Kinetic formulation and uniqueness for scalar conservation laws with discontinuous flux
aa r X i v : . [ m a t h . A P ] S e p KINETIC FORMULATION AND UNIQUENESS FOR SCALARCONSERVATION LAWS WITH DISCONTINUOUS FLUX
GRAZIANO CRASTA, VIRGINIA DE CICCO, AND GUIDO DE PHILIPPIS
Abstract.
We prove a uniqueness result for BV solutions of scalar conservation lawswith discontinuous flux in several space dimensions. The proof is based on the no-tion of kinetic solution and on a careful analysis of the entropy dissipation along thediscontinuities of the flux. Introduction
This paper is concerned with multidimensional scalar conservation laws with discon-tinuous flux, namely(1) u t + div A ( x, u ) = 0 , ( t, x ) ∈ (0 , + ∞ ) × R n , where the flux function A : R n × R → R n is possibly discontinuous in the first variable andsatisfies some structural assumptions listed in Section 2. These type of equations haveattracted a lot of attention in the last years since they naturally arise in several models(for instance models of traffic flow, flow in porous media, sedimentation processes, etc.),see [5, 18, 25] and references therein for a more detailed account on the theory.It is well known that, even for smooth fluxes, in general the Cauchy problem(2) ( u t + div A ( x, u ) = 0 , ( t, x ) ∈ (0 , + ∞ ) × R n ,u (0 , x ) = u ( x ) , x ∈ R n , does not admit classical solutions. On the other hand, the notion of distributionalsolution is too weak in order to achieve well-posedness, in particular it does not provideuniqueness of the solution. In this context, the notion of entropy solution turns out tobe the right one, as it has been shown by Vol’pert [28] in the BV setting and by Kruzkov[20] in the L ∞ framework.The classical Kruzkov’s approach [20] completely solves the problem of well-posednessin the case of smooth fluxes. Moreover, as it has been shown in recent years, using aclever change of variables and the concept of adapted entropies , this approach also worksfor a restricted class of discontinuous fluxes (see [7, 8, 11, 23, 25]).The case of more general discontinuous fluxes in one space variable has been exten-sively studied (see for example [2, 5, 6, 7, 9, 16, 18, 19, 23, 24] and references therein).In particular, it has been pointed out ([2, 5, 18]) that many different admissibility cri-teria generate continuous semigroups of solutions, and the choice of the right criterionmay depend on the physics of the problem under consideration. Indeed, in addition to Date : September 12, 2014.2010
Mathematics Subject Classification.
Primary: 35L65; Secondary: 26A45.
Key words and phrases.
Chain rule, BV functions, conservation laws with discontinuous flux. the classical entropy criteria, one has to impose some conditions on the behavior of thesolutions on the discontinuities of the flux. Roughly speaking different conditions giverise to different criteria. These different conditions are coded, for example, in the notionof germ (see [5]) or of well-posed Riemann solver (see [18]). One of the most studiedadmissibility criterion is based on the notion of vanishing viscosity solution. This willactually be the criterion that we are going to use in this paper, see Remark 3.3 below.In spite of the intensive study concerning conservation laws with discontinuous fluxes,in the multidimensional case there are very few results available in the literature. Avery general existence result has been obtained by Panov [25]. On the other hand, awell-posedness result for a restricted class of fluxes (having only one regular hypersurfaceof discontinuity) has been recently proved by Mitrovic [22].In this paper we propose an entropy criterion, modeled on the ones introduced inthe one-dimensional case in [6, 16, 24], which allows to prove uniqueness of BV entropysolutions of the Cauchy problem (2) under mild assumptions on the flux A . Moreprecisely our main result is the following: Theorem.
Let A ∈ L ∞ ( R n × R ; R n ) satisfies ( H – ( H in Section 2 below and let u and u be two BV entropy solutions of (1) (see Definition 3.2), then Z R n | u ( T, x ) − u ( T, x ) | dx ≤ Z R n | u (0 , x ) − u (0 , x ) | dx. Let us mention that our assumptions are satisfied in the particular case of fluxes of theform A ( x, u ) = b A ( k ( x ) , u ) where k is in SBV ( R n ; R N ) ∩ L ∞ ( R n ; R N ) and b A is smooth,see Remark 2.3.On the other hand we do not investigate the issue of existence of BV entropy solutions.Let us point out that, unlike the case of fluxes with smooth dependence on x , when A is discontinuous there is no general known result concerning existence of BV solutions,see however Remark 3.4.Our method of proof is based on the notion of kinetic solution introduced by Lions,Perthame and Tadmor (see [21, 26, 27]). This technique has been successfully applied byDalibard to the case of nonautonomous smooth fluxes (see [13, 14]). More precisely, weshow the equivalence between the entropic and the kinetic formulation of (1). Relying onthe nonautonomous chain rule for BV functions proved in [3] and on a careful analysisof the entropy dissipation along the discontinuities of the flux, we are then able to provethat the semigroup of solutions is contractive in the L norm.The plan of the paper is the following. In Section 2 we state the assumptions on theflux (see (H1)–(H7) below) and we discuss some preliminary results. In Section 3 weintroduce our notions of entropy and kinetic solution, we prove their equivalence (seeTheorem 3.9), and we state our main uniqueness result (see Theorem 3.5). In Section 4we prove some preliminary estimates for kinetic solutions, and, finally, in Section 5 weprove the uniqueness of kinetic solutions (see Theorem 5.1). Acknowledgments.
The authors would like to thank Luigi Ambrosio and GianlucaCrippa for some useful discussions, and two referees for having improved, with their com-ments, the presentation of the paper. The authors have been supported by the Gruppo
ONSERVATION LAWS WITH DISCONTINUOUS FLUX 3
Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni (GNAMPA) ofthe Istituto Nazionale di Alta Matematica (INdAM).2.
Assumptions on the vector field and the chain rule
In this section we first survey some useful facts about BV function that we need inthe sequel, we state our main structural hypotheses on the vector field (assumptions(H1)-(H7) below) and prove some consequences of these assumptions.2.1. BV functions.
Let us start recalling our main notation and preliminary facts on BV and SBV functions. A general reference is Chapter 3 of [4], and occasionally wewill give more precise references therein.We denote by L n the Lebesgue measure in R n and by H k the k -dimensional Hausdorffmeasure. By Radon measure we mean a nonnegative Borel measure finite on compactsets. If µ is a Radon measure on X and ν is a Radon measure on Y we denote by µ × ν the the product measure on X × Y , sometimes we will also write µ x × ν y . Given a Borelset D we will denote by µ D the Radon measure given by µ D ( B ) = µ ( B ∩ D ). A setΣ ⊂ R n is said to be countably H n − rectifiable if H n − -almost all of Σ can be coveredby a sequence of C hypersurfaces. Let us recall that if Σ ⊂ R n is countably H n − rectifiable, then ( a, b ) × Σ ⊂ R × R n is countably H n rectifiable and H n ( a, b ) × Σ = L ( a, b ) × H n − Σ , see [17, Theorem 3.2.23].A function u ∈ L ( R n ) belongs to BV ( R n ) if its derivative in the sense of distributionsis representable by a vector-valued measure Du = ( D u, . . . , D n u ) whose total variation | Du | is finite, i.e. Z R n u ∂φ∂x i dx = − Z φ D i u ∀ φ ∈ C ∞ c ( R n ) , i = 1 , . . . , n and | Du | ( R n ) < ∞ . Approximate continuity and jump points.
We say that x ∈ R n is an approximatecontinuity point of u if, for some z ∈ R , it holdslim r ↓ − Z B r ( x ) | u ( y ) − z | dy = 0 . The number z is uniquely determined at approximate continuity points and denoted by˜ u ( x ), the so-called approximate limit of u at x . The complement of the set of approximatecontinuity points, the so-called singular set of u , will be denoted by S u .Analogously, we say that x is a jump point of u , and we write x ∈ J u , if there existsa unit vector ν ∈ S n − and u + , u − ∈ R satisfying u + = u − andlim r ↓ − Z B ± ( x,r ) | u ( y ) − u ± | dy = 0 , where B ± ( x, r ) := { y ∈ B r ( x ) : ±h y − x, ν i ≥ } are the two half balls determined by ν .At points x ∈ J u the triplet ( u + , u − , ν ) is uniquely determined up to a permutation of( u + , u − ) and a change of sign of ν ; for this reason, with a slight abuse of notation, wedo not emphasize the ν dependence of u ± and B ± ( x, r ). Since we impose u + = u − , itis clear that J u ⊂ S u , moreover, for every u ∈ BV loc , H n − ( S u \ J u ) = 0 and J u is H n − countably rectifiable (see [4, Theorem 3.78]). Finally, we define the precise representativeas(3) u ∗ ( x ) = ˜ u ( x ) x ∈ R n \ S u (cid:0) u + ( x ) + u − ( x ) (cid:1) / x ∈ J u H n − ( { e u = u ∗ } \ J u ) = 0, in particular, since | Du | << H n − , | e Du | ( { e u = u ∗ } ) = 0. Decomposition of the distributional derivative.
For any oriented and countably H n − -rectifiable set Σ ⊂ R n we have(4) Du Σ = ( u + − u − ) ν Σ H n − Σ . For any u ∈ BV ( R n ), we can decompose Du as the sum of a diffuse part, that weshall denote e Du , and a jump part, that we shall denote by D j u . The diffuse part ischaracterized by the property that | e Du | ( B ) = 0 whenever H n − ( B ) is finite, while thejump part is concentrated on a set σ -finite with respect to H n − . The diffuse part canbe then split as e Du = D a u + D c u where D a u is the absolutely continuous part with respect to the Lebesgue measure,while D c u is the so-called Cantor part. The density of Du with respect to L n can berepresented as follows D a u = ∇ u d L n , where ∇ u is the approximate gradient of u , see [4, Proposition 3.71 and Theorem 3.83].Note also that | e Du | ( { e u = u ∗ } ) = 0. The jump part can be easily computed by takingΣ = J u (or, equivalently, S u ) in (4), namely D j u = Du J u = ( u + − u − ) ν J u H n − J u . We will say that u ∈ SBV ( R n ) if D c u = 0, i.e. if Du = ( u + − u − ) ν J u H n − J u + ∇ u L n . All these concepts and results extend, mostly arguing component by component, tovector-valued BV functions, see [4] for details. Sets of finite perimeter and coarea formula.
