Regularity Results for Eikonal-Type Equations with Nonsmooth Coefficients

Regularit y Results for Eik onal-T yp e Equations with Nonsmo oth Co efficien ts Piermarco Cannarsa ∗ & Pierre Cardaliaguet † June 9, 2018 Abstract Solutions of the Hamilton-Jacobi equation H ( x, − Du ( x )) = 1 , with H ( · , p ) Hölder co ntin uous and H ( x, · ) c onv ex and p ositiv ely homogeneous of degree 1 , are shown to b e locally semiconca ve with a p o wer-lik e mod- ulus. An essentia l step of the pro of is th e C 1 ,α -regularit y of the extremal tra jectories associated with the multif u nction generated by D p H . Key w ords : viscosity solutions, semiconcav e functions, differen tial inclus io ns, extremal tra jectories MSC Sub ject class ifications : 49L25 3 4 A60 26B25 4 9N60 1 In tro duction The impor tance of semiconcavit y for the study of Ha milton-Jacobi equations and optimal con tro l problems is b y now widely ackno wledged. Indeed, such a qualitative pro p erty ensures the upp er semicontin uity and quasi-monotonicity of the super differ e ntial, provides u pper b ounds for the set where differen tiability fails pro viding, at the same time, criteria for the propag ation of singularities, and leads to stro nger optimality conditions than the ones holding for a contin uous (or Lipschit z con tinuous) function, se e, for instance, [4] and the r eferences therein. Typically , a r eal-v alued function u is semiconcav e o n the conv ex set D ⊂ R N if there exists a mo dulus (i.e., no ndecreasing upp er semicontin uous function, v a nis hing at 0 ) ω : [0 , ∞ ) → [0 , ∞ ) such that u ( λx + (1 − λ ) y ) ≥ λu ( x ) + (1 − λ ) u ( y ) − C λ (1 − λ ) | x − y | ω ( | x − y | ) for a ll x, y ∈ D and λ ∈ [0 , 1 ] . Semiconcavit y results with a linear mo dulus ho ld for viscosit y solutions of Hamilton-Jacobi equations with conv ex Hamiltonians which are sufficien tly smo oth with respe c t to the space v a riables, as w ell as v a lue functions o f opti- mal control problems with smo oth dynamics and running cost (see, e.g., [8], [7], ∗ Dipartimen to di Matematica, Università di Roma “T or V ergata”, Via della Ricerca Scien- tifica 1, 00133 Roma (Italy) , e-mai l : cannarsa@mat.uniroma2.it † Unive rsi té de Brest, UMR 6205, 6 A v. Le Gorgeu, BP 809, 29285 Brest (F rance); e-mail: Pierre.Cardaliague t@univ-brest.fr 1 [3]; see also [4]). Known genera lizations allow for Lipsc hitz contin uous dep en- dance with resp ect to spa ce, provided the Ha miltonian is strictly conv ex and supe r linear in the gradient v ar ia bles (see [5], [9]). In this pap er we sha ll study the Dirichlet problem  H ( x, − D u ( x )) = 1 in Ω u ( x ) = 0 on ∂ Ω (1) where Ω is an o p e n subset o f R N , H ( x, · ) is conv ex and p o sitively ho mo geneous of degree 1, and H ( · , p ) is just Hölder contin uous. Consequently , (1 ) fits none of the aforementioned settings. Nevertheless, our main result—Theor em 5.1 below—guara nt ees that the solution u of (1) is loca lly semiconcav e in Ω with the p ower-lik e mo dulus ω ( t ) = C t θ , for some θ > 0 depending on H . The metho d o f pro of relies on the repres entation of u ( x ) as the minimum time needed to reach ∂ Ω along a tra jectory o f the differen tial inclusion ( x ′ ( t ) ∈ F ( x ( t )) t ≥ 0 a.e. x (0) = x , (2) where F ( x ) = co { D p H ( x, p ) : p ∈ R N \{ 0 }} ∀ x ∈ R N . An esse ntial step of the analysis is the C 1 ,α -regular it y o f the ex tr emal tra jectories of (2), see Theorem 4.1. F or time-dep endent a nd isotropic Hamiltonia ns ( H = a ( t, x ) | p | ), suc h a r egularity pro p erty—in teresting in its own right—has already bee n observed in [2] for N = 2 , and [6] for general N . Howev er, the unexp ected connection be t ween Theor em 4 .1 and the semiconcavit y o f the solution of (1) is, to our best knowledge, e ntirely new. The main technical to o ls we b o rrow from co nv ex a nalysis are r ecalled in detail in section 3, which makes this paper essentially se lf- contained. 