1 Introduction

Existence of strong traces for quasi-solutions ofmultidimensional scalar conservation lawsE.Yu. Panove-mail: pey@novsu.ac.ruAbstractAbstract. We consider a conservation law in the domain Ω ⊂ R n+1with C 1 boundary ∂Ω. For the wide class of functions including generalizedentropy sub- and super-solutions we prove existence of strong traces for normalcomponents of the entropy fluxes on ∂Ω. No non-degeneracy conditions onthe flux are required.1 IntroductionIn the open domain Ω ⊂ R n+1 we consider a first-order conservation lawdiv x f(u) = 0, (1.1)u = u(x), x = (x 0 , . . .,x n ) ∈ Ω; f = (f 0 , . . .,f n ) ∈ C(R, R n+1 ), where the flux functionsare supposed to be only continuous: f i (u) ∈ C(R), i = 0, . . .,n. We assumethat ∂Ω is a C 1 boundary that is for each point x ∈ ∂Ω there exists its neighborhoodU and a C 1 -diffeomorphism ζ : U ¯Ω → W rh , where ¯Ω = Cl Ω and W rh = [0, h)×V r isa cylinder in R n+1 , V r = { y = (y 1 , . . .,y n ) ∈ R n | |y| 2 = y 2 1 +· · ·+y2 n < r2 } is an openball, r, h > 0. We shall also assume that derivatives of ζ and ζ −1 are bounded ( thisassumption can be always achieved by passing to a less neighborhood ). Clearly,boundaries of the sets W rh and U ∩ ¯Ω must correspond each other under the map ζ:ζ(∂Ω ∩ U) = {0} × V r . We shall refer to triples (U, ζ, W rh ) as boundary charts onΩ. The corresponding neighborhood U will be called a coordinate neighborhood.For any point ¯x ∈ ∂Ω we can choose the boundary chart (U, ζ, W rh ), ¯x ∈ U in thefollowing special way. As is rather well-known, ∂Ω locally is a graph of some C 1function, upon rotating and shift of coordinate axes if necessary. More precisely,making a shift of coordinate axes, we can assume that ¯x = 0. Then there exists aC 1 function g = g(y) defined in some open ball V r and a neighborhood U of thepoint ¯x = 0 such that, after possible rotating of coordinate axes¯Ω ∩ U = { x = (x 0 , x ′ ) ∈ R n+1 | 0 ≤ x 0 − g(x ′ ) < h, x ′ ∈ V r }with some h > 0. In particular, ∂Ω ∩ U is a graph x 0 = g(x ′ ), x ′ ∈ V r . Takingζ(x) = (t(x), y(x)) with t(x) = x 0 −g(x ′ ), y(x) = x ′ we obtain the chart (U, ζ, W rh ),which will be called canonical. Observe that the Jacobian of the map ζ equalsidentically 1, therefore ζ is a measure preserving map.Denote by ⃗ν(x) the outward normal vector at the point x ∈ ∂Ω. Let (U, ζ, W rh )be a chart, ζ(x) = (t(x), y(x)) then ∂Ω ∩ U is the level set t(x) = 0 and t(x) > 0 for1

x ∈ Ω ∩ U. Therefore, ⃗ν(x) = −∇t(x). Of course, this vector could be normalizedto be unit but this is not necessary for the sequel. Notice also that if t(x) = τ > 0then ⃗ν(x) = −∇t(x) is the normal vector to the hyper-surface t(x) = τ.Now, let us recall the definition of a generalized entropy solution of (1.1) in thesense of S.N. Kruzhkov [11].Definition 1.1. A bounded measurable function u = u(x) ∈ L ∞ (Ω) is called a generalizedentropy solution ( briefly - g.e.s. ) of the equation (1.1) in the domain Ω if∀k ∈ Rdiv x [sign (u − k)(f(u) − f(k))] ≤ 0 (1.2)in the sense of distributions on Ω ( in D ′ (Ω) ).In the case when (1.2) holds with equality sign: for all k ∈ Rdiv x [sign (u − k)(f(u) − f(k))] = 0 in D ′ (Ω) (1.3)a g.e.s. u(x) is called an isentropic solutions (briefly - i.s.)Condition (1.2) means that for any test function h = h(x) ∈ C0(Ω), 1 h ≥ 0∫{sign (u − k)(f(u) − f(k), ∇ x h)}dtdx ≥ 0.ΩHere and below we use the notation (·, ·) for the scalar product in a finite-dimensionalEuclidean space. We also denote by | · | the corresponding Euclidean norm.Setting in (1.2) k = ±‖u‖ ∞ , one can easily derive that div x f(u) = 0 in D ′ (Ω)and u(x) satisfies (1.1) in the sense of distributions, i.e. u(x) is a generalized solution( g.s. ) of this equation.If we replace the condition (1.2) by one of the following conditionsdiv x [sign + (u − k)(f(u) − f(k))] ≤ 0 in D ′ (Ω), (1.4)div x [sign − (u − k)(f(u) − f(k))] ≤ 0 in D ′ (Ω), (1.5)where sign + (u − k) = max(sign (u − k), 0); sign − (u − k) = min(sign (u − k), 0), weobtain notions of a generalized entropy sub-solution (g.e.sub-s.) and a generalizedentropy super-solution (g.e.super-s.) respectively ( see [12, 4, 20] ). Obviously, afunction u(x) is a g.e.s. of (1.1) if and only if u(x) is a g.e.sub-s. and g.e.super-s ofthis equation simultaneously.The theory of g.e.sub-s. (g.e.super-s.) plays an important role in the study ofconservation laws, especially in the case when the flux vector is only continuous( see [12, 4, 20] ).Denote by ¯M loc (Ω) the space of Borel measures γ on Ω that are locally finite upto the boundary ∂Ω, i.e. for any compact set K ⊂ R n+1Varγ(K ∩ Ω) < ∞. (1.6)2

Here Varγ(A) is the variation of γ on a Borel set A, so Varγ is a nonnegative Borelmeasure on Ω. We will consider ¯Mloc (Ω) as a locally convex space with topologygenerated by semi-norms p K (γ) = Varγ(K ∩ Ω).Now we introduce the notion of a quasi-solution of (1.1).Definition 1.2. A function u = u(x) ∈ L ∞ (Ω) is called a quasi-solution ( a quasi-s.for short ) of equation (1.1) if for some dense set F ⊂ R of values kdiv x ψ k (u) = −γ k in D ′ (Ω), (1.7)where γ k = γ k (x) ∈ ¯M loc (Ω), and ψ k (u) = sign (u−k)(f(u)−f(k)) is the Kruzhkov’sentropy flux.From (1.7) with k > ‖u‖ ∞ it follows thatdiv x f(u) = −γ in D ′ (Ω), (1.8)γ ∈ ¯M loc (Ω). We underline that functions satisfying (1.7) generally are not g.s. oforiginal equation (1.1).Without loss of generality we can assume that the set F from Definition 1.2 iscountable.Remark 1.1. Definitions 1.1, 1.2 (as well as the definitions of g.e.sub-s. and g.e.supers.)remain valid without changes also for the case when the flux vector in (1.1)depends on variables x ∈ Ω, f = f(x, u), and satisfies the assumptiondiv x f(x, k) = 0 in D ′ (Ω) ∀k ∈ R. (1.9)As we will demonstrate below, the class of quasi-solutions includes g.e.s. of (1.1)as well as g.e.sub-s. and g.e.super-s. of this equation.Denote by H n the n-dimensional Hausdorff measure on ∂Ω. The main our resultis the following theoremTheorem 1.1. Suppose a function u(x) is a quasi-s. of (1.1). Then there exists afunction u 0 = u 0 (x) ∈ L ∞ (∂Ω) such that for each k ∈ R the normal component offlux vector (ψ k (u(x)),⃗ν(x)) has the strong trace (ψ k (u 0 (x)),⃗ν(x)) at the boundary∂Ω. This means that for any chart (U, ζ, W rh )ess lim(ψ k (u(t, y)),⃗ν(t, y)) = (ψ k (u 0 (y)),⃗ν(y)) in L 1 (V r ), (1.10)t→0where u(t, y) = u(ζ −1 (t, y)), ⃗ν(t, y) = ⃗ν(ζ −1 (t, y)), and u 0 (y) = u 0 (ζ −1 (0, y)), ⃗ν(y) =⃗ν(ζ −1 (0, y)).Besides, if for H n -a.e. x ∈ ∂Ω the function u → (f(u),⃗ν(x)) is not constant onnon-degenerate intervals then u 0 (x) is the strong trace of u(x).3

Remark, that the first result on existence of strong traces for g.e.s. of the equationu t +f(u) x = 0 on initial line t = 0 was established in [7] under the condition thatthe flux function f(u) ∈ C 1 (R) is not affine on non-degenerate intervals. In multidimensionalcase existence of the strong traces for g.e.s. of equation u t +div x f(u) = 0was later proved in [26] under the assumptions that the flux f(u) ∈ C 3 (R, R n )and satisfies the following non-degeneracy condition: ∀ξ ∈ R n \ {0} the functionu → (ξ, f ′ (u)) is not constant on sets of positive Lebesgue measure. In [19] theexistence of strong traces on initial hyperplane t = 0 was proved for quasi-solutionsof the equation u t + div x ϕ(u) = 0 without any restrictions on only continuous fluxvector ϕ(u) = (ϕ 1 (u), . . .,ϕ n (u)). Recently in paper [13] the authors prove existenceof strong traces for g.e.s. of the equation u t + f(u) x = 0 on the boundary ∂Ω of aplane domain Ω ⊂ R 2 without non-degeneracy restrictions but under the regularityassumption f(u) ∈ C 2 (R).In the present paper we extend results of paper [19] to the case of an arbitraryboundary ∂Ω. We keep the main scheme of the paper [19], utilizing the techniquesof H-measures and induction on the spatial dimension. Recall that the flux vectoris assumed to be only continuous. Our main result remains valid for more generalcase of a locally Lipschitz-deformable boundary ∂Ω ( in the sense of [6] ) with thesame proof. We consider the case of C 1 boundaries only to simplify the notations.We underline that Theorem 1.1 allows to formulate boundary value problemsfor (1.1) in the sense of Bardos, LeRoux & Nédélec [1]. For instance the Dirichletproblem u| ∂Ω = u b can be understood in the sense of the inequality: ∀k ∈ R(sign (u − k) + sign (k − u b ))(f(u) − f(k),⃗ν) ≥ 0 on ∂Ω,which is well defined due to existence of the strong traces ofsign (u − k)(f(u) − f(k),⃗ν) and (f(u),⃗ν).1.1 Existence of the weak tracesLet ⃗v(x) ∈ L ∞ (Ω, R n ) be a bounded vector field on Ω. This field is called a divergencemeasure field if div⃗v = γ ∈ ¯M loc (Ω) in D ′ (Ω). From results by G.-Q. Chen& H. Frid [6] it follows existence of the weak trace for normal component (⃗ν,⃗v) ofthe field ⃗v at the boundary ∂Ω.Proposition 1.1. Let ⃗v be a divergence measure field on Ω. Then there exists afunction v 0 (x) ∈ L ∞ (∂Ω) such that for any chart (U, ζ, W rh )ess lim(⃗v(t, y),⃗ν(t, y)) = v 0 (y) weakly-∗ in L ∞ (V r ),t→0where ⃗v(t, y) = ⃗v(ζ −1 (t, y)), ⃗ν(t, y) = ⃗ν(ζ −1 (t, y)), and v 0 (y) = v 0 (ζ −1 (0, y)).The function v(x) will be called a weak normal trace of ⃗v. Applying this Theoremto the fields ψ k (u), where u = u(x) is a quasi-s. of (1.1), we derive the followingresult.4

Theorem 1.2. Let u(x) be a quasi-s. of (1.1). Then for each k ∈ R there exists afunction v 0k (x) ∈ L ∞ (∂Ω) being the weak normal trace of the field ψ k (u(x)).Proof. Let u = u(x) be a quasi-s. of (1.1), M = ‖u‖ ∞ . If k ∈ F then by Definition1.2 ψ k (u(x)) is a divergence measure field and by Proposition 1.1 there existsits weak normal trace v 0k (x) ∈ L ∞ (∂Ω) that is for any boundary chart (U, ζ, W rh )(ψ k (u(t, y)),⃗ν(t, y)) → v 0k (y) weakly-∗ in L ∞ (V r ) (1.11)as t → 0 running a set E k ⊂ (0, h) of full measure. Here u(t, y) = u(ζ −1 (t, y)),⃗ν(t, y) = ⃗ν(ζ −1 (t, y)), v 0k (y) = v 0k (ζ −1 (0, y)). Let E = ⋂ k∈FE k . Since F is countablethe set E has full measure in (0, h) and limit relations (1.11) hold as t → 0, t ∈ Efor all k ∈ F simultaneously.As easy to see, ‖ψ k (u) − ψ k ′(u)‖ ∞ ≤ 2ω(|k − k ′ |), whereω(δ) = sup{ |f(u 1 ) − f(u 2 )| | u 1 , u 2 ∈ [−M, M], |u 1 − u 2 | < δ }is a continuity modulus of the vector f(u) on the segment [−M, M]. This impliesthat for any k, k ′ ∈ R|(ψ k (u(t, y)),⃗ν(t, y)) − (ψ k ′(u(t, y)),⃗ν(t, y))| ≤2ω(|k − k ′ |)|⃗ν(t, y)| ≤ Cω(|k − k ′ |), (1.12)C = const. Relations (1.11), (1.12) implies in the limit as t → 0, t ∈ E that‖v 0k (y) − v 0k ′(y)‖ ∞ ≤ Cω(|k − k ′ |) ∀k, k ′ ∈ F (1.13)and since ω(δ) → 0 as δ → 0, we find that the map k → v 0k (y) ∈ L ∞ (V r ) is uniformlycontinuous on F. Taking into account that F is dense in R the map k → v 0k isuniquely extended as a continuous map on the whole R. Thus, the functions v 0k (y)are defined for all real k and relation (1.13) remains valid already for all real k, k ′ .From limit relation (1.11) and estimates (1.12), (1.13) it readily follows that (1.11) issatisfied for all k ∈ R. Clearly, the weak limit v 0k (x) ≡ v 0k (ζ(x)) does not depend onthe choice of a boundary chart. This allows to define v 0k (x) on the whole boundary∂Ω and by the construction v 0k (x) is the weak normal trace of the field ψ 0k (u(x)).The proof is complete.1.2 Localization of the problem.Since our result has local character we reformulate the problem in a coordinate neighborhoodU of a boundary point ¯x ∈ ∂Ω. More precisely, suppose that (U, ζ, W rh ) isa canonical boundary chart, i.e. after rotating and shift of coordinate axes if necessary¯x = 0 and (t, y) = ζ(x) = (t(x), x ′ ), t(x) = x 0 − g(x ′ ), x ′ = (x 1 , . . .,x n ), whereg ∈ C 1 (V r ), |∇g| ≤ const. Let u = u(x) ∈ L ∞ (U ∩ Ω), and u(t, y) ∈ L ∞ ((0, h) ×V r )be the same function written in the coordinates (t, y) that is u(t, y) = u(ζ −1 (t, y)).We have the following statement.5

