One-dimensional transport equations with discontinuous
coefficients
François Bouchut∗
François James†
November 24, 2009
Abstract
We consider one-dimensional linear transport equations with bounded but possibly
discontinuous coefficient a. The Cauchy problem is studied from two different points of
view. In the first case we assume that a is piecewise continuous. We give an existence
result and a precise description of the solutions on the lines of discontinuity. In the second
case, we assume that a satisfies a one-sided Lipschitz condition. We give existence,
uniqueness and general stability results for backward Lipschitz solutions and forward
measure solutions, by using a duality method. We prove that the flux associated to these
measure solutions is a product by some canonical representative b
a of a.
Key-words. Linear transport equations, discontinuous coefficients, weak stability, duality,
product of a measure by a discontinuous function, nonnegative solutions.
1991 Mathematics Subject Classification. Primary 35F10, 35B35, 34A12.
To appear in Nonlinear Analysis, TMA
∗
Département de Mathématiques et Applications, UMR CNRS 8553, École Normale Supérieure et CNRS,
45 rue d’Ulm, 75230 Paris Cedex 05, France, francois.bouchut@ens.fr
†
Mathématiques, Applications et Physique Mathématique d’Orléans, UMR CNRS 6628, Université
d’Orléans, 45067 Orléans Cedex 2, France, james@math.cnrs.fr
1
Summary
1 Introduction
2 Notations and preliminaries
2.1 Some notations
2.2 From conservative to nonconservative by integration
2.3 Uniqueness when the coefficient is continuous with respect to x
2.4 Uniqueness for bounded nonnegative solutions
3 Transport equations with piecewise continuous coefficient
3.1 Two examples of non-uniqueness
4 Transport equations with coefficient satisfying a one-sided Lipschitz condition
4.1 Lipschitz solutions, backward problem
4.1.1 General properties
4.1.2 Limit solutions
4.1.3 Reversible solutions
4.1.4 The conservative case
4.1.5 The generalized backward flow
4.2 Duality solutions, forward problem
4.2.1 The nonconservative case
4.2.2 The conservative case
4.3 Flux and universal representative
4.3.1 Flux of a conservative duality solution
4.3.2 Characteristics in Filippov’s sense
4.3.3 Reversibility and renormalization
4.4 On viscous problems
4.4.1 Duality for viscous problems
4.4.2 Backward problem with discontinuous final data
Equations de transport unidimensionnelles à coefficients discontinus
Résumé
Nous considérons des équations de transport linéaires en une dimension
avec coefficient a borné mais éventuellement discontinu. Le problème de Cauchy
est abordé par deux points de vue différents. Dans le premier cas nous supposons que a est continu par morceaux. Nous donnons un résultat d’existence
et une description précise des solutions sur les lignes de discontinuité. Dans
le deuxième cas, nous supposons que a vérifie une condition de Lipschitz d’un
seul côté. Nous donnons des résultats d’existence, d’unicité et de stabilité faible
générale pour les solutions rétrogrades lipschitziennes et les solutions directes
mesures par une méthode de dualité. Nous montrons que le flux associé à ces
b de a.
solutions mesures est un produit par un représentant canonique a
Mots-clés. Equations de transport linéaires, coefficients discontinus, stabilité faible, dualité,
produit d’une mesure par une fonction discontinue, solutions positives.
2
1
Introduction
This paper is devoted to one-dimensional homogeneous linear transport equations
∂t u + a(t, x)∂x u = 0
in
]0, T [×R,
(1.1)
with T > 0 and a a given bounded coefficient. This equation will be referred as the nonconservative problem. By differentiating (1.1) with respect to x, we obtain the conservative
problem
∂t µ + ∂x (a(t, x)µ) = 0 in ]0, T [×R,
(1.2)
with µ = ∂x u. This type of equations appears naturally in the study of some systems of
conservation laws where solutions belong to some measure space, as in H.C. Kranzer and
B.L. Keyfitz [17], P. Le Floch [18], D. Tan, T. Zhang and Y. Zheng [25], Y. Zheng and A.
Majda [26]. It is especially the case for the system of pressureless gases, see F. Bouchut
[2], E. Grenier [14], Y. Brenier and E. Grenier [4], W. E, Y.G. Rykov and Y.G. Sinai [11].
Therefore, we are interested in solutions µ to (1.2) that are measures in x (for a fixed time
t), or solutions u to (1.1) that are functions of bounded variation in x (for a fixed time t).
Notice that we have an a priori estimate on the total mass of µ since by multiplying (1.2) by
sgn µ we get
∂t |µ| + ∂x (a|µ|) = 0,
(1.3)
and thus
d
dt
Z
|µ(t, dx)| = 0.
(1.4)
R
This computation is only formal, but in the cases of interest we can prove that (1.3) is actually
true with an inequality (Theorem 4.3.6).
Both equations (1.1) and (1.2) also appear in problems of identification or control of the
non-linearity of a scalar conservation law
∂t v + ∂x f (v) = 0,
(1.5)
where the coefficient is given by a = f 0 (v), see F. James and M. Sepúlveda [16]. Eq. (1.1)
can actually be seen as a linearized version of (1.5). It should be noted that there is a very
extended literature on non-linear problems such as (1.5), although very few papers consider
linear equations. It is due to the very specific difficulty of (1.2) which lies in the product of
the measure µ by the possibly discontinuous coefficient a. In this situation, the theory of
R.J. DiPerna and P.-L. Lions [10] (see also I. Capuzzo Dolcetta and B. Perthame [5]) does
not apply. In [9], G. Dal Maso, P. Le Floch and F. Murat have introduced a definition for
the product a∂x u when a is a function of u (and for vector valued functions). This condition
happens to be well-suited to treat certain non-linear systems.
In this paper, we are interested in defining the product aµ when no a priori relation is
assumed between a and µ. We actually only use the implicit relation induced by the fact
that µ solves (1.2). In order to handle this product, we use two different approaches.
In the first case, we use the classical product of a measure by a bounded Borel function,
which is well-defined provided that a is everywhere defined. We assume that a is piecewise
3
continuous, and we define admissibility conditions ((AC1)-(AC2) in Section 3) which limit
the possible behaviors of a on discontinuity lines. In this context, we have an existence result
for BVloc solutions to (1.1) with initial data u◦ ∈ BVloc (R). Moreover, we can give a very
precise description of the solutions along discontinuity lines (Proposition 3.2). There is no
uniqueness for a general piecewise smooth a. However, we prove that under the one-sided
Lipschitz condition
∂x a ≤ α(t)
in
α ∈ L1 (]0, T [),
]0, T [×R,
(1.6)
uniqueness holds. We have similar results for (1.2) with initial data µ◦ ∈ Mloc (R).
In the second case, we assume that
a ∈ L∞ (]0, T [×R)
(1.7)
only satisfies the one-sided Lipschitz condition (1.6). In the literature, this condition is often
written under the equivalent form
a.e. (t, x, y) ∈]0, T [×R × R (a(t, x) − a(t, y))(x − y) ≤ α(t)(x − y)2 ,
(1.8)
with the same α ∈ L1 (]0, T [). We prove that the problems (1.1) and (1.2) are well-posed for
Cauchy data u◦ ∈ BVloc (R) and µ◦ ∈ Mloc (R) respectively. These solutions are understood
in the duality sense. This means that we consider locally Lipschitz solutions p to
∂t p + a∂x p = 0
in
]0, T [×R,
(1.9)
with final data pT ∈ Liploc (R). Then, a formal computation shows that
∂t (pµ) + ∂x (apµ) = 0,
(1.10)
and hence
d
hµ, pi = 0.
(1.11)
dt
This last formula makes up the definition of duality solutions µ (see Definition 4.2.4). It is
well-known that (1.6) ensures the existence of Lipschitz solutions to (1.9), see O.A. Oleinik
[21], E.D. Conway [7], D. Hoff [15], E. Tadmor [23], P. Le Floch and Z. Xin [19]. However,
there is no uniqueness. The corner stone of this paper is the introduction of the notion
of reversible solutions p to (1.9), a class for which there is existence and uniqueness for
the backward Cauchy problem. These reversible solutions can be characterized by various
properties: support properties (Definition 4.1.4), monotonicity properties (Theorem 4.1.9),
total variation properties (Proposition 4.1.7), entropy inequality or equality (Theorem 4.3.13).
Then, only reversible solutions are taken into account in (1.11) for the definition of duality
solutions (see Definition 4.2.4).
Finally, we prove that duality solutions µ actually satisfy in the distribution sense
bµ) = 0
∂t µ + ∂x (a
in
]0, T [×R
(1.12)
b of a (Theorem 4.3.4), and this result answers the question
for some canonical representative a
b can be computed explicitly.
raised by the product a × µ. In the piecewise continuous case, a
4
Then, we have a general stability result (Theorem 4.3.2) which states that a bounded
sequence of duality solutions µn to (1.2) with coefficient an satisfying the one-sided Lipschitz
condition (1.6) with αn bounded in L1 (]0, T [) will converge weakly to the duality solution µ
bn µn → a
bµ.
associated to the L∞ − w∗ limit a of (an ). We also obtain that a
Notice that the above cited authors only used existence for the backward problem in order
to get uniqueness for some non-linear problems. In [14], E. Grenier proves that the one-sided
Lipschitz condition holds for the system of pressureless gases, with α(t) = 1/t. Therefore,
our stability results apply to that system.
In the classical theory (i.e. for a smooth a), transport equations such as (1.1), (1.2) or
(1.9) are closely related to characteristics. The flow X(s, t, x) is defined to be the solution to
the ODE

