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Analysis of a nonlinear fluid–structure interaction problem in velocity–displacement formulation
Nonlinear Analysis 35 (1999) 561 – 587
Analysis of a nonlinear uid–structure interaction problem in velocity–displacement formulation Fabien Flori∗ , Pierre Orenga Centre de Mathematique et de Calcul Scienti que, URA 2053, Universite de Corse, Quartier Grossetti, 20250 Corte, France Received 15 March 1997; accepted 20 April 1997
Keywords: Compressible Navier–Stokes equations; Fluid–structure interaction; Problem velocity–displacement formulation
in
0. Introduction In the present study, we propose an existence and uniqueness theorem for the solution of a bidimensional nonlinear problem of uid–structure coupling as well a method of numerical resolution. If u; w and p represent the displacement within the structure, the velocity and the pressure in the uid, respectively, the problem (P) to be studied is the following: Fluid equations:
(F)
∗
wt + 12 ∇w2 + curl w ∧ w − w + ∇p = 0 pt + div w = 0 in Qf curl w = 0 on ×]0; T [ w · n = 0 on f ×]0; T [ w(t = 0) = w0 in f p(t = 0) = p0 in f
Corresponding author.
0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 7 1 7 - 7
in Qf
(I)
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Plate equations: 2 utt − utt + u + p − div w = f (S) u = ∇u · n = 0 on p ×]0; T [ u(t = 0) = u0 ; ut (t = 0) = u1 in p Coupling condition: @u (C) − =w · n @t
in Qp (II)
(III)
in Qp
with Qp = p ×]0; T [; Qf = f ×]0; T [ where p is an open domain of R and f is an open domain of R2 simply connected with boundary = f ∪ p of class C 1; 1 . We classically denote by v the viscosity of the uid. In a previous study we determined the existence conditions of the linearized problem (P). This last study was carried out in order to point out the problems inherent to the coupling which arise in particular from the meaning given to the trace of p − div w in the plate equation. Here, the nonlinearity of the momentum equation accentuates these diculties and requires the introduction of smoothness results. We show that problem (P) admits a xed point such that w · n=−
@u ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )): @t
Sections 1–5 are entirely devoted to the proof of the theorem and in Section 6 we present the implementation of the numerical method. 1. An existence and uniqueness result Let us begin by providing several elements necessary to state the theorem. We consider the following functional space W = {w ∈ H 1 ( f )2 ; w · n = 0 on
f}
which we can decompose in the following manner [15]: W = V + W˜ with V = {v = grad h; h ∈ H 2 ( f ); grad h · n = 0 on
} ⊕ Rot(H 2 ( f ) ∩ H01 ( f ))
and ˜ grad h˜ · n = 0 on ˜ h˜ ∈ H 2 ( f ); h˜ = h; W˜ = {w˜ = grad h;
f }:
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563
Thus, all functions of W are represented in the following unique manner: ˜ w = v + w˜ = vq + vh + w˜ = Curl q + grad h + grad h; where q; h are the solutions to the following scalar problems: q = curl vq in f ; (Q) q=0 on and
( (H)
h = div vh grad h · n = 0
in f ; on
:
In order to mute the problem (F) homogeneous, we then introduce the function hg such that hg; t − hg = 0 in Qf ; grad hg · n = 0 on f ×]0; T [; (R) grad hg · n = g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) on Qp ; hg (0) = 0 in f : We easily establish that wg = grad hg ∈ W 1; 4 (0; T ; W 2; 4 ( f )2 ); div wg ∈ L4 (0; T ; L∞ ( f )); wg · n = g
on Qp ;
curl wg = 0
in Qf :
