Existence and uniqueness of solutions for linear
conservations laws with velocity field in L∞ .
Olivier Besson and Jérôme Pousin
Preprint (May 2005)
1
EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR LINEAR
CONSERVATIONS LAWS WITH VELOCITY FIELDS IN L∞ .
OLIVIER BESSON
∗ AND
JÉRÔME POUSIN
†
Abstract. A Space-Time Integrated Least Squares (STILS) method is derived for solving the linear conservation
law with a velocity field in L∞ . An existence and uniqueness result is given for the solution of this equation. A
maximum principle is established and finally a comparison with renormalized solution is presented.
Key words. Conservation laws, transport equation, space-time least squares methods.
1. Introduction. Many works are dedicated to linear conservation laws, and according
to the regularity of datum, different points of view have been used. The semi-group approach
first developed in [7] requires a C 1 regularity for the velocity field. Moreover this vector field
has to be extendable by zero outside a neighborhood of the spatial domain Ω. The characteristic flow generated by the velocity u can be defined for less regular fields. In [18] for
a velocity field in L1 (0, T ; W 1,1 (Ω)) with div u ∈ L1 (0, T ; L∞(Ω)) the notion of renormalized solution is introduced allowing to handle initial conditions with very low regularity.
When the velocity field u belongs to H 1/2 with a divergence free existence and uniqueness
has been proved in [14] and when the velocity field u belongs to BV , results of existence and
uniqueness of solutions in L∞ is provided in [3, 13], see also [9]. For domains Ω included
in R2 and for a time independent velocity field u in L2loc with a divergence free a solution to
the linear conservation laws is presented in [23] and compare to renormalized solutions. The
question of uniqueness for weak solutions in L∞ to linear conservation laws is discussed in
[16] for a velocity field u in L∞ with a divergence free and a domain included in R2 .
In this paper, the question of existence and uniqueness is addressed for linear conservation
laws on a domain Ω with Lipchitz boundary that satisfy the cone property. In our case the
velocity field u is only bounded. i.e. u ∈ L∞ and div u ∈ L∞ . The proposed method does
not deal with the characteristic flow generated by the velocity field, but uses the functional
setting of anisotropic Sobolev’s spaces in the same way as in [22] combined with a formulation of the problem in the time-space least squares sense in the same spirit as in [20] and [2].
In [18], the velocity field is required to be more regular than in our formulation (u ∈ L∞ ,
and div u ∈ L∞ ). This allow them to handle boundary conditions with very few regularity.
In our method we must assume that the boundary conditions have some regularity.
The least squares method is widely used to solve partial differential equations, see [19] and
[20] for elasticity and fluid mechanics problems. Few general mathematical results have
been obtained for this method in the case of first order time dependent conservation laws.
It seems that the STILS method (Space-Time Integrated Least Squares) is originated to [11]
and [26]. In [11, 28], a least squares method is used to solve a 2D stationary first order
conservation equation with regularity assumptions on the advection velocity. Other results
have been obtained in [4, 5, 6]. In this paper, a general mathematical analysis of this method
is given for the linear conservation law when the advection velocity u has low regularity, more
precisely when u ∈ L∞ , and div u ∈ L∞ . The solution obtained in this way is compared
with renormalized solutions [18].
In section 2 a description of the problem is given. In section 3 a variational formulation of the
problem is given. The section 4 is dedicated to the proof of the existence and uniqueness of
∗ Université de Neuchâtel; Institut de Mathématiques; 11, rue E. Argand; 2007 Neuchâtel, Switzerland. e-mail:
[email protected]
† INSA Lyon, Laboratoire MAPLY, UMR-CNRS 5585; Bâtiment Léonard de Vinci; 20 av. Albert Einstein;
69621 Villeurbanne, France. e-mail: [email protected]
2
solutions to the variational formulation described in section 3. Moreover a comparison with
renormalized solution is given.
2. The problem description. Let Ω ⊂ Rd be a domain with a Lipschitz boundary ∂Ω
satisfying the cone property. If T > 0 is given, set Q = Ω×]0, T [. Consider an advection
velocity u : Q → Rd and f : Q → R a given source term. In all this paper, the velocity u has
the following regularity
u ∈ L∞ (Q)d and div u ∈ L∞ (Q).
