BLOW-UP FOR THE TWO-COMPONENT CAMASSA–HOLM SYSTEM 4 1 KATRINGRUNERT 0 2 Abstract. Followingconservativesolutionsofthetwo-componentCamassa– n Holmsystemut−utxx+3uux−2uxuxx−uuxxx+ρρx=0,ρt+(uρ)x =0along a characteristics, we determine if wave breaking occurs inthe nearby future or J not, forinitialdatau0∈H1(R)andρ0∈L2(R). 5 2 ] 1. Introduction P A Overthelastfewyearsalotofgeneralizationsofthewell-studiedCamassa–Holm . (CH) equation, [2], h t (1.1) u u +3uu 2u u uu =0 a t− txx x− x xx− xxx m have been introduced. Among them the two-component Camassa–Holm (2CH) [ system, 1 (1.2) ut utxx+3uux 2uxuxx uuxxx+ρρx =0 − − − v ρ +(uρ) =0, 9 t x 6 which we want to consider here. It has been derived as a modell for shallow water 5 by Constantin and Ivanov [4] and has been studied intensively over the last few 6 years, see e.g. [7, 8, 12, 13, 14] and the references therein. . 1 The great interest in the CH equation and its generalizations, see e.g. [3, 5, 0 6, 10, 11, 18, 19] relies on the fact that even smooth initial data might lead to 4 classical solutions which only exist locally. In the case of the 2CH system, this 1 : is due to the fact that wave breaking can occur within finite time, that is ux v becomesunboundedfrombelowpointwisewhile u and ρ remainbounded, i k kHR1 k kL2R X see e.g. [12, 13, 14, 15, 20, 21] and the references therein. More precisely the r functionu2+ρ2 turnsintoapositive,finiteRadonmeasureatbreakingtime,which a x means that energy concentrates on sets of measure zero. Thus in order to describe weak solutions, one does not only consider pairs (u,ρ), such that u H1(R) and ρ L2(R), but triplets (u,ρ,µ) , where µ is a positive, finite Ra∈don measure ∈ ∈ D with µ =(u2 +ρ2)dx, see Definition 2.1. ac x However,oneisstillfacingtheproblem,whatwouldbeanaturalwaytoprolong the solution beyond wave breaking. One possibility is to reformulate the 2CH system in Lagrangian coordinates (y,U,h,r) (see Definition 2.2 and 2.5). A ∈ F big advantage of this approach is that the measure µ in Eulerian coordinates is mapped to the function h in Lagrangiancoordinates, 2010 Mathematics Subject Classification. Primary: 35Q53,35B35;Secondary: 35B44. Key words and phrases. Two-componentCamassa–Holmsystem,blowup,regularization. ResearchsupportedbytheAustrianScienceFund(FWF)underGrantNo.J3147. 1 2 K.GRUNERT y(ξ)=sup y µ(( ,y))+y <ξ , { | −∞ } h(ξ)=1 y (ξ). ξ − Moreover, this new setting allows to identify where wave breaking takes place by identifying the points where y (ξ)=0 and in particular, all the involved functions ξ (y,U,h,r) are bounded. The associated system of differential equations (2.10), whichdescribesthetimeevolution,providesusthenwithonepossibilitytocontinue the solution after wave breaking. However, dependent on how h is manipulated at breaking time, one obtains different kinds of solutions. The most prominent ones aretheconservativesolutions,wheretheenergycontainedremainsunchangedwith respect to time, see [7] and the dissipative ones, where a sudden drop of energy