We will say that a measurable set E is of finite perimeter if its characteristic function χ E belongs to BV ( R n ). In this casewe denote by ∂ ∗ E = J χ E its reduced boundary (note that this is slightly larger than what it is usually called reducedboundary, however it coincides with it up to a H n − negligible set, see [4, Chapter 3]).Then, Dχ E = D j χ E = − ν ∂ ∗ E H n − ∂ ∗ E, ONSERVATION LAWS WITH DISCONTINUOUS FLUX 5 in particular χ E ∈ SBV . We also recall the coarea formula: if u ∈ BV ( R n ), then for L almost every v ∈ R , the characteristic function of the set { u > v } belongs to BV andwe have the following equalities between measures: Du = Z R Dχ u>v dv | Du | = Z R | Dχ { u>v } | dv = Z R H n − ( ∂ ∗ { u > v } ) dv . The following lemma relates the pointwise behavior of χ { u>v } to the pointwise behaviorof u , see [15, Lemma 2.2] for the proof. Lemma 2.1.
For any function u : R n → R and any v ∈ R , let ϕ u,v ( x ) = χ { u>v } ( x ) . If u ∈ BV ( R n ) , then for L -a.e. v ∈ R there exists a Borel set N v ⊂ R n , with H n − ( N v ) =0 , such that the following relation holds: ϕ u ± ,v ( x ) = ϕ ± u,v ( x ) , ∀ x ∈ R n \ N v . Structural assumptions on the vector field.
Let A ∈ L ∞ ( R n × R ; R n ) be suchthat:(H1) There exists a set C A with L n ( C A ) = 0 such that A ( x, · ) ∈ C ( R ) for every x ∈ R n \ C A and A ( · , v ) ∈ SBV ( R n ) for every v ∈ R n .(H2) There exists a constant M such that | ∂ v A ( x, v ) | ≤ M ∀ x ∈ R n \ C A , v ∈ R . (H3) There exists a modulus of continuity ω such that | ∂ v A ( x, u ) − ∂ v A ( x, w ) | ≤ ω ( | u − w | ) ∀ x ∈ R n \ C A , u , w ∈ R . (H4) There exists a function g ∈ L ( R n ) such that |∇ x A ( x, u ) − ∇ x A ( x, w ) | ≤ g ( x ) | u − w | ∀ x ∈ R n \ C A , u , w ∈ R , where ∇ x A ( x, v ) denotes the approximate gradient of the map x A ( x, u ).(H5) The measure σ := _ u ∈ R | D x A ( · , u ) | satisfies σ ( R n ) < ∞ . Here D x A ( · , u ) is the distributional gradient of the map x A ( x, u ) (which is a measure since A ( · , u ) ∈ BV ) and W denotes the leastupper bound in the space of nonnegative Borel measures, see [4, Definition 1.68].2.3. Chain rule and fine properties of A . Assumptions (H1)-(H5) imply that A satisfies the hypothesis of [3]. Let us summarize some consequence of this fact. If wedefine N = n x ∈ R n : lim inf r → σ ( B r ( x )) r n − > o , then N is a H n − rectifiable set. In the sequel we shall assume that:(H6) H n − ( N ) < + ∞ . G. CRASTA, V. DE CICCO, AND G. DE PHILIPPIS
In our context, Theorem 2.2 in [3] reads as follows: For every u ∈ BV ( R n ; R ) thecomposite function w ( x ) = A ( x, u ( x )) belongs to BV ( R n ; R n ) with(5) | Dw | ≤ σ + M | Du | and e Dw = ∇ x A ( x, e u ( x )) L n + ∂ v A ( x, e u ( x )) ⊗ e Du (6) D j w = (cid:2) A + ( x, u + ( x )) − A − ( x, u − ( x )) (cid:3) ⊗ ν J u ∪N H n − ( J u ∪ N ) . (7)Here the functions A ± ( x, v ) are defined for H n − almost every x ∈ N ∪ J u and every v ∈ R as(8) A ± ( x, v ) = lim r → − Z B ± r ( x ) A ( y, v ) dy, in particular A + ( x, v ) = A − ( x, v ) H n − -a.e. x ∈ J u \ N and every v ∈ R .Note that by interchanging the derivative with the integral, applying Ascoli-Arzel`aTheorem and taking into account (H2) and (H3), we can deduce as in [3, Section 3]that the map v A ± ( x, v ) is C for almost every x ∈ N with derivative given by ∂ v A ± ( x, v ) = ( ∂ v A ( x, v )) ± (it is part of the statement the fact that this last quantity iswell defined). In particular for H n − -a.e. x ∈ N and all u , w ∈ R we have(9) | ( ∂ v A ) ± ( x, u ) − ( ∂ v A ) ± ( x, w ) | ≤ ω ( | u − w | ) . In the same way, see [3, Section 3], for H n − almost every point of R n \ N and every v ∈ R there exists the limit e A ( x, v ) = lim r → − Z B r ( x ) A ( y, u ) dy, and the map v e A ( x, v ) is continuously differentiable with ∂ v e A ( x, v ) = g ∂ v A ( x, v ).Moreover for L n -a.e. x and all u , w ∈ R (10) | g ∂ v A ( x, u ) − g ∂ v A ( x, w ) | ≤ ω ( | u − w | ) . We conclude this section with the following simple remark. Thanks to (H4) and theprevious discussion, the functions B h ( x, v ) = A ( x, v + h ) − A ( x, v ) h satisfy(11) | B h ( x, v ) | ≤ M, | D x B h ( · , v ) | ≤ g ( x ) L n + M H n − N , where the first inequality follows from (H2). Using (H3) one can show that B h ( x, v ) → ∂ v A ( x, v ) for almost every x and every v ∈ R , see [3, Section 3] for similar arguments.From (11) we deduce that ∂ v A ( · , v ) ∈ SBV . Let us now consider the decomposition D x ∂ v A ( · , v ) = ∇ x ∂ v A ( · , v ) L n + (cid:0) ( ∂ v A ) + ( · , v ) − ( ∂ v A ) − ( · , v ) (cid:1) ⊗ ν N H n − N . ONSERVATION LAWS WITH DISCONTINUOUS FLUX 7
According to the discussion below equation (8) we have ( ∂ v A ) ± = ∂ v ( A ± ) for H n − N almost every x and every v . Moreover, by (H4) for every v the family h
7→ ∇ x A ( x, v + h ) − ∇ x A ( x, v ) h , h = 0 , is weakly compact in L and every one of its cluster points has to coincide with ∇ x ∂ v A ( x, v ),i.e. ∇ x ∂ v A ( x, v ) = w L − lim h → ∇ x A ( x, v + h ) − ∇ x A ( x, v ) h . Let us now define for every Borel and bounded function ϕ the maps h ( v ) := Z ϕ ( x ) ∇ x A ( x, v ) dx, h ( v ) := Z N ϕ ( x )( A + ( x, v ) − A − ( x, v )) ⊗ ν N d H n − ( x ) . By the previous discussion we then see that h , h are Lipschitz continuous and every-where differentiable with derivatives given by dh ( v ) dv = Z ϕ ( x ) ∇ x ∂ v A ( x, v ) dxdh ( v ) dv = Z N ϕ ( x ) (cid:16) ∂ v A + ( x, v ) − ∂ v A − ( x, v ) (cid:17) ⊗ ν N d H n − . (12)However in the sequel we will also need the following assumption, ensuring the continuityof the map v → dh /dv :(H7) There exist a L function g and a modulus of continuity ω (which we can assumewithout loss of generality to be equal to the one appearing in (H3)) such that (cid:12)(cid:12) ∇ x ∂ v A ( x, u ) − ∇ x ∂ v A ( x, w ) | ≤ g ( x ) ω ( | u − w | ) ∀ u, w ∈ R . With this assumption we have
Lemma 2.2.
Let A satisfy (H1)–(H7), then there exists a set e C A with L n ( e C A ) = 0 suchthat every x ∈ R n \ e C A is a Lebesgue point for x
7→ ∇ x A ( x, v ) , x
7→ ∇ x ∂ v A ( x, v ) andfor any such x the map v ] ∇ x A ( x, v ) is C with derivative given by ^ ∇ x ∂ v A ( x, v ) Proof.
Let U ⊂ R be a countable dense set and let e C A = [ u ∈ U R n \ S A ( · ,u ) ∪ ( R n \ S g ) ∪ ( R n \ S g ) , which clearly satisfies L n ( e C A ) = 0. By arguing as in [3, Section 3] and using (H4) and(H7) we see that for every x ∈ R n \ e C A the limits ] ∇ x A ( x, v ) = lim r → − Z B r ( x ) ∇ x A ( y, v ) dy ^ ∇ x ∂ v A ( x, v ) = lim r → − Z B r ( x ) ∇ x ∂ v A ( y, v ) dy exist for every v ∈ R . By the first equality in (12) the map v h r ( v ) = − Z B r ( x ) ∇ x A ( y, v ) dy are differentiable with derivative given by h ′ r ( v ) = − R B r ( x ) ∇ x ∂ v A ( y, v ) dy . Since x is aLebesgue point for g , thanks to (H7) this is a family of equi-continuous functions in G. CRASTA, V. DE CICCO, AND G. DE PHILIPPIS v converging to ^ ∇ x ∂ A ( x, v ). It is now a standard argument to see that ] ∇ x A ( x, v ) =lim r h r ( v ) is C with derivative given by lim r h ′ r ( v ) = ^ ∇ x ∂ v A ( x, v ). (cid:3) Remark . Let us point out that our hypotheses include (and actually are modeled on)the case A ( x, u ) = b A ( k ( x ) , u ) where k ∈ SBV ( R n ; R N ) ∩ L ∞ ( R n ; R N ), H n − ( J k ) < + ∞ and b A ∈ C ( R N × R , R n ) ∩ Lip( R N × R , R n ).3. Formulation of the problem
Entropic formulation.
