Gazzola F., Secchi P. - Inflow-outflow problems for Euler equations in a rectangular...

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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 ina (nonsmooth) bounded cylinder admit a regular solution. The problemsconsidered are symmetric hyperbolic systems with partly characteristic andpartly noncharacteristic boundary; for such problems, no general theory isavailable. Therefore, we introduce particular spaces of functions satisfyingsuitable additional boundary conditions which allow to determine a regularsolution by means of a reflection technique.

1 Introduction

Let be an open bounded cylinder in IR3 having a rectangular section: moreprecisely, let be the cartesian product of an open rectangle R with a boundedinterval (a, b) ( =R (a, b)) and let 1 = R {a, b}and 0 = R [a, b] so thatthe piecewise smooth boundary may be characterized by = 1

0. In the

cylinder we consider the following initial-boundary value problem for the Eulerequations for a barotropic inviscid compressible fluid

t + (v) = 0 in [0, T]

(tv+ (v )v f) + p= 0 in [0, T]

M(, v) = G on (0, T) 1

v = 0 on (0, T) 0

(0, x) = 0(x) in

v(0, x) = v0(x) in ,

(1)

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where t = /t, Mis a matrix which depends on the particular inflow-outflowproblem considered, see (6), (9), and denotes the unit outward normal to ,

when it exists; the density = (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 spacevariablex . In (1) the density is assumed to be positive for physical reasons;moreover, andp are related by the equation of state p= p() wherep: IR+ IR+is a given smooth function such that p(s) > 0 for all s > 0. The external forcefield f = f(t, x), the boundary data G and the initial data v0 and 0 are knownfunctions.

Usually, the Euler system is studied under the slip boundary condition v =0 on 0, see [1, 2, 3, 9]. In this paper we study the existence and uniqueness ofa regular solution for some inflow-outflow problems for (1): we assume that thefluid flows in on i1 = R {a}, that it flows out on

o1 = R {b} and satisfies

the slip condition v = 0 on 0. On 1 = i1 o1 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 inthe boundary conditions. The existence of a regular solution of (1) is not at allobvious: first because the domain is nonsmooth, second because the boundarymatrix does not have constant rank. Nevertheless, if suitable functional spacesare considered, the presence of a nonsmooth boundary avoids in some sense theproblems arising from boundary points where the boundary matrix changes rankand enables us to prove the existence of a regular solution. In next section wewrite (1) as a symmetric hyperbolic system [4]; with our assumptions it becomespartly characteristic and partly noncharacteristic: more precisely, the boundary1 is noncharacteristic while the boundary 0 is characteristic of rank two. To ourknowledge no general theory is available for these problems even if some partialresults may be found in [6, 7, 8, 11, 13].

Our method consists in introducing a class of functional spaces of functionshaving some vanishing traces; these spaces allow, by a reflection technique, toreduce the linearized problem associated to (1) to a noncharacteristic problem.Then, standard existence results apply and we obtain a solution of the linearizedproblem in 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 theexistence and uniqueness of a regular local in time solution (, v) of (1) for boththe supersonic and the subsonic cases: we assume that the fluid flows in i1 eithersupersonic or subsonic and analogously on the outflow part o1. 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 conditionshave to be considered.

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2 Notations and results

For simplicity, we take

= (0, 2) (0, 1) (0, 1), (2)

we denoteQT = (0, T) , T = [0, T] and we define the sets

i1 = {0} [0, 1] [0, 1] o1 = {2} [0, 1] [0, 1] 1=

i1

o1

2 = [0, 2] {0, 1} [0, 1] 3 = [0, 2] [0, 1] {0, 1} 0 = 2

3 ;

then, the piecewise smooth boundary is given by = 1

2

3.Let Hm() be the usual Sobolev space of order m (m IN) and let m

denote its norm: the L2()-norm is simply denoted by . We also introduce thefollowing spaces of functions having some (even or odd) traces vanishing on parts

of the boundary

H3e ={ H3(); = 0 on 2, 3= 0 on 3, 22= 0 on 2}

H3o = { H3(); = 0 on 3, 2= 0 on 2,

23= 0 on 3}

H3 ={ H3(); = 0 on 0} ;

clearly, these are closed subspaces ofH3() and contain H30 (): here, =/,i =/xi,

2i =

2/x2i and the traces are well-defined even if the domain isnonsmooth, see [5]. Similarly, we define the spaces H3(1), H

3o(1) and H

3e(1).

For any vector function defined on (a subset of) QT we denote by i its i-thcomponent,i = 1, 2, 3. LetB be a Banach space and letT >0: thenC(0, T; B) andL(0, T; B) denote respectively the space of continuous and essentially boundedfunctions defined on [0, T] and taking values in B . Define the space

CT(H3

) = C(0, T; H3) C(0, T; H3) C(0, T; H3e) C(0, T; H3o) .

Consider now the spaces

CT(H3) =

3k=0

Ck(0, T; H3k()) L(H3) =

3k=0

Wk,(0, T; H3k())

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

|||u|||3,T= sup[0,T]

|||u(t)|||3 , |||u(t)|||23 =

3

k=0

ktu(t)23k .

