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Nonlinear World 3 (1996), pp. 567587 Nonlinear World

c Walter de Gruyter

Berlin New York 1996

On the initial value problem for a linear model of well-reservoir coupling

Thierry Cazenave

1

and Flavio Dickstein

2

1 Analyse Numerique, URA CNRS 189, Universite Pierre et Marie Curie, 4, place Jussieu,

75252 Paris, Cedex 05, France, [email protected] Instituto de Matematica, Universidade Federal do Rio de Janeiro, Caixa Postal 68530,

21944 Rio de Janeiro, R.J., Brazil, [email protected]

Received June 19, 1995

Abstract. We consider a linearized model of well-reservoir coupling for a monophasic flow,

with boundary conditions corresponding to oil production at either a given pressure or at a

given flow rate. By using the semigroup theory, we show that the initial value problem is

well-posed and we study the asymptotic behavior of the solutions.

AMS clsssification scheme numbers:35Q35, 35B10, 35B35, 76S05

Key words: Flow in porous media, initial value problem, semigroup theory, asymptotic be-

havior of solutions.

1. Introduction

Porous medium flow is usually modelled through Darcys law, which relates

fluid velocity u and pressure p. For a monophasic flow, it is given by the

equation

u= k

(p gz), (1.1)

where k is the permeability tensor, is the fluid viscosity, g is the gravity

constant andz is the vertical coordinate. Similar relation holds for multiphase

flow.

567


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568 Thierry Cazenave and Flavio Dickstein

An important field of application of flow in porous media is in oil recovery. Oil

exploitation is performed through the drilling of wells, which can be considered

as long and thin tubes, into the reservoir. Therefore, oil reservoir models

usually treat the wells as Dirac mesures over an interval length. Moreover, the

well-reservoir coupling is modelled under some quite simplified assumptions,

in particular by considering that the well pressure distribution is hydrostatic.

It is recognized, however, that in many situations, such as those related to

non-vertical wells, these simplifications do not take into account some relevant

physical aspects of the coupled flow. In order to overcome this difficulty, more

complex models have been proposed (see for example [3, 1]). For a monophasic

flow, A. Bourgeat [3] has considered the coupling of a non-linear hyperbolic

system for the well to a non-linear parabolic equation for the reservoir.In this work we consider a linearization of the model presented in [3]. Two

different boundary conditions, oil production at a given flow rate or at a given

pressure, are discussed. For technical reasons, they are treated separately.

However, they both lead to evolution equations of the form

U(t) + AU(t) =F(t),

U(0) =U0,(1.2)

in appropriate functional spaces. In Section 2 we introduce the set of equations

to be solved and we discuss in Section 3 the properties of the operator A.

We show, in particular, that A is m-accretive for the boundary conditions

considered. It follows that, under appropriate regularity assumptions, (1.2) is

well-posed. The asymptotic behavior of (T(t))t0, the semigroup generated by

A, is studied in Section 4. Taking convenient perturbed energy functionals,

we show that T(t) exponentially decays to zero. Finally, in Section 5 we prove

the main results of existence and asymptotic behavior, Theorems 2.1, 2.2, 2.3

and 2.4 below.

2. The coupled model

Let = (0, 1) (0, 1), 0 = {0} [0, 1] and = \ 0. Given q0, w0 :

(0, 1) R,Q0: R and : (0, T) R, we consider the problem of finding


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On the initial value problem for a linear model of well-reservoir coupling 569

q, w: (0, T) (0, 1) R and Q: (0, T) Rsuch that:

qt+ wy+

Q

n |0 = 0 in (0, T) (0, 1),

wt+ qy = 0 in (0, T) (0, 1),

q(0, y) =q0(y), w(0, y) =w0(y) for y (0, 1),

(2.1)

and

Qt Q= 0 in (0, T) ,

Q(t,x,y) =q(t, y) for (t,x,y) (0, T) 0,Q

n(t,x,y) = 0 for (t,x,y) (0, T) ,

Q(0, x , y) =Q0(x, y) for (x, y) .