We consider the following scalar conservation law(13) u t + div A ( x, u ) = 0 , where A : R n × R → R n satisfies the structural assumption (H1)–(H7). Definition 3.1 (Convex entropy pair) . We say that ( S, η ) is a convex entropy pair if S ∈ C ( R ) is a convex function, and η = ( η , . . . , η n ) is defined by(14) η i ( x, v ) := Z v ∂ v A i ( x, w ) S ′ ( w ) dw , i = 1 , . . . , n. In the above definition and in the sequel, A i = A · e i , are the components of A .Note that according to the discussion in Section 2.3, η ( · , v ) ∈ SBV ( R n ; R n ) for every v ∈ R and its distributional derivative is given by D x η ( · , v ) = (cid:18)Z v ∇ x ∂ v A ( x, w ) S ′ ( w ) dw (cid:19) L n + (cid:18)Z v (cid:0) ∂ v A + ( x, w ) − ∂ v A − ( x, w ) (cid:1) S ′ ( w ) dw (cid:19) ⊗ ν N d H n − N . Definition 3.2 (Entropy solutions) . A function u ∈ C ([0 , T ]; L ( R n )) ∩ L ∞ ((0 , T ) × R n ) ∩ L ((0 , T ); BV ( R n ))is an entropy solution of (13) if u is a solution to (13) in the sense of distributions, andthere exists a (everywhere defined) Borel representative ˆ u of u with | ˆ u ( t, x ) | ≤ k u k ∞ such that, for every convex entropy pair ( S, η ), one has ∂ t S ( u ) + div (cid:0) η ( x, u ) (cid:1) − div η ( x, v ) (cid:12)(cid:12)(cid:12) v =ˆ u ( t,x ) + S ′ (ˆ u ) div A ( x, v ) (cid:12)(cid:12)(cid:12) v =ˆ u ( t,x ) ≤ A ( x, v ) (cid:12)(cid:12) v =ˆ u ( t,x ) we mean the measure whoseaction on a bounded and Borel function ϕ = ϕ ( t, x ) is given by(16) n X i =1 Z T dt Z R n ϕ ( t, x ) ∇ i A i ( x, ˆ u ( t, x )) dx + n X i =1 Z T dt Z N ϕ ( t, x ) (cid:0) A + i ( x, ˆ u ( t, x )) − A + i ( x, ˆ u ( t, x )) (cid:1) ν i N d H n − ( x ) , ONSERVATION LAWS WITH DISCONTINUOUS FLUX 9 and the same for div η ( x, v ) (cid:12)(cid:12) v =ˆ u ( t,x ) . Remark . Some comments about our definition of entropy solution are in order. Weshall see in Section 3.2 that, if x ∈ N , then our entropy condition characterizes the(closure of the) so-called vanishing viscosity germ defined in equation (5.4) of [5] (see(24) below). This characterization has also been used by Diehl (see Condition Γ in [16]),Mitrovic [24] and Andreianov & Mitrovic [6]. In this sense, despite the appearance ofa somewhat arbitrary Borel representative ˆ u , this definition seems to be the naturalextension to our general framework of the conditions cited above. We also note that,although in Definition 3.2 we require ˆ u to be defined everywhere, it will be clear by ourarguments below, that it is enough to define ˆ u L × ( L n + σ )-a.e. Remark . By definition, if u is an entropy solution to (13), then u ( t, · ) ∈ BV ( R n ) forevery t . For regular fluxes with respect to the space variables, it is well known that theentropy solution to the Cauchy problem with initial data u ∈ BV ( R n ) remains in BV for all times (see [20, 10]). On the other hand, it is not clear when such regularity has tobe expected for entropy solutions when the flux is discontinuous in the space variables.A relevant situation for which we can expect BV -regularity of solutions is the case n = 1,at least for a class of fluxes widely studied in the literature (see for example [1, 18] andreferences therein). For instance, if we assume that the flux is piecewise constant andthat, at every point of discontinuity, it satisfies an appropriate version of the crossingcondition, one can show the existence of a BV solution of the Cauchy problem (see [1,Theorem 2.13] and [18, Lemma 9]).With these definitions at hand we can now restate our main result: Theorem 3.5.
Let A ∈ L ∞ ( R n × R ; R n ) satisfies ( H – ( H and let u and u be twoentropy solutions of (13) , then (17) Z R n | u ( T, x ) − u ( T, x ) | dx ≤ Z R n | u (0 , x ) − u (0 , x ) | dx. Remark . Since we are dealing with bounded solutions, one can suitably localizeassumptions (H1)–(H7) in the “vertical” variable v , see for instance [3]. Furthermoreone can also localize in the space variable x by just requiring for instance that A ( · , v )belongs to SBV loc and similarly for u . Since this will not add any new ideas to the proofbelow, we leave this generalization to the interested reader. Moreover by exploitingstandard techniques in the context of hyperbolic equation the following localized versionof (17) can be easily obtained from the proof of Theorem 3.5: Z B R (0) | u ( T, x ) − u ( T, x ) | dx ≤ Z B R + V T (0) | u (0 , x ) − u (0 , x ) | dx ∀ R ≥ , where V := k A k ∞ .Since u ∈ L (cid:0) (0 , T ); BV ( R n ) (cid:1) , by the discussion in Section 2.3, see (5) in particular,we have that A ( x, u ( t, x )) ∈ L (cid:0) (0 , T ); BV ( R n ; R n )) . By arguing for instance as in [12, Theorem 4.3.1] we then deduce the following:
Lemma 3.7.
Let u be an entropy solution of (13) , then u ∈ BV ((0 , T ) × R n ) . Obviously we can think of A ( · , v ) as a function in SBV ((0 , T ) × R n ; R n ) constant intime, hence equations (13) and (7) give the following Rankine-Hugoniot conditions (cid:0) u + − u − ) ν t + (cid:0) A + ( x, u + ) − A − ( x, u − ) (cid:1) · ν x = 0for H n -a.e. ( t, x ) ∈ ((0 , T ) × N ) ∪ J u ,where ν = ( ν t , ν x ) is the normal to the H n rectifiable set ((0 , T ) × N ) ∪ J u ⊂ (0 , T ) × R n .In particular, since for H n almost every ( t, x ) in ((0 , T ) × N ) ∩ J u we have ν = (0 , ν N ),we obtain(18) A + ( x, u + ) = A − ( x, u − ) for H n -a.e. ( t, x ) ∈ ((0 , T ) × N ) ∩ J u ,where we have introduced the notation(19) A ± ( x, v ) = A ± ( x, v ) · ν N v ∈ R . Analysis of the entropy condition on discontinuities.
Let u be an entropysolution of (13). Thanks to (7) and (16), on (0 , T ) × N the entropy inequality (15) readsas(20) η + ( x, u + ) − η + ( x, ˆ u ) + A + ( x, ˆ u ) S ′ (ˆ u ) ≤ η − ( x, u − ) − η − ( x, ˆ u ) + A − ( x, ˆ u ) S ′ (ˆ u ) , H n almost everywhere on (0 , T ) × N and we have dropped the dependence on ( t, x ) from u ± and ˆ u in order to simplify the notations. Here we understand that u ± = e u if x / ∈ J u and we are using the short hand notations (19) and η ± ( x, v ) := η ± ( x, v ) · ν N ( x ) . Thanks to the definition of η , (14), and (12), condition (20) can be rewritten as(21) A + ( x, u + ) S ′ ( u + ) − Z u + ˆ u A + ( x, v ) S ′′ ( v ) dv ≤ A − ( x, u − ) S ′ ( u − ) − Z u − ˆ u A − ( x, v ) S ′′ ( v ) dv . By a standard approximation argument, we can plug in (21) the Kruzkov–type entropies S ( v ) := | v − c | , η i ( v ) := Z v ∂ v A i ( x, w ) sign( w − c ) dw, where c ∈ R is a constant. We then obtain H n almost everywhere on (0 , T ) × N (22) A + ( x, u + ) sign( u + − c ) − u + − ˆ u ) A + ( x, c ) (ˆ u,u + ) ( c ) ≤ A − ( x, u − ) sign( u − − c ) − u − − ˆ u ) A − ( x, c ) (ˆ u,u − ) ( c ) . Here and in the sequel, for a, b ∈ R the symbol ( a,b ) will denote the characteristicfunction of the open interval I ( a, b ) with endpoints a and b , i.e. I ( a, b ) = ( a, b ) if a < b or I ( a, b ) = ( b, a ) if b < a .We shall now derive as a consequence of (22) some inequalities which will be usefulin the last part of the proof of Theorem 3.5. To this end, let us now fix a point ( t, x ) ∈ ONSERVATION LAWS WITH DISCONTINUOUS FLUX 11 (0 , T ) × N for which (22) and (18) are valid and let I ( u − , u + ) be the open interval withendpoints u − ( t, x ) and u + ( t, x ). Let us consider the following two cases:(a) ˆ u ∈ I ( u − , u + )(b) ˆ u / ∈ I ( u − , u + ) .In the case (a), taking into account the Rankine–Hugoniot condition (18), by (22) weget(23) (cid:2) sign( u + − c ) − sign( u − − c ) (cid:3) A + ( u + ) ≤ u + − u − ) (cid:2) A + ( c ) (ˆ u,u + ) ( c ) + A − ( c ) (ˆ u,u − ) ( c ) (cid:3) where we have dropped the dependence on x from A . This condition gives informationonly for c ∈ I ( u − , u + ):(24) sign( u + − u − ) A + ( u + ) ≤ sign( u + − u − ) (cid:2) A + ( c ) (ˆ u,u + ) ( c ) + A − ( c ) (ˆ u,u − ) ( c ) (cid:3) , c ∈ I ( u − , u + ) . As particular cases, taking c ր ˆ u and c ց ˆ u we getsign( u + − u − ) A + ( u + ) ≤ sign( u + − u − ) A ± (ˆ u ) . Similarly, in the case (b), taking again into account the Rankine–Hugoniot condition(18), by (22) we get(25) (cid:2) sign( u + − c ) − sign( u − − c ) (cid:3) A + ( u + ) ≤ u + − ˆ u ) (cid:2) A + ( c ) (ˆ u,u + ) ( c ) − A − ( c ) (ˆ u,u − ) ( c ) (cid:3) , c I ( u − , u + ) . We will now analyze conditions (23) and (25) in all the possible cases of different posi-tions of u − , u + , ˆ u and c (see [24] for a similar analysis). We list all the cases for reader’sconvenience. First of all we remark that if c ≥ max { u + , u − , ˆ u } or c ≤ min { u + , u − , ˆ u } ,then, by the Rankine–Hugoniot condition (18), condition (22) does not give any infor-mation. Therefore we list all other possible cases:Case 1: u + ≤ u − .Subcase 1a: u + ≤ u − ≤ ˆ u (26) ( i ) u + ≤ u − ≤ c ≤ ˆ u ⇒ A + ( c ) ≤ A − ( c )(27) ( ii ) u + ≤ c ≤ u − ≤ ˆ u ⇒ A + ( c ) ≤ A + ( u + ) . Subcase 1b: u + ≤ ˆ u ≤ u − (28) ( iii ) u + ≤ ˆ u ≤ c ≤ u − ⇒ A − ( c ) ≤ A − ( u − )(29) ( iv ) u + ≤ c ≤ ˆ u ≤ u − ⇒ A + ( c ) ≤ A + ( u + ) . Subcase 1c: ˆ u ≤ u + ≤ u − (30) ( v ) ˆ u ≤ u + ≤ c ≤ u − ⇒ A + ( c ) ≤ A + ( u + )(31) ( vi ) ˆ u ≤ c ≤ u + ≤ u − ⇒ A − ( c ) ≤ A + ( c ) . Case 2: u − ≤ u + .Subcase 2a: u − ≤ u + ≤ ˆ u (32) ( i ) u − ≤ u + ≤ c ≤ ˆ u ⇒ A + ( c ) ≤ A − ( c )(33) ( ii ) u − ≤ c ≤ u + ≤ ˆ u ⇒ A − ( u − ) ≤ A − ( c ) . Subcase 2b: u − ≤ ˆ u ≤ u + (34) ( iii ) u − ≤ ˆ u ≤ c ≤ u + ⇒ A + ( u + ) ≤ A + ( c )(35) ( iv ) u − ≤ c ≤ ˆ u ≤ u + ⇒ A − ( u − ) ≤ A − ( c ) . Subcase 2c: ˆ u ≤ u − ≤ u + (36) ( v ) ˆ u ≤ u − ≤ c ≤ u + ⇒ A + ( u + ) ≤ A + ( c )(37) ( vi ) ˆ u ≤ c ≤ u − ≤ u + ⇒ A − ( c ) ≤ A + ( c ) . Kinetic formulation.
