Inflow-outflow problems for Euler equations

in a rectangular cylinder

FILIPPO GAZZOLA
Dipartimento di Scienze T.A. - via Cavour 84, 15100 Alessandria, Italy

PAOLO SECCHI
Dipartimento di Matematica - via Valotti 9, 25133 Brescia, Italy

Abstract
We prove that some inflow-outflow problems for the Euler equations in
a (nonsmooth) bounded cylinder admit a regular solution. The problems
considered are symmetric hyperbolic systems with partly characteristic and
partly noncharacteristic boundary; for such problems, no general theory is
available. Therefore, we introduce particular spaces of functions satisfying
suitable additional boundary conditions which allow to determine a regular
solution by means of a “reflection technique”.

1 Introduction
Let Ω be an open bounded cylinder in IR3 having a rectangular section: more
precisely, let Ω be the cartesian product of an open rectangle R with a bounded
interval (a, b) (Ω = R × (a, b)) and let Γ1 = R × {a, b} and Γ0 = ∂R × [a,
S b] so that
the piecewise smooth boundary ∂Ω may be characterized by ∂Ω = Γ1 Γ0 . In the
cylinder Ω we consider the following initial-boundary value problem for the Euler
equations for a barotropic inviscid compressible fluid



 ∂t ρ + ∇ · (ρv) = 0 in [0, T ] × Ω


ρ(∂t v + (v · ∇)v − f ) + ∇p = 0 in [0, T ] × Ω






on (0, T ) × Γ1

 M (ρ, v) = G
(1)


 v·ν =0 on (0, T ) × Γ0





 ρ(0, x) = ρ0 (x) in Ω



 v(0, x) = v (x)
0 in Ω ,

1
where ∂t = ∂/∂t, M is a matrix which depends on the particular inflow-outflow
problem considered, see (6), (9), and ν denotes the unit outward normal to ∂Ω,
when it exists; the density ρ = ρ(x, t), the velocity field v = v(t, x) = (v1 , v2 , v3 )
and the pressure p = p(t, x) are unknown functions of time t ∈ (0, T ) and space
variable x ∈ Ω. In (1) the density ρ is assumed to be positive for physical reasons;
moreover, ρ and p are related by the equation of state p = p(ρ) where p : IR+ → IR+
is a given smooth function such that p0 (s) > 0 for all s > 0. The external force
field f = f (t, x), the boundary data G and the initial data v0 and ρ0 are known
functions.
Usually, the Euler system is studied under the slip boundary condition v · ν =
0 on Γ0 , see [1, 2, 3, 9]. In this paper we study the existence and uniqueness of
a regular solution for some inflow-outflow problems for (1): we assume that the
fluid flows in on Γi1 = R × {a}, that it flows out on Γo1 = R × {b} and satisfies
the slip condition v · ν = 0 on Γ0 . On Γ1 = Γi1 ∪ Γo1 the flow is assumed to be
either supersonic or subsonic. As far as we are aware, inflow-outflow problems for
(1) have only been studied in [12] provided that there is no “discontinuity” in
the boundary conditions. The existence of a regular solution of (1) is not at all
obvious: first because the domain Ω is nonsmooth, second because the boundary
matrix does not have constant rank. Nevertheless, if suitable functional spaces
are considered, the presence of a nonsmooth boundary avoids in some sense the
problems arising from boundary points where the boundary matrix changes rank
and enables us to prove the existence of a regular solution. In next section we
write (1) as a symmetric hyperbolic system [4]; with our assumptions it becomes
partly characteristic and partly noncharacteristic: more precisely, the boundary
Γ1 is noncharacteristic while the boundary Γ0 is characteristic of rank two. To our
knowledge no general theory is available for these problems even if some partial
results may be found in [6, 7, 8, 11, 13].
Our method consists in introducing a class of functional spaces of functions
having some vanishing traces; these spaces allow, by a “reflection technique”, to
reduce the linearized problem associated to (1) to a noncharacteristic problem.
Then, standard existence results apply and we obtain a solution of the linearized
problem in Ω by a partition of unity. Finally, the solution of the nonlinear prob-
lem (1) is obtained by a fixed point argument. With this method, we prove the
existence and uniqueness of a regular local in time solution (ρ, v) of (1) for both
the supersonic and the subsonic cases: we assume that the fluid flows in Γi1 either
supersonic or subsonic and analogously on the outflow part Γo1 . Even if the under-
lying abstract framework is similar, these two cases turn out to be quite different:
the subsonic case is slightly more delicate because nonlinear boundary conditions
have to be considered.

2
2 Notations and results
For simplicity, we take

Ω = (0, 2) × (0, 1) × (0, 1) , (2)

we denote QT = (0, T ) × Ω, ΣT = [0, T ] × ∂Ω and we define the sets

Γi1 = {0} × [0, 1] × [0, 1] Γo1 = {2} × [0, 1] × [0, 1] Γ1 = Γi1 Γo1
S
S
Γ2 = [0, 2] × {0, 1} × [0, 1] Γ3 = [0, 2] × [0, 1] × {0, 1} Γ0 = Γ2 Γ3 ;
S S
then, the piecewise smooth boundary ∂Ω is given by ∂Ω = Γ1 Γ2 Γ3 .
Let H m (Ω) be the usual Sobolev space of order m (m ∈ IN) and let k · km
denote its norm: the L2 (Ω)-norm is simply denoted by k · k. We also introduce the
following spaces of functions having some (even or odd) traces vanishing on parts
of the boundary

He3 = {φ ∈ H 3 (Ω); φ = 0 on Γ2 , ∂3 φ = 0 on Γ3 , ∂22 φ = 0 on Γ2 }

Ho3 = {φ ∈ H 3 (Ω); φ = 0 on Γ3 , ∂2 φ = 0 on Γ2 , ∂32 φ = 0 on Γ3 }
H3 = {φ ∈ H 3 (Ω); ∂ν φ = 0 on Γ0 } ;

clearly, these are closed subspaces of H 3 (Ω) and contain H03 (Ω): here, ∂ν = ∂/∂ν,
∂i = ∂/∂xi , ∂i2 = ∂ 2 /∂x2i and the traces are well-defined even if the domain Ω is
nonsmooth, see [5]. Similarly, we define the spaces H3 (Γ1 ), Ho3 (Γ1 ) and He3 (Γ1 ).
For any vector function φ defined on (a subset of) Q̄T we denote by φi its i-th
component, i = 1, 2, 3. Let B be a Banach space and let T > 0: then C(0, T ; B) and
L∞ (0, T ; B) denote respectively the space of continuous and essentially bounded
functions defined on [0, T ] and taking values in B. Define the space

CT (H∗3 ) = C(0, T ; H3 ) × C(0, T ; H3 ) × C(0, T ; He3 ) × C(0, T ; Ho3 ) .

Consider now the spaces

3
\ 3
\
CT (H 3 ) = C k (0, T ; H 3−k (Ω)) L(H 3 ) = W k,∞ (0, T ; H 3−k (Ω))
k=0 k=0

with the norm given by (∂tk = ∂ k /∂tk ):

3
X
|||u|||3,T = sup |||u(t)|||3 , |||u(t)|||23 = k∂tk u(t)k23−k .
[0,T ] k=0

We seek solutions (ρ, v) of (1) in the closed subspace of [CT (H 3 )]4 defined by

KT = CT (H∗3 ) ∩ [CT (H 3 )]4 .

