This is the accepted version of the journal article: Clop, Albert; Jylhä, Heikki; Orobitg i Huguet, Joan; [et al.]. «Well-posedness for the continuity equation for vector fields with suitable modulus of continuity». Journal of Functional Analysis, Vol. 276, Issue 1 (January 2019), p. 45-77. DOI 10.1016/j.jfa.2018.10.001 This version is available at https://ddd.uab.cat/record/288144 under the terms of the license WELL-POSEDNESS FOR THE CONTINUITY EQUATION FOR VECTOR FIELDS WITH SUITABLE MODULUS OF CONTINUITY ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG Abstract. We prove well-posedness of linear scalar conservation laws using only assumptions on the growth and the modulus of continuity of the velocity field, but not on its divergence. As an application, we obtain uniqueness of solutions in the atomic Hardy space, H 1 , for the scalar conservation law induced by a class of vector fields whose divergence is an unbounded BM O function. 1. Introduction The scalar continuity equation ( d ρ + div(b ρ) = 0 dt (1) ρ(0, ·) = ρ0 appears in many conservation phenomena in nature. Among the most rep- resentative ones, we mention the so called aggregation equation (see for instance [8]), which describes the time evolution of an active scalar ρ(t, ·) with initial state ρ0 , under a velocity field b(t, ·) = K ∗ ρ(t, ·). The dependence of b with respect to the unknown ρ(t, ·) makes the aggrega- tion equation nonlinear, while the analytic nature of the kernel K = ∇N (N is the Newtonian potential) gives it a gradient flow structure. In the present paper, though, we omit both nonlinearities and gradient flow structure, and focus our attention on much simpler linear case. It is known by the so-called superposition principle (see for instance [3]) that if a vector field b admits a unique flow φ of trajectories t 7→ φt (x) then non-negative solutions to (1) are uniquely determined by ρ0 . For this reason, in the present paper we focus on the much more subtle question of signed solutions. Having a mass conservation structure, one may think that continuity equa- tions have L1 as the natural function space for solutions. It is then reason- able to think its adjoint equation, also known as transport equation, ( d dt ω + b · ∇ω = 0, ω(0, ·) = ω0 . Date: January 25, 2017. 1 2 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG in the L∞ setting. Indeed, the classical theory by DiPerna and Lions [16], together with Ambrosio’s more recent developments [1], provides an ex- istence, uniqueness and stability theory for solutions u ∈ L∞ (0, T ; L∞ ) for any Sobolev vector field b which, among other assumptions, satisfies div(b) ∈ L∞ . There is, though, an increasing interest in removing both boundedness assumptions on the divergence [11, 12, 13, 15, 21] as well as on the datum [14]. In this setting, the Euler system of equations deserves a special mention. Its scalar formulation in the plane, dω i (2) + v · ∇ω = 0, where v(t, ·) = ∗ ω(t, ·), dt 2πz has the structure of a nonlinear transport equation, together with the initial i condition v(0, ·) = 2πz ∗ ω(0, ·). After Yudovich’s proof of global existence and uniqueness of solutions ω ∈ L∞ (0, T ; L∞ ) for ω0 ∈ L1 ∩ L∞ [28], there has been many attempts to understand the case of unbounded vorticities. Particular attention is devoted to spaces that stay close to BM O, the space of functions of bounded mean oscillation, which arises naturally since it con- tains the image of L∞ under any Calderón-Zygmund singular integral op- erator. It is conjectured that well-posedness of (2) fails in BM O. Also, strong ill-posedness has been proven in certain Sobolev spaces included in BM O [9]. However, there exist classes of unbounded vorticities for which (2) is well-posed [6, 7]. In a similar way, the present paper deals with the well-posedness of (1) in the atomic Hardy space H 1 , which is the predual of BM O, and which consists of L1 functions having L1 image under any Calderón-Zygmund operator. Cauchy problems for the linear continuity equation with non-smooth ve- locity fields were successfully treated with the DiPerna-Lions scheme and the notion of renormalized solution, as well as the more recent extensions by Ambrosio in the BV setting. In both approaches, the starting point is the classical Cauchy-Lipschitz theory, which allows to write the solution ρ = ρ(t, x) of ( d (3) dt ρ + div(b ρ) = 0 ρ(0, ·) = ρ0 as the adjoint composition operator, (4) ρ(t, x) = ρ0 ◦ φ−1 −1 t (x) J(x, φt ) where φt : Rn → Rn is the flow generated by the velocity field b, ( φ̇t (x) = b(t, φt (x)), (5) φ0 (x) = x, and J(x, φ−1t ) denotes the jacobian determinant of the inverse flow φt −1 at the point x, at least for smooth enough b. Abusing of the classical mass transport notation, one could equivalently write ρ(t, ·) = (φt )] ρ0 . Towards finding explicit solutions ρ ∈ L∞ (0, T ; H 1 (Rn )) of the problem (3) for a given ρ0 ∈ H 1 (Rn ), there are two things to be analyzed. First, describing the class Q of diffeomorphisms φt under which (4) defines a bounded operator in H 1 (Rn ). Second, describing the class of velocity fields b such that (5) has a WELL-POSEDNESS FOR CONTINUITY EQUATION 3 solution φt that falls into Q. Both questions drive us naturally to Reimann, who partially solved them both in two papers in the 70’s [23, 22]. In the first one, quasiconformality was found to be the fundamental notion. In the second, uniform bounds for the anticonformal part of Db were proven to be enough. Let us recall that the anticonformal part of Db is, by definition, Db(t, x) + DT b(t, x) divb(t, x) SA b(t, x) = − In , 2 n where In denotes the n-dimensional identity matrix. 1,1 Theorem 1. Let b : [0, T )×Rn → Rn be a vector field, b ∈ L1 (0, T ; Wloc (Rn )), such that SA b ∈ L1 (0, T ; L∞ (Rn )). Assume also that |b(t, x)| ∈ L1 (0, T ; L∞ (Rn )). 