Hamilton-Jacobi Equations

Transcription

1 Universiy Of Helsinki Deparmen Of Mahemaics and Saisics Hamilon-Jacobi Equaions Maser s Thesis Nikia Skoura supervised by Pr. Kari Asala May 31, 2017

2 HELSINGIN YLIOPISTO HELSINGFORS UNIVERSITET UNIVERSITY OF HELSINKI Tiedekuna/Osaso Fakule/Sekion Faculy Laios Insiuion Deparmen Maemaais-luonnonieeellinen Tekijä Förfaare Auhor Nikia Skoura Työn nimi Arbees iel Tile Maemaiikan ja ilasoieeen laios Hamilon-Jacobi Equaions Oppiaine Läroämne Subjec Maemaiikka Työn laji Arbees ar Level Aika Daum Monh and year Sivumäärä Sidoanal Number of pages Pro gradu-ukielma Toukokuu s. Tiiviselmä Refera Absrac This hesis presens he heory of Hamilon-Jacobi equaions. I is firs shown how he equaion is derived from he Lagrangian mechanics, hen he radiional mehods for searching for he soluion are presened, where he Hopf-Lax formula along wih he appropriae noion of he weak soluion is defined. Laer he flaws of his approach are remarked and he new noion of viscosiy soluions is inroduced in connecion wih Hamilon-Jacobi equaion. The imporan properies of he viscosiy soluion, such as consisency wih he classical soluion and he sabiliy are proved. The inroducion ino he conrol heory is presened, in which he Hamilon-Jacobi-Bellman equaion is inroduced along wih he exisence heorem. Finally muliple numerical mehods are inroduced and aligned wih he heory of viscosiy soluions. The knowledge of he heory of parial differenial analysis, calculus and real analysis will be helpful. Avainsana Nyckelord Keywords Hamilon-Jacobi Equaion, PDE, Classical Mechanics, Viscosiy Soluions Säilyyspaikka Förvaringssälle Where deposied Kumpulan iedekirjaso Muia ieoja Övriga uppgifer Addiional informaion

3 Conens Inroducion 2 1 Hamilon-Jacobi Equaions. Derivaion and Basic Properies Hamilon s Principle Hamilon s ODE Canonical Transformaions Hamilon-Jacobi Equaion and Is Properies Mehod of Characerisics Applicaions Eikonal Equaion Schrödiner Equaion Tradiional Approach Separaion Of Variables Legendre Transform Classical Hopf-Lax Formula Weak Soluions Viscosiy Soluions Mehod Of Vanishing Viscosiy Definiion and Basic Properies Hopf-Lax Formula as a Viscosiy Soluion Uniqueness Opimal Conrol Theory Inroducion Hamilon-Jacobi-Bellman Equaion Numerical Approach Firs Order Monoone Schemes Time Discreizaion Higher Order Schemes and Furher Reading References 70 1

4 Inroducion The opic of his hesis is he Hamilon-Jacobi equaion. This equaion is an example of a parial differenial equaions ha arises in many differen areas of physics and mahemaics, and hence has an umos imporance in science. Originally, i was derived by William Rowen Hamilon in his essays [15] and [16] as a formulaion of classical mechanics for a sysem of moving paricles. Hamilon exended he Lagrangian formulaion, using he calculus of variaions, hus deriving he following equaion: ( S S + H,..., S ),q 1,..., q d =0. q 1 q d Here H was a Hamilonian - a descripor of he sysem, defined by Hamilon, and q i are he generalized coordinaes. The soluion o he equaion he classical acion S = (U + T )ds, also called he principal funcion, where U is a poenial and T 0 is a kineic energy of he sysem. However, mahemaically Hamilon s argumens conained some flaws. For example, Hamilon assumed ha he equaion would hold only in case of conservaive sysems (saisfying he canonical Hamilon s equaions). Jacobi showed laer in his papers, ha such an assumpion is no necessary and also provided mahemaically meaningful adjusmens o he equaion. Tha s why he modern name for he equaion is Hamilon-Jacobi equaion. Laer, anoher version of he derivaion of he equaion was presened. I showed ha he principal funcion S can be seen as a generaing funcion for a canonical ransformaion of coordinaes. In his hesis we ll follow his approach. In mahemaics, he similar equaion appeared in conrol heory in 50s, when Richard Bellman developed a dynamic programming mehod in [2] wih his colleagues. This so-called Hamilon-Jacobi-Bellman equaion defines a broader class of equaions, han he version from he classical mechanics. Mahemaicians have been working on solving his equaions for many years. I appeared, ha his equaion in general has no classical soluions, hence firs, some noions of weak soluion was developed by Hopf ([17]) and Lax([21]). Their invesigaions led o he imporan Hopf-Lax formula, ha can be used for calculaing he soluion or sudying is properies. Finally, in he 80s, he new noion of viscosiy soluion was developed for he Hamilon-Jacobi equaion and exended o a broader class of parial differenial equaions. The firs ideas abou his weaker noion of soluions appeared in Evans paper [10] by applying he vanishing viscosiy mehod o he equaion, however he exac definiion was given laer by Crandall and Lions in [9] along wih some very imporan properies. 2

5 The purpose of his hesis is o presen he heory of Hamilon-Jacobi equaions from muliple sides. In he firs chaper we will show how his equaion is derived by using he canonical ransformaion argumen. In he second chaper he radiional approach o he exisence and uniqueness heory is presened along wih Hopf-Lax formula. And finally, in he hird chaper we ll inroduce he viscosiy soluions from scrach: we ll firs show how he applicaion of he mehod of vanishing viscosiy o he Hamilon-Jacobi equaion leads o he definiion, and hen define i sricly and show, ha such weak soluions are consisen and sable. Also we give a shor inroducion ino he conrol heory and dynamic programming, hus also deriving he Hamilon-Jacobi-Bellman equaion. This hesis uses Chapers 3.3 and 10 from [11] as a skeleon. The prior knowledge of heory of parial differeniable equaions, real analysis is assumed. 1 Hamilon-JacobiEquaions. Derivaionand Basic Properies The purpose of his hesis is o sudy he iniial value problem for he following Hamilon-Jacobi equaion: { u + H(D x u, x) =0 (1.1) u(x, 0) = g. Here u : R d [0, ) R,u= u(x, ) is an unknown funcion, H : R d R d R,H = H(p, x) is a Hamilonian (we ll define i laer) and g : R d R,g = g(x) is he iniial condiion. We denoe by D x u a spaial gradien, i.e. D x u =( u x 1,..., u x d ). In his secion we ll show how his equaion is derived from Lagrangian mechanics. We ll firs derive he formulaion of Lagrangian mechanics, using he principle of leas acion, hen we ll show how Hamilon s ODE are derived and finally we ll show how Hamilon-Jacobi equaion comes in place. Remark. The noaion and definiions presened in his paper may be differen from he ones in mechanics books. I was chosen as i beer suis he mahemaical PDE heory, ha we re sudying in his hesis. You can read more abou Hamilon s principle and Hamilon s ODEs from he physical perspecive in 2 and 40 in [20], or Chaper 9 in [1]. 3