Theorem 1.3. If the function u(x) is a g.e.s, an i.s., a g.e.sub-s., a g.e.super-s, aquasi-s. of equation (1.1) in Ω then the function u(t, y) is, respectively, a g.e.s., ani.s., a g.e.sub-s, a g.e.super-s, a quasi-s. of the equationϕ 0 (y, u) t + div y ϕ(u) = 0 (1.14)in the domain W = (0, h) × V r , where ϕ(u) = ¯f(u) = (f 1 (u), . . ., f n (u)) ∈ R n andϕ 0 (y, u) = f 0 (u) − (∇g(y), ¯f(u)) = −(⃗ν(y), f(u)). (1.15)Proof. Let h(x) ∈ C0 1 (U ∩Ω) be a test function, h(t, y) be the same function writtenin the variables (t, y) ∈ W. Then h(t, y) ∈ C0(W) 1 and by the chain ruleh xi (x) = − ∂g∂x i(y)h t (t, y) + h yi (t, y), i = 1, . . ., n,h x0 (x) = h t (t, y), (t, y) = ζ(x).Using this relations and making the measure preserving change of variables (t, y) =ζ(x), we obtain the identity∫∫(ψ k (u(x)), ∇ x h(x))dx= sign (u(x)−k)(f(u)−f(k), ∇ x h(x))dx=Ω∫Ωsign (u(t, y) − k)[(ϕ 0 (y, u(t, y)) − ϕ 0 (y, k))h t (t, y) +W rh(ϕ(u(t, y)) − ϕ(k), ∇ y h(t, y))]dtdy. (1.16)Clearly, the functions h(t, y) runs the whole space C0 1(W) while h(x) ∈ C1 0 (U ∩ Ω).Therefore, u = u(x) is g.e.s of (1) if and only if for all k ∈ R∂∂t [sign (u − k)(ϕ 0(y, u(t, y)) − ϕ 0 (y, k))] + div y [sign (u − k)(ϕ(u(t, y)) − ϕ(k))] ≤ 0in D ′ (W), i.e. ( see Remark 1.1 ) u(t, y) is a g.e.s. of (1.14). Observe, that thecondition (1.9) from this Remarkϕ 0 (y, k) t + div y ϕ(k) = 0 in D ′ (W) ∀k ∈ Ris evidently satisfied. If u(x) is an i.s. of (1.1) then the same arguments yield thatu(t, y) is an i.s. of (1.14).Finally, if u(x) is a quasi-s. of (1.1) then, by relation (1.16) again, ∀k ∈ F,h(t, y) ∈ C0 1(W)∫sign (u(t, y) − k)[(ϕ 0 (y, u(t, y)) − ϕ 0 (y, k))h t (t, y) +W∫∫(ϕ(u(t, y)) − ϕ(k), ∇ y h(t, y))]dtdy = h(x)dγ k (x) = h(t, y)d˜γ k (t, y),ΩW6

˜γ k (t, y) being the image ζ ∗ γ k | U under the map ζ. Thus, for all k ∈ F in D ′ (W)∂∂t [sign (u−k)(ϕ 0(y, u(t, y))−ϕ 0 (y, k))]+div y [sign (u−k)(ϕ(u(t, y))−ϕ(k))] = −˜γ k .Clearly, ˜γ k ∈ ¯M loc (W) and therefore u(t, y) is a quasi-s. of (1.14).From identity (1.16) with sign replaced by sign + , sign − we derive that u(t, y)is a g.e.sub-s or a g.e.super-s. together with u(x). The proof is complete.For our aims it is only essential behavior of quasi-s. near the boundary ∂Ω. Inparticular, the space ¯M loc (W) appearing in the definition of quasi-s. can be definedas the space of Borel measures on W that are locally finite up to the ”essential part”of the boundary laying in the hyperplane t = 0. Thus, the condition (1.6) has theform:Varγ(K ∩ W) < ∞ (1.17)for each compact set K ⊂ W rh = [0, h) × V r .Remark also that the function ϕ 0 (y, u) = −(f(u),⃗ν(y)), where ⃗ν(y) is the normalvector to the hyper-surface t(x) = t written in the coordinate (t, y) ( it does notdepend on t ). Thus, to prove Theorem 1.1 we have to establish that the functionsψ 0k (u(t, y)) = sign (u(t, y) − k)(ϕ 0 (y, u(t, y)) − ϕ 0 (y, k)), k ∈ R,where u(t, y) being a quasi-s. of (1.14), have strong traces at t = 0.For equation (1.14) relation (1.7) has the formwhereψ 0k (y, u) t + div y ψ k (u) = −γ k in D ′ (W), (1.18)ψ 0k (t, y, u) = sign (u − k)(ϕ 0 (y, u) − ϕ 0 (y, k)),ψ k (u) = sign (u − k)(ϕ(u) − ϕ(k)) ∈ R n ,and u = u(t, y). Observe also that the analog of (1.8) is the relationϕ 0 (y, u) t + div y ϕ(u) = −γ in D ′ (W). (1.19)Now we demonstrate that g.e.s. as well as g.e.sub-s. and g.e.super-s. of (1.14) arequasi-s. of this equation. Let us firstly show that a g.e.s. u(t, y) satisfies (1.18)for all k ∈ R, moreover in this case γ k ≥ 0. Indeed, as it follows from the knownrepresentation of nonnegative distributions, ψ 0k (y, u) t +div y ψ k (u) = −γ k in D ′ (W),where γ k is a nonnegative locally finite measure on W. We show that this measure islocally finite up to the initial hyperplane t = 0, i.e. γ k ∈ ¯M loc (W). For this, choose7

a function β(t) ∈ C 1 0 (R) such that supp β ⊂ [0, 1], β(t) ≥ 0, ∫for ν ∈ Nβ(t)dt = 1 and setδ ν (t) = νβ(νt), θ ν (t) =∫ t0δ ν (s)ds, (1.20)so that θ ν ′ (t) = δ ν(t). Clearly, the sequence δ ν (t) converges as ν → ∞ to the Diracδ-measure in D ′ (R) while the sequence θ ν (t) converges pointwise to the Heavisidefunction{0 , t ≤ 0,θ(t) =1 , t > 0.Let ρ(t, y) ∈ C0 1(W rh), ρ(t, y) ≥ 0. Applying the equality ψ 0k (y, u) t + div y ψ k (u) =−γ k to the test function ρ(t, y)θ ν (t − t 0 ) ∈ C0(W), 1 where t 0 > 0, we derive that∫∫ρ(t, y)θ ν (t − t 0 )dγ k (t, y) = ψ 0k (y, u)ρ(t, y)δ ν (t − t 0 )dtdy +W∫W[ψ 0k (y, u)ρ t (t, y) + (ψ k (u), ∇ y ρ)]θ ν (t − t 0 )dtdy.WSupposing that t 0 is a Lebesgue point of the function∫t → ψ 0k (y, u(t, y))ρ(t, y)dy,V rwe obtain, in the limit as ν → ∞, that∫∫ρ(t, y)dγ k (t, y) = ψ 0k (y, u(t 0 , y))ρ(t 0 , y)dy +(t 0 ,h)×V r V∫r[ψ 0k (y, u)ρ t (t, y) + (ψ k (u), ∇ y ρ)]dtdy ≤(t 0 ,h)×V r∫|ψ 0k (y, u(t 0 , y))|ρ(t 0 , y)dy +V∫r[|ψ 0k (y, u)| · |ρ t | + |ψ k (u)| · |∇ y ρ|]dtdy ≤ C ρ ,Wwhere the constant C ρ does not depend on t 0 since u ∈ L ∞ (W).Passing to the limit as t 0 → 0 we derive the estimate∫ρ(t, y)dγ k (t, y) ≤ C ρ ,Wwhich implies, in view of arbitrariness of the nonnegative test function ρ(t, y) ∈C 1 0 (W rh), that the measure γ k satisfies (1.17).8

The condition (1.18) is also satisfied for g.e.sub-s. (and g.e.super-s.) u(t, y).Indeed, assume for definiteness that u = u(t, y) is a g.e.sub-s. of (1.14). Then, inthe same way as for g.e.s., we find that ∀k ∈ R[sign + (u − k)(ϕ 0 (y, u) − ϕ 0 (y, k))] t + div y [sign + (u − k)(ϕ(u) − ϕ(k))] = −β k ,where β k ∈ ¯M loc (W), β k ≥ 0.In particular, taking in this relation k < −‖u‖ ∞ , we obtain thatϕ 0 (y, u) t + div y ϕ(u) = (ϕ 0 (y, u) − ϕ 0 (y, k)) t + div y (ϕ(u) − ϕ(k)) = −βin D ′ (W), where β ∈ ¯M loc (W), β ≥ 0. Taking into account thatψ 0k (t, y, u) = (2sign + (u − k) − 1)(ϕ 0 (y, u) − ϕ 0 (y, k)),ψ k (u) = (2sign + (u − k) − 1)(ϕ(u) − ϕ(k)),we derive the equality ψ 0k (y, u) t + div y ψ k (u) = −(2β k − β) that is (1.18) is satisfiedfor all k ∈ R.Extending ∇g(y) from a ball |y| ≤ ρ < r as a continuous bounded nonzerovector on the whole space R n , we can consider equation (1.14) in the half spaceΠ = R + × R n . If u = u(t, y) is a quasi-s. of (1.14) in W 1 = (0, h) × V ρ then we canextend this function as a quasi-s. in Π. For instance, one can set u(t, y) = 0 for(t, y) /∈ W 1 . Then, as easily follows from the generalized Gauss-Green formula ( see[6] ), for each k ∈ F, g = g(t, y) ∈ C0 1(Π)∫∫∫[ψ 0k (y, u)g t (ψ k (u), ∇ y g)]dtdy = g(t, y)dγ k (t, y) + α k (t, y)g(t, y)dH n (t, y),ΠS ′where S ′ = ((0, h) × S ρ ) ∪ ({h} × V ρ ), S ρ = ∂V ρ is a sphere |y| = ρ, α k (t, y) =α 1k (t, y)−α 0k (t, y), where α 1k (t, y) is a weak normal trace of the divergence measurefield ⃗ ψ(y, u) = (ψ 0k (y, u), ψ k (u)), u = u(t, y) at S ′ ( in the sense of [6] ), whileα 0k (t, y) = ( ⃗ ψ(y, 0),⃗n(t, y)), ⃗n being the outward unit normal vector on S ′ . Thisrelation shows that for all k ∈ F(ψ 0k (y, u)) t + div y ψ k (u) = −γ k − β k ,where the measure β k , defined by the relation 〈g, β k 〉 = ∫ S ′ α k (t, y)g(t, y)dH n (t, y),evidently lays in ¯M loc (Π), as well as the measure γ k .Hence the extended functions u(t, y) is a quasi-s. of (1.14) in the whole half-spaceΠ.We have reduced our problem to the model case of equation (1.14) in Π. In thiscase our main result have the following form9

Theorem 1.4. If u(t, y) ∈ L ∞ (Π) is a quasi-s. of (1.14) then there exists a functionu 0 (y) ∈ L ∞ (R n ) such that for all k ∈ Ress lim ψ 0k (y, u(t, y)) = ψ 0k (y, u 0 (y)) in L 1 loc(R n ).t→0By Theorem 1.2 there exist the weak traces v 0k (y) of ψ 0k (y, u(t, y)) on the initialhyperplane t = 0.In the simple case of the plane boundary the proof of existence of the weak tracescan be simplified. For completeness, we give below the proof, independent of resultsof Proposition 1.1.Suppose u = u(t, y) ∈ L ∞ (Π) is a quasi-s. of (1.14). We denoteE = { t ∈ R + | (t, y) is a Lebesgue point ofu(t, y) for a.e. y ∈ R n }. (1.21)Obviously, E is the set of full measure in R + . We have the following proposition:Proposition 1.2. There exists functions v 0k (y) ∈ L ∞ (R n ) such that ψ 0k (·, u(t, ·)) →v 0k weakly-∗ in L ∞ (R n ) as t → 0, t ∈ E.Proof. Fix k ∈ F. Clearly, ∀t ∈ E ‖ψ 0k (·, u(t, ·))‖ ∞ ≤ const and we can find asequence t m ∈ E, m ∈ N, t m → 0 and a function v 0k (y) ∈ L ∞ (R n ) such thatm→∞ψ 0k (y, u(t m , y)) → v 0k (y) weakly-∗ in L ∞ (R n ) as m → ∞.If τ ∈ E then t m < τ for large enough m. We fix such t m < τ and set χ ν (t) =θ ν (t − t m ) − θ ν (t − τ), where the sequence θ ν (t), ν ∈ N was defined above in (1.20).Let h(y) ∈ C0(R 1 n ). Then for large enough ν ∈ N the function h(y)χ ν (t) ∈C0 1 (Π). Applying relation (1.18) to this test function, we obtain, after simple transforms,that∫ +∞(∫)ψ 0k (y, u(t, y))h(y)dy δ ν (t − τ)dt −0 R∫ n +∞(∫)ψ 0k (y, u(t, y))h(y)dy δ ν (t − t m )dt =0 R∫n ∫(ψ k (u), ∇ y h)χ ν (t)dtdy − h(y)χ ν (t)dγ k (t, y). (1.22)ΠΠPassing in (1.22) to the limit as ν → ∞∫and taking into account that t m , τ ∈ E areLebesgue points of the function I(t) = ψ 0k (y, u(t, y))h(y)dy, and also that theR nsequence χ ν (t) converges pointwise to the indicator function of the interval (t m , τ]and is bounded, we derive the equality∫∫I(τ) − I(t m ) = (ψ k (u), ∇ y h)dtdy − h(y)dγ k (t, y),(t m,τ]×R n (t m,τ]×R n10

