Newton s problem of the body of minimal resistance under a single-impact assumption

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1 Newton s problem of the body of minimal resistance under a single-impact assumption M. Comte T. Lachand-Robert Abstract We consider the problem of the body of minimal resistance as formulated in [2], section 5: minimize F (u) := Ω dx/( + u(x) 2 ), where Ω is the unit disc of R 2, in the class of radial functions u : Ω [, M] satisfying a geometrical property (), corresponding to a single-impact assumption (M > is a given parameter). We prove the existence of a critical value M of M. For M M, there exist a unique local minimizer of the functional. For M < M, the set of local minimizers is not compact in H, though they all achieve the same value of the functional. Résumé Nous étudions le problème de résistance minimale proposé dans [2], section 5 : minimiser F (u) := Ω dx/( + u(x) 2 ), où Ω est le disque unité de R 2, dans la classe des fonctions à symétrie radiale u : Ω [, M] satisfaisant une propriété géométrique (), correspondant à une contrainte de choc unique et élastique (M > est un paramètre donné). Nous montrons l existence d une valeur critique de M au-delà de laquelle il existe un unique minimiseur local de la fonctionnelle, et endeçà de laquelle l ensemble des minimiseurs locaux n est pas compact dans H, bien que ceux-ci donnent la même valeur pour la fonctionnelle. Université Pierre et Marie Curie, Laboratoire d Analyse Numérique, Paris Cedex 5, France. comte@ann.jussieu.fr; lachand@ann.jussieu.fr;

2 Introduction Sir Isaac Newton formulated in his Principia Mathematica one of the oldest problems in the calculus of variations: find the body of minimal resistance moving with constant velocity through a gas. The gas is assumed to be very dilute, so particles do not interact with each other. Under these assumption, the shape of the front part of the required body can be described by the graph of a function u : Ω R, where Ω R 2 is a given bounded domain. This problem is explained with more details in [2]; see also [4] for an historical introduction. In order to avoid degeneracies (such as very long and pointed bodies), we will consider here only functions whose oscillation osc u := max u min u does not exceed a given constant M > ; since the resistance does not change by adding a constant to u (translation of the body), it is assumed without loss of generality that u takes values in [, M]. Even with these restrictions, the computation of the effective resistance of the body could be very complicated. Some particles hit the body, transmitting momentum; this quantity depends only on the impact angle and therefore on u at the contact point. But after reflection, some particles may hit the body a second time; therefore it is necessary to restrict the class of admissible functions in order to exclude this event. If each particle hit the body at most once, then the effective resistance of the body can be expressed (in appropriate units) by: F (u) = Ω dx + u(x) 2. (Throughout the paper, u is assumed to be differentiable almost everywhere; the domain of u will be noted dom( u).) One of the simplest way to ensure the single-impact assumption is to require that u is concave. This is indeed the case considered first by Newton, and we will use it for comparisons in this paper. However, this is very restrictive and appears to be mathematical artifact. Therefore, many authors tried to remove this limitation. It was already noticed by Legendre in 786 [7] that the solution of Newton does not minimize the functional among all radial functions; for, if you consider wildly oscillating functions, the value of the functional becomes arbitrarily small. However, many authors objected that it does not make sense from the physical statement of the problem: indeed for these functions, the resistance is not expressed by F (u), since the single impact assumption is not taken into account. Some different approaches are explained in [2]. But the least restrictive one, for this expression of the resistance, is to require only that any particle hitting the graph of u at x with vertical velocity does not hit it again, 2

3 u(r) M u (r )= u (r +) = t N.88 u (+)= 2 r - innerparabola Eulerpart r ru (+ u 2 ) 2 = const. Figure : A local minimizer and reflected particles (thin dashed lines). as explained in section 5 of [2]. This leads to the following (complicated) condition: x dom( u), τ >, such that x τ u(x) Ω, u(x τ u(x)) u(x) ( u(x) 2 ). () τ 2 An example of a radial function (with Ω the unit disc of R 2 ) satisfying this condition is pictured in Figure. The main goal of this paper is to exhibit all radially symmetric local minimizers satisfying this condition. It turns out that they are different from the radial function given by Newton; some examples are pictured in Figure 2. As proved in [2], the condition () implies u W, loc (Ω); for convenience, the proof is recalled in Section 2.. This is not restrictive enough, since functions in W, loc may have a wildly oscillating gradient; indeed, the set of those functions satisfying (5) is not closed (see [3]). On the other hand, it is obvious, from a physical point of view, that the gradient must have some sort of continuity property. The assumption u C is too restrictive, since the physical problem makes sense with, for 3

4 .8 M = M = Figure 2: Comparison between the Newton concave minimizer and our local minimizer (thicker line) for M = 4/5 and M = 2/5. 4

5 instance, polyhedral functions. In addition, Newton himself considered concave functions, and these functions are not necessarily of class C, but are more regular than general functions in W, loc. In this paper, we will consider only the case of radial functions on a disc Ω of unit radius. We will give necessary conditions for u to be a minimizer. The existence of a minimizer in the more general case is to be studied in [3]. For radial functions, we will restrict ourselves to those having finite or infinite left and right derivatives at every point (noted respectively u (r ), u (r+)); it is also assumed that u (r ) is left continuous on (, ], and u (r+) is right continuous on [, ). We recall that every concave or convex function (and obviously any C function) does satisfy these conditions. For radial functions on the unit disc of R 2, the corresponding set will be noted W(, ) := {u W, (, ) ; u is left and right differentiable, loc u (r+) is right continuous, u (r ) is left continuous }. For these functions, the Newton functional can be written in the form: F (u) := We study the following problem : where r dr + u (r) 2 min F (u) (2) u C M C M = {u W(, ) ; u M, u( x ) satisfies ()}. (3) Let us mention here that C M is not convex: see Remark 2.A. It is easy to check that, for any p [, ], C ([, ]) W(, ) W,p (, ), and that W(, ) is closed for the (strong) topology of W,p (, ). Also, we recall that F is bounded and continuous in W,p. In the whole paper, we consider local minimizers of F in W,p C M, with p <, in the following sense: a function u W,p C M is called a local minimizer for F is there exists δ > such that u v W,p < δ implies F (v) F (u). The existence of global minimizers is considered in [3]; we prove in this other paper that all local minimizers are actually global minimizers. We will prove in this paper that there exists a unique local minimizer for M large. The graph of this minimizer (pictured in Figure ) has a unique angular point in (, ). On the other hand, if M is smaller than a critical value M , there is a variable part in the graphs of minimizers allowing the construction of an infinite number of them (as pictured in Figure 3): see Theorem 2. for precise statements. 5

