Existence of minimizers for the Newton s problem of the body of minimal resistance under a single impact assumption M. Comte T. Lachand-Robert 30th September 999 Abstract We prove that the infimum of the Newton s functional of minimal resistance F (u := Ω dx/( + u(x 2, where Ω R 2 is a strictly convex domain, is not attained in a wide class of functions satisfying a single-impact assumption, proposed in []. On the other hand, we prove that the infimum is attained in the subclass of radial functions; hence the minimizers are the local minimizers already described in [3]. Résumé Nous montrons que l infimum de la fonctionnelle de résistance minimale de Newton F (u := Ω dx/( + u(x 2, où Ω R 2 est un domaine strictement convexe, n est pas atteint dans une classe assez large de fonctions satisfaisant une contrainte de choc unique et élastique suggérée dans []. En revanche, si on se limite aux fonctions à symétrie radiale, nous montrons que l infimum est atteint et par conséquent les minimiseurs coïncident avec les minimiseurs locaux étudiés dans [3]. Université Pierre et Marie Curie, Laboratoire d Analyse Numérique, 75252 Paris Cedex 05, France. comte@ann.jussieu.fr; lachand@ann.jussieu.fr; www.ann.jussieu.fr.
Introduction We consider the existence question for the problem dx minimize F (u := + u(x 2 where Ω R 2 is a strictly convex domain, and the unknown function u is to be found in the class of maps u : Ω [0, M] (M > 0 a given parameter satisfying the geometrical condition: x dom( u, τ > 0, such that x τ u(x Ω, u(x τ u(x u(x ( 2 u(x ( τ 2 Here dom( u is merely the dense subset of Ω where u is differentiable; we will define it more precisely later. As explained in [] and [3], this is a mathematical formulation for the Newton s problem of minimal resistance: find a body with cross-section Ω having minimal resistance to a flux of particles. The body here is represented by the hypograph of u: H u := {(x, z Ω R; z u(x} The assumption ( means that every particle coming downward at x, hitting the body at (x, u(x, and reflected in an elastic shock at this point, does not hit the body again. This single-impact assumption ensures that the resistance of the body is correctly computed (in appropriate units by the functional F. More details on this assumption can be found in [] and [3]. The natural topology for this problem is induced by the constraint (: it is the topology of W, loc (Ω C0 (Ω. In fact, it is proved in [] that any function satisfying ( must be locally Lipschitz; we recall the proof in Section 2 for convenience. As already explained in [3], the physical meaning of the problem requires more regularity for the gradient than usually found for functions of W, loc (Ω. So we have to define a smaller vector space to work with. We consider Ω P(Ω := {u C 0 (Ω ; u polyhedral} where polyhedral means that u is a finite sequence of minimum or maximum operations on affine functions. The vector space of interest here is W(Ω := P(Ω 2
where the closure operation is with respect to the topology of W, loc (Ω C 0 (Ω. It is easy to check that C (Ω C 0 (Ω W(Ω W, loc (Ω C0 (Ω and it is a well-known property that W(Ω contains all convex or concave functions of C 0 (Ω. The set of minimization is: and we consider the problem: C M = {u W(Ω ; 0 u M, u satisfies (}. inf F (u where F (u := u C M Ω dx + u(x 2. (2 In the original problem of Newton, Ω is the unit ball of R 2 ; we will consider this case for radial functions in Section 3. In general we will assume here that Ω is any strictly convex bounded domain of R 2. In this paper, we will prove in Section 2 the following result: Theorem. Let u C M be regular on the boundary (in the sense of Definition 2. hereafter. Then u is not a minimizer for (2. We strongly believe that all functions in C M are regular on the boundary (hence, the infimum of F is not attained but we don t have a general proof. On the other hand, the class C M is restrictive enough to avoid degeneracy of the infimum of the functional: Lemma. We have inf u CM F (u > 0 for all values of M > 0. The proof of this lemma is given in section 2.. In view of these results, it is natural to ask if some radial minimizer exists when Ω is a ball. We prove in Section 3 that it is indeed true. If Ω is a ball, W r is the set of radial functions in W, we have: Theorem.2 There exists u W r such that F (u = min F(v. v W r C M In this case, the nature of the minimizing set has been fully characterized in [3]; in particular, for M large enough, there exists a unique radial minimizer. 3