(2.2)

qand w satisfy one of the following boundary conditions. Either

q(t, 0) =(t) and w(t, 1) = 0, (2.3)

or

w(t, 0) =(t) and w(t, 1) = 0. (2.4)

We can interpret (2.1)-(2.2) as follows. is the reservoir region, 0 the well-

bore, q the well pressure, w the well velocity and Q the reservoir pressure.

Considering the fluid state equation relating density and pressure as (q) =q,

the first equation in (2.1) expresses mass conservation. Then, Q

n |0 is the

well-reservoir mass flow exchange, given by Darcys law (2.1) with k

= 1 and

g = 0. Frequently (see [2, 6]) the diffusion equation has been considered as

a simplified model for the reservoir flow. The second equation in (2.2) is a

continuity condition for the pressure at the wellbore. No-flow is assumed at

the reservoir boundary. We consider production at a given pressure (2.3) or at

a given mass flow rate (2.4).

We show under appropriate assumptions on that the initial value prob-

lems (2.1)-(2.2)-(2.3) and (2.1)-(2.2)-(2.4) are well posed for (q0, w0, Q0)

L2(0, 1) L2(0, 1) L2(). We also establish some regularity properties and

determine the asymptotic behavior of the solutions.We first consider the problem (2.1)-(2.2)-(2.3). In order to get homogeneous

boundary conditions we make the following change of variables.p(t, y) =q(t, y) (t)(y) for (t, y) (0, T) (0, 1),

v(t, y) =w(t, y) for (t, y) (0, T) (0, 1),

P(t,x,y) =Q(t,x,y) (t)(y) for (t,x,y) (0, T) .


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Here, C([0, 1]) is such that (0) = 1 and (1) = (0) = (1) = 0, for

example (y) =1 + cos(y)

2 . System (2.1)-(2.2)-(2.3) is equivalent to

pt+ vy+ P

n|0 =

(t)(y) in (0, T) (0, 1),

vt+py =(t)(y) in (0, T) (0, 1),

p(0, y) =p0(y), v(0, y) =v0(y) for y (0, 1),

p(t, 0) =v(t, 1) = 0 for t (0, T),

(2.5)

and

Pt P =(t)(y) (t)(y) in (0, T) ,

P(t,x,y) =p(t, y) for (t,x,y) (0, T) 0,

Pn

(t,x,y) = 0 for (t,x,y) (0, T) ,

P(0, x , y) =P0(x, y) for (x, y) ,

(2.6)

where p0(y) =q0(y) (0)(y),

v0(y) =w0(y),

P0(x, y) =Q0(x, y) (0)(y).

We write the system (2.5)-(2.6) in the form

dU

dt + AU=F(t),U(0) =U0,

(2.7)

with

U0= (p0, v0, P0).

Here, the operator Ais defined on the Hilbert space

H=L2(0, 1) L2(0, 1) L2(),

by

D(A) = (p, v, P) H1(0, 1)2 H1(); p(0) =v(1) = 0,

P L2(),P

n|= 0, P|0 =p

(2.8)


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and

A(p, v, P) = (vy+ P

n |0

, py, P), (2.9)

for (p, v, P) D(A). F(t) : (0, T) His defined by

F(t) = ((t)(y), (t)(y), (t)(y) (t)(y)). (2.10)

We will show in the next section that Ais well defined and that Aism-accretive

inH, so that Agenerates a semigroup of contractions in Hwhich we denote

by (T(t))t0. Therefore, by Duhamels formula, the solution Uof (2.7) is given

by

U(t) =T(t)U0+

t

0T(t s)F(s) ds. (2.11)

Conversely,Ugiven by (2.11) is called the weak solution of (2.7). We have the

following results.

Theorem 2.1. Let A be defined by (2.8)-(2.9), let T > 0 and let

W1,1(0, T).