1 + |x| log(e + |x|) For every h0 ∈ H 1 (Rn ), the problem ∂t h + div(bh) = 0 in (0, T ) × Rn  (6) h(0, ·) = h0 in Rn . admits a unique weak solution h ∈ L∞ (0, T ; H 1 (Rn )). The essential point for existence is the fact that quasiconformal maps trans- port boundedly measures with H 1 (Rn ) density into themselves. For this, we rely on previous results by Reimann [23, 14] and Fefferman’s H 1 − BM O duality. In the proof of uniqueness in Theorem 1, most of the available literature cannot be used, due to the unboundedness of div(b). However, we remark two exceptions. First, a work by Ambrosio - Bernard [2], where uniqueness is obtained from diagonal modulus of continuity assumptions on b, together with global boundedness. Unfortunately, this was not enough for proving Theorem 1, which includes many unbounded vector fields. We found a sec- ond exception in a paper by Seis [25], which provides a detailed stability estimate for solutions u ∈ L1 (0, T ; Lq (Rn )) of continuity equations with a 1,p velocity field b ∈ Wloc (Rn ; Rn ), p1 + 1q = 1, p ∈ (1, ∞]. It turns out that a modification of the argument in [25] brings the following uniqueness result for solutions in L1 (0, T ; L1 (Rn )) (actually a more general one, see Section 4 for details). Proposition 2. Let b : [0, T ] × Rn → Rn be a vector field. Assume that (A1) |b(t, x)| ≤ C(1 + |x|), and that (A2) |b(t, x) − b(t, y)| ≤ C |x − y| log(e + 1/|x − y|), of all x, y ∈ Rn and almost every t ∈ [0, T ]. If ρ0 ∈ L1 (Rn ), then the Cauchy problem ∂t ρ + div(bρ) = 0 in (0, T ) × Rn  ρ(0, ·) = ρ0 in Rn has at most one weak solution ρ ∈ L1 (0, T ; L1 (Rn )). 4 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG The above result still does not provide a complete proof for uniqueness in Theorem 1. Indeed, there are vector fields b admissible for Theorem 1 for which (A1) and (A2) may both fail at the same time. It is worth mention- ing too that existence of solutions may fail in the setting of Proposition 2, since flows arising from the above velocity fields b do not preserve Lebesgue measurable sets in general. See [11, 12, 13] for optimal conditions in this direction. In particular, these flows could move an initial datum ρ0 away from L1 (Rn ) and transform it into a non absolutely continuous measure. It is then natural to look at the case where ρ(t, ·) belongs to the space of signed Borel measures on Rn with finite total variation, which is denoted by M(Rn ). It turns out that this is the right choice for a full well-posedness theorem. Theorem 3. Let b : [0, T ]×Rn → Rn be a vector field such that the following two conditions hold: There exists ´ ∞ adscontinuous and nondecreasing function G : [0, ∞) → (0, ∞) satisfying r G(s) = ∞ for some r > 0 such that |b(t, x)| (B1) sup ∈ L∞ (0, T ), x∈Rn G(|x|) ´ r dsa continuous and nondecreasing function ω : [0, ∞) → [0, ∞) and there exists satisfying 0 ω(s) = ∞ for some r > 0 such that |b(t, x) − b(t, y)| (B2) sup ∈ L∞ (0, T ). x,y∈B(0,R) ω(|x − y|) for any radius R > 0. Then for any ρ0 ∈ M(Rn ) there exists a unique solution (as defined in Definition 4) ρ ∈ L1 (0, T ; M(Rn )) for the Cauchy problem = 0 in (0, T ) × Rn  ∂t ρ + div(bρ) (7) ρ(0, ·) = ρ0 in Rn . Moreover, this solution is of the form ρ(t, ·) = (φt )] ρ0 , where φt is the flow of the vector field b. It is worth mentioning that Theorem 3 holds under slightly more general- ity. Namely, if one further knows that ρ ∈ L∞ (0, T ; M(Rn )), then unique- ness still holds if one replaces L∞ (0, T ) by L1 (0, T ) in both (B1) and (B2). The argument for obtaining existence in Theorem 3 is straightforward, be- cause of to the measure-valued setting and the continuity of b. Concerning uniqueness, the proof is much trickier. Exactly as in Proposition 2, the proof expands even further the methods used by Seis in [25] using opti- mal transport. Among the consequences, uniqueness in Theorem 1 now follows from Theorem 3. Moreover, it also provides uniqueness of solutions in L1 (0, T ; L1 (Rn )) for the continuity equation in these cases where one can guarantee the preservation of Lebesgue measurable sets under the flow. This is the case for instance if   L div(b) ∈ Exp , log L log log L · · · WELL-POSEDNESS FOR CONTINUITY EQUATION 5 see also [13]. Nevertheless, conditions (B1), (B2) are neither a consequence of, nor a reason for any condition on the divergence of b. Some counterex- amples in this direction are given at the end of Section 5. The paper is structured as follows. In Section 2 we give some definitions and basic results for continuity equations. In Section 3 we overview some results from optimal transport theory. In Section 4 we prove a slightly more general version of Proposition 2, which allows us to present the ideas behind the proof of Theorem 3 without getting hung up on all the details. In Section 5 we prove Theorem 3 and give some counterexamples for the optimality of its assumptions. In Section 6 we prove Theorem 1. All along the paper we will be abusing notation, so that we identify ρ(t, ·) = ρt and φ(t, ·) = φt . Acknowledgements. A.C., J.M. and J. O. were partially supported by re- search grants 2014SGR75 (Generalitat de Catalunya) and MTM2016-75390- P (spanish government). All named authors were partially supported by the research grant FP7-607647 (European Union). 2. Basic properties of continuity equations Let us first introduce our notation. By M+ (Rn ) we denote the set of finite nonnegative Borel measures on Rn . We denote by M(Rn ) the space of signed Borel measures on Rn with finite total variation, that is, µ ∈ M(Rn ) if µ is a signed Borel measure and nX o kµkT V := sup |µ(Aj )| : Aj ∈ B(Rn ), Ai ∩ Aj = ∅ for i 6= j < ∞, j∈N where B(Rn ) is the Borel σ-algebra. For µ ∈ M(Rn ) we denote by |µ| the total variation measure of µ. We use the Jordan decomposition of measures: Any µ ∈ M(Rn ) can be written as µ = µ+ − µ− for some mutually singular measures µ+ , µ− ∈ M+ (Rn ) Given µ ∈ M(Rn ) and a homeomorphism φ : Rn → Rn we define the push-forward φ] µ by φ] µ(A) := µ(φ−1 (A)) for any A ∈ B(Rn ). Equivalently, one can define φ] µ by duality with the space C0 (Rn ) of con- tinuous functions vanishing at infinity, ˆ hφ] µ, ϕi := ϕ(φ(x)) dµ(x), ∀ϕ ∈ C0 (Rn ). Then it is easy to see that φ] µ ∈ M(Rn ) with kφ] µkT V ≤ kµkT V . Definition 4. A measure valued map ρ ∈ L1 (0, T ; M(Rn )) is a weak solu- tion of the Cauchy problem (7) for the continuity equation if the equality ˆ Tˆ ˆ   ∂t ϕ(t, x) + hb(t, x), ∇ϕ(t, x)i dρt (x) dt + ϕ(0, x) dρ0 (x) = 0 0 Rn Rn holds for all functions ϕ ∈ Cc∞ ([0, T ) × Rn ). Let’s point out some simple properties of solutions, which we will prove for completeness. First is related to the weak continuity of solutions. 