6 innerparabola variable part Eulerpart ru (+ u 2 ) 2 = const Figure 3: Differents minimizers for M =.44 < M (shifted for clarity). 2 Statement of the problem We will note f(t) := + t 2 the function appearing in the functional; we note here, for future use, that f (t) = 2t and f (t) = 2(3t2 ) ( + t 2 ) 2 ( + t 2 ). (4) 3 2. Constraints for radial functions We recall that every function is assumed to satisfy (); this can also be expressed as follows: if x dom( u) is given, such that u(x), let us note d x = sup{t > ; x t u(x) Ω}; then u(x) x dom( u), t (, d x ], u(x t u(x) u(x) ) u(x) t ( ) 2 u(x) u(x) (5) As proved in [2], the fact that osc u M and that Ω is bounded imply that the left hand side of (5) is bounded from below by M/ dist(x, Ω). 6

7 Therefore u(x) M + M 2 + dist(x, Ω) 2 dist(x, Ω) for all x dom( u). Hence u W, loc (Ω). Note that by (6) we have, for radial functions, (6) r [, ) dom u, u (r) ψ(r) := M + M 2 + ( r) 2, (7) r and a similar bound holds for u (r ) and u (r+). In particular, both directional derivatives are bounded, except possibly u ( ). For the radial functions, the constraint (5) can be written as follows: for all r (, ) dom(u ), with u (r) <, s (r, ), u(s) u(r) s r 2 and for all r (, ) dom(u ), with u (r) >, s [, r), u( s ) u(r) s r 2 ( u (r) ) u (r) ( u (r) ) u (r) (Here we have to consider also s < because the trajectory of the reflected particle at r crosses the axis r =, and is assumed to remain above the symmetrical part of u.) In particular, this last relation implies (by setting s = r). (8) (9) r (, ), u (r) () Remark 2.A. Let us prove that C M is not convex by an example: let u and v satisfy (8) with r =, s =, u() =, u (+) =, v() = 9/5, v (+) = 2, and u() = v() =. The similar condition for w = (u + v)/2 would lead to the inequality 2/5 5/2, which is not true. It is always possible to find functions u, v C M satisfying these conditions, by considering for instance arcs of parabolas: see section Main statements We will now give the main statement of this paper. We first recall some wellknown facts about the minimizer of F in the class of concave radial functions; it has been constructed by Newton from the Euler equation associated with the problem (rf (u (r))) =. () 7

8 F N M Figure 4: Value of the functional for our local minimizers (thicker line), compared to the value for Newton s concave minimizers (thin dashed line). This leads to rf (u (r)) = c = const. It turns out that for f(t) = /( + t 2 ), this equation can be solved explicitly; we recall the formulas hereafter. For c, the solution is a strictly concave decreasing (c > ) or convex increasing (c < ) C function. The concave minimizer v is constant (v M) in an interval [, r c (M)] (where r c depends on M through a formula given later); it satisfies () in (r c (M), ] and v (r c +) = (see Figure 2). We now turn our attention to the minimizers of problem (2), the minimization of F in the class C M. Our main result is: Theorem 2. There exists a number M such that, if M M, then there is exactly one local minimizer of F ; if M < M, there is an infinite number of local minimizers for F, and the corresponding set is not compact in W,p. For each value of M >, there exists a unique number r (M) (, ) and a unique concave C function η : [r, ] [, M] solution of the Euler equation () satisfying η(r ) = M, η() = and η < in (r, ), such that all solutions u of (2) satisfy u η in [r, ]. If M M, {(r, u(r)} r [,r ] is an arc of the parabola whose focus is ( r, M) and containing the point (r, M). 8

9 Figure 5: Unique local minimizer for M = M If M < M, {(r, u(r)} r [,4M/3] is an arc of the parabola whose focus is ( 4M/3, M) and containing the point (, ); and in (4M/3, r (M)), the graph of every minimizer is composed of arcs of parabolas. For a given value of M, all local minimizers achieve the same value of F. The proof of this theorem will be given throughout the rest of the paper. Some examples of the shape of u are pictured in figures 5 7. Remark 2.B. It will appear in the proof that u (r +) > for all minimizers; hence u and the concave minimizer does not coincide in [r, ]. On the other hand, we will obtain the relation u [ 2, ] in [, r ]. Therefore the well known property for concave minimizer v: v / (, ) (which has been generalized to non radial functions in the form v / (, ), see [2]) is not true for u. Remark 2.C. It is easy to check out that for M < M, the set of minimizers is not compact in W,p topology (p < ). Indeed it is possible to choose a sequence of minimizers (u n ) as pictured in Figure 3, such that osc u n in the variable part; this implies that (u n ) converges weakly to a constant function in this interval, but since u n /2, there is no strong convergence. More generally, the class C M is not compact. Hence the existence of a global minimizer of (2) is not obvious. This problem is investigated, for the radial functions and for the more general class of functions satisfying (5), in [3]. 3 Properties of admissible functions If u satisfies (8 9), but is not differentiable at some point r (, ), we have 9

10 Figure 6: Unique local minimizer for M =.73 > M Figure 7: An example of local minimizer for M =.44 < M.

11 . If u (r+) <, u satisfies (8) at r with u (r+) replacing u (r). 2. If u (r+) >, u satisfies (9) at r with u (r+) replacing u (r). Similar properties hold for u (r ). This follows from the continuity properties of the derivatives. As a consequence, we have: Lemma 3. Let u satisfy (8 9). If, at some point r (, ), the left and right derivative do not coincide, then: And similarly: Also, at r =, we have: If u (r+) >, 2u (r+) u (r ) u (r+) 2. (2) If u (r ) <, 2u (r+) u (r ) u (r ) 2. (3) If u (+) >, 2u (+) 2 u (+) 2 u (+) 3. (4) (Note that, if u (r ) = u (r+), these relations reduce to u (r) 2, which is not interesting.) Proof. It suffices to pass to the limit s r in (8); this yields (2). Similarly (3) can be deduced from (9). The case r = can be considered similarly, but here we have to replace u( ) by u(+). Definition 3. Let r [, ] be given. We define { Σ + (r) := s J + (r) ; where u( s ) u(r) s r = 2 [, r) if u (r+) > J + (r) := if u (r+) = (r, ] if u (r+) < ( u (r+) u (r+) and Σ + (r) := Σ + (r) {r} if u (r+) > and equality is attained in (2), Σ + (r) := Σ + (r) otherwise. We define similarly Σ (r) and J (r) with u (r ), for any r (, ], and Σ (r) := Σ (r) {r} if u (r ) < and equality is attained in (3), Σ (r) := Σ (r) otherwise. Finally, we define Σ(r) := Σ (r) Σ + (r). (5) As before, we shall say that a function u saturates the constraint Σ at r = r if Σ(r ) is not empty. We shall also say that s is a contact point for r [, ] if s Σ(r). )}