We seek solutions (, v) of (1) in the closed subspace of [CT(H3)]4 defined by

KT =CT(H3

) [CT(H3)]4 .

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Finally, consider the space

H3(QT) = { [H3(QT)]3; (t) H3 H3e H3o for a.e. t [0, T]}

normed by

[]23,T =

T0

|||(t)|||23 dt ;

similarly, we define H3

(T) and the norm []3,T .

By introducing the sound velocity c() =

p(), the equations in problem(1) become

1

(t + v ) + v= 0 in [0, T]

c2() (tv+ (v )v f) + = 0 in [0, T] ,

(3)

to which we associate the initial conditions

(0, x) = 0(x) in

v(0, x) = v0(x) in ,(4)

for0 andv0 in a suitable functional space.We may write (3) as a quasilinear symmetric hyperbolic system, that is in

the form

A0(u)tu +3

j=1

Aj(u)ju= F(u) (5)

where u = (, v), the Aj (j = 0,..., 3) are symmetric and A0 is positive definite.

The boundary matrix is

A=

1v T

c2()

v I3

;

therefore, the matrix A10 Ahas eigenvalues

v , v , v + c() , v c().

Recall that a part of the boundary is said noncharacteristic if detA= 0 on ;moreover, if we are given homogeneous boundary conditions M u= 0, then kerMis said to be maximally non-negative forA if

(Au, u) 0 on (0, T) u kerM ,

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

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integrals when one looks for a priori estimates. This condition corresponds torequiring that the number of boundary conditions equals the number of negative

eigenvalues ofA. Let us first consider the supersonic case: on the inflow part ofthe boundary i1 (which is noncharacteristic) the four eigenvalues ofA

10 A are

negative and the whole state of the fluid must be prescribed while on the outflowpart o1 the boundary matrix A

10 Ais positive definite and no condition has to

be assigned. For the subsonic case we also have that 1 is noncharacteristic: on i1

three eigenvalues ofA10 A are negative and three conditions must be prescribedwhile on o1 the boundary matrix A

10 Aonly has a negative eigenvalue and one

condition has to be assigned. Finally, the impermeable part 0 is characteristicof rank two with A10 Ahaving one negative eigenvalue and only the conditionv = 0 is given.

So, if we fix T0 > 0 and we take into account the shape of in (2), for thesupersonic case we require that

(, v) = (r, g) on (0, T0) i1

v2 = 0 on (0, T0) 2

v3 = 0 on (0, T0) 3

(6)

while no conditions should be imposed on o1; however, to ensure that the fluidflows out supersonic at least in a small interval of time, we require that the initialflow satisfies such condition: we assume that

0 > 0 in

(v0)1 > c(0) on 1

r >0 on [0, T0] i1

g1 > c(r) on [0, T0] i1 .

(7)

Finally, to find regular solutions one needs to impose some necessary compatibilityconditions between the boundary data and the initial values; denote by kt0 andktv0the functions obtained by formally takingk 1 time derivatives of (3), solvingfor kt and

ktv and evaluating at time t = 0. Then the compatibility conditions

in the supersonic case read

kt0 = ktr(0),

ktv0 =

ktg(0) on

i1

ktv0 = 0 on 0 k= 0, 1, 2 ;

(8)foru0 = (0, v0) we set|||u0|||

2m =

mk=1(||

kt0||

2mk+ ||

ktv0||

2mk).

Our main result in the supersonic case states

Theorem 1 Let IR3 be as in (2), letT0 > 0 and assume that:(i) the boundary datar, g H3((0, T0) i1) satisfy

(r, g) H3(i1) H3(i1) H

3e(

i1) H

3o(

i1) for a.e.t [0, T0] ;

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(ii) the initial data satisfy(0, v0) H3 H3 H3e H

3o;

(iii) the external force satisfiesfH3

(QT0);

(iv) (7) and (8) hold.Then, there existsT >0 such that (3)-(4)-(6) admits a unique solution(, v) KTsatisfyingv1 > c() on[0, T] o1.

Remark.Some of the compatibility conditions (8) are hidden in the assumptions(i) - (iii); further regularity may be obtained by using Sobolev spacesHm of higherorder (m 4): in such case, one must obviously strengthen (8) with supplementaryconditions.

In the subsonic case we assign the tangential velocity of the fluid on the inflowpart and the normal velocity on the outflow part; we require that

v2= g2 , v3 = g3 , v1+ c() = on (0, T0) i1

v1= g1 on (0, T0) o1

v2= 0 on (0, T0) 2

v3= 0 on (0, T0) 3

(9)

and we assume that

0 > 0 in

c(0) 0 on [0, T0] o1 .

(10)

In order to ensure that the fluid flows in subsonic on i1 one should require thatc() < + c() < 0 on [0, T0]

i1 while to ensure that the fluid flows out

subsonic on o1one should require thatg1 < c() on [0, T0]o1but these conditions

depend on the solution; nevertheless, we will prove in Theorem 2 below that theyhold in some interval of time [0, T], T >0. We also refer to [12] for the derivationof (9) and for other possible boundary conditions.

