotheses on the LagrangianΛ(t,x,v)are weakened somewhat: we suppose in this chapter that the gradientsΛx,Λvexist and, together withΛ, are continuous functions of(t,x,v). Note that under these hypotheses, the functional b J(x) =Λt,x(t),x(t)dt a is well-defined (as a Lebesgue integral) for everyx∈Lip[a,b]. It can be shown (see 2 Exer.??) that a minimum overC[a,b]is also a minimum over Lip[a,b], so the extension of the classical theory to Lipschitz functions is a faithful one.
10.1 The integral Euler equation
The first order of business is to see what becomes of the basic necessary con-dition, the Euler equation, when the basic problem (P) is posed over the class Lip[a,b]. The definition of weak local minimumx∗is essentially unchanged: the minimum is relative to the admissible functionsxthat satisfy|x(t)−x∗(t)|εfor allt∈ [a,b], and|x(t)−x∗(t)|εalmost everywhere on[a,b]. Next, we must concede the regrettable fact that the proof of Theorem 9.2failswhenx∈Lip[a,b]. The t,x(t) problem is that the functiont→Λv∗,x∗(t)is no longer differentiable, so the integration by parts that was used before can no longer be justified. There is, however, a good way to extend the Euler equation to our new situation.
10.2 Theorem. (duBois-Reymond 1879)Letx∗∈Lip[a,b]be a weak local mini-mum for the basic problem(P). Thenx∗satisfies theintegral Euler equation:for n some constantc∈R, we have t Λt,x(t),x(t) =c+Λs,x(s)ds,t v∗ ∗x∗(s),x∗∈[a,b]a.e. a
10.1 The integral Euler equation
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Proof.We defineg(λ)as we did in the proof of Theorem 9.2, where now the variationylies in Lip[a,b]. The difference quotient whose limit isg(0)is given by b Λ(t,x∗+λy,x+λy)−Λ(t,x∗,x∗) ∗ dt. aλ SinceΛis locally Lipschitz, and since all the functions insideΛare bounded, there is a constantKsuch that, for allλnear 0, the integrand is bounded by K|(y(t),y(t))|.It now follows from Lebesgue’s dominated convergence theorem thatg(0)exists and b • • g(0) =α(t)y(t) +β(t)y(t)dt=0, a where (as before) α(t) =Λxt,x∗(t),x(t),β(t) =Λvt,x∗(t),x(t) ∗ ∗
are essentially bounded functions. We apply integration by parts, but now to thefirst term in the integral; this yields b t β(t)−α(s)ds y(t)dt=0. • a a n For anyc∈R, we also have b t • β(t)−c−α(s)ds y(t)dt=0. a a n This holds for all Lipschitz variationsyand all constantsc∈R. We now choosec such that the function t t y(t) =β(t)−c−α(s)ds dt a a
defines a variation (that is, such thaty(b) =0). With this choice ofy, we dis-cover b t 2 β(t)−c−α(s)dsdt=0, a a which implies the desired conclusion. 2 Whenx∗∈C[a,b]andn=1, it is evident that the integral Euler equation is equiv-alent to the original one given by (1) (which can be thought of as the “differentiated form”). So we have lost nothing in this new formulation. Note, however, that when n>1, we are now dealing with asystemof equations, since the gradientsΛxand Λvare vector-valued. In other words, there arenunknown functions involved, then component functions ofx.
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10.3 Example. (Geodesics)The problem of finding curves of shortest length (geo-desics) is a venerable one in the subject. Finding the shortest curve joining two given points in the plane(a,A)and(b,B)is most readily expressed in terms of two unknown functionsx(t),y(t)parametrized on a given interval, which we can always take to be[0,1]. Then the functional to be minimized is 1 1/2 22 x(t) +y(t)dt, 0 subject to the given boundary conditions x(0),y(0() = a,A),x(1),y(1) = (b,B).
The integral Euler equation (a system of two differential equations) asserts the exis-tence of constants(c,d)satisfying 1/2 1/2 22 22 c=x/x+y,d=y/x+y.