 dX
= a(s, X),
(1.13)
ds
 X(t)
= x.
In this formulation, t and x are parameters (the initial time and initial position), and s is the
only variable. Then, the solution p to (1.9) with data p(s, .) = ps (for a fixed s) is given by
p(t, x) = ps (X(s, t, x)).
(1.14)
For discontinuous a, this equivalence between characteristics and transport equations is
no longer valid. However, the theory of A.F. Filippov [12] ensures that (1.13) has got a
solution in a generalized sense for bounded a. Moreover, if a satisfies the one-sided Lipschitz
condition (1.6), this solution is unique on the right (i.e. for s > t). This theory is used to
study differential equations and inclusions, see J.-P. Aubin and A. Cellina [1], A.F. Filippov
[13], and also in non-linear hyperbolic problems, see C.M. Dafermos [8], E. Tadmor and T.
Tassa [24]. As noticed by E.D. Conway [7], this situation leads to the following paradox:
when a satisfies (1.6), uniqueness holds for Filippov solutions to the forward problem (1.13)
(for s > t), although the corresponding backward problem (1.9) with final data ps (solved for
t < s) can have many solutions. Similarly, the backward problem (1.13) (for s < t) can have
many solutions, although duality can only provide uniqueness for the corresponding forward
problem (1.1) with initial data us (solved for t > s). In that case existence was an open
problem.
In this paper, we give the following answer: the formula (1.14) with X the Filippov flow
(for s > t) actually gives the unique reversible backward solution to (1.9). As stated above,
it is the only one that is stable by approximation of a. Notice that partial results in that
direction were already obtained by Conway in [7]. By duality, we then obtain existence for
the forward problem (1.1). Incidentally, we also obtain a stability result for forward Filippov
solutions to (1.13) with the assumption that the sequence of coefficients an satisfies (1.6) with
αn bounded in L1 .
The paper is organized as follows. In Section 2, we introduce some notations and prove
our main lemma which states that conservative and nonconservative equations are equivalent.
We also prove two sharp results of uniqueness. Section 3 is concerned with the case of a piecewise continuous a. Finally, Section 4 is devoted to the case where a satisfies the one-sided
Lipschitz condition. In 4.1 we study the backward problem and Lipschitz solutions, in 4.2 we
define duality solutions for the forward problem, 4.3 contains more sophisticated results and
5
the relation with the generalized Filippov flow, and 4.4 is devoted to some comments about
viscous problems.
The main results of this paper were announced in [3].
2
Notations and preliminaries
This section is devoted to notations that are used throughout the paper and to general results
concerning equations (1.1) and (1.2).
2.1
Some notations
The following sets will often be used.
• B(Y, Z): set of bounded functions from Y to Z,
• C(Y, Z): set of continuous functions from Y to Z,
• Cc (Y, Z): set of continuous functions with compact support from Y to Z,
• Lip(Y, Z): set of Lipschitz continuous functions from Y to Z,
• M(R): space of Borel measures on R.
An index c as for Cc will always refer to functions of compact support, and an index loc
will refer to the corresponding local spaces. For example, Mloc (R) denotes the space of local
Borel measures on R, i.e. the space of distributions of order 0. In the above sets, a missing
Z (as in C(Y )) means that Z = R.
In the study of (1.1) and (1.2) we fix T > 0, and set
Ω =]0, T [×R.
(2.1.1)
We introduce the four following spaces
\
SBV = C([0, T ], L1loc (R)) B([0, T ], BVloc (R)),
SM = C([0, T ], Mloc (R) − σ(Mloc (R), Cc (R))),
SLip = Liploc ([0, T ] × R),
∞
1
SL∞ = C([0, T ], L∞
loc (R) − σ(Lloc (R), Lc (R))).
(2.1.2)
In (2.1.2), we denote by σ(Y, Z) the weak topology on Y generated by the natural bilinear
form on Y × Z.
We will sometimes use the space L∞ (]0, T [, M(R)), which is defined as follows. First, a
function f :]0, T [→ M(RN ) is said to be measurable if for any ϕ ∈ C0 (RN ), t 7→ hf (t), ϕi is a
Borel function, where C0 (RN ) denotes the space of functions in C(RN ) that tend to 0 at ∞.
Then, M (]0, T [, M(RN )) is defined to be the quotient of the space of measurable functions
]0, T [→ M(RN ) by the subspace of functions that vanish a.e. in ]0, T [. Finally,
(
L∞ (]0, T [, M(RN )) =
)
µ ∈ M (]0, T [, M(RN )) ; ess sup kµ(t)kM(RN ) < ∞ .
t∈]0,T [
6
(2.1.3)
It is a Banach space with the obvious norm. Let us now state without proof a few properties
related to this space.
1. If f :]0,R T [→ M(RN ) is measurable, and ϕ is a bounded Borel function ]0, T [×RN → R,
then t 7→ RN ϕ(t, x)f (t, dx) is a Borel function.
2. If f :]0, T [→ M(RN ) is measurable, then t 7→ |f (t, .)| is also measurable. Therefore, for
any µ ∈ L∞ (]0, T [, M(RN )), |µ| ∈ L∞ (]0, T [, M(RN )).
3. If µ ∈ L∞ (]0, T [, M(RN )), we define the measure dtµ(t, dx) ∈ Mloc (]0, T [×RN ) by
Z T
Z
Z
dt
dtµ(t, dx) =
RN
0
A
1IA (t, x) µ(t, dx).
Then for any bounded Borel function ϕ :]0, T [×RN → R
Z T
Z
]0,T [×RN
Z
dt
ϕ(t, x) dtµ(t, dx) =
ϕ(t, x) µ(t, dx).
RN
0
Moreover, we have |dtµ(t, dx)| = dt|µ(t, dx)|.
4. If µn is a bounded sequence in L∞ (]0, T [, M(RN )), then there exists a measure µ ∈
L∞ (]0, T [, M(RN )) and a subsequence µn0 such that µn0 → µ in the distribution sense in
]0, T [×RN .
5. Let u ∈ C(]0, T [, L1 (RN )). Then u ∈ Lip(]0, T [, L1 (RN )) if and only if ∂t u ∈ L∞ (]0, T [, M(RN )).
Moreover, we have kukLip(]0,T [,L1 (RN )) = k∂t ukL∞ (]0,T [,M(RN )) .
2.2
From conservative to nonconservative by integration
Let u ∈ SBV solve (1.1). Then it is obvious that µ = ∂x u ∈ SM and solves (1.2). We here
prove that conversely, it is possible to go from (1.2) to (1.1).
Lemma 2.2.1 Let a :]0, T [×R → R a bounded Borel function, and assume that µ ∈ SM
solves (1.2). Then there exists u ∈ SBV solving (1.1) such that µ = ∂x u. Moreover, u is
unique up to an additive constant.
Here, (1.1) and (1.2) are understood in the usual distribution sense, and the product a × µ
is the usual product of a measure by a bounded Borel function.
Remark 2.2.1 If µ has more regularity, we automatically get more regularity for u. For
example, if µ ∈ SL∞ , then u ∈ SLip .
Proof of Lemma 2.2.1. Uniqueness up to a constant is obvious. For existence, denote by
v(t, .) the primitive of µ(t, .) such that v(t, 0+) = 0. Then v ∈ SBV and ∂x v = µ,
∂x (∂t v + a∂x v) = 0
in
Ω.
Therefore, there exists ψ ∈ D0 (]0, T [) such that
∂t v + a∂x v = ψ
in
Ω.
Let ϕ ∈ Cc∞ (R). Then for any χ ∈ Cc∞ (]0, T [),
Z
(
ϕ)hψ, χi = h∂t v + a∂x v, χ ⊗ ϕi
=−
ZZ
Z
0
v(t, x)χ (t)ϕ(x) dtdx +
7
Z
χ(t) dt
a(t, x)ϕ(x)∂x v(t, dx),
and thus
Z
(
Z
Z
ϕ)ψ(t) = ∂t
v(t, x)ϕ(x) dx +
a(t, x)ϕ(x)∂x v(t, dx),
so that ψ = ∂t η with η ∈ C([0, T ]). We finally define u = v −η(t) which solves the problem.
2.3
Uniqueness when the coefficient is continuous with
respect to x
Lemma 2.3.1 Let a ∈ L∞ (]0, T [×R) such that
a.e. t ∈]0, T [
a(t, .) ∈ C(R).
(2.3.1)
Assume that v ∈ SL∞ solves
(
∂t v + ∂x (av) = 0
v(0, .) = 0.
in
]0, T [×R,
(2.3.2)
Then v = 0.
Proof. By Lemma 2.2.1 (and Remark 2.2.1), there exists u ∈ SLip solving (1.1) such that
∂x u = v. Since u is defined up to a constant and ∂x u(0, .) = v(0, .) = 0, we can assume
that u(0, .) = 0. Let ρn be a smoothing sequence in R2 , ρn (t, x) = ρtn (t)ρxn (x), with suppρtn ⊂
] − ∞, 0[. Let us define un = ρn ∗ u ∈ Liploc ([0, T − ε] × R) (for a fixed ε ∈]0, T [ such that
t,x
1/n ≤ ε), and an = ρn ∗ a ∈ L∞ (]0, T − ε[×R). We have
t,x
∂t un + ρn ∗ (a∂x u) = 0
in ]0, T − ε[×R,
t,x
so that
∂t un + an ∂x un = fn
]0, T − ε[×R,
in
(2.3.3)
with
fn = an ∂x un − ρn ∗ (a∂x u).
t,x
(2.3.4)
Since an , un , fn ∈ C ∞ ([0, T − ε] × R) we can consider the flow Xn (s, t, x) of an , and for
0 ≤ t ≤ T − ε we have
Z t
un (t, x) = un (0, Xn (0, t, x)) +
fn (s, Xn (s, t, x)) ds.
(2.3.5)
0
Therefore for any 0 ≤ t ≤ T − ε and x1 < x2
kun (t, .)kL∞ (]x1 ,x2 [) ≤ kun (0, .)kL∞ (]x1 −kak∞ t,x2 +kak∞ t[)
Z t
+
0
kfn (s, .)kL∞ (]x1 −kak∞ t,x2 +kak∞ t[) ds.
Now since un → u locally uniformly in [0, T − ε] × R, it is enough to prove that for any
x1 < x2
Z
T −ε
0
kfn (s, .)kL∞ (]x1 ,x2 [) ds → 0.
8
(2.3.6)
We have
fn = (an − a)∂x un + a(ρn ∗ ∂x u) − ρn ∗ (a∂x u) ≡ fn1 + fn2 .
t,x
t,x
(2.3.7)
In order to treat the first term fn1 , we bound ∂x un by a constant, and an − a → 0 in
L1 (]0, T − ε[, L∞ (]x1 , x2 [)) because by (2.3.1), a can be approached in L1 (]0, T [, L∞
loc (R)) by
C ∞ functions. For the second term,
fn2 (t, x) =
ZZ
[a(t, x) − a(t − τ, x − y)]∂x u(t − τ, x − y)ρn (τ, y) dτ dy,
kfn2 kL1 (]0,T −ε[,L∞ (]x1 ,x2 [)) ≤ k∂x ukL∞ (]0,T [×]x1 −1,x2 +1[)
× sup ka(t, x)−a(t−τ, x−y)kL1 (]0,T −ε[t ,L∞ (]x1 ,x2 [x ))
−1/n<τ <0
|y|<1/n
which tends to 0 by the same approximation argument. Therefore, (2.3.6) is proved.
2.4
Uniqueness for bounded nonnegative solutions
We prove here a uniqueness result for a bounded coefficient, by a slight modification of the
method of T.-P. Liu and M. Pierre [20].
Lemma 2.4.1 Let a ∈ L∞ (]0, T [×R) and v ∈ SL∞ solve
∂t v + ∂x (av) = 0
in
]0, T [×R,
(2.4.1)
with v(0, .) = 0. Assume moreover that there exists ve ∈ SL∞ some nonnegative solution to
(2.4.1) such that
|v| ≤ ve.
(2.4.2)
Then v = 0.
In particular, notice that condition (2.4.2) is fulfilled if v = v1 − v2 with v1 , v2 ∈ SL∞ some
nonnegative solutions to (2.4.1) (take ve = v1 + v2 ). Therefore, if v1 , v2 ∈ SL∞ are two nonnegative solutions to (2.4.1) and have the same initial data, then they are equal.
Proof of Lemma 2.4.1. By Lemma 2.2.1 (and Remark 2.2.1), there exists u ∈ SLip solving
(1.1) such that ∂x u = v. Since u is defined up to a constant and ∂x u(0, .) = v(0, .) = 0, we
can assume that u(0, .) = 0. Let us define ven = ρn ∗ ve and bn = ρn ∗ (ave) (with ρn (t, x) a
t,x
t,x
smoothing sequence in R2 ), so that
∂t ven + ∂x bn = 0
in
]ε, T − ε[×R
(2.4.3)
as soon as 1/n < ε. Since u is locally Lipschitz continuous, we can multiply (1.1) by sgn u to
get
∂t |u| + a∂x |u| = 0.
(2.4.4)
Now (2.4.3) and (2.4.4) yield
∂t (ven |u|) + ∂x (bn |u|) = (bn − aven )∂x |u| in
9
]ε, T − ε[×R.
(2.4.5)
Since ve ≥ 0 we have ven ≥ 0 and |bn | ≤ kakL∞ ven , and by standard integration we deduce that
for ε < t1 ≤ t2 < T − ε and x1 < x2 ,
Zx2
ven (t2 , x)|u(t2 , x)| dx ≤
x1
x
Z2 +kak∞ (t2 −t1 )
ZZ
ven (t1 , x)|u(t1 , x)| dx
+
(bn − aven )∂x |u| dtdx.
t1 <t<t2
x1 −kak∞ (t2 −t)<x<x2 +kak∞ (t2 −t)
x1 −kak∞ (t2 −t1 )
By letting n → ∞ we get
Zx2
x
Z2 +kak∞ (t2 −t1 )
ve(t2 , x)|u(t2 , x)| dx ≤
x1
ve(t1 , x)|u(t1 , x)| dx,
x1 −kak∞ (t2 −t1 )
for any x1 < x2 and 0 < t1 ≤ t2 < T . Then we let t1 → 0 and obtain for 0 < t2 < T
Z x2
x1
ve(t2 , x)|u(t2 , x)| dx ≤ 0,
so that by the nonnegativity of ve
∀ 0 < t < T,
a.e. x ve(t, x)|u(t, x)| = 0.
Now since |∂x u| = |v| ≤ ve, we get u∂x u = 0 a.e., and (1.1) gives also u∂t u = 0 a.e. Therefore,
∂x u2 = 0 and ∂t u2 = 0 so that u2 = cst = 0, and hence u = 0, v = ∂x u = 0.
3
Transport equations with piecewise continuous coefficient
In this section a : Ω =]0, T [×R → R is a bounded Borel function (everywhere defined),
and we consider solutions u ∈ SBV to (1.1), where the product is understood in the usual
sense since ∂x u ∈ L∞ (]0, T [, Mloc (R)) ⊂ Mloc (Ω). We assume in all Section 3 that a satisfies
the following piecewise continuity condition: Ω = C ∪D∪S with C, D, S some disjoint sets and
(PC1) C is open, a is continuous in C,
(PC2) S is locally finite in Ω,
(PC3) D is a one-dimensional submanifold of class C 1 of Ω, such that a has got limit values
at (t0 , x0 ) ∈ D on both sides of D. These limits are denoted by a− (t0 , x0 ) and a+ (t0 , x0 ), and
it is assumed that they can be locally chosen continuous with respect to (t0 , x0 ) ∈ D.
Since the values of a on D cannot be arbitrary, we have to prescribe them where the measure dt∂x u(t, dx) can concentrate. We thus assume that the following admissibility conditions
are fulfilled for any (t0 , x0 ) ∈ D.
(AC1) a(t0 , x0 ) ∈ [a− (t0 , x0 ), a+(t0, x0 )],
(AC2) if (dt)D (t0 , x0 ) 6= 0 and dx
dt (t0 , x0 ) ∈ [a− (t0 , x0 ), a+ (t0 , x0 )] then
a(t0 , x0 ) =
D
dx
dt D(t0 , x0 ).
Remark 3.1 Notice that C ∪ D is an open set.
10
Remark 3.2 The values of a on the set S do not matter, neither in the assumptions nor in
Eq. (1.1).
Remark 3.3 Any function a ∈ Cb (Ω) (continuous and bounded) verifies all the above hypotheses.
Remark 3.4 For a given function a, the sets C, D and S are not uniquely determined. For
example, it is possible to add artificial lines of discontinuity. Then since Eq. (1.1) does not
depend on this choice it is a way to get informations on the behavior of u along any curve in
C (by Lemma 3.1 and Proposition 3.2 below).
With (PC1)-(PC3) and (AC1)-(AC2), we are able to characterize the solutions to (1.1)
(Proposition 3.2) and to prove an existence result (Theorem 3.4). Let us begin by a lemma.
Lemma 3.1 If u ∈ SBV solves (1.1), then the measure dt∂x u(t, dx) does not concentrate in
the set of unconsistency
Duc = (t0 , x0 ) ∈ D ; (dt)D (t0 , x0 ) 6= 0
Z
More precisely, it means that
and
a(t0 , x0 ) 6=
dx
(t0 , x0 ) .
dt D
(3.1)
|dt∂x u(t, dx)| = 0.
Duc
(t0 , x0 )
Proof of Lemma 3.1. Let
∈ D such that (dt)D (t0 , x0 ) 6= 0. Then D is defined by a
curve x = ξ(t) in a neighborhood V of (t0 , x0 ), with ξ ∈ C 1 . Let us denote by ζε (x) a cutoff
function, ζε (x) = ζ1 (x/ε), ζ1 ∈ Cc∞ (] − 1, 1[), 0 ≤ ζ1 ≤ 1, ζ1 (x) = 1 if |x| ≤ 1/2. Then we
have
ζε (x − ξ(t)) (∂t u + a∂x u) = 0 in V.
(3.2)
Since by Lebesgue’s theorem ζε (x − ξ(t))a(t, x)∂x u → 1ID a∂x u when ε → 0, and similarly
ζε (x − ξ(t))∂t u = ∂t (ζε (x − ξ(t))u) + ζε0 (x − ξ(t))ξ 0 (t)u
= ∂t (ζε (x − ξ(t))u) + ∂x ζε (x − ξ(t))ξ 0 (t)u − ζε (x − ξ(t))ξ 0 (t)∂x u
→ ∂t (1ID u) + ∂x (1ID ξ 0 (t)u) − 1ID ξ 0 (t)∂x u,
(3.3)
ε→0
we get by letting ε → 0 in (3.2) and by using that meas(D) = 0
1ID a(t, x) −
dx
(t, x) ∂x u = 0
dt D
in V.
(3.4)
The result follows easily.
Notice that the above proof does not use the admissibility conditions.
Proposition 3.2 (Characterization of solutions) Let u ∈ SBV . Then u solves (1.1) if
and only if the three following conditions are satisfied
(i)
u ∈ Lip([0, T ], L1loc (R)),
(3.5)
(ii) ∂t u + a∂x u = 0 in C,
(3.6)
(iii) the measure dt∂x u(t, dx) does not concentrate in the set of transversality
dx