The compatibility conditions between the initial and boundary data in thesubsonic case read

(ktv0)2 = ktg2(0), (

ktv0)3 =

ktg3(0), (

ktv0)1+

ktc0 =

kt(0) on

i1

(ktv0)1 = ktg1(0) on

o1

ktv0 = 0 on 0 k= 0, 1, 2 ,

(11)wherektc0 denotes thek

th time derivative att = 0 ofc(). Then, for the subsoniccase we prove the following

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Theorem 2 Let IR3 be as in (2), letT0 > 0 and assume that:(i) , g2, g3 H

3((0, T0) i1), g1 H

3((0, T0) o1) satisfy

g2 H3e(

i1), g3 H

3o(

i1), H

3(i1), g1 H3(o1) for a.e.t [0, T0] ;

(ii) the initial data satisfy(0, v0) H3 H3 H3e H3o;

(iii) the external force satisfiesfH3

(QT0);(iv) (10) and (11) hold;(v) the functions c(s) is concave.Then, there exists T > 0 such that (3)-(4)-(9) admits a solution (, v) KTsatisfying

0< v1 < c() on[0, T] 1 .

Moreover, if the functions c(s)satisfiesc(s)> 0 for alls >0, then there exists >0 such that if max

[0,T]i1

|v1|< then(, v) is the unique solution of (3)-(4)-(9).

Remark. Ifp() = R with > 1 then c() is concave for 3; for commonisentropic gases (1, 53 ) and therefore (v) is physically meaningful. Moreover, if >1 then c(s)> 0 for all s >0.

Remark.Uniqueness in Theorem 2 is certainly ensured in some time interval [0, T](T > 0) if (0) c(0) is sufficiently small on i1 (recall that by (10) we have(0) c(0)> 0).

3 The linearized problems in space sectors

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

3.1 The supersonic case

In this section we first consider the case where

= IR3++ := IR+ IR+ IR , 1 := {0} IR+ IR , 2 := IR+ {0} IR ,

and we study the linearized problem

1

(t + w ) + v= h in [0, T]

c2() (tv+ (w )v) + = k in [0, T]

(12)

together with initial conditions (4) and boundary conditions

(, v) = (r, g) on (0, T) 1

v2 = 0 on (0, T) 2 .(13)

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In this case 3 = and the conditions on 3 in the definitions of the spacesH3 disappear; in particular, we have H3o =H

3. Consider the set ST of functions

(, w) KT such that

3

2sup

0 1

2inf

0 > 0 in QT,

w1+ c()1

2sup1

(v0)1+ c(0)

0 . (14)

Let us point out that the definition of ST gives some restrictions on the initialdata0, v0 since some inequalities of the kind of (7) are required.

The following result holds:

Proposition 1 Let = IR3++, letT >0 and assume that:(i) the boundary datar, g H3((0, T) 1) satisfy

(r, g) H3(1) H3(1) H

3e(1) H

3o(1) for a.e. t [0, T] ;

(ii) the initial data satisfy(0, v0) H3 H3 H3e H3o;

(iii) the right hand side of (12) satisfies (h, k) H3(QT) H3

(QT), h(t) H3

for a.e.t [0, T];(iv) the data satisfy (7) and the compatibility conditionskt(0) =

ktr(0),

ktv(0) =

ktg(0)on1, ktv(0) = 0 on0, k = 0, 1, 2, where

kt(0),

ktv(0)are thekth

time derivatives of, v at time t= 0 obtained from (12), (4).Then, for all (, w) ST there exists a unique (, v) KT solving (12)-(4)-(13). Moreover, there exist two constantsC1, C2 > 0 (C1 depends increasingly on|||(, w)|||2,T whileC2 depends increasingly on|||(, w)|||3,T) such that

|||(, v)|||23,T

C1(1 + |||(, w)|||23,T)|||(0, v0)|||

23+

+C1[(r, g)]23,T + C2[(h, k)]23,T

e(C1+C2)T ,

(15)

where= () is defined in (14) and ( 12 , 1).

Proof. Let = IR+ IR IR and = . For all x = (x1, x2, x3) IR3 we

denote x= (x1, x2, x3) and for all continuous function defined on we define

the functions and on by

(x) =

(x) ifx2 0

(x) ifx2 < 0(x) =

(x) ifx2 0

(x) ifx2 < 0 ;

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this definition implies that the functions and are, respectively, even and oddwith respect to x2: to be precise, is odd only if(x1, 0, x3) = 0 for all x1, x3.

Consider now the auxiliary problem

1

(t + w ) + v=h in [0, T]

c2() (tv+ ( w )v) + =

k in [0, T]

= r v= g on (0, T)

(0, x) = 0(x) in

v(0, x) = v0(x) in ,

(16)

where we have set

= g= (g1, g2, g3) = (g1, g2, g3) r= r h=h

k = (k1, k2, k3) = (k1, k2, k3) w= ( w1, w2, w3) = ( w1, w2, w3)

0 = 0 v0 = ((v0)1, (v0)2, (v0)3) = ((v0)1, (v0)2, (v0)3) :

for simplicity, we have omitted to underline the dependence on t.Since (, w) KT, it is not difficult to verify that the reflected functions

, wi(i= 1, 2, 3) belong toCT(H3()); moreover, (14) holds for as () = ().

