The modified Camassa-Holm equation in Lagrangian coordinates

In this paper, we study the modified Camassa-Holm (mCH) equation in Lagrangian coordinates. For some initial data $m_0$, we show that classical solutions to this equation blow up in finite time $T_{max}$. Before $T_{max}$, existence and uniqueness of classical solutions are established. Lifespan for classical solutions is obtained: $T_{max}\geq \frac{1}{||m_0||_{L^\infty}||m_0||_{L^1}}.$ And there is a unique solution $X(\xi,t)$ to the Lagrange dynamics which is a strictly monotonic function of $\xi$ for any $t\in[0,T_{max})$: $X_\xi(\cdot,t)>0$. As $t$ approaching $T_{max}$, we prove that classical solution $m(\cdot ,t)$ in Eulerian coordinate has a unique limit $m(\cdot,T_{max})$ in Radon measure space and there is a point $\xi_0$ such that $X_\xi(\xi_0,T_{max})=0$ which means $T_{max}$ is an onset time of collision of characteristics. We also show that in some cases peakons are formed at $T_{max}$. After $T_{max}$, we regularize the Lagrange dynamics to prove global existence of weak solutions $m$ in Radon measure space.


Introduction
In this work, we consider the following nonlinear partial differential equation in R: x )m] x = 0, m = u − u xx , x ∈ R, t > 0, (1.1) subject to an initial condition m(x, 0) = m 0 (x). (1.2) This equation is referred to as the modified Camassa-Holm(mCH) eqaution with cubic nonlinearity, which was introduced as a new integrable system by several different researchers [14,16,24,25]. It has a bi-Hamiltonian structure [18,24] and a Lax-pair [25]. Equation (1.1) also has solitary wave solutions of the form [18]: u(x, t) = pG(x − x(t)), m(x, t) = pδ(x − x(t)), and x(t) = 1 6 where p is a constant representing the amplitude of the soliton and G(x) = 1 2 e −|x| is the fundamental solution for the Helmholtz operator 1 − ∂ xx . With this fundamental solution G, we have the following relation between functions u and m: u(x, t) = G * m = Moreover, global existence of N -peakon weak solutions of the following form was obtained in [17]: In the present paper, we study local well-posedness for classical solutions and global weak solutions to (1.1) in Lagrangian coordinates. Below we introduce the Lagrange dynamics for the mCH equation. To this end, we first review the Lagrange dynamics for incompressible 2D Euler equation: where the velocity u is determined from the vorticity ω by the Biot-Savart law involving the kernel K 2 (x) = (2π|x| 2 ) −1 (−x 2 , x 1 ). Assume X(ξ, t) is the flow map generated by the velocity field u(x, t): ßẊ (ξ, t) = u(X(ξ, t), t), ξ ∈ R 2 , t > 0, X(ξ, 0) = ξ.
(1. 6) When m 0 ∈ L 1 (R), the following useful properties can be easily obtained: In the rest of this paper, we assume the initial m 0 satisfying supp{m 0 } ⊂ (−L, L) for some constant L > 0. Next, we summarize our main results in four theorems. Here, BV (R) is the space of functions with bounded variation (see definition 5.1).
(iv) There exists a unique m(·, T max ) ∈ M(R) (Radon measure space on R) such that m(·, t) * m(·, T max ) in M(R), as t → T max .
(a) and (b) tells us that T max is an onset time of collisions of characteristics. (1.9) implies that the supports for classical solutions will not change.
Our another main theorem is about finite time blow-up behaviors and the lifespan of classical solutions. Let T max (m 0 ) be the maximum existence time of the classical solution to the mCH equation subject to an initial condition m 0 . Then we have the following theorem about lifespan for classical solutions. (1.10) (ii) If there exists ξ ∈ [−L, L] such that m 0 (ξ)∂ x u 0 (ξ) < 0, |m 0 (ξ)|(∂ x u 0 (ξ)) 2 > 1 2 ||m 0 || 3 L 1 , (1.11) then the classical solution to the mCH equation will blow up in finite time. Moreover, for any > 0 we have (1. 12) This theorem implies that there are smooth initial data with arbitrary small support and arbitrary small C k (R)-norm, k ∈ N, for which the classical solution does not exist globally.
Next, we give a theorem to show the formation of peakons at finite blow-up time T max . From Theorem 1.1, we know there is a point ξ 0 ∈ [−L, L] such that X ξ (ξ 0 , T max ) = 0. Set For any x ∈ F Tmax , because X ξ (·, T max ) ≥ 0, we know that X −1 (x, T max ) is either a single point or a closed interval. Denote " F Tmax := {x ∈ F Tmax : X −1 (x, T max ) = [ξ 1 , ξ 2 ] for some ξ 1 < ξ 2 }.