Dtr = (t0 , x0 ) ∈ D; (dt)D (t0 , x0 ) 6= 0 and
(t0 , x0 ) ∈
/ [a− (t0 , x0 ), a+ (t0 , x0 )] .
dt D
11
(3.7)
Proof. First, if u solves (1.1), then (ii) is obvious, (i) is deduced from the equation (1.1) and
Property 5 in § 2.1, and (iii) results from Lemma 3.1 and (AC1).
Let us now assume that (i), (ii), (iii) are verified. By (i) and Property 5 in § 2.1, we have
∂t u ∈ L∞ (]0, T [, Mloc (R)). Since u ∈ SBV , we have also ∂x u ∈ L∞ (]0, T [, Mloc (R)) and thus
∂t u and ∂x u do not concentrate on points. Then since S is locally finite we just have to prove
that ∂t u + a∂x u = 0 in the open set C ∪ D. But (ii) indicates that it holds in C. Therefore,
we take (t0 , x0 ) ∈ D and we just have to prove that ∂t u + a∂x u = 0 in a neighborhood V of
(t0 , x0 ). We distinguish two cases.
1st case: (dt)D (t0 , x0 ) 6= 0.
As in the proof of Lemma 3.1, we have
ζε (x − ξ(t)) (∂t u + a∂x u) → 1ID a(t, x) −
ε→0
dx
(t, x) ∂x u in V.
dt D
(3.8)
Since by (ii) [1 − ζε (x − ξ(t))](∂t u + a∂x u) = 0 in V , we obtain by addition
∂t u + a∂x u = 1ID a(t, x) −
dx
(t, x) ∂x u in V.
dt D
(3.9)
Then by (iii) and (AC2) the right-hand side vanishes and we get the result.
2nd case: (dt)D (t0 , x0 ) = 0.
Let V be a small open set containing (t0 , x0 ), and define
N = {(t, x) ∈ V ∩ D ; (dt)D (t, x) = 0} .
(3.10)
By the first case, we know that the measure ∂t u + a∂x u vanishes in the open set V \N . It remains to prove that λ = ∂t u+a∂x u does not concentrate in N . Since λ ∈ L∞ (]0, T [, Mloc (R))
we have
Z
Z
ZZ
|dtλ(t, dx)| = dt 1IN (t, x)|λ(t, dx)|,
(3.11)
N
and by Sard’s lemma |{t ∈ R; ∃x ∈ R (t, x) ∈ N }| = 0. This proves that the right-hand side
of (3.11) vanishes, and the proof is complete.
Proposition 3.3 (Renormalization) If u ∈ SBV solves (1.1) then for any S : R → R
Lipschitz continuous, S(u) ∈ SBV solves (1.1).
Before proving Proposition 3.3, let us state a simple estimate, which proof is easy: if
v ∈ BVloc (R) and S : R → R is Lipschitz continuous, then S(v) ∈ BVloc (R) and |∂x [S(v)]| ≤
Lip(S)|∂x v| in the sense of measures.
Proof of Proposition 3.3. By the above mentioned estimate, we have S(u) ∈ SBV ,
|∂x [S(u)]| ≤ Lip(S)|∂x u|. By Proposition 3.2, we have to prove that S(u) verifies conditions (i), (ii), (iii). Since u verifies these properties, it is obvious with the above estimate
that (i) and (iii) hold. It remains to prove (ii), and since a is continuous in C, it is easy to
reduce the problem to the case where S ∈ C 1 .
Therefore, let us prove that ∂t [S(u)] + a∂x [S(u)] = 0 in C with S ∈ C 1 . Consider an open
subset ω of C with compact closure in C, and aε (t, x) a sequence of bounded C 1 functions
such that aε → a locally uniformly in C. Since u solves (1.1) we have
∂t u + aε ∂x u = (aε − a)∂x u = µε
12
(3.12)
with µε a locally bounded measure, and µε → 0 strongly in ω. Let us define un = ρn (x) ∗ u,
x
with ρn a standard smoothing sequence in R. By using an estimate from the paper of R.J.
DiPerna and P.-L. Lions [10], we have
in L1loc (Ω),
ρn ∗(aε ∂x u) − aε ∂x un = rε,n → 0
x
n→∞
(3.13)
and since
∂t un + aε ∂x un = ρn ∗ µε − rε,n ,
x
we have
∂t [S(un )] + aε ∂x [S(un )] = S 0 (un )ρn ∗ µε − S 0 (un )rε,n .
x
Then we let n → ∞ and obtain by (3.13)
∂t [S(u)] + aε ∂x [S(u)] = λS,ε ,
with λS,ε a locally bounded measure, such that λS,ε → 0 in ω strongly when ε → 0. We
finally let ε → 0, and using that aε → a uniformly in ω we obtain
∂t [S(u)] + a∂x [S(u)] = 0
in
ω,
which ends the proof.
Theorem 3.4 (Existence) For any u0 ∈ BVloc (R) there exists u ∈ SBV solving (1.1) with
u(0, .) = u0 , and such that for any x1 < x2 and t ∈ [0, T ]
T VI (u(t, .)) ≤ T VJ (u0 ),
(3.14)
ku(t, .)kL∞ (I) ≤ ku0 kL∞ (J)
(3.15)
with I =]x1 , x2 [ and J =]x1 − kak∞ t, x2 + kak∞ t[.
Proof. Let aε (t, x) a sequence of bounded C 1 functions such that aε → a locally uniformly in
C (for example a convolution of a by a smoothing kernel in (t, x)), and u0ε ∈ C 1 (R) a bounded
sequence in BVloc (R) such that u0ε → u0 . Let us define uε as the (classical) solution to
∂t uε + aε ∂x uε = 0
in
Ω,
uε (0, .) = u0ε .
(3.16)
We are going to prove that for suitable aε and u0ε (such as convolutions of a and u0 ), and
after extraction of a subsequence, uε tends to a solution.
We have uε (t, x) = u0ε (Xε (0, t, x)) with Xε (s, t, x) the solution to
∂s Xε = aε (s, Xε ),
Xε (t, t, x) = x,
(3.17)
and thus ∂x uε = ∂x u0ε (Xε (0, t, x))∂x Xε (0, t, x). We have
∂s ∂xZ
Xε = ∂x aε (s, Xε )∂x Xε , s
∂x Xε = exp
∂x aε (τ, Xε (τ, t, x)) dτ > 0,
tZ s
Xε = x +
(3.18)
aε (τ, Xε (τ, t, x)) dτ,
t
|Xε − x| ≤ kak∞ |s − t|.
13
Therefore, |∂x uε | = |∂x u0ε |(Xε (0, t, x))∂x Xε (0, t, x), and
Z x2
|∂x uε | dx =
Z Xε (0,t,x2 )
Xε (0,t,x1 )
x1
|∂x u0ε |(x) dx ≤
Z x2 +kak∞ t
x1 −kak∞ t
|∂x u0ε |(x) dx.
(3.19)
Hence, we conclude that uε is bounded in SBV . Then, since ∂t uε = −aε ∂x uε is bounded
in C([0, T ], L1loc (R)), we get by Property 5 in § 2.1 that uε is bounded in Lip([0, T ], L1loc (R)).
Therefore after extraction of a subsequence
uε → u in C([0, T ], L1loc (R)),
(3.20)
with u ∈ SBV ∩ Lip([0, T ], L1loc (R)). It is easy to check that u(0, .) = u0 and that u verifies
(3.14)-(3.15) (if the variation of u0ε is not too large, for example u0ε = ρε ∗ u0 ). Since aε → a
locally uniformly in C, we obtain
∂t u + a∂x u = 0
in C.
By Proposition 3.2, it now remains to prove (iii). We are going to prove that property by a
dispersion estimate.
Let us consider an open set V 3 (t0 , x0 ) with (t0 , x0 ) ∈ D, V as small as we want, such
that for any (t, x) ∈ V ∩ D
(dt)D (t, x) 6= 0
and
dx
(t, x) ∈
/ [a− (t, x), a+ (t, x)].
dt D
(3.21)
Then D is defined in V by a curve x = ξ(t), ξ ∈ C 1 . Let s1 < t0 < s2 such that (t, ξ(t)) ∈ V
for s1 ≤ t ≤ s2 , χ ∈ Cc (]s1 , s2 [) a given test function, and define
ZZ
I=
χ(t)1I(t,x)∈V ∩D dt∂x u(t, dx).
(3.22)
We have to prove that I = 0. We have I = lim Iη , Iη = lim Iηε , with
ε
η→0
Z
Iη =
Z
χ(t) dt
D
D
E
ζη (x − ξ(t))∂x u(t, dx) = ∂x u, χ(t)ζη (x − ξ(t)) ,
E
Iηε = ∂x uε , χ(t)ζη (x − ξ(t)) =
Z
Z
χ(t) dt
(3.23)
ζη (x − ξ(t))∂x uε (t, x) dx.
Here ζη is the same cutoff function as in Lemma 3.1. In order to use a duality formula, let
us define the test function ϕηε = ζη (Xε (s, t, x) − ξ(s)), which is compactly supported in the
x variable and solves
ϕηε (s, s, x) = ζη (x − ξ(s)).
∂t ϕηε + aε (t, x)∂x ϕηε = 0,
Since ∂x uε (t, x) = ∂x uε (s, Xε (s, t, x))∂x Xε (s, t, x), we have
Z
ϕηε (s, t, x)∂x uε (t, x) dx
Z
=
Z
=
ζη (Xε (s, t, x) − ξ(s))∂x uε (s, Xε (s, t, x))∂x Xε (s, t, x) dx
ζη (x − ξ(s))∂x uε (s, x) dx,
14
(3.24)
which is independent of t. Then
Z
Iηε =
Z
χ(s) ds
Z
=
ζη (x − ξ(s))∂x uε (s, x) dx
Z
χ(s) ds
ϕηε (s, t0 , x)∂x uε (t0 , x) dx
Z
=
(3.25)
Z
∂x uε (t0 , x) dx
χ(s)ϕηε (s, t0 , x) ds
Z
=
∂x uε (t0 , x)ψηε (x) dx,
with
Z
ψηε (x) =
χ(s)ϕηε (s, t0 , x) ds
Z
=
(3.26)
χ(s)ζη (Xε (s, t0 , x) − ξ(s)) ds.
Since suppψηε remains in a compact set independent of η and ε, and because of (3.25), it is
enough to prove that kψηε kL∞ ≤ Cη.
By (3.21) and by reducing V if necessary, we can assume that for some σ ∈ {−1, 1} and
α>0
∀(t, x) ∈ V− σ(a− (t, x) − ξ 0 (t)) ≥ α,
(3.27)
∀(t, x) ∈ V+ σ(a+ (t, x) − ξ 0 (t)) ≥ α.
Then by reducing V and α again, for small enough ε and for a well chosen aε (such as a
convolution in t, x of a), we get
∀(t, x) ∈ V
σ(aε (t, x) − ξ 0 (t)) ≥ α.
(3.28)
Let us denote δε,x (s) = Xε (s, t0 , x) − ξ(s). The set Jηε (x) = {s ∈]s1 , s2 [; |δε,x (s)| < η} is
open, and for small η we have
∀s ∈ Jηε (x)
(s, Xε (s, t0 , x)) ∈ V,
and thus by (3.28)
∀s ∈ Jηε (x)
0
σδε,x
(s) = σ aε (s, Xε (s, t0 , x)) − ξ 0 (s) ≥ α.
This implies that Jηε (x) is an interval, and δε,x is a diffeomorphism from Jηε (x) onto its
image. We have
Z
Z
ψηε (x) =
χ(s)ζη (δε,x (s)) ds =
Jηε (x)
χ(s)ζη (y)
δε,x (Jηε (x))
dy
0 (s)|
|δε,x
,
with δε,x (s) = y, and thus
|ψηε (x)| ≤
1
kχk∞
α
Z
ζη (y) dy ≤
2
kχk∞ η.
α
Therefore we get Iη → 0 = I when η → 0 as announced and the proof is complete.
Remark 3.5 The above proof actually provides a stability result for a sequence of solutions
associated to a smooth approximate coefficient aε . The limit u is a solution associated to
a, provided that aε satisfies a convexity assumption ensuring (3.28). This result has to be
compared with the stability result of A.F. Filippov [12] for differential equations. Notice that
in both cases, the convexity assumption compensates for non uniqueness.
15
Remark 3.6 In the above proof, we obtain condition (iii) of Proposition 3.2 by a dispersion
estimate in a neighborhood of (t0 , x0 ) ∈ Dtr . This estimate means that u is ”approximately
continuous” through Dtr , which comes from the transversality expressed in (3.7). That
transversality property has been used by N. Caroff [6] and P. Serfati [22] to prove some
continuity estimates for u when only allowing discontinuity lines x = cst for a, and assuming
ess inf a > 0.
In general, solutions to (1.1) are not unique. Some examples of non-uniqueness are shown
in § 3.1 and at the beginning of Section 4 (Example 4.1.1). However, when a satisfies the
one-sided Lipschitz condition, we are able to prove uniqueness by the method of P. Le Floch
[18].
Theorem 3.5 (Uniqueness) Let us assume that a satisfies the one-sided Lipschitz condition (1.6). Then any solution u ∈ SBV to (1.1) verifies
∂t |u| + ∂x (a|u|) ≤ α(t)|u|
in
]0, T [×R,
(3.29)
|u(0, x)| dx.
(3.30)
and for any x0 ∈ R, R > 0, 0 ≤ t ≤ T we have
Z
|u(t, x)| dx ≤ e
Rt
0
|x−x0 |<R
Z
α
|x−x0 |<R+kak∞ t
Therefore, under condition (1.6), there exists at most one solution u ∈ SBV to (1.1) with
Cauchy data u0 ∈ BVloc (R).
Proof. The estimate (3.30) and the uniqueness property can be deduced from (3.29) by
standard integration. Therefore, we only prove that (3.29) holds in distribution sense.
First, Proposition 3.3 ensures that
∂t |u| + a∂x |u| = 0
in
Ω.
(3.31)
Then let us define an = ρn ∗ a (with ρn a smoothing sequence in R), which is a bounded Borel
x
function. We have by (3.31)
∂t |u| + ∂x (an |u|) = (∂x an )|u| + (an − a)∂x |u|
≤ α(t)|u| + (an − a)∂x |u|,
(3.32)
because ∂x an ≤ α(t) by (1.6). Therefore, since an → a locally uniformly in C,
∂t |u| + ∂x (a|u|) ≤ α(t)|u| in C.
(3.33)
It now remains to study what happens around S and D. Let us consider the nonnegative
measure
µn = (α(t) − ∂x an )|u| = α(t)|u| + an ∂x |u| − ∂x (an |u|).
(3.34)
After extraction of a subsequence, µn * µ ≥ 0. Then for any χ ∈ Cc∞ (]0, T [) and ψ ∈ Cc∞ (R),
ψ ≥ 0,
hµ, χ ⊗ ψi = limhµ
n , χ ⊗ ψi
ZZ
= lim
α(t)|u|χ(t)ψ(x) dtdx + han ∂x |u|, χ ⊗ ψi +
16
ZZ
0
an |u|χ(t)ψ (x) dtdx .
Therefore for a fixed ψ we get
|hµ, χ ⊗ ψi| ≤ C
Z T
(1 + |α(t)|)|χ(t)| dt,
(3.35)
0
and hence hµ, 1It=t0 ψ(x)i = 0. Since ∂x (an |u|) → ∂x (a|u|), we conclude that ∂x (a|u|) is a local
measure in (t, x) in Ω, and does not concentrate in points. Then since S is locally finite, it
remains to prove that (3.29) holds in a small neighborhood V of (t0 , x0 ) ∈ D. As in the proof
of Proposition 3.2 we distinguish two cases.
1st case: (dt)D (t0 , x0 ) 6= 0.
We use the same approach as in Lemma 3.1. By (3.33) we have
h
ih
i
1 − ζε (x − ξ(t)) ∂t |u| + ∂x (a|u|) − α(t)|u| ≤ 0
in V,
and since
ζε (x − ξ(t))∂x (a|u|) → 1ID ∂x (a|u|),
ε→0
ζε (x − ξ(t))∂t |u| → −1ID
ε→0
dx
(t, x)∂x |u|,
dt D
we get
∂t |u| + ∂x (a|u|) + 1ID
dx
∂x |u| − 1ID ∂x (a|u|) ≤ α(t)|u| in V.
dt D
Therefore, it is enough to prove that
1ID
dx
∂x |u| − 1ID ∂x (a|u|) ≥ 0
dt D
in V.
(3.36)
We have
1ID ∂x (a|u|) = lim ζε (x − ξ(t))∂x (a|u|)
ε→0 = lim ∂x (ζε (x − ξ(t))a|u|) − ζε0 (x − ξ(t))a|u|
ε→0
= lim −ζε0 (x − ξ(t))a|u|,
ε→0
and thus for any ϕ ∈
Cc∞ (V
)
h1ID ∂x (a|u|), ϕ(t, x)i = lim −
ε→0
Z
=
Z
Z
dt
dx ϕ(t, x)a(t, x)|u(t, x)|ζε0 (x − ξ(t))
"
#
dt ϕ(t, ξ(t)) a(t, ξ(t)+)u(t, ξ(t)+) − a(t, ξ(t)−)u(t, ξ(t)−) .
Similarly
h1ID
dx
∂x |u|, ϕ(t, x)i = lim h−ζε0 (x − ξ(t))ξ 0 (t)|u|, ϕi
ε→0
dt D
"
Z
=
#
dt ϕ(t, ξ(t))ξ (t) u(t, ξ(t)+) − u(t, ξ(t)−) .
0
Therefore, the inequality to be proved (3.36) can be written
a.e. t,
a(t, ξ(t)+)u(t, ξ(t)+) − a(t, ξ(t)−)u(t, ξ(t)−)
#
≤ ξ (t) u(t, ξ(t)+) − u(t, ξ(t)−) .
"
0
17
(3.37)
Let us notice that (1.6) is equivalent to a.e. t ∂x [a(t, .)] ≤ α(t) in R, and thus
∀x ∈ R a(t, x+) ≤ a(t, x−).
a.e. t
(3.38)
Now, by property (iii) of Proposition 3.2 (applied to |u|), for any t out of a set of Lebesgue
measure zero, one of the two following cases occurs.
a. ξ 0 (t) ∈
/ [a− (t, ξ(t)), a+ (t, ξ(t))] and u(t, ξ(t)+) = u(t, ξ(t)−). Then (3.38) gives (3.37).
b. ξ 0 (t) ∈ [a− (t, ξ(t)), a+ (t, ξ(t))]. Choosing a+ (t, x) = a(t, x+) and a− (t, x) = a(t, x−), we
get
a+ (t, ξ(t))u(t, ξ(t)+) ≤ ξ 0 (t)u(t, ξ(t)+),
−a− (t, ξ(t))u(t, ξ(t)−) ≤ −ξ 0 (t)u(t, ξ(t)−),
and (3.37) follows by addition.
2nd case: (dt)D (t0 , x0 ) = 0.
Let us define
N = {(t, x) ∈ V ∩ D ; (dt)D (t, x) = 0} .
By the first case, we know that the measure ∂t |u| + ∂x (a|u|) − α(t)|u| is nonpositive in the
open set V \N . In order to prove that it is nonpositive in V , it is enough to prove that it
does not concentrate in N . Since α(t)|u| and ∂t |u| do not, it remains to check that ∂x (a|u|),
or equivalently µ, does not concentrate in N .
Let ψ ∈ Cc∞ (R), ψ ≥ 0. We define the measure λ on ]0, T [ by
ZZ
ψ(x) µ(dt, dx).
λ(E) =
t∈E
Then for any χ ∈ Cc∞ (]0, T [)
Z
ZZ
χ(t) λ(dt) =
χ(t)ψ(x) µ(dt, dx),
Z
Z T
χ(t) λ(dt) ≤ C
(1 + |α(t)|)|χ(t)| dt
0
because of (3.35). Therefore, λ is absolutely continuous with respect to the Lebesgue measure,
and λ({t ∈]0, T [; ∃y ∈ R (t, y) ∈ N }) = 0. Thus, µ({(t, x); ∃y ∈ R (t, y) ∈ N }) = 0 and
µ(N ) = 0.
By using Lemma 2.2.1, we can also give existence and uniqueness results for (1.2). The
following result is a straightforward consequence of Theorems 3.4 and 3.5.
Theorem 3.6 (Conservative Cauchy problem) For any µ0 ∈ Mloc (R), there exists µ ∈
SM solution to (1.2) with initial datum µ0 , and such that for any x1 < x2 and t ∈ [0, T ]
Z
|µ(t, dx)| ≤
I
Z
|µ0 (dx)|,
(3.39)
J
with I =]x1 , x2 [ and J =]x1 − kak∞ t, x2 + kak∞ t[. Moreover, if a satisfies the one-sided
Lipschitz condition (1.6), then this solution is unique.
Notice that by Proposition 3.2, any solution µ satisfies
ZZ
dt|µ(t, dx)| = 0.
Dtr
18
(3.40)
3.1
Two examples of non-uniqueness
In this paragraph we show how non-uniqueness can occur. We also refer to Section 4 for the
sgn function example.
1. Non-uniqueness when (1.6) is not satisfied
Let al < ar two real numbers, and define
a(t, x) =
x
≤ al ,
t
x
if al ≤ ≤ ar ,
t
x
if
≥ ar .
t


