Hamilton- J acobi Equation: Explicit Formulas In this lecture we try to apply the method of characteristics to the Hamilton-Jacobi equation: u t
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1 M ah Fall L ecure 1 0 O c. 7, Hamilon- J acobi Equaion: Explici Formulas In his lecure we ry o apply he mehod of characerisics o he Hamilon-Jacobi equaion: u + H D u, x = 0 in R n 0, 1 u = g on R n { = 0. 2 To avoid confusion, we use he following noaion: x D u x, u z, u p = p p n Then we can re-wrie he equaion o where F D u, u, u, x, = 0 4 F p, z, x p n+ 1 + H p, x. 5 The characerisics ODEs hen are x D = x = D p F = p H, 6 1 D ż = D p F p = p H p = p 1 p n+ 1 + D p H p = D p H p H p, x, 7 n+ 1 ṗ Dx H = p = D ṗ z F p D x F =. 8 n Mehod of characerisics. We ry o solve he characerisic ODEs. Firs noice ha, since = 1, we can simply use as he parameer insead of s. Thus he equaions become x = D p H, 9 ż = D p H p H p, x, 1 0 ṗ = D x H, 1 1 p n+ 1 = p n+ 1 = I is clear he all we need o do is o solve he firs 3 equaions. Losing a bi rigor, we assume for now only H is differeniable and sricly convex. We also assume H grows super-linearly a infiniy: H p lim = +, 1 3 p p Noicing ha We can define p 0 = argmax p R n { D p H p 0, x p H p, x. 1 4 q D p H p, x 1 5 which also give p as a funcion of q as D p 2 H is non-singular due o he convexiy of H. We wrie As a consequence he z equaion becomes where q saisfies L q, x sup p R n { q p H p, x. 1 6 ż = L q, x 1 7 q = D p H p, x. 1 8
2 Therefore he soluion u is given by u x = u x 0 + where x and x 0 are relaed by where X solves To furher simplify he sysem, we noice ha implies which implies ha q, x minimizes 0 L q τ, x τ dτ. 1 9 x = X s 20 d d X = D ph = q, X 0 = x wih x 0, x fixed. See Evans for deails. To see his, wrie L q, x = q p q, x H p q, x, x, and compue x = D p H, ṗ = D x H 22 d ds D ql + D x L = L q τ, x τ dτ 24 D q L = p + q D q p D p H D q p = p, 25 D x L = q D x p D x H D p H D x p = D x H, 26 where we have used q = 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 q. Below we will see ha in a special case, his can indeed be done in some sense. 2. The Hopf-Lax formula. This special case is when H is independen of x, ha is H = H D u. The characerisic equaions can hen be furher simplified o x = D p H, 27 ż = D p H p H p = L q, 28 ṗ = D x H = 0, 29 p n+ 1 = p n+ 1 = We see ha p is a consan vecor along he characerisic curve, and as a consequence x = D p H is a consan vecor, and herefore he characerisics x are sraigh lines. Furhermore we know ha he velociy q = x is consan. Thus if x 0 = y and x = x, we mus have As a consequence d z = L q = L d q =. 31 z = z 0 + L = g y + L. 32 Now he only problem is ha y is no known. Now hink of g y as no merely an iniial funcion, bu as an inermediae record. In oher words, insead of saring a = 0, imagine our sysem sars from =, say, 1. We consider all possible rajecories emanaing from some poin a = 1, passing y a = 0, and finally reach ime a x. Think of g y as he record of work done from = 1 o = 0. Obviously he correc rajecory should be he one ha is he minimizer among hem all.