We note by CS , the imbedding constant of H 1 in L2 , and by CG , Gagliardo– Nirenberg’s optimal constant of inequality kvk2L4 ≤ CG kvkL2 kvkH 1 . Thus, if we consider the real numbers ∈]0; 1[; ¿ 12 CG CS 1 ¿0, 2 ¡ 12 , and the functions f ∈ H 1 (0; T ; H 1 ( p ));
(1.1)
g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ));
(1.2)
w0 ∈ H 2 ( f )2 ;
(1.3)
p0 ∈ H 2 ( f );
(1.4)
u0 ∈ H 4 ∩ H02 ( p );
(1.5)
u1 ∈ H 3 ∩ H02 ( p );
(1.6)
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we assume kv0 k2f + kp0 k2f +
CG CS 2 2 A2 kwg k2L4 (0; T ; H 1 ( f )3 ) + khg; t k2L1 (0; t1 ; L2 ( p )) ¡ 4 2 1 2 CG
(1.7)
with A= −
CG CS 1 − 2CS CG kwg (t)kL∞ (0; T ; L2 ( f )2 ) ¿0; 2
(1.8)
we obtain the existence and uniqueness result shown below. Theorem 1.1. Under the conditions (1.1)–(1.8), the trace of p − div w is de ned in L2 (0; T ; H 3=2 ( )) and the problem (P) admits a unique xed point (w; u) such that w · n = −ut ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )):
(1.9)
The proof of this theorem is made possible using Kakutani’s xed point theorem. To this end, we prescribe w · n = g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) and we solve problems (F) and (S). The main diculty is to give a meaning to the trace of p − div w on p . We will see that problem (P) admits a xed point such that w · n = −ut ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )) and with this regularity the solution (w; p) of (F) is such that
(p − div w) ∈ L2 (0; T ; H 3=2 ( )):
2. Study of problem (F) By using the orthogonal decomposition introduced in the previous section, Eq. (I) are transformed in order to obtain a homogeneous problem denoted as (Fh ). By setting w = v + wg , we obtain the following problem (Fh ) vt + 12 ∇w 2 + curl v (v + wg ) − v + ∇p = 0 pt + div v = −div wg in Qf ; v · n = 0 on ; curl v = 0 on ; v(t = 0) = w0 − wg (t = 0) = v0 in f ; p(t = 0) = p0 in f :
in Qf ;
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The problem (Fh ) is then solved using the following weak formulation (VFh ): Find (v; p) ∈ L∞ (0; T ; L 2 ( f ) 2 ) ∩ L 2 (0; T ; V ) × L∞ (0; T ; L 2 ( f )) such that (vt ; ’)f − 12 (v 2 ; div ’)f − (vwg ; div ’)f + (curl v (v); ’)f +(rot v (wg ); ’)f + a(v; ’) − (p; div ’)f = 12 (wg2 ; div ’)f v(t = 0) = v0 ; ( (pt ; )f + (div v; )f = −(div wg ; )f ∀ ∈ L2 ( f ) p(t = 0) = p ; 0
∀’ ∈ V
where a(v; ’) is the bilinear form Z
Z a(v; ’) =
f
div v div ’ +
f
curl v curl ’:
If v0 ∈ L2 ( f ) 2 and p0 ∈ L 2 ( f ) then we obtain Theorem 2.1. If Eqs. (1.7) and (1.8) hold and if g ∈ H 1 (0; T ; H 1=2 ( p )); then the problem (VFh ) admits a unique solution such that (v; p) ∈ L2 (0; T ; V ) ∩ L 2 (0; T ; L∞ ( f ) 2 ) × L∞ (0; T ; L∞ ( f )); in addition; the solution is such that kv(t)kL∞ (0; T ; L 2 ( f ) 2 ) ≤ 2 kvkL22 (0; T ; V ) ≤ 2
A ; CG
2 A ; 1 − CG2
(2.1) kp(t)kL2∞ (0; T ; L 2 ( f ) 3 ) ≤
4 2 A : − 22 )
CG2 (1
(2.2)
For the proof of this theorem, we refer to the results developed in [12]. It should be noted that the main diculty in obtaining an estimate is in the estimation of the following term 1 2 (v ; div v)f ; 2 which is not necessarily bounded. To overcome this diculty we must build a stability space. These results show that w = v + wg ∈ L 2 (0; T ; W ) and that p ∈ L∞ (0; T ; L 2 ( f )) and with this regularity we are unable to give a meaning to the trace of p − div w in the plate equation. We therefore introduce a series of smoothness results in Section 3 which will allow us, based on several additional assumptions, to prove that (p − div w) has a meaning in L 2 (0; T ; H 3=2 ( )). This result is sucient to solve the plate equation.