12 Remark 3.A. It should be noticed that there is essentially three ways to saturate the constraints at some point r (, ): first we can have u(r ) = or u(r ) = M; second, we can have equality in (8) or (9), or more generally Σ(r ) ; third, we can have r Σ(s) for some s, that is Σ (r ). Consequently, some small variations of the function u near r could lead to non-admissible functions. In this paper, we will mainly use three types of small perturbations of a local minimizer u:. If u does not saturate any of the constraints in a subinterval (a, b) [, ], then for any function v C c (a, b), there exists ε > such that, for ε ( ε, ε ), u + εv is admissible; hence F (u).v =. If v is not differentiable at some points of (a, b), we must check the compatibility conditions given in Lemma 3. for u + εv at these points. 2. If Σ(a, b) = and u > in (a, b), u + εv is admissible for v C c (a, b), v and ε < small. Indeed u + εv and Σ u+εv = for ε < small enough, since u + εv(s) u(s) for all s (, ), as one can see in the expression of the constraints (8 9). 3. If Σ (a, b) = and u < M in (a, b), u+εv is admissible for v C c (a, b), v and ε > small. Lemma 3.2 The set S u := {r [, ] ; Σ(r) } is closed in [, ]. Proof. Let (r n ) S u with limit r [, ]. Restricting to subsequences if necessary, we can assume that for all n, Σ + (r n ) (the case with Σ is similar), and that (r n ) is a monotone sequence. Let us first assume that for an infinite number of indices n, we have r n Σ + (r n ), i.e. 2u (r n )u (r n +) = u (r n +) 2. This contradicts the assumption on the continuity of the derivatives: for, if (r n ) is increasing, for instance, lim u (r n ) = u (r ) = lim u (r n +); then we get, passing to the limit, 2u (r ) 2 = u (r ) 2, which is impossible. Therefore, for an infinite number of indices n, there exists s n Σ + (r n ) (or in Σ (r n ), but this gives something similar), i.e. u( s n ) u(r n ) s n r n = 2 ( u (r n +) u (r n +) ) (6) Extracting again a subsequence, we can assume that (s n ) converges to some limit s. If s r, and (r n ) is decreasing, then u (r n +) converges to u (r+), and we get (passing to the limit) u( s ) u(r) s r = 2 ( u (r+) ) u (r+) 2

13 Therefore, s Σ + (r) and r S u. If s r, and (r n ) is increasing, then u (r n +) converges to u (r ); hence s Σ (r) and r S u. If s = r, and (r n ) is decreasing, we must have u (r n +) > for n large; for, if not, we would have s n (r, ] and u( sn ) u(rn) s n r n would converge to u (r+). Therefore, passing to the limit in (6), we get u (r+) 2 =, which is impossible. Now the fact that s n < r for n large implies, passing to the limit in (6): u (r ) = ( u (r+) ) 2 u (r+) Hence (2) is satisfied and r Σ(r), implying r S u. A similar proof leads to the same conclusion if s = r and (r n ) is increasing. Remark 3.B. Let us say that r (, ) is a concavity point for u if u (r ) > u (r+). Then, conditions of Lemma 3. are always satisfied at these points. This can be easily checked as follows: there is no condition if u (r ) u (r+); otherwise, if u (r+) > for instance, then 2u (r+) u (r ) < 2u (r+) 2 < u (r+) 2, hence (2) is satisfied. Remark 3.C. Conversely, let us say that s (, ) is a convexity point for u if u (s ) < u (s+). One interesting fact about convexity points is that they cannot be contact points. This can be proved as follows. Assume that there exists r < s with u (r) <, such that equality holds in (8) for instance; chosing s = s + ε for ε >, we see that 2 ( u (r) u (r) ) = u(s) u(r) s r u(s + ε) u(r). s + ε r This implies after multiplying by the denominators, and subtracting (s r) (u(s) u(r)): ε (u(s) u(r)) (s r) (u(s + ε) u(s)) Dividing by ε and letting ε leads to (u(s) u(r))/(s r) u (s+); similarly, u (s ) (u(s) u(r))/(s r). This cannot happen if s is a convexity point. This observation can even be strengthened as follows, using the continuity of left and right derivatives of u: there must exist a neighborhood V of s such that σ V, the set Σ (σ) := {r [, ] ; s Σ(r)} is empty. 3

14 4 A priori properties of local minimizers We now state some useful a priori properties of local minimizers (in W,p topology, p [, + )), in subintervals where the constraints are not completely saturated. Theorem 4. Let u be a radial local minimizer of (2). Assume that there is an interval (a, b) (, ) such that Σ(r) = for all r (a, b) ( i.e. u does not saturate the constraint in (a, b)). Then u is differentiable in (a, b), u (r) in (a, b), and u satisfies the Euler equation () in the sense of distributions in (a, b). It is important to remark that it is not assumed that Σ (r) = for r (a, b); this property appears to be true for minimizers, but that follows from different considerations. Proof. Step. Let us choose r (a, b) dom(u ). We will prove first that u (r ) / 3. We restrict here to those points where u > ; the general case will then follow from the continuity properties of the directional derivatives. Since u (r ), we have u(r ) (, M). Let α, β, γ be given in (, ], and consider the function v := min (u, u + α(r r ) + + β(r r ) γ). Note that v u, except in (r, r 2 ) := (r γ, r β + γ ) where v < u. For γ α small enough, we may assume that this interval is included in (a, b). We will first prove that v satisfies the constraint for appropriate values of α, β, γ. We see that v u, u v W, is small for α, β, γ small. The points r = r, r = r γ, r = r β + γ do not belong to α dom(v ). For almost all values of α, β, γ, the last two points belong to dom(u ). In that case, v is not continuous at these points; but they are concavity points for v (cf. Remark 3.B): for instance, v (r ) = u (r ) > v (r +) = u (r ) β. At the point r, the situation is different; here, v is discontinuous, and has left derivative equal to u (r ) β, and right derivative equal to u (r )+α. For α, β small enough, we always have 2(u (r ) β)(u (r ) + α) < max{(u (r ) β) 2, (u (r ) + α) 2 } (7) since 2u (r ) 2 u (r ) 2 < u (r ) 2. Hence v satisfies the conditions of Lemma 3.. 4

15 Note that all previous conditions can be written in the form α+β+γ < ε for some ε small; if this is satisfied, v satisfies the constraints in [a, b], as explained in Remark 3.A. On the other hand, v satisfies the constraints in (, )\[a, b] (where v u), because u does and v u. Since u is a local minimizer, we must have F (v) F (u) for α, β, γ small. This is equivalent to: r + γ α r γ β r dr r + u (r) 2 r γ β r dr r + γ + (u (r) β) + α 2 r r dr + (u (r) + α) 2 (8) Let us divide this inequality by r γ >, and let γ tend to. We get: (/α) + (/β) + u (r ) 2 This can also be written φ(α) φ( β), /β + (u (r ) β) + /α (9) 2 + (u (r ) + α) 2 where φ(t) := t [ ] + (u (r ) + t). 2 + u (r ) 2 Recall that this must hold for any α, β in some interval (, ε ). Since φ is of class C (R), this implies φ (), i.e. φ () = 2 f (u (r )) = 3u (r ) 2 ( + u (r ) 2 ) 3 ; hence, we must have u (r ) 3, as claimed. Step 2. We will prove that u is differentiable on (a, b), except possibly at some c satisfying u (c+) 3 < 3 u (c ) and Σ (c). (2) Assume that there is some r (a, b) such that u (r +) u (r ). Note that we must have u (r +) / 3 and u (r ) / 3 (2) from the previous step and the continuity of the left and right derivatives. Using the same argument, we can still obtain (8), for α >, β >, γ > small enough. Taking the limit γ leads to /β + u (r ) 2 + /α + u (r +) 2 /β + (u (r ) β) 2 + /α + (u (r +) + α) 2. 5

