Finite time stability for the Riemann problem with extremal shocks for a large class of hyperbolic systems
FFINITE TIME STABILITY FOR THE RIEMANN PROBLEM WITHEXTREMAL SHOCKS FOR A LARGE CLASS OF HYPERBOLICSYSTEMS
SAM G. KRUPA
Abstract.
In this paper on hyperbolic systems of conservation laws in one space di-mension, we give a complete picture of stability for all solutions to the Riemann problemwhich contain only extremal shocks. We study stability of the Riemann problem amongsta large class of solutions. We show stability among the family of solutions with shocksfrom any family. We assume solutions verify at least one entropy condition. We have nosmall data assumptions. The solutions we consider are bounded and satisfy a strong tracecondition weaker than BV loc . We make only mild assumptions on the system. In par-ticular, our work applies to gas dynamics, including the isentropic Euler system and thefull Euler system for a polytropic gas. We use the theory of a-contraction (see Kang andVasseur [ Arch. Ration. Mech. Anal. , 222(1):343–391, 2016]), and introduce new ideasin this direction to allow for two shocks from different shock families to be controlledsimultaneously. This paper shows L stability for the Riemann problem for all time. Ourresults compare to Chen, Frid, and Li [ Comm. Math. Phys. , 228(2):201–217, 2002] andChen and Li [
J. Differential Equations , 202(2):332–353, 2004], which give uniqueness andlong-time stability for perturbations of the Riemann problem – amongst a large classof solutions without smallness assumptions and which are locally BV . Although, theseresults lack global L stability. Contents
1. Introduction 22. Hypotheses on the system 113. Overview of the proofs of Theorem 6.1 and Theorem 6.2 144. Technical Lemmas 155. Construction of the shift 275.1. Proof of Proposition 5.1 295.2. Proof of Lemma 5.4 356. Proofs of Theorem 6.1 and Theorem 6.2 39
Date : May 10th, 2019.2010
Mathematics Subject Classification.
Primary 35L65; Secondary 76N15, 35L45, 35A02, 35B35,35D30, 35L67, 35Q31, 76L05, 35Q35, 76N10.
Key words and phrases.
System of conservation laws, compressible Euler equation, Euler system, isen-tropic solutions, Riemann problem, rarefaction wave, Rankine–Hugoniot discontinuity, shock, stability,uniqueness.This work was partially supported by NSF Grant DMS-1614918. a r X i v : . [ m a t h . A P ] M a y KRUPA L Stability for the Riemann Problem with Extremal ShocksVerifying Strong Form of Lax’s E-condition 396.2. Theorem 6.2: L Stability for the Riemann Problem with Extremal Shocksbut No Rarefactions 45References 491.
Introduction
We consider the following n × n system of conservation laws in one space dimension: (cid:40) ∂ t u + ∂ x f ( u ) = 0 , for x ∈ R , t > ,u ( x,
0) = u ( x ) , for x ∈ R . (1.1)For a fixed T > T = ∞ ), the unknown is u : R × [0 , T ) → M n × . Thefunction u : R → M n × is in L ∞ ( R ) and is the initial data . The function f : M n × → M n × is the flux function for the system. We assume the system (1.1) is endowed with a strictlyconvex entropy η and associated entropy flux q . Note the system will be hyperbolic on thestate space where η exists. We assume the functions f, η , and q are defined on an openconvex state space V ⊂ R n . We assume f, q ∈ C ( V ) and η ∈ C ( V ). By assumption, theentropy η and its associated entropy flux q verify the following compatibility relation: ∂ j q = n (cid:88) i =1 ∂ i η∂ j f i , ≤ j ≤ n. (1.2)By convention, the relation (1.2) is rewritten as ∇ q = ∇ η ∇ f, (1.3)where ∇ f denotes the matrix ( ∂ j f i ) i,j .For u ∈ V where η exists, the system (1.1) is hyperbolic, and the matrix ∇ f ( u ) isdiagonalizable, with eigenvalues λ ( u ) ≤ . . . ≤ λ n ( u ) , (1.4)called characteristic speeds .We consider both bounded classical and bounded weak solutions to (1.1). A weaksolution u is bounded and measurable and satisfies (1.1) in the sense of distributions. I.e.,for every Lipschitz continuous test function Φ : R × [0 , T ) → M × n with compact support,(1.5) T (cid:90) ∞ (cid:90) −∞ (cid:34) ∂ t Φ u + ∂ x Φ f ( u ) (cid:35) dxdt + ∞ (cid:90) −∞ Φ( x, u ( x ) dx = 0 . We only consider solutions u which are entropic for the entropy η . That is, they satisfythe following entropy condition: ∂ t η ( u ) + ∂ x q ( u ) ≤ , (1.6) TABILITY FOR THE RIEMANN PROBLEM 3 in the sense of distributions. I.e., for all positive, Lipschitz continuous test functions φ : R × [0 , T ) → R with compact support:(1.7) T (cid:90) ∞ (cid:90) −∞ (cid:34) ∂ t φ (cid:0) η ( u ( x, t )) (cid:1) + ∂ x φ (cid:0) q ( u ( x, t )) (cid:1)(cid:35) dxdt + ∞ (cid:90) −∞ φ ( x, η ( u ( x )) dx ≥ . For u L , u R ∈ R n , the function u : R × [0 , ∞ ) → R n defined by u ( x, t ) := (cid:40) u L if x < σt,u R if x > σt (1.8)is a weak solution to (1.1) if and only if u L , u R , and σ satisfy the Rankine-Hugoniot jumpcompatibility relation: f ( u R ) − f ( u L ) = σ ( u R − u L ) , (1.9)in which case (1.8) is called a shock solution.Moreover, the solution (1.8) will be entropic for η (according to (1.7)) if and only if, q ( u R ) − q ( u L ) ≤ σ ( η ( u R ) − η ( u L )) . (1.10)In this case, ( u L , u R , σ ) is an entropic Rankine–Hugoniot discontinuity .For a fixed u L , we consider the set of u R which satisfy (1.9) and (1.10) for some σ . Fora general n × n strictly hyperbolic system of conservation laws endowed with a strictlyconvex entropy , we know that locally this set of u R values is made up of n curves (see forexample [25, p. 140-6]).The present paper concerns the finite-time stability of Riemann problem solutions to(1.1), working in the L setting. We work in a very general setting. Our techniques arebased on the theory of shifts in the context of the relative entropy method as developed byVasseur (see [34]). We consider systems of the form (1.1), with minimal assumptions onthe shock families. We ask that the extremal shock speeds (1-shock and n-shock speeds)are separated from the intermediate shock families. If we want to consider solutions tothe Riemann problem with a 1-shock, we ask that the 1-shock family satisfy the Liuentropy condition (shock speed decreases as the right-hand state travels down the 1-shockcurve), and we ask that the shock strength increase in the sense of relative entropy (an L requirement) as the right-hand state travels down the 1-shock curve. If we want toconsider n-shocks, we ask for similar requirements on the n-shock family.The intermediate wave families have far fewer requirements. The intermediate shockcurves might not even be well-defined and characteristic speeds might cross.Systems we have in mind include the isentropic Euler system and the full Euler systemfor a polytropic gas (both in Eulerian coordinates).We study the stability of solutions ¯ v to the Riemann problem. We study the stability anduniqueness of these solutions among a large class of weak solutions u which are bounded,measurable, entropic for at least one strictly convex entropy, and verify a strong tracecondition (weaker than BV loc ). We require the solution ¯ v contain shocks of only the KRUPA extremal families (1-shocks and n-shocks), if it contains shocks at all. However, the roughersolutions u which we compare to ¯ v may have shocks of any type or family.Previous results in this direction include Chen, Frid, and Li [7] where for the full Eulersystem, they show uniqueness and long-time stability for perturbations of Riemann initialdata among a large class of entropy solutions (locally BV and without smallness conditions)for the 3 × x . For an extension to the relativistic Euler equations, seeChen and Li [8]. However, these papers do not give L stability results for all time.We will occasionally use a strong form of Lax’s E-condition, saying we want a shock tobe compressive but not overcompressive [13, p. 359-60], for a shock with left state u L , right state u R , and shock speed σ ,there exists i ∈ { , . . . , n } such that λ i − ( u L/R ) < λ i ( u R ) ≤ σ ≤ λ i ( u L ) < λ i +1 ( u L/R ) , where u L/R denotes u L or u R and λ := −∞ and λ n +1 := ∞ . (1.11)The condition (1.11) is a type of separation between the characteristic speeds. In par-ticular, note that for any strictly hyperbolic system of conservation laws, the first and lastinequalities in (1.11) will hold whenever | u L − u R | is sufficiently small. Furthermore, forhyperbolic systems where the characteristic speeds are completely separated in value, anyshock ( u L , u R ) will trivially satisfy (1.11). For example, for the full Euler system for gasdynamics in Lagrangian coordinates the first characteristic speed is always negative, themiddle one is always zero, and the third characteristic speed is always positive.We will also occasionally consider systems of the form (1.1) (endowed with the entropy η ) and verifying the additional sign condition, (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) ¯ v T ( x, t ) (cid:33) ∇ η (¯ v ( x, t )) f ( u | ¯ v ( x, t )) ≥ , for every rarefaction wave solution ¯ v of (1.1) and for every u ∈ R n ,(1.12)where f ( ·|· ) denotes the relative flux, f ( a | b ) := f ( a ) − f ( b ) − ∇ f ( b )( a − b ) , for a, b ∈ M n × .In particular, the system of isentropic gas dynamics verifies the property (1.12). For aproof of this fact, see [34]. The full Euler system also satisfies (1.12) in the case of onespace dimension and multiple space dimensions (see [15] for proof of this in multiple spacedimensions). TABILITY FOR THE RIEMANN PROBLEM 5
Fix
T >
0. For u L , u R ∈ R n , assume there exists a (potentially weak) solution ¯ v ∈ L ∞ ( R × [0 , T )) entropic for the entropy η , with the initial data¯ v ( x,
0) = (cid:40) u L if x < u R if x > . (1.13)In other words, ¯ v solves the Riemann problem (1.13).Assume ¯ v has the following standard form for a solution to the Riemann problem, con-stant on lines through the origin in the x-t plane: ¯ v is made up of n + 1 constant states u L = ¯ v , . . . , ¯ v n +1 = u R , where if ¯ v i (cid:54) = ¯ v i +1 ,then ¯ v i is joined to ¯ v i +1 by either an i-shock or an i-rarefaction fan.Otherwise ¯ v i = ¯ v i +1 and we do not need a shock or a rarefaction to connect ¯ v i to ¯ v i +1 .(1.14)Before we can present our stability and uniqueness results, for a fixed ¯ v as in (1.14), wedefine the Property ( D ).We say a function Ψ ¯ v : R × [0 , T ) → R n verifies property ( D ) if( D ): • If ¯ v contains at least one rarefaction wave, and if there are any shocks in ¯ v theyare either a 1-shock verifying (1.11) or an n-shock verifying (1.11), then – If ¯ v contains a 1-shock verifying (1.11) for i = 1, but no other shocks, thenthere exists a Lipschitz continuous function h : [0 , T ) → R with h (0) = 0and verifying h ( t ) < λ (¯ v ) t (1.15) for all t ∈ [0 , T ) such that Ψ ¯ v : R × [0 , T ) → R n verifies,Ψ ¯ v ( x, t ) := ¯ v if x < h ( t )¯ v if h ( t ) < x < λ (¯ v ) t ¯ v ( x, t ) if λ (¯ v ) t < x. (1.16) – If ¯ v contains an n-shock verifying (1.11) for i = n , but no other shocks, thenthere exists a Lipschitz continuous function h n : [0 , T ) → R with h n (0) = 0,and verifying λ n − (¯ v n ) t < h n ( t )(1.17) for all t ∈ [0 , T ) such that Ψ ¯ v : R × [0 , T ) → R n verifies,Ψ ¯ v ( x, t ) := ¯ v ( x, t ) if x < λ n − (¯ v n ) t ¯ v n if λ n − (¯ v n ) t < x < h n ( t )¯ v n +1 if h n ( t ) < x. (1.18) KRUPA – If ¯ v contains a 1-shock and an n-shock verifying (1.11) for i = 1 and i = n , respectively, but no other shocks, then there exists Lipschitz continuousfunctions h , h n : [0 , T ) → R with h (0) = h n (0) = 0, where h verifies (1.15)and h n verifies (1.17) such that Ψ ¯ v : R × [0 , T ) → R n verifies,Ψ ¯ v ( x, t ) := ¯ v if x < h ( t )¯ v if h ( t ) < x < λ (¯ v ) t ¯ v ( x, t ) if λ (¯ v ) t < x < λ n − (¯ v n ) t ¯ v n if λ n − (¯ v n ) t < x < h n ( t )¯ v n +1 if h n ( t ) < x. (1.19) – If ¯ v contains no shocks, then Ψ ¯ v : R × [0 , T ) → R n verifies,Ψ ¯ v ( x, t ) := ¯ v ( x, t )(1.20) for all ( x, t ) ∈ R × [0 , T ). • If ¯ v does not contain any rarefactions , and if ¯ v contains any shocks, they are eithera 1-shock or an n-shock, then – If ¯ v contains a 1-shock, but no other shocks, then there exists a Lipschitzcontinuous function h : [0 , T ) → R with h (0) = 0 such that Ψ ¯ v : R × [0 , T ) → R n verifies, Ψ ¯ v ( x, t ) := (cid:40) ¯ v if x < h ( t )¯ v if h ( t ) < x. (1.21) – If ¯ v contains an n-shock, but no other shocks, then there exists a Lipschitzcontinuous function h n : [0 , T ) → R with h n (0) = 0 such that Ψ ¯ v : R × [0 , T ) → R n verifies, Ψ ¯ v ( x, t ) := (cid:40) ¯ v n if x < h n ( t )¯ v n +1 if h n ( t ) < x. (1.22) – If ¯ v contains a 1-shock and an n-shock, but no other shocks, then there existsLipschitz continuous functions h , h n : [0 , T ) → R with h (0) = h n (0) = 0,and verifying h ( t ) ≤ h n ( t )(1.23) for all t ∈ [0 , T ) such that Ψ ¯ v : R × [0 , T ) → R n verifies,Ψ ¯ v ( x, t ) := ¯ v if x < h ( t )¯ v if h ( t ) < x < h n ( t )¯ v n +1 if h n ( t ) < x. (1.24) – If ¯ v contains no shocks, then ¯ v must be a constant function and Ψ ¯ v : R × [0 , T ) → R n verifies, Ψ ¯ v ( x, t ) := ¯ v (1.25) TABILITY FOR THE RIEMANN PROBLEM 7 for all ( x, t ) ∈ R × [0 , T ).Let u ∈ L ∞ ( R × [0 , T )) be any weak solution to (1.1), entropic for the entropy η (assumealso that u has strong traces (Definition 2.1)). With the definition of Property ( D ) outof the way, we present our main and most important theorem regarding L -type stabilityand uniqueness results between u and ¯ v . The hypotheses ( H ) and ( H ) ∗ in the theoremdepend only on the system (1.1) and the Riemann problem solution ¯ v . The hypothesesare related to conditions on 1-shocks and n-shocks and in particular are satisfied by theisentropic Euler and full Euler systems. They are with small modifications related to thehypotheses in [27]. These hypotheses are explained in detail in Section 2. The theoremgives a general overview of the results in this paper: Theorem 1.1 (Main theorem – L Stability for the Riemann Problem with ExtremalShocks) . Fix
T > . Assume u, ¯ v ∈ L ∞ ( R × [0 , T )) are solutions to the system (1.1) .Assume that u and ¯ v are entropic for the entropy η . Further, assume that u has strongtraces (Definition 2.1).Assume also that ¯ v is a solution to the Riemann problem (1.13) and has the form (1.14) .If ¯ v contains a 1-shock, assume the hypotheses ( H ) hold. Likewise, if ¯ v contains an n-shock,assume the hypotheses ( H ) ∗ hold.Assume (1.12) holds. Further, assume the system (1.1) has at least two conserved quan-tities ( n ≥ ).If ¯ v contains at least one rarefaction wave , assume that if there are any shocks in ¯ v theyare either a 1-shock verifying (1.11) or an n-shock verifying (1.11) .If ¯ v does not contain any rarefactions, and if ¯ v contains any shocks, assume they areeither a 1-shock or an n-shock (and we do not require (1.11) ).Then there exists a Ψ ¯ v with Property ( D ) , and verifying the following stability estimate: R (cid:90) − R (cid:12)(cid:12) u ( x, t ) − Ψ ¯ v ( x, t ) (cid:12)(cid:12) dx ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (1.26) for a constant µ > , and for all t , R > verifying t ∈ (0 , R ) and R > max i { Lip [ h i ] } t , (1.27) where the max runs over the i-shock families contained in ¯ v (1-shocks and/or n-shocks)and the h i are in the context of Property ( D ) .We also have the following L -type control on the shift functions h i : t (cid:90) (cid:88) i (cid:12)(cid:12)(cid:12) σ i (¯ v i , ¯ v i +1 ) − ˙ h i ( t ) (cid:12)(cid:12)(cid:12) dt ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (1.28) where the sum runs over the i-shock families contained in ¯ v (1-shocks and/or n-shocks).Remark. KRUPA • Note that H¨older’s inequality and (6.3) give control on the shifts in the form of1 t t (cid:90) (cid:12)(cid:12)(cid:12) σ i (¯ v i , ¯ v i +1 ) − ˙ h i ( t ) (cid:12)(cid:12)(cid:12) dt ≤ √ µ √ t (cid:13)(cid:13)(cid:13) u ( · ) − ¯ v ( · , (cid:13)(cid:13)(cid:13) L ( − R − rt ,R + rt ) . (1.29) • The relative entropy method can handle the occurance of vacuum states in theweak, entropic solution u (where u is in the context of Theorem 1.1). In particular,the method of relative entropy can be extended to allow for vacuum states in thefirst slot of the relative entropy η ( ·|· ). For simplicity, in the present article wedo not consider these generalizations to vacuum states. However, our results andarguments would be the same even if we considered vacuum states in the solution u . For details, see [34], [17, p. 346-7], and [27, p. 277-8].For more details on the results in Theorem 1.1, see Theorem 6.1 and Theorem 6.2 below.Our method is the relative entropy method, a technique created by Dafermos [11, 10] andDiPerna [14] to give L -type stability estimates between a Lipschitz continuous solutionand a rougher solution, which is only weak and entropic for a strictly convex entropy (theso-called weak-strong stability theory). For a system (1.1) endowed with an entropy η , thetechnique of relative entropy considers the quantity called the relative entropy , defined as η ( u | v ) := η ( u ) − η ( v ) − ∇ η ( v ) · ( u − v ) . (1.30)Similarly, we define relative entropy-flux, q ( u ; v ) := q ( u ) − q ( v ) − ∇ η ( v ) · ( f ( u ) − f ( v )) . (1.31)Remark that for any constant v ∈ R n , the map u (cid:55)→ η ( u | v ) is an entropy for the system(1.1), with associated entropy flux u (cid:55)→ q ( u ; v ). Furthermore, if u is a weak solution to(1.1) and entropic for η , then u will also be entropic for η ( ·| v ). This can be calculateddirectly from (1.1) and (1.6) – note that the map u (cid:55)→ η ( u | v ) is basically η plus a linearterm.Moreover, by virtue of η being strictly convex, the relative entropy is comparable to the L distance, in the following sense: Lemma 1.2.
