2. Nonlinear Conservation Law Equations

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1 . Nonlinear Conservaion Law Equaions One of he clear lessons learned over recen years in sudying nonlinear parial differenial equaions is ha i is generally no wise o ry o aack a general class of nonlinear problems bu o focus insead on specific problems, preferably ones which are physically moivaed. Consider hen he siuaion involving a quaniy of some kind conained in a region U in R n. We will refer o his quaniy only as Q and we suppose he amoun of Q conained in U need no be consan bu can change wih ime. However, we suppose he amoun of change is due only o he flow of Q across he boundary of U plus any losses or gains due o inernal sources or sinks inside U. These assumpions hen provide he basis for deriving a conservaion of Q equaion. The densiy of Q a posiion x a ime is a scalar valued funcion which will be denoed by ux,. Similarly, F Fx,,u will be used o denoe he vecor flux field for Q and S Sx,, u will denoe he scalar source densiy field. Then a ime, he amoun of Q in an arbirary ball B in U is given by he expression B ux, dx amoun of Q in B a ime Similarly, he ouflow hrough he boundary of he ball during he ime inerval, is given by B Fx,,u n dsd ouflow hrough he boundary of he ball during heimeinerval, where n denoes he ouward uni normal o B, he boundary he surface of he ball. Then we can formulae he following he conservaion law equaion B ux, dx B ux, dx B Fx,,u n dsd B Sx,,u dxd. expressing he fac ha he amoun of Q in he ball a ime equals he amoun of Q in he ball a ime, less ha amoun of Q ha has flowed ou of he ball hrough he boundary plus any amoun ha has been creaed or desroyed inside he ball during he ime inerval. Then (.) is he inegral form of he conservaion law equaion. We can use he fundamenal heorem of calculus o wrie B ux, dx B ux, dx B ux, dxd and he divergence heorem o wrie B Fx,,u n ds B div F x,,u dx Then (.) impllies

2 ux, div F x,,u Sx,,u dx d 0. B Since his mus hold for every ball B conained in U, and every ime inerval,, i follows ha if all of he funcions here are assumed o be smooh hen, ux, div F x,,u Sx,,u a all x U, and all. If we suppose ha F F u, hen his becomes where ux, a u x u... a nu xn u Sx,,u a all x U, and all.. a ju F j,jn. u Equaion (.) is he differenial form of he conservaion law. We refer o (.) as a scalar conservaion law equaion. Laer we will consider problems where he unknown is a vecor valued funcion. Then we will be faced wih a sysem of equaions of he form (.). Shock Type Soluions Equaion (.) is a special case of a quasilinear firs order pde and, as we have seen in he examples, such equaions can have soluions wih sponaneous singulariies. Then since global classical soluions do no exis, in general, we mus consider a weaker noion of soluion. We will begin by considering he simples example of equaion (.). In he case of one space variable, n, he Cauchy problem for equaion (.) wih Su 0, reduces o ux, x Fu 0, ux,0 0x, x R, 0..3 Le R x R, 0 and R x R, 0. Then a classical soluion of (.3) is a funcion in C R CR. As examples in he previous secion have shown, i may be he case ha no classical soluion for (.3) exiss. We will herefore define a weak soluion for.3, o mean a funcion u L loc R, saisfying 0 R ux, x, x x,dxd R u 0x x,0 dx 0 C c R..4 I is eviden ha (.4) implies ha u saisfies he equaions of (.3) in he sense of disribuions. Then.4 and (.3) have meaning in he disribuional sense even when he funcion u is disconinuous. Wha is no eviden is ha hese disconinuiies in he soluion are subjec o some consrains. Problem. Show ha if u L loc R saisfies (.4) hen u solves (.3) in he sense of disribuions, and if u C R,saisfies (.4), hen u solves (.3) in he classical sense.