occurs at breaking time, which have been studied in [8]. Rather recently a new class of solutions has been introduced, the so-called α-dissipative solutions, which provide a continuous interpolation between conservative and dissipative solutions, [9]. The aim of this note is to show when wave breaking can occur for initial data (u ,ρ ,µ ) . All our considerations will be based on the description of conser- 0 0 0 ∈D vative solutions in Lagrangian coordinates. Therefore we summarize in Section 2 the interplay between Eulerian and Lagrangian coordinates and how the underlying system of ordinary differential equations, which describes global conservative solutionsofthe2CHsystem,lookslike. Whiledoingso,wewillfocusonwhatwave breakingmeansindifferentformulationsandwhatwecanreadofffromthe system of differential equations for the behavior before and after breaking time. Section 3 then focuses on the prediction of wave breaking by following u and ρ x along characteristics y, which solve (1.4) y (t,ξ)=u(t,y(t,ξ)) t for some given initial data y (ξ). We will prove the following result. 0 Theorem1.1. Given(u ,ρ ,µ ) anddenoteby(u(t),ρ(t),µ(t)) theglobal 0 0 0 ∈D ∈D conservative solution of the 2CH system at time t. Moreover, let C = 2(ku0k2L2R + µ (R)), then the following holds 0 (i) If ρ (x)=0 and u (x)< √2C for some x R, then wave breaking will 0 0,x − ∈ occur within the time interval [0,T], where T denotes the solution of u (x)+√2C (1.5) 0,x =exp( √2CT). u (x) √2C − 0,x − (ii) If ρ (x) = 0 and u (x) > √2C for some x R, then wave breaking 0 0,x ∈ occured within the time interval [T,0], where T denotes the solution of u (x)+√2C (1.6) 0,x =exp( √2CT). u (x) √2C − 0,x − (iii) Assumeρ (x)=0forsomex R. Letξ Rbesuchthaty (ξ)=x, where 0 0 6 ∈ ∈ y (ξ) denotes the initial characteristic curve defined by (2.6a). Denote by 0 y(t,ξ) the solution of y (t,ξ)=u(t,y(t,ξ)), then u (t,y(t,ξ)) cannot blow t x up. BLOW-UP FOR THE 2CH SYSTEM 3 2. Eulerian vs. Lagrangian coordinates The aim of this section is to present the interplay between Eulerian and La- grangiancoordinates and how the underlying system of ordinary differential equations in Lagrangian coordinates, which describes global conservative solutions of the 2CH system, looks like. Since the results stated in this section have already been proved earlier, we refer the interested reader for more details to [7], while we here focus on pointing out properties related to wave breaking. LetusfirstfocusontheinterplaybetweenEulerianandLagrangiancoordinates. Itiswell-knownthatsolutionsofthe2CHsystemmightenjoywavebreakingwithin finitetime,whichmeansthatpositiveamountsofenergycanconcentrateonsetsof measure zero. Thus the set of possible initial data, will not consist of pairs (u,ρ), where u and ρ are functions, but of triplets (u,ρ,µ), where µ is a positive, finite Radonmeasure,whichdescribesthe concentrationofenergy. The admissiblesetof Eulerian coordinates is then given by Definition2.1(Euleriancoordinates). Theset iscomposedofalltriplets(u,ρ,µ) such