Uniqueness of conservative solutions for the Hunter-Saxton equation

We show that the Hunter-Saxton equation $u_t+uu_x=\frac14\big(\int_{-\infty}^x d\mu(t,z)- \int^{\infty}_x d\mu(t,z)\big)$ and $\mu_t+(u\mu)_x=0$ has a unique, global, weak, and conservative solution $(u,\mu)$ of the Cauchy problem on the line.


Introduction
The Hunter-Saxton (HS) equation [14] reads Here u is an H 1 (R) function for each time t, and µ(t) is a non-negative Radon measure. Derived in the context of modeling liquid crystals, the HS equation has turned out to have considerable interest mathematically. It has, e.g., a geometric interpretation [17,19,20,21,18], convergent numerical methods [9,12], and a stochastic version [11], in addition to numerous extensions and generalizations [24,25], too many to mention here. The first comprehensive study appeared in [15,16]. While the HS equation was originally derived on differential form where, in the case of smooth functions, µ equal to u 2 x will automatically satisfy the second equation, we prefer to work on the integrated version. Note that there are several ways to integrate this equation, say u t + uu x = 1 2 x 0 u 2 x (t, y)dy, for which the uniqueness of conservative solutions on the half-line has been established in [5], but we prefer the more symmetric form. For us it is essential to introduce a measure µ(t) on the line such that for almost all times dµ = dµ ac = u 2 x dx. The times t when dµ = u 2 x dx will precisely be the times when uniqueness can break down. Our task here is to analyze this situation in detail, and restore uniqueness by carefully selecting particular solutions called conservative solutions. The aim of this paper is to show the following uniqueness result (Theorems 3.6 and 5.10): For any initial data (u 0 , µ 0 ) ∈ D the Hunter-Saxton equation has a unique global conservative weak solution (u, µ) ∈ D. Here D is given in Definition 2.2.
In the case of the so-called dissipative solutions, where energy is removed exactly at the times when the measure µ ceases to be absolutely continuous, the uniqueness question has been addressed in [7] by showing uniqueness of the characteristics.
The problem at hand can be illustrated by the following explicit example [6]. Consider the trivial case u 0 = 0, which clearly has u(t, x) = 0 as one solution. However, as can be easily verified, also is a solution for any α ≥ 0, with µ(0) = αδ 0 and dµ(t) = 4t −2 I (−αt 2 /8,αt 2 /8) (x)dx for t = 0. Here I A is the indicator (characteristic) function of the set A. Thus the initial value problem is not well-posed without further constraints.
Furthermore, it turns out that the solution u of the HS equation may develop singularities in finite time in the following sense: Unless the initial data is monotone increasing, we find (1.2) inf(u x ) → −∞ as t ↑ t * = 2/ sup(−u ′ 0 ). Past wave breaking there are at least two different classes of solutions, denoted conservative (energy is conserved) and dissipative (where energy is removed locally) solutions, respectively, and this dichotomy is the source of the interesting behavior of solutions of the equation (but see also [8,10]). We will in this paper consider the so-called conservative case where the associated energy is preserved.
However, in this case the function u = u(t, x) will in general only be Hölder and not Lipschitz continuous. This is the crux of the problem. Thus we cannot expect uniqueness of solutions of this equation. Indeed, it is precisely in the case where uniqueness fails that the HS equation encounters singularities. See [3,5,7,26,27,28]. We will reformulate the HS equation in new variables, the aim being to identify variables where the singularities disappear.
Furthermore, to study stability questions for conservative solutions the coordinates from [6] seem to be favorable, but not for investigating the uniqueness. The main difficulty stems from the fact that for each t ∈ R, the function F (t, x) = µ(t, (−∞, x)), where µ denotes a positive, finite Radon measure, is increasing but not necessarily strictly increasing. This means, in particular, that its spatial inverse χ(t, η) might have jumps. Time evolution of increasing functions with possible jumps can lead to the same problems as for conservation laws. What happens to jumps as time evolves? Do they satisfy some kind of Rankine-Hugoniot condition or do they behave more like rarefaction waves? In [6] this issue has been resolved by using the system (1.4) to show that any jump preserves position and height. Thus the associated system for (χ(t, η), U(t, η)) cannot be treated using classical ODE theory, but only in a weak sense with some additional constraints. Hence these new variables would not simplify the study of uniqueness questions.
In contrast to u(t, x), the function U (t, ξ) is Lipschitz continuous and hence the above system can be solved uniquely using the method of characteristics, which is sufficient to ensure the uniqueness of conservative solutions. In particular, it can be shown that by applying the method of characteristics the above system turns into (1.4), see Remark 3.5. Although the uniqueness question is successfully addressed, the above system has one main drawback: The definition of the function y(t, ξ) is far from unique. On the other hand, the above system can be used to find other equivalent formulations of the Hunter-Saxton equation, which might be advantageous for addressing, e.g., stability questions. As an illustration, we here introduce a novel set of coordinates, which can be studied on its own, without relying on special properties of solutions to (1.4) and which avoids the formation of jumps but requires to impose additional moment conditions. The main idea is to introduce an auxiliary measure ν, such that G(t, x) = ν(t, (−∞, x)) is strictly increasing for each t ∈ R. To that end define the auxiliary function (the power n to be fixed later) which will be a smooth function for all Radon measures µ and let U(t, η) = u(t, χ(t, η)), P(t, η) = p(t, χ(t, η)).
Provided (u, µ) is a weak, conservative solution of the HS equation, which satisfies an additional moment condition, see (5.2), we show, cf. Theorem 5.9, that the triplet (χ(t, η), U(t, η), P(t, η)) satisfies In particular, h(t, η) is continuous w.r.t. time and Lipschitz continuous w.r.t. space, so that the above system has a unique solution and can be solved by applying the method of characteristics. This is sufficient to ensure the uniqueness of conservative solutions that satisfy an additional moment condition, cf. Theorem 5.10.