For any fixed compact set V ⊂ V , there exists c ∗ , c ∗∗ > such that for all u, v ∈ V , c ∗ | a − b | ≤ η ( u | v ) ≤ c ∗∗ | a − b | . (1.32) The constants c ∗ , c ∗∗ depend on V and bounds on the second derivative of η . This lemma follows from Taylor’s theorem; for a proof see [27, 34].Now that we have defined the relative entropy, we remark that what we prove in thisarticle is actually stronger than Theorem 1.1. We get more than the L stability estimate(1.26). In fact, we get a contraction in a properly defined pseudo-distance . For simplicity,here in the introduction we define the pseudo-distance only when ¯ v (in the context ofTheorem 1.1) contains two shocks. The definition of the pseudo-distance is very similar TABILITY FOR THE RIEMANN PROBLEM 9 for the case of one shock or no shock. Then: for u , ¯ v , Ψ ¯ v as in the context of Theorem 1.1and α, β >
0, we define the pseudo-distance(1.33) E (cid:16) u ( · , t ); Ψ ¯ v ( · , t ); α ; β (cid:17) := (cid:34) h ( t ) (cid:90) − R η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx + α h n ( t ) (cid:90) h ( t ) η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx + β R (cid:90) h n ( t ) η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx (cid:35) , and where R > u onlylocally. The h and h n used in the definition (1.33) are from the Property ( D ) which Ψ ¯ v verifies.The pseudo-distance (1.33) is a technical tool we use in the proof of Theorem 1.1 (andin particular Theorem 6.1 and Theorem 6.2). By Lemma 1.2, it gives us the L stabilityestimate (1.26). The constants α and β we choose do not depend on the weak, entropicsolution u . Our use of the pseudo-distance (1.33) is based on the work [17].Given a Lipschitz continuous solution ¯ u to (1.1), and weak solution u to (1.1) which isentropic for at least one entropy, the method of relative entropy can be used to determineestimates on the growth in time of (cid:13)(cid:13) ¯ u ( · , t ) − u ( · , t ) (cid:13)(cid:13) L ( R ) . (1.34)To estimate the growth of this quantity, consider ∂ t (cid:82) η ( u | ¯ u ) dx . By (1.2), we get estimatesof the L -type (1.34). The point is that due the entropy inequality (1.6), it is more naturalto consider the quantity (cid:82) η ( u | ¯ u ) dx than to consider the L norm itself.However, the relative entropy method breaks down if a discontinuity is introduced intothe otherwise smooth solution ¯ u . In fact, simple examples for the scalar conservation lawsshow that when ¯ u has a discontinuity, there is no L stability in the same sense as in theclassical weak-strong estimates.In order to recover L stability in the sense of the classical weak-strong estimates, wemust allow the discontinuity in ¯ u to be moved (‘shifted’) with an artificial velocity whichdepends on the weak solution u . This is the theory of shifts. Within the context of therelative entropy method, this idea was devised by Vasseur [34]. Since then, this techniquehas been the subject of intense study by Vasseur and his team, and has yielded new results.The first result was for the scalar conservation laws in one space dimension. Further workconsidered the scalar viscous conservation laws in one space dimension [19] and multiplespace dimensions [20]. To handle systems, which allow for shocks from differing wavefamilies, the technique of a-contraction is used [17, 33, 35, 31, 27]. Recent work for scalar[22] has also allowed for many discontinuities to exist in the otherwise classical solution ¯ u which the method of relative entropy considers. By adding more and more discontinuitiesto the otherwise classical solution ¯ u , the method of relative entropy and the theory of shiftscan be used to show uniqueness for solutions which are entropic for at least one strictly convex entropy. For a general overview of theory of shifts and the relative entropy method,see [32, Section 3-5]. The theory of stability up to a shift has also been used to studythe asymptotic limits when the limit is discontinuous (see [9] for the scalar case, [36] forsystems). There are many other results using the relative entropy method to study theasymptotic limit. However, without the theory of shifts these results can only considerlimits which are Lipschitz continuous (see [28, 30, 4, 1, 37, 2, 5, 16] and [34] for a survey).The present paper is another step in this program of stability up to a shift.We use the construction of shifts based on the generalized characteristic introduced in[23, 21]. In this paper, we are able to handle shocks from two different wave familiesin the same solution, which is necessary for handling the Riemann problem with shocksfrom extremal wave families. As mentioned in [23, 21], the generalized-characteristic-basedshifts are an improvement over previous shift constructions partly because they are verysimple, and thus amenable to analysis and control. In particular, to do stability estimatesfor a solution to the Riemann problem with two extremal shocks, we need two shifts –one for each shock. Using prior constructions of the shifts, it was impossible to tell if thetwo shifts necessary for the Riemann problem would interact in a bad way or not. Usinggeneralized-characteristic-based shifts, this analysis is easy: due to the separation of shockspeeds, and the fact that generalized-characteristic-based shifts travel at characteristic-likespeed (for the characteristic of the shock they are shifting), we know immediately that theshifts for a 1-shock will stay to the left of the shifts for an n-shock. See Theorem 6.2 andProposition 5.1.In order to control the two shifts, one for the 1-shock and one for the n-shock in a solutionto the Riemann problem, we needed to extend the theory of Filippov flows to construct thetwo shifts in the sense of Filippov flows, while still maintaining control on their ordering(keeping the 1-shock shift to the left of the n-shock shift). The need to control the orderingof two different Filippov flows arose in the first paper where the theory of shifts in thecontext of the relative entropy method was used (see [26, Proposition 2]). However, ourresult (Lemma 5.4) is more general and has a significantly simpler proof.In [21], for a solution ¯ u to (1.1) which is Lipschitz continuous on both sides of one singleshock curve in space-time, to maintain L stability between ¯ u and another solution u whichis weak and entropic for at least one entropy, the solution ¯ u is translated artificially in space,instead of simply moving only the discontinuity itself. However, if ¯ u is a solution to theRiemann problem, it might contain rarefactions, which have a blow up in the derivative at t = 0. This causes tremendous entropy production if the rarefaction is artificially translatedin space. Moreover, ¯ u could easily contain two shocks – making it impossible to artificiallytranslate ¯ u in such a way that each discontinuity is moving at the velocity necessary tomaintain L stability against the solution u . Both of these concerns, ¯ u containing twoshocks and the blowup of rarefactions at t = 0, are addressed in the present paper. SeeSection 3 for a related discussion.For hyperbolic systems of conservation laws in one space dimension, one difficulty toshowing stability and uniqueness of (entropic) solutions is that many systems admit onlya single nontrivial entropy. The best well-posedness theory to date has been the L -basedtheory of Bressan, Crasta, and Piccoli [6]. However, this work only considers solutions with TABILITY FOR THE RIEMANN PROBLEM 11 small total variation. It would be interesting to study the stability of these solutions in alarger class. In fact, existence of solutions to the 2 × L -type. We use the relative entropy method and the related theories of shifts and a-contraction. Due to these theories not being perturbative, we are able to prove resultswithout small data limitations. Furthermore, because we use techniques based on therelative entropy method, we only use a single entropy and require only a single entropycondition.The outline of the paper is as follows: in Section 2, we give our hypotheses on the system.In Section 3, we present an overview of the proof of the main theorem (Theorem 1.1), whichis actually proved in two parts: Theorem 6.1 and Theorem 6.2. In Section 4, we presenttechnical lemmas. In Section 5, we construct the shift. Finally, in Section 6 we proveTheorem 6.1 and Theorem 6.2, which make up the main theorem Theorem 1.1.2. Hypotheses on the system
We will consider the following structural hypotheses ( H ), ( H ) ∗ on the system (1.1),(1.6) regarding the 1-shock and n-shock curves (they are closely related to hypotheses in[27, 21, 17]). For a fixed i-shock ( v L , v R ) ( i = 1 or i = n ): • ( H r > u ∈ B r ( v L ), there is a 1-shock curve (issuing from u ) S u : [0 , s u ) → V (possibly s u = ∞ ) parameterized by arc length. Moreover, S u (0) = u and theRankine-Hugoniot jump condition holds: f ( S u ( s )) − f ( u ) = σ u ( s )( S u ( s ) − u ) , (2.1) where σ u ( s ) is the velocity function. The map u (cid:55)→ s u is Lipschitz on V . Further,the maps ( s, u ) (cid:55)→ S u ( s ) and ( s, u ) (cid:55)→ σ u ( s ) are both C on { ( s, u ) | s ∈ [0 , s u ) , u ∈V} , and the following conditions are satisfied:(a) (Liu entropy condition) dd s σ u ( s ) < , σ u (0) = λ ( u ) , (b) (shock “strengthens” with s ) dd s η ( u | S u ( s )) > , for all s > , (c) (the shock curve cannot wrap tightly around itself)For all R >
0, there exists ˜
S > (cid:110) S u ( s ) (cid:12)(cid:12)(cid:12) s ∈ [0 .s u ) , | u | ≤ R and (cid:12)(cid:12)(cid:12) S u ( s ) (cid:12)(cid:12)(cid:12) ≤ R (cid:111) ⊆ (cid:110) S u ( s ) (cid:12)(cid:12)(cid:12) | u | ≤ R and s ≤ ˜ S (cid:111) . • ( H u L , u R ) is an entropic Rankine-Hugoniot discontinuity with shock speed σ , then σ > λ ( u R ). • ( H u L , u R ) (with u L ∈ B r ( v L )) is an entropic Rankine-Hugoniot discontinu-ity with shock speed σ verifying σ ≤ λ ( u L ) , (2.2) then u R is in the image of S u L . In other words, there exists s u R ∈ [0 , s u L ) such that S u L ( s u R ) = u R (and by implication, σ = σ u L ( s u R )).Similarly, we will consider the following structural hypotheses ( H ) ∗ on the system (1.1),(1.6) regarding the n-shock curves: • ( H ∗ : (Family of n-shocks verifying the Liu condition) There exists r > u ∈ B r ( v R ), there is an n-shock curve (issuing from u ) S nu : [0 , s u ) → V (possibly s u = ∞ ) parameterized by arc length. Moreover, S nu (0) = u and theRankine-Hugoniot jump condition holds: f ( S nu ( s )) − f ( u ) = σ nu ( s )( S nu ( s ) − u ) , (2.3) where σ nu ( s ) is the velocity function. The map u (cid:55)→ s u is Lipschitz on V . Further,the maps ( s, u ) (cid:55)→ S nu ( s ) and ( s, u ) (cid:55)→ σ nu ( s ) are both C on { ( s, u ) | s ∈ [0 , s u ) , u ∈V} , and the following conditions are satisfied:(a) (Liu entropy condition) dd s σ nu ( s ) > , σ nu (0) = λ n ( u ) , (b) (shock “strengthens” with s ) dd s η ( u | S nu ( s )) > , for all s > , (c) (the shock curve cannot wrap tightly around itself)For all R >
0, there exists ˜
S > (cid:110) S nu ( s ) (cid:12)(cid:12)(cid:12) s ∈ [0 .s u ) , | u | ≤ R and (cid:12)(cid:12) S nu ( s ) (cid:12)(cid:12) ≤ R (cid:111) ⊆ (cid:110) S nu ( s ) (cid:12)(cid:12)(cid:12) | u | ≤ R and s ≤ ˜ S (cid:111) . • ( H ∗ : If ( u R , u L ) is an entropic Rankine-Hugoniot discontinuity with shock speed σ , then σ < λ n ( u L ). • ( H ∗ : If ( u R , u L ) (with u R ∈ B r ( v R )), is an entropic Rankine-Hugoniot disconti-nuity with shock speed σ verifying σ ≥ λ n ( u R ) , (2.4) then u L is in the image of S nu R . In other words, there exists s u L ∈ [0 , s u R ) such that S nu R ( s u L ) = u L (and by implication, σ = σ nu R ( s u L )). Remark.