At last, we give a theorem to show global existence of weak solutions (see Definition 6.2). Theorem 1.1 (iv) tells that classical solutions become Radon measures when blow-up happens. After the blow-up time T max , we can extend our solution m(x, t) globally in the Radon measure space. We have: Theorem 1.4. Let m 0 ∈ M(R) with compact support. Then there exists a global weak solution to the mCH equation satisfying: Now, we compare the mCH equation with the Camassa-Holm (CH) equation: The CH equation was established by Camassa and Holm [6] to model the unidirectional propagation of waves at free surface of a shallow layer of water (u(x, t) representing the height of water's free surface above a flat bottom). It is also a complete integrable system which has a bi-Hamiltonian structure and a Lax pair [6]. There are some different properties between the CH equation and the mCH equation.
• Classical solutions and blow-up criteria. For a large class of initial data, classical solutions to the CH equation blow up in finite time (see [2] and references in it). Moreover, the only way that a classical solution of the CH equation fails to exist globally is that the wave breaks [10] in the sense that the solution u remains bounded while the spatial derivative u x becomes unbounded. For the mCH equation, blow-up behaviors also happen for a large class of initial data (see [8,18,22]). However, u xx (hence m) becomes unbounded when blow-up happens, while u and u x remain bounded.
• Lifespan for classical solutions. Comparing with (1.12), the lower bound for lifespan of strong solutions to the CH equation with initial data u 0 (x) is given by [13,23]: • N -peakon weak solutions. Trajectories for N -peakon weak solutions to the CH equation never collide [7,9] provided that the initial datum m 0 (x) = N i=1 p i δ(x − c i ) satisfies p i > 0 and c i = c j for i = j. However, the trajectories for N -peakon solutions of the mCH equation may collide in finite time even if m 0 ≥ 0 [17]. Moreover, for the CH equation, when blow-up happens at finite time T max , we have lim inf t→Tmax u x (x, t) = −∞ (see [10,23]). Peakon solutions u and its derivative u x are in BV space, which are bounded functions. Hence, peakon solutions can not be formed when blow-up happens (comparing with Theorem 1.3) for the CH equation.
For more results about local well-posedness and blow up behavior of strong solutions to the Cauchy problem (1.1)-(1.2), one can refer to [8,15,18,22]. For weak solutions, one can refer to [17,29].
The rest of this article is organized as follows. In Section 2, we use contraction mapping theorem to prove local existence and uniqueness of solutions X(ξ, t) to the Lagrange dynamics (1.6). Then, we use X(ξ, t) to give (u(x, t), m(x, t)) and prove that it is a unique classical solution to the mCH equation (1.1)-(1.2). Besides, when sup t∈[0,T ) ||m(·, t)|| L ∞ is finite, we can extend this classical solution in time. In Section 3, we show some blow-up criteria for classical solutions. In Section 4, we prove that for some initial data classical solutions blow up in a finite time and the estimates for blow-up rates are given. For small initial data, almost global existence of classical solutions is obtained. In Section 5, we study classical solutions at blow-up time T max . u(·, T max ) and u x (·, T max ) are BV functions while m(·, t) has a unique limit m(·, T max ) in Radon measure space as t → T max . Moreover, we prove that in some cases peakons are formed at T max . In the last section, we use regularized Lagrange dynamics to prove global existence of weak solutions in Radon measure space.

Lagrange dynamics and short time classical solutions
In this section, we study the existence and uniqueness of solutions to Lagrange dynamics (1.6). Then, we prove (u(x, t), m(x, t)) defined by (1.8) is a unique classical solution to First, let's introduce the spaces for solutions. For nonnegative integers k, n and real number T > 0, we denote and the function space Similarly, we can define C k n (R × [0, T ]). We will present the results of this section in three subsections as follows.
1. In Subsection 2.1, when m 0 ∈ C k c (−L, L), we prove local existence and uniqueness of a solution X ∈ C k+1 we can extend the classical solution in time.