3
Finally, consider the space

H∗3 (QT ) = {φ ∈ [H 3 (QT )]3 ; φ(t) ∈ H3 × He3 × Ho3 for a.e. t ∈ [0, T ]}

normed by
Z T
[φ]23,T = |||φ(t)|||23 dt ;
0

similarly, we define H∗3 (ΣT ) and the norm [·]3,ΣT .

p
By introducing the sound velocity c(ρ) = p0 (ρ), the equations in problem
(1) become

 1 (∂ ρ + v · ∇ρ) + ∇ · v = 0 in [0, T ] × Ω
ρ t
(3)
 ρ (∂ v + (v · ∇)v − f ) + ∇ρ = 0 in [0, T ] × Ω ,
c2 (ρ) t

to which we associate the initial conditions


 ρ(0, x) = ρ (x) in Ω
0
(4)
 v(0, x) = v (x) in Ω ,
0

for ρ0 and v0 in a suitable functional space.

We may write (3) as a quasilinear symmetric hyperbolic system, that is in
the form
X3
A0 (u)∂t u + Aj (u)∂j u = F (u) (5)
j=1

where u = (ρ, v), the Aj (j = 0, ..., 3) are symmetric and A0 is positive definite.
The boundary matrix is
 
1 T
v · ν ν
Aν =  ρ  ;
ρ
ν c2 (ρ) v · νI3

therefore, the matrix A−1

0 Aν has eigenvalues

v·ν , v·ν , v · ν + c(ρ) , v · ν − c(ρ) .

Recall that a part Γ of the boundary is said noncharacteristic if detAν 6= 0 on Γ;

moreover, if we are given homogeneous boundary conditions M u = 0, then kerM
is said to be maximally non-negative for Aν if

(Aν u, u) ≥ 0 on ∂Ω × (0, T ) ∀u ∈ kerM ,

and kerM is not properly contained in any other subspace having this property;
the maximal non-negativity is useful because it allows to neglect some boundary

4
integrals when one looks for a priori estimates. This condition corresponds to
requiring that the number of boundary conditions equals the number of negative
eigenvalues of Aν . Let us first consider the supersonic case: on the inflow part of
the boundary Γi1 (which is noncharacteristic) the four eigenvalues of A−1 0 Aν are
negative and the whole state of the fluid must be prescribed while on the outflow
part Γo1 the boundary matrix A−1 0 Aν is positive definite and no condition has to
be assigned. For the subsonic case we also have that Γ1 is noncharacteristic: on Γi1
three eigenvalues of A−10 Aν are negative and three conditions must be prescribed
while on Γo1 the boundary matrix A−1 0 Aν only has a negative eigenvalue and one
condition has to be assigned. Finally, the impermeable part Γ0 is characteristic
of rank two with A−1 0 Aν having one negative eigenvalue and only the condition
v · ν = 0 is given.
So, if we fix T0 > 0 and we take into account the shape of Ω in (2), for the
supersonic case we require that



 (ρ, v) = (r, g) on (0, T0 ) × Γi1

v2 = 0 on (0, T0 ) × Γ2 (6)


on (0, T0 ) × Γ3

 v =0
3

while no conditions should be imposed on Γo1 ; however, to ensure that the fluid
flows out supersonic at least in a small interval of time, we require that the initial
flow satisfies such condition: we assume that

ρ0 > 0 in Ω̄
(v0 )1 > c(ρ0 ) on Γ1
(7)
r>0 on [0, T0 ] × Γi1
g1 > c(r) on [0, T0 ] × Γi1 .

Finally, to find regular solutions one needs to impose some necessary compatibility
conditions between the boundary data and the initial values; denote by ∂tk ρ0 and
∂tk v0 the functions obtained by formally taking k−1 time derivatives of (3), solving
for ∂tk ρ and ∂tk v and evaluating at time t = 0. Then the compatibility conditions
in the supersonic case read

∂tk ρ0 = ∂tk r(0), ∂tk v0 = ∂tk g(0) on Γi1 ∂tk v0 · ν = 0 on Γ0 k = 0, 1, 2 ;

Pm (8)
for u0 = (ρ0 , v0 ) we set |||u0 |||2m = k=1 (||∂tk ρ0 ||2m−k + ||∂tk v0 ||2m−k ).
Our main result in the supersonic case states
Theorem 1 Let Ω ⊂ IR3 be as in (2), let T0 > 0 and assume that:
(i) the boundary data r, g ∈ H 3 ((0, T0 ) × Γi1 ) satisfy
(r, g) ∈ H3 (Γi1 ) × H3 (Γi1 ) × He3 (Γi1 ) × Ho3 (Γi1 ) for a.e. t ∈ [0, T0 ] ;

5
(ii) the initial data satisfy (ρ0 , v0 ) ∈ H3 × H3 × He3 × Ho3 ;
(iii) the external force satisfies f ∈ H∗3 (QT0 );
(iv) (7) and (8) hold.
Then, there exists T > 0 such that (3)-(4)-(6) admits a unique solution (ρ, v) ∈ KT
satisfying v1 > c(ρ) on [0, T ] × Γo1 .

Remark. Some of the compatibility conditions (8) are “hidden” in the assumptions
(i) - (iii); further regularity may be obtained by using Sobolev spaces H m of higher
order (m ≥ 4): in such case, one must obviously strengthen (8) with supplementary
conditions. 2
In the subsonic case we assign the tangential velocity of the fluid on the inflow
part and the normal velocity on the outflow part; we require that



 v2 = g2 , v3 = g3 , v1 + c(ρ) = ψ on (0, T0 ) × Γi1


on (0, T0 ) × Γo1

 v =g
1 1
(9)


 v2 = 0 on (0, T0 ) × Γ2


on (0, T ) × Γ

 v =0
3 0 3

and we assume that

ρ0 > 0 in Ω̄
−c(ρ0 ) < −ψ(0) + c(ρ0 ) < 0 on Γi1
(10)
0 < g1 (0) < c(ρ0 ) on Γo1
g1 > 0 on [0, T0 ] × Γo1 .