For useful remarks on these hypotheses, see [21, 17, 27]. We include the remarks here forcompleteness.
TABILITY FOR THE RIEMANN PROBLEM 13 • Note that the system (1.1) verifies the hypotheses ( H H
3) on the 1-shock familyif and only if the system (cid:40) ∂ t u − ∂ x f ( u ) = 0 , t > ,u ( x,
0) = u ( x ) for x ∈ R . (2.5) verifies the properties ( H ∗ -( H ∗ for the n-shock family. It is in this way that( H H
3) are dual to ( H ∗ -( H ∗ . • On top of the Liu entropy condition (Property (a) in ( H S u L when measured via the pseudo-distance of the relative entropy (recall that the map( u, v ) (cid:55)→ η ( u | v ) measures L -distance somehow – see Lemma 1.2). This growthcondition arises naturally in the study of admissibility criteria for systems of con-servation laws. In particular, Property (b) ensures that Liu admissible shocks areentropic for the entropy η even for moderate-to-strong shocks (see [12, 24, 29]).In [3], Barker, Freist¨uhler, and Zumbrun show that stability and in particularcontraction fails to hold for the full Euler system if we replace Property (b) withdd s η ( S u ( s )) > , s > . (2.6) This shows that it is better to measure shock strength using the relative entropyrather than the entropy itself. • Recall the famous Lax E-condition for an i-shock ( u L , u R , σ ), λ i ( u R ) ≤ σ ≤ λ i ( u L ) . (2.7) The hypothesis ( H
2) is implied by the first half of the Lax E-condition along withthe hyperbolicity of the system (1.1). In addition, we do not allow for right 1-contact discontinuities. • The hypothesis ( H
3) is a statement about the well-separation of the 1-shocks fromall other Rankine-Hugoniot discontinuities entropic for η ; the 1-shocks do not inter-fere with any other shocks. In particular, ( H
3) will hold for any strictly hyperbolicsystem in the form (1.1) if all Rankine-Hugoniot discontinuities ( u L , u R , σ ) entropicfor η lie on an i-shock curve for some i and the extended Lax admissibility conditionholds: λ i − ( u L ) ≤ σ ≤ λ i +1 ( u R ) , (2.8) where λ := −∞ and λ n +1 := ∞ . Moreover, we only use the first inequality in (2.8)and the fact that λ ( u ) ≤ λ i − ( u ) for all u ∈ V and for all i > any strictly hyperbolic system in the form (1.1), if u R and u L live in a fixed compact set, then there exists δ > | u R − u L | ≤ δ . Similarly, for any strictly hyperbolic system endowed witha strictly convex entropy, all Rankine-Hugoniot discontinuities ( u L , u R , σ ) entropicfor η will locally be in the form S iu L ( s ) = u R for some s >
0, and where S iu L is thei-shock curve issuing from u L . See [25, Theorem 1.1, p. 140] and more generally [25, p. 140-6]. For the full Euler system, ( H
3) will hold regardless of the size of theshock ( u L , u R ). • Note that due to the map ( s, u ) (cid:55)→ S u ( s ) being Lipschitz, we have (cid:12)(cid:12)(cid:12) S u ( s ) − u (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) S u ( s ) − S u (0) (cid:12)(cid:12)(cid:12) ≤ Lip (cid:104) ( s, u ) (cid:55)→ S u ( s ) (cid:105) s, (2.9) for all u ∈ B r ( I − ) and all s ∈ [0 , s u ). Equivalently,1Lip (cid:104) ( s, u ) (cid:55)→ S u ( s ) (cid:105) (cid:12)(cid:12)(cid:12) S u ( s ) − u (cid:12)(cid:12)(cid:12) ≤ s. (2.10) • On the state space V where the strictly convex entropy η is defined, the system(1.1) is hyperbolic. Further, by virtue of f ∈ C ( V ), the eigenvalues of ∇ f ( u )vary continuously on the state space V . Further, if the eigenvalue λ ( u ) ( λ n ( u )) issimple for u ∈ V (such as when the system (1.1) is strictly hyperbolic), the map u (cid:55)→ λ ( u ) ( u (cid:55)→ λ n ( u )) will be in C ( V ) due to the implicit function theorem.We study solutions u to (1.1) among the class of functions verifying a strong traceproperty (first introduced in [27]): Definition 2.1.
Fix
T >
0. Let u : R × [0 , T ) → R n verify u ∈ L ∞ ( R × [0 , T )). We say u has the strong trace property if for every fixed Lipschitz continuous map h : [0 , T ) → R ,there exists u + , u − : [0 , T ) → R n such thatlim n →∞ t (cid:90) ess sup y ∈ (0 , n ) (cid:12)(cid:12) u ( h ( t ) + y, t ) − u + ( t ) (cid:12)(cid:12) dt = lim n →∞ t (cid:90) ess sup y ∈ ( − n , (cid:12)(cid:12) u ( h ( t ) + y, t ) − u − ( t ) (cid:12)(cid:12) dt = 0(2.11)for all t ∈ (0 , T ).Note that for example a function u ∈ L ∞ ( R × [0 , T )) will satisfy the strong trace propertyif for each fixed h , the right and left limitslim y → + u ( h ( t ) + y, t ) and lim y → − u ( h ( t ) + y, t )(2.12)exist for almost every t . In particular, a function u ∈ L ∞ ( R × [0 , T )) will have strong tracesaccording to Definition 2.1 if u has a representative which is in BV loc . However, the strongtrace property is weaker than BV loc .3. Overview of the proofs of Theorem 6.1 and Theorem 6.2
Within the context of the relative entropy method, the theory of shifts often worksby moving shocks with an artificial velocity, as opposed to the velocity dictated by theRankine-Hugoniot jump condition. One difficulty in applying the theory of shifts to solvingthe Riemann problem is, what to do if the graph of a x = h ( t ) shift function (in the x-tplane) for a particular shock intersects one of the rarefaction fans? At this point, it isnot guaranteed that the states to the left and right of the shift function are an entropic TABILITY FOR THE RIEMANN PROBLEM 15 discontinuity (they might not even satisfy Rankine-Hugoniot) – and this prevents analysis.But this is again solved using generalized-characteristic-based shifts. For example, thegeneralized-characteristic-based shifts for a 1-shock in ¯ v will either travel at characteristic-like speed of u , or they will travel to the left very quickly (super-characteristic speed).When the generalized-characteristic-based shift (for a 1-shock) is traveling to the left veryfast, we do not have to worry about it intersecting with a rarefaction fan, which will spreadout with characteristic speed. When the generalized-characteristic-based shift is travelingwith characteristic speed, then we must control the speed of generalized characteristic of u versus the speed the rarefaction fans in ¯ v are spreading out. Heuristically, the function u goes into the first slot η ( ·| ) of the relative entropy, and ¯ v goes into the second η ( |· ), andthere is little connection between the two slots of the relative entropy. However, throughthe strong form of Lax’s E-condition (1.11), we can connect these two worlds of the firstand second slot of the relative entropy and show that the generalized characteristic of u will not intersect the rarefaction fans in the x-t plane. In fact, the analysis will depend onthe quantity ( λ i +1 (¯ v i +1 ) − λ i (¯ v i )) if ¯ v has an i-shock (¯ v i , ¯ v i +1 ). For example, for hyperbolicsystems of conservation laws where the characteristics speeds are completely separated invalue, any shock will satisfy (1.11). Furthermore, for such systems it is clear that a shiftfunction traveling at the speed of a generalized characteristic for one wave family cannotintersect the rarefaction fan of a different wave family. See Theorem 6.1.If ¯ v does not contain any rarefactions, then we do not have to compare the shifts tothe rarefactions to make sure they are not interacting. Instead, we only need to preventthe shifts corresponding to a 1-shock from interacting with the shifts corresponding to ann-shock. We want the two shifts to stay away from each other, because if they touch andstick together then the left and right hand states to the left and right of the (now single)shift will in general not make an entropic shock. Without rarefactions in between thesetwo shifts to separate them, we cannot use the arguments from Theorem 6.1. We insteadstudy the two shifts directly. See Theorem 6.2.4. Technical Lemmas
For use throughout this paper, we define the relative flux f ( a | b ) := f ( a ) − f ( b ) − ∇ f ( b )( a − b ) , (4.1)for a, b ∈ M n × . Further, for a, b ∈ M n × , we define the relative ∇ η : ∇ η ( a | b ) := ∇ η ( a ) − ∇ η ( b ) − [ a − b ] T ∇ η ( b ) . (4.2) Lemma 4.1.
Fix
B > . Then there exists a constant C > depending on B such thatthe following holds:If u L , u R ∈ V with | u L | , | u R | ≤ B , then whenever α, θ ∈ (0 , verify α < θ C , (4.3) then R a := { u | η ( u | u L ) ≤ aη ( u | u R ) } ⊂ B θ ( u L ) for all < a < α .Remark. The set R a is compact. The proof of Lemma 4.1 is found in the proof of Lemma 4.3 in [17].The following Lemma gives us the entropy dissipation caused by changing the domainof integration and translating the piecewise-smooth solution ¯ u in x (by a function X ( t )). Lemma 4.2 (Local entropy dissipation rate) . Let u, ¯ u ∈ L ∞ ( R × [0 , T )) be weak solutionsto (1.1) . We assume that u, ¯ u are entropic for the entropy η . Assume that ¯ u is Lipschitzcontinuous on { ( x, t ) ∈ R × [0 , T ) | x < s ( t ) } and on { ( x, t ) ∈ R × [0 , T ) | x > s ( t ) } , where s : [0 , T ) → R is a Lipschitz function . Assume also that u verifies the strong trace property(Definition 2.1). Let T, t , t ∈ R verify ≤ t < t < T . Let h , h , X : [0 , T ) → R beLipschitz continuous functions with the property that h ( t ) − h ( t ) > for all t ∈ ( t , t ) .We also require that if t (cid:54) = 0 , then h ( t ) = h ( t ) . Further assume that for all t ∈ [ t , t ] , s ( t ) − X ( t ) is not in the open set ( h ( t ) , h ( t )) .Then, (4.4) t (cid:90) t (cid:20) q ( u ( h ( t )+ , t ); ¯ u (( h ( t ) + X ( t ))+ , t )) − q ( u ( h ( t ) − , t ); ¯ u (( h ( t ) + X ( t )) − , t ))+˙ h ( t ) η ( u ( h ( t ) − , t ) | ¯ u (( h ( t ) + X ( t )) − , t )) − ˙ h ( t ) η ( u ( h ( t )+ , t ) | ¯ u (( h ( t ) + X ( t ))+ , t )) (cid:21) dt ≥ h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx − h ( t ) (cid:90) h ( t ) η ( u ( x ) | ¯ u ( x )) dx + t (cid:90) t h ( t ) (cid:90) h ( t ) (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ∇ η (¯ u ( x, t )) (cid:33) f ( u ( x, t ) | ¯ u ( x + X ( t ) , t ))+ (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ¯ u T ( x, t ) ˙ X ( t ) (cid:33) ∇ η (¯ u ( x + X ( t ) , t ))[ u ( x, t ) − ¯ u ( x + X ( t ) , t )] dxdt. Remark. If t (cid:54) = 0, then h ( t ) = h ( t ) and h ( t ) (cid:90) h ( t ) η ( u ( x ) | ¯ u ( x )) dx = 0 . (4.5) Remark.
Lemma 4.2, and in particular (4.4), are not true if h ( t ) = h ( t ) for all t in someopen interval. TABILITY FOR THE RIEMANN PROBLEM 17
To see this, consider the following simple example: Let ¯ u := v for some constant state v ∈ R n . Let ( u L , u R , σ ( u L , u R )) be a shock entropic for the entropy η . Define u ( x, t ) := (cid:40) u L if x < σ ( u L , u R ) tu R if x > σ ( u L , u R ) t. (4.6)Choose h ( t ) := h ( t ) := σ ( u L , u R ) t .With these choices, the right hand side of (4.4) vanishes.The left hand side of (4.4) becomes t (cid:90) t (cid:20) q ( u R ; v ) − q ( u L ; v ) − σ ( u L , u R )( η ( u R | v ) − η ( u L | v )) (cid:21) dt. (4.7)Note that because u is entropic for the entropy η , u is also entropic for the entropy u (cid:55)→ η ( u | v ) . (4.8)This follows because the map (4.8) is simply the function η ( u ) plus a term (affine) linearin u .Thus, the shock ( u L , u R , σ ( u L , u R )) is entropic for (4.8). This implies that q ( u R ; v ) − q ( u L ; v ) − σ ( u L , u R )( η ( u R | v ) − η ( u L | v )) ≤ . (4.9)By choosing a shock ( u L , u R , σ ( u L , u R )) such that (4.9) is strictly negative, we haveshown that (4.4) does not hold (recall (4.7)).Intuitively, why does (4.4) fails to hold when h ( t ) = h ( t )? This is because when h ( t ) (cid:54) = h ( t ) the function h thinks that if it moves to the right (or left), it is reducing (orcreating more of) the entropy in the integral h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx, (4.10)by contracting (or expanding) the domain of integration. And similarly for h . However,if for a positive amount of time h ( t ) = h ( t ), then (4.10) is always zero and no mass iscreated or destroyed.However, as long as h ( t ) = h ( t ) only for brief moments, a version of Lemma 4.2 stillholds. See Corollary 4.3. Proof of Lemma 4.2.