16 Again letting α and β tend to yields f (u (r )) f (u (r +)) (22) where f is given in (4). Note that f is negative on [ 3, + ), positive on (, 3 ], and increasing in both intervals. Therefore, taking (2) into account, we must have u (r ) > > u (r +) or u (r +) > u (r ) > or u (r ) < u (r +) <. Let us first prove that neither of the two last possibilities can hold. In fact, in either cases, r is a convexity point for u: cf. Remark 3.C. As observed in this remark, there exists a neigborhood V of r such that s V, Σ (s) =. Therefore, if we consider the function w := max (u, u α(r r ) + β(r r ) + γ), for α, β, γ positive but small enough, this function still satisfies the constraint. Using again the fact that F (w) F (u) and letting γ, then α and β, we obtain relations similar to the proceding ones but with reversed signs: f (u (r )) f (u (r +)) (23) Hence, using (22), we get u (r +) = u (r ), a contradiction. Similarly, if we have u (r ) > > u (r +) with Σ (r ) =, we have again (23), and this is a contradiction since the sign of f (t) is opposite to the sign of t, for any t R. Hence, if u is discontinuous at some point c (a, b), we must have (2). Note that this can happen only once in the interval, because c is now a local maximum, but no local minimum can exist (u = is excluded by the fact that u 3 and u (r ) < < u (r+) by (22)). Step 3. Let us prove that for r (a, b), we have ( rf (u (r)) ) ( ) 2ru (r) = [ + u (r) 2] 2. (24) in distributional sense. Let us first observe that u > in (a, b); indeed u(r) = for r (a, c) (c, b) would imply u (r) =, in contradiction to the result of Step ; and u(c) = contradicts (2). Indeed, since Σ(r) = for all r (a, b), if v is any regular function with compact support in (a, b), then u εv satisfies the constraints for ε > small enough, as explained in Remark 3.A. 6

17 u u u µ + Figure 8: Construction of u ε. Hence F (u εv) F (u), and this implies b a f (u (r))v (r)r dr = b a ( rf (u (r)) ) v(r)r dr (25) Since this is true for any regular v, we find (24). Step 4. We now prove that ( rf (u (r)) ) =. We fix here a compact subinterval [α, β] (a, b) and v Cc (α, β) such that v. We consider the continuous function u ε defined as follows: u(r) ε for r [α, β] u ε (r) = u(r) for r > β + εν ε or r < α εµ ε an affine function in [α εµ ε, α] and [β, β + εν ε ] Here, µ ε > and ν ε > are chosen such that u ε satisfies the constraints. We will first explain how it is possible to find these numbers. (The construction of u ε is illustrated in Figure 8.) Indeed, since Σ(α) =, and u is continuous, there exists p < u (α ) such that s J (α), u( s ) u(α) < ( 2 (s α) p ). (26) p 7

18 We fix u ε p in (α εµ ε, α); since u ε is continuous, this implicitly defines µ ε as the smallest positive solution of the equation u(α εµ ε ) = u ε (α εµ ε ), i.e. u(α εµ ε ) = u(α) ε p εµ ε. Note that µ ε µ := /(u (α ) p ) as ε by a Taylor expansion; in particular, µ ε > exists when ε is small enough. Similarly, we fix u ε p 2 in (β, β + εν ε ), for an appropriate value of p 2, and this defines ν ε. We claim that u ε satisfies the constraints if ε > is small enough. Indeed, if r < α εµ ε or r > β + εν ε, then u u ε in a neigborhood of r, and u ε u in the whole domain; hence it is clear that the constraint is satisfied here. If r (α εµ ε, α), u ε(r) = p ; hence, since u u ε ε everywhere, u ε (s) u ε (r) ( 2 (s r) u ε(r) ) u ε(r) = u(s) u(α) ( 2 (s α) p ) + O(ε). p Therefore, the left-hand side is negative for small ε (by (26)), and the constraint is satisfied at this point. A similar property holds for r (β, β + εν ε ). Finally, for r (α, β), we recall that Σ(r) = ; thus γ + (r) := min s J + (r) [ (s r) 2 ( u (r+) u (r+) ) ] u(s) + u(r) > for all r [c, β] (note that u (r+) = u (r) for r c). Moreover, γ + (r) is a continuous function of r (since u is differentiable in (c, β], and therefore continuously differentiable by the continuity properties of the directional derivatives). Similarly, defining γ with the left derivatives, we get a continuous function in [α, c]. (It is possible that u does not change sign in [α, β], i.e. no point c exists in this interval; in that case, γ + = γ are defined in [α, β].) Hence if ( ) ε < max max γ (r), max γ + (r), r [α,c] r [c,β] then u ε satisfies the constraint in (α, β). Hence, we have proved that u ε satisfies the constraints. Moreover, if ε is small enough, S uε (a, b) =. Therefore u ε + εv is admissible and F (u ε + εv) F (u) for small ε; this implies F (u) F (u ε ) F (u ε + εv) F (u ε ) = ε 8 β α f (u ε)v r dr = ε β α f (u )v r dr.