In order to ensure that the fluid flows in subsonic on Γi1 one should require that
−c(ρ) < −ψ + c(ρ) < 0 on [0, T0 ] × Γi1 while to ensure that the fluid flows out
subsonic on Γo1 one should require that g1 < c(ρ) on [0, T0 ]×Γo1 but these conditions
depend on the solution; nevertheless, we will prove in Theorem 2 below that they
hold in some interval of time [0, T ], T > 0. We also refer to [12] for the derivation
of (9) and for other possible boundary conditions.
The compatibility conditions between the initial and boundary data in the
subsonic case read

(∂tk v0 )2 = ∂tk g2 (0), (∂tk v0 )3 = ∂tk g3 (0), (∂tk v0 )1 + ∂tk c0 = ∂tk ψ(0) on Γi1
(∂tk v0 )1 = ∂tk g1 (0) on Γo1 ∂tk v0 · ν = 0 on Γ0 k = 0, 1, 2 ,
(11)
where ∂tk c0 denotes the k th time derivative at t = 0 of c(ρ). Then, for the subsonic
case we prove the following

6
Theorem 2 Let Ω ⊂ IR3 be as in (2), let T0 > 0 and assume that:
(i) ψ, g2 , g3 ∈ H 3 ((0, T0 ) × Γi1 ), g1 ∈ H 3 ((0, T0 ) × Γo1 ) satisfy

g2 ∈ He3 (Γi1 ), g3 ∈ Ho3 (Γi1 ), ψ ∈ H3 (Γi1 ), g1 ∈ H3 (Γo1 ) for a.e. t ∈ [0, T0 ] ;

(ii) the initial data satisfy (ρ0 , v0 ) ∈ H3 × H3 × He3 × Ho3 ;

(iii) the external force satisfies f ∈ H∗3 (QT0 );
(iv) (10) and (11) hold;
(v) the function s 7→ c(s) is concave.
Then, there exists T > 0 such that (3)-(4)-(9) admits a solution (ρ, v) ∈ KT
satisfying
0 < v1 < c(ρ) on [0, T ] × Γ1 .
Moreover, if the function s 7→ c(s) satisfies c0 (s) > 0 for all s > 0, then there exists
δ > 0 such that if max i |v1 | < δ then (ρ, v) is the unique solution of (3)-(4)-(9).
[0,T ]×Γ1

Remark. If p(ρ) = Rργ with γ > 1 then c(ρ) is concave for γ ≤ 3; for common
isentropic gases γ ∈ (1, 53 ) and therefore (v) is physically meaningful. Moreover, if
γ > 1 then c0 (s) > 0 for all s > 0. 2
Remark. Uniqueness in Theorem 2 is certainly ensured in some time interval [0, T ]
(T > 0) if ψ(0) − c(ρ0 ) is sufficiently small on Γi1 (recall that by (10) we have
ψ(0) − c(ρ0 ) > 0). 2

3 The linearized problems in space sectors

In order to simplify notations, in the sequel we denote by IR+ both the open
interval (0, +∞) and its closure, depending on the context.

3.1 The supersonic case

In this section we first consider the case where

Ω = IR3++ := IR+ × IR+ × IR , Γ1 := {0} × IR+ × IR , Γ2 := IR+ × {0} × IR ,

and we study the linearized problem


 1 (∂ ρ + w · ∇ρ) + ∇ · v = h in [0, T ] × Ω
π t
(12)
 π (∂ v + (w · ∇)v) + ∇ρ = k in [0, T ] × Ω
c2 (π) t

together with initial conditions (4) and boundary conditions


 (ρ, v) = (r, g) on (0, T ) × Γ1
(13)
 v =0
2 on (0, T ) × Γ2 .

7
In this case Γ3 = ∅ and the conditions on Γ3 in the definitions of the spaces
H3 disappear; in particular, we have Ho3 = H3 . Consider the set ST of functions
(π, w) ∈ KT such that
3 1
sup ρ0 ≥ π ≥ inf ρ0 > 0 in Q̄T ,
2 Ω̄ 2 Ω̄
1

−w1 + c(π) ≤ sup − (v0 )1 + c(ρ0 ) < 0 on (0, T ) × Γ1 ,
2 Γ1
∂tk π(0) = ∂tk ρ0 , ∂tk w(0) = ∂tk v0 k = 0, 1, 2,
where ∂tk ρ0 , ∂tk v0 are the same of (8), obtained from (3), (4). From the previous
assumption  
1 π
γ(π) := min inf , inf 2 >0. (14)
Q̄T π Q̄T c (π)
Let us point out that the definition of ST gives some restrictions on the initial
data ρ0 , v0 since some inequalities of the kind of (7) are required.
The following result holds:
Proposition 1 Let Ω = IR3++ , let T > 0 and assume that:
(i) the boundary data r, g ∈ H 3 ((0, T ) × Γ1 ) satisfy
(r, g) ∈ H3 (Γ1 ) × H3 (Γ1 ) × He3 (Γ1 ) × Ho3 (Γ1 ) for a.e. t ∈ [0, T ] ;
(ii) the initial data satisfy (ρ0 , v0 ) ∈ H3 × H3 × He3 × Ho3 ;
(iii) the right hand side of (12) satisfies (h, k) ∈ H 3 (QT ) × H∗3 (QT ), h(t) ∈ H3
for a.e. t ∈ [0, T ];
(iv) the data satisfy (7) and the compatibility conditions ∂tk ρ(0) = ∂tk r(0), ∂tk v(0) =
∂tk g(0) on Γ1 , ∂tk v(0) · ν = 0 on Γ0 , k = 0, 1, 2, where ∂tk ρ(0), ∂tk v(0) are the k−th
time derivatives of ρ, v at time t = 0 obtained from (12), (4).
Then, for all (π, w) ∈ ST there exists a unique (ρ, v) ∈ KT solving (12)-(4)-
(13). Moreover, there exist two constants C1 , C2 > 0 (C1 depends increasingly on
|||(π, w)|||2,T while C2 depends increasingly on |||(π, w)|||3,T ) such that
n
γ|||(ρ, v)|||23,T ≤ C1 (1 + |||(π, w)|||2ε 2
3,T )|||(ρ0 , v0 )|||3 +
o (15)
+C1 [(r, g)]23,ΣT + C2 [(h, k)]23,T e(C1 +C2 )T ,

where γ = γ(π) is defined in (14) and ε ∈ ( 12 , 1).

Proof. Let Ω0 = IR+ × IR × IR and Γ0 = ∂Ω0 . For all x = (x1 , x2 , x3 ) ∈ IR3 we

denote x̄ = (x1 , −x2 , x3 ) and for all continuous function φ defined on Ω̄ we define
the functions φ̃ and φ̂ on Ω0 by
 
 φ(x) if x ≥ 0  φ(x) if x2 ≥ 0
2
φ̃(x) = φ̂(x) =
 φ(x̄) if x < 0  −φ(x̄) if x < 0 ;
2 2

8
this definition implies that the functions φ̃ and φ̂ are, respectively, even and odd
with respect to x2 : to be precise, φ̂ is odd only if φ(x1 , 0, x3 ) = 0 for all x1 , x3 .
Consider now the auxiliary problem




1
π̄ (∂t ρ + w̄ · ∇ρ) + ∇ · v = h̄ in [0, T ] × Ω0


 c2π̄(π̄) (∂t v + (w̄ · ∇)v) + ∇ρ = k̄ in [0, T ] × Ω0




ρ = r̄ v = ḡ on (0, T ) × Γ0 (16)


in Ω0

 ρ(0, x) = ρ̄0 (x)




in Ω0 ,

 v(0, x) = v̄0 (x)

where we have set

π̄ = π̃ ḡ = (ḡ1 , ḡ2 , ḡ3 ) = (g̃1 , ĝ2 , g̃3 ) r̄ = r̃ h̄ = h̃

k̄ = (k̄1 , k̄2 , k̄3 ) = (k̃1 , k̂2 , k̃3 ) w̄ = (w̄1 , w̄2 , w̄3 ) = (w̃1 , ŵ2 , w̃3 )
ρ̄0 = ρ̃0 v̄0 = ((v̄0 )1 , (v̄0 )2 , (v̄0 )3 ) = ((ṽ0 )1 , (v̂0 )2 , (ṽ0 )3 ) :

for simplicity, we have omitted to underline the dependence on t.