6 1.1 Hamilon s Principle Hamilonian mechanics is one of he several formulaions of classical mechanics, ha uses generalised coordinaes x =(x 1,..., x d )asoneofheparameers. Weknowfrom he Newonian formulaion, ha o define he posiion of a sysem of N moving paricles 1 in hree-dimensional Euclidian space, we need o specify N vecors, which requires a oal pf 3N coordinaes. This number of coordinaes is also called he number of degrees of freedom. The generalised coordinaes are some number of independen quaniies, ha are needed o describe he moion of sysem of paricles (3N in he example above). Noe, ha hese quaniies mus no necessarily be Caresian coordinaes and he opimal choice ofen depends on he given problem. In his paper, we are looking a he moion law for a sysem of paricles wih d degrees of freedom, which is mahemaically described as a pah x : [0, ] R d, i.e. x j = x j (),j =1...d. Our firs goal is o derive he Euler-Lagrange equaions for his law. Before we sae he Hamilon s principle and is consequences, le s give some definiions firs: Definiion 1.1. The Lagrangian of he sysem of moving paricles is a definie smooh funcion L : R d R d R, denoedasl = L(q, x), hacompleelycharacerizes he condiions of he movemen. Lagrangian is he cenral noion of he so-called Lagrangian mechanics. In general, here s no single expression o calculae he Lagrangian, and any funcion, ha generaes he physically correc moion equaions, can be aken as a Lagrangian. There re however some wide classes of problems, where he Lagrangian can be easily consruced. For example, in non-relaivisic case, he Lagrangian for he moving sysem of paricles can be defined as L = T V,whereT is he oal kineic energy of he sysem and V is he poenial energy. More abou i can be found in [24]. Definiion 1.2. The acion funcional is he following inegral: I[x( )] = 0 L(ẋ(s),x(s))ds, (1.2) defined for funcions x(s) =(x 1 (s),...x d (s)), belongingoheadmissibleclassa: A = {u : [0,] R d,u C 2 ([0,]) u(0) = y, u() =x} (1.3) 1 In his paper we mean by a paricle a physical body whose dimensions may be negleced in describing is moion. 4

7 The Hamilon s principle in mechanics saes, ha he paricle moves beween poins y and x in a way, ha he acion inegral akes he leas possible value. This principle is also called he principle of leas acion. According o his, o find he law of moion of he paricle, we have he following calculus of variaion problem: we wan o find a C 2 -curve, ha minimizes he acion funcional among all admissible curves w A: I[x( )] = min w A 0 L(ẇ(s),w(s))ds, (1.4) To approach his minimizaion problem, we can solve he corresponding Euler- Lagrange equaions of he variaional problem. In our case i s a sysem of d second order ordinary differenial equaions. Theorem 1.1. Assume here exiss a funcion x A,hasolvesheminimizaion problem (1.4). Then his funcion saisfies he following Euler-Lagrange equaion: d ds (D ql(ẋ(s),x(s))) + D x L(ẋ(s),x(s)) = 0 (1.5) Proof. Le s choose a smooh funcion v : [0,] R d,suchhav(0) = v() =0. For τ R we define a funcion: w( ) := x( )+τv( ) (1.6) Clearly, w Aand by definiion of x, we have ha I[x( )] I[w( )]. Hence, he real valued funcion i, defined as: i(τ) := I[x( )+τv( )], (1.7) aains a minimum a τ = 0. Hence, if i (τ) exiss, we have ha i (τ) = 0. Le s calculae his derivaive explicily. i(τ) = Using Leibniz inegral rule: Nex, le τ =0: i (τ) = 0=i (0) = L(ẋ(s)+τ v(s),x(s)+τv(s))ds, (D q L(ẋ(s)+τ v(s),x(s)+τv(s)) v(s) + D x L(ẋ(s)+τ v(s),x(s)+τv(s)) v(s))ds. (D q L(ẋ(s),x(s)) v(s)+d x L(ẋ(s),x(s)) v(s))ds (1.8) 5

8 Finally, noice ha we can simplify he firs par of he inegrand wih inegraion by pars, aking ino accoun he ha v vanishes a he inegraion limis: 0 D q L(ẋ(s),x(s)) v(s)ds = = = d i=1 0 0 d i=1 0 (D qi L(ẋ(s),x(s)) v i (s)ds ( d ds D q i L(ẋ(s),x(s))v i (s)ds ( d ds D ql(ẋ(s),x(s)) v(s)ds Now passing his resul back o (1.8), we obain: 0= 0 ( d ds (D ql(ẋ(s),x(s)) + D x L(ẋ(s),x(s))) v(s)ds (1.9) Due o he fac, ha v was an arbirary smooh funcion, we mus have, ha: d ds (D ql(ẋ(s),x(s))) + D x L(ẋ(s),x(s)) = 0 Euler-Lagrange equaions wih he corresponding iniial values consiue he formulaion of he classical mechanics, ha uses generalized coordinaes and generalized velociies as parameers. Remark. Noe, ha no every soluion o Euler-Lagrange equaion is aminimizer of he acion funcional. The soluion o he Euler-Lagrange equaioniscalleda criical poin of he corresponding funcional. So, wha he proven heorem says,is ha every minimizer is a criical poin of I[x( )], buacriicalpoinneed nobea minimizer. This is similar o Ferma s heorem in calculus, which saes ha a any poin, where he differenial funcion aains a local exremum, is derivaive is zero. Remark. Theorem (1.1) doesn prove he exisence of he minimizer. However, under some assumpions, we can ensure he exisence. According o he heory of calculus of variaion, he minimizer exiss, if he Lagrangian L is coercive and convex in q. More abou his fac can be found in [11], Chaper

9 1.2 Hamilon s ODE As he second sep of he consrucion, we conver he sysem of Euler-Lagrange equaions ino Hamilon s equaions. We assume ha x is a C 2 funcion, ha is a criical poin of he acion funcional. We define firs he generalized momenum as a funcion p : [0,] R d, ha will serve as he second parameer for he Hamilonian formulaion of mechanics: p(s) := D q L(ẋ(s),x(s)). (1.10) Definiion 1.3. Assume, ha for all x, p R d he following equaion: p = D q L(q, x) (1.11) can be uniquely solved for q as a smooh funcion of p, x, q = q(p, x). The Hamilonian, associaedwihhelagrangianl is he funcion H : R d R d R, defined as: H(p, x) := p q(p, x) L(q(p, x),x), (1.12) where q is defined implicily by (1.11) Example 1.1. Remember wha we old before, ha for a moving paricle Lagrangian is L = T V,or,aferexpanding: L(q, x) = 1 2 m q 2 V (x). Now, since we are using he Caresian coordinaes, we know, ha in his case he momenum is p = mq, sowecancalculaehehamilonianbyheabovedefiniion: H(p, x) = 1 m p p L(q, x) = 1 m p 2 1 2m p 2 + V (x) = 1 2m p 2 + V (x) The firs erm in he sum is jus he kineic energy, wrien via hemomenum. So in his case he Hamilonian equals he oal energy of he paricle. However, his is no always he case, and i happens only if he Caresian coordinaes can be wrien in erms of he generalized coordinaes wihou ime dependence and vice versa. Oherwise he expression ges more complicaed. Now ha we ve defined he Hamilonian and generalized momenum in erms of Lagrangian and generalized velociy, we are ready o derive he Hamilon s ODEs. 7