19 On the other hand, u u ε in [, α εµ ε ) (α, β) (β+εν ε, ); and u u ε = O(ε) by the construction of u ε. Thus [ F (u) F (u ε ) = f (u ) f (u ε) ] r dr = O(ε 2 ). (α εµ ε,α) (β,β+εν ε) Hence we have (also using (25)): β α f (u )v r dr O(ε) Hence (rf (u )) = in (α, β) since v is arbitrary in C c (α, β). This proves that () holds in (a, b). Step 5. We prove now that u < in (a, b); in particular, there is no point c satisfying (2). Assume by contradiction that u > in (a, c). Since u >, we get u > from the Euler equation, that is, u is strictly convex. Let α > be a small parameter; we consider a positive function φ C 2 ([a+α, c]), such that φ(a+α) = (u(a + α) u(a))/α and φ(c) = φ (c) =. We define a function v by u(r) for r [, a) (c, ) v(r) = u(r) αφ(r) for r (a + α, c) u(a) for r (a, a + α) We claim that, for α > small enough, v satisfies the constraint. This follows from the fact that Σ u (r) = : hence Σ v (r) = for r (a + α, r ) for small α, since u v C (a+α,c) is small. It is also obvious that Σ v(r) = for r (a + α, c) (because v = on this set). Finally, for small α, v is convex in (a, c) since u is strictly convex here; therefore, no contact point can exist in (a, c). Let us compute F (v) F (u), assuming α is small; we get, integrating by parts: F (v) F (u) = = a+α a a+α a [ f() f(u (r)) ] r dr + c a+α [f(u αφ ) f(u )]r dr ru (r) 2 c + u (r) dr α rf (u (r)) φ (r) dr + O(α 2 ) 2 a+α = α au (a) 2 + u (a) 2 + αφ(a + α)(a + α)f (u (a + α)) + O(α 2 ) 9

20 since rf (u ) = const. = (a + α)f (u (a + α)) from the Euler equation. Therefore, using αφ(a + α) = u(a + α) u(a) = αu (a) + O(α 2 ): F (v) F (u) = α au (a) 2 + u (a) 2 + αu (a)af (u (a)) + O(α 2 ) = αau (a) 2 ( + u (a) 2 ) 2 ( + u (a) 2 2 ) + O(α 2 ) This is negative for α > small, since < u (a) <, as follows from (9) with s = c, r = a. Hence, u is not a local minimizer; this is a contradiction. Step 6. We now have to prove that u. Recall that we already know, from the previous steps, that u / 3. Let us assume, to force a contradiction, that u (r ) > for some r (a, b). Let ε > small be given; we perturb u in a neighbordhood of r as follows: u ε u in [, r ε] [r + ε, ]; u ε const = u(r + ε) in [r, r + ε]; and u ε is affine in [r ε, r ], such that u ε is continuous on [, ]. In particular u > in [r ε, r ], for small enough ε; hence, the point r is a convexity point for u ε which satisfies condition (3). Since Σ(r ) =, u ε is admissible. Since u is assumed to be a local minimizer, we have, taking into account that u ε = u + O(ε) in (r ε, r ): F (u ε ) F (u) = = r r ε r r ε r[f(u ε(r)) f(u (r))] dr + r +ε r r[f() f(u (r))] dr rf (u (r))[u ε(r) u (r)] dr + o(ε) + εr [f() f(u (r ))] = r f (u (r ))[u ε (r ) u(r )] + o(ε) + εr [f() f(u (r ))] integrating by parts and taking into account that u(r ε) = u ε (r ε) and using (). Since u ε (r ) u(r ) = u(r + ε) u(r ) = εu (r ) + o(ε), we get, to the first order in ε: f (u (r )) u (r ) + f() f(u (r )). This inequality is never satisfied for u (r ) (, 3 ]. Indeed, we have pf (p) + f() f(p) = p2 (p 2 ) ( + p 2 ) 2. This concludes the proof of the Theorem. 2

21 Lemma 4. Let u be a radial minimizer of (2). Then u() = and max u(r) = M. r [,] Proof. If u(), then u() > since u C M. Since u (r) for all r from (), either u is bounded in (, ), or there exists a sequence (r n ) (, ) such that u (r n ) and r n as n +. Let us first assume that u is bounded, say by a constant C. Let ε (, ) be given, and consider the function { u(r) if r < ε v(r) = ( r) u( ε) if r [ ε, ] ε It is clear that v C M, and F (v) F (u) = ε r[f(v ) f(u )] dr < if v = u( ε)ε > C. It is possible to find ε > satisfying this condition, since u() > and u is continuous. This is a contradiction since u was supposed to be a minimizer. Let us now assume that there exists a sequence (r n ) (, ) such that u (r n ) and r n as n +. We may assume that u (r) u (r n ) for r (, r n ), for instance by choosing r n as a point where u is maximal in [, ]. Let us define the function v n n by u(r) if r (, t n ] v n (r) = u(r n ) 2 u() + u (r n )(r r n ) if r (t n, r n ] u(r) u() if r (r 2 n, ) where t n is the smallest number in the set {r [, r n ) ; u(r) u(r n ) 2 u() + u (r n )(r r n )} Note that this set contains at least an interval (r n ε, r n ) since u() >. We then have v n(r) u (r) for r (t n, r n ), and v = u elsewhere. Hence F (v n ) < F (u) except if u is constant in (t n, r n ), which cannot happen for all n. This is again a contradiction with the fact that u is a minimizer. We now prove that max u = M. Suppose that m := max u < M; then û := u+m m C M, and F (û) = F (u); so û is a minimizer. This contradicts the previous result, since û() = u() + M m >. 2

22 5 Construction of a local minimizer We now explain how a local minimizer u of F can be constructed from a priori properties; it will appear in the following process that this minimizer is unique for large M. 5. Separation into a Euler part and a constrained part We will now prove that any minimizer should have a Euler part near the boundary, that is, some neighborhood of the boundary where the constraints are not saturated and the Euler equation () is satisfied. In the remaining part, the constraints must be saturated: Theorem 5. Let r := sup S u (where S u is defined in Lemma 3.2). Then r <, and u (r ) > > u (r +). Moreover, u is strictly convex and differentiable in any subinterval of S u where u has constant sign. Proof. Step. We first prove that r <. We recall that S u is a closed set (Lemma 3.2), hence r S u. Therefore, if r = we must have Σ(), that is, Σ () (since Σ + () is empty by definition). From the definition, this implies that there exists s such that equality occurs in the constraint, but in particular we must have u ( ) >. On the other hand, we know from Lemma 4. that u() =. Hence, we find u < near r =, and this contradicts the assumption that u. Step 2. We now prove that u (r +) < < u (r ). As explained in the introduction, the solution of the Euler equation () is a function of class C whose derivative does not vanish. Since u() = and u, we must have u < in (r, ). In particular, u (r +) < by the right continuity of the right derivative. This implies Σ + (r ) = ; since r S u, we then have Σ (r ). Hence u (r ). Suppose u (r ) < ; then, there exists s > r such that s Σ(r ). This implies u (s) = (u(s) u(r ))/(s r ) since u is differentiable at s. On the other hand, u is strictly concave in [r, s]; therefore u (s) > (u(s) u(r ))/(s r ). This is a contradiction. Step 3. Let [a, b] S u such that u has constant sign in (a, b); we claim that u is differentiable and convex in (a, b). We assume u > for instance; the other case is similar. (Use (8) instead of (9) in (a, b) if u <.) Let r (a, b); assume that u (r ) u (r+). If for instance u (r ) > u (r + ) then, for all s [, r): ( u (r ) ) > ( u (r+) ) u( s ) u(r) 2 u (r ) 2 u (r+) s r 22