Since (π, w) ∈ KT , it is not difficult to verify that the “reflected” functions
π̄, w̄i (i = 1, 2, 3) belong to CT (H 3 (Ω0 )); moreover, (14) holds for π̄ as γ(π̄) = γ(π).
Note also that since (π, w) ∈ ST we have
 
inf 0 w1 − c(π) > 0 ,
(0,T )×Γ

so that Γ0 is noncharacteristic. Moreover we observe that (ρ̄0 , v̄0 ) ∈ H 3 (Ω0 ), (h̄, k̄) ∈
H 3 ((0, T ) × Ω0 ) and that (r̄, ḡ) ∈ H 3 ((0, T ) × Γ0 ). Finally we observe that the
compatibility conditions of order 2 hold on Γ0 . We apply Theorem A.1 in [12]
(also valid for halfspaces) in the case m = s = 3 and find a unique solution
(ρ, v) ∈ [CT (H 3 (Ω0 ))]4 of (16); furthermore, from the estimates (A.5) and (A.7) in
[12] and by reasoning as in the proof of Lemma 3.4 in [12] we arrive at

γ|||(ρ, v)(t)|||23 ≤ C1 (1 + |||(π, w)|||2ε 2

3,T )|||(ρ0 , v0 )|||3 +

+C1 [(r, g)]23,ΣT + C2 [(h, k)]23,T + (C1 + C2 )[(ρ, v)]23,t :

then, (15) follows by the Gronwall Lemma.

We claim that ρ, v1 and v3 are even with respect to x2 while v2 is odd with
respect to x2 . To this end, define the functions

ρ̄(t, x) = ρ(t, x̄) v̄1,3 (t, x) = v1,3 (t, x̄) v̄2 (t, x) = −v2 (t, x̄)

for x ∈ Ω0 ; it is not difficult to verify that the couple (ρ̄, v̄) satisfies (16) as well:
then, by uniqueness of the solution of (16), we infer that ρ̄ ≡ ρ and v̄ ≡ v in Ω0 ,

9
which proves the claim. Therefore, for all t ∈ [0, T ] the functions ρ, v1 , v3 belong
to the space Ho3 while v2 ∈ He3 ; in particular, we have v2 = 0 on Γ2 × (0, T ): this
proves that (ρ, v) ∈ KT solves (12)-(4)-(13). 2
Next we consider the case where

Ω = IR3+++ := [IR+ ]3 , Γ1 := {0} × IR+ × IR+ ,

Γ2 := IR+ × {0} × IR+ , Γ3 := IR+ × IR+ × {0} ,

and we study (12)-(4) together with the boundary conditions




 (ρ, v) = (r, g) on (0, T ) × Γ1

v2 = 0 on (0, T ) × Γ2 (17)


on (0, T ) × Γ3 :

 v =0
3

by using the same “reflection technique” as in the proof of Proposition 1 we prove

existence and uniqueness of a solution:

Proposition 2 Let Ω = IR3+++ , let T > 0 and assume that (i)-(iv) of Proposition
1 hold. Then, for all (π, w) ∈ ST there exists a unique (ρ, v) ∈ KT solving (12)-(4)-
(17). Moreover, there exist two constants C1 , C2 > 0 (as in Proposition 1) such
that (15) holds.
0
Proof. LetS Ω0 = IR+ × IR+ × IR, Γ01 = {0} × IR+ × IR and Γ02 = IR+ × {0} × IR so
that Γ1 Γ2 = ∂Ω . For all x = (x1 , x2 , x3 ) ∈ IR3 we define x̄ = (x1 , x2 , −x3 ) and
0 0

for all function φ defined on Ω̄ we define the functions φ̃ and φ̂ on Ω0 by

 
 φ(x) if x ≥ 0  φ(x) if x3 ≥ 0
3
φ̃(x) = φ̂(x) =
 φ(x̄) if x < 0  −φ(x̄) if x < 0 ;
3 3

then, the functions φ̃ and φ̂ are, respectively, even and odd with respect to x3 .
Consider now the auxiliary problem

1
 π̄ (∂t ρ + w̄ · ∇ρ) + ∇ · v = h̄

 in [0, T ] × Ω0


π̄
c2 (π̄) (∂t v + (w̄ · ∇)v) + ∇ρ = k̄ in [0, T ] × Ω0






on (0, T ) × Γ01

 ρ = r̄ v = ḡ
(18)
 v =0

 2 on (0, T ) × Γ02


in Ω0



 ρ(0, x) = ρ̄0 (x)


in Ω0 ,

 v(0, x) = v̄ (x)
0

10
where we have set

π̄ = π̃ ḡ = (ḡ1 , ḡ2 , ḡ3 ) = (g̃1 , g̃2 , ĝ3 ) r̄ = r̃ h̄ = h̃

k̄ = (k̄1 , k̄2 , k̄3 ) = (k̃1 , k̃2 , k̂3 ) w̄ = (w̄1 , w̄2 , w̄3 ) = (w̃1 , w̃2 , ŵ3 )
ρ̄0 = ρ̃0 v̄0 = ((v̄0 )1 , (v̄0 )2 , (v̄0 )3 ) = ((ṽ0 )1 , (ṽ0 )2 , (v̂0 )3 ) .

Again, it is not difficult to verify that the reflected functions (π̄, w̄) belong to
CT (H∗3 (Ω0 )) and the properties (i) − (iv) of Proposition 1 for the reflected data;
by Proposition 1, problem (18) admits a unique solution (ρ, v) ∈ [CT (H 3 (Ω0 ))]4
which satisfies the estimate (15). By proceeding as in the proof of Proposition 1 we
obtain that ρ, v1 and v2 are even with respect to x3 while v3 is odd with respect
to x3 . Therefore, (ρ, v) ∈ KT : in particular, we have v3 = 0 on (0, T ) × Γ3 and
(ρ, v) solves (12)-(4)-(17). 2

3.2 The subsonic case

We repeat the same arguments of the previous section and we maintain the same
notations. Again, we first consider the case where Ω = IR3++ and we study the
linearized problem (12) together with the initial conditions (4) (ρ0 and v0 satisfy
some of the inequalities in (10) due to the definition of ST below) and the (linear)
boundary conditions

 v = g , v = g , v + ρ c(π) = ψ on (0, T ) × Γ1
2 2 3 3 1 π
(19)
 v =0
2 on (0, T ) × Γ , 2

where π is a given function; for these linearized boundary conditions the compat-
ibility conditions read

∂tk v2 (0) = ∂tk g2 (0), ∂tk v3 (0) = ∂tk g3 (0), ∂tk v1 (0) + ∂tk (ρ c(π) k
π )(0) = ∂t ψ(0) on Γ1