that u H1(R), ρ L2(R) and µ is a poDsitive, finite Radon measure whose ∈ ∈ absolutely continuous part, µ , satisfies ac (2.1) µ =(u2 +ρ2)dx. ac x The weak formulation of the 2CH system (2.2a) u +uu +P =0, t x x (2.2b) ρ +(uρ) =0, t x where 1 1 (2.3) P(t,x)= e−|x−z|u2(t,z)dz+ e−|x−z|dµ(t,z), 2ZR 4ZR hints that it could be possible to describe solutions of the 2CH system with the help of the method of characteristics. Indeed, one defines the time evolutionof the characteristic y(t,ξ) to be (2.4) y (t,ξ)=u(t,y(t,ξ)), t for some given initial data y (ξ)=y(0,ξ), and introduces the functions 0 (2.5) U(t,ξ)=u(t,y(t,ξ)) and r(t,ξ)=ρ(t,y(t,ξ))y (t,ξ), ξ whose time evolution is given through (2.2). However, this ansatz has two draw- backs, on the one hand we do not know yet how µ(t,x) evolves with respect to time, and on the other hand how to combine µ(t,x) and y(t,ξ). The last issue will be resolved next by introducing a suitable mapping from Eulerian coordinates D to Lagrangiancoordinates . F Definition 2.2. For any (u,ρ,µ) in , let D (2.6a) y(ξ)=sup y µ(( ,y))+y <ξ , { | −∞ } (2.6b) h(ξ)=1 y (ξ), ξ − (2.6c) U(ξ)=u y(ξ), ◦ (2.6d) r(ξ)=ρ y(ξ)y (ξ). ξ ◦ Then (y,U,h,r) . We denote by L: the mapping which to any element ∈F D →F (u,ρ,µ) associates X =(y,U,h,r) given by (2.6). ∈D ∈F 4 K.GRUNERT Since µ is a positive, finite Radon measure, we have that the function F(x) = µ(( ,x)) is increasing and lower semi-continuous. We therefore obtain, as an −∞ immediate consequence, that y is an increasing function, such that we can find to any x R at least one ξ R such that y(ξ) = x. In particular, we have that if ∈ ∈ a positive amount of energy concentrates at some point x, i.e. µ( x ) = c > 0, { } then(2.6a)impliesthatthepointxwillbemappedtoaninterval[ξ ,ξ ],suchthat 1 2 ξ ξ = c and y(ξ) = x for all ξ [ξ ,ξ ]. Moreover, y is Lipschitz continuous 2 1 1 2 − ∈ with Lipschitz constant at most one, and hence differentiable almost everywhere by Rademacher’s theorem. Thus y (ξ) = 0 for almost every ξ [ξ ,ξ ], which is ξ 1 2 ∈ equivalent to h(ξ) = 1 for almost every ξ [ξ ,ξ ]. Thus µ( x ) = ξ2h(ξ)dξ. In ∈ 1 2 { } ξ1 general, one has that R ξ R y (ξ)=0 = ξ R y(ξ) supp(µ ) , ξ s { ∈ | } { ∈ | ∈ } byBesicovitch’sderivationtheorem,see e.g. [1, Theorem2.22],andµ(R)= h . k kL1R The above considerations show that wave breaking in Lagrangian coordinates means that y (ξ) equals 0, and the concentrated energy can be read of from the ξ function h(ξ) integrated over ξ R y (ξ) = 0 . In particular, one can view, ξ { ∈ | } roughlyspeaking, h, whichis a function, asthe relabeledversionofthe measureµ. Therefore,(2.1) implies that hy =U2+r2 for almostevery ξ R. Since y and h ξ ξ ∈ ξ are bounded, we have that y (ξ)=0 implies U (ξ)=r(ξ)=0. ξ ξ Before introducing the set of Lagrangian coordinates, which provides us with a list of all properties inherited by (y,U,h,r) necessary for global conservative solutions to exist, we will characterize relabeling functions. These functions will not only play a key