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3. Smoothness of the solutions of (F) 3.1. Preliminary results The results presented in this paragraph are demonstrated in detail in [3]. We search w = vh + vq + wg under the approximated forms [15] w=
N X
Aj (t) Curl qj +
j=1
N X
Bi (t)∇hi +
i=1
M X
Cl (t)∇h˜ l
l=1
in which hi and qj are solutions of the following spectral problems: ( ( −qj = j qj in f −hi = i hi in f and ∇hi ·n = 0 on qj = 0 on and where ˜ (H)
˜ ˜ hl = hl in Qf grad h˜ l · n = 0 on f ×]0; T [ grad h˜ l · n = l on f ×]0; T [
2 with { l }∞ l=1 a basis of H0 ( p ) associated to the motion of the plate. In addition, as the continuity equation must be veri ed, if we set
Bi (t) =
dbi (t) dt
and
Cl (t) =
dcl (t) ; dt
then we look for the pressure as p=
N X
i bi (t)hi −
i=1
M X
cl (t)h˜ l
l=1
if bi (0) and cl (0) are chosen such that p(0) =
N X i=1
i bi (0)hi −
M X
cl (0)h˜ l :
l=1
The projection for the scalar product of L 2 ( f ) 2 in the momentum equation on the eld of rotational vectors allows us to write vq; t − Rot rot vq + P(rot vq (w)) = 0; where P is the projection operator on the subspace of the rotational eld vectors. In the same way, the projection onto the eld of gradient vectors leads to ! vq2 (h + hg ) 2 ∇ ht − h + + vq ∇(h + hg ) + + + p = 0; 2 2
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
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that is to say, if we set ˜ = (h + hg ), ht − h +
vq2 ˜2 + vq ∇˜ + + + p= 2 2
(3.1)
with such that ∇ represents the projection, for the scalar product L 2 ( f ) 2 , of curl vq (w) on the eld gradient vectors and a function which is only dependent on time. We will now give some results dealing with the regularity of vq and . Lemma 3.1. If we assume that w0 ∈ H 1 ( f ) 2 ; then vq ∈ L 4 (0; T ; W 1; 4 ( f ) 2 ) ∩ L∞ (0; T ; H 1 ( f ) 2 );
(3.2)
vq; t ∈ L 2 (Qf ):
(3.3)
If; in addition w0 ∈ H 2 ( f ) 2 and wh; t = ∇˜t ∈ L 2 (Qf ); we obtain vq; t ∈ L∞ (0; T ; L 2 ( f ) 2 );
(3.4)
curl vq; t ∈ L 2 (Qf );
(3.5)
vq ∈ W 1; 4 (Qf ):
(3.6)
And nally; if wh = L 2 (0; T ; W 1; 4 ( f ) 2 ); then vq ∈ L 4 (0; T ; W 2; 4 ( f ) 2 );
(3.7)
curl vq ∈ L 4 (0; T ; L∞ ( f ) 2 ):
(3.8)
Lemma 3.2. Given that represents the function de ned above; then veri es ∈ L 4 (Qf );
(3.9)
k∇ kL 2 ( f ) ≤ C(1 + kvq; t kL 2 ( f ) 2 )
a.e. on [0; T ]:
(3.10)