Local existence and uniqueness of solutions to Lagrange dynamics
In this subsection, we use the contraction mapping theorem to prove short time existence and uniqueness of solutions to the Lagrange dynamics (1.6), which is equivalent to the following integral equation: For constants C 2 > C 1 > 0 and t 1 > 0, we define Obviously, Q t1 (C 1 , C 2 ) is a closed subset of C(U t1 ). We will look for suitable constants C 1 , C 2 , t 1 and then use the contraction mapping theorem in the set Q t1 (C 1 , C 2 ). Before presenting the existence and uniqueness theorem, we give two useful lemmas.
Lemma 2.1. Assume g ∈ L ∞ (−L, L) and X(ξ, t) ∈ Q t1 (C 1 , C 2 ) for some constants C 2 > C 1 > 0 and t 1 > 0. Let Proof. According to (2.4), X(ξ, t) is monotonic about ξ. For given (x, t) ∈ U t1 , we separate the proof into three parts. Step For (y, s) closed to (x, t) and because X ∈ C(U t1 ) is monotonic, we can assume y > X(θ, s) for θ ∈ (−L, L). A direct estimate gives Therefore, according to the uniform continuity of X, A is continuous at (x, t). The proof of the case x < X(−L, t) is similar.
Due to the continuity of X, for (y, s) closed to (x, t), there exists η ∈ [−L, L] such that X(η, s) = y. Without lose of generality, we assume ξ > η.
Then, the monotonicity of X(θ, t) implies that From the definition of Q t1 (C 1 , C 2 ), we have For (y, s) closed to (x, t), we have two cases. When y > X(L, s), we can use Step 1. When there exists ξ ∈ (−L, L) such that y = X(ξ, s), we can use Step 2.
This is the end of the proof.
We have the following existence and uniqueness theorem.

9)
and such that (2.2) has a unique solution X ∈ C k+1 Moreover, for any ∈ N, 0 ≤ ≤ k + 1, there exists a constant " C (depending on ||m 0 || C k , ||m 0 || L 1 and t 1 ) such that Proof. We separate this proof into two parts. Part I.(Existence and Uniqueness) We use the contraction mapping theorem to prove the existence of a unique solution X ∈ C 0 1 (U t1 ) to (2.2). Step 1. When 0 < t 1 <
At last, by the contraction mapping theorem, the system (2.2) (or (1.6)) has a unique solution in C(U t1 ).
On the other hand, using Lemma 2.1 we can see Hence, X ∈ C 0 1 (U t1 ) and Part I is finished. Part II. (Regularity) We show X obtained in the first part belongs to C k+1 1 (U t1 ). From the first part, we can see solution X belongs to C 0 1 (U t1 ). For this solution we have the following properties On the other hand, G(x) = 1 2 e −|x| satisfies: Hence, We obtain (2.21) Because X(ξ, t) is monotonic about ξ, its derivative exists for a.e. ξ ∈ [−L, L]. Differentiating with respect to ξ shows that for a.e. ξ ∈ [−L, L], Due to (2.18) and (2.19), the sum of the last two terms in (2.22) is zero, which leads to (2.24) Differentiating (2.23) with respect to ξ shows that Hence, we obtain X ξξ ∈ C(U t1 ) and We have X ∈ C 2 1 (U t1 ). Similarly, taking derivative about ξ for k times on both sides of (2.23) gives that and (2.12) holds.
Remark 2.1. Monotonicity of X(·, t) plays an important role in our proof. Without monotonicity, the vector field for the Lagrange dynamics may be not Lipschitz. From (2.23), we know supp{X ξ (·, t) − 1} ⊂ (−L, L). Hence, we can continuously extend X ξ (·, t) globally as