(2.1)
Let
Γ− = {x ∈ ∂Ω : ( u(x, t) | n(x) ) < 0}
where n(x) is the outer normal to ∂Ω at point x. For the sake of the presentation, it is assumed
that Γ− do not depend on t.
The problem consists in finding a function c : Q → R satisfying the following partial differential equation
∂t c + div (c u) = f
in Q,
(2.2)
for x in Ω
for x on Γ− .
(2.3)
(2.4)
and the initial and inflow boundary conditions
c(x, 0) = c0 (x)
c(x, t) =c1 (x, t)
As usual, when c1 , c0 , and u are sufficiently regular, changing the source term f if necessary,
one can assume that c1 = 0 on Γ− , and c0 = 0 on Ω. A similar result will be given later,
using a suitable trace theorem.
3. Functional Setting. In this section the functional setting for a variational formulation
of the problem (2.2-2.4) will be settled, (see also [4, 5, 6]). Moreover a trace operator is given
in this context.
3.1. The Hilbert spaces. For u ∈ L∞ (Q)d , with div u ∈ L∞ (Q), define u
e as
u
e = (1, u1 , u2 , . . . , ud )t ∈ L∞ (Q)d+1
and for a sufficiently regular function ϕ defined on Q, set
and
e =
∇ϕ
∂ϕ
∂ϕ ∂ϕ ∂ϕ
,
,
,...,
∂t ∂x1 ∂x2
∂xd
t
,
d
X ∂
f u ϕ) = ∂ϕ +
div(e
(ui ϕ).
∂t
∂xi
i=1
Finally n
e denotes the outward unit vector on ∂Q. The following theorem is proved in [12].
T HEOREM 3.1. Under the assumption u ∈ L∞ (Q)d , and div u ∈ L∞ (Q), the normal trace
of u, ( u
e|n
e ) is in L∞ (∂Q).
3
Let now
∂Q− = {(x, t) ∈ ∂Q, ( u
e|n
e ) < 0}
= Γ− × (0, T ) ∪ Ω × {0},
and set
cb (x, t) =
c0 (x)
c1 (x, t)
if
if
(x, t) ∈ Ω × {0}
(x, t) ∈ Γ− × (0, T ).
(3.1)
We will assume that
cb ∈ L2 (∂Q− ).
For ϕ ∈ D(Q), consider the norm
kϕkH(u,Q) =
kϕk2L2 (Q)
2
f
+ div(e
u ϕ)
L2 (Q)
+
Z
2
∂Q−
|(u
e|n
e ) |ϕ de
σ
!1/2
,
(see also [4, 5, 6, 8]) and then define the space H(u, Q) as the closure of D(Q) for this norm:
H(u,Q)
H(u, Q) = D(Q)
If u is regular enough, it can be seen that
n
o
f u ρ) ∈ L2 (Q), ρ|∂Q ∈ L2 (∂Q− , | ( u
H(u, Q) = ρ ∈ L2 (Q), div(e
e|n
e ) | de
σ)
−
(see e.g. [25, 22]). We now give a trace result for functions belonging to H(u, Q). Let us
start with the well known normal trace operator γ defined from H(div, Q) with values in
1
H − 2 (∂Q) (see [21, 10])
v 7→ ( n
e | v ) |∂Q ,
∀v ∈ H(div, Q), with the associated Green formula:
Z
f
e
div(v)ψ
+ v | ∇ψ
dx dt =< ( v | n
e),ψ >
Q
1
∀ψ ∈ H 1 (Q). Plugging v = u
eρ in the previous formula, we have:
Z
f uρ)ψ + u
e
div(e
e | ∇ψ
ρ dx dt =< ρ ( u
e|n
e ) , ψ > − 12
H
Q
1
H − 2 (∂Q);H 2 (∂Q)
1
,
(∂Q);H 2 (∂Q)
,
∀ψ ∈ H 1 (Q). Let us now consider the bilinear form L : D(Q) × D(Q) ⊂ H(u, Q) ×
H(u, Q) −→ R defined for all ϕ, ψ ∈ D(Q) by:
Z
Z
f uϕ)ψ + u
e
L(ϕ, ψ) =
div(e
e | ∇ψ
ϕ dx dt +
|(u
e|n
e ) |ϕψ de
σ.