Background
In this section we introduce the concept of weak conservative solutions for the Hunter-Saxton equation. Afterwards we show that there indeed exists at least one weak conservative solution to every admissible initial data. We use C ∞ c to denote smooth functions with compact support and C ∞ 0 to denote smooth functions that vanish at infinity.
As a starting point we introduce the spaces we work in.
Definition 2.1. Let E be the vector space defined by and consider the corresponding partition of unity ψ + and ψ − , i.e., ψ + and ψ − belong to C ∞ (R), and the linear mapping The mappings R 1 and R 2 are linear, continuous, and injective. Accordingly introduce E 1 and E 2 , the images of H 1 1 (R) and H 1 2 (R) by R 1 and R 2 , respectively, i.e., ) and E 2 = R 2 (H 1 2 (R)). The corresponding norms are given by Note that the mappings R 1 and R 2 are also well-defined for all (f , a) ∈ L 2 With these spaces in mind, we can define next the admissible set of initial data.
Definition 2.2 (Eulerian coordinates). The space D consists of all pairs (u, µ) such that x dx, where M + (R) denotes the set of positive, finite Radon measures on R.
A weak conservative solution is not only a weak solution of the Hunter-Saxton equation, but has to satisfy several additional conditions, which make it possible to single out a unique, energy preserving solution.
Definition 2.3. We say that (u, µ) is a weak conservative solution of the Hunter-Saxton equation with initial data (u(0, · ), There exists a set N ⊂ R with meas(N ) = 0 such that for every t ∈ N the measure µ(t) is absolutely continuous and has density u 2 x (t, · ) w.r.t. the Lebesgue measure. Note that the family {µ(t) | t ∈ R} provides a measure-valued solution w to the linear transport equation w t + (uw) x = 0. Thus one has that µ(t, R) = µ(0, R) for all t ∈ R.
In [4] weak conservative solutions in D have been constructed. A closer look at their construction reveals that the following theorem holds.
Theorem 2.4. For any initial data (u 0 , µ 0 ) ∈ D the Hunter-Saxton equation has a global conservative weak solution (u, µ) in the sense of Definition 2.3.
In other words, all the properties stated in Definition 2.3 are satisfied for the conservative solutions constructed in [4]. However, some of them are better hidden than others. This is especially true for (iv) and (vi), which we show here.
We start by recalling the set of Lagrangian coordinates.
Definition 2.5 (Lagrangian coordinates). The set F consists of all triplets (y,Ǔ ,Ȟ) such that •y ξ ≥ 0,Ȟ ξ ≥ 0 a.e., • there exists c > 0 such thaty ξ +Ȟ ξ ≥ c > 0 a.e., •Ǔ 2 ξ =y ξȞξ a.e.. Note that there cannot be a one-to-one correspondence between Eulerian and Lagrangian coordinates. Instead, one has that each element in Eulerian coordinates corresponds to an equivalence class in Lagrangian coordinates. Furthermore, all elements belonging to one and the same equivalence class can be identified using so-called relabeling functions. Definition 2.6 (Relabeling functions). We denote by G the group of homeomorphisms f from R to R such that where Id denotes the identity function.
Let X 1 = (y 1 ,Ǔ 1 ,Ȟ 1 ) and X 2 = (y 2 ,Ǔ 2 ,Ȟ 2 ) in F . Then X 1 and X 2 belong to the same equivalence class if there exists a relabeling function f ∈ G such that Then F 0 contains exactly one representative of each equivalence class in F . Note that if X = (y,Ǔ ,Ȟ) ∈ F 0 and f ∈ G, then one haš This implies that for each X = (y,Ǔ ,Ȟ) ∈ F one has thaty +Ȟ ∈ G.
Whether or not a function is a relabeling function, can be checked using the following lemma, which is taken from [13].
In [4], one rewrites the Hunter-Saxton equation, with the help of a generalized method of characteristics as a linear system of differential equations, cf. (1.4), where C = µ(0, R) = µ(t, R). This system of differential equations does not preserve F 0 , but respects equivalence classes. It can be integrated to yielď with initial data determined as introduced next in (2.8).
The connection between the pairs (u, µ) ∈ D and the triplet (y,Ǔ ,Ȟ) ∈ F is given by the following definitions.
Definition 2.8. Let the mapping L : D → F 0 be defined by L(u, µ) = (y,Ǔ ,Ȟ), wherey Now we can finally focus on showing that the weak conservative solutions constructed in [4] satisfy Definition 2.3 (iv) and (vi).