This implies thatxandyare constant: any shortest curve is affine in in each com-ponent (unsurprisingly). Whena<b, we can perhaps accept that the class of curves of interest may be restricted to those which can be parametrized in the formt,y(t), in which case we return to the case of a single unknown function. In general, however, several unknown functions will be involved, as in the general problem of finding geodesics 3 on a given surfaceSinR, whereSis described by the equationϕ(x,y,z) =0. Then the problem becomes that of minimizing 1 2 2 2 1/2 ϕxx+ϕyy+ϕzz dt 0 subject to not only the boundary conditions, but also to a pointwise constraint ϕx(t),y(t),z(t) =0∀t∈[0,1].
We shall discuss this type of problem later, for which a multiplier rule exists, in §11.3. Note the distinction with the isoperimetric problem: there are an infinite number of equality constraints here, one for eacht. There exist situations when the surface constraint can be solved for to some extent, and in which the problem takes a simpler form. One such case is that of the cylinder 2 2 3 Sdefined byx+y=1 inR. Then any curve onSmay be parametrized in the form cosθ(t),sinθ(t),z(t),0t1 for certain functionsθandz. Suppose (without loss of generality) that the initial point is given by(θ,z) = (0,0), and the final point by(θf,zf), withθf∈(−π,π]. The geodesic problem then reduces to the minimization of
10.1 The integral Euler equation 1 1/2 22 θ(t) +z(t)dt 0
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subject to (θ,z)(0) = (0,0),(θ,z)(1) = (θf+2πk,zf),k∈Z, where the term 2πkreflects the possibility of winding around the cylinder several times. For fixedk, this is the same problem as that of finding geodesics in the(θ,z)plane, for which we know that the extremals are affine: θ(t),z(t) =tθf+2πk,t zf. 2 1/2 2 The resulting cost is given byθf+2πk+z,which is minimized rela-f 1 tive tokby the choicek=0. Thus the geodesic describes ahelixθ(t),z(t) = tθf,t zfhaving at most a half turn around thez-axis. We may question this analysis on the basis that we do not know that a shortest curve actually exists. (Only a mathematician would worry about this; we mean that as a compliment.) This is why the need for existence theory makes itself felt. Or, failing that, we would require sufficient conditions of a global nature. We return to this point later (see Example 10.9).
The costate.In the context of Theorem 10.2, let us define t p(t) =c+Λxs,x∗(s),x(s)ds. ∗ a
n Then the functionp:[a,b]→Rbelongs to Lip[a,b]and satisfies p(t),p(t) =∇x,vΛt,x∗(t),x(t)a.e. ∗
This is a convenient way to write the integral Euler equation of Theorem 10.2.
In classical mechanics,pis referred to as thegeneralized momentum; another syn-onym:adjoint variable. In optimal control, wherepfigures in the celebrated Pon-tryagin Maximum Principle,pis often called thecostate; we shall retain this termi-nology.
The costate is also convenient for expressing the appropriate Transversality condi-tion when the values ofxat the boundary are not fully prescribed. In the context of Theorem 9.14, for example, the condition becomes −p(b) =x∗(b)
1 ¯ Whenθ=π, the final point is antipodal to the initial point, and there is another (equally good) possibility:k=−1.
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These facts suggest using(x,p)as new coordinates, a procedure which Legendre was the first to employ. In classical mechanics, thephase coordinates(x,p)can be used to rewrite the integral Euler equation inHamiltonian formas follows: −p(t) =Hxt,x(t),p(t),x(t) =Hpt,x(t),p(t),
where the functionHis theHamiltonianobtained from the LagrangianΛvia the Legendre transformation (the ancestor of the Fenchel transform in convex analysis). Details are given in Exer.??.
10.2 Regularity of Lipschitz solutions
Aregularity theoremis one which affirms that, under some additional hypotheses, solutions of the basic problem initially known to lie merely in the class in which (P) is posed (currently, Lip[a,b]) actually lie in a smaller class of more regular 1 functions (below,C[a,b]). This is a way of having your cake and eating it too: we allow nonsmooth functions in formulating the problem, yet we obtain smooth solutions.