6 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG Lemma 5. Let b ∈ L∞ (0, T ; L∞ n 1 n loc (R )). Let ρ ∈ L (0, T ; M(R )) be a solution for the Cauchy problem (7). Then we have the convergence ˆ ˆ t→0 ϕ(x) dρt (x) −→ ϕ(x) dρ0 (x) Rn Rn for every ϕ ∈ ∞ n Cc (R ). Proof. Let ϕ ∈ Cc∞ (Rn ). Fix t0 ∈ (0, T ) and ε ∈´(0, t0 ). Choose a nonnega- tive function ηt0 ,ε ∈ Cc∞ (t0 − ε, t0 + ε) such that ηt0 ,ε (s) ds = 1, and define ´t ψt0 ,ε (t) := 1 − 0 ηt0 ,ε (s) ds. Then using ψt0 ,ε ϕ as a test function we get ˆ T ˆ ˆ T ˆ 0 ψt0 ,ε (t) ϕ dρt dt + ψt0 ,ε (t) b(t, x), ∇ϕ(x) dρt (x) dt Rn Rn 0 ˆ 0 + ϕ dρ0 = 0. Rn Taking ε → 0 we obtain (for a.e. t0 , if we are precise) ˆ ˆ t0 ˆ ˆ (8) − ϕ dρt0 + b(t, x), ∇ϕ(x) dρt (x) dt + ϕ dρ0 = 0. Rn 0 Rn Rn Thus we can deduce that ˆ ˆ ˆ t0 ˆ ϕ dρt0 − ϕ dρ0 ≤ b(t, x), ∇ϕ(x) dρt (x) dt → 0, Rn Rn 0 Rn as t0 → 0, which is what we wanted to prove.  Before the next Lemma we need to introduce speciffic cut-off functions that will be used throughout the paper. Let k ∈ N and consider the function ( 1, if |x| ≤ k χ̃k (x) = ´ |x| dr max{0, 1 − k G(r) }, if |x| > k. Modifying this function we find a smooth, compactly supported function χk ∈ Cc∞ (Rn ) such that χk = 1 in B(0, k), χk = 0 in Rn \ B(0, Rk ) for some k < Rk < ∞ (9) 2 and |∇χk (x)| ≤ . G(|x|) ´ ∞ ds Here the existence of Rk follows from the assumption r G(s) = ∞ for some (and thus any) r > 0. The second important property of solutions is the conservation of mass, or from the point of view of this paper, the conservation of mass balance. Lemma 6. Suppose the vector field b satisfies the condition (B1) and that ρ ∈ L1 (0, T ; M(Rn )) is a weak solution for the Cauchy problem  ∂t ρ + div(bρ) = 0 (10) ρ(0, ·) = 0 ´ Then for a.e. t ∈ [0, T ] it holds that ρt (Rn ) = Rn dρt = 0, or in other words: − ρ+ n n t (R ) = ρt (R ). WELL-POSEDNESS FOR CONTINUITY EQUATION 7 Proof. Use a test function ϕ(t, x) = ψ(t)χk (x), where ψ ∈ Cc∞ (0, T ). This gives ˆ T ˆ ˆ T ˆ 0 ψ (t) χk (x) dρt (x) dt = − ψ(t) b(t, x), ∇χk (x) dρt (x). 0 Rn 0 Rn Since this works for every ψ ∈ ∞ Cc (0, T ), we obtain ˆ  ˆ ∂t χk (x) dρt (x) = b(t, x), ∇χk (x) dρt (x). Rn Rn Combining this with Lemma 5 we get ˆ n |ρt (R )| = lim χk (x) dρt (x) k→∞ Rn ˆ t ˆ ≤ lim sup b(s, x), ∇χk (x) dρs (x) ds k→∞ 0 Rn ˆ tˆ |b(t, x)| ≤2 lim sup d|ρs |(x)ds = 0, G(|x|) L∞ ((0,T )×Rn ) k→∞ 0 Rn \B(0,k) and thus conclude the proof.  In the smooth case, the solution to the continuity equation can be found using the method of characteristics. Actually, this works also in our non- smooth case. Proof of the existence part of Theorem 3. Let φ be the flow of b. This flow exists by the classical Cauchy-Lipschitz theory, and moreover it satisfies ˆ t b s, φ(s, x) ds for any x ∈ Rn .  (11) φ(t, x) = x + 0 We need to show that ρt := φ(t, ·)] ρ0 is a solution to the Cauchy problem (7). First, the properties of the push-forward ensure that we actually have ρt ∈ L∞ (0, T ; M(Rn )), since kρt kT V is bounded by kρ0 kT V . Let ϕ ∈ Cc∞ ([0, T ) × Rn ). First notice that ϕ is Lipschitz and t 7→ φ(t, x) is absolutely continuous for any x ∈ Rn . This implies that the function fx (t) := ϕ(t, φ(t, x)) is absolutely continuous for any x ∈ Rn and ˆ T (12) fx0 (t) dt = fx (T ) − fx (0) = −ϕ(0, x). 0 On the other hand if we fix x ∈ Rn , then for L1 -a.e. t ∈ [0, T ) we have fx0 (t) = ∂t ϕ(t, φ(t, x)) + ∇ϕ(t, φ(t, x)), ∂t φ(t, x) (13) = ∂t ϕ(t, φ(t, x)) + b(t, φ(t, x)), ∇ϕ(t, φ(t, x)) . Now we can check that ρt is a solution. Using the definition of the push- forward measure and applying Fubini’s Theorem, (13) and (12) we calculate ˆ Tˆ ˆ   ∂t ϕ(t, x) + b(t, x), ∇ϕ(t, x) dρt (x) dt + ϕ(0, x) dρ0 (x) 0 Rn Rn ˆ T ˆ ˆ   = ∂t ϕ(t, φ(t, x)) + b(t, φ(t, x)), ∇ϕ(t, φ(t, x)) dρ0 (x) dt + ϕ(0, x) dρ0 (x) 0 Rn Rn ˆ ˆ T ˆ = fx0 (t) dt dρ0 (x) + ϕ(0, x) dρ0 (x) = 0. Rn 0 Rn 8 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG Since ϕ ∈ Cc∞ ([0, T )×Rn ) was arbitrary, φ(t, ·)] ρ0 is a solution to the Cauchy problem (7).  3. Useful results from the theory of optimal transport In this section we present the part of the theory of optimal transport which we will use in the uniqueness proof. For further interest to this topic we refer to the books [27], [26] by Villani, as well as [24] by Santambrogio and [4] by Ambrosio, Gigli and Savaré. We want to transport mass in Rn but our use of cut-off functions might lead to a difference between the initial and final mass. This could be a problem. The typical way (see e.g. [17], [10] and [19]) to avoid this mass inbalance is to add an isolated point ♦ to Rn . We write R̂n := Rn ∪ {♦}. We also ”extend” the euclidean distance to R̂n by setting |x − ♦| = |♦ − x| = ∞ whenever x ∈ Rn and obviously |♦ − ♦| = 0. Given two Borel measures µ, ν ∈ M+ (R̂n ) with µ(R̂n ) = ν(R̂n ) we de- note the set of transport plans between µ and ν by Π(µ, ν). In other words, a measure λ ∈ M+ (R̂n × R̂n ) belongs to Π(µ, ν), if λ(A × R̂n ) = µ(A) and λ(R̂n × A) = ν(A) for all Borel sets A ⊂ R̂n . Equivalently, λ ∈ Π(µ, ν) if ˆ ˆ ˆ  u(x) + v(y) dλ(x, y) = u(x) dµ(x) + v(y) dν(y) R̂n ×R̂n R̂n R̂n for all u ∈ L1 (µ), v ∈ L1 (ν). If in addition to the measures µ, ν ∈ M+ (R̂n ) we are given a continuous cost function c : R̂n × R̂n → [0, ∞], we can study the optimal transport problem ˆ inf c(x, y) dλ(x, y). λ∈Π(µ,ν) R̂n ×R̂n The existence of minimizers, called optimal transport plans, can be proved with the direct method in the calculus of variations (see e.g. [26, Theorem 4.1]). In our case, we are interested in cost functions of the form c(x, y) = c(|x − y|). With our notation, this means that c(x, ♦) = c(♦, x) = c(∞) is a constant among all x ∈ Rn . We assume that c : [0, ∞] → [0, ∞] satisfies the following conditions:    c(0) = 0, c(s) > 0 for every s > 0 and c is nondecreasing c is bounded and continuous  (14)   c is concave c is Lipschitz w.r.t. the euclidean distance in [0, ∞).  