(i) Given U0 = (p0, v0, P0) H, there exists a unique weak solution U =

(p, v, P) C([0, T], H) of (2.7). In addition, PL2((0, T), H1()).

(ii) If W2,1(0, T) and U0 D(A), then U C([0, T], D(A))

W1,1((0, T), H) and U solves the equation (2.7) for almost all t

(0, T). If furthermore C

2

([0, T]), thenUC

1

([0, T], H).

Theorem 2.2. Assume W1,1loc(0, ).

(i) Ifsupt0

t+1t

(|(s)|+|(s)|) ds


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System (2.1)-(2.2)-(2.4) is then equivalent to

pt+ vy+

P

n |0 =(t) in (0, T) (0, 1),

vt+py = (1 y)(t) in (0, T) (0, 1),

p(0, y) =p0(y) v(0, y) =v0(y) for y (0, 1),

v(t, 0) =v(t, 1) = 0 for t (0, T),

(2.12)

and

Pt P = 0 in (0, T) ,

P(t,x,y) =p(t, y) for (t,x,y) (0, T) 0,P

n(t,x,y) = 0 for (t,x,y) (0, T) ,

P(0, x , y) =P0(x, y) for (x, y) ,

(2.13)

where p0(y) =q0(y),

v0(y) =w0(y) (1 y)(0),

P0(x, y) =Q0(x, y).

System (2.12)-(2.13) can be written as (2.7), where now the operator A is

defined on the Hilbert space H by

D(A) =

(p, v, P) H; p H1(0, 1), v H10 (0, 1),

P L2(), Pn

| = 0, P|0 =p, (2.14)and

A(p, v, P) =

vx+ P

n |0, px, P

, (2.15)

for (p, v, P) D(A). F(t) : (0, T) His defined by

F(t) = ((t), (1 y)(t), 0). (2.16)

We have the following results.

Theorem 2.3. Let A be defined by (2.14)-(2.15), let T > 0 and let

W

1,1

(0, T).(i) Given U0 = (p0, v0, P0) H, there exists a unique weak solution U =

(p, v, P) C([0, T], H) of (2.7). In addition, PL2((0, T), H1()).


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(ii) If W2,1(0, T) and U0 D(A), then U C([0, T], D(A))

W1,1((0, T), H) and U solves the equation (2.7) for almost all t

(0, T). If furthermore C2([0, T]), thenUC1([0, T], H).

Theorem 2.4. Assume W1,1loc(0, ) and that t0

(s) ds,

is bounded as t .

(i) Ifsupt0

t+1t

(|(s)|+|(s)|) ds


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Step 1. Existence. We observe that if v H1(), then v L2((0, 1),

H1(0, 1)) H1((0, 1), L2(0, 1)). Therefore, v C([0, 1], L2(0, 1)), so that v|0is well defined, v|0 L

2(0, 1). It follows that

V ={v H1(); v|0 = 0},

is a closed subspace of H1(). Furthermore, vV = vL2() is a norm

on Vwhich is equivalent to the H1 norm, as follows easily by integrating the

identity

v(x, y) =

x0

v

x(s, y) dy.

As usual, we write the equation (3.1) in a weak form, i.e. we say that u H1()

is a solution of (3.1) ifu|0 =p and

u v=

f v, for allv V. (3.3)

Setting

u(x, y) =u(x, y) +p(y),solving (3.1) is then equivalent to solving the following problemu V,

u v=

f v

pyvy, for allv V.

Existence and uniqueness for this last problem follows immediately from Lax-

Milgrams lemma, since the right hand side is clearly a linear continuous func-tional ofv V. Continuity of the mapping (f, p)u fromL2()H1(0, 1)

H1() follows.

Step 2. Properties of u

n|0 in the case f = 0. Suppose that f = 0, and

consider first the case p(y) = cos(my) for some integer m 0. Then one

verifies by a direct calculation that

u(x, y) =cosh(m(x 1))

cosh(m) cos(my);

and sou

n |0(y) =m tanh(m)cos(my).


