A note on traveling wave solutions to the two component Camassa-Holm equation

A note on tra v eling w a v e solutions to the t w o comp onen t C amassa-Holm equati on K.Moha jer Departmen t of Mathematics and Statistics Univ ersit y of Sask atc hew an 106 Wiggins Road Sask ato on, SK, S7N 5E6 CANAD A moha jer@math.usask.ca Octob er 27 , 2018 Abstract In this paper w e show that non-smo oth functions whic h are dis tri- butional tra veling wa ve solutions to the tw o comp onen t Camass a-Holm equation are distributional tra vel ing w av e solutions to the Camassa-Holm equation pro vided that the set u − 1 ( c ), where c is the sp eed of the wa ve, is of measure zero. In particular there are no new p eak on or cuspon solutions b ey ond those al ready satisfying the Camassa-Holm equation. How ever, the tw o component Camassa-Holm equation has distinct from Camassa-Holm equation smooth trav eling w a ve solutions as w ell as n ew distributional solutions when the measure of u − 1 ( c ) is n ot zero. W e pro- vide examples of suc h solutions. Mathematics Sub ject Classification. 35Q35, 35Q53 Keywords. Camassa-Holm equation, T ra veling W av es, Peak ons The Camass a-Holm equation [1] u t + κu x − u xxt + 3 uu x = 2 u x u xx + uu xxx , (0.1) arises as a model fo r the unidirectiona l pro pagation of shallo w water wa ves ov er a flat b ottom, u ( x , t ) represe n ting the w ater’s fre e s urface, and κ ∈ R b eing a parameter related to the critica l shallow water sp eed. Camassa and Holm [1] discov ered that the equation ha s non-smo oth solitary wa ves that r etain their individual characteristics throug h the int era ction and even tually emerge with their origina l shap es and speeds. The trav eling w av e solutions of the Camassa- Holm equation ha ve b e en cla ssified by J. Lenells [4]. An alterna tiv e, and useful for genera lizations form of this equation is m t + um x + 2 mu x = 0 , (0.2) 1 where m = u − u xx + 1 2 κ . One such a genera lization has b een introduced b y M. Chen, S. Liu and Y. Zhang [2]: ( m t + um x + 2 mu x − ρρ x = 0 , ρ t + ( ρu ) x = 0 , (0.3) The traveling wa v e so lutions are obtained by setting u = u ( x − ct ) and ρ = ρ ( x − ct ). In this case, ea sy manipulations show that (0.3) can be written as follows ( − 2 c ( u ′ − u ′′′ ) + 2 κ u ′ + 3( u 2 ) ′ + (( u ′ ) 2 ) ′ − ( u 2 ) ′′′ = ( ρ 2 ) ′ , − cρ ′ + ( ρu ) ′ = 0 . (0.4) These equations are v alid in the sense o f distributions, if u ∈ H 1 loc ( R ) and ρ ∈ L 2 loc ( R ). Indeed, for a given function ρ , if ( ρ 2 ) ′ ∈ D ′ ( R ), then ρ ∈ L 2 loc ( R ). Since ev ery dis t ribution has a primitive whic h is a distribution (see [3]), we can int egr a te and then rewrite ( ( v 2 ) ′′ = ( v ′ ) 2 + p ( v ) − ρ 2 , ρv = B 1 . (0.5) where v = u − c and p ( v ) = 3 v 2 + (2 κ + 4 c ) v + K for so me constants K and B 1 . Definition 0.1 . A p air of functions ( u, ρ ) wher e u ∈ H 1 loc ( R ) and ρ ∈ L 2 loc ( R ) , is c al le d a tr aveling wave solution for (0.3) if u and ρ satisfy (0.5) in t h e sense of distributions. The following Lemma is due to J . Lene lls [4]. Lemma 0.1. L et p ( v ) b e a p olynomial with r e al c o efficient. Assume that v ∈ H 1 loc ( R ) satisfies ( v 2 ) ′′ = ( v ′ ) 2 + p ( v ) in D ′ ( R ) . (0.6) Then v k ∈ C j ( R ) for k ≥ 2 j . (0.7) In our ca s e, we hav e the following g eneralization: Lemma 0.2. L et p ( v ) b e a p olynomial with r e al c o efficients. Assume that v ∈ H 1 loc ( R ) and ρ ∈ L 2 loc ( R ) satisfy the fol lowing system in D ′  R  : ( ( v 2 ) ′′ = ( v ′ ) 2 + p ( v ) − ρ 2 , ρv = B 1 . (0.8) Then v k ∈ C j  R  for k ≥ 2 j and j ≥ 0 . (0.9) 2 Pr o of. Since v ∈ H 1 loc ( R ) and ρ ∈ L 2 loc ( R ), (0.8) implies that ( v 2 ) ′′ ∈ L 1 loc ( R ). Therefore, ( v 2 ) ′ is a bsolutely contin uous and v 2 ∈ C 1 ( R ). Also, since v ∈ H 1 loc ( R ), then v is abs olutely con tinu ous and we can claim ( v k ) ′ = k 2  v k − 2 ( v 2 ) ′  for k ≥ 3 . T o see why the claim is tr ue, we fir s t note that in fact, it is obviously tr ue if k is an even num ber . Also, note that s ince the first deriv a tiv e of an absolutely contin uous function exists almost everywhere, in taking the fir st deriv ative of the pro duct o f t wo absolutely con tinuous functions w e can use the Leibniz Rule almost everywhere. Now, if k is an odd num ber, let’s say k = 2 n + 1, then we can write ( v k ) ′ = ( v 2 n v ) ′ = v ( v 2 n ) ′ + v ′ v 2 n = v ( nv 2( n − 1) )( v 2 ) ′ + 1 2 ( v 2 ) ′ v 2 n − 1 = k 2 v k − 2 ( v 2 ) ′ . Thu s, we ha ve ( v k ) ′′ = k 2  v k − 2 ( v 2 ) ′  ′ = k 2  ( v k − 2 ) ′ ( v 2 ) ′ + v k − 2 ( v 2 ) ′′  = k ( k − 2) v k − 2 ( v ′ ) 2 + k 2 v k − 2 ( v 2 ) ′′ for k ≥ 3 . Substituting from (0 .8) w e hav e ( v k ) ′′ = k ( k − 2) v k − 2 ( v ′ ) 2 + k 2 v k − 2  ( v ′ ) 2 + p ( v ) − ρ 2  = k ( k − 3 2 ) v k − 2 ( v ′ ) 2 + k 2 v k − 2 p ( v ) − k 2 B 1 v k − 3 ρ. (0.10) F or k ≥ 3 the right hand side o f the ab o ve e q uation b elongs to L 1 loc ( R ). There- fore v k ∈ C 1  R  for k ≥ 2 . (0.11) Thu s, the a ssertion holds for j = 1. W e pr oceed by induction on j . Supp ose v k ∈ C j − 1  R  for k ≥ 2 j − 1 and j ≥ 2 . Then for k ≥ 2 j we ha ve v k − 2 ( v ′ ) 2 = 1 2 j − 1 (2 j − 1 v 2 j − 1 − 1 v ′ ) 1 k − 2 j − 1 (( k − 2 j − 1 ) v k − 2 j − 1 − 1 v ′ ) = 1 2 j − 1 ( k − 2 j − 1 ) ( v 2 j − 1 ) ′ ( v k − 2 j − 1 ) ′ ∈ C j − 2 ( R ) . (0.12) 3 Also, we ha ve v k − 2 p ( v ) ∈ C j − 1 ( R ) and v k − 3 ρ = B 1 v k − 4 ∈ C j − 2 ( R ). Therefore the right hand side of equation (0.10) b elongs to C j − 2 ( R ). Hence, v k ∈ C j  R  for k ≥ 2 j . Remark. Lemma (0.2) implies tha t v ′ is po ssibly discontin uous only at po in ts where v = 0. In fact, a muc h stro nger r esult is true: Corollary 0.1. If v ∈ H 1 loc ( R ) and ρ ∈ L 2 loc ( R ) satisfy (0.8) in D ′  R  , then v ∈ C ∞  R \ v − 1 (0)  and ρ ∈ C ∞  R \ v − 1 (0)  . Pr o of. Supp ose k ≥ 2 . The n v k ∈ C 1 ( R ). T her efore k v k − 1 v ′ = ( v k ) ′ ∈ C ( R ) . This implies that v ′ ∈ C  R \ v − 1 (0)  . Thus, v ∈ C 1  R \ v − 1 (0)  . Now, a ssume that v ∈ C j  R \ v − 1 (0)  for j ≥ 1 . F or k ≥ 2 j +1 , we hav e v k ∈ C j +1 ( R ). Therefor e k v k − 1 v ′ = ( v k ) ′ ∈ C j ( R ) . This shows that v ′ ∈ C j  R \ v − 1 (0)  . Hence, v ∈ C j +1  R \ v − 1 (0)  . Thus, u is in the desired space. No w the statemen t for ρ follows fro m the second equation of (0.5). Remark. Since v = u − c , Co rollary (0.1) shows that u ∈ C ∞  R \ u − 1 ( c )  . Since R \ u − 1 ( c ) is an o pen set, we hav e R \ u − 1 ( c ) = ∞ [ i =1 ( a i , b i ) . So, u is smo oth in ev ery interv al ( a i , b i ) where the following Lemma holds (b elo w ( a i , b i ) = ( a, b )): Lemma 0.3. L et ( u, ρ ) b e a tr aveling wave solut i on to (0.3) . Supp ose u is smo oth in t h e interval ( a, b ) . Then in the interval ( a, b ) , u satisfies the fol lowing e qu a tion: ( u − c ) 2 u ′ 2 = P ( u ) , (0.13) wher e P ( u ) = ( u 2 + κu + A )( u − c ) 2 + C ( u − c ) + B , (0.14) and A , B and C ar e some c onst a nts. 4 Pr o of. Since b oth u and ρ are s mo oth in ( a, b ) we use standard c alculus r ule s . By the first equatio n o f (0.5), we ha ve 2( v ′ ) 2 + 2 v v ′′ = ( v ′ ) 2 + p ( v ) − ρ 2 . Therefore, ( v ′ ) 2 + 2 v v ′′ = p ( v ) − ρ 2 . Multiplying by v ′ we have ( v ′ ) 3 + v  ( v ′ ) 2  ′ = v ′ p ( v ) − v ′ ρ 2 . Thu s,  v ( v ′ ) 2  ′ = v ′ p ( v ) − v ′ ρ 2 . Hence,  v ( v ′ ) 2  ′ = (3 v 2 + (2 κ + 4 c ) v + K ) v ′ − B v ′ v 2 , where B = B 2 1 . Int egr a tion yields v ( v ′ ) 2 = v 3 + ( κ + 2 c ) v 2 + K v + B v + C. Now, multiplying this equa tion by v we get v 2 ( v ′ ) 2 =  v 2 + ( κ + 2 c ) v + K  v 2 + C v + B . Substituting v = u − c and simplifying, w e hav e ( u − c ) 2 ( u ′ ) 2 = ( u 2 + κu + A )( u − c ) 2 + C ( u − c ) + B , for some consta nt A. Theorem 0.1. Supp ose ( u, ρ ) is a non-smo oth tr aveling wave solution t o (0.3) . If u − 1 ( c ) is a set of me asur e zer o, then u is a solution t o the Camassa-Holm e qu a tion. Pr o of. Supp ose ξ ∈ R \ u − 1 ( c ). Since, u − 1 ( c ) 6 = ∅ , there exists an η ∈ u − 1 ( c ) such that either ξ > η or ξ < η . Without loss of ge ne r alit y , assume that ξ < η . Let η 0 = inf { η ∈ u − 1 ( c ) : η > ξ } . Since u − 1 ( c ) is a clos e d set, η 0 ∈ u − 1 ( c ). So, ( ξ , η 0 ) ⊆ R \ u − 1 ( c ). Thus, we hav e proved tha t there ex ists an η ∈ u − 1 ( c ) such tha t either ( ξ , η ) ⊆ R \ u − 1 ( c ) or ( η , ξ ) ⊆ R \ u − 1 ( c ). Now, consider the equation (0.1 3 ) and s et F ( u ) = P ( u ) ( u − c ) 2 . W e cla im that B in (0.13) equals 0. Suppo se B 6 = 0 . Since B = B 2 1 , w e have B > 0. Then (0.14) implies that 1 p F ( u ) = 1 √ B | u − c | + O  ( u − c ) 2  u → c. 5 On the other hand, w e hav e dξ du = ± 1 p F ( u ) . Since u ∈ C ( R ), for ξ close e no ugh to η , integration yields | ξ − η | = 1 2 √ B ( u − c ) 2 + O  ( u − c ) 3  u → c. (0.15) Therefore, | ξ − η | = 1 2 √ B ( u − c ) 2  1 + O  u − c )  u → c. So, | ξ − η | 1 2 = 1 p 2 √ B | u − c | q  1 + O  u − c )  u → c. Thu s, | ξ − η | 1 2 = 1 p 2 √ B | u − c |  1 + O  u − c )  u → c. Hence, | ξ − η | 1 2 = 1 p 2 √ B | u − c | + O  ( u − c ) 2  u → c. This implies that ( u − c ) = O  ( ξ − η ) 1 2  ξ → η . Therefore, ( u − c ) 2 = O ( ξ − η ) ξ → η . Thu s, we ha ve | u − c | = q 2 √ B | ξ − η | 1 2 + O ( ξ − η ) ξ → η . (0.16) Hence, | ξ − η | − 1 2 − q 2 √ B | u − c | − 1 = O  | ξ − η | 1 2 u − c  = O (1 ) ξ → η . So, | u − c | − 1 = 1 p 2 √ B | ξ − η | − 1 2 + O (1) ξ → η . (0.17) On the other hand, from (0.14) we ha ve | u ′ | = √ B ( u − c ) − 1 + O (1) ξ → η . (0.18) 6 Now co m bining (0.18) and (0.1 