REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ORDINARY DIFFERENTIAL EQUATIONS. Bernard Dacorogna. Alessandro Ferriero

Mnuscript submitted to AIMS Journls Volume X, Number 0X, XX 200X Website: http://aimsciences.org pp. X XX REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ORDINARY DIFFERENTIAL EQUATIONS Bernrd Dcorogn Section de Mthémtiques, EPFL 1015 Lusnne, Switzerlnd Alessndro Ferriero Deprtmento de Mtemátics, Universidd Autónom de Mdrid Cmpus de Cntoblnco, 28049 Mdrid, Spin (Communicted by the ssocite editor nme) This rticle is dedicted to Polo Mrcellini on the occsion of his sixtieth birthdy Abstrct. Implicit Ordinry or Prtil Differentil Equtions hve been widely studied in recent times, essentilly from the existence of solutions point of view. One of the min issues is to select meningful solution mong the infinitely mny ones. The most celebrted principle is the viscosity method. This selection principle is well dpted to convex Hmiltonins, but it is not lwys pplicble to the non-convex setting. In this work we present n lterntive selecting principle tht singles out the most regulr solutions (which do not lwys coincide with the viscosity ones). Our method is bsed on generl regulrity theorem for Implicit ODEs. We lso provide severl exmples. Introduction. Let f be function from [, b] R R to R. An Implicit Ordinry Differentil Eqution, s defined by Dcorogn-Mrcellini [7], is n eqution of the form f(t, u(t), u(t)) = 0, for.e. t in (, b), (1) where solutions u re serched in W 1, 0 (, b). As it is well known, under suitble comptibility conditions between the dt nd the function f, this problem dmits priori infinitely mny solutions [7]. The purpose of the present work is to determine criteri to select, mong these infinitely mny solutions, some of them. At the moment there is no generl criterion tht permits to select unique solution in ll contexts. Depending on the eqution some of them re more dpted thn the others. Here re some of the most nturl criteri. 1) The most widely used criterion is the viscosity method introduced in systemtic wy by Crndll-Lions [5, 10]. However powerful nd generl is the method, it essentilly pplies only when the Hmiltonin f is convex in the lst vrible. 2) Another nturl criterion is the one selecting the mximl solution. In mny instnces, the pointwise supremum of solutions is however not solution. 2000 Mthemtics Subject Clssifiction. Primry: 34A60, 34H05, 35F30, 49J45, 49L25. Key words nd phrses. Implicit ODEs, Differentil Inclusions, Viscosity Solutions, Regulrity. 1

2 BERNARD DACOROGNA AND ALESSANDRO FERRIERO 3) We will here emphsize new criterion selecting the most regulr solution. Since we re deling with first order ode s coupled with Dirichlet dt, the rule is tht there re no C 1 solutions nd only Lipschitz ones re to be expected. Therefore it is nturl to choose mong the mny solutions, those tht exhibit the smllest number of discontinuities of their derivtives. In order to mke the lst principle precise, we first estblish generl regulrity result. It sys tht, under only mild nd nturl ssumptions on the boundry dt nd for lrge clss of f, the solutions in W 1, 0 (, b) to (1) hve piecewise continuous derivtive (Theorem 1), with finite number of discontinuities. We therefore define the most regulr solutions s those which hve derivtives with the lest number of discontinuity points. By using this result we show (Theorem 14) tht there exist finite number, tht we provide explicitly, of solutions with t most one discontinuity point of the derivtive. Furthermore, we show n essentil uniqueness result (Theorem 17), tht is, in the strictly convex cse (f strictly convex with respect to u), there exist two solutions, one positive nd one negtive, with t most one discontinuity point of the derivtive. We would like to conclude this introduction pointing out tht we look t Problem (1) essentilly s differentil inclusion. In other words, from our stndpoint, Problem (1) is equivlent to find u in W 1, 0 (, b) tht solves (t, u(t), u(t)) Z f, where Z f is the set of zeros of f, nmely for.e. t in (, b), Z f = {(t, u, ξ) [, b] R R : f (t, u, ξ) = 0}. Therefore, no other properties of f, besides the ones relted to its set of zeros, will ply ny role in our work. Nevertheless, we del throughout the pper with our problem minly in form (1), since this form is the most commonly studied. The pper is orgnized s follows. In Section 1 we present the regulrity result for the derivtive of the solutions to our problem (1); in Section 2 we ddress briefly whether the comptibility condition on the boundry dt re lso necessry for the existence. In Sections 3.1 nd 3.2 we recll the viscosity solution method nd the mximl solution selecting principle nd finlly, in Section 3.3, we present our criterion bsed on choosing the most regulr solutions nd lso the uniqueness result. 1. Regulrity. In this section we present our regulrity results. We prove tht for lrge clss of f, under comptibility condition in order to gurntee existence, the problem (1) dmits solution ū in W 1, 0 (, b) with piecewise continuous derivtive. We denote by C piec ([, b]) the set of piecewise continuous functions, with finite number of discontinuities, nd such tht the left nd the right limits exist nd re finite t every points of [, b]. We lso introduce the set C k piec ([, b]) := {u C piec ([, b]) : u hs t most k continuity intervls}, for given k in N, tht is the set of functions in C piec ([, b]) with t most k 1 discontinuity points. We denote lso by S f the set of solutions to (1), i.e. S f := {u W 1, 0 (, b) : f(t, u, u) = 0}. nd we sy tht function F (t, u) : [, b] R R is continuous in u nd piecewise continuous in t if there exist finite prtition of [, b] in intervls with extreme

REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ODES 3 points j < b j, j = 1,, K, such tht F restricted to ny ( j, b j ) R is uniformly continuous. We strt proving the Theorem tht hs centrl role in the present work. Theorem 1. Assume tht Z f contins the zeros of the grph of positive nd negtive function, i.e. Z f contins the zeros of P + (t, u, ξ) := ξ f + (t, u), P (t, u, ξ) := ξ f (t, u), where f +, f : [, b] [0, δ] R re continuous in u, for certin δ > 0, piecewise continuous in t nd f + > 0 > f. Then, (1) dmits W 1, 0 (, b) solution ū + > 0, with ū + in C piec ([, b]). Anlogously, ssume tht Z f contins the zeros of the grph of positive nd negtive function, i.e. Z f contins the zeros of P + (t, u, ξ) := ξ g + (t, u), P (t, u, ξ) := ξ g (t, u), where g +, g : [, b] [ δ, 0] R re continuous in u, for certin δ > 0, piecewise continuous in t nd g + > 0 > g. Then, (1) dmits W 1, 0 (, b) solution ū < 0, with ū in C piec ([, b]). Proof. We cn suppose tht f + nd f re continuous in t on the entire intervl [, b]. The generl cse follows by proceeding s below for every continuity intervls [ j, b j ], j = 1,, K. Let R > 0 be the mximum between f + nd f on [, b] [0, δ]. Consider the Cuchy problem { u(t) = f+ (t, u(t)), u() = 0. t >, (2) By the Cuchy-Peno Theorem, this problem dmits locl solution ū l in C 1 ([, 1 ]), with 1 >. We lso hve tht ny solution to (2) is incresing since f + > 0. By the continuity of ū l nd ū l () = 0, we cn choose 1 such tht ū l (t) is smller thn δ, for ny t in [, 1 ]. We then consider nother Cuchy problem defined by { u(t) = f (t, u(t)), t > 1, (3) u( 1 ) = ū l ( 1 ). Agin this new problem dmits locl solution ū l in C 1 ([ 1, 2 ]) which is now decresing since f < 0. The bound on f implies tht, defining 2 by 2 := 1 + ūl( 1 ) 2R, ū l (t) ū l ( 1 )/2, for ny t in [ 1, 2 ]. Indeed, for ny t in [ 1, 2 ], by definition of R, t ū l (t) = ū l ( 1 ) + f (τ, ū l (τ))dτ ū l ( 1 ) R(t 1 ) 1 Hence, ū l ( 2 ) is greter thn ū l ( 1 )/2 nd is smller thn ū l ( 1 ) < δ. Denoting { f+ if n is even f ( ) n := if n is odd, f by itertion, we cn define ū l on the entire intervl [, b] where ū l is solution to { u(t) = f( ) n(t, u(t)), t > n, u( n ) = ū l ( n ).