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Now if

p(y) =

m=0 amcos(my),we obtain

u(x, y) =

m=0

amcosh(m(x 1))

cosh(m) cos(my);

and so

u

n |0(y) =

m=0

mamtanh(m) cos(my).

Elementary calculations yield

py

2

L2

=

1

2

m=1 m22a2m,and u

n

2L2(0)

=1

2

m=1

m22a2m(tanh(m))2,

from which we deduce un

L2(0)

py2L2 p

2H1 .

Since (cos(my))m0 is the set of eigenvectors ofd2

dy2 in H1(0, 1) with Neu-

mann boundary conditions, it is a complete orthonormal system of H1(0, 1),

and the result follows for general p H1(0, 1). (Note that when f= 0 we have

u C(\0). This can be seen either on the Fourier series expansion, or else

by extending u to a harmonic functionuon the strip (0, 2) R by symmetry,see Step 3.)

Step 3. Properties of u

n |0 in the case p = 0. Suppose that p = 0, and

define the functionuon R2 as follows.

u(x, y) =u(x, y) for (x, y) ,u(x, y) =

u(x, y) for (x, y) ,

u(x, y) =u(x, y) for (x, y) ,u(x + 2m, y+ 2) = (1)mu(x, y) for (x, y) R2 and m, Z.


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Define

f by

f(x, y) =f(x, y) for (x, y) ,f(x, y) = f(x, y) for (x, y) ,f(x, y) =f(x, y) for (x, y) ,f(x + 2m, y+ 2) = (1)m f(x, y) for (x, y) R2 and m, Z.

It follows easily thatu H1loc(R2),fL2loc(R2), and thatu=f inD (R2).

Therefore,

u H2loc(R

2). It follows in particular that u H2(), from which

we deduce (see Step 1) u

n

|0 L2(0) and

u

n L2(0) CfL2().Step 4. Conclusion. The properties of

u

n|0 in the general case p

H1(0, 1), f L2() follow from Steps 2 and 3. Greens formula (3.2) is im-

mediate in the case p(y) =

m=0

amcos(my), 0 such that AUH (pH1(0,1) +vH1(0,1) +PH1()) for allU= (p, v, P) D(A).


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(iv) A is closed. In particular, D(A) equipped with the norm U2D(A) =

U2H+ AU2H is a Hilbert space. AUH is an equivalent norm on

D(A). Moreover, the embeddingD(A)H is compact.

(v) A ism-accretive.

Proof. (i) We have

(AU,U)H=

10

(pvy+ pyv) +

10

pP

n |0

PP

=

10

pP

n|0

PP.

On the other hand, (3.2) gives

PP =

|P|2 10

p Pn |0

,

hence property (i).

(ii) It is clear from (i) that A 0. Let now ( ,,f ) H. Set

p(y) =

y0

(s) ds for y (0, 1).

We have p H1(0, 1) and p(0) = 0. Let PH1() be the solution of

P =f in ,P

n

= 0 in ,

P =p in 0,

given by Lemma 3.1. Since P

n|0 L

2(0, 1), if

v(y) =

1y

P

n|0+

1y

,

then v H1(0, 1) and v(1) = 0. Therefore U = (p, v, P) D(A), AU =

(,,f) and so, R(A) =H.

(iii) Let U = (p, v, P) D(A). We have

AU2H= 10vy+ Pn |02 + 10 p2y+ |P|2.


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Since p(0) = 0, we have

pH1 CpyL2 ,

for some C independent ofU. Next, it follows from Lemma 3.1 that

PH1()+P

n |0L2(0)

C(PL2()+ pH1(0,1)).

Finally, since v(1) = 0, we have

vH1 CvyL2 CP

n|0

L2(0)

+ Cvy+ P

n|0

L2(0)

.

Combining the above inequalities we obtain (iii).