Classical solutions to the mCH equation
Next, we prove the short time existence and uniqueness of the classical solutions to (1.1)-(1.2).
The following lemma shows that we can construct classical solutions to the mCH equation (1.1)-(1.2) from the solutions to the Lagrange dynamics (1.6). Moreover, we show that the support of m(·, t) will not change.
Next, we present a useful lemma which is similar to Lemma 2.1.
However, in order to make no confusion, we still use [X(−L, t), X(L, t)] in this proof. By using the inverse function theorem, for any t ∈ [0, t 1 ], there is a continuously differ- Moreover, we have 1 Changing variable and using the property of Dirac measure, we have Next, we separate the proof into three parts, which is similar to the proof of Lemma 2.1. Step In this case, we have A(x, t) = 0. For any (y, s) closed to (x, t) and because X ∈ C(U t1 ), we can assume y ≥ X(L, s). Because g(·, s) ∈ C c (−L, L), we have A(y, s) = 0. Hence, A is continuous at (x, t).
Step 2. Continuity at ( Due to the continuity of X, for (y, s) closed enough to (x, t), we can assume y ∈ [X(−L, s), X(L, s)]. In other words, there exists η ∈ [−L, L] such that X(η, s) = y. Because we only have to prove Z and Z x are continuous at (x, t). (2.5) shows that Hence, A(x, t) is continuous at (x, t).
Step 3. x = X(L, t). The case x = X(−L, t) is similar. For (y, s) closed to (x, t), we have two cases. When y > X(L, s), we can use Step 1. When there exists ξ ∈ (−L, L) such that y = X(ξ, s), we can use Step 2.
Part I. Regularity.
Step 3. If k > 2, we can keep using integration by parts and Lemma 2.4 and obtain Step 4. Because m = u−u xx , from the above steps, we already know ∂ k Taking derivative of both sides of (2.31), we have and d dt From the above proof (or Lemma 2.3), we can see that u(x, t), m(x, t) is a classical solution to (1.1)-(1.2).
Part II. Uniqueness of the classical solution to (1. ). We prove that u 1 (x, t) can also be defined by the solution X(ξ, t) to (1.6), which means (2.34) By standard ODE theory, we can obtain a solution Y ∈ C k+1 Taking derivative with respect to ξ shows thaṫ (2.35) This implies Hence, we can see Step 2. We prove Y (ξ, t) = X(ξ, t). From (2.37), we obtaiṅ which means that Y (ξ, t) is also a solution to (1.6). From Theorem 2.1 we know that the strictly monotonic solution to (1.6) is unique. Therefore, to prove Y (ξ, t) = X(ξ, t), we only have to prove Y (·, t) is strictly monotonic for t ∈ [0, t 1 ].
Combining (2.35) and (2.36) gives that , the minimum and maximum of ( and The strictly monotonic property of X plays an crucial role in the proof of the above Theorem. Whenever X is strictly monotonic, we can use integration by parts to obtain the regularity of u(x, t). Conversely, if m(x, t) is a classical solution, then the characteristics for the mCH equation is strictly monotonic.
For the convenience of the rest proof, we summarize the results in the proof of Part II of Theorem 2.2 and give a corollary. and Proof. The proof for (2.39) and (2.40) is the same as the proof for uniqueness in Theorem 2.2.
Remark 2.4. From (2.40), we know that m(X(θ, t), t) does not change sign for any t ∈ [0, T ]. We present a precise argument here. Set Hence, For t ∈ [0, T ], denote and Due toẊ This can also be obtained by (2.23).

Solution extension
In this subsection, we will show that as long as classical solutions to (1.1)-(1.2) satisfying ||m(·, t)|| L ∞ < ∞ we can extend the solutions X and m in time. ||m(·, t)|| L ∞ < +∞, then there exists T 0 > T 0 such that is a solution to (1.6), and is a solution to (1.1)-(1.2).
Proof. There exists a constant M ∞ satisfies From Lemma 2.3, we know m(·, t) has a uniform (in t) support. Hence, there exists a constant › M 1 such that sup . Our target is to prove that the classical solution can be extend to T 0 : > T 0 . We will show this in two steps.
Step 1. In this step we consider a dynamic system from time T 1 .