Q
∂Q−
Accounting for theorem 3.1 we have
f
kψkL2 (Q)
uϕ) 2
|L(ϕ, ψ)| ≤ div(e
L (Q)
f
f u)ψ kϕkL2 (Q)
uψ) − div(e
+ div(e
2
L (Q)
+ kϕkL2 (∂Q− ,|( ue | ne )|deσ) kψkL2 (∂Q− ,|( ue | ne )| deσ) .
4
And the following estimate holds true
|L(ϕ, ψ)| ≤ (1 + kdiv(u)kL∞ (Q) ) kϕkH(u,Q) kψkH(u,Q) .
2
Since it is straightforward to check that L(ϕ, ϕ) = kϕkL2 (∂Q+ ,|( ue | ne )|deσ) , if we extend by
continuity the bilinear form L to H(u, Q) × H(u, Q) we have:
P ROPOSITION 3.2. Under the assumption u ∈ L∞ (Q)d , and div u ∈ L∞ (Q) there exists a
linear continuous trace operator
γne : H(u, Q) −→ L2 (∂Q, | ( u
e|n
e ) |de
σ)
ϕ 7→ γne ϕ = ϕ|∂Q ,
which can be localized as:
γne± : H(u, Q) −→ L2 (∂Q± , | ( u
e|n
e ) |de
σ)
ϕ 7→ γne± ϕ = ϕ|∂Q± .
Finally define the space
H0 = H0 (u, Q, ∂Q−) = {ρ ∈ H(u, Q), ρ = 0 on ∂Q− }
= H(u, Q) ∩ Ker γne− .
3.2. Curved Poincaré inequality. We now give an extension of the curved Poincaré
inequality obtained in [4, 5].
T HEOREM 3.3. If u ∈ L∞ (Q)d and div u ∈ L∞ (Q), the semi-norm on H(u, Q) defined by
|ρ|1,u =
Z
Q
f uρ) dx dt +
(div(e
2
Z
2
∂Q−
is a norm, equivalent to the norm given on H(u, Q).
|(u
e|n
e ) |ρ de
σ
!1/2
(3.2)
Proof. We have to show that there is a constant C such that
kϕkL2 (Q) ≤ C · |ϕ|1,u
for all ϕ ∈ D(Q). We have
Z
Z h
i
f u ϕ) · ξ + ϕ · u
e
div(e
e | ∇ξ
dx dt −
∂Q−
Q
ξϕ ( u
e|n
e ) de
σ=
Z
∂Q+
ξϕ ( u
e|n
e ) de
σ
(3.3)
for all regular enough function ξ. For α : (0, T ) → R, choose ξ = α · ϕ, then
∂ξ
∂ϕ
∂ϕ
0
+( u | ∇ξ ) = α·
+ ( u | ∇ϕ ) +α ϕ = α·
+ div (u ϕ) − ϕ div u +α0 ϕ.
∂t
∂t
∂t
Let v ∈ L∞ (0, T ) be defined by
v(t) = sup | div(u(t, x))|.
x∈Ω
5
With the above choices, equation (3.3) has the form
Z h
i
f u ϕ) dx dt−
(α0 + αv − α(v + div u)) ϕ2 + 2αϕ · div(e
Q
Z
Z
2
αϕ ( u
e|n
e ) de
σ=
∂Q−
∂Q+
Let α be the solution of the differential equation
α0 + αv = −2,
αϕ2 ( u
e|n
e ) de
σ
(3.4)
α(T ) = 0.
An easy computation gives
α(t) = 2e−w(t)
Z
t
Rt
T
ew(s) ds ≥ 0,
v(s)
with w(t) = 0 e ds. Introducing this value in equation (3.4) we obtain
Z h
i
f u ϕ) dx dt−
−2ϕ2 − α(v + div u)ϕ2 + 2αϕ div(e
Q
Z
Z
αϕ2 ( u
e|n
e ) de
σ=
αϕ2 ( u
e|n
e ) de
σ ≥ 0.