∂tk v(0) · ν = 0 on Γ0 k = 0, 1, 2 ,
(20)
where ∂tk ρ(0), ∂tk v(0) are the k-th time derivatives of ρ, v at time t = 0, obtained
from (12), (4). Consider now the set ST of functions (π, w) ∈ KT which satisfy
3 1
sup ρ0 ≥ π ≥ inf ρ0 > 0 in Q̄T
2 Ω̄ 2 Ω̄
1  
−w1 + c(π) ≥ inf − (v0 )1 + c(ρ0 ) > 0 on (0, T ) × Γ1
2 Γ1
1
w1 ≥ inf (v0 )1 > 0 on (0, T ) × Γ1
2 Γ1
∂tk π(0) = ∂tk ρ0 , ∂tk w(0) = ∂tk v0 k = 0, 1, 2.
The following result holds:

11
Proposition 3 Let Ω = IR3++ , let T > 0 and assume that:
(i) ψ, g2 , g3 ∈ H 3 ((0, T ) × Γ1 ) satisfy
g2 ∈ He3 (Γ1 ) , g3 ∈ Ho3 (Γ1 ) , ψ ∈ H3 (Γ1 ) for a.e. t ∈ [0, T ] ;
(ii) the initial data satisfy (ρ0 , v0 ) ∈ H3 × H3 × He3 × Ho3 ;
(iii) the right hand side of (12) satisfies (h, k) ∈ H 3 (QT ) × H∗3 (QT ), h(t) ∈ H3
for a.e. t ∈ [0, T ];
(iv) the data satisfy (10)1 , (10)2 and the compatibility conditions (20) of order 2.
Then, for all (π, w) ∈ ST there exists a unique (ρ, v) ∈ KT solving (12)-(4)-(19).
Moreover, there exist three constants C1 , C2 , CM > 0 (C1 depends increasingly on
|||(π, w)|||2,T , C2 depends increasingly on |||(π, w)|||3,T , CM depends increasingly
on [(ψ, g2 , g3 )]3,ΣT ) such that

|||(ρ, v)(t)|||23 + [(ρ, v)]23,Σt ≤ C1 (1 + |||(π, w)|||2ε 2

3,T )|||(ρ0 , v0 )|||3 +
(21)
+C1 [(ψ, g2 , g3 )]23,Σt + C2 [(h, k)]23,t + (C1 CM + C2 )[(ρ, v)]23,t

for all t ∈ [0, T ]; here, ε ∈ ( 21 , 1).

Proof. Let Ω0 = IR+ × IR × IR, Γ0 = ∂Ω0 and consider the auxiliary problem




1
π̄ (∂t ρ + w̄ · ∇ρ) + ∇ · v = h̄ in [0, T ] × Ω0


π̄
(∂ v + (w̄ · ∇)v) + ∇ρ = k̄ in [0, T ] × Ω0


 c2 (π̄) t


v2 = ḡ2 , v3 = ḡ3 , v1 + ρ c(π̄)π̄ = ψ̄ on (0, T ) × Γ0 (22)


in Ω0



 ρ(0, x) = ρ̄0 (x)


in Ω0 ,

 v(0, x) = v̄ (x)
0

where we have set

π̄ = π̃ (ḡ2 , ḡ3 ) = (ĝ2 , g̃3 ) ψ̄ = ψ̃ h̄ = h̃

k̄ = (k̄1 , k̄2 , k̄3 ) = (k̃1 , k̂2 , k̃3 ) w̄ = (w̄1 , w̄2 , w̄3 ) = (w̃1 , ŵ2 , w̃3 )
ρ̄0 = ρ̃0 v̄0 = ((v̄0 )1 , (v̄0 )2 , (v̄0 )3 ) = ((ṽ0 )1 , (v̂0 )2 , (ṽ0 )3 ) .

Note that Γ0 is noncharacteristic and by Theorem A.1 in [12] we find a unique

solution (ρ, v) ∈ [CT (H 3 (Ω0 ))]4 of (22); moreover, we get (21) by (A.5) and (A.7)
in [12] and by arguing as in the proof of Lemma 3.4 in the same paper.
Since also ρ(t, x̄), v1,3 (t, x̄), −v2 (t, x̄) solves (22), by uniqueness of the solution
of (22) we infer that ρ, v1 and v3 are even with respect to x2 while v2 is odd with
respect to x2 : then (ρ, v) ∈ KT solves (12)-(4)-(19). 2
Remark. In (21) we do not highlight the dependence of the constants C1 , C2 , CM
on γ (defined in (14)) because γ is uniformly bounded from below and above when
(π, w) vary in ST . 2

12
 Next, consider the case where Ω = IR3+++ and the problem (12)-(4) together
with the boundary conditions

 v = g2 , v3 = g3 , v1 + ρ c(π)π =ψ on (0, T ) × Γ1
 2


v2 = 0 on (0, T ) × Γ2 (23)


on (0, T ) × Γ3 ;

 v =0
3

then, we obtain the following:

Proposition 4 Let Ω = IR3+++ , let T > 0 and assume that (i)-(iv) of Proposition
3 hold.
Then, for all (π, w) ∈ ST there exists a unique (ρ, v) ∈ KT solving (12)-(4)-(23).
Moreover, there exist three constants C1 , C2 , CM > 0 (as in Proposition 3) such
that (21) holds for all t ∈ [0, T ].

Proof. It follows by the same arguments used in the proof of Proposition 2. 2

Finally, we also need a similar result for the subsonic outflow problem: here
we take Ω = (−∞, 2) × IR+ × IR+ instead of IR3+++ so that we do not have
to change sign to the functions defined on ∂Ω. Let Γ1 = {2} × IR+ × IR+ , Γ2 =
(−∞, 2]×{0}×IR+ , Γ3 = (−∞, 2]×IR+ ×{0} and consider the boundary conditions

 v = g1 on (0, T ) × Γ1
 1


v2 = 0 on (0, T ) × Γ2 (24)


on (0, T ) × Γ3 .

 v =0
3

Consider the same set ST as for Proposition 3; then, we obtain the following:

Proposition 5 Let Ω = (−∞, 2) × IR+ × IR+ , let T > 0 and assume that:
(i) g1 ∈ H 3 ((0, T ) × Γ1 ) satisfies g1 ∈ H3 (Γ1 ) for a.e. t ∈ [0, T ];
(ii) the initial data satisfy (ρ0 , v0 ) ∈ H3 × H3 × He3 × Ho3 ;
(iii) the right hand side of (12) satisfies (h, k) ∈ H 3 (QT ) × H∗3 (QT ), h(t) ∈ H3
for a.e. t ∈ [0, T ];
(iv) the data satisfy (10)1 , (10)3 , (10)4 and the compatibility conditions of order
2.
Then, for all (π, w) ∈ ST there exists a unique (ρ, v) ∈ KT solving (12)-(4)-(24).
Moreover, there exist three constants C1 , C2 > 0 (as in Proposition 3) and CM > 0
(depending increasingly on [g1 ]3,ΣT ) such that (21) holds (with [g1 ]3,Σt instead of
[(ψ, g2 , g3 )]3,Σt ) for all t ∈ [0, T ].