3.2. Smoothness of w and P The diculty in the following estimates is determining the proper smoothness for g. Indeed, this smoothness must permit the obtention of estimates which allow a meaning to be given to the trace of p − div w. We present here only the obtention of a priori estimates as the passage to the limit is of no great diculty. Let us introduce a rst lemma on the smoothness of . Lemma 3.3. The function de ned above satis es ∈ L∞ (0; T ) t mes( f ) =
Z
f
wwt −
1
(3.11)
Z
f
hg; t :
(3.12)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Proof. By integrating Eq. (3.1) with respect to f , we obtain Z Z Z Z Z @ 1 h− h + w2 + + p: (t)· mes( f ) = @t f 2 f
f
f
f First of all, it should be noted that Z h = 0:
f
In addition, as h and can be expressed as the sum of a scalar function and the sum of a constant, we can always choose a constant such that Z Z h= = 0:
f
f
This yields mes ( f ) =
1 2
Z
f
w2 +
Z p:
f
(3.13)
In this way, as p∈L∞ (0; T ; L2 ( f )) and w2 ∈L∞ (0; T ; L1 ( f )), we obtain Eq. (3.11). The result (3.12) is then obtained by di erentiating Eq. (3.13) with respect to time and by observing that Z Z Z 1 Pt = − hg = − hg; t : v f
f
f Lemma 3.4. If p0 ∈ H 1 ( f ); w0 ∈ H 2 ( f )2 then ˜t ∈ L4 (Qf );
(3.14)
p ∈ L4 (Qf );
(3.15)
and if; in addition; we consider that g ∈ H 1 (0; T ; W 1=2; 4 ( p )) we obtain div wh = ˜ ∈ L4 (Qf ):
(3.16)
Proof. Let us begin by proving Eq. (3.14). Di erentiating Eq. (3.1) with respect to time, multiplying it by 2ht and integrating it with respect to Qf , we obtain kht (t)k2L∞ (0; T ; L2 ( f )) + 2vk∇ht k2L2 (0; T ; L2 ( f )2 ) +
Z tZ 0
f
wwt ht + 2
Z tZ 0
f
t ht + 2
Z tZ 0
f
pt ht = kht (0)k2f + 2
Z tZ 0
f
t ht : (3.17)
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569
The di erent terms are estimated in the following manner: Z tZ Z tZ wwt ht = w(∇ht + ∇hg; t + vq; t )ht
f
f 0 0 Z Z 3 t ≤ C + k∇ht k2L2 (0; T ; L2 ( f )2 ) + |w|2 |ht |2 ; 0 f Z tZ
2
0
2
Z t ht ≤ 2
f
t
0
k∇ t kL1 kht kf ≤ C + kht k2L∞ (0; T ; L2 ( f )) ;
Z tZ
1 pt ht = kdiv wk2L1 (0; T ; L2 ( f )) + kht k2L∞ (0; T ; L2 ( f ))
f
0
≤ C + kht k2L∞ (0; T ; L2 ( f )) : The last term is estimated to be ! Z # Z t" Z Z Z tZ 2 t ht ≤ |wwt | + |hg; t | · |ht | ; 2 mes( f ) 0
f
f
f
f 0 however Z t Z 0
f
Z ≤
t
0
!
Z |wwt |
f
|ht | Z
kwkf kvq; t kf kht kL1 +
t
0
Z kwkf k∇ht kf kht kL1 +
t
0
kwkf k∇hg; t kf kht kL1
Z t 3 2 2 2 ≤ C + mes( f ) kwkf kht kf + k∇ht kL2 (0; T ; L2 ( f )2 ) 0 and Z
f
0
hence 2
Z tZ 0
f
!