Blow-up criteria
In this section, we give some criteria on finite time blow-up of classical solutions to the mCH equation.
Let T max > 0 be the maximal existence time of classical solution to the mCH equation. In other words, T max satisfies Next lemma shows that the solution to Lagrange dynamics (1.6) can be extended to the blow-up time T max . Then we have Proof. Let t go to T max in (2.21) and we obtain X(ξ, T max ). Using (2.24) and Lipschitz property of G(x) = 1 2 e −|x| , we can obtain that Let t go to T max in (2.23) and (2.25). Similarly, combining (2.12) gives Keep doing like this and we can see At last, let t go to T max in (2.20) and combining (1.6), we have ∂ t X ∈ C([−L, L] × [0, T max ]).
We have the following blow up criteria.
is the solution to Lagrange dynamics (1.6). Assume T max < +∞ is the maximum existence time for the classical solution to (1.1)-(1.2). Then, the following equivalent statements hold.
Proof. We follow the following lines to prove this theorem, Step 1. We prove ( This is a contradiction to (3.2).
Hence, we can see that (3.8) and (3.5) are equivalent.
Step 6. At last, we prove   (3.3) shows that there is a ξ 0 such that X ξ (ξ 0 , T max ) = 0. This means T max is an onset time of collision of characteristics. Now, we can conclude that if m(x, t) blows up in finite time T max , then we have ). The blow-up criterion (3.5) can also be found in [18]. Besides, (3.7) is similar to the well known blow-up criterion for smooth solutions to 3D Euler equation [1].  Another proof for (3.7). By Theorem 2.1 and Theorem 2.2, we know m ∈ C k 1 (R×[0, T max )). From (1.1), we obtain Multiplying both sides by m x and taking integral show that Integration by parts for the last term implies that On the other hand, we have

Finite time blow up and almost global existence of classical solutions
In the rest of this paper, we assume m 0 ∈ C 1 c (−L, L). In this section, we show that for some initial data solutions to the mCH equation blow up in finite time. Some blow-up rates are obtained. Moreover, for any > 0 and initial data m 0 (x) ∈ C 1 c (R), we prove that the lifespan of the classical solutions satisfies where C is a constant depends on m 0 (x). Our finite time blow-up results are similar to the blow-up results in [8,18,22] but with some subtle differences. All these three papers apply the idea from transport equation and focus on the derivative of u 2 − u 2 x which is 2mu (4.1) (ii) We have the following lower bound for blow-up time Proof. (i) The mCH equation (1.1) can be rewritten as Therefore, we have d dt m(X(ξ, t), t) = −2(m 2 u x )(X(ξ, t), t).

In view of equation (4.3), the most natural way to study blow-up behavior is following the characteristics. This method was used for the Burgers equation and the CH equation. Equality (4.5) reminds us the proof for finite time blow-up of Burgers equation:
u t + uu x = 0, for x ∈ R, t > 0. (4.6) Consider its characteriesẊ(x, t) = u(X(x, t), t) and we have d dt u(X(x, t), t) = 0.
Taking derivative of (4.6) gives Then we have Hence, if there exists x 0 ∈ R such that u 0x (x 0 ) < 0, then u x goes to −∞ in finite time.  Proof. From (1.1), we obtain Taking derivative to (4.9) with respect to x yields After some calculation we obtain For the last two terms on the right side, integration by parts shows that Hence Young's inequality and (1.7) give that which implies (4.8).
Next, we state and prove our main results in this section.
Moreover, when T max = t * , we have the following estimate of the blow-up rate for m: for t ∈ [0, T max ), (4.12) and for X ξ we have Proof.
Similarly, we have the following theorem.
Moreover, when T max = t * , we have the following estimate of the blow-up rate for m(x, t): and for X ξ we have where (1.7) was used. Together with (1.10) we can obtain (1.12).