23 hence Σ (r) = Σ (r) = ; similarly, Σ + (r) = if u (r ) < u (r + ). We recall that S u [a, b], hence Σ(ρ) for all ρ in a neighborhood of r; on the other hand, the fact that either Σ (r) = or Σ + (r) = implies that Σ(ρ) = in a right or left neighborhood of r, from the continuity of u and its directional derivatives. This is a contradiction. We now prove that u is strictly convex in (a, b). In order to do that, we just have to prove that u is increasing. Let us first prove that for all r (a, b), Σ(r) (a, b) =. For if not, there exists r (a, b), and s Σ(r) (a, b). Since u (r) > thus s < r and 2 ( u (r) u (r) ) = u(r) u(s) r s = u (s) from (9) and the fact that u is differentiable at s. Since s (a, b) S u, there exists σ < s such that ( u (s) ) u(s) u(σ) =. 2 u (s) s σ ) Substituting u(s) = u(r) (u (r s) (r) in this relation, we get: 2 u (r) u (s) u(σ) = 2u(r) r s ( u (r) ) u (s) s σ s σ u (r) Since σ < r, we can apply (9) with r and σ, and replace in the last relation: u (s) ( u (s) u (r) ) [ r σ u (r) s σ r s ] = u (r) s σ u (r) = 2u (s) Hence u (s) + /u (s), in contradiction to the relation u (s) >. Now we can prove that u is increasing in (a, b). Let us consider r, ρ in (a, b), with ρ < r. Let us fix s Σ(r); note that s a < ρ from the previous remark. Therefore ( u (ρ) ) u(ρ) u(s) 2 u (ρ) ρ s from (9) (applied at ρ). On the other hand, since ρ < r, we have: from (9) u(ρ) < u(r) ( 2 (r ρ) u (r) ) u (r) (inequality is strict since ρ / Σ(r)); substituting in the previous equation, we get: u (ρ) u(s) < 2u(r) r ρ ( u (r) ) u (ρ) ρ s ρ s u (r) 23

24 Since s Σ(r), we have 2(u(r) u(s)) = (r s) hence a new substitution leads to u (ρ) u (ρ) < [ r s ρ s r ρ ρ s ( u (r) ) u (r) ] ( u (r) ) = u (r) u (r) u (r) This implies u (ρ) < u (r) since t t /t is increasing in (, + ). This ends the proof of the theorem. 5.2 Optimal computation Hence, we need to construct some function in two parts: a constrained part in some interval [, r ], and an Euler part in (r, ]. We will see in the next section that the first one is indeed composed from pieces of parabolas, but we shall first study an abstract problem, where the shape of the constrained part is imposed. For any given numbers ρ (, ), µ >, there exists at most one decreasing concave function η ρ,µ : [ρ, ] [, µ] satisfying the Euler equation () in [ρ, ], and boundary conditions η ρ,µ (ρ) = µ and η ρ,µ () =. This is a well-known fact, since this function is needed for the construction of the minimizer in the classical Newton s problem; since we need the precise value of η ρ,µ, we will prove this fact again in the proof of the next theorem, also giving a condition on µ, ρ in order to ensure the existence of η ρ,µ. Actually, the Euler equation with these boundary conditions can be solved only if µ µ(ρ) defined by: µ(ρ) := g ( ) (ρ) ρ 3 ρ g ( ) (ρ) 3 dτ g(τ) (27) where g(t) := f (t)/f (/ 3) for t / 3 is decreasing, and g ( ) is its inverse map. It turns out that the map ρ µ(ρ) is decreasing, satisfies lim ρ µ(ρ) = + and µ() = (cf. proof of Theorem 5.2 hereafter). Theorem 5.2 Let φ : [, ] R be a given Lipschitz function satisfying φ() = ; we note m φ := max t [,] φ(t), and we assume that m φ < M. For any ρ (, ) such that µ := M m φ ρ µ(ρ), let u ρ be the continuous function defined as follows: { ρφ(r/ρ) + µ for r [, ρ] u ρ (r) = (28) η ρ,µ (r) for r [ρ, ] 24

25 Then, the minimal value of F (u ρ ) for all ρ (, ) is attained for a unique value ρ depending only on m φ and J φ := sf(φ (s)) ds. The optimal value F (u ρ ) also depends only on m φ and J φ. Moreover, if φ, φ 2 are two functions satisfying the conditions given for φ, with J φ = J φ2 but m φ > m φ2, then we have F φ > F φ2. It should be noticed that, by construction, max u ρ = M; on the other hand, min u ρ = min(, ρ min φ + M m φ ρ); hence u ρ has values in [, M] if and only if osc φ M/ρ. Also, we don t take care here of the constraints: it is not required that u ρ belongs to W. Remark 5.A. It will appear in the proof that, in the particular case m φ =, the value of ρ is given by ρ = f (a φ ) (29) f (t φ ) and we have also where F (u ρ ) = F φ := f(a φ ) (a φ M)f (a φ ) (3) t φ = 3 4J φ + 9 6J φ 4J φ (3) and a φ is the unique solution in (t φ, + ) of the equation M = a φ t φf (a φ ) f (t φ ) f (a φ ) aφ t φ dτ f (τ) = a φ 4a 2 φ + 3a4 φ 4 ln(a φ) 4t 2 φ 3t4 φ + 4 ln(t φ) 4( + a 2 φ )2. (32) Since this particular case will prove important in the following, we will need these relations later. Proof of Theorem 5.2. We first explain the construction of η ρ,µ (we write just η here). It can be constructed in a parametrized form as follows. We recall that the Euler equation can be written rf (η (r)) = const. We will use as a new parameter t = η (r); since η is assumed to be concave decreasing, then t > and η is bijective. We note a = η (), which is the maximal value of t. Hence, 25

26 the Euler equation reduces to rf (t) = f (a) (since f is even, so f is odd), or: r = r e (t) = f (a) f (t). As a consequence, we have: dη dt = dη dr Since by assumption, η() =, we get η(r e (t)) = f (a) dr e dt = tf (a)f (t) f (t) 2. a t τf (τ) f (τ) dτ 2 = a tf (a) f (t) f (a) a t dτ f (τ) from an integration by parts. In order to find η e, we need to find a > t 3 satisfying two conditions: ρ = r e (t ) = f (a) f (t ) and a µ = η(r e (t )) = a t ρ f dτ (a) t f (τ). (33) These equations can be uniquely solved if µ µ(ρ). Indeed, let us first recall that f (t) = 2t/( + t 2 ) 2 is negative increasing in [ 3, + ). Hence, if a > 3 and ρ (, ) are given, there exist a unique t = t (a) satisfying f (t ) = f (a)/ρ. Hence: d da η(r e(t (a))) = d [ a ] a t (a)ρ f dτ (a) da f (τ) = ρ dt = f (a) [ da f (a) a t (a) dτ f (τ) >. t (a) f (a) f (t ) dt da ] a f (a) t (a) dτ f (τ) Hence, if a solution of (33) exists, it is unique. Also we have lim a η(r e(t (a))) = + (34) as one can prove by writing f (t) = 2t 3 + O(t 4 ) as t ; hence, t (a) = ρ /3 a + o(a), and η(r e (t (a))) = a ρ 4/3 a a 3 a t (a) 26 τ 3 dτ + o(a) = 3 4 ( ρ4/3 )a + o(a).