We set ˆ Cc (µ, ν) := min c(|x − y|) dλ(x, y) λ∈Π(µ,ν) R̂n ×R̂n We will study the total transport cost Cc (µ, ν) for a certain family of cost functions and then compare the estimates for these costs to the special case ˆ W (µ, ν) := min min{|x − y|, 1} dλ(x, y), λ∈Π(µ,ν) R̂n ×R̂n WELL-POSEDNESS FOR CONTINUITY EQUATION 9 which is used as a reference cost in our considerations. The key comparison between transportation costs Cc (µ, ν) and W (µ, ν) is given by the following Lemma, which is a generalization of [25, Lemma 5] . Lemma 7. Let µ, ν ∈ M+ (R̂n ) with µ(R̂n ) = ν(R̂n ) and let c : [0, ∞] → [0, ∞] be continuous and strictly increasing. Given any ε > 0 we have the upper bound   −1 Cc (µ, ν) Cc (µ, ν) W (µ, ν) ≤ c µ(R̂n ) + ε + , ε c(1) if Cc (µ,ν)  ε ∈ c [0, ∞] . Proof. Define the following sets in R̂n × R̂n :   Cc (µ, ν) K1 = |x − y| ≤ 1 : c(|x − y|) ≤ , ε   Cc (µ, ν) K2 = |x − y| ≤ 1 : c(|x − y|) > , ε  and K3 = |x − y| > 1 . Let λ be an optimal plan for Cc (µ, ν). Using λ as a test plan for W (µ, ν) we see that ˆ W (µ, ν) ≤ min{|x − y|, 1} dλ(x, y) R̂n ×R̂n ˆ ˆ = min{|x − y|, 1} dλ(x, y) + min{|x − y|, 1} dλ(x, y) K1 K2 ˆ + min{|x − y|, 1} dλ(x, y) K3   −1 Cc (µ, ν) ≤c λ(K1 ) + λ(K2 ) + λ(K3 ). ε Here it is easy to see that ˆ ˆ c(|x − y|) ε λ(K2 ) = dλ(x, y) ≤ c(|x − y|) dλ(x, y) ≤ ε K2 c(|x − y|) Cc (µ, ν) K2 and similarly ˆ c(|x − y|) Cc (µ, ν) λ(K3 ) = dλ(x, y) ≤ , K3 c(|x − y|) c(1) which concludes the proof.  If c satisfies the conditions (14), then c(|x − y|) gives a metric in R̂n . For this kind of cost functions the optimal transport problem admits a dual formulation due to the celebrated Kantorovich Duality Theorem (see e.g. [26, Theorem 5.10]). 10 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG Theorem 8. Let µ, ν ∈ M+ (R̂n ) with µ(R̂n ) = ν(R̂n ) > 0. Suppose c : [0, ∞] → [0, ∞) satisfies conditions (14). Then we have the duality: ˆ  min c(|x − y|) dλ(x, y) : λ ∈ Π(µ, ν) R̂n ×R̂n (ˆ ˆ ) |v(x) − v(y)| = max v dµ − v dν : sup ≤ 1, v(♦) = 0 . R̂n R̂n x6=y c(|x − y|) In addition, the maximizers of the dual problem, called Kantorovich poten- tials, satisfy v(x) − v(y) = c(|x − y|) for every (x, y) ∈ suppλ, where λ ∈ Π(µ, ν) is an optimal plan for Cc (µ, ν). Since it is important for us, we emphasize that in our case any test func- tion v for the dual problem is bounded and Lipschitz in the euclidean dis- tance: (15) |v(x) − v(y)| |c(t) − c(s)| sup |v(x)| ≤ c(∞) and sup ≤ sup , x∈R̂n x,y∈Rn |x − y| t,s∈[0,∞) |t − s| In addition, information about the gradient of the Kantorovich potentials can also be extracted from the Duality Theorem. Corollary 9. Suppose c : [0, ∞] → [0, ∞) is C 1 and satisfies (14). Let µ, ν ∈ M+ (R̂n ) with µ(R̂n ) = ν(R̂n ). Let λ ∈ Π(µ, ν) be an optimal plan for Cc (µ, ν) and let v be a corresponding Kantorovich potential. Then there exists a set E ⊂ Rn such that Ln (Rn \ E) = 0 and for any (x, y) ∈ suppλ ∩ (E × E) such that x 6= y we have x−y ∇v(x) = ∇v(y) = c0 (|x − y|) |x − y| In addition, ∇v(x) = 0, if x ∈ E and (x, ♦) ∈ suppλ or (♦, x) ∈ suppλ. Proof. Since v is Lipschitz in the euclidean distance, we can apply the Rademacher Theorem to find a set E ⊂ Rn such that Ln (Rn \ E) = 0 and v is differentiable in E. Now, given (x, y) ∈ suppλ ∩ (E × E) we have by the Duality Theorem c(|x − y|) + v(y) = v(x) = 0inf n {c(|x − y 0 |) + v(y 0 )}. y ∈R Thus, if x 6= y we can differentiate the function f (y 0 ) = c(|x − y 0 |) + v(y 0 ) at its minimum point y ∈ E, which gives y−x x−y c0 (|x − y|) + ∇v(y) = 0, i.e. ∇v(y) = c0 (|x − y|) . |x − y| |x − y| Similarly, the Duality Theorem implies c(|x − y|) − v(x) = −v(y) = 0inf n {c(|x0 − y|) − v(x0 )}, x ∈R from which we obtain by differentiation x−y ∇v(x) = c0 (|x − y|) . |x − y| WELL-POSEDNESS FOR CONTINUITY EQUATION 11 If x ∈ E and (x, ♦) ∈ suppλ, then using the Duality Theorem we see that c(∞) − v(x) = 0 = 0inf n {c(∞) − v(x0 )}, x ∈R and again differentiation gives ∇v(x) = 0. The case where (♦, x) ∈ suppλ can be handled similarly.  In the uniqueness part of the proof on Theorem 3 we will deal with certain integrals involving the gradient of a Kantorovich potential. These integrals can be estimated easily using optimal transport theory: Corollary 10. Suppose c : [0, ∞] → [0, ∞] is C 1 and satisfies (14). Let µ, ν ∈ M+ (R̂n ) with µ(R̂n ) = ν(Rn ) and let v : R̂n → R be a Kantorovich potential for Cc (µ, ν). Assume that µ and ν are mutually singular, µ, ν << Ln and b ∈ L1 (µ) ∩ L1 (ν). Then ˆ hb, ∇vid(µ − ν) Rn x−y ≤ min{µ(Rn ), ν(Rn )} sup c0 (|x − y|) b(x) − b(y), , x,y∈F |x − y| where F := (suppµ ∪ suppν) ∩ Rn . Proof. Let λ ∈ Π(µ, ν) be an optimal plan for Cc (µ, ν). Since µ and ν are singular, there exists a Borel set A ⊂ R̂n such that µ(A) = 0 and ν(Rn \ A) = 0. Then λ is concentrated on the set (Rn \ A) × A. Define functions f, g : R̂n → R by  hb(x), ∇v(x)i, x ∈ E f (x) = 0, x ∈ R̂n \ E  −hb(y), ∇v(y)i, y ∈ E g(y) = 0, y ∈ R̂n \ E Then f ∈ L1 (µ) and g ∈ L1 (ν), which means that ˆ ˆ ˆ  (16) f dµ + g dν = f (x) + g(y) dλ(x, y). R̂n R̂n R̂n ×R̂n Here ˆ ˆ ˆ ˆ f dµ + g dν = hb, ∇vidµ − hb, ∇vidν, R̂n R̂n Rn Rn since µ, ν << Ln and Ln (Rn \ E) = 0. Thus we only need to estimate the right side of (16). To do this, we notice that λ is concentrated on the set Λ := suppµ ∩ (E ∪ {♦}) ∩ R̂n \ A × suppν ∩ (E ∪ {♦}) ∩ A .   12 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG For any (x, y) ∈ Λ ∩ suppλ we can further apply Corollary 9. This gives ˆ  f (x) + g(y) dλ(x, y) n n ˆR̂ ×R̂  = b(x), ∇v(x) − b(y), ∇v(y) dλ(x, y) Λ∩suppλ∩(E×E) ˆ  + f (x) + g(y) dλ(x, y) Λ∩suppλ∩((E×{♦})∪({♦}×E)) ˆ = b(x) − b(y), ∇v(x) dλ(x, y) Λ∩suppλ∩(E×E) ˆ x−y = b(x) − b(y), c0 (|x − y|) dλ(x, y). Λ∩suppλ∩(E×E) |x − y| Now the proof can be completed by taking the absolute value inside the integral and noticing that suppλ ⊂ F × F and λ(Λ ∩ suppλ ∩ (E × E)) ≤ min{µ(Rn ), ν(Rn )}.  