4 BERNARD DACOROGNA AND ALESSANDRO FERRIERO on [ n, n+1 ], n+1 := n + min{ū l ( 1 )/(2R), b n } nd 0 :=. Observe tht the itertion process stops fter integer prt of 2R(b )/ū l ( 1 ) steps. Furthermore, ū l (b) ūl( 1 ) > 0. 2 Anlogously, strting from b nd proceeding bckwrd in t, we cn define function ū r on [, b] solution to { u(t) = f( ) n+1(t, u(t)), t < b n, u(b n ) = ū r (b n ). on [b n, b n+1 ], b n+1 := b n min{ū r (b 1 )/(2R), b n } nd b 0 := b, ū r (b 0 ) := 0. The itertion process stops fter integer prt of 2R(b )/ū r (b 1 ) steps. Furthermore, ū r () ūr(b 1 ) > 0. 2 Define the function ū + := min{ū l, ū r }. It is continuous on [, b] with piecewise continuous derivtive, ū + () = ū + (b) = 0 nd it stisfies [ ū + (t) f + (t, ū + (t))][ ū + (t) f (t, ū + (t))] = 0. We therefore conclude tht f(t, ū + (t), ū + (t)) = 0 for.e. t in (, b). Anlogously, we cn prove the existence of ū < 0 in cse of g + nd g defined on [, b] [ δ, 0]. We refer to the ssumptions of Theorem 1 s the comptibility condition. Theorem 1 is shrp result (see Section 2). Even if the ssumptions re expressed in term of the set of zeros of f nd not directly on f, function f which does not stisfy such ssumptions must be built on purpose in n rtificil wy. However, in wht follows we propose n explicit generl non-convex clss of functions f which stisfy the needed conditions on its set of zeros. We sy tht function h(ξ) : R R is piecewise strictly monotone incresing (respectively decresing) if there exists loclly finite prtition of intervls of R such tht h is strictly monotone incresing (respectively decresing) on ny of such intervls. Further, function q(t, u) : [, b] R R is loclly monotone incresing (respectively decresing) t u = 0 if there exists δ > 0 such tht q(t, u) q(, 0) (respectively q(t, u) q(, 0)), for ny t in (, b), u in (0, δ). We sy tht q is piecewise loclly monotone t u = 0 if there exits finite prtition of intervls of [, b] such tht q is loclly monotone t u = 0 on ny of such intervls. Theorem 2. Let q(t, u) : [, b] R R be function continuous with respect to u nd piecewise continuous with respect to t nd let h(ξ) : R R be continuous. Suppose tht either h is piecewise strictly monotone, or h is piecewise monotone nd q is loclly piecewise monotone t u = 0, f(t, u, ξ) := q(t, u) + h(ξ), h(0) < q(t, 0), for ny t in [, b], nd h(ξ) +, s ξ. Then, problem (1) dmits two W 1, 0 (, b) solutions ū + > 0 > ū, with ū + nd ū in C piec ([, b]). Proof. Whitout loss of generlity, we cn suppose tht q is continuous with respect to t in the whole intervl [, b]. The generl cse follows by proceeding s below for every continuity intervls [ j, b j ], j = 1,, K. We split the proof in two cses. Cse 1. Suppose tht h is piecewise strictly monotone.

REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ODES 5 Let {I j } j N be prtition of (0, + ) such tht h restricted to the interior int(i j ) of I j is strictly monotone nd the I j s re mximl. By the strict monotonicity of h int(ij), the inverse functions h 1 j exist nd re continuous from h(int(i j )) to I j. Furthermore, h(int(i j )) = ( inf h, + ). (0,+ ) j N By the piecewise strict monotonicity of h, there exists prtition 0 := < 1 < < n 1 < b of [, b] nd n integer numbers j 1,, j n such tht h(int(i jk )) { q(t, 0) : t [ k 1, k )}, for k = 1,, n 1, h(int(i jn )) { q(t, 0) : t [ n 1, b]}. Therefore, by continuity, there exists lso δ > 0 such tht h(int(i jk )) { q(t, u) : t [ k 1, k ), u [0, δ)}, for k = 1,, n 1, h(int(i jn )) { q(t, u) : t [ n 1, b], u [0, δ)}. Define { h 1 j f + (t, u) := k ( q(t, u)), (t, u) [ k 1, k ) [0, δ), k = 1,, n 1, h 1 j n ( q(t, u)), (t, u) [ n 1, b] [0, δ), or equivlently in closed form n 1 f + (t, u) := h 1 j k ( q(t, u))χ [k 1, k ) [0,δ)(t, u) + h 1 j n ( q(t, u))χ [n 1,b] [0,δ)(t, u), k=1 which is function from [, b] [0, δ) with vlues in (0, ). Similrly, strting from the prtition of (, 0) of intervls of strict monotonicity for h, we cn define f (t, u) : [, b] [0, δ) (, 0) nd, considering the left neighbourhood of u = 0 insted of the right one, we define g + (t, u) : [, b] ( δ, 0] (0, ) nd g (t, u) : [, b] ( δ, 0] (, 0) nlogously to f + nd f. By construction, if t, u nd ξ re such tht [ξ f + (t, u)][ξ f (t, u)][ξ g + (t, u)][ξ g (t, u)] = 0, then f(t, u, ξ) = 0. Moreover, one cn verify tht f ±, g ± re continuous in u nd piecewise continuous in t. Therefore, by pplying Theorem 1, we conclude tht (1) dmits two W 1, 0 (, b) solutions ū + > 0 > ū, with ū + nd ū in C piec ([, b]). Cse 2. Suppose tht h is piecewise monotone nd q is loclly piecewise monotone t u = 0. Let {I j } j N be prtition of (0, + ) s in Cse 1. Differently thn in the previous cse, since h is not priori strictly monotone, we hve j N h(int(i j )) = ( inf (0,+ ) h, + ) \ {z i} i N, where {z i } i N is set of isolted points of R. By the locl piecewise monotonicity t u = 0 of q nd the piecewise monotonicity of h, there exists prtition 0 := < 1 < < n := b of [, b] nd n integer numbers j 1,, j n such tht h(int(i jk )) { q(t, 0) : t ( k 1, k )}, for k = 1,, n. Therefore, by continuity, there exists lso δ > 0 such tht h(int(i jk )) { q(t, u) : t ( k 1, k ), u [0, δ)}, for k = 1,, n. Defining f + s in Cse 1, with f + ( k, u) := lim h 1 t + j k ( q(t, u)), k