27 Therefore, if µ is large, a is large, too. On the other hand, one can note that f (t) f (/ 3) = 3 3/8, so f (a) 3 3ρ/8; hence, for ρ small and µ large, there is no solution to (33). As a conclusion, we see that for ρ (, ) given, there exists µ > such that for all µ [, µ], the equations (33) can be uniquely solved. The (complicated) map ρ µ(ρ) is decreasing; since the smaller value of ρ is attained for t = / 3, it can be parametrized with respect to a as follows: µ = η(r e ( )) = a f / 3 (a) 3 f (/ 3) f (a) ρ = f (a) f (/ 3). a 3 dτ f (τ) This gives (27) using the auxiliary function g and its inverse map. Let us now compute the contribution of η to the functional, namely: F e (ρ, µ) := ρ rf(η ρ,µ(r)) dr Using the former parametrization, this can be written: F e (ρ, µ) = a t = 2 f(a) 2 f (a) 2 f (τ)f(τ) f (τ) 3 dτ f(t )f (a) 2 f (t ) 2 2 f (a) 2 a integrating by parts again. In order to compute F (u ρ ), we need to add the value of F φ (ρ) := Hence we have ρ t dτ f (τ). rf(ρ d dr φ(r/ρ)) dr = ρ2 sf(φ (s)) ds = ρ 2 J φ. F (u ρ ) = F φ (ρ) + F e (ρ, M m φ ρ) = ρ 2 J φ + F e (ρ, M m φ ρ). We have to minimize this function of ρ; on the other hand, since F e cannot be expressed in a simple way with respect to ρ, we use again the parameters t, a (but we will write t instead of t to shorten notations). The function to be minimized is Ψ(t, a) := F φ (ρ) + F e (ρ, M m φ ρ) = f (a) 2 f (t) J 2 φ + 2 f(a) f(t)f (a) 2 2 f (t) 2 2 f (a) 2 27 a t dτ f (τ).

28 under the constraint that µ = M m φ ρ, or f (a) M = m φ f (t) + a tf (a) a f (t) f dτ (a) t f (τ) We have, from a careful computation: t Ψ(t, a) = f (a) 2 f (t) f (t)(f(t) 2J 3 φ ) a Ψ(t, a) = f (a) a f (t) f (a) (f(t) + f (t) 2 2 t t Ξ(t, a) = f (a) f (t) f (t)(t m 2 φ ) ( a Ξ(t, a) = a f (t) f (a) t + f (t) t =: Ξ(t, a) (35) ) dτ f (τ) 2J φ ) dτ f (τ) m φ Therefore, the Lagrange condition Ψ Ξ a t = Ψ Ξ reduces to t a a ) (f(t) + f (t) 2 dτ t f (τ) 2J φ (t m φ ) = ( a ) = (f(t) 2J φ ) t + f dτ (t) f (τ) m φ since other terms do not vanish. This again can be simplified in 2J φ = f(t) f (t)(t m φ ) = + 3t2 2m φ t ( + t 2 ) 2 (36) (This equation can be easily solved if m φ =, giving (3).) It should be noticed that this equation does not depend on a: hence, the optimal value t φ for t must solve (36). This equation has at most one solution in [ 3, + ) if m φ / 3, since d dt [f(t) f (t)(t m φ )] = f (t)(t m φ ) If m φ > / 3, then (36) may have two solutions, one of them in [ 3, m φ ), the other one in (m φ, + ). It will appear soon that only the last one gives a minimum for Ψ, and this one will be noted t φ ; the other one we note t φ for a while. t 28

29 Let us note Ξ(a) := Ξ(t φ, a). We note that t φ > m φ implies a > Ξ(a) from (35); moreover, we have ( d da Ξ(a) = ) a f (t φ ) f (a) t φ + f dτ (t φ ) f (τ) m φ and = f (a) f (a) [ a Ξ(a) ] > lim Ξ(a) = mφ and lim Ξ(a) = + a t φ a (since f (a) and a f (a) a t the equation dτ f (τ) t φ = 3a/4+O(/a) as a ). Therefore, M = Ξ(a) (37) has a unique solution a φ (t φ, + ) if M > m φ. If m φ > / 3, we recall that another solution t φ of (36) can exist; the corresponding a should be less than Ξ(t φ, a) from (35) again. However, the equation M = Ξ(t φ, a) do not have any solution in (t φ, M) since M > m φ by assumption and d da Ξ(t φ, a) = f (a) [ a Ξ(t f (a) φ, a) ] < and lim Ξ(t φ, a) = m φ. a t φ Let us substitute (36) and (35) in the expression of Ψ(a) := Ψ(t φ, a): 2 Ψ(a) = f (a) 2 f (t) 2 [f(t) f (t)(t m φ )] + f(a) f(t)f (a) 2 = f(a) (a Ξ(a))f (a). a f (a) 2 f (t) 2 t dτ f (τ) Hence, the optimal value for Ψ is f(a φ ) (a φ M)f (a φ ); this gives (3). (The unique optimal value ρ is given by (29).) We now prove the last assertion of the theorem, with different functions φ. So let us assume that we change smoothly the function φ, with J φ = const.. It is convenient here to use m := m φ as a parameter. Since J φ is constant, t φ changes accordinly to (36) and a φ accordingly to (37). Differentiating this last one with respect to the parameter m, we get = d dm Ξ(t φ, a φ ) = Ξ t (t φ, a φ ) dt φ dm + Ξ a (t φ, a φ ) da φ dm + Ξ m (t φ, a φ ). 29

30 The optimal value F φ = Ψ(t φ, a φ ) changes with derivative: d dm Ψ(t φ, a φ ) = Ψ t (t φ, a φ ) dt φ dm + Ψ a (t φ, a φ ) da φ dm (since Ψ does not depend explicitly on m, and J φ is assumed to be constant). We recall that, from (36), we have for t = t φ : Ψ t (t φ, a φ ) = f (a φ ) Ξ t (t φ, a φ ) and Ψ a (t φ, a φ ) = f (a φ ) Ξ a (t φ, a φ ). Therefore, we get: { d Ξ dm Ψ(t φ, a φ ) = f (a φ ) t (t φ, a φ ) dt φ dm + Ξ a (t φ, a φ ) da } φ dm = f (a φ ) Ξ m (t φ, a φ ) = f (a φ ) 2 f (t φ ) >. Hence, any increase in m φ will increase F φ, and the proof of the theorem is complete. Let us give some important special cases for φ. The special case φ = leads to the classical Newton s problem (u φ is concave). We get here from (36) (with m φ =, J φ = /2) the equation = f(t) tf (t), whose only solution is t = ; this proves the well-known fact that the derivative is at the discontinuity point in this problem []. The case (t + )2 φ(t) = 4 whose graph is a parabola of focus (, ) will prove important in the following. In that case, the graph of u φ, in the constrained part, is a parabola of focus ( ρ, M) (note that m φ = here). We get here J φ = γ N := sf(φ (s)) ds = 4s 4 + (s + ) 2 ds = 6 ln(2) π 2 2 ln(5) + 2 arctan (38) and from (3): t φ = t N := 3 4γ N + 9 6γ N 4γ N.884. (39) 3