which implies in the limit as m → ∞ that∫I(τ) − I(0+) = ψ 0k (y, u(τ, y))h(y)dy −R∫n ∫∫v 0k (y)h(y)dy =R n(ψ k (u), ∇ y h)dtdx −(0,τ]×R n h(y)dγ k (t, y) → 0,(0,τ]×R n τ→0(1.23)i.e. ∀h(x) ∈ C 1 0(R n )∫∫ψ 0k (y, u(τ, y))h(y)dy → v 0k (y)h(y)dy.R n τ→0R nThis, together with boundedness of u(τ, ·), τ ∈ E and density of C 1 0(R n ) in L 1 (R n ),implies that for each k ∈ Fψ 0k (·, u(t, ·)) → v 0k weakly-∗ in L ∞ (R n ) as t → 0, t ∈ E. (1.24)Since the map k → ψ 0k (y, u) ∈ C(R n × I) is continuous, where I = [−M, M], M =‖u‖ ∞ , then repeating the arguments from the proof of Theorem 1.2, we concludethat (1.24) remains valid for all k ∈ R. The proof is complete.2 Preliminaries2.1 Measure valued functionsBelow we will frequently utilize results of the theory of measure valued functions.Recall ( see [9, 24] ) that a measure valued function on a domain Ω ⊂ R N isa weakly measurable map x → ν x of the set Ω into the space of Borel probabilitymeasures having compact supports on R. Weak measurability of ν x means that forany continuous function f(λ) the function∫x → 〈f(λ), ν x (λ)〉 = f(λ)dν x (λ)is Lebesgue measurable on Ω.A measure valued function ν x is said to be bounded if there is a constant M > 0such that supp ν x ⊂ [−M, M] for a.e. x ∈ Ω. The minimal of such M will bedenoted by ‖ν x ‖ ∞ .Finally, measure valued functions of the kind ν x (λ) = δ(λ − u(x)), where δ(λ −u) = δ u (λ) is the Dirac measure at the point u, are called regular and they areidentified with the corresponding functions u(x). Thus, the set MV(Ω) of boundedmeasure valued functions on Ω includes the space L ∞ (Ω).Remark 2.1. As was demonstrated in [15], for ν x ∈ MV(Ω) measurability of thefunctions x → 〈f(λ), ν x (λ)〉 remains valid for all Borel functions f(λ).11

We shall use also the following notions:Definition 2.1. Suppose ν m x ∈ MV(Ω), m ∈ N; ν x ∈ MV(Ω). We shall say that1) the sequence ν m x is bounded if for some constant M > 0 ‖νm x ‖ ∞ ≤ M ∀m ∈ N;2) the sequence ν m x converges to ν x weakly if∀f(λ) ∈ C(R) 〈f(λ), ν m x (λ)〉 →m→∞〈f(λ), ν x (λ)〉3) the sequence ν m x converges to ν x strongly if∀f(λ) ∈ C(R) 〈f(λ), ν m x (λ)〉 →m→∞〈f(λ), ν x (λ)〉weakly-∗ in L ∞ (Ω);in L 1 loc(Ω).Remark 2.2. If a bounded sequence νxm converges to ν x weakly (strongly) then〈f(x, λ), νx m(λ)〉 → 〈f(x, λ), ν x(λ)〉 weakly-∗ in L ∞ (Ω) (respectively, - in L 1 loc (Ω)m→∞for each function f(x, λ) ∈ C(Ω × R), which is bounded on sets Ω × I for eachsegment I ⊂ R. This easily follows from the fact that f(x, λ) can be uniformly onl∑any compact approximated by functions of the form α i (x)β i (u), α i (x) ∈ C(Ω),β i (u) ∈ C(R), i = 1, . . .,l.Measure valued functions naturally arise as weak limits of bounded sequences inL ∞ (Ω) in the sense of the following theorem of Tartar ( see [24] ).Theorem 2.1. Let u m (x) ∈ L ∞ (Ω), m ∈ N be a bounded sequence. Then thereexist a subsequence u r (x) and a measure valued function ν x ∈ MV(Ω) such thatu r (x) → ν x weakly. This means that∀f(λ) ∈ C(R)f(u r ) →m→∞〈f(λ), ν x (λ)〉i=1weakly-∗ in L ∞ (Ω)that is the sequence of regular measure valued functions νx r = δ(λ −u r (x)), identifiedwith u r , converges weakly to ν x .Besides, ν x is regular, i.e. ν x = δ(λ − u(x)) if and only if u r (x) → u(x) inr→∞L 1 loc (Ω) that is νr x → ν x strongly.More generally, the following statement on weak precompactness of bounded setof measure valued functions is valid ( see [16] ).Theorem 2.2. Let νxm be a bounded sequence of measure valued functions. Thenthere exist a subsequence νx(x) r and a measure valued function ν x ∈ MV(Ω) suchthat νx r → ν x weakly as r → ∞.Corollary 2.1. Suppose a bounded sequence νxm ∈ MV(Ω) weakly converges toν x ∈ MV(Ω), and a function p(x, λ) ∈ C(Ω × R), bounded on sets Ω × I for eachsegment I ⊂ R, is such that p(x, λ) ≡ u m (x) on supp νxm for a.e. x ∈ Ω. Thenu m (x) → u(x) = 〈p(x, λ), ν x (λ)〉 weakly-∗ in L ∞ (Ω), and u m (x) → u(x) in L 1 loc (Ω)m→∞if and only if p(x, λ) ≡ const = u(x) on supp ν x for a.e. x ∈ Ω.12

Proof. Clearly, u m (x) = 〈p(x, λ), ν m x (λ)〉 ∈ L∞ (Ω). Denote by p(x, ·) ∗ ν x the imageof measure ν x under the map λ → p(x, λ). Obviously, x → p(x, ·) ∗ ν x is a boundedmeasure valued function on Ω. From the relation ( see Remark 2.2 )f(u m (x)) = 〈f(p(x, λ)), ν m x (λ)〉 →m→∞ 〈f(p(x, λ)), ν x(λ)〉 = 〈f(λ), p(x, ·) ∗ ν x (λ)〉∀f(λ) ∈ C(R) it follows that the the sequence u m (x) converges weakly to the measurevalued function p(x, λ) ∗ ν x , and in particular, u m (x) → u(x) = 〈p(x, λ), ν x (λ)〉weakly-∗ in L ∞ (Ω). By Theorem 2.1 this convergence is strong if and only ifp(x, ·) ∗ ν x (λ) = δ(λ − u(x)) for a.e. x ∈ Ω. Since the latter is equivalent to thecondition p(x, λ) ≡ u(x) on supp ν x , the proof is complete.We also need the following simple resultLemma 2.1. Let νx m , ˜ν x m , m ∈ N be bounded sequences on Ω, which converge weaklyto the measure valued functions ν x , ˜ν x , respectively. Then, for each p(λ, µ) ∈ C(R 2 )∫∫∫∫p(λ, µ)dνx m (λ)d˜νm y (µ) → p(λ, µ)dν x (λ)d˜ν y (µ) (2.1)m→∞weakly-∗ in L ∞ (Ω × Ω).Proof. Suppose firstly that p(λ, µ) = p 1 (λ)p 2 (µ), p 1 , p 2 ∈ C(R). Since the variablesx, y are independent we have∫∫p(λ, µ)dνx m (λ)dνy m (µ) = 〈p 1 (λ), νx m (λ)〉〈p 2 (µ), ˜ν y m (µ)〉 →m→∞∫∫〈p 1 (λ), ν x (λ)〉〈p 2 (µ), ˜ν y (µ)〉 = p(λ, µ)dν x (λ)d˜ν y (µ)weakly-∗ in L ∞ (Ω × Ω), as required. Certainly, this relation remains true also forthe functions p(λ, µ) represented as linear combinations of functions of the kindp 1 (λ)p 2 (µ). In particular (2.1) is valid for polynomials p(λ, µ). Since polynomialsare dense in C(R × R) we conclude that relation (2.1)is satisfied for general p(λ, µ).2.2 Some properties of isentropic solutionsFix some segment I = [−M, M], M > 0 and denote by L ⊂ C(I) the subspace ofconstant functions on I, and by C(I)/L the corresponding factor space. Consideralso the space BV (I) consisting of continuous from the left functions, which havelocally bounded variation. If η(u) ∈ BV (I) then η ′ (u) = dη(u) in the sense ofdistributions, where dη(u) is a Radon measure of finite variation on I ( the Stieltjesmeasure generated by η(u) ), and Varη = Var dη(I). Let us define the linear13

operator T η : C(I)/L → C(I)/L such that, up to an additive constant ( i.e. in thespace C(I)/L )∫T η f(u) = f(u)η(u) − f(s)dη(s) =[−M,u)∫(f(u) − f(s))dη(s) + f(u)η(−M). (2.2)[−M,u)From the representation T η f = ∫ (f(u) − f(s))dη(s) + f(u)η(−M) it easily[−M,u)follows that the function T η f is continuous and T η f 1 − T η f 2 = const wheneverf 1 − f 2 = const. Thus, the operator T η is well-defined on C(I)/L. It is easy to seethat‖T η f(u)‖ ∞ ≤ ‖f(u)‖ ∞ [|η(−M)| + Varη],which shows that the operator T η is continuous. In the case f(u) ∈ C 1 (I), integratingby parts in (2.2), we obtain that T η f is uniquely defined in C(I)/L by theequality(T η f) ′ (u) = η(u)f ′ (u) in D ′ ((−M, M)). (2.3)If η 1 , η 2 ∈ BV (I) when the product η 1 η 2 ∈ BV (I) as well andT η1 T η2 = T η1 η 2. (2.4)Indeed, if f ∈ C 1 (I) when the equality T η1 T η2 f = T η1 η 2f directly follows from (2.3).Since C 1 (I) is dense in C(I) and operators T η are continuous we conclude that thisequality holds for all f ∈ C(I) ( modulo L, of course ) and identity (2.4) is satisfied.Taking η = θ(u) ={ −1 , u ≤ k,1 , u > k with k ∈ I we derive that T ηf = f k (u) =sign (u − k)(f(u) − f(k)). The following Lemma shows that for general η ∈ BV (I)the function T η f(u) is generated by the functions f k (u).Lemma 2.2. Let η(u) ∈ BV (I). Then ∀f(u) ∈ C(I)/LT η f(u) = 1 ∫sign (u − k)(f(u) − f(k))dη(k) + Af(u), (2.5)2[−M,M)A = const = (η(M) + η(−M))/2on the segment [−M, M], up to an additive constant.Proof. Revealing the integral in the right-hand side of (2.5) and taking into account14

epresentation (2.2), we obtain∫sign (u − k)(f(u) − f(k))dη(k) =∫[−M,M)∫(f(u) − f(k))dη(k) − (f(u) − f(k))dη(k) =∫[−M,u)∫[u,M)2 (f(u) − f(k))dη(k) − (f(u) − f(k))dη(k) =∫[−M,u)[−M,M)2 (f(u) − f(k))dη(k) − (η(M) − η(−M))f(u) + const =[−M,u)and (2.5) follows.2T η f(u) − 2Af(u) + constLemma 2.3. Let I = [−M, M], ϕ(u) ∈ C(R), ψ k (u) = sign (u − k)(ϕ(u) − ϕ(k)),v k ∈ R, k ∈ R. Then the setC = { u ∈ I | ∀k ∈ R ψ k (u) = v k }is closed and connected. In particular if C ≠ ∅ then C is a segment [a, b] ⊂ I.Proof. Since all functions ψ k (u) are continuous the set C is closed. To prove thatthis set is connected, take points u 1 , u 2 ∈ C, u 1 < u 2 . We have to show that[u 1 , u 2 ] ⊂ C. Since for fixed a ≤ −M ϕ(u) = ψ a (u) + ϕ(a) we see that ϕ(u 1 ) =ϕ(u 2 ) = v = v a + ϕ(a). Then ∀k ∈ [u 1 , u 2 ]v k = ψ k (u 1 ) = ϕ(k) − ϕ(u 1 ) = ψ k (u 2 ) = ϕ(u 2 ) − ϕ(k),which immediately implies that ϕ(k) = v ∀k ∈ [u 1 , u 2 ]. Thus, ϕ(u) ≡ v on [u 1 , u 2 ].From this it follows that ψ k (u) ≡ ψ k (u 1 ) = v k on the segment [u 1 , u 2 ] for all k ∈ R.Hence, [u 1 , u 2 ] ⊂ C and the set C is connected.Now, we consider the equationϕ 0 (u) t + div y ϕ(u) = 0, u = u(t, y), (t, y) ∈ Π = R + × R n . (2.6)Here ϕ 0 (u) ∈ C(R), ϕ(u) = (ϕ 1 (u), . . ., ϕ n (u)) ∈ C(R, R n ).We are going to prove that the statement of Theorem 1.4 is valid for isentropicsolutions of (2.6), more precisely - for measure valued i.s.Definition 2.2. A bounded measure valued function ν t,y on the half-space Π iscalled a measure valued isentropic solution of (2.6) if for all k ∈ R∂∂t 〈ψ 0k(λ), ν t,y (λ)〉 + div y 〈ψ k (λ), ν t,y (λ)〉 = 0 in D ′ (Π), (2.7)where ψ 0k (λ) = sign (λ −k)(ϕ 0 (λ) −ϕ 0 (k)), ψ k (λ) = sign (λ −k)(ϕ(λ) −ϕ(k)) ∈ R nis the Kruzhkov entropy flux.15