Let us define the function χ : R → R ,(38) χ ( v, u ) := v < u, / v = u, u < v. Note that if u ∈ BV ((0 , T ) × R n ) then for almost every v ∈ R the function ( t, x ) χ ( v, u ( t, x )) belongs to SBV loc .Let A satisfy conditions (H1)–(H7), and let us define a i ( x, v ) := ∂ v A i ( x, v ) ∀ i = 1 , . . . , na n +1 ( · , v ) := − n X i =1 ∂ i A i ( · , v ) . We remark that, for every v ∈ R , a i ( · , v ) is a BV function, while a n +1 ( · , v ) is a Radonmeasure, moreover according to (H5)(39) sup v | a n +1 ( · , v ) | ( R n ) ≤ σ ( R n ) < + ∞ . Let us denote by a ( · , v ) := ( a ( · , v ) L nx , . . . , a n ( · , v ) L nx , a n +1 ( · , v )) . Note that a is a Radon measure and that div x,v a = 0 . Definition 3.8 (Kinetic solutions) . A function u ∈ C ([0 , T ]; L ( R n )) ∩ L ∞ ((0 , T ) × R n ) ∩ L ((0 , T ); BV ( R n ))is a kinetic solution of (13) if u is a solution to (13) in the sense of distributions, andthere exists a (everywhere defined) Borel representative ˆ u of u with | ˆ u ( t, x ) | ≤ k u k ∞ and a positive measure m ( t, x, v ) with m ((0 , T ) × R n +1 ) < + ∞ such that the function( t, x, v ) χ ( v, ˆ u ( t, x )) satisfies(40) ∂ t χ ( v, ˆ u ( t, x )) + div x,v [ a ( x, v ) χ ( v, ˆ u ( t, x ))] = ∂ v ( m ( t, x, v )) ONSERVATION LAWS WITH DISCONTINUOUS FLUX 13 in the sense of distributions.Our first results establishes the equivalence between Definitions 3.2 and 3.8.
Theorem 3.9.
Let u ∈ C ([0 , T ); L ( R n )) ∩ L ∞ ((0 , T ) × R n ) ∩ L ((0 , T ); BV ( R n )) . Then u is an entropy solution to (13) if and only if it is a kinetic solution to (13) .Proof. We divide the proof in two steps.
Step 1 . Let u be a kinetic solution and let S ∈ C ∞ c ( R ). By testing (40) with φ ( t ) ϕ ( x ) S ′ ( v ),we then obtain Z (0 ,T ) × R n × R φ ′ ( t ) ϕ ( x ) S ′ ( v ) χ ( v, ˆ u ( t, x )) dtdxdv + n X i =1 Z (0 ,T ) × R n × R φ ( t ) ∂ i ϕ ( x ) S ′ ( v ) a i ( x, v ) χ ( v, ˆ u ( t, x )) dtdxdv + Z (0 ,T ) × R n × R φ ( t ) ϕ ( x ) S ′′ ( v ) χ ( v, ˆ u ( t, x )) da n +1 ( x, v ) dvdt = Z (0 ,T ) × R n × R S ′′ ( v ) ϕ ( x ) φ ( t ) dm ( t, x, v ) . (41)Now, since S is compactly supported,(42) Z R S ′ ( v ) χ ( v, ˆ u ( t, x )) dv = Z ˆ u ( t,x ) −∞ S ′ ( v ) dv = S (ˆ u ( t, x ))and, for i = 1 , . . . , n ,(43) Z R S ′ ( v ) a i ( x, v ) χ ( v, ˆ u ( t, x )) dv = Z ˆ u ( t,x ) −∞ S ′ ( v ) ∂ v A i ( x, v ) dv = η i ( x, ˆ u ( t, x )) , where η ( x, v ) := Z v −∞ S ′ ( w ) ∂ v A ( x, w ) dw = η ( x, v ) + Z −∞ S ′ ( w ) ∂ v A ( x, w ) dw by the definition of η in (14). Moreover Z R n × R ϕ ( x ) S ′′ ( v ) χ ( v, ˆ u ( t, x )) da n +1 ( x, v ) dv = − n X i =1 Z R n × R ϕ ( x ) S ′′ ( v ) χ ( v, ˆ u ( t, x )) ∇ i A i ( x, v ) dv dx − Z N × R ϕ ( x ) S ′′ ( v ) χ ( v, ˆ u ( t, x )) (cid:0) A + ( x, v ) − A − ( x, v ) (cid:1) dv d H n − ( x ) , (44)where we have used the short hand notation (19). Now by the discussion in Section 2.3,the map v A ± ( x, v ) is C for H n − almost every x ∈ N with derivative given by v ∂ v A ± ( x, v ), hence for any such x Z R S ′′ ( v ) χ ( v, ˆ u ( t, x )) (cid:16) A + ( x, v ) − A − ( x, v ) (cid:17) dv = Z ˆ u ( t,x ) −∞ S ′′ ( v ) (cid:0) A + ( x, v ) − A − ( x, v ) (cid:1) dv = S ′ (ˆ u ( t, x )) (cid:0) A + ( x, ˆ u ( t, x )) − A − ( x, ˆ u ( t, x )) (cid:1) − Z ˆ u ( t,x ) −∞ S ′ ( v ) (cid:0) ∂ v A + ( x, v ) − ∂ v A − ( x, v ) (cid:1) dv = S ′ (ˆ u ( t, x )) (cid:0) A + ( x, ˆ u ( t, x )) − A − ( x, ˆ u ( t, x )) (cid:1) − (cid:0) η + ( x, ˆ u ( t, x )) − η − ( x, ˆ u ( t, x )) (cid:1) (45)where we are using for η the same convention (19) used for A and η . In the same wayby Lemma 2.2, for almost every x ∈ R n the map v
7→ ∇ i A i ( x, v ) is C with derivativegiven by ∇ i ∂ v A i ( x, v ), hence for any such x Z R S ′′ ( v ) χ ( v, ˆ u ( t, x )) ∇ i A i ( x, v ) dv = S ′ (ˆ u ( t, x )) ∇ i A ( x, ˆ u ( t, x )) − Z ˆ u ( t,x ) −∞ S ′ ( v ) ∇ i ∂ v A i ( x, v ) dv = S ′ (ˆ u ( t, x )) ∇ i A ( x, ˆ u ( t, x )) − ∇ i η i ( x, ˆ u ( t, x )) . (46)Combining (41), (42), (43), (44), (45) and (46) we deduce that if u is a kinetic solutionof (13), then for every function S ∈ C ∞ c ( R ) we have ∂ t S ( u ) + div (cid:0) η ( x, u ) (cid:1) − div η ( x, v ) (cid:12)(cid:12)(cid:12) v =ˆ u + S ′ (ˆ u ) div A ( x, v ) (cid:12)(cid:12)(cid:12) v =ˆ u = − Z S ′′ ( w ) dm ( · , · , w )in the sense of distribution. We now note that η − η is a function of the x variable only,hence the above equation implies that ∂ t S ( u ) + div (cid:0) η ( x, u ) (cid:1) − div η ( x, v ) (cid:12)(cid:12)(cid:12) v =ˆ u + S ′ (ˆ u ) div A ( x, v ) (cid:12)(cid:12)(cid:12) v =ˆ u = − Z S ′′ ( w ) dm ( · , · , w )(47)for every S ∈ C ∞ c ( R ). Using the very same approximation argument of the second stepof the proof of Theorem 3 in [13], we conclude that (47) holds for every convex function S of class C . Since m ≥
0, this fact implies that u is an entropy solution of (13). Step 2 . Let u be an entropy solution of (13), let us define the distribution m ( t, x, v ) = ∂ t Z v χ ( w, u ( t, x )) dw + n X i =1 ∂ i (cid:26)Z v a i ( x, w ) χ ( w, u ( t, x )) dw (cid:27) + a n +1 ( x, v ) χ ( v, ˆ u ( t, x )) . (48)Clearly (40) is satisfied in the sense of distributions, hence to conclude that u is a kineticsolution we only have to show that m is a positive measure with m ((0 , T ) × R n × R ) < ∞ . ONSERVATION LAWS WITH DISCONTINUOUS FLUX 15
First note that by testing (48) with φ ( t, x ) ψ ( v ) with spt ψ ∩ [ −k u k ∞ , k u k ∞ ] = ∅ andrecalling that | ˆ u | ≤ k u k ∞ we obtain that, as a distribution, m ( t, x, v ) = ∂ t u + div A ( x, u ) = 0 outside [0 , T ] × R n × [ −k u k ∞ , k u k ∞ ] , that is spt m ⊂ [0 , T ] × R n × [ −k u k ∞ , k u k ∞ ]. We want to show that m is a positivemeasure, to this end note that by the same computations of Step 1, u satisfies (47) forevery S ∈ C ∞ c ( R ). If ψ ( v ) ≥ m vanishes outside[0 , T ] × R n × [ −k u k ∞ , k u k ∞ ] we can construct a function S ∈ C ∞ c which is convex onthe range of ˆ u and for which Z ψ ( v ) dm ( t, x, v ) = Z S ′′ ( v ) dm ( t, x, v ) . By (48) and since u is an entropy solution we then deduce that Z ψ ( v ) dm ( t, x, v ) ≥ ∀ ψ ≥ , hence m is a positive measure. Moreover choosing φ = 1 on [ −k u k ∞ , k u k ∞ ] so that S ( v ) = v / u and integrating (47) we get Z (0 ,T ) × R n × R dm ( t, x, v ) ≤ Z R n | u (0 , x ) | dx + Z (0 ,T ) × R n × R d | a n +1 | dt . By (39) we finally conclude. (cid:3) Preliminary estimates
In this section we shall prove some preliminary estimates for approximate solutionsthat will be used in Section 5.Given an entropy solution ˆ u , let us define the function f ( t, x, v ) := χ ( v, ˆ u ( t, x )) andlet us consider a regularization of f with respect to the v variable. More precisely, let ϕ ∈ C ∞ c ([ − / , / ϕ ( w ) = ϕ ( − w ), ϕ ≥ R ϕ = 1. Let ϕ ǫ ( v ) = 1 ǫ ϕ (cid:16) vǫ (cid:17) denote the standard family of mollifiers. If we define f ǫ ( t, x, v ) := ( f ( t, x, · ) ∗ ϕ ǫ )( v ) = (cid:0) χ ( · , u ) ∗ ϕ ǫ (cid:1) ( v ) (cid:12)(cid:12)(cid:12) u =ˆ u ( t,x ) , then we have the following Proposition 4.1.