6 BERNARD DACOROGNA AND ALESSANDRO FERRIERO for k = 1,, n 1, f + (b, u) := lim t b h 1 j n ( q(t, u)), nd nlogously f, g ±. By pplying Theorem 1, we conclude tht (1) dmits two W 1, 0 (, b) solutions ū + > 0 > ū, with ū + nd ū in C piec ([, b]). We now give severl exmples to discuss the optimlity of Theorem 2. In these exmples but the lst one there exist solutions in W 1, 0 (, b), none of them hving piecewise continuous derivtive, ech one violting different hypothesis. Exmple 3. Let h : R R be the continuous function defined by ξ 2, ξ [0, 5/2) [3, ), h(ξ) := 2(1 2 2n ) 1, ξ [3 2 2n+1, 3 2 2n ), n 1, 3 ξ 3 + 2 2n + 2(1 2 2n ) 1, ξ [3 2 2n, 3 2 2n 1 ), n 1, nd, for ξ < 0, h(ξ) := h( ξ). Consider, for ny t in [0, 1] nd for ny ξ in R, f(t, ξ) := t 2 + h(ξ). Then, ny solution to (1) in W 1, 0 (0, 1) hs infinitely mny discontinuities close to the point t = 1. The function h does not stisfy the ssumption of piecewise monotonicity. Exmple 4. Let h : R R be the continuous function defined by ξ 2, ξ [0, 2), h(ξ) := 0 ξ [2, 3), ξ 3, ξ [3, ), nd, for ξ < 0, h(ξ) := h( ξ). Consider, for ny u nd ξ in R, f(u, ξ) := u sin 1 u + h(ξ). Then, ny solution to (1) in W 1, 0 (, b) hs infinitely mny discontinuities close to the points t = nd t = b. Anlogously, if f(t, ξ) := t sin 1 t + h(ξ), then, ny solution to (1) in W 1, 0 ( 1, 1) hs infinitely mny discontinuities close to the point t = 0. In both cses the function h is piecewise (but not strictly) monotone but q does not verify the hypothesis of locl piecewise monotonicity t u = 0. Exmple 5. Let f(t, u, ξ) := mx{u 2 t 4, 0} + (1 ξ 2 ) 2. A solution ū in W 1, 0 (0, 1) to (1) with derivtive in C piec ([0, 1]) must stisfy ū(t) t 2 nd ū(t) = 1 (or ū(t) = 1) t every t in right neighbourhood of 0. Therefore, by the bsolute continuity of ū, for ny t smll enough, we obtin t 2 ū(t) = t 0 ū = t, reching contrdiction.

REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ODES 7 Notice tht there exists solution in W 1, 0 (0, 1). Indeed, one verifies tht [ (n+1 + n ) 2 ū(t) := 4 t ] n+1 + n 2 χ (n+1, n](t) n=1 [ + 1 ( 1 + 3) 4 t 3 ] 1 + 1 4 χ (1,( 1+1)/2](t) [ + 1 ( 1 + 3) 4 t ] 1 + 3 4 χ ((1+1)/2,1](t), where n+1 := 1 + 4 n (1 + n ), 0 := 1, is function in W 1, 0 (0, 1) tht solves (1). The function f does not verify the hypothesis f(t, 0, 0) < 0, for every t in [, b]. Exmple 6. Here is n exmple where there exists solution in W 1, 0 ( 1, 1), but its derivtive does not belong to C piec ([ 1, 1]). Let { tξ 2t 2 sin π f(t, ξ) := t + πt cos π t, t 0, 0, t = 0. A solution ū in W 1, 0 ( 1, 1) to (1) is given by ū(t) := t 2 sin(π/t). Its derivtive does not belong to C piec ([ 1, 1]), becuse lim t 0 ū (t) is not well defined. The function f does neither verify f(t, 0) < 0, for every t in ( 1, 1), nor f(t, ξ) +, s ξ, for every t in ( 1, 1). Exmple 7. We give n exmple of non-existence of solutions in W 1, 0 (0, 1). Let f(u, ξ) := u 2 ξ 2 1. A solution ū in W 1, 0 (0, 1) to (1), if it exists, should stisfy ū(t) ū(t) = 1 nd, therefore, ū(t) 0, for.e. t in (0, 1). Since ū(0) = 0, by the continuity of ū, we obtin tht ū is unbounded in neighbourhood of 0. The function f(0, ξ) does not tend to + s ξ. 2. On The Existence. In this section, we would like to ddress briefly whether the comptibility condition on the boundry dt stted in Theorem 1, tht is sufficient to gurntee the existence of solutions to problem (1), is lso necessry. We discuss the specil cses in which f does not depend either on t or on u. We prove tht the comptibility condition is indeed lso necessry condition for the existence of positive nd negtive solution to (1) in W 1, 0 (, b) with C piec ([, b]) derivtive. More precisely: Theorem 8. Let ū + > 0 > ū in W 1, 0 (, b), with ū in C piec ([, b]), be solutions to (1). Assume tht f is of the form f(t, ξ) or f(u, ξ). Then, there exist two functions f + (t, u), f (t, u) : [, b] [ δ, δ] R continuous in u, for certin δ > 0, piecewise continuous in t nd f + > 0 > f such tht the set Z f of zeros of f contins the zeros of P + (t, u, ξ) := ξ f + (t, u), P (t, u, ξ) := ξ f (t, u). Proof. Cse 1. Suppose tht f(t, u, ξ) = f(t, ξ) does not depend on u. We cn define two piecewise continuous functions f + nd f by f + (t) := ū + (t), f (t) := ū (t),