10 Theorem 1.2. The funcions x and p (defined as in (1.10)), saisfy Hamilon s ODE sysem: { ṗ(s) = D x H(p(s),x(s)) (1.13) ẋ(s) =D p H(p(s),x(s)) for 0 s. Moreover,hemappings H(p(s),x(s) is consan. This is he sysem of 2d firs-order differenial equaions. Proof. Remember, in he definiion (1.3), he generalized velociy q is uniquely defined by he equaion p = D q L(q(p, x),x). Then from (1.10) i follows ha ẋ(s) = q(p(s),x(s)). We calculae he corresponding gradiens of H from is definiion (1.12): D x H(p, x) =p D x q(p, x) D q L(q(p, x),x) D x q(p, x) D x L(q(p, x),x) =(p D q L(q(p, x),x)) D x q(p, x) D x L(q(p, x),x) (1.11) = D x L(q(p, x),x)= D x L(ẋ(s),x(s)) Now if we remember he Euler-Lagrange equaion (1.5) and he above, we ge: D x H(p(s),x(s)) = D x L(ẋ(s),x(s)) = = d ds D ql(ẋ(s),x(s)) = ṗ(s) Thus we obains he firs equaion from (1.13). The second one is obained in he same way: D p H(p, x) =q(p, x)+p D p q(p, x) D q L(q(p, x),x) D p q(p, x) = q(p, x)+(p D q L(q(p, x),x)) D p q(p, x)) (1.11) = q(p, x) =ẋ(s) Finally, using his, we can easily prove ha he mapping s H(p(s),x(s) is consan: d ds H(p(s),x(s)) = D ph(p(s),x(s)) ṗ(s)+d x H(p(s),x(s)) ẋ(s) = D p H(p(s),x(s)) ( D x H(p(s),x(s))) + D x H(p(s),x(s)) D p H(p(s),x(s)) =0 8

11 Unlike Lagrangian formulaion, he Hamilonian ODEs are coupled and he order of he equaions is one, no wo, hence mahemaically i can be easier o solve hem in some cases. Bu here s sill a quesion, where he Hamilonian formalism can be useful and why i was inroduced in he firs place? Le us lis some of he reasons: 1. Hamilonian is a physical quaniy, namely a oal energy of he sysem, as described in he example (1.1) above, and herefore all pars of Hamilon s equaions have a physical inerpreaion. 2. Geomerically, Hamilon s equaions say, ha flowing in ime is equivalen o flowing along a vecor field in phase space. Tha gives a geomerical picure of ime evoluions in dynamic Hamilonian sysems. 3. In Hamilonian mechanics one can use canonical ransformaions which allow o change coordinaes in phase space which can make i easier o solve problem. (This will be he opic of he nex chaper). 4. And probably he mos imporan feaure of Hamilon s equaions is ha i gives he easy way o quanize classical sysems, hence i s possible o use in quanum mechanics. Moreover, i can be used in hermodynamics, saisical physics and many oher branches, where Lagrangian formalism fails or becomes incomprehensible. Laer we ll show, how one can derive he Schrödinger equaion direcly from he Hamilonian formalism. 1.3 Canonical Transformaions As we showed above, he Hamilonian mechanics uses 2d parameers o describe he sysem of moving paricles. Someimes, however, i makes sense o perform a change of variables o use differen 2d parameers. This change in is mos general form is described mahemaically in he following way: (p(),x()) (P (p, x, ),X(p, x, )). (1.14) There are several requiremens we wan o pu on his ransformaion. Firs, i should be inverible, so ha we can obain he soluion o he original Hamilon s equaions (p(),x()), once we know he soluion (P (),X()) o he ransformed equaion. Also, we wan o have he equaions for new parameers o be similar o he original Hamilon s equaions. Definiion 1.4. Amapg : R d R d R R d R d is called a canonical ransformaion, ifipreserveshecanonicalformofhehamilon sequaions, i.e. if g =(P, X), 9

12 where P = P (p, x, ) and X = X(p, x, ), andhehamilonianinnewcoordinaes is denoed as K(P, X) =H(p(P, X),x(P, X)), henhehamilon sequaionsinnew coordinaes have he following form: { P () = D X K(P (),X()) (1.15) Ẋ() =D P K(P (),X()) Now le us remind ourselves abou Hamilon s principle. The pah x() minimizes he acion funcional, hus solving he variaional problem (1.4). In erms of calculus of variaion, i means, ha x() is a saionary poin of he acion funcional, meaning ha δi[x( )] = δ 0 L(ẋ(s),x(s))ds =0, where δ denoes he funcional derivaive. In physics his version is ofen referred o as he Hamilon s principle in configuraion space. We wan o rewrie i via he Hamilonian. Le p(s) =D q L(ẋ(s),x(s)). Then from he definiion 1.3 i follows, ha H(p(s),x(s)) = p(s) ẋ(s) L(ẋ(s),x(s)). We can rewrie he variaional relaion above now: δi[x( )] = δ 0 [p(s) ẋ(s) H(p(s),x(s))]ds =0. (1.16) This version is called he Hamilon s principle in phase space. Nex we wan o apply he canonical ransformaion and expec he Hamilon s principle o hold in he new coordinaes, i.e. here exiss some pah X(s) in new coordinaes, such ha δi[x( )] = δ 0 [P (s) Ẋ(s) K(P (s),x(s))]ds =0, (1.17) where K(P, X) is he Hamilonian in new coordinaes. The expressions under he inegral inside (1.16) and (1.17) need no be equal, however, due o he fac, ha we have fixed endpoin values for x(),x(0) and hence p(),p(0), so here s a zero variaion a endpoins, we mus have he following relaion beween he inegrands in old and new coordinaes: α(p(s) ẋ(s) H(p(s),x(s)) = P (s) Ẋ(s) K(P (s),x(s)) + du d =s, (1.18) where u is some good enough funcion wih zero variaion on endpoins (his will ensure he simulaneous saisfacion of boh (1.16) and (1.17)) and α is a scaling consan. From now on we consider α = 1, as i s no relevaion for our discussion. This funcion F is called ageneraingfuncion, depending on he variables, on which i depends. Goldsein in [14] inroduces four generic ypes of generaing funcions: 10

13 1. u 1 := u 1 (x, X, ). 2. u 2 := u 2 (P, x, ) X P. 3. u 3 := u 3 (p, X, )+x p. 4. u 4 := u 4 (p, P, )+x p X P. We are ineresed in ransformaion laws ha appear in case of he generaing funcion ype-2. Le u 2 = u(p, x, ) X P.Weremark,haheerm X P is added o ge rid of Ẋ in (1.17). Le us see wha happens (we ll omi he parameer s for he sake of clearer calculaions): d(u(p, x, ) X P ) p ẋ H(p, x) =P Ẋ K(P, X)+ d =s = P Ẋ K(P, X)+u + D P u P + d x u ẋ P Ẋ P X = K(P, X)+u + D P u P + D x u ẋ X P. Or, rewriing i: (p D x u) ẋ +(X D P u) P = H(p, x) K(P, X)+u. (1.19) Separaely he new and old coordinaes are independen of each oher, hen he whole equaion can be idenically rue, if he corresponding coefficiens for ẋ and P vanish, i.e. if he following is rue: p(s) =D x u(p (s),x(s),s) X(s) =D P u(p (s),x(s),s). And ha means, ha he res of he equaion should also be equal o zero, hus giving us he formula for he Hamilonian in he new coordinaes and hus proving he following heorem. Theorem 1.3. If (p, x, ) (P (p, x, ),X(p, x, ),) is a canonical ransformaion, generaed by a funcion u, henhehamilonianinhenewcoordinaesis: K(P, X, ) =H(p, x)+ u(p, x, ). (1.20) Generally he process of applying he ransformaion is defined by several seps: 1. We sar wih he funcion u : R d R d R R, u := u(p, x, ). 2. We define he coordinaes p = D x u, X = D P u. 3. Then (p(),x()) (P (),X()) is a canonical ransformaion. 11