al



x
if

t



 ar
Then a ∈ Cb (]0, ∞[×R). Given ul , ur ∈ R we consider the initial datum
0
u (x) =
ul
ur
if x < 0,
if x > 0.
For any ϕ ∈ BV (]al , ar [) we define for t > 0


ul



 x
u(t, x) = ϕ( )

t



 ur
x
< al ,
t
x
if al < < ar ,
t
x
> ar .
if
t
if
Then u ∈ SBV (for any T > 0) and solves (1.1) in ]0, ∞[×R. To see this, just add artificial
discontinuity lines and apply Proposition 3.2. Therefore, there is no uniqueness, although
∂x a ≤ 1/t.
2. Non-uniqueness for a nonlinear problem
Let ul , ur ∈ R and consider the initial datum
0
u (x) =
ul
ur
if x < 0,
if x > 0.
Given v ∈ [ul , ur ] we define
u(t, x) =

 ul
v

ur
if x < vt,
if x = vt,
if x > vt.
Then u is a piecewise continuous admissible coefficient in the sense of (PC1)-(PC3) and
(AC1)-(AC2), and u ∈ SBV (for any T > 0) at the same time. By using Proposition 3.2 we
easily obtain that
∂t u + u∂x u = 0 in ]0, ∞[×R.
Therefore, if ul 6= ur this problem has infinitely many solutions. Notice that if ul > ur , we
have ∂x u ≤ 0 so that (1.6) is satisfied with α = 0.
19
4
Transport equations with coefficient satisfying a one-sided
Lipschitz condition
In all Section 4 we consider a coefficient a ∈ L∞ (Ω), Ω =]0, T [×R, satisfying the one-sided
Lipschitz condition
∂x a ≤ α(t) in Ω,
α ∈ L1 (]0, T [).
(4.1)
Notice that in contrast with Section 3, a is now defined almost everywhere.
Condition (4.1) implies some regularity for a. In fact, it can be seen by appropriate
smoothing that (4.1) is equivalent to
a.e. t ∈]0, T [
∂x [a(t, .)] ≤ α(t)
in R.
(4.2)
Therefore, it is quite easy to obtain that for almost every t ∈]0, T [, a(t, .) ∈ BVloc (R) and for
any x1 < x2
T V[x1 ,x2 ] (a(t, .)) ≤ 2 (|α(t)|(x2 − x1 ) + kakL∞ ) .
(4.3)
Notice also that since a is bounded, (4.2) implies that α ≥ 0.
In Section 4.1 we consider the backward problem, in Section 4.2 the forward problem by
the duality method, in Section 4.3 we prove some sharp properties for both problems by using
the generalized flow, and in Section 4.4 we make some comments about viscous problems.
4.1
Lipschitz solutions, backward problem
We are here interested in solutions p ∈ SLip to
∂t p + a∂x p = 0
in
Ω.
(4.1.1)
The space of these solutions will be denoted by L. We study solutions to the backward
problem, consisting of all p ∈ L such that
p(T, .) = pT ,
(4.1.2)
for a given pT ∈ Liploc (R).
Remark 4.1.1 One can of course transform (4.1.1)-(4.1.2) in a direct problem, by setting
e(t, x) = −a(T − t, x), and u(t, x) = p(T − t, x). Condition (4.1) then becomes ∂x a
e ≥
a
−α(T − t).
In general, the problem (4.1.1)-(4.1.2) does not have a unique solution, even in the class
of Lipschitz continuous functions. Let us consider the sign example, already studied by E.D.
Conway [7].
Example 4.1.1 Take
a(t, x) = − sgn x =