2 Notation and assumpti ons Let N be a p ositive in teg er. Denote by h· , ·i and | · | the Euclidean sca lar pro duct and nor m in R N , respectively , and set B = { x ∈ R N : | x | ≤ 1 } . More gener a lly , for all x ∈ R N and ρ > 0 , B ( x, ρ ) stands for the closed ball o f radius ρ centered at x , that is, B ( x, ρ ) = x + ρB . Let H : R N × R N → R b e a co ntin uous function satisfying the following assumptions for some pos itive consta nt s C 0 , r, R , with r < R , a nd α ∈ (0 , 1 / 2) . Standing Assumptions (SA): • F or all p ∈ R N , the function x 7→ H ( x, p ) is 2 α -Hölder contin uous, and | H ( x, p ) − H ( y , p ) | ≤ C 0 | x − y | 2 α | p | ∀ x, y ∈ R N . (3) • F or a ll x ∈ R N , the function p 7→ H ( x, p ) is conv ex o n R N , p ositively homogeneous of degree o ne, and has linear gr owth, i.e., r | p | ≤ H ( x, p ) ≤ R | p | ∀ p ∈ R N . (4) 2 • F or all x ∈ R N , the function p 7→ H ( x, p ) is contin uously differentiable o n R N \{ 0 } , and, for all p, q ∈ R N \{ 0 } , − 1 2 r | D p H ( x, q ) − D p H ( x, p ) | 2 ≤ h D p H ( x, q ) − D p H ( x, p ) , p | p | i ≤ − 1 2 R | D p H ( x, q ) − D p H ( x, p ) | 2 , (5) where D p H ( x, p ) denotes the gradient of H in the p -v ar iables at ( x, p ) . Hereafter, b y a universal constant—briefly , a constant —we mean a p ositive real num b er that only dep ends on the parameters N , α, r, R, and C 0 in tro duced ab ov e. Generic constants app ea ring in co mputations will be denoted by C . A subscript ( C 1 , C 2 , . . . ) will b e added when necessary fo r future reference. Example 2.1. Let H ( x, p ) = | A ( x ) p | , where A : R N → R N × N is α -Hölder contin uous on R N , A ( x ) is in vertible for all x ∈ R N , and | A ( x ) | ≤ C and | A ( x ) − 1 | ≤ C ∀ x ∈ R N for s ome co nstant C . Then H sa tisfies (SA) for a suitable c ho ic e of co ns tants. 3 Prelimina ry results Let H : R N × R N → R b e a contin uous function satisfying our Standing As- sumptions with fixed constant s α, r , R, and C 0 . F o r all p 6 = 0 , set f p ( x ) = D p H ( x, p ) and define F ( x ) = co { f p ( x ) : p ∈ R N \{ 0 }} ∀ x ∈ R N , (6) where ‘co ’ sta nds for co nv ex hull. Note that, fo r all ( x, p ) ∈ R N × ( R N \{ 0 } ) , H ( x, p ) = max v ∈ F ( x ) h v , p i and f p ( x ) = ar gmax v ∈ F ( x ) h v , p i . (7) W e b egin by recov er ing some prop erties of F ( · ) that follow dir ectly from (SA). Lemma 3.1. The set -value d map F is 2 α -Hölder c ontinuous, i. e., F ( x ) ⊂ F ( y ) + C 0 | x − y | 2 α B ∀ x, y ∈ R N , (8) and satisfies, for al l x ∈ R N , the curv ature estimates B  f p ( x ) − r p | p | , r  ⊂ F ( x ) and F ( x ) ⊂ B  f p ( x ) − R p | p | , R  ∀ p 6 = 0 , (9) as wel l as the co nt rollability c ondition B (0 , r ) ⊂ F ( x ) ⊂ B (0 , R ) . (10) 3 Remark 3.2. The first inclusio n in (9)—which can b e interpreted a s an uppe r bo und for the curv a tur e of ∂ F ( x ) —is equiv alent to the inequality    v − f p ( x ) + r p | p |    ≥ r ∀ v ∈ ∂ F ( x ) , which in turn can b e recast as follows − 1 2 r | v − f p ( x ) | 2 ≤ D v − f p ( x ) , p | p | E ∀ v ∈ ∂ F ( x ) . (11) The seco nd inclusion in (9)—a lower b ound for the cur v atur e —c a n b e r ephrased as    v − f p ( x ) + R p | p |    ≤ R ∀ v ∈ F ( x ) , which is equiv alent to D v − f p ( x ) , p | p | E ≤ − 1 2 R | v − f p ( x ) | 2 ∀ v ∈ F ( x ) , or, since F ( x ) is conv ex, D v − f p ( x ) , p | p | E ≤ − 1 2 R | v − f p ( x ) | 2 ∀ v ∈ ∂ F ( x ) . (12) Pr o of. Note that, since H ( x, · ) is the suppor t function o f F ( x ) , inequality (3) directly implies (8) while (4) entails (10). Let us now chec k that the regularity condition (5) implies (9). F or this we just have to no te that, for any v ∈ ∂ F ( x ) , there is some q 6 = 0 such tha t v = f q ( x ) , so that (5) b ecomes − 1 r | v − f p ( x ) | 2 ≤ D v − f p ( x ) , p | p | E ≤ − 1 R | v − f p ( x ) | 2 ∀ v ∈ ∂ F ( x ) . These tw o equalities are equiv alent to (11 ) and (12). ✷ Next, we derive a r egularity r esult for f p ( · ) , which is a ctually a co nsequence of the Hölder con tinuit y of F in (8) co mbin ed with the low er curv a ture b o und in (5). Lemma 3.3. F or al l p ∈ R N \{ 0 } we have | f p ( x ) − f p ( y ) | ≤ ( C 0 + p 2 C 0 R ) | x − y | α ∀ x, y ∈ R N . Pr o of. Let x, y ∈ R N . In v ie w of (8) there are v y ∈ F ( x ) and v x ∈ F ( y ) s uch that | f p ( y ) − v y | ≤ C 0 | x − y | 2 α and | f p ( x ) − v x | ≤ C 0 | x − y | 2 α . Then, b y (12), D v y − f p ( x ) , p | p | E ≤ − 1 2 R | v y − f p ( x ) | 2 and D v x − f p ( y ) , p | p | E ≤ − 1 2 R | v x − f p ( y ) | 2 . A dding up the ab ov e tw o inequalities yields | v y − f p ( x ) | 2 + | v x − f p ( y ) | 2 ≤ − 2 R D v y − f p ( x )+ v x − f p ( y ) , p | p | E ≤ 2 C 0 R | x − y | 2 α . 