(iv) It follows from (iii) that AUH UH. Since R(A) = H, the

closedness of A follows immediately. Next, the mapping U (AU,U) is an

isometryD(A) H H, whose image is G(A). SinceG(A) is closed, G(A) isa Hilbert space, and so is D(A). Compactness of the embeddingD(A) H

follows from (iii) and the compactness of the embeddings H1(0, 1)L2(0, 1)

and H1()L2().

(v) SinceA 0, we need only show that R(I+A) =H. LetV R(I+A).

We have

(V, U+ AU)H = 0 for all UD(A).

LetWD(A) be such that V =AW. ChoosingU=Win the above identity,

we obtain

0 = (AW, W+ AW)H= (AW, W)H+ AW2

H AW2

HW2

H;and so W= 0, thus V = 0. Therefore R(I+ A) =H, and we need only show

that R(I+A) is closed. Suppose Zn n

Z in H, where Zn = Un+AUn.

ThenZn is a Cauchy sequence in H. Since A is accretive, Un is also a Cauchy

sequence. Therefore,Un UH. Hence,AUn Z U. By the closedness

ofA we deduce that Z=U+ AU, which proves the closedness ofR(I+ A).

It follows from Lemma 3.3 that A generates a semigroup of contractions

in H, which we denote by (T(t))t0.

We now consider the operator A on H defined by (2.14)(2.15). It follows

easily from Lemma 3.1 that A is well defined D(A) H. On the other hand,Ahas a nontrivial nullspace. More precisely, it can be easily seen that

N(A) ={U= (p, v, P) H; c R, p c and Pc}.


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In other words, the nullspace of A is the (one-dimensional) subspace of H

spanned by the vector (1(0,1), 0, 1). It is convenient to introduce the restriction

ofAto N(A). Therefore, we consider the Hilbert space

H=N(A) = (p, v, P) L2(0, 1)2 L2(); 10

p +

P = 0

,

and we define the operatorAonH byD( A) = (p, v, P)H; p H1(0, 1), v H10 (0, 1),

PL2(), P

n|= 0, P|0 =p

, (3.5)

and

A(p, v, P) = (vx+ Pn |0 , px, P), (3.6)for (p, v, P) D( A). Clearly, if (p, v, P) D( A), then 1

0

vy+

P

n|0

+

(P) =

10

P

n|0+

(P) = 0,

where the last identity follows from (3.2) applied with u = P and v = 1.

Therefore,AU H; and so,A is an operator onH. We describe in thefollowing lemma the main properties ofA.Lemma 3.4. If

Ais defined by (3.5)(3.6), then the following properties hold.

(i) ( AU,U) eH= |P|2, for allUD( A).(ii)A 0 andR( A) =H.

(iii) There exists > 0 such that AU eH (pH1(0,1) +vH1(0,1) +PH1()) for allUD( A).

(iv)A is closed. In particular, D( A) equipped with the norm U2D(eA)

=

U2eH

+ AU2eH

is a Hilbert space. AU eH is an equivalent norm onD( A). Moreover, the embeddingD( A)H is compact.

(v)

A ism-accretive.

(vi) There exists a constant >0 such that

( AU,U) eH P2L2()+ p2L2(0,1) ,


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for allUD(

A).

Proof. The proof of property (i) is the same as the proof of Lemma 3.3 (i).We next prove (vi), which we will use in the proof of (iii). We argue by

contradiction and we assume that there exists (Un)n0 D( A) such thatpnL2(0,1)+ PnL2()= 1, and

|Pn|2 n

0.

By the compact embedding H1() L2(), there exists a subsequence

(nk)k0 and a constant C such that Pnk k

C in H1(). It follows that

pnk = Pnk |0 k

C in L2(0, 1). Since (Un)n0 H, we deduce that C = 0;and so pnkL2(0,1)+ PnkL2()

k0, which is absurd.