2 The two-dimensional case 2. On the geometrical constraint As explained in [], the facts that osc u M and Ω is bounded imply u(x ψ(x := M + M 2 + dist(x, Ω 2 dist(x, Ω (3 for all x where u is differentiable. Hence u W, loc (Ω. This estimate allows us to prove Lemma.. Proof of Lemma.. Since any function u C M satisfies (3, we have dx F (u c ψ := + ψ(x. 2 The function /( + ψ(x 2 is bounded, positive and continuous on Ω; thus c ψ > 0. Indeed, for a unit ball we have: c ψ = πm 2 + π 2 + πm ln( M 2 + πm ln M + πm M 2 +. This ends the proof of the lemma. We recall that a locally Lipschitz function is almost everywhere differentiable. In particular, the constraint (7 applies almost everywhere. Moreover, since u W(Ω, u(x is also a geometrical gradient : the vector ( u(x, R 3 is normal to a tangent plane to the hypograph H u at (x, u(x. That follows from an important property of functions u W(Ω: Ω If u is differentiable at x, u is continuous at x. (4 This property is proved for convex functions in [4, Theorems 25.4 5]. A similar proof applies here. In view of these remarks, and since u C 0 (Ω, it is natural to take also into account the points on the boundary. In fact, if x Ω is such that u is not differentiable at x, but H u as a vertical tangent plane at (x, u(x, we shall say that x is a regular point for u. We extend the notation dom( u as follows: dom( u = {x Ω; u is differentiable at x} {x Ω; x is a regular point for u} 4
Formally, u(x = for regular points on the boundary, but the property (4 still holds for these points in the following sense: (x n dom( u Ω, lim x n = x, u(x n and u(x n u(x n has a limit in S. In this paper, we will restrict ourselves to those functions u C M such that the non-regular points are not dense on the boundary: Definition 2. A function u C M is said regular on the boundary if Γ u Ω, nonempty and open in Ω, Γ u dom( u. (5 The non-regular functions yield to great difficulties, some of them are detailed in Section 2.3. The constraint ( can also be written as follows: if x dom( u is given, such that u(x 0, let us note { d x = sup t > 0 ; x t u(x u(x Ω }. (6 Then t (0, d x ], u(x t u(x u(x u(x t ( 2 u(x u(x (7 In particular, if x Ω is regular with u(x =, we get in the limit u(x tζ for all t > 0, where ζ = lim u(x n u(x n for any sequence (x n dom( u Ω with limit x. That means that ζ must be an inward direction for Ω: Ω (x R + ζ = {x}. We first state a useful lemma on functions satisfying the constraints: Lemma 2. If u and v are in C M, then w := min(u, v is in C M too. Proof. It is obvious that w(x [0, M] for all x, and that w W(Ω. Hence, we have merely to prove ( for w. Let us first notice that x dom( w if and only if one of the three following cases occurs:. u(x < v(x and x dom( u (then w(x = u(x; 5
2. u(x > v(x and x dom( v (then w(x = v(x; 3. u(x = v(x, x dom( u dom( v and u(x = v(x. In the first case, we have for all τ > 0 (since w u and w(x = u(x: w(x τ w(x w(x τ u(x τ u(x u(x τ 2 ( u(x 2 = 2 ( w(x 2. The other cases are similar. 2.2 Main Theorem in two dimensions We are now in position to state precisely our result on the problem dx inf F (u where F (u := u C M + u(x 2. (2 We recall here the statement of our Theorem: Theorem. Let u C M be regular (in the sense of Definition 2. above. Then u is not a minimizer for (2. Proof of Theorem.. Assume by contradiction that there exists a regular minimizer u C M. We will construct a sequence (u n n such that F(u n < F (u for n large enough, giving a contradiction. We first have to prove some a priori properties on u. Step. We claim that for all x Ω dom( u, we have either u(x = 0 or u(x =. (This property is due to G. Buttazzo. Indeed assume that there exists x 0 Ω dom( u such that u(x < and u(x 0 > 0. Then u is bounded in a neighborhood V of x 0 in Ω, since it is continuous at x 0. Let ξ S be an internal direction of Ω at x 0 and φ(x := ξ (x x 0. For λ > sup x ω u large enough, the set ω := {x Ω ; λφ(x < u(x} is included in V ; indeed, from the strict convexity of Ω, we see that u/φ is continuous on Ω \ {x 0 }. On the other hand, the hypothesis u(x 0 > 0 and the fact that φ(x 0 = 0 implies that ω is an open neighborhood of x 0. 6 Ω