Proof. It can be obtained by “double reflection” as for Propositions 1 and 2. 2

13
4 Proof of Theorem 1
Let Ω be the bounded cylinder defined in (2), let ST be the subset of KT introduced
in Section 3.1; we prove existence and uniqueness for the linearized problem in
(0, T ) × Ω:

 1 (∂ ρ + w · ∇ρ) + ∇ · v = 0 in [0, T ] × Ω
π t
(25)
 π (∂ v + (w · ∇)v − f ) + ∇ρ = 0 in [0, T ] × Ω
c2 (π) t

together with initial conditions (4) and boundary conditions (6).

Proposition 6 Let Ω ⊂ IR3 be as in (2), let T > 0 and assume that (i)-(iv) of
Theorem 1 hold (with T0 replaced by T ). Then, for all (π, w) ∈ ST there exists
a unique (ρ, v) ∈ KT solving (25)-(4)-(6). Moreover, there exist two constants
C1 , C2 > 0 (C1 depends increasingly on |||(π, w)|||2,T while C2 depends increasingly
on |||(π, w)|||3,T ) such that

γ|||(ρ, v)|||23,T ≤
n o
≤ C1 (1 + |||(π, w)|||2ε
3,T )|||(ρ 0 , v 0 )|||2
3 + C 1 [(r, g)]2
3,ΣT + C2 [f ]2
3,T e
(C1 +C2 )T
,
(26)
where γ = γ(π) is defined in (14) and ε ∈ ( 12 , 1).

Proof. We first prove the result in the unbounded cylinder

Ω0 = (0, +∞) × (0, 1) × (0, 1)

and in the time interval [0, T ]. We localize (12) in Ω0 by using a partition of unity
for Ω̄0 so that the problem reduces to four problems considered in the previous
section: cover Ω̄0 by a family of 4 open sets {Ui } (i = 1, ..., 4) so that each one
of them contains one (and only one) of the infinite edges of Ω0 , namely the lines
{x2 = x3 = 0}, {x2 = 1, x3 = 0}, {x2 = 0, x3 = 1}, {x2 = x3 = 1}; moreover,
each Ui should not intersect the two faces of Ω0 which are not adjacent to the edge
contained in Ui . Take a partition of unity {χi }4i=1 subordinate to the covering
P4
{Ui } such that i=1 χi = 1, χi ≥ 0, χi ∈ H3 . We multiply (25) by χi , i = 1, . . . , 4.
After a suitable change of variables, we obtain the following equations in IR3+++
for (ρi , vi ) = (χi ρ, χi v):

 1 (∂ ρ + w · ∇ρ ) + ∇ · v = 1 w · ∇χ ρ + v · ∇χ
π t i i i π i i
(27)
π π

c2 (π) (∂ v + (w · ∇)v − χ f ) + ∇ρ =
t i i i i (w · ∇χ )v + ρ∇χ .
c2 (π) i i

14
In view of a fixed point π = ρ, w = v, instead of (27) we consider the problems




1
π (∂t ρi + w · ∇ρi ) + ∇ · vi = 2w · ∇χi in [0, T ] × IR3+++


π
c2 (π) (∂t vi + (w · ∇)vi − χi f ) + ∇ρi =






= c2π(π) (w · ∇χi )w + π∇χi in [0, T ] × IR3+++





(ρi , vi ) = (χi r, χi g) on [0, T ] × Γ1 (28)


vi · ν = 0 on [0, T ] × Γ0






in IR3+++



 ρi (0, x) = χi (x)ρ0 (x)


in IR3+++ .

 v (0, x) = χ (x)v (x)
i i 0

We verify that the data of problem (28) satisfy (i) − (iv) of Proposition 2; in
particular (iii) follows from (π, w) ∈ CT (H∗3 ) and χi ∈ H3 , the compatibility
conditions (iv) follow from ∂tk π(0) = ∂tk ρ0 , ∂tk w(0) = ∂tk v0 , k = 0, 1, 2, and (8). By
Proposition 2, (28) admits a unique solution (ρi , vi ) ∈ KT . By adding together the
functions (ρi , vi ) we obtain a solution (ρ, v) in (0, T ) × Ω0 of (25),(4) and the first
boundary condition on (0, T ) × Γi1 of (6). As regards the boundary conditions on
Γ2 and Γ3 , if vi · ν 6≡ 0 on P = {x2 = 1} ∪ {x3 = 1} for some i, after adding
the solutions (ρi , vi ) we could obtain v2 6= 0 on Γ2 or v3 6= 0 on Γ3 . To overcome
this point, we proceed in two steps. First, since each (ρi , vi ) has initially a support
which does not intersect P, by using the finite speed of propagation we show that
(ρi , vi ) vanishes on P for each t ∈ (0, T 0 ), for P
a sufficiently small T 0 > 0. It follows
that (after the inverse change of variables) i vi · ν = 0 on Γ0 . Thus we have
found a unique solution (ρ, v) of (25),(4),(6) defined on [0, T 0 ] × Ω0 . We verify that
T 0 depends only on ||w||L∞ and on the extension of each support, namely on the
functions χi . We take t = T 0 as a new initial time, decompose the data by means
of the χi ’s and by the same arguments as above find a solution defined on [T 0 , 2T 0 ].
We proceed by this continuation argument up to when the solution is extended to
the whole interval [0, T ]. Moreover, since all the (ρi , vi ) satisfy (15), (ρ, v) satisfies
(26). 2
For all given r, g, ρ0 , v0 , f satisfying the assumptions of Theorem 1 is defined
a map Λ such that Λ(π, w) = (ρ, v); we achieve the proof of Theorem 1 by showing
that Λ admits a fixed point:

Lemma 1 Assume that r, g, ρ0 , v0 , f satisfy the assumptions of Theorem 1; then,

for sufficiently small T there exists a compact subset S of C(0, T, L2 (Ω)) such that
the map Λ defines a contraction in S.

Proof. For the moment take T ∈ (0, T0 ): if needed, later on we will take a smaller
value of T . Let u0 = (π, w) and let u = (ρ, v) = Λu0 ; as for (5), we may write (25)

15
as a linear symmetric hyperbolic system, that is in the form
3
X
L(u0 )u = A0 (u0 )∂t u + Aj (u0 )∂j u = F (u0 ) ,
j=1

together with the initial and boundary conditions (4) and (6) which we write as

u(0, x) = u0 (x) in Ω
Mu = G on (0, T ) × ∂Ω ;

in this case we have

M = Id , G = (r, g) on (0, T ) × Γi1 ,

M =0, G = 0 on (0, T ) × Γo1
M = (0, 0, 1, 0) , G = 0 on (0, T ) × Γ2 ,
M = (0, 0, 0, 1) , G = 0 on (0, T ) × Γ3 .

STEP 1: definition of the set S.

