Continuous and discontinuous piecewise linear solutions of the linearly forced inviscid Burgers equation

Con tin uous and discon tin uous piecewis e linear solutions of the linearly forced i n viscid Burgers equation Hans Lundmark ∗ Jacek Szmigielski † Marc h 31, 20 08 Abstract W e study a class of piecewise linear solutions t o the inviscid Burgers equa- tion driv en by a linear forcing term. Inspired by the analogy with peakons, w e think of these solutions as b eing made up of solitons situated at the breakp oints. W e derive and solv e ODEs go verning the soliton dyn am- ics, first for con tinuous solutions, and then for more general sho ck wa ve solutions with d iscon tinuities. W e show t hat triple collisions of soli tons cannot take place for contin uous solutions, but giv e an example of a triple collisio n in the presence of a shock. 1 In tro duc tion The sub ject of this pape r is piecew is e linear solutions o f the PDE ( u t + u u x ) xx = 0 , (1.1) which w e earlier [1] have ca lled the derivative Bur gers e quation . This name refers o f course to the w ell-known Burgers equatio n u t + uu x = ν u xx and its sp ecial case the inviscid Burg e rs equatio n u t + uu x = 0, which is the pr ototype equation for studying sho ck w av e solutions of h yp erb olic conser v ation laws. In some applications one co nsiders also for c e d Burgers equatio ns with terms of the form F ( x, t ) on the right-hand side, often written as F = − ∂ V /∂ x with a po tent ial V . Since equation (1.1) is equiv a lent to u t + u u x = A ( t ) x + B ( t ), it is per haps mor e appro priate to ta lk a b o ut it as a for c e d inviscid Bur gers e quation with line ar fo r c e (or quadratic p otential). Mor eov er, the latter equatio n can be rewritten as u t + 1 2 ( u 2 ) x = A ( t ) x + B ( t ) , (1.2) ∗ Departmen t of Mathematics, Link¨ oping U nive rsi t y , SE-581 83 Link¨ oping, SWEDEN; halun@mai.liu.se † Departmen t of Mathematics a nd Sta tistics, Universit y of Sask atc hewan, 106 Wiggins Road, Sask atoon, Sa sk atc hewan, S7N 5E6, CANADA; szmigiel@math.usask.ca 1 which mak es sense for a m uch larg er class of functions than just u ∈ C 1 ( R ). F or exa mple, if u ∈ L 2 lo c ( R ) w e can interpret (1.2) to hold in the sense of dis- tributions. One could w o rk with distributions acting on test f unctions ψ ( x, t ) ∈ D ( R 2 ), but the following simpler interpretation is sufficie nt for our purp os e s here: we view u ( x, t ) as a mapping that takes a r eal n umber t to a function u ( · , t ) ∈ L 2 lo c ( R ) which we can iden tify with a distribution in D ′ ( R ). The deriv a tive with resp ect to x is then the distributional deriv ative defined b y its action on a test function ψ ( x ) ∈ D ( R ) in the usual wa y , h u x , ψ i = −h u, ψ x i , while the der iv ative with resp ect to t is the limit of a difference quotient. If equation (1.2) is satisfied in D ′ ( R ) for each t , then we then say that it holds in a w eak sense and that u is its w eak solution. W e w ere led to the Burgers eq uation by our previo us w ork on p e akon and sho ckp e akon so lution of the Degasp er is –Pro cesi (DP) equatio n u t − u txx + 4 uu x = 3 u x u xx + u u xxx , (1.3) an integrable wa ve equation disc overed a few years ago [2, 3]. Indeed, the problems trea ted in this paper are to some ex tent “ toy problems” , but w e hop e that they might provide s o me guidance a nd intuition for the fu ture study o f the DP equation. Equation (1.1) can b e obtained for ma lly from the DP eq uation by substi- tuting x 7→ εx , t 7→ εt , and then letting ε → 0. This “high-frequency limit” is a natural thing to try on the DP equa tion, since it is the same pro cedur e that takes the c e le brated integrable Cama ssa–Holm (CH) shallow water equation [4 ], u t − u txx + 3 uu x = 2 u x u xx + u u xxx , (1.4) to the Hunter–Saxton (HS) equation for nematic liquid crystals [5, 6], ( u t + u u x ) xx = u x u xx . (1.5) The CH a nd DP equations b oth a dmit p e akon solutions, whic h ar e m ulti- soliton solutions of the form u ( x, t ) = n X k =1 m k ( t ) exp  − | x − x k ( t ) |  , (1.6) where the functions x k ( t ) and m k ( t ) (p ositions and momenta of the individual pea k-shap ed s olitons) are requir ed to satisfy a certain system of 2 n ODEs in