Consider the function v = min(λφ, u; it follows from Lemma 2. that v C M. We have v = u on Ω \ ω and v = λ on ω; hence using λ > sup x ω u, we get: F(v F(u = and therefore u is not a minimizer. ω + λ 2 + u 2 < 0 Step 2. We now prove that u(x > 0 for x Ω dom( u. Indeed assume that there exists x Ω dom( u such that u(x = 0 (and then u(x = 0 from the previous step. Hence there exists a neighborhood V of x such that u(y < / 3 for all y V ; we can choose V large enough in order to have sup V u > 0. Let us consider a given point x V such that u(x > 0, and the function v defined by v(y := y x 3. By choosing x near x, we can assume that the set ω := {y V ; u(y > v(y} is compactly embedded in V (taking into account that u < / 3 = v in V ; this set is open and nonempty since u(x > 0. It is easy to check out that v satisfies (7; hence the function w defined by w := min(u, v in V, w := u in Ω \ V belongs to C M from Lemma 2.. Using the fact that t /( + t 2 is strictly decreasing with respect to t 0 and u < / 3 = v in V, we have F(u F (w = This is a contradiction since u is a minimizer. ω [ + u(x 2 ] + dx > 0. 3 Step 3. We claim that u is nonincreasing in a neighborhood of Ω dom( u. More precisely, if x Ω dom( u, and (x n Ω dom( u such that x n x, then lim sup d xn = 0 (8 where d x is defined in (6. Indeed assume that there exists c > 0 such that d xn > c for all n. Since u satisfies the constraints, we have in particular M c u(y n u(x n ( d xn 2 u(x n u(x n 7
where y n := x n d xn ζ n and ζ n = u(x n / u(x n. Since ζ n S (the unit circle, we can assume, extracting subsequences if necessary, that ζ n converges to ζ S, and d xn δ > 0. That implies that u(x n converges to some finite value: u(x <. Using the first step, we deduce that u(x = 0, and u(x > 0 from the previous step. Hence u(x is a inner vector at x, since y n y = x β u(x with β = δ/ u(x > 0 belongs to the convex set Ω. Since u(y 0 = u(x for all y Ω, we have This contradicts u(x > 0. Step 4. 0 lim t 0+ u(x t u(x u(x t = u(x 2 We claim that there exists an open subset ω Ω such that and a number c > 0 such that x ω, t (0, d x ], c + ω Ω Ω dom( u (9 u(x t u(x u(x u(x t ( 2 u(x u(x. (0 Indeed assume that there exists sequences (x n dom( u, (ε n R +, and (t n R +, satisfying t n d xn, x n converges to x dom( u Ω, ε n 0, and ε n + u(y n u(x n = ( t n 2 u(x n u(x n ( where y n := x n t n ζ n and ζ n = u(x n / u(x n. Since ζ n S (the unit circle, we can assume, extracting subsequences if necessary, that ζ n converges to ζ S. From the previous step, we have lim sup d xn = 0. Hence we must have lim t n = 0. From Step and 2, we know that u(x > 0. If u(x <, we have ( u(x 2 u(x = lim 2 ( u(x n u(x n = lim u(x n u(x n t n ζ n t n = lim u(x n ζ n = u(x 8
taking into account that u is continuous at x dom( u, since u W(Ω. This is impossible. Hence we get u(x =. Dividing ( by u(x n, we get η n + u(y n u(x n θ n = 2 ( u(x n 2 (2 where η n := ε n / u(x n 0 and θ n := t n u(x n. If lim sup θ n > 0, we get, passing to the limit in (2 for a subsequence, 0 = /2 since y n x and x n x. Hence θ n 0. From a Taylor expansion near x n, we have, ξ S and t > 0, u(x n tξ u(x n = t u(x n ξ + tρ n (t, ξ. Since u is continuous at x, there exists ρ : R + R such that ρ n (t, ξ < ρ(t for all t > 0, and lim t 0+ ρ(t = 0. Hence u(y n u(x n = + ρ n(t n, ζ n θ n u(x n as n. Passing to the limit in (2, we get = /2, again a contradiction. Step 5. There exists an open ball B ω such that u(x (0, M for all x B. For if not, we would have u 0 or u M in ω from the continuity of u. That implies u 0, in contradiction to step 2. We claim that u div ( + u 2 = 0 in B (3 2 in the sense of distributions. Indeed, if v Cc (B is given, there exists ε 0 > 0 such that (u + εv C M for all ε ( ε 0, ε 0, as follows clearly from (0 and the fact that u(x (0, M for all x B. Therefore, F(u + εv F (u implies F (u v = 0, that is u v 0 = ( + u 2 2 B and this gives (3 using Green s formula. As a consequence, the curl of the vector field u/( + u 2 2 vanishes identically; hence, it can be written as a gradient. In other words, we can find w W, loc (Ω such that u w = 0 in B. (4 9