This proof is based on a similar argument in [23].Step 1We first show that for all positive, Lipschitz continuous test functions φ : R × [0 , T ) → R with compact support and that vanish on the set { ( x, t ) ∈ R × [0 , T ) | x = s ( t ) − X ( t ) } , we have(4.11) T (cid:90) ∞ (cid:90) −∞ [ ∂ t φη ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) + ∂ x φq ( u ( x, t ); ¯ u ( x + X ( t ) , t ))] dxdt + ∞ (cid:90) −∞ φ ( x, η ( u ( x ) | ¯ u ( x )) dx ≥ T (cid:90) ∞ (cid:90) −∞ φ (cid:34)(cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ∇ η (¯ u ( x, t )) (cid:33) f ( u ( x, t ) | ¯ u ( x + X ( t ) , t ))+ (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ¯ u T ( x, t ) ˙ X ( t ) (cid:33) ∇ η (¯ u ( x + X ( t ) , t ))[ u ( x, t ) − ¯ u ( x + X ( t ) , t )] (cid:35) dxdt. Note that (4.11) is the analogue in our case of the key estimate used in Dafermos’s proof ofweak-strong stability, which gives a relative version of the entropy inequality (see equation(5.2.10) in [13, p. 122-5]). The proof of (4.11) is based on the famous weak-strong stabilityproof of Dafermos and DiPerna [13, p. 122-5]. We then modify the Dafermos and DiPernaproof as in [23] to allow for the translation of the solution ¯ u by the function X and toaccount for the additional entropy this creates.Note that on the complement of the set { ( x, t ) ∈ R × [0 , T ) | x = s ( t ) } , ¯ u is smooth andso we have the exact equalities, ∂ t (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) ¯ u ( x, t ) (cid:1) + ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) f (¯ u ( x, t )) (cid:1) = 0 , (4.12) ∂ t (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) η (¯ u ( x, t )) (cid:1) + ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) q (¯ u ( x, t )) (cid:1) = 0 . (4.13)Thus for any Lipschitz continuous function X : [0 , T ) → R with X (0) = 0 we have onthe complement of the set { ( x, t ) ∈ R × [0 , T ) | x = s ( t ) − X ( t ) } ,(4.14) ∂ t (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) ¯ u ( x + X ( t ) , t ) (cid:1) + ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) f (¯ u ( x + X ( t ) , t )) (cid:1) = (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) (cid:0) ¯ u ( x, t ) (cid:1)(cid:33) ˙ X ( t ) , and(4.15) ∂ t (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) η (¯ u ( x + X ( t ) , t )) (cid:1) + ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x,t ) (cid:0) q (¯ u ( x + X ( t ) , t )) (cid:1) = ∇ η (¯ u ( x + X ( t ) , t )) (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) (cid:0) ¯ u ( x, t ) (cid:1)(cid:33) ˙ X ( t ) . TABILITY FOR THE RIEMANN PROBLEM 19
We can now imitate the weak-strong stability proof in [13, p. 122-5], using (4.14) and(4.15) instead of (4.12) and (4.13). This gives (4.11). For more details, the reader can referto [21], where the computation is done under the additional assumption that the system(1.1) has a source term G . Due to considering the source term G , the work [21] assumesthat the entropy η ∈ C ( V ), but the computations go through unchanged if we take G ≡ η ∈ C ( V ).Step 2We will now test (4.11) with some particular test functions. The rest of the proof ofLemma 4.2 is decomposed into two cases: Case 1 t = 0, h ( t ) < h ( t )and Case 2 h ( t ) = h ( t )We start with Case 1 . Case 1 t = 0, h ( t ) < h ( t )Choose t ∗ ∈ ( t , t ).Define δ := inf t ∈ (cid:2) t ,t ∗ + t − t ∗ (cid:3) ( h ( t ) − h ( t )) . (4.16)Note δ > < (cid:15) < min { δ, t − t ∗ } .We apply the test function ω ( t ) χ ( x, t ) to (4.11), where ω ( t ) := ≤ t < t ∗ (cid:15) ( t ∗ − t ) + 1 if t ∗ ≤ t < t ∗ + (cid:15) t ∗ + (cid:15) ≤ t. (4.17)and χ ( x, t ) := x < h ( t ) (cid:15) ( x − h ( t )) if h ( t ) ≤ x < h ( t ) + (cid:15) h ( t ) + (cid:15) ≤ x ≤ h ( t ) − (cid:15) − (cid:15) ( x − h ( t )) if h ( t ) − (cid:15) < x ≤ h ( t )0 if h ( t ) < x. (4.18)The function ω is modeled from [13, p. 124]. The function χ is from [26, p. 765]. We receive,(4.19) t ∗ (cid:90) (cid:34) − h ( t )+ (cid:15) (cid:90) h ( t ) (cid:15) ˙ h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx + h ( t )+ (cid:15) (cid:90) h ( t ) (cid:15) q ( u ( x, t ); ¯ u ( x + X ( t ) , t )) dx + h ( t ) (cid:90) h ( t ) − (cid:15) (cid:15) ˙ h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx − h ( t ) (cid:90) h ( t ) − (cid:15) (cid:15) q ( u ( x, t ); ¯ u ( x + X ( t ) , t )) dx (cid:35) dt + h (0) (cid:90) h (0) η ( u ( x ) | ¯ u ( x )) dx − t ∗ + (cid:15) (cid:90) t ∗ (cid:15) h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dxdt + O ( (cid:15) ) ≥ t ∗ (cid:90) h ( t ) (cid:90) h ( t ) RHS dxdt, where RHS represents everything being multiplied by φ in the integral on the right handside of (4.11).Recall the convexity of η . Furthermore, remark that for weak solutions u to (1.1), themap t (cid:55)→ u ( · , t ) is continuous in L ∞ weak-* . Thus, from these two facts we have thefollowing lower-semicontinuity property for r ∈ [0 , T ): h ( r ) (cid:90) h ( r ) η ( u ( x, r ) | ¯ u ( x + X ( r ) , r )) dx ≤ lim inf s → r h ( s ) (cid:90) h ( s ) η ( u ( x, s ) | ¯ u ( x + X ( s ) , s )) dx. (4.20)Let (cid:15) → u satisfies the strong trace property (Definition 2.1). This yields, TABILITY FOR THE RIEMANN PROBLEM 21 (4.21) t ∗ (cid:90) t (cid:20) q ( u ( h ( t )+ , t ); ¯ u (( h ( t ) + X ( t ))+ , t )) − q ( u ( h ( t ) − , t ); ¯ u (( h ( t ) + X ( t )) − , t ))+ ˙ h ( t ) η ( u ( h ( t ) − , t ) | ¯ u (( h ( t ) + X ( t )) − , t )) − ˙ h ( t ) η ( u ( h ( t )+ , t ) | ¯ u (( h ( t ) + X ( t ))+ , t )) (cid:21) dt + h (0) (cid:90) h (0) η ( u ( x ) | ¯ u ( x )) dx ≥ h ( t ∗ ) (cid:90) h ( t ∗ ) η ( u ( x, t ∗ ) | ¯ u ( x + X ( t ∗ ) , t ∗ )) dx + t ∗ (cid:90) t h ( t ) (cid:90) h ( t ) RHS dxdt, where we used (4.20) to take the limit of the term t ∗ + (cid:15) (cid:90) t ∗ (cid:15) h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dxdt (4.22)for every t ∗ and not just almost every t ∗ .We let t ∗ → t in (4.21). Recall the dominated convergence theorem, and again use(4.20) to handle the term h ( t ∗ ) (cid:90) h ( t ∗ ) η ( u ( x, t ∗ ) | ¯ u ( x + X ( t ∗ ) , t ∗ )) dx (4.23)This yields (4.4). Case 2 h ( t ) = h ( t )Choose t ∗ , t ∗∗ ∈ ( t , t ) with t ∗∗ < t ∗ .Define δ := inf t ∈ (cid:2) t ∗∗ ,t ∗ + t − t ∗ (cid:3) ( h ( t ) − h ( t )) . (4.24)Note δ > < (cid:15) < min { δ, t − t ∗ } . We repeat the above calculations, but instead of using ω , we use ω : ω ( t ) := ≤ t < t ∗∗ (cid:15) ( t − t ∗∗ ) if t ∗∗ ≤ t < t ∗∗ + (cid:15) t ∗∗ + (cid:15) ≤ t < t (cid:15) ( t − t ) + 1 if t ≤ t < t + (cid:15) t + (cid:15) ≤ t. (4.25)The function ω is from [26, p. 765].The function χ (4.18) is used exactly as it is.We test (4.11) with the test function ω ( t ) χ ( x, t ). This gives us,(4.26) t ∗ (cid:90) t ∗∗ (cid:34) − h ( t )+ (cid:15) (cid:90) h ( t ) (cid:15) ˙ h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx + h ( t )+ (cid:15) (cid:90) h ( t ) (cid:15) q ( u ( x, t ); ¯ u ( x + X ( t ) , t )) dx + h ( t ) (cid:90) h ( t ) − (cid:15) (cid:15) ˙ h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx − h ( t ) (cid:90) h ( t ) − (cid:15) (cid:15) q ( u ( x, t ); ¯ u ( x + X ( t ) , t )) dx (cid:35) dt + t ∗∗ + (cid:15) (cid:90) t ∗∗ (cid:15) h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dxdt − t ∗ + (cid:15) (cid:90) t ∗ (cid:15) h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dxdt + O ( (cid:15) ) ≥ t ∗ (cid:90) t ∗∗ h ( t ) (cid:90) h ( t ) RHS dxdt.
Note that we can estimate (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t ∗∗ + (cid:15) (cid:90) t ∗∗ (cid:15) h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dxdt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C sup t ∈ [ t ∗∗ ,t ∗∗ + (cid:15) ] (cid:12)(cid:12) h ( t ) − h ( t ) (cid:12)(cid:12) , (4.27)for some constant C >
TABILITY FOR THE RIEMANN PROBLEM 23 (4.28) t ∗ (cid:90) t ∗∗ (cid:34) − h ( t )+ (cid:15) (cid:90) h ( t ) (cid:15) ˙ h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx + h ( t )+ (cid:15) (cid:90) h ( t ) (cid:15) q ( u ( x, t ); ¯ u ( x + X ( t ) , t )) dx + h ( t ) (cid:90) h ( t ) − (cid:15) (cid:15) ˙ h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx − h ( t ) (cid:90) h ( t ) − (cid:15) (cid:15) q ( u ( x, t ); ¯ u ( x + X ( t ) , t )) dx (cid:35) dt + C sup t ∈ [ t ∗∗ ,t ∗∗ + (cid:15) ] (cid:12)(cid:12) h ( t ) − h ( t ) (cid:12)(cid:12) + O ( (cid:15) ) − t ∗ + (cid:15) (cid:90) t ∗ (cid:15) h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dxdt ≥ t ∗ (cid:90) t ∗∗ h ( t ) (cid:90) h ( t ) RHS dxdt.
Let (cid:15) → u satisfies the strong trace property (Definition 2.1). This yields,(4.29) t ∗ (cid:90) t ∗∗ (cid:20) q ( u ( h ( t )+ , t ); ¯ u (( h ( t ) + X ( t ))+ , t )) − q ( u ( h ( t ) − , t ); ¯ u (( h ( t ) + X ( t )) − , t ))+ ˙ h ( t ) η ( u ( h ( t ) − , t ) | ¯ u (( h ( t ) + X ( t )) − , t )) − ˙ h ( t ) η ( u ( h ( t )+ , t ) | ¯ u (( h ( t ) + X ( t ))+ , t )) (cid:21) dt + C (cid:12)(cid:12) h ( t ∗∗ ) − h ( t ∗∗ ) (cid:12)(cid:12) ≥ h ( t ∗ ) (cid:90) h ( t ∗ ) η ( u ( x, t ∗ ) | ¯ u ( x + X ( t ∗ ) , t ∗ )) dx + t ∗ (cid:90) t ∗∗ h ( t ) (cid:90) h ( t ) RHS dxdt, where we used (4.20) to take the limit of the term t ∗ + (cid:15) (cid:90) t ∗ (cid:15) h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dxdt (4.30)for every t ∗ and not just almost every t ∗ .We let t ∗∗ → t +0 in (4.29), recalling the dominated convergence theorem. This gives,(4.31) t ∗ (cid:90) t (cid:20) q ( u ( h ( t )+ , t ); ¯ u (( h ( t ) + X ( t ))+ , t )) − q ( u ( h ( t ) − , t ); ¯ u (( h ( t ) + X ( t )) − , t ))+ ˙ h ( t ) η ( u ( h ( t ) − , t ) | ¯ u (( h ( t ) + X ( t )) − , t )) − ˙ h ( t ) η ( u ( h ( t )+ , t ) | ¯ u (( h ( t ) + X ( t ))+ , t )) (cid:21) dt ≥ h ( t ∗ ) (cid:90) h ( t ∗ ) η ( u ( x, t ∗ ) | ¯ u ( x + X ( t ∗ ) , t ∗ )) dx + t ∗ (cid:90) t ∗∗ h ( t ) (cid:90) h ( t ) RHS dxdt.
Finally, we let t ∗ → t − in (4.31). We recall again the dominated convergence theoremand (4.20).We receive (4.4).This completes the proof of Lemma 4.2. (cid:3) Corollary 4.3.
Let u, ¯ u ∈ L ∞ ( R × [0 , T )) be weak solutions to (1.1) . Assume that u and ¯ u are entropic for the entropy η . Assume that ¯ u is Lipschitz continuous on { ( x, t ) ∈ R × [0 , T ) | x < s ( t ) } and on { ( x, t ) ∈ R × [0 , T ) | x > s ( t ) } , where s : [0 , T ) → R is aLipschitz function . Assume also that u verifies the strong trace property (Definition 2.1).Let T, t ∈ R verify < t < T . Let h , h , X : [0 , T ) → R be Lipschitz continuous. Werequire that • h ( t ) ≤ h ( t )(4.32) for all t ∈ [0 , T ) , • if h ( t ) = h ( t ) for some t ∈ [0 , T ) , and h and h are both differentiable at t , then ˙ h ( t ) < ˙ h ( t ) . (4.33) Assume also that for all t ∈ [0 , t ] , s ( t ) − X ( t ) is not in the open set ( h ( t ) , h ( t )) . TABILITY FOR THE RIEMANN PROBLEM 25
Figure 1.
Corollary 4.3 allows us to consider shift functions h and h which occasionally touch as shown. Then, (4.34) t (cid:90) (cid:20) q ( u ( h ( t )+ , t ); ¯ u (( h ( t ) + X ( t ))+ , t )) − q ( u ( h ( t ) − , t ); ¯ u (( h ( t ) + X ( t )) − , t ))+ ˙ h ( t ) η ( u ( h ( t ) − , t ) | ¯ u (( h ( t ) + X ( t )) − , t )) − ˙ h ( t ) η ( u ( h ( t )+ , t ) | ¯ u (( h ( t ) + X ( t ))+ , t )) (cid:21) dt ≥ h ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ u ( x + X ( t ) , t )) dx − h (0) (cid:90) h (0) η ( u ( x ) | ¯ u ( x )) dx + t (cid:90) h ( t ) (cid:90) h ( t ) (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ∇ η (¯ u ( x, t )) (cid:33) f ( u ( x, t ) | ¯ u ( x + X ( t ) , t ))+ (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ¯ u T ( x, t ) ˙ X ( t ) (cid:33) ∇ η (¯ u ( x + X ( t ) , t ))[ u ( x, t ) − ¯ u ( x + X ( t ) , t )] dxdt. Remark.
This corollary says that the dissipation rate formula (4.4) holds if h ( t ) = h ( t )for only a small number of t values (see Figure 1). Proof.
