https www.ljll.math.upmc.fr ~cazenave 52
TRANSCRIPT
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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.
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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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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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On the initial value problem for a linear model of well-reservoir coupling 575
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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580 Thierry Cazenave and Flavio Dickstein
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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582 Thierry Cazenave and Flavio Dickstein
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,
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584 Thierry Cazenave and Flavio Dickstein
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.
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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
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586 Thierry Cazenave and Flavio Dickstein
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.
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