role when defining , but also when identifying equivalence F classesinLagrangiancoordinates,whichturnup sinceweintroduced4Lagrangian coordinates in contrast to 3 Eulerian coordinates, and hence a degree of freedom, which could cause trouble, when considering questions concerning the uniqueness of solutions. Definition 2.3 (Relabeling functions). We denote by G the subgroup of the group of homeomorphisms from R to R such that (2.7a) f Id and f−1 Id both belong to W1,∞(R), − − (2.7b) f 1 belongs to L2(R), ξ − where Id denotes the identity function. Given κ > 0, we denote by G the subset κ G of G defined by κ (2.8) G = f G f Id + f−1 Id κ . κ { ∈ |k − kW1,∞ − W1,∞ ≤ } (cid:13) (cid:13) At first sight it seems difficult to check if(cid:13)a function(cid:13)belongs to G or not, the following lemma simplifies this task considerably. Lemma 2.4 ([16, Lemma 3.2]). Let κ 0. If f belongs to G , then 1/(1+κ) κ ≥ ≤ f 1+κ almost everywhere. Conversely, if f is absolutely continuous, f Id ξ W1≤,∞(R), f satisfies (2.7b) and there exists d 1 such that 1/d f d −almos∈t ξ ≥ ≤ ≤ everywhere, then f G for some κ depending only on d and f Id . ∈ κ k − kW1,∞ WearenowreadytointroducethesetofLagrangiancoordinates,whichprovides uswithalistofallpropertiesof(y,U,h,r),whicharenecessaryforglobalsolutions to exist. BLOW-UP FOR THE 2CH SYSTEM 5 Definition 2.5 (Lagrangian coordinates). The set is composed of all X = F (ζ,U,h,r), such that (2.9a) (ζ,U,h,r,ζ ,U ) L∞(R) [L∞(R) L2(R)]5, ξ ξ ∈ × ∩ (2.9b) y 0,h 0,y +h>0 almost everywhere, ξ ξ ≥ ≥ (2.9c) y h=U2+r2 almost everywhere, ξ ξ (2.9d) y+H G, ∈ ξ where we denote y(ξ)=ζ(ξ)+ξ and H(ξ)= h(η)dη. −∞ R Note that (2.9c), together with (2.4) and (2.5), provides us with a possibility to define h (t,ξ) for all ξ such that y (t,ξ) = 0. Since we want to obtain solutions, t ξ 6 which are continuous with respect to time, the time evolution of the 2CH system in Lagrangiancoordinates is given through (2.10a) ζ =U, t (2.10b) ζ =U , ξ,t ξ (2.10c) U = Q, t − 1 (2.10d) U = h+(U2 P)y , ξ,t ξ 2 − (2.10e) h =2(U2 P)U , t ξ − (2.10f) r =0, t where 1 (2.11) P(t,ξ)= e−|y(t,ξ)−y(t,η)|(2U2y +h)(t,η)dη, 4ZR ξ and 1 (2.12) Q(t,ξ)= sign(ξ η)e−|y(t,ξ)−y(t,η)|(2U2y +h)(t,η)dη. −4ZR − ξ One can show that both P and Q belong to H1(R). Thissystem,whichyieldsforanyinitialdataX auniqueglobalsolutionin 0 ∈F , describes the conservative solutions. By conservative we mean that the energy, F which is given in the case of the 2CH system through (2.13) E(t)=ZR(U2yξ+h)(t,ξ)dξ =ku(t,.)k2L2R +µt(R), is preserved for all times, i.e. E(t) = E(0) for all t R. Note that E(t) is well- ∈ defined for all (y,U,h,r) . Moerover, it is not only possible to consider (2.10) ∈ F forwardin time, but also backward. We already saw that wave breaking at time t for some ξ R means that b ∈ y (t ,ξ) = U (t ,ξ) = r(t ,ξ) = 0. Then we get from (2.10) that h (t ,ξ) = ξ b ξ b b t b y (t ,ξ) = r (t ,ξ) = 0, while U (t ,ξ) = 1h(t ,ξ). Applying Lemma 2.4 to ξ,t b t b ξ,t b 2 b y + H G, we get that h(t ,ξ) = h(t ,ξ) + y (t ,ξ) > 0, which means that b b ξ b ∈ U (t,ξ) changes from negative to positive at breaking time t . We will see some ξ b consequences of this observation in the next section. Althoughrelabelingwillnotplayanimportantroleinourfurtherinvestigations, wewillstatethemostimportantresultshere,forthesakeofcompleteness. Forany X =(y,U,h,r) andanyfunctionf Gwe denote (y f,U f,h ff ,r ff ) ξ ξ ∈F ∈ ◦ ◦ ◦ ◦ 6 K.GRUNERT by X f. We say that X and X′ are equivalent, if there exists a relabeling ◦ ∈ F function f G such that X′ =X f. Let ∈ ◦ (2.14) = X y+H =Id , 0 F { ∈F | } then contains exactly one element of each equivalence class. In particular, we 0 F have that L is a mapping from to and therefore different elements in are 0 D F D mapped to different elements in , which is why the mapping L is well-defined. 0 F Moerover,letus denotebyS (X),the solutionsof (2.10)attime twithinitialdata t X, then (2.15) S (X f)=S (X) f. t t ◦ ◦ SinceX doesnotimplyS (X) forallt R,itwouldbeahugeadvantage 0 t 0 ∈F ∈F ∈ if we couldfind a mapping fromLagrangiancoordinates to Euleriancoordinates F , such that any two elements in belonging to the same equivalence class are D F mapped to the same element in Eulerian coordinates. Theorem 2.6. Given any element X = (y,U,h,r) . Then, the measure ∈ F y (r(ξ)dξ) is absolutely continuous, and we define (u,ρ,µ) as follows # (2.16a) u(x)=U(ξ) for any ξ such that x=y(ξ), (2.16b) µ=y (h(ξ)dξ), # (2.16c) ρ(x)dx=y (r(ξ)dξ), # We have that (u,ρ,µ) belongs to . We denote by M: the mapping which D F →D toanyX in associates theelement(u,ρ,µ) asgiven by (2.16). Inparticular, F ∈D the mapping M is invariant under relabeling. Finallywerecallthatthepush-forwardofameasureν byameasureablefunction f is the measure f ν(B) = ν(f−1(B)). Thus Theorem 2.6 implies that if there # exists an interval of positive length [ξ ,ξ ] such that y(ξ) = x for all ξ [ξ ,ξ ], 1 2 1 2 ∈ then µ( x )>0 and ρ(x)=0. { } 3. Wave breaking or not? After recalling the interplay between Eulerian and Lagrangian coordinates and presenting the system of differential equations describing the time evolution in Lagrangian coordinates, we want to investigate when wave breaking might occur, that means when u can become unbounded from below pointwise. x Proof of Theorem 1.1. Given some initial data (u ,ρ ,µ ) , we can find to 0 0 0 almost every x R, a ξ R such that x=y(0,ξ) and y (0,ξ∈)>D0. Let us denote ξ ∈ ∈ (3.1) = ξ R y(0,ξ),U(0,ξ) are differentiable and y (0,ξ)>0 . ξ B { ∈ | } Then the set y( ) has full measure. Indeed, we have, after a change of variables B (3.2) meas(y( )c)= dx= y dξ =0. B Z Z ξ y(B)c Bc Then for all ξ , the variables ∈B U r ξ (3.3) α(0,ξ)= (0,ξ) and β(0,ξ)= (0,ξ), y y ξ ξ BLOW-UP FOR THE 2CH SYSTEM 7 Figure 1. Plot of the vectorfield for the functions (α(t),β(t)) (with (U2 P) replaced by the constant value 5). − are well-defined. Since U =u yy and r =ρ yy , we have ξ x ξ ξ ◦ ◦ U (0,ξ) ξ (3.4) u (0,x)=u (0,y(0,ξ))= =α(0,ξ), x x y (0,ξ) ξ and r(0,ξ) (3.5) ρ(0,x)=ρ(0,y(0,ξ))= =β(0,ξ). y (0,ξ) ξ From (2.10), we get α = Uξ,tyξ−Uξyξ,t = 21hyξ+(U2−P)yξ2−Uξ2 = 1β2 1α2+(U2 P) t y2 y2 2 − 2 − ξ ξ and ry rU ξ,t ξ β = = = αβ. t − y2 − y2 − ξ ξ Thus the system under consideration will be 1 1 (3.6a) α = β2 α2+(U2 P), t 2 − 2 − (3.6b) β = αβ, t − and we want to find out when α can become unbounded from below, or in other words when wave breaking occurs. Here it is important to note, that for u H1(R) and ρ L2(R), the energy 0 0 u(t) +µ (R), is preserved (cf. (2.13)), ∈since we are cons∈idering conservative k kL2R t solutions. Thus P and P U2 can be bounded uniformly (with respect to time) − by a constant only dependent on u 2 and µ (R). In particular, we have k 0kL2R 0 (3.7) P(t,.) u 2 +µ (R), k kL∞R ≤k 0kL2R 0 and (3.8) U2(t,.) P(t,.) 2( u 2 +µ (R)). (cid:13) − (cid:13)L∞R ≤ k 0kL2R 0 We will distingui(cid:13)sh two cases for fi(cid:13)xed ξ : ∈B 8 K.GRUNERT (i): If β(0,ξ)=0, (3.6b) implies that β(t,ξ)=0 for all t 0 and(3.6a) reduces ≥ to 1 (3.9) α = α2+(U2 P). t −2 − Thus we see that α (t,ξ) 0 for α2(t,ξ) 4( u 2 +µ (R)). t ≤ ≥ k 0kL2R 0 Assume that α(0,ξ) 2 u 2 +µ (R), then we can find to any constant ≤ − qk 0kL2R 0 C <α(0,ξ)<−2qku0k2L2R +µ0(R), a time tC, such that α(tC,ξ)=C, and α(t,ξ) is strictly decreasing on the interval [0,t ]. C LetC beadecreasingsequenceofconstantssuchthatC <α(0,ξ)foralln N n n ∈ and C . Then we can associate to each C , as before, a time t , such that n n n →−∞ α(t ,ξ)=C and α(t,ξ) is strictly decreasing on the interval [0,t ]. Thus n n n (3.10) α(t ,ξ)=C , as n . n n →−∞ →∞ Unfortunately, this argument has one drawback so far, we do not know if the sequence t is convergent. However what we do know, is that the sequence t is n n increasing. This means if we can show that the sequence t is uniformly bounded n then there exists a positive time t such that b (3.11) t t , as n , n b ↑ →∞ and t is the time where wave breaking occurs, since α(t,ξ) becomes unbounded b from below. Therefore consider the differential equation 1 (3.12) γ (t)= γ2(t)+C, t −2 whereC =2( u 2 +µ (R)),withinitialdataγ(0)=α(0,ξ). Thenγ2(0) 2C >0 k 0kL2R 0 − and γ(t) is strictly decreasing. Moreover, γ(t) is negative and an upper bound for α(t,ξ), i.e. α2(t,ξ) γ2(t) > 2C, as long as α(t,ξ) is well-defined. A qualitative ≥ analysisof (3.12)showsthatγ(t),willblowup. We claimthatthishappens within finite time. If so the blow up time of γ(t) provides us with an upper bound for the sequence t . n The solution of (3.12) is given by √2Cγ(0)+2C+(√2Cγ(0) 2C)exp( √2Ct) (3.13) γ(t)= − − . γ(0)+√2C (γ(0) √2C)exp( √2Ct) − − − Acloselookrevealsthatboththe numerator,whichremainsnegativeforalltimes, and the denominator, which is positive at initial time, are absolutely bounded. Thus γ2(t) can only blow up within finite time if the denominator equals zero, i.e. γ(0)+√2C (3.14) =exp( √2Ct). γ(0) √2C − − Solving this very last equation for t, provides us with an upper bound on the sequence t and proves (3.11). n AlthoughourargumentisrigorousforaparticularsequenceC ,itislefttoshow