14 1.4 Hamilon-Jacobi Equaion and Is Properies Now we are searching for a specific canonical ransformaion, ha will make he resuling Hamilonian K o be 0. Then he Hamilon s equaions in he new coordinaes will be rivial: { P () =0 (1.21) Ẋ() =0 So we are looking for a generaing funcion u(p, x, ), ha will produce such a ransformaion. By heorem (1.3), K(P, X, ) =H(p, x)+ u(p, x, ), and by he definiion of a ype-2 generaing funcion, p = D x u. So, in order for he new Hamilonian K o be zero, we mus have: u(p, x, )+H(D x u(p, x, ),x)=0. However, P doesn play any mahemaical role in his equaion, so for furher invesigaion, we assume ha u has no explici dependence on P and he equaion becomes he Hamilon-Jacobi equaion: u(x, )+H(D x u(x, ),x)=0. (1.22) The funcion u, as a soluion o he Hamilon-Jacobi equaion, is also called he Hamilon principle funcion. We ll show, ha i s acually closely relaed o he acion inegral. Since u is a generaing funcion, hen D x u = p. Also ẋ = q. Using hese facs and he Hamilon-Jacobi equaion, we can calculae he ime derivaive of u: du d = D xu dx d + u = p ẋ + u = p ẋ H(p, x) =p q H(p, x) =L(p, q) (1.23) So acually, u is a classical acion plus some consan Mehod of Characerisics There s anoher imporan connecion of Hamilon-Jacobi equaion and Hamilon s ODEs: apparenly Hamilon s ODEs are characerisic equaions of Hamilon-Jacobi PDE. We ll prove his fac laer in his chaper. Le us firs recall he mehodof characerisics. Assume, we have a nonlinear firs-order PDE wih boundary condiions in he se Ω R d : { F (Du, u, x) = 0 in Ω, (1.24) u = g on Γ Ω. 12

15 We assume, ha F = F (p, z, x) andg are given smooh funcions. The idea of he mehod of characerisics if he following: we assume, ha u solves he equaion (1.24) and we fix x Ωandx 0 Γ. We know he value of u in x 0 because of he boundary condiion and hen wan o find he value u(x) by finding a curve inside Ω, ha connecs x and x 0 and along which we can calculae he value of u. The curve mahemaically is described as x(s) =(x 0 (s),..., x d (s)). We define z(s) = u(x(s)) and p(s) = Du(x(s)). Le s firs differeniae: ṗ i (s) = n u xi x j ẋ j (s). (1.25) j=1 Second, we differeniae he original equaion (1.24) wih respec o x i : n j=1 Now we calculae his expression a x = x(s): F u xj x p i + F j z u x i + F = 0 (1.26) x i n F (p(s),z(s),x(s))u xj x p i (x(s)) j j=1 = F z (p(s),z(s),x(s))u x i (x(s) F x i (p(s),z(s),x(s)). (1.27) We wan o ge rid of he second derivaives in (1.25), so we require: Now we can pass hese resuls back o (1.25): ẋ j (s) = F p j (p(s),z(s),x(s)). (1.28) ṗ i (s) = F z (p(s),z(s),x(s))p i(s) F x i (p(s),z(s),x(s)) (1.29) Finally, we calculae he derivaive of z(s): ż(s) = n j=1 u x j (x(s))ẋ j (s) = n j=1 ṗ j (s) F p j (p(s),z(s),x(s)) = D p F (p(s),z(s),x(s)) p(s). (1.30) 13

16 Equaions (1.28), (1.29) and (1.30) are he characerisic lines we were looking for. Le s rewrie hem in he vecor noaion: ṗ(s) = D z F (p(s),z(s),x(s))p(s) D x F (p(s),z(s),x(s)) ż(s) =D p F (p(s),z(s),x(s)) p(s) (1.31) ẋ(s) =D p F (p(s),z(s),x(s)). I should be noed, ha for hese equaions o be used, appropriae iniial condiions should be found. More abou i can be found in Chaper 3.2 in [11]. We, however, reurn o he Hamilon-Jacobi PDE. Theorem 1.4. Hamilon s equaions are characerisic equaions of Hamilon-Jacobi PDE. Proof. In he noaion inroduced above, we have he following siuaion: F ((D x u, u ),u,(x, )) = u + H(D x u, x) =0. Here we have F (P, z, X) =q + H(p, x), where P =(p, q) andx =(x, ). Now we calculae: D P F =(D p H, 1). D P F P =(D p H, 1) (p, q) =D p H p + q = D p H p H(p, x). D z F =0. D X F =(D x H, 0). Now we can calculae hese derivaives a (y(s),z(s),y(s)) and use (1.31) and we ge he following sysem: P (s) = D x H(p(s),x(s)) ż(s) =D p H(p(s),x(s)) H(p(s),x(s)) Ẋ(s) =D p H(p(s),x(s)). The equaions for x(s) andp(s) are exacly he Hamilon s ODEs, as we have derived hem from he Lagrangian mechanics. Noice also, ha he equaion forz(s) is rivial, once we find x(s) andp(s), because he funcion H doesn direcly depend on z, so we can immediaely inegrae he equaion for z(s). 14

17 1.5 Applicaions In his chaper we ll show some useful examples, ha shows he relaions of he Hamilon-Jacobi equaion o such fields of physics as opics and quanum mechanics Eikonal Equaion If we consider he saic process, hen he Hamilon-Jacobi equaion urns o he following form: H(D x u(x),x)=0. (1.32) One example of such a saic Hamilon-Jacobi equaion arises in geomerical opics and is called he eikonal equaion. Since here s no ime dependency, he equaion is usually considered in he bounded domain Ω R d and supplied wih he boundary condiions: { D x u(x) = n(x) in Ω (1.33) u(x) =0on Ω This equaion can be derived from he Maxwell s equaions of elecromagneism and i provides a link beween wave opics and ray opics. Or i can also be derived from he Ferma s principle of leas ime, which says ha ligh ravels beween wo poins along he pah ha requires he leas ime, as compared o oher nearby pahs. Physically, he funcion u here describes he shores ime, needed for ligh o ravel from he boundary Ω o he poin x Ω, and n(x) is he refracive index of a medium a poin x Schrödiner Equaion We have shown previously, ha u is a classical acion. If we know u, henaany ime we can deermine he isosurfaces 2 u(x 0,). The moion of his isosurface is deermined by he moions of he paricles, belonging o he sysem, ha sar a a poin x 0. According o he Feynman pah inegral formulaion of quanum mechanics, here re wo posulaes: 1. The probabiliy ha a paricle has a pah lying in a cerain region of spaceime is he absolue square of a sum of conribuions, one from each pah in his region. 2. The pahs conribue equally in magniude, bu he phase of he conribuions is he classical acion. 2 An isosurface represens poins of consan acion wihin some volume of space. 15