31 In that case, we get ρ = ρ N (M) := f (a (M))/f (t N ) (4) where a (M) is the only solution in [t N, + ) of M = a 4a2 + 3a 4 4 ln(a) 4t 2 N 3t4 N + 4 ln(t N) 4( + a 2 ) 2. (4) 5.3 Saturation of constraints Theorem 5.3 With the notations of Theorem 5., we have u(r ) = M and S u = [, r ]. Proof of Theorem 5.3. Step. We first prove that u(r ) = M. Let φ : [, ] R be defined by φ(t) = [u(tr ) u(r )]/r ; then u = u ρ (as defined in Theorem 5.2) if ρ = r. Since by assumption u is a local minimizer, we must have r = ρ (φ). Note also that φ satisfies the constraints in [, ], except that we can have φ < somewhere. Assume by contradiction that u(r ) < M; then m φ = [M u(r )]/r >. If we can find ψ : [, ] R satisfying the assumptions for φ in Theorem 5.2 and such that u ψ is admissible, with J ψ = J φ and m ψ < m φ, then from Theorem 5.2 we get F (u ψ ) < F (u φ ) and we have a contradiction. We will construct ψ in two steps as pictured in figure 9. First, we construct an intermediate function η satisfying m η = m φ but J η > J φ. In order to do that, we observe that there exists a finite or infinite sequence r 2 > r 3 > r 4 >, with r 2 < r such that u has constant sign in (r i, r i ) (i > 2) and u (r i +)u (r i ) <. For each value of i, r 2i is a local minimum for u, and r 2i is a local maximum. We note t i := r i /r. Since m φ >, there exists i such that φ(t 2i+ ) = m φ and φ(t 2i+ ) > φ(t 2i ). As a consequence, the constraints are saturated in the left and in the right of r 2i and u (r 2i ) > u (r 2i +) >. (42) If ε > is small, we define η as a continuous function as follows: η φ in [, t 2i ε] [t 2i + ε, ]; η is affine in (t 2i ε, t 2i ) and (t 2i, t 2i + ε); and the value η(t 2i ) is chosen such that η (t 2i ) = φ (t 2i +) + O(ε), η (t 2i +) = φ (t 2i ) + O(ε), and the convexity points t 2i ε, t 2i + ε for η are admissible. Hence η is admissible, and J η > J φ from (42) as claimed. Now we construct ψ from η as follows: ψ η in [t 2i ε, ]; ψ η δ in [, t 2i+ ], where δ > is a small number to be chosen in a while; and ψ is 3

32 t 2i+ t 2i Figure 9: Construction of η, ψ. any C decreasing function near η in (t 2i+, t 2i ε) giving for ψ a continuous admissible function. Usually, this implies that ψ < η in this interval, so J ψ < J η ; however, by chosing appropriately δ with respect to ε, we can assume J ψ = J φ. On the other hand, m ψ = m η δ = m φ δ < m φ, as claimed. (Note that it is possible that ψ becomes negative somewhere in [, t 2i+ ); in that case, we can add a δr to u ψ in order to have an admissible function.) Step 2. Let us prove that S u = [, r ]. Assume by contradiction that there exists r [, r ) such that r / S u. Since S u is closed, there is a neighborhood (a, b) of r such that S u (a, b) =. From Theorem (4.), we get u in (a, b) and u is strictly decreasing as a consequence of (). In particular, u (r ) < ; from the constraint (8), we get u(s) < u(r ) for all s > r. Using this relation with s = r, and the fact that u(r ) = M, we get u(r ) > M, contradicting the assumption on u. 5.4 Proof of the main theorem We now turn on the proof of Theorem 2.. Proof of Theorem 2.. Step. Recall that u (r ) > ; hence there exists a left neighborhood of r where u >. We define recursively for j integer: r 2j := inf{r (, r 2j ) ; u (ρ+) >, ρ (r, r 2j )} (43) r 2j+ := inf{r (, r 2j ) ; u (ρ ) <, ρ (r, r 2j )} (44) 32

33 We first prove that this is a finite decreasing sequence, that is, for some finite k integer, we must have r k =. Indeed, let us remark that the directional derivatives at r i must have opposite signs from the definition; more precisely, we have u (r i ) and u (r i +) since Σ(r) for r near r i. Hence, if the sequence would satisfy r i > for all i, then the right derivative at r = would vanish (which cannot happen since S u ) or not be continuous, contrarily to our assumptions. Note also that u has constant sign in (r i+, r i ), and this interval is a subset of [, r ] = S u ; hence, from Theorem 5., u is stricly convex and differentiable in these intervals. Step 2. We now prove that Indeed, if s Σ(r), then r S u, Σ(r) {r, r 3,..., r +2 }. (45) k 2. s / (r, ] [, r ). Indeed, if we would have s (r, ] for instance, then (since u is differentiable at s left differentiable for s = ) the point (r, u(r)) belongs to the tangent line to the graph of u at s; since u is concave in (r, ], that implies u(r) > u(r ) = M, a contradiction. 2. s / (r i+, r i ) since u is strictly convex in this interval. 3. s r 2j for any j, since r 2j is a convexity point for u (because u (r 2j ) < < u (r 2j+ +) by construction), see Remark 3.C. Hence s = r 2j+ for some j, and (45) is proved. Step 3. Let us now prove that the graph of u is a parabola for r (r i+, r i ). We recall that, from the geometric definition of the parabola, if Σ(r) = {s } for all r (α, β), then the graph of u in (α, β) is a parabola with axis r = s and focus at (s, u(s )). Hence we have to prove that Σ(r) contains only one point if r (r i+, r i ), and is constant in (r i+, r i ). Assume by contradiction that for some r (r 2, r ) (for instance), Σ(r ) contains two different points, say r 2i+ and r 2j+ (with i < j); we can choose i, j such that r 2i+ = max Σ(r ) and r 2j+ = min Σ(r ). Note that the three points (r 2i+, u(r 2i+ )), (r 2j+, u(r 2j+ )) and (r, u(r )) are aligned in R 2. Therefore, the set of points like r, where Σ(r) contains more than one point, must be finite. In particular, r is isolated and Σ(r) contains only one point s(r) for all points r r in a neighborhood of r. 33