Note also that since (, w) STwe have

inf(0,T)

w1 c()

>0 ,

so that is noncharacteristic. Moreover we observe that (0, v0) H3(), (h, k)

H3

((0, T)

) and that (r, g) H3

((0, T)

). Finally we observe that thecompatibility conditions of order 2 hold on . We apply Theorem A.1 in [12](also valid for halfspaces) in the case m = s = 3 and find a unique solution(, v) [CT(H

3())]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

|||(, v)(t)|||23 C1(1 + |||(, w)|||23,T)|||(0, v0)|||

23+

+C1[(r, g)]23,T + C2[(h, k)]23,T+ (C1+ C2)[(, v)]

23,t :

then, (15) follows by the Gronwall Lemma.We claim that, v1 andv3 are even with respect to x2 while v2 is odd with

respect to x2. To this end, define the functions

(t, x) = (t,x) v1,3(t, x) = v1,3(t,x) v2(t, x) = v2(t,x)

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

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which proves the claim. Therefore, for all t [0, T] the functions , v1, v3 belongto the space H3o while v2 H

3e; in particular, we have v2 = 0 on 2 (0, T): this

proves that (, v) KTsolves (12)-(4)-(13).

Next we consider the case where

= IR3+++ := [IR+]3 , 1 := {0} IR+ IR+ ,

2 := IR+ {0} IR+ , 3 := IR+ IR+ {0},

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

(, v) = (r, g) on (0, T) 1

v2 = 0 on (0, T) 2

v3 = 0 on (0, T) 3 :

(17)

by using the same reflection technique as in the proof of Proposition 1 we proveexistence and uniqueness of a solution:

Proposition 2 Let = IR3+++, letT >0 and assume that (i)-(iv) of Proposition1 hold. Then, for all(, w) STthere exists a unique(, v) KTsolving (12)-(4)-(17). Moreover, there exist two constants C1, C2 > 0 (as in Proposition 1) suchthat (15) holds.

Proof. Let = IR+ IR+ IR,

1 = {0} IR+ IR and

2 = IR+ {0} IR sothat 1

2 =

. For all x = (x1, x2, x3) IR3 we define x= (x1, x2, x3) and

for all function defined on we define the functions and on by

(x) =

(x) ifx3 0

(x) ifx3 < 0(x) =

(x) ifx3 0

(x) ifx3 < 0 ;

then, the functions and are, respectively, even and odd with respect to x3.Consider now the auxiliary problem

1

(t + w ) + v=h in [0, T]

c2() (tv+ ( w )v) + =

k in [0, T]

= r v= g on (0, T) 1

v2 = 0 on (0, T) 2

(0, x) = 0(x) in

v(0, x) = v0(x) in ,

(18)

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where we have set

= g= (g1, g2, g3) = (g1, g2, g3) r= r h=hk = (k1, k2, k3) = (k1, k2, k3) w= ( w1, w2, w3) = ( w1, w2, w3)

0 = 0 v0 = ((v0)1, (v0)2, (v0)3) = ((v0)1, (v0)2, (v0)3) .

Again, it is not difficult to verify that the reflected functions (, w) belong toCT(H3(

)) and the properties (i) (iv) of Proposition 1 for the reflected data;by Proposition 1, problem (18) admits a unique solution (, v) [CT(H

3())]4

which satisfies the estimate (15). By proceeding as in the proof of Proposition 1 weobtain that , v1 andv2 are even with respect to x3 while v3 is odd with respectto x3. Therefore, (, v) KT: in particular, we have v3 = 0 on (0, T) 3 and(, v) solves (12)-(4)-(17).

3.2 The subsonic case

We repeat the same arguments of the previous section and we maintain the samenotations. Again, we first consider the case where = IR3++ and we study thelinearized problem (12) together with the initial conditions (4) (0 and v0 satisfysome of the inequalities in (10) due to the definition ofSTbelow) and the (linear)boundary conditions

v2 = g2 , v3 = g3 , v1+

c()

= on (0, T) 1

v2 = 0 on (0, T) 2 ,(19)

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

ktv2(0) =ktg2(0),

ktv3(0) =

ktg3(0),

ktv1(0) +

kt(

c()

)(0) =kt(0) on 1

ktv(0) = 0 on 0 k= 0, 1, 2,

(20)where kt(0),

ktv(0) are the k-th time derivatives of, v at time t = 0, obtained

from (12), (4). Consider now the set STof functions (, w) KTwhich satisfy

3

2sup

0 1

2inf

0 > 0 in QT

w1+ c()1

2inf1

(v0)1+ c(0)

>0 on (0, T) 1

w1

1

2inf1 (v0)1 > 0 on (0, T) 1

kt(0) =kt0,

ktw(0) =

ktv0 k = 0, 1, 2.