Step 1Note that (4.32) and (4.33) imply that h ( t ) = h ( t ) will not occur for t values whereboth h and h are differentiable. Thus the set { t | h ( t ) = h ( t ) } (4.35) is measure zero because Lipschitz continuous functions are differentiable almost everywhere.Step 2Remark that { t ∈ (0 , t ) | h ( t ) (cid:54) = h ( t ) } (4.36)is an open subset of R .Thus, we can write (4.36) as a union of at most countably many disjoint open intervals: { t ∈ (0 , t ) | h ( t ) (cid:54) = h ( t ) } = (cid:91) i ∈ Λ ( x i , y i ) , (4.37)where Λ is an at most countable index set, and x i , y i ∈ R , x i (cid:54) = y i .We now show the following two claims: If h ( t ) (cid:54) = h ( t ), then there exists i ∈ Λ such thatthe open interval ( x i , y i ) is equal to the open interval ( x i , t ).Further, if h ( y i ) (cid:54) = h ( y i ) for some i ∈ Λ, then y i = t .(4.38)and If h (0) (cid:54) = h (0), then there exists i ∈ Λ such thatthe open interval ( x i , y i ) is equal to the open interval (0 , y i ).Further, if h ( x i ) (cid:54) = h ( x i ) for some i ∈ Λ, then x i = 0.(4.39)The proofs of (4.38) and (4.39) are similar. We will only show (4.38):If h ( t ) (cid:54) = h ( t ), then by continuity of h , h there exists α ∈ [0 , t ) such that h ( t ) (cid:54) = h ( t ) for all t in the open interval ( α, t ), with either h ( α ) = h ( α ) or α = 0. By (4.37),we must have ( α, t ) ⊆ (cid:91) i ∈ Λ ( x i , y i ) . (4.40)Consider i ∈ Λ such that ( x i , y i ) ∩ ( α, t ) (cid:54) = ∅ . Then if y i < t , we have a contradictionto the fact that the open intervals ( x i , y i ) are disjoint. This proves the first part of (4.38).Assume now that h ( y i ) (cid:54) = h ( y i ) for some i ∈ Λ. By definition (4.37), y i ≤ t . If y i < t ,then by continuity of h , h , there exists (cid:15) > h ( t ) (cid:54) = h ( t ) for all t ∈ [ y i , y i + (cid:15) ).This contradicts that the open intervals ( x i , y i ) are disjoint. Recall also that x i (cid:54) = y i for all i . Thus, we conclude that y i = t .This proves (4.38).Step 3For each i ∈ Λ, we apply (6.21) to the time interval ( x i , y i ). Note that we can do thisbecause by (4.39), h ( x i ) = h n ( x i ) whenever x i (cid:54) = 0. This gives, TABILITY FOR THE RIEMANN PROBLEM 27 (4.41) y i (cid:90) x i (cid:20) q ( u ( h ( t )+ , t ); ¯ u (( h ( t ) + X ( t ))+ , t )) − q ( u ( h ( t ) − , t ); ¯ u (( h ( t ) + X ( t )) − , t ))+ ˙ h ( t ) η ( u ( h ( t ) − , t ) | ¯ u (( h ( t ) + X ( t )) − , t )) − ˙ h ( t ) η ( u ( h ( t )+ , t ) | ¯ u (( h ( t ) + X ( t ))+ , t )) (cid:21) dt ≥ h ( y i ) (cid:90) h ( y i ) η ( u ( x, y i ) | ¯ u ( x + X ( y i ) , y i )) dx − h ( x i ) (cid:90) h ( x i ) η ( u ( x ) | ¯ u ( x )) dx + y i (cid:90) x i h ( t ) (cid:90) h ( t ) (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ∇ η (¯ u ( x, t )) (cid:33) f ( u ( x, t ) | ¯ u ( x + X ( t ) , t ))+ (cid:32) ∂ x (cid:12)(cid:12)(cid:12)(cid:12) ( x + X ( t ) ,t ) ¯ u T ( x, t ) ˙ X ( t ) (cid:33) ∇ η (¯ u ( x + X ( t ) , t ))[ u ( x, t ) − ¯ u ( x + X ( t ) , t )] dxdt, where when we write the term h ( x i ) (cid:90) h ( x i ) η ( u ( x ) | ¯ u ( x )) dx (4.42)we have again used that h ( x i ) = h ( x i ) for x i (cid:54) = 0, in which case this term vanishes.We then sum both sides of the inequality (4.41) over all i ∈ Λ. Recall that the set { t | h ( t ) = h ( t ) } (4.43)has measure zero. Recall also (4.38) and (4.39). Further, recall that the intervals ( x i , y i )are disjoint. Lastly, note that terms of the form h ( t ) (cid:90) h ( t ) η ( u | ¯ u ) dx (4.44)equal zero when h ( t ) = h ( t ).This proves (4.34). (cid:3) Construction of the shift
In this section, we prove
Proposition 5.1 (Existence of the shift functions) . Fix
T > . Assume u is a weaksolution to (1.1) . Assume u is entropic for the entropy η , and u has strong traces (Defini-tion 2.1). Let ( u L, , u R, , σ ( u L, , u R, )) be a 1-shock verifying the hypotheses ( H ) and let ( u L,n , u
R,n , σ n ( u L,n , u
R,n )) be an n-shock verifying the hypotheses ( H ) ∗ .Assume also that there exists ρ > such that r i > ρ, (5.1) for i = 1 and i = n and where r i satisfies S u L,i ( r i ) = u R,i .Then, there exist positive constants a , ∗ , a n, ∗ such that for all a ∈ (0 , a , ∗ ) and all a n ∈ ( a n, ∗ , ∞ ) , there are Lipschitz continuous maps h , h n : [0 , T ) → R with h (0) = h n (0) = 0 such that for almost every t , (5.2) a (cid:0) q ( u ; u R, ) − ˙ h ( t ) η ( u | u R, ) (cid:1) − q ( u − ; u L, ) + ˙ h ( t ) η ( u − | u L, ) ≤− c (cid:12)(cid:12)(cid:12) σ ( u L, , u R, ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) , and (5.3) 1 a n (cid:0) q ( u n + ; u R,n ) − ˙ h n ( t ) η ( u n + | u R,n ) (cid:1) − q ( u n − ; u L,n ) + ˙ h n ( t ) η ( u n − | u L,n ) ≤− c n (cid:12)(cid:12)(cid:12) σ n ( u L,n , u
R,n ) − ˙ h n ( t ) (cid:12)(cid:12)(cid:12) , where u i ± := u ( u ( h i ( t ) ± , t ) for i = 1 , n . The constants c i > depend on (cid:107) u (cid:107) L ∞ , ρ , (cid:12)(cid:12) u R,i (cid:12)(cid:12) , (cid:12)(cid:12) u L,i (cid:12)(cid:12) , and a i . The constants a i, ∗ depend on (cid:107) u (cid:107) L ∞ , (cid:12)(cid:12) u L,i (cid:12)(cid:12) , (cid:12)(cid:12) u R,i (cid:12)(cid:12) , and (cid:12)(cid:12) u R,i − u L,i (cid:12)(cid:12) , for i = 1 , n .For each t ∈ [0 , T ) either ˙ h ( t ) < inf λ or ( u , u − , ˙ h ) is a 1-shock with u − ∈ { u | η ( u | u L, ) ≤ a η ( u | u R, ) } (possibly u = u − and ˙ h = λ ( u ± ) ). Similarly, for each t ∈ [0 , T ) either ˙ h n ( t ) > sup λ n or ( u n + , u n − , ˙ h n ) is an n-shock with u n + ∈ { u | η ( u | u R,n ) ≤ a n η ( u | u L,n ) } (pos-sibly u n + = u n − and ˙ h n = λ n ( u n ± ) ).Moreover, h ( t ) ≤ h n ( t )(5.4) for all t ∈ [0 , T ) . The proof of Proposition 5.1 is based on the following Lemma proved in [21].
Lemma 5.2 (from [21]) . Assume the hypotheses ( H ) hold.Let B, ρ > . Then there exists a constant a ∗ ∈ (0 , depending on B and ρ such thatthe following is true:For any a ∈ (0 , a ∗ ) , there exists a constant c depending on B , ρ , and a such that (5.5) a (cid:0) q ( S u ( s ); S u L ( s R )) − σ u ( s ) η ( S u ( s ) | S u L ( s R )) (cid:1) − q ( u ; u L ) + σ u ( s ) η ( u | u L ) ≤− c (cid:12)(cid:12)(cid:12) σ u L ( s R ) − σ u ( s ) (cid:12)(cid:12)(cid:12) , for all u L ∈ V with | u L | ≤ B , all u ∈ { u | η ( u | u L ) ≤ aη ( u | S u L ( s R )) } , any s ∈ [0 , B ] , and any s R ∈ [ ρ, B ] . TABILITY FOR THE RIEMANN PROBLEM 29
Moreover, (5.6) a (cid:0) q ( u ; S u L ( s R )) − λ ( u ) η ( u | S u L ( s R )) (cid:1) − q ( u ; u L ) + λ ( u ) η ( u | u L ) ≤ − c , for all u ∈ { u | η ( u | u L ) ≤ aη ( u | S u L ( s R )) } and for the same constant c . Lemma 5.2 follows from the proof of Lemma 4.3 in [17], but the proof of Lemma 5.2 (asproved in [21]) keeps careful track of the dependencies on the constants and makes sure inthe calculations to leave some extra negativity in the entropy dissipation lost at the shock( u L , u R , σ L,R ) (thus we have a negative right hand side in our (5.5) and (5.6)). The ideaof creating negative entropy dissipation is related to the previous works [18, 23, 21].The proof of Lemma 5.2 is powered by Lemma 5.3:
Lemma 5.3 (from [21]) . Assume the system (1.1) satisfies the hypothesis ( H . Fix B, ρ > . Then there exists k, δ > depending on B and ρ such that for any δ ∈ (0 , δ ] , u ∈ V with | u | ≤ B and for any s ∈ ( ρ, B ) and s ≥ , (5.7) q ( S u ( s ); S u ( s )) − σ u ( s ) η ( S u ( s ) | S u ( s )) ≤ − k (cid:12)(cid:12)(cid:12) σ u ( s ) − σ u ( s ) (cid:12)(cid:12)(cid:12) , for | s − s | < δ,q ( S u ( s ); S u ( s )) − σ u ( s ) η ( S u ( s ) | S u ( s )) ≤ − kδ (cid:12)(cid:12)(cid:12) σ u ( s ) − σ u ( s ) (cid:12)(cid:12)(cid:12) , for | s − s | ≥ δ. The formula (5.7) is a modification on a key lemma due to DiPerna [14]. The proof ofLemma 5.3 in [21] is based on the proof of a very similar result in [17, p. 387-9]. The proofin [21] modifies the proof in [17, p. 387-9] – being careful to keep the constants k and δ uniform in s and u .5.1. Proof of Proposition 5.1.