4 So, | f p ( y ) − f p ( x ) | ≤ | f p ( y ) − v y | + | v y − f p ( x ) | ≤ ( C 0 + p 2 C 0 R ) | x − y | α , and the pro of is co mplete. ✷ Describing the way how f p ( x ) dep ends on p is the ob ject of our next r e sult. Lemma 3.4. F or every x ∈ R N we have 1 R | f p ( x ) − f q ( x ) | ≤     p | p | − q | q |     ≤ 1 r | f p ( x ) − f q ( x ) | , ∀ p, q ∈ R N \{ 0 } . Pr o of. Let x ∈ R N and let p, q ∈ R N \{ 0 } . Let us sta r t with the first inequality . Recalling the seco nd c ondition in (9) in its equiv a lent form (12), we hav e, since f q ( x ) ∈ ∂ F ( x ) , D f q ( x ) − f p ( x ) , p | p | E ≤ − 1 2 R | f q ( x ) − f p ( x ) | 2 . In a symmetric w ay we also ha ve D f p ( x ) − f q ( x ) , q | q | E ≤ − 1 2 R | f p ( x ) − f q ( x ) | 2 . A dding the t wo inequalities easily gives the first inequa lity . W e now pr ove the second inequalit y , which is slightly more subtle. Recalling the fir st inclusion in (9 ) and the definition of f p ( x ) , w e conclude that D q , f p ( x ) − r p | p | + rb E ≤ h q , f q ( x ) i , ∀ b ∈ B . Hence, − r D q | q | , p | p | E + r ≤ D f q ( x ) − f p ( x ) , q | q | E . Thu s, exchanging p and q , − r D p | p | , q | q | E + r ≤ D f p ( x ) − f q ( x ) , p | p | E . A dding the abov e inequalities together leads to r    p | p | − q | q |    2 = 2 r  1 − D p | p | , q | q | E ≤ D f p ( x ) − f q ( x ) , p | p | − q | q | E . (13) Since f p ( x ) and f q ( x ) are b oundary points, (1 1) yields D f p ( x ) − f q ( x ) , p | p | E ≤ 1 2 r | f q ( x ) − f p ( x ) | 2 . and D f q ( x ) − f p ( x ) , q | q | E ≤ 1 2 r | f q ( x ) − f p ( x ) | 2 . Therefore, D f p ( x ) − f q ( x ) , p | p | − q | q | E ≤ 1 r | f p ( x ) − f q ( x ) | 2 . (14) The conclusion follows from (13) and (14). ✷ 5 Let us now consider the p olar of H , namely the function H 0 defined by H 0 ( x, q ) := max  h p, q i : H ( x, p ) ≤ 1  ∀ ( x, q ) ∈ R N × R N . It is well-kno wn that, for all ( x, q ) ∈ R N × R N , H 0 ( x, q ) ≤ 1 ⇐ ⇒ q ∈ F ( x ) , (15) and H 0  x, D p H ( x, p )  = H 0  x, f p ( x )  = 1 ∀ ( x, p ) ∈ R N × ( R N \{ 0 } ) . (16) The dualit y b etw een H and H 0 brings similar qualitative prop erties for these t wo functions. F or instance, on account o f (1 0), we hav e | q | R ≤ H 0 ( x, q ) ≤ | q | r ∀ ( x, q ) ∈ R N × R N . (17) Moreov e r , H 0 is als o Hö lder contin uous with r esp ect to x , with the same exp o- nen t a s H . Lemma 3.5. F or al l q ∈ R N , | H 0 ( x, q ) − H 0 ( y , q ) | ≤ C 0 r 2 | q || x − y | 2 α ∀ x, y ∈ R N . (18) Pr o of. Let x, y , q ∈ R N . T ake p ∈ R N , with H ( x, p ) ≤ 1 , such that H 0 ( x, q ) = h p, q i . Then, by (4), | p | ≤ 1 /r . Also, by (3), H ( y , p ) ≤ 1 + C 0 r | x − y | 2 α . So, H 0 ( y , q ) ≥ D p 1 + C 0 r | x − y | 2 α , q E = H 0 ( x, q ) 1 + C 0 r | x − y | 2 α . On the other hand, in view o f (17), H 0 ( x, q ) = H 0 ( x, q ) 1 + C 0 r | x − y | 2 α + C 0 r | x − y | 2 α 1 + C 0 r | x − y | 2 α H 0 ( x, q ) ≤ H 0 ( x, q ) 1 + C 0 r | x − y | 2 α + C 0 r 2 | q || x − y | 2 α . Thu s, H 0 ( y , q ) ≥ H 0 ( x, q ) − C 0 r 2 | q || x − y | 2 α . Hence, w e obtain the conclusion e x changing the roles of x and y . ✷ W e now turn to the analysis of the level s et F 0 ( x ) =  p ∈ R N : H ( x, p ) ≤ 1  x ∈ R N . 