7 ), we ha ve | u ′ | = 4 √ B √ 2 | ξ − η | − 1 2 + O (1) ξ → η . (0.19) Hence, u ′ / ∈ L 2 loc ( R ). This co n tradiction shows that B = 0. Therefore, the second equation of (0.5) implies that ρ = 0 a lmost ev erywhere. Now, we provide an example of a smo oth s olution of (0 .3) tha t is not a solution of Camass a -Holm equa t ion. Example. Let P ( u ) b e as in the previous Theorem. Observe that P ( u ) = ( u − G ) 2 ( u − L ) 2 if and only if          κ = 2  c − ( L + G )  , A = 2 cκ − c 2 + ( L + G ) 2 + 2 LG, C = 2 c A − κc 2 − 2 LG ( L + G ) , B = C c − Ac 2 + L 2 G 2 . (0.20) Suppo se | u | < 1 and c > 1. Therefore, if G = − 1 and L = 1, in tegration yields (1 − u ) 1 − c (1 + u ) 1+ c = e 2( ξ − ξ 0 ) . (0.21) Let’s say c = 2 a nd ξ 0 = 0. W e obser v e that the equation (1 + u ) 3 1 − u = e 2 ξ , provides a smo oth solution of (0 .3) which is not a solutio n of Cama s sa-Holm equation. See figure 1. 3 2 1 0 −1 −2 −3 4 −0.5 −4 −5 −6 1.0 0.5 0.0 −1.0 5 6 3 2 1 0 −1 −2.5 −2 −3 −4 −5 −6 −1.0 −1.5 −2.0 −3.0 6 5 4 Figure 1: ( u on the left a nd ρ on the righ t) A smo oth solution of (0.3) whic h is not a solutio n o f Cama ssa-Holm equation. The following Lemma pr o vides necessa ry and sufficient conditions for a piece- wise smo oth function to b e a distributional solutio n to (0.3) . 7 Lemma 0.4 . Supp ose u is a pie c ewise sm o oth fun ctio n. The p air ( u, ρ ) is a distributional solution to (0.3) in the sense of definition 0.1 if and only if al l of the fol lo wing c onditions hold: 1. u ∈ H 1 loc ( R ) and ρ ∈ L 2 loc ( R ) . 2. ( u − c ) 2 ∈ W 2 , 1 loc ( R ) . 3. u and ρ satisfy t h e e quation (0.5) lo c al ly with the same c onstant K on every interval wher e u is smo oth. Pr o of. The part ( ⇒ ) is easy . F or the conv erse ( ⇐ ), we note that since ( u − c ) 2 ∈ W 2 , 1 loc ( R ), then (( u − c ) 2 ) ′ is absolutely co n tin uous and has no jumps. Therefor e, (( u − c ) 2 ) ′′ defines a regular dis tribution [3] .Thus, every term in the equation (0.5) can b e represented b y an int egr a l that defines a distribution on the space of test functions a nd we are allow ed to write each in tegral as a finite sum of int egr a ls ov er lo cal interv als and use condition 3 to prov e that u and ρ satisfy (0.5) in the se ns e o f distributions. Remark. W e note that if the measure of u − 1 ( c ) is not zero, then the equation (0.5) implies that ρ 2 = K on u − 1 ( c ). How ev er in the Camassa-Holm equation if the mea sure of u − 1 ( c ) is not zero, then K = 0 beca use ρ = 0 . This implies that solutions of the form given in the following example cannot arise from the Camas s a-Holm eq ua tion. Example. Set κ = 0 . The pa ir of funct ions ( u, ρ ) g iv en b y u ( x ) = ( ce 1 −| x | if | x | > 1 , c if | x | < 1 , ρ ( x ) = ( c if | x | < 1 , 0 if | x | > 1 , is a solution to (0.3) but u is not a solution of Camassa-Holm equation. T o see this, observe that the left hand side deriv ative o f u at − 1 and the r igh t hand side deriv a tiv e of u at 1 ar e no n-zero and finite in contrast with the Camassa -Holm equation for which Lenells [4] show ed that if the measure of u − 1 ( c ) is not zero, then these limits cannot be finite. See figur e 2 . 