Obviously, a regular measure valued function ν t,y (λ) = δ(λ−u(t, y)) is a measurevalued i.s. of (2.6) if and only if the function u(t, y) is a ”usual” i.s. of this equation.Remark also that, as follows from (2.7) with k > ‖ν t,y ‖ ∞ , any measure valued i.s.ν t,y satisfies the equality∂∂t 〈ϕ 0(λ), ν t,y (λ)〉 + div y 〈ϕ(λ), ν t,y (λ)〉 = 0 in D ′ (Π). (2.8)Proposition 2.1. If ν t,y ∈ MV(Π) is a measure valued i.s. of (2.6), ‖ν t,y ‖ ∞ ≤ Mthen it is a measure valued i.s. of equations(T η ϕ 0 (u)) t + div y T η ϕ(u) = 0 ∀η(u) ∈ BV ([−M, M]).Proof. Let I = [−M, M]. By Lemma 2.2 ∀λ ∈ IT η ϕ 0 (λ) = 1 ∫ψ 0k (λ)dη(k) + Aϕ 0 (λ),2T η ϕ(λ) = 1 2∫[−M,M)[−M,M)ψ k (λ)dη(k) + Aϕ(λ),where A = const. Taking into account (2.7), (2.8), we see that in the sense ofdistributions∫[−M,M)〈T η ϕ 0 (λ), ν t,y (λ)〉 t + div y 〈T η ϕ(λ), ν t,y (λ)〉 ={〈ψ0k (λ), ν t,y (λ)〉 t + div y 〈ψ k (λ), ν t,y (λ)〉 } dη(k) +A { 〈ϕ 0 (λ), ν t,y (λ)〉 t + div y 〈ϕ(λ), ν t,y (λ)〉 } = 0 (2.9)for each η(u) ∈ BV (I). Let η(u) ∈ BV (I), k ∈ R. Denoteψ 0 (λ) = T η ϕ 0 (λ), ψ(λ) = T η ϕ(λ) ∈ R n .By identity (2.4), we have that, up to additive constants,sign (λ − k)(ψ 0 (λ) − ψ 0 (k)) = T θη ϕ 0 (λ), sign (λ − k)(ψ(λ) − ψ(k)) = T θη ϕ(λ),{ −1 , u ≤ k,where θ(u) =1 , u > k . Therefore∂∂t 〈sign (λ − k)(ψ 0(λ) − ψ 0 (k)), ν t,y (λ)〉 + div y 〈sign (λ − k)(ψ(λ) − ψ(k)), ν t,y (λ)〉 =〈T θη ϕ 0 (λ), ν t,y (λ)〉 t + div y 〈T θη ϕ(λ), ν t,y (λ)〉 = 0 in D ′ (Π)in view of (2.9). Since k ∈ R is arbitrary, ν t,y is a measure valued i.s. of the equationas was to be proved.ψ 0 (u) t + div y ψ(u) = 0,16

No we can prove the following assertion on existence of strong traces for measurevalued i.s.Theorem 2.3. Let ν t,y be a measure valued i.s. of (2.6) such that for a.e. (t, y) ∈ Πand all k ∈ R functions ψ 0k (λ) ≡ v k (t, y) on supp ν t,y . Then there exist tracefunctions v 0k (y) ∈ L ∞ (R n ) such thatess lim v k (t, ·) = v 0k in L 1 loc(R n ) ( strongly ).t→0Proof. We deduce from Lemma 2.3 that ψ 0k (λ) ≡ v k (t, y) also on the convex hullCo supp ν t,y of supp ν t,y . Define the set of full measureE = { t ∈ R + | (t, y) is a Lebesgue point of the functions(t, y) → 〈p(λ), ν t,y 〉 for a.e. y ∈ R n and all p(λ) ∈ C(R) }.From the fact that the space C(R) is separable it easily follows that the set ofcommon Lebesgue points of the functions (t, y) → 〈p(λ), ν t,y 〉 has full measure onΠ, which directly implies that E ⊂ R + is a set of full measure, as required. Takingp(λ) = ϕ 0 (λ), ψ 0k (λ), we conclude that for t ∈ E for a.e. y ∈ R n points (t, y) areLebesgue points of functions v(t, y) = 〈ϕ 0 (λ), ν t,y (λ)〉, v k (t, y) = 〈ψ 0k (λ), ν t,y (λ)〉,k ∈ R. From the relationv t + div y w = 0 in D ′ (Π), where w = w(t, y) = 〈ϕ(λ), ν t,y (λ)〉 ∈ R nit follows, in correspondence with Proposition 1.1, that the function v(t, y) have theweak trace v 0 (y) ∈ L ∞ (R n ), namely v(t, ·) → v 0 weakly-∗ in L ∞ (R n ) as t → 0,t ∈ E.Obviously, for each τ ∈ E the measure valued function ν τ y = ν τ,y ∈ MV(R n ) iswell-defined, due to the relation〈p(λ), ν τ y(λ)〉 = 〈p(λ), ν τ,y (λ)〉on the set of full measure consisting of y ∈ R n such that (τ, y) is a common Lebesguepoint of all functions (t, y) → 〈p(λ), ν t,y 〉. Moreover, ‖νy τ ‖ ∞ ≤ M = ‖ν t,y ‖ ∞ . ByTheorem 2.2 we can choose a sequence t m ∈ E, m ∈ N such that t m → 0 andthe bounded sequence of measure valued functions νym = ν t m,y converges weakly asm → ∞ to some measure valued function ν 0,y ∈ MV(R n ).We have to prove that the trace v 0 is strong, i.e. v(t, ·) → v 0 in L 1 loc (Rn ) ast → 0, t ∈ E. We firstly establish that v(t m , ·) → v 0 in L 1 loc (Rn ) as m → ∞.To do this, we apply the measure valued variant of the Kruzhkov method ofdoubling variables ( like in [15], Theorem 2.3 ).By relations (2.7) for all µ ∈ R∫∂∂t [sign (v−µ)(ϕ 0(v)−ϕ 0 (µ))]+div y sign (λ−µ)(ϕ(λ)−ϕ(µ))dν t,y (λ) = 0 in D ′ (Π).17

Integrating this relation applied to a test function g = g(t, y; s, z) ∈ C0 1 (Π × Π)firstly over the measure ν s,z (µ) and then over (s, z), we readily obtain thatdiv y∫∫∂∂t [sign (v(t, y) − v(s, z))(ϕ 0(v(t, y)) − ϕ 0 (v(s, z)))] +sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν s,z (µ) = 0 in D ′ (Π × Π). (2.10)Here we take into account thatsign (v − µ)(ϕ 0 (v) − ϕ 0 (µ)) ≡ sign (v − v(s, z))(ϕ 0 (v) − ϕ 0 (v(s, z)))on supp ν s,z (µ) for a.e. (s, z) ∈ Π.Analogously, changing the places of the variables λ and µ, (t, y) and (s, z), wefind∂∂s [sign (v(t, y) − v(s, z))(ϕ 0(v(t, y)) − ϕ 0 (v(s, z)))] +div z∫∫sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν s,z (µ) = 0 in D ′ (Π × Π). (2.11)Combining (2.10) and (2.11) we arrive at the relation: in D ′ (Π × Π)( ∂∂t ∂s)+ ∂ [sign (v(t, y) − v(s, z))(ϕ 0 (v(t, y)) − ϕ 0 (v(s, z)))] +∫∫(div y + div z ) sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν s,z (µ) = 0. (2.12)Now∫we choose a function β(t) ∈ C0(R) 1 such that supp β ⊂ [0,∫1], β(t) ≥ 0,tβ(t)dt = 1 and set, as in (1.20), for ν ∈ N δ ν (t) = νβ(νt), θ ν (t) = δ ν (s)ds, sothat θ ν ′ (t) = δ ν(t). Recall that the sequence δ ν (t) converges as ν → ∞ to the Diracδ-measure in D ′ (R) while the sequence θ ν (t) converges pointwise to the Heavisidefunction. Let f(t, y, z) ∈ C0(R 1 + × R n × R n ), g(t, y; s, z) = δ ν (s − t)f(t, y, z), ν ∈ N.It is clear that g(t, y; s, z) ∈ C0 1 (Π×Π). Applying relation (2.12) to this test functionand making simple transformations, we find∫ [sign (v(t, y) − v(s, z))(ϕ 0 (v(t, y)) − ϕ 0 (v(s, z)))f t +Π×Π(∫∫)]sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν s,z (µ), (∇ y + ∇ z )f dydz ×δ ν (s − t)dsdt = 0018

Passing in this relation to the limit as ν → ∞ we derive[sign (v(t, y) − v(t, z))(ϕ 0 (v(t, y)) − ϕ 0 (v(t, z)))f t +∫R + ×R n ×R(∫∫n )]sign (λ−µ)(ϕ(λ)−ϕ(µ))dν t,y (λ)dν t,z (µ), (∇ y +∇ z )f dydzdt=0 (2.13)for each f(t, y, z) ∈ C 1 0(R + × R n × R n ), i.e. in D ′ (R + × R n × R n )∂∂t [sign (v(t, y) − v(t, z))(ϕ 0(v(t, y)) − ϕ 0 (v(t, z)))] +∫∫(div y + div z ) sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν t,z (µ) = 0.Let h(y, z) ∈ C0 1(Rn × R n ), τ ∈ E, and f(t, y, z) = χ ν (t)h(y, z), where χ ν (t) =θ ν (t − t m ) − θ ν (t − τ), ν, m ∈ N, t m < τ.Taking in (2.13) the test function f, we obtain the relation∫sign (v(t, y) − v(t, z))(ϕ 0 (v(t, y)) − ϕ 0 (v(t, z)))h(y, z)dydz ×∫R + R n ×R n (∫∫(δ ν (t − t m ) − δ ν (t − τ))dt +sign (λ − µ)×∫R + ×R n ×R n )(ϕ(λ) − ϕ(µ))dν t,y (λ)dν t,z (µ), (∇ y + ∇ z )h χ ν (t)dydzdt = 0Passing in this relation to the limit as ν ∫→ ∞ and taking into account that t ∈ Eare Lebesgue points of the function t → sign (v(t, y) − v(t, z))(ϕ 0 (v(t, y)) −R n ×R nϕ 0 (v(t, z)))h(y, z)dydz and that χ ν (t) converges point-wise to the indicator functionof (t m , τ], we obtain∫sign (v(t m , y)−v(t m , z))(ϕ 0 (v(t m , y))−ϕ 0 (v(t m , z)))h(y, z)dydzR n ×R∫n− sign (v(τ, y) − v(τ, z))(ϕ 0 (v(τ, y)) − ϕ 0 (v(τ, x)))h(y, z)dydzR n ×R n (∫∫+sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν t,z (µ),∫(t m,τ]×R n ×R n)(∇ y + ∇ z )h dydzdt = 0. (2.14)19

By Lemma 2.1sign (v(t m , y) − v(t m , z))(ϕ 0 (v(t m , y)) − ϕ 0 (v(t m , z))) =∫∫sign (λ − µ)(ϕ 0 (λ) − ϕ 0 (µ))dνy m (λ)dνm z (µ)∫∫→ sign (λ − µ)(ϕ 0 (λ) − ϕ 0 (µ))dν 0y (λ)dν 0z (µ)m→∞weakly-∗ in L ∞ (R n × R n ) and from (2.14) it follows, in the limit as m → ∞, that(∫∫)sign (λ − µ)(ϕ 0 (λ) − ϕ 0 (µ))dν 0y (λ)dν 0z (µ) h(y, z)dydz∫R n ×R∫n= sign (v(τ, y) − v(τ, z))(ϕ 0 (v(τ, y)) − ϕ 0 (v(τ, z)))h(y, z)dydzR n ×R n (∫∫−sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν t,z (µ),∫(0,τ]×R n ×R n)(∇ y + ∇ z )h dydzdt. (2.15)Now we choose the function h in the form h(y, z) = ρ(y)¯δ ν (y − z), with ¯δ ν (y − z) =n∏δ ν (z i − y i ), ρ(y) ∈ C0 1(Rn ). Taking this function in (2.15), we findi=1∫R n ×R n (∫∫∫)sign (λ − µ)(ϕ 0 (λ) − ϕ 0 (µ))dν 0y (λ)dν 0z (µ) ρ(y)¯δ ν (y − z)dydz =sign (v(τ, y) − v(τ, z))(ϕ 0 (v(τ, y)) − ϕ 0 (v(τ, z)))ρ(y)¯δ ν (y − z)dydz −R n ×R n (∫∫)sign (λ−µ)(ϕ(λ)−ϕ(µ))dν t,y (λ)dν t,z (µ), ∇ y ρ ¯δ ν (y−z)dydzdt.∫(0,τ]×R n ×R nPassing to the limit as ν → ∞, we derive that for all τ ∈ E(∫∫)sign (λ − µ)(ϕ 0 (λ) − ϕ 0 (µ))dν 0y (λ)dν 0y (µ) ρ(y)dy =∫R n (∫∫)− sign (λ − µ)(ϕ(λ) − ϕ(µ))dν t,y (λ)dν t,y (µ), ∇ y ρ dydt∫(0,τ]×R nand in the limit at τ → 0 this implies that(∫∫)sign (λ − µ)(ϕ 0 (λ) − ϕ 0 (µ))dν 0y (λ)dν 0y (µ) ρ(y)dy = 0.∫R nAs ρ(y) ∈ C 1 0 (Rn ) is arbitrary, we conclude that for a.e. y ∈ R n∫∫sign (λ − µ)(ϕ 0 (λ) − ϕ 0 (µ))dν 0y (λ)dν 0y (µ) = 0. (2.16)20

In view of Proposition 2.1 one could replace equation (2.6) by the equation(T η ϕ 0 (u)) t + div y T η ϕ(u) = 0, η(u) ∈ BV ([−M, M])and relation (2.16) yields∫∫sign (λ − µ)(T η ϕ 0 (λ) − T η ϕ 0 (µ))dν 0y (λ)dν 0y (µ) = 0 for a.e. y ∈ R n . (2.17)Since the space C(I × I) is separable and integrands in (2.17) generate a subspaceof C(I × I), we can choose a set of full measure D ⊂ R n such that for y ∈ Dsupp ν 0y ⊂ [−M, M] and (2.17) is satisfied for all η(u) ∈ BV (I). Fix y ∈ D anddenote ν = ν 0y , [a, b] = Co supp ν. We are going to show that ϕ 0 (λ) ≡ const on[a, b]. Assuming the contrary, we can find a point c ∈ (a, b) such that ϕ 0 (λ) ≢ conston any segment [c, d], c < d < b ( otherwise, ϕ 0 (I) is at most countable andtherefore ϕ 0 (λ) ≡ const on I ). Hence, there exists a sequence c r , r ∈ N such thatc < c r+1 < c r < b ∀r ∈ N, c r → c as r → ∞, and|ϕ 0 (c r ) − ϕ 0 (c)| = max |ϕ 0 (u) − ϕ 0 (c)| > 0. (2.18)u∈[c,c r]Denote h r = ϕ 0 (c r ) − ϕ 0 (c) and set η r (u) = 1 h rχ (c,cr](u), where χ (c,cr](u) is theindicator function of (c, c r ]. It is clear that η r ∈ BV (I) and⎧⎨ 0 , u ≤ c,ψ r (u) = T ηr ϕ 0 (u) = (ϕ 0 (u) − ϕ 0 (c))/h r , c < u ≤ c r ,⎩1 , u > c r .(2.19)Observe that, up to an additive constant, ψ r (u) = (ψ 0c (u) − ψ 0cr (u))/(2h r ). Asfollows from (2.18), |ψ r (u)| ≤ 1 and obviously the sequence ψ r converges point-wiseto the Heaviside function ψ(u) = θ(u − c). Taking in (2.17) η = η r and passing tothe limit as r → ∞, we find∫∫0 = sign (λ − µ)(ψ(λ) − ψ(µ))dν(λ)dν(µ) =∫∫∫∫dν(λ)dν(µ) + dν(λ)dν(µ) = 2ν([a, c])ν((c, b]).λ≤c