6 Lemma 3.6. L et x ∈ R N . Then, for every p, p ′ ∈ R N with H ( x, p ) = H ( x, p ′ ) = 1 , | p − p ′ | ≤ C    p ′ | p ′ | − p | p |    (19) for some c onst ant C . Pr o of. First of a ll, the r eader b e w arned that, as x plays no role in this pro o f, the x − dep endence in H will b e omitted. F or all θ , θ ′ ∈ S N − 1 , w e ha ve    θ H ( θ ) − θ ′ H ( θ ′ )    ≤ | θ − θ ′ | H ( θ ) + | H ( θ ) − H ( θ ′ ) | H ( θ ) H ( θ ′ ) . Since H is Lipschitz contin uous by (10), recalling r ≤ H ( θ ) , H ( θ ′ ) ≤ R we conclude that | θ − θ ′ | H ( θ ) + | H ( θ ) − H ( θ ′ ) | H ( θ ) H ( θ ′ ) ≤ C | θ − θ ′ | for s ome co nstant C . Therefor e,    θ H ( θ ) − θ ′ H ( θ ′ )    ≤ C | θ − θ ′ | . Now, obs erve that the ma p θ 7→ θ /H ( θ ) is a bijection betw ee n the unit s phere S N − 1 and ∂ F 0 ( x ) . So, a pplying the above inequa lit y to θ , θ ′ ∈ S N − 1 chosen such that p = θ /H ( θ ) and p ′ = θ ′ /H ( θ ′ ) we obtain the conclusion. ✷ Lemma 3.7 (Lower curv ature estimate for F 0 ) . Ther e is a c onstant R ′ such that F 0 ( x ) satisfies the lower curvatur e estimate of r adius R ′ for al l x ∈ R N , i.e., F 0 ( x ) ⊂ B  p H ( x, p ) − R ′ f p ( x ) | f p ( x ) | , R ′  ∀ x, p ∈ R N , p 6 = 0 or, e quivalently, D p ′ − p, f p | f p | E ≤ − 1 2 R ′ | p ′ − p | 2 ∀ p, p ′ ∈ ∂ F 0 (20) Pr o of. Again, we shall dr o p the x -dep endence in all the for mu las below s ince it is of no int er est for this pro of. Recalling Remar k 3.2 we conclude that it suffices to prove inequality (20) for some constan t R ′ . Let then p, p ′ ∈ ∂ F 0 . Since H is po sitively ho mogeneous of degree 1, we hav e H ( p ′ ) − H ( p ) − h D p H ( p ) , p ′ − p i = h D p H ( p ′ ) , p ′ i − h D p H ( p ) , p i − h D p H ( p ) , p ′ − p i = h D p H ( p ′ ) − D p H ( p ) , p ′ i where D p H ( p ) = f p and D p H ( p ′ ) = f p ′ . F rom the lower curv ature estimate on F given in (12) it follows that D f p − f p ′ , p ′ | p ′ | E ≤ − 1 2 R | f p ′ − f p | 2 . Thu s, co mbining the above inequality with the previous identit y , and using the fact that H ( p ) = H ( p ′ ) = 1 , h f p , p ′ − p i ≤ − | p ′ | 2 R | f p ′ − f p | 2 . 7 Now, a pply Lemma 3.4 to obtain h f p , p ′ − p i ≤ − r 2 | p ′ | 2 R    p ′ | p ′ | − p | p |    2 . (21) Finally , let C b e the co nstant g iven by Lemma 3.6. Then, (19) and (21) yield D f p | f p | , p ′ − p E ≤ − r 2 | p ′ | 2 R 2    p ′ | p ′ | − p | p |    2 ≤ − r 2 2 C 2 R 3 | p ′ − p | 2 . whence the conclusion follo ws with R ′ = C 2 R 3 /r 2 . ✷ In particular, Lemma 3.7 ensures F 0 ( x ) is a strictly con vex set for any x ∈ R N . Thus, since H 0 ( x, · ) is the supp o r t function o f F 0 ( x ) , D q H 0 ( x, q ) exis ts for a ny x, q ∈ R N with q 6 = 0 (see, for instance, [4, Theo rem A.1.20]). In fact, we shall so o n prove a stronge r prop erty: the map q → D q H 0 ( x, q ) is lo cally Lipschit z contin uous in R N \{ 0 } . Before doing this, let us collect some tech nical remarks on the link b etw een H 0 and H and their deriv atives. Lemma 3.8. W e hav e, for any p, q ∈ R N \{ 0 } ,  q ∈ ∂ F ( x ) and p = D q H 0 ( x, q )  ⇔  p ∈ ∂ F 0 ( x ) and q = D p H ( x, p )  (22) In p articular, D q H 0  x, f p ( x ) | f p ( x ) |  = p ∀ p ∈ ∂ F 0 ( x ) . (23) Pr o of. W e just need to show the implication  q ∈ ∂ F ( x ) and p = D q H 0 ( x, q )  ⇒  p ∈ ∂ F 0 ( x ) and q = D p H ( x, p )  bec a use H 00 = H . Let q ∈ ∂ F ( x ) and p = D q H 0 ( x, q ) . Note that H ( x, p ) = H 0 ( x, q ) = 1 a nd, in particular, p ∈ ∂ F 0 ( x ) . B y definition, we hav e H 0 ( x, q ′ ) H ( x, p ′ ) ≥ h p ′ , q ′ i ∀ p ′ , q ′ ∈ R N . (24) This inequalit y beco mes a n equality for ( p ′ , q ′ ) = ( p, q ) because h p, q i = h D q H 0 ( x, q ) , q i = H 0 ( x, q ) = 1 = H 0 ( x, q ) H ( x, p ) . T a king the deriv ative in (24) with resp ect to p then gives H 0 ( x, q ) D p H ( x, p ) = D p H ( x, p ) = q . Next we turn to the pr o of o f (23). Reca ll first that f p ( x ) = D p H ( x, p ) for any p 6 = 0 . So, if p ∈ ∂ F 0 ( x ) , then (22) implies that D q H 0  x, f p ( x ) | f p ( x ) |  = D q H 0  x, f p ( x )  = p , since D q H 0 ( x, · ) is 0 − homogeneous. ✷ 8 Lemma 3.9. Ther e is a c onst ant C such t hat, for every x ∈ R N ,   D q H 0 ( x, q ) − D q H 0 ( x, q ′ )   ≤ C | q | ∨ | q ′ | | q − q ′ | ∀ q , q ′ ∈ R N \ { 0 } . Pr o of. Let us fix p, p ′ ∈ ∂ F 0 ( x ) . Owing to Lemma 3 .7 in its equiv a lent fo rm (20), we deduce that D p ′ − p, f p ( x ) | f p ( x ) | E ≤ − 1 2 R ′ | p ′ − p | 2 and h p − p ′ , f p ′ ( x ) | f p ′ ( x ) | i ≤ − 1 2 R ′ | p ′ − p | 2 . A dding up the last tw o inequalities, | p ′ − p | 2 ≤ R ′ D p ′ − p, f p ′ ( x ) | f p ′ ( x ) | − f p ( x ) | f p ( x ) | E ≤ R ′ | p ′ − p |    f p ′ ( x ) | f p ′ ( x ) | − f p ( x ) | f p ( x ) |    . (25) Now, recall that the map p 7→ f p ( x ) / | f p ( x ) | is a bijection from ∂ F 0 ( x ) to S N − 1 to deduce that for all q , q ′ ∈ S N − 1 there are p, p ′ ∈ ∂ F 0 ( x ) such that q = f p ( x ) / | f p ( x ) | and q ′ = f p ′ ( x ) / | f p ′ ( x ) | . Then, combining (23) and (25),   D q H 0 ( x, q ′ ) − D q H 0 ( x, q )   =    D q H 0  x, f p ′ ( x ) | f p ′ ( x ) |  − D q H 0  x, f p ( x ) | f p ( x ) |     = | p ′ − p | ≤ R ′    f p ′ ( x ) | f p ′ ( x ) | − f p ( x ) | f p ( x ) |    = R ′ | q ′ − q | . This is the desired estimate for q , q ′ ∈ S N − 1 . Next, let q , q ′ ∈ R N \{ 0 } . Then, since D q H 0 ( x, · ) is homogeneous of degree 0 ,   D q H 0 ( x, q ′ ) − D q H 0 ( x, q )   ≤ R ′    q ′ | q ′ | − q | q |    . Finally , o bserve that    q ′ | q ′ | − q | q |    ≤    q ′ | q ′ | − q | q ′ |    +    q | q ′ | − q | q |    = | q ′ − q | | q ′ | + | q | | q || q ′ | | | q ′ | − | q | | ≤ 2 | q ′ − q | | q ′ | to co mplete the pro of. ✷ 4 Regularit y of extremal tra jectories In this section, we shall pr ov e a r egularity r esult for the extremal tr a jectories of the differe ntial inclusion x ′ ( t ) ∈ F ( x ( t )) t ≥ 0 , (26) where F is the multif unction introduced in (6), a nd H is a given function satisfy- ing (SA). Alternatively , this a nalysis co uld b e addres s ed to differential inclusions asso ciated with a multifunct ion F : R N ⇒ R N that sa tisfies (8), (9), a nd (1 0) 9 as standing ass umptions, in which case the Hamiltonia n H should b e defined a s in (7). A tr aje ctory of the ab ove differen tia l inclusion is a lo cally a bs o lutely co ntin - uous arc x ( · ) : [0 , ∞ ) → R N that satisfies (26) for a.e. t ≥ 0 . Given a clo sed subset K of R N , we denote b y R ( t ) , t ≥ 0 , the r e achable set (from K ) in time t , that is, R ( t ) = { x ( t ) : x ( · ) is a tra jectory of (26) with x (0) ∈ K } . A tra jectory ¯ x ( · ) of (26) is ca lled extremal on the time interv al [0 , t ] if ¯ x ( t ) ∈ ∂ R ( t ) . In this c a se, o ne ca n show that in fact ¯ x ( s ) ∈ R ( s ) for every s ∈ [0 , t ] . Due to the sp ecial structure of F , descr ibed by the prop erties (8), (9), a nd (10), we will b e able to sho w that all extrema l tra jector ies ar e C 1 ,α/ 2 -smo oth. More precisely , w e hav e the following re s ult. Theorem 4.1. Assume (SA) and let ¯ x b e an extr emal tr aje ctory of (26) on some time interval [0 , T ] .Then | ¯ x ′ ( t 2 ) − ¯ x ′ ( t 1 ) | ≤ C ( t 2 − t 1 ) α/ 2 ∀ t 1 , t 2 ∈ [0 , T ] (27) for some c onst ant C . Pr o of. Let ¯ x be an extremal tra jectory on [0 , T ] . Then, by extremality , x ′ ( t ) ∈ ∂ F ( ¯ x ( t )) for almost all t ∈ [0 , T ] , so that w e can set ¯ p ( t ) = D q H 0 ( ¯ x ( t ) , ¯ x ′ ( t )) a.e. in [0 , T ] . Using Lemma 3.8, w e obtain the following r elation betw een ¯ x and ¯ p : ¯ x ′ ( t ) = D p H  ¯ x ( t ) , ¯ p ( t )  for a .e. t ∈ [0 , T ] . Step 1. W e first claim that, for any 0 ≤ t 1 < t 2 ≤ T we hav e t 2 − t 1 ≤ H 0  ¯ x ( t 2 ) , Z t 2 t 1 D p H  ¯ x ( t 2 ) , ¯ p ( t )  dt  + C ( t 2 − t 1 ) 1+ α . (28) Pr o of of (28): Let us set q = ¯ x ( t 2 ) − ¯ x ( t 1 ) | ¯ x ( t 2 ) − ¯ x ( t 1 ) | . Let λ : [ t 1 , t 2 ] → R b e a so lution of the Cauch y problem    λ ′ ( t ) = 1 H 0 ( ¯ x ( t 1 ) + λ ( t ) q , q ) , t ∈ [ t 1 , t 2 ] λ ( t 1 ) = 0 . Then x ( t ) := ¯ x ( t 1 ) + λ ( t ) q is a tra jectory of (26) s ince, o wing to (15), H 0  x ( t ) , x ′ ( t )  = H 0  x ( t ) , λ ′ ( t ) q  = H 0  ¯ x ( t 1 ) + λ ( t ) q, q  H 0  ¯ x ( t 1 ) + λ ( t ) q, q  = 1 for all t ∈ [ t 1 , t 2 ] . Therefor e , since ¯ x is a n extremal tra jectory , the p oint ¯ x ( t 1 ) + λ ( t 2 ) q b elo ng s to the s egment [ ¯ x ( t 1 ) , ¯ x ( t 2 )] . So, λ ( t 2 ) − λ ( t 1 ) ≤ | ¯ x ( t 2 ) − ¯ x ( t 1 ) | . 