The following result holds:

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Proposition 3 Let = IR3++, letT >0 and assume that:(i) , g2, g3 H

3((0, T) 1) satisfy

g2 H3e(1) , g3 H

3o(1) , H

3(1) for a.e. t [0, T] ;


3o;


(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(, w) ST there exists a unique(, v) KT solving (12)-(4)-(19).Moreover, there exist three constantsC1, C2, CM>0 (C1 depends increasingly on|||(, w)|||2,T, C2 depends increasingly on |||(, w)|||3,T, CMdepends increasinglyon[(, g2, g3)]3,T ) such that

|||(, v)(t)|||23+ [(, v)]23,t

C1(1 + |||(, w)|||23,T)|||(0, v0)|||

23+

+C1[(, g2, g3)]23,t + C2[(h, k)]23,t+ (C1CM+ C2)[(, v)]23,t(21)

for all t [0, T]; here, ( 12 , 1).

Proof. Let = IR+ IR IR, = and consider the auxiliary problem

1

(t + w ) + v=h in [0, T]

c2() (tv+ ( w )v) + =

k in [0, T]

v2= g2 , v3 = g3 , v1+ c()

= on (0, T)

(0, x) = 0(x) in

v(0, x) = v0(x) in ,

(22)

where we have set

= (g2, g3) = (g2, g3) = h=h

k = (k1, k2, k3) = (k1, k2, k3) w= ( w1, w2, w3) = ( w1, w2, w3)

0 = 0 v0 = ((v0)1, (v0)2, (v0)3) = ((v0)1, (v0)2, (v0)3) .

Note that is noncharacteristic and by Theorem A.1 in [12] we find a uniquesolution (, v) [CT(H

3())]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(t,x), v1,3(t,x), v2(t,x) solves (22), by uniqueness of the solutionof (22) we infer that , v1 andv3 are even with respect tox2 whilev2 is odd withrespect to x

2: then (, v) KTsolves (12)-(4)-(19).

Remark. In (21) we do not highlight the dependence of the constants C1, C2, CMon(defined in (14)) becauseis uniformly bounded from below and above when(, w) vary in ST.

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Next, consider the case where = IR3+++ and the problem (12)-(4) togetherwith the boundary conditions

v2 = g2 , v3 = g3 , v1+ c()

= on (0, T) 1

v2 = 0 on (0, T) 2

v3 = 0 on (0, T) 3 ;

(23)

then, we obtain the following:

Proposition 4 Let = IR3+++, letT >0 and assume that (i)-(iv) of Proposition3 hold.Then, for all(, w) ST there exists a unique(, v) KT solving (12)-(4)-(23).Moreover, there exist three constantsC1, C2, CM > 0 (as in Proposition 3) suchthat (21) holds for all t [0, T].

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

Finally, we also need a similar result for the subsonic outflow problem: herewe take = (, 2) IR+ IR+ instead of IR

3+++ so that we do not have

to change sign to the functions defined on . Let 1 = {2} IR+ IR+, 2 =(, 2]{0}IR+, 3 = (, 2]IR+{0} and consider the boundary conditions

v1 = g1 on (0, T) 1

v2 = 0 on (0, T) 2

v3 = 0 on (0, T) 3 .

(24)

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

Proposition 5 Let = (, 2) IR+ IR+, letT >0 and assume that:(i) g1 H

3((0, T) 1) satisfiesg1 H3(1) for a.e. t [0, T];


3o;


(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 order2.Then, for all(, w) ST there exists a unique(, v) KT solving (12)-(4)-(24).Moreover, there exist three constantsC1, C2> 0 (as in Proposition 3) andCM>0(depending increasingly on [g1]3,T) such that (21) holds (with [g1]3,t instead of[(, g2, g3)]3,t) for allt [0, T].

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

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4 Proof of Theorem 1

Let be the bounded cylinder defined in (2), let STbe the subset ofKTintroducedin Section 3.1; we prove existence and uniqueness for the linearized problem in(0, T) :

1

(t + w ) + v= 0 in [0, T]

c2() (tv+ (w )v f) + = 0 in [0, T]

(25)

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

Proposition 6 Let IR3 be as in (2), letT > 0 and assume that (i)-(iv) ofTheorem 1 hold (with T0 replaced by T). Then, for all (, w) ST there existsa unique (, v) KT solving (25)-(4)-(6). Moreover, there exist two constants

C1, C2 > 0 (C1 depends increasingly on|||(, w)|||2,TwhileC2 depends increasinglyon|||(, w)|||3,T) such that

|||(, v)|||23,T

C1(1 + |||(, w)|||23,T)|||(0, v0)|||23+ C1[(r, g)]

23,T

+ C2[f]23,T

e(C1+C2)T ,

(26)where= () is defined in (14) and ( 12 , 1).

Proof. We first prove the result in the unbounded cylinder

= (0, +) (0, 1) (0, 1)

and in the time interval [0, T]. We localize (12) in by using a partition of unityfor so that the problem reduces to four problems considered in the previoussection: cover by a family of 4 open sets {Ui} (i = 1, ..., 4) so that each oneof them contains one (and only one) of the infinite edges of , namely the lines{x2 = x3 = 0}, {x2 = 1, x3 = 0}, {x2 = 0, x3 = 1}, {x2 = x3 = 1}; moreover,eachUi should not intersect the two faces of

which are not adjacent to the edgecontained in Ui. Take a partition of unity {i}

4i=1 subordinate to the covering

{Ui}such that4

i=1 i = 1, i 0, i H3. We multiply (25) byi, i = 1, . . . , 4.