order for u ( x, t ) to satisfy the PDE in a w eak sense. In sho rthand notatio n these O DE s a re ˙ x k = u ( x k ), ˙ m k = − ( b − 1) u x ( x k ), where b = 2 for the CH equation and b = 3 for the DP equation. One can think of this as an in tegrable mechanical system of n pa r ticles on the real line, simililar to , for exa mple, the op en T o da lattice. If follows from the rapid decay of e −| x | that ˙ x k = u ( x k ) ≈ m k when all distances | x i − x j | are large, so it agrees with intuition to rega rd m k as the mo ment um of the k th par ticle. Asy mpo tically (when t → ±∞ ) the particles will spread apa r t, ea ch moving with its own (near ly) constan t v elo city which is 2 nonzero and distinct from the other particles’ velocities. The latter is a highly nontrivial fact for the DP equation [7, Theorem 2.4]. There is a n analogous class of so lutions of the HS equation and the forced Burgers equation (1.2), na mely the piecewise linear solutions u ( x, t ) = n X k =1 m k ( t ) | x − x k ( t ) | . (1.7) In the s horthand no ta tion used a bove, the g ov erning ODEs take exactly the same form again: ˙ x k = u ( x k ), ˙ m k = − ( b − 1) u x ( x k ), where b = 2 for the HS equation and b = 3 for the f or ced Burge rs equation. How ever, for p eakons the term m k e −| x k − x k | usually dominates the other terms in the equatio n ˙ x k = u ( x k ), while her e w e instea d have the term m k | x k − x k | which is zero while all other terms are large. Th us, in contrast to pea kons wher e the int era ction is strongly lo calized, these piecewise linear solitons influence eac h other more strongly the more sepa rated they are. Although it is a bit hard to dev elop a useful in tuition ab out these ODEs as a “mech anica l” system (p erhaps one can think of so me kind of expanding ga s with long-r a nge correlations ), the ana logy with p e a kons still makes it natural to think of the piecewise linear solutions a s being comp os e d of some kind of s o litons situated at the br eakp oints x k . (But w e hav e not b een able to make sense of the idea that the piecewise linear solutions are somehow high-frequency limits of p ea kons). In all four ca ses men tioned ab ov e, the ODEs gov erning the so liton dynamics can be explicitly solved using in verse spectra l methods [8, 9, 7, 10, 1, 11]. In the forced Burgers case the O DEs are also easily s olved direc tly by e lementary metho ds, as w e will see. In the Degas p er is–Pr o cesi equa tion (but not in the Cama ssa–Holm equation) there a lso app ea rs a more complicated phenomenon, namely discontin uous so- lutions of the for m u ( x, t ) = n X k =1  m k ( t ) − s k ( t ) sgn  x − x k ( t )   exp  − | x − x k ( t ) |  . (1.8) Such sho ckp e akons [12] are gov erned b y 3 n ODEs for po s itions x k , momen ta m k , and shock strengths s k . Even if one starts with the usual p eakon ansatz (1.6), sho ck s olutions of the form (1.8) can for m a fter finite time when a p ea kon with m k > 0 collide s with a n antip e akon with m k +1 < 0 moving in the o pp osite direction. (In the CH equation, such collisions give rise to “zer o -strength shocks” where u x momentarily blows up but u r emains contin uous, still b eing of the form (1.6) after the collision [8 ], and a similar thing occurs for the HS equation [6].) The shockpea kon ODEs ha ve so far only been solved in the trivia l ca se n = 1 and in a very particular sub case when n = 2. T he problem is that the La x pair for the DP equation, whic h w as crucial for deriving the p ea kon solution formulas, do es not ma ke sense for the weak for mulation of the DP equa tion that is used when working with discontin uous solutions. 3 The forced Burgers equation (1.2) admits an ana logous class o f solutions, given by the discontin uous piecewise linear a nsatz u ( x, t ) = n X k =1  m k ( t ) | x − x k ( t ) | − s k ( t ) sgn  x − x k ( t )   . (1.9) Such solutions with sho cks can form a fter finite time, ev en if the initial profile is co nt inuous. Unlike the Degasp eris– Pro ces i ca se, it turns out here that the extra generality of ha ving jumps in u ca n be handled without problems. The outline of the pap er is simple: we derive and so lve the ODEs gover- ing piecewise linear solutions of the forced Burgers equation (1.2), first in the simpler case (1.7) of contin uous solutions (using elemen tary metho ds and, for compariso n, inv erse sp ectr al metho ds), then in the general case (1 .9) of discon- tin uous solutions (b y r eduction to the previous case). W e conclude with a few examples. 