Hamilon Jacobi equaions Inoducion o PDE The rigorous suff from Evans, mosly. We discuss firs u + H( u = 0, (1 where H(p is convex, and superlinear a infiniy, H(p lim p p = + This by comes by inegraion from special hyperbolic sysems of he form (n = m v + F j (v j v = 0 when here exiss a ponenal for F j, i.e. F j = j H(v, and when we seek soluions as gradiens v = u. Typical example (n = 1 is Burgers equaion v + vv x = 0 which by inegraion v = u x gives, omiing consans, u + 1 2 u2 x = 0. We know soluions of Burgers or conservaion laws conserve L norms and produce shocks. So we expec soluions of HJ equaions o be Lipschiz, and loose second derivaives. Tha hey do. We recall ha he Legendre ransform of a convex superlinear-a-infiniy funcion L is L (p = sup q R n [p q L(q] Noe ha L is convex and superlinear a infiniy and (L = L. 1
Le H = L. Noe ha, if L is smooh (and i is, if H is sricly convex, hen a a maximum (hey are aained becuse of superlineariy q L(q = p, so, solving for q gives a funcion of p. (This is unique if L is sricly convex. So, H(p = p q(p L(q(p Differeniaing his, we have p H(p = q(p + p p q q L(q(p p q = q(p This can be undersood: he Legendre ransform of L is given by H(p = p q(p L(q(p where q solves q L = p, or by he equaion H(p = p p H L( p H(p which looks horrible, and iself i is a seady Hamilon-Jacobi equaion. Similarly, H (q = L(q is compued by solving p H(p = q and seing L(q = p(q q H(p(q. Of course, if L is no sricly convex we do no have necessarily a unique soluion, and he minimizaion is necessary. Now we have a magical soluion of (1, u(x, = min y { L ( x y } + u 0 (y where u 0 is he value of u a = 0. This is ermed he Hopf-Lax formula in Evans. The magic is hen enveloped in mysery: {ˆ } u(x, = min L(ẇ(sds + u 0 (w(0 A where 0 A = {w C 1 ([0, ], R n w( = x} and ẇ(s = dw. This is easier o prove han i looks (see Evans, and makes ds conac wih minimum acion principles, or conrol, bu does no illuminae (2 2
he Hopf Lax formula by even a single lumen. By wha divinaion process could hey have arrived a such an amazing formula? Le us differeniae our equaion (1, moivaed by where i came from. We obain v + ( p H(v v = 0 where v = u. So, le us inroduce characerisics, dx d = ph(v(x,, x(0 = x 0. Denoe for a second x(, x 0 he characerisic issued from x 0. On i, v is consan, v(x(, x 0, = v(x 0, 0. Bu v = u, which is he sole argumen of p H. So he characerisics are sraigh lines and x(, x 0 = x 0 + q, u(x, = p, q = p H(p for x = x 0 + p H(p. Now le us look a u on he characerisic. We differeniae d ds u(x 0 + sq, s = s u(x 0 + sq, s + q u(x 0 + sq, s Bu we are on he characerisic, so u = p, and q = p H(p. Using he equaion (1 we also have s u(x 0 + sq, s = H(p. The righ-hand side is a consan, p q H(p. Tha we know how o inegrae in ime: u(x, u(x 0, 0 = (p q H(p We recall ha q = p H(p. Wai a minue, ha means ha p q H(p = L(q. So u(x, = L(q + u 0 (x 0 Now we express q in erms of x and x 0 : From x = x 0 + q, i follows ha q = x x 0. So, wihou magic, we arrived a ( x x0 u(x, = L + u 0 (x 0. Now, we assumed sric convexiy, and smooh iniial daa, and his would work only for shor ime. Remarkably, he variaional formula is rue for all ime and gives he good weak soluion even afer u develops shocks. When hey do, x 0 is no uniquely deermined on characerisics by u(x,, x an. We sar wih a semigroup propery: 3