2.1.
On the Hölder continuity in the definition of weak conservative solutions. In [4] a generalized method of characteristics was used to construct weak conservative solutions as outlined above. This ansatz yields solutions u that are Hölder continuous with respect to space and time, but not Lipschitz continuous. Indeed, assume we are given a solution (u, µ) with corresponding Lagrangian coordinates (y,Ǔ ,Ȟ) satisfying (2.6). Choose two points (t 1 , x 1 ) and (t 2 , x 2 ). Then we can find ξ 1 and ξ 2 such that (2.10)y(t 1 , ξ 1 ) = x 1 andy(t 2 , ξ 2 ) = x 2 .
Moreover, one has, in general, that and hence every time wave breaking occurs, the Lipschitz continuity is lost.

2.2.
On the Lipschitz continuity in the definition of weak conservative solutions. In [4] a generalized method of characteristics was used to construct weak conservative solutions. The same approach has been used in [23], see also [6], in the case of the two-component Hunter-Saxton system, which generalizes the HS equation. However, there is a slight, but important difference in the solution spaces.
The one used in [23] is bigger, since one only assumes u(t, · ) ∈ L ∞ (R) and F (t, · ) ∈ L ∞ (R) instead of u(t, · ) ∈ E 0 2 and F (t, · ) ∈ E 0 1 . Thus one would expect that the mapping t → u(t, · ) is Lipschitz continuous from [0, T ] into L ∞ (R). Yet, a closer look at reveals that u t (t, · ) cannot be uniformly bounded in L ∞ (R), since u x (t, · ) does not belong to L ∞ (R) and hence t → u(t, · ) is not Lipschitz continuous from [0, T ] into L ∞ (R). The smaller solution space used in [4] and here, on the other hand, is the correct choice, since the right-hand side of (2.15) belongs to E 0 2 and hence the mapping On the continuity in the topology of weak convergence of measures in the definition of weak conservative solutions. In [4] a generalized method of characteristics was used to construct weak conservative solutions as outlined above. This ansatz yields measures µ, such that the mapping t → µ(t, · ) is locally Lipschitz continuous if we equip the set of positive Radon measures with the Kantorovich-Rubinstein norm, which generates the weak topology [1].
Denote by BL(R) the space of all bounded and Lipschitz continuous functions equipped with the norm Then the Kantorovich-Rubinstein norm of µ ∈ M + (R) is given by Given a solution (u, µ) with corresponding Lagrangian coordinates (y,Ǔ ,Ȟ), which satisfy (2.6). Let φ ∈ BL(R) such that φ ≤ 1. Then we have for all φ ∈ BL(R) such that φ ≤ 1, and, in particular, which proves the local Lipschitz continuity, since Ǔ (s, · ) L ∞ can be uniformly bounded on any bounded time interval. Note that we cannot expect global Lipschitz continuity in time due to the last term in the above inequality.