2.3 One criterion of existence of the strong tracesAs was shown in Proposition 1.2, there exist weak traces v 0k (y) ∈ L ∞ (R n ) ofψ 0k (y, u(t, y)) = sign (u − k)(ϕ 0 (y, u) − ϕ 0 (y, k)), where u = u(t, y) is a quasi-s. of(1.14). In the following Theorem we give a simple criterion for v 0k to be the strongtraces. We formulate this theorem even for the following more general situation.Theorem 2.4. Let V be a bounded open set in R n , W = (0, h) × V ⊂ Π, 0 < h

are the minimal and respectively the maximal among such functions u 0 (y). Let usshow that u ± 0 (x) are measurable on V . This follows from evident relations: for allλ ∈ R{ y ∈ V | u − 0 (y) ≥ λ } = ⋂ ⋃{ y ∈ V | ψ 0k (0, y, µ) ≠ v 0k (y) },µ∈(−M,λ)∩Q k∈Q{ y ∈ V | u + 0 (y) ≤ λ } = ⋂µ∈(λ,M)∩Q k∈Q⋃{ y ∈ V | ψ 0k (0, y, µ) ≠ v 0k (y) }and measurability of v 0k (y). Hence, u ± 0 ∈ L∞ (V ), ‖u ± 0 ‖ ∞ ≤ M.Conversely, suppose that there exists a bounded function u 0 (y) such thatv 0k (y) = ψ 0k (0, y, u 0 (y)) a.e. on V for each k ∈ R. We take M ≥max{‖u‖ ∞ , sup |u 0 (y)|} and set I = [−M, M]. Since maps k → ψ 0k (0, y, ·) ∈ C(I),k → v 0k (y) are continuous, there is a set Y ⊂ V of full measure such thatv 0k (y) = ψ 0k (0, y, u 0 (y)) for all k ∈ R if y ∈ Y (see the arguments for theset Y in the first part of the proof). We should prove that for all k ∈ Rψ 0k (t, y, u(t, y)) → v 0k (y) = ψ 0k (0, y, u 0 (y)) in L 1 (V ) as t → 0, t ∈ E, wherethe set E ⊂ (0, h) of full measure is defined by (2.20). As was shown above, this isequivalent to the convergenceψ 0k (0, y, u(t, y)) → ψ 0k (0, y, u(t, y)) in L 1 (V ) ∀k ∈ R. (2.22)t→0,t∈EAssuming that (2.22) fails, we can find k ∈ R and a sequence t m ∈ E, t m → 0, suchthat for some δ > 0∫lim inf |ψ 0k (0, y, u(t m , y)) − v 0k (y)|dy > δ. (2.23)m→∞VWithout loss of generality we can suppose that u(t m , y) converges as m → ∞ tosome measure valued function ν y . Clearly, ‖ν y ‖ ≤ M, and by the weak-∗ convergenceψ 0k (0, y, u(t m , y)) → v 0k (y) in L ∞ (V ) we have the relationsm→∞∫ψ 0k (0, y, u 0 (y))=v 0k (y)=〈ψ 0k (0, y, λ), ν y (λ)〉= ψ 0k (0, y, λ)dν y (λ) ∀k ∈ R. (2.24)Clearly, the set of full measure Y 1 ⊂ Y of values y, for which this relation holds, canbe chosen common for all k ∈ R. This easily follows from the fact that both sidesof (2.24) depends continuously on k. Taking in (2.24) k > M, we derive∫ϕ 0 (0, y, u 0 (y)) = v 0 (y) = 〈ϕ 0 (0, y, λ), ν y (λ)〉 = ϕ 0 (0, y, λ)dν y (λ). (2.25)Fix y ∈ Y 1 and denote ϕ 0 (u) = ϕ 0 (0, y, u), u 0 = u 0 (y), ν = ν y . Let η(u) ∈ BV (I).Integrating relation (2.24) over the measure dη(k) and using Lemma 2.2 with f(u) =ϕ 0 (u), we find that for all η ∈ BV (I)〈T η ϕ 0 (λ), ν(λ)〉 = T η ϕ 0 (u 0 ). (2.26)23

Here we also take into account relation (2.25). Now, we are going to derive from(2.26) that ϕ 0 (u) ≡ const on [a, b] = Co supp ν. Assuming the contrary, we canchoose a value c ∈ (a, b) and sequences c r , h r , r ∈ N such that the sequence ψ r =T ηr ϕ 0 (u) defined by (2.19) is bounded, and converges point-wise as r → ∞ to theHeaviside function θ(u − c). Taking in (2.26) η = η r and passing to the limit asr → ∞ we find that ν((c, b]) = θ(u 0 − c) ∈ {0, 1}, which contradicts to the factthat [a, b] is the minimal segment containing supp ν and therefore 0 < ν((c, b]) < 1.Thus, ϕ 0 (u) = ϕ 0 (u 0 ) for all u ∈ [a, b]. From the obtained relation it follows thatψ 0k (u) ≡ ψ 0k (u 0 ) on [a, b]. Hence for a.e. y ∈ V ψ 0k (0, y, u) = ψ 0k (0, y, u 0 (y)) =v 0k (y) on supp ν y and by Corollary 2.1 for all k ∈ R ψ 0k (0, y, u(t m , y)) convergesas m → ∞ to v 0k (y) in L 1 (V ). But this contradicts (2.23). Therefore, (2.22) issatisfied, as was to be proved.If, in addition, the function ϕ 0 (0, y, ·) is not constant on non-degenerate intervalsfor a.e. y ∈ V then we take a sequence t m ∈ E, t m → 0 such that u(t m , y) convergesweakly to a measure valued function ν y . Repeating the above arguments we find thatfor a.e. y ∈ V supp ν y ⊂ [u − (y), u + (y)], u 0 (y) ∈ [u − 0 (y), u+ 0 (y)], and ϕ 0(0, y, u) ≡ϕ 0 (0, y, u 0 (y)) on [u − 0 (y), u + 0 (y)]. By our assumption the functions ϕ 0 (0, y, ·) are notconstant on non-degenerate intervals for a.e. fixed y ∈ V . Therefore, for a.e. y ∈ Vu 0 (y) = u − 0 (y) = u+ 0 (y), and ν y(λ) = δ(λ − u 0 (y)) is a regular measure valuedfunction. By Theorem 2.1 the sequence u(t m , y) → u 0 (y) in L 1 (V ) as m → ∞ andsince the limit function u 0 (y) does not depend on the choice of a vanishing sequencet m ∈ E, we conclude that u(t, ·) → u 0 in L 1 (V ) as t → 0, t ∈ E. The proof iscomplete.Corollary 2.2. Suppose u(t, y) ∈ L ∞ (Π) is a quasi-s. of (1.14), E ⊂ R +is defined by (1.21), and there is a sequence t m ∈ E such that t m → 0, andm→∞ψ 0k (·, u(t m , ·)) → v 0k in L 1 loc (Rn ). Then ψ 0k (·, u(t, ·)) → v 0k in L 1 loc (Rn ) as t → 0,m→∞t ∈ E.Proof. Extracting a subsequence if necessary, we can assume that u(t m , y) convergesweakly to a measure valued function ν y ∈ MV(R n ). In view of strong convergenceψ 0k (0, ·, u(t m , ·)) → v 0k from Corollary 2.1 it follows that for all k ∈ R ψ 0k (0, y, λ) ≡v 0k (y) on supp ν y for all y ∈ Y , where Y ⊂ R n is a set of full measure, which doesnot depend on k ( see the proof of Theorem 2.4 ). Thus, ψ 0k (0, y, u 0 (y)) = v 0k (y)for each y ∈ Y , where u 0 (y) ∈ supp ν y . By Theorem 2.4 we conclude that v 0k (y) arethe strong traces for ψ 0k (t, y, u(t, y)) for all k ∈ R, as was to be proved.3 The ”blow-up” procedureLet u(t, y) ∈ L ∞ (Π) be a quasi-s. of (1.14). We consider a sequence ε m > 0, m ∈ Nsuch that ε m → 0 as m → ∞ and setu m = u m (t, y; z) = u(ε m t, ε m y + z),24

where z ∈ R n is treated as a fixed parameter. By Proposition 1.2 there exists theweak traces v 0k (y) ∈ L ∞ (Π) such that ψ 0k (y, u(t, y)) → v 0k (y) weakly-∗ in L ∞ (R n )as t → 0, t ∈ E. We set also v m 0k (y; z) = v 0k(ε m y + z).The following lemma was essentially proved in [26] ( Lemma 3 ) and in [19]( Lemma 3.1 ).Lemma 3.1. There exists a subsequence of ε m ( we keep the notation ε m ) such thatfor a.e. z ∈ R n and all k ∈ R v m 0k (y; z) → v 0k(z) = const in L 1 loc (Rn ) as m → ∞.For completeness we give the proof.Proof. Let ρ(y) = e −|y| ,∫Ik m (z) = |v 0k (ε m y + z) − v 0k (z)|ρ(y)dy.R nWe integrate this equality over z. Changing the order of integrating we obtain that(∫)Ik ∫R∫R m (z)ρ(z)dz = |v 0k (ε m y + z) − v 0k (z)|ρ(z)dz ρ(y)dy. (3.1)n n R nBy continuity of the average, for all y ∈ R nJ m k (y) = ∫R n |v 0k (ε m y + z) − v 0k (z)|ρ(z)dz →m→∞0,and also ‖Jk m(y)‖ ∞ ≤ const. Using the Lebesgue theorem on dominated convergence,we conclude that the right-hand side of (3.1) converges to zero. Then, by(3.1) ∫Ik m (z)ρ(z)dz → 0.R n m→∞Therefore, after possible extraction of a subsequence (we keep the previous notationfor it), Ik m (z) → 0 as m → ∞ for a.e. z and all k ∈ Q. This means that for suchvalues of y v0k m(y; z) → v 0k(z) in L 1 loc (Rn ). Since the above functions depends on kcontinuously and Q is dense, we see that this limit relation is satisfied for all k ∈ R.The proof is complete.Evidently, u m (t, y; z) satisfies equalities like (1.18), namely in D ′ (Π)ψ 0k (ε m y + z, u m ) t + div y ψ k (u m ) = −S m z (γ k) = −ε m γ k (ε m t, ε m y + z). (3.2)The operator Sz m is defined on the space ¯Mloc (Π) by the equality Sz m(γ) =ε m γ(ε m t, ε m y + z) understood in the sense of distributions. This means that∀h(t, y) ∈ C 0 (Π)∫〈Sz m (γ), h〉 = (ε m ) −n h(t/ε m , (y − z)/ε m )dγ(t, y).ΠThe following lemma can be proved in just the same way as Lemma 2 in [26] ( seealso Lemma 3.2 in [19] ).25

Lemma 3.2. If γ ∈ ¯M loc (Π) then, after possible extraction of a subsequence, fora.e. z ∈ R n Sz m (γ) → 0 in ¯M loc (Π).m→∞Proof. Firstly, observe that VarSz m(γ) = Sm z (Var γ). Therefore, without loss ofgenerality we may assume that γ ≥ 0. Let r > 0. Denote by B r the ball {y | |y| ≤r}, and by χ(t, y) the indicator function of the cylinder (0, r] × B r . We set∫I m (z) = Sz m (γ)((0, r] × B r) = (ε m ) −n χ(t/ε m , (y − z)/ε m )dγ(t, y).Now we integrate this equality over z ∈ B R , where R > 0. Changing the order ofintegrating and making the change x = (y − z)/ε m we derive that∫ ∫I∫B m (z)dz = (ε m ) −n χ(t/ε m , (y − z)/ε m )dzdγ(t, y) =R Π B∫ ∫Rχ(t/ε m , x)dxdγ(t, y) ≤ C r γ((0, ε m r] × B R+εmr), (3.3)Π|y−ε mx|≤Rwhere C r is Lebesgue measure of the ball B r . As m → ∞and in view of (3.3), for all R > 0γ((0, ε m r] × B R+εmr) → 0,limm→∞∫ΠB RI m (z)dz = 0.Therefore, we can choose a subsequence of ε m (as usual, we keep the notation ε m )such that I m (z) → 0 a.e. on R n . Using the diagonal extraction, we can chose theindicated subsequence and the set of full measure z ∈ R n , for which I m (z) → 0,being common for all rational values of r. Then, for such zas required. The proof is complete.S m z (γ) →m→∞ 0 in ¯M loc (Π),Applying Lemma 3.2 to the measures γ k , k ∈ F, we see that there exists asubsequence of the sequence ε m ( we keep the notation ε m for this subsequence)and a set Z ⊂ R n of full measure such that S m z (γ k) → 0 in ¯M loc (Π) as m → ∞∀z ∈ Z. Recall that the dense set F ⊂ R is supposed to be countable. Then, usingthe standard diagonal extraction, we can choose the indicated subsequence beingcommon for all k ∈ F.Thus, we can assume that the sequence ε m and the set Z ⊂ R n of full measureare chosen in such way that the assertions of Lemma 3.1 and Lemma 3.2 for eachk ∈ F and γ = γ k hold.Our interest to the sequence u m (t, y; z) is stipulated by the following theorem:26