2 Con tin uous piecewise linear solutions Theorem 1. Th e c ontinuous p ie c ewise line ar ansatz (1.7) , u = P m k | x − x k | , is a we ak solution to the line arly for c e d inviscid Bur gers e quation (1 .2 ) if and only if ˙ x k = n X i =1 m i | x k − x i | , ˙ m k = 2 m k n X i =1 m i sgn( x i − x k ) , (2.1) for k = 1 , . . . , n . F or t his class of solutions, e quation (1.2) takes the form u t + 1 2 ( u 2 ) x = M 2 x − M M + , (2.2) wher e M = P n k =1 m k and M + = P n k =1 m k x k ar e c onstants of motion. Pr o of. This is a sp ecial ca se (all s k = 0) o f The o rem 7 which is proved later. One c a n assume that all m k 6 = 0, since it follows from (2.1) tha t any v anishing m k remains iden tically zero . If we think of x k and m k as p o sitions and masses of particles on a line, then the total ma ss M and the cen ter of mass M + / M (if M 6 = 0) a r e c o nserved. Note that when M = 0 we hav e the unforced Burger s equation. There are some additiona l constants of motion M 2 , . . . , M n that come together with M 1 = M fro m the Lax pair presen ted in the next section, but w e will not need them here [1]. The pr esence of absolute v alues and the sign function in (2.1) natur ally divides the p o sition space R n int o sector s. More precisely , to an y permutation σ = σ 1 σ 2 . . . σ n of the num b er s { 1 , 2 , . . . , n } o ne can assign the sector X σ = { ( x 1 , x 2 , . . . , x n ) ∈ R n | x σ 1 < x σ 2 < · · · < x σ n } . W e will concentrate on the sector X e corres p o nding to the identit y permutation e =12 . . . n , since there is 4 no loss of generality in assuming that the initial p ositions x k (0) are sorted in increasing order: X e = { ( x 1 , x 2 , . . . , x n ) ∈ R n | x 1 < x 2 < · · · < x n } . (2.3) F or positions in X e the ODEs (2.1) take the fo r m ˙ x k = n X i =1 m i ( x k − x i ) sgn( k − i ) , ˙ m k = 2 m k n X i =1 m i sgn( i − k ) . (2.4) The follo wing theorem solves this system completely . Theorem 2. Given any initial data { x k (0) , m k (0) } n k =1 (with the x k (0) ’s or der e d or not), t he solution of the ODEs (2.4) is given by the formulas b elow, wher e M = P m k and M + = P m k x k as b efor e, and wher e the empty s ums P 0 1 and P n n +1 in F 0 and F n ar e t o b e interpr ete d as zer o (so that F 0 ( t ) = e − M t and F n ( t ) = e M t ). • Wh en M 6 = 0 the solution of (2.4) is x k ( t ) = M + M + e M t M   X j k  x k (0) − x j (0)  m j (0)   , m k ( t ) = m k (0) F k − 1 ( t ) F k ( t ) , (2.5) for k = 1 , . . . , n , wher e F k ( t ) = e M t M   k X j =1 m j (0)   + e − M t M   n X j = k +1 m j (0)   . (2.6) • Wh en M = 0 the solution of (2.4) is x k ( t ) = x k (0) + t   X j k  x k (0) − x j (0)  m j (0)   , m k ( t ) = m k (0) F k − 1 ( t ) F k ( t ) , (2.7) for k = 1 , . . . , n , wher e F k ( t ) = 1 + t   k X j =1 m j (0) − n X j = k +1 m j (0)   . (2.8) 5 • L etting l k = x k +1 − x k for k = 1 , . . . , n − 1 , we have in b oth c ases l k ( t ) = l k (0) F k ( t ) . (2.9) The pro of is presen ted at the end of this section. As an immediate corollary we obtain information abo ut the origina l O DE s (2.1). Theorem 3. Gi ven initial d ata { x k (0) , m k (0) } n k =1 to the ODEs (2.1) such that x 1 (0) < x 2 (0) < · · · < x n (0) (that is, with the p ositions in t he se ctor X e of R n ), the solution is given lo c al ly (ar ound t = 0 ) by the f ormulas of The or em 2, and this solution is valid as long as the p ositions x k ( t ) r emain in X e . A loca l so lution that starts in X e hits the b ounda ry of X e whenever x k = x k +1 for a t lea st one k , an even t which we refer to as a c ol lisi on . It is clea r from (2.9) that a collis ion o ccurs when some F k bec omes zero, a t whic h time m k and m k +1 blow up. The lo ca l so lutio n is v a lid up un til the time o f the first collision. In general a shock will then for m, and the con tinuous ansa tz (1.7) will no t be able to desc r ib e the s olution b eyond the p oint of collis ion. W e will return to this in the