10 Note that, owing to (18 ), the ab ove inequality , a nd the bo undedness of F ,    1 H 0 ( ¯ x ( t 1 ) + λ ( t ) q , q ) − 1 H 0 ( ¯ x ( t 2 ) , q )    ≤ C | ¯ x ( t 2 ) − ¯ x ( t 1 ) | 2 α ≤ C ( t 2 − t 1 ) 2 α for a ll t ∈ [ t 1 , t 2 ] and some constant C . Hence, λ ( t 2 ) − λ ( t 1 ) = Z t 2 t 1 dt H 0 ( ¯ x ( t 1 ) + λ ( t ) q , q ) ≥ t 2 − t 1 H 0 ( ¯ x ( t 2 ) , q ) − C ( t 2 − t 1 ) 1+2 α . So, a pp ea ling to Lemma 3.3, t 2 − t 1 ≤ H 0 ( ¯ x ( t 2 ) , q ) | ¯ x ( t 2 ) − ¯ x ( t 1 ) | + C ( t 2 − t 1 ) 1+2 α = H 0 ( ¯ x ( t 2 ) , ¯ x ( t 2 ) − ¯ x ( t 1 )) + C ( t 2 − t 1 ) 1+2 α = H 0  ¯ x ( t 2 ) , R t 2 t 1 D p H ( ¯ x ( t ) , ¯ p ( t )) dt  + C ( t 2 − t 1 ) 1+2 α ≤ H 0  ¯ x ( t 2 ) , R t 2 t 1 D p H ( ¯ x ( t 2 ) , ¯ p ( t )) dt  + C ( t 2 − t 1 ) 1+ α , where the ab ov e constants may c hang e from line to line. W e hav e thus proved (28). Step 2. Let us fix 0 ≤ t 1 < t 2 ≤ T and let ¯ t b e such that H 0 ( ¯ x ( t 2 ) , ¯ x ( ¯ t ) − ¯ x ( t 1 )) = H 0 ( ¯ x ( t 2 ) , ¯ x ( t 2 ) − ¯ x ( ¯ t )) . (29) Define a = ¯ x ( ¯ t ) − ¯ x ( t 1 ) and b = ¯ x ( t 2 ) − ¯ x ( ¯ t ) . (30) W e claim that H 0 ( ¯ x ( t 2 ) , a ) + H 0 ( ¯ x ( t 2 ) , b ) ≤ H 0 ( ¯ x ( t 2 ) , a + b ) + C ( t 2 − t 1 ) 1+ α (31) and | 2 ¯ t − t 1 − t 2 | ≤ C ( t 2 − t 1 ) 1+ α . (32) Pr o of of (31) and (32) : A g ain by Lemma 3.3, and then using Jensen’s inequal- it y , we obtain H 0 ( ¯ x ( t 2 ) , a ) = H 0  ¯ x ( t 2 ) , R ¯ t t 1 D p H ( ¯ x ( s ) , ¯ p ( s )) ds  ≤ H 0  ¯ x ( t 2 ) , R ¯ t t 1 D p H ( ¯ x ( t 2 ) , ¯ p ( s )) ds  + C ( t 2 − t 1 ) 1+ α ≤ R ¯ t t 1 H 0  ¯ x ( t 2 ) , D p H ( ¯ x ( t 2 ) , ¯ p ( s ))  ds + C ( t 2 − t 1 ) 1+ α ≤ ¯ t − t 1 + C ( t 2 − t 1 ) 1+ α . (33) Applying (28) b etw een t 1 and ¯ t gives ¯ t − t 1 ≤ H 0  ¯ x ( ¯ t ) , Z ¯ t t 1 D p H ( ¯ x ( ¯ t ) , ¯ p ( t )) dt  + C ( ¯ t − t 1 ) 1+ α . (34) Now, in order to b ound the ab ove r ight-hand side observe that   D p H ( ¯ x ( ¯ t ) , ¯ p ( t )) − D p H ( ¯ x ( t ) , ¯ p ( t ))   ≤ C | ¯ x ( ¯ t ) − ¯ x ( t ) | α ≤ C ( ¯ t − t 1 ) α 11 in view of Lemma 3.3, and   H 0  ¯ x ( ¯ t ) , a  − H 0  ¯ x ( t 2 ) , a    ≤ C | a | | ¯ x ( ¯ t ) − ¯ x ( t 2 ) | 2 α ≤ C ( ¯ t 2 − t 1 ) 1+2 α owing to Lemma 3.5. Therefore, (34) lea ds to ¯ t − t 1 ≤ H 0  ¯ x ( ¯ t ) , Z ¯ t t 1 D p H ( ¯ x ( t ) , ¯ p ( t )) dt | {z } a  + C ( ¯ t − t 1 ) 1+ α ≤ H 0 ( ¯ x ( t 2 ) , a ) + C ( t 2 − t 1 ) 1+ α . (35) On a c c ount of (33) and (35), we have H 0 ( ¯ x ( t 2 ) , a ) − C ( t 2 − t 1 ) 1+ α ≤ ¯ t − t 1 ≤ H 0 ( ¯ x ( t 2 ) , a ) + C ( t 2 − t 1 ) 1+ α . In the same w ay , H 0 ( ¯ x ( t 2 ) , b ) − C ( t 2 − t 1 ) 1+ α ≤ t 2 − ¯ t ≤ H 0 ( ¯ x ( t 2 ) , b ) + C ( t 2 − t 1 ) 1+ α . Combining the ab ov e tw o inequalities with the c ho ice of ¯ t made in (29) gives (32). Moreov e r , adding up the ab ov e inequa lities and recalling (28), we get H 0 ( ¯ x ( t 2 ) , a )+ H 0 ( ¯ x ( t 2 ) , b ) ≤ ( t 2 − t 1 )+ C ( t 2 − t 1 ) 1+ α ≤ H 0 ( ¯ x ( t 2 ) , a + b )+ C ( t 2 − t 1 ) 1+ α , which yields (31). Step 3. W e no w claim that, for an y 0 ≤ t 1 < t 2 ≤ T , we hav e    ¯ x  t 1 + t 2 2  − ¯ x ( t 2 ) + ¯ x ( t 1 ) 2    ≤ C ( t 2 − t 1 ) 1+ α . (36) Pr o of of (36): Ha v ing fixed 0 ≤ t 1 < t 2 ≤ T , we will use the sa me no tation for ¯ t , a , and b a s in (29) and (30). Moreov er, since x ( t 2 ) is fixed in the r e a soning below, as we often did befor e we will o mit the x ( t 2 ) -depe nda nce of H 0 and all the o ther ma ps app ea ring in this proo f. Let us set, for a ny q ∈ R N \{ 0 } , g q = D q H 0 ( q ) . W e use b elow r ep etitiv ely the fo llowing remar k: for any q 6 = 0 , if p = g q , then f p = q /H 0 ( q ) . Indeed, since q /H 0 ( q ) ∈ ∂ F and p = D q H 0 ( q /H 0 ( q )) (b ecause D q H 0 is 0 − homog e neo us), Lemma 3 .8 implies that q /H 0 ( q ) = D p H ( p ) = f p . W e first sho w that 1 2 R ′ | g q − g q ′ | 2 ≤ h g q − g q ′ , q | q | i ∀ q , q ′ ∈ R N \{ 0 } , (37) where R ′ is the consta nt app ear ing in Lemma 3.7. F or this, let us co nsider the low er curv ature estimate (20) in Lemma 3.7 with p = g q and p ′ = g q ′ : b eca us e of the remark ab ove and since p, p ′ ∈ ∂ F 0 , we hav e D g q ′ − g q , q | q | E ≤ − 1 2 R ′ | g q ′ − g q | 2 12 which is exactly (37). Next we note that | a − b | ≤ C H 0 ( a ) | g a − g b | (38) Indeed let us apply the first inequality in Lemma 3.4 to p = g a and q = g b . Since f p = a/H 0 ( a ) and f q = b/H 0 ( b ) and since H 0 ( a ) = H 0 ( b ) by (29), we hav e | a − b | ≤ RH 0 ( a )     g a | g a | − g b | g b |     ≤ C H 0 ( a ) | g a − g b | . In order to estimate the right-hand side of inequality (38), let us obser ve that, in view of (31), 0 ≤ H 0 ( a + b ) − H 0 ( a ) − H 0 ( b ) + C ( t 2 − t 1 ) 1+ α = h g a + b , a + b i − h g a , a i − h g b , b i + C ( t 2 − t 1 ) 1+ α so tha t 0 ≤ h g a + b − g a , a i + h g a + b − g b , b i + C ( t 2 − t 1 ) 1+ α . Plugging ineq ua lit y (37) in to this inequalit y leads to | a | | g a + b − g a | 2 + | b | | g a + b − g b | 2 ≤ C ( t 2 − t 1 ) 1+ α , i.e., since | a | ≥ ( ¯ t − t 1 ) /C and | b | ≥ ( t 2 − ¯ t ) /C and (32) holds, | g a + b − g a | ≤ C ( t 2 − t 1 ) α/ 2 , | g a + b − g b | ≤ C ( t 2 − t 1 ) α/ 2 . So, r e c a lling tha t H 0 ( a ) ≤ C ( ¯ t − t 1 ) ≤ C ( t 2 − t 1 ) , we get from (38), | a − b | ≤ C H 0 ( a )( | g a + b − g a ( x ) | + | g a + b − g b ( x ) | ) ≤ C ( t 2 − t 1 ) 1+ α/ 2 . F r om the definition of a and b , this means that | 2 ¯ x ( ¯ t ) − ¯ x ( t 1 ) − ¯ x ( t 2 ) | ≤ C ( t 2 − t 1 ) 1+ α/ 2 . Using a g ain (3 2) a nd the Lipschit z contin uity of ¯ x then easily yields to (36). Conclusion. In view of (36), Theorem 2.1.10 of [4] states that ea ch comp onent of ¯ x is semi-conv ex and semi-concave with a mo dulus m of the form m ( ρ ) = C ρ α/ 2 . Then, from Theorem 3.3.7 of [4], ¯ x is of class C 1 ,α/ 2 and (27) holds. ✷ 5 The semiconca vit y result Let H : R N × R N → R b e a contin uous function satisfying our Standing As- sumptions with constants α, r , R, and C 0 , and let Ω ⊂ R N be an open set. In this sec tio n, we will apply the previous analys is to study the r egularity of the s olution to the Dirichlet pr oblem  H ( x, − D u ( x )) = 1 in Ω u ( x ) = 0 on ∂ Ω (39) The e x istence, uniqueness, and Lipschit z c ontin uit y of the viscosit y s olution u of the ab ov e problem is w ell- known, as w ell as the r epresentation formula u ( x ) = inf  t ≥ 0 : ∃ x ( · ) tra jectory of (26) with x (0) = x , x ( t ) ∈ ∂ Ω  (40) 13 (see, e.g., [1]). W e reca ll that a function v : Ω → R is lo cally θ -semiconcave, with θ ∈ (0 , 1 ] , if for every c o mpact co nv ex set O ⊂ Ω ther e is a constant C O such that v ( λx + (1 − λ ) y ) ≥ λv ( x ) + (1 − λ ) v ( y ) − C O λ (1 − λ ) | x − y | 1+ θ for a ll x, y ∈ O and λ ∈ [0 , 1] . W e are no w ready for our ma in res ult. Theorem 5.1. Assume (SH) . Then t he solution u of (39 ) is lo c al ly θ -semic onc ave in Ω for every θ ∈ (0 , α 4+ α ) . Pr o of. The strategy of the pro of is the following. Fix β ∈  2 2 + α , 4 4 + α  , (41) and obs erve that 0 < θ := β 2 + α 2 − 1 < α 4 + α . Let O ⊂⊂ Ω be an op en conv ex set. W e are going to show that u ( x + h ) + u ( x − h ) − 2 u ( x ) ≤ C | h | β (2+ α ) / 2 (42) for all h ∈ R N sufficient ly small (in this pro of, C denotes a generic c o nstant depending only on α, r , R, C 0 , and O ). Since u is contin uous, owing to [4, The- orem 2.1.10] the ab ove inequality implies that u is lo cally θ -semico ncave in Ω . Step 1. Let ¯ x ∈ O and let ¯ x ( · ) b e a solution of the minimization problem in (40)—an optimal tra jector y for short. Since ¯ x ( · ) is extremal on [0 , u ( ¯ x )] , Theorem 4.1 implies that ¯ x ( · ) is of class C 1 ,α/ 2 and sa tisfies | ¯ x ′ ( t 2 ) − ¯ x ′ ( t 1 ) | ≤ C | t 2 − t 1 | α/ 2 ∀ t 1 , t 2 ∈ [0 , u ( ¯ x )] . Setting ¯ v = ¯ x ′ (0) , from the above inequality we obtain | ¯ x ( t ) − ¯ x − t ¯ v | ≤ C t 1+ α/ 2 ∀ t ∈ [0 , u ( ¯ x )] . (43) Step 2. Let h ∈ R N be small enough, and set ¯ t = | h | β . W e will now build a tra jectory x + ( · ) suc h that      x + (0) = ¯ x + h , x + ( t ) ∈ [ ¯ x + h , ¯ x ( ¯ t )] ∀ t ∈ [0 , τ + ] , x + ( t ) = ¯ x ( t + ¯ t − τ + ) ∀ t ∈ [ τ + , u ( ¯ x ) + τ + − ¯ t ] . Notice that x + ( u ( ¯ x ) + τ + − ¯ t ) = ¯ x ( u ( ¯ x )) ∈ R N \ Ω , so that u ( ¯ x + h ) ≤ u ( ¯ x ) + τ + − ¯ t . (44) Pr o of of S tep 2. In or der to construct the line segment part, let us set q + = ¯ x ( ¯ t ) − ( ¯ x + h ) and o bserve that, in view of (43), q + = ¯ t ¯ v + O ( ¯ t 1+ α/ 2 ) − h . (45) 14 Then, q + 6 = 0 since ¯ t >> | h | and | ¯ v | ≥ 1 / r . Let λ ( · ) b e a so lution of the Cauch y problem    λ ′ ( t ) = 1 H 0 ( ¯ x + h + λ ( t ) q + , q + ) , t ≥ 0 λ (0) = 0 . Since λ ( · ) is strictly increa sing there