8 x 1.5 5.0 0.0 −5.0 1.0 0.5 −2.5 2.5 2.0 0.0 Figure 2: u ( x ) is a solution of (0.3) but it is not a solution of Ca massa-Holm equation. Definition 0 .2. Supp ose f is a c ontinuous function on R . 1. We say f has a p e ak at x if f is smo oth lo c al ly on b oth sides of x and 0 6 = lim y ↓ x f ′ ( y ) = − lim y ↑ x f ′ ( y ) 6 = ±∞ . T r aveling wave solutions of (0.3) with p e aks ar e c al le d p e akons. 2. We say f has a cusp at x if f is smo oth lo c al ly on b oth sides of x and lim y ↓ x f ′ ( y ) = − lim y ↑ x f ′ ( y ) = ±∞ . T r aveling wave solutions of (0.3) with cusps ar e c al le d cusp ons. 3. We say that f has a stu mp if ther e is an interval [ a, b ] on which f is a c onst a nt and f is smo oth lo c al ly t o the left of a and to the right of b and 0 6 = lim x ↑ a f ′ ( x ) = − lim x ↓ b f ′ ( x ) . T r aveling wave solutions of (0.3 ) with stumps ar e c al le d stump ons. Note that, in the definition of a stump the limits c an b e either finite or infinite. Theorem 0.1 limits the exis tence of new distributiona l p eak on or cusp on solutions to the (0.3) . Corollary 0 .2. Every p e akon or cusp on t r aveling wave solution t o (0.3) is a tr aveling wave solution to the Camassa-Holm e quation. Finally we would like to comment on the p eak ed so lut ion rep orted in [2]. F or reasons explained below, that solutio n is not a distributional solution. First, we note that by Corollar y (0.1) the non-smo oth points of a dis tr ibut ional solution 9 u c an only appea r when u = c . Also, Lemma (0.2) s ho ws that if ( u, ρ ) is a trav eling wa ve s olution to (0.3), then ( u − c ) 2 ∈ C 1 ( R ). Now, consider the pea k ed function (see [2]) u = χ + p χ 2 − c 2 , ρ = p − cK 1  1 + r χ + c χ − c  , χ = − ( c + K 1 ) cos h( x − ct ) + K 1 , where K 1 = − 1 4 κ , K 1 < 0 and c > | K 1 | > 0 . Aw ay from it’s non-smo oth po in t, u is a solution to (0.3). How ever, It is clear that u is not smo oth at ξ = 0 and u (0) = − c . F urthermor e, ( u − c ) 2 / ∈ C 1 ( R ) b ecause lim ξ ↓ 0  ( u − c ) 2  ′ − lim ξ ↑ 0  ( u − c ) 2  ′ = − 8 c p c ( c + K 1 ) , where ξ = x − ct . Ther e fo re, u is not a distributional solutio n to (0.3) even though it sup erficially lo oks lik e a p eak on solution (see fig ure 3). −1.5 −5 x 10 0.0 5 −0.5 −1.0 0 −2.0 −10 2.5 1.5 2.75 2.25 2.0 1.75 x 10 5 0 −5 −10 Figure 3: ( u o n the left and ρ on the right) This pair is not a distributional trav eling w av e solution of (0.3). c = 2 and K 1 = − 1. Ac kno wledgment. I would like to thank professor J. Szmigielski for sug- gesting the pro blem a nd tre mendo usly helpful commen ts. References [1] Rob erto Camassa and Darryl D.Holm, An inte gr able shal low water e quation with p e ake d solitons , arXiv:patt-so l/ 9305002 v1 13 May 1993 [2] M. Cheng, S. Liu, Y. Zhang A Two-c omp onent Gener alization of the Camassa-Holm Equation and Its Solutions , Letters in Ma thema t ical Physics (2006) 75 :1-15 [3] I. M. Gel’fand a nd G. E. Shilov, Gener alize d F u nctions , Aca demic P ress, New Y ork and London, 19 64 10 [4] J. Lenells, T r aveling Wave Solutions of the Camassa-Holm Equation , J . Dif- ferential Equations, 217 (2005) 393- 430 11