Solutions at blow-up time and formation of peakons
In this section, we study the behavior of classical solutions at blow-up time T max . First, we show that u and u x are uniformly BV function for t ∈ [0, T max ] (including the blow-up time T max ) and m(·, t) has a unique limit in Radon measure space as t approaching T max .
Let us recall the concept of the space BV (R).
(ii) (Equivalent definition for one dimension case) A function f belongs to BV (R) if for any {x i } ⊂ R, x i < x i+1 , the following statement holds:  (1.6). Then, the following assertions hold: (i) There exists a function u(x, T max ) such that and Proof. We use three steps to prove (i) and (ii) together.
Step 3. We prove (5.1) and show that u(x, T max ) satisfies (5.2). The first part of (5.1) is deduced by u ∈ C(R × [0, T max ]). To prove the second part, we have to do a little more job.
Combining (1.7), step 2, and [3, Theorem 2.3], we know that there exists a consequence {t k }(→ T max ) and two BV functions u(x), v(x) such that for every x ∈ R and we obtain This is the end of the proof.
Next we give a theorem to prove that m(·, t) has a unique limit in Radon measure space M(R) as t approaching T max . Before this, let's recall the definition A + t and A − t in Remark 2.4 and denote Because X(ξ, T max ) may not be strictly monotonic, it is not obvious to see that A + Tmax and A − Tmax are open sets. We give a lemma to show this. Because m 0 (ξ 0 ) > 0, we know η 1 < ξ 0 < η 2 and m 0 (ξ) > 0 for ξ ∈ (η 1 , η 2 ). Hence Because X(ξ, T max ) is nondecreasing, we obtain 2. If ξ 1 < ξ 2 , we have By definition we know m 0 (ξ i ) ≥ 0 for i = 1, 2. When m 0 (ξ i ) = 0 for i = 1 or i = 2, from Remark 2.4 we know X ξ (ξ i , T max ) = 1. This implies that X(ξ, T max ) is strictly monotonic in a neighborhood of ξ i which is a contradiction with (5.3). Hence, we have m 0 (ξ i ) > 0, for i = 1, 2.
Because u x (·, T max ) is a BV function, its derivative u xx (·, T max ) is a Radon measure. We know m(·, T max ) = u(·, T max ) − u xx (·, T max ) is a Radon measure and for any test This proves (5.4).
For φ ∈ C ∞ c (R) and φ ≥ 0, by the definition of A + we have Hence, m + Tmax is a positive Radon measure. With the same argument, we can see that m − Tmax is a negative Radon measure.
On the other hand, by using (5.9), we have which implies (5.6).
For any test function φ ∈ C ∞ c (R), we have This ends the proof.
Remark 5.2. In Section 6, we will prove the global existence of weak solutions to the mCH equation when initial data m 0 belongs to M(R). Hence, we can extend m globally in time after blow up time. Similar results can be found in [17], where a sticky particle method was used.
Step 1. We first consider the cases when x ∈ (X(−L, T max ), X(L, T max )). Because m 0 (L) = 0, from Remark 2.4 we know X ξ (L, T max ) = 1, which means X(ξ, T max ) is strictly monotonic in a small neighborhood of L. Hence, From this we know, if x ≥ X(L, T max ), we have x − X(ξ, T max ) > 0 for ξ ∈ (−L, L). From Theorem 5.1, we know Thus
Step 2. We only left the case for Using X η (η, T max ) > 0, we can obtain Hence, Taking derivative again shows that Because m 0 ∈ C 1 c (R) and X ξ (·, T max ) ∈ C 1 (−L, L), which implies Together with Step 1 and Step 2, we obtain Step where (5.8) was used. Because φ is arbitrary and u(·, T max ) ∈ C 3 (R \ F Tmax ), we obtain From Theorem 5.2, we know m(·, T max ) has compact support in (−L, L). Hence, Because the Radon measure m(·, T max ) has finite total variation, we obtain From (5.13), we know where x ∈ O Tmax and X(η, T max ) = x. This means (2.40) holds in the set O: This finishes our proof.
Because u(·, T max ) and u x (·, T max ) are BV functions, their discontinuous points are countable. We give a proposition to show discontinuous points of u x (·, T max ). First, let us introduce two subsets of F Tmax .
Step 1. Assume y ∈ ‹ F Tmax and we prove u x (·, T max ) is continuous at y. By definition of F Tmax , we know there is only one point ξ 0 ∈ F , such that X(ξ 0 , T max ) = y. Due to (5.11), there exist two sequence {y n } and { y n } such that the following hold: Because y n ∈ O Tmax , there is a unique ξ n ∈ O such that X(ξ n , T max ) = y n . Similarly, we have a unique ξ n ∈ O such that X( ξ n , T max ) = y n . (Uniqueness is because X(ξ, T max ) is strictly monotonic in O.) Moreover, we have ξ n < ξ 0 < ξ n , and lim n→+∞ ξ n = ξ 0 = lim n→+∞ ξ n .
Because formula (5.12) holds for x ∈ O Tmax , we know Let n goes to infinity and we obtain Similarly, we have and This implies u x (y−, T max ) = u x (y+, T max ). For any y ∈ ‹ F Tmax , define Then using similar argument for any sequence R \ " F Tmax y n → y, we know Step 2. Assume y ∈ " F Tmax and we prove u x (·, T max ) is discontinuous at y. Set ξ 1 = min{ξ ∈ F : X(ξ, T max ) = y} and ξ 2 = max{ξ ∈ F : X(ξ, T max ) = y}.
From (5.12), we know Let n go to +∞ and we obtain Similarly, we also have Hence, using the above claim, we have m 0 (θ)dθ = 0 (5. 16) which shows that u x (·, T max ) is not continuous at y.
Next, we prove Theorem 1.3. Let's give some notations first.
. From the proof (5.16), we know that for each 1 ≤ i ≤ N there exist ξ i1 < ξ i2 such that Proof of Theorem 1.3. For any text function φ ∈ C ∞ c (R), we have Because u x (·, T max ) ∈ C k+2 (R \ F Tmax ), integration by parts leads to Because u x (·, T max ) is continuous at x i for i ≥ N + 1, combining (5.14) and (5.17) gives that This theorem tells us that peakons are exactly the points in the set " F Tmax . Hence, a peakon is formulated when some Lagrangian labels in a interval [ξ 1 , ξ 2 ] aggregate into one point at T max and the weight of the peakon is the integration of m 0 (x) on [ξ 1 , ξ 2 ].