The proof of (5.2) and (5.3) is based on the work [21].The result (5.4) is a novel contribution.Proof of (5.2)We will use Lemma 5.2. The 1-shock ( u L, , u R, , σ ( u L, , u R, )) in Proposition 5.1 willplay the role of ( u L , S u L ( s R )) in Lemma 5.2. Take R := max {(cid:107) u (cid:107) L ∞ , (cid:12)(cid:12) u L, (cid:12)(cid:12) } and then takethe ˜ S corresponding to this R as in Property (c) of ( H B in Lemma 5.2 to be B := max { R, ˜ S, (cid:12)(cid:12) u R, (cid:12)(cid:12) } . Then, we have that for all ( u − , u + , σ ) 1-shock with u − , u + < R ,there exists s ∈ (0 , B ) such that u + = S u − ( s ). Further, note that B depends on (cid:107) u (cid:107) L ∞ and (cid:12)(cid:12) u L, (cid:12)(cid:12) .Then, we will have a constant 0 < a , ∗ < a , ∗ is playing therole of the a ∗ in Lemma 5.2. Then, as in the statement of Proposition 5.1, we choose any a ∈ (0 , a , ∗ ).Throughout this proof, c denotes a generic constant that depends on (cid:107) u (cid:107) L ∞ , ρ , (cid:12)(cid:12) u R, (cid:12)(cid:12) , (cid:12)(cid:12) u L, (cid:12)(cid:12) , and a .Step 1We now show that for any γ > η ( u | u L ) − a η ( u | u R ) ≥ c γ (5.8) for a constant c >
0, where the infimum runs over all ( u, u L , u R ) such that dist( u, { w | η ( w | u L ) ≤ a η ( w | u R ) } ) ≥ γ and | u L | , | u R | ≤ B . Here, B is from Lemma 5.2 and the distancedist( x, A ) between a point x and a set A is defined in the usual way,dist( x, A ) := inf y ∈ A | x − y | . (5.9)Consider any triple ( u, u L , u R ) such that dist( u, { w | η ( w | u L ) ≤ a η ( w | u R ) } ) ≥ γ and | u L | , | u R | ≤ B .By Lemma 5.2, the set { w | η ( w | u L ) ≤ a η ( w | u R ) } is compact. Thus, there exists w ∈{ w | η ( w | u L ) ≤ a η ( w | u R ) } such that | u − w | = dist( u, { w | η ( w | u L ) ≤ a η ( w | u R ) } ) . (5.10)We Taylor expand the functionΓ( u ) := η ( u | u L ) − a η ( u | u R )(5.11)around the point w :Γ( u ) = Γ( w ) + ∇ Γ( w )( u − w ) + (cid:90) (1 − t )( u − w ) T ∇ Γ( w + t ( u − w ))( u − w ) dt. (5.12)By definition of w , we must have Γ( w ) = 0 and ∇ Γ( w )( u − w ) ≥ ∇ Γ = (1 − a ) ∇ η . Thus, by strict convexity of η and because 0 < a < ∇ Γ ≥ cI for some constant c > (cid:90) (1 − t )( u − w ) T ∇ Γ( w + t ( u − w ))( u − w ) dt (5.13) ≥ . (cid:90) (1 − t )( u − w ) T ∇ Γ( w + t ( u − w ))( u − w ) dt, (5.14)where we have changed the limits of integration. Continuing, ≥ . c | u − w | ≥ . cγ , (5.15)where the last inequality comes from dist( u, { w | η ( w | u L ) ≤ a η ( w | u R ) } ) ≥ γ . This proves(5.8).We choose γ := c L ∗ , (5.16)where c is from Lemma 5.2 and L ∗ is the Lipschitz constant of the map( u, u L , u R ) (cid:55)→ a (cid:0) q ( u ; u R ) − λ ( u ) η ( u | u R ) (cid:1) − q ( u ; u L ) + λ ( u ) η ( u | u L ) . (5.17)Step 2 TABILITY FOR THE RIEMANN PROBLEM 31
Define V ( u ) := λ ( u ) − C ∗ , { u | a η ( u | u R, ) <η ( u | u L, ) } ( u ) , (5.18)where C ∗ , > C ∗ , := 1 c γ (cid:32) sup u,u L ,u R ∈ B B (0) (cid:12)(cid:12) aq ( u ; u R ) − q ( u ; u L ) (cid:12)(cid:12) + 1 (cid:33) + 2 sup u ∈ B B (0) (cid:12)(cid:12) λ ( u ) (cid:12)(cid:12) , (5.19)where c is from (5.8).We solve the following ODE in the sense of Filippov flows, (cid:40) ˙ h ( t ) = V ( u ( h ( t ) , t )) h (0) = 0 , (5.20)The existence of such an h comes from the following lemma, Lemma 5.4 (Existence and ordering of Filippov flows) . For i = 1 , let V i ( u, t ) : R n × [0 , ∞ ) → R be bounded on R n × [0 , ∞ ) , upper semi-continuous in u , and measurable in t .Let u be a weak solution to (1.1) , entropic for the entropy η , and that takes values in acompact set K . Assume also that u verifies the strong trace property (Definition 2.1). Let x ∈ R . Then for i = 1 , we can solve (cid:40) ˙ g i ( t ) = V i ( u ( g i ( t ) , t ) , t ) g i (0) = x , (5.21) in the Filippov sense. That is, there exist Lipschitz functions g i : [0 , ∞ ) → R such thatLip [ g i ] ≤ (cid:107) V i (cid:107) L ∞ , (5.22) g i (0) = x , (5.23) ˙ g i ( t ) ∈ I [ V ( u i + , t ) , V ( u i − , t )] , (5.24) for almost every t , where u i ± := u ( g i ( t ) ± , t ) and I [ a, b ] denotes the closed interval withendpoints a and b .Moreover, for almost every t , f ( u i + ) − f ( u i − ) = ˙ g i ( u i + − u i − ) , (5.25) q ( u i + ) − q ( u i − ) ≤ ˙ g i ( η ( u i + ) − η ( u i − )) , (5.26) which means that for almost every t , either ( u i + , u i − , ˙ g i ) is an entropic shock (for η ) or u i + = u i − .Furthermore, if there exists µ > such that for all v ∈ K we have V ( v ) − V ( v ) ≥ µ, (5.27) then g and g satisfy g ( t ) ≥ g ( t ) for all t ∈ [0 , T ) . (5.28) The proof of (5.22), (5.23), and (5.24) is very similar to the proof of Proposition 1 in[27].It is well known that (5.25) and (5.26) are true for any Lipschitz continuous function g : [0 , ∞ ) → R when u is BV. When instead u is only known to have strong traces(Definition 2.1), then (5.25) and (5.26) are given in Lemma 6 in [27]. We do not prove(5.25) and (5.26) here; their proof is in the appendix in [27].The result (5.28) is a new result about Filippov flows novel to this article.The proof of (5.28) is in Section 5.2. Moreover, for completeness, the proofs of (5.22),(5.23) and (5.24) are also in Section 5.2.Note that V (see (5.18)) is upper semi-continuous in u because indicator functions ofopen sets are lower semi-continuous and the negative of a lower semi-continuous functionis upper semi-continuous.Step 3Let u ± := u ( u ( h ( t ) ± , t ).Note that by Lemma 5.4,˙ h ( t ) ∈ I (cid:34) λ ( u ) − C ∗ , { u | a η ( u | u R, ) <η ( u | u L, ) } ( u ) , (5.29) λ ( u − ) − C ∗ , { u | a η ( u | u R, ) <η ( u | u L, ) } ( u − ) (cid:35) . (5.30)We are now ready to show (5.2).For each fixed time t , we have 4 cases to consider to prove (5.2): Case 1 a η ( u − | u R, ) < η ( u − | u L, ) , (5.31) a η ( u | u R, ) < η ( u | u L, ) . (5.32) Case 2 a η ( u − | u R, ) < η ( u − | u L, ) , (5.33) a η ( u | u R, ) ≥ η ( u | u L, ) . (5.34) Case 3 a η ( u − | u R, ) ≥ η ( u − | u L, ) , (5.35) a η ( u | u R, ) < η ( u | u L, ) . (5.36) Case 4 a η ( u − | u R, ) ≥ η ( u − | u L, ) , (5.37) a η ( u | u R, ) ≥ η ( u | u L, ) . (5.38)Note that we allow for u = u − .We start with Case 1
TABILITY FOR THE RIEMANN PROBLEM 33
In this case, by (5.24), (5.19), and (5.29) we know that(5.39) ˙ h ( t ) ≤ − c γ (cid:32) sup u,u L ,u R ∈ B B (0) (cid:12)(cid:12) aq ( u ; u R ) − q ( u ; u L ) (cid:12)(cid:12) + 1 (cid:33) − sup u ∈ B B (0) (cid:12)(cid:12) λ ( u ) (cid:12)(cid:12) < inf u ∈ B B (0) λ ( u ) . If u (cid:54) = u − , then we have (5.25) and (5.26). But then (5.39) contradicts ( H u = u − .Let v := u = u − .If dist( v, { w | η ( w | u L, ) ≤ a η ( w | u R, ) } ) ≥ γ , then(5.40) a (cid:18) q ( u ; u R, ) − ˙ h ( t ) η ( u | u R, ) (cid:19) − q ( u − ; u L, ) + ˙ h ( t ) η ( u − | u L, )= a (cid:18) q ( v ; u R, ) − ˙ h ( t ) η ( v | u R, ) (cid:19) − q ( v ; u L, ) + ˙ h ( t ) η ( v | u L, )= aq ( v ; ¯ u + ( t )) − q ( v ; ¯ u − ( t )) − ˙ h ( t ) (cid:0) aη ( v | ¯ u + ( t )) − η ( v | ¯ u − ( t )) (cid:1) ≤ − , because of (5.39) and (5.8). Because the term (cid:12)(cid:12)(cid:12) σ ( u L, , u R, ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) on the right handside of (5.2) is bounded due to (5.22), we have proven (5.2) by choosing c sufficiently small.If on the other hand, dist( v, { w | η ( w | u L, ) ≤ a η ( w | u R, ) } ) < γ , then(5.41) a (cid:18) q ( u ; u R, ) − ˙ h ( t ) η ( u | u R, ) (cid:19) − q ( u − ; u L, ) + ˙ h ( t ) η ( u − | u L, )= a (cid:18) q ( v ; u R, ) − ˙ h ( t ) η ( v | u R, ) (cid:19) − q ( v ; u L, ) + ˙ h ( t ) η ( v | u L, )= aq ( v ; ¯ u + ( t )) − q ( v ; ¯ u − ( t )) − ˙ h ( t ) (cid:0) aη ( v | ¯ u + ( t )) − η ( v | ¯ u − ( t )) (cid:1) ≤ a (cid:18) q ( v ; u R, ) − λ ( v ) η ( v | u R, ) (cid:19) − q ( v ; u L, ) + λ ( v ) η ( v | u L, ) , because η ( v | u L, ) − a η ( v | u R, ) ≥ h ≤ − sup u ∈ B B (0) (cid:12)(cid:12) λ ( u ) (cid:12)(cid:12) . Continuing,we get ≤ − c , from (5.6), the definition of γ (5.16), the assumption that dist( v, { w | η ( w | u L, ) ≤ a η ( w | u R, ) } ) <γ and the assumption that r ≥ ρ . Again because the term (cid:12)(cid:12)(cid:12) σ ( u L, , u R, ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) onthe right hand side of (5.2) is bounded due to (5.22), we have proven (5.2) by choosing c sufficiently small. Note c will depend on ρ . Case 2
In this case, we must have u − (cid:54) = u . Recall also that (1.1) is hyperbolic. Further-more, we have from (5.24) that ˙ h ∈ (cid:34) − c γ (cid:32) sup u,u L ,u R ∈ B B (0) (cid:12)(cid:12) aq ( u ; u R ) − q ( u ; u L ) (cid:12)(cid:12) +1 (cid:33) − sup u ∈ B B (0) (cid:12)(cid:12) λ ( u ) (cid:12)(cid:12) , λ ( u ) (cid:35) . However, this implies that ( u , u − , ˙ h ) is a right 1-contact discontinuity (see [13, p. 274]). This contradicts the hypothesis ( H
2) on the shock( u + , u − , ˙ h ), which is entropic for η because of (5.25) and (5.26). The hypothesis ( H Case 2 ) cannotactually occur.
Case 3
In this case, we have from (5.24) that˙ h ∈ (cid:34) − c γ (cid:32) sup u,u L ,u R ∈ B B (0) (cid:12)(cid:12) aq ( u ; u R ) − q ( u ; u L ) (cid:12)(cid:12) + 1 (cid:33) − sup u ∈ B B (0) (cid:12)(cid:12) λ ( u ) (cid:12)(cid:12) , λ ( u − ) (cid:35) . (5.42)By the hypothesis ( H u , u − , ˙ h ) must be a1-shock. Also, u − verifies a η ( u − | u R, ) ≥ η ( u − | u L, ). Thus, we can apply Lemma 5.2.Recall that r > ρ (see (5.1)). We receive (5.2). Case 4
In this case, we have from (5.24) that ˙ h ∈ I [ λ ( u ) , λ ( u − )]. Then, by the hypothesis( H I [ λ ( u + ) , λ ( u − )] = ( λ ( u − ) , λ ( u + ))(5.43)because then (5.43) would imply that ( u , u − , ˙ h ) is a right 1-contact discontinuity. How-ever, ( H
2) prevents right 1-contact discontinuities. Recall ( H u , u − , ˙ h )is a 1-shock. Moreover, u − verifies a η ( u − | u R, ) ≥ η ( u − | u L, ). We can now applyLemma 5.2. Recall that r > ρ (see (5.1)). This gives (5.2).Proof of (5.3)To prove (5.3), note that if v ( x, t ) solves (1.1), then v ( − x, t ) will solve v t + ( − f ( v )) x = 0 , (5.44)where we have replaced the flux f with − f .The n th characteristic family of (1.1) corresponds to the first characteristic family of(5.44). Thus to prove (5.3) we simply apply (5.2) to the system (5.44).Define V n ( u ) := λ n ( u ) + C ∗ ,n { u | a n η ( u | u L,n ) <η ( u | u R,n ) } ( u ) . (5.45) TABILITY FOR THE RIEMANN PROBLEM 35
Note that the shift function h n (from (5.3)) will solve the following ODE in the sense ofFilippov flows, (cid:40) ˙ h n ( t ) = V ( u ( h n ( t ) , t )) h n (0) = 0 , (5.46)for a large constant C ∗ ,n > λ n as usual refers to the n th characteristic familyof (1.1).Let K be a compact set which contains the range of u (note by assumption u is bounded).Then due to the strict hyperbolicity of (1.1), there is θ > λ n ( v ) − λ ( v ) ≥ θ (5.47)for all v ∈ K .Then, (5.47) along with (5.18) and (5.45) imply that V and V n satisfy (5.27) for some µ . Then (5.28) implies (5.4).This completes the proof of Proposition 5.1.5.2. Proof of Lemma 5.4.
Proof of (5.22) , (5.23) , and (5.24)The following proof of (5.22), (5.23), and (5.24) is based on the proof of Proposition 1in [27], the proof of Lemma 2.2 in [31], and the proof of Lemma 3.5 in [22]. We do notprove (5.25) or (5.26) here; these properties are in Lemma 6 in [27], and their proofs arein the appendix in [27].For i = 1 , v i,n ( x, t ) := (cid:90) V i (cid:18) u ( x + yn , t ) , t (cid:19) dy. (5.48)Let g i,n be the solution to the ODE: (cid:40) ˙ g i,n ( t ) = v i,n ( g i,n ( t ) , t ) , for t > g i,n (0) = x . (5.49)The v i,n are uniformly bounded in n because by assumption V i is bounded ( (cid:13)(cid:13) v i,n (cid:13)(cid:13) L ∞ ≤(cid:107) V i (cid:107) L ∞ ). The v i,n are measurable in t , and due to the mollification by n are also Lipschitzcontinuous in x . Thus (5.49) has a unique solution in the sense of Carath´eodory.The g i,n are Lipschitz continuous with Lipschitz constants uniform in n , due to the v i,n being uniformly bounded in n . Thus, by Arzel`a–Ascoli the g i,n converge in C (0 , T ) for anyfixed T > g i (passing to a subsequence if necessary).Note that ˙ g i,n converges in L ∞ weak* to ˙ g i .We define V i, max( t ) := max { V ( u i − , t ) , V ( u i + , t ) } , (5.50) V i, min( t ) := min { V ( u i − , t ) , V ( u i + , t ) } , (5.51)where u i ± := u ( g i ( t ) ± , t ).To show (5.24), we will first prove that for almost every t > n →∞ [ ˙ g i,n ( t ) − V i, max( t )] + = 0 , (5.52) lim n →∞ [ V i, min( t ) − ˙ g i,n ( t )] + = 0 , (5.53)where [ · ] + := max(0 , · ).The proofs of (5.52) and (5.53) are similar; we only show the first one.[ ˙ g i,n ( t ) − V i, max( t )] + (5.54) = (cid:34) (cid:90) V i (cid:18) u ( g i,n ( t ) + yn , t ) , t (cid:19) dy − V i, max( t ) (cid:35) + (5.55) = (cid:34) (cid:90) V i (cid:18) u ( g i,n ( t ) + yn , t ) , t (cid:19) − V i, max( t ) dy (cid:35) + (5.56) ≤ (cid:90) (cid:104) V i (cid:18) u ( g i,n ( t ) + yn , t ) , t (cid:19) − V i, max( t ) (cid:105) + dy (5.57) ≤ ess sup y ∈ (0 , n ) (cid:104) V i (cid:18) u ( g i,n ( t ) + y, t ) , t (cid:19) − V i, max( t ) (cid:105) + (5.58) ≤ ess sup y ∈ ( − (cid:15) i,n ,(cid:15) i,n ) (cid:104) V i (cid:18) u ( g i ( t ) + y, t ) , t (cid:19) − V i, max( t ) (cid:105) + , (5.59)where (cid:15) i,n := (cid:12)(cid:12) g i,n ( t ) − g i ( t ) (cid:12)(cid:12) + n . Note (cid:15) i,n → + .Fix a t ≥ u has a strong trace in the sense of Definition 2.1. Then becausethe map u (cid:55)→ V i ( u, t ) is upper semi-continuous,lim n →∞ ess sup y ∈ (0 , n ) (cid:104) V i (cid:18) u ( g i ( t ) ± y, t ) , t (cid:19) − V i (cid:0) u i ± , t (cid:1)(cid:105) + = 0 , (5.60)where u i ± := u ( g i ( t ) ± , t ). Recall that the map u (cid:55)→ V i ( u, t ) being upper semi-continuous atthe point u means that lim sup u → u V i ( u, t ) ≤ V i ( u , t ) . (5.61)From (5.60), we getlim n →∞ ess sup y ∈ (0 , n ) (cid:104) V i (cid:18) u ( g i ( t ) ± y, t ) , t (cid:19) − V i, max( t ) (cid:105) + = 0 . (5.62) TABILITY FOR THE RIEMANN PROBLEM 37
We can control (5.59) from above by the quantity(5.63) ess sup y ∈ ( − (cid:15) i,n , (cid:104) V i (cid:18) u ( g i ( t ) + y, t ) , t (cid:19) − V i, max( t ) (cid:105) + +ess sup y ∈ (0 ,(cid:15) i,n ) (cid:104) V i (cid:18) u ( g i ( t ) + y, t ) , t (cid:19) − V i, max( t ) (cid:105) + . By (5.62), we have that (5.63) goes to 0 as n → ∞ . This proves (5.52).Recall that ˙ g i,n converges in L ∞ weak* to ˙ g i . Thus, due to the convexity of the function[ · ] + , T (cid:90) [ ˙ g i ( t ) − V i, max( t )] + dt ≤ lim inf n →∞ T (cid:90) [ ˙ g i,n ( t ) − V i, max( t )] + dt. (5.64)By the dominated convergence theorem and (5.52),lim inf n →∞ T (cid:90) [ ˙ g i,n ( t ) − V i, max( t )] + dt = 0 . (5.65)We conclude, T (cid:90) [ ˙ g i ( t ) − V i, max( t )] + dt = 0 . (5.66)From a similar argument, T (cid:90) [ V i, min( t ) − ˙ g i ( t )] + dt = 0 . (5.67)This proves (5.24). Proof of (5.28)Let us first explain the idea behind the proof of (5.28). We use the fact that, for a fixed t , according to (5.48) and (5.49), the value of ˙ g i,n ( t ) is based on the value of u ( x, t ) for x ∈ [ g i,n ( t ) , g i,n ( t ) + n ]. Then, if the values of g ,n ( t ) and g ,n ( t ) are close enough together(see (5.70) below), the domain of u ( · , t ) used to calculate ˙ g ,n ( t ) and the domain of u ( · , t )used to calculate ˙ g ,n ( t ) (according to (5.48) and (5.49)) will have some overlap. On thisoverlap, the estimate (5.27) says that V ( u ( · , t )) − V ( u ( · , t )) > µ. (5.68)Thus, when the value of g ,n ( t ) and g ,n ( t ) are close enough together, the estimate (5.68)allows us to compensate for the lack of control we have for the parts of the domain of u ( · , t )which are not overlapping, and we find that whenever g ,n ( t ) and g ,n ( t ) are close enoughtogether, the difference ˙ g ,n − ˙ g ,n must be strictly positive (see (5.73)). This means thatwhenever g ,n and g ,n get close together, they start being pushed apart. This, combined Figure 2.