The proof of the properties (ii) to (v) is almost the same as the proof ofLemma 3.3. We only indicate the modifications. To show that R( A) =H, let(,,f) H. Using Lemma 3.3, we find U = (p, v, P) H1(0, 1)2 H1()such that

vy+P

n0 =,

py = ,

P =f,

with the boundary conditions

p(0) =v(1) = 0,Pn |

= 0 and P|0 =p.

But (,,f)H impliesv(0) =

10

vy =

10

0

P

n =

10

+

f= 0,

that is, v H10 (0, 1). We now consider

c=1

2

10

p +1

2

10

P,

and we setU = Uc(1(0,1), 0, 1). Then, it is clear thatU D( A) andAU= (,,f).


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The proof of (iii) is modified as follows. We have

PH1()+ Pn |0L2(0) C(PL2()+ pH1) C( AU eH+ pL2).Since v H10 (0, 1), we have

vH1 CvyL2 CP

n|0

L2(0)

+ C vy+ P

n0

L2(0)

C( AU eH+ pL2).Therefore,

pH1(0,1)+ vH1(0,1)+ PH1() C( AU eH+ pL2).The desired estimate follows by applying estimate (vi) and Cauchy-Schwarz

inequality.

Corollary 3.5. Consider the operator A on H defined by (2.14)(2.15) and

the operatorA onH defined by (3.5)-(3.6).(i) A ism-accretive and(AU,U)H=

|P|2.

(ii) If(T(t))t0 is the semigroup of contractions inHgenerated byAand

(T(t))t0 is the semigroup of contractions inHgenerated by A, thenfor everyU0= (p0, v0, P0) H, we have

T(t)U0= c +

T(t)

U0,

where c = 12 1

0p0+

P0, andU0 H is defined byU0 = (p0

c, v0, P0 c).

Proof. Property (i) follows from Lemma 3.4 and the propertyH = N(A).Property (ii) is immediate, sinceU0 is the orthogonal projection of U0 onH. 4. Asymptotic behavior

We consider first the case where the wellhead pressure is known, that is, letAbe given by (3.5)-(3.6) and let ( T(t))t0 be the semigroup of contractions inH generated by A. We defineK:HH byK(U) = A1U, U,


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where the scalar product , is defined by

U1, U2= 10

p1p2+ 2 10

v1v2+

P1P2,

ifUj = (pj, vj, Pj). We have the following results.

Lemma 4.1. There exists c0 > 0 such that ifU0 D( A) andU =T(t)U0,then

d

dtK(U) c0 1

0p2 1

2

10v2

P2, (4.1)for all t 0.

Proof. Indeed,

ddtK(U(t)) = A1 dU

dt,U + A1 U ,d U

dt = A1 AU ,U A1 U ,AU

= U ,U A1 U ,AU.Set (p, v, P) =A1 U. We have

vy+P

n|0 =p,

py = v,P =P .

(4.2)

Next,

A1 U ,AU= 10

pvy+ P

n|0

+ 2

10

vpy

P P=

10

pyv 2 10

vyp

PP+ 10

P

n |0p

=

10v2 2 1

0p2 + 3 1

0p P

n|0 +

P2.Hence,

d

dt

K(

U(t)) =

10

p2

10

v2 2

P2 3

10

pP

n|0 . (4.3)


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On the other hand, it follows from (3.4) that

Pn |0L2 C(PL2+ pyL2) =C( PL2 + vL2). (4.4)(4.1) now follows from (4.3), (4.4) and Cauchy-Schwarz inequality.

Remark 4.2. Let A be the operator in H defined by (2.8)-(2.9), and let

(T(t))t0 be the semigroup generated by A. Let K(U) = A1U, U for

UD(A), where, is as above, but in Hinstead ofH. The same argumentas in Lemma 4.1 shows that ifU0 D(A) and U=T(t)U0, then

d

dtK(U(t)) c0

10

p2 1

2

10

v2

P2,

for all t 0, where c0 is the same constant as in (4.1).Lemma 4.3. IfT(t))t0 is the semigroup generated by A, then there exists >0 such that

T(t)L(eH)

C et,

for all t 0.