∂Q−
Hence
so
∂Q+
Z
f u ϕ) dx dt − 1
αϕ2 ( u
e|n
e ) de
σ≤
αϕ · div(e
2 ∂Q−
Q
Q
Z
Z
Z
1
1
1
2
2f
2
αϕ2 ( u
e|n
e ) de
σ
ϕ dx dt +
α div(e
u ϕ) dx dt −
2 Q
2 Q
2 ∂Q−
Z
ϕ2 dx dt ≤
Z
Q
2
Z
ϕ dx dt ≤
2
Z
Q
f u ϕ)2 dx dt −
α · div(e
2
Z
∂Q−
If A = max(kαkL∞ , kαkL∞ ), we get
Z
Z
Z
f u ϕ)2 dx dt −
div(e
ϕ2 dx dt ≤ A
Q
∂Q−
Q
and the theorem is proved.
αϕ2 ( u
e|n
e ) de
σ.
2
!
ϕ (u
e|n
e ) de
σ ,
Henceforth the space H(u, Q) is equipped with the norm |ϕ|1,u .
R EMARK 1. a) Using the above result, if cb = 0, the semi-norm
Z
1/2
|ρ|1,u =
(Aρ)2 dx dt
Q
in a norm on H0 which is equivalent to the usual norm on H(u, Q).
b) As an easy consequence of the above arguments, for any ρ ∈ H(u, Q), the norm defined
by:
!1/2
Z
1
2
2
(u
e|n
e ) (T − t)ρ de
σ
|||ρ||| = kρkL2 (Q) +
2 ∂Q+
6
verifies
kρkL2 (Q) ≤ |||ρ||| ≤
√
A |ρ|1,u .
3.3. A weak formulation. In L2 (Q)), a solution of equation (2.2) corresponds to a zero
of the following convex, positive functional
!
Z
Z 2
1
2
f
(c − cb ) ( u
e|n
e ) de
σ
div(e
u c) − f dx dt −
J(c) =
2
∂Q−
Q
The Gâteau derivative of J is
Z
Z f u c) − f div(e
f u ϕ) dx dt −
div(e
DJ(c)ϕ =
∂Q−
Q
(c − cb )ϕ ( u
e|n
e ) de
σ
So a sufficient condition to get the least squares solution of (2.2 - 2.4) is the following weak
formulation: Find c ∈ H(u, Q) such that
Z
Q
f u c) · div(e
f u ϕ) dx dt −
div(e
Z
∂Q−
Z
Q
c · ϕ(u
e|n
e ) de
σ=
f u ϕ) dx dt −
f · div(e
Z
∂Q−
for all ϕ ∈ H(u, Q) (see [4, 5, 6, 8, 15, 17]).
cb · ϕ ( u
e|n
e ) de
σ
(3.5)
3.4. Stampacchia’s theorems. In this section, we assume that the domain Ω is bounded.
Later we will use the following versions of Stampacchia’s theorems (see [27, 24]).
T HEOREM 3.4. Let ρ ∈ H(u, Q), then
f u ρ) = 0 a.e. on the set {(x, t) ∈ Q; ρ(x.t) = 0}
div(e
T HEOREM 3.5. Let f : R → R be a Lipschitz continuous function.
a) If ρ ∈ H(u, Q), then f (ρ) ∈ H(u, Q).
b) If f is differentiable except at a finite number of points, say {z1 , . . . , zn }, then
f u ρ) if ρ(x, t) ∈
f 0 (ρ) div(e
/ {z1 , . . . , zn }
f
div(e
u f (ρ)) =
0 elsewere.
(3.6)
(3.7)
For the proof of these theorems the following lemma is used.
L EMMA 3.6. Let ρ ∈ H(u, Q) then |ρ| ∈ H(u, Q) and
where
f u |ρ|) = sgn(ρ) · div(e
f u ρ)
div(e

 +1
0
sgn(ρ(x, t)) =

−1
7
if
if
if
ρ(x, t) > 0
ρ(x, t) = 0
ρ(x, t) < 0
(3.8)
Proof. (see [24]). For ε > 0, let fε (t) =
and
√
t2 + ε. If ρ ∈ H(u, Q), then fε (ρ) ∈ H(u, Q)
f u fε (ρ)) = p ρ
f u ρ).
div(e
div(e
ρ2 + ε
We have
Z
2
Q
|fε (ρ)| dx dt = εT |Ω| +
Z
Q
2
|ρ|2 dx dt → kρkL2 (Q) if ε → 0
and
Z Q
f u
fε0 (ρ) div(e
Z
2
ρ2 f
ρ) dx dt =
div(e
u ρ)2 dx dt
2
Q ρ +ε
Z
f u ρ)2 dx dt if ε → 0
div(e
→
Q
So the set {fε (ρ)}ε>0 is bounded in H(u, Q) and there exists a sequence (εn ) → 0 such that
fεn (ρ) * η in H(u, Q).