Notice that w is differentiable on dom( u. Let ω be relatively compact open subset of B, and χ C c (ω ; [0, ] be given. For any n, we set u n = u n 3 χ sin2 (n 2 w (5 Since u(x (0, M for x B, we see that u n (x [0, M] for all x Ω if n is large enough. We have dom( u n dom( u and in dom( u, u n = u n 3 sin(n2 w χ n χ sin(2n2 w w. (6 Therefore u n converges uniformly to u in dom( u. x dom( u: converges to u n (x t u n(x u n (x u n(x t u(x u(x t u(x u(x t 2 In particular for ( 2 u n (x u n(x ( u(x u(x uniformly in dom( u B. From (0, it follows that u n satisfies (7 on B dom( u. For x dom( u, x / B, we have u n (x = u(x. Hence for any t (0, d x ], we get x t := x t u n(x u n (x = x t u(x u(x and u n (x t u n (x ( t 2 u n (x u n(x = u(x t u(x t tn χ 3 sin2 (n 2 w(x t ( 2 u(x u(x tn χ 3 sin2 (n 2 w(x t 0. Hence u n C M for n large enough. On the other hand, u n 2 = u 2 + n 2 χ2 sin 2 (2n 2 w w 2 + O ( n 3 (7 using (4. Hence, u n u, and this inequality is strict almost everywhere in {x; χ(x > 0} (for otherwise, we would have w 0 in B and this is not possible as explained before. We conclude that F (u n < F (u and u n C M, a contradiction. 0
2.3 Points of non-differentiability We would like to explain here what happens to points x / dom( u for some u C M. For any convex function w : Ω R {+ }, it is well-known (cf. [4] that w w(x + ty w(x (x; y := lim (8 t 0+ t exists in R {+ } for all x dom(w, and for all y R 2 (if x is in the interior of dom(w, or all y in the tangent cone to the interior of dom(w at x (if x is in the boundary of dom(w. Clearly, if x dom( w, then w (x; y = w(x y; however, for all x Ω, w (x; y is a convex, homogeneous function with respect to y; that implies that it is a continuous function with respect to y (not with respect to x. Similar properties hold for concave functions, with R {+ } changed to R { }. It is easy to check out that similar properties hold for any function in W(Ω. Hence, passing to the limit in (, we will get the following: x Ω, τ > 0, z u(x, such that x τz Ω, u(x τz u(x ( z 2 (9 τ 2 where u(x := {z; (x n dom( u; lim x n = x and lim u(x n = z}. Notice that we can pass to the limit τ 0 again in (9: x Ω, z u(x \ {0}, u (x; z 2 ( z 2. (20 (The similar limit with ( leads to u(x 2 2 ( u(x 2 which is always true. We will say that u saturates the constraint at x if equality occurs in (20 for some z u(x or in (9 for some τ > 0 and some z u(x such that x τz Ω. We will note S u Ω the corresponding set of saturation points. Contrarily to basic intuition, we have: Lemma 2.2 There exists some functions u C M such that S u is not closed in Ω. The existence of such functions prevents to extend the proof of Theorem. to the whole class C M : the reader can compare with the proof in the radial case (Section 3, where the similar set S u is always closed.
Proof. We just have to give a counter-example. Let 0 be an interior point of Ω. We consider the function defined as follows: x if x 2 x 0 0 if x 2 x 0 u(x, x 2 = c + 2x 2 x if x 0 and x 2 x 2x + 2x 2 if x 0 and x 2 x The constant c > 0 is chosen such that u 0 in Ω; we assume that M is large enough to have M in Ω. It is easy to check out that u C M. Moreover we have S u = {x = (x, x 2 Ω ; x = 0, x 2 > 0}. Indeed if x = (0, x 2, with x 2 > 0, e = (, 0 belongs to u(x. If τ [0, x 2 ], we have u(x τe u(x τ = 0 = 2 ( e 2. It is easy to check that x = (x, x 2 with x 0 or x 2 < 0 implies that ( is not saturated. Let us now consider the point (0, 0. We have On the other hand we have: u(0 = {0, e, 2e 2 e, 2e + 2e 2 }. u (0; e = lim t 0+ u( te t u (0; 2e 2 + e = lim t 0+ u(t, 2t t u (0; 2e 2 2e = lim t 0+ u( 2t, 2t t Hence 0 / S u, and S u is not closed. = 2 < ( e 2 = 0 2 = 5 < ( 2e2 e 2 = 2 2 = 8 < 2 ( 2e + 2e 2 2 = 7 2 As apparent in the proof herebefore, the very reason for which this happens is related to the relative speed of convergence of x n x and τ n 0 in the quantity u(x n τ n u(x n u(x n τ n. The limit can be any combination in the form u (x; z λu (x; ξ where z = lim u(x n, λ [0, and ξ = lim(x n x/ x n x. In that case, taking the limit in the constraint ( does not give any useful information. 2