Finally, we want to know that the total transport cost behaves well under convergence of measures. Lemma 11. Suppose c : [0, ∞] → [0, ∞] satisfies conditions (14). Let ρk , ρ ∈ M(R̂n ), k ∈ N, be measures such that ρk (R̂n ) = ρ(R̂n ) = 0 for all k ∈ N. Assume that ρk → ρ weakly as k → ∞, that is ˆ ˆ ϕ dρk → ϕ dρ for every ϕ ∈ C 0 (R̂n ). R̂n R̂n Then − Cc (ρ+ , ρ− ) ≤ lim inf Cc (ρ+ k , ρk ). k→∞ Proof. Let v : R̂n → R be such that v(♦) = 0 and |v(x) − v(y)| ≤ c(|x − y|) for all x, y ∈ R̂n . Then v is continuous and bounded, so it follows from weak convergence and the Duality Theorem that ˆ ˆ − v dρ = lim v dρk ≤ lim inf Cc (ρ+ k , ρk ). R̂n k→∞ R̂n k→∞ Taking the supremum over v we get the claim.  4. Proof of Proposition 2 In this section we prove the following more general version of Proposition 2. Proposition 12. Let b : [0, T ] × Rn → Rn be a vector field. Assume that there exists´ a continuous and nondecreasing function G : [0, ∞) → (0, ∞) ∞ ds satisfying r G(s) = ∞ for some r > 0 such that |b(t, x)| (B1) sup ∈ L∞ (0, T ). x∈Rn G(|x|) WELL-POSEDNESS FOR CONTINUITY EQUATION 13 Assume also that there exists ´ r ads continuous and nondecreasing function ω : [0, ∞) → [0, ∞) satisfying 0 ω(s) = ∞ for some r > 0 such that |b(t, x) − b(t, y)| (B2’) Cω := ess supt∈(0,T ) sup < ∞. x,y∈Rn ω(|x − y|) If u ∈ L1 (0, T ; L1 (Rn )) is a weak solution of the Cauchy problem ∂t u + div(b u) = 0 in (0, T ) × Rn  (17) u(0, ·) = 0 in Rn , then u(t, ·) = 0 as L1 (Rn ) functions, for each t ∈ [0, T ]. In order to prove Proposition 12, we are given a function u ∈ L1 (0, T ; L1 (Rn )), weak solution of the Cauchy problem (17). Associated to this solution u, we define µk (t) := χk u+ (t, ·)Ln + m+ k (t)δ♦ νk (t) := χk u− (t, ·)Ln + m− t (t)δ♦ , ´ ± where m±  k (t) := − Rn χk (x)u(t, x) dx , where the superscripts denote the positive and negative parts of a number, and χk are as in (9) . Then µk (t)(R̂n ) = νk (t)(R̂n ), and so we can consider the optimal transport prob- lem ˆ Dδ,k (t) := inf cδ (|x − y|) dλ(x, y), λ∈Π(µk (t),νk (t)) R̂n ×R̂n where the cost function is cδ : [0, ∞] → [0, ∞], ˆ r ds cδ (r) = . 0 ω(s) + δ Since we can increase ω(s) when ´ ∞ sds> 0 is large without affecting condition (B2’), we may assume that 1 ω(s) < ∞. This ensures that cδ is C 1 , strictly increasing and satisfies conditions (14), so that all the results from the previous section are at our disposal. Our strategy is to estimate the total cost Dδ,k and then use Lemma 7 as k → ∞ and δ → 0. Lemma 13. If k and δ are fixed, then Dδ,k (t) → 0 as t → 0. Proof. By the Duality Theorem it is sufficient to prove that as t → 0 ˆ (18) Dδ,k (t) = sup v(x)χk (x)u(t, x) dx → 0, v∈F Rn where |v(x) − v(y)| F := {v ∈ C(R̂n ) : sup ≤ 1, v(♦) = 0}. x6=y cδ (|x − y|) Since u ∈ L1 (0, T ; L1 (Rn ))we can use Lipschitz functions in the definition of solutions of (17). In particular, we can follow the arguments in the proof of Lemma 5 in the case where ϕ = vχk , v ∈ F. Thus we get the following version of (8): ˆ ˆ t0 ˆ − v(x)χk (x)u(t0 , x) dx + b(t, x), ∇(vχk )(x) u(t, x) dx dt = 0. Rn 0 Rn 14 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG 1 2cδ (∞) Using the fact that |∇(vχk )| ≤ δ + G(k) we arrive at the estimate ˆ 1 ˆ t0 ˆ 2cδ (∞)  v(x)χk (x)u(t0 , x) dx ≤ Cb + |u(t, x)| dx dt, Rn δ G(k) 0 Rn where Cb := kbkL∞ (0,T ;L∞ (B(0,Rk )) < ∞. Since the bound on the right hand side is independent of v ∈ F and goes to zero as t0 → 0 , we obtain (18).  When we study the weak derivative of Dδ,k the Kantorovich potentials come into play. To that end we denote by vk (t, ·) the Kantorovich potential for Dδ,k (t). Lemma 14. The total cost Dδ,k has a weak derivative ˆ ∂t Dδ,k (t) = b(t, x), ∇vk (t, x) χk (x)u(t, x) dx Rn ˆ + vk (t, x) b(t, x), ∇χ(x) u(t, x) dx. Rn Proof. Our possible derivative is clearly in L1 (0, T ), so integrability is not an issue. Using the duality and the assumption that vk (t, ♦) = 0 we get for any h ∈ R Dδ,k (t) − Dδ,k (t − h) ˆ ˆ (19) ≤ vk (t, x)χk (x)u(t, x) dx − vk (t, x)χk (x)u(t − h, x) dx. Rn Rn Let ψ ∈ Cc∞ (0, T ), ψ ≥ 0, be a test function and h ∈ R so small that supp(ψ) ⊂ (|h|, T − |h|). It follows from a change of variables that ˆ T ˆ T ψ(t)(Dδ,k (t) − Dδ,k (t − h)) dt = (ψ(t) − ψ(t + h))Dδ,k (t) dt 0 0 In particular we see that ˆ T ˆ T 1 (20) lim ψ(t)(Dδ,k (t) − Dδ,k (t − h)) dt = − ψ 0 (t)Dδ,k (t) dt. h→0 h 0 0 Similarly it follows from a change of variables that ˆ T ˆ ψ(t)vk (t, x) χk (x)(u(t, x) − u(t − h, x)) dx dt 0 Rn ˆ T ˆ = (ψ(t)vk (t, x) − ψ(t + h)vk (t + h, x))χk (x)u(t, x) dx dt 0 Rn ˆ T ˆ ˆ t  = ∂t ψ(s)vk (s, x) ds χk (x)u(t, x) dx dt. 0 Rn t+h WELL-POSEDNESS FOR CONTINUITY EQUATION 15 If we have a smooth function vε instead of vk above, we can use the fact that u is a solution to the continuity equation to get ˆ Tˆ ˆ t  ∂t ψ(s)vε (s, x) ds χk (x)u(t, x) dx dt 0 Rn t+h ˆ T ˆ ˆ t (21) =− hb(t, x), ψ(s)∇vε (s, x) dsiχk (x)u(t, x) dx dt 0 Rn t+h ˆ T ˆ ˆ t − ψ(s)vε (s, x) dshb(t, x), ∇χk (x)iu(t, x) dx dt. 0 Rn t+h Since vk is Lipschitz in the euclidean distance, we can approximate it with smooth functions vε so that ˆ Tˆ lim ψ(t)vε (t, x) χk (x)(u(t, x) − u(t − h, x)) dx dt ε→0 0 Rn ˆ Tˆ = ψ(t)vk (t, x) χk (x)(u(t, x) − u(t − h, x)) dx dt 0 Rn and ˆ T ˆ ˆ t lim hb(t, x), ψ(s)∇vε (s, x) dsiχk (x)u(t, x) dx dt ε→0 0 Rn t+h ˆ T ˆ ˆ t  + ψ(s)vε (s, x) dshb(t, x), ∇χk (x)iu(t, x) dx dt 0 Rn t+h ˆ T ˆ ˆ t = hb(t, x), ψ(s)∇vk (s, x) dsiχk (x)u(t, x) dx dt 0 Rn t+h ˆ T ˆ ˆ t + ψ(s)vk (s, x) dshb(t, x), ∇χk (x)iu(t, x) dx dt. 0 Rn t+h Thus we get ˆ Tˆ ψ(t)vk (t, x) χk (x)(u(t, x) − u(t − h, x)) dx dt 0 Rn ˆ T ˆ ˆ t =− hb(t, x), ψ(s)∇vk (s, x) dsiχk (x)u(t, x) dx dt 0 Rn t+h ˆ T ˆ ˆ t − ψ(s)vk (s, x) dshb(t, x), ∇χk (x)iu(t, x) dx dt, 0 Rn t+h and furthermore, if we divide by h and let h → 0 we obtain ˆ ˆ 1 T lim ψ(t)vk (t, x) χk (x)(u(t, x) − u(t − h, x)) dx dt h→0 h 0 Rn ˆ Tˆ (22) = ψ(t)hb(t, x), ∇vk (t, x)iχk (x)u(t, x) dx dt 0 Rn ˆ T ˆ + ψ(t)vk (t, x)hb(t, x), ∇χk (x)iu(t, x) dx dt. 0 Rn Now we want to combine this with (19) and (20). The limits (20) and (22) do not depend on the sign of h, but the inequality (19) changes depending 16 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG on whether we divide the sides by positive or negative h. Thus by combining (19), (20) and (22) we have in fact proved that ˆ T − ψ 0 (t)Dδ,k (t) dt 0 ˆ T ˆ (23) = ψ(t)hb(t, x), ∇vk (t, x)iχk (x)u(t, x) dx dt 0 Rn ˆ T ˆ + ψ(t)vk (t, x)hb(t, x), ∇χk (x)iu(t, x) dx dt. 0 Rn To conclude the proof we need to show (23) for general test functions ψ ∈ Cc∞ (0, T ). This is done by approximating ψ + and ψ − by smooth functions for which (23) holds.  