Proof. LetU0 D( A), and setU(t) = (p(t), v(t),P(t)) =T(t)U0. Then,d

dtU2eH= 2(Ut,U) eH=2( AU ,U) eH=2

| P|2. (4.5)

For >0 small enough

|U|2 = U2eH

+ K(U),defines an equivalent norm onU, sinceA1 L( H). It follows from (4.1)and (4.5) that

d

dt|U|2 2

| P|2

2

10v2

P2 + c0 10p2.

Applying properties (i) and (vi) of Lemma 3.4, we obtain by choosing small

enough,

d

dt

|U|2 | P|2

2 1

0 v2 1U2eH 2|U|2,


18/21


where1 and 2 are positive constants independent of

U0. Therefore,

|U(t)| e2t|U0|. (4.6)Since 2 is independent ofU0, (4.6) holds for everyU0 H, and the resultfollows.

We now consider again the operator A given by (2.8)-(2.9). For the study

of the asymptotic behavior of the semigroup (T(t))t0 generated by A, we

will use the following lemma.

Lemma 4.4. For any >0, there existsC such that

P2L2()+ p2L2(0,1) C

|P|2 +

|P|2

+ v2L2(0,1), (4.7)

for everyU= (p, v, P) D(A). Here, U = (p, v, P) =A1(U).

Proof. Let >0. If (4.7) does not hold, then there exists a sequence (Un)n0

D(A) such that

Pn2L2()+ pn

2L2(0,1)= 1, (4.8)

||vn||2L2(0,1)

1

, (4.9)

Pn2L2 + Pn

2L2 n

0. (4.10)

It follows from (4.8) and (4.10) (cf. the proof of Lemma 3.4 (vi)) that there

exists a constant C1 such that Pn C1 in H1() and pn C1 in L2(0, 1).

Moreover, sinceA1 is compact, Pn has a limit in L2(). By (4.10), this limit

is a constant C2, and Pn C2 in H1(). Therefore,Pn 0 in H

1();

and so C1= 0, since Pn= Pn. This contradicts (4.8).

Lemma 4.5. If(T(t))t0 is the semigroup generated byA, then there exists

>0 such that

T(t)L(H) C et,

for all t 0.

Proof. With the notation of Remark 4.2, let > 0 be small enough so that|U|= (U2H+K(U))

1

2 be equivalent to . Let U0 D(A) and U(t) =

T(t)U0. It follows from Remark 4.2 (see the proof of Lemma 4.3) that

d

dt|U|2 2

|P|2

2

10

v2

P2 + c0

10

p2.


19/21


Let now U= (p, v, P) =A1U. In particular, U(t) =T(t)U(0); and so

d

dt U2

=2(AU,U)H=2 |P|2.We define the norm |U|1 =

|U|2 + A1U2H

1/2, which is equivalent to

the | | norm. It follows from the two above relations that

d

dt|U|21+ 2P

2L2+ 2P

2L2+

2v2L2+ P

2L2 c0p

2L2 0. (4.11)

On the other hand, taking = 1

8c0in (4.7), we find a constant C such that

2c0p2L2

1

4v2L2 C(P

2L2 + P

2L2). (4.12)

On multiplying (4.12) by and summing up with (4.11), we obtain

d

dt|U|21+ P

2L2+ 2P

2L2+

4v2L2+ P

2L2+ c0p

2L2

C(P2L2 + P2L2).

Now, if we choose small enough so that C1, we find >0 such that

d

dt|U|21+ |U|

21 0,

from which we obtain the desired result.

5. Proofs of Theorems 2.1, 2.2, 2.3 and 2.4

Theorems 2.1 and 2.3 are immediate consequences of the results of Section 3

and the semigroup theory (see Pazy [5], Chapter 4, Corollaries 2.2, 2.10

and 2.5), except for the property P L2((0, T), H1()) for U0 H. This

last property is a consequence of the identity (AU,U)H = U2L2 . More

precisely, we have the following result.