Since kfεn (ρ)kH(u,Q) → kρkH(u,Q) if n → ∞ and fε (t) → |ρ| if ε → 0, we have η = |ρ|
and |ρ| ∈ H(u, Q).
Let now ϕ ∈ D(Q), then
Z
Z
f u fε (ρ)) ϕ dx dt
f u ρ) ϕ dx dt =
div(e
fε0 (ρ) div(e
Q
Q
Z
f u |ρ|) ϕ dx dt
div(e
→
Q
But
and
f u ρ) ϕ → sgn(ρ) div(e
f u ρ) ϕ a.e.
fε0 (ρ) div(e
f u ρ) ϕ| ≤ sgn(ρ) div(e
f u ρ) ϕ
|fε0 (ρ) div(e
and we get the second result.
Proof of theorem 3.4. This is a consequence of lemma 3.6. Indeed when ρ ≥ 0 then |ρ| = ρ
and
f u |ρ|) = div(e
f u ρ) = sgn(ρ) div(e
f u ρ).
div(e
f u ρ) = 0 a.e. on the subset {(x, t) ∈ Q, ρ(x, t) = 0}.
If ρ = 0, then sgn(ρ) = 0, so div(e
Let now ρ ∈ H(u, Q), then ρ+ = 21 (|ρ| + ρ) ∈ H(u, Q), ρ− = 21 (|ρ| − ρ) ∈ H(u, Q), and
ρ = ρ+ − ρ− . But
{(x, t) ∈ Q, ρ(x, t) = 0} = (x, t) ∈ Q, ρ+ (x, t) = 0 ∩ (x, t) ∈ Q, ρ− (x, t) = 0 ,
so
f u ρ+ ) = 0
div(e
on
(x, t) ∈ Q, ρ+ (x, t) = 0 ,
8
and
f u ρ− ) = 0
div(e
and we get
on
f u ρ) = 0
div(e
(x, t) ∈ Q, ρ− (x, t) = 0 ,
on {(x, t) ∈ Q, ρ(x, t) = 0} ,
and the theorem is proved.
The proof of theorem 3.5 is similar to the proof given in [24].
4. Study of the least squares formulation . This section is devoted to the study of
equation (3.5). More precisely, an existence and uniqueness theorem for the solution of equation (3.5) is given. Then a maximum principle is deduced from the Stampacchia’s theorems.
With the notations and hypothesis of section 3 we have
T HEOREM 4.1. For a fixed u ∈ L∞ (Q)d with div u ∈ L∞ (Q), the problem (3.5):
Z
Q
f u c) · div(e
f u ϕ) dx dt −
div(e
Z
∂Q−
c · ϕ(u
e|n
e ) de
σ=
Z
Z
f u ϕ) dx dt −
f · div(e
∂Q−
Q
for all ϕ ∈ H(u, Q), has a unique solution. Moreover
|c|1,u =
2
f
u c)
div(e
2
L (Q)
−
Z
2
∂Q−
c (u
e|n
e ) de
σ
!1/2
cb · ϕ ( u
e|n
e ) de
σ
≤ kf kL2 (Q) .
This solution is the space-time least squares solution of (2.2).
Proof. This assertion is a consequence of the Curved Poincaré inequality (theorem 3.3) and
of the Lax-Milgram theorem (see also [4, 5]).
R EMARK 2. For the numerical solution of equation (3.5), a time marching approach can be
used to avoid the consideration of the all of Q (see e.g. [8, 15, 17]).
C OROLLARY 4.2. The solution c of equation (3.5) belongs to the space
X = L2 (Q) ∩ L2 (∂Q+ , ( u
e|n
e ) de
σ)
equipped with the norm |||c|||.