After a suitable change of variables, we obtain the following equations in IR3+++for (i, vi) = (i, iv):

1

(ti+ w i) + vi = 1

w i + v i

c2

()

(tvi+ (w )vi if) + i=

c2

()

(w i)v+ i.(27)

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In view of a fixed point = , w= v, instead of (27) we consider the problems

1

(ti+ w i) + vi= 2w i in [0, T] IR3+++

c2() (tvi+ (w )vi if) + i =

= c2() (w i)w+ i in [0, T] IR

3+++

(i, vi) = (ir, ig) on [0, T] 1

vi = 0 on [0, T] 0

i(0, x) = i(x)0(x) in IR3+++

vi(0, x) = i(x)v0(x) in IR3+++ .

(28)

We verify that the data of problem (28) satisfy (i) (iv) of Proposition 2; in

particular (iii) follows from (, w) CT(H

3

) and i H

3

, the compatibilityconditions (iv) follow from kt(0) =kt0,

ktw(0) =

ktv0, k= 0, 1, 2, and (8). By

Proposition 2, (28) admits a unique solution (i, vi) KT. By adding together thefunctions (i, vi) we obtain a solution (, v) in (0, T) of (25),(4) and the firstboundary condition on (0, T) i1 of (6). As regards the boundary conditions on2 and 3, ifvi 0 on P = {x2 = 1} {x3 = 1} for some i, after addingthe solutions (i, vi) we could obtain v2 = 0 on 2 or v3 = 0 on 3. To overcomethis point, we proceed in two steps. First, since each (i, vi) has initially a supportwhich does not intersect P, by using the finite speed of propagation we show that(i, vi) vanishes on Pfor each t (0, T

), for a sufficiently small T >0. It followsthat (after the inverse change of variables)

i vi = 0 on 0. Thus we have

found a unique solution (, v) of (25),(4),(6) defined on [0, T] . We verify thatT depends only on ||w||L and on the extension of each support, namely on the

functionsi. We take t = T

as a new initial time, decompose the data by meansof theis and by the same arguments as above find a solution defined on [T, 2T].We proceed by this continuation argument up to when the solution is extended tothe whole interval [0, T]. Moreover, since all the (i, vi) satisfy (15), (, v) satisfies(26).

For all givenr, g,0, v0, fsatisfying the assumptions of Theorem 1 is defineda map such that (, w) = (, v); we achieve the proof of Theorem 1 by showingthat admits a fixed point:

Lemma 1 Assume thatr, g,0, v0, fsatisfy the assumptions of Theorem 1; then,for sufficiently smallTthere exists a compact subsetSofC(0, T , L2())such thatthe map defines a contraction inS.

Proof. For the moment take T (0, T0): if needed, later on we will take a smallervalue ofT. Let u = (, w) and let u = (, v) = u; as for (5), we may write (25)

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as a linear symmetric hyperbolic system, that is in the form

L(u)u= A0(u)tu +

3j=1

Aj(u)ju= F(u

) ,

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

u(0, x) = u0(x) in

M u= G on (0, T) ;

in this case we have

M=Id , G= (r, g) on (0, T) i1 ,

M= 0 , G= 0 on (0, T) o

1

M= (0, 0, 1, 0) , G= 0 on (0, T) 2 ,

M= (0, 0, 0, 1) , G= 0 on (0, T) 3 .

STEP 1: definition of the set S.Choose K1 > |||u0|||3: if needed, later on we will take a larger value ofK1; choosealsoK2 > |||u0|||2. Consider the set Sof functions u = (, w) KT such that

3

2sup

0 1

2inf

0 > 0 in QT

w1+ c()1

2max1

(v0)1+ c(0)|||u0|||3 there exists TK1 >0 such that for all T TK1 we have S =.STEP 3: proof that Sis compact in X=C(0, T; L2()).Consider a sequence{uk} Sand note that the set Sis bounded inKT: in partic-ular,S is bounded in C1(0, T; H2()). Then, by the Ascoli-Arzela Theorem andthe compact imbedding H2() L2() we can extract a subsequence converginginXto some u X. Furthermore, u satisfies the inequalities which characterizeS: indeed, the pointwise inequalities in QTand on (0, T) 1 are satisfied by the

compact imbedding KT C(QT) while the bounds on the norms are satisfied bythe lower semicontinuity of the norms under weak* convergence. Hence, u SandS is compact.STEP 4: proof that (S) S.