10.4 Theorem.Letx∗∈Lip[a,b]satisfy the integral Euler equation, where, for almost everyt∈[a,b], the functionv→Λ(t,x∗(t),v)is strictly convex. Thenx∗∈ 1 C[a,b].
Proof.The goal is to find a continuous functionv¯ on[a,b]which agrees withx ∗ almost everywhere. For such a function, we have t t s=x(a) +v¯(s)ds∀t∈[a,b], x∗(t) =x∗(a) +x∗(s)d∗ a a
1 from which it follows thatx∗∈C[a,b]. We define Ω=t∈(a,b):x(t)exists andp(t) =Λvt,x∗(t),x(t). ∗ ∗
ThenΩis of full measure. Fix anyτ∈[a,b], and let{si}and{ti}be any two sequences inΩconverging toτ, and such that
s:=limx(si),t:=limx(ti) ∗ ∗ i→∞i→∞ both exist. Passing to the limit in the equationp(s) =Λs,x(s)yields i v i∗(si),x∗i p(τ) =Λv(τ,x∗(τ),s). Similarly, we derivep(τ) =Λv(τ,x∗(τ),t). It follows that the strictly convex functionv→Λ(τ,x∗(τ),v)has the same gradient atsandt, whences=t(see Exer. 4.18).
10.2 Regularity of Lipschitz solutions
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Let us now define a functionv¯ on[a,b]as follows: take any sequence{si}inΩ converging toτand such that(s)exists (such sequences exist becauseΩ limix∗i is of full measure, and becausex v(τ) =lim . In view of is bounded); set ¯ix∗(si) ∗ the preceding remarks,v¯ is well defined. Butv¯ is continuous by construction, as is easily seen, and agrees withxonΩ. ∗
10.5 Exercise.As an application of Theorem 10.4, we prove the necessity of Ja-cobi’s condition; the setting is that of Theorem 9.10. We reasonad absurdum, by supposing that a conjugate pointτ∈(a,b)exists. a) Show that, for everyy∈Lip[a,b]vanishing ataandb, we have b 2 2 I(y):=P(t)y(t) +Q(t)y(t)dt0. a [Hint: see the proof of Theorem 9.7, and invoke Exer.??.] b) By assumption there is a nontrivial solutionuof Jacobi’s equation vanishing ata andτ. Observe thatu(τ)=0, and that c 2 2 P(t)u(t) +Q(t)u(t)dt=0. a
c) Extenduto[a,b]by setting it equal to 0 betweenτandb. It follows thatu minimizes the functionalI(y)above relative to the arcsy∈Lip[a,b]satisfying y(a) =y(b) =0. Apply Theorem 10.4 to obtain a contradiction.
Higher regularity.It is possible to go further in the regularity theory, and show that, under suitable conditions, the solutions of the basic problem inherit the full regularity of the Lagrangian.
10.6 Theorem. (Hilbert-Weierstrass c. 1875)Letx∗∈Lip[a,b]satisfy the inte-m gral Euler equation, whereΛbelongs toC(m2)and satisfies, for someε>0,
n t∈[a,b],|x−x∗(t)|<ε,v∈R=⇒Λvv(t,x,v)>0 (positive definite).
m Thenx∗belongs toC[a,b].
Proof.The hypotheses imply the strict convexity of the functionv→Λ(t,x∗(t),v) 1 for eacht. It follows from Theorem 10.4 thatx∗belongs toC[a,b]. We deduce 1 that the costatep(∙)belongs toC[a,b], since we have (by the integral Euler equa-tion) (t) (continuous on[a,b] ). p(t) =Λxt,x∗(t),x∗ Note that for fixedt, the unique solutionvof the equationp(t) =Λv(t,x∗(t),v) x t)It follows from the Implicit Function theorem that(s 0). isv=∗(inceΛvv> 1 1m−1 2 x∈C[a,b], sincep ∗isCandΛvisCwithm2. Thusx∗lies inC[a,b].