18 Mahemaically, his means, ha he ampliude for a given pah x(s) is Ae iu(x,), where A is some real funcion and u is a classical acion, performed by he sysem beween imes 1 and 2. And he oal ampliude of a sysem is K 21 = n K n 21 = n A n 21 e iun (1.34),whereA n 21 and u n correspond o he pah x n (). This funcion K is called a spaceime propagaor and i s relaed o he quanum-mechanical wavefuncion. Now, o esablish he connecion beween Hamilon-Jacobi equaion and Schrödinger equaion, we ll consider he following simplified wavefuncion: ψ(x, ) := e iu(x,) (1.35) is a Planck consan. This funcion has all he properies of wave funcion of wave mechanics. Invering his, we ge: u(x, ) = i ln ψ(x, ). (1.36) Now consider he Hamilonian from he example (1.1) H(p, x) = 1 2m p 2 + V (x). We wan wrie a Hamilon-Jacobi equaion for his H and u in erms of he wave funcion ψ. Now, le s derive some relaions for his: u x i = i ψ ψ ψ = iψ x i x i Now, as we assume, ha u is a generaing funcion ype-2 for he canonical ransformaion, hence D x u = p, andhap is a momenum, so p = mq = mx and hus: 2 u = p i = d x i =0. 2 x i x i d x i Now, we can calculae: Which is equivalen o: 2 ψ = i ψ u + iψ 2 x i x i x i u x i 2 u = ψ ( ) 2 u 2 x i 2 x i H(D x u, ) = 1 2m D x 2 + V (x) = 2 2mψ D2 xψ + V (x). (1.37) 16

19 Finally, he ime derivaive is: u = i ψ ψ And now we can rewrie he Hamilon-Jacobi equaion in erms of ψ: 2 2m D2 xψ + V (x)ψ i ψ = 0 (1.38) This is he ime-dependen non-relaivisic Schrödiner equaion, which is a fundamendal mahemaical descripion of behavior of quanum sysems. Alhough we gave only he brief physical moivaion of our guess, ha u can be views as an ampliude of he wave, he derivaion above suggess a deep link beween Hamilonian formulaion of classical mechanics and quanum mechanics. More abou his connecion can be found in [12]. 2 Tradiional Approach Now ha we have esablished all he connecions beween he Hamilon-Jacobi PDE, Hamilon s ODEs and calculus of variaion problems, we can move on o finding he soluion o he problem (1.1). We ll demosrae how o solve he HJE by separaion of variables and hen show ha in classical sense, he soluion may be found only in some resriced cases, and his soluion migh no even be unique. 2.1 Separaion Of Variables In cerain cases we can be lucky and he Hamilon-Jacobi equaion can be easily solved using he separaion of variables mehod in an addiive way. Noe, ha in his case, solving his equaion is acually easier han using he Hamilon s mehod direcly. Assume, ha he Hamilonian has he following dependence on u and p: So he HJE has he following form: H = H(f 1 (x 1,p 1 ),..., f d (x d,p d )) u + H(f 1 (x 1, u x 1 ),..., f d (x d, u x d )) = 0 (2.1) Now we separae he firs variable and look for he soluion u in he following form: u = u 1 (x 1 )+u (x 2,..., x d,) (2.2) 17

20 Passing i back o (2.1), we ge he following equaion: u + H(f 1 (x 1, u 1 x 1 ),..., f d (x d, u x d )) = 0 (2.3) Assume now, ha we have already found a soluion u in he form (2.2). Then (2.3) becomes an ideniy for any value of x 1.Buwhenwechangex 1, only he funcion f 1 is affeced. Hence i mus be consan. Thus, we obain wo equaions: f 1 (x 1, u 1 x 1 )=α 1, (2.4) u + H(α 1,f 2 (x 2, u )..., f d (x d, u )) = 0 (2.5) x 2 x d α 1 is an arbirary consan. Assume also, ha f 1 is inverible in he second variable in a sense, ha here exiss g(x 1,y 1 ), such ha for each (x 1,p 1 )andeachy 1 R, f(x 1,g(x 1,y 1 )) = y 1. For our case his means, ha u 1 x 1 = g(x 1,α 1 ). Then afer simple inegraion we find ou u 1 : u 1 (x 1 )= g 1 (x 1,α 1 )dq 1 + C 1, (2.6) where C 1 is he inegraion consan. Now we make he similar separaion seps consecuively for oher variables, hence he soluion has he following form: u = u 1 (x 1 )+u 2 (x 2 ) u d (x d )+u 0 (), (2.7) where each u k is calculaed via he inverse funcion of f k,denoedasg k : u k (x k )= g k (x k,α 1,..., α k )dq k + C k, (2.8) Finally, he equaion (2.1) akes he simple form: u 0 + H(α 1,...α d ) = 0 (2.9) Afer inegraing by and combining all inegraion consans ino one, we have he complee inegral of he equaion u: d u = H(α 1,...α d ) + g k (x k,α 1,..., α k )dq k + C, (2.10) k=1 The separabiliy of he equaion depends no only on he Hamilonian iself, bu on he generalized coordinaes oo. 18

21 Example 2.1. We ll demonsrae his mehod on he very rivial Hamilon-Jacobi PDE for he moion of a paricles on a line (one dimension). u +(u x ) 2 =0. (2.11) Then, following he seps of he mehod, we have ha H(x, p) =f(x, p) =p 2,hence g(x, α) = α.now,wearelookingforasoluioninaformu(x, ) =u 0 ()+u 1 (x). We can calculae u 1 (x) by (2.6): and ge he final soluion by (2.10): u 1 (x) = αx + C, u(x, ) = α + αx + C. To specify α and C, weneedosupplyheiniialcondiions. Forexample,ifhe iniial condiion is u(x, 0) = 0, henobviouslyα =0and hence also C =0. Furher, more complicaed examples of he usages of his mehod can be found in 48 in [20]. 2.2 Legendre Transform Before we move on o sudying oher mehods for solving he HJE, we inroduce anoher imporan noion, connecing Lagrangian and Hamilonian, namely he Legendre ransform. Firs, we inroduce he general noion of wha i is, and hen we proceed wih is relaions o he classical mechanics. Definiion 2.1. Le f : R d R, f = f(q) be a convex funcion. Is Legendre ransform is he following funcion: The funcion f is called he convex conjugae of f. f (p) := sup q R d {p q f(q)}. (2.12) Remark. Hisorically, he Legendre ransfrom is defined as above, for convex funcion only, however i can be generalized o non-convex funcions, and his generalizaion is called he Legendre-Fenchel ransformaion. The convexiy of he funcions, o which he Legendre ransform is applied, is imporan due o he propery, called convex dualiy, whichsimplypu,means,ha f = f, 19

22 We ll prove his fac laer in his chaper for he case of Lagrangian and Hamilonian. Moreover, his equaliy above is in fac, a charaerizaion of a convex lower semiconinuous funcion. This fac is know as a Fenchel-Moreau heorem. In case of f being non-convex, he biconjugae (he ransformaion applied wice) will no be he same as he original funcion, bu i will be he largeslowersemiconinuous convex funcion wih f = f, whichiscalledhe closed convex hull. Remark. There is a good geomerical inerpreaion of wha a Legendre ransform is. Assume ha f = f(q) is a sricly convex smooh funcion. Consider a poin q 0 and a supporing plane o he graph of he funcion f a his poin. I will have he form p q + b and has o saisfy wo condiions: f(q 0 )=p q 0 + b, p = D q f(q 0 ). For sricly convex funcion, each componen of he gradien is sricly monoone, so he second equaion can be uniquely solved for q 0 = q 0 (p), whichgivesusawayo calculae b as a funcion of p: b = f(q 0 (p)) p q 0 (p) = f (p). Therefore, a family of supporing planes o he graph of f can be given by: y = p q f (p) So he Legendre ransform maps he graph of he funcion o he familyofheangen planes of his graph. To simplify he discussion, we assume, ha he Lagrangian L and is corresponding Hamilonian H don explicily depend on x, so we wrie L = L(q) andh = H(p). From now on in his chaper we also assume ha Lagrangian L is convex and saisfies he following condiion: L(q) lim =+ (2.13) q q This condiion is called superlineariy. Firs we ll show ha he Legendre ransform of such a Lagrangian is acually a corresponding Hamilonian. Theorem 2.1. Hamilonian H, associaed wih he Lagrangian L is is Legendre ransform, or: H(p) =L (p) (2.14) 20