3 Jump Condiions Suppose ux, L loc R is a weak soluion for (.3) such ha u has a jump disconinuiy across a curve C x, R : x s. This means ha u has (finie) one sided limis as C is approached from eiher side bu a any poin of C, he one sided limi from he lef does no equal he one sided limi from he righ. Le U denoe an open se in R and suppose ha C passes hrough U, dividing U ino wo disjoin pars, U U x s and U U x s. Noe ha a weak soluion u, of (.3) is a disribuional soluion of he pde in U. Suppose furher, ha u saisfies he pde in he classical poinwise sense in each of he ses U and U. Noe ha ux, x Fu, x u Fu div u Fu. In addiion, for any C c U, div div u Fu u, Fu x and for k,, div div u Fu u, Fu x. Now he assumpion ha u saisfies he equaion in (.3) in he classical poinwise sense in U and U leads o div u Fu u x Fu 0 for k,. Furhermore, (assuming ha he disribuional soluion u is regular enough o permi he use of he divergence heorem) he divergence heorem gives div n k. where n k denoes he ouer normal o U k. Then n k. u, Fu x 3

4 and, by adding he equaion wih k o he equaion wih k we obain U n. U n. U u, Fu x, Since u is a weak soluion of (.3) and C c U, he las inegral on he righ vanishes. Also, since supp lies inside U i follows ha vanishes on U. However he boundaries of he ses U k are no simply he wo pieces of U, bu are composed of he curve C ogeher wih porions of U. Then here is a par of U k where is no zero, namely on he disconinuiy curve, C. I is imporan o noe, however, ha he orienaion of C as par of he boundary of U is opposie o he orienaion of C when i is considered as par of he boundary of U ; in paricular his means, n n. Then he las equaion reduces o, C U n. C U n. 0 for all C c U. Here he subscrips and indicae ha limiing values have been obained by approaching C from wihin U or U. u(x,) is disconinuous across C Now suppose he curve C along which u is disconinuous is given paramerically by, s C :, x xs. and as s varies from s a o s b, he par of he boundary of U ha lies on C is covered. Then he angen o C can be wrien as T s,x s, and, N x s, s is easily seen o be normal o T. The ouward uni normal o C U is hen 4

5 n s x s, s (he parameer is scaled o make his a uni vecor). This leads o where Then, a b x s, s k p lim qp qu k ds a b x s, s q q a b sx su sfu Fu ds 0, u k Fu k. ds 0 where u and u denoe respecively he lef and righ hand limiing values of u a poins along C, he curve of disconinuiies for ux,. Since his inegral vanishes for all es funcions, we can conclude ha su x sfu Fu 0, or equivalenly, dx Fu Fu d u Fu u.5 where we use he noaion u o denoe he jump in he value of u as he curve C is crossed from lef o righ. Noe ha x is equal o he speed wih which he curve C propagaes. Then (.5) assers ha Fu u, and his is referred o as he Rankine-Hugonio relaion for he conservaion law (.3). A soluion for (.3) which is disconinuous across a curve C bu is smooh on eiher side of he curve and in addiion saisfies (.5), is referred o as a shock ype soluion for he conservaion law. The curve C is called a shock curve. The Rankine-Hugonio relaion assers ha disconinuiies in he soluion o a conservaion law do no propagae along arbirary curves bu can be consisen wih he conservaion law only when hey propagae along curves which are soluion curves for (.5). Noe ha if Fu ax,x,, hen he conservaion law is a linear pde and (.5) reduces o x ax,; i.e., he shock curve C is jus a base characerisic for he linear pde. The conservaion law is said o be genuinely nonlinear if F u 0. (i.e, if Fu ax,x, hen F u ax,, and F u 0 ). When he equaion is genuinely nonlinear, he shocks are no base characerisics. Example.. Consider he Cauchy problem ux, x, x ux, 0, ux,0 if x 0 0 if x 0 Here, he iniial condiion has a jump disconinuiy a x 0 hence we expec a soluion which is a weak soluion having a jump disconinuiy across a curve C which originaes a 5

6 he poin 0,0. Of course his curve will hen have o conform o he Rankine-Hugenio condiions. Noe ha ux, x ux, x u x, x Fu. This equaion is known as Burger s equaion and in his case, we have u 0, and Fu Fu Fu u u. Then Fu u u u u u 0, and he Rankine-Hugenio condiions imply ha he shock curve is a soluion curve for x /. Since he disconinuiy is locaed a x 0 when 0, we have he iniial condiion, x0 0; i.e., x /. Then he shock curve is he sraigh line x /, and he shock soluion for his iniial value problem is he disconinuous funcion, ux, if x / 0 if x / 0. Of course since he soluion is disconinuous, his is only a soluion in he weak sense.. Now consider he Cauchy problem ux, x, x ux, 0, ux,0 0 if x 0 if x 0 As in he previous example, we can show ha x / is a shock curve for his problem and a shock soluion is given by he disconinuous funcion u x, 0 if x / if x / 0. On he oher hand, i is easy o check ha he coninuous funcion u x, 0 if x 0 x if 0 x, if x 0 is also a soluion o he iniial value problem. This coninuous soluion is an example of wha is called an expansion fan soluion. We shall have o appeal o physical principles in order o decide which (if eiher is) is he physically relevan soluion o he problem. 3.The previous example provided our firs encouner wih loss of uniqueness. We consider now a more exreme case of lack of uniqueness. Suppose ux, solves, 6