Solutions after blow-up.
At the blow up time, the solution to the mCH equation m becomes a Radon measure. In this section, we assume initial data m 0 belongs to the Radon measure space M(R) and use the Lagrange dynamics to prove that weak solution to (1.1)-(1.2) exists globally in Radon measure space.  (ii) For each > 0, set With this definition, we define Hence, G ∈ C k (R) for k ≥ 2. By Young's inequality we have and On the other hand, because G xx ∈ C[− , ], there is a constant > 0 such that Hence, G x (x) is a global Lipschitz function. For any measurable function X(ξ, t), we define and m (·, t) := X (·, t)#m 0 (·). (6.4) By the definition, we have Hence, we have the following relation between m and m m ( In the following of this paper, we denote Hence, we haveẊ (ξ, t) = U (X (ξ, t), t). We have the following Lemma about u .
Lemma 6.1. Let m 0 ∈ M(R) satisfy (6.1). For > 0, u (x, t) is defined by (6.3). Then, the following statements hold:  Moreover, for any T > 0, there exist subsequences of u , u x (also denoted as u , u x ) and two functions u, u x ∈ BV (R × [0, T )) such that and u, u x satisfy all the properties in (i), (ii) and (iii).
Proof. (i) From (6.2) and the definition of u , we can easily obtain (i).
Similarly, we can obtain The rest results can be obtained by using [3, Theorem 2.4,2.6].

Weak consistency and convergence theorem
In this subsection, we show that u defined by (6.3) is weak consistent with the mCH equation (1.1)-(1.2). We rewrite (1.1) as equation of u, Now, we introduce the definition of weak solution in terms of u. To this end, for φ ∈ C ∞ c (R × [0, T )), we denote the functional Then, the definition of the weak solution to (1.1) in terms of u(x, t) is given as follows.  On the other hand, combining the definition (6.3) and (6.8) gives Combining the last two equalities, we define We now state the main result of this section.
Proposition 6.1. We have the following estimate |E | ≤ C .
The constant C is independent of .
Proof. By the definition of m and m , the first term in (6.10) can be estimated as For the second term of (6.10), because ρ is an even function, by the definition of U we can obtain U m − U m , φ x This ends the proof.
Next, we state our main theorem in this section, which contains Theorem 1.4. Proof.
For each φ ∈ C ∞ c (R × [0, +∞)), there exists T = T (φ) such that φ ∈ C ∞ c (R × [0, T )) We now consider convergence for each term of L(u , φ), For the first term, because supp{φ} is compact, we can see The second term can be estimated as follows Similarly, we obtain Combining the above estimates and Proposition 6.1 gives This proves that u is a global weak solution to the mCH equation.
For any test function φ ∈ C 1 c (R × [0, T )), integrating by parts and using the relationship m = (1 − ∂ xx )u imply that Taking → 0, the right hand side of the above equality converges to Hence, m * m in M(R × [0, T )). This ends the proof.
The weak solution is unique when u ∈ L ∞ (0, ∞; W 2,1 (R)). Moreover, examples about nonuniqueness of peakon weak solutions can also be found in [17]. Notice that peakon solutions are not in the solution class W 2,1 (R).