Now we are ready to prove Proposition 12. Note that Proposition 2 immedi- ately follows by the linearity of the equation, since (A1) and (A2) are partic- ular cases of (B1) (with G(t) = 1 + t) and (B2’) (with ω(s) = s log(e + 1s )). Proof of Proposition 12. Fix δ > 0 and k ∈ N. We note that using Lemmas 13 and 14 we see that ˆ Tˆ Dδ,k (t) ≤ b(t, x), ∇vk (t, x) χk (x)u(t, x) dx dt 0 Rn (24) ˆ Tˆ + vk (t, x) b(t, x), ∇χk (x) u(t, x) dx dt . 0 Rn The first term on the left hand side can be estimated by Corollary 10, where we take into account that 1 1 c0δ (|x − y|) = ≤ . ω(|x − y|) + δ ω(|x − y|) This gives ˆ Tˆ ˆ T b(t, x), ∇vk (t, x) χk (x)u(t, x) dx dt ≤ Cω ku(t, ·)kL1 (Rn ) dt, 0 Rn 0 where Cω < ∞ by assumption (B2’). For the second term we only need the bounds on vk and ∇χk : ˆ Tˆ vk (t, x) b(t, x), ∇χk (x) u(t, x) dx dt 0 Rn ˆ T ˆ ≤ 2CG cδ (∞) |u(t, x)| dx dt, 0 Rn \B(0,k) where |b(t, x)| <∞ CG := ess supt∈(0,T ) sup G(|x|) x∈Rn by assumption (B1). Thus we obtain Dδ,k (t) ≤ Cδ,k , where ˆ T ˆ Tˆ Cδ,k := Cω ku(t, ·)kL1 (Rn ) dt + 2CG cδ (∞) |u(t, x)| dx dt. 0 0 Rn \B(0,k) ´ Now we denote µ(t) := u+ (t, ·)Ln and ν(t) := u− (t, ·)Ln . Since Rn u(t, x) dx = 0 by Lemma 6, we see that µk (t) − νk (t) → µ(t) − ν(t) weakly. This enables WELL-POSEDNESS FOR CONTINUITY EQUATION 17 us to apply first Lemma 11 and then Lemma 7, which gives for any ε > 0 and δ > 0 small enough W (µ(t), ν(t)) ≤ lim inf W (µk (t), νk (t)) k→∞ −1 Cδ,k Cδ,k   (25) ≤ lim sup cδ ku(t, ·)kL1 (Rn ) + ε + k→∞ ε cδ (1) C  C = c−1 δ ku(t, ·)kL1 (Rn ) + ε + , ε cδ (1) where ˆ T C := lim Cδ,k = Cω ku(t, ·)kL1 (Rn ) dt < ∞. k→∞ 0 As δ → 0, our assumption on ω implies that c−1 C δ ( ε ) → 0 and cδ (1) → ∞. Thus it follows from (25) that W (µ(t), ν(t)) ≤ ε. Since this holds for any ε > 0, we see that W (µ(t), ν(t)) = 0, which implies µ(t) = ν(t). In other words, u = 0.  5. Proof of Theorem 3 Let ρ ∈ L1 (0, T ; M(Rn )) be a solution of (10). We want to repeat the argu- ments in the previous section, but we run into trouble if we try to integrate the gradient of a Lipschitz function over general measure ρt . So we regu- larize the measures ρt , and take into account the fact that the regularized measures might not be solutions of the same equation. Lemma 15. Let ρ ∈ L1 (0, T ; M(Rn )) be a solution of (10) and η ∈ Cc∞ (Rn ). Define ρηt := η ∗ ρt . Then for every ϕ ∈ Cc∞ ([0, T ) × Rn ) we have ˆ Tˆ ∂t ϕ(t, x) + b(t, x), ∇ϕ(t, x) ρηt (x) dx dt   0 Rn (26) ˆ Tˆ ˆ = (b(t, y) − b(t, x)), ∇ϕ(t, y) η(x − y) dy dρt (x) dt. 0 Rn Rn Proof. Let ϕ ∈ Cc∞ ([0, T ) × Rn ). Then we use the definition of ρηt and Fubini’s Theorem to calculate ˆ Tˆ ∂t ϕ(t, x) + b(t, x), ∇ϕ(t, x) ρηt (x) dx dt   0 Rn ˆ T ˆ ˆ   = ∂t ϕ(t, x) + hb(t, x), ∇ϕ(t, x)i η(x − y) dρt (y) dx dt 0 Rn Rn (27) ˆ T ˆ ˆ = ∂t ϕ(t, x)η(x − y) dx dρt (y) dt 0 Rn Rn ˆ Tˆ ˆ + b(t, x), ∇ϕ(t, x) η(x − y)dx dρt (y) dt. 0 Rn Rn ´ Now, if we define ψ(t, y) := Rn ϕ(t, x)η(x−y) dx, then ψ ∈ Cc∞ ((0, T )×Rn ), ˆ ∂t ψ(t, y) = ∂t ϕ(t, x)η(x − y) dx Rn ˆ ˆ and ∇ψ(t, y) = − ϕ(t, x)∇η(x − y) dx = ∇ϕ(t, x)η(x − y) dx, Rn Rn 18 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG where the last equality follows from integration by parts. Thus we may use the fact that ρ solves (10) to get ˆ T ˆ ˆ ∂t ϕ(t, x)η(x − y) dx dρt (y) dt 0 Rn Rn ˆ T ˆ = ∂t ψ(t, y) dρt (y) dt 0 Rn ˆ T ˆ =− hb(t, y), ∇ψ(t, y)idρt (y) dt 0 Rn ˆ T ˆ ˆ =− b(t, y), ∇ϕ(t, x) η(x − y) dx dρt (y) dt. 0 Rn Rn Combining this with (27) gives the claim.  Now we can proceed similarly to the previous section. For fixed β, δ > 0 we define cβ,δ : [0, ∞] → [0, ∞] by ˆ r ds cβ,δ (r) := β . 0 ω(s) + δ Then´cβ,δ is C 1 , strictly increasing and satisfies conditions (14), if we assume ∞ ds that r ω(s) < ∞. As before, we may assume this, because modifying ω for large values does not affect the condition (B2). ´ For 0 < α < 1 we choose a mollifier ηα ∈ Cc∞ (Rn ) such that ηα ≥ 0, Rn ηα (x) dx = 1 and suppηα ⊂ B(0, α). Let µα,k (t) := χk (ηα ∗ ρt )+ Ln + m+ α,k (t) να,k (t) := χk (ηα ∗ ρt )− Ln + m− α,k (t), ´ ± where m± α,k (t) := − Rn χk (x)(ηα ∗ ρt )(x) dx , and χk as in (9). Then µα,k (R̂n ) = να,k (R̂n ). We denote ˆ  Dα,β,δ,k (t) := min cβ,δ (|x − y|) dλ(x, y) : λ ∈ Π(µt , νt ) , R̂n ×R̂n and let vα,β,δ,k (t, ·) be a Kantorovich potential for Dα,β,δ,k (t). As before, we try to obtain estimates for Dα,β,δ,k and then apply these estimates to our reference cost via Lemma 7. But first, we need a counterpart to Lemma 13: Lemma 16. Dα,β,δ,k (t) → 0, as t → 0. Proof. Just repeat the proof of Lemma 13, but instead of using the defini- tion of a solution to (17) use Lemma 15. This gives an extra term, which fortunately causes no problems.  Combining the proof of Lemma 14 with Lemma 15 we get the following result. WELL-POSEDNESS FOR CONTINUITY EQUATION 19 Lemma 17. The cost Dα,β,δ,k has a weak derivative: ˆ ∂t Dα,β,δ,k (t) = b(t, x), ∇vα,β,δ,k (t, x) χk (x)(ηα ∗ ρt )(x) dx Rn ˆ + vα,β,δ,k (t, x) b(t, x), ∇χ(x) (ηα ∗ ρt )(x) dx Rn ˆ ˆ + b(t, y) − b(t, x), ∇ϕα,β,δ,k (t, x) ηα (y − x) dy dρt (x) dt, Rn Rn where ϕα,β,δ,k (t, x) = vα,β,δ,k (t, x)χk (x). Proof. Repeat the proof of Lemma 14, but for equality (21) use Lemma 15, which gives the extra term.  