Lemma 5.1. Let A be the operator inH defined by (2.8)-(2.9) (respectively,

by (2.14)-(2.15)) and let(T(t))t0be the semigroup of contractions inHgener-

ated byA. Let W1,1

(0, T)and letF(t)be defined by (2.10) (respectively,by (2.16)). IfU0 = (p0, v0, P0) H andU= (p, v, P) is given by (2.11), then


20/21


PL2((0, T), L2()) and

U(t)2H+ t0

|P(t,x,y)|2 dx dy dt CU02H+ ||||2W1,1(0,t) , (5.1)

for all t [0, T]. Here, Cis a constant independent ofT, andU0.

Proof. By density and continuous dependence, we need only show (5.1) when

C2([0, T]) andU0 D(A). In this case,UC([0, T], D(A))C1([0, T], H)

is the solution of (2.7). Taking the scalar product of the equation with U and

applying property (i) of Lemma 3.3 (respectivey, property (i) of Corollary 3.5),

we find1

2

d

dtU(t)2H+

|P|2 = (F(t), U(t))H.

Since

(F(t), U(t))H F(t)HU(t)HC(|(t)| + |(t)|) ||U(t)||H,

it follows that

1

2

d

dt

U(t)2H+ 2

t0

|P|2

C(|(t)| + |(t)|) ||U(t)||H.

The above differential inequality yields||U(t)||2H+ 2

t0

|P(t,x,y)|2 dx dy dt

12

||U0||H+ C||||W1,1(0,t),

from which (5.1) follows.

Theorem 2.2 follows from the exponential decay of the semigroup (T(t))t0

(see Section 4) and the results of Haraux [4] (see Theorem 7 p. 154).

For proving Theorem 2.4, we write U(t) =U(t) + V(t), where V(t) =(a(t)1(0,1), 0, a(t)1) with

a(t) = 1

2

10

p0+

P0+

t0

(s)

.

Note that is -periodic and that

t0

(s) dsis bounded, so that has mean

value 0. Therefore V is -periodic. Furthermore, we have

U(t) =T(t)U0+ t0

T(t s) F(s) ds,


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where

U0= (p0 c, v0, P0 c) with

c=1

2 10 p0+ P0 ,so thatU0H, andF(t) = 1

2((t), (1 y)(t), (t))H.

Therefore, the result follows from the exponential decay of the semigroup

(T(t))t0 (see Section 4) and the same argument as above.Acknowledgments.We thank A. Haraux for his helpful suggestions concerning this

work. Part of this work was done while T. Cazenave was visiting the Universidade

Federal do Rio de Janeiro; he thanks the Instituto de Matematica for its invitationand hospitality.

References

[1] J. Alvestad, K. Holing, K. Christoffersen, O. Langeland, O. Stava, Interactive

Modelling of multiphase inflow performance of horizontal and highly deviated wells, in

European Petroleum Computer Conference, Aberdeen, 1994.

[2] K. Aziz, A. Settari, Petroleum Reservoir Simulation, Applied Science Publishers,

London, 1979.

[3] A. Bourgeat,Simulating gas-liquid flow in a well-reservoir system,Numerical Methods

in Engeneering and Applied Sciences, H. Alder, J.C. Heinrich, S. Lavanchy, E. Onateand B. Suarez (eds.), CIMNE, Barcelona, 1992.

[4] A. Haraux, Nonlinear evolution equations: Global behavior of solutions, Lecture Notes

in Math. 841, Springer, New York, 1981.

[5] A. Pazy, Semi-groups of linear operators and applications to partial differential equa-

tions, Applied Math. Sciences 44, Springer, New-York, 1983.

[6] D.W. Peaceman, Fundamentals of Numerical Reservoir Simulation, Elsevier Science

Publishers, Amsterdam, 1977.

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