Uniqueness of weak conservative solutions via Lagrangian coordinates
The main goal of this section is to present the proof of Theorem 3.6. To be a bit more precise, we will show that the characteristic equatioň y t (t, ξ) = u(t,y(t, ξ)) has a unique solution and thereby establish rigorously that each weak conservative solution satisfies the system of ordinary differential equations (2.6) in Lagrangian coordinates. The pair (u, µ) will be a solution in the sense of Definition 2.3. In particular, this means that the function u(t, x) is Hölder continuous in (t, x), and the map t → u(t, · ) is Lipschitz continuous from [0, T ] into E 0 2 , the set of locally square integrable functions with possible non-vanishing asymptotics at ±∞. The measure µ(t) is finite, µ(t, R) = C, absolutely continuous and has density u 2 x (t, · ) w.r.t. the Lebesgue measure, except on a set N of zero measure. Furthermore, no moment condition is assumed on the measure here.
3.1. The differential equation satisfied by the characteristics y(t, ξ). Recall that u(t, x) is a weak solution to and hence we obtain, by computing u(t, · ) along characteristics, that Thus for every characteristic x(s) given by Recall that due to the Hölder continuity of u the equation (3.7) will in general not have a unique solution. This estimate together with the Hölder continuity of the weak conservative solution, helps us to refine the estimate for |x(t) − x(s)|. Indeed, by assumption we know that there exists a constant D such that Thus every characteristic x(t) given through (3.7), satisfies Recalling (3.8) we end up with Integration then yields for all s and t in [0, T ] that Here M denotes some positive constant, which is independent of s and t. Furthermore, using (3.8), there exists a positive constant N such that We are now ready to turn our attention towards the equation In the case of a classical solution, one has that F (t, x(t)) = F (s, x(s)). In our more general case, one has We will show this estimate in the next lemma. For simplicity we let s = 0 and only consider the right inequality.
Lemma 3.2. In the above notation, we have the following result Proof. We have that µ(t, R) = C for all t. We will first show that for a given ε > 0, we can find an M > 0 such that To that end we first observe that for t = 0 we can find anM 0 > 0 such that Let δ be a small positive number to be decided later. Since u(t, · ) ∈ E 2 for all t ∈ [0, T ], we can find u ±∞ (t), x δ , andx δ such that We then find Furthermore, using that (u, µ) is a weak solution and ψ is a test function, we get Direct computations yield This implies for all t ∈ [0, T ], using (3.14), By (3.13) the terms in the next to last line can be made arbitrarily small by increasing M , so that If we choose b ≥ u a.e., then For any t ∈ R, introduce the strictly increasing function L(t, · ) : R → R given by which satisfies, cf. (3.2) and (3.4), Then , and choosing x(s) = y(s, ξ), we get Recalling that L(t, · ) is strictly increasing we end up with Since ξ → y(t, ξ) is Lipschitz with Lipschitz constant at most one, it follows that y(t, ξ) is Lipschitz and hence differentiable almost everywhere in [0, T ] × R.