23 Proof. Remember he definiion 1.3 of he Hamilonian. In he view of condiion (2.13), for each p here exiss a poin q(p) R d, such ha he mapping q p q L(q) has a maximum a q(p) and: L (p) =p q(p) L(q(p)). (2.15) Now i follows ha D q L(q(p)) = p, which is a solvable for q(p). follows, ha (2.15) gives us he definiion of he Hamilonian. I immediaely Now we urn o some properies of he Legendre ransform, ha make i so valuable o he discussion. Theorem 2.2. Assume, he convex Lagrangian L saisfies he condiion (2.13). Define H = L.Thenhefollowingfacsarerue: 1. H is a convex map, 2. lim p H(p) p 3. L = H. Proof. =+, These properies are called he convex dualiy of Hamilonian and Lagrangian. 1. Take τ [0, 1) and p 1,p 2 R d.wehave: H(τp 1 +(1 τ)p 2 )=L (τp 1 +(1 τ)p 2 )=sup{(τp 1 +(1 τ)p 2 ) q L(q)} q τ sup q {p 1 q L(q)} +(1 τ)sup{p 2 q L(q)} = τh(p 1 )+(1 τ)h(p 2 ) Hence H is a convex map by definiion. 2. Take p 0andλ>0. For esimae below we ll jus use q 0 = λ p p.wehave: H(p) =sup{p q L(q)} q λ p L(λ p p ) λ p max L(q). q B(0,λ) Now he laer erm is fixed, so i s no affeced by he change of p, sowecan deduce ha: H(p) lim inf λ. p p Since λ was an arbirary consan, we proved he second propery. 21 q

24 3. Since H = L, hen from he definiion of he Legendre ransform i follows, ha for all p, q R d : H(p)+L(q) p q. In paricular, he following is rue: On he oher hand: L(q) sup{p q H(p)} = H (q). p H (q) =L (q) =sup{p q L (p)} =sup{p q sup{p r L(r)}} p p r =supinf {p (q r)+l(r)}} (2.16) p r Since L is a convex funcion, hen by supporing hyperplane heorem (Theorem 1 in Appendix B.1 in [11]), here exiss s R d,suchha: L(r) L(q)+s (r q) So we can choose p = s in (2.16), and hen we ge: H (q) inf{s (q r)+l(r)} L(q) (2.17) r 2.3 Classical Hopf-Lax Formula Now we ry o solve he simplified iniial value problem (1.1), where he Hamilonian H(p) is convex and saisfies he condiion lim p = +. The problem will have he p following form: { u + H(D x u)=0 (2.18) u(x, 0) = g. We suppose ha he funcion g is Lipschiz coninuous. Remember, ha Hamilon-Jacobi equaion has characerisic lines, which are exacly he Hamilon s ODEs. Furhermore, he Hamilon s ODEs are direcly relaed o he variaional problem (1.4). Nex we ll ry o build a connecion beween calculus of variaion and he Hamilon-Jacobi equaion bypassing he characerisic equaions. Given x R d and >0, we ry o minimize he classical acion: 0 L(ẇ(s))ds 22

25 over all w : [0,] R d,suchha,w() =x. We also need o ensure ha he iniial condiion is ake ino accoun, so we ll add a erm g(w(0)). So, we have he following variaional problem: { } u(x, ) := inf L(ẇ(s))ds + g(w(0)) w() = x (2.19) 0 We emphasize again, ha his seemingly random guess comes from he relaions of he Hamilon s equaions o he Lagrange s variaional problem and he fac ha he Hamilon s equaions are in fac he characerisic equaions of he HJE. Nex we simplify he expression (2.19). Take y R d and define w(s) := y + s (x y). Then by definiion of u, wehave ( ) x y u(x, ) L(ẇ(s))ds + g(y) =L + g(y), and hence also 0 ( ) x y u(x, ) inf {L + g(y)}. y R d Nex, if w is a C 1 funcion, he Jensen s inequaliy yields, ha: ( 1 ) L ẇ(s)ds 1 L(ẇ(s)ds. If we again wrie y = w(0), we obain: ( ) x y L + g(y) and hence also: We have jus shown, ha: 0 { ( ) x y inf L y R d u(x, ) = inf y R d { L 0 0 L(ẇ(s))ds + g(y), } + g(y) u(x, ). ( ) x y } + g(y). (2.20) This looks already beer, han he original form of u(x, ). Noice now, ha he funcion ψ(y) =L ( ) x y + g(y) is coninuous, due o he coninuiy of boh componens by assumpion. We wan o show, ha he infimum in he above formula is in 23

26 fac a minimum. Laer we denoe by Lip(g) he Lipschiz consan of g, which is { g(x) g(y) Lip(g) = x y sup x,y R d,x y }. (2.21) In he case of Lipschiz funcions Lip(g) <. We make an esimae, using he Lipschiz coninuiy: g(y) g(x) Lip(g) y x 0. Then we add his erm o he definiion of ψ o esimae i from below: ( ) x y ψ(y) L + g(y) + g(y) g(x) Lip(g) y x ( ) x y L + g(y)+ g(y) g(x) Lip(g) y x ( ) x y L g(x) Lip(g) y x ( ( L x y ) ) = x y g(x) x y Lip(g) x y We claim, ha he RHS of his inequaliy ends o +, asy.indeed, L ( ) x y y +, x y by our assumpion (2.13). Obviously, g(x) 0as y.thaproves,ha: x y lim ψ(y) =+, (2.22) y which means by definiion, ha for any x, here exiss R>0, such ha if y / B(0,R)(by B(0,R) we denoe a closed ball of radius R), hen ψ(y) >u(x, )+ϵ for some fixed ϵ>0. Bu by exreme value heorem inside his closed ball ψ aains is minimum a some poin y 0 B(0,R). By his consrucion, inf y R ψ(y) ψ(y 0 ). Also by definiion of he infimum, here exiss a poin y B(0,R), such ha ψ(y ) <u(x, ) + ϵ. From which i follows, 2 ha: inf ψ(y) >u(x, )+ϵ>u(x, )+ ϵ y >R 2 >ψ(y ) ψ(y 0 ). 24