The function f ǫ satisfies the following equation: (49) ∂ t f ǫ ( t, x, v ) + div x,v [ a ( x, v ) f ǫ ( t, x, v )] = ∂ v ( m ǫ ( t, x, v )) + r ǫ , where m ǫ ( t, x, v ) := ( m ( t, x, · ) ∗ ϕ ǫ )( v ) , and the commutator r ǫ := − div x,v (cid:0) ( a f ) ∗ ϕ ǫ (cid:1) + div x,v (cid:0) a f ǫ (cid:1) is a measure on (0 , T ) × R n × R such that (50) lim ǫ → | r ǫ | (cid:0) (0 , T ) × R n × ( c, d ) (cid:1) = 0 for every c, d ∈ R .Proof. The fact that f ǫ satisfies (49) is evident, hence we only have to verify the lastpart of the statement. To this end note let us write, with obvious notations, r ǫ as r ε = div x ( ∂ v A f ǫ ) − div x (( ∂ v A f ) ǫ )+ ∂ v ( a n +1 f ǫ ) − ∂ v (( a n +1 f ) ǫ ) := r ,ǫ + r ,ǫ and let us show that both r ,ǫ and r ,ǫ are measure satisfying (50). • Computation of r ,ǫ : If we test the distribution r ,ǫ with functions φ ( t, x, v ) = φ ( t ) φ ( x ) φ ( v ) and we applyFubini Theorem we obtain h r ,ǫ , φ i = − Z R Z T dvdt φ ( t ) φ ( v ) Z dw ϕ ( w ) × Z R n dx D x φ ( x ) · (cid:0) ∂ v A ( x, v ) − ∂ v A ( x, v − ǫw ) (cid:1) f ( t, x, v − ǫw )= Z R Z T dvdt φ ( t ) φ ( v ) Z dw ϕ ( w ) × Z R n dx φ ( x ) div x h(cid:0) ∂ v A ( x, v ) − ∂ v A ( x, v − ǫw ) (cid:1) f ( t, x, v − ǫw ) i . (51)Recall that, by the coarea formula, for L almost every ( v, w ) the set (cid:8) ( t, x ) ∈ (0 , T ) × R n : u ( t, x ) > v − ǫw (cid:9) is of finite perimeter in (0 , T ) × R n and that denoting by J v,w = ∂ ∗ (cid:8) ( t, x ) ∈ (0 , T ) × R n : u ( t, x ) > v − ǫw (cid:9) we have D x f ( t, x, v − ǫw ) = D jx f ( t, x, v − ǫw ) = ν J v,w H n − J v,w . Exploiting the Leibniz rule in BV , [4, Exercise 3.97], we then find that for L almostevery ( v, w )div h(cid:0) ∂ v A ( x, v ) − ∂ v A ( x, v − ǫw ) (cid:1) f ( t, x, v − ǫw ) i = n X i =1 (cid:0) ∇ i ∂ v A i ( x, v ) − ∇ i ∂ v A i ( x, v − εw ) (cid:1) f ( t, x, v − εw ) L t × L nx + (cid:0) ∂ v A + ( x, v ) − ∂ v A + ( x, v − εw ) (cid:1) f + ( t, x, v − εw ) L t × H n − N− (cid:0) ∂ v A − ( x, v ) − ∂ v A − ( x, v − εw ) (cid:1) f − ( t, x, v − εw ) L t × H n − N + (cid:0) g ∂ v A ( x, v ) − g ∂ v A ( x, v − εw ) (cid:1) · ν J v,w × (cid:0) f + ( t, x, v − εw ) − f − ( t, x, v − εw ) (cid:1) H n J v,w \ ((0 , T ) × N ) , (52)where we are using the notation (19). According to (H3), (H7) and (9), (10) we thenobtain from (51) and (52) that r ,ǫ is a measure and that | r ,ǫ | ((0 , T ) × R n × ( c, d )) ≤ T ω ( ǫ )( d − c ) k g k L ( R n ) + 2 T ω ( ǫ )( d − c ) H n − ( N )+ ω ( ǫ ) Z ϕ ( w ) dw Z R H n ( J v,w ) dv. (53) ONSERVATION LAWS WITH DISCONTINUOUS FLUX 17
Clearly the first two terms on the right hand side of (53) go to zero as ǫ goes to zero.For what concerns the last one we notice that by the coarea formula Z ϕ ( w ) dw Z R H n ( J v,w ) dv ≤ Z ϕ ( w ) dw Z R H n (cid:0) ∂ ∗ (cid:8) u ( t, x ) > v (cid:9)(cid:1) dv ≤ | Du | ((0 , T ) × R n ) . Hence also the third term in right hand side of (53) goes to zero. • Computation of r ,ǫ : A computation similar to the previous one shows that r ,ǫ is given by the followingmeasure: r ,ǫ = n n X i =1 Z ∇ i ∂ v A i ( x, v ) f ( t, x, v − ǫw ) ϕ ( w ) dw − n X i =1 Z (cid:0) ∇ i A i ( x, v ) − ∇ i A i ( x, v − εw ) (cid:1) f ( t, x, v − ǫw ) ϕ ′ ( w ) ǫ dw o L t × L nx × L v + h Z (cid:0) ∂ v A + ( x, v ) − ∂ v A − ( x, v ) (cid:1) f ( t, x, v − ǫw ) ϕ ( w ) dw − Z (cid:0) A + ( x, v ) − A + ( x, v − ǫw ) (cid:1) f ( t, x, v − ǫw ) ϕ ′ ( w ) ǫ dw + Z (cid:0) A − ( x, v ) − A − ( x, v − ǫw ) (cid:1) f ( t, x, v − ǫw ) ϕ ′ ( w ) ǫ dw i L t × H n − N × L v , (54)where we are again using the convention (19). Note that by (H2) and the discussionin Section 2.3, | A ± ( x, v ) − A ± ( x, v − ǫw ) | ≤ M ǫ | w | for H n − -a.e. x ∈ N and that, by(H4), (cid:12)(cid:12) ∇ i A i ( x, v ) − ∇ i A i ( x, v − εw ) (cid:12)(cid:12) ≤ g ( x ) ǫ | w | . Hence if we can show that the term incurly brackets (respectively in square bracket) in (54) goes to zero as ǫ goes to zero for L t × L nx × L v almost every ( t, x, v ) (respectively for L t × H n − N × L v almost every( t, x, v )), an application of the Lebesgue dominated convergence theorem applied withrespect to the measure L t × L nx × L v (respectively L t × H n − N × L v ) will imply that r ,ǫ satisfies (50). To this end let us write Z ∇ i ∂ v ∇ A i ( x, v ) f ( t, x, v − ǫw ) ϕ ( w ) dw − Z (cid:0) ∇ i A i ( x, v ) − ∇ i A i ( x, v − εw ) (cid:1) f ( t, x, v − ǫw ) ϕ ′ ( w ) ǫ dw = Z (cid:8) ∇ i ∂ v A i ( x, v ) ϕ ( w ) − ∇ i ∂ v A i ( x, v ) wϕ ′ ( w ) (cid:9) f ( t, x, v − ǫw ) dw − Z ( ∇ i A i ( x, v ) − ∇ i A i ( x, v − εw ) ǫ − w ∇ i ∂ v A ( x, v ) ) f ( t, x, v − ǫw ) ϕ ′ ( w ) dw. (55)Since by Lemma 2.2, the map v → ∇ i A i ( x, v ) is differentiable for almost every x withderivative given by ∇ i ∂ v A i ( x, v ), by the fundamental theorem of calculus and (H7) for every such x we can estimate the second integral in the right hand side of (55) by ω ( ǫ ) g ( x ) Z | wφ ′ ( w ) | dw, which goes to zero as ǫ →
0. For what concerns the first term we notice that f ( t, x, v − εw ) → f ( t, x, w ) for almost every ( t, x, v ) since L × L n (cid:16) { ( t, x ) : ˆ u ( t, x ) = v } (cid:17) = 0for all but countably many v . Hence the first term in the right hand side of (55) goes to ∇ i ∂ v A i ( x, v ) f ( t, x, v ) Z [ ϕ ( w ) − wϕ ′ ( w )] dw = 0 . The term in square bracket in (54) can be treated in the same way this time using that v A ± ( x, v ) is differentiable for H n − almost every x ∈ N with derivative given by ∂ v A ± ( x, v ) and that L × H n − N (cid:16) { ( t, x ) : ˆ u ( t, x ) = v } (cid:17) = 0for all but countably many v . (cid:3) The next step concerns the derivation of the evolution equation satisfied by f ǫ := ( f ǫ ) .In order to simplify the notation, we define the differential operator L by Lg ( t, x, v ) := ∂ t g ( t, x, v ) + div x,v [ a ( x, v ) g ( t, x, v )] . Lemma 4.2.
The function f ǫ satisfies the following equation Lf ǫ = 2 f ∗ ǫ Lf ǫ + R [ f ǫ ] that is (56) ∂ t f ǫ ( t, x, v ) + div x,v [ a ( x, v ) f ǫ ( t, x, v )]= 2( f ǫ ( t, x, v )) ∗ [ ∂ v ( m ǫ ( t, x, v )) + r ǫ ] + R [ f ǫ ] where f ∗ ǫ is defined as in (3) and R [ f ǫ ] := n [( ∂ v A ) + − ( ∂ v A ) − ] (cid:0) f + ǫ − f ǫ (cid:1)(cid:0) f ǫ − f − ǫ (cid:1) + (cid:0) f + ǫ − f ǫ + f − ǫ − f ǫ (cid:1) [ A + − A − ] ∂ v f ǫ o L t × H n − N × L v . (57) Here we are using the convention (19) .Proof.
