Homogenization of nonlinear scalar conservation laws
aa r X i v : . [ m a t h . A P ] J un Homogenization of nonlinear s alar onservation lawsAnne-Laure DalibardO tober 24, 2018Abstra tWe study the limit as ε → of the entropy solutions of the equation ∂ t u ε + div x ˆ A ` xε , u ε ´˜ = 0 .We prove that the sequen e u ε two-s ale onverges towards a fun tion u ( t, x, y ) , and u is the uniquesolution of a limit evolution problem. The remarkable point is that the limit problem is not a s alar onservation law, but rather a kineti equation in whi h the ma ros opi and mi ros opi variablesare mixed. We also prove a strong onvergen e result in L lo .1 Introdu tionThis arti le is on erned with the asymptoti behavior of the sequen e u ε ∈ C ([0 , ∞ ) , L lo ( R N )) , as theparameter ε vanishes, where u ε is the entropy solution of the s alar onservation law ∂u ε ( t, x ) ∂t + N X i =1 ∂∂x i A i (cid:16) xε , u ε ( t, x ) (cid:17) = 0 t ≥ , x ∈ R N , (1) u ε ( t = 0) = u (cid:16) x, xε (cid:17) . (2)The fun tions A i = A i ( y, v ) ( y ∈ R N , v ∈ R ) are assumed to be Y -periodi , where Y = Π Ni =1 (0 , T i ) is the unit ell, and u is also assumed to be periodi in its se ond variable.Under regularity hypotheses on the (cid:29)ux, namely A ∈ W , ∞ per,lo ( R N +1 ) , and when the initial data u ε ( t = 0) belongs to L ∞ , it is known that there exists a unique entropy solution u ε of the above systemfor all ε > given (see [4, 16, 17, 26℄). The study of the homogenization of su h hyperboli s alar onservation laws has been investigated by several authors, see for instan e [9, 10, 11℄, and in the linear ase [14, 15℄. In dimension one, there is also an equivalen e with Hamilton-Ja obi equations whi h allowsto use the results of [18℄. In general, the results obtained by these authors an be summarized as follows:there exists a fun tion u = u ( t, x, y ) su h that u ε − u (cid:16) t, x, xε (cid:17) → in L lo ((0 , ∞ ) × R N ) . (3)The fun tion u ( t, x, y ) satis(cid:28)es a mi ros opi equation, alled ell problem, and an evolution equation,whi h is a s alar onservation law in whi h the oe(cid:30) ients depend on the mi ros opi variable y . Ingeneral, there is no (cid:16)de oupling(cid:17) of the ma ros opi variables t, x , and the mi ros opi variable y : theaverage of u with respe t to the variable y is not the solution of an (cid:16)average(cid:17) onservation law.To our knowledge, there are no results as soon as the dimension is stri tly greater than one whenthe (cid:29)ux does not satisfy a stru tural ondition of the type A ( y, ξ ) = a ( y ) g ( ξ ) . Here, we investigate thebehavior of the family u ε for arbitrary (cid:29)uxes. We prove that (3) still holds in some sense whi h will bepre ised later on, and the fun tion u is a solution of a mi ros opi ell problem. Pre isely, we provethat even though there is no simple evolution equation satis(cid:28)ed by the fun tion u itself, the fun tion f ( t, x, y, ξ ) = ξ
0) = 0 for all y ∈ R N .However, it is easily proved, using the identity f ε = g ε + ξ< , that f ε satis(cid:28)es (4), even when a N +1 ( y, does not vanish.We now de(cid:28)ne the limit system, whi h is reminis ent of equation (4) :De(cid:28)nition 1. Let f ∈ L ∞ ([0 , ∞ ) × R N × Y × R ) , u ∈ L ∞ ( R N × Y ) . We say that f is a generalized kineti solution of the limit problem, with initial data ξ su h that Supp
M ⊂ [0 , ∞ ) × R N × Y × [ − M, M ] , (6) f ( t, x, y, ξ ) = 1 if ξ < − M, (7) f ( t, x, y, ξ ) = 0 if ξ > M. (8)2. Mi ros opi equation for f : there exists a non-negative measure m ∈ M ((0 , ∞ ) × R N × Y × R ) su h that f is a solution in the sense of distributions of div y,ξ ( a ( y, ξ ) f ( t, x, y, ξ )) = ∂ ξ m, (9)and Supp m ⊂ [0 , ∞ ) × R N × Y × [ − M, M ] .3. Evolution equation: the ouple ( f, M ) is a solution in the sense of distributions of ∂ t f + N X i =1 a i ( y, ξ ) ∂ x i f = M ,f ( t = 0 , x, y, ξ ) = ξ
4. Conditions on f : there exists a non-negative measure ν ∈ M per ([0 , ∞ ) × R N × Y × R ) su
h that ∂ ξ f = − ν, (11) ≤ f ( t, x, y, ξ ) ≤ almost everywhere. (12)And for all
ompa
t set K ⊂ R N , τ Z τ || f ( s ) − f || L ( K × Y × R ) ds −→ τ → . (13)5. Condition on M : de(cid:28)ne the set G := { ψ ∈ L ∞ lo ( Y × R ) , ∂ ξ ψ ≥ , and ∃ µ ∈ M per ( Y × R ) , ∃ C > , ∃ α − ∈ R , div y,ξ ( aψ ) = − ∂ ξ µ, Supp µ ⊂ Y × [ − C, C ] , µ ≥ ,ψ ( y, ξ ) = α − if ξ < − C } . Then for all ϕ ∈ D ([0 , ∞ ) × R N ) su
h that ϕ ≥ , the fun
tion M ∗ t,x ϕ belongs to C ([0 , ∞ ) × R N , L ( Y × R )) , and ∀ ( t, x ) ∈ [0 , ∞ ) × R N , ∀ ψ ∈ G , Z Y × R ( M ∗ t,x ϕ ) ( t, x, · ) ψ ≤ . (14)We now state an existen
e and uniqueness result for solutions of the limit problem :Theorem 1. Let A ∈ W , ∞ per,lo ( Y × R ) .1. Existen