Lemma 1. If H is convex and superlinear a infiniy and if u is given by (2, hen { ( } x y u(x, = min ( sl + u(y, s (3 y s for 0 s <. Proof. Take z so ha u(y, s = sl ( y z s + u0 (z and use convexiy of L o show ( x z ( L 1 s ( x y L + s ( y z s L s Then u(x, L ( x z = ( sl ( x y + u0 (z ( s L ( x y s ( + sl y z s + u0 (z s + u(y, s. On he oher hand, pick w so ha u(x, = L ( x w + u0 (w and ake y = sx + (1 sw. Then x y s = x w = y w s and consequenly ( sl ( ( x y s + u(y, s ( sl x w ( + sl y w s + u0 (w = L ( x w + u0 (w = u(x,. Proposiion 1. If u 0 is Lipschiz, hen he funcion u(x, given in (2 is Lipschiz coninuous and coninuous up o = 0. Proof. Fix > 0. Pick x, x 1. Find y so ha ( x y u(x, = L + u 0 (y Then { ( u(x 1, u(x, = min z L x1 z + u0 (z } L ( x y u0 (y g(x 1 x + y g(y C x x 1. Then we swich he roles of x 1 and x. For he coninuiy a = 0, on one hand we have u(x, L(0 + u 0 (x 4
direcly from (2 by aking y = x. On he oher hand, using u 0 (y u 0 (x C x y in he definiion (2 we have { ( u(x, u 0 (x + min y C x y + L x y } = u 0 (x max w {C w L(w} = u 0 (x max z C max w {z w L(w} = u 0 (x max z C H(z We also have u(x, u(x, s C s using he semigroup propery and he bounds above. Proposiion 2. Le u 0 be Lipschiz and ake a poin (x, where he funcion u(x, defined by (2 is differeniable. (This happens a.e. by Rademacher s heorem and he resul above. Then u(x, + H( u(x, = 0. Proof. le q R n, h > 0. By he semigroup propery { ( u(x + hq, + h = min y hl x+hq y } h + u(y, hl(q + u(x,, which implies This is valid for all q, so q u(x, + u(x, L(q u(x, + H( u(x, = u(x, + min {q u(x, L(q} 0. q On he oher hand, choose z so ha u(x, = L ( x z + u0 (z. Take h > 0, pu s = h, y = sx + (1 s x z z. Then = y z and so, s u(x, u(y, s L ( x z + u0 (z { sl ( y z s + u0 (z } = ( sl ( x z. This means 1 h Leing h 0 ( (( u(x, u 1 h x + h z, h L u(x, + x z ( x z u(x, L 5 ( x z
Then u(x, + H( u(x, = u(x, + max w {q u(x, L(q} u(x, + x z u(x, L ( x z 0. This concludes his verificaion. Now i urns ou ha an addiional propery holds, semi-concaviy. Lemma 2. If here exiss a consan such ha u 0 (x + z + u 0 (x z 2u 0 (x C z 2 holds for all x, z, hen he soluion (2 saisfies, wih he same C u(x + z, + u(x z, 2u(x, C z 2 Proof. Le y so ha u(x, = L ( x y + u0 (y and use his y + z and y z in he definiions of u(x + z,, u(x z,. This will cancel he L erms and give he resul. Semi-concaviy of u implies ha u ɛ = φ ɛ u where φ ɛ is a sandard mollifier saisfies u ɛ CI I urns ou ha if H is sricly convex, hen for > 0 soluions become semiconcave, even if u 0 was no, wih bounds ha explode a = 0. Lemma 3. If H is sricly convex ( p1 + p 2 H 1 2 2 H(p 1 + 1 2 H(p 2 c p 1 p 2 2 hen he soluion (2 saisfies u(x + z, + u(x z, 2u(x, C z 2. Proof. Using definiions, i urns ou ha 1 2 L(q 1 + 1 ( 2 L(q q1 + q 2 2 L + C q 1 q 2 2 2 Then, choosing y o have u(x, = L ( x y + u0 (y and using he same y for u(x + z, and u(x z, he u 0 erms cancel ou, and he resul emerges. 6
Theorem 1. Le H C 2, sricly convex, superlinear a infiniy. Assume u 0 is Lipschiz. Then here is only one Lipschiz coninuous funcion u saisfying u(x, 0 = u 0 (x and u + H( u(x, = 0, a.e. u(x + z, + u(x z, 2u(x C(1 + 1 z 2 Proof. Taking he difference u of wo soluions u 1, u 2 we arrive a where v is bounded, and given by v(x, = ˆ 1 0 u + v u = 0 We mollify he soluions u 1, u 2 and define v ɛ (x, = ˆ 1 0 p H(λ u 1 (x, + (1 λu 2 (x, dλ p H(λ u 1ɛ (x, + (1 λu 2ɛ (x, dλ where u jɛ = u j φ ɛ. The divergence saisfies a one-sided bound divv ɛ C(1 + 1. This used he semiconcaviy of boh soluions. Now because boh soluions are Lipschiz u(x, c, u C. We wrie u + v ɛ u = (v ɛ v u We ake a funcion w = f(u wih f smooh and nonnegaive. This sill solves w + v ɛ w = (v ɛ v w We inegrae on a fasly shrinking ball ˆ e( = x x 0 V ( 0 7 wdx
wih V > b L. Then de d C(1 + 1 e + ˆ x x 0 V ( 0 (v v ɛ w We le ɛ 0: de d C(1 + 1 e Now we selec f o vanish for u cɛ, posiive oherwise. Then e = 0 for ɛ and from hen on, by Gronwall e( = 0. I follows ha u cɛ, bu ɛ was arbirary. 8