The following theorem is a maximum principle for the solution of problem (3.5).
T HEOREM 4.3. Assume that the domain Ω is bounded, that the function f = 0 in equation
(3.5) and that the function cb ∈ L∞ (∂Q− ). If div u = 0, the solution of equation (3.5)
satisfies
inf cb ≤ c ≤ sup cb .
Proof. The solution c verifies
Z
Z
f
f
div(e
u c) · div(e
u ϕ) dx dt −
Q
∂Q−
c · ϕ(u
e|n
e ) de
σ=−
9
Z
∂Q−
cb · ϕ ( u
e|n
e ) de
σ
for all ϕ ∈ H(u, Q). Let
M = sup cb
∂Q−
and put
ϕ = (c − M )+ , Q1 = (x, t) ∈ Q, c − M > 0 , Σ1 = ∂Q− ∩ Q1
Then, using the Stampacchia’s lemma,
Z
Q1
f u c) · div(e
f u (c − M )) dx dt −
div(e
Z
Σ1
c · (c − M ) ( u
e|n
e ) de
σ=
Z
−
cb · (c − M ) ( u
e|n
e ) de
σ.
Σ1
Since div u = 0, we get
Z
Q1
f u (c − M )))2 dx dt −
(div(e
Z
((c − M ))2 ( u
e|n
e ) de
σ=
Σ1
Z
−
(cb − M ) · (c − M ) ( u
e|n
e ) de
σ ≤ 0.
Σ1
Hence, using theorem 3.3, the set Q1 has a zero measure, so c ≤ M . We show in the same
way that c ≥ inf cb .
Fianlly, let us do the following
R EMARK 3. The reduction of problem (2.2-2.4) to an homogeneous Dirichlet problem on
∂Q− , i.e. assume that cb = 0 can be done as follows.
Let c be the solution of (2.2):
Z
Q
f u c) · div(e
f u ϕ) dx dt −
div(e
Z
∂Q−
c · ϕ(u
e|n
e ) de
σ=
Z
Z
f u ϕ) dx dt −
f · div(e
Q
∂Q−
for all ϕ ∈ H(u, Q), and let η the solution of
Z
Z
Z
f u η) · div(e
f u ϕ) dx dt −
η · ϕ(u
e|n
e ) de
σ=−
div(e
Q
∂Q−
∂Q−
cb · ϕ ( u
e|n
e ) de
σ,
cb · ϕ ( u
e|n
e ) de
σ
for all ϕ ∈ H(u, Q). Then ρ = c − η is the unique solution of
Z Z
f
f
f u η) · div(e
f u ψ) dx dt
div(e
u ρ) · div(e
u ψ) dx dt =
f − div(e
Q
(4.1)
Q
for all ψ ∈ H0 = H0 (u, Q, ∂Q−). Moreover the solution of problem (4.1) is equivalent to
the solution of (2.2).
Therefore, modifying the source term if necessary, it is sufficient to only deal with homogeneous Dirichlet boundary conditions on ∂Q− .
10
5. Comparison with renormalized solutions.. This section is devoted to the comparison between the least squares solution of equation (2.2-2.4) and the renormalized solution of
these equations in the sense of [18].
More precisely, let u ∈ L∞ (Q)d , with div u = 0, and ( u | n ) = 0 on ∂Ω. Let φ ∈ L∞ (Q)
be the space-time least square solution of
∂t φ + div (φ u) = 0
in Q,
with the boundary condition
φ = φb
on ∂Q− ,
and φb ∈ L∞ (∂Q− ).
f u φ) ∈ L2 (Q); as we have seen in remark 3, the space time least squares
Now let ψ = −div(e
∞
solution ρ ∈ L (Q) of
∂t ρ + div (ρ u) = ψ
ρ=0
in Q,
(5.1)
on ∂Q− ,
(5.2)
gives an equivalent solution to the previous problem.
D EFINITION 5.1. [18] For u ∈ L∞ (Q)d , div u ∈ L∞ (Q), cb ∈ L∞ (∂Q− ), and f ∈ L2 (Q),
the function c ∈ L∞ (Q) is a renormalized solution of
f u c) = f
div(e
with
c = cb
on ∂Q−
if for any β ∈ C 1 (R), β(0) = 0, β(c) is a weak solution of
and
f u β(c)) = β 0 (c)f,
div(e
β(c) = β(cb ) on ∂Q− .
where the equations are understood in distributions sense.