3 The radial case We consider here the problem of minimizing F in the set W r C M, where: W r := {u W(Ω; u is radial } where Ω is the unit disc of R 2. For these functions, it is more convenient to write them as functions from [0, ] to R, or equivalently, as even functions from [, ] to R. Taking this convention into account, we will prove in Lemma 3. that W r = {u W, loc (0, C0 ([0, ] ; u has a right-limit and left-limit everywhere in (0, }. (Unless otherwise specified, u is the distributional derivative of u. yields easily W r = {u W, loc (0, C0 ([0, ] ; u is left and right differentiable, u (r+ is right continuous, u (r is left continuous }. (2 This (22 For radial functions in the unit disc of R 2, the Newton functional can be written in the form: F (u := 0 r dr + u (r 2 It is already known, from [3], that there exists a critical value M such that, for M M, only one minimizer can exist. On the other hand, for M < M there exists a noncompact set of local minimizers, all of them achieving the same value of the functional. This phenomenon is due to the constraint u 0. If this constraint is removed, except at the boundary, only one minimizer can exist, whatever the value of M. This leads us to study first the following problem : where min F (u (23 u EM r E r M = {u W r ; u( 0, u M, u( x satisfies (} (24 Notice that E r M is larger than C M W r, since we do not require that u 0 in [0,. We recall that E r M and C M W r are not convex (cf. [3]. 3
Theorem 3. There exists u W r (0, such that F (u = min F (v. v EM r The proof is given in section 3.3. Since the set of local minimizers is known from [3], and E r M C M W r, we deduce from Theorem 3. that the minimum of F on C M W r is attained, too. Hence it is just equal to the set of local minimizers described in [3]. In this section, we will note f(t = + t 2 the function appearing in the fonctionnal; we notice here, for further usage, that f (t = 2t and f (t = 2(3t2 (25 ( + t 2 2 ( + t 2 3 3. Constraints for radial functions For the radial functions, the constraint (7 can be written as follows: for all r (0, dom(u, with u (r < 0, u(s u(r s (r,, ( u (r (26 s r 2 u (r and for all r (0, dom(u, with u (r > 0, s [, r, u( s u(r s r 2 ( u (r u (r (27 (Here we have to consider also s < 0 because the trajectory of the reflected particle at r crosses the axis r = 0, and is assumed to remain above the symmetrical part of u. In particular, this last relation implies u (r (28 (by setting s = r. Therefore, any function u EM r u M. is bounded from below: 3.2 Minimizing sequences We consider here minimizing sequences for F, i.e. sequences (u n n N E r M such that F(u n m F := min F (v. (29 v EM r 4
The usual method in calculus of variations is to prove that such a sequence is relatively compact. On the other hand, in this particular problem, it is impossible to do so, since we know that (for some values of the parameter M there exists a non-compact set of local minimizers in C M W r (cf. [3]. So, in our proofs, we will change the minimizing sequence (u n as follows:. First extract subsequence (still noted (u n. 2. Second, find a new sequence (ũ n such that as n. u n ũ n L (0, 0 and F (u n F (ũ n 0 (30 Under these conditions, we shall say that (ũ n is a slight change of (u n. The new sequence (ũ n is then noted (u n again, and the whole process iterated eventually. It is useful to consider minimizing sequences more regular than usual functions in EM r. In particular, let us consider affine by parts functions (that is, function whose derivative is constant in all parts of a finite partition of [0, ]. It follows from the definition of W(Ω that the set is dense in W r. On the other hand, if we define: A := {u C 0 ([0, ], affine by parts } E A M := E r M A it is not obvious that EM A is dense in E M r, due to additional constraints. But this appears to be true: Lemma 3. The set W r is equal to the set given in (2, and EM A is dense in EM r,, for the W ([0, topology. Proof. loc Step. Let u C 0 ([0, ] be a function having a derivative u admitting a right and left limit at every point r (0,. Using a classical topological result [2], we know that, on any compact subset [a, b] (0,, u is a uniform limit of a sequence of step functions (v n n. We define v n (r := 0 if r / [a, b], and we introduce u n (r := u(a + 5 r a v n (s ds.