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By Proposition 6 (and (26)) there exist two constants C1, C2 > 0 (C1 dependsincreasingly on |||u|||2,T whileC2 depends increasingly on |||u

|||3,T) such that

|||u|||23,T

C1(1 + |||u|||23,T)|||u0|||

23+ C1[G]

23,T

+ C2[f]23,T

e(C1+C2)T ;

here, the data are fixed, so thatis given and can be included inC1, C2. The proofthen follows by reasoning as in Lemma 3.4 in [12], the only differences being thatwe deal with theH3 norm instead of theH4 norm: in particular, by the imbeddingH2() C() we have thatS C1(0, T; C()) so thatu satisfies the inequalitiesthat characterize Son some interval of time [0, T], T >0.STEP 5: proof that is a contraction in S.We argue as in [12]. Takeu1 = (1, w1),u

2= (2, w2) inSand letui = (i, vi) =ui fori = 1, 2: then, u1 u2 satisfies

L(u1)(u1 u2) = [L(u

2) L(u

1)](u2) + F(u

1) F(u

2) in [0, T]

M(u1 u2) = 0 on (0, T)

(u1 u2)(0, x) = 0 in ;

now, setH= [L(u2) L(u

1)](u2) + F(u

1) F(u

2), multiply the first equation byu1 u2 and integrate by parts over to obtain

d

dt

A0(u

1)(u1 u2), u1 u2

+

A(u

1)(u1 u2), u1 u2

=

B(u1)(u1 u2), u1 u2

+ 2

(H, u1 u2),

where B = tA0+ jjAj . Note that u1 u2 kerMso that by the maximallynon-negativity assumption the boundary integral is non negative; moreover, byHolder inequality we get the estimate

B(u1)(u1 u2), u1 u2

C|||u1 u2|||

20,T :

hence, again Holder inequality yields

d

dt

A0(u

1)(u1 u2), u1 u2

C|||u1 u2|||20,T+ 2H(t) u

1(t) u

2(t).

Finally, integrate over [0, t], note that [H]0,t C[u

1 u

2]0,t, take into account thepositive definiteness ofA0 (sayA0 ) and use Young inequality to obtain

|||u1 u2|||20,T C1T|||u1 u2|||

20,T+ C2T|||u

1 u

2|||20,T ;

now take T 2C1 so that the previous inequality becomes

|||u1 u2|||20,T CT|||u

1 u

2|||20,T :

17


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if we takeC T c() on (0, T)

o1.

5 Proof of Theorem 2

The proof follows the same lines as that of Theorem 1; nevertheless, it is slightlymore delicate since nonlinear boundary conditions are involved. Let be as in (2)and consider the boundary conditions

v2 = g2 , v3 = g3 , v1+ c() = on (0, T) i1

v1 = g1 on (0, T) o1

v2 = 0 on (0, T) 2

v3 = 0 on (0, T) 3 .

(29)

Consider the same setSTas for Proposition 3; by reasoning as for Proposition6, we obtain

Proposition 7 Let IR3 be as in (2), letT >0 and assume that:(i) , g2, g3 H

3((0, T) i1), g1 H3((0, T) o1) satisfy

g2 H3e(

i1), g3 H

3o(

i1), H

3(i1), g1 H3(o1) for a.e. t [0, T] ;

(ii) the initial data satisfy(0, v0) H3 H3 H3e H3o;(iii) the external force satisfiesfH3

(QT);

(iv) the data satisfy (10) and the compatibility conditions

(ktv0)2 = ktg2(0), (

ktv0)3 =

ktg3(0), (

ktv0)1+

kt(0

c((0))(0) ) =

kt(0) on

i1

(ktv0)1 = ktg1(0) on

o1

ktv0 = 0 on0 k = 0, 1, 2 .

Then, for all(, w) ST there exists a unique(, v) KT solving (25)-(4)-(29).Moreover, there exist three constantsC1, C2, CM>0 (C1 depends increasingly on|||(, w)|||2,T, C2 depends increasingly on |||(, w)|||3,T, CMdepends increasinglyon[(, g1, g2, g3)]3,T) such that

|||(, v)(t)|||

2

3+ [(, v)]

2

3,t C1(1 + |||(, w)|||

2

3,T)|||(0, v0)|||

2

3++C1[(, g1, g2, g3)]23,t + C2[f]

23,t+ (C1CM+ C2)[(, v)]

23,t

(30)

for all t [0, T]; here, ( 12 , 1).

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Proof. The partition of unity is slightly different from that in Proposition 6: cover 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 , 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, eachUi should not intersect the three faces of which are not adjacent to the vertexcontained 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 fori= 5, ..., 8 it reduces to a subsonic outflow problem that we solve by Proposition5. Then, the proof follows.

For all given , g, 0, v0, fsatisfying the assumptions of Proposition 7 is de-fined a map such that (, w) = (, v); in order to prove a result similar toLemma 1 in the subsonic case we need to take into account the boundary condi-tions:

Lemma 2 Assume that , g, 0, v0, f satisfy the assumptions of Proposition 7;then, for sufficiently smallTthere exists a compact subsetS ofC(0, T , L2()) L2(T) such that maps continuouslyS into itself.

Proof. To start, just take T (0, T0]: if necessary, later on we will take a smallervalue ofT.

Letu = (, w) and let u = (, v) = u; we write (25) in the form

L(u)u= A0(u)tu +

3j=1

Aj(u)ju= F(u

) ,

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

u(0, x) = u0(x) in

M(u)u= G on (0, T) ;

in this case we have

M(u) =

0 0 1 0

0 0 0 1c()

1 0 0

, G= (g2, g3, ) on (0, T) i1 ,

M= (0, 1, 0, 0), G= g1 on (0, T) o1 ,

M= (0, 0, 1, 0) , G= 0 on (0, T) 2 ,

M= (0, 0, 0, 1) , G= 0 on (0, T) 3 .