27 All of he above means, ha u(x, ) = inf y R d ψ(y) ψ(y 0 ). I immediaely follows from he definiion of he infimum, ha: { ( ) } x y u(x, ) = min L + g(y). (2.23) y R d We have jus proven he following heorem, which originally was consruced by Hopf in [17] and Lax in [21]: Theorem 2.3 (Hopf-Lax). If x R d and >0, henhesoluionu = u(x, ) of he minimizaion problem (2.19) has he form (2.23), which is called he Hopf-Lax formula. Remark. Acually, he Hopf-Lax formula can be moivaed in a differen way. Noice, ha for any y, z R d aken as a parameer, he funcion F (x, ) =(x y) z H(z)+g(y) solves he Hamilon-Jacobi equaion from he iniial value problem (2.18). Indeed, D x F (x, ) =z, andhencef (x, ) = H(z) = H(D x F (x, )), whichproveshis claim. Then he Hopf-Lax formula can be obained from his F (x, ) by he wo seps envelope soluion: u(x, ) = inf sup{(x y) z H(z)+g(y)} y R d z R { d ( ) } x y = inf H + g(y). y R d Which is exacly he Hopf-Lax formula. The fac ha inf in his version is min can be proved in he same way as in he reasoning above. Moreover, Hopf in [17] consruced a second formula, by swiching he inf and sup in he above envelope formula, afer which we can wrie a Legendre ransform of g: u(x, ) =sup inf {(x y) z H(z)+g(y)} y R d z R d =sup z R d {x z g (z) H(z)}. Before showing, how he Hopf-Lax formula can be used o define some reasonable soluion o he Hamilon-Jacobi equaion, le us firs show some of he useful properies of he funcion u, defined by he Hopf-Lax formula. 25

28 Lemma 2.1. The funcion u, definedbyhopf-laxformula(2.23),whereg is Lipschiz coninuous, has he following properies: 1. u is Lipschiz coninuous. 2. u = g on R d { =0}. 3. For each x R d and 0 s<we have he following funcional ideniy for u(x, ), definedby(2.23): { ( ) } x y u(x, ) = min ( s)l + u(y, s) (2.24) y R d s Proof. 1. Fix >0, x, ˆx R d and choose y R d,suchha ( ) x y u(x, ) =L + g(y). Then { ( ) } ( ) ˆx z x y u(ˆx, ) u(x, ) = inf L + g(z) L g(y) z R d = [le z =ˆx x + y for he expression inside he inf] ( ) ( ) x y x y L + g(ˆx x + y) L g(y) g(ˆx x + y) g(y) Lip(g) ˆx x. The similar resul can be archieved by inerchanging roles of x and ˆx in he above argumen, from where i follows ha: u(ˆx, ) u(x, ) Lip(g) ˆx x. (2.25) This shows he Lipschiz coninuiy of x u(x, ). 2. Le 0 <s<. The formula (2.24), ha we need o prove, means is ha once we know he value of u(x, s), we can us i as a iniial condiion for calculaing u(x, ). Fix y R d and 0 <s<and choose z R d,suchha ( ) y z u(y, s) =sl + g(z). s 26

29 Noice, ha: x z = I follows by convexiy of L, ha: ( ) x z L ( 1 s ) x y s + s ( 1 s ) L y z. s ( ) x y + s ( ) y z s L. s Therefore ( ) ( ) ( ) x z x y y z u(x, ) L + g(z) ( s)l + sl + g(z) s s ( ) x y =( s)l + u(y, s). s This is rue for all y R d. We have already shown he Lipschiz coninuiy, and hence he coninuiy of a map y u(y, ). So i follows, ha: { ( ) } x y u(x, ) min ( s)l + u(y, s) y R d s Now we can choose w, such,ha ( ) x w u(x, ) =L + g(w) (2.26) and se y = sx +(1 s )w. Then x y s = x w = y w. s And consequenly, using he convexiy of L again: ( ) ( ) ( ) x y x w y w ( s)l + u(y, s) ( s)l + sl + g(w) s s ( ) x w = L + g(w) =u(x, ). This ends he proof of he ideniy (2.24). 27

30 3. We now need o show ha here is also Lipschiz coninuiy wih respec o he variable. Le us again fix >0andx R d and choose y = x in he Hopf-Lax formula (2.23). Then u(x, ) L(0) + g(x). (2.27) Moreover, using he Lipschiz condiion on g, we can esimae, g(x) g(y) Lip(g) x y g(y) g(x) Lip(g) x y. u(x, ) = min y R d { L g(x) + min y R d ( ) x y { L } + g(y) ) ( x y } Lip(g) x y [now se z = x y ] = g(x) + min {L(z) Lip(g) z } z R d [nex we use min{f(x)} = max{ f(x)} = g(x) max {Lip(g) z L(z)} z R d = g(x) max max{w z L(z)} w B(0,Lip(g)) z R d = g(x) max w B(0,Lip(g)) L (w) =g(x) From his resul and (2.27) i follows, ha where max H(w) w B(0,Lip(g)) u(x, ) g(x) C, (2.28) C := max{ L(0), max H(w)}. w B(0,Lip(g)) This inequaliy immediae yields he iniial condiion for u. If = 0,hen obviously u(x, 0) = g(x). Finally, by using he (2.24) and he fac, ha Lip(u(,) Lip(g), proven above, we can repea he above calculaions. Choose 0 <s<and x = y in (2.24). Then: u(x, ) ( s)l(0) + u(x, s). 28

31 And, similarly o when s=0: u(x, ) u(x, s) + min y R d = u(x, s) C( s) { ( s)l This proves he Lipschiz coninuiy of u(,). ( ) } x y Lip(u(,)) x y s Theorem 2.4. Le u be defined by Hopf-Lax formula (2.23). Then u is differeniable a.e. in R d (0, ) and solves he iniial value problem (2.18) in he poins of differeniabiliy. Proof. The differeniabiliy a.e. follows immediaely from he Lipschiz coninuiy by he Radamacher heorem. Now le (x, ), where x R d, >0, be a poin, where u is differeniable. Fix q R d, h>0. By (2.24): hence u(x + hq, + h) = min y R d Noice, ha u(x + hq, + h) u(x, ) h { hl hl(q)+u(x, ). Combined wih (2.29) we obain: ( ) x + hq y h u(x + hq, + h) u(x, ) h } + u(y, ) u(x + hq, + h) u(x + hq, ) = + h q D x u(x, )+u (x, ) ash 0 + Since q is arbirary, hen, in pariculal, we have: [inside min le y = x] L(q). (2.29) u(x + hq, ) u(x, ) h q D x u(x, )+u (x, ) L(q). (2.30) 0 u (x, )+max q R d {q D x u(x, ) L(q)} = u (x, )+L (D x u(x, )). (2.31) To ge he reversed inequaliy, we sar by choosing z, suchha: ( ) x z u(x, ) =L + g(z) 29

32 and fix h>0. We se s = h and y = s x +(1 s x z )z. Then ( ) ( ( ) x z y z u(x, ) u(y, s) L + g(z) L s ( ) x z =( s)l. From where i follows, ha: u(x, ) u(x h x z, h) h ( x z L By leing h 0 +, we obain: ( ) x z x z D x u(x, )+u (x, ) L ). = y z.therefore s ) + g(z) Finally 0 u (x, )+ x z ( ) x z D x u(x, ) L u (x, )+max q R d {q D x u(x, ) L(q)} = u (x, )+L (D x u(x, )). Taking ino accoun he fac ha H = L, his ends he proof. One may conclude from his, ha he Lipschiz coninuous funcion, ha solves he HJE a.e. and agrees wih g, when = 0, migh be a good definiion for he weak soluion of (2.18), bu unforunaely, such a soluion is no unique, ha can be illusraed by he following example: Example 2.2. Le s look a he eikonal equaion in one dimension: { u (x) =1on [ 1, 1] u( 1) = u(1) = 0. (2.32) We can noice, ha he funcions u 1 (x) := 1 x and u 2 (x) := x 1 solve he equaion almos everywhere (excep for x =0,as he derivaive doesn exis here). Moreover, since we only wan a.e. soluion, we can noice, ha any funcion, ha has he piecewise propery u (x) =1and agrees wih he boundary condiions, will 30