 −1
0

1
20
if x > 0,
if x = 0,
if x < 0.
For any h ∈ Lip([0, T ]) such that h(0) = pT (0), we set for (t, x) ∈ [0, T ] × R,
p(t, x) =
T
p (x − (T − t) sgn x)
h(T − t − |x|)
if T − t ≤ |x|,
if T − t ≥ |x|.
(4.1.3)
Then p ∈ SLip solves (4.1.1)-(4.1.2), while p(t, 0) = h(T − t) is arbitrary. Actually any
Lipschitz continuous solution to (4.1.1)-(4.1.2) is of the form (4.1.3) for some h.
Notice that there is a ”canonical choice” for the above h, namely h ≡ pT (0). The aim of
this section is to give a criterion which selects this solution.
In all the following, (ρn )n will stand for a sequence of standard Cc∞ mollifiers on R.
4.1.1
General properties
We first give some properties of Lipschitz solutions to (4.1.1).
Lemma 4.1.1 For any p ∈ L we have the following properties.
(i) Forward stability. Let an ∈ C 1 ([0, T ] × R) be a bounded sequence in L∞ ([0, T ] × R), such
that an → a in L1loc (]0, T [×R), and ∂x an ≤ αn (t), αn → α in L1 (]0, T [). Let p0n ∈ C 1 (R),
p0n → p(0, .) in L1loc (R). Then, denoting by pn the (classical) solution to
∂t pn + an ∂x pn = 0
in
Ω,
pn (0, .) = p0n ,
we have pn → p in C([0, T ], L1loc (R)).
(ii) Decrease of the total variation. The application
t 7→
Z
|∂x p(t, x)| dx ∈ [0, ∞]
R
is nonincreasing in [0, T ].
(iii) Let x0 ∈ R. Then, ∀t ∈ [0, T ], ∃x ∈ [x0 − kak∞ (T − t), x0 + kak∞ (T − t)] such that
p(t, x) = p(T, x0 ).
Proof. (i) We write
∂t (p − pn ) + an ∂x (p − pn ) = (an − a)∂x p,
so that by the Lipschitz functions calculus
∂t |p − pn | + ∂x (an |p − pn |) = (an − a) sgn(p − pn )∂x p + |p − pn |∂x an
≤ |an − a||∂x p| + αn |p − pn |.
Then, denoting qn = |p − pn | exp −
Z t
αn
we get
0
∂t qn + ∂x (an qn ) ≤ |an − a||∂x p|e−
Rt
0
αn
,
and by standard integration on a cone
Z
qn (t, x) dx ≤
|x−x0 |<R
Z t Z
Z
qn (0, x) dx +
0
|x−x0 |<R+kan k∞ t
21
|an − a||∂x p|e−
|x−x0 |<R+kan k∞ (t−τ )
Rτ
0
αn
dτ dx.
−
Rτ
αn
Z
L1loc .
is bounded uniformly, an → a in
Thus
qn (t, x) dx → 0 uniformly
But |∂x p|e 0
for t ∈ [0, T ].
|x−x0 |<R
(ii) Take an as in (i), and set p0n = ρn ∗ p(0, .), so that for t ∈ [0, T ]
Z
|∂x pn (t, x)| dx =
Z
R
|∂x pn (0, x)| dx ≤
Z
R
|∂x p(0, x)| dx.
R
Then by (i)
Z
|∂x p(t, x)| dx ≤ lim
Z
|∂x pn (t, x)| dx ≤
Z
|∂x p(0, x)| dx.
We get the result by a similar proof on any subinterval of [0, T ].
(iii) Assume on the contrary that for some t ∈ [0, T [, for any x ∈ [x0 − kak∞ (T − t), x0 +
kak∞ (T − t)], p(t, x) 6= p(T, x0 ) ≡ y. Then, since p is continuous, we have for some ε > 0 and
η>0
∀x ∈ [x0 − kak∞ (T − t) − η, x0 + kak∞ (T − t) + η], |p(t, x) − y| ≥ ε.
Now we approximate a by an as in (i) with kan k∞ ≤ kak∞ , we set ptn = ρn ∗ p(t, .), and we
call pn the solution to ∂t pn + an ∂x pn = 0, pn (t, .) = ptn . By (i), pn → p in C([t, T ], L1loc (R)).
Now since ptn → p(t, .) locally uniformly, we have for n large enough
∀x ∈ [x0 − kak∞ (T − t) − η, x0 + kak∞ (T − t) + η],
|ptn (x) − y| ≥ ε/2.
Going forward up to T by characteristics, we obtain |pn (T, x) − y| ≥ ε/2 for any x ∈ [x0 −
η, x0 + η]. Since pn (T, .) → p(T, .) in L1loc (R), we get |p(T, x) − y| ≥ ε/2, for a.e. x ∈
[x0 − η, x0 + η]. But this contradicts p(T, x0 ) = y, since p is continuous.
Remark 4.1.2 Properties (i) and (ii) reflect some kind of irreversibility of the solution, in the
sense that the inequality ∂x a ≤ α gives some control on p(t, .) and ∂x p(t, .) in terms of p(0, .)
and ∂x p(0, .), and not the converse. It remains true when replacing 0 by any t1 , for t > t1 .
Remark 4.1.3 Property (iii) asserts that the final data is somewhat transported by the backward equation. Going back to Example 4.1.1, this means that for any t < T , ∃x ∈ [t−T, T −t]
such that p(t, x) = pT (0). A ”natural” solution is then to transport the constant pT (0) in
the whole fan {(t, x); |x| ≤ T − t, t ∈ [0, T ]}.
The remainder of this section is devoted to the introduction of some criteria selecting the
”natural” solutions. As we shall see, these solutions are ”reversible” in a sense to be precised,
and we have existence and uniqueness of a reversible solution to (4.1.1)-(4.1.2). However,
before that, we have to build up such solutions, by a completely different technique.
4.1.2
Limit solutions
The limit solutions to (4.1.1) are obtained by testing against some ”pseudo-solutions” to
the direct problem (1.1) with initial data u0 ∈ BVloc (R). This looks very much as a duality
argument, and will appear finally to be actually one. Let us first build the ”pseudo-solutions”.
Let an → a as in Lemma 4.1.1 with kan k∞ ≤ kak∞ , u0 ∈ BVloc (R) and set u0n = ρn ∗ u0 .
Some classical estimates along the characteristics prove that the solution un to
∂t un + an ∂x un = 0,
22
un (0, .) = u0n
(4.1.4)
is bounded in SBV . Since ∂t un = −an ∂x un , un is also bounded in Lip([0, T ], L1loc (R)),
and therefore compact in C([0, T ], L1loc (R)). Up to a subsequence, we get that un → u in
C([0, T ], L1loc (R)) with u ∈ SBV ∩ Lip([0, T ], L1loc (R)), and u(0, .) = u0 . We wish now to
extract a subsequence for which convergence holds for any u0 ∈ BVloc (R).
By a diagonal process, we can choose a subsequence for which convergence holds for a
countable subset of initial data A ⊂ C 1 (R), dense in L1loc (R). Then we deduce from the
estimate
Z
Z
Rt
αn
0
|un (t, x) − vn (t, x)| dx ≤ e
|u0n (x) − vn0 (x)| dx
(4.1.5)
|x−x0 |<R
|x−x0 |<R+kan k∞ t
C([0, T ], L1loc (R))
that the Cauchy criterion is verified in
for any u0 ∈ BVloc (R) (and actually
for any u0 ∈ L1loc (R)). One can also deduce that for another approximation u0n ∈ C 1 (R),
bounded in BVloc (R), such that u0n → u0 in L1loc (R), the corresponding solution also converges,
to the same limit.
Now this subsequence is fixed once for all, and we denote by U (u0 ) the limit u. It is quite
easy to check that U (u0 ) ≥ 0 whenever u0 ≥ 0; U (u0 ) is compactly supported in x if u0 is;
and U (u0 v 0 ) = U (u0 )U (v 0 ).
Definition 4.1.2 Let p ∈ L. We say that p is a limit solution to (4.1.1) if
∀u0 ∈ BVloc (R),
∂t U (u0 )∂x p + ∂x a U (u0 )∂x p = 0
in
Ω.
(4.1.6)
Notice that the operator U and the notion of limit solution depend on the sequence of
approximate coefficients (an ).
Remark 4.1.4 As a consequence of the whole construction we shall achieve, we shall see that
1. They actually not depend on the choice of the sequence (an ), and it is not necessary to
extract any subsequence,
2. U (u0 ) is actually solution to (1.1), in the sense of duality which will be defined later.
Proposition 4.1.3 Let pT ∈ Liploc (R). Then there exists a unique p ∈ L limit solution to
(4.1.1) with final data p(T, .) = pT .
Proof. Uniqueness. It follows from a duality argument.
Integrate (4.1.6) with respect to
R
x, for a compactly supported u0 . We obtain that t 7→ U (u0 )∂x p dx is constant in [0, T ].
Assuming that p(T, .) = 0, it follows that
Z
0=
Z
0
U (u )(T, x)∂x p(T, x) dx =
R
u0 (x)∂x p(0, x) dx.
R
Since this holds for any u0 , we have ∂x p(0, .) = 0. But by Lemma 4.1.1(ii) the total variation
is nonincreasing, so that ∂x p(t, .) = 0 for any t. Using the equation, we find that ∂t p = 0 and
p = 0.
Existence. Let pTn → pT , pTn bounded in Liploc (R), with pTn ∈ C 1 (R), and pn solve
∂t pn + an ∂x pn = 0,
23
pn (T, .) = pTn .
(4.1.7)
If Xn is the flow corresponding to an , one has
0 ≤ ∂x Xn (T, t, x) ≤ e
RT
t
αn
.
Since pn (t, x) = pTn (Xn (T, t, x)), we easily deduce from this inequality a L∞
loc bound on ∂x pn
and, using the equation, on ∂t pn . Thus (pn ) is bounded in Liploc ([0, T ]× R), and therefore has
a subsequence converging in C([0, T ] × [−R, R]) to some p ∈ Liploc ([0, T ] × R), which satisfies
(4.1.1)-(4.1.2). Now for any n, we multiply (4.1.4) and ∂x (4.1.7) respectively by ∂x pn and un
to obtain
∂t (un ∂x pn ) + ∂x (an un ∂x pn ) = 0.
When n → ∞, ∂x pn → ∂x p weakly, but an → a and un → U (u0 ) strongly so we recover
(4.1.6).
4.1.3
Reversible solutions
We wish to prove now that the limit solution is indeed the ”natural” solution mentioned in
Remark 4.1.3.
Definition 4.1.4 (Nonconservative reversible solutions) (i) We call exceptional solution any function pe ∈ L such that pe (T, .) = 0. We denote by E the vector space of exceptional
solutions.
(ii) We call domain of support of exceptional solutions the open set
n
o
Ve = (t, x) ∈]0, T [×R; ∃pe ∈ E pe (t, x) 6= 0 .
(iii) Any p ∈ L is called reversible if p is locally constant in Ve .
In the case of Example 4.1.1 a(t, x) = − sgn x, the exceptional solutions are given by pe (t, x) =
h((T − t − |x|)+ ) with h ∈ Lip([0, T ]), h(0) = 0, and we have Ve = {(t, x) ∈ Ω; |x| < T − t}.
Remark 4.1.5 1. The vector space of reversible solutions to (4.1.1) will be denoted by R. It
is a subspace of L, containing constants.
2. If p is a reversible solution, then S(p) is also a reversible solution, for any S : R → R
Lipschitz continuous. Also, the product of two reversible solutions is a reversible solution.
3. It is easy to see that R ∩ E = {0}. For a proof see Corollary 4.1.6.
4. We have the equivalence
E = {0} ⇔ Ve = ◦/ ⇔ any p ∈ L is reversible.
For example, it is the case if a.e. t a(t, .) ∈ C(R), see Lemma 2.3.1. Another example is
a(t, x) = 1Ix<0 .
Theorem 4.1.5 (Nonconservative backward Cauchy problem) Let be given a final datum pT ∈ Liploc (R). Then there exists a unique p ∈ L reversible solution to (4.1.1) such that
p(T, .) = pT . Moreover we have for any x1 < x2 and t ∈ [0, T ]
kp(t, .)kL∞ (I) ≤ kpT kL∞ (J) ,
k∂x p(t, .)kL∞ (I) ≤ e
RT
t
α
k∂x pT kL∞ (J) ,
with I =]x1 , x2 [ and J =]x1 − kak∞ (T − t), x2 + kak∞ (T − t)[.
24
Proof. Uniqueness follows from Remark 4.1.5(3). For existence, we actually prove that the
limit solution p obtained in Proposition 4.1.3 is reversible.
Let pe ∈ E. By Lemma 4.1.1(i), we have pe = U (pe (0, .)). Then since p is a limit solution,
∂t (pe ∂x p) + ∂x (ape ∂x p) = 0.
Now by Lemma 2.2.1 (and Remark 2.2.1), there exists q ∈ L such that ∂x q = pe ∂x p. Then,
by the formula U (u0 v 0 ) = U (u0 )U (v 0 ), we easily see that q is a limit solution. Now by
uniqueness of limit solutions (Proposition 4.1.3) and since ∂x q(T, .) = 0, we find that q is a
constant and pe ∂x p = ∂x q = 0. Since this holds for any pe ∈ E, we get ∂x p = 0 in Ve , and
(4.1.1) gives also ∂t p = 0 in Ve . The estimates are left to the reader.
Remark 4.1.6 The above proof shows that the notions of limit and reversible solutions actually coincide.
Corollary 4.1.6 Recall that L, R, E are the vector spaces of Lipschitz, reversible and exceptional solutions respectively. Then we have
L = R ⊕ E.
Moreover, for any pr ∈ R, pe ∈ E and t ∈ [0, T ], ∂x pr (t, .) and ∂x pe (t, .) are supported by
disjoint sets.
Proof. The decomposition of L follows from Theorem 4.1.5. Then, let pr ∈ R, pe ∈ E
and t ∈]0, T [, and define A = {x ∈ R; (t, x) ∈ Ve }, which is open. Since pr is reversible,
pr (t, .) is locally constant in A. Therefore, a.e. x ∈ A ∂x pr (t, x) = 0. Now, by the definition of Ve , we have ∀x ∈ Ac pe (t, x) = 0, and thus a.e. x ∈ Ac ∂x pe (t, x) = 0. This
proves the support property if 0 < t < T . If t = T , we just take A = ◦/. If t = 0, take
A = {x ∈ R; ∃qe ∈ E qe (0, x) 6= 0}. The proof is then similar to the above one.
We now come to more handable characterizations of reversible solutions.
Proposition 4.1.7 Let p ∈ L.
Z Then
(i) If p is reversible then t 7→
|∂x p(t, x)| dx is constant in [0, T ].
R
(ii) If the above function is constant and finite, then p is reversible.
We recall that for any p ∈ L, the total variation is in general a nonincreasing function of t
(Lemma 4.1.1(ii)).
Proof of Proposition 4.1.7. In order to prove (i), let us assume that p is reversible. Then
by Theorem 4.1.5 and by construction, p is a limit solution, obtained as the limit of pn
solution to ∂t pn + an ∂x pn = 0, pn (T, .) = ρn ∗ pT , pT = p(T, .). Therefore
Z
|∂x pn (0, x)| dx =
Z
|∂x pTn (x)| dx
≤
Z
|∂x pT (x)| dx.
Thus, by lower semicontinuity we get
Z
|∂x p(0, x)| dx ≤
Z
25
|∂x p(T, x)| dx,
and we conclude since we already know that the total variation is nonincreasing. The converse (ii) is easily obtained by (i) and the decomposition of Corollary 4.1.6.
The most useful criterion for reversibility (Theorem 4.1.9) is closely related to monotone
solutions. We first state a lemma.
Lemma 4.1.8 Let p ∈ L. We have
(i) if p ∈ E and ∂x p ≥ 0 then p = 0,
(ii) if ∂x p ≥ 0 then p is reversible,
(iii) if p is reversible and ∂x p(T, .) ≥ 0 then ∂x p ≥ 0.
Proof. (i) Since p(T, .) = 0, by Lemma 4.1.1(iii), for any t ∈ [0, T ] and
any R > 0, there
R
exists x1 < −R and x2 > R such that p(t, x1 ) = p(t, x2 ) = 0. Thus xx12 ∂x p(t, x) dx = 0,
and by nonnegativity ∂x p(t, x) = 0 a.e. x ∈]x1 , x2 [, and therefore for a.e. x ∈ R since R is
arbitrary. The result easily follows.
(ii) By Corollary 4.1.6 we have p = pr +pe with pr reversible and pe exceptional, and such that
∂x pr (t, .) and ∂x pe (t, .) have disjoint supports. Therefore, the condition ∂x p ≥ 0 is equivalent
to ∂x pr ≥ 0 and ∂x pe ≥ 0. Then by (i) pe = 0, and thus p = pr is reversible.
(iii) This property easily follows from the construction of reversible solutions by an approximate problem.
Theorem 4.1.9 (Main criterion for reversibility) Let p ∈ L. Then p is reversible if
and only if there exists p1 , p2 ∈ L such that ∂x p1 ≥ 0, ∂x p2 ≥ 0 and p = p1 − p2 .
Proof. If p is reversible, we write p(T, .) = pT1 − pT2 with pTi ∈ Liploc (R) and ∂x pTi ≥ 0. Then
p = p1 − p2 with p1 , p2 the reversible solutions with final data pT1 , pT2 , and Lemma 4.1.8(iii)
ensures that ∂x pi ≥ 0. The converse follows from Lemma 4.1.8(ii) and from the fact that R
is a vector space.
Remark 4.1.7 1. The above decomposition was already used by D. Hoff [15].
2. From the characterization in Theorem 4.1.9 we deduce that the time restriction of a
reversible solution is also reversible. This property was not obvious in Definition 4.1.4
3. If we choose the above property as a definition for reversible solutions, then there is a
direct proof of Theorem 4.1.5. Existence is easily obtained by any reasonable approximation,
and uniqueness follows from Lemma 2.4.1. However, the basic renormalization property of
Remark 4.1.5(2) is not obvious with this definition.
We next give an important feature of reversible solutions, namely the stability with respect
to coefficient and final data.
Theorem 4.1.10 (Stability) Let (an ) be a bounded sequence in L∞ (]0, T [×R), with an → a
in L∞ (]0, T [×R) − w∗. Assume ∂x an ≤ αn (t), where (αn ) is bounded in L1 (]0, T [), ∂x a ≤
α ∈ L1 (]0, T [). Let (pTn ) be a bounded sequence in Liploc (R), pTn → pT , and denote by pn the
reversible solution to
(
∂t pn + an ∂x pn = 0 in ]0, T [×R,
pn (T, .) = pTn .
26
Then pn → p in C([0, T ] × [−R, R]) for any R > 0, where p is the reversible solution to
(
∂t p + a∂x p = 0
p(T, .) = pT .
in
]0, T [×R,
Proof. By the bounds of Theorem 4.1.5 and the equation on pn , we easily get that pn is
bounded in SLip . Therefore, we can extract a subsequence pn0 , such that
pn0 → p
in C([0, T ] × [−R, R]),
p ∈ Liploc ([0, T ] × R),
p(T, .) = pT .
Let us first prove that p satisfies the equation. Let µn = αn (t) − ∂x an ≥ 0. Since µn is
a measure, so is ∂x an . We have ∂x an → ∂x a in D0 (Ω), and, up to a subsequence, αn0 * λ
in M(]0, T [) − w∗, so that µn0 → µ = λ − ∂x a in D0 (Ω), and µ ≥ 0. The convergence holds
therefore weakly in the space of measures. Going back to ∂x an , we have now ∂x an0 * ∂x a
locally weakly in the space of measures. Thus we can pass to the limit in the product pn0 ∂x an0 ,
so that an0 ∂x pn0 = ∂x (an0 pn0 ) − pn0 ∂x an0 → ∂x (ap) − p∂x a = a∂x p. This yields ∂t p + a∂x p = 0.
It remains to prove that p is reversible. By Theorem 4.1.9 we have pn = pn,1 − pn,2 ,
∂x pn,i ≥ 0, where pn,i are solutions corresponding to an and pTn,i chosen bounded in Liploc (R).
Up to a subsequence, pTn0 ,i → pTi ∈ Liploc (R), and we can pass to the limit for pn0 ,i as for