n that t is independent of the sequence C we choose. Assume the opposite, that is b n there exist two sequences C and C˜ as above, with n n (3.15) lim t =t 0 (the case β(0,ξ)<0 is similar), we get from (3.6b) that t (3.17) β(t,ξ)=β(0,ξ)exp( α(s,ξ)ds), −Z 0 and hence β(t,ξ) 0 for all t 0. In particular, we see that β(t,ξ) increases if ≥ ≥ α(t,ξ) is negative and β(t,ξ) decreases if α(t,ξ) is positive. Note additionally that β(t,ξ) remains finite on some interval [0,T], if α(t,ξ) is bounded from below on [0,T]. We now want to show that no wave breaking can occur in this case. Let us assume the opposite. Namely, α(t,ξ) becomes unbounded from below within finite time for some fixed ξ R. Denote by t the first time when wave breaking occurs, b ∈ then (3.18) α(t,ξ) as t t . b →−∞ → Moreover,(3.17) implies that either β(t ,ξ) is positive and finite or b (3.19) β(t,ξ) as t t . b →∞ → Since the righthans side of (3.6) depends onα2(t,ξ), β2(t,ξ) and the uniformly boundedfunctionU2(t,ξ) P(t,ξ),wearegoingtoinvestigatetheinterplaybetween − α2(t,ξ) and β2(t,ξ). Therefore consider the function α2(t,ξ) (3.20) , β2(t,ξ) 10 K.GRUNERT which is bounded from below by 0. According to (3.6), we have α2 α (3.21) = (α2+β2+2(U2 P)). (cid:16)β2(cid:17)t β2 − Sinceα(t,ξ) ast t ,thereexistsatime 0 t˜ t ,suchthatα (t˜,ξ)<0, b b t →−∞ → ≤ ≤ that is (3.22) α2(t˜,ξ) β2(t˜,ξ)+2(U2(t˜,ξ) P(t˜,ξ)), ≥ − and (3.23) α2(t,ξ) 8( u 2 +µ (R)) 4 U2(t,.) P(t,.) , ≥ k 0kL2R 0 ≥ (cid:13) − (cid:13)L∞R (cid:13) (cid:13) for all t [t˜,t ). Note that the last inequality implies that α(t,ξ) < 0 for all b t [t˜,t ).∈Furthermore, α2(t˜,ξ) isfiniteandstrictlypositivedueto(3.22)and(3.23). ∈ b β2(t˜,ξ) Combining now (3.21) and (3.23), yields that the function α2(t,ξ) is decreasing on β2(t,ξ) the interval [t˜,t ), since α(t,ξ) is negative for all t [t˜,t ). This implies, since b b ∈ α2(t,ξ) α2(t˜,ξ) (3.24) 0 for all t [t˜,t ), ≤ β2(t,ξ) ≤ β2(t˜,ξ) ∈ b that lim α2(t,ξ) exists. Additionally, we have by (3.21) and (3.23), that t↑tb β2(t,ξ) (3.25) t˜ t˜ α2(s,ξ) 2(U2(s,ξ) P(s,ξ)) 0 α(s,ξ)ds α(s,ξ)+α(t,ξ) + − ds ≤Ztb ≤Ztb (cid:16) (cid:16)β2(s,ξ) β2(s,ξ) (cid:17)(cid:17) α2(t˜,ξ) α2(t ,ξ) b = < . β2(t˜,ξ) − β2(tb,ξ) ∞ Thus, according to (3.17), lim β(t,ξ) exists and t↑tb tb (3.26) limβ(t,ξ)=β(t˜,ξ)exp α(s,ξ)ds . t↑tb (cid:16)−Zt˜ (cid:17) Moerover,we get α2(t,ξ) α2(t ,ξ) (3.27) limα2(t,ξ)=limβ2(t,ξ)lim =β2(t ,ξ) b < , t↑tb t↑tb t↑tb β2(t,ξ) b β2(tb,ξ) ∞ and in particular α(t ,ξ) is finite and no blow up occurs, which contradicts our b assumption. Note, that the above considerations imply that if α (t,ξ) < 0, then β2(t,ξ) is t growing faster than α2(t,ξ). This is the main reason why wave breaking cannot occur. (cid:3) Last but not least we want to point out that if β(0,ξ) = 0, then u cannot x 6 tend to along characteristics. The proof follows the same lines as the one of ∞ Theorem 1.1 (iii), with slight modifications.

Blow-up for the two-component Camassa-Holm system

Similar Blow-up for the two-component Camassa-Holm system