Note that f ε ( t, x, v ) is a BV loc function with respect to all its variables. By thechain rule e ∂ t f ǫ ( t, x, v ) = 2 f ∗ ǫ ( t, x, v ) e ∂ t f ǫ ( t, x, v ) , f D x f ǫ ( t, x, v ) = 2 f ∗ ǫ ( t, x, v ) f D x f ǫ ( t, x, v )for H n +1 almost every ( t, x, v ) ∈ R n +2 \ J f ǫ . Since v f ε ( t, x, v ) is C we also have(recall that ˆ u ( t, x ) is everywhere defined) ∂ v f ǫ ( t, x, v ) = 2 f ǫ ( t, x, v ) ∂ v f ǫ ( t, x, v ) , ONSERVATION LAWS WITH DISCONTINUOUS FLUX 19 for every ( t, x, v ). Moreover χ ( · , u ) ∗ ϕ ǫ = 1 ǫ Z u −∞ ϕ (cid:16) v − wǫ (cid:17) dw is a smooth function of u . According to this J f ǫ = J f ǫ = J u × R . Hence, recalling that(19) is in force, Lf ǫ − f ǫ ) ∗ Lf ǫ = n(cid:0) ∂ v A + ( f ǫ ) + − ∂ v A − ( f ǫ ) − (cid:1) − f ∗ ǫ (cid:0) ∂ v A + f + ǫ − ∂ v A − f − ǫ (cid:1)o L t × H n − N × L v + g ∂ v A · ν J fǫ n(cid:0) ( f ǫ ) + − ( f ǫ ) − (cid:1) − f ∗ ǫ (cid:0) f + ǫ − f − ǫ (cid:1)o H n +1 t,x,v ( J f ǫ \ [(0 , T ) × N × R ])+ n(cid:0) A + − A − (cid:1)(cid:0) f ǫ − f ∗ ǫ (cid:1) ∂ v f ǫ o L t × H n − N × L v + n(cid:0) ( ∂ v A ) + − ( ∂ v A ) − (cid:1)(cid:0) f ǫ − f ∗ ǫ (cid:1) f ǫ o L t × H n − N × L v , (58)where we have used that ∂ v ( A ± ) = ( ∂ v A ) ± for H n − almost every x in N and that f ∗ ǫ = f ǫ , for L n + | g D t,x f ǫ | almost every x , see Section 2.3. Since ( f ǫ ) ± = ( f ± ǫ ) we have( f ǫ ) + − ( f ǫ ) − − f ∗ ǫ (cid:0) f + ǫ − f − ǫ (cid:1) = 0 , hence the second line in the right hand side of (58) vanishes. A simple algebraic com-putation now shows that (58) reduces to (57). (cid:3) Let us now consider, for
R >
0, the following test function ψ R ∈ C ∞ c ( R n +1 ), ψ R ≥ ψ R ( x, v ) = θ (cid:16) xR (cid:17) φ R ( v ) , with θ ∈ C ∞ c ( R n ) , φ R ∈ C ∞ ( R ) , φ R ( v ) = 1 if | v | ≤ R , φ R ( v ) = 0 if | v | ≥ R + 1 , θ ( x ) = 1 if | x | ≤ | φ ′ R | ≤ Lemma 4.3. If u is a kinetic solution of (13) , then lim R → + ∞ lim ǫ → Z T Z R n +1 ( ν ∗ ϕ ǫ ) ψ R dm ǫ = − lim R → + ∞ lim ǫ → Z T Z R n +1 ψ R ( x, v ) d R [ f ǫ ]= Z T Z N Q ( u ) d H n − dt , (60) where (recall (19) ) Q ( u ) := χ ( u − , u + ) A + ( u − ) + χ ( u + , u − ) A + ( u + ) − χ ( u + , u − ) A − ( u + ) − χ ( u − , u + ) A − ( u − ) − χ ( u + , ˆ u ) A + ( u + ) + χ ( u + , ˆ u ) A − ( u + ) − χ ( u − , ˆ u ) A + ( u − ) + χ ( u − , ˆ u ) A − ( u − ) , (61) and ν ( t, x, v ) := δ u + ( t,x ) ( v ) + δ u − ( t,x ) ( v ) = − ∂ v f ∗ . Proof.
Let η δ ∈ C c (0 , T ) with η = 1 on [ δ, T − δ ]. Since u ∈ C ([0 , T ]; L ( R n )), we havethat f ε ∈ C ([0 , T ]; L ( R n +1 )), hence by testing (56) and (49) with η δ ( t ) ψ R ( x, v ) andletting δ → Z R n +1 [ f ǫ ( T, x, v ) − f ǫ ( T, x, v )] ψ R ( x, v ) dx dv (62) = Z R n +1 [ f ǫ (0 , x, v ) − f ǫ (0 , x, v )] ψ R ( x, v ) dx dv (63) + Z T Z R n +1 ( f ǫ − f ǫ ) ∇ x,v ψ R · d a ( x, v ) dt (64) + Z T Z R n +1 [ − ∂ v f ∗ ǫ ] ψ R dm ǫ (65) + Z T Z R n +1 [1 − f ∗ ǫ ] ∂ v ψ R dm ǫ (66) + Z T Z R n +1 [2 f ∗ ǫ − ψ R dr ǫ (67) + Z T Z R n +1 ψ R d R [ f ǫ ]( x, v ) dt . (68)As ǫ goes to 0, the integral (62) tends to Z R n +1 (cid:2) f ( T, x, v ) − f ( T, x, v ) (cid:3) ψ R ( x, v ) dx dv = 0 , since f = f . In the same way the integral (63) tends to 0. Moreover, recalling thedefinition of a we obtain that Z T Z R n +1 ( f ǫ − f ǫ ) ∇ x,v ψ R ( x, v ) · d a ( x, v ) dt = Z T Z R n +1 ( f ǫ − f ǫ ) ∇ x ψ R · ∂ v A dx dv dt + Z T Z R n +1 ( f ǫ − f ǫ ) ∂ v ψ R da n +1 dt . We now note that, arguing as the end of the proof of Lemma 4.1, f ǫ → f = f , for L t × ( L n + σ ) x × L v almost every point. This and the fact that | a n +1 | << L n + σ , see(39), imply that the integral (64) tends to 0, as ǫ → ∂ v f ∗ ǫ = − ν ∗ φ ǫ = −
12 [ δ u − ( t,x ) ( v ) + δ u + ( t,x ) ( v )] ∗ φ ǫ . Therefore the integral (65) is equal to Z T Z R n +1 ( ν ∗ φ ǫ ) ψ R dm ǫ . ONSERVATION LAWS WITH DISCONTINUOUS FLUX 21
Let us now estimate the integral (66). Recalling the choice of the test function made in(59), we have that (cid:12)(cid:12)(cid:12)(cid:12)Z T Z R n +1 m ǫ ( t, x, v )[1 − f ∗ ǫ ] ∂ v ψ R ( x, v ) dt dx dv (cid:12)(cid:12)(cid:12)(cid:12) ≤ m ǫ ([0 , T ] × R n × I R ) , where I R = { v : R ≤ | v | ≤ R + 1 } . Hence, by letting ǫ → R → + ∞ , the integral(66) tends to 0 since m ([0 , T ] × R n × R ) < + ∞ .By (50), the integral in (67) tends to 0 as ǫ →
0. Gathering all the information above,the first equality in (60) is proved.It remains to compute − lim R → + ∞ lim ǫ → Z T Z R n +1 ψ R ( x, v ) d R [ f ǫ ]and to show that the second equality in (60) holds. We recall that the explicit form of R [ f ǫ ] is given in Lemma 4.2. We have that − Z T Z R n +1 ψ R ( x, v ) d R [ f ǫ ]= − Z T Z R n +1 ( f + ǫ − f ǫ )( f − ǫ − f ǫ )( A + − A − ) ∂ v ψ R dv d H n − ( x ) dt − Z T Z R n +1 (cid:2) ( f + ǫ − f ǫ ) ∂ v ( f − ǫ − f ǫ ) + ( f − ǫ − f ǫ ) ∂ v ( f + ǫ − f ǫ )+ ( f + ǫ − f ǫ + f − ǫ − f ǫ ) ∂ v f ǫ (cid:3) ( A + − A − ) ψ R dv d H n − ( x ) dt =: I ( ǫ, R ) + I ( ǫ, R ) . Again, the first integral I ( ǫ, R ) tends to 0 as ǫ → R → + ∞ thanks to the choiceof the test function ψ R . In order to compute the I ( ǫ, R ), let us recall that ∂ v ( f ± ǫ − f ǫ ) = [ δ ˆ u ( v ) − δ u ± ( v )] ∗ ϕ ǫ ( v ) , ∂ v f ǫ = − δ ˆ u ( v ) ∗ ϕ ǫ ( v ) . A straightforward computation based on Lemma 4.4 below, now gives that(69) lim R → + ∞ lim ǫ → I ( ǫ, R ) = Z T Z N Q ( u ) d H n − ( x ) dt, where Q ( u ) := (cid:2) − χ ( u + , ˆ u ) + χ ( u + , u − ) (cid:3) [ A + ( u + ) − A − ( u + )]+ (cid:20) − χ (ˆ u, u − ) + 12 − χ (ˆ u, u + ) (cid:21) [ A + (ˆ u ) − A − (ˆ u )]+ (cid:2) − χ ( u − , ˆ u ) + χ ( u − , u + ) (cid:3) [ A + ( u − ) − A − ( u − )]+ (cid:2) − χ (ˆ u, u − ) + χ (ˆ u, u + ) (cid:3) [ A + (ˆ u ) − A − (ˆ u )] , (70)which, after some computations is easily seen to coincide with (61). The second equalityin (60) now follows form this together with (69) and (70). (cid:3) We conclude this section with the following technical lemma that we have used in theproof of (70), for later use we state the lemma in a slightly more general setting.
Lemma 4.4.
