Existence of strong traces for quasisolutions of scalar conservation - PDF document

Existence of strong traces for quasisolutions of scalar conservation laws E.Yu. Panov (Velikiy Novgorod, Russia) Introduction. Consider a scalar conservation law div x ϕ ( u ) = 0 , (1) where u = u ( x ), x ∈ Ω, Ω ⊂ R n is a domain with C 1 -smooth boundary ∂ Ω; the flux vector ϕ ( u ) = ( ϕ 1 ( u ) , . . . , ϕ n ( u )), ϕ i ( u ) ∈ C ( R ) ∩ BV loc ( R ). Definition 1 (Pa,2005). A function u = u ( x ) ∈ L ∞ (Ω) is called a quasisolution of (1) if ∀ k ∈ F , where F ⊂ R is a dense set (in the sequel we suppose it being countable) there exists a Borel measure γ k ∈ ¯ M loc (Ω) such that div ψ k ( u ) = − γ k in D ′ (Ω) , ψ k ( u ) = sign ( u − k )( ϕ ( u ) − ϕ ( k )) . (i) G.e.s. as well as g.e.sub-s., g.e.super-s. are quasis.; (ii) If a, b ∈ F , a < b , and u = u ( x ) is a quasis. then v = max( a, min( u, b )) is also a quasis. Our main result is the following There exists a function u 0 ( y ) ∈ L ∞ ( ∂ Ω) (the trace) Theorem 1. such that ∀ k ∈ R ( ψ k ( u ( x )) , ν ) → ( ψ k ( u 0 ( y )) , ν ) as x → y ∈ ∂ Ω in L 1 loc . Here ν = ν ( y ) is the normal vector at y ∈ ∂ Ω . In particular the normal flux component ( ϕ ( u ( x )) , ν ) has the strong trace. Corollary. If for a.e. y ∈ ∂ Ω functions u → ( ϕ ( u ) , ν ( y )) are not constant on nondegenerate intervals then u ( x ) → u 0 ( y ) as x → y strongly. 1

The above result allows to formulate boundary value problems for (1) in the sense of Bardos, LeRoux & N´ ed´ elec. For instance the Dirichlet problem u | ∂ Ω = u b is understood in the set of the inequality: ∀ k ∈ R (sign ( u − k ) + sign ( k − u b ))( ϕ ( u ) − ϕ ( k ) , ν ) ≥ 0 on ∂ Ω , which is well defined due to existence of strong traces of sign ( u − k )( ϕ ( u ) − ϕ ( k ) , ν ) and ( ϕ ( u ) , ν ). Reformulation of the problem. The assertion of the main Th.1 has the local character. Thus, we can assume that for some f ( u ) ∈ C 1 ( R n − 1 ) Ω = { x ∈ R n | x 1 > f ( x 2 , . . . , x n ) } . Making the change t = x 1 − f ( x 2 , . . . , x n ), x i = x i , i = 2 , . . . , n we transform (1) to n ( ϕ 0 ( x, u )) t + div x ˜ ϕ ( u ) = ( ϕ 0 ( x, u )) t + i =2 ϕ i ( u ) x i = 0 , (2) � u = u ( t, x ), ( t, x ) ∈ Π = R + × R n − 1 ; here ˜ ϕ ( u ) = ( ϕ 2 ( u ) , . . . , ϕ n ( u )), i =2 ϕ i ( u ) ∂f n ϕ 0 ( x, u ) = ϕ 1 ( u ) − = ( ϕ ( u ) , ν ( x )) . � ∂x i Clearly u ( x ) is a quasis. of (1) iff u ( t, x ) is a quasis. of (2). For equa- tion (2) Th.1 is formulated as follows. There exists a function u 0 ( y ) ∈ L ∞ ( R n − 1 ) such that Theorem 1. ∀ k ∈ R ψ 0 k ( x, u ( t, x )) = ψ 0 k ( x, u 0 ( x )) in L 1 loc ( R n − 1 ) . ess lim t → 0 Here ψ 0 k ( x, u ) = sign ( u − k )( ϕ 0 ( x, u ) − ϕ 0 ( x, k )) = sign ( u − k )( ϕ ( u ) − ϕ ( k ) , ν ( x )) , u ∈ R , x = ( x 2 , . . . , x n ) ∈ R n − 1 . 2

Remark. To prove Th. 1 it is sufficient to establish strong convergence: ess lim ψ 0 k ( x, u ( t, x )) = v k ( x ). Indeed, there exists the measure valued t → 0 trace ν x so that ∀ p ( λ ) ∈ C ( R ) � p ( λ ) dν x ( λ ) weakly- ∗ in L ∞ ( R n − 1 ) . p ( u ( t m , x )) → Here t m is an arbitrary vanishing sequence, which consists of ”essen- tial points”. By the strong convergence of ψ 0 k ( x, u ( t m , x )) we see that ψ 0 k ( x, λ ) is constant on supp ν x for a.e. x ∈ R n − 1 . This easily implies that supp ν x contains in segments where ϕ 0 ( x, λ ) is constant. We can � λdν x ( λ ). Then for a.e. x ∈ R n − 1 define u 0 ( x ) = � v k ( x ) = ψ 0 k ( x, λ ) dν x ( λ ) = ψ 0 k ( x, u 0 ( x )) , as required. Reduction of the dimension. Suppose one of the component, say ϕ n ( u ) ≡ const. x n the function u ( t, x ′ ) = Proposition 1 (Pa,2005). For a.e. u ( t, x ′ , x n ) is a quasis. of reduced equations (2) n − 1 i =2 ϕ i ( u ) = 0 , ( t, x ′ ) ∈ Π ′ = R + × R n − 2 . ϕ 0 ( x, u ) t + � Weak traces. From the condition ( ψ 0 k ( x, u )) t + div x ˜ ψ k ( u ) = − γ k ∈ ¯ M loc (Π) , k ∈ F it follows (see Chen & Frid, 1999) existence of the weak traces: ψ 0 k ( x, u ( t, x )) = v k ( x ) weakly- ∗ in L ∞ ( R n − 1 ) ess lim t → 0 for all k ∈ R (we also take into account that ψ 0 k ( x, u ) is equicontinuous w.r.t. k ). Let u ( t, x ) be a quasis. of (2), y ∈ R n − 1 . ”Blow-up” procedure. Introduce the sequences u ε ( t, x ; y ) = u ( εt, y + εx ) ∈ L ∞ (Π), ε = ε m → 0; v ε k ( x ; y ) = v k ( y + εx ) ∈ L ∞ ( R n − 1 ), 3

and the sequences of measures γ ε k ( t, x ) = εγ k ( εt, y + εx ). The latter equality is understood in the sense of distributions, i.e. < γ ε k , f > = ε 1 − n � f ( t/ε, ( x − y ) /ε ) dγ k ( t, x ) ∀ f ∈ C 0 (Π) . Evidently, ( ψ 0 k ( y + εx, u ε )) t + div x ψ k ( u ε ) = − γ ε k ; (3) ψ 0 k ( y + εx, u ε ) | t =0 = v ε k ( x ; y ) in the weak sense. (4) Theorem 2 (Vasseur, 2001; Panov, 2005). (i) There exist a sequence ε = ε m → 0 such that for a.e. y ∈ R n − 1 k → 0 in ¯ v ε k ( x ; y ) → v k ( y ) = const in L 1 loc ( R n − 1 ); γ ε M loc (Π); (ii) Existence of the strong trace ψ 0 k ( x, u (0 , x )) = v k ( x ) holds iff there exists a sequence ε = ε m → 0 such that for a.e. y ∈ R n − 1 the sequence ψ 0 k ( y, u ε ( t, x ; y )) is strictly convergent. Besides, in this case the limit function is v k ( y ) = const for a.e. y . In order to prove (ii) (the inverse implication) we use some property of isentropic solutions (i.s.) of conservation laws. Choose a sequence ε = ε m → 0 and the set of full measure Y ⊂ R n − 1 of values y , for which (i) is satisfied. Suppose that y ∈ Y and ∀ k ∈ F ψ 0 k ( y + εx, u ε ( t, x ; y )) → w k ( t, x ; y ) in L 1 loc (Π) . (5) We can assume in addition that the sequence u ε ( t, x ; y ) weakly converges to some measure valued function ν t,x , i.e. ∀ p ( u ) ∈ C ( R ) p ( u ε ( t, x ; y )) → p ( λ ) dν t,x ( λ ) weakly- ∗ in L ∞ (Π) . � � From (5) it follows that ψ 0 k ( y, λ ) dν t,x ( λ ) = w k ( t, x ; y ), moreover supp ν t,x contains in an interval where ϕ 0 ( y, λ ) = const. Passing to the limit as m → ∞ in (3), we obtain that ∀ k ∈ F ∂ � � ψ k ( λ ) dν t,x ( λ ) = 0 in D ′ (Π) . ψ 0 k ( y, λ ) dν t,x ( λ ) + div x (6) ∂t 4

Since F is dense and the functions ψ 0 k , ψ k are equicontinuous in k we conclude that (6) holds for all k and ν t,x is an isentropic m.v.s. of the equation ϕ 0 ( u ) t + div x ˜ ϕ ( u ) = 0 (7) with some measure valued initial data ν 0 ,x = ν x . (8) Here and below we simplify the notations dropping the dependence on y : ϕ 0 ( u ) = ϕ ( y, u ), ψ 0 k ( u ) = ψ 0 k ( y, u ), w k ( t, x ) = w k ( t, x ; y ). From (4) in the limit as m → ∞ it follows that also w k (0 , x ) = v k ( y ) = const in the weak sense. In particular, taking k sufficiently large, we have w (0 , x ) = v ( y ) = const, where � w ( t, x ) = ϕ 0 ( λ ) dν t,x ( λ ) . Theorem 3. Let ν t,x be an isentropic m.v.s. of problem (7), (8) such that for for a.e. ( t, x ) ∈ Π supp ν t,x contains in a segment where ϕ 0 ( u ) is constant. Then � ψ 0 k ( λ ) dν t,x ( λ ) have the strong (i) ∀ k ∈ R the functions w k ( t, x ) = � ψ 0 k ( λ ) dν x ( λ ) . traces v k ( x ) = (ii) If v k = const then w k ( t, x ) ≡ v k . To prove Th.3, observe that ν t,x is an isentropic m.v.s. of any equation η ( u ) t + div x ψ ( u ) = 0, where η ′ ( u ) = ϕ ′ 0 ( u ) f ′ ( u ), ψ ′ ( u ) = ˜ ϕ ′ ( u ) f ′ ( u ). Taking f ′ ( u ) = α ( u ), where α is defined by the relation µ ( u ) = α ( u ) d | µ | ( u ), µ = ϕ ′ 0 we obtain that η ′ ( u ) = | µ | ≥ 0, i.e. η ( u ) in- creases and it is constant on the same intervals as ϕ 0 ( u ). So the problem is reduced to the case when ϕ 0 ( u ) increases. This case is treated similarly to the model case ϕ 0 ( u ) = u based on the key Kruzhkov relation for two isentropic m.v.s. ν t,x , ˜ ν t,x � � | ϕ 0 ( u ) − ϕ 0 ( v ) | t +div x sign ( λ 1 − λ 2 )( ˜ ϕ ( λ 1 ) − ˜ ϕ ( λ 2 )) dν t,x ( λ 1 ) d ˜ ν t,x ( λ 2 ) = 0 , � ϕ 0 ( λ ) dν t,x ( λ ), v = � ϕ 0 ( λ ) d ˜ u = ν t,x ( λ ). In particular, we apply this relation with ˜ ν t,x = ν t,x + h . 5