is a unique time τ + such that λ ( τ + ) = 1 . Now, set x + ( t ) = ¯ x + h + λ ( t ) q + t ∈ [0 , τ + ] . Then x + ( · ) is a solution of the differen tial inclusion (26) on [0 , τ + ] b ecause H 0 ( x + ( t ) , x ′ + ( t )) = 1 ∀ t ∈ [0 , τ + ] . Moreov e r x + ( τ + ) = ¯ x ( ¯ t ) . Thus, defining x + ( t ) = ¯ x ( t + ¯ t − τ + ) ∀ t ∈ [ τ + , u ( ¯ x ) + τ + − ¯ t ] completes the construction of x + ( · ) . Step 3. W e will now prove the es timate τ + ≤ ¯ t +  D q H 0 ( ¯ x , ¯ t ¯ v ) , q + − ¯ t ¯ v  + C | h | 2 ¯ t . (46) Pr o of of Step 3. T o b egin with, let us note that a ny ξ ∈ [ ¯ t ¯ v , q + ] sa tisfies | ξ | ≥ ¯ t/C . Indeed, if ξ = µq + + (1 − µ ) ¯ t ¯ v for some µ ∈ [0 , 1 ] , then, b y (45), ξ = ¯ t ¯ v + O ( ¯ t 1+ α/ 2 ) − h with | h | = ¯ t 1 /β and β ∈ (0 , 1) . So, for | h | s ma ll enoug h, we hav e the desired cla im: | ξ | ≥ ¯ t/C . Then, by Lemma 3.9 we c o nclude that the map ξ 7→ D q H 0 ( ¯ x , ξ ) is Lipschitz on [ ¯ t ¯ v, q + ] with co ns tant C / ¯ t . So, for all t ∈ [0 , τ + ] , H 0 ( x + ( t ) , q + ) ≤ H 0 ( ¯ x, q + ) + C | x + ( t ) − ¯ x | 2 α | q + | ≤ H 0 ( ¯ x, ¯ t ¯ v ) + h D q H 0 ( ¯ x , ¯ t ¯ v ) , q + − ¯ t ¯ v i + ( C / ¯ t ) | q + − ¯ t ¯ v | 2 + C | x + ( t ) − ¯ x | 2 α | q + | . Since x + ( t ) ∈ [ ¯ x + h , ¯ x ( ¯ t )] , w e hav e | x + ( t ) − ¯ x | ≤ max {| h | , | ¯ x ( ¯ t ) − ¯ x |} ≤ C ¯ t . Also, o n account of (45) and (41), | q + | ≤ C ¯ t and | q + − ¯ t ¯ v | = | O ( ¯ t 1+ α/ 2 ) − h | ≤ C | h | . Noting that H 0 ( ¯ x , ¯ v ) = 1 b eca use ¯ v ∈ ∂ F ( ¯ x ) , the a b ove ineq uality yield, by the homogeneity o f H 0 ( ¯ x, · ) , H 0 ( x + ( t ) , q + ) ≤ ¯ t +  D q H 0 ( ¯ x, ¯ t ¯ v ) , q + − ¯ t ¯ v  + C  | h | 2 ¯ t + ¯ t 1+2 α  , where | h | 2 / ¯ t > ¯ t 1+2 α bec a use ¯ t = | h | β with β > 1 / (1 + α/ 2) . So, H 0 ( x + ( t ) , q + ) ≤ ¯ t +  D q H 0 ( ¯ x , ¯ t ¯ v ) , q + − ¯ t ¯ v  + C | h | 2 / ¯ t . 15 Then 1 = Z τ + 0 λ ′ ( t ) dt = Z τ + 0 dt H 0 ( x + ( t ) , q + ) ≥ τ + ¯ t +  D q H 0 ( ¯ x, ¯ t ¯ v ) , q + − ¯ t ¯ v  + C | h | 2 / ¯ t , which in turn yields (46). Conclusion. Rep eating the a b ov e reasoning with q − = ¯ x ( ¯ t ) − ( ¯ x − h ) , we can build a solution x − ( · ) to (26) suc h that x − (0) = ¯ x − h , x − ( t ) ∈ [ ¯ x − h, ¯ x ( ¯ t )] on the time interv al [0 , τ − ] , and x − ( t ) = ¯ x ( t + ¯ t − τ − ) on [ τ − , u ( ¯ x ) + τ − − ¯ t ] . Therefore, u ( ¯ x − h ) ≤ u ( ¯ x ) + τ − − ¯ t , (47) where τ − can b e estimated as a bove: τ − ≤ ¯ t +  D q H 0 ( ¯ x , ¯ t ¯ v ) , q − − ¯ t ¯ v  + C | h | 2 ¯ t . (48) Hence, b y (44), (4 7 ), (46), (48), and (43) we o btain u ( ¯ x + h ) + u ( ¯ x − h ) − 2 u ( ¯ x ) ≤ τ + − ¯ t + τ − − ¯ t ≤ 2  D H 0 ( ¯ x, ¯ t ¯ v ) , ¯ x ( ¯ t ) − ¯ x − ¯ t ¯ v  + C | h | 2 ¯ t ≤ C | h | β (2+ α ) / 2 since β < 4 / (4 + α ) . W e hav e thus a ttained (42), which completes the pro of. ✷ References [1] M. Bardi and I. Capuzzo Dolcetta. Optimal c ontr ol and visc osity solutions of Hamilton-Jac obi-Bel lman e quations . Springer , 1 997. [2] B.Su and M.Burger. Global weak solutions of non-isotherma l front propa g a- tion problem. Ele ctr on. R es. A nnoun c. A mer. Math. So c. , 13:46–5 2, 20 0 7. [3] P . Ca nnarsa a nd H. F ra nko wsk a. Some characterizatio ns of optimal tra jec- tories in control theory . SIA M J. Contr ol Optim. , 29:132 2 –134 7, 1991 . [4] P . Cannarsa and C. Sinestrari. Semic onc ave functions, Hamilton-Jac obi e quations and optimal c ont r ol . Bir khäuser, 2 004. [5] P . Cannarsa and H.M. Soner. Generalized one-sided estima tes for solutions of hamilton-jacobi equations and applications. Nonline ar A n al. , 1 3(3):305 – 323, 1 989. [6] P . Car daliaguet, O.Ley , and A. Monteillet. Viscosity solutions for a p olymer crystal growth mo del. Indiana Univ. Math. J. (to app e ar) . [7] H. Ishii. Uniqueness of un bo unded viscosity so lutions of hamilton-jacobi equations. In diana Univ. Math. J. , 33:721– 7 48, 198 4. [8] P .L. Lions . Gener alize d solutions of Hamilton-Jac obi e quations . Pitman, 1982. [9] C. Sinestra r i. Semiconcavity of solutions of stationary hamilton-ja c o bi equa- tions. Nonline ar A nal. , 24:132 1–13 26, 1995. 16