p. Thus pn0 ,i → pi (again up to a subsequence), pi is solution to (4.1.1), ∂x pi ≥ 0, and
p = p1 − p2 . Thus p is reversible. Finally, uniqueness ensures that it is not necessary to
extract any subsequence.
4.1.4
The conservative case
Let us now consider the conservative problem. Let π ∈ SL∞ solve
∂t π + ∂x (aπ) = 0
in
Ω.
(4.1.8)
Then by the integration lemma 2.2.1 (and Remark 2.2.1), there exists a unique (up to an
additive constant) p ∈ SLip which satisfies
∂x p = π,
∂t p + a∂x p = 0
in
Ω.
Definition 4.1.11 We say that π ∈ SL∞ solving (4.1.8) is a reversible solution to (4.1.8) if
p is reversible.
From Definition 4.1.4 and Theorem 4.1.9, we have the following characterization for reversible
conservative solutions.
Theorem 4.1.12 (Conservative reversible solutions) The following three properties are
equivalent for π ∈ SL∞ solution to (4.1.8).
(i) π is reversible,
(ii) π = 0 in Ve ,
(iii) π = π1 − π2 , for some πi ∈ SL∞ solutions to (4.1.8), such that πi ≥ 0.
Obviously, the set of reversible conservative solutions is a vector space. Notice also that the
time restriction of a reversible conservative solution is reversible as well. From the existence
and uniqueness theorem 4.1.5 for the nonconservative Cauchy problem, we have immediately
27
Theorem 4.1.13 (Conservative backward Cauchy problem) Let π T ∈ L∞
loc (R). Then
there exists a unique π ∈ SL∞ reversible solution to (4.1.8) such that π(T, .) = π T . This
solution satisfies for any x1 < x2 and t ∈ [0, T ]
kπ(t, .)kL∞ (I) ≤ e
RT
t
α
kπ T kL∞ (J) ,
where I =]x1 , x2 [ and J =]x1 − kak∞ (T − t), x2 + kak∞ (T − t)[.
Moreover, π ≥ 0 if π T ≥ 0.
4.1.5
The generalized backward flow
We can define in a natural way a backward flow associated to a.
Definition 4.1.14 Let Db = {(s, t) ∈ R2 ; 0 ≤ t ≤ s ≤ T }. We define the backward flow
X ∈ Lip(Db × R) in the following way.
If 0 < s ≤ T , X(s, ., .) is the unique reversible SLip solution to
(
∂t X + a(t, x)∂x X = 0
X(s, s, x) = x.
in
]0, s[×R,
If s = 0, we simply set X(0, 0, x) = x.
Remark 4.1.8 If a ∈ C 1 ([0, T ] × R), we recover the usual flow, solution to the ordinary
differential equation ∂s X = a(s, X), X(t, t, x) = x.
The following Lipschitz estimates are easily proved by approximation:
k∂s Xk
◦
L∞ (Db ×R)
≤ kak∞ ,
∀(s, t) ∈ Db ,
k∂t Xk
◦
L∞ (Db ×R)
≤ kak∞ e
k∂x X(s, t, .)kL∞ (R) ≤ e
Rs
t
α
RT
0
α
,
(4.1.9)
.
We have also |X(s, t, x) − x| ≤ kak∞ (s − t), and for any t ∈ [0, T [ ∂s ∂x X ≤ α(s)∂x X
in ]t, T [s ×Rx . Notice also that since the final data is increasing, the reversibility reads as
∂x X ≥ 0 (by Lemma 4.1.8). From this we deduce that x 7→ X(s, t, x) is onto. We have also
the usual composition formula: for 0 ≤ s1 ≤ s2 ≤ s3 ≤ T and x ∈ R
X(s3 , s2 , X(s2 , s1 , x)) = X(s3 , s1 , x).
Example 4.1.2 (i) a(t, x) = − sgn x. For s ≥ t ≥ 0, X(s, t, x) = (|x| − (s − t))+ sgn x and
∂x X(s, t, x) = 1I|x|≥s−t .
(ii) a(t, x) = 1Ix<0 . For s ≥ t ≥ 0, X(s, t, x) = x+ − (x + (s − t))− and ∂x X(s, t, x) =
1Ix∈]−(s−t),0[
. Here, a and ∂x X are discontinuous on the line x = 0.
/
From the above Lipschitz estimates we easily deduce the following stability result.
Theorem 4.1.15 Let an and a be as in Theorem 4.1.10, and denote by Xn and X their
respective backward flows. Then Xn → X in C(Db × [−R, R]) for any R > 0.
28
Now, as a simple consequence of renormalization for nonconservative reversible solutions
(Remark 4.1.5(2)), we can rewrite all reversible solutions in terms of the flow. Differentiating
this formula, we obtain a similar representation for conservative solutions.
Proposition 4.1.16 Let s ∈]0, T ]. We have
(i) If ps ∈ Liploc (R), the reversible solution p to
(
∂t p + a∂x p = 0
p(s, .) = ps ,
in
]0, s[×R,
is given by p(t, x) = ps (X(s, t, x)) for any 0 ≤ t ≤ s and x ∈ R.
(ii) If π s ∈ L∞
loc (R), the reversible solution π to
(
∂t π + ∂x (aπ) = 0
π(s, .) = π s ,
in
]0, s[×R,
is given by π(t, x) = π s (X(s, t, x))∂x X(s, t, x), for any 0 ≤ t ≤ s and a.e. x ∈ R.
4.2
4.2.1
Duality solutions, forward problem
The nonconservative case
We turn now to the study of the forward problem, in nonconservative form to begin with
∂t u + a∂x u = 0,
(4.2.1)
with u ∈ SBV . Recall that a still satisfies the assumption (4.1), i.e. ∂x a is upper-bounded
by some L1t function α. This corresponds to some compressive case, by analogy to fluid
mechanics.
Definition 4.2.1 (Nonconservative duality solutions) We say that u ∈ SBV is a duality solution to (4.2.1) if for any 0 < τ ≤ T , and any reversible
solution π with compact
Z
support in x, to ∂t π + ∂x (aπ) = 0 in ]0, τ [×R, the function t 7→
u(t, x)π(t, x) dx is constant
R
on [0, τ ].
Remark 4.2.1 1. A duality solution is not a solution in the sense of distributions. We do not
intend yet to give a meaning to the product a∂x u.
2. The set of duality solutions is a vector space which contains the constants. Conversely, a
duality solution u satisfying ∂x u = 0 is constant.
3. The time restriction of a duality solution is a duality solution also.
Theorem 4.2.2 (Nonconservative forward Cauchy problem) Let be given u0 ∈ BVloc (R).
Then there exists a unique u ∈ SBV duality solution to (4.2.1), such that u(0, .) = u0 . This
solution satisfies, for any x1 < x2 and any t ∈ [0, T ],
T VI (u(t, .)) ≤ T VJ (u0 ),
(4.2.2)
ku(t, .)kL∞ (I) ≤ ku0 kL∞ (J) ,
(4.2.3)
with I =]x1 , x2 [ and J =]x1 − kak∞ t, x2 + kak∞ t[. Moreover, u ∈ Lip([0, T ], L1loc (R)).
29
Proof. Uniqueness. Let u be a duality solution with u(0, .) = 0. For any compactly
supported reversible π, we have
Z
∀t ∈ [0, τ ],
Z
u(t, x)π(t, x) dx =
u(0, x)π(0, x) dx = 0.
Since the choice of π(τ, .) in L∞
c is arbitrary, we obtain u(τ, .) = 0, for any 0 < τ ≤ T .
Existence. Consider the pseudo-solution U (u0 ) we built up in § 4.1.2. Then it is actually
a duality solution. Indeed, if π is a reversible conservative solution, by definition π = ∂x p,
with p reversible nonconservative solution. Thus p is also a limit solution (by Remark 4.1.6).
Therefore (since we can restrict the operator U on [0, τ ]),
∂t U (u0 )∂x p + ∂x a U (u0 )∂x p = 0
in
]0, τ [×R.
If π = ∂x p is compactly supported, we can integrate this equality and obtain
d
dt
Z
U (u0 )(t, x)π(t, x) dx = 0
in
]0, τ [.
R
Since the integral is continuous in t, it is a constant and we can take u = U (u0 ). The bounds
on u are obtained easily.
Remark 4.2.2 The uniqueness result proves that there is actually no need to extract any
subsequence in the construction of U (see Remark 4.1.4).
Remark 4.2.3 If u ∈ SBV , u is a duality solution to (4.2.1) in ]0, T [×R if and only if for any
T1 ∈]0, T [, u is a duality solution to (4.2.1) in ]T1 , T [×R. This last formulation can be taken
as a definition of duality solutions when a satisfies the one-sided Lipschitz condition with
only α ∈ L1loc (]0, T ]), such as α(t) = 1/t.
One can easily check that, if S : R → R is any Lipschitz continuous function, then
S(U (u0 )) = U (S(u0 )). Since the duality solution is given by U (u(0, .)), we can rewrite this
as
Proposition 4.2.3 (Renormalization) Let u ∈ SBV be a duality solution to (4.2.1), and
S : R → R a Lipschitz continuous function. Then S(u) is also a duality solution.
Remark 4.2.4 If p ∈ SLip solves (4.1.1), then p is a duality solution to (4.2.1), even if p is not
reversible. This results from Lemma 4.1.1(i).
4.2.2
The conservative case
Definition 4.2.4 (Conservative duality solutions) We say that µ ∈ SM is a duality
solution to
∂t µ + ∂x (aµ) = 0 in ]0, T [×R
(4.2.4)
if for any 0 < τ ≤ T , and any reversible
solution p with compact support in x, to ∂t p+a∂x p =
Z
0 in ]0, τ [×R, the function t 7→
p(t, x)µ(t, dx) is constant on [0, τ ].
R
30
Remark 4.2.5 The set of conservative duality solutions is obviously a vector space, and the
time restriction of a conservative duality solution is also a conservative duality solution.
Theorem 4.2.5 (Conservative forward Cauchy problem) Given µ0 ∈ Mloc (R), there
exists a unique µ ∈ SM duality solution to (4.2.4), such that µ(0, .) = µ0 . This solution
satisfies for any x1 < x2 and t ∈ [0, T ]
Z
|µ(t, dx)| ≤
[x1 ,x2 ]
Moreover, t 7→
Z
Z
|µ0 (dx)|.
(4.2.5)
[x1 −kak∞ t,x2 +kak∞ t]
|µ(t, dx)| (which has values in [0, ∞]) is nonincreasing on [0, T ].
R
Remark 4.2.6 It is easy to prove that µ0 ≥ 0 implies that µ ≥ 0.
Proof of Theorem 4.2.5. Uniqueness is obtained similarly as in Theorem 4.2.2. For existence, just define µ = ∂x u, with u obtained by Theorem 4.2.2 with initial data u0 ∈ BVloc (R)
such that ∂x u0 = µ0 . Then, regarding to Definition 4.1.11, it is easy to obtain that µ is a
duality solution, and (4.2.5) follows from (4.2.2). The final property is obtained by restriction
on a subinterval and by letting x1 → −∞, x2 → ∞.
Two easy consequences of the above proofs are the following properties of derivation and
integration, which generalize Lemma 2.2.1.
Proposition 4.2.6 (i) Let u ∈ SBV be a duality solution to ∂t u + a∂x u = 0. Then µ =
∂x u ∈ SM is a duality solution to ∂t µ + ∂x (aµ) = 0.
(ii) Let µ ∈ SM be a duality solution to ∂t µ + ∂x (aµ) = 0. Then there exists u ∈ SBV duality
solution to ∂t u + a∂x u = 0, such that µ = ∂x u. Moreover, u is unique up to an additive
constant.
The following result justifies the terminology duality solutions.
Proposition 4.2.7 (i) Let π ∈ SL∞ be a reversible solution to ∂t π+∂x (aπ) = 0, and u ∈ SBV
a duality solution to ∂t u + a∂x u = 0. Then πu ∈ SL∞ is a reversible solution to
∂t (πu) + ∂x (aπu) = 0.
(ii) Let p ∈ SLip be a reversible solution to ∂t p + a∂x p = 0, and µ ∈ SM a duality solution to
∂t µ + ∂x (aµ) = 0. Then pµ ∈ SM is a duality solution to
∂t (pµ) + ∂x (apµ) = 0.
Proof. For (i) the assertion solution is contained in the proof of Theorem 4.2.2. The
reversibility follows from the cancellation in Ve (Theorem 4.1.12).
For (ii), use the definition, and the fact that if p1 , p2 are reversible, then so is p1 p2 .
31
Theorem 4.2.8 Let µ0 ∈ Mloc (R). Then the duality solution µ ∈ SM obtained in Theorem
4.2.5 is given by
∀t ∈ [0, T ],
∀ϕ ∈ Cc (R),
Z
Z
ϕ(x) µ(t, dx) =
R
ϕ(X(t, 0, x)) µ0 (dx),
(4.2.6)
R
where X is the backward flow (Definition 4.1.14).
Proof. Consider first ϕ ∈ Lipc (R), and let p be the reversible solution to
(
∂t p + a∂x p = 0
p(τ, .) = ϕ,
in
]0, τ [×R,
for a given τ ∈]0, T ]. By Proposition 4.1.16, p(t, x) = ϕ(X(τ, t, x)), so that
Z
Z
ϕ(x) µ(τ, dx) =
Z
p(τ, x) µ(τ, dx) =
Z
p(0, x) µ(0, dx) =
ϕ(X(τ, 0, x)) µ0 (dx),
and the identity is proved for that ϕ. The general case ϕ ∈ Cc (R) follows by uniform
approximation.
Remark 4.2.7 The identity (4.2.6) means that µ(t, dx) is the image of µ0 (dx) by the application x 7→ X(t, 0, x), which is well-known for a smooth a. Therefore, (4.2.6) also holds for
ϕ bounded, measurable with compact support. From this we deduce that for 0 ≤ t ≤ T ,
|µ(t, .)| ≤ λ(t, .), where λ is the duality solution with initial data |µ0 |.
Example 4.2.1 For µ0 = δ(x − x0 ), x0 ∈ R, one has µ(t, .) = δ(x − X(t, 0, x0 )). This gives the
trajectory of a particle with initial position x0 .
Let us close this section by a lower estimate, which ensures that vacuum cannot appear
if not present at time 0. An easy proof consists in smoothing a and µ0 .
Proposition 4.2.9 Let µ ∈ SM be a duality solution to (4.2.4), and assume that µ(0, dx) ≥
δ dx for some δ > 0, in the sense of measures on R. Then for any t ∈ [0, T ]
−
µ(t, dx) ≥ δe
4.3
4.3.1
Rt
0
α
dx.
Flux and universal representative
Flux of a conservative duality solution
Definitions 4.2.1 and 4.2.4 of duality solutions are a priori not written in the distribution
sense. We give here a simple way to define the product of the coefficient a by the measure
µ, thus recovering a distribution equation. This product is in a way stable with respect to
perturbations of a and initial data. Notice that it is not defined for an arbitrary measure µ,
but only for duality solutions. That is the main reason why very general weak stability can
hold. It cannot be true for more general definitions such as in G. Dal Maso, P. Le Floch, F.
Murat [9] where the equation is not taken into account.
32
Definition 4.3.1 (Generalized flux) Let µ ∈ SM be a duality solution to (4.2.4). We
define the flux corresponding to µ by
a ∆ µ = −∂t u,
(4.3.1)
where µ = ∂x u and u ∈ SBV is a duality solution to the nonconservative problem (cf Proposition 4.2.6(ii)). We have therefore
∂t µ + ∂x (a ∆ µ) = 0
in
D0 (Ω).
(4.3.2)
The application µ 7→ a ∆ µ is of course linear, and since u ∈ Lip([0, T ], L1loc (R)), we have by
Property 5 in § 2.1 a ∆ µ ∈ L∞ (]0, T [, Mloc (R)), and for any x1 < x2 ,
Z
ka ∆ µkL∞ (]0,T [,M(]x1 ,x2 [)) ≤ kak∞
|µ(0, dx)|.
]x1 −kak∞ T,x2 +kak∞ T [
Remark 4.3.1 The flux corresponding to a time restriction is the time restriction of the flux.
Theorem 4.3.2 (Weak stability) Let (an ) be a bounded sequence in L∞ (]0, T [×R), with
an → a in L∞ (]0, T [×R) − w∗. Assume ∂x an ≤ αn (t), where (αn ) is bounded in L1 (]0, T [),
∂x a ≤ α ∈ L1 (]0, T [). Consider a sequence (µn ) ∈ SM of duality solutions to
∂t µn + ∂x (an µn ) = 0
in
Ω,
such that µn (0, .) is bounded in Mloc (R), and µn (0, .) * µ0 ∈ Mloc (R).
Then µn → µ in SM , where µ ∈ SM is the duality solution to
∂t µ + ∂x (aµ) = 0
in
Ω,
µ(0, .) = µ0 .
Moreover, an ∆ µn * a ∆ µ weakly in Mloc (Ω).
Proof. The sequence (µn ) is bounded in SM by (4.2.5). By Proposition 4.2.6(ii), we can write
µn = ∂x un , with un ∈ SBV duality solution to ∂t un +an ∂x un = 0, such that un (0, .) is bounded
in BVloc (R). Then un is bounded in Lip([0, T ], L1loc (R)), and thus (µn ) is equicontinuous in t.
Therefore, up to a subsequence, µn → µ ∈ SM .
Let us prove that µ is a duality solution. Let p be a compactly supported reversible
solution to ∂t p + a∂x p = 0 in ]0, τ [×R. From the stability theorem 4.1.10, the sequence
(pn ) of reversible solutions corresponding
to an and pn (τ, .) = p(τ, .) converges to p. Since
R
p
is also compactly supported, t 7→ pn (t, x)µn (t, dx) is constant in [0, τ ], and so is t 7→
Rn
p(t, x)µ(t, dx) by passing to the limit. Therefore, µ is a duality solution. Then uniqueness
ensures that it is not necessary to extract any subsequence.
The convergence of the flux follows in a similar way from the convergence of un .
We now come to a more precise study of the flux.
33
Lemma 4.3.3 Let µ ∈ SM a conservative duality solution, and ϕ ∈ Cc (Ω). Then
ha ∆ µ, ϕi =
Z
!
Z T
ϕ(t, X(t, 0, x))∂s X(t, 0, x) dt
R
µ0 (dx).
(4.3.3)
0
Notice that the expression between brackets is continuous with respect to x, because x 7→
∂s X(., 0, x) is weakly continuous.
Proof of Lemma 4.3.3. We know that µ = ∂x u, with u ∈ SBV a nonconservative duality
solution. Let us first assume that ϕ ∈ Cc∞ (Ω). We have by definition of the flux
ha ∆ µ, ϕi = −h∂t u, ϕi =
Z Z T
u(t, x)∂t ϕ(t, x) dtdx.
R 0
By the definition of duality solutions and by Proposition 4.1.16(ii), we obtain
Z
Z
u(t, x)∂t ϕ(t, x) dx =
R
u0 (x)∂t ϕ(t, X(t, 0, x))∂x X(t, 0, x) dx,
R
Z
so that ha ∆ µ, ϕi = −
u0 (x)I(x) dx, where I is the following L∞ (R) function:
I(x) = −
Z T
∂t ϕ(t, X(t, 0, x))∂x X(t, 0, x) dt.
0
Let us define
Z x
Φ(t, x) =
ϕ(t, y) dy,
ψ(x) = −
Z T
−∞
∂t Φ(t, X(t, 0, x)) dt
0
∈ Liploc (R),
so that ψ 0 = I. We have
Z T
ψ(x) =
Z0T
d
− [Φ(t, X(t, 0, x))] + ∂x Φ(t, X(t, 0, x))∂s X(t, 0, x)
dt
dt
ϕ(t, X(t, 0, x))∂s X(t, 0, x) dt,
=
0
thus ψ ∈ Lipc (R) and
ha ∆ µ, ϕi = −
Z
u0 (x)ψ 0 (x) dx =
R
Z
ψ(x)µ0 (dx),
R
and we are done for a smooth ϕ. For ϕ ∈ Cc (Ω), the result follows by uniform approximation.
We now state the main result of this section.
b :
Theorem 4.3.4 (Universal representative) There exists a bounded Borel function a
]0, T [×R → R such that for any conservative duality solution µ, one has
bµ.
a∆µ=a
We call such a function a universal representative of a.
34
(4.3.4)
Proof. Define a Borel set A by
X(t + ε, t, x) − x
A = (t, x) ∈ [0, T [×R;
has a limit when ε → 0+ ,
ε
and let
b(t, x) =
a