Theorem 3.1. Let v k (t, y) = ψ 0k (y, u(t, y)), vk m(t, y; z) = v k(ε m t, ε m y + z) =ψ 0k (ε m y + z, u m (t, y; z)). Then(i) Existence of the strong traces ess lim ψ 0k (y, u(t, y)) = v 0k (y) in L 1 loct→0(Rn ) impliesthat, after possible extraction of a subsequence, for a.e. z ∈ R n the sequencesvk m(t, y; z) converge as m → ∞ to some functions v k(t, y; z) in L 1 loc (Π);(ii) Conversely, suppose that for a.e. z ∈ R n and each k ∈ F there is a subsequenceof ε m such that the sequence vk m (t, y; z) is strongly convergent. Then thereexist strong traces ess lim ψ 0k (y, u(t, y)) = v 0k (y) in L 1 loct→0(Rn ).Proof. Suppose ess lim v k (t, y) = v 0k (y) in L 1 loct→0(Rn ). We denote ρ(z) = e −|z| . Thenfor a.e. (t, y) ∈ Π∫|vk m (t, y; z) − v 0k(ε m y + z)|ρ(z)dz =∫|v k (ε m t, ε m y + z) − v 0k (ε m y + z)|ρ(z)dz =∫|v k (ε m t, x) − v 0k (x)|ρ(x − ε m y)dx → 0.m→∞Integrating this relation over (t, y) weighted p(t, y) = e −t−|y| , and using the Lebesguedominated convergence theorem we derive the relation∫ (∫)|v k (ε m t, ε m y + z) − v 0k (ε m y + z)|p(t, y)dtdy ρ(z)dz → 0.m→∞ΠFrom this it follows that, after possible extraction of a subsequence, vk m (t, y; z) −v0k m(y; z) → 0 as m → ∞ in L1 loc (Π) for a.e. z ∈ Z. By Lemma 3.1 we see also thatv0k m(y; z) → v 0k(z) in L 1 loc (Π) and we can conclude that for a.e. z vm k (t, y; z) → v 0k(z)in L 1 loc (Π), as required. Observe also that the subsequence ε m and the set of fullmeasure of the values z, for which the above limit relation holds can be chosento be common for all k ∈ R. This easily follows from continuity vk m (t, y; z) =ψ 0k (ε m y + z, u m (t, y; z)) with respect to the parameter k.Conversely, suppose that ∀z ∈ Z 1 , where Z 1 ⊂ R n is a set of full measure, thereis a subsequence of ε m such that vk m(·; z) converges in L1 loc (Π) to some functionv k (·; z) ∈ L ∞ (Π). Clearly, this subsequence can be chosen common for all k ∈ R.Without loss of generality we can assume that Z 1 ⊂ Z. Fix z ∈ Z 1 and simplifythe notations u m (t, y) = u m (t, y; z), vk m(t, y) = vm k (t, y; z), v k(t, y) = v k (t, y; z).In correspondence with Theorem 2.1 we can suppose that the sequence u m (t, y)converges weakly to a measure valued function ν t,y ∈ MV(Π). Clearly, ‖ν t,y ‖ ∞ ≤M = ‖u‖ ∞ . Since the sequence vk m(t, y) = ψ 0k(ε m y + z, u m (t, y)) converges stronglywhile the function ψ 0k (y, u) is continuous, we see that the sequence ψ 0k (z, u m (t, y))converges strongly to the limit v k (t, y) = 〈ψ 0k (z, λ), ν t,y (λ)〉. By Corollary 2.1 wesee that the functions ψ 0k (z, λ) ≡ v k (t, y) on supp ν t,y for a.e. (t, y) ∈ Π.27

Passing to the limit as m → ∞ in (3.2) and taking into account that for k ∈ FS m z (γ k ) → 0 in D ′ (Π), we obtain that ∀k ∈ F(〈ψ 0k (z, λ), ν t,y (λ)〉) t + div y 〈ψ k (λ), ν t,y (λ)〉 = 0 in D ′ (Π). (3.4)Since the left-hand side of this equality is continuous with respect to k in D ′ (Π),while F ⊂ R is dense we conclude that (3.4) holds for all k ∈ R, i.e. ν t,y is a measurevalued i.s. of the equation (2.6) with ϕ 0 (u) = ϕ 0 (z, u). As was shown above, thisi.s. satisfies the assumptions of Theorem 2.3 and, therefore, the functions v k (t, y)have strong traces at t = 0.By the relation like (1.23) we see that for a.e. τ > 0 ∀h(y) ∈ C0(R 1 n )∫∫Ik m (τ) − Im k (0+) = vk m (τ, y)h(y)dy − v0k m (y)h(y)dx =∫R n ∫R n(ψ k (u m ), ∇ y h)dtdy − h(x)dγk m (t, y),(0,τ]×R n (0,τ]×R nwhere γ m k = Sm z (γ k). From this equality we derive the estimate|I m k (τ) − I m k (0+)| ≤ C h (τ + |γ m k |((0, T] × K)), (3.5)C h = const, K = supp h,which holds for a.e. τ ∈ (0, T).By our assumptions the conclusion of Lemma 3.2 holds, that is |γk m |((0, T]×K) →0 as m → ∞. Further, for a.e. τ > 0 vk m(τ, y) → v k(τ, y) as m → ∞ in L 1 loc (Rn )(after possible extraction of a subsequence), and also v0k m(y) → v 0k(z) = const inL 1 loc (Rn ) (by Lemma 3.1). Therefore, from (3.5) it follows, in the limit as m → ∞,that for a.e. τ > 0, ∀h(y) ∈ C0 1(Rn )∫∫∣ v k (τ, y)h(y)dy − v 0k h(y)dy∣ ≤ C hτ → 0,R n R ni.e. ess lim v k (t, ·) = v 0k = v 0k (z) weakly-∗ in L ∞ (R n ).t→0Since the traces for v k (t, y) must be strong we findess lim〈ψ 0k (z, λ), ν t,y (λ)〉 = v 0k (z) in L 1 loct→0(Rn ) (3.6)for each k ∈ F. Taking into account that the set F is dense and both sides in (3.6)are continuous with respect to k in L 1 loc (Rn ), we derive that (3.6) holds for all k ∈ R.As in the proof of Theorem 2.3 we can choose the sequence t r → 0 such that thecorresponding sequence of measure valued functions νy r = ν t r,y ∈ MV (R n ) is welldefined,converges weakly as r → ∞ to a measure valued function ν y , ‖ν y ‖ ∞ ≤ M.In correspondence with (3.6) 〈ψ 0k (z, λ), νy(λ)〉 r → v 0k (z) in L 1r→∞loc (Rn ). Therefore, byCorollary 2.1, ψ 0k (z, λ) ≡ v 0k (z) on supp ν y for all k ∈ R and almost all y ∈ R n . Fix28

such y and set u 0 ∈ supp ν y . Then, u 0 = u 0 (z) satisfies the properties: |u 0 | ≤ M andψ 0k (z, u 0 ) = v 0k (z). Since here z ∈ Z 1 is arbitrary, we obtain the bounded functionu 0 (z) such that ψ 0k (z, u 0 (z)) = v 0k (z) a.e. on R n . By Theorem 2.4 we conclude thatthe traces v 0k (y) are strong, which completes the proof.4 Reduction of the space dimensionSuppose that u(t, y) is a quasi-s. of (1.14) and one of the components of the fluxvector, say ϕ n , is absent, i.e. equation (1.14) has the form∑n−1ϕ 0 (y, u) t + ϕ i (u) yi = ϕ 0 (y, u) t + div y ′ϕ(u) = 0,i=1y ′ = (y 1 , . . .,y n−1 ), ϕ(u) = (ϕ 1 (u), . . .,ϕ n−1 (u)) ∈ R n−1 .It is natural to consider the equation for the fixed variable y n = λϕ 0 (y ′ , λ, u) t + div y ′ϕ(u) = 0, u = u(t, y ′ ) (4.1)in the half-space Π ′ = R + × R n−1 with reduced space dimension.Below, to justify the inductive jump in the proof of Theorem 1.4, we will needthe following result.Theorem 4.1. For a.e. y n ∈ R the function ũ(t, y ′ ) = u(t, y ′ , y n ) is a quasi-s. ofequation (4.1).In the case ϕ 0 (y, u) = u Theorem 4.1 was proved in [19]. To prove this theoremwe shall need the following lemma established in [19] ( Lemma 4.2 ). Forcompleteness we reproduce the proof.Lemma 4.1. Let N ∈ N and γ be a locally finite measure on R N × R. Then fora.e. λ ∈ R there exist a locally finite measure γ λ on R N such that ∀h(y) ∈ C 0 (R N )∫∫lim h(y)δ ν (s − λ)dγ(y, s) = h(y)dγ λ (y).ν→∞R N+1 R NProof. We denote |γ| = Varγ and assume firstly that|γ|(R N × [a, b]) < +∞ for any segment [a, b] ⊂ R.(∗)Then we can define a projection m = pr ∗ |γ|, which is the image of |γ| under themap pr(y, s) = s. It is clear that m is a locally finite nonnegative measure on R.Further, for any function h(y) ∈ C 0 (R n ) the correspondence∫g → h(y)g(s)dγ(y, s), g ∈ C 0 (R)R N+129

defines a bounded linear functional l h on C 0 (R), moreover∫|〈l h , g〉| ≤ ‖h‖ ∞ |g(s)|d|γ|(y, s) = ‖h‖ ∞ |g(s)|dm(s). (4.2)R∫RN+1In particular,|〈l h , g〉| ≤ ‖h‖ ∞ m(K)‖g(s)‖ ∞ , K = supp g,which implies that the functional l h is in fact continuous. By the known representationtheorem l h is given as the integral∫〈l h , g〉 = g(s)dµ h (s)Rover some locally finite measure µ h , and, as it follows from (4.2), its variation |µ h | ≤‖h‖ ∞ · m. The latter implies that the measures µ h are absolutely continuous withrespect to the measure m and by the Radon-Nikodym theorem µ h = α h (s)m, whereα h (s) are Borel functions, |α h | ≤ ‖h‖ ∞ .Now, we consider the Lebesgue decomposition m = β(s)ds + σ, where β(s) ∈L 1 loc (R), β(s) ≥ 0, and σ is a singular measure. By known properties of measures( see for instance [23] ) there exists a set of full Lebesgue measure A ⊂ R, on whichthe measure σ has the null derivative over the Lebesgue measure ds that is ∀λ ∈ A∫limν→∞δ ν (s − λ)dσ(s) = 0. (4.3)We choose a countable dense set G ⊂ C 0 (R N ) and consider the set A ′ ⊂ A consistingof common Lebesgue points for the function β(s) and the countable family offunctions ρ h (s) = α h (s)β(s) with h ∈ G. Since|ρ h1 (s) − ρ h2 (s)| = |α h1 −h 2(s)|β(s) ≤ ‖h 1 − h 2 ‖ ∞ β(s) ∀h 1 , h 2 ∈ C 0 (R N )and G is dense in C 0 (R N ) we conclude that λ ∈ A ′ are Lebesgue points of allfunctions ρ h (s), h ∈ C 0 (R N ). For λ ∈ A ′ we have∫∫h(y)δ ν (s − λ)dγ(y, s) = δ ν (s − λ)α h (s)dm(s) =R N+1 ∫R∫δ ν (s − λ)ρ h (s)ds + δ ν (s − λ)α h (s)dσ(s). (4.4)RSince λ is a Lebesgue point of the functions ρ h (s) then∫δ ν (s − λ)ρ h (s)ds → ρ h (λ) as ν → ∞.RR30

Further, in view of (4.3)∫∫∣ δ ν (s − λ)α h (s)dσ(s)∣ ≤ ‖α h‖ ∞ δ ν (s − λ)dσ(s) ≤RR∫‖h‖ ∞ δ ν (s − λ)dσ(s) → 0ν→∞and from (4.4) it follows that∫RR N+1 h(y)δ ν (s − λ)dγ(y, s) → ρ h (λ) as ν → ∞. (4.5)For a fixed λ ∈ A ′ the correspondence h → ρ h (λ) determines a bounded linearfunctional on C 0 (R N ) and, taking into account the estimate |ρ h (λ)| = |α h (λ)β(λ)| ≤β(λ)‖h‖ ∞ , we have the representation∫ρ h (λ) = h(y)dγ λ (y), (4.6)R Nwhere γ λ is a finite Borel measure such that |γ λ |(R N ) ≤ β(λ), λ ∈ A ′ . From (4.5),(4.6) it directly follows the conclusion of the Lemma.In the general case of arbitrary locally finite measure γ we introduce measuresγ l = γ| Bl ×R, where B l is a ball |y| ≤ l, l ∈ N. We see that the measures γ l can beconsidered on the whole space R N × R and satisfy the condition (*). As we havealready proved, there are sets A l ⊂ R of full Lebesgue measure and finite Borelmeasures γλ l such that ∀λ ∈ A l∫∫limν→∞R N+1 h(y)δ ν (s − λ)dγ l (y, s) =R N h(y)dγ l λ(y). (4.7)It is clear that the set A = ∩ l∈N A l has full measure and for λ ∈ A relation (4.7)holds for all l ∈ N. This, in particular, implies that measures γλ l are compatible:γλ l′ | B l= γλ l for l′ > l. Therefore, there exists a locally finite measure γ λ on R N suchthat γ λ | Bl = γλ l ∀l ∈ N. From (4.7) it follows that for λ ∈ A ∀h(y) ∈ C 0(R N )∫∫lim h(y)δ ν (s − λ)dγ(y, s) = h(y)dγ λ (y),ν→∞R N+1 R Nas was to be proved.Proof of Theorem 4.1. Remark firstly that the measures γ k from (1.18) can be extendedon the whole space R n+1 , so that the extended measure of a Borel setA ⊂ R n+1 equals to γ k (A ∩ Π). As it is easy to see, the extended measures arelocally finite and by Lemma 4.1 there exists a set A ⊂ R of full measure such thatfor λ ⊂ A the following limit relation holds: ∀k ∈ F, h(t, y ′ ) ∈ C 0 (Π ′ )∫∫limν→∞Πh(t, y ′ )δ ν (s − λ)dγ k (t, y ′ , s) =31Π ′ h(t, y ′ )dγ k,λ (t, y ′ ), (4.8)

where γ k,λ (t, y ′ ) ∈ ¯M loc (Π ′ ). We also utilize the fact that the set F is countable,which allows us to choose the set A being common for all k ∈ F.Let A 1 ⊂ A be a set of λ ∈ A such that for a.e. (t, y ′ ) ∈ Π ′ (t, y ′ , λ) is a Lebesguepoint of the function u(t, y) = u(t, y ′ , s). Obviously, A 1 is a set of full measure on R.Let λ ∈ A 1 , h(t, y ′ ) ∈ C0 1(Π′ ). We choose a test function g(t, y) = h(t, y ′ )δ ν (y n − λ).By condition (1.18), applying to the test function g, ∀k ∈ F∫∫[ψ 0k (y, u)h t +(ψ k (u), ∇ y ′h)]δ ν (y n −λ)dtdy ′ dy n = h(t, x ′ )δ ν (y n −λ)dγ k (t, y ′ , y n ).ΠPassing in this equality to the limit as ν → ∞ and taking into account (4.8), wefind that∫∫[ψ 0k (y ′ , λ, u(·, λ))h t + (ψ k (u(·, λ)), ∇ y ′h)]dtdy ′ = f(t, y ′ )dγ k,λ (t, y ′ ).Π ′ Π ′Since h(t, y ′ ) ∈ C 1 0 (Π′ ) is arbitrary, we obtain that ∀k ∈ F∂∂t ψ 0k(·, λ, u(·, λ)) + div y ′ψ k (u(·, λ)) = −γ k,λ in D ′ (Π ′ ),i.e. the function ũ = u(·, λ) is a quasi-s. of equation (4.1). The proof is complete.Π5 H-measures associated with sequences of measurevalued functionsThe notion of H-measure was introduced in [25, 10] and was further extended in[16] for the case of sequences of bounded measure valued functions.Let νx m ∈ MV(Ω), m ∈ N be a bounded sequence of measure valued functionsweakly convergent to a measure valued function ν x ∈ MV(Ω). For x ∈ Ω and p ∈ Rwe setV m (x, p) = νx m ((p, +∞)), V 0 (x, p) = ν x ((p, +∞)).As was shown in [16], for m ∈ N ∪ {0}, p ∈ R the functions V m (x, p) ∈ L ∞ (Ω) and0 ≤ V m (x, p) ≤ 1. Let{}P = P(ν x ) = p 0 ∈ R | V 0 (x, p) → V 0 (x, p 0 ) in L 1 loc (Ω) .p→p0The following lemma was proved in [16]:Lemma 5.1. The complement CP = R \ P is at most countable and for p ∈ PV m (x, p) →m→∞V 0 (x, p) weakly-∗ in L ∞ (Ω).32