Then u n A, and u n (r u(r r a v n (s u (s ds and the result follows from the uniform convergence of (v n. Hence u W r. Conversely if u W r, u is the limit of a sequence of step functions, in L loc (0,. From the reciprocal of the same topological result, we know that u has a right and left limit at every point. This ends the proof of (2. Step 2. We now prove that EM A is dense in E M r. Let u E M r, ε > 0 and [a, b] [0, be given, and v A such that u v W, (a,b < ε. (3 Obviously, we can assume that v M in [0, and v( 0, but in general v does not satisfy the constraints (26 27. On the other hand, we recall that u satisfies these constraints. We claim that for all r [a, b] such that v (r < 0, we have: s (r, ], v(s v(r s r K u ε + 2 ( v (r v (r. (32 with K u > 0 depends only on u. This relation holds obviously for all pair (r, s satisfying (26 for v. Let r [a, b] with v (r < 0, such that u and v are both differentiable at r but (26 is not satisfied for v, that is, there exists s (r, ] with: v(s v(r s r > 2 ( v (r v (r. (33 Since the left-hand side in bounded above by v L ([a,b] < u L ([a,b] +ε < C u := u L ([a,b] +, this implies: v (r > + C 2 u C u > 0. Therefore, if ε is less than ε u := [ + Cu 2 C u]/2, we have v (r > 2ε u > 2ε, and this implies u (r < ε u < 0. Then we have from (26: u(s u(r s (r, ], ( u (r. (34 s r 2 u (r From (3, the Lipschitz constant of (u v is bounded by ε, hence, for all s, r: u(s u(r v(s v(r s r s r ε 6
Let us note g : t 2 (t /t; notice that g is increasing in (, 0 with a derivative bounded by g (ε u in (, ε u. Hence g(v (r g(u (r g (ε u ε. Therefore using (34, we get (32 with K u := + g (ε u. This ends the proof of (32 for all r [a, b] such that v (r < 0. Similarly we have, for all r such that v (r > 0: and s [, r, v( s v(r s r K u ε + 2 For ε > 0 small enough, we can find λ satisfying ε 0 λ 2 λ 2 2εK u(ε + u L (0,. ( v (r v (r (35 (36 Since u L (0, + ε v L (a,b, we get, for r [a, b] such that v (r < 0, 2εK u v (r, and this implies λ 2 λk u ε + λ ( v (r ( w (r 2 v (r 2 w (r where w := λv. Taking (32 into account, we see that w satisfies (26. Similarly, from (35, we see that w satisfies (27. Hence w EM A ; from (36, we have also u w W, (a,b ε. for ε small enough. This concludes the proof of the lemma. Consequently, we can restrict from now on, to minimizing sequences in EM A. We will prove that any minimizing sequence can be changed to another one enjoying better properties. Our first objective is to reduce the (possibly large number of sign changes for u n. More specifically, let us define, for any u EM A, the set Z u of points r 0 [0, where u (r 0 and u (r 0 + have opposite signs. Lemma 3.2 Let (u n n N EM A be a minimizing sequence. Then, for a subsequence still noted (u n, there exists k N, ε > 0, and for all n, v n EM A such that F(v n F (u n 0, v n > ε, and the cardinal of Z v n equals k. 7
Proof. The proof consists en several slight changes and subsequences extractions of the given minimizing sequence (u n. Step. We first note that for a subsequence, there exists ε > 0 and a slight change (v n of (u n such that v n > ε. That is just obvious, since f attains a strict maximum at 0. Step 2. Hence, changing notations, we can restrict to minimizing sequences satisfying u n > ε. We prove that we can extract subsequence and change slightly such a sequence (u n in order to have Card Z un < C for some constant C. Let us assume that no subsequence of the given sequence does satisfy this condition. Hence Card Z un + as n ; renumbering the sequence, we can assume for instance that Card Z un = n. Let us note 0 z n < zn 2 < < zn n the elements of Z u n ; we also note z0 n := 0, zn n+ :=. From the definition of Z un, u n has constant sign in every subinterval (zn i, zn i+. We can assume for instance that u n > 0 in (zn 2j, zn 2j, and u n < 0 in (zn 2j, zn 2j+. Let J N be the set of indices j N satisfying lim sup n (z2j+ n zn 2j = 0. (37 Notice that N \ J is finite, since the sequences (zi n n belong to the bounded set [0, ]. We note δj n := (zn 2j+ zn 2j for short. For any given n N, let j N [0, n] be such that δj n δk n, for all k j. We recall that u n is affine by parts; there is no loss of generality in assuming that u is constant in the small intervals A j := (zn 2j δn j, zn 2j, A 2 j := (z2j, n z2j, n A 3 j := (z2j, n z2j+ n and A 4 j := (z2j+, n z2j++δ n j n for all j J, since this can be obtained by a slight change in u n, taking into account (37 and the fact that δj n δk n, for all k j. We note pk j, k =,..., 4, the constant value of u in A k j, and αk j the length of A k. We recall that p j < 0, p 2 j > 0, p3 j < 0, p4 j > 0 and α j = α4 j = δn j = α2 j + α3 j. Let (ε n be a sequence with limit zero, to be chosen in a while. Let us define a function v n EM A by: v n (t = u n (t for all t [0, z n 2j δn j (zn 2j+ + δn j, ]. and p 3 j + ε n for all t (z2j n δn j, zn 2j δn j + α3 j. v n (t = p j + ε n for all t (z2j n δn j + α3 j, ζn j. p 4 j ε n for all t (ζj n, z2j+ n + δj n α 2 j. p 2 j ε n for all t (z2j+ n + δn j α2 j, zn 2j+ + δn j. 8