Choose K1 > |||u0 |||3 : if needed, later on we will take a larger value of K1 ; choose
also K2 > |||u0 |||2 . Consider the set S of functions u0 = (π, w) ∈ KT such that
3 1
sup ρ0 ≥ π ≥ inf ρ0 > 0 in Q̄T
2 Ω̄ 2 Ω̄

1

−w1 + c(π) ≤ max − (v0 )1 + c(ρ0 ) < 0 on (0, T ) × Γ1
2 Γ1
∂tk u0 (0) = ∂tk u0 in Ω̄ k = 0, 1, 2
|||u0 |||3,T ≤ K1 |||u0 |||2,T ≤ K2 .
STEP 2: proof that S 6= ∅.
By reasoning as in the proof of Lemma 3.2 in [12] we obtain that for all K1 >
|||u0 |||3 there exists TK1 > 0 such that for all T ≤ TK1 we have S 6= ∅.
STEP 3: proof that S is compact in X = C(0, T ; L2 (Ω)).
Consider a sequence {u0k } ⊂ S and note that the set S is bounded in KT : in partic-
ular, S is bounded in C 1 (0, T ; H 2 (Ω)). Then, by the Ascoli-Arzela Theorem and
the compact imbedding H 2 (Ω) ⊂ L2 (Ω) we can extract a subsequence converging
in X to some u0 ∈ X. Furthermore, u0 satisfies the inequalities which characterize
S: indeed, the pointwise inequalities in Q̄T and on (0, T ) × Γ1 are satisfied by the
compact imbedding KT ⊂ C(Q̄T ) while the bounds on the norms are satisfied by
the lower semicontinuity of the norms under weak* convergence. Hence, u0 ∈ S
and S is compact.
STEP 4: proof that Λ(S) ⊆ S.

16
By Proposition 6 (and (26)) there exist two constants C1 , C2 > 0 (C1 depends
increasingly on |||u0 |||2,T while C2 depends increasingly on |||u0 |||3,T ) such that
n o
|||u|||23,T ≤ C1 (1 + |||u0 |||2ε
3,T )|||u 0 |||2
3 + C1 [G]2
3,ΣT + C2 [f ]2
3,T e
(C1 +C2 )T
;

here, the data are fixed, so that γ is given and can be included in C1 , C2 . The proof
then follows by reasoning as in Lemma 3.4 in [12], the only differences being that
we deal with the H 3 norm instead of the H 4 norm: in particular, by the imbedding
H 2 (Ω) ⊂ C(Ω) we have that S ⊂ C 1 (0, T ; C(Ω̄)) so that u satisfies the inequalities
that characterize S on some interval of time [0, T ], T > 0.
STEP 5: proof that Λ is a contraction in S.
We argue as in [12]. Take u01 = (π1 , w1 ), u02 = (π2 , w2 ) in S and let ui = (ρi , vi ) =
Λu0i for i = 1, 2: then, u1 − u2 satisfies



 L(u01 )(u1 − u2 ) = [L(u02 ) − L(u01 )](u2 ) + F (u01 ) − F (u02 ) in [0, T ] × Ω

M (u1 − u2 ) = 0 on (0, T ) × ∂Ω


 (u − u )(0, x) = 0

in Ω ;
1 2

now, set H = [L(u02 ) − L(u01 )](u2 ) + F (u01 ) − F (u02 ), multiply the first equation by
u1 − u2 and integrate by parts over Ω to obtain
Z   Z 
d 
A0 (u01 )(u1 − u2 ), u1 − u2 + Aν (u01 )(u1 − u2 ), u1 − u2
dt Ω ∂Ω
Z   Z
= B(u01 )(u1 − u2 ), u1 − u2 + 2 (H, u1 − u2 ) ,
Ω Ω
P
where B = ∂t A0 + j ∂j Aj . Note that u1 − u2 ∈kerM so that by the maximally
non-negativity assumption the boundary integral is non negative; moreover, by
Hölder inequality we get the estimate
Z  
B(u01 )(u1 − u2 ), u1 − u2 ≤ C|||u1 − u2 |||20,T :
Ω

hence, again Hölder inequality yields

Z 
d 
A0 (u01 )(u1 − u2 ), u1 − u2 ≤ C|||u1 − u2 |||20,T + 2kH(t)k ku01 (t) − u02 (t)k .
dt Ω
Finally, integrate over [0, t], note that [H]0,t ≤ C[u01 − u02 ]0,t , take into account the
positive definiteness of A0 (say A0 ≥ γ) and use Young inequality to obtain

γ|||u1 − u2 |||20,T ≤ C1 T |||u1 − u2 |||20,T + C2 T |||u01 − u02 |||20,T ;

γ
now take T ≤ 2C1 so that the previous inequality becomes

|||u1 − u2 |||20,T ≤ CT |||u01 − u02 |||20,T :

17
if we take CT < 1, then Λ is a contraction in S with respect to the C(0, T ; L2 (Ω))
norm. 2
We are now ready to give the
Proof of Theorem 1. If T is small enough, Lemma 1 states that Λ has a unique
fixed point (ρ, v) ∈ S: by definition of S and by Proposition 6, (ρ, v) ∈ KT solves
(3)-(4)-(6) and satisfies v1 > c(ρ) on (0, T ) × Γo1 . 2

5 Proof of Theorem 2
The proof follows the same lines as that of Theorem 1; nevertheless, it is slightly
more delicate since nonlinear boundary conditions are involved. Let Ω be as in (2)
and consider the boundary conditions



 v2 = g2 , v3 = g3 , v1 + ρ c(π)π =ψ on (0, T ) × Γi1


on (0, T ) × Γo1

 v =g
1 1
(29)


 v2 = 0 on (0, T ) × Γ2


on (0, T ) × Γ .

 v =0
3 3

Consider the same set ST as for Proposition 3; by reasoning as for Proposition

6, we obtain
Proposition 7 Let Ω ⊂ IR3 be as in (2), let T > 0 and assume that:
(i) ψ, g2 , g3 ∈ H 3 ((0, T ) × Γi1 ), g1 ∈ H 3 ((0, T ) × Γo1 ) satisfy
g2 ∈ He3 (Γi1 ), g3 ∈ Ho3 (Γi1 ), ψ ∈ H3 (Γi1 ), g1 ∈ H3 (Γo1 ) for a.e. t ∈ [0, T ] ;
(ii) the initial data satisfy (ρ0 , v0 ) ∈ H3 × H3 × He3 × Ho3 ;
(iii) the external force satisfies f ∈ H∗3 (QT );
(iv) the data satisfy (10) and the compatibility conditions

(∂tk v0 )2 = ∂tk g2 (0), (∂tk v0 )3 = ∂tk g3 (0), (∂tk v0 )1 + ∂tk (ρ0 c(π(0)) k i
π(0) ) = ∂t ψ(0) on Γ1

(∂tk v0 )1 = ∂tk g1 (0) on Γo1 ∂tk v0 · ν = 0 on Γ0 k = 0, 1, 2 .

