Vanishing Viscosiy Mehod. There are anoher insrucive and perhaps more naural disconinuous soluions of he conservaion law (1 u +(q(u x 0, he so called vanishing viscosiy mehod. This mehod consiss in viewing equaion (25 as he limi for 0 + of he equaion (2 u +(q(u x u xx, which corresponds o choosing he flux funcion (3 q(u q(u u x, where is a small posiive number. Alhough we recognize u xx as a diffusion erm, his kind of model arises mosly in fluid dynamics where u is he fluid velociy and is viscosiy, from which comes he name of he mehod. There are several good reasons in favor of his approach. Firs of all, a small amoun of diffusion or viscosiy makes he mahemaical model more realisic in mos applicaions. Noe ha u xx becomes relevan only when u xx is large, ha is in a region where u x changes rapidly and a shock occurs. For insance in our model of raffic dynamics, i is naural o assume ha drivers would slow down when hey see increased (relaive densiy ahead. Thus, an appropriae model for heir velociy is ṽ(ρ v(ρ ρ x ρ which corresponds o q(ρ ρv(ρ ρ x for he flow rae of cars. Anoher reason comes from he fac ha shocks consruced by he vanishing viscosiy mehod are physical shocks, since hey saisfy he enropy inequaliy. As for he hea equaion, in principle, we expec o obain smooh soluions even wih disconinuous iniial daa. On he oher hand, he nonlinear erm may force he evoluion owards a shock wave. This phenomenon is clearly seen in he imporan case of viscous Burger s equaion. Typese by AMS-TEX 1
2 The Viscous Burger Equaion. Burger s Shock Soluion. Le us deermine a soluion of he viscous Burger s equaion (4 u + uu x u xx. The viscous Burger equaion is one of he mos celebraed examples of nonlinear diffusion equaion. I arose (Burger, 1948 as a simplified form of he Navier-Sokes equaion, in an aemp o sudy some aspecs of urbulence. I appears also in gas dynamics, in he heory of sound waves and in raffic flow modelling. I consiues a basic example of compeiion beween dissipaion (due o linear diffusion and seepening (shock formaion due o he nonlinear ranspor erm uu x. The success of Burger s equaion is in large par due o he raher surprising fac ha he iniial value problem can be solved analyically. In fac, via he so-called Hopf-Cole ransformaion, Burger s equaion is convered ino he hea equaion. Namely, wrie he equaion in he form u + ( 1 x 2 u2 u x 0. Then he planar vecor field ( u, 1 2 u2 u x is curl-free and herefore here exiss a poenial ψ ψ(x, such ha Thus, ψ solves he equaion ψ x u and ψ 1 2 u2 u x. (5 ψ 1 2 ψ2 x ψ xx 0. Now we ry o ge rid of he quadraic erm by leing wih w o be chosen. We have w φ(ψ, w φ (ψψ, w x φ (ψψ x, w xx φ (ψ(ψ x 2 + φ (ψψ xx. Subsiuing ino (5 we find ( w φ (ψψ φ (ψ ψ xx + 1 2 ψ2 x [ w xx φ (ψ+ 1 ] 2 φ (ψ (ψ x 2.
3 If we choose φ(s s 2, i.e. (6 ( w exp ψ, 2 or ψ 2 log w, which is he Hopf-Cole ransformaion, hen and we are lef wih φ (ψ+ 1 2 (φ (ψ 2 0 (7 w w xx 0. From u ψ x we obain (8 u 2 w x w An iniial daa u(x, 0 g(x ransforms ino an iniial daa of he form { x } g(z (9 w 0 (x exp 2 dz, (a R. If 1 x x 2 g(zdz 0 as z, a he iniial value problem (7, (9 has a unique smooh soluion in he half-plane >0, given by w(x, 1 4π w 0 (y exp ( (x y2 dy. 4 The soluion is coninuous ogeher wih is x-derivaive up o 0 a any coninuiy poin of g. Consequenly, by (8, problem (4 has a unique smooh soluion in he half-plane >0, coninuous up o 0 a any coninuiy poin of u 0, given by (10 u (x, 2 w x w ( x y w 0 (y (x y2 4 w 0(y (x y2 4.