Figure : Construction of v n (dashed line based on u n (plain line, here with ε n = 0. (The construction of v n is pictured in Figure. Here ζj n is the unique number permitted for v n to be continuous. If we assume that ε n is small, then ζ n j = O(ε n + z n 2j δn j + α3 j + α j = O(ε n + z n 2j+ + δn j α2 j α4 j. In other words, v n achieves almost the same values than u n, on intervals of almost the same length; but they are not taken in the same order. Consequently, we have: F(v n F(u n < cε n δ n j for some constant c > 0. Moreover, ε n > 0 can be chosen such that v n satisfies the constraints. Indeed we just have to check them at r = ζj n, r = zn 2j δn j +α3 j, r = zn 2j+ + δj n α2 j. At r = ζn j the left derivative of v n is p j +ε n < 0; since by assumption the constraint (26 is satisfied by u n on the left at r = z2j n = ζn j + O(δn j, we see that choosing ε n small, of order δj n, will imply that v n satisfies the constraint on the left of ζj n. Similarly we can ensure that the constraints are satisfied on the right of ζj n and also at r = zn 2j δn j +α3 j, r = zn 2j+ +δn j α2 j. From the last inequality and the fact that ε n is of order δj n, we see that 9
there exists a constant c > 0 such that F(v n F (u n < c (δ n j 2 Also, from our construction, v n has constant sign in (zn 2j, ζn j and (ζn j, zn 2j+, hence Z vn = n 2. Now, for all n large enough, we can repeat the same process for all indices j J, giving a new affine function w n such that F (w n F (u n < c j J (δ n j 2 and the cardinal of Z wn is bounded. Since j δn j, and max j J δ n j 0 as n by assumption, the sum j J (δn j 2 converges to zero as n. Hence F (w n F (u n 0 and w n is the required minimizing sequence. Step 3. We now have Card Z wn C for some sequence (w n E A M with F(w n F (u n 0. Hence there exists k N [0, C] and a subsequence (v n of (w n such that Card Z vn = k. This ends the proof of the lemma. 3.3 Existence of a minimizer Proof of Theorem 3.. We consider a minimizing sequence (u n EM A, with Card Z u n = const. = k. We will prove that there is another minimizing sequence with similar properties, and relatively compact in W, loc (0,. This obviously implies that the minimum of F is attained. Step. We claim that we can change (u n slightly in order to have Z un = Z = {z,..., z k } and u n (z j = m j, n, j [0, k + ] N. (38 where by convention z 0 = 0 and z k+ =. Note that (38 implies also that all u n have the same sign in (z j, z j+, for j = 0,..., k. Indeed, we have Z un = {z n,..., zk n}, and u(zn j [ M, M] for all j; we define z0 n := 0 and zn k+ = for all n. Extracting subsequences, we can assume that zj n and u n (zj n converge to some limits z j, m j, for all j [0, k + ] N. Using an argument similar to the one in the second step of the proof of Lemma 3., we can find a sequence (y n EM A, such that u n y n W, (K 0 in all compact subsets K [0,, and (y n satisfies (38. This new sequence is called (u n again in the next steps. 20
Step 2. We claim that there exists another minimizing sequence (v n with the same properties than those of (u n, and also satisfying: v n is increasing in all intervals where v n > 0. Assume that there is some n N and some r (0, such that u n (r > u n (r + > 0. Since u n is a step function, u n p 0 := u n (r in some interval (r 0, r, and u n p := u n (r + in (r, r 2. We define ρ := r 0 +r 2 r and a new function y n as follows: y n is continuous, y n u n in [0, ]\(r 0, r 2, y n p in (r 0, ρ and y n p 0 in (ρ, r 2. Obviously y n A and y n satisfies (38 (in particular y n M, y n ( 0. We claim that we have also y n EM A and F(y n < F (u n. Let us first observe that u n is concave and y n is convex in [r 0, r 2 ]; since they both achieve the same limit values, y n u n in this interval (then in [0, ], since they are just equal otherwise. Hence, for all r [0, ] \ [r 0, r 2 ], y n satisfies the constraints (26 27 since u n does. Hence we just have to check (27 for y n and for r = r 0 + and r = ρ + since y n const. > 0 in (r 0, ρ and (ρ, r 2. We recall that 0 < y (r 0 + = p < p 0 = u (r 0 +; hence ( y n 2 (r 0+ < ( u y n(r n 0 + 2 (r 0+ u n(r 0 + u n( s u n (r 0 = y n( s y n (r 0 s r 0 s r 0 for all s < r, since u n satisfies (27; hence y n satisfies (27 as well for r = r 0 +. We now check for r = ρ +. We recall that y n (ρ + = p 0 (0, ] (taking (28 into account. Hence p 0 p 0 > 0 and then, for s < r 0 : y n ( s = u n ( s u n (r 0 + ( 2 (r 0 s p 0 p 0 y n (ρ + ( 2 (ρ s p 0 p 0 since y n (ρ > y n (r 0 = u n (r 0 and u n satisfies (27. Obviously, this relation is also true for s [r 0, ρ since y n is affine here. Thus y n satisfies (27 and we have proved that y n EM A. We now prove that F(y n < F (u n. Indeed we have F (y n F (u n = r2 r 0 [f(y n f(u n ]r dr = 2 (ρ2 r2 0 f(p + 2 (r2 2 ρ2 f(p 0 2 (r2 r 2 0f(p 0 2 (r2 2 r 2 f(p = [f(p 0 f(p ](r 2 r (r r 0 < 0 2