In this section we will show that ρ is a renormalized solution of equations (5.1)- (5.2).
Let us first prove the following result which is a sort of Meyers-Serrin theorem.
f u ρ) ∈
T HEOREM 5.2. Let u ∈ H01 (Q)d with div u = 0, and let ρ ∈ L∞ (Q) with div(e
2
L (Q), and ρ(x, 0) = 0. Then ρ ∈ H0 (u, Q).
Proof. Let > 0 be given and sufficiently small. Set
Q = Ω × (, T ) D = Ω × (0, ),
and define ρ = ρ|Q ∗ω , where ω is the usual mollifier. Then ρ ∈ D(Q), and ρ (x, 0) = 0.
Let us show that
→0
f
f u ρ )
u ρ) − div(e
−→ 0.
2
div(e
L (Q)
We have
f u ρ) − div(e
f u ρ ) = div(e
f u ρ) − div(e
f u ρ) ∗ ω
div(e
f u ρ) ∗ ω − div(e
f u ρ ∗ ω )
+ div(e
f u ρ ∗ ω ) − div(e
f u ρ ).
+ div(e
11
It is clear that
f
f u ρ) ∗ ω u ρ) − div(e
div(e
→0
L2 (Q)
−→ 0;
moreover, from [18, 9], we have
f
f u ρ ∗ ω )
u ρ) ∗ ω − div(e
div(e
→0
−→ 0.
L2 (Q)
Finally, let
f u ρ ∗ ω ) − div(e
f u ρ )
h = div(e
f u (ρ − ρ|Q ) ∗ ω ).
= div(e
If B is the ball of center (x, t) and radius in Rd+1 , then
Z
e (x − y, t − s) dy ds.
e(x, t) | ∇ω
ρ(y, s)1D (y, s) u
h (x, t) =
B
Since
Z
e (x − y, t − s) dy ds = 0,
ρ(y, s)1D (y, s) u
e(y, s) | ∇ω
Z
e (x − y, t − s) dy ds.
ρ(y, s)1D (y, s) (e
u(x, t) − u
e(y, s))/ | ∇ω
B
we get
h (x, t) =
B
Hence, using the Cauchy-Schwartz inequality we get
2
2
kh kL2 (Q) ≤ kρkL2 (Q) ·
Z Z
2
2 e
u(x, t) − u
e(y, s))/k ∇ω
1D (y, s) k(e
(x
−
y,
t
−
s)
dy ds dx dt,
Q
B
e is bounded, we get, as in [18, 9]
since ∇ω
2
2
2
kh kL2 (Q) ≤ M 2 |Ω| kρkL2 (Q) k∇ukL2 (Q)d2 .
Let um be the following regularization of the velocity field u given by the unique solution of
the following space-time stationary Stokes problem.
−
1 e m
∆u + um + ∇p = u in Q
m
f um = 0 in Q
div
um = 0 on ∂Q
with
2
e = ∆+ ∂ .
∆
∂t2
12
Then um ∈ H01 (Q) ∩ L∞ (Q), and um → u in L2 (Q) when m → ∞.
Let now η m be the renormalized solution of
∂t η m + div (η m um ) = ψ
in Q,
with the initial condition
η m (x, 0) = 0.
m η m ) ∈ L2 (Q). As η m ∈ L∞ (Q), we deduce from
f uf
Since ψ ∈ L2 (Q) we have div(
m
m
theorem 5.2 that η ∈ H0 (u , Q). Therefore η m is also the space-time least squares solution
m η m ) = ψ in H (um , Q), and we have
f uf
of div(
0
f f
= kψkL2 (Q)
div(um η m ) 2
L (Q)
kη m kL2 (Q) ≤ C kψkL2 (Q)
Hence, extracting a subsequence if necessary, one can assume that
ηm * η
in L2 (Q),
m ηm ) * κ
f uf
div(
in L2 (Q).