Then, for all (π, w) ∈ ST there exists a unique (ρ, v) ∈ KT solving (25)-(4)-(29).
Moreover, there exist three constants C1 , C2 , CM > 0 (C1 depends increasingly on
|||(π, w)|||2,T , C2 depends increasingly on |||(π, w)|||3,T , CM depends increasingly
on [(ψ, g1 , g2 , g3 )]3,ΣT ) such that

|||(ρ, v)(t)|||23 + [(ρ, v)]23,Σt ≤ C1 (1 + |||(π, w)|||2ε 2

3,T )|||(ρ0 , v0 )|||3 +
(30)
+C1 [(ψ, g1 , g2 , g3 )]23,Σt + C2 [f ]23,t + (C1 CM + C2 )[(ρ, v)]23,t

for all t ∈ [0, T ]; here, ε ∈ ( 21 , 1).

18
Proof. The partition of unity is slightly different from that in Proposition 6: cover
Ω̄ by a family of 8 open sets {Ui } (i = 1, ..., 8) so that each one of them contains
one (and only one) of the vertices of Ω, namely the points V1 (0, 0, 0), V2 (0, 1, 0),
V3 (0, 0, 1), V4 (0, 1, 1), V5 (2, 0, 0), V6 (2, 1, 0), V7 (2, 0, 1), V8 (2, 1, 1); moreover, each
Ui should not intersect the three faces of Ω̄ which are not adjacent to the vertex
contained in Ui . Then, for i = 1, ..., 4, (25)-(4)-(29) reduces to the problem (28)
with subsonic linearized boundary conditions, solved by Proposition 4, while for
i = 5, ..., 8 it reduces to a subsonic outflow problem that we solve by Proposition
5. Then, the proof follows. 2
For all given ψ, g, ρ0 , v0 , f satisfying the assumptions of Proposition 7 is de-
fined a map Λ such that Λ(π, w) = (ρ, v); in order to prove a result similar to
Lemma 1 in the subsonic case we need to take into account the boundary condi-
tions:

Lemma 2 Assume that ψ, g, ρ0 , v0 , f satisfy the assumptions of Proposition 7;

then, for sufficiently small T there exists a compact subset S of C(0, T, L2 (Ω)) ∩
L2 (ΣT ) such that Λ maps continuously S into itself.

Proof. To start, just take T ∈ (0, T0 ]: if necessary, later on we will take a smaller
value of T .
Let u0 = (π, w) and let u = (ρ, v) = Λu0 ; we write (25) in the form
3
X
L(u0 )u = A0 (u0 )∂t u + Aj (u0 )∂j u = F (u0 ) ,
j=1

together with the initial and boundary conditions (4) and (29) which we write as

u(0, x) = u0 (x) in Ω
M (u0 )u = G on (0, T ) × ∂Ω ;

in this case we have

 
0 0 1 0
0
M (u ) =  0 0 0 1  , G = (g2 , g3 , ψ) on (0, T ) × Γi1 ,
 
c(π)
π 1 0 0

M = (0, 1, 0, 0) , G = g1 on (0, T ) × Γo1 ,

M = (0, 0, 1, 0) , G = 0 on (0, T ) × Γ2 ,
M = (0, 0, 0, 1) , G = 0 on (0, T ) × Γ3 .
STEP 1: definition of the set S.

19
Choose K1 > 0 so that K1 > |||u0 |||3 : if needed, later on we will take a larger value
of K1 ; choose also K2 > |||u0 |||2 . Consider the set S of functions u0 = (π, w) ∈ KT
such that
3 1
sup ρ0 ≥ π ≥ inf ρ0 > 0 in Q̄T
2 Ω̄ 2 Ω̄
1

−w1 + c(π) ≥ min − (v0 )1 + c(ρ0 ) > 0 on (0, T ) × Γ1
2 Γ1
1
w1 ≥ min(v0 )1 > 0 on (0, T ) × Γ1
2 Γ1
∂tk u0 (0) = ∂tk u0 in Ω̄ k = 0, 1, 2
0 0
[u ]3,ΣT ≤ K1 |||u |||3,T ≤ K1 |||u0 |||2,T ≤ K2 .
STEP 2: proof that S 6= ∅.
This can be obtained as in step 2 in the proof of Lemma 1.
STEP 3: proof that S is compact in X = C(0, T ; L2 (Ω)) ∩ L2 (ΣT ).
Consider a sequence {u0k } ⊂ S: by step 3 in the proof of Lemma 1 we know that
it converges in C(0, T ; L2 (Ω)) to some u ∈ S, up to a subsequence. Moreover,
uk → u0 in L2 (ΣT ) and hence S is compact.
STEP 4: proof that Λ(S) ⊆ S.
By Proposition 7 (and (30)) there exist three constants C1 , C2 , CM > 0 (C1 de-
pends increasingly on |||u0 |||2,T , C2 depends increasingly on |||u0 |||3,T , CM depends
increasingly on [u0 ]3,ΣT ) such that

|||u(t)|||23 + [u]23,Σt ≤
≤ C1 (1 + |||u0 |||2ε 2 2 2 2
3,T )|||u0 |||3 + C1 [G]3,Σt + C2 [f ]3,t + (C1 CM + C2 )[u]3,t

for all t ∈ [0, T ]. The proof then follows by reasoning as in Lemma 3.4 in [12] with
the H 4 norms replaced by the H 3 norms.
STEP 5: proof that Λ is continuous in S.
We argue as in the proof of Lemma 3.5 in [12]. Let u01 , u02 ∈ S and let ui = Λu0i
(i = 1, 2); consider the two problems corresponding to ui and u0i (i = 1, 2), subtract
them and multiply by u1 − u2 : then, with an integration by parts on Ω we get
Z   Z 
d 
A0 (u01 )(u1 − u2 ), u1 − u2 + Aν (u01 )(u1 − u2 ), u1 − u2
dt Ω ∂Ω
Z   Z
= B(u01 )(u1 − u2 ), u1 − u2 + 2 (H, u1 − u2 ) ,
Ω Ω

where H = [L(u02 )−L(u01 )](u2 )+F (u01 )−F (u02 ) and B = ∂t A0 + j ∂j Aj . Integrate
P
the previous equality on [0, t] and take into account the estimates

[M (u02 )u2 − M (u01 )u2 ]0,Σt ≤ c[u01 − u02 ]0,Σt [H]0,t ≤ c[u01 − u02 ]0,t

20
to obtain

k(u1 − u2 )(t)k2 + [u1 − u2 ]20,Σt ≤ c([u01 − u02 ]20,Σt + [u01 − u02 ]20,t + [u1 − u2 ]20,t ) ;

the continuity of Λ follows by applying the Gronwall Lemma and from the estimate
[u01 − u02 ]20,T ≤ T |||u01 − u02 |||20,T . 2
We are now ready to give the
Proof of Theorem 2. As ρ 7→ c(ρ) is concave, the set S found in Lemma 2 is convex;
therefore, by the Schauder fixed point Theorem, the map Λ admits a fixed point
(ρ, v) ∈ S; by definition of S and by Proposition 7, (ρ, v) ∈ KT solves (3)-(4)-(9)
and satisfies 0 < v1 < c(ρ) on [0, T ] × Γ1 .
Assume now that there exists δ > 0 such that max[0,T ]×Γi1 |v1 | < δ; then
uniqueness follows by arguing as in the proof of Theorem 2 in [12]. 2
Acknowledgement. The authors wish to thank the referee for pointing out the
uncorrectness of the proof in the first version of the paper.

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