Let Um p (x) = V m(x, p) − V 0 (x, p). By Lemma 5.1 for p ∈ Pm → ∞ weakly-∗ in L ∞ (Ω). We introduce the following notations:F(u)(ξ), ξ ∈ R N is the Fourier transform of u(x) ∈ L 2 (R N );S = S N−1 = { ξ ∈ R N | |ξ| = 1 } denotes the unit sphere in R N ;u → ū, u ∈ C, is complex conjugation.Um p (x) → 0 asProposition 5.1 (see [16]). 1) There exists a family of complex-valued locally finiteBorel measures {µ pq } p,q∈P on Ω × S and a subsequence U r (x) = {U p r (x)} p∈P suchthat ∀Φ 1 (x), Φ 2 (x) ∈ C 0 (Ω), ψ(ξ) ∈ C(S)〈µ pq , Φ 1 (x)Φ 2 (x)ψ(ξ)〉 =∫lim F(Φ 1 Ur p )(ξ)F(Φ 2 Ur)(ξ)ψ(ξ/|ξ|)dξ.qr→∞R N2) The map (p, q) → µ pq is continuous as a map from P × P into the spaceM loc (Ω × S) of locally finite Borel measures on Ω × S.Definition 5.1. The family {µ pq } p,q∈P is called the H-measure corresponding tothe subsequence ν r x .In the following lemma some important properties of the H-measure are collected(see [16]):Lemma 5.2. 1) ∀p ∈ P µ pp ≥ 0; 2) ∀p, q ∈ P µ pq = µ qp ; 3) for p 1 , . . .,p l ∈ Pand for any bounded Borel set C ⊂ Ω × S the matrix a ij = µ p ip j(C), i, j = 1, . . .,lis positively definite; 4) µ pq = 0 ∀p, q ∈ P if and only if the sequence ν r x is stronglyconvergent as r → ∞.Remark that, as it directly follows from 3), ∀p, q ∈ P(µ pq (C) )2 ≤ µ pp (C)µ qq (C) (5.1)for any bounded Borel set C ⊂ Ω × S.We will need also the following result.Lemma 5.3. Let {µ pq } p,q∈P be the H-measure corresponding to a bounded sequenceν r x ∈ MV(Ω) and C ⊂ P be a closed set such that µpp = 0 for all p ∈ C. If s(u) isa continuous from the left nondecreasing function, which is constant on connectedcomponents of R \ C, then the sequence s ∗ ν r x converges strongly as r → ∞.Proof. Suppose ν r x → ν x as r → ∞. Choose M > sup ‖ν r x ‖ ∞ and defines −1 (µ) = inf{ λ ∈ [−M, M] | s(λ) > µ }.We agree, as usual, that s −1 (µ) = M in the case when s(λ) ≤ µ on [−M, M]. Let usdemonstrate that λ = s −1 (µ) ∈ C ∪ {−M, M}. Indeed, if this is not true then λ ∈33

(a, b), where (a, b) is some connected component of the complement (−M, M) \ C.Clearly s(u) > µ in the interval u ∈ (λ, b). But by our assumption s(u) is constanton (a, b), which implies that s(u) > µ on (a, b). Therefore, λ = s −1 (µ) ≤ a. Theobtained contradiction yields the required relation λ = s −1 (µ) ∈ C ∪ {−M, M}.Observe that p = ±M ∈ P and for these values µ pp = 0 ( because Ur p (x) ≡ 0 forp = ±M ). Hence, we can suppose that ±M ∈ C. Taking into account that thefunction s(λ) is continuous from the left, we find s ∗ νx((µ, r +∞)) = νx((s r −1 (µ), +∞)),r ∈ N; s ∗ ν x ((µ, +∞)) = ν x ((s −1 (µ), +∞)), which implies the relationŨ µ r (x) = s∗ ν r x ((µ, +∞)) − s∗ ν x ((µ, +∞)) = U λ m (x) = νr x ((λ, +∞)) − ν x((λ, +∞)),where λ = s −1 (µ). Hence, the H-measure {˜µ pq } p,q∈R corresponding the sequences ∗ νx r is well defined and ˜µpp = µ qq , q = s −1 (p). Since in this relation q = s −1 (p) ∈ Cwe conclude that ˜µ pp = 0 for all p. By (5.1) we find ˜µ pq ≡ 0 and in correspondencewith Lemma 5.2,4) the sequence s ∗ νx r is strongly convergent.The proof is complete.Corollary 5.1. Let {µ pq } p,q∈P be the H-measure corresponding to a bounded inL ∞ (Ω) sequence u r (x), r ∈ N (these functions are considered as regular measurevalued functions). Suppose u r (x) converges weakly to a measure valued functionν x and [a(x), b(x)] = Co supp ν x . Then for a.e. x ∈ Ω µ pp ≠ 0 for all p ∈(a(x), b(x)) ∩ P.Proof. Clearly for λ ∈ R{ x ∈ Ω | b(x) ≤ λ } = { x ∈ Ω | ν x ((λ, +∞)) = 0 },{ x ∈ Ω | a(x) ≥ λ } = { x ∈ Ω | ν x ((−∞, λ)) = 0 }and since functions x → ν x ((λ, +∞)), x → ν x ((−∞, λ)) are measurable on Ω, byRemark 2.1, we see that a(x), b(x) are measurable as well. Hence a(x), b(x) ∈L ∞ (Ω). Let x be a common Lebesgue point of the functions a(x), b(x). Let usshow that µ pp ≠ 0 for p ∈ (a(x), b(x)) ∩ P. Assuming the contrary, take a valuep ∈ P ∩ (a(x), b(x)) such that µ pp = 0 and define s(u) = sign + (u − p). Thisfunction satisfies the assumptions of Lemma 5.3 with C = {p} and we concludethat the sequence s(u r ) strongly converges to s ∗ ν x . By Theorem 2.1 s ∗ ν x is regular.Therefore p /∈ (a(x), b(x)) for a.e. x ∈ Ω. But this contradicts to the fact thatinequality a(x) < p < b(x) remains valid on a set of positive measure. The proof iscomplete.Now, let ϕ(u) ∈ C(R, R N ) be a continuous vector function. Suppose that thesequence νx r satisfies the condition:(C) ∀p ∈ R the sequence of distributions∫L p r = div x (ϕ(λ) − ϕ(p))dνx r (λ)(p,+∞)34

is precompact in H −1loc (Ω).Here, as usual, the space H −1loc(Ω) consists of distributions u(x) such that for allf(x) ∈ C0 ∞ (Ω) the product fu ∈ H−1 2 . The topology in H−1loc(Ω) is generated bysemi-norms ‖fu‖ H−1 , f ∈ C∞ 2 0 (Ω).From the results of [17] (see Lemma 2 with q = p 0 and the proof of Theorem 4)it directly follows the statement, which plays a key role in the proof of our mainTheorems 1.4 and 1.1.Theorem 5.1. Suppose that {µ pq } p,q∈P is the H-measure corresponding to the sequenceν r x, which satisfies condition (C), and µ pp ≠ 0 for some p = p 0 ∈ P. Thenthere exists a non-degenerate interval I = (p 0 −δ, p 0 +δ) such that (ξ, ϕ(u)) = conston I.In particular from Theorem 5.1 it follows that under the non-degeneracy condition∀ξ ∈ S the function u → (ξ, ϕ(u)) is not constant on non-degenerate intervalsµ pq ≡ 0 and, therefore, the sequence ν r x is strongly convergent.In [18] this result was generalized to the case when the flux vector ϕ dependsalso on the spatial variables x.Observe that the Vasseur’s result in [26] directly follows from the indicatedprecompactness property, with no regularity restrictions on the flux vector and theweaker non-degeneracy assumption.6 Proof of Theorems 1.4 and 1.1We shall use the method of mathematical induction on the spatial dimension n.If n = 0 (the base) then by (1.18) ψ 0k (u) ′ = −γ k ∈ ¯M loc (R + ). This implies thatψ 0k (u(t)) has bounded variation on any interval (0, T), T > 0. Therefore, we canfind v 0k ∈ R such that lim ψ 0k (u(t)) = v 0k . Clearly, v 0k = ψ 0k (u 0 ), where u 0 is ant→0+arbitrary limit point of u(t) at t = 0. We see that the statement of Theorem 1.4 isfulfilled.Assuming that the conclusion of Theorem 1.4 is true for n − 1 space variables,we prove it for dimension n. So, suppose that the function u = u(t, y) is a quasi-s.of equation (1.14).We introduce the set I of segments I = [a, b], a, b ∈ F, such that for somenonzero vector ξ = (ξ 0 , ξ ′ ) ∈ R n+1 the function u → (f(u), ξ) is constant on I. Heref(u) is the flux vector of original equation (1.1). Since the set F is assumed to becountable, the set I is also countable (or empty).Let I = [a, b] ∈ I. We set u I = u I (t, y) = max(a, min(u, b)) so that u I (t, y) ∈ I.Let us show that the function u I (t, y) is also a quasi-s. of (1.14). Take k ∈ F. Then35

k ′ = max(a, min(k, b)) ∈ F and one can easily verify thatψ 0k (y, u I ) = sign (u I − k)(ϕ 0 (y, u I ) − ϕ 0 (y, k)) =ψ 0k ′(y, u) − (ψ 0a (y, u) + ψ 0,b (y, u))/2 + (ψ 0a (y, k) + ψ 0b (y, k))/2,which implies that in D ′ (Π)ψ k (u I ) = sign (u I − k)(ϕ(u I ) − ϕ(k)) =ψ k ′(u) − (ψ a (u) + ψ b (u))/2 + (ψ a (k) + ψ b (k))/2,ψ 0k (y, u I ) t + div y ψ k (u I ) = (γ a + γ b )/2 − γ k ′ ∈ ¯M loc (Π)for all k ∈ F. Thus, u I (t, y) satisfies (1.18), i.e. it is a quasi-s. of (1.14).Now, recall that (f(u), ξ) = const for u ∈ I. Consider the following two cases.1) ξ 0 = 0. In this case ξ ′ = (ξ 1 , . . .,ξ n ) ≠ 0 and there exists a linear changez = z(y), y = (y 1 , . . .,y n ) ∈ R n such that z n = (ξ ′ , y). After this change (1.14)reduces to the formϕ 0 (z, u) t + div z ¯ϕ(u) = 0, (6.1)where ¯ϕ n = (ξ ′ , ϕ(u)) = (ξ ′ , ¯f(u)) = const on I. We consider the function u I as afunction of new variables (t, z). By Theorem 4.1, u I (t, z ′ , z n ) is a quasi-s. of reducedequation ( defined on the domain Π ′ = R + × R n−1 )u t +∑n−1i=1˜ϕ i (u) zi = 0 (6.2)for a.e. z n .As is rather well-known ( see for instance [23] ), the setM = { (t, z) = (t, z ′ , z n ) ∈ Π | (t, z) is a Lebesgue point of u,(t, z ′ ) is a Lebesgue point of u(·, z n ) }has full measure. Therefore, for a set E 1 of full measure of values t > 0 the sectionsM t = {z ∈ R n | (t, z) ∈ M } have full measure on R n . Observe that E 1 ⊂ E, wherethe set E is defined by (1.21). Now, we choose a sequence t r ∈ E 1 , t r → 0, andr→∞introduce the sets A r ⊂ R consisting of values s ∈ R such that (z ′ , s) ∈ M tr for a.e.z ′ ∈ R n−1 . Clearly, A r have full measure, consequently A = ∩ r∈N A r is also a set offull measure. We see also that for s ∈ A all t r ∈ E(s), where E(s) is defined by(1.21) for the function u(·, s) on Π ′ = R + × R n−1 .As we have already established, for a.e. z n ∈ A u I (·, z n ) is a quasi-s. of equation(6.2) on Π ′ . By the inductive assumption for all k ∈ R ψ 0k (·, z n , u I (t r , ·, z n )) →v 0k (·, z n ) in L 1 loc (Rn−1 ) as r → ∞, where v 0k = v 0k (z ′ , z n ) ∈ L ∞ (R n ). Usingthe Lebesgue dominated convergence theorem, we conclude that as r → ∞ also36