The idea for the proof of (5.28).with the the identical starting values g ,n (0) = g ,n (0) = x , yields (5.28) in the n → ∞ limit. See Figure 2.We now give the proof.Fix n ∈ N .Define M := max i ∈{ , } (cid:107) V i (cid:107) L ∞ . (5.69)Assume that for some t ∗ , (cid:12)(cid:12) g ,n ( t ∗ ) − g ,n ( t ∗ ) (cid:12)(cid:12) < µn ( µ + 4 M ) . (5.70)Recall that due to g ,n and g ,n solving (5.49) in the sense of Carath´eodory, for i = 1 , g i,n ( t ) = v i,n ( g i,n ( t ) , t ) , (5.71)for almost every t . TABILITY FOR THE RIEMANN PROBLEM 39
Then if g ,n and g ,n also satisfy the differential equation (5.71) at this time t ∗ , then wehave(5.72)˙ g ,n ( t ∗ ) − ˙ g ,n ( t ∗ )= v ,n ( g ,n ( t ∗ ) , t ∗ ) − v ,n ( g ,n ( t ∗ ) , t ∗ )= (cid:90) V (cid:18) u ( g ,n ( t ∗ ) + yn , t ∗ ) , t ∗ (cid:19) dy − (cid:90) V (cid:18) u ( g ,n ( t ∗ ) + yn , t ∗ ) , t ∗ (cid:19) dy. Then from (5.27), (5.69) and (5.70), we have(5.73) (cid:90) V (cid:18) u ( g ,n ( t ∗ ) + yn , t ∗ ) , t ∗ (cid:19) dy − (cid:90) V (cid:18) u ( g ,n ( t ∗ ) + yn , t ∗ ) , t ∗ (cid:19) dy ≥ M µµ + 4
M .
Thus, from (5.72) and (5.73) and the fundamental theorem of calculus for W , functionswe get that for any t , t ∈ [0 , T ) verifying t < t and such that condition (5.70) holds forall t ∗ ∈ [ t , t ], g ,n ( t ) − g ,n ( t ) < g ,n ( t ) − g ,n ( t ) . (5.74)Because g ,n (0) = g ,n (0) = x , (5.74) implies that g ,n ( t ) ≤ g ,n ( t )(5.75)for all t ∈ [0 , T ).Thus in the n → ∞ limit, from (5.75) we in fact get (5.28).6. Proofs of Theorem 6.1 and Theorem 6.2
Theorem 6.1: L Stability for the Riemann Problem with Extremal ShocksVerifying Strong Form of Lax’s E-condition.Theorem 6.1 ( L Stability for the Riemann Problem with Extremal Shocks VerifyingStrong Form of Lax’s E-condition) . Fix
T > . Assume u, ¯ v ∈ L ∞ ( R × [0 , T )) are solutionsto the system (1.1) . Assume that u and ¯ v are entropic for the entropy η . Further, assumethat u has strong traces (Definition 2.1).Assume also that ¯ v is a solution to the Riemann problem (1.13) and that ¯ v has the form (1.14) . If ¯ v contains a 1-shock, assume the hypotheses ( H ) hold. Likewise, if ¯ v containsan n-shock, assume the hypotheses ( H ) ∗ hold.Assume that ¯ v contains at least one rarefaction wave, and if there are any shocks in ¯ v they are either a 1-shock verifying (1.11) or an n-shock verifying (1.11) .Also assume (1.12) holds. Further, assume the system (1.1) has at least two conservedquantities ( n ≥ ). Then there exists Ψ ¯ v with Property ( D ) and verifying the following stability estimate: R (cid:90) − R (cid:12)(cid:12) u ( x, t ) − Ψ ¯ v ( x, t ) (cid:12)(cid:12) dx ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (6.1) for all t , R > verifying t ∈ (0 , R ) and R > max i { Lip [ h i ] } t , (6.2) where the max runs over the i-shock families contained in ¯ v (1-shocks and/or n-shocks)and the h i are in the context of Property ( D ) .We also have the following L -type control on the shift functions h i : t (cid:90) (cid:88) i (cid:12)(cid:12)(cid:12) σ i (¯ v i , ¯ v i +1 ) − ˙ h i ( t ) (cid:12)(cid:12)(cid:12) dt ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (6.3) where the sum runs over the i-shock families contained in ¯ v (1-shocks and/or n-shocks).If ¯ v contains an i-shock, the constants µ , µ > depend on (cid:107) u (cid:107) L ∞ , | ¯ v i | , | ¯ v i +1 | , and | ¯ v i − ¯ v i +1 | . Further, µ , and µ depend on bounds on the second derivative of η on therange of u and ¯ v . In addition, if ¯ v contains an i-shock, µ depends on sup (cid:107)∇ λ i (cid:107) (wherethe supremum runs over the range of u and ¯ v ), and ( λ i +1 (¯ v i +1 ) − λ i (¯ v i )) .Proof. We assume that in ¯ v there is both a 1-shock and an n-shock (in addition to ararefaction fan). The three cases when there is only a 1-shock, only an n-shock, or noshocks at all are all very similar and are left to the reader.We first focus on the shock connecting ¯ v to ¯ v . Label the shock speed σ (¯ v , ¯ v ).Let j be such that the leftward most rarefaction wave in ¯ v is a j-rarefaction wave.Then the j-rarefaction wave must be joining ¯ v j = ¯ v and ¯ v j +1 . We can write thisj-rarefaction wave as, ¯ v j if xt < λ j (¯ v j ) V j ( xt ) if λ j (¯ v j ) < xt < λ j (¯ v j +1 )¯ v j +1 if λ j (¯ v j +1 ) < xt , (6.4)for a function V j : R → R n .From (1.11), we get that λ (¯ v ) < λ (¯ v ). Then, from strict hyperbolicity of the system(1.1), we get λ (¯ v ) < λ (¯ v ) ≤ λ j (¯ v ) (recall it is possible j = 2). Define(6.5) L := sup | u |≤ B (cid:13)(cid:13) ∇ λ ( u ) (cid:13)(cid:13) ,(cid:15) := λ (¯ v ) − λ (¯ v )2 L , where B verifies (cid:107) u (cid:107) L ∞ , (cid:107) ¯ v (cid:107) L ∞ ≤ B . Note L exists by the remarks after the hypotheses( H ) and ( H ) ∗ . TABILITY FOR THE RIEMANN PROBLEM 41
From Proposition 5.1, we can find a positive a such that a < (cid:15) C , (6.6)where C is the constant from Lemma 4.1, and from (5.2), we have a shift function h :[0 , T ) → R such that h (0) = 0 and(6.7) a (cid:0) q ( u ; ¯ v ) − ˙ h ( t ) η ( u | ¯ v ) (cid:1) − q ( u − ; ¯ v ) + ˙ h ( t ) η ( u − | ¯ v ) ≤− c (cid:12)(cid:12)(cid:12) σ (¯ v , ¯ v ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) , for all t ∈ [0 , T ) and where u ± := u ( u ( h ( t ) ± , t ).Note that from Lemma 4.1 and (6.6) we know that { u | η ( u | ¯ v ) ≤ a η ( u | ¯ v ) } ⊂ B (cid:15) (¯ v ).Furthermore, for each t ∈ [0 , T ) either ˙ h ( t ) < inf λ or ( u , u − , ˙ h ) is a 1-shock with u − ∈ { u | η ( u | ¯ v ) ≤ a η ( u | ¯ v ) } (possibly u = u − and ˙ h = λ ( u ± )). From (6.5) and (6.6),we have that λ ( u − ) ≤ λ (¯ v ) + λ (¯ v ) − λ (¯ v )2 < λ (¯ v ) . (6.8)Then, due to the hypothesis ( H h ( t ) ≤ λ ( u − ). Then because of (6.8),˙ h ( t ) ≤ λ ( u − ) < λ (¯ v )(6.9)for all t ∈ [0 , T ). Finally, recalling that λ (¯ v ) ≤ λ j (¯ v ) due to strict hyperbolicity of (1.1),we get ˙ h ( t ) ≤ λ ( u − ) < λ (¯ v ) ≤ λ j (¯ v )(6.10)for all t ∈ [0 , T ).We now consider the n-shock connecting ¯ v n to ¯ v n +1 .Let k be such that the rightward most rarefaction wave in the solution ¯ v is a k-rarefactionwave.Note first that the k -rarefaction wave joins ¯ v k and ¯ v k +1 = ¯ v n . Note k ≤ n −
1. We canwrite this k-rarefaction wave as, ¯ v k if xt < λ k (¯ v k ) V k ( xt ) if λ k (¯ v k ) < xt < λ k (¯ v n )¯ v n if λ k (¯ v n ) < xt , (6.11)for a function V k : R → R n .Following the same argument as above for the 1-shock, we get from (5.3), a function h n : [0 , T ) → R such that h n (0) = 0 and(6.12) 1 a n (cid:0) q ( u n + ; ¯ v n +1 ) − ˙ h n ( t ) η ( u n + | ¯ v n +1 ) (cid:1) − q ( u n − ; ¯ v n ) + ˙ h n ( t ) η ( u n − | ¯ v n ) ≤− c n (cid:12)(cid:12)(cid:12) σ n (¯ v n , ¯ v n +1 ) − ˙ h n ( t ) (cid:12)(cid:12)(cid:12) , for all t ∈ [0 , T ) and where u n ± := u ( u ( h n ( t ) ± , t ) and 0 < a n < t ∈ [0 , T ) either ˙ h n ( t ) > sup λ n or ( u n + , u n − , ˙ h n ) is an n-shock with u n + ∈ { u | η ( u | ¯ v n +1 ) ≤ a n η ( u | ¯ v n ) } (possibly u n + = u n − and ˙ h n = λ n ( u n ± )).We get (as an analogue of (6.10)),˙ h n ( t ) ≥ λ n ( u n + ) > λ n − (¯ v n ) ≥ λ k (¯ v n )(6.13)for all t ∈ [0 , T ).Define Ψ ¯ v : R × [0 , T ) → R n ,Ψ ¯ v ( x, t ) := ¯ v if x < h ( t )¯ v if h ( t ) < x < λ j (¯ v ) t ¯ v ( x, t ) if λ j (¯ v ) t < x < λ k (¯ v n ) t ¯ v n if λ k (¯ v n ) t < x < h n ( t )¯ v n +1 if h n ( t ) < x. (6.14)Note that h and h satisfy (1.15) and (1.17), respectively, and Ψ ¯ v is well-defined, dueto (6.10), (6.13), and the fundamental theorem of calculus for W , functions.Choose R > max { Lip[ h ] , Lip[ h n ] } t . (6.15)Define(6.16) h left ( t ) := − R + r ( t − t ) h right ( t ) := R − r ( t − t ) , where r > (cid:12)(cid:12) q ( a ; b ) (cid:12)(cid:12) ≤ rη ( a | b ) , (6.17)for a, b within the range of u and ¯ v . Note that r > η ( a | b ) and q ( a ; b ) bothbeing locally quadratic in a − b and η being strictly convex.Then, we use Lemma 4.2 three times with X ≡ h left and h and with the constant function¯ v (6.18)playing the role of the function ¯ u in Lemma 4.2. Note that h ( t ) − h left ( t ) > t dueto (6.15).We use Lemma 4.2 again with h and h n and with ¯ v if x < λ (¯ v ) t ¯ v ( x, t ) if λ (¯ v ) t < x < λ n − (¯ v n ) t ¯ v n if λ n − (¯ v n ) t < x (6.19)playing the role of ¯ u . TABILITY FOR THE RIEMANN PROBLEM 43
Note that h ( t ) − h n ( t ) > t >
0, because of (6.10) and (6.13). Note also that thefact that the solution ¯ v to the Riemann problem (1.13) exists means that λ (¯ v ) < λ n − (¯ v n ).Recall also the fundamental theorem of calculus for W , functions.Further, remark that (6.19) is Lipschitz continuous on R × (0 , ∞ ), due to the form ofthe rarefaction waves (6.4) and(6.11).Finally, we use Lemma 4.2 a third time, with h n and h right . Note that h right ( t ) − h n ( t ) > t due to (6.15). The constant function¯ v n +1 (6.20)plays the role of ¯ u .We now take a linear combination of the three applications of Lemma 4.2. Recall thespace derivative of constant states is zero, and otherwise we have (1.12). This yields,(6.21) t (cid:90) (cid:20) q ( u ( h left ( t )+ , t ); ¯ v ) − ˙ h left ( t ) η ( u ( h left ( t )+ , t ) | ¯ v ) − a a n (cid:0) q ( u ( h right ( t ) − , t ); ¯ v n +1 ) − ˙ h right ( t ) η ( u ( h right ( t ) − , t ) | ¯ v n +1 ) (cid:1) + a a n (cid:0) q ( u n + ; ¯ v n +1 ) − ˙ h n ( t ) η ( u n + | ¯ v n +1 ) (cid:1) − a (cid:0) q ( u n − ; ¯ v n ) + ˙ h n ( t ) η ( u n − | ¯ v n ) (cid:1) + a (cid:0) q ( u ; ¯ v ) − ˙ h ( t ) η ( u | ¯ v ) (cid:1) − q ( u − ; ¯ v ) + ˙ h ( t ) η ( u − | ¯ v ) (cid:21) dt ≥ (cid:34) h ( t ) (cid:90) h left ( t ) η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx + a h n ( t ) (cid:90) h ( t ) η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx + a a n h right ( t ) (cid:90) h n ( t ) η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx (cid:35) − (cid:34) h (0) (cid:90) h left (0) η ( u ( x ) | Ψ ¯ v ( x, dx + a h n (0) (cid:90) h (0) η ( u ( x ) | Ψ ¯ v ( x, dx + a a n h right (0) (cid:90) h n (0) η ( u ( x ) | Ψ ¯ v ( x, dx (cid:35) . Recall (6.7), (6.12), (6.16) and (6.17). In particular, note that ˙ h left = r and ˙ h right = − r .Then, we get from (6.21), (6.22) − t (cid:90) (cid:20) c n (cid:12)(cid:12)(cid:12) σ n (¯ v n , ¯ v n +1 ) − ˙ h n ( t ) (cid:12)(cid:12)(cid:12) + c (cid:12)(cid:12)(cid:12) σ (¯ v , ¯ v ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) (cid:21) dt + (cid:34) h (0) (cid:90) − R − rt η ( u ( x ) | Ψ ¯ v ( x, dx + a a n R + rt (cid:90) h n (0) η ( u ( x ) | Ψ ¯ v ( x, dx (cid:35) ≥ (cid:34) h ( t ) (cid:90) − R η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx + a h n ( t ) (cid:90) h ( t ) η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx + a a n R (cid:90) h n ( t ) η ( u ( x, t ) | Ψ ¯ v ( x, t )) dx (cid:35) . Note that we have also used that a h n (0) (cid:90) h (0) η ( u ( x ) | Ψ ¯ v ( x, dx = 0(6.23)due to h (0) = h n (0) = 0.Note that for i = 1 , n the constant c i depends on (cid:107) u (cid:107) L ∞ , a i , | ¯ v i − ¯ v i +1 | (by (2.10)), and | ¯ v i | , | ¯ v i +1 | .Recall Lemma 1.2, which says that due to the strict convexity of η , there exist constants c ∗ , c ∗∗ > c ∗ | a − b | ≤ η ( a | b ) ≤ c ∗∗ | a − b | , (6.24)for all a, b in a fixed compact set. Note that c ∗ , c ∗∗ depend on bounds on the secondderivative of η on the range of a and b .Recall also that a depends on (cid:107) u (cid:107) L ∞ , | ¯ v | , | ¯ v | , and | ¯ v − ¯ v | . Further, a n depends on (cid:107) u (cid:107) L ∞ , | ¯ v n | , | ¯ v n +1 | , and | ¯ v n − ¯ v n +1 | . Recall the relation (2.10).Thus, from (6.22) we can write, R (cid:90) − R (cid:12)(cid:12) u ( x, t ) − Ψ ¯ v ( x, t ) (cid:12)(cid:12) dx ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − Ψ ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (6.25)where the constant µ > (cid:107) u (cid:107) L ∞ , | ¯ v i | , | ¯ v i +1 | , and | ¯ v i − ¯ v i +1 | for both i = 1 and i = n . Further, µ depends on bounds on the second derivative of η on the range of u and¯ v . Recall also that due to (6.6), for both i = 1 and i = n , µ depends on sup (cid:107)∇ λ i (cid:107) (wherethe supremum runs over the range of u and ¯ v ), and ( λ i +1 (¯ v i +1 ) − λ i (¯ v i )).This gives us (6.1). Note Ψ ¯ v ( x,
0) = ¯ v ( x, t ). TABILITY FOR THE RIEMANN PROBLEM 45
We also get from (6.22), t (cid:90) (cid:20)(cid:12)(cid:12)(cid:12) σ n (¯ v n , ¯ v n +1 ) − ˙ h n ( t ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) σ (¯ v , ¯ v ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) (cid:21) dt ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − Ψ ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (6.26)where the constant µ > (cid:107) u (cid:107) L ∞ , | ¯ v i | , | ¯ v i +1 | , and | ¯ v i − ¯ v i +1 | for both i = 1 and i = n . Moreover, µ depends on bounds on the second derivative of η on the range of u and ¯ v . Note again Ψ ¯ v ( x,
0) = ¯ v ( x, t ). (cid:3) Theorem 6.2: L Stability for the Riemann Problem with Extremal Shocksbut No Rarefactions.