Next we aim at computing y t (t, ξ), using (3.10), and deriving the differential equation for y(t, ξ). One has, combining (3.10) and the analysis used to derive (3.12), We can derive this estimate as follows. For simplicity let s = 0 and consider only the right estimate. Lemma 3.3. In the above notation, we have the following result Proof. Given an ε > 0, we can find, as in the proof of Lemma 3.2, an M > 0 such that By (3.18) the terms in the next to last line can be made arbitrarily small by in- Letx ∈ R and consider the Hölder continuous function g(t, x) = u(0,x) + D(t + |x −x|) 1/2 , which satisfies g ≥ u a.e. by (3.9). Furthermore, let ξ = ξ(t, z) solve is a strictly increasing function. To see this, observe that one has if and thus ξ(t, · ) and φ(t, · ) are strictly increasing functions.
Since g ≥ u a.e., we have It remains to estimate ξ(t, z) − ξ(0, z). Integrating the differential equation for ξ(t, z) we find which by the Gronwall inequality implies Plugging this estimate into the integral representation of the solution, we find , which means thatM depends linearly on |ξ(0, z) −x|. On the other hand, one has for z = ξ(0, z) =x, Since the above argument holds for any choice ofx ∈ R we end up with Recalling (3.15), (3.16), and choosing x(s) = y(s, ξ), we get and, applying (3.16) once more, Since L(t, · ) is strictly increasing we end up with Note that the above inequality implies that lim s→t y(t, ξ + u(s, y(s, ξ))(t − s)) − y(s, ξ) t − s = u(t, y(t, ξ)), since combining (3.9) and (3.17) yields Thus, one has and it is left to compute Recalling (3.21), the above equality (3.22) holds since y(t, · ) is Lipschitz continuous with Lipschitz constant at most one. Moreover, note that for u(t, y(t, ξ)) = 0, one has This result also remains valid in the case u(t, y(t, ξ)) = 0. Hence we conclude that y(t, ξ) satisfies Furthermore, recalling (3.2), direct computations yield 3.2. The differential equation satisfied by U (t, ξ). To begin with we have a closer look at the system of differential equations, given by (3.23) and (3.24), which reads, using (3.3) This systems of equations can be solved (uniquely) by the method of characteristics, if the differential equation has a unique solution and k ζ (t, · ) is strictly positive for all t ∈ [0, T ]. According to classical ODE theory, (3.26) has for each fixed ζ a unique solution if the function U (t, ξ) = u(t, y(t, ξ)) is continuous with respect to time and Lipschitz with respect to space. The continuity with respect to time is an immediate consequence of (3.21). To establish the Lipschitz continuity with respect to space is a bit more involved. A closer look at (3.2) and (3.4) reveals that one has . This means especially, given ξ ∈ R, there exist ξ − ≤ ξ ≤ ξ + such that are Lipschitz continuous in space with Lipschitz constant at most one. Thus (3.26) has a unique solution. Furthermore, if k(0, ζ) = ζ for all ζ ∈ R and ζ 1 < ζ 2 , we have, as long as the function k(t, · ) remains nondecreasing that Thus k(t, · ) not only remains strictly increasing, it is also Lipschitz continuous with Lipschitz constant e t and hence according to Rademacher's theorem differentiable almost everywhere. In particular, one has that