Moreover, since um → u in L2 (Q), we obtain
m η m ) → div(e
f uf
f u η) in H −1 (Q),
div(
f u η) ∈ L2 (Q).
and so κ = div(e
Let us now prove that
m η m ) → div(e
f uf
f u η)
div(
For this it is sufficient to show that
f f
div(um η m )
L2 (Q)
but
2
f
u η) 2
div(e
L (Q)
=
Z Q
Z
Q
=
Q
Q
=
f
u η)
→ div(e
L2 (Q)
,
m η m ) div(e
f u η) − div(
f uf
f u η) dx dt +
div(e
m η m )div(e
f uf
f u η) dx dt
div(
Z Z
in L2 (Q).
m η m ) div(e
f u η) − div(
f uf
f u η) dx dt +
div(e
f u η) dx dt
ψ div(e
Z m η m ) div(e
f u η) − div(
f uf
f u η) dx dt +
div(e
Q
Z
m η m )2 dx dt,
f uf
div(
lim
m→∞
Q
13
and so
f f
div(um η m )
f
→ div(e
u η)
L2 (Q)
L2 (Q)
.
f u η) = ψ, so
As a consequence of the above results, η is the least squares solution of div(e
η ∈ H0 (u, Q). Indeed, let w ∈ D(Q) with w = 0 on ∂Q− , then for all m, w ∈ H0 (um , Q),
and we have
Z
Z
m w) dx dt.
m η m ) · div(
m w) dx dt =
f uf
f uf
f uf
ψ · div(
div(
Q
Q
So, passing to the limit
Z
Z
f u η) · div(e
f u w) dx dt =
f u w) dx dt,
div(e
ψ · div(e
Q
Q
and this equation remains true for all w ∈ H0 (u, Q). Therefore η ∈ H0 (u, Q), and is the
f u η) = ψ, and so η = ρ ∈ L∞ (Q).
least squares solution of div(e
Now we can show that the sequence η m strongly converges to η. Since div um = div u = 0
and η m ∈ H0 (um , Q), with the same idea as in theorem 3.3, we have
Z
Z h
i
m η m ) · ξ + η m · div(
m ξ) dx dt =
f uf
f uf
ξη m ( u
e|n
e ) de
σ
div(
∂Q+
Q
m
for all regular enough function ξ. If we choose ξ = (T − t)η , then
Z
Z
m 2
m η m ) η m (T − t)dx dt.
f uf
div(
(η ) dx dt = 2
But
Z
Q
(5.3)
Q
Q
m η m ) η m (T − t)dx dt →
f uf
div(
Z
Q
f u η) η(T − t)dx dt =
div(e
Z
η 2 dx dt,
Q
when m → ∞, since ( u | n ) = 0 on ∂Ω; so η m → η when m → ∞.
Finally, let us show that η is a renormalized solution. Let β ∈ C 1 (R), β(0) = 0, then
β(η m ) → β(η). Moreover
m | ∇η
m β(η m )) = β 0 (η m ) · u
f uf
e m ,
f
div(
but
β 0 (η m ) → β 0 (η) in Lp (Q)
and
so
m | ∇η
e m
uf
→
e
u
e | ∇η
m β(η m )) → div(e
f uf
f u β(η))
div(
in L2 (Q),
in L1 (Q),
thus η is a renormalized solution. We have proved the following
T HEOREM 5.3. If u ∈ L∞ (Q)d , div u = 0, and cb ∈ L∞ (∂Q− ), the least squares solution
of
f u c) = 0 with
div(e
is a renormalized solution.
14
c = cb
on ∂Q−
(5.4)
6. Conclusions and remarks . We have shown in this paper that the conservation law
∂t c + div (c u) = f
in Q,
(6.1)
can be solved for irregular vector fields, using a very simple approach compared to the methods like e.g. [18, 3, 9].
Our method leads to some numerical schemes which are much simpler to use than the usual
one (like the stream-line diffusion method, the characteristics method, the discontinuous finite
element method with flux limiter, etc . . . ). Some numerical examples are presented in [15, 8,
17].
In [7], it is proved that the solution of equation 6.1 gives an isomorphism on L2 (Q) when
u ∈ C 1 (Ω) is independent of t. Our method still gives an isomorphism, but in some unusual
spaces. It allows to solve equations 6.1 with an irregular velocity field u. This situation where
not known until now.
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