section ab out discontin uous solutions. If all m k (0)’s ha ve the same sign, then (2.6) shows that there are no c ollisions, so the solution is global. In the case when all ar e positive, the asymptotic behaviour of this global so lution as t → + ∞ is tha t x 1 → M + / M and m 1 → M , while x k → + ∞ and m k → 0 for all k > 1. When the m k (0)’s hav e mixed signs, co llisions ma y or may not o ccur for t > 0 . F o r example, in the case n = 2 a co llision takes place when F 1 ( t ) = ( m 1 (0) e M t + m 2 (0) e − M t ) / M b ec o mes zero, whic h happ ens when m 2 (0) /m 1 (0) < 0 and t = (2 M ) − 1 ln | m 2 (0) /m 1 (0) | . Consideratio n of cas es s hows that this v a lue of t is p ositive iff m 1 (0) < 0 < m 2 (0). The even t when l k − 1 = l k = 0 is ca lled a triple co llision, since thre e particles come together at one point. The absence of triple collisions in the CH equation is a nontrivial r esult [8 , 13], but for the linear ly forced Burgers equatio n it is muc h simpler. (Note, how ever, that triple collisions ar e p oss ible for discontin uous piecewise linear solution; s ee the examples a t the end of the pa pe r .) Theorem 4. Col li sions o c curing in c ontinuous p ie c ewise line ar solutions of the line arly for c e d B urg ers e quation (1 .2) c annot b e t riple c ol lisions. Pr o of. A triple collis io n w ould occur if l k − 1 ( t 0 ) = 0 = l k ( t 0 ) for so me t 0 , whic h amounts to F k − 1 ( t 0 ) = 0 = F k ( t 0 ) b y (2.9). F ro m the definition of F k , it is obvious that this is imp ossible in the case M = 0, since we are assuming m k 6 = 0. In the case M 6 = 0, it is also impossible, a lthough less ob vious ; F k ( t 0 ) = 0 iff t 0 = 1 2 M log − P j>k m j (0) m k (0)+ P jk m j (0), whic h requires that ( m k + A )( m k + B ) = AB , and hence m k ( A + m k + B ) = 0. But this is ruled out by m k and A + m k + B = M both being nonzero . 6 W e finish this section with the p ostp oned pro o f of the main theorem. Pr o of of The or em 2. Assume to be g in with that x 1 (0) < · · · < x n (0). Then (2.4) is equiv ale nt to (2 .1), and we ca n attack the problem by trying to find x k ( t ) and m k ( t ) such that the cor resp onding piecewise linear u ( x, t ) given by (1.7) satisfies the PDE (2 .2). The x k ’s divide the real line in to n + 1 in terv als which w e num b er by k = 0 , . . . , n . In each such interv al u takes the form u ( x, t ) = a k ( t ) x + b k ( t ). Inserting this in to (2 .2) yields ˙ a k + a 2 k = M 2 and ˙ b k + b k a k = − M M + , from whic h (in the ca se M 6 = 0) a k ( t ) = M a k (0) cos h( M t ) + M sinh( M t ) a k (0) sinh( M t ) + M cosh( M t ) (2.10) is found immediately , and by making an ansatz for b k with the same denominator as a k one also obtains b k ( t ) = a k (0) M +  1 − cosh( M t )  + M  b k (0) − M + sinh( M t )  a k (0) sinh( M t ) + M c o sh( M t ) . (2.11) Now x k ( t ) and m k ( t ) are r e cov ered from the r elations m k = 1 2 ( a k − a k − 1 ) and x k = − ( b k − b k − 1 ) / ( a k − a k − 1 ). Beca use of the algebr aic na ture of the formulas th us o btained, they satisfy the O DE s (2.4) identically , whic h shows that the assumption x 1 < . . . < x n is immaterial and can b e removed. (This will b e impo rtant later; see the commen ts after Theor em 8.) The simpler case M = 0 (unforced Burgers) is e ntirely similar, except that a k ( t ) = a k (0) ta k (0) + 1 , b k ( t ) = b k (0) ta k (0) + 1 . (2.12) (The solution for M = 0 ca n als o b e obtained b y expanding e ± M t = 1 ± M t + O ( M 2 ) in the s o lution for M 6 = 0 and letting M → 0.) 3 In v erse sp ectral construction of solutions The Lax pa ir − ∂ 3 x φ = z mφ, (3.1) φ t =  z − 1 ∂ 2 x + c + u x − u ∂ x  φ, (3.2) with c an arbitrary constant, is compatible iff m t + m x u + 3 mu x = 0 and m x = u xxx , under the as sumption of s ufficient smo othness needed to justify the cross- differ entiation. In pa rticular, it is compatible if u evolves a ccording to the deriv a tive Burger s eq ua tion (1.1), which can b e wr itten as m t + m x u + 3 mu x = 0 with m = u xx . T o obtain the linearly forced Bur gers equatio n (1.2) from equation (1.1) the rule ( u 2 ) x = 2 uu x is used. It is no t ob vio us if all these formal c alculations hav e any relev ance to weak solutions, where the