33 solve he equaion. For example, he funcion, ha has he following graph, also solves he equaion a.e., along wih u 1 and u 2 : also solves he iniial value problem almos everywhere and i s Lipschiz coninuous. Moreover, here re infiniely many funcion, ha saisfy his problem almos everywhere. So as a conclusion, we need o pu more resricions on he funcions g and H, andrehinkourdefiniionofheweaksoluion. 2.4 Weak Soluions We ll sar wih looking ino some more properies of he Hopf-Lax formula, esablishing more connecions beween funcion u, defined by he formula and he given iniial value problem. Firs, we give a definiion of semiconcaviy. Definiion 2.2. Areal-valuedfunciong is called semiconcave, if here exiss a consan C, suchhaforallx, z R d,hefollowinginequaliyholds g(x + z) 2g(x)+g(x z) C z 2 (2.33) Remark. An exensive overview of he semiconcave funcions and is applicaions is given in [5]. The erm semi -concaviy comes from some of he properies of he funcions. Corollary in [5] proves, ha any semiconcave funcion can be represened as a sum of a smooh funcion and a concave funcion. Hence, o he surprise, a sricly convex funcion, as x 2 can be seen as an example of a semiconcave funcion, which is no concave. Lemma 2.2. Le g be semiconcave. Then u, definedby(2.23),isalsosemiconcave w.r.. x, meaninghaforallx, z R d,and>0: u(x + z, ) 2u(x, )+u(x z, ) C z 2. (2.34) Proof. Take y R d a minimizer in Hopf-Lax formula, i.e.: ( ) x y u(x, ) =L + g(y). 31

34 Now, we can esimae u(x+ z, ) and u(x z, ) by he Hopf-Lax formula from above, by choosing respecively y +z and y z as corresponding y-variable from he formula. I follows ha: ( ( ) ) x + z y z u(x + z, ) 2u(x, )+u(x z, ) L + g(y + z) ( ( ) ) x y 2 L + g(y) + ( L = g(y + z) 2g(y)+g(y z) C z 2. ( ) ) x z y + z + g(y z) The second sep will be o esablish he connecion beween H and u. Wedrop he semiconcaviy assumpion on g and consider now he uniformly convex H. I happens, ha his propery is enough o ensure he semiconcaviy of he soluion u for any fixed >0. Again, we firs give a definiion. Definiion 2.3. A C 2 funcion H : R d R is called srongly convex wih consan θ>0, ifhefollowinginequaliyisrueforallp, ξ R d : ξ T 2 H(p)ξ θ ξ 2. (2.35) Remark. This definiion is acually a special case of a uniformly convex funcion, which means ha here exiss a funcion φ, whichisincreasingandvanishesonlya zero, such ha: ξ T Dx 2 H(p) ξ φ( ξ ) (2.36) Obviously, he case of φ(α) =θα 2 gives he above definiion of a srongly convex funcion. Lemma 2.3. Le H be a srongly convex Hamilonian wih consan θ, L = H be he corresponding Lagrangian, and u is defined by he Hopf-Lax formula (2.23). Then u is semiconcave for any >0, inparicular,forallx, z R d : u(x + z, ) 2u(x, )+u(x z, ) 1 θ z 2. Proof. Le p 1,p 2 R d. Then, by Taylor s formula, we have he following ideniies: ( ) ( ) ( ) p1 + p 2 p1 + p 2 p1 p 2 H(p 1 )=H + D x H ( ) T ( ) T p1 p 2 Dx H(ˆp p1 p 2 1) 2 32

35 ( ) ( ) ( ) p1 + p 2 p1 + p 2 p2 p 1 H(p 2 )=H + D x H ( ) T ( ) T p2 p 1 Dx H(ˆp p2 p 1 2) 2 ˆp 1 and ˆp 2 are he mid-poins, corresponding o he remainder erm in Lagrange form. Now, summing i up and esimaing each of he remainder erms by (2.35), we obain: ( ) p1 + p 2 H(p 1 )+H(p 2 ) 2H + θ 2 4 p 1 p 2 2 (2.37) Nex, le q 1,q 2 R d be arbirary. We choose p 1,p 2 be such, ha H(p 1 )=p 1 q 1 L(q 1 ) and H(p 2 )=p 2 q 2 L(q 2 ). The exisence and uniqueness is ensured by he relaions beween Hamilonian and Lagrangian. Then we can esimae: L(q 1 )+L(q 2 )=p 1 q 1 + p 2 q 2 H(p 1 ) H(p 2 ( ) p1 + p 2 p 1 q 1 + p 2 q 2 2H θ 2 4 p 1 p 2 2 Now, by definiion of he Legendre ransform, we have ( ) p1 + p 2 H 1 ( ) 2 4 (p q1 + q p 2 ) (q 1 + q 2 ) L 2 Combining i wih he previous inequaliy, we obain: L(q 1 )+L(q 2 ) p 1 q 1 + p 2 q 2 1 ( ) 2 (p q1 + q p 2 ) (q 1 + q 2 )+2L θ 2 4 p 1 p ( ) 2 (p q1 + q 2 1 p 2 ) (q 1 q 2 )+2L θ 2 4 p 1 p 2 2 (2.38) Finally, we noice, ha: θ 4 p 1 p (p 1 p 2 ) (q 1 q 2 )+ 1 4θ q 1 q 2 2 ( ) θ θ 4 p 1 p p 1 p 2 = ( ) θ 2 p 1 p θ q 1 q ( ) 1 2 θ q 1 q θ q 1 q 2 2

36 This will help o ge rid of he p i from (2.38). We obain: ( ) q1 + q 2 L(q 1 )+L(q 2 ) 2L θ q 1 q 2 2 (2.39) Finally, we choose y o be a minimizer in he Hopf-Lax formula for u(x, ), and hen using he same value for esimaing u(x + z, ) andu(x z, ), we obain: ( ( ) ) x + z y u(x + z, ) 2u(x, )+u(x z, ) L + g(y) ( ( ) ) ( ( ) ) x y x z y 2 L + g(y) + L + g(y) ( ( ) ( ) ( )) x + z y x y x z y L 2L + L 1 2 2z 4θ = 1 θ z 2. Here we applied he inequaliy (2.39) here wih q 1 = x+z y and q 2 = x z y. Now we can combine he resuls of boh he lemmas o creae a suiable definiion of he weak soluion of he HJE iniial value problem (2.18). Definiion 2.4. ALipschizconinuousfuncionu : R d [0, ) R is called he weak soluion of he iniial value problem (2.18), provided: 1. u(x, 0) = g(x), whenx R d, 2. u (x, )+H(D x u(x, )) = 0 for a.e. (x, ) R d (0, ). 3. u(x + z, ) 2u(x, )+u(x z, ) C(1 + 1 ) z 2 for some consan C 0 and all x, z R d,>0. The previous lemmas (2.2) and (2.3) direcly lead us o he exisence heorem of he weak soluion, under some resricions on he funcion H or g. Theorem 2.5 (Hopf-Lax formula as a weak soluion). If H is a C 2 superlinear convex funcion, L = H be a corresponding Lagrangian and g is Lipschiz coninuous, hen u, definedbyhehopf-laxformula(2.23)isaweaksoluionofhe iniial value problem for he HJE (2.18), if eiher g is semiconcave or H is srongly convex. 34