4 Example. Iniial Pulse. Consider problem (4, (10 wih he iniial condiion g(x Mδ(x where δ denoes he Dirac densiy a he origin. We have, choosing a 1, { x } { g(y 1 x>0 ϕ 0 (x exp 2 dy M 2 x<0. Formula (10 gives, afer some rouine calculaions, u(x, 4 1 2 π x2 4 M/(2 1 + 1 2 [ π erf( x 4 ] where is he error funcion. erf(x x 0 e z2 dz Remark. Consider he nonlinear PDE (11 { u a u + b u 2 0, in R n (0,, u f, in R n { 0}, where a>0, b R. (This PDE arises in sochasic conrol heory. Assuming for he momen ha u is a smooh soluion of (11, we se w φ(u, where φ : R R is a smooh funcion o be seleced. We have w φ (uu, w xi φ (uu xi, w xix i φ (u(u xi 2 + φ (uu xix i. Subsiuing ino (11 we find w φ (uu φ (u (a u b Du 2 w xx [ aφ (u+bφ (u ] Du 2. Hence, if we choose φ(s bs a, i.e. (12 w bs a or u a b log w which is he Hopf-Cole ransformaion, hen aφ (u+b(φ (u 2 0 and we are lef wih (13 w a w 0.
5 The iniial daa u(x, 0 f(x ransforms ino an iniial daa ( (14 w 0 (x exp bf. a The iniial value problem (13, (14 has a unique smooh soluion in he half-plane >0, given by w(x, 1 x y 2 exp ( bf dy. 4πa 4a a The soluion is coninuous ogeher wih is x-derivaive up o 0 a any coninuiy poin of u 0. Consequenly, from (12, problem (11 has a unique smooh soluion in he half-plane >0, coninuous up o 0 a any coninuiy poin of u 0, given by u(x, a [ b log 1 + x y 2 exp ( bf ] dy. 4πa 4a a Laplace s Mehods Laplace s mehod concerns he asympoics as 0 of inegrals involving expressions of he form e I/, where I denoes some given funcions. In view of (8 and (9, we see ha (15 where u (x, (16 K(x, y, ( x y w 0 (y (x y2 4 w 0(y (x y2 4 ( x y ( x y (x y2 4 (x y2 4 (x y2 4 K(x,y, 2 ( K(x,y, 2 h(y 2 h(y 2, + h(y, wih h(x x 0 g(y. In consideraion of (4, leing F (u u2 2, we have F (q q2 2, and (16 becomes ( x y (17 K(x, y, F + h(y. We expec ha he vanishing viscosiy echnique should allow us o recover Lax formula for he weak soluion u of (4; namely, ( x y(x, x y(x, lim 0 u (x, G u(x,, for all bu counably many x, where y(x, is he unique poin a which K(x, y, aains is minimum. For his purpose, i suffices o esablish he following.
6 Lemma. Suppose ha k, l : R R are coninuous, l grows a mos linearly, and k grows a lea quadraically. Assume also here exiss a unique poin y 0 R such a k(y 0 min k(y. y R Then lim 0 Proof. Consider he funcion k(y l(y l(y k(y 0. µ (y k(y0 k(y k(y0 k(z dz, { µ 0, which saisfies µ 1, µ y 0, as 0, exponenially fas for y y 0. and µ δ(y 0, as 0; indeed, we may suppose he exisence of consans δ 0, C 1 and C 2 such ha C 1 z 2+δ k(z k(y C 2 z 2+δ and for any coninuous funcion l(y which grows a mos linearly, we have l(y 0 l(yµ (y µ (y[l(y 0 l(y]dy Consequenly, C 6 k(y0 k(y k(y0 k(z C1 y y0 2+δ C2 y y0 2+δ dz [l(y 0 l(y]dy dz [l(y 0 l(y]dy C 3 2+δ s 2+δ C 4 2+δ + s2+δ ds [l(y 0 l(y 0 + 1 2+δ s]ds C 5 s 2+δ [l(y 0 l(y 0 + 1 2+δ s]ds 1 2+δ s s 2+δ ds C 7 1 2+δ. lim 0 k(y0 k(y l(y k(y0 k(y lim 0 l(yµ (y l(y 0.