Mah 527 Lecure 6: Hamilon-Jacobi Equaion: Explici Formulas Sep. 23, 2 Mehod of characerisics. We r o appl he mehod of characerisics o he Hamilon-Jacobi equaion: u +Hx, Du = in R n, u = g on R n =. 2 To avoid confusion, we use he following noaion: x, u z, x u p p =. 3 Du p Then we can re-wrie he equaion o where The characerisics ODEs hen are = x = D p F = ẋ D p H ż = D p F p = D p H ṗ ṗ Fx, z, p = 4 Fx, z, p 6 p + Hx, p. 5, 6 = p = D z F p D x F = p p = p + D p H p =D p H p Hp, x, 7. 8 D x H Now we r o solve he characerisic ODEs. Firs noice ha, since =, we can simpl use as he parameer insead of s. Thus he equaions become ẋ = D p H, 9 ż = D p H p Hx, p, N ṗ = D x H, p = p =. 2 I is clear he all we need o do is o solve he firs 3 equaions. Losing a bi rigor, we assume for now onl H is differeniable and sricl convex. We also assume H grows super-linearl a infini: Hx, p lim =+, 3 p ր p Now ake Legendre ransform: The z equaion hen becomes where q saisfies Therefore he soluion u is given b ux=ux + where x and x are relaed b Lx, v 6 sup p R n v p Hx, p. 4 ż = Lx, v 5 q =D p Hx, p. 6 Lxτ, vτ dτ. 7 ẋ = D p H =v, x =x. 8
2 Mah 527 Lecure 6: Hamilon-Jacobi Equaion: Explici Formulas To furher simplif he ssem, we noice ha implies which implies ha q, x minimizes wih x, x fixed. To see his, wrie Lx, v=v px, v Hx, px, v, and compue ẋ = D p H, ṗ = D x H 9 d d D vl+d x L = 2 Lxτ, vτ dτ 2 D v L= p + v D q p D p H D v p = p, 22 D x L=v D x p D x H D p H D x p = D x H, 23 where we have used v =D p H. Now he equaion ṗ = D x H gives wha we wan. Thus we see ha he Hamilon-Jacobi equaion can be solved as soon as we find ou he rajecories x and v. Below we will see ha in a special case, his can indeed be done in some sense. The Hopf-Lax formula. This special case is when H is independen of x, ha is H = HDu. The characerisic equaions can hen be furher simplified o ẋ = D p H, 24 ż = D p H p Hp=Lv, 25 ṗ = D x H =, 26 p n+ = p n+ N =. 27 We see ha p is a consan vecor along he characerisic curve, and as a consequence ẋ =D p H is a consan vecor, and herefore he characerisics x are sraigh lines. Furhermore we know ha he veloci q = ẋ is consan. Thus if x= and x=x, we mus have As a consequence d d z =Lv=L v =. 28 z=z+l = g+l. 29 Now he onl problem is ha is no known. Now hink of g as no merel an iniial funcion, bu as an inermediae record. In oher words, insead of saring a =, imagine our ssem sars from =, sa,. We consider all possible rajecories emanaing from some poin a =, passing a =, and finall reach ime a x. Think of g as he record of work done from = o =. Obviousl he correc rajecor should be he one ha is he minimizer among hem all. Remark. Noe ha he above explanaion means ha he rajecor ma no sa C as i crosses =. This should be expeced. Because in general he given g canno be he resul of a dnamical ssem wih H = HDu, ha is free paricle. The H producing g has o be dependen on x or even u. Thus here is no surprise ha his sudden change leads o a sudden change of direcion in he rajecor v. Following his idea, we reach he following Hopf-Lax formula: u, x=z= inf L + g. 3 R n Remark 2. I can be shown ha L grows superlinearl a infini. As a consequence, if we assume g o be Lipschiz coninuous, hen he infimum is acuall a minimum.
Sep. 23, 2 3 Remark 3. Noe ha convex funcions are coninuous. The proof can go roughl as follows. Firs one can show ha f he convex funcion is bounded, le he bound be denoed M. Then using he definiion of convexi we have, for an fixed x,, Leing α we see ha u + α u +α ux u u +2αM. 3 On he oher hand, for an x n x we have, b convexi This gives limsupux n ux. 32 x n x ux 2 [ux n+u2x x n ]. 33 ux 2 liminf[ux n +u2x x n ]. 34 x n x Coninui hen follows. One can in fac prove ha an convex funcion is Lipschiz coninuous, see e.g. B. Dacorogna Direc Mehods in he Calculus of Variaions, 2nd ed., Springer, 28, 2.3. Soluion of he H-J equaion. Now we show ha he Hopf-Lax formula ux, = inf L R n indeed solves he Hamilon-Jacobi equaion, albei onl almos everwhere. + g. 35 Remark 4. I is eas o see ha in general one canno expec he exisence of classical soluions due o possible inersecions of characerisics. There are hree hings o show.. u = g on R n =, 2. u, Du exis almos everwhere, 3. u + HDu= a.e. We show hem one b one.. u = g on R n =. Recall he formula: Taking = x we have u, x=min L u, x gx+l On he oher hand, we compue u, x = min L = gx+min gx max = gx max z = gx max + g. 36 limsupu, x gx. 37 ց + g + g gx L Lipg x L Lipg z Lz max w z Lz w B Lipg z = gx max w B Lipg Hw. 38
4 Mah 527 Lecure 6: Hamilon-Jacobi Equaion: Explici Formulas As H is coninuous, we have Thus ends he proof. 2. u, Du exis almos everwhere. I suffices o show ha u is Lipschiz wih respec o x and o. liminfu, x gx. 39 ց u is Lipschiz w.r.. x. We esimae u, xˆ u, x. Choose such ha u, x=l + g. 4 Then xˆ z u, xˆ u, x = min L + gz L g. 4 Taking z =xˆ x+ such ha xˆ z = we have u, xˆ u, x gxˆ x+ g Lipg xˆ x. 42 Similarl we can show u, x u, xˆ Lipg xˆ x. 43 The Lipschiz coninui of u hen follows. u is Lipschiz w.r... This follows from he following proper of he Hopf-Lax formula: u, x= min sl +us,. 44 R n s Tha his should hold is inuiivel ver clear following our derivaion of he formula. For a proof see Evans p. 26. Using his formula, we see ha esimaing u, x us, x is no differen han esimaing ux, gx. Thus a similar argumen as in Sep. gives u, x us, x C s. 45 3. u + HDu= a.e. Fix an q R n, we compue x+hq u + h, x+hq = min h L + u, h hlq+u, x. 46 This implies for all q R n. Therefore and u + q Du Lq u Du q Lq 47 u max Du q Lq = HDu 48 q u + HDu. 49 For he oher direcion ha is u + HDu, we onl need o find one q such ha or more specificall where is in he direcion of q. u + q Du Lq 5 u, x us, s Lq 5
Sep. 23, 2 5 As u is a minimum, o ge u, x us, somehing, we ge rid of he minimum in u, x. Take z such ha x z u, x=l + gz. 52 Now ha q = Then we have This gives As we ge and finishes he proof. is alread chosen, has o be on he line segmen connecing x and z. Thus we ake s = h, = s x + s z. 53 = z s = q 54 [ ] z u, x us, L + gz s L + gz s = sl. 55 u, x us, s u + x z u + Du 56 x z Du L 57