X(t + ε, t, x) − x
ε
 ε→0+
anything
lim
if (t, x) ∈ A,
(4.3.5)
(4.3.6)
if (t, x) ∈
/ A.
b has the desired property for µ0 = δ(x − x0 ), x0 ∈ R. We know in
Let us first prove that a
this case that µ(t) = δ(x − X(t, 0, x0 )) (Example 4.2.1). On the one hand, by Lemma 4.3.3,
we have for any ϕ ∈ Cc (Ω)
ha ∆ µ, ϕi =
Z T
ϕ(t, X(t, 0, x0 ))∂s X(t, 0, x0 ) dt,
0
and on the other hand
Z T
bµ, ϕi =
ha
0
b(t, X(t, 0, x0 )) dt.
ϕ(t, X(t, 0, x0 ))a
We are thus led to prove
b(t, X(t, 0, x0 ))
∂s X(t, 0, x0 ) = a
a.e. in t.
(4.3.7)
But since X is Lipschitz continuous, we have
X(t + ε, 0, x0 ) − X(t, 0, x0 )
ε→0+
ε
∂s X(t, 0, x0 ) = lim
a.e. in t.
Pick a t for which this formula holds true, and set y = X(t, 0, x0 ). Since X(t+ε, 0, x0 ) = X(t+
ε, t, X(t, 0, x0 )), we have ∂s X(t, 0, x0 ) = lim X(t+ε,t,y)−y
, so that (t, y) ∈ A and ∂s X(t, 0, x0 ) =
ε
b(t, y) = a
b(t, X(t, 0, x0 )). This concludes the proof for Dirac masses.
a
For a general duality solution µ, we have by Lemma 4.3.3 for any ϕ ∈ Cc (Ω),
ha ∆ µ, ϕi =
Z
ψ(x)µ0 (dx),
Z T
ψ(x) =
Z T
ϕ(t, X(t, 0, x))∂s X(t, 0, x) dt =
0
0
b(t, X(t, 0, x)) dt
ϕ(t, X(t, 0, x))a
by (4.3.7). Using Remark 4.2.7 we get
ha ∆ µ, ϕi =
Z T
Z
dt
Z0
=
ZR
T
dt
0
R
b(t, X(t, 0, x)) µ0 (dx)
ϕ(t, X(t, 0, x))a
b(t, x)µ(t, dx)
ϕ(t, x)a
bµ, ϕi.
= ha
b1 and a
b2 are two universal representatives, then
Remark 4.3.2 By definition, if a
b1 = a
b2 a.e. dtµ(t, dx).
a
35
b can take any value on Ac , we have
Remark 4.3.3 In the above construction, since a
Z
dt|µ(t, dx)| = 0
for any duality solution µ.
Ac
From this we deduce that for any duality solution µ
X(t + ε, t, x) − x
b(t, x) dtµ(t, dx) a.e. in Ω.
→ a
ε→0+
ε
(4.3.8)
b1 be a universal representative of a, and a
b2 : Ω → R a bounded
Proposition 4.3.5 (i) Let a
b
measurable function. Then a2 is a universal representative if and only if
∀x ∈ R
a.e. s ∈]0, T [
b2 (s, X(s, 0, x)) = a
b1 (s, X(s, 0, x)).
a
(4.3.9)
(ii) One has for any (t, x) ∈ [0, T [×R,
a.e. s ∈]t, T [
b(s, X(s, t, x)).
∂s X(s, t, x) = a
(4.3.10)
b1 and a
b2 be two universal representatives of a. Then
(iii) Let a
b1 = a
b2
a
a.e. in Ω.
(4.3.11)
b2 is a universal representative, and take for µ0 a Dirac mass at x0 .
Proof. (i) Assume a
Then by (4.3.7), (4.3.9) holds for x0 . Conversely, if (4.3.9) holds, then the definition property
(4.3.4) is satisfied for Dirac masses. The general case follows as in the proof of Theorem 4.3.4.
(ii) The case t = 0 was treated in the proof of Theorem 4.3.4 (see (4.3.7)). In the general
case, there exists x0 ∈ R such that x = X(t, 0, x0 ). Thus, if s ∈ [t, T ],
X(s, t, x) = X(s, t, X(t, 0, x0 )) = X(s, 0, x0 ).
Differentiating in s and using (4.3.7) we obtain the result.
0
R(iii) Take µ = dx, it satisfies the assumption of Proposition 4.2.9! Thus dtµ(t, dx) ≥
e−
t
0
α
b1 and a
b2 coincide dtµ(t, dx) a.e., so we are done.
dtdx. By Remark 4.3.2, a
b. Indeed, if a
b1 is another
Remark 4.3.4 Formula (4.3.10) does not depend on the choice of a
universal representative, we have by (4.3.9), for any t ∈ [0, T [ and x ∈ R,
a.e. s ∈]t, T [
b(s, X(s, t, X(t, 0, x))) = a
b1 (s, X(s, t, X(t, 0, x))).
a
But x 7→ X(t, 0, x) is onto, so that
∀t ∈ [0, T [, ∀y ∈ R, a.e. s ∈]t, T [,
b(s, X(s, t, y)) = a
b1 (s, X(s, t, y)).
a
Remark 4.3.5 We give in Theorem 4.3.10 a more precise version of (iii), stating that actually
b = a a.e. in Ω.
a
36
Remark 4.3.6 Property (ii) shows that the ”hat” operator and time restriction commute.
We are now able to prove an important feature for applications.
Theorem 4.3.6 (Entropy inequality) Let µ ∈ SM be a conservative duality solution.
Then
b|µ|) ≤ 0 in D 0 (Ω).
∂t |µ| + ∂x (a
(4.3.12)
Proof. Let ϕ ∈ Cc∞ (R), ϕ ≥ 0, t0 ∈]0, T [, and ε > 0 such that t0 + ε < T . Consider the
conservative duality solution λ on [t0 , T ], such that λ(t0 , dx) = |µ(t0 , dx)|. Making use of
Remark 4.2.7 and of the representation formula (4.2.6), we obtain
Z
ϕ(x)|µ(t0 + ε, dx)| ≤
Z
Z
ϕ(x)λ(t0 + ε, dx) =
ϕ(X(t0 + ε, t0 , x))|µ(t0 , dx)|.
Therefore
≤
1
Zε
Z
ϕ(x)|µ(t0 + ε, dx)| −
Z
ϕ(x)|µ(t0 , dx)|
X(t0 + ε, t0 , x) − x
|µ(t0 , dx)|
ϕ0 (x)
ε
Z h
i
1
+
ϕ(X(t0 + ε, t0 , x)) − ϕ(x) − ϕ0 (x)(X(t0 + ε, t0 , x) − x) |µ(t0 , dx)|.
ε
The second term in the right-hand side tends to 0 when ε → 0, uniformly in t0 , and the first
one, by Remark 4.3.3, satisfies for a.e. t0
Z
ϕ0 (x)
X(t0 + ε, t0 , x) − x
|µ(t0 , dx)| −→
ε→0+
ε
Z
b(t0 , x)|µ(t0 , dx)|.
ϕ0 (x)a
Thus we obtain in the sense of D0 (]0, T [), for any ϕ ∈ Cc∞ (R), ϕ ≥ 0,
d
dt
Z
ϕ(x)|µ(t, dx)| ≤
Z
b(t, x)|µ(t, dx)|,
ϕ0 (x)a
or equivalently for any ϕ ∈ Cc∞ (R), ϕ ≥ 0, and χ ∈ Cc∞ (]0, T [), χ ≥ 0,
b|µ|), χ ⊗ ϕi ≤ 0.
h∂t |µ| + ∂x (a
Now we take a mollifier ρn = χn ⊗ ϕn and obtain
b|µ|) ≤ 0
ρn ∗ ∂t |µ| + ∂x (a
t,x
pointwise.
Therefore, we get (4.3.12) by letting n → ∞.
We end this section by a geometric description in the piecewise continuous case.
Theorem 4.3.7 (Piecewise continuous coefficient) Let us assume that in addition to
the hypothesis (4.1) used in all Section 4, a is piecewise continuous in the sense of (PC1)b which coincides with a in C, and which
(PC3) in Section 3. Then there exists a function a
is admissible in the sense of (AC1)-(AC2).
b, µ ∈ SM is a duality solution to (4.2.4) if and only if ∂t µ + ∂x (a
bµ) = 0 in
With this a
bµ.
D0 (Ω). Then a ∆ µ = a
b is a universal representative of a.
In particular, a
37
Proof. By Theorems 4.2.5 and 3.6, there exists unique solutions to the Cauchy problem for
both kinds of solutions (duality solutions and distributional solutions). Since they are built
by the same procedure (approximation of a), they do coincide. Therefore, both notions are
identical. The remaining assertions are obvious.
4.3.2
Characteristics in Filippov’s sense
Lemma 4.3.8 Assume that a.e. t ∈]0, T [, a(t, .) ∈ C(R). Then for any conservative duality
solution µ ∈ SM ,
a ∆ µ = aµ.
(4.3.13)
Proof. Consider the same mollifier ρn as in Lemma 2.3.1, let an = ρn ∗ a and µ0n = ρxn ∗ µ0 ,
x
t,x
and denote by µn the (classical) solution to
∂t µn + ∂x (an µn ) = 0
in ]0, T − ε[×R,
µn (0, .) = µ0n .
The stability result (Theorem 4.3.2) ensures that µn → µ in C([0, T −ε], Mloc (R)−σ(Mloc (R), Cc (R)))
and an µn → a ∆ µ in D0 (]0, T − ε[×R). On the other hand, an µn = aµn + (an − a)µn , and
since a is continuous in x, aµn * aµ σ(Mloc (R), Cc (R)) for a.e. t ∈]0, T − ε[. Integrating
in t and using Lebesgue’s theorem leads therefore to aµn → aµ in D0 (]0, T − ε[×R). Next,
as in the proof of Lemma 2.3.1, an → a in L1 (]0, T − ε[, L∞
loc (R)) and thus (an − a)µn → 0
in L1loc (]0, T − ε[×R). Therefore an µn → aµ in D0 (]0, T − ε[×R), which gives a ∆ µ = aµ in
]0, T − ε[×R, for any ε > 0.
We now prove that the backward flow actually coincides with the generalized forward
solution introduced by A.F. Filippov [12]. Recall that g ∈ Lip([t, T ]) solves g 0 (s) = a(s, g(s))
in the Filippov sense if
a.e. s ∈]t, T [
lim
ess inf a(s, y) ≤ g 0 (s) ≤ lim ess sup a(s, y).
δ→0+ |y−g(s)|<δ
δ→0+ |y−g(s)|<δ
In our case, since a satisfies the one-sided Lipschitz condition (4.1), a(t, .) ∈ BVloc (R) for a.e.
t so that a(t, x+) = lim a(t, y) and a(t, x−) = lim a(t, y) are well-defined for any x ∈ R.
y→x+
y→x−
Moreover, again by (4.1), a(t, x+) ≤ a(t, x−). Therefore Filippov’s condition reads as
a.e. s ∈]t, T [ g 0 (s) ∈ [a(s, g(s)+), a(s, g(s)−)] .
(4.3.14)
Proposition 4.3.9 (Filippov’s flow) For any (t, x) ∈ [0, T [×R, s 7→ X(s, t, x) is the
unique function g ∈ Lip([t, T ]) which satisfies (4.3.14) and g(t) = x.
Proof. Uniqueness is contained in Filippov’s paper [12]. In order to prove that g(s) ≡
X(s, t, x) is a solution, take ρn a smoothing sequence in R, and set an = ρn ∗ a, which
x
bn = an , and we have by (4.3.10)
satisfies ∂x an ≤ α. By Lemma 4.3.8, we can choose a
a.e. s ∈]t, T [,
∂s Xn (s, t, x) = an (s, Xn (s, t, x)),
which means that Xn is solution in Filippov’s sense. By the stability theorem 4.1.15, Xn → X
uniformly on bounded subsets of Db × R. Therefore, by Filippov’s stability theorem [12], X
verifies (4.3.14).
38
Remark 4.3.7 Theorem 4.1.15 can be interpreted as a weak stability result for Filippov’s
solutions.
b of a such that
Theorem 4.3.10 One can choose a universal representative a
b(t, x) ∈ [a(t, x+), a(t, x−)] .
a.e. t ∈]0, T [, ∀x ∈ R, a
(4.3.15)
In particular, we have
b(t, x) = a(t, x) = a(t, x+) = a(t, x−)
a
a.e. in ]0, T [×R.
(4.3.16)
Notice that by Proposition 4.3.5(iii), (4.3.16) therefore holds for any universal representative
b of a.
a
Proof of Theorem 4.3.10. By changing a if necessary on a set of t of measure zero, we
can assume that a(t, .) ∈ BVloc (R) for any t. Then the set
X(t + ε, t, x) − x
exists and belongs to [a(t, x+), a(t, x−)]
ε→0+
ε
B = (t, x) ∈ Ω; lim
is a Borel set, and we can define
b(t, x) =
a