7 ux, x, x ux, 0, ux,0 if x 0 if x 0 We can show ha his problem has an infinie family of shock ype soluions and, in addiion, i has one expansion fan soluion. We do no ye have any way of selecing he correc soluion ou of his collecion. Since he iniial condiion here is disconinuous a x 0, if a shock ype soluion exiss, i mus originae a x 0, 0. In fac, here are an infinie number of shocks ha could originae from his poin. For each choice of a parameer, a,0 a, here are hree shocks originaing a 0,0, all of which saisfy he R-K condiion. For an arbirary a, 0 a, he corresponding shock ype soluions for he pde are given by ux, if x a, 0 a if a x 0, 0 a if 0 x a, 0 if a x, 0 The firs shock, S, separaes he region where u from a region where u a hence i saisfies, x u a, x0 0; i.e., x a. The second shock, S, separaes he region where u a from a region where u a hence his shock saisfies, x u a a 0, x0 0; i.e., x 0. Finally he las shock, S3, separaes he region where u a from a region where u so i mus saisfy, x u a, x0 0; i.e., x 3 a Since his locally consan funcion saisfies he pde in he disribuional sense, and since each of he disconinuiy lines saisfies he R-H condiions, his is a mahemaically accepable shock ype soluion for he Cauchy problem. Visually, he soluion can be represened as follows: 7

8 A 3 Shock Soluion In addiion o he shock soluion, his problem has anoher soluion of he expansion fan ype. The following, coninuous and piecewise differeniable funcion can be shown o solve he Cauchy problem ux, This soluion looks as follows, if x, 0 x if x, 0 if x, 0 A Fan Type Soluion We shall have o awai he developmen of selecion principles o deermine which of hese soluions is he physically correc one. 4. Consider he Cauchy problem ux, x, x ux, 0, ux,0 fx if x 0 x if 0 x 0 if x Here he iniial condiion is coninuous so we expec a non-shock weak soluion for he iniial value problem, and we will use he mehod of characerisics o consruc his soluion. Since he pde is homogeneous, he soluion is consan along characerisics. Then he coefficien in he equaion, u x,, is consan on each characerisic (alhough i is no 8

9 necessarily he same consan on each characerisic). This leads o he resul ha he characerisics are sraigh lines. The equaion for he base characerisics is x, and i follows ha x 0 x 0 is he equaion for he characerisic originaing a he poin x 0,0, where u 0 fx 0. There are hree cases o consider. If x 0 0 hen u 0 fx 0, hence he characerisics ha originae on he lef half line 0 are given by x x 0 and he soluion is given by ux, if x 0. If x 0 hen u 0 fx 0 0, and he characerisics ha originae on he half line are given by x x 0. Then he soluion is given by ux, 0 if x, 0. If 0 x 0 hen u 0 fx 0, and he soluion is given by u fx x. Solving his equaion for u in erms of x and leads o, ux, x, for x, 0. Noe ha a, he value u carried along he characerisic x, collides wih he value u 0 being carried along he characerisic, x (no o menion he values u p x p ha move along he lines x p x p, which all collide a he poin,). Alhough he soluion o he pde is coninuous up o he ime, a he poin,, x, a disconinuiy appears and he classical soluion no longer exiss. For he soluion mus be a weak soluion originaing from a disconinuous iniial condiion a ; i.e., ux, x, x ux, 0, ux, if x 0 if x In he same way we did in he firs example, hen we find he shock curve is he sraigh line,x /,, and a weak soluion is provided by a funcion which is consan on eiher side of he shock curve, u x, if x / 0 if x /, Then u is he weak soluion of he iniial value problem originaing a, and a complee weak soluion o he original problem is given by ux, if x, x if x, 0 if x, if x /, 0 if x /, 0 All we know a his poin is ha his is he only possible shock ype soluion for his problem. 9