ψ 0k (z, u I (t r , z)) → v 0k (z) in L 1 loc (Rn ). This easily implies the same limit relationunder original variables y: ψ 0k (y, u I (t r , y)) → v 0k (y) in L 1 loc (Rn ). By Corollary2.2 ψ 0k (y, u I (t, y)) have the strong traces v 0k (y) on the hyperspace t = 0, i.e.ψ 0k (y, u I (t, y)) → v 0k (y) in L 1 loc (Rn ) as t → 0, t ∈ E.2) ξ 0 ≠ 0.Let C ⊂ R n be the set of y such that ξ ′ + ξ 0 ∇g(y) = 0. Since by (1.15)f 0 (u) = ϕ 0 (y, u) + (∇g(y), ϕ(u)) we have the relationξ 0 ϕ 0 (y, u) + (ξ ′ + ξ 0 ∇g(y), ϕ(u)) = const ∀u ∈ I.Thus, for y ∈ C ϕ 0 (y, u) = v 0 (y) = const on I and consequently ψ 0k (y, u) areconstant on I as well: ψ 0k (y, u) = v 0k (y). Therefore, for u 0 ∈ I ψ 0k (y, u 0 ) = v 0k (y)for every y ∈ C.If y /∈ C then in a vicinity of the point (0, y) ∈ ∂Ω we can make the linear changeof original variables x: z 0 = ξ 0 x 0 +(ξ ′ , x ′ ) ( keeping the variables x ′ = (x 1 , . . .,x n ) ),which removes the component f 0 from equation (1.1). Clearly, the boundary ∂Ωis locally represented as the graph z 0 = ˜g(x ′ ) = ξ 0 g(x ′ ) + (ξ ′ , x ′ ) and since ∇˜g =ξ ′ + ξ 0 ∇g(x ′ ) ≠ 0 in a vicinity of our boundary point, we can express some variablex i , i = 1, . . .,n as a function of remaining variables on ∂Ω. The correspondingcanonical boundary chart transforms our equation to the form (6.2). As in the case1), we deduce from the inductive assumption the strong trace property for ψ 0k (·, u I ).By Theorem 2.4 this implies that for some u 0 (y) ∈ I v 0k (y) = ψ 0k (y, u 0 (y)) fora.e. y /∈ C. Combining the cases y ∈ C, y /∈ C, we find that the representationv 0k (y) = ψ 0k (y, u 0 (y)) holds for a.e. y ∈ R n and by Theorem 2.4 again we concludethat the strong trace property is satisfied on the whole hyperplane t = 0.Thus, in the both cases the functions ψ 0k (y, u I ) have strong traces at the hyperplanet = 0.By Theorem 3.1 we can find a subsequence of the sequence ε m such that thecorresponding sequences ψ 0k (ε m y+z, u m I (t, y; z)), with um I (t, y; z) = u I(ε m t, ε m y+z),converge strongly ( that is in L 1 loc (Π) ) for a.e. z ∈ Rn .The established property remains valid for any choice of the segment I ∈ I. SinceI is countable then, using the diagonal extraction, we can choose the subsequenceε m and the set Z ⊂ R n of full measure to be common for all I ∈ I. More exactly,we can assume that the conclusions of Lemma 3.1 and Lemma 3.2 ( for all γ = γ k ,k ∈ F ) hold, and ∀I ∈ I, z ∈ Z and k ∈ R the functions ψ 0k (ε m y + z, u m I (t, y; z))are strongly convergent as m → ∞.Now, we fix z ∈ Z and k ∈ R. According to Theorem 2.1 we can extract a subsequenceε r such that the corresponding sequence u r (t, y) = u r (t, y; z) = u(ε r t, ε r y+z)converges as r → ∞ weakly to a measure valued function ν t,y , and for this sequencethe H-measureµ pq = µ pq (t, y, ξ) ∈ M loc (Π × S), p, q ∈ P37

is determined, where S = S n = { ξ = (ξ 0 , ξ ′ ) ∈ R n+1 | |ξ| = 1 } is the unit sphere,and the set P ⊂ R is defined as in the previous section.We consider the sequence of regular measure valued functions νt,y r (λ) = δ(λ −u r (t, y; z)) and introduce, as in the previous section,V r (t, y, p) = ν r t,y((p, +∞)) = sign + (u r (t, y) − p),V 0 (t, y, p) = ν t,y ((p, +∞)), U p r (t, y) = V r(t, y, p) − V 0 (t, y, p),here (t, y) ∈ Π, p ∈ P. Let us show that the sequence νt,y r satisfies to condition (C)applied to the vector (ϕ 0 (z, u), ϕ(u)) ∈ R n+1 . Indeed,L p r = ∂ ∫∫(ϕ 0 (z, λ) − ϕ 0 (z, p))dνt,y r ∂t(λ) + div y (ϕ(λ) − ϕ(p))dνt,y r (λ) =(p,+∞)(p,+∞)∂∂t [sign + (u r − p)(ϕ 0 (z, u r ) − ϕ 0 (z, p))] + div y [sign + (u r − p)(ϕ(u r ) − ϕ(p))] =∂∂t [sign + (u r − p)(ϕ 0 (z + ε r y, u r ) − ϕ 0 (z + ε r y, p))] +wherediv y [sign + (u r − p)(ϕ(u r ) − ϕ(p))] + F p r ,F p r = ∂ ∂t {sign + (u r − p)[ϕ 0 (z, u r ) − ϕ 0 (z + ε r y, u r ) + ϕ 0 (z + ε r y, p) − ϕ 0 (z, p)]}and from the identitiessign + (u − p)(ϕ 0 (y, u) − ϕ 0 (y, p)) = (ψ 0p (y, u) + ϕ 0 (y, u) − ϕ 0 (y, p))/2,sign + (u − p)(ϕ(u) − ϕ(p)) = (ψ p (u) + ϕ(u) − ϕ(p))/2,relations (1.18), (1.19), and (3.2) it follows that for p ∈ FL p r = F p r − Sr z (β p), where β p = (γ p + γ)/2.By Lemma 3.2 the sequence Sz(β r p ) → 0 as r → ∞ in ¯M loc (Π) and, using, forinstance, Murat’s lemma ( [14] ), we conclude that Sz r(β p) → 0 in H −1loc. As easy tosee, Fr p → 0 in H−1loc as well and we conclude that Lp r → 0 in H−1loc. Further, if h(t, y)is a function in the Sobolev space H2(Π) 1 having compact support K ⊂ Π then, asis easily verified, for p > q|〈L p r − Lq r , h〉| ≤ C K( max |ϕ 0 (z, u) − ϕ 0 (z, q)| · ‖h t ‖ 2 +u∈[q,p]max |ϕ(u) − ϕ(q)| · ‖∇ y h‖ 2 ) ≤u∈[q,p]C K ( max |ϕ 0 (z, u) − ϕ 0 (z, q)| + max |ϕ(u) − ϕ(q)|)‖h‖ H 1u∈[q,p]u∈[q,p]2,which implies that the sequence L p r is equicontinuous over the parameter p in H−1loc (Π)and, since L p r → 0 in H −1locfor the dense set p ∈ F, this limit relation remains valid for38

all p. Thus, condition (C) is satisfied and we can use the conclusion of Theorem 5.1.By this Theorem, if p ∈ P and µ pp ≠ 0 then p ∈ I for some I ∈ I ( we also take hereinto account that the correspondence between vectors (ϕ 0 (z, u), ϕ(u)) and f(u) isgiven by a linear isomorphism of R n+1 ). Therefore, the set C = R \ ⋃ I∈II satisfiesthe property:∀p ∈ C ∩ P µ pp = 0. (6.3)Now, suppose that [a(t, y), b(t, y)] is the convex hull of supp ν t,y . Since the sequencesψ 0k (z, u r I (t, y; z)) are strongly convergent we see that for a.e. (t, y) ∈ Π all the functionsψ 0k (z, u) must be constant on [a(t, y), b(t, y)] ∩ I, I ∈ I. By Corollary 5.1and (6.3) we see that for a.e. (t, y) ∈ Π the set [a(t, y), b(t, y)] ∩ C lays in thecomplement R \ P and, therefore, is at most countable. Hence, the sets of valuesof the continuous functions ψ 0k (z, u) on [a(t, y), b(t, y)] are at most countableand we conclude that these functions must be constant on [a(t, y), b(t, y)] for a.e.(t, y) ∈ Π. In correspondence with Corollary 2.1 this yields the strong convergence ofψ 0k (z, u r (t, y; z)) and by Theorem 3.1,(ii) we conclude that there exist strong tracesess lim ψ 0k (y, u(t, y)) = v 0k (y) in L 1 loct→0(Rn ). The proof of Theorem 1.4 is complete.To prove Theorem 1.1 remark that by Theorem 1.4 relation (1.10) has alreadyproved for canonical boundary charts (U, ζ, W rh ). Since the corresponding neighborhoodsU cover the boundary ∂Ω we derive from Theorem 2.4 that weak tracesv 0k (x) = (ψ k (u 0 (x)),⃗ν(x)) for some function u 0 (x) ∈ L ∞ (∂Ω). Taking an arbitraryboundary chart (U, ζ, W rh ), we have u = u(t, y) ∈ L ∞ ((0, h) × V r ), andess lim ψ 0k (t, y, u(t, y)) = v 0k (y) = ψ 0k (0, y, u 0 (y)) weakly-∗ in L ∞ (V r ), wheret→0ψ 0k (t, y, u) = sign (u − k)(ϕ 0 (t, y, u) − ϕ 0 (t, y, k)), ϕ 0 (t, y, u) =(⃗ν(x), f(u)) = (∇t(x), f(u)), x = ζ −1 (t, y); v 0k (y) = v 0k (ζ −1 (0, y)).Again by Theorem 2.4 we conclude that ess lim ψ 0k (t, y, u(t, y)) = v 0k (y) in L 1 (V r ),t→0i.e. (1.10) is satisfied for any boundary charts. This completes the proof.Remark 6.1. The presented results remain valid for more general case of unboundedquasi-s., which are understood in the following sense.Definition 6.1. A measurable function u(x) is called a quasi-s. of (1.1) if thereexists a dense set F ⊂ R such that for each pair a, b ∈ F, a < b the cut-off functionu a,b (x) = max(a, min(u(x), b)) satisfies the relationdiv x f(u a,b (x)) = −γ a,b in D ′ (Ω) (6.4)with a measure γ a,b ∈ ¯M loc (Ω).39

Observe that renormalized entropy sub- and super-solutions of equation (1.1) inthe sense of [3] are quasi-s. of this equation.Let us demonstrate that Definition 6.1 is compatible with Definition 1.2. Indeed,if in Definition 6.1 u(x) ∈ L ∞ (Ω) and M = ‖u‖ ∞ then taking in (6.4) a < −M, b >M we derive div x f(u) = −γ, γ ∈ ¯M loc (Ω). Further, taking a = k, b > max(k, M),we finddiv x ψ k (u) = div x [2f(u k,b ) − f(u) − f(k)] = −(2γ k,b − γ) in D ′ (Ω).Thus, condition (1.7) is satisfied with γ k = 2γ k,b − γ ∈ ¯M loc (Ω), and u(x) is aquasi-s. of equation (1.1) in the sense of Definition 1.2. As is easily follows fromthe definition, if u(x) is a quasi-s. of (1.1) then u a,b (x) is a bounded quasi-s. of thisequation for each a, b ∈ F. Taking also into account that the set F is dense, we derivefrom Theorem 1.1 that for each a, b ∈ R, a < b the functions (f(u a,b (x)),⃗ν(x)) havestrong traces at the boundary ∂Ω. Moreover, as follows from Theorem 2.4, thesetraces can be represented as (f(u 0 a,b (x)),⃗ν(x)), where u0 a,b (x) = max(a, min(u0 (x), b))and u 0 (x) is a measurable function on ∂Ω with values from R∪{±∞}. Among suchfunctions u 0 (x) there are the unique minimal u 0 − (x) and the maximal u0 + (x) and(f(u),⃗ν(x)) = const on the intervals (u 0 −(x), u 0 +(x)) for a.e. x ∈ ∂Ω. In particular,under the assumption that for a.e. x ∈ ∂Ω the function u → (f(u),⃗ν(x)) ≠ conston non-degenerate intervals, u 0 −(x) = u 0 +(x) = u 0 (x) a.e. on ∂Ω and u 0 (x) is thestrong trace of the function u(x) itself ( in the sense of strong convergence of thecut-off functions ).Remark 6.2. For the general equationdivf(x, u) = 0 (6.5)existence of the strong trace for quasi-s. u(x) can be proved under the nondegeneracycondition:for H n -a.e. x ∈ ∂Ω and all ξ ∈ R n+1 \ {0} the function u → (ξ, f(x, u)) is notconstant on non-degenerate intervals.Indeed, under this condition Theorem 5.1 yields the strong convergence ( for a.e.z ) of the sequences u m (t, y; z) generated by the blow-up procedure, and existenceof the strong trace follows from Theorem 3.1. Moreover, in view of locality of thisresult it remains true also in the case of equation (6.5) on a manifold Ω ( in thesense of [21], see also forthcoming paper [2] ).Without non-degeneracy conditions existence of strong normal traces for entropysolutions of (6.5) is not generally true. For instance there are numerous examples( see [5, 8, 22] ) of linear transport equations div(a(x)u) = 0 with divergence freefields a(x) ( i.e. diva(x) = 0 in the sense of distributions ) which admits generalizedsolutions such that the vector a(x)u has no strong normal traces on some hyperplane.40

Figure 1:7 Example of a weak solution without a strongtraceExistence of strong traces is not valid for g.s. of equation (1.1), which do not satisfyentropy condition (1.7). We confirm this fact by the following simple example. Letn = 1. We consider the Burgers equation u t + (u 2 ) x = 0. To construct the desiredg.s. u(t, x) we introduce the function⎧⎨ 0 , |x| + |t − 6| ≤ 1w(t, x) = −sign x , |x| + |t − 6| > 1, t ∈ (6, 8] ,⎩sign (1 − x)sign x , |x| + |t − 6| > 1, t ∈ (4, 6)defined in the square t ∈ (4, 8], −2 ≤ x < 2. We extend this function in thewhole layer t ∈ (4, 8], as a 4-periodic function v over the variable x so that fory = x − 4[x/4] − 2 (here [a] denotes the integer part of a) v(t, x) = u(t, y). Inthe half-space Π we define the piecewise constant function u(t, x) = v(2 k t, 2 k x) ift ∈ (4 · 2 −k , 8 · 2 −k ], k ∈ Z, see fig. 1. As is easily verified, on the discontinuitylines the Rankine-Hugoniot condition is satisfied and therefore u(t, x) is a g.s. ofthe Burgers equation. As t → 0 u(t, ·) → 0 weakly-∗ in L ∞ (R) but there is nostrong limit of u(t k , x) for any choice of a sequence t k → 0. Remark also that forthe constructed g.s. condition (1.18) is satisfied with the measures γ k ∈ M loc (Π)∀k ∈ R, but certainly γ k /∈ ¯M loc (Π).AcknowledgmentsThis work was carried out under financial support of Russian Foundation for BasicResearch (grant No. 06-01-00289) and DFG project 436 RUS 113/895/0-1.41

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