When ¯ v contains no rarefactions, then we do not require (1.11)or (1.12). We get the following stability result, Theorem 6.2 ( L Stability for the Riemann Problem with Extremal Shocks but No Rar-efactions) . Fix
T > . Assume u, ¯ v ∈ L ∞ ( R × [0 , T )) are solutions to the system (1.1) .Assume that u and ¯ v are entropic for the entropy η . Further, assume that u has strongtraces (Definition 2.1).Assume also that ¯ v is a solution to the Riemann problem (1.13) and that ¯ v has the form (1.14) . If ¯ v contains a 1-shock, assume the hypotheses ( H ) hold. Likewise, if ¯ v containsan n-shock, assume the hypotheses ( H ) ∗ hold.Assume ¯ v does not contain any rarefactions, and if ¯ v contains any shocks, they are eithera 1-shock or an n-shock.Assume the system (1.1) has at least two conserved quantities ( n ≥ ).Then there exists Ψ ¯ v with Property ( D ) and verifying the following stability estimate: R (cid:90) − R (cid:12)(cid:12) u ( x, t ) − Ψ ¯ v ( x, t ) (cid:12)(cid:12) dx ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (6.27) for all t , R > verifying t ∈ (0 , R ) and R > max i { Lip [ h i ] } t , (6.28) where the max runs over the i-shock families contained in ¯ v (1-shocks and/or n-shocks)and the h i are in the context of Property ( D ) .Moreover, there is L -type control on the shift functions h i : t (cid:90) (cid:88) i (cid:12)(cid:12)(cid:12) σ i (¯ v i , ¯ v i +1 ) − ˙ h i ( t ) (cid:12)(cid:12)(cid:12) dt ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (6.29) where the sum runs over the i-shock families contained in ¯ v (1-shocks and/or n-shocks).If ¯ v contains an i-shock, the constant µ > depends on (cid:107) u (cid:107) L ∞ , | ¯ v i | , | ¯ v i +1 | , and | ¯ v i − ¯ v i +1 | .Further, µ depends on bounds on the second derivative of η on the range of u and ¯ v . Proof.
There are three cases to consider: • ¯ v contains a 1-shock and an n-shock • ¯ v contains a 1-shock but no n-shock or ¯ v contains an n-shock but no 1-shock • ¯ v does not contain any shocksWe begin with the first case, Case ¯ v contains a 1-shock and an n-shock.From (5.2), we have a ∈ (0 ,
1) and a shift function h : [0 , T ) → R such that h (0) = 0and(6.30) a (cid:0) q ( u ; ¯ v ) − ˙ h ( t ) η ( u | ¯ v ) (cid:1) − q ( u − ; ¯ v ) + ˙ h ( t ) η ( u − | ¯ v ) ≤ − c (cid:12)(cid:12)(cid:12) σ (¯ v , ¯ v ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) , where u ± := u ( u ( h ( t ) ± , t ).Similarly, for the n-shock, we get from (5.3), the existence of an a n ∈ (0 ,
1) and a function h n : [0 , T ) → R such that h n (0) = 0 and(6.31) 1 a n (cid:0) q ( u n + ; ¯ v n +1 ) − ˙ h n ( t ) η ( u n + | ¯ v n +1 ) (cid:1) − q ( u n − ; ¯ v n ) + ˙ h n ( t ) η ( u n − | ¯ v n ) ≤ − c n (cid:12)(cid:12)(cid:12) σ n (¯ v n , ¯ v n +1 ) − ˙ h n ( t ) (cid:12)(cid:12)(cid:12) , where u n ± := u ( u ( h n ( t ) ± , t ). Note that by virtue of there not being any rarefactions,¯ v n = ¯ v .Note that from Proposition 5.1 we know that the constant a i depends on (cid:107) u (cid:107) L ∞ , | ¯ v i | , | ¯ v i +1 | ,and | ¯ v i − ¯ v i +1 | , for i = 1 , n . For i = 1 , n , the constant c i > (cid:107) u (cid:107) L ∞ , | ¯ v i | , | ¯ v i +1 | , | ¯ v i − ¯ v i +1 | , and a i .Choose R > max { Lip[ h ] , Lip[ h n ] } t . (6.32)Define(6.33) h left ( t ) := − R + r ( t − t ) h right ( t ) := R − r ( t − t ) , where r > (cid:12)(cid:12) q ( a ; b ) (cid:12)(cid:12) ≤ rη ( a | b ) , (6.34)for a, b within the range of u and ¯ v . Note that r > η ( a | b ) and q ( a ; b ) bothbeing locally quadratic in a − b and η being strictly convex.We use Lemma 4.2 once with X ≡ h left , and h and with the constant function¯ v (6.35)playing the role of the function ¯ u in Lemma 4.2. Note that h ( t ) − h left ( t ) > t dueto (6.32). TABILITY FOR THE RIEMANN PROBLEM 47
We also use Corollary 4.3 with h and h n and with¯ v (6.36)playing the role of ¯ u . Note that we can apply Corollary 4.3 with h and h n because byProposition 5.1, we know that for each t ∈ [0 , T ) either ˙ h ( t ) < inf λ or ( u , u − , ˙ h ) is a1-shock (including possibly u = u − and ˙ h = λ ( u ± )). Similarly, for each t ∈ [0 , T ) either˙ h n ( t ) > sup λ n or ( u n + , u n − , ˙ h n ) is an n-shock (including possibly u n + = u n − and ˙ h n = λ n ( u n ± )).By the hypotheses ( H ) and ( H ) ∗ , the assumptions necessary to apply Corollary 4.3 with h and h n are satisfied. The hypotheses ( H ) and ( H ) ∗ say that the speeds of 1-shocks andn-shocks and the characteristic speeds of the 1-family and n-family are well-separated.Furthermore, by virtue of Proposition 5.1 we know that h ( t ) ≤ h n ( t ) for all t . Inparticular, this gives (1.23).Finally, we use Lemma 4.2 a second time, with X ≡ h n , and h right . Note that h right ( t ) − h n ( t ) > t due to (6.32). The constant function¯ v n +1 (6.37)plays the role of ¯ u .We now take a linear combination of the two applications of Lemma 4.2 and the oneapplication of Corollary 4.3. Note the space derivative of constant states in ¯ v is zero. Thisyields,(6.38) t (cid:90) (cid:20) q ( u ( h left ( t )+ , t ); ¯ v ) − ˙ h left ( t ) η ( u ( h left ( t )+ , t ) | ¯ v ) − a a n (cid:0) q ( u ( h right ( t ) − , t ); ¯ v n +1 ) − ˙ h right ( t ) η ( u ( h right ( t ) − , t ) | ¯ v n +1 ) (cid:1) + a a n (cid:0) q ( u n + ; ¯ v n +1 ) − ˙ h n ( t ) η ( u n + | ¯ v n +1 ) (cid:1) − a (cid:0) q ( u n − ; ¯ v n ) + ˙ h n ( t ) η ( u n − | ¯ v n ) (cid:1) + a (cid:0) q ( u ; ¯ v ) − ˙ h ( t ) η ( u | ¯ v ) (cid:1) − q ( u − ; ¯ v ) + ˙ h ( t ) η ( u − | ¯ v ) (cid:21) dt ≥ (cid:34) h ( t ) (cid:90) h left ( t ) η ( u ( x, t ) | ¯ v ) dx + a h n ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ v ) dx + a a n h right ( t ) (cid:90) h n ( t ) η ( u ( x, t ) | ¯ v n +1 ) dx (cid:35) − (cid:34) h (0) (cid:90) h left (0) η ( u ( x ) | ¯ v ) dx + a h n (0) (cid:90) h (0) η ( u ( x ) | ¯ v ) dx + a a n h right (0) (cid:90) h n (0) η ( u ( x ) | ¯ v n +1 ) dx (cid:35) . Recall (6.30), (6.31), (6.33) and (6.34). In particular, note that ˙ h left = r and ˙ h right = − r .Then, we get from (6.38), (6.39) − t (cid:90) (cid:104) c (cid:12)(cid:12)(cid:12) σ (¯ v , ¯ v ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) + c n (cid:12)(cid:12)(cid:12) σ n (¯ v n , ¯ v n +1 ) − ˙ h n ( t ) (cid:12)(cid:12)(cid:12) (cid:105) dt + (cid:34) h (0) (cid:90) − R − rt η ( u ( x ) | ¯ v ) dx + a a n R + rt (cid:90) h n (0) η ( u ( x ) | ¯ v n +1 ) dx (cid:35) ≥ (cid:34) h ( t ) (cid:90) − R η ( u ( x, t ) | ¯ v ) dx + a h n ( t ) (cid:90) h ( t ) η ( u ( x, t ) | ¯ v ) dx + a a n R (cid:90) h n ( t ) η ( u ( x, t ) | ¯ v n +1 ) dx (cid:35) . Note that we have also used that a h n (0) (cid:90) h (0) η ( u ( x ) | ¯ v ) dx = 0(6.40)due to h (0) = h n (0) = 0.The reader will recall Lemma 1.2: by virtue of the strict convexity of η , there existconstants c ∗ , c ∗∗ > c ∗ | a − b | ≤ η ( a | b ) ≤ c ∗∗ | a − b | , (6.41)for all a, b in a fixed compact set. Note that c ∗ , c ∗∗ depend on bounds on the secondderivative of η on the range of a and b .Recall also that a depends on (cid:107) u (cid:107) L ∞ , | ¯ v | , | ¯ v | , and | ¯ v − ¯ v | (via the relation (2.10)).Similarly, a n depends on (cid:107) u (cid:107) L ∞ , | ¯ v n | , | ¯ v n +1 | , and | ¯ v n − ¯ v n +1 | (via the relation (2.10)).Thus, from (6.39) we can write, R (cid:90) − R (cid:12)(cid:12) u ( x, t ) − Ψ ¯ v ( x, t ) (cid:12)(cid:12) dx ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − Ψ ¯ v ( x, (cid:12)(cid:12)(cid:12) dx, (6.42)for a constant µ >
0. This gives us (6.27). Note Ψ ¯ v ( x,
0) = ¯ v ( x, t (cid:90) (cid:34)(cid:12)(cid:12)(cid:12) σ (¯ v , ¯ v ) − ˙ h ( t ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) σ n (¯ v n , ¯ v n +1 ) − ˙ h n ( t ) (cid:12)(cid:12)(cid:12) (cid:35) dt ≤ µ R + rt (cid:90) − R − rt (cid:12)(cid:12)(cid:12) u ( x ) − Ψ ¯ v ( x, (cid:12)(cid:12)(cid:12) dx. (6.43)This is (6.29). Note again Ψ ¯ v ( x,
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