e: let u ∈ L lo ( R N ; C per ( Y )) ∩ L ∞ ( R N ) su
h that there exists a non-negative measure m = m ( x, y, ξ ) su
h that f ( x, y, ξ ) = ξ su
h that for all t > , for all R, R ′ > , || f ( t ) − g ( t ) || L ( B R × Y × R ) ≤ e Ct + R (cid:16) || u − v || L ( B R ′ × Y ) + e − R ′ (cid:17) . (17)As a
onsequen
e, for all u ∈ L ∞ ( R N × Y ) ∩ L lo ( R N , C per ( Y )) satisfying (15) and (16), there exists aunique generalized kineti
solution f ∈ L ∞ ([0 , ∞ ) × R N × Y × R ) of the limit problem.Remark 1. Noti
e that for any fun
tion v ∈ L ∞ ( Y ) , v is an entropy solution of the
ell problem div y A ( y, v ( y )) = 0 if and only if there exists a non-negative measure m ∈ M per ( Y × R ) su
h that the equation div y,ξ ( a ( y, ξ ) ξ
M, C ) ; re all that M and C are su h that Supp f ⊂ [0 , ∞ ) × R N × Y × [ − M, M ] , and ψ ( y, ξ ) = α − if ξ < − C . Integrating on [0 , ∞ ) × R N × [ − R, R ] , we obtain Z ∞ Z R N Z R − R f ε ( t, x, ξ ) h ∂ t θ ( t, x ) + a i (cid:16) xε , ξ (cid:17) ∂ x i θ ( t, x ) i ψ (cid:16) xε , ξ (cid:17) dx dξ dt − ε Z ∞ Z R N Z R − R f ε ( t, x, ξ ) ∂µ∂ξ (cid:16) xε , ξ (cid:17) θ ( t, x ) dx dξ dt + α − Z ∞ Z R N ε a N +1 (cid:16) xε , − R (cid:17) θ ( t, x ) dt dx = Z ∞ Z R N Z R − R m ε ( s, z, ξ ) ∂ ξ ψ (cid:16) xε , ξ (cid:17) dz dξ ds − Z R N Z R − R ξ
12e have (cid:20)Z ∞ Z R N ϕ ( t − s, x − z ) θ ( t, x ) dt dx (cid:21) K ( ξ ) ∂ ξ ψ δ (cid:16) zε , ξ (cid:17) ≥ . Moreover, thanks to (7), (8), and the assumptions on ψ and K , we have ∂ ξ K = 0 on Supp m ε , and (cid:20)Z ∞ Z R N ϕ ( t − s, x − z ) θ ( t, x ) dt dx (cid:21) ψ δ (cid:16) zε , ξ (cid:17) ∂ ξ K ( ξ ) f ε ( s, z, ξ )= α − (cid:20)Z ∞ Z R N ϕ ( t − s, x − z ) θ ( t, x ) dt dx (cid:21) ∂ ξ K ( ξ ) . Hen e, we obtain, for all ε, δ > , − Z f ε ( s, z, ξ ) ( ∂ t ϕ ( t − s, x − z ) + N X i =1 a i (cid:16) zε , ξ (cid:17) ∂ i ϕ ( t − s, x − z ) ) ×× ψ δ (cid:16) zε , ξ (cid:17) θ ( t, x ) dξ dx dz ds dt + 1 ε Z f ε ( s, z, ξ ) a (cid:16) zε , ξ (cid:17) · ∇ y,ξ ψ δ (cid:16) zε , ξ (cid:17) ϕ ( t − s, x − z ) θ ( t, x ) K ( ξ ) dt dx ds dz dξ + α − ε Z ϕ ( t − s, x − z ) θ ( t, x ) ∂ ξ K ( ξ ) a N +1 (cid:16) zε , ξ (cid:17) dt dx ds dz dξ ≥ . Following the formal al ulations above, we have to investigate the sign of the term Z f ε ( s, z, ξ ) a (cid:16) zε , ξ (cid:17) · ∇ y,ξ ψ δ (cid:16) zε , ξ (cid:17) ϕ ( t − s, x − z ) θ ( t, x ) K ( ξ ) dt dx ds dz dξ. Sin e div y,ξ ( aψ ) = − ∂ ξ µ , we have div y,ξ ( aψ δ ) = − ∂µ δ ∂ξ + r δ where µ δ = µ ∗ y ϕ δ ∗ ξ ϕ δ . Then − Z ∞ Z R N +1 f ε ( s, z, ξ ) ∂µ δ ∂ξ (cid:16) xε , ξ (cid:17) (cid:20)Z ∞ Z R N ϕ ( t − s, x − z ) θ ( t, x ) dt dx (cid:21) ds dz dξ = − Z ∞ Z R N +1 δ ( ξ = u ε ( t, x )) µ δ (cid:16) xε , ξ (cid:17) (cid:20)Z ∞ Z R N ϕ ( t − s, x − z ) θ ( t, x ) dt dx (cid:21) ds dz dξ ≤ . Hen e, we have to prove that as δ → , r δ → in L lo ( Y × R ) . The proof is quite lassi al. We have r δ ( y, ξ ) = a ( y, ξ ) ψ ∗ (cid:0) ∇ y,ξ ϕ δ ϕ δ (cid:1) − [ a ( y, ξ ) ψ ] ∗ (cid:0) ∇ y,ξ ϕ δ ϕ δ (cid:1) = N X i =1 Z [ a i ( y, ξ ) − a i ( y , ξ )] ψ ( y , ξ ) ∂ y i ϕ δ ( y − y ) ϕ δ ( ξ − ξ ) dy dξ + Z [ a N +1 ( y, ξ ) − a N +1 ( y , ξ )] ψ ( y , ξ ) ϕ δ ( y − y ) ∂ xi ϕ δ ( ξ − ξ ) dy dξ Thus, we ompute, for ( y, y , ξ, ξ ) ∈ R N +2 , ≤ i ≤ N + 1 , a i ( y, ξ ) − a i ( y , ξ ) = ( y − y ) · Z ∇ y a i ( τ y + (1 − τ ) y , τ ξ + (1 − τ ) ξ ) dτ +( ξ − ξ ) · Z ∂ ξ a i ( τ y + (1 − τ ) y , τ ξ + (1 − τ ) ξ ) dτ. ≤ k, i ≤ N , y ∈ R N , ξ ∈ R , φ k,i ( y, ξ ) = y k ∂ϕ ∂y i ( y ) ϕ ( ξ ) , φ k,N +1 ( y, ξ ) = y k ∂ϕ ∂ξ ( ξ ) ϕ ( y ) ,ζ i ( y, ξ ) = ξ ∂ϕ ∂y i ( y ) ϕ ( ξ ) , ζ N +1 ( y, ξ ) = ξ ∂ϕ ∂ξ ( ξ ) ϕ ( y ) . Noti e that Z R N +1 φ k,i = − δ k,i , Z R N +1 ζ i = − δ N +1 ,i . Then r δ ( y, ξ ) = N +1 X i =1 N X k =1 Z ∂a i ∂y k ( τ y + (1 − τ ) y , τ ξ + (1 − τ ) ξ ) ψ ( y , ξ ) φ δk,i ( y − y , ξ − ξ ) dy dξ dτ + N +1 X i =1 Z ∂a i ∂ξ ( τ y + (1 − τ ) y , τ ξ + (1 − τ ) ξ ) ψ ( y , ξ ) ζ δi ( y − y , ξ − ξ ) dy dξ dτ. Hen e as δ → , r δ onverges to − div y,ξ ( a ( y, ξ )) ψ ( y, ξ ) = 0 in L p lo ( R N +1 ) for any p < ∞ and for all ( t, x ) ∈ [0 , ∞ ) × R N . We now pass to the limit as δ → , with ε (cid:28)xed, and we obtain − Z f ε ( s, z, ξ ) n ∂ t ϕ ( t − s, x − z ) + a i (cid:16) zε , ξ (cid:17) ∂ i ϕ ( t − s, x − z ) o ψ (cid:16) zε , ξ (cid:17) θ ( t, x ) dξ dx dz ds dt − α − Z θ ( t, x ) ∂ ξ K ( ξ ) A (cid:16) zε , ξ (cid:17) · ∇ x ϕ ( t − s, x − z ) dt dx ds dz dξ ≥ . Passing to the limit as ε vanishes, we are led to − Z f ( s, z, y, ξ ) { ∂ t ϕ ( t − s, x − z ) + a i ( y, ξ ) ∂ i ϕ ( t − s, x − z ) } ψ ( y, ξ ) θ ( t, x ) dξ dx dz ds dy dt − α − Z θ ( t, x ) ∂ ξ K ( ξ ) A ( y, ξ ) · ∇ x ϕ ( t − s, x − z ) dt dx ds dy dz dξ ≥ . Sin e Z θ ( t, x ) ∇ x ϕ ( t − s, x − z ) dt dx ds dz = − (cid:18)Z θ ( t, x ) dt dx (cid:19) (cid:18)Z ∇ z ϕ ( s, z ) ds dz (cid:19) = 0 , we dedu e that Z f ( s, z, y, ξ ) ( ∂ t ϕ ( t − s, x − z ) + N X i =1 a i ( y, ξ ) ∂ i ϕ ( t − s, x − z ) ) ψ ( y, ξ ) θ ( t, x ) dξ dx dz ds dy dt ≤ , whi h means that f satis(cid:28)es ondition (14). There only remains to he k the strong ontinuity of f attime t = 0 .2.4 Strong ontinuity at time t = 0 The ontinuity property for f is inherited from uniform ontinuity properties at time t = 0 for thesequen e f ε . This is strongly linked to the well-preparedness of the initial data ( ondition (9)), that is,the fa t that for all x ∈ R N , u ( x, · ) is an entropy solution of the ell problem div y A ( y, u ( x, y )) = 0 . g δn = f ∗ x ρ n ∗ y ϕ δ ∗ ξ ϕ δ . with ρ n a onvolution kernel ( n ∈ N ), δ > , and ϕ δi de(cid:28)ned as in the previous subse tion. Then we anwrite N X i =1 a i (cid:16) xε , ξ (cid:17) · ∂∂x i h g δn (cid:16) x, xε , ξ (cid:17)i + 1 ε a N +1 (cid:16) xε , ξ (cid:17) ∂∂ξ g δn (cid:16) x, xε , ξ (cid:17) = 1 ε a (cid:16) xε , ξ (cid:17) · (cid:0) ∇ y,ξ g δn (cid:1) (cid:16) x, xε , ξ (cid:17) + N X i =1 a i (cid:16) xε , ξ (cid:17) (cid:18) ∂∂x i g δn (cid:19) (cid:16) x, xε , ξ (cid:17) (30) := r εn,δ . Noti e that ||∇ x g δn || L ∞ ( R N × Y × R ) ≤ ||∇ x ρ n || L ( R N ) , and a ( y, ξ ) ∇ y g δn ( x, y, ξ ) = ∂ ξ m δn + r δn , where m δn = m ∗ x ρ n ∗ y ϕ δ ∗ ξ ϕ δ ,r δn ( x, y, ξ ) = a ( y, ξ ) ∇ y g δn ( x, y, ξ ) − [ af ∗ x ρ n ] ∗ y,ξ ∇ y,ξ ϕ δ ( y ) ϕ δ ( ξ ) . Then for all n ∈ N , for all x ∈ R N , r δn vanishes as δ → in L lo ( Y × R ) and almost everywhere. Theproof of this fa t is exa tly the same as in the pre eding subse tion, and thus, we leave the details to thereader. As a onsequen e, r εn,δ ( x, ξ ) = 1 ε ∂ ξ m δn (cid:16) x, xε , ξ (cid:17) + R εn,δ ( x, ξ ) , and there exists a onstant C n , independent of ε , su h that for all n ∈ N , for all ε > , and for almostevery x, ξ lim sup δ → | R εn,δ ( x, ξ ) | ≤ C n . Moreover,
Supp R εn,δ ⊂ R N × [ − R − , R + 1] if δ < .Now, we multiply (4) by − g δn ( x, x/ε, ξ ) , and (30) by − f ε ( t, x, ξ ) . Setting h εn,δ ( t, x, ξ ) := f ε ( t, x, ξ ) h − g δn (cid:16) x, xε , ξ (cid:17)i + g δn (cid:16) x, xε , ξ (cid:17) [1 − f ε ( t, x, ξ )]= (cid:12)(cid:12)(cid:12) f ε ( t, x, ξ ) − g δn (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) + g δn (cid:16) x, xε , ξ (cid:17) − (cid:12)(cid:12)(cid:12) g δn (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) , we obtain ∂∂t h εn,δ ( t, x, ξ ) + N X i =1 a i (cid:16) xε , ξ (cid:17) ∂ x i h εn,δ ( t, x, ξ ) + 1 ε a N +1 (cid:16) xε , ξ (cid:17) ∂ ξ h εn,δ ( t, x, ξ ) == ∂m ε ∂ξ h − g δn (cid:16) x, xε , ξ (cid:17)i + 1 ε ∂ ξ m δn (cid:16) x, xε , ξ (cid:17) [1 − f ε ( t, x, ξ )] + R εn,δ ( x, ξ ) [1 − f ε ( t, x, ξ )] . (31)Noti
e that ∂ ξ [1 − f ε ( t, x, ξ )] = 2 δ ( ξ = u ε ( t, x )) ,∂∂ξ (cid:16) − g δn (cid:16) x, xε , ξ (cid:17)(cid:17) = 2 ν n,ε,δ ( x, ξ ) , where ν n,ε,δ is a non-negative fun
tion in C ∞ ( R N +1 ) , with support in R N × [ − M − , M + 1] if δ < .Noti
e also that f ε ( t, x, ξ ) − g δn ( x, x/ε, ξ ) = 0 if | ξ | is large enough ( | ξ | > M + 1 ). Take a
ut-o(cid:27) fun
tion15 = ζ ( x ) su
h that ζ ( x ) = e −| x | when | x | ≥ , and e ≤ ζ ( x ) ≤ for | x | ≤ . Then there exists a
onstant C su
h that |∇ x ζ ( x ) | ≤ Cζ ( x ) ∀ x ∈ R N . Hen
e, mutliplying (31) by ζ ( x ) and integrating on R N +1 , we obtain a bound of the type ddt Z R N +1 h εn,δ ( t, x, ξ ) ζ ( x ) dx dξ ≤ C Z R N +1 h εn,δ ( t, x, ξ ) ζ ( x ) dx dξ + Z R N +1 (cid:12)(cid:12) R εn,δ ( x, ξ ) (cid:12)(cid:12) | − f ε ( t, x, ξ ) | ζ ( x ) dx dξ. Using Gronwall's lemma and passing to the limit as δ → with ε and n ∈ N (cid:28)xed, we retrieve, for all t ≥ , Z R N +1 (cid:12)(cid:12)(cid:12) f ε ( t, x, ξ ) − g n (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) ζ ( x ) dx dξ ≤ e Ct Z R N +1 (cid:12)(cid:12)(cid:12) f (cid:16) x, xε , ξ (cid:17) − g n (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) ζ ( x ) dx dξ + e Ct Z R N +1 (cid:20) g n (cid:16) x, xε , ξ (cid:17) − (cid:12)(cid:12)(cid:12) g n (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) (cid:21) ζ ( x ) dx dξ + C n ( e Ct − , where the
onstant C n does not depend on ε , and g n = f ∗ x ρ n . And for all n ∈ N , ε > , we have Z R N +1 (cid:12)(cid:12)(cid:12) f (cid:16) x, xε , ξ (cid:17) − g n (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) ζ ( x ) dx dξ ≤ Z R N +1 Z R N (cid:12)(cid:12)(cid:12) f (cid:16) x, xε , ξ (cid:17) − f (cid:16) x ′ , xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) ρ n ( x − x ′ ) ζ ( x ) dx dx ′ dξ ≤ Z R N Z R N (cid:12)(cid:12)(cid:12) u (cid:16) x, xε , ξ (cid:17) − u (cid:16) x ′ , xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) ρ n ( x − x ′ ) ζ ( x ) dx dx ′ ≤ Z R N Z R N sup y ∈ Y | u ( x, y, ξ ) − u ( x ′ , y, ξ ) | ρ n ( x − x ′ ) ζ ( x ) dx dx ′ . The right-hand side of the above inequality vanishes as n → ∞ be
ause u ∈ L lo ( R N , C per ( Y )) . Similarly, Z R N +1 (cid:20) g n (cid:16) x, xε , ξ (cid:17) − (cid:12)(cid:12)(cid:12) g n (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) (cid:21) ζ ( x ) dx dξ ≤ Z R N +1 h g n (cid:16) x, xε , ξ (cid:17) − f (cid:16) x, xε , ξ (cid:17)i ζ ( x ) dx dξ + Z R N +1 (cid:20) f (cid:16) x, xε , ξ (cid:17) − g n (cid:16) x, xε , ξ (cid:17) (cid:21) ζ ( x ) dx dξ ≤ Z R N +1 (cid:12)(cid:12)(cid:12) g n (cid:16) x, xε , ξ (cid:17) − f (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) ζ ( x ) dx dξ ≤ Z R N Z R N sup y ∈ Y | u ( x, y, ξ ) − u ( x ′ , y, ξ ) | ρ n ( x − x ′ ) ζ ( x ) dx dx ′ . Hen
e, we dedu
e that there exists a fun
tion ω : [0 , ∞ ) → [0 , ∞ ) , independent of ε and satisfying lim t → ω ( t ) = 0 , su
h that Z R N +1 (cid:12)(cid:12)(cid:12) f ε ( t, x, ξ ) − f (cid:16) x, xε , ξ (cid:17)(cid:12)(cid:12)(cid:12) ζ ( x ) dx dξ ≤ ω ( t ) for all t > .Then, we prove that the same property holds for the fun
tion f . Indeed, we write (cid:12)(cid:12)(cid:12) f ε ( t, x, ξ ) − ξ , Z ∞ Z R N +1 h f ε − f ε ξ arbitrary, we dedu
e that τ Z τ | f ( t ) − χ ( ξ, u ( x, y )) | ζ ( x ) dt dx dy ≤ τ Z τ ω ( t ) dt, and the left-hand side vanishes as τ → . Thus the
ontinuity property is satis(cid:28)ed at time t = 0 .Hen
e, we have proved that any two-s
ale limit of the sequen
e f ε is a solution of the limit system.Thus the existen
e result in Theorem 1 is proved, as well as the
onvergen
e result of Theorem 2. Wenow ta
kle the proof of the uniqueness and rigidity results of Theorem 1. The strong
onvergen
e resultof Theorem 1 will follow from the rigidity.3 Uniqueness of solutions of the limit evolution problemIn this se
tion, we prove the se
ond and the third point in Theorem 1, that is, if f is any solution of thelimit evolution problem, then there exists a fun
tion u ∈ L ∞ ([0 , ∞ ) × R N × Y ) su
h that f ( t, x, y, ξ ) = ξ arbitrary, and let θ ∈ D ( R ) , θ ∈ D ( R N ) su
h that θ ≥ , θ ≥ , Z R θ = Z R N θ = 1 , Supp θ ⊂ [ − , and θ (0) = 0 . We set, for ( t, x ) ∈ R N +1 θ δ ( t, x ) = 1 δ N +1 θ (cid:18) tδ (cid:19) θ (cid:16) xδ (cid:17) . Set f δ := f ∗ t,x θ δ , M δ := M ∗ t,x θ δ . Then f δ is a solution of ∂f δ ∂t + N X i =1 a i ( y, ξ ) ∂f δ ∂x i = M δ . f δ satis(cid:28)es the following properties ≤ f δ ≤ , (32) div y,ξ ( a ( y, ξ ) f δ ) = ∂ ξ m ∗ t,x θ δ , (33) ∂ ξ f δ = − ν ∗ t,x θ δ , (34)whereas M δ satis(cid:28)es M δ ∈ C ((0 , T ) × R N , L ( Y × R )) ∩ L ∞ ([0 , ∞ ) × R N × Y × R ) , (35) M δ ( · , ξ ) = 0 if | ξ | > M, f δ ( · , ξ ) = 0 if ξ > M, f δ ( · , ξ ) = 1 if ξ < − M, (36) Z Y × R M δ ψ ≤ ∀ ψ ∈ G . (37)In parti
ular, noti
e that (1 − f δ ( t, x )) ∈ G for all t, x , and f δ ( t, x, y, ξ ) − f δ ( t, x, y, ξ ) = 0 if | ξ | > M .Let ζ ∈ C ∞ ( R N ) be a
ut-o(cid:27) fun
tion as in the previous subse
tion. We multiply the equation on f δ by (1 − f δ ) ζ ( x ) , and we integrate over R N × Y × R . We obtain ddt Z R N × Y × R (cid:0) f δ − | f δ | (cid:1) ζ − Z R N × Y × R a i ( y, ξ ) ∂ i ζ ( x ) (cid:0) f δ − | f δ | (cid:1) = Z R N × Y × R M δ (cid:0) − f δ (cid:1) ζ ≤ . We then dedu
e su
essively, using Gronwall's lemma, ddt Z R N × Y × R (cid:0) f δ − | f δ | (cid:1) ζ ≤ C Z R N × Y × R (cid:0) f δ − | f δ | (cid:1) ζ, Z R N × Y × R (cid:0) f δ ( t ) − | f δ ( t ) | (cid:1) ζ ≤ e Ct Z R N × Y × R (cid:0) f δ ( t = 0) − | f δ ( t = 0) | (cid:1) ζ ∀ t > , Z T Z R N × Y × R (cid:0) f δ − | f δ | (cid:1) ζ ≤ e CT − C Z R N × Y × R (cid:0) f δ ( t = 0) − | f δ ( t = 0) | (cid:1) ζ, (38)with a
onstant C depending only on || a || L ∞ ( Y × [ − R,R ]) .Let us now
he
k that f δ ( t = 0) strongly
onverges towards ξ Z R N × Y × R (cid:12)(cid:12) f δ ( t = 0) − f ∗ x θ δ (cid:12)(cid:12) ζ ( x ) dx dy dξ ≤ Z R N × Y × R Z R N +1 | f ( s, z, y, ξ ) − f ( z, y, ξ ) | ζ ( x ) θ δ ( − s, x − z ) dx dy dξ ds dz ≤ Z R || f ( s ) − f || L ( R N × Y × R ,ζ ( x ) dx dy dξ ) δ θ (cid:18) − sδ (cid:19) ds dx dy dξ + 2 R | Y | || ζ − ζ ∗ ˇ θ δ || L ( R N ) ≤ Cδ Z δ || f ( s ) − f || L ( R N × Y × R ,ζ ( x ) dx dy dξ ) ds + 2 R | Y | || ζ − ζ ∗ ˇ θ δ || L ( R N ) . δ → , and thus f δ ( t = 0) onverges towards f as δ → in L ( R N × Y × R , ζ ( x ) dx dy dξ ) , and hen
e also in L ( R N × Y × R , ζ ( x ) dx dy dξ ) . Consequently, Z R N × Y × R (cid:0) f δ ( t = 0) − | f δ ( t = 0) | (cid:1) ζ → as δ → . Above, we have used the fa
t that f = ξ , Z T Z R N × Y × R (cid:0) f − | f | (cid:1) ϕ ≤ . Sin
e the integrand in the left-hand side is non-negative, we dedu
e that | f | = | f | almost everywhere.The rigidity property follows.3.2 Contra
tion prin
ipleLet f , f be two generalized kineti
solutions of the limit problem; we denote by M , M , and M , M ,the
onstants and distributions asso
iated to f , f , respe
tively. Without loss of generality, we assumethat M ≤ M . A
ording to the rigidity result, there exist fun
tions u , u ∈ L ∞ ([0 , ∞ ) × R N × Y ) ∩ L ∞ ([0 , ∞ ) , L ( R N × Y )) su
h that f i = ξ , and thus Z R N × Y × R g δ ( t, x, y, ξ ) ζ ( x ) dx dy dξ ≤ e Ct Z R N × Y × R g δ ( t = 0 , x, y, ξ ) ζ ( x ) dx dy dξ. f δi ( t = 0) derived in the previous se
tion, we
an pass tothe limit as δ → . We infer that for almost every t > , Z R N × Y × R | f ( t, x, y, ξ ) − f ( t, x, y, ξ ) | ζ ( x ) dx dy dξ ≤ e Ct Z R N × Y × R | f ( t = 0 , x, y, ξ ) − f ( t = 0 , x, y, ξ ) | ζ ( x ) dx dy dξ. (40)This
ompletes the derivation of the
ontra
tion prin
iple for the limit system. Uniqueness of solutionsof the limit system follows. In parti
ular, we dedu
e that the whole sequen
e f ε of solutions of (4) two-s
ale
onverges towards f .3.3 Strong
onvergen
e resultHere, we explain why the strong
onvergen
e result stated in Theorem 2 holds, that is, we prove (21).This fa
t is rather
lassi
al, and is a dire
t
onsequen
e of the fa
t that ξ n . Then we write (cid:12)(cid:12)(cid:12) ξ , Z C Z R ξ , for all
ompa
t set C ⊂ [0 , ∞ ) × R N , lim ε → (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u ε ( t, x ) − u δ (cid:16) t, x, xε (cid:17)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L ( C ) = || u − u δ || L ( C × Y ) . Statement (21) then follows from standard
onvolution results.3.4 Appli
ation: proof of the
onvergen
e result for kineti
solutionsIn this subse
tion, we prove Theorem 3; this result is in fa
t an easy
onsequen
e of the
onvergen
eresult stated in Theorem 2 for entropy solutions, and of the
ontra
tion prin
iple for the limit system.Assume that a N +1 ≡ , and let u ε be a kineti
solution of equation (1), with an initial data u ( x, x/ε ) su
h that u ∈ L ( R N , C per ( Y )) and N X i =1 ∂∂y i ( a i ( y, ξ ) χ ( ξ, u ( x, y ))) = 0 (41)in the sense of distributions.For n ∈ N , let u n := sgn( u ) inf( | u | , n ) . Then for all n ∈ N , u n belongs to L ∞ ( R N × Y ) and u n → u as n → ∞ in L ( R N , C per ( Y )) . Moreover, χ ( ξ, u n ) = χ ( ξ, u ) ξ
7→ || f λ ( t ) || L ( R N × Y × R ) is nonin reasingon [0 , T ] . The equality Z R N × Y × R | χ ( ξ, u ( x, y )) | dx dy dξ = Z R N × Y × R | χ ( ξ, u ( x, y )) | dx dy dξ = Z R N × Y | u ( x, y ) | dx dy then yields the desired result.Third step. Compa t support in ξ . Let us prove now that f λ ( · , ξ ) = 0 if | ξ | > M : let ϕ ∈ D ( R ) be anarbitrary test fun tion su h that ϕ ( ξ ) = 0 when | ξ | ≤ M . Then P ( f λ ) ϕ = 0 sin e P ( f λ ) ∈ K , and thus f λ ϕ is a solution of ∂∂t ( f λ ϕ ) + a · ∇ x ( f λ ϕ ) + λ ( f λ ϕ ) = 0 , ( f λ ϕ ) ( t = 0 , x, y, ξ ) = 0 . Hen e ( f λ ϕ )( t, x, y, ξ ) = 0 for almost every t, x, y, ξ , and f λ ( · , ξ ) = 0 if | ξ | > M .Fourth step. Sign property. We now prove the sign property, namely sgn( ξ ) f λ = | f λ | ≤ a.e.This relies on the following fa t: if g ∈ K , then sgn( ξ ) g ( y, ξ ) ∈ [0 , for almost every y, ξ . Indeed, g ( · , ξ ) = 0 if ξ < − M , and thus if − M < ξ < , g ( y, ξ ) = − Z ξ − M ν ( y, ξ ′ ) dξ ′ ≤ . Hen e g ( y, · ) is non positive and non in reasing on ( −∞ , . Similarly, g ( y, · ) is non negative and nonde reasing on (0 , ∞ ) . And if ξ < < ξ ′ , then g ( y, ξ ′ ) − g ( y, ξ ) = 1 − Z ξ ′ ξ ν ( y, w ) dw ≤ . Hen e the sign property is true for fun tions in K .Multiplying (47) by sgn( ξ ) , we are led to ∂∂t (sgn( ξ ) f λ ) + a ( y, ξ ) · ∇ x (sgn( ξ ) f λ ) + λ (sgn( ξ ) f λ ) = λ P ( f λ ) ∈ [0 , λ ] . And at time t = 0 , sgn( ξ ) f λ ( t = 0) = | χ ( ξ, u ) | ∈ [0 , . Thus, using a maximum prin iple for this lineartransport equation, we dedu e that the sign property is satis(cid:28)ed for f λ .Fifth step. Uniform ontinuity at time t = 0 . Let δ > be arbitrary, and let f δ := f ∗ x θ δ , with θ δ astandard molli(cid:28)er. Then f δ ( x ) ∈ K for all x ∈ R N , and thus f λ − f δ is a solution of the equation ∂∂t (cid:0) f λ − f δ (cid:1) + a ( y, ξ ) · ∇ x (cid:0) f λ − f δ (cid:1) + λ (cid:0) f λ − f δ (cid:1) = λ (cid:0) P ( f λ ) − P ( f δ ) (cid:1) − a ( y, ξ ) · (cid:0) f ∗ x ∇ θ δ (cid:1) . Multiply the above equation by (cid:0) f λ − f δ (cid:1) and integrate on R N × Y × R . Using on e more the Lips hitz ontinuity of the proje tion P , we obtain ddt (cid:12)(cid:12)(cid:12)(cid:12) f λ − f δ (cid:12)(cid:12)(cid:12)(cid:12) L ( R N × Y × R ) ≤ || a || L ∞ ( Y × ( − M,M )) || f λ − f δ || L ( R N × Y × R ) || f || L ( R N × Y × R ) ||∇ θ δ || L ddt (cid:12)(cid:12)(cid:12)(cid:12) f λ − f δ (cid:12)(cid:12)(cid:12)(cid:12) L ( R N × Y × R ) ≤ Cδ .
25s a onsequen e, we obtain the following estimate, whi h holds for all t > , λ > and δ > (cid:12)(cid:12)(cid:12)(cid:12) f λ ( t ) − f δ (cid:12)(cid:12)(cid:12)(cid:12) L ( R N × Y × R ) ≤ Ctδ + (cid:12)(cid:12)(cid:12)(cid:12) f − f δ (cid:12)(cid:12)(cid:12)(cid:12) L ( R N × Y × R ) . Hen e the uniform ontinuity property is true, with ω ( t ) := inf δ> (cid:18) Ctδ + 2 (cid:12)(cid:12)(cid:12)(cid:12) f − f δ (cid:12)(cid:12)(cid:12)(cid:12) L ( R N × Y × R ) (cid:19) . Sixth step. Inequality for M λ . Inequality (50) is merely a parti ular ase of the inequality hP ( f ) − f, P ( f ) − g i E ≤ whi h holds for all f ∈ E , for all g ∈ K .4.2 The hydrodynami limitIn this subse tion , we prove the following result :Proposition 3. Let ( f λ ) λ> be the family of solutions of the relaxation model (47), and let f ( t ) = χ ( ξ, u ) be the unique solution of the limit system (43)-(46) with initial data χ ( ξ, u ( x, y )) . Then as λ → ∞ , f λ → f in L ((0 , T ) × R N × Y × R ) . The above Proposition relies on an inequality of the type ddt Z R N × Y × R | f λ − f | ≤ r λ ( t ) , with r λ ( t ) → as λ → ∞ . The al ulations are very similar to those of the ontra tion prin iple in theprevious se tion; the only di(cid:27)eren e lies in the fa t that f λ and f are not solutions of the same equation.Before ta kling the proof itself, let us derive a few properties on the weak limit of the sequen e f λ .Sin e the sequen e f λ is bounded in X T ⊂ L ((0 , T ) × R N × Y × R ) , we an extra t a subsequen e, whi hwe relabel f λ , and (cid:28)nd a fun tion g ∈ L ((0 , T ) × R N × Y × R ) su h that f λ weakly onverges to g in L .Moreover, the sequen e P ( f λ ) is bounded in L ((0 , T ) × R N × Y × R ) , for all T > . Hen e, extra tinga further subsequen e if ne essary, we an (cid:28)nd a fun tion h ∈ L ((0 , T ) × R N × Y × R ) su h that P ( f λ ) weakly onverges towards h as λ → ∞ . Noti e that the onvex set K is losed for the weak topology in L . Consequently, h ( t, x ) ∈ K for almost every t, x . At last, P ( f λ ) − f λ = O (cid:18) λ (cid:19) , where the O is meant in the sense of distributions. Hen e, g = h , and in parti ular, we dedu e that g ( t, x ) ∈ K for almost every ( t, x ) .We are now ready to prove the ontra tion inequality; onsider a mollifying sequen e θ δ as in theprevious se tion, and set f δ = f ∗ t,x θ δ , f δ ′ λ = f λ ∗ t,x θ δ ′ . Then ∂ t f δ + a ( y, ξ ) · ∇ x f δ = M δ ,∂ t f δ ′ λ + a ( y, ξ ) · ∇ x f δ ′ λ = M δ ′ λ . Let us multiply the (cid:28)rst equation by sgn( ξ ) − f δ ′ λ , the se ond by f δ ′ λ − f δ ) , and add the two identitiesthus obtained; setting F δ,δ ′ λ = sgn( ξ ) f δ + | f δ ′ λ | − f δ f δ ′ λ , we have ∂ t F δ,δ ′ λ + a ( y, ξ ) · ∇ x F δ,δ ′ λ = M δ (cid:16) sgn( ξ ) − f δ ′ λ (cid:17) + 2 M δ ′ λ ( f δ ′ λ − f δ ) .
26e integrate over (0 , t ) × R N × Y × R and obtain Z R N × Y × R F δ,δ ′ λ ( t, x, y, ξ ) dx dy dξ = Z t Z R N × Y × R M δ (cid:16) sgn( ξ ) − f δ ′ λ (cid:17) +2 Z t Z R N × Y × R M δ ′ λ ( f δ ′ λ − f δ )+ Z R N × Y × R F δ,δ ′ λ ( t = 0 , x, y, ξ ) dx dy dξ. We now pass to the limit as δ ′ → , with all the other parameters (cid:28)xed. Noti e that lim δ ′ → Z t Z R N × Y × R M δ ′ λ ( f δ ′ λ − f δ ) = Z t Z R N × Y × R M λ ( f λ − f δ )= − λ Z t Z R N × Y × R ( f λ − P ( f λ )) + Z t Z R N × Y × R M λ ( P ( f λ ) − f δ ) ≤ , sin e f δ ( t, x ) ∈ K for all t, x . The passage to the limit in F δ,δ ′ λ ( t = 0) does not rise any di(cid:30) ulty be auseof the strong ontinuity of the fun tions f λ at time t = 0 . Hen e, we retrieve Z R N × Y × R (cid:8)(cid:0) | f δ ( t ) | − | f δ ( t ) | (cid:1) + | f δ ( t ) − f λ ( t ) | (cid:9) ≤ Z t Z R N × Y × R M δ (sgn( ξ ) − f λ )+ Z R N × Y × R (cid:8)(cid:0) | f δ ( t = 0) | − | f δ ( t = 0) | (cid:1) + | f δ ( t = 0) − χ ( ξ, u ) | (cid:9) , and thus, integrating on e again this inegality for t ∈ [0 , T ] , Z T Z R N × Y × R (cid:8)(cid:0) | f δ | − | f δ | (cid:1) + | f δ ( t ) − f λ | (cid:9) ≤ Z T dt (cid:20)Z t Z R N × Y × R M δ ( s ) (sgn( ξ ) − f λ ( s )) ds (cid:21) + T Z R N × Y × R (cid:8)(cid:0) | f δ ( t = 0) | − | f δ ( t = 0) | (cid:1) + | f δ ( t = 0) − χ ( ξ, u ) | (cid:9) . We now pass to the limit as λ → ∞ , with δ > (cid:28)xed. Then lim inf λ →∞ || f λ − f δ || L ((0 ,T ) × R N × Y × R ≥ || g − f δ || L ((0 ,T ) × R N × Y × R , and lim λ →∞ Z T dt (cid:20)Z t Z R N × Y × R M δ ( s ) (sgn( ξ ) − f λ ( s )) ds (cid:21) = Z T dt (cid:20)Z t Z R N × Y × R M δ ( s ) (sgn( ξ ) − g ( s )) ds (cid:21) ≤ . Thus, we obtain, for all δ > || g − f δ || L ((0 ,T ) × R N × Y × R ≤ T Z R N × Y × R (cid:8)(cid:0) | f δ ( t = 0) | − | f δ ( t = 0) | (cid:1) + | f δ ( t = 0) − χ ( ξ, u ) | (cid:9) . We have already proved in the previous se tion that the family f δ ( t = 0) strongly onverges towards χ ( ξ, u ) as δ vanishes, due to the ontinuity assumption at time t = 0 . Hen e, we obtain in the limit || g − f || L ((0 ,T ) × R N × Y × R ≤ , and onsequently, g = f . Hen e the result is proved.27 The separate ase : identi(cid:28) ation of the limit problemThis se tion is devoted to the proof of Proposition 1. Thus we fo us on the limit system in the asewhere the (cid:29)ux A an be written as A ( y, ξ ) = a ( y ) g ( ξ ) , with div y a = 0 . The interest of this ase lies in the spe ial stru ture of the limit system; indeed, we shall prove thatthe fun tion u , whi h is the two-s ale limit of the sequen e u ε , is the solution of the s alar onservationlaw (19). In view of Theorem 1, we wish to emphasize that Proposition 1 implies in parti ular that theentropy solution of (19) satis(cid:28)es the onstraint equation div y ( a ( y ) g ( u ( t, x ; y ))) for almost every t > , x ∈ R N ; this fa t is not ompletely obvious when g = Id . We will prove in thesequel that u ( t, x ) a tually belongs to the onstraint spa e K for a.e. t, x .Before ta kling the proof of the theorem, let us mention that the limit problem (19) is not the onewhi h is expe ted from a vanishing vis osity approa h. Pre isely, for any given δ > , let u εδ be thesolution of ∂ t u εδ + div x A (cid:16) xε , u εδ (cid:17) − εδ ∆ x u εδ = 0 , with the initial data u εδ ( t = 0 , x ) = u ( x, x/ε ) . Then for all ε > , u εδ → u ε as δ → ; moreover, thebehavior of u εδ as ε → is known for ea h δ > (see [5, 6℄). In the divergen e-free ase, for all δ > , lim ε → u εδ = ¯ u ( t, x ) in L lo , where ¯ u is the entropy solution of ∂ t ¯ u + div x ( h a i g (¯ u )) = 0 , with initial data ¯ u ( t = 0 , x ) = h u ( x, · ) i . Hen e, it ould be expe ted that the limits ε → and δ → an be ommuted, that is lim ε → lim δ → u εδ = lim δ → lim ε → u εδ , whi h would entail lim ε → u ε = ¯ u. In general, this equality is false, even in a weak sense: a generi ounter-example is the one of shear(cid:29)ows (see for instan e the al ulations in [9℄). In that ase, we have N = 2 and A ( y, ξ ) = ( a ( y ) ξ, ,and the equation (19) be omes ∂ t u + a ( y ) ∂ x u = 0 , with the initial ondition u ( t = 0 , x, y ) = u ( x , x , y ) . It is then easily he ked that in general, theaverage of u over Y is not the solution of the transport equation ∂ t ¯ u + h a i ∂ x ¯ u = 0 . We now turn to the proof of Proposition 1. In view of Theorem 1, it is su(cid:30) ient to prove that theentropy solution of (19) belongs to K for a.e. t, x , or in other words, that K is invariant by the semi-group asso iated to equation (19). We prove this result in the slightly more general ontext of kineti solutions. The ore of the proof lies in the followingProposition 4. Let u ∈ L ( R N , L ∞ ( Y )) su h that u ( x, · ) ∈ K for almost every x ∈ R N .Let v = v ( t, x ; y ) ∈ C ([0 , ∞ ); L ( R N × Y )) be the kineti solution of (cid:26) ∂ t v ( t, x ; y ) + div x (˜ a ( y ) g ( v ( t, x ; y ))) = 0 , t > , x ∈ R N , y ∈ Y,v ( t = 0 , x ; y ) = u ( x, y ) , i.e. f ( t, x, y, ξ ) := χ ( ξ, v ( t, x ; y )) is a solution in the sense of distributions of (cid:26) ∂ t f + ˜ a ( y ) · ∇ x f g ′ ( ξ ) = ∂ ξ m, t > , x ∈ R N , y ∈ Y, ξ ∈ R ,f ( t = 0 , x, y, ξ ) = χ ( ξ, u ( x, y )) , (51)and m is a non-negative measure on [0 , ∞ ) × R N × Y × R .Then for a.e. t > , x ∈ R N , u ( t, x ) ∈ K . 28roof. First, let us re all (see [23, 24℄) that for all T > , f = lim λ →∞ f λ in C ([0 , T ]; L ( R N × Y × R )) , where f λ = f λ ( t, x, y, ξ ) ( λ > ) is the unique solution of the system ∂ t f λ + ˜ a ( y ) · ∇ x f λ g ′ ( ξ ) + λf λ = λχ ( ξ, u λ ) ,u λ ( t, x, y ) = R R f λ ( t, x, y, ξ ) dξ,f λ ( t = 0) = χ ( ξ, u ) . (52)Moreover, for every λ > , u λ is the unique (cid:28)xed point of the ontra tant appli ation φ λ : C ((0 , T ); L ( R N × Y )) → C ((0 , T ); L ( R N × Y )) u u where u = R ξ f and f is the solution of ∂ t f + ˜ a ( y ) · ∇ x f g ′ ( ξ ) + λf = λχ ( ξ, u ) ,f ( t = 0) = χ ( ξ, u ) . (53)Thus, the whole point is to prove that the spa e { u ∈ C ([0 , T ]; L ( R N × Y )); u ( t, x ) ∈ K a.e } is invariant by the appli ation φ λ .First, let us stress that for all u ∈ L ( Y ) , u ∈ K ⇐⇒ div y ( a ( y ) χ ( ξ, u )) = 0 in D ′ ( Y × R ) . (54)Indeed, if u ∈ K , then for all δ > , set u δ = u ∗ θ δ , with θ δ a standard molli(cid:28)er. The fun tion u δ is asolution of div y ( a u δ ) = r δ , and the remainder r δ vanishes strongly in L ( Y ) (see the al ulations in the previous se tions). Sin ethe fun tion u δ is smooth, if G ∈ C ( R N ) , we have div y ( a G ( u δ )) = G ′ ( u δ ) r δ . Passing to the limit as δ vanishes, we infer div y ( a G ( u )) = 0 for all G ∈ C ( R N ) . At last, taking asequen e of smooth fun tions approa hing χ ( ξ, u ) , we dedu e that div y ( a χ ( ξ, u )) = 0 in D ′ per ( Y × R ) .Conversely, assume that div y ( a χ ( ξ, u )) = 0 ; then integrating this equation with respe t to ξ yields u ∈ K . Hen e (54) is proved.Now, let u ∈ C ([0 , T ]; L ( R N × Y )) su h that u ( t, x ) ∈ K a.e. Then div( a χ ( ξ, u ) = 0) . Let f bethe solution of (53); sin e ˜ a ∈ K , the distribution div y ( a f ) satis(cid:28)es the transport equation ∂ t (div( a f )) + g ′ ( ξ )˜ a ( y ) · ∇ x (div( a f )) + λ div( a f ) = 0 , and div( a f )( t = 0) = 0 be ause u ( x ) ∈ K a.e. Hen e div y ( a f ) = 0 ; integrating this equation withrespe t to ξ gives u ∈ K a.e.Consequently, u λ ( t, x ; · ) ∈ K a.e. Passing to the limit, we dedu e that v ( t, x ; · ) ∈ K a.e.Let us now re-write equation (51): setting b ( y ) = a ( y ) − ˜ a ( y ) , we have ∂ t f + a ( y ) ∇ x f g ′ ( ξ ) = ∂ ξ m − b ( y ) ∇ x f g ′ ( ξ ) =: M . If u ∈ L ∞ ( R N ) , then v ∈ L ∞ ([0 , ∞ ) × R N × Y ) , and it is easily he ked that f and M satisfy the ompa t support assumptions. A ording to the above Proposition, f also satis(cid:28)es (43), and thanks tothe stru ture of the right-hand side, the distribution M satis(cid:28)es (46). Thus f is the unique solutionof the limit system, and Proposition 1 is proved. 29 Further remarks on the notion of limit systemHere, we have gathered, by way of on lusion, a few remarks around the limit evolution system introdu edin de(cid:28)nition 1. The main idea behind this se tion is that the limit system is not unique (although its so-lution always is), and thus several other relevant equations an be written instead of (10). Unfortunately,there does not seem to be any rule whi h would allow to de ide between two limit systems.Let us illustrate these words by a (cid:28)rst series of examples : assume that the (cid:29)ux is divergen e free,and let K := { f ∈ L lo ( Y × R ) , N X i =1 ∂ y i ( a i f ) = 0 in D ′ } . We denote by P the proje tion on K in L lo ( Y × R ) . Pre isely, onsider the dynami al system X ( t, y ; ξ ) de(cid:28)ned by (cid:26) ˙ X ( t, y ; ξ ) = a ( X ( t, y ; ξ ) , ξ ) , t > X ( t = 0 , y ; ξ ) = y. Then for all ξ ∈ R , the Lebesgue measure on Y is invariant by the semi-group X ( t ; ξ ) be ause of thehypothesis div y a ( y, ξ ) = 0 . Hen e by the ergodi theorem, for all f ∈ L lo ( Y × R ) , there exists a fun tionin L lo ( Y × R ) , denoted by P ( f )( y, ξ ) , su h that P ( f )( y, ξ ) = lim T →∞ T Z T f ( X ( t, y ; ξ ) , ξ ) dt, and the limit holds a.e. in y, ξ and in Y × ( − R, R ) for all R > .Set ˜ a := P ( a ) . Then if f is a solution of the limit system, f also satis(cid:28)es ∂ t f + ˜ a ( y, ξ ) · ∇ x f = ˜ M and f , ˜ M satisfy (9) and (11) - (14). Indeed, ˜ M = M + [˜ a ( y, ξ ) − a ( y, ξ )] · ∇ x f and the term [˜ a ( y, ξ ) − a ( y, ξ )] ·∇ x ( f ∗ x ϕ )( t, x, y, ξ ) belongs to K ⊥ for all t, x . Of ourse, uniqueness holdsfor this limit system (the proof is exa tly the same as the one in se tion 3), and thus this onstitutes aslegitimate a limit system as the one in de(cid:28)nition 1. In fa t, in the separate ase, Proposition 1 indi atesthat the above system seems to be the relevant one, rather than the one in de(cid:28)nition 1. Noti e that thedistribution ˜ M satis(cid:28)es the additional property ˜ M ∗ t,x φ ( t, x ) ∈ K ⊥ ∀ t, x. Let us now go a little further: let θ ∈ C ( R ) su h that ≤ θ ≤ , and let a θ ( y, ξ ) = θ ( ξ ) a ( y, ξ ) + (1 − θ ( ξ ))˜ a ( y, ξ ) . Then f is a solution of ∂ t f + ˜ a θ ( y, ξ ) · ∇ x f = M θ , for some distribution M θ satisfying (14). Thus this still onstitutes a limit system whi h has the samestru ture as the one of de(cid:28)nition 1. Hen e the limit system is highly non unique, and it must be seenas a way of identifying the two-s ale limit of the sequen e f ε , rather than as a kineti formulation of agiven onservation law, for instan e. We wish to emphasize that if the (cid:29)ux A is not (cid:16)separated(cid:17), that is,if the hypotheses of Proposition 1 are not satis(cid:28)ed, then in general, the fun tion u su h that f = ξ
3. For all p ∈ [ p , p ] , h v ( · , p ) i Y = 0 ;4. The distribution ∂ p v is a nonnegative fun tion in L ( Y × [ p , p ]) ; this implies in parti ular thatfor all ouples ( p, p ′ ) ∈ [ p , p ] su h that p ≥ p ′ , for almost every y ∈ Y , v ( y, p ) ≥ v ( y, p ′ ) . Under these onditions, one an onstru t a kineti formulation for equation (1), based on the family v ( x/ε, p ) of stationary solutions of (1), rather than on the family of Kruzkov's inequalities. This kind of onstru tion was a hieved in [7℄ in a paraboli setting, following an idea developed by Emmanuel Audusseand Benoît Perthame in [3℄; these authors de(cid:28)ne a new notion of entropy solutions for a heterogeneous onservation law in dimension one, based on the omparison with a family of stationary solutions. Letus explain brie(cid:29)y how the kineti formulation for entropy solutions of (1) is derived: let u ε be an entropysolution of (1). De(cid:28)ne the distribution m ε ∈ D ′ ((0 , ∞ ) × R N × ( p , p )) by m ε ( t, x, p ) := − (cid:26) ∂∂t (cid:16) u ε − v (cid:16) xε , p (cid:17)(cid:17) + + ∂∂y i h v ( xε ,p )
M ⊂ [0 , ∞ ) × R N × Y × [ p ′ , p ′ ]; f ( t, x, y, p ) = 1 if p < p < p ′ , f ( t, x, y, p ) = 0 if p ′ < p < p .
2. Mi ros opi equation for f : f is a solution in the sense of distributions on Y × ( p , p ) of div y (˜ a ( y, p ) f ( t, x, y, p )) = 0 . (58)3. Evolution equation: the ouple ( f, M ) is a solution in the sense of distributions on [0 , ∞ ) × R N × Y × ( p , p ) of (cid:26) ∂ t ( v p ( y, p ) f ) + ˜ a ( y, p ) · ∇ x f = M ,f ( t = 0 , x, y, p ) = v ( y,p )
4. Conditions on f : there exists a nonnegative measure ν ∈ M per ([0 , ∞ ) × R N × Y × R ) su h that ∂ p f = − ν, (60) ≤ f ( t, x, y, ξ ) ≤ a.e. , (61) τ Z τ || f ( s ) − f || L ( R N × Y × ( p ,p ) ds −→ τ → . (62)5. Condition on M : for all ϕ ∈ D ([0 , ∞ ) × R N su h that ϕ ≤ , the fun tion M ∗ t,x ϕ belongs to C ([0 , ∞ ) × R N , L ( Y × R )) , and (cid:26) R Y × R ( M ∗ t,x ϕ ) ( t, x, · ) ψ ≤ , ∀ ψ ∈ L ∞ lo ( Y × R ) , div y (˜ aψ ) = 0 , and ∂ ξ ψ ≥ . (63)We now state without proof a result analogue to Theorems 1, 2 :Proposition 5. Let A ∈ W , ∞ per,lo ( Y × R ) . Assume that a ∈ C per ( Y × R ) and that ˜ a ∈ W , ( Y × ( p , p )) .Let u ∈ L ∞ ( R N × Y ) ∩ L lo ( R N , C per ( Y )) su h that u ( x, · ) is an entropy solution of the ell problemfor almost every x ∈ R N . Assume furthermore that there exists p ′ < p ′ in ( p , p ) su h that v ( y, p ′ ) ≤ u ( x, y ) ≤ v ( y, p ′ ) , and let f ( x, y, p ) := v ( y,p )