X(t + ε, t, x) − x


 lim
if (t, x) ∈ B,
ε

a(t, x+) + a(t, x−)


2
ε→0+
if (t, x) ∈
/ B.
By Proposition 4.3.9, we have for any x ∈ R
a.e. t ∈]0, T [,
∂s X(t, 0, x) ∈ [a(t, X(t, 0, x)+), a(t, X(t, 0, x)−)].
b is a universal representative of a.
Then, following the proof of Theorem 4.3.4, a
4.3.3
Reversibility and renormalization
Let us now come back to the backward problem.
Lemma 4.3.11 (Strong continuity) Let π ∈ SL∞ be a solution to ∂t π + ∂x (aπ) = 0 in Ω.
Then π ∈ C([0, T ], L1loc (R)).
Notice that π is not assumed to be reversible.
Proof of Lemma 4.3.11. We use the same method as in R.J. DiPerna and P.-L. Lions
[10]. Consider ρn a mollifier in R, and define πn = ρn ∗ π ∈ C([0, T ] × R), and an = ρn ∗ a
x
x
∈ L∞ (Ω). We notice that πn ∈ Liploc ([0, T ] × R), since it satisfies
∂t πn + ∂x (ρn ∗(aπ)) = 0
x
39
in
Ω.
Writing
∂t πn + ∂x (an πn ) = ∂x (an πn − ρn ∗(aπ)) ∈ L∞
loc ,
x
we multiply this equality by πn , to obtain
∂t
Let ϕ ∈
π2
π2
πn2
+ ∂x (an n ) = − n ∂x an + πn ∂x (an πn − ρn ∗(aπ)) ≡ fn
x
2
2
2
Cc∞ (R).
d
dt
Z
πn2
ϕ dx is continuous on [0, T ], and satisfies
2
Z
The function t 7→
πn2
ϕ dx =
2
∈ L∞
loc .
Z
an
πn2 0
ϕ dx +
2
Z
fn ϕ dx in
]0, T [.
In view of the regularity of a (4.3), we write
fn = −
h
i
πn2
ρn ∗ ∂x a + πn (an − a)ρ0n ∗ π + πn ρn ∗ ∂x a + aρ0n ∗ π − ρ0n ∗(aπ) .
x
x
x
x
x
2
Notice now that
an − a =
aρ0n ∗ π − ρ0n ∗(aπ) =
x
x
Z
[a(t, x − y) − a(t, x)] ρn (y) dy,
ZR
[a(t, x) − a(t, x − y)] π(t, x − y)ρ0n (y) dy.
R
By using (4.3), we obtain for a.e. t ∈]0, T [
Z x2
|a(t, x − y) − a(t, x)| dx ≤ 2|y| |α(t)|(x2 − x1 + |y|) + kak∞ ,
x1
and setting M =
ess sup
|π(t, x)|,
0<t<T
x1 −1/n<x<x2 +1/n
Z x2
1
2
|α(t)|(x2 − x1 + ) + kak∞ ,
n n
Z x2x1
1
0
0
|aρn ∗ π − ρn ∗(aπ)| dx ≤ C M |α(t)|(x2 − x1 + ) + kak∞ .
x
x
n
x1
|an (t, x) − a(t, x)| dx ≤
Since kρ0n ∗ πkL∞ (]x1 ,x2 [) ≤ C M n, we obtain, for a.e. t ∈]0, T [,
x
Z x2
|fn (t, x)| dx ≤ C M
x1
2
1
|α(t)|(x2 − x1 + ) + kak∞ .
n
Putting things together, and choosing x1 < x2 such that suppϕ ⊂]x1 , x2 [, we get that for a.e.
t ∈]0, T [,
Z
d
πn2
1
M2
ϕ dx ≤ C M 2 |α(t)|(x2 − x1 + ) + kak∞ kϕk∞ +
kak∞ kϕ0 k1 .
dt
2
n
2
!
πn2
The family t 7→
ϕ dx
is therefore relatively compact in C([0, T ]). Since it converges
2
n
pointwise, the convergence is thus uniform, and the limit
Z
t 7→
Z
π(t, x)2
ϕ(x) dx
2
40
is continuous in [0, T ]. The result follows since it is true for any ϕ ∈ Cc∞ (R).
In a similar way, one can prove
Proposition 4.3.12 The backward flow X satisfies ∂x X ∈ C(Db , L1loc (R)), with
Db = {(s, t) ∈ R2 ; 0 ≤ t ≤ s ≤ T }.
We can now prove the following characterization for reversibility by renormalization.
Theorem 4.3.13 (Reversibility by renormalization) Let π ∈ SL∞ be a distributional
solution to ∂t π + ∂x (aπ) = 0 in Ω. Then
∂t |π| + ∂x (a|π|) ≤ 0
in
Ω,
and equality holds if and only if π is reversible.
Proof. By Remark 4.2.4 π is a duality solution. Thus Theorem 4.3.6 gives the inequality
b = a a.e. by (4.3.16).
since a
Concerning equality, first assume that π is reversible. Then by Proposition 4.1.16(ii)
π(t, x) = π(T, X(T, t, x))∂x X(T, t, x), and thus |π(t, x)| = |π(T, X(T, t, x))|∂x X(T, t, x) is
the reversible solution with final data |π(T, .)|. Conversely, if |π| solves the equation with
equality, we notice that by Lemma 4.3.11 |π| ∈ SL∞ . But |π| ≥ 0, so |π| is reversible by
Theorem 4.1.12(iii). Therefore, |π| = 0 in Ve by Theorem 4.1.12(ii), and π = 0 in Ve . The
same criterion (ii) thus ensures that π is reversible.
4.4
4.4.1
On viscous problems
Duality for viscous problems
The duality method can be used to treat viscous problems. Let us show formally how it
works for a constant viscosity ε. The forward conservative problem is written in this case
2
∂t µ + ∂x (aµ) − ε∂xx
µ = 0,
µ(0, .) = µ0 ,
(4.4.1)
and the backward nonconservative problem is
2
∂t p + a∂x p + ε∂xx
p = 0,
p(T, .) = pT .
(4.4.2)
For both problems, we have the same a priori estimates as for the non-viscous equations.
Notice also that the decomposition of p as a difference of monotone solutions is valid. The
duality between (4.4.1) and (4.4.2) can be seen by multiplying (4.4.1) by p, (4.4.2) by µ and
by adding the results. This yields
∂t (pµ) + ∂x (apµ) + ε∂x (µ∂x p − p∂x µ) = 0.
(4.4.3)
Therefore,
d
hµ, pi = 0
(4.4.4)
dt
as in the non-viscous case. This shows that p and µ tend respectively to the reversible and
duality solutions to the non-viscous problems when ε → 0.
41
4.4.2
Backward problem with discontinuous final data
Let us again consider the backward nonconservative problem
∂t p + a∂x p = 0,
p(T, .) = pT .
(4.4.5)
To any pT ∈ Liploc (R) we can associate the reversible solution p ∈ Liploc ([0, T ] × R) to
(4.4.5). Since this operator is linear and continuous for the topology of uniform convergence
on compact sets (by Theorem 4.1.5), it extends in a unique linear continuous operator from
C(R) to C([0, T ] × R). Actually this extension can be expressed in terms of the flow (see
Proposition 4.1.16(i)).
We want here to show that there is no natural way to solve (4.4.5) if pT is discontinuous,
even if pT ∈ BVloc (R).
Let us first choose a sequence pTn ∈ Liploc (R) such that pTn → pT in L1loc (R), with pTn
bounded in BVloc (R). Then the corresponding reversible solution pn is also a duality solution
by Remark 4.2.4, which is bounded in SBV since pn is reversible. Therefore, after extraction of
a subsequence, pn (0, .) → p0 ∈ BVloc , and by Theorem 4.3.2 pn → p the duality solution with
initial data p0 . Moreover, p is locally constant in Ve since pn is. However, the values of p in Ve
strongly depend on the sequence of approximations (pTn ). For example, if a(t, x) = − sgn x,
this value is lim pTn (0), which can be anything if pT is discontinuous at 0.
However, it may be possible to determine what is p in Ve by selecting a specific approximation procedure, see F. James and M. Sepúlveda [16]. For example, let us solve
2
∂t pε + a∂x pε + ε∂xx
pε = 0,
pε (T, .) = pT .
(4.4.6)
In the case where a = − sgn x,
pε → p
and
p = (pT (0+) + pT (0−))/2
in Ve .
(4.4.7)
uε (0, .) = u0 = pT .
(4.4.8)
Actually by setting uε (t, x) = pε (T − t, x) we are led to
2
∂t uε + (sgn x)∂x uε − ε∂xx
uε = 0,
The solution is given by
Z
uε (t, x) =
Gε (t, x, y)u0 (y) dy,
(4.4.9)
R
1
e−|y|/ε
|x| + |y| − t
2
√
Gε (t, x, y) = √
e−(x−t sgn(x)−y) /4tε−11xy<0 |y|/ε +
erf
,
2ε
4πtε
2tε
Z ∞
1
2
√ e−z /2 dz,
erf (y) =
2π
y
and we easily get (4.4.7).
42
(4.4.10)
(4.4.11)
References
[1] J.-P. Aubin and A. Cellina, Differential Inclusions, Springer-Verlag, 1984.
[2] F. Bouchut, On Zero Pressure Gas Dynamics, Advances in Kinetic Theory and Computing, Selected Papers, Series on Advances in Mathematics for Applied Sciences vol.
22, 171-190, World Scientific, 1994.
[3] F. Bouchut and F. James, Equations de Transport Unidimensionnelles à Coefficients
Discontinus, C.R. Acad. Sci. Paris t. 320, Serie I, 1097-1102 (1995).
[4] Y. Brenier and E. Grenier, On the Model of Pressureless Gases with Sticky Particles,
UCLA Computational and Applied Math. report no 94-45 (1994).
[5] I. Capuzzo Dolcetta and B. Perthame, On Some Analogy Between Different Approaches
to First Order PDE’s with Non Smooth Coefficients, to appear in Advances in Math.
Sci. and Appl.
[6] N. Caroff, Generalized Solutions of Linear Partial Differential Equations with Discontinuous Coefficients, preprint, CEREMADE Paris (1991).
[7] E.D. Conway, Generalized Solutions of Linear Differential Equations with Discontinuous
Coefficients and the Uniqueness Question for Multidimensional Quasilinear Conservation
Laws, J. of Math. Anal. and Appl. 18, 238-251 (1967).
[8] C.M. Dafermos, Generalized Characteristics and the Structure of Solutions of Hyperbolic
Conservation Laws, Indiana Univ. Math. J. 26, 1097-1119 (1977).
[9] G. Dal Maso, P. Le Floch and F. Murat, Definition and Weak Stability of Nonconservative Products, J. Math. Pures Appl. 74, 483-548 (1995).
[10] R.J. DiPerna and P.-L. Lions, Ordinary Differential Equations, Transport Theory and
Sobolev Spaces, Invent. Math. 98, 511-547 (1989).
[11] W. E, Y.G. Rykov and Y.G. Sinai, Generalized Variational Principles, Global Weak
Solutions and Behavior with Random Initial Data for Systems of Conservation Laws
Arising in Adhesion Particle Dynamics, Comm. Math. Phys. 177, 349-380 (1996).
[12] A.F. Filippov, Differential Equations with Discontinuous Right-Hand Side, A.M.S.
Transl. (2) 42, 199-231 (1964).
[13] A.F. Filippov, Differential Equations with Discontinuous Righthand Sides, Coll. Mathematics and Its Applications, Kluwer Academic Publishers Dordrecht/Boston/London,
1988.
[14] E. Grenier, Existence Globale pour le Système des Gaz sans Pression, C.R. Acad. Sci.
Paris t. 321, Serie I, 171-174 (1995).
[15] D. Hoff, The Sharp Form of Oleinik’s Entropy Condition in Several Space Variables,
Trans. of the A.M.S. 276, 707-714 (1983).
43
[16] F. James and M. Sepúlveda, Convergence Results for the Flux Identification in a Scalar
Conservation Law, Preprint Université d’Orléans (1994).
[17] H.C. Kranzer and B.L. Keyfitz, A Strictly Hyperbolic System of Conservation Laws
Admitting Singular Shocks, Nonlinear Evolution Equations that Change Type, IMA
Volumes in Mathematics and its Applications, vol. 27, Springer-Verlag (1990).
[18] P. Le Floch, An Existence and Uniqueness Result for Two Nonstrictly Hyperbolic Systems, Nonlinear Evolution Equations that Change Type, IMA Volumes in Mathematics
and its Applications, vol. 27, Springer-Verlag (1990).
[19] P. Le Floch and Z. Xin, Uniqueness via the Adjoint Problems for Systems of Conservation
Laws, Comm. on Pure and Applied Maths., XLVI (11), 1499-1533 (1993).
[20] T.-P. Liu and M. Pierre, Source-solutions and Asymptotic Behavior in Conservation
Laws, J. Diff. Eq. 51, 419-441 (1984).
[21] O.A. Oleinik, Discontinuous Solutions of Nonlinear Differential Equations, Amer. Math.
Soc. Transl. (2) 26, 95-172 (1963).
[22] P. Serfati, Thesis Université Paris 6 (1992).
[23] E. Tadmor, Local Error Estimates for Discontinuous Solutions of Nonlinear Hyperbolic
Equations, SIAM J. Numer. Anal. 28, 891-906 (1991).
[24] E. Tadmor and T. Tassa, On the Piecewise Smoothness of Entropy Solutions to Scalar
Conservation Laws, Comm. in P.D.E. 18, 1631-1652 (1993).
[25] D. Tan, T. Zhang and Y. Zheng, Delta-Shock Waves as Limits of Vanishing Viscosity
for Hyperbolic Systems of Conservation Laws, J. Diff. Eq. 112, 1-32 (1994).
[26] Y. Zheng and A. Majda, Existence of Global Weak Solutions to One-Component VlasovPoisson and Fokker-Planck-Poisson Systems in One Space Dimension with Measures as
Initial Data, Comm. on Pure and Appl. Math. 47, 1365-1401 (1994).
44