8 BERNARD DACOROGNA AND ALESSANDRO FERRIERO for ny t in [, b]. By definition, ū + is solution to u(t) f + (t) = 0 nd ū is solution to u(t) f (t) = 0, for lmost every t in (, b). Furthermore, by ssumption, f(t, ū + (t)) = 0, f(t, ū (t)) = 0. for t in [, b]. We conclude tht, if ξ f + (t) = 0 or ξ f (t) = 0, then f(t, ξ) = 0, s stted. Cse 2. Suppose tht f(t, u, ξ) = f(u, ξ) does not depend on t. From the ssumed regulrity on ū := ū +, there exists 1 < b 1 in (, b) such tht ū belongs to C 1 ([, 1 ] [b 1, b]) nd ū is respectively incresing on [, 1 ] nd decresing on [b 1, b]. We cn define two continuous functions f + nd f by f + (u) := ū(ū 1 [, 1] (u)), f (u) := ū(ū 1 [b 1,b] (u)), for ny u in [0, min{ū( 1 ), ū(b 1 )}]. By definition, ū is solution to u(t) f + (u(t)) = 0, for t in [, 1 ], u(t) f (u(t)) = 0, for t in [b 1, b]. Tht implies, by the chnges of vrible t = ū 1 [, 1] (u) or t = ū 1 [b (u), 1,b] f(u, f + (u)) = 0, f(u, f (u)) = 0, for ny u in [0, min{ū( 1 ), ū(b 1 )}]. We infer tht, if ξ f + (u) = 0 or ξ f (u) = 0, then f(u, ξ) = 0. To conclude, we proceed nlogously s bove with ū := ū in order to extend the definition of f + nd f to the left of 0. 3. Selecting Principles. In this section we present three selecting principles for solutions to (1). Nmely, the viscosity method developed by Crndll-Lions, the mximl solutions method nd the most regulr solutions method. Before presenting these methods, we would like to recll here simple exmple of Georgy [8] showing tht these three methods select, in generl, different solutions. However if the function f is convex in the vrible ξ nd incresing in the vrible u, then the three methods select the sme solution. Consider the continuous function f given by (ξ + 1) f(t, ξ) := t + ξ in (, 3/2] ξ + 2 ξ in ( 3/2, 1] ξ ξ in ( 1, 1] (ξ 2) ξ in (1, 3/2] ξ 1 ξ in (3/2, ). Let β R nd u 0 (t) = tβ/2. We then look for solutions u u 0 + W 1, 0 (0, 2) of f(t, u(t)) = 0, for.e. t in (0, 2). (4) In [8], the following results re proved. (i) Eqution (4) hs solution u u 0 +W 1, 0 (0, 2) if nd only if β [ 7/2, 7/2]. Moreover, in the intervl of existence β [ 7/2, 7/2], there lwys exists mximl (nd miniml) solution nd lso solution with t most two points of discontinuities of the derivtive. (ii) In β [ 7/2, 2] [2, 7/2], there exists solution with t most one point of discontinuity. (iii) There re no C 1 solutions. (iv) There exists viscosity solution if nd only if β [ 3, 7/2]. However, in this intervl if, moreover, β 3, 2, 7/2, then ny viscosity solution hs t lest two points of discontinuities of its derivtive. Therefore, the viscosity solution is not

REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ODES 9 lwys the most regulr solution. It cn lso be seen tht the viscosity solutions cn be different from the mximl one s we dd other wells to the function f. 3.1. The Viscosity Solutions. In this section, we would like to recll very briefly few fcts on the viscosity solutions. For more detils, we refer to [1, 5, 10]. The nme viscosity refers to the so-clled vnishing viscosity term ɛü (or ɛ u in the multidimensionl cse). Indeed solutions u of (1) re obtined s the limit of the sequence u ɛ which re solutions of the second order eqution ɛü ɛ (t) + f(t, u ɛ (t), u ɛ (t)) = 0, t (, b). This method is, by now, more generl nd does not refer nymore to the bove limit procedure nd the following definition is now the stndrd one. Definition 9. A continuous function u is viscosity solution to (1) if nd only if for ll φ in C 1 (, b) such tht u φ hs locl mximum t t 0 in (, b), then f(t 0, u(t 0 ), φ(t 0 )) 0, nd for ll φ in C 1 (, b) such tht u φ hs locl minimum t t 0 in (, b), then f(t 0, u(t 0 ), φ(t 0 )) 0. For the existence nd uniqueness of the viscosity solutions in W 1, 0 (, b) to (1), the bsic ssumptions re, s lredy sid, the convexity with respect to ξ nd the incresing monotonicity with respect to u of f(t, u, ξ) (see [10]). The convexity is essentilly necessry condition not only for the uniqueness of viscosity solutions but lso for the existence, s the bove exmple of Georgy shows (for β in [ 7/2, 3)). 3.2. The Pointwise Mximl Solution. In the cse where f does not depend explicitly on the vrible u, it cn esily be proved s in [8] tht there exists mximl solution. The result is however flse, in generl, if the Hmiltonin depends explicitly on u s will be seen in Exmple 12. Theorem 10. If Z f [, b] R is closed set, bounded with respect to ξ nd { } S f = u W 1, 0 (, b) : (t, u) Z f, then ū S f where ū(t) := sup u(t), for ny t in [, b], Remrk 11. In prticulr, if we ssume tht f(t, ξ) is continuous, f(t, 0) < 0 nd f(t, ξ) +, s ξ, for ny t [, b], then the set verifies the ssumptions of the theorem. Z f = {(t, ξ) [, b] R : f(t, ξ) = 0} Proof. Since, by ssumption, Z f is bounded with respect the lst vrible, there exists ρ > 0 such tht Z f [, b] [ ρ, ρ]. Hence, Observing tht u L (,b) ρ, for ny u in S f. sup u u = 0 inf u,

10 BERNARD DACOROGNA AND ALESSANDRO FERRIERO by continuity it follows tht there exists point t in [, b] such tht t sup u + t inf u = 0. (If there exist such two points t 1 < t 2, then u = sup ū(t) := sup u, t [, t], inf u, t ( t, b]. u = inf u in ( t 1, t 2 ).) Set By definition, ū(t) := t ū belongs to W 1, 0 (, b). Furthermore, for ny t in [, t] ū(t) = t sup u t u = u(t), nd, s t belongs to ( t, b], ū(t) = inf u u = u(t), t t for ny u in S f. Hence, ū sup{u : u S f }. We re left to prove tht u is solution. In fct, by the fct tht Z f is closed, we hve (t, ū(t)) = (t, sup u(t)), t [, t], (t, inf u(t)), t ( t, b], nd hence (t, ū(t)) Z f = Z f. We conclude tht ū = sup u is solution to (4). We now emphsize tht the result is not true in cse Z f depends on u. In fct, consider the following exmple. Exmple 12. The pointwise mximl, mong ll solutions, function might not be solution in cse f depends explicitly on u. If we let then does not belong to S f. f(u, ξ) := mx{u 2 1, 0} + (ξ 2 1) 2, = 0 nd b = 4, ū(t) = sup u(t) = u S Zf t, t [0, 1], 1, t (1, 3], 4 t, t (3, 4], In the theorem below, we recll result contined in [8] giving necessry nd sufficient condition for the existence of (mximl) solution to (1). Theorem 13. Let f : [, b] R R, f = f(t, ξ), be continuous function. If the set E(t) := {ξ R : f(t, ξ) = 0} is non-empty nd bounded for ny t in [, b], then, there exists mximl solution ū W 1, 0 (, b) to (1) if nd only if min{ξ E(t)}dt 0 mx{ξ E(t)}dt.

REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ODES 11 3.3. The Most Regulr Solutions. In this section we present our selecting principle bsed on choosing the most regulr solution. Furthermore, in Theorem 17 we give uniqueness result under convexity ssumption. To illustrte our purpose, we strt with n elementry exmple, nmely the eikonl eqution u(t) = 1, for.e. t in (0, 1). (5) As it is well known, this problem dmits infinitely mny solutions. For instnce, for ny integer j, the function { t i2 ū j (t) := j, t [i2 j, (i + 2 1 )2 j ], i = 0, 1,, 2 j 1 (i + 1)2 j t, t [(i + 2 1 )2 j, (i + 1)2 j ], i = 0, 1,, 2 j 1 is solution of (5). Even if we look t the solutions with fixed number of discontinuity points for the derivtive, we obtin infinitely mny functions. Indeed, if k = 3, the functions ū α (t) := tχ [0,α) (t) + (α t)χ [α,(α+1)/2) (t) + (t 1)χ [(α+1)/2,1] (t) re solutions to (5), for ny rel α (0, 1/2). Anlogously, we cn find infinitely mny solutions for every k > 3. Nevertheless, the solutions with exctly one discontinuity point for the derivtive re just two, nmely ū + (t) := 2 1 t 2 1, ū (t) := 2 1 + t 2 1, nd there re no C 1 solutions. Incidentlly the function ū + is the unique viscosity solution of (5). In the theorem below, under simple ssumptions on the set of zeros of f, we cn gurntee the existence of finite number of solutions whose derivtive dmits t most one discontinuity point. More precisely, we show tht if Z f is the union of n + positive nd n negtive grphs of continuous functions f i +(t, u), f i (t, u) : [, b] R R, with f i 0 f i +, then there re solutions to (1) with t most one point of discontinuity of the derivtive, nd there exist t most 2n + n such solutions. If we ssume tht the zeros do not intersect the plne ξ = 0, mening tht f i < 0 < f i +, we cn then prove tht the solutions of (1) with miniml number of discontinuity points hve exctly one singulrity, nd there exist t most 2n + n solutions of this type. Note tht, we cn lwys suppose, without chnging the problem (1), tht f i + nd f i re defined over the entire spce [, b] R (but clerly without supposing tht they re continuous on [, b] R in generl) by considering f i +(t, u) := mx{f i +(t, u), 0} nd f i (t, u) := min{f i (t, u), 0}. Theorem 14. Assume tht the set Z f of zeros of f is finite union of n + positive nd n negtive continuous grphs, nmely where Z f = n + i=1 P i +(t, u, ξ) := ξ f i +(t, u), Z P i + n Z P i i=1 P i (t, u, ξ) := ξ f i (t, u),

12 BERNARD DACOROGNA AND ALESSANDRO FERRIERO f i +, f i : [, b] R R re loclly Lipschitz with respect to u, continuous with respect to t, f i 0 f i + nd one of them hs sub-liner growth, i.e. f i +(t, u) (or f i (t, u) ) m u + q, for certin m > 0, q R nd for ny (t, u) [, b] R. Then, the solutions to (1) with miniml number of discontinuity points hve derivtive in C 2 piec ([, b]), i.e. the derivtive hs t most one singulrity, nd there exist t lest 2 nd t most 2n + n solutions of this type. Remrk 15. Theorem 14 sttes the existence of t most 2n + n solutions. The fct tht we might lose some solutions is due to the structure of the zeros of f. If, for instnce, we ssume tht the functions f i + nd f i do not intersect one with echother, then, we do not lose ny solution nd therefore we obtin exctly 2n + n distinct solutions. Proof. We first prove the existence of solution to (1). For ny i = 1,, n +, consider the Cuchy problem { u(t) = f i +(t, u(t)), t >, (6) u() = 0. By the Cuchy-Lipschitz Theorem nd the growth ssumption on f+ i this problem dmits globl solution ū i l C 1 ([, b]), with ū i l () = 0. We lso hve tht ūi l is incresing since f + 0. Anlogously, for ny j = 1,, n, considering the Cuchy problem { u(t) = f j (t, u(t)), t < b, (7) u(b) = 0, we hve C 1 solution ū j r in right neighbourhood of t = b, with ū j r(b) = 0 nd ū j r + t the left extreme of tht neighbourhood, nd ū j r is decresing. Defining ū i,j (t) := min{ū i l(t), ū j r(t)}, t [, b], we obtin tht ū i,j belongs to W 1, 0 (, b) nd ū i,j belongs to the spce C 2 piec ([, b]). Indeed, by the monotonicity of ū i l, ūj r, there exists exctly one point t i,j such tht ū i,j (t) = ū i l(t)χ [,ti,j)(t) + ū j r(t)χ [ti,j,b](t). Note tht, if ū i l (t i,j) = ū j r(t i,j ), then both derivtives ū i l (t i,j), ū j r(t i,j ) must be equl to 0 nd, hence, ū i,j hs continuous derivtive on the whole [, b]. (If we hd considered f j sub-liner insted of f i +, proceeding s bove we would hve obtined the sme result.) Now, if we consider problem (6), for every i = 1,, n +, nd problem (7), for every j = 1,, n, we define t most 2n + n solutions to (1) nd no more thn 2n + n, by the locl Lipschitzinity of f i + nd f j. If we suppose tht f i +, f i do not ssume the vlue zero, i.e. f i < 0 < f i + on [, b] R, then, from the proof of Theorem 14, it follows tht the solutions to (1) with miniml number of discontinuity points hve derivtive with exctly one singulrity. The eikonl eqution belongs to this clss. We therefore obtin the following corollry.