Z
Z
t
|hg; t |
f
|ht |
≤ C + mes( f )kht k2L∞ (0; T ; L2 ( f ))
t ht ≤ C + 2kht (t)k2L∞ (0; T ; L2 ( f ))
+2kht k2L2 (0; T ; H 1 ( f )) +
6
Z 0
t
kwk2f kht k2f :
Finally, if the value of is chosen small enough, then Eq. (3.17) becomes Z t kwk2f kht k2f ; kht (t)k2L∞ (0; T ; L2 ( f )) + k∇ht k2L2 (0; T ; L2 ( f )2 ) ≤ C1 + C2 0
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
w, however, is bounded into L2 (0; T ; L2 ( f )2 ), therefore, by applying Gronwall’s lemma to kht (t)kL∞ (0; T ; L2 ( f )) we obtain kht (t)kL∞ (0; T ; L2 ( f )) ≤ C and, subsequently k∇ht k2L2 (0; T ; L2 ( f )2 ) ≤ C: Therefore ht ∈ L4 (Qf ) and as hg; t ∈ L4 (Qf ) then ˜t ∈ L4 (Qf ). By multiplying the continuity equation by 4|p|2 p we deduce dp4 + 4˜ · |p|2 p = 0: dt As a result, by expressing ˜ using Eq. (3.1), we obtain dp4 + 4p4 = −4˜t · |p|2 p − 2w2 · |p|2 p − 4 · |p|2 p + 4 · |p|2 p: (3.18) dt Then, by integrating Eq. (3.18) with respect to Qf , we get the following estimate:
kp(t)k4L∞ (0; T ; L4 ( f )) + 4kpk4L4 (Qf ) Z t Z t ≤ kp0 k4L4 + 4 k˜t kL4 kp3 kL4=3 + 4 k kL4 kp3 kL4=3 Z +4
0
0
t
kkL4 kp3 kL4=3 + 4
Z 0
0
t
kw2 kL4 kp3 kL4=3 :
The right-hand side terms are estimated as follows: Z t 3 4 k˜t kL4 kp3 kL4=3 ≤ k˜t k4L4 (Qf ) + 3kpk4L4 (Qf ) ; 0 Z t 3 4 k kL4 kp3 kL4=3 ≤ k k4L4 (Qf ) + 3kpk4L4 (Qf ) ; 0 Z t 3 kkL4 kp3 kL4=3 ≤ kk4L4 (Qf ) + 3kpk4L4 (Qf ) ; 4 0 Z t Z t 2 3 4 kw kL4 kp kL4=3 ≤ 4 kw2 kL4 kpk3L4 ≤ 4kw2 kL1 (0; T ; L4 ( f )) kPk3L∞ (0; T ; L4 ( f )) 0
0
1 ≤ kw2 k4L1 (0; T ; L4 ( f )) + kPk4L∞ (0; T ; L4 ( f )) : As a result, if is small enough and if the fact that p0 ∈ H 1 ( f ), , , and ˜t ∈ L4 (Qf ) is taken into consideration, then kp(t)k4L∞ (0; T ; L4 ( f )) + kpk4L4 (Qf ) ≤ C(1 + kw2 k4L1 (0; T ; L4 ( f )) ): Moreover, w ∈ L2 (0; T ; H 1 ( f )2 )⊂L2 (0; T ; L8 ( f )2 ), therefore w2 ∈L1 (0; T ; L2 ( f )) and so p ∈ L4 (Qf ).