since p 0 > p > 0 implies f(p 0 < f(p. Repeating a similar process a finite number of steps, we can find, for each fixed n, a function v n EM A such that v n satisfies (38, v n is increasing where v n > 0 and F (v n < F (u n. (Note that this is not a slight change on the sequence (u n, since p 0, p are not assumed to be close together, and the interval (r 0, r 2 is not assumed to be small; hence u n v n W, is possibly large. Step 3. There exists another minimizing sequence (w n with the same properties, and also satisfying: w n / 3 in all intervals where w n < 0. In order to prove that, we fix n here (and we drop indices n for short, and we consider a given j [0, k] N such that v < 0 in (z j, z j+. Since v is a step function, there exists t 0 = z j < t < < t l+ = z j+ such that v is constant in (t i, t i+ for all i = 0,..., l. We consider the set { Q := y EM A ; y v in [0, ] \ (z j, z j+, } y 0 has at most l discontinuities in (z j, z j+ We claim that Q is compact for the topology W,p (z j, z j+. Indeed, if (y n is a given sequence in Q, let z j = τ0 n < τ n τ 2 n τ l n < τ l+ n = z j+ be the discontinuities of y n, ηn i the constant value of y n n in (τi, τ i+ n (i [0, l] N. By assumption we have l i=0 (τ n i+ τ n i ηn i = zj+ z j y n = y n(z j+ y n (z j = v(z j+ v(z j = const. Extracting subsequences, we can assume that (τi n converge to some limit τ i as n, and (ηi n converge to some limit η i. We have in the limit l (τ i+ τ i η i = v(z j+ v(z j i=0 hence the continuous function y A defined by y = v in [0, ] \ (z j, z j+, y = η i in (τ i, τ i+ belongs to W r and is the W,p -limit of (y n. We now prove that y EM A. We just have to check the constraints (26 27. Since y n satisfies these relations for all n, y n (t y(t for all t [0, ], and y n (t y (t for all t / {τ 0,..., τ l+ }, these relations follow by passing pointwise to the limit. From this compactness property and the continuity of F for the W,p topology, we deduce that F has a minimizer y Q. If we still note τ 0 = z j < 22
τ < < τ l+ = z j+ the discontinuities of y and η i the values of y, we have F (y = C + 2 l (τi+ 2 τi 2 f(η i i=0 where C is the constant part of the integral in [0, ] \ (z j, z j+. We have also the additional constraint G(y := l (τ i+ τ i η i = const. = v(z i+ v(z i. i=0 Since y is assumed to be a minimizer of F, we get in particular, for some Lagrange multiplier λ: 0 2 (F + λg η 2 i = 2 (τ 2 i+ τ 2 i f (η i, i [0, l] N Hence f (η i 0, which implies η i / 3 for all i (recall that y 0 by assumption. Hence we get that y / 3 in (z j, z j+. Repeating the same process for each fixed n and for each j such that v n < 0 in (z j, z j+, we obtain a function w n satisfying w n EM A, w n v n in each [z j, z j+ ] where v > 0, F (w n F (v n, w n / ( / 3, 0 as claimed. Step 4. Such a sequence w n is relatively compact in EM r,, for the Wloc topology. Indeed, if j [0, k] N is such that w n > 0 in (z j, z j+, we know from step 2 that w n is increasing, that is, w n is convex. As a consequence, the restriction of w n in (z j, z j+ is relatively compact for the W, loc topology. On the other hand, if j [0, k] N is such that w n < 0 in (z j, z j+, we know from step 3 that w n / 3 in this interval. Since f > 0 in (, / 3, the functional F j (w := zj+ z j f(w (rr dr is strictly convex with respect to w. Since the restriction of (w n in (z j, z j+ is a minimizing sequence for this functional, it converges to the unique minimizer of F j in the class { L j := w EM; r w satisfies (38, w } in (z j, z j+. 3 This ends the proof of the Theorem. 23
References [] G. Buttazzo, V. Ferone, B. Kawohl, Minimum Problems over Sets of Concave Functions and Related Questions, Math. Nachrichten, 73 (993, pp. 7 89. [2] G. Choquet, Cours d analyse. Topologie, Masson 964. [3] M. Comte, T. Lachand-Robert, Newton s problem of the body of minimal resistance under a single-impact assumption, to appear. Prépublications du Laboratoire d Analyse Numérique; http://www.ann.jussieu.fr/~lachand/publications.html. [4] R. T. Rockafellar, Convex Analysis, Princeton University Press (970. 24