STEP 1: definition of the set S.

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ChooseK1 > 0 so thatK1 > |||u0|||3: if needed, later on we will take a larger valueofK1; choose also K2 > |||u0|||2. Consider the setSof functionsu

= (, w) KTsuch that

3

2sup

0 1

2inf

0 > 0 in QT

w1+ c()1

2min1

(v0)1+ c(0)

>0 on (0, T) 1

w1 1

2min1

(v0)1 > 0 on (0, T) 1

ktu(0) =ktu0 in k= 0, 1, 2

[u]3,T K1 |||u|||3,T K1 |||u

|||2,T K2 .

STEP 2: proof that S =.

This can be obtained as in step 2 in the proof of Lemma 1.STEP 3: proof that Sis compact in X=C(0, T; L2()) L2(T).Consider a sequence{uk} S: by step 3 in the proof of Lemma 1 we know thatit converges in C(0, T; L2()) to some u S, up to a subsequence. Moreover,uk u

inL2(T) and hence S is compact.STEP 4: proof that (S) S.By Proposition 7 (and (30)) there exist three constants C1, C2, CM > 0 (C1 de-pends increasingly on |||u|||2,T,C2depends increasingly on |||u

|||3,T,CMdependsincreasingly on [u]3,T) such that

|||u(t)|||23+ [u]23,t

C1(1 + |||u|||23,T)|||u0|||

23+ C1[G]

23,t

+ C2[f]23,t+ (C1CM+ C2)[u]

23,t

for allt [0, T]. The proof then follows by reasoning as in Lemma 3.4 in [12] withtheH4 norms replaced by the H3 norms.STEP 5: proof that is continuous in S.We argue as in the proof of Lemma 3.5 in [12]. Let u1, u

2 Sand let ui = u

i

(i= 1, 2); consider the two problems corresponding to uiand u

i(i= 1, 2), subtractthem and multiply by u1 u2: then, with an integration by parts on we get

d

dt

A0(u

1)(u1 u2), u1 u2

+

A(u

1)(u1 u2), u1 u2

=

B(u1)(u1 u2), u1 u2

+ 2

(H, u1 u2),

whereH= [L(u

2)L(u

1)](u2)+F(u

1)F(u

2) andB = tA0+

jjAj . Integratethe previous equality on [0, t] and take into account the estimates

[M(u2)u2 M(u

1)u2]0,t c[u

1 u

2]0,t [H]0,t c[u

1 u

2]0,t

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to obtain

(u1 u2)(t)2 + [u1 u2]20,t c([u

1 u

2]20,t + [u

1 u

2]20,t+ [u1 u2]20,t) ;

the continuity of follows by applying the Gronwall Lemma and from the estimate[u1 u

2]20,T T|||u

1 u

2|||20,T.

We are now ready to give the

Proof of Theorem 2.As c() is concave, the set Sfound in Lemma 2 is convex;therefore, by the Schauder fixed point Theorem, the map admits a fixed point(, v) S; by definition ofSand by Proposition 7, (, v) KT solves (3)-(4)-(9)and satisfies 0< v1 < c() on [0, T] 1.

Assume now that there exists > 0 such that max[0,T]i1

|v1| < ; then

uniqueness follows by arguing as in the proof of Theorem 2 in [12].

Acknowledgement. The authors wish to thank the referee for pointing out theuncorrectness of the proof in the first version of the paper.

References

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[2] H. Beirao da Veiga, On the barotropic motion of compressible perfect fluids, Ann.Sc. Norm. Sup. Pisa 8, 1981, 317-351

[3] H. Beirao da Veiga, Perturbation theorems for linear hyperbolic mixed problems andapplications to the compressible Euler equations, Comm. Pure Appl. Math. 46, 1993,221-259

[4] K.O. Friedrichs,Symmetric positive linear differential equations, Comm. Pure Appl.Math. 11, 1958, 333-418

[5] P. Grisvard, Elliptic problems in nonsmooth domains, Pitman Advanced PublishingProgram, 1985

[6] T. Nishitani, M. Takayama, A characteristic initial boundary value problem for asymmetric positive system, Hokkaido Math. J. 25, 1996, 167-182

[7] T. Nishitani, M. Takayama, Regularity of solutions to characteristic boundary valueproblem for symmetric systems, preprint

[8] J. Rauch, Boundary value problems with nonuniformly characteristic boundary, J.Math. Pures Appl. 73, 1994, 347-353

[9] S. Schochet, The compressible Euler equations in a bounded domain: existence ofsolutions and the incompressible limit, Comm. Math. Phys. 104, 1986, 49-75

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[10] P. Secchi,Well-posedness of characteristic symmetric hyperbolic systems, Arch. Rat.Mech. Anal. 134, 1996, 155-197

[11] P. Secchi, A symmetric positive system with nonuniformly characteristic boundary,Diff. Int. Eq. 11, 1998, 605-621

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