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
571
Eq. (3.16) remains to be proven. It has been shown in [3] that h ∈ L4 (Qf ) if ht ∈ L4 (Qf ). In addition, if g ∈ H 1 (0; T ; W 1=2; 4 ( p )) then wg ∈ H 1 (0; T ; H 2 ( f )) and as the embedding of H 2 ( f ) is compact in W 1; 4 ( f ) in dimension 2, we obtain div wg = hg ∈ H 1 (0; T ; L4 ( f )). So ˜ = (h + hg ) ∈ L4 (Qf ). As a continuation of the proof, we will now present a regularity and fundamental trace result. Lemma 3.5. Under the assumptions of Lemma 3.4, we have p ∈ H 1 (0; T ; H 1 ( f )):
(3.20)
Proof. By multiplying the momentum equation by ∇div vh and by integrating with respect to Qf , we obtain 1 1 kdiv vh (t)k2L∞ (0; T ; L2 ( f )) + k∇vh k2L2 (0; T ; L2 ( f )2 ) + k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) 2 2 Z Z Z tZ Z tZ 1 t + ∇p∇div wg − ∇w2 vh − curl vq (w)vh 2 0 f
f
f 0 0 =
1 1 kdiv vh (0)k2f + k∇p(0)k2f : 2 2
(3.21)
To estimate the non-linear terms we proceed as follows: Z tZ
curl vq (w)vh ≤
f
0
3 kcurl vq (w)k2L2 (Qf ) + kvh k2L2 (0; T ; L2 ( f )2 ) : 3
Let us show that curl vq (w) is bounded into L2 (Qf ). To this end we observe that vq ∈ L4 (0; T ; W 1; 4 ( f )2 ) ⇒ rot vq ∈ L4 (Qf ) and that vq ∈ L4 (0; T ; W 1; 4 ( f )3 ) wh = ∇˜ ∈ L4 (0; T ; W 1; 4 ( f )3 )
) ⇒ w ∈ L4 (0; T ; W 1; 4 ( f )2 ):
As a result kcurl vq (w)k2L2 (Qf ) ≤ kcurl vq k2L4 (Qf ) k (w)k2L4 (Qf ) ≤ C: In the same way, as w ∈ L4 (0; T ; w1; 4 ( f )2 ) we prove that ∇w2 is bounded into L (Qf ) and, consequently, that 2
1 2
Z tZ 0
∇w2 vh ≤ C + kvh k2L2 (0; T ; L2 ( f )2 ) : 3
f
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Finally, since the solution of problem (R) satis es kdiv wg (t)k2L∞ (0; T ; L2 ( f )) + kwg k2L2 (0; T ; L2 ( f )2 ) ≤ C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ) (3.22) we obtain Z tZ 0
f
∇p∇ div wg ≤ k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) + C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ):
(3.23)
Consequently, if is small enough, then Eq. (3.21) becomes kdiv vh (t)k2L∞ (0; T ; L2 ( f )) + kvh k2L2 (0; T ; L2 ( f )2 ) +k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) ≤ C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ):
(3.24)
However, by adding the inequalities (3.22), (3.24) and if we recall that pt = −div vh − div wg , then we easily prove that kpt (t)k2L∞ (0; T ; L2 ( f )) + k∇pt k2L2 (0; T ; L2 ( f )2 ) +k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) ≤ C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ):
(3.25)
In addition, thanks to Theorem 2.1 we know that p ∈ L∞ (0; T ; L2 ( f )). As a result, the inequality (3.25) shows that p ∈ L∞ (0; T ; H 1 ( f )) and that pt ∈ L2 (0; T ; H 1 ( f )) which leads to Eq. (3.20). Also, by using the continuity equation, this result demonstrates that div wh ∈ L2 (0; T ; H 1 ( f )) and that the trace of p − div wh has a meaning in L2 (0; T ; H 1=2 ( )). Lemma 3.6. If w0 ∈ H 2 ( f ) 2 and if g ∈ H 1 (0; T ; H 1=2 ( p )); then we obtain vh; t ∈ L4 (Qf ):