3 Following his idea, we reach he following Hopf-Lax formula: { u x, = z = inf L + g y. 33 y R n Remark 1. I can be shown ha L grows superlinearly a infiniy. As a consequence, if we assume g o be Lipschiz coninuous, hen he infimum is acually a minimum. Remark 2. The relaion L q = H q sup p R n { q p H p 34 is called Legendre ransform and is very useful. I can be shown ha he following heorem holds Evans p Theorem 3. Le H = H p be convex, and saisfies hen i. H q is also convex, ii. H = H. H p lim = +, 35 p p H q lim = +, 36 q q Inspecing he proof, one sees ha i sill holds even if H is no convex, bu convexiy is necessary for ii If H is no convex, hen i canno be he same as H, which is convex by i. Remark 4. Noe ha convex funcions are coninuous. The proof can go roughly as follows. Firs one can show ha f he convex funcion is bounded, le he bound be denoed M. Then using he definiion of convexiy we have, for any fixed x, y, Leing α 0 we see ha u y + α u y + α u x u y u y + 2 α M. 37 On he oher hand, for any x n x we have, by convexiy This gives limsup u x n u x. 38 x n x u x 1 2 [ u x n + u 2 x x n ]. 39 u x 1 2 liminf [ u x n + u 2 x x n ]. 40 x n x Coninuiy hen follows. One can in fac prove ha any convex funcion is Lipschiz coninuous, see e. g. B. Dacorogna Direc Mehods in he Calculus of Variaions, 2nd ed., Springer, 2008, Soluion of he Hamilon-Jacobi equaion. Now we show ha he Hopf-Lax formula { u x, = inf L y R n indeed solves he Hamilon-Jacobi equaion, albei only almos everywhere. + g y. 41 Remark 5. I is easy o see ha in general one canno expec he exisence of classical soluions due o possible inersecions of characerisics.
4 There are hree hings o show. 1. u = g on R n { = 0, 2. u, D u exis almos everywhere, 3. u + H D u = 0 a. e. We show hem one by one. 1. u = g on R n { = 0. Recall he formula: Taking y = x we have { u x, = min L y + g y. 42 u x, g x + L 0 limsup u x, g x On he oher hand, we compue { u x, = min L + g y y { = g x + min L + g y g x y { g x max Lip g y x L y = g x max { Lip g z L z z { = g x max max { w z L z w B L i p g z = g x max H w. 44 w B L i p g As H is coninuous, we have Thus ends he proof. 2. u, D u exis almos everywhere. I suffices o show ha u is Lipschiz wih respec o x and o. liminf u x, g x u is Lipschiz w. r.. x. We esimae u xˆ, u x,. Choose y such ha u x, = L + g y. 46 Then { xˆ z u xˆ, u x, = min L + g z L g y. 47 Taking z = xˆ x + y such ha xˆ z = we have Similarly we can show u xˆ, u x, g xˆ x + y g y Lip g xˆ x. 48 The Lipschiz coninuiy of u hen follows. u x, u xˆ, Lip g xˆ x. 49 u is Lipschiz w. r... This follows from he following propery of he Hopf-Lax formula: { u x, = min s L + u y, s. 50 y R n s Tha his should hold is inuiively very clear following our derivaion of he formula. For a proof see Evans p
5 Using his formula, we see ha esimaing u x, u x, s is no differen han esimaing u x, g x. Thus a similar argumen as in Sep 1. gives u x, u x, s C s u + H D u = 0 a. e. Fix any q R n, we compue u x + h q, + h = { x + h q y min h L + u y, h h L q + u x,. 52 This implies for all q R n. Therefore and u + q D u L q u D u q L q 53 u max { D u q L q = H D u 54 q u + H D u For he oher direcion ha is u + H D u 0, we only need o find one q such ha or more specifically u + q D u L q 56 u x, u y, s s L q 57 where is in he direcion of q. As u is a minimum, o ge he u x, u y, s somehing, we ge rid of he minimum in u x,. Take z such ha x z u x, = L + g z. 58 Now ha q = x z is already chosen, y has o be on he line segmen connecing x and z. Thus we ake s = h, y = s x + 1 s z. 59 Then we have x z As we ge and finish he proof. = q. We compue x z u x, u y, s L x z = s L = y z s u x, u y, s s u + x z ] y z s L + g z s. 60 [ + g z u + x z D u, 61 x z D u L 62
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