10 Selecion Principles We will describe now he simples of he so called selecion principles by means of which we can choose he physically relevan soluion from he collecion of all possible soluions o a nonlinear conservaion law equaion. The principle can be saed as he condiion ha a shock can exis only if he nearby characerisics carry informaion oward he shock no away from he shock. This is consisen wih he examples we have seen in which he shock occurs when characerisics carrying conflicing values collide. When characerisics carrying conflicing informaion diverge from one anoher, leaving an empy wedge conaining no characerisics, a shock soluion can be consruced bu his soluion is nonphysical in he sense ha i violaes he principle of causaliy and, in he conex of gas dynamics, he soluion can be shown o produce a decrease in enropy across he shock, violaing he second law of hermodynamics. For his reason, he condiion we are abou o discuss is called he enropy condiion. Oher selecion principles leading o he same resuls will be given laer. If he equaion is given by ux, x Fu ux, au x u 0, hen he enropy condiion akes he form, Recalling earlier experiences wih he linear ranspor equaion, we can view au F u as he wave speed in a ranspor equaion. Then he condiion of admissibiliy for a shock soluion is ha he wave behind he shock ravels faser han he wave ahead of he shock; i.e., he shock resuls when a fas moving wave runs up he back of a slow moving wave. On he oher hand, if he slow moving wave is behind he faser wave, hen he fas wave jus pulls away from he slower wave. In his case no shock develops bu insead we have a fan ype soluion. Applying he selecion principle o example 4, we observe he following: s shock, S- F u x a a F u condiion no saisfied nd shock, S- F u a x 0 a F u condiion is saisfied 3rd shock, S3- F u a x a F u condiion no saisfied 0

11 Then his shock soluion, while mahemaically correc, fails o saisfy he selecion condiion and i is herefore no a physically relevan soluion. The fan ype soluion is herefore he one ha is physically relevan. Expansion Fans Expansion fans are also called rarefacion wave soluions as hey represen he physical siuaion where a fas wave runs away from slower railing waves, leaving an area of rarefacion in beween. In order o moivae his ype of soluion, consider he following example for some L 0, ux, x, x ux, 0, ux,0 fx 0 if x 0 x/l if 0 x L if x L As we have seen in previous examples, he base characerisics are given by x 0 x 0 ; i.e., his is he equaion of he characerisic originaing a x 0,0 where ux 0,0 0. Now here are hree cases o consider, if x 0 0 hen u 0 0 so x x 0 for all characerisics ha originae o he lef of zero if x 0 L hen u 0 so x x 0 for all characerisics ha originae o he righ of x L if 0 x 0 L, hen ux, fx and ux 0,0 x 0 /L fx 0 so ux, x L This hird case leads o he soluion ux, x L 0 x L, 0. Coninuous soluion for L 0

12 Leing L end o zero causes his picure o become i.e., L0, Rarefacion Wave Soluion ux, 0 if x 0 x L if 0 x L, 0 if x L, 0 0 if x 0 x if 0 x, 0 if x, 0 This las soluion is he rarefacion wave soluion. To furher moivae he expansion fan soluion, consider he iniial value problem ux, x Fu 0, ux,0 u L if x x 0 u R if x x 0 where we suppose F u 0, au F u and u L R so he equaion is genuinely nonlinear and he shock soluion is no admissible. The characerisics in his case are, i) x au L x p if x p x 0 and u L for x x 0 au L, 0 ii) x au R x p if x p x 0 and u R for x x 0 au L, 0 Since u L R, here are no characerisics in he wedge, au L x x 0 au R

13 Noe ha he family of rays, x x 0, fill he wedge as varies over he inerval, au L au R. This suggess ha we ry a soluion of he form, ux, G x x 0 he wedge. This leads o, in and ux, G x x 0 ux, x Fu G x x 0 x x 0, x ux, G x x 0 x x 0 a G x x 0, 0. If G x x 0 0, hen a G x x 0 x x 0 ; Since F u 0, F u au is inverible, and i follows ha G a. Then he following is a soluion of he pde in all hree regions of he half plane, ux, u L if x au L x 0 G x x 0 if au L x x 0 au R u R if x au L x 0 and, in addiion, he soluion is coninuous (bu no differeniable) across he boundaries of he wedge; i.e., uau p x 0, G x x 0 Gau p p, p L,R. In he case of Burger s equaion, Fu au ideniy, so we have G a ideniy, Gu. 3

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