Now it is easy to estimate the actual transport cost. Lemma 18. We have the estimate ˆ T Dα,β,δ,k (t) ≤βCω,k |ρt |(Rn ) dt 0 ˆ ∞ ˆ T ds + 2βCG |ρt |(Rn \ B(0, k − 1)) dt 0 ω(s) + δ 0 ˆ ∞ ˆ β β ds  T + Cω,k ω(α) + |ρt |(Rn ), δ G(k) 0 ω(s) + δ 0 where |b(t, x) − b(t, y)| Cω,k := ess supt∈(0,T ) sup x,y∈B(0,Rk +1) ω(|x − y|) |b(t, x)| and CG := ess supt∈(0,T ) sup . x∈Rn G(|x|) Proof. Just integrate Lemma 17 while taking Lemma 16 into account. Lemma 17 gives three terms, which we estimate. First, it follows from Corollary 10 that ˆ Tˆ b(t, x), ∇vα,β,δ,k (t, x) χk (x)(ηα ∗ ρt )(x) dx 0 Rn (28) ˆ T ≤ Cω,k |ρt |(Rn ) dt. 0 For the second term recall that 2 b(t, x), ∇χk (x) ≤ |b(t, x)| ≤ 2CG . G(|x|) Thus it is easy to see that ˆ Tˆ vα,β,δ,k (t, x) b(t, x), ∇χ(x) (ηα ∗ ρt )(x) dx 0 Rn (29) ˆ T ≤ 2CG cβ,δ (∞) |ρt |(Rn \ B(0, k − 1)) dt. 0 Finally, we note that for ϕα,β,δ,k (t, x) = vα,β,δ,k (t, x)χk (x) we have β 2cβ,δ |∇ϕα,β,δ,k | ≤ + , δ G(k) 20 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG which implies (30) ˆ T ˆ ˆ b(t, y) − b(t, x), ∇ϕα,β,δ,k (t, x) ηα (y − x) dy dρt (x) dt 0 Rn Rn  β 2c (∞)  ˆ T β,δ ≤ Cω,k ω(α) + |ρt |(Rn ) dt. δ G(k) 0 Combining the estimates (28), (29) and (30) and writing out cβ,δ (∞) gives the claim.  Now we can prove the uniqueness part of Theorem 3. Proof of uniqueness in Theorem 3. It suffices to prove that if ρ ∈ L1 (0, T ; M(Rn )) solves (10), then ρ = 0. Fix k ≥ 2, and choose parameters α, β and δ de- pending on k. For ease of notation we denote ˆ T ˆ T  1 I := |ρt |(Rn ) dt and Ik := max |ρt |(Rn \ B(0, k − 1)) dt, . 0 0 k 1 Now we choose βk := Cω,k I+1 , δk > 0 such that ˆ ∞ ds Cω,k I + 1 = 0 ω(s) + δk 2(CG + 1)Ik and 0 < αk < 1 such that 1 ω(αk ) ≤  ´∞  . βk βk ds δk + G(k) 0 ω(s)+δk (Cω,k I + 1) Now we set µk (t) := µαk ,k (t), νk (t) := ναk ,k (t) and ck := cβk ,δk . We consider the optimal transport problem Dk (t) := Dαk ,βk ,δk ,k (t). Using Lemma 18 we see that Dk (t) ≤ 3. Lemma 7 then gives for any ε > 0 the inequality 3 (31) W (µk (t), νk (t)) ≤ c−1 k (3/ε) + ε + . ck (1) Since ρt (Rn ) = 0 by Lemma 6, we see that µk (t) − νk (t) → ρt weakly. Thus we get from Lemma 11 − W (ρ+ t , ρt ) ≤ lim inf W (µk (t), νk (t)). k→∞ Combining this with (31) we see that to prove the Theorem it suffices to show that ck (r) → ∞ for any r > 0 (which also implies that c−1 k (r) → 0 for any r > 0). So let’s do just that. Because of our choice of βk and δk we have ˆ ∞ ds ck (∞) = βk → ∞, as k → ∞. 0 ω(s) + δk On the other hand, it holds that ˆ ∞ ˆ r ˆ ∞ ds ds ds βk ≤ βk + βk , 0 ω(s) + δk 0 ω(s) + δ k r ω(s) WELL-POSEDNESS FOR CONTINUITY EQUATION 21 where we can use the bound βk ≤ 1 to see that ˆ ∞ ˆ ∞ ds ds βk ≤ < ∞. r ω(s) r ω(s) ´ r ds Thus we obtain ck (r) = βk 0 ω(s)+δ k → ∞.  As mentioned in the introduction, the existence of solutions to the Cauchy problem for the continuity equation (7) in L1 (0, T ; L1 (Rn )) may fail in gen- eral. Indeed, given a vector field b satisfying (B1) and (B2), it admits a well defined flow map φt : Rn → Rn solving ( ˙ = b(t, φt (x)) φt (x) φ0 (x) = x but it is far from clear wether φt will (or will not) preserve the set of abso- lutely continuous measures. In [12] (see also [13]) this situation is analyzed in terms of div(b). In particular, a smooth vector field b is provided for which the flow does not preserve Lebesgue measurable sets (both backward and forward in time). As shown in [13], and modulo other conditions, a certain sub-exponential degree of integrability for div(b) suffices to guaran- tee the preservation of Lebesgue measurable sets. Below, we present some examples to show that the exponential integrability of the divergence div(b) and the Osgood modulus of continuity (B2) are independent conditions. By simplicity we write the details in the plane. It is easy to find vector fields b satisfying the Osgood continuity condition (B2) but with div(b) ∈/ L∞ . For instance, let f1 , f2 : R2 → R be BM O func- tions with compact suport (in the next section we provide more information about the BM O space). Then, define the vector field b(x) := (I1 ∗ f1 (x), I1 ∗ f2 (x)), where I1 (x) = c/|x| is the Riesz potential of order 1. Then b belongs to the Zygmund class, and so satifies an Osgood condition. Clearly, b has derivatives locally in Lp for all p < ∞. Likewise, the divergence is div(b) = R1 ∗ f1 + R2 ∗ f2 , where Rj denotes the j-th Riesz transform. Clearly, divb ∈ / L∞ in general, even though it is exponentially integrable because div(b) ∈ BM O. In order to get examples of vector fields b satisfying (B2) but with divergence not exponentially integrable we have to be more precise. Let b a vector field with compact support, smooth in R2 \ {(0, 0)} and such that in a convenient neighbourhood of the origin is defined by b(x, y) = (xf (x, y), yf (x, y)), whith f (x, y) = log(x2 + y 2 )(log(− log(x2 + y 2 ))). When δ is small, the modulus of continuity ω(δ) of b is like δ log(δ −2 ) log(log(δ −2 )) and so, b satisfies the Osgood condition. Besides, in a neighbourhood of the origin, div(b) = 2 f (x, y) + log(− log(x2 + y 2 )) + 1) 22 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG and then div(b) is not exponentially integrable because f does not admit any BM O majorant. It is worth mentioning, though, that div(b) is subex- ponentially integrable. A slight modification gives an example of a vector field b with exponentially integrable div(b) and not satisfying the Osgood condition (B2). Consider the vector field b with compact support, smooth in R2 \ {(0, 0)} and such that in a convenient neighbourhood of the origin is defined by b(x, y) = (xg(x, y), −yg(x, y)), whith g(x, y) = log(x2 + y 2 )(log(− log(x2 + y 2 )))2 . In this case, the modulus of continuity ω(δ) of b is as δ log(δ −2 ) (log(log(δ −2 ))2 when δ is small, thus Osgood condition does not hold. However, div(b) is exponentially integrable because in a neighbourhood of the origin x2 − y 2  2 2 2 2 2  div(b) = 2 (log(− log(x + y ))) + 2 log(− log(x + y )) , x2 + y 2 that is, |div(b)| ≤ C| log(x2 + y 2 )| and so div(b) has a BM O majorant. 