Introducingȳ
(t, ζ) = y(t, k(t, ζ)) andH(t, ζ) =H(t, k(t, ζ)), we have from (3.25)ȳ In particular, one hasH (t, ζ) =H(s, ζ) and (3.2) turns into Furthermore, note thatȳ(t, ζ) is a characteristic due to (3.28a). Introducinḡ The above system can be extended to the system (2.6), which has been introduced in [4] and which describes conservative solutions in the sense of Definition 2.3, if we can show that As an immediate consequence, one then obtains the uniqueness of global weak conservative solutions. The proof of (3.30) is based on an idea that has been used in [2]. According to the definition of a weak solution, one has for all Note that in the above equality, one can replace and hence

Direct calculations then yield
Introduce the function and note that F ac (τ, · ) is absolutely continuous. Moreover, one has For the last two terms, observe that for every τ ∈ N , i.e., for almost every τ Thus, the dominated convergence theorem yields In particular, i.e.,Ū (t, ζ) is Lipschitz continuous with respect to time, and Moreover, one has, using (3.27) and (3.29) i.e.,Ū (t, ζ) is Lipschitz continuous with respect to space. The final step is to derive the differential equation for U (t, ξ) from (3.31). Recall that we have the relation k(t, ζ)).
Since k(t, · ) is continuous and strictly increasing, there exists a continuous and strictly increasing function o(t, · ) such that Now, given ξ ∈ R, there exists a unique ζ ∈ R such that k(t, ζ) = ξ, and thus k(t, ζ)).
Following closely [4] one can show that for each time t the triplet (y, U,H) belongs to F 0 . Furthermore, the system given by (3.25) and (3.37) can be uniquely solved in F 0 with the help of the method of characteristics. Thus we have shown the following result.  3), respectively, satisfy the following system of differential equations which can be solved uniquely in F 0 with the help of the method of characteristics. In particular, applying the method of characteristics yields the system of ordinary differential equations (2.6), which describes the weak conservative solutions constructed in [4].
Remark 3.5. Note that the above system of differential equations (3.38) is related to the system (1.4) by relabeling. Indeed, denote by (y,Ǔ ,Ȟ) the classical Lagrangian solution with initial data in F 0 , which solves (1.4), as introduced in [4].
Applying the method of characteristics to (3.38), will yield a unique solution, since the characteristic equation can be solved uniquely and k(t, · ) is strictly increasing, see the beginning of this subsection. Thus we are led to investigating the system While the last three equations coincide with the ones in (1.4), which suggests the role of the first equation has to be clarified to prove (3.40). By construction one has that (y +H)(t, ξ) = ξ for all (t, ξ), which implies that k(t, ζ) = (y +H)(t, k(t, ζ)) = (ŷ +Ĥ)(t, ζ) and in particular, k(t, · ) is the relabeling function connecting (y, U,H)(t, · ) ∈ F 0 with (ŷ,Û ,Ĥ)(t, · ) ∈ F for every t ∈ R. Furthermore, since k(t, ζ) can be recovered usingŷ(t, ζ) andĤ(t, ζ), the system (3.39) can be reduced to (1.4) and hence one obtains the weak conservative solutions constructed in [4].
We have proved the following theorem.

Introduction of an auxiliary function
As a preparation for rewriting the Hunter-Saxton equation in a set of coordinates, which shares the essential features with the Lagrangian coordinates, while at the same time avoiding equivalence classes, we introduce an, at the moment, auxiliary function p(t, x).
According to the definition of a weak solution, one has for all φ ∈ C ∞ c ([0, ∞)× R) that Furthermore, recall that F (t, x) = µ(t, (−∞, x)). If we would use the change of variables from [6], which is based on the pseudo inverse of F , one difficulty turns up immediately. The function F (t, · ), might have intervals, where it is constant and that would especially mean that its inverse would have jumps. The classical method of characteristics implies that these intervals, where F (t, · ) is constant, will change their position (i.e., they move to the right or to the left), but their length remains unchanged. This would imply that one has to deal with jumps in the inverse, and hence the involved change of variables does not simplify the problem we are interested in. Therefore, a change of variables, for proving the existence and uniqueness of conservative solutions of the HS equation, while at the same time avoiding equivalence classes, should not be based on the inverse of F , but the inverse of a strictly increasing and bounded function. This is where p(t, x) will come into the play. Let n ∈ N. Introduce the non-negative function We note the following elementary result.
Lemma 4.1. Let n ∈ N and K n be given by (4.3).
The jth derivative K (j) n satisfies where q n,j (x) denotes a polynomial with degree not exceeding 2 j+1 n − 2n − j.
Remark 4.2. We will drop the subscript n in the notation, and the value of n will only be fixed later. Thus we write What can we say about the time evolution of the function p(t, x)? We have two main ingredients: On the one hand the definition of p(t, x). Consider ψ ∈ C ∞ c (R) such that supp(ψ) ⊂ (−1, T + 1), ψ ≡ 1 on [0, T ], and 0 ≤ ψ ≤ 1. If we define φ(t, x) = ψ(t)K(x), then φ(t, x) ∈ C ∞ ([0, ∞) × R) and it can be approximated by admissible test functions. On the other hand we have the definition of a weak solution (4.1), which implies can be uniformly bounded. Therefore observe that Furthermore, since u(t, x) is a weak solution to we have that Ct for all t ≥ 0.