(3.26)
Proof. Let us di erentiate Eq. (3.1) with respect to time, multiply it by vh; t and integrate it with respect to Qf . We obtain 1 kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) + kdiv vh; t k2L2 (0;T ; L2 ( f )) 2 Z tZ Z tZ Z tZ 1 = kvh; t (0)k2f + wwt div vh; t − ∇ t vh; t + pt div vh; t : 2
f
f
f 0 0 0 Thanks to the properties of the special basis, the trace terms are cancelled. In addition, we estimate the right-hand side terms in the following manner: Z tZ 0
f
wwt div vh; t ≤ kwkL∞ (0;T ; L4 ( f ) 2 ) kwt kL2 (0;T ; L4 ( f ) 2 ) kdiv vh; t kL2 (0;T ; L2 ( f ))
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
573
≤ CG kwkL∞ (0;T ; L4 ( f ) 2 ) kwt k1=2 L2 (0;T ; L2 ( f ) 2 ) × kcurlvq; t + div wg; t k1=2 kdiv vh; t kL2 (0;T ; L2 ( f )) L2 (0;T ; L2 ( f ) 2 ) + CG kwkL∞ (0;T ; L4 ( f ) 2 ) kwt kt=2 kdiv vh; t k3=2 L2 (0;T ; L2 ( f ) 2 ) L2 (0;T ; L2 ( f ) 2 ) ≤
CG kwk2L∞ (0;T ; L4 ( f ) 2 ) kvq; t + vh; t + wg; t kL2 (0;T ; L2 ( f ) 2 ) Z CG t × kcurlvq; t + div wg; t kL2 (0;T ; L2 ( f ) 2 ) + kwk4L4 kwt k2L2 0 + 2 CG kdiv vh; t k2L2 (0;T ; L2 ( f ))
≤
with
Z 0
t
CG kwk2L∞ (0;T ; L4 ( f ) 2 ) kvq; t + wg; t kL2 (0;T ; L2 ( f ) 2 ) CG × kcurlvq; t + div wg; t kL2 (0;T ; L2 ( f ) 2 ) + 3 kwk4L∞ (0;T ; L4 ( f ) 2 ) 2 × kcurlvq; t + div wg; t kL2 (Qf ) + CG kvh; t k2L2 (0;T ; L2 ( f ) 2 ) Z CG t + kwk4L4 kwt k2L2 + 2 CG kdiv vh; t k2L2 (0;T ; L2 ( f )) (3.27) 0
kwk4L4 kwt k2L2 ≤ kwk4L∞ (0;T ; L4 ( f ) 2 ) kvq; t + wg; t k2L2 (Qf ) 2 kwk8L∞ (0;T ; L4 ( f ) 2 ) kvq; t + wg; t k2L2 (Qf ) 2 Z t + 22 T 2 kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) + kwk4L4 kvh; t k2L2 : +
0
(3.28)
Moreover, pt = −(div vh + div wg ) ∈ L∞ (0; T ; L2 ( f )) and wg ∈ H 1 (0; T ; H 1 ( f ) 2 );
(3.29)
hence vh ∈ L∞ (0; T ; H 1 ( f ) 2 ):
(3.30)
In addition, as w0 ∈ H 2 ( f ) 2 , Lemma 3.1 and the results (3.29) and (3.30) show that w = vq + vh + wg ∈ L∞ (0; T ; L4 ( f ) 2 );
(3.31)
2
vq; t + wg; t ∈ L (Qf );
(3.32)
curl vq; t + div wg; t ∈ L2 (Qf ):
(3.33)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
and, nally, (3.27) and (3.28) lead to Z tZ wwt div vh; t ≤ C + 2CG (T 2 kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) + kdiv vh; t k2L2 (Qf ) ) 0
f
CG +
Z 0
t
kwk4L4 |vh; t |2L2 :
The remaining terms are estimated as follows: Z tZ 1 pt div vh; t ≤ kpt k2L2 (0;T ; L2 ( f )) + kdiv vh; t k2L2 (0;T ; L2 ( f ))
f 0 ≤ C + kdiv vh; t k2L2 (0;T ; L2 ( f )) ; Z tZ 0
f
t div vh; t ≤
1
≤
C
Z
t
k t k2L2 + kdiv vh; t k2L2 (0;T ; L2 ( f ))
0
Z
t
0
k∇ t k2L1 + kdiv vh; t k2L2 (0;T ; L2 ( f ))