6. Proof of Theorem 1 In this section, we prove Theorem 1. In order to do so, we need to recall some fundamentals on quasiconformality. We say that a sense-preserving homeomorphism φ : Rn → Rn is K-quasiconformal if φ belongs to the local 1,n Sobolev space Wloc (Rn ; Rn ) and its distributional differential Dφ(x) satisfies the distortion inequality, |Dφ(x)|n ≤ K J(x, φ) at almost every point x ∈ Rn . Here |Dφ(x)| denotes the operator norm of the differential matrix Dφ(x), and J(x, φ) denotes the jacobian deter- minant det Dφ(x). The inverse of a K-quasiconformal map is a K n−1 - quasiconformal map. The composition of a K1 -quasiconformal map and a K2 -quasiconformal map is a K1 K2 -quasiconformal map. For more about this topic, we refer the interested reader to the monograph [20]. In classical harmonic analysis and Calderón-Zygmund theory, one introduces H 1 (Rn ) as the space of L1 (Rn ) functions h : Rn → R such that all the Riesz xj transforms Rj h = cn P.V. |x|n+1 ∗ h , j = 1, . . . , n belong to L1 (Rn ) as well. One indeed can define n X khkH 1 (Rn ) := khkL1 (Rn ) + kRj hkL1 (Rn ) . j=1 The Hardy space H 1 (Rn ) is the appropriate substitute in harmonic analysis for L1 (Rn ), on which Calderón-Zygmund theory fails. It also appears as the topological predual of BM O(Rn ), H 1 (Rn )∗ = BM O(Rn ) WELL-POSEDNESS FOR CONTINUITY EQUATION 23 by Fefferman Theorem [18]. Let us recall here that BM O(Rn ) consists of functions b : Rn → R with bounded mean oscillation, that is, ˆ 1 kukBM O(Rn ) = sup inf |u(x) − c| dx < ∞. B⊂Rn c∈R |B| B The dual pairing hh, bi, h ∈ H 1 (Rn ) and b ∈ BM O(Rn ), is given by an abso- lutely converging integral only in some particular cases, but not in general. It is also of interest for us the space V M O(Rn ) of functions with vanishing mean oscillation, which is the closure of Cc (Rn ) = {f : Rn → R; f is continuous and has compact support} under the k · kBM O -topology. It is a proper, closed subspace of BM O(Rn ), whose topological dual is precisely H 1 (Rn ), V M O(Rn )∗ = H 1 (Rn ). In Calderón-Zygmund theory, BM O(Rn ) is the appropiate substitute for L∞ (Rn ), while V M O(Rn ) is the one for Cc (Rn ). The following theorem was proven by Reimann in [22]. Theorem 19 (Reimann, [22]). Given K ≥ 1 and an integer n ≥ 2, there exists a constant C(n, K) with the following property. If φ : Rn → Rn is K- quasiconformal, and b ∈ BM O(Rn ), then b ◦ φ−1 ∈ BM O(Rn ). Moreover, one has 1 kbkBM O ≤ kb ◦ φ−1 kBM O ≤ C(n, K)kbkBM O C(n, K) for each b ∈ BM O(Rn ). Reimann’s result is optimal. More precisely, if φ : Rn → Rn is a sense- preserving homeomorphism for which Cφ (b) = b◦φ induces a topological iso- morphism on BM O(Rn ), then necessarily φ is K-quasiconformal, for some K ≥ 1 depending only on n and kCφ kBM O , see [5]. If we want to dualize the above statement, then the composition operator Cφ must be replaced by the transport operator Tφ , hTφ h, gi = hh, Cφ gi. The above equality may not make sense in general, even in the cases when the dual pairing h·, ·i is a true integral. However, quasiconformal maps are known to preserve Lebesgue null sets, and so the change of variables formula holds. As a consequence, if h ∈ L1loc then the transport operator Tφ maps the absolutely continuous measure dµ = h(x) dx to another absolutely continuous measure Tφ (dµ) = φ] (dµ) whose density comes from the change of variables formula, Tφ (dµ) = h(φ−1 (x)) J(x, φ−1 ) dx. By abbusing notation, one could simply write Tφ (h) = φ] h = h(φ−1 ) J(·, φ−1 ). The following corollary looks at this fact from the H 1 (Rn ) perspective. 24 ALBERT CLOP, HEIKKI JYLHÄ, JOAN MATEU, AND JOAN OROBITG Corollary 20. Let φ : Rn → Rn be K-quasiconformal. Then the transport operator Tφ maps measures with H 1 (Rn ) density boundedly into themselves. That is, if h ∈ H 1 (Rn ) then Tφ h = h(φ−1 ) J(·, φ−1 ) is an H 1 (Rn ) function, and kTφ hkH 1 (Rn ) ≤ C(n, K) khkH 1 (Rn ) , for each h ∈ H 1 (Rn ). Proof. We will prove that Tφ : H 1 (Rn ) → H 1 (Rn ) is the adjoint of the composition operator Cφ : V M O(Rn ) → V M O(Rn ), Cφ b = b ◦ φ. The rest of the proof follows by Schauder’s Theorem, kTφ kH 1 →H 1 = kCφ∗ kH 1 →H 1 ≤ kCφ kV M O→V M O ≤ C(n, K) together with Reimann’s previous result. Given h ∈ H 1 (Rn ), it suffices to see that if h ∈ H 1 (Rn ) and g ∈ V M O(Rn ), then (32) hTφ h, gi = hh, Cφ gi where h·, ·i denotes the V M O − H 1 duality. By density, we can assume that g ∈ Cc (Rn ). But then (32) becomes an equality of absolutely converging integrals, ˆ ˆ Tφ h(x) g(x) dx = h(y) g(φ−1 (y)) dy Rn Rn which is easily seen to be true due to the change of variables formula. Thus, the claim follows.  We are now in position of proving Theorem 1. Proof of Theorem 1. According to [14, Theorem 5] (see also [23]), the Cauchy problem ( φ̇t (x) = b(t, φt (x)) φ0 (x) = x admits a unique flow φt : Rn → Rn of Kt -quasiconformal maps, where  ˆ t  Kt ≤ exp (n − 1) 2kSA b(s, ·)k∞ ds , 0 ≤ t ≤ T. 0 By Corollary 20, we deduce that the transport operator Tφt maps measures with H 1 (Rn ) density boundedly into themselves. In particular, at every fixed time t ∈ [0, T ] we can define an H 1 (Rn ) function as follows, h(t, ·) = Tφt h0 , or more explicitly h(t, x) = h0 (φ−1 −1 t (x)) J(x, φt ), with norm kh(t, ·)kH 1 (Rn ) ≤ C(n, Kt ) kh0 kH 1 (Rn ) . In particular, h ∈ L∞ (0, T ; H 1 (Rn )), since 1 ≤ Kt ≤ KT for all 0 ≤ t ≤ T . It is an easy exercise to prove that h(t, ·) solves the Cauchy problem (6). For uniqueness, we simply note that b is in the assumptions of Theorem 3, which forces (6) to have a unique measure valued solution.  As in the BM O setting, quasiconformality can also be characterized in H 1 terms. WELL-POSEDNESS FOR CONTINUITY EQUATION 25 Corollary 21. Let φ : Rn → Rn be a sense-preserving homeomorphism. Assume that the transport operator Tφ maps absolutely continuous measures with H 1 (Rn ) density boundedly into themselves. That is, there is a constant C0 with the following property. If dµ = h dx with h ∈ H 1 (Rn ), then Tφ (dµ) is absolutely continuous and d Tφ (dµ) ≤ C0 khkH 1 (Rn ) . dx H 1 (Rn ) Then φ must be K-quasiconformal, for some K = K(C0 , n) ≥ 1. Proof. By assumptin, given any h ∈ H 1 (Rn ), the measure Tφ (h dx) is abso- lutely continuous with H 1 (Rn ) density. Thus, for each b ∈ BM O(Rn ) we can define a linear functional (Tφ b)∗ : H 1 (Rn ) → R as follows, h(Tφ b)∗ , hi := hb, Tφ hi. By our assumption, (Tφ b)∗ is well defined and bounded, with norm k(Tφ b)∗ kBM O(Rn ) ≤ C(n) C0 kbkBM O(Rn ) where C(n) comes from Fefferman’s duality. 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