Let ϕ ǫ : R → R be a family of symmetric mollifiers and let h h : R → R be two bounded and uniformly continuous functions, then for every u , ˆ u ∈ R (71) lim ǫ → Z h ( v ) ( δ u ∗ ϕ ǫ )( v ) [ h ( · ) χ ( · , ˆ u )] ∗ ϕ ǫ ( v ) dv = h ( u ) h ( u ) χ ( u, ˆ u ) . Here χ ( u, ˆ u ) is defined according to (38) .Proof. We start noticing that I := Z h ( v ) ( δ u ∗ ϕ ǫ )( v ) [ h ( · ) χ ( · , ˆ u )] ∗ ϕ ǫ ( v ) dv = Z Z h ( v ) ϕ ǫ ( v − u ) h ( w ) χ ( w, ˆ u ) ϕ ǫ ( v − w ) dw dv = Z Z h ( v ) ϕ ǫ ( v − u )[ h ( w ) − h ( u )] χ ( w, ˆ u ) ϕ ǫ ( v − w ) dw dv + h ( u ) Z Z h ( v ) ϕ ǫ ( v − u ) χ ( w, ˆ u ) ϕ ǫ ( v − w ) dw dv =: I + h ( u ) I . By exploiting the uniform continuity of h we obtain for some modulus of continuity ω ,that | I | ≤ Z Z | h ( v ) | ϕ ǫ ( v − u ) ω ( | v − w | ) ϕ ǫ ( v − w ) dw dv ≤ k h k ∞ Z (cid:18) ω ( | s | ) ϕ ǫ ( s ) Z ϕ ǫ ( s + w − u ) dw (cid:19) dt ≤ k h k ∞ Z ω ( | s | ) ϕ ǫ ( s ) ds (72)and the last integral tends to 0 as ǫ →
0. On the other hand(73) lim ǫ → I = h ( u ) χ ( u, ˆ u ) . Indeed, from the estimate | I − h ( u ) χ ( u, ˆ u ) | ≤ max v ∈ [ u − ǫ,u + ǫ ] | h ( v ) − h ( u ) | + | h ( u ) | (cid:12)(cid:12)(cid:12)(cid:12)Z Z ϕ ǫ ( v − u ) χ ( w, ˆ u ) ϕ ǫ ( v − w ) dwdv − χ ( u, ˆ u ) (cid:12)(cid:12)(cid:12)(cid:12) and the uniform continuity of h it is enough to show that(74) lim ǫ → Z Z ϕ ǫ ( v − u ) χ ( w, ˆ u ) ϕ ǫ ( v − w ) dwdv = χ ( u, ˆ u ) . If u = ˆ u , then for ǫ small enough we have that χ ( w, ˆ u ) = χ ( u, ˆ u ) for w ∈ ( u − ǫ, u + 2 ǫ ),so that χ is constant in the integral. Then Z Z ϕ ǫ ( v − u ) χ ( w, ˆ u ) ϕ ǫ ( v − w ) dwdv = χ ( u, ˆ u ) Z Z ϕ ǫ ( v − u ) ϕ ǫ ( v − w ) dwdv = χ ( u, ˆ u ) ONSERVATION LAWS WITH DISCONTINUOUS FLUX 23 and (74) follows. If u = ˆ u , then the integral in (74), for ǫ > Z Z ϕ ǫ ( v − u ) χ ( w, u ) ϕ ǫ ( v − w ) dwdv = Z + ∞−∞ ϕ ǫ ( v − u ) Z u −∞ ϕ ǫ ( v − w ) dw dv = Z + ∞−∞ ϕ ǫ ( v − u ) (cid:20)
12 + Z uv ϕ ǫ ( v − w ) dw (cid:21) dv = 12 , where we have exploited that ϕ ǫ ( s ) = ϕ ǫ ( − s ). Equation (74) now follows form thedefinition of χ , (38). Hence (71) follows from (72) and (73). (cid:3) Uniqueness
In this Section we prove Theorem 3.5, to this end let us fix some notation. For u , u two entropy solutions of (13), with corresponding everywhere defined Borel representa-tives ˆ u , ˆ u , we will set f i ( t, x, v ) := χ ( v, ˆ u i ( t, x )) and m i , i = 1 ,
2, for the correspondingfunctions and measures appearing in the kinetic formulation. We will also set f ǫ ( t, x, v ) := ( f ( t, x, · ) ∗ ϕ ǫ )( v ) , f ǫ ( t, x, v ) := ( f ( t, x, · ) ∗ ϕ ǫ )( v ) . and m ǫ ( t, x, v ) := ( m ( t, x, · ) ∗ ϕ ǫ )( v ) , m ǫ ( t, x, v ) := ( m ( t, x, · ) ∗ ϕ ǫ )( v ) , where ϕ ǫ denotes the standard family of mollifiers.The following theorem is the main result of this section and, by Cavalieri’s principle,it immediately implies Theorem 3.5. Theorem 5.1.
Let u , u be two entropy solution of (13) , with corresponding everywheredefined Borel representatives ˆ u , ˆ u . Setting f i ( t, x, v ) := χ ( v, ˆ u i ( t, x )) , i = 1 , , we havethat Z R n +1 | f − f | ( T, x, v ) dx dv ≤ Z R n +1 | f − f | (0 , x, v ) dx dv . In order to prove Theorem 5.1 we need some preliminary results concerning the inter-action of two kinetic solutions:
Proposition 5.2.
With the notation above, for every test function ψ ( x, v ) we have that Z R n +1 | f − f | ( T, x, v ) ψ ( x, v ) dx dv ≤ Z R n +1 | f − f | (0 , x, v ) ψ ( x, v ) dx dv − lim sup ǫ → Z T Z R n +1 ∂ v [( f ǫ − f ǫ ) ∗ ψ ] d ( m ǫ − m ǫ )( t, x, v )+ lim sup ǫ → Z T Z R n +1 ψ d R [ f ǫ − f ǫ ]( t, x, v ) − Z T Z R n +1 | f − f | ∇ x ψ · ∂ v A ( x, v ) dx dv dt + Z T Z R n +1 | f − f | ∂ v ψ da n +1 ( x, v ) dt , (75) where R [ f ǫ − f ǫ ] is defined according to (57) . Proof.
We recall that Lf ǫ = ∂ v m ǫ + r ǫ , Lf ǫ = ∂ v m ǫ + r ǫ . Since f ǫ − f ǫ is a solution of (56), by Lemma 4.2 we see that ∂ t ( f ǫ − f ǫ ) + div x,v [ a ( f ǫ − f ǫ ) ]= 2( f ǫ − f ǫ ) ∗ L ( f ǫ − f ǫ ) + R [ f ǫ − f ǫ ] . Hence, by arguing as in Lemma 4.3, for every test function ψ ( x, v ) we have that Z R n +1 ( f ǫ − f ǫ ) ( T, x, v ) ψ ( x, v ) dx dv − Z R n +1 ( f ǫ − f ǫ ) (0 , x, v ) ψ ( x, v ) dx dv = − Z T Z R n +1 ( f ǫ − f ǫ ) ∇ x,v ψ ( x, v ) · d a ( x, v ) dt + 2 Z T Z R n +1 ( f ǫ − f ǫ ) ∗ L ( f ǫ − f ǫ ) ψ ( x, v ) dx dv dt + Z T Z R n +1 ψ ( x, v ) d ( R [ f ǫ − f ǫ ])=: I + I + I . (76)Noticing that ( f ǫ − f ǫ ) → ( f − f ) , that the equality ( f − f ) = | f − f | holdsfor L t × ( L n + σ ) × L v almost every point, and that | a | << ( L n + σ ) × L v , we can passto the limit in I obtaining the last two integrals in the right hand side of (75). In thesame way, the left hand side of (76) tends, as ǫ →
0, to Z R n +1 | f − f | ( T, x, v ) ψ ( x, v ) dx dv − Z R n +1 | f − f | (0 , x, v ) ψ ( x, v ) dx dv . Let us now consider I and I . Since, by Proposition 4.1, Lf ǫ = ∂ v m ǫ + r ǫ , Lf ǫ = ∂ v m ǫ + r ǫ we infer that inequality (75) holds. (cid:3) Proposition 5.3.
Let u , u be two entropy solution of (13) , with corresponding repre-sentatives ˆ u , ˆ u . Setting f i ( t, x, v ) := χ ( v, ˆ u i ( t, x )) , i = 1 , , we have that Z R n +1 | f − f | ( T, x, v ) dx dv ≤ Z R n +1 | f − f | (0 , x, v ) dx dv + Z T Z N W ( u , u ) d H n − dt , (77) where W ( u , u ) := A + ( u +1 ) (cid:2) − χ ( u +1 , u +2 ) + 2 χ ( u − , u − ) (cid:3) + A + ( u +2 ) (cid:2) − χ ( u +2 , u +1 ) + 2 χ ( u − , u − ) (cid:3) . (78) ONSERVATION LAWS WITH DISCONTINUOUS FLUX 25
Proof.
Let us start from inequality (75) using the test function ψ = ψ R defined in (59).It is easy to show that, as R → + ∞ , the last two integrals in (75) goes to 0. Hence, weonly need to estimate the contribution of the two terms I ( ǫ, R ) := − Z T Z R n +1 ∂ v (cid:2) ( f ǫ − f ǫ ) ψ R (cid:3) d ( m ǫ − m ǫ )( t, x, v )and H ( ǫ, R ) := Z T Z R n +1 ψ R d ( R [ f ǫ − f ǫ ])( t, x, v )for ǫ → R → + ∞ . We start noticing that I ( ǫ, R ) = − Z T Z R n +1 ∂ v [( f ǫ − f ǫ ) ∗ ] ψ R d ( m ǫ − m ǫ )( t, x, v ) − Z T Z R n +1 f ǫ − f ǫ ) ∗ ∂ v ψ R d ( m ǫ − m ǫ )( t, x, v ) , (79)and, reasoning as in the final part of the proof of Lemma 4.3, the last integral goes to 0as ǫ → R → + ∞ . Let us compute the first term in (79): − Z T Z R n +1 ∂ v [( f ǫ − f ǫ ) ∗ ] ψ R d ( m ǫ − m ǫ )= − Z T Z R n +1 [ δ u +2 ( v ) + δ u − ( v ) − δ u +1 ( v ) − δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R d ( m ǫ − m ǫ )= Z T Z R n +1 [ δ u +1 ( v ) + δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R dm ǫ + Z T Z R n +1 [ δ u +2 ( v ) + δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R dm ǫ − Z T Z R n +1 [ δ u +2 ( v ) + δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R dm ǫ − Z T Z R n +1 [ δ u +1 ( v ) + δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R dm ǫ =: I ( ǫ, R ) + I ( ǫ, R ) + I ( ǫ, R ) + I ( ǫ, R ) . By Lemma 4.3, lim sup R →∞ lim sup ǫ → I ( ǫ, R ) + I ( ǫ, R ) ≤ Z T Z N [ Q ( u ) + Q ( u )] d H n − dt , where Q ( u ) is defined in (61). In order to estimate I , we note that I = − Z T Z R n +1 [ δ u +2 ( v ) + δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R dm ǫ ( t, x, v ) ≤ − Z T Z R n +1 [ δ u +2 ( v ) + δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R dm s ǫ ( t, x, v ) , where m s ǫ ( t, · , v ) is the restriction to N of the singular part of the measure m ǫ ( t, · , v ),namely m s ǫ ( t, · , v ) = m s ( t, · , v ) ∗ ϕ ǫ ( v ), with m s ( t, x, v ) = Z v (cid:2) ( ∂ v A ) + ( x, w )( χ ( w, u +1 ( t, x ))) − ( ∂ v A ) − ( x, w )( χ ( w, u − ( t, x ))) (cid:3) d H n − dw − (cid:0) A + ( x, v ) − A − ( x, v ) (cid:1) χ ( v, ˆ u ( t, x )) d H n − . Hence, the contribution given by I to (77) can be estimated by − lim sup R → + ∞ lim sup ǫ → Z T Z R n +1 [ δ u +2 ( v ) + δ u − ( v )] ∗ ϕ ǫ ( v ) ψ R dm s ǫ . By the explicit expression of m s ǫ and by also taking into account Lemma 4.4, we get I ≤ Z T Z N h − χ ( u +2 , u +1 ) A + ( u +2 ) − χ ( u +1 , u +2 ) A + ( u +1 )+ χ ( u +2 , u − ) A − ( u +2 ) + χ ( u − , u +2 ) A − ( u − ) − χ ( u +2 , ˆ u ) A − ( u +2 ) + χ ( u +2 , ˆ u ) A + ( u +2 ) i d H n − + Z T Z N h − χ ( u − , u +1 ) A + ( u − ) − χ ( u +1 , u − ) A + ( u +1 )+ χ ( u − , u − ) A − ( u − ) + χ ( u − , u − ) A − ( u − ) − χ ( u − , ˆ u ) A − ( u − ) + χ ( u − , ˆ u ) A + ( u − ) i d H n − . In the same way we can estimate the contribution of I . The contribution of H canthen be estimated with the aid of Lemma 4.4 by arguing as in the final part of the proofof Lemma 4.3. Therefore, by summing up all the estimates we obtain that (77) holdswith the following choice of W ( u , u ) where, for the reader convenience, the terms aregrouped according to their provenience: W ( u , u )= χ ( u − , u +1 ) A + ( u − ) + χ ( u +1 , u − ) A + ( u +1 ) − χ ( u +1 , u − ) A − ( u +1 ) − χ ( u − , u +1 ) A − ( u − ) − χ ( u +1 , ˆ u ) A + ( u +1 ) + χ ( u +1 , ˆ u ) A − ( u +1 ) − χ ( u − , ˆ u ) A + ( u − ) + χ ( u − , ˆ u ) A − ( u − ) ) I + χ ( u − , u +2 ) A + ( u − ) + χ ( u +2 , u − ) A + ( u +2 ) − χ ( u +2 , u − ) A − ( u +2 ) − χ ( u − , u +2 ) A − ( u − ) − χ ( u +2 , ˆ u ) A + ( u +2 ) + χ ( u +2 , ˆ u ) A − ( u +2 ) − χ ( u − , ˆ u ) A + ( u − ) + χ ( u − , ˆ u ) A − ( u − ) ) I − χ ( u +1 , u +2 ) A + ( u +1 ) − χ ( u − , u +2 ) A + ( u − ) − χ ( u +2 , u +1 ) A + ( u +2 ) − χ ( u +2 , u − ) A + ( u +2 )+ χ ( u +1 , u − ) A − ( u +1 ) + χ ( u − , u − ) A − ( u − ) + χ ( u − , u +1 ) A − ( u − ) + χ ( u − , u − ) A − ( u − )+ χ ( u +1 , ˆ u ) A + ( u +1 ) − χ ( u +1 , ˆ u ) A − ( u +1 ) + χ ( u − , ˆ u ) A + ( u − ) − χ ( u − , ˆ u ) A − ( u − ) I − χ ( u +2 , u +1 ) A + ( u +2 ) − χ ( u − , u +1 ) A + ( u − ) − χ ( u +1 , u +2 ) A + ( u +1 ) − χ ( u +1 , u − ) A + ( u +1 )+ χ ( u +2 , u − ) A − ( u +2 ) + χ ( u − , u − ) A − ( u − ) + χ ( u − , u +2 ) A − ( u − ) + χ ( u − , u − ) A − ( u − )+ χ ( u +2 , ˆ u ) A + ( u +2 ) − χ ( u +2 , ˆ u ) A − ( u +2 ) + χ ( u − , ˆ u ) A + ( u − ) − χ ( u − , ˆ u ) A − ( u − ) I − [ χ ( u +1 , u − ) − χ ( u +1 , ˆ u ) − χ ( u +1 , u − ) + χ ( u +1 , ˆ u )] (cid:2) A + ( u +1 ) − A − ( u +1 ) (cid:3) + [ χ ( u +2 , u − ) − χ ( u +2 , ˆ u ) − χ ( u +2 , u − ) + χ ( u +2 , ˆ u )] (cid:2) A + ( u +2 ) − A − ( u +2 ) (cid:3) − [ χ ( u − , u +1 ) − χ ( u − , ˆ u ) − χ ( u − , u +2 ) + χ ( u − , ˆ u )] (cid:2) A + ( u − ) − A − ( u − ) (cid:3) + [ χ ( u − , u +1 ) − χ ( u − , ˆ u ) − χ ( u − , u +2 ) + χ ( u − , ˆ u )] (cid:2) A + ( u − ) − A − ( u − ) (cid:3) . H We now note that all terms in A + ( u − j ) and A − ( u + j ), j = 1 ,
2, cancel out. Using theRankine-Hugoniot conditions (18), we can sum up the terms A + ( u + j ) with A − ( u − j ),obtaining (78). (cid:3) We are now ready to prove Theorem 5.1.
Proof of Theorem 5.1.
By Proposition 5.3, it remains to prove that(80) A + ( u +1 ) (cid:2) − χ ( u +1 , u +2 ) + χ ( u − , u − ) (cid:3) + A + ( u +2 ) (cid:2) − χ ( u +2 , u +1 ) + χ ( u − , u − ) (cid:3) ≤ . Since u , u are entropy solutions, recalling that χ ( a, b ) + χ ( b, a ) = 1 ∀ a, b ∈ R , in order to prove (80), we have to show that(81) W ( u , u ) = ( χ + − χ − ) (cid:2) A + ( u +2 ) − A + ( u +1 ) (cid:3) ≤ , where we have set χ ± = χ ( u ± , u ± ).If sign( u +2 − u +1 ) = sign( u − − u − ), then χ + = χ − , and (81) is satisfied. Hence we shallrestrict our attention to the case 1 = sign( u +2 − u +1 ) = − sign( u − − u − ), i.e.(82) u +1 < u +2 , u − < u − , so that χ + = 1, χ − = 0, since the case − u +2 − u +1 ) = − sign( u − − u − ) can behandled in a similar way.Since (82) is in force in order to prove (81) we need to show that(83) A + ( u +2 ) − A + ( u +1 ) ≤ A − ( u − ) − A − ( u − ) ≤ u +1 < u +2 < u − < u − , (84) u +1 < u − < u +2 < u − , (85) u − < u +1 < u +2 < u − , (86) u − < u +1 < u − < u +2 , (87) u − < u − < u +1 < u +2 , (88) u +1 < u − < u − < u +2 . (89)Here, for simplicity, we have considered only strict inequalities, but the equality casescan be handled as well. Namely, if u − = u − or u +1 = u +2 , then (83) trivially holds, while the other equality cases can be proved using a continuity argument, since the quantities χ ± do not change.The analysis will be the same as the one performed in [24]. For the reader’s conve-nience, we briefly report here how to use the inequalities (26)–(37) in order to prove(83) in each of the six cases listed above. When not explicitly stated, the inequalities(26)–(37) are supposed to be used with u = u . • Case (84). We have the following possibilities: • u +1 < u +2 < u − < u − < ˆ u : choose c = u +2 in (27). • u +1 < u +2 < ˆ u < u − : choose c = u +2 in (29). (We remark that the relativeposition of ˆ u and u − is not relevant, so that this case includes the two subcases u +1 < u +2 < ˆ u < u − < u − and u +1 < u +2 < u − < ˆ u < u − ; similar considerationswill apply also in some of the following cases without further reference.) • u +1 < ˆ u < u − < u − : choose c = u − in (28). • ˆ u < u +1 < u +2 < u − < u − : choose c = u +2 in (30). • Case (85). We have the following possibilities: • u +1 < u − < u +2 < u − < ˆ u : choose c = u +2 in (27). • u +1 < u − < u +2 < ˆ u < u − : choose c = u +2 in (29). • u +1 < ˆ u < u − < u +2 < u − : choose c = u − in (28). • ˆ u < u +1 < u − < u +2 < u − : choose c = u +2 in (30). • u +1 < u − < ˆ u < u +2 < u − : this case is slightly more involved because we needto consider also the position of ˆ u . We have the following subcases:(1) u +1 < u − < ˆ u < u +2 < ˆ u : apply (33) to u with c = ˆ u and then (28) to u again with c = ˆ u , obtaining A − ( u − ) ≤ A − (ˆ u ) ≤ A − ( u − ) . (2) u +1 < u − < ˆ u < ˆ u < u +2 < u − : apply (35) to u with c = ˆ u and then (28)to u again with c = ˆ u obtaining A − ( u − ) ≤ A − (ˆ u ) ≤ A − ( u − ) . (3) u +1 < u − < ˆ u < ˆ u < u +2 < u − : apply (34) to u with c = ˆ u and then (29)to u again with c = ˆ u obtaining A + ( u +2 ) ≤ A + (ˆ u ) ≤ A + ( u +1 ) . (4) ˆ u < u − < ˆ u < u +2 < u − : apply (36) to u with c = ˆ u and then (29) to u again with c = ˆ u obtaining A + ( u +2 ) ≤ A + (ˆ u ) ≤ A + ( u +1 ) . • Case (86). We have the following possibilities: • u − < u +1 < u +2 < u − < ˆ u : choose c = u +2 in (27). • u − < u +1 < u +2 < ˆ u < u − : choose c = u +2 in (29). • ˆ u < u +1 < u +2 < u − : choose c = u +2 in (30). • u − < u +1 < ˆ u < u +2 < u − : this case is slightly more involved because we needto consider also the position of ˆ u . We have the following subcases: ONSERVATION LAWS WITH DISCONTINUOUS FLUX 29 (1) u − < u +1 < ˆ u < u +2 < ˆ u : apply (33) to u with c = ˆ u and then (28) to u again with c = ˆ u obtaining A − ( u − ) ≤ A − (ˆ u ) ≤ A − ( u − ) . (2) u − < u +1 < ˆ u < ˆ u < u +2 < u − : apply (35) to u with c = ˆ u and then (28)to u again with c = ˆ u obtaining A − ( u − ) ≤ A − (ˆ u ) ≤ A − ( u − ) . (3) u − < ˆ u < ˆ u < u +2 < u − : apply (34) to u with c = ˆ u and then (29) to u again with c = ˆ u obtaining A + ( u +2 ) ≤ A + (ˆ u ) ≤ A + ( u +1 ) . (4) ˆ u < u +1 < ˆ u < u +2 < u − : apply (36) to u with c = ˆ u and then (29) to u again with c = ˆ u obtaining A + ( u +2 ) ≤ A + (ˆ u ) ≤ A + ( u +1 ) . • Case (87). This case is symmetric to (85). It is enough to replace ˆ u with ˆ u and toapply (32)–(37) instead of (26)–(31) or conversely. • Case (88). This case is symmetric to (84). It is enough to replace ˆ u with ˆ u and toapply (32)–(37) instead of (26)–(31). • Case (89). This case is symmetric to (86). It is enough to replace ˆ u with ˆ u and toapply (32)–(37) instead of (26)–(31) or conversely. (cid:3) References [1] Adimurthi, Rajib Dutta, Shyam Sundar Ghoshal, and G. D. Veerappa Gowda,
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ONSERVATION LAWS WITH DISCONTINUOUS FLUX 31
Dipartimento di Matematica “G. Castelnuovo”, Univ. di Roma I, P.le A. Moro 2 – I-00185Roma (Italy)
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