smo othness assumptions may b e violated. T o inv estigate this, let us say that (3.1) and (3.2 ) 7 constitute a we ak L ax p air if they a re satisfied in the weak sense discuss ed in the int ro duction (thus φ , lik e u , is a D ′ ( R )-v alued function o f t , and the equations hold in the space of distributions D ′ ( R )). Solutions u of the form (1.7), u = P m k | x − x k | , do a dmit a weak Lax pair with m = u xx = 2 P n k =1 m k δ x k , and φ is in this cas e a co ntin uous f unction (in f act, it is piecewise a quadratic po lynomial in x with t -dep endent co efficients). The pro duct mφ in (3.1) is well- defined since the distr ibution m can b e multiplied by the contin uo us function φ . W e hope to treat w eak Lax pair s in mor e depth in future papers. Here w e just state a theorem which can b e v erified by careful use o f the calculus of distributions. Theorem 5. The fol lowing ar e e quivalent c onditions on a function u of the form (1.7) , u = P m k | x − x k | : 1. u is a we ak solution to the line arly for c e d Bu rge rs e quation (1.2) , and { x k , m k } satisfy e quations (2.1) . 2. u has a we ak L ax p air (3.1) , (3.2) . When u = P m k | x − x k | , a solution to equation (3.1) with the asy mpto tic condition φ ( x, t ; z ) = 1 for x < x 1 ( t ) will be consisten t with the time ev olu- tion g iven by (3.2) provided that w e c ho ose th e constant c = − M . Such a solution ev a luated at x > x n ( t ) will ta ke the form φ ( x, t ; z ) = A ( t ; z ) 1 2 ( x − x n ) 2 + B ( t ; z )( x − x n ) + C ( t ; z ), where all three coe fficie nt s are p olynomials in z , which , b y e q uation (3.2), satisfy ˙ A = 0, ˙ B = M B , and ˙ C = A z + 2 M C (see [1]). Th us it is consistent with equatio ns (3.1) a nd (3.2) to imp ose the condition A ( t ; z ) = 0, which together with φ = 1 for x < x 1 amounts to the b oundar y conditions φ x ( −∞ ) = φ xx ( −∞ ) = φ xx ( ∞ ) = 0. With these boundar y condi- tions in place, the problem of solving the ODEs (2.1) b ecomes an isosp ectr al deformation pr oblem which can be so lved if one knows how to solve the in verse problem for equa tion (3.1 ). This is exac tly the inv erse problem that was studied in [1] under the additional assumption that all m k (0) > 0. W e no w give a brief summary of results fro m that pa pe r . Theorem 6. The “Neumann -like discr ete cubic string” b oundary value pr oblem − ∂ 3 x φ = z mφ, φ x ( −∞ ) = φ xx ( −∞ ) = φ xx ( ∞ ) = 0 , wher e m = 2 P n k =1 m k δ x k with al l m k > 0 , has a sp e ctru m of the form { 0 = z 0 < z 1 < z 2 < · · · < z n − 1 } . Ther e is a o ne-t o-one (up t o tr anslations of m along the x axis) and onto sp e ctra l map m 7→ { M , µ } , wher e M = P m k > 0 and µ is a me asur e of the form µ = P n − 1 j =1 b j δ z j , with b j > 0 for j = 1 , . . . , n − 1 (se e details in [1]). The inverse pr oblem of r e c overing the discr ete me asur e m fr om { M , µ } has the ex plicit solution m n − k = C k D k 2 A k +1 A k , x n − k +1 − x n − k ≡ l n − k = − 2 A k D ′ k . (3.3) in terms of determinants of bimoment matric es c onstru ct e d out of the me asur e µ and the c onstant M (se e b elow). 8 W e recall the following definitions from [1]. Given a measur e µ , let β j = Z z j dµ ( z ) , I ij = I j i = Z Z z i w j z + w dµ ( z ) dµ ( w ) . (3.4) Let A 0 = B 0 = C 0 = D 0 = 1, A 1 = I 00 + 1 2 M , D ′ 1 = β 0 , and for other v a lues of k let A k =            I 00 + 1 2 M I 01 · · · I 0 ,k − 1 I 10 I 11 · · · I 1 ,k − 1 I 20 I 21 · · · I 2 ,k − 1 . . . . . . . . . I k − 1 , 0 I k − 1 , 1 · · · I k − 1 ,k − 1            , B k =          I 00 I 01 · · · I 0 ,k − 1 I 10 I 11 · · · I 1 ,k − 1 . . . . . . . . . I k − 1 , 0 I k − 1 , 1 · · · I k − 1 ,k − 1          , C k =          I 11 I 12 · · · I 1 k I 21 I 22 · · · I 2 k . . . . . . . . . I k 1 I k 2 · · · I kk          , (3.5 ) D k =          I 10 I 11 · · · I 1 ,k − 1 I 20 I 21 · · · I 2 ,k − 1 . . . . . . . . . I k 0 I k 1 · · · I k,k − 1          , D ′ k =          β 0 I 10 · · · I 1 ,k − 2 β 1 I 20 · · · I 2 ,k − 2 . . . . . . . . . β k − 1 I k 0 · · · I k,k − 2          . In all these cases, the index k a g rees with the size k × k of the deter minant. Note that A k = B k + 1 2 M C k − 1 for k ≥ 1. Let us analyze the form ula (3 .3) fo r l k in or der to compare it with (2.9) obtained earlier. First, (3.2) implies tha t the linea r ly f or ced Bur gers equation induces a very simple evolution of the measure µ , namely µ ( z ; t ) = e M t µ ( z ; 0). Because of this it is easy to factor out the time dependence from all the deter- minants inv olved in (3 .3). This elemen tar y exercise leads to l k ( t ) = l k (0) F k ( t ), where F k ( t ) = B n − k (0) e M t + 1 2 M C n − k − 1 (0) e − M t B n − k (0) + 1 2 M C n − k − 1 (0) . (3.6) This is in full agre ement with (2.6) and (2 .9). The formula fo r m k can b e chec ked in a s imilar w ay . 