REGULARITY AND SELECTING PRINCIPLES FOR IMPLICIT ODES 13 Corollry 16. Assume tht the set Z f of zeros of f is finite union of n + positive nd n negtive continuous grphs, nmely where Z f = n + i=1 P i +(t, u, ξ) := ξ f i +(t, u), Z P i + n Z P i i=1 P i (t, u, ξ) := ξ f i (t, u), f i +, f i : [, b] R R re loclly Lipschitz with respect to u, continuous with respect to t, f i < 0 < f i + nd one of them hs sub-liner growth. Then, the solutions to (1) with miniml number of discontinuity points hve derivtive with exctly one singulrity, nd there exist t lest 2 nd t most solutions of this type. 2n + n Proof. It follows directly from Theorem 14 tht, since ū i r < 0 < ū j l on [, b] nd ū i,j () = ū i,j (b) = 0, the function ū i,j cnnot hve continuous derivtive on [, b]. As consequence of Theorem 14, we obtin tht, for n + = n = 1, there exist exctly two solutions, one positive nd one negtive, to (1) with derivtive in 1, C 2 piec ([, b]). For instnce, the problem of finding u in W0 (, b) solution to u(t) = f(t, u(t)), for.e. t in (, b), with f loclly Lipschitz with respect to u nd f 0, for ny (t, u) [, b] R, dmits exctly positive nd negtive solution in W 1, 0 (, b) with derivtive with t most one discontinuity. In the sme spirit of the observtion bove, if we ssume the strict convexity of f with respect to ξ, we cn prove the essentil uniqueness of the most regulr solution. Theorem 17. Assume tht f is continuously differentible function, strictly convex with respect to ξ, f(t, u, 0) < 0 nd f(t, u, ξ) χ [,b] {(u,ξ) R2 : ξ m u +q}(t, u, ξ), for certin m > 0, q in R nd for ny (t, u) [, b] R. Then, the solutions of (1) with miniml number of discontinuity points hve derivtive in C 2 piec ([, b]) nd there exist exctly 2 solutions of this type, one positive nd one negtive. Proof. From the Implicit Function Theorem, there exists two functions f + (t, u) nd f (t, u) in C 1 ([, b] R) such tht f(t, u, f ± (t, u)) = 0. Therefore, the result follows from Corollry 16. The eikonl eqution (5) stisfies the ssumptions of Theorem 17 nd, in fct, it dmits exctly 2 solutions, one positive nd one negtive, with one discontinuity point of the derivtives, nd it does not dmit C 1 solutions. A prticulr cse of Theorem 14 is provided by function f which is polynomil in ξ of even degree.

14 BERNARD DACOROGNA AND ALESSANDRO FERRIERO Corollry 18. Let f i : [, b] R R, i = 0,, 2N, be continuous functions, loclly Lipschitz with respect to u, such tht f 0 (t, 0) < 0, for ny t in [, b]. Let f(t, u, ξ) := 2N i=0 ξ i f i (t, u). Then, the solutions of (1) with miniml number of discontinuity points hve derivtive in C 2 piec ([, b]) nd there exist t lest 2 nd t most 2n + n solutions of this type, where n + is the number of ξ > 0 such tht f(t, 0, ξ) = 0 nd n is the number of ξ < 0 such tht f(t, 0, ξ) = 0. Proof. The proof follows directly from Theorem 14. We now discuss the optimlity of our Theorem 14 through two simple exmples. Exmple 19. If we drop the Lipschitz ssumption with respect to u, we might end up with more solutions thn the ones stted by Theorem 14. Indeed, consider f(u, ξ) := (ξ 1 u 2 )(ξ + 1 u 2 ), for.e. t in (0, 2π). (8) Eqution (1) dmits exctly four solutions with continuous derivtive, tht re ū 1 (t) := sin t, ū 2 (t) := sin t, ū 3 (t) := sin(t)χ [0,π/2) (t) + χ [π/2,3π/2] (t) sin(t)χ (3π/2,2π] (t) nd ū 4 (t) := ū 3 (t), while if f + nd f were Lipschitz in u, the solutions would be only two (cf. lso Exmple 21). Exmple 20. The ssumption on the zeros of f to be the grph of continuous functions defined for ny u in R is optiml. In fct, consider the modified eikonl eqution u(t) 3 8 χ { ξ u 8/3}(u(t), u(t)) + 3 8 u(t) 1 χ { ξ > u 8/3}(u(t), u(t)) = 0,.e. t (0, 1). This problem dmit infinitely mny solutions with derivtive in C k piec ([0, 1]), for ny k 3, but it does not dmit solutions for k = 1, 2. Observe tht f(u, ξ) = u 3 8 χ { ξ u 8/3}(u, ξ) + 3 8 ξ 1 χ { ξ > u 8/3}(u, ξ) is convex in ξ (nd lso in u). The set Z f is the rectngle with sides prllel to the xis u nd ξ of coordintes ξ = 3/8, 3/8 nd u = 1, 1. Finlly we conclude with nother exmple showing tht there is non-uniqueness even mong smooth solutions. Exmple 21. Consider Let α i C ([0, 1]), i = 1,, n, f (t, ξ) = with 1 n [ξ α i (t)], i=1 0 α i (t) dt = 0.

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