≤ C + kdiv vh; t k2L2 (0;T ; L2 ( f )) : Therefore, if is small enough, we obtain the following estimate: Z t kwk4L4 kvh; t k2L2 ; kvh; t (t)k2L2 ( f ) 2 + kdiv vh; t k2L2 (0;T ; L2 ( f )) ≤ C1 + C2 0
(3.35)
as w ∈ L∞ (0; T ; L4 ( f ) 2 ) we can apply Gronwall’s lemma to kvh; t (t)k2L2 ( f ) 2 . In this way, we obtain kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) ≤ C and consequently, kdiv vh; t k2L2 (0;T ; L2 ( f ) 2 ) ≤ C which shows that vh; t ∈ L4 (Qf ). Lemma 3.7. If p0 ∈ W 1; 4 ( f ) and if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) then vh ∈ L4 (0; T ; W 2; 4 ( f )):
(3.36)
Proof. Let us apply the operator ∇ to the mass equation and multiply it by 4∇p|∇p|2 . By integrating with respect to f , we obtain 1 1 d k∇pk4L4 ≤ kvh |4L4 + kwg k4L4 + 2k∇pk4L4 : dt
(3.37)
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
575
It should be noted that if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) then wg ∈ W 1; 4 (0; T ; W 2; 4 ( f ) 2 ). As a result Eq. (3.37) becomes d 1 k∇pk4L4 ≤ C + kvh k4L4 + 2k∇pk4L4 : dt
(3.38)
Let us provide an estimate of vh . To this end, the momentum equation is multiplied by 4vh |vh |2 and integrated on f . 1 kvh k4L4 ≤ (kvh; t k4L4 + k∇w2 k4L4 + k∇ k4L4 + k∇pk4L4 ) + 4kvh k4L4 :
(3.39)
First, we recall that div vh and vh; t ∈ L4 (Qf ); thus vh ∈ W 1; 4 (Qf ):
(3.40)
Moreover, we have wg ∈ W 1; 4 (0; T ; W 2; 4 ( f ) 2 );
vq ∈ W 1; 4 (Qf );
which imply that w = vq + wh = vq + vh + wg ∈ W 1; 4 (Qf ): Using Sobolev’s inequality we obtain ∇w2 ∈ W 1; 4 (Qf ) and then ∇w2 ∈ L4 (Qf ):
(3.41)
In addition, using inequality (3.35) we have vh ∈ H 1 (0; T ; H 1 ( f ) 2 ); moreover vq ∈ L∞ (0; T ; H 1 ( f ) 2 ) and wg ∈ H 1 (0; T ; H 1 ( f ) 2 ), thus w ∈ L∞ (0; T ; H 1 ( f ) 2 ): Finally, as curl vq ∈ L4 (0; T ; W 1; 4 ( f ) 2 ) we deduce that ∇ ∈ L4 (0; T ; W 1; 4=3 ( f ) 2 ) ⊂ L4 (Qf ):
(3.42)
Then, if in (3.39) is small enough, these estimates show that kvh k4L4 ≤ C(1 + k∇pk4L4 ):
(3.43)
By combining Eqs. (3.38) and (3.43) we obtain d k∇pk4L4 ≤ C1 + C2 k∇pk4L4 dt
(3.44)
and as p0 ∈ W 1; 4 ( f ); we arrive at k∇pk4L∞ (0;T ; L4 ( f ) 2 ) ≤ C;
(3.45)
so by Eq. (3.43), vh ∈ L4 (Qf ). Finally, we have vh ∈ L4 (Qf ); vh ∈ L4 (Qf ) and vh · n = 0 which implies that vh ∈ L4 (0; T ; W 2; 4 ( f ) 2 ).
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Lemma 3.8. If vh0 ∈ H 2 ( f ) 2 and if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )); then wh ∈ L∞ (0; T ; L2 ( f ) 2 );
(3.46)
(p − div w) ∈ L2 (0; T ; H 3=2 ( p )):
(3.47)
Proof. Let us apply the divergence operator to the conservation equation and multiply by div vh; t . After integrating on Qf , we deduce kdiv vh; t k2L2 (0; T ; L2 ( f )) + kvh k2L∞ (0; T ; L2 ( f )2 ) 2 Z Z Z tZ 1 t =− p div vh; t − