l1 solutions to rst order hyperbolic equations in bounded...
TRANSCRIPT
L1 solutions to first order hyperbolic equations in bounded
domains
Alessio Porretta∗ Julien Vovelle†
April 21, 2009
Abstract
We consider L1 solutions to the initial-boundary value problem for first order hyperbolic equations.We detail the interactions between the boundary condition and the flux function and the way theyinfluence the integrability of the solution. Following [BCW00], we define a notion of renormalizedentropy solution in order to ensure the integrability of the non-linear term in the entropy inequalities.Existence, uniqueness and stability of such a solution is established.
Keywords: L1 solution, hyperbolic equation, initial-boundary value problem, renormalized entropysolution
MSC: 35L60 (35D05 35F25 35F30 35L65)
1 Introduction
We are interested here in studying the initial boundary value problem for first order hyperbolic equationsin a bounded domain Ω ⊂ IRd
ut(x, t) + divxf(u(x, t)) = g , (x, t) ∈ Ω× (0, T ) ,
u(x, 0) = u0(x) , x ∈ Ω ,
u(x, t) = u(x, t) , (x, t) ∈ ∂Ω× (0, T ) ,
(1.1)
in case of unbounded data g, u0, u.In a recent work ([BCW00]) P. Benilan, J. Carrillo and P. Wittbold consider the Cauchy problem in
the whole space (x ∈ IRd) with data g ∈ L1 and u0 ∈ L1. In order to deal with unbounded solutionsthey introduce the definition of renormalized entropy solutions which extends the classical definition ofentropy solutions ([Vol67], [Kru70]) used in the L∞ framework. They prove existence and uniquenessof such generalized solutions which are also proved to coincide with the mild solutions constructed viasemigroup theory (see [Ben72], [Cra72]).
We deal here with the problem in a bounded domain. As it is well known in the study of boundaryvalue problems for first order hyperbolic equations, the correct meaning of the boundary condition has tobe precised since this condition may be not assumed pointwise but should be read as an entropy conditionat the boundary. This latter condition (known as the BLN condition) has been defined in [BlRN79] forsolutions u having bounded variation and data u ∈ C2(∂Ω× (0, T )). Such a condition makes no sense ifu and u are merely in L∞, consequently a new formulation of the problem was needed in the L∞ setting.It has been given by F. Otto ([Ott96]) who introduced an integral formulation of the boundary condition∗Dipartimento di Matematica, Universita’ di Roma ”Tor Vergata”, Via della Ricerca Scientifica 1, 00133 Roma -Italy,
[email protected]†Universite de Provence, CMI, F-13453 Marseille, [email protected]
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and proved that (1.1) is well posed. Further results for bounded domains in case of boundary data u inL∞ have been obtained in [Val00] and in [CW].
We focus our attention here on the role of the boundary data u in order to extend the previous resultsto an L1 framework, namely to L1 solutions of (1.1) in presence of possibly unbounded boundary datau. To this aim it is quite natural to make use of the concept of renormalized entropy solution and try toadapt the definition given in [Ott96] to this setting. The main difficulty is related to understanding theinfluence of the boundary condition, while trying to find the assumptions the function u should satisfy inorder to have suitable L1–estimates on the solution. In fact, as far as the Cauchy problem is concerned,the L1 norm of the solution is immediately estimated, thanks to the conservation of mass, in terms ofthe L1 norm of the initial datum. In a bounded domain Ω, the L1 norm of the solution also depends onthe normal trace at the boundary of the flux f(u), which may be different from f(u).
Actually, we give a quite general and simple example, constructed together with J.L. Vazquez, showingthat requiring u in L1(∂Ω× (0, T )) is not enough in general to ensure solutions in L1(Ω× (0, T )). Thisfeature had already been observed in [AM99], where the authors study the unidimensional case for astrictly convex and superlinear flux f (with g = 0), using the explicit formula given by P. Lefloch [LeF88]and constructing an example for the Burgers equation.
While in most cases the more natural assumption would be to have f(u) in L1(∂Ω × (0, T )), thishypothesis may be not enough yet to have solutions in L1, and we give here (in Section 2) an exampleshowing that this situation may actually occur. Thus, we are led to introduce a suitable assumption onu allowing solutions of finite mass; actually we define a sort of “maximal effective flux”(the maximal fluxwhich may be seen by the equation) as
f(s) = sup |f(t)| , t ∈ [−s−, s+] ,
and we assume that u is a measurable function almost everywhere finite on Σ such that
f(u) ∈ L1(∂Ω× (0, T )) . (1.2)
If the function f is monotone at infinity then this assumption reduces to ask that f(u) is in L1(∂Ω×(0, T )),and if the function f is concave, our assumption is even weaker than assuming u to be in L1. It isinteresting to note that a condition of similar kind on u, but slightly stronger than ours, has also beengiven in [NOV99] to work in an L1 framework from a kinetic point of view. Under hypothesis (1.2)we prove existence, uniqueness and continuous dependence of renormalized entropy solutions of (1.1).We leave to the next section the precise definition of solutions, further comments and main examplesconcerning assumption (1.2).
2 Statement of the results, remarks and counterexamples.
2.1 Definition of renormalized entropy solution
Let us fix some notations. For fixed T > 0, we set Q = Ω× (0, T ) and Σ = ∂Ω× (0, T ). We assume thatΩ is a bounded strongly Lipschitz subset of IRd, that
g ∈ L1(Q), (2.3)
u0 ∈ L1(Ω), (2.4)
and f : IR → IR is a locally Lipschitz continuous function. We introduce the modified flux
f(s) = sup |f(t)| , t ∈ [−s−, s+] ,
and we assume that u is a measurable function satisfying
u is finite almost everywhere on Σ, and f(u) ∈ L1(Σ) . (2.5)
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Actually, the flux f can be seen as the “maximal effective flux”which may affect the equation, as it canbe observed by solving local Riemann problems, for which solutions are determined by convex or concaveenvelopes of the flux; this will be made more clear in our Example 2.6, which motivates assumption (2.5).
For real numbers a, b, we denote (u⊥a) = min(a, b) and a>b = max(a, b), and we define the truncationfunction Tl(s) = (u⊥l)>(−l). The Lipschitz constant of f on [−l, l] will be denoted as
Lipl := Lip(f |[−l,l]).
Given a bounded measure µ on Q, we denote by ||µ|| its total mass, that is to say the norm of µ viewedas an element of the dual space C(Q)′, or, still, ||µ|| = µ(Q) if the measure µ is non-negative. Hereis the definition of renormalized entropy solution, which adapts the one introduced in [BCW00] to theformulation for boundary value problems given in [Ott96] (see also [Vov02]).
Definition 2.1 A function u of L1(Q) is said to be a renormalized entropy solution of problem (1.1) ifthere exist some non-negative bounded measures µl and νl on Q whose masses vanish with l, that is tosay
||µl|| , ||νl|| →l→+∞
0 ,
and such that the following entropy inequalities are satisfied: for all κ ∈ IR, for all l ∈ IR such thatl ≥ |κ|, for all ϕ ∈ C∞c (Ω× [0, T ); IR+),∫ T
0
∫Ω
((u⊥l − κ)+ϕt + Φ+(u⊥l, κ) · ∇ϕ
)dx dt+
∫Ω
(u0⊥l − κ)+ϕ(x, 0) dx
+Lipl
∫ T
0
∫∂Ω
(u⊥l − κ)+ϕ(x, t) dγ(x) dt +∫ T
0
∫Ω
g sgn+(u⊥l − κ)ϕdx dt ≥ −〈µl, ϕ〉 ,(2.6)
and, similarly,∫ T
0
∫Ω
((u>(−l)− κ)−ϕt + Φ−(u>(−l), κ) · ∇ϕ
)dx dt+
∫Ω
(u0>(−l)− κ)−ϕ(x, 0) dx
+Lipl
∫ T
0
∫∂Ω
(u>(−l)− κ)−ϕ(x, t) dγ(x) dt+∫ T
0
∫Ω
g sgn−(u>(−l)− κ)ϕdx dt ≥ −〈νl, ϕ〉 ,(2.7)
where we denoted by 〈·, ·〉 the duality between C(Q)′ and C(Q) and by Φ±(u, κ) the entropy-flux function
Φ±(u, κ) = sgn±(u− κ)(f(u)− f(κ)) ,
having defined sgn±(s− κ) = 0 if s = k.
Note that, as in the purpose of renormalization, all integrals appearing in (2.6) and (2.7) are restrictedto sets where u and u are bounded, so as to make sense. On the other hand, the conditions on theasymptotic behavior of the measures µl, νl allow to recover informations on the zone where u is unbounded.
Remark 2.2 In case of bounded data u, g, u0, choosing ϕ = T − t, |k| great enough and letting l go toinfinity in (2.6) and (2.7), one easily checks that any renormalized entropy solution is bounded and thencoincides with the unique entropy solution u in the sense of [Ott96], which of course can be seen as arenormalized entropy solution where the measures µl, νl are taken to be zero for l large enough.
2.2 Statement of the results
For such a notion of renormalized solution, we prove existence and uniqueness, as described in thefollowing theorem.
Theorem 2.3 Let assumptions (2.3), (2.4), (2.5) hold true. Then a renormalized entropy solution uexists and is unique. Moreover, u ∈ C0([0, T ];L1(Ω)).
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Besides, we also prove stability of solutions with respect to convergence of the data.
Theorem 2.4 Let (ui) be a sequence of renormalized entropy solutions of Problem (1.1) with data(ui0, u
i, gi). Assume thatgi → g strongly in L1(Q),
ui0 → u0 strongly in L1(Ω),
andui → u a.e. on Σ and f(ui)→ f(u) strongly in L1(Σ).
Then the sequence (ui) strongly converges in C0([0, T ];L1(Ω)) to the unique solution u of Problem (1.1)with data (u0, u, g).
The main step in the proof of Theorems 2.3 and 2.4 is proving an L1–contraction result for renormal-ized solutions only depending on the L1 norm of f(u); to this aim we follow the basic lines of the proofgiven in [Ott96] improving the estimates with respect to the boundary data and taking into accountthe presence of the extra terms given by measures µl and νl. The contraction argument allows us toobtain uniqueness and continuous dependence, which also implies compactness of bounded approximatedsolutions corresponding to truncated data. The existence is then obtained by passing to the limit in theproblem with truncated data; here the definition of the measures µl and νl appears quite naturally as in[BCW00] and our main task is proving the estimate on the masses of those measures in terms of possiblysingular boundary data.
2.3 Comment on the assumption on u.
Example 2.5 This example, which was constructed together with J. L. Vazquez, shows that assumingthe boundary datum u in L1(Σ) is not enough in order to have a priori estimates in L1(Q), so that onecan have an integrable boundary datum but solutions with infinite mass, in sharp contrast with whathappens for the Cauchy problem, where there is conservation of the mass. Let here f(s) = sp+1 withp > 1, Ω = (0, 1), u0 = 0. First of all we build a solution u which blows up at a given fixed time T in anypoint x; this is already enough to see that a priori estimates on the L1(Ω) norm of solutions at time t arenot available. In order to build such a solution u we fix the line x = ζ(t) = Tα−(T−t)α
Tα as a line of shockfor the solution, obtain the value of u from the Rankine–Hugoniot condition and transport this value
through the characteristics in the domain x < ζ(t). To this purpose, for α ∈ (0, 1), let u(s) = α1p (T−s)
α−1p
Tαp
be the limit value, from the left, of the solution u on the line x = ζ(s). For any fixed s ∈ (0, T ) thecharacteristic line starting from the point (ζ(s), s) is defined by the equation x = f ′(u(s))(t− s) + ζ(s).It is easy to check that, since the flux is convex, these characteristic lines do not meet one another inthe domain x < ζ(t); in fact for any s and x < ζ(s) there is only one value t < s such that the point(x, t) belongs to the characteristic line starting at (ζ(s), s). Rephrasing we can also say that the equationf ′(u(s))(t − s) + ζ(s) − x = 0 implicitly defines s = s(x, t) as a function of (x, t), hence the function
u(x, t) = u(s(x, t)) = α1p (T−s(x,t))
α−1p
Tαp
is well defined for x < ζ(t). Since u is defined through the methodof characteristics, it is a classical solution in the domain x < ζ(t). Indeed, a straightforward calculationgives, for x < ζ(t),
ut + f(u)x = α1p
(1− αp
)(T − s)
α−1p −1
Tαp
[∂s
∂t+ (p+ 1)α
(T − s)α−1
Tα∂s
∂x
],
and computing the partial derivatives of s(x, t) in terms of the function A(x, t, s) = f ′(u(s))(t−s)+ζ(s)−xand the equation A(x, t, s) = 0, we obtain[
∂s
∂t+ (p+ 1)α
(T − s)α−1
Tα∂s
∂x
]= −
(∂A
∂s
)−1 [f ′(u(s))− (p+ 1)α
(T − s)α−1
Tα
]= 0 .
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Define u = 0 for x > ζ(t); since by construction u satisfies the Rankine–Hugoniot condition on x = ζ(t)and since the entropy condition is ensured from the fact that the flux is convex and the shock decreasing,we have that u is an entropy solution of the problem ut + f(u)x = 0 in (0, 1)× (0, T ),
u(x, 0) = 0 in (0, 1),u(0, t) = u(s(0, t)) , u(1, t) = 0 in (0, T ).
(2.8)
As far as the boundary datum is concerned, note that since the flux is an increasing function the solutioncoincides with the boundary datum at x = 0, while any boundary datum is admissible at x = 1 so thatwe have taken a trivial choice. Computing the L1(0, T ) norm of the boundary datum gives∫ T
0
u(0, t) dt = α1p
∫ T
0
(T − s)α−1p
Tαp
[p
p+ 1+ζ(s)(1− α)
(p+ 1)αTα
(T − s)α
],
and since p > 1 and α ∈ (0, 1) we get ∫ T
0
u(0, t) dt ≤ C T 1− 1p , (2.9)
where the constant C does not depend on T . Thus the boundary datum has a finite L1 norm, but thesolution blows up at t = T and this is due to the fact that f(u(0, t)) is not in L1(0, T ), as easily checked.
The solution u previously found blows up at t = T but still belongs to L1(Q). However, the examplecan be slightly modified in order to build a solution which has infinite mass on (0, 1)× (0, T ).
Let τn = 1
n2pp−1
; define time intervals En = (en−1, en) such that e0 = 0, e2n − e2n−1 =√τn
p+1 and
e2n−1− e2n−2 = τn, for n ≥ 1. We denote by u(x, t;T ) the solution of (2.8) previously constructed in the
interval (0, T ), and we define u(0, t) =∞∑n=1
u(0, t − e2n−2; τn)χE2n−1 . Thus in all odd intervals we have
iterated the example built before and in all even intervals we have put u(0, t) = 0 and we can use that atthe initial time of the even intervals the solution is infinite. Precisely, let v be any entropy solution of vt + f(v)x = 0 in (0, 1)× (0, T ∗),
v(x, 0) = 0 in (0, 1),v(0, t) = u(t) , v(1, t) = 0 in (0, T ∗),
(2.10)
where T ∗ = limn→∞ en =∑∞n=1
(1
n2pp−1
+ 1p+1
1
npp−1
), which is finite since p > 1. Clearly v is nonnegative
so that, in each interval (e2n−2, e2n−1), v is greater than u(x, t− e2n−2; τn), which is the solution of ut + f(u)x = 0 in (0, 1)× (e2n−2, e2n−1),u(x, e2n−2) = 0 in (0, 1),u(0, t) = u(t) , u(1, t) = 0 in (e2n−2, e2n−1).
Since u(x, t − e2n−2; τn) blows up at e2n−1, in particular this implies that v(x, e2n−1) ≥ n1p−1 , hence in
each interval (e2n−1, e2n) we have that v is greater than the solution ofwt + f(w)x = 0 in (0, 1)× (e2n−1, e2n),w(x, e2n−1) = n
1p−1 in (0, 1),
w(0, t) = 0 , w(1, t) = 0 in (e2n−1, e2n).
This function w is obtained by solving a Riemann problem with data 0 and n1p−1 , and standard considera-
tions (the flux is convex) imply that w = n1p−1 if x > f ′(n
1p−1 )(t−e2n−1) = (p+1)n
pp−1 (t−e2n−1). We de-
duce that v ≥ n1p−1 for x > (p+1)n
pp−1 (t−e2n−1) and t ∈ (e2n−1, e2n) (note that e2n−e2n−1 = 1
(p+1)npp−1
).
We conclude that
‖v‖L1((0,1)×(0,T∗)) ≥1
2(p+ 1)
∞∑n=1
1n
= +∞ ,
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so that any entropy solution of (2.10) has infinite mass, whereas the boundary datum u belongs to L1(0, T )since, by (2.9),
‖u‖L1(0,T ) ≤ C∞∑n=1
τ1− 1
pn = C
∞∑n=1
1n2
<∞ .
The previous example proves that the L1 norm of the boundary datum u does not control, in general,the L1 norm of the solution, but an assumption concerning the L1 norm of f(u) is needed. In somecases, this can even lead to weaker assumptions; for instance, for all positive even concave flux functionsf , we have |f(u)| ≤ |f(u)| ≤ C|u|, in this case the L1 norm of u controls the L1 norm of the solutionbut assuming (2.5) we still obtain L1 solutions under a weaker assumption (as an example one can takef(s) =
√1 + |s|, (2.5) only requires |u| 12 , and not u, to be integrable).
It is also worth pointing out that for a large class of functions f , our hypothesis (2.5) reduces to askthat f(u) ∈ L1(Σ); for example, if f is monotone for |s| ≥ s0, then clearly
f(u) ≤ max[−s0,s0]
|f | + |f(u)| , a.e. in Σ,
and since by definition |f(u)| ≤ f(u) it follows that the two assumptions are equivalent. However, in thefollowing example we show that assuming f(u) ∈ L1(Σ) may be not enough to ensure L1 solutions, sothat requiring the stronger hypothesis (2.5) seems to be necessary.
Example 2.6 This example underlines the accuracy of the hypothesis f(u) ∈ L1(Σ) in order to haveL1(Q) solutions; more precisely, for a certain flux function f , we construct a solution of the problem ut + f(u)x = 0 in (0, 1)× (0, T ),
u(x, 0) = 0 in (0, 1),u(0, t) = u(t) , u(1, t) = u(t) in (0, T ),
(2.11)
such that u, u are almost everywhere finite and
f(u) = f(u) = 0 , f(u) /∈ L1(0, T ) , and u /∈ L1((0, 1)× (0, T )) . (2.12)
The function u is estimated by using a principle of comparison and the resolution of a sequence of RiemannProblems, as in Example 2.5. We consider the flux function f defined in the following way: let (Sn) and(Yn) be the sequences defined by Sn = (n+ 1)2 and Yn = (n+ 1)4. Set S+
n = Sn + 1, S−n = Sn − 1, and
f(s) =
Yn(s− Sn−) if S−n ≤ s ≤ Sn ,Yn(S+
n − s) if Sn ≤ s ≤ S+n ,
0 if S+n ≤ s ≤ S−n+1 .
(2.13)
Notice that the function f is defined as being piecewise affine in order to make the several resolutionsof Riemann problems more readable. A regular approximation of this flux could be considered, withoutaffecting the main lines of this counter-example.
Let an be the slope of the straight line joining the point (Sn, f(Sn)) to the point (Sn+1, f(Sn+1)) anddefine time intervals En = (en−1, en) such that e0 = 0, e2n − e2n−1 = 1
Ynand e2n−1 − e2n−2 = 1
an−1, for
n ≥ 1. Notice that
|E2n| = O(1n4
) and |E2n+1| = O(1n2
) (2.14)
so that T =∞∑n=1
|En| is well defined. We set u(t) =∞∑n=1
S+n χE2n−1 and u(t) =
∞∑n=1
S−n−1 χE2n−1 , and in
all odd intervals E2n−1 we solve the Riemann problem(v2n−1)t + f(v2n−1)x = 0 in (0, 1)× E2n−1,v2n−1(x, e2n−2) = S−n−1 in (0, 1),v2n−1(0, t) = S+
n , v2n−1(1, t) = S−n−1 in E2n−1
(2.15)
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starting from n = 1. The solution v2n−1 consists of two shocks with positive slopes 1/an−1 > 1/Yn−1 inthe (x, t)-plane, separating three constant states which are (turning counter-clockwise): S−n−1, Sn−1 andSn. Note that the time e2n−1 is defined so that the line t = e2n−2 + x
an−1touches the boundary x = 1 at
t = e2n−1. In all even intervals, we solve the Riemann problem (v2n)t + f(v2n)x = 0 in in (0, 1)× E2n,v2n(x, e2n−1) = v2n−1(e2n−1) in (0, 1),v2n(0, t) = 0 , v2n(1, t) = 0 in E2n
(2.16)
with n ≥ 1. Since v2n−1(e2n−1) = Sn, it appears, after examination of the convex and concave hulls ofthe function f between the points (0, 0) and (Sn, f(Sn)), that, cutting the rectangle (0, 1)×E2n along thediagonale with positive slope (equal to 1/Yn in the (x, t)-plane), the function v2n consists of two constantstates: S−n above the diagonale and Sn below. Thus v2n(e2n) = S−n , which is consistent in order to iteratethe construction and solve repeatedly the Riemann problems (2.15) and (2.16). We deduce that, for everyn ∈ IN, the solution u of Problem (2.11) is greater than Sn+1 on half the domain (0, 1) × E2n+1. Thenusing the estimate (2.14), it is easy to see that u does not belong to L1(Q). Moreover, since
|u > l ∪ u > l| ≤ C∑n>√l
|E2n−1| → 0 as l→∞,
we have that u and u are almost everywhere finite in virtue of (2.14). Clearly we also have f(u) = f(u) = 0and again (2.14) implies that f(u) /∈ L1(0, T ), so that (2.12) is indeed satisfied.
3 Proof of the results
3.1 Localization at the boundary
Here, we set the tools used to study different terms related to the boundary, which are localization andthe introduction of a cut-off function.
Localization. The set Ω being a strong Lipschitz open subset of IRd, there exists a finite open cover(Bα)0,N of Ω and a partition of unity (λα)0,N on Ω subordinate to (Bα)0,N such that, up to a changeof coordinates represented by an orthogonal matrix Aα, the set ∂Ω ∩ Bα is the graph of a Lipschitzcontinuous function hα, that is to say:
Ω ∩Bα = x ∈ Bα ; (Aα x)d > hα(Aα x)
and∂Ω ∩Bα = x ∈ Bα ; (Aα x)d = hα(Aα x) ,
where y stands for (yi)1,d−1 if y ∈ IRd. The problem will be localized with the help of the functionsλα. We drop the index α and, for the sake of clarity, suppose that the change of coordinates is trivial:A = Id. We set Π = x ; x ∈ B which, clearly, is an open subset of IRd−1. If a function θ is defined on∂Ω ∩B, then we denote by θ the function defined on Π by θ(x) = θ(x, h(x)). Then we have∫
∂Ω∩Bθ(x)dγ(x) =
∫Π
θ(x)√
1 + |∇h(x)|2 dx, (3.17)
where ∇ denotes the gradient operator in IRd−1. Finally, when the function θ appears in volume integralson Ω ∩ B, it will be meant that we use the constant prolongation of θ inside Ω, that is (with a slightabuse of notation) θ(x, xd) = θ(x, h(x)) for all (x, xd) ∈ Ω ∩B.
Cut-off function. Let (ρn) be a sequence of mollifiers on IR defined by
ρn(t) = nρ(nt) (3.18)
7
where ρ is a non-negative function of C∞c (−1, 0) such that∫ 0
−1
ρ(t) dt = 1. The function h is extended
outside Π so as to get a Lipschitz continuous function. For ε a positive number, we also denote by ρε thefunction t 7→ ρ1/ε(t) and define the cut-off function ωε by
ωε(x) =∫ 0
h(x)−xdρε(σ) dσ . (3.19)
On Ω ∩ B, the function ωε vanishes in a neighbourhood of ∂Ω and equals 1 if dist(x, ∂Ω) > ε. Ifψ ∈ H1(Ω)N then∫
Ω
λψ · ∇ωε dx = −∫
Ω
div(λψ)ωε dx→ −∫
Ω
div(λψ) dx = −∫∂Ω
λψ · ndγ(x) , (3.20)
Also notice that an estimate on the L1 norm of the gradient of ωε (uniform with respect to ε) holds.Indeed, from the definition (3.19), we have
∇ωε(x) = ρε(h(x)− xd)(−∇h(x)
1
), (3.21)
and, in particular:
∀x ,∫xd>h(x)
|∇ωε(x)| dxd ≤√
1 + |∇h(x)|2 . (3.22)
Eventually, if ψ is a smooth function defined in a neighbourhood V of ∂Ω ∩ B then we have, using also(3.22), ∫
V ∩Ω
ψ(x) · ∇ωε(x) dx =∫V ∩Ω
ψ(x) · ∇ωε(x) dx+ o(1) [ε→ 0] . (3.23)
3.2 L1–contraction principle.
Here, we aim at proving the following result.
Theorem 3.1 Let u and v be two renormalized entropy solutions of Problem (1.1) with data (u0, u, g)and (v0, v, g) respectively. Then we have∫
Ω
|u− v|(t) ≤∫ t
0
∫∂Ω
S(u, v) dγ(x) ds+∫ t
0
∫Ω
|g − g| dx ds+∫
Ω
|u0 − v0| dx , (3.24)
for almost every t ∈ (0, T ), where
S(u, v)(t, x) = sup 2 |f(σ)− f(τ)| ; min(u(t, x), v(t, x)) ≤ σ , τ ≤ max(u(t, x), v(t, x)) , (t, x) ∈ Σ .
This theorem extends the known contraction result in case the boundary data are not bounded, andthe main novelty with respect to previous results is the possibility to obtain this estimate depending onf(u) and f(v) at the boundary. Its proof however is an adaptation of the proof given by Otto in case ofbounded data, proof which lies itself on the tecnhique of doubling variables of Kruzkov-Kutnetzov. As in[Ott96], preliminary results are required, in order to deal with the boundary terms in an accurate way.To expose them, we localize again and refer to the notations previously introduced.
Lemma 3.2 Let u ∈ L1(Q) be a renormalized entropy solution of the problem (1.1), let v be a measurablefunction on Σ and ϕ ∈ C∞c (Ω× [0, T )). Then
1. for every l ≥ |κ|, the following limit is well defined and, if the test function ϕ is nonnegative,bounded as follows:
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ>Tl(u)) · ∇ωε λϕdx dt ≤∫ T
0
∫∂Ω
λϕdµl , (3.25)
where, in the right hand-side, we still denoted by µl the restriction of the measure µl to ∂Ω× [0, T ].
8
2. Similarly, for every l ≥ 0 the following limit
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, Tl(v)) · ∇ωε λϕdx dt
exists and only depends on the value ϕ = ϕ|∂Ω×[0,T ). Actually, this limit defines a linear continuousform on L1(Σ, λ dγ dt), which is represented by a function K+
l (u, v) of L∞(Σ, λ dγ dt), that is wehave
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, Tl(v)) · ∇ωε λϕdx dt =∫ T
0
∫∂Ω
K+l (u, v)λϕdγ(x) dt . (3.26)
Proof: Let us set
L(ϕ) = ∫ T
0
∫Ω
((u⊥l − κ)+ϕt + Φ+(u⊥l, κ) · ∇ϕ
)dx dt+
∫Ω
(u0⊥l − κ)+ϕ(x, 0) dx
+ Lipl
∫ T
0
∫∂Ω
(u⊥l − κ)+ϕ(x, t) dγ(x) dt +∫ T
0
∫Ω
g sgn+(u⊥l − κ)ϕdx dt+ 〈µl, ϕ〉 .
By definition of entropy solution we have that L is a linear nonnegative operator. Letting ωε be definedin (3.19), and ϕ ≥ 0, since (1 − ωε)ϕλ is a nonincreasing sequence we deduce that L((1 − ωε)ϕλ) isnonincreasing and nonnegative, hence admits a limit as ε tends to zero. In particular this implies thefollowing inequality
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕdx dt ≤ Lipl
∫ T
0
∫∂Ω
(u⊥l − κ)+λϕ(x, t) dγ(x) dt + limε→0〈µl, (1− ωε)λϕ〉 .
Using the monotone convergence theorem we have
limε→0〈µl, (1− ωε)λϕ〉 =
∫ T
0
∫∂Ω
λϕdµl .
Moreover, since Φ+(u⊥l, κ) is bounded for l fixed and k ≥ −l, reasoning as in (3.23) we easily get that
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕdx dt = limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕdx dt ,
so that we obtain
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕdx dt ≤ Lipl
∫ T
0
∫∂Ω
(u⊥l − κ)+λϕdγ(x) dt +∫ T
0
∫∂Ω
λϕdµl . (3.27)
Our main task is now replacing ϕ with ϕχB in (3.27) for any borelian subset B ⊂ ∂Ω × [0, T ). Indeed,given such a B, for any δ > 0 there exists a function ψδ ∈ C∞c (Ω× [0, T )) such that
‖ψδ − χB‖L1(Σ) + ‖ψδ − χB‖L1(Σ,dµl) ≤ δ , (3.28)
where ψδ = ψδ|∂Ω×[0,T ). To construct such a ψδ, it is enough to use Uryshon lemma applied to an openset U and to a compact set K ⊂ U such that γ(U \ K) + µl(U \ K) ≤ δ (recall that γ is the d − 1dimensional Lebesgue measure), the existence of U and K being given by the regularity of µl and of theLebesgue measure on Σ. Define B = (x, xd) ∈ Ω : (x, h(x)) ∈ B, which is a thickening of B inside Ω.Since we have, by (3.22),∣∣∣∣∣
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕ(χB − ψδ) dx dt
∣∣∣∣∣ ≤ C(l)‖ϕ‖L∞(Σ)
∫ T
0
∫∂Ω
|χB − ψδ|λ dγ(x) dt ,
9
we deduce from (3.27) and (3.28)
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕχB dx dt
≤ limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕψδ dx dt+ C δ
≤ Lipl
∫ T
0
∫∂Ω
(u⊥l − κ)+λϕψδ dγ(x) dt +∫ T
0
∫∂Ω
λϕψδ dµl + Cδ ,
and then, as δ tends to zero, thanks to (3.28)
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ) · ∇ωε λϕχB dx dt ≤ Lipl
∫ T
0
∫∂Ω
(u⊥l − κ)+λϕχB dγ(x) dt
+∫ T
0
∫∂Ω
λϕχB dµl .
(3.29)
Let now sn =n∑i=1
wiχBi be a sequence of step functions strongly converging to Tl(v) in L1(Σ), with
|wi| ≤ l. Choosing B = Bi, replacing κ with κ>wi in (3.29) and summing over i we get
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ>sn) · ∇ωε λϕdx dt ≤ Lipl
∫ T
0
∫∂Ω
(u⊥l − κ>sn)+λϕdγ(x) dt
+∫ T
0
∫∂Ω
λϕdµl .
(3.30)
Using again the following estimate (uniform on ε)∫ T
0
∫Ω
∣∣[Φ+(u⊥l, κ>sn)− Φ+(u⊥l, κ>Tl(v))]· ∇ωε λϕ
∣∣ dx dt≤ Lipl ‖ϕ‖L∞(Σ)
∫ T
0
∫∂Ω
|sn − Tl(v)|λ dγ(x) dt ,
we obtain from (3.30):
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, κ>Tl(v)) · ∇ωε λϕdx dt ≤ Lipl
∫ T
0
∫∂Ω
(u⊥l − κ>Tl(v))+λϕdγ(x) dt
+∫ T
0
∫∂Ω
λϕdµl .
(3.31)
Applying this result with v = u (we use that (u⊥l − k>Tl(u))+ = 0) proves (i). The proof of (ii) isanalogous in what concerns the fact that there exists
limε→0
∫ T
0
∫Ω
Φ+(u⊥l, Tl(v)) · ∇ωε λϕ dx dt = limε→0
∫ T
0
∫Ω
Φ+(u⊥l, Tl(v)) · ∇ωε λϕdx dt .
Now, define
Λε(ϕ) =∫ T
0
∫Ω
Φ+(u⊥l, Tl(v)) · ∇ωε λϕdx dt ,
10
if ϕ ∈ C∞c (Ω × [0,+∞); IR+). Since the function Φ+(u⊥l, Tl(v)) is bounded a.e., say by Ml, we have,according to the estimate (3.22) on the gradient of the cut-off function ωε,
Λε(ϕ) =∫ T
0
∫Π
∫xd>h(x)
Φ+(u⊥l, Tl(v))(x, t) · ∇ωε(x)λ(x)ϕ(x, t) dxd dx dt
≤ Ml
∫ T
0
∫Π
√1 + |∇h(x)|2λ(x, t)|ϕ|(x) dx dt
= Ml
∫ T
0
∫∂Ω
λϕdγ dt .
(3.32)
Therefore, the function Λε is well defined on L1(Σ, λ dγ dt) and induces a linear form on this space. Itsnorm is bounded independently of ε and it is pointwise converging. Therefore, Λ : β 7→ lim
ε→0Λε(β) is a con-
tinuous linear form on L1(Σ, λ dγ dt) which can be represented by an element K+l (u, v) of L∞(Σ, λ dγ dt).
Remark 3.3 The two conclusions of Lemma 3.2 together imply the existence, in a weak sense, of thenormal trace of φ+(u⊥l, k>Tl(u)) at the boundary, this normal trace being represented by the L∞ functionK+l (u, k>Tl(u)). Note that requiring, from the definition of renormalized solution, that ‖µl‖ goes to zero
as l tends to infinity implies that K+l (u, k>Tl(u)) converges to zero in L1(Σ) as l tends to infinity for any
|k| ≤ l.
Working on the ”negative” semi Kruzkov entropy functions u 7→ (u− κ)−, analogous results to thoseof Lemma 3.2 can be proved, that is:
1. for every l ≥ |κ|, if the test function ϕ is nonnegative, there holds
limε→0
∫ T
0
∫Ω
Φ−(u>(−l), κ⊥Tl(u)) · ∇ωε λϕdx dt ≤∫ T
0
∫∂Ω
λϕdνl . (3.33)
2. There exists K−l (u, v) ∈ L∞(Σ, λ dγ dt) such that for every l ≥ 0 and for every ϕ ∈ C∞c (Ω× [0, T )),
limε→0
∫ T
0
∫Ω
Φ−(u>(−l), Tl(v)) · ∇ωε λϕdx dt =∫ T
0
∫∂Ω
K−l (u, v)λϕd γ(x) dt . (3.34)
Gathering these two kind of results, we obtain an entropic formulation for the truncations of the solutions.
Proposition 3.4 Let u ∈ L1(Q) be a renormalized entropy solution of the problem (1.1), let v be ameasurable function on Σ and ϕ ∈ C∞c (Ω× [0, T )). Then
1. There exists a function Kl(u, v) ∈ L∞(Σ, λ dγ dt) such that
limε→0
∫ T
0
∫Ω
Φ(Tl(u), Tl(v)) · ∇ωε λϕdx dt =∫ T
0
∫∂Ω
Kl(u, v)λϕd γ(x) dt (3.35)
for every l and every ϕ ∈ C∞c (Ω× [0, T )).
2. For every k ∈ IR, l ∈ IR+, with |k| ≤ l, and for every ϕ ∈ C∞c (Ω× [0, T ),∫ T
0
∫Ω
|Tl(u)− κ|(λϕ)t + Φ(Tl(u), κ) · ∇(λϕ) dx dt+∫
Ω
|Tl(u0)− κ|λ(x)ϕ(x, 0) dx
−∫ T
0
∫∂Ω
(Kl(u, u)− Φ(κ, Tl(u)) · n)λϕd γ(x) dt
+∫ T
0
∫Ω
g sgn(Tl(u)− κ)λϕdx dt ≥ −2〈µl + νl, λ ϕ〉 .
(3.36)
11
Proof: In view of the identity
(u⊥l − κ)+ + (u>(−l)− κ)− = |Tl(u)− κ| , (3.37)
which is valid for every u, l, κ ∈ IR with l ≥ |κ|, (3.35) follows from (3.26) and (3.34) by simply definingthe function Kl(u, v) by
Kl(u, v) = K+l (u, v) +K−l (u, v) .
Moreover, once the limit in (3.35) has been defined thanks to the function Kl(u, v), we have the followingestimate: for all κ ∈ IR, for all ϕ ∈ C∞c (Ω× [0, T )),
limε→0
∫ T
0
∫Ω
Φ(Tl(u), κ) · ∇ωε λϕdx dt
≤ −∫ T
0
∫∂Ω
(Kl(u, u)− Φ(κ, Tl(u)) · n)λϕd γ(x) dt+ 2∫ T
0
∫∂Ω
λϕd(µl + νl) .
(3.38)
The proof of (3.38) lies on the manipulation of algebraic formula. Indeed, defining as in [MNRR96] thefunction F by
F(s, κ, w) = 2(Φ+(s, κ>w) + Φ−(s, κ⊥w)
),
we haveΦ(Tl(u), κ) = F(Tl(u), κ, Tl(u))− Φ(Tl(u), Tl(u)) + Φ(κ, Tl(u))
which yields, by (3.35),
limε→0
∫ T
0
∫Ω
Φ(Tl(u), κ) · ∇ωε λϕdx dt
≤ limε→0
∫ T
0
∫Ω
F(Tl(u), κ, Tl(u)) · ∇ωε λϕdx dt−∫ T
0
∫∂Ω
(Kl(u, u)− Φ(κ, Tl(u)) · n)λϕd γ(x) dt .
Moreover, one can check that, for l ≥ |κ|, one has
F(Tl(u), κ, Tl(u)) = 2(Φ+(u⊥l, κ>Tl(u)) + Φ−(u>(−l), κ⊥Tl(u))
),
and deduce from (3.25) and (3.33) the estimate
limε→0
∫ T
0
∫Ω
F(Tl(u), κ, Tl(u)) · ∇ωε λϕdx dt ≤ 2∫ T
0
∫∂Ω
λϕd(µl + νl).
Estimate (3.38) then follows. It is used to derive the entropy inequalities for the truncations Tl(u).Indeed, adding entropy inequalities (2.6) and (2.7) where the function ωε λϕ has been chosen as a testfunction yields, thanks to formula (3.37): for l ≥ |κ|,
∫ T
0
∫Ω
(|Tl(u)− κ|(λϕ)t + Φ(Tl(u), κ) · ∇(λϕ)) dx dt+∫
Ω
|Tl(u0)− κ|λ(x)ϕ(x, 0) dx
+∫ T
0
∫Ω
Φ(Tl(u), κ) · ∇ωε λϕdxdt+∫ T
0
∫Ω
g sgn(Tl(u)− κ)λϕdxdt
≥ −∫ T
0
∫Ω
ωε λϕd(µl + νl) + o(1) .
Here o(1) is a quantity which group together the terms in which the function 1 − ωε is integrated and,therefore, tends to zero with ε. Consequently, passing to the limit [ε→ 0], using that ωε ≤ 1 and takinginto account estimate (3.38) yields the entropy inequality (3.36).
12
We prove the fundamental L1–contraction principle by using the doubling variable technique ofKruzkov-Kutnetsov.
Proof of Theorem 3.1.Let u and v ∈ L1(Q) be two entropy solutions of Problem (1.1) associated to the data (u0, u, g) and
(v0, v, g) respectively and let ρp,q,r be the function defined on IRd+1 by
ρp,q,r(t, x) = ρp(t)ρq(x)ρr(xd)
where the mollifying function ρn is defined by (3.18). To compare u with v, choose κ = Tl(v(y, s)) ininequality (3.36) and (x, t) 7→ ρp,q,r(t− s, x− y)ψ(x, t) as a test function, where ψ ∈ C∞c (Ω× [0, T ); IR+),then integrate with respect to (y, s). This yields the inequality
A1 +A2 +A3 +A4 +A5 +A6 +A7 ≥ A8 (3.39)
where
A1 =∫ T
0
∫Ω
∫ T
0
∫Ω
|Tl(u(x, t))− Tl(v(y, s))|(λψ)t ρp,q,r(t− s, x− y)
A2 =∫ T
0
∫Ω
∫ T
0
∫Ω
|Tl(u(x, t))− Tl(v(y, s))|λψ (∂tρp,q,r)(t− s, x− y) dx dt dy ds
A3 =∫ T
0
∫Ω
∫ T
0
∫Ω
Φ(Tl(u(x, t)), Tl(v(y, s)))∇(λψ) ρp,q,r(t− s, x− y) dx dt dy ds
A4 =∫ T
0
∫Ω
∫ T
0
∫Ω
Φ(Tl(u(x, t)), Tl(v(y, s)))λψ (∇ρp,q,r)(t− s, x− y) dx dt dy ds
A5 =∫ T
0
∫Ω
∫Ω
|Tl(u0)− Tl(v(y, s))|(λψ)(x, 0) ρp,q,r(−s, x− y) dx dy ds
A6 = −∫ T
0
∫Ω
∫ T
0
∫Π
(Kl(u, u)(x, t)
√1 + |∇h(x)|2 − Φ(Tl[v(y, s)], Tl[u(x, t)]) ·
(−∇h(x)
1
))×λψ(x, t)ρp(t− s) ρq(x− y)ρr(h(x)− yd) dx dt dy ds
A7 =∫ T
0
∫Ω
∫ T
0
∫Ω
g sgn(Tl(u(x, t))− Tl(v(y, s)))λψ ρp,q,r(t− s, x− y) dx dt dy ds
A8 = −2 < µl + νl, λ ψ > .
Notice that, in A8, the mollifiers have been integrated with respect to (y, s) and that Identity (3.17) hasbeen used to write the term A6.
The function v also satisfies the entropy inequality∫ T
0
∫Ω
|Tl(v)− κ|θs + Φ(Tl(v), κ) · ∇θ dy ds+∫
Ω
|Tl(v0(y))− κ|θ(y, 0)∫ T
0
∫Ω
∫ T
0
∫Ω
f(u)(s, y) sgn(Tl(v(y, s))− κ) θ dy ds ≥ −∫ T
0
∫Ω
θ d(µl + νl)(3.40)
for every θ ∈ C∞c (Ω × [0, T ); IR+) and l ≥ |κ|. In (3.40), choose κ = Tl(u(x, t)) and θ : (y, s) 7→ρp,q,r(t − s, x − y)λ(x)ψ(x, t) as a test function (notice that for y ∈ ∂Ω × [0, T ] one has xd − yd ≥h(x)−h(y) ≥ −Liph |x−y|, hence, for q small enough with respect to r, the function θ vanishes, because
13
the function ρ, through which the mollifier ρn is defined, is supported on the left of zero). Integratingthe result with respect to (x, t) yields the inequality
−A2 −A4 +B7 ≥ −∫ T
0
∫Ω
θ d(µl + νl)
which, added to (3.39) gives
A1 +A3 +A5 +A6 +A7 +B7 ≥ A8 −∫ T
0
∫Ω
θ d(µl + νl) , (3.41)
where
B7 =∫ T
0
∫Ω
∫ T
0
∫Ω
g(s, y) sgn(Tl(v(y, s))− Tl(u(x, t)))λψ ρp,q,r(t− s, x− y) dx dt dy ds .
To deal with the term A5 , let us denote by Rp the function defined by
Rp(s) =∫ −s−∞
ρp(z) dz
(notice that (Rp) is bounded and converges to zero everywhere on (0,+∞)) and let us choose κ =Tl(u0(x)), θ(y, s) = Rp(s)ρq,r(x − y)λ(x)ψ(x, 0) in the entropy inequality (3.40). Integrating the resultwith respect to x yields the following upper bound on A5:
A5 ≤ C2 + C5 + C7 + C8
where
C2 = −∫
Ω
∫ T
0
∫Ω
Φ(Tl(v(y, s)), Tl(u0(x))) · (∇ρq,r)(x− y)Rp(s)(λψ)(x, 0)dx dy ds ,
C5 =∫
Ω
∫Ω
|Tl(v0(y))− Tl(u0(x))|ρq,r(x− y)(λψ)(x, 0)dx dy ,
C7 =∫
Ω
∫ T
0
∫Ω
g sgn(Tl(v(y, s))− Tl(u0(x)))λ(x)ρq,r(x− y)Rp(s)ψ(x, 0)dx dy ds ,
C8 =∫
Ω
∫ T
0
∫Ω
λ(x)ρq,r(x− y)Rp(s)ψ(x, 0) d(µl + νl)(y, s) dx .
The theorem of convergence in means ensures limp→+∞
C2 = 0 and limp→+∞
C7 = 0 (because the sequence
(Rp) converges to 0 in L1) and |C8| ≤ ‖ψ‖L∞(Q)(‖µl‖+ ‖νl‖).Besides, owing to the continuity of the translations in L1, we have
B7 =∫ T
0
∫Ω
∫ T
0
∫Ω
g(x, t) sgn(Tl(v(y, s))− Tl(u(x, t)))λψ ρp,q,r(t− s, x− y) dx dt dy ds+ ηp,q,r
where ηp,q,r tends to zero when p, q, and r tend to ∞. Since sgn(−a) = −sgn(a), a ∈ IR, there holds:
A7 +B7 ≤ F7 =∫ T
0
∫Ω
|g − g|λψ dx dt+ ηp,q,r .
Therefore, first passing to the limit p→ +∞, then q → +∞ in inequality (3.41) yields
A1 +A3 + C5 +A6 + F7 + ηr ≥ A8 − ‖ψ‖L∞(Q)(‖µl‖+ ‖νl‖) (3.42)
14
where limr→+∞
ηr = 0 and
A1 =∫
Ω
∫ T
0
∫yd>h(x)
|Tl(u(x, t))− Tl(v(x, yd, t))|(λψ)t ρr(xd − yd)dx dt dyd
A3 =∫
Ω
∫ T
0
∫yd>h(x)
Φ(Tl(u(x, t)), Tl(v(x, yd, t)))∇(λψ) ρr(xd − yd) dx dt dyd
A6 = −∫ T
0
∫Π
∫yd>h(x)
(Kl(u, u)(x, t)
√1 + |∇h(x)|2 − Φ(Tl[v(x, yd, t)], Tl[u(x, t)]) ·
(−∇h(x)
1
))×λψ(x, t)ρr(h(x)− yd) dx dt dyd
C5 =∫
Ω
∫yd>h(x)
|Tl(v0(x, yd))− Tl(u0(x))|ρr(xd − yd)(λψ)(x, 0)dx dyd .
Replacing yd by xd in the term A6 and using Formula (3.21) and Identity (3.17), we have
A6 = −∫ T
0
∫∂Ω
Kl(u, u)λψdγ(x) dt+∫ T
0
∫Ω
Φ(Tl[v(x, xd, t)], Tl[u(x, t)]) · ∇ω1/r(x)λψ dx dt ,
so that
limr→+∞
A6 = −∫ T
0
∫∂Ω
(Kl(u, u)−Kl(v, u)
)λψdγ(x) dt .
From the theorem of convergence in means we derive the limit as r tends to +∞ of A1, A3 and C5
respectively to deduce from inequality (3.42) the following result:∫Ω
∫ T
0
|Tl(u)− Tl(v)|(λψ)t + Φ(Tl(u), Tl(v))∇(λψ) dx dt+∫
Ω
|Tl(u0)− Tl(v0)|λψ(0) dx
−∫ T
0
∫∂Ω
(Kl(u, u)−Kl(v, u)
)λψdγ(x) dt
+∫ T
0
∫Ω
|g − g|λψ dx dt ≥ −C‖ψ‖L∞(Q) (‖µl‖+ ‖νl‖+ ‖µl‖+ ‖νl‖) .
(3.43)
Changing the role of the solutions u and v in (3.43), then adding the result (side by side) to (3.43) anddividing the result by 2, recalling that the function λ is actually one of the functions of localizations(λα)0,N and summing the previous result over α ∈ 0, . . . , N we get:∫
Ω
∫ T
0
|Tl(u)− Tl(v)|ψt + Φ(Tl(u), Tl(v))∇ψ dx dt+∫
Ω
|Tl(u0)− Tl(v0)|ψ(0) dx
−∫ T
0
∫∂Ω
(Kl(u, u)−Kl(v, u) +Kl(v, v)−Kl(u, v)
)ψ dγ(x) dt
+∫ T
0
∫Ω
|g − g|ψ dx dt ≥ −C2‖ψ‖L∞(Q) (‖µl‖+ ‖νl‖+ ‖µl‖+ ‖νl‖) .
(3.44)
Coming back to the definition of Kl(u, v), we have∫ T
0
∫∂Ω
(Kl(u, u)−Kl(v, u) +Kl(v, v)−Kl(u, v)
)ψ dγ(x) dt
= limε→0
∫ T
0
∫Ω
(Φ(Tl(u), Tl(u))− Φ(Tl(v), Tl(u)) + Φ(Tl(v), Tl(v))− Φ(Tl(u), Tl(v))
)∇ωεψ dx dt .
15
Discussing the respective positions of u, v, u, v, we have a pointwise estimate:∣∣Φ(Tl(u), Tl(u))− Φ(Tl(v), Tl(u)) + Φ(Tl(v), Tl(v))− Φ(Tl(u), Tl(v))∣∣ ≤ S(u, v) ,
where
S(u, v)(t, x) = sup 2 |f(σ)− f(τ)| ; min(u(t, x), v(t, x)) ≤ σ , τ ≤ max(u(t, x), v(t, x)) , (t, x) ∈ Σ .
Then we deduce, using also (3.22),
∣∣ ∫ T
0
∫∂Ω
(Kl(u, u)−Kl(v, u) +Kl(v, v)−Kl(u, v)
)ψ dγ(x) dt
∣∣ ≤ ∫ T
0
∫∂Ω
S(u, v)ψ dγ(x) dt . (3.45)
Taking also into account the estimate (3.45) and passing to the limit [l→ +∞] we get from (3.44)
−∫
Ω
∫ T
0
|u− v|ψt ≤∫ T
0
∫∂Ω
S(u, v)ψ dγ(x) dt+∫ T
0
∫Ω
|g − g|ψ dx dt+∫
Ω
|u0 − v0|ψ(0) dx ,
for every nonnegative ψ ∈ C∞c [0, T ). Since S(u, v) ∈ L1(Σ) by (2.5), and since |g− g| ∈ L1(Q), |u0−v0| ∈L1(Ω), standard arguments allow to obtain (3.24).
3.3 Existence and uniqueness
For n ∈ IN, we define the truncated data gn = Tn(g), un0 = Tn(u0) and un = Tn(u). There exists a uniqueentropy solution un ∈ L∞(Q) of the problem (1.1) with data gn, un0 and un [Ott96], moreover |un| ≤ nand un satisfies the following entropy inequalities: for every κ ∈ IR, for every ϕ ∈ C∞c (Ω× [0, T ); IR+),∫ T
0
∫Ω
((un − κ)±ϕt + Φ±(un, k) · ∇ϕ) dx dt+∫
Ω
(un0 − κ)±ϕ(x, 0) dx
+Lipn
∫ T
0
∫∂Ω
(un − κ)±ϕ(x, t) dγ(x) dt +∫ T
0
∫Ω
gn sgn±(un − κ)ϕdx dt ≥ 0.(3.46)
Remark 3.5 Ineq, (3.46) remains true if the test-function ϕ does not vanish on T , that is ϕ ∈ C∞(Ω×[0, T ]; IR+). This can be seen by considering ϕ ξα as a test function in that case, the function ξα being acut-off function defined by ξα(t) =
∫ 0
t−T ρα for example, and by noticing that the additional term∫ T
0
∫Ω
(un − κ)± ϕ ξ′α(t) dx dt
is non-positive for every α > 0.
First, we will reformulate the entropy inequalities in order to prove the convergence of the sequence (un)to a solution of Problem (1.1). To that purpose, let us set (for l ∈ IR and ϕ ∈ C∞(Ω× [0, T ]), with λ thecut–off function defined in Section 3.1)
T +λ (un, l, ϕ) =
∫ T
0
∫Ω
((un − l)+ λϕt + Φ+(un, l) · ∇(ϕλ) dx dt+∫
Ω
(un0 − l)+ϕ(x, 0)λ dx
+∫ T
0
∫Ω
gn sgn+(un − l)ϕλdx dt ,
(3.47)and
< µλl,n, ϕ >= T +λ (un, l, ϕ) + lim
ε→0
∫ T
0
∫Ω
(Φ+(un, l⊥un)− Φ+(un, un)
)· ∇ωε ϕλdx dt . (3.48)
16
Finally, recalling that the function λ is actually an element of a partition of unit (λα)0,N , we globallydefine the functional
< µl,n, ϕ >=N∑α=0
< µλαl,n, ϕ > . (3.49)
Lemma 3.6 The linear functional µl,n is well defined; it is a non-negative measure on Ω × [0, T ] and,for positive l ≤ n, its mass can be estimated as follows:
||µl,n|| ≤∫
Ω
(u0 − l)+ dx+ 2∫ T
0
∫∂Ω
f(u) 11u>l dγ dt+∫ T
0
∫Ω
11un>l |g| dx dt . (3.50)
Proof: Reasoning as in Lemma 3.2 we deduce that the functional µλl,n is well defined. Indeed, from(3.46), applied with κ = l ∈ [−n, n], and Remark 3.5, we have that the linear functional
ϕ 7→ T +λ (un, l, ϕ) + Lipn
∫ T
0
∫∂Ω
(un − l)+ϕλdγ(x) dt
is non-decreasing on C∞(Ω× [0, T ]). Consequently, if ψ ∈ C∞(Ω× [0, T ]; IR+), then
ε 7→ T +λ (un, l, ψ(1− ωε)) + Lipn
∫ T
0
∫∂Ω
(un − l)+ψλdγ(x) dt
is a nonincreasing sequence of non-negative reals and, therefore, admits a limit as ε → 0, which is non-negative. Since lim
ε→0(1 − ωε) = 0, the limit considered in the following inequality is well defined and we
have:
limε→0
∫ T
0
∫Ω
Φ+(un, l) · ∇ωε ψλdx dt ≤ Lipn
∫ T
0
∫∂Ω
(un − l)+ψλdγ(x) dt .
Moreover, this result is still true if l is replaced with l>un (see [Ott96], [Vov02], but the result is alsocontained in Lemma 3.2 if µl = 0 ). This proves that the following limit exists and is non positive:
limε→0
∫ T
0
∫Ω
Φ+(un, l>un) · ∇ωε ψλdx dt ≤ 0 . (3.51)
Using equation(3.46) with ϕ = ψλωε as a test function we get, as ε tends to zero,
T +λ (un, l, ψ) + lim
ε→0
∫ T
0
∫Ω
Φ+(un, l) · ∇ωε ψλdx dt ≥ 0 . (3.52)
Combining this last inequality with Inequation (3.51) and the algebraic identity
Φ+(un, l) = Φ+(un, l>un) + Φ+(un, l⊥un)− Φ+(un, un)
we obtain that µλl,n is well defined and it is a non-negative measure on Ω× [0, T ]; besides it has a finitemass. Indeed we have
||µλl,n|| =< µλl,n, 1 >=∫
Ω
(un0 − l)+λ dx+∫ T
0
∫Ω
Φ+(un, l) · ∇λ dx dt
+∫ T
0
∫Ω
gn sgn+(un − l)λ dx dt
+ limε→0
∫ T
0
∫Ω
(Φ+(un, l⊥un)− Φ+(un, un)
)· ∇ωε λ dx dt .
(3.53)
Suppose 0 < l ≤ n. If l ≥ un, then Φ+(un, l⊥un)− Φ+(un, un) = 0. If l < un then
Φ+(un, l⊥un)− Φ+(un, un) =f(un)− f(l) if l ≤ un ≤ un ,f(un)− f(l) if un < un ,
17
and in both cases one has |Φ+(un, l⊥un)− Φ+(un, un)| ≤ 2 f(u). Therefore we have
||µλl,n|| ≤∫
Ω
(un0 − l)+λ dx+ 2 limε→0
∫ T
0
∫Ω
f(u) 11u>l|∇ωε|λ dx dt
+∫ T
0
∫Ω
11un>l |gn|λ dx dt+∫ T
0
∫Ω
Φ+(un, l) · ∇λ dx dt ,
and taking into account the estimate (3.22) and the fact that un0 = Tn(u0), gn = Tn(g), we get
||µλl,n|| ≤∫
Ω
(u0 − l)+λ dx+ 2∫ T
0
∫∂Ω
f(u) 11u>l λ dγ dt
+∫ T
0
∫Ω
11un>l |g|λ dx dt+∫ T
0
∫Ω
Φ+(un, l) · ∇λ dx dt ,
Since ||∑Nα=0 µ
λαl,n|| ≤
∑Nα=0 ||µ
λαl,n||, we end up with the estimate (3.50).
Proposition 3.7 Let un be the solution of problem (1.1) with data gn, un0 and un. Then, for all κ, l,with n ≥ l ≥ |κ|, for all non-negative functions ϕ in C∞c (Ω× [0, T )), the function un satisfies the followingentropy inequality:∫ T
0
∫Ω
((un⊥l − κ)+ϕt + Φ+(un⊥l, κ) · ∇ϕ) dx dt+∫
Ω
(un0⊥l − κ)+ϕ(x, 0) dx
+Lipl
∫ T
0
∫∂Ω
(un⊥l − κ)+ϕ(x, t) dγ(x) dt +∫ T
0
∫Ω
gn sgn+(un⊥l − κ)ϕdx dt ≥ − < µl,n, ϕ > ,
(3.54)where µl,n is defined in (3.49).
Proof: the proof is quite directly deduced from the formula
(u⊥l − κ)+ = (u− κ)+ − (u− l)+ .
Indeed, this implies∫ T
0
∫Ω
(un⊥l − κ)+λϕt + Φ+(un⊥l, κ) · ∇(ϕλ) dx dt+∫
Ω
(un0⊥l − κ)+ϕ(x, 0)λ dx
+∫ T
0
∫Ω
gn sgn+(un⊥l − κ)ϕλdx dt = T +λ (un, κ, ϕ)− T +
λ (un, l, ϕ) . (3.55)
By definition of µλl,n (Eq.(3.48)), we have
T +λ (un, κ, ϕ)− T +
λ (un, l, ϕ) =< µλκ,n, ϕ > − < µλl,n, ϕ > +δε(l, κ) (3.56)
where δε(l, κ) = − limε→0
(∫ T
0
∫Ω
(Φ+(un, κ⊥un)− Φ+(un, l⊥un)
)· ∇ωε ϕλdx dt
).
Since |κ| ≤ l, we have Φ+(un, κ⊥un)− Φ+(un, l⊥un) = 0 if un ≤ κ and, therefore,
|Φ+(un, κ⊥un)− Φ+(un, l⊥un)| ≤ Lipl(l⊥un − k)+ ,
so that |δε(l, κ)| ≤ Lipl
∫ T
0
∫∂Ω
(un⊥l − κ)+ϕλdγ dt . Also noticing that < µλκ,n, ϕ >≥ 0 (see Lemma
3.6), we get from (3.55) and (3.56)∫ T
0
∫Ω
(un⊥l − κ)+λϕt + Φ+(un⊥l, κ) · ∇(ϕλ) dx dt+∫
Ω
(un0⊥l − κ)+ϕ(x, 0)λ(x) dx
+Lipl
∫ T
0
∫∂Ω
(un⊥l − κ)+ϕλdγ(x) dt +∫ T
0
∫Ω
gn sgn+(un⊥l − κ)ϕλdx dt ≥ − < µλl,n, ϕ > .
18
Summing the previous inequality over α ∈ 0, . . . , N and recalling the definition of µl,n in (3.49) we get(3.54).
We are now in the position to conclude the proof of the existence of a solution, and then to proveTheorem 2.3.
Proof of Theorem 2.3.We prove first the existence of a renormalized entropy solution as limit of the sequence un of bounded
solutions satisfying (3.46). Since bounded entropy solutions are in particular renormalized entropy solu-tions (see Remark 2.2), we can apply Theorem 3.1 to un and um, obtaining:∫
Ω
|un − um|(t) ≤∫ T
0
∫∂Ω
S(Tn(u), Tm(u)) dγ(x) ds
+∫ T
0
∫Ω
|Tn(g)− Tm(g)| dx ds+∫
Ω
|Tn(u0)− Tm(u0)| dx ,(3.57)
for all t ∈ [0, T ], where
S(Tn(u), Tm(u)) = sup 2 |f(σ)− f(τ)| ; min(Tn(u), Tm(u)) ≤ σ , τ ≤ max(Tn(u), Tm(u)) ,
SinceS(Tn(u), Tm(u)) ≤ 2|f(u)| 11|u|>n a.e. on Σ,
using that u is almost everywhere finite on Σ and f(u) belongs to L1(Σ) we deduce that the quantityS(Tn(u), Tm(u)) converges to zero in L1(Σ) as n and m tend to infinity. Similarly since g and u0 are inL1 the remaining terms in the right hand side of (3.57) converge to zero, so that we conclude from (3.57)that un is a Cauchy sequence in C0([0, T ];L1(Ω)) and strongly converges in C0([0, T ];L1(Ω)) towards afunction u as n tends to infinity. This implies that we can pass to the limit as n tends to infinity inthe left hand side of the entropy inequality (3.54) by simply using Lebesgue’s Theorem. Moreover, since0 ≤ 11un>l ≤ 1, the estimate (3.50) shows that the sequence of measures (µn,l)n is bounded independentlyof n. Therefore, there exists a subsequence still denoted by (µn,l)n and µl a bounded positive measure onQ such that µn,l −→ µl in C(Q)′ for the weak–∗ topology. Then passing to the limit in (3.54) we obtainthat u satisfies the renormalized entropy inequality (2.6). Finally, since (un) converges to u in L1(Q)then, in view of the estimate (3.50), we have
||µl|| ≤ lim infn→+∞
||µn,l||
≤∫
Ω
(u0 − l)+ dx+ 2∫ T
0
∫∂Ω
f(u) 11u>l dγ dt+∫ T
0
∫Ω
11u>l |g| dx dt ,
which ensures liml→+∞
||µl|| = 0. Clearly, reasoning as in Lemma 3.6 and Proposition 3.7, one also proves
that un satisfies, for all |k| ≤ l < n,∫ T
0
∫Ω
((un>(−l)− κ)−ϕt + Φ−(un>(−l), κ) · ∇ϕ
)dx dt+
∫Ω
(un0>(−l)− κ)−ϕ(x, 0) dx
+ Lipl
∫ T
0
∫∂Ω
(un>(−l)− κ)−ϕ(x, t) dγ(x) dt
+∫ T
0
∫Ω
gn sgn−(un>(−l)− κ)ϕdx dt ≥ − < νl,n, ϕ > ,
where νl,n is a nonnegative bounded measure on Q whose mass is estimated as follows
||νl,n|| ≤∫
Ω
(u0 + l)− dx+ 2∫ T
0
∫∂Ω
f(u) 11u<−l dγ dt+∫ T
0
∫Ω
11un<−l |g| dx dt .
19
Then passing to the limit we conclude that u also satisfies (2.7) with νl satisfying liml→+∞
||νl|| = 0. This
concludes the proof of the existence of a renormalized entropy solution.The uniqueness of the renormalized entropy solution of Problem (1.1) is a straightforward consequence
of Theorem 3.1.
3.4 Continuous dependence
Proof of Theorem 2.4 Let u be the unique renormalized solution of (1.1). We use (3.24) with v = ui,v0 = ui0, v = ui, g = gi so as to have∫
Ω
|ui − u|(t) ≤∫ T
0
∫∂Ω
S(ui, u) dγ(x) dt+∫ T
0
∫Ω
|gi − g| dx dt+∫
Ω
|ui0 − u0| dx , (3.58)
for all t ∈ [0, T ], where, for (t, x) ∈ Σ
S(ui, u)(t, x) = sup
2 |f(σ)− f(τ)| ; min(ui(t, x), u(t, x)) ≤ σ , τ ≤ max(ui(t, x), u(t, x)).
Since ui almost everywhere converges to u on Σ we have that S(ui, u) converges to zero almost everywhereon Σ. Moreover
|S(ui, u)| ≤ 2(|f(ui)|+ |f(u)|) .
Since the sequence f(ui) is strongly converging in L1(Σ) we deduce, from Vitali theorem, that S(ui, u)strongly converges to zero in L1(Σ). Since (ui0) also converges to u0 in L1(Ω) and (gi) converges to g inL1(Q), we can pass to the limit in (3.58) and deduce that the sequence (ui) strongly converges to u inC0([0, T ];L1(Ω)) (in particular in L1(Q)).
Acknowledgment We are grateful to Juan Luis Vazquez who encouraged our work, and suggestedand constructed with us Example 2.5 during his visit in Rome.
The first author also thanks Thierry Gallouet for the very warm hospitality given during his stayingat Universite de Provence de Marseille, where part of this work was done. Italian Government MURST60% research funds have also supported this work.
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