Recalling (2.6), direct calculations yield
Note that combining (4.5), (3.6), and Lemma 4.1, one has For later use, note that, we have where B is independent of time. A closer look reveals that Furthermore, Lemma 4.1 implies that p x (t, ·) ∈ C ∞ 0 (R) and (4.11) |p x (t, x)| ≤ np(t, x).

Uniqueness in a new set of coordinates
In this section we will rewrite the Hunter-Saxton equation in a set of coordinates, which shares the essential features with the Lagrangian coordinates, while at the same time avoids equivalence classes. However, there is a price to pay: we have to impose an additional moment condition.
Given γ > 2 and a weak conservative solution (u, µ) in the sense of Definition 2.3, such that Using the reformulation of the Hunter-Saxton equation in Lagrangian coordinates, whose time evolution is given by (2.6) and that f γ (x) = |x| γ is convex for γ > 2, it follows that We can see this as follows.
Thus we obtain the same function χ(t, η) independent of which representative in the corresponding equivalence class we choose. This is in contrast to some of the following steps, where relabeling will play a crucial role. In particular, it is then of great importance, which representative we choose from the equivalence class.

5.1.
Differentiability of χ(t, η) with respect to time. Next, we want to study the time evolution of χ(t, η). On the one hand, we will see that χ(t, η) behaves like a characteristic. On the other hand, we expect χ t (t, · ) ∈ L γ ([0, B + C]), since one has To begin with we aim at showing that χ(t, η) is differentiable with respect to time in the following sense: We establish that χ( · , η) is Lipschitz continuous and show that for each t ∈ [0, T ] one has that t − s exists for almost every η ∈ [0, B + C]. The dominated convergence theorem then implies that χ t (t, · ) ∈ L γ ([0, B + C]).
An important consequence of the above equality is, that we can choose whether we want to study the differentiability of χ(t, η) =l(t, η)− η or ofl(t, η) with respect to time. Sincel(t, · ) is the inverse toĤ(t, · ), it seems advantageous to studyl(t, η) in detail. The basis will be a good understanding of the relabeling function v(t, η) and its inverse w(t, η).
Applying (3.2) the above inequality reads As the following lemma shows a similar estimate holds for H(t, ξ). The proof relies on a detailed investigation of the function g(t, ξ) and can be found in Lemma A.1.
Furthermore, note that the above inequality implies that η(s) → η as s → t and hence ℓ(s, η) → ℓ(t, η) as s → t by (5.32). Thus we have y(t, ℓ(s, η))), if all the above limits exist. The first limit is of the form Moreover, ℓ(t, · ) is strictly increasing and continuous and hence differentiable almost everywhere. Thus, the above limit exists for almost every η ∈ R and equals ℓ η (t, η).
Combining (5.34)-(5.39) we have shown that for a given t, one has for almost every η ∈ [0, B + C] that where h(t, η) is given by (5.38). This completes the computation of ℓ t (t, η).
To summarize, we showed with the help of the dominated convergence theorem that Actually we showed that which is a limiting process in L γ ([0, B + C]). Thus the correct Banach space to work in, is L γ ([0, B + C]). This might be surprising at first sight since only χ t (t, · ) belongs to L γ ([0, B + C]) but not χ η (t, · ). On the other hand, one has that the function g(t, ℓ(t, · )) is non-decreasing and hence differentiable almost everywhere. Furthermore, (5.9) and (5.11) imply that g(t, ℓ(t, · )) is Lipschitz continuous with Lipschitz constant at most one and for almost every η ∈ [0, B + C] it follows that using (5.43), which is finite, cf. (4.10) and which implies that hχ η (t, · ) ∈ L γ ([0, B + C]). Thus we will not study (5.42) pointwise, but as a differential equation in L γ ([0, B + C]).
Theorem 5.3. The function χ(t, · ) satisfies for almost all t the following differential equation in L γ ([0, B + C]) Remark 5.4. The above computations are based on error estimates. This idea is also used when proving the chain rule in the classical setting for functions of several variables. However, one cannot use the chain rule here. Namely, writing one is tempted to look at χ(t, η) as a composition of two functions, since v(t, ξ) is differentiable almost everywhere with respect to both time and space. However, v t (t, ξ) and v ξ (t, ξ) are not continuous with respect to time and space, and this fact prevents us from using the following splitting together with a limiting process based on the existence of the partial derivatives of v. Furthermore, ℓ(t, η) is not differentiable almost everywhere, we only know that for fixed t, the derivatives ℓ t (t, η) and ℓ η (t, η) exist for almost every η.
for any test function φ ∈ C ∞ c ([0, ∞) × R). Using (5.5) and changing the coordinates according to (5.7) and (5.36), we end up with First of all note that using integration by parts (for the integral with respect to η) in the triple integral we obtain Furthermore, the right-hand side of (5.46) can be rewritten as Combining the last two equation with (5.46), we have Now we are ready to read off the differential equation for χ(t, η), since the above equality is equivalent to where h(t, η) is given by (5.38). Since the above equality must hold for any test function φ, we end up with χ t (t, η) + h(t, η)χ η (t, η) = U(t, η).