4 Discon tin uous piecewise linear solutions Theorem 7. The disc ont inuous pie c ewise line ar ansatz (1.9) , u = P ( m k | x − x k |− s k sgn( x − x k )) , is a we ak solution of the line arly for c e d inviscid Bu r gers e quation 9 (1.2) if and only if ˙ x k = n X i =1  m i | x k − x i | + s i sgn( x i − x k )  , ˙ m k = 2 m k n X i =1 m i sgn( x i − x k ) , ˙ s k = s k n X i =1 m i sgn( x i − x k ) , (4.1) for k = 1 , . . . , n . F or t his class of solutions, e quation (1.2) takes the form u t + 1 2 ( u 2 ) x = M 2 x − M ( M + + S ) , (4.2) with M = P m k and M + = P m k x k as b efor e, and with S = P s k . The quantities M and M + + S ar e c onstants of motion, and so is s 2 k /m k for k = 1 , . . . , n (p r ovide d that m k 6 = 0 ). Pr o of. W e will rep eatedly use the following distributional form ula v alid for an a rbitrary piecewise differe nt iable function f with p oints of discontin uity at x 1 , x 2 , . . . , x n : f x = { f x } + P n k =1 [ f ] k δ x k , where { f x } means the ordinar y deriv a- tive tak en awa y from discontin uities and [ f ] k = f ( x + k ) − f ( x − k ) deno tes the jump at x k . Mor eov er, u t = P k  ˙ m k | x − x k | − ( m k ˙ x k + ˙ s k ) sgn( x − x k ) + 2 s k ˙ x k δ x k  . Now the left-hand side of (1.2), u t + 1 2 ( u 2 ) x , must b e a function since the righ t- hand side is a function; hence a ll Dirac deltas m ust cancel out. Similarly , there m ust be no Dirac deltas in the first or second x deriv atives o f u t + 1 2 ( u 2 ) x , The s e conditions give, in turn, 0 = 2 s k ˙ x k + 1 2 [ u 2 ] k , 0 = − 2( m k ˙ x k + ˙ s k ) + 1 2 [ { ( u 2 ) x } ] k , 0 = 2 ˙ m k + 1 2 [ { ( u 2 ) xx } ] k . (4.3) An elementary computation o f jumps for the case of piecewise contin uous func- tions no w produces (4.1). The co efficients of the for cing term in the PDE are ident ified from the smooth part of the term 1 2 ( u 2 ) x , while the consta nt s of mo- tion follo w from (4.1). W eak solutions to a n initial v a lue problem ar e usua lly not unique unless the PDE is supplemented with a so-called en tropy condition that pic ks out the “physical” solution. In the case of the Burge rs equation this condition requires u to jump down, not up, at discon tinuities. This is satisfied by the a nsatz (1.9 ) if all shock str e nght s k are nonneg ative, s o we will assume s k ≥ 0 from now o n. When considering the initial v alue problem for the ODEs (4.1) w e can as- sume without loss of g enerality that x 1 (0) < x 2 (0) < . . . < x n (0). Th us on a sufficiently sma ll time in terv al we w ill still have x 1 ( t ) < x 2 ( t ) < . . . < x n ( t ); in other words, the po sitions stay in the sector X e (see (2.3)). In X e the equations 10 (4.1) can b e written as ˙ x k = n X i =1  m i sgn( k − i )( x k − x i ) − s i sgn( k − i )  , ˙ m k = 2 m k n X i =1 m i sgn( i − k ) , ˙ s k = s k n X i =1 m i sgn( i − k ) , (4.4) for k = 1 , . . . , n . These equations can b e solved explicitly , since the simple change of v ariable s in the follo wing theo rem reduces them to the O DEs alre ady solved in Theorem 2. Theorem 8. If { x k , m k , s k } n k =1 satisfy (4.4) , if al l m k (0) 6 = 0 , and if y k = x k + s k /m k , (4.5) then { y k , m k } n k =1 satisfy (2.4 ) (with y k taking the p lac e o f x k everywher e). Pr o of. Straightf or ward calculation. Note that the initital v alues y k (0) will not necess arily b e distinct or sor ted in increasing order even thoug h the x k (0)’s are, but this does not matter since the solution formulas of Theorem 2 are v alid for any initial conditions. So Theorem 2 gives us y k ( t ) a nd m k ( t ) (note that P m k y k = P ( m k x k + s k ) = M + + S replaces M + in the