5.2.
Differentiability of U(t, η) with respect to time. To begin with we have a closer look at the differential equation (5.44), which reads where h(t, η) is given by (5.45). This equation can be solved (uniquely) by the method of characteristics, if the differential equation has a unique solution and m θ (t, · ) is strictly positive for all t ∈ [0, T ]. According to classical ODE theory, (5.47) has for each fixed θ a unique solution, if the function h(t, η) is continuous with respect to time and Lipschitz with respect to space. This is the result of the next lemma, whose proof can be found in Lemma A.2. Hence (5.47) has a unique solution.
The proof of (5.53) is again based on an idea that has been used in [2]. According to the definition of a weak solution, one has, see (2.3a) and (5.5) (5.55) In the above equality one can replace φ ∈ C ∞ c ([0, ∞) × R) by φ(t, x) such that φ(t, · ) ∈ C ∞ c (R) for all t and φ( · , x) ∈ C 1 (R).
As a closer look reveals the differential equation ( The differential equation for U(t, η). In view of (5.44) it remains to derive the differential equation for U(t, η) from (5.57). Recall that we have by (5.52) the relationŪ (t, θ) = U(t, m(t, θ)).

5.3.
Differentiability of P(t, η) with respect to time. To close the system of differential equations (5.44) and (5.60), i.e., it remains to derive the differential equation satisfied by P(t, η), given by (5.36).
Theorem 5.8. The function P(t, · ), given by (5.36), satisfies for almost all t the following differential equation in L γ ([0,

Summary.
We have shown the following result.
Theorem 5.9. The functions χ(t, η), U(t, η), and P(t, η) defined by (5.7) and (5.36), respectively, satisfy the following system of equations: which can be uniquely solved using the method of characteristics in L γ ([0, We remark that Definition 2.3, which lists all properties that have to be satisfied by weak conservative solutions, requires (5.69) u(t, · ) ∈ E 2 and µ(t, (−∞, · )) = F (t, · ) ∈ E 0 1 . In our new coordinates these conditions can be formulated in terms of Uχ η (t, η) and χ η (t, η), whose time evolution cannot be described with the help of (5.66). On the other hand, the imposed moment condition (5.2), which is preserved with respect to time, implies (5.69). Indeed by the definition of E 2 . Here we only consider one integral, since all of them can be investigated using similar ideas. One has which is finite if γ > 2 and (5.2) is satisfied.
For the second condition, recall that by the definition of E 1 0 . Again we only consider one integral, since all of them can be considered using similar ideas. One has which is finite if γ > 2 and (5.2) is satisfied. We have proved the following theorem.
Next we will show that each of the terms II 1 , II 2 , and II 3 is of order |t − s| 3/2 . Therefore it is important to keep in mind that 0 ≤ y ξ ,H ξ ≤ 1.
Then for any n ∈ N such that n ≥ γ R (1 + |x| γ )p(t, x)dx < ∞ for all t ∈ R.