solution form ula (2.5)), and w e can then r ecov er s k ( t ) from the fact that s 2 k /m k is co nstant for each k ; this g ives s k ( t ) = s k (0) / p F k − 1 ( t ) F k ( t ), a nd allows us to also r ecov er x k ( t ) = y k ( t ) − s k ( t ) /m k ( t ). This solution { x k , m k , s k } to (4.4) is also the so lution to (4.1) , at le ast lo cally in some time int er v al ar ound t = 0 (so that the x k ’s remain in the sector X e ). F or illustration, here is the gener al solution with sho c ks in the ca se n = 2, when M 6 = 0 , m 1 (0) 6 = 0, m 2 (0) 6 = 0: m 1 ( t ) = m 1 (0) F 0 ( t ) F 1 ( t ) , s 1 ( t ) = s 1 (0) p F 0 ( t ) F 1 ( t ) , m 2 ( t ) = m 2 (0) F 1 ( t ) F 2 ( t ) , s 2 ( t ) = s 2 (0) p F 1 ( t ) F 2 ( t ) , x 1 ( t ) = M + + S − K m 2 (0) e − M t M − s 1 (0) m 1 (0) p F 0 ( t ) F 1 ( t ) , x 2 ( t ) = M + + S + K m 1 (0) e M t M − s 2 (0) m 2 (0) p F 1 ( t ) F 2 ( t ) , F 0 ( t ) = e − M t , F 1 ( t ) = m 1 (0) e M t + m 2 (0) e − M t M , F 2 ( t ) = e M t , K = x 2 (0) − x 1 (0) + s 2 (0) m 2 (0) − s 1 (0) m 1 (0) . (4.6) In the contin uo us case (2.1) w e as sumed all m k 6 = 0, but for (4.1) it do es make sense to have m k = 0 pro vided that the corr esp onding s k is nonzero. If 11 x − ξ ( t ) ξ ( t ) 0 σ ( t ) 2 µ ( t ) ξ ( t ) x x 0 Figure 1: Left/middle: W a ve profile u ( x, t ) as given by (4.7) a t tw o different times t < t coll , with ξ ( t ) decreasing tow ar ds z e r o at a constan t ra te. Righ t: Stationary profile after co llision ( t ≥ t coll ). m k (0) = 0, then clearly m k ( t ) = 0 for a ll t , and the ab ov e solutio n procedure do es not work. But this is ea sily fixed: just write down the g eneral solution obtained for a nonzer o initial v alue m k (0) = a , and let a → 0 there. W e will finish with a few examples that sho w ho w to deal with the solution when it hits the b oundary of the sector X e . Example. A par ticular antisymmetric s olution of (4.1) with n = 3 is given by − x 1 = x 3 ≡ ξ > 0, x 2 = 0, − m 1 = m 3 ≡ µ > 0, m 2 = 0 , s 1 = s 3 = 0, s 2 ≡ σ ≥ 0, wher e ξ ( t ) = ξ (0) F ( t ) − σ (0 ) t , µ ( t ) = µ (0) /F ( t ), σ ( t ) = σ (0) /F ( t ), with F ( t ) = 1 − 2 µ (0) t . (These formulas are obtained either b y reducing ( 4.1 ) to ODEs for ξ , µ , σ a nd so lv ing them directly; or by assuming m 2 (0) = a 6 = 0, changing v aria bles to y 1 = x 1 , y 2 = x 2 + s 2 /m 2 , y 3 = x 3 , wr iting down the general solution using Theorems 8 and 2, and letting a → 0; or simply by noting that M = 0 so that w e a re dealing with the unf or ced Burge rs equatio n whose so lutio n can be found in the textb o ok w ay using characteris tics.) Since M + + S = 2 µξ + σ is constant in time, the w av e profile (see Figure 1) is u ( x, t ) = − µ ( t ) | x + ξ ( t ) | + µ ( t ) | x − ξ ( t ) | − σ ( t ) sgn( x ) =                2 µ (0) ξ (0) + σ (0) , x < − ξ ( t ) , − 2 µ ( t ) x + σ ( t ) , − ξ ( t ) ≤ x < 0 , 0 , x = 0 , − 2 µ ( t ) x − σ ( t ) , 0 < x ≤ ξ ( t ) , −  2 µ (0) ξ (0) + σ (0 )  , ξ ( t ) < x. (4.7) If σ (0) = 0 then this is a shockless so lutio n (with n = 2 r eally , since there is neither mass nor sho ck at the site x 2 = 0). It is defined unt il ξ ( t ) = ξ (0) F ( t ) bec omes zero at time t coll =  2 µ (0)  − 1 . Then x 1 and x 3 collide at x = 0 while m 1 and m 3 blow up to −∞ and + ∞ , resp e ctively . H ow ever, u r emains bo unded, and tends to a sho ck pr ofile: u ( x, t ) → − 2 µ (0) ξ (0) sgn( x ) as t ր t coll . This illustrates that sho cks can form naturally even if they are not pres ent in the initia l w ave profile. The profile will b e statio nary after th e collis io n, bec ause its con tinued e volution is g iven by the n = 1 case of (4.1) ( ˙ x 1 = m 1 , ˙ m 1 = ˙ s 1 = 0 ) with x 1 = 0, m 1 = 0, s 1 = 2 µ (0) ξ (0). Conse quently , u ( x, t ) = − 2 µ (0) ξ (0) sgn( x ) for all t ≥ t coll . 12 x x 1 (0) = − 2 x 2 (0) = 0 x 3 (0) = 1 Figure 2: Solid: Contin uous initial w av e profile u ( x, 0 ). Dashed/ dotted: u ( x, t ) at times t = ln 4 3 and t = ln 5 3 , respectively . If σ (0) > 0 there is a s ho ck waiting at the o rigin betw een the t wo appr oaching particles (as in Figure 1). The solution hits the bounda ry of the secto r X e when ξ ( t ) b e comes zero at time t coll =  2 µ (0) + σ (0) /ξ (0)  − 1 . Then x 1 = x 2 = x 3 = 0, which illustra tes that triple collisions ma y o ccur when sho c ks are present. Since the collision o ccurs earlier than in the sho ckless case, F ( t ) has not yet r eached zero at the time o f collision; hence m 1 and m 3 do not blo w up in this case. Again, u tends to a s tationary shock profile: u ( x, t ) = −  2 µ (0) ξ (0) + σ (0)  sgn( x ) for all t ≥ t coll . Example. Consider now the sho ckless ODEs (2.1) with n = 3 and initial data m 1 (0) = 2 3 , m 2 (0) = − 1 and m 3 (0) = 4 3 , so that M = 1. W e assume x 1 (0) < x 2 (0) < x 3 (0) but lea ve them otherwise unspecified. Since u = ± ( M x − M + ) as x → ±∞ , and since the s lop e u x jumps by 2 m k at each x k , the initial profile u ( x, 0) consists of line segments with slo p e − 1, 1 3 , − 5 3 and 1, jo ined at the p oints ( x k , u ( x k , 0)). Figure 2 illustrates this for the par ticular v alues x 1 (0) = − 2, x 2 (0) = 0, x 3 (0) = 1. Note t hat if the lines u = ± ( M x − M + ) to the left a nd to the right are c o ntin ued, they intersect on the x a xis a t the center of mass x = M + / M (= 0 in the figure), whic h is a cons ta nt of motion. Recall that l 1 = x 2 − x 1 and l 2 = x 3 − x 2 . F ro m (2.6) we obtain F 0 ( t ) = e − t , F 1 ( t ) = 2 3 e t + 1 3 e − t , F 2 ( t ) = − 1 3 e t + 4 3 e − t , and F 3 ( t ) = e t . Equations (2.5) a nd (2.9) give x 1 ( t ) = x 1 (0) + (1 − e − t )  1 3 l 1 (0) + 4 3 l 2 (0)  , x 2 ( t ) = x 1 ( t ) + l 1 (0) F 1 ( t ), and x 3 ( t ) = x 2 ( t ) + l 2 (0) F 2 ( t ). The r e is a collision betw een x 2 and x 3 when F 2 ( t ) bec omes zero, which ha pp e ns a t time t = t coll = ln 2 when e t = 2. A t that time we hav e F 0 = 1 2 , F 1 = 3 2 , F 2 = 0, F 3 = 2, hence by (2 .5) m 1 = m 1 (0) /F 0 F 1 = 8 9 , 13 x x 1 ( t coll ) = − 1 x 2 ( t coll ) = x 3 ( t coll ) = 2 Figure 3: Solid: Disco ntin uo us w av e profile u ( x, t ) formed at the instant of collision t = t coll = ln 2 . Dashed: u ( x, t ) at time t = t coll + 1 2 . Dotted: No more collisions occur , and u ( x, t ) → | x | as t → + ∞ . m 2 = −∞ , m 3 = + ∞ . As fo r the w av e pr ofile u , we ha ve u ( x 1 ( t ) , t ) = m 2 l 1 + m 3 ( l 1 + l 2 ) = ( M − m 1 ) l 1 + m 3 l 2 = ( F 1 − m 1 (0) /F 0 ) l 1 (0) + m 3 (0) l 2 (0) /F 3 → 1 6 l 1 (0) + 2 3 l 2 (0) , as t ր t coll , (4.8) and u ( x 2 ( t ) , t ) − u ( x 3 ( t ) , t ) = ( m 1 l 1 + m 3 l 2 ) − ( m 1 ( l 1 + l 2 ) + m 2 l 2 ) = ( m 3 − m 1 − m 2 ) l 2 =  m 3 (0) /F 3 − m 1 (0) F 2 /F 0 F 1 − m 2 (0) /F 1  l 2 (0) → 4 3 l 2 (0) , as t ր t coll . (4.9) Thu s the limiting w av e profile at t = t coll consists of a line segmen t with slop e − 1, joined to a line s egment with slop e − 1 + 2 · 8 9 = 7 9 at x = x 1 ( t coll ) = x 1 (0) + 1 2  1 3 l 1 (0) + 4 3 l 2 (0)  and height u = 1 6 l 1 (0) + 2 3 l 2 (0); the profile ju mps down by 4 3 l 2 (0) at x = x 2 ( t coll ) = x 3 ( t coll ) = x 1 ( t coll ) + 3 2 l 1 (0), and contin ues from there with slop e 1. See Fig ure 3. The contin ued evolution of the profile for t ≥ t coll is illustrated in Figure 3; it is given by the shock ODEs (4.1) with n = 2, using a new set of v aria bles whos e initial v alues a t t = t coll are ˜ x 1 = x 1 ( t coll ), ˜ x 2 = x 2 ( t coll ), ˜ m 1 = 8 9 , ˜ m 2 = 1 9 , ˜ s 1 = 0, and ˜ s 2 = 2 3 l 2 (0). In terms o f the new tim e v ar iable τ = t − t coll ≥ 0 one finds from the g eneral solution (4 .6) that, for exa mple, ˜ x 2 ( τ ) − ˜ x 1 ( τ ) =  ˜ x 2 (0) − ˜ x 1 (0) + ˜ s 2 (0) ˜ m 2 (0)  ˜ F 1 ( τ ) − ˜ s 2 (0) ˜ m 2 (0) q ˜ F 1 ( τ ) ˜ F 2 ( τ ) , (4.10) 14 where ˜ F 1 ( τ ) = 8 9 e τ + 1 9 e − τ and ˜ F 2 ( τ ) = e τ . W riting this express io n as ˜ x 2 − ˜ x 1 = ( A + B ) ˜ F 1 − B p ˜ F 1 ˜ F 2 , we se e that it is zero if F 1 ( τ ) = 0, which can never happen, or if ( A + B ) 2 F 1 = B 2 F 2 , whic h is the same a s e − 2 τ = 9(( A + B ) /B ) 2 − 8 that can’t happ en either since the right-hand side is > 1 and the left-hand side is ≤ 1 for τ ≥ 0. The conclusio n is that, in this example, ˜ x 2 ( τ ) − ˜ x 1 ( τ ) remains po sitive for a ll τ > 0 , so there a re no mor e collis ions. Instead, as τ (or t ) → + ∞ , we hav e ˜ x 1 → 0, ˜ x 2 → + ∞ , ˜ m 1 → M , ˜ m 2 → 0, and ˜ s 2 → 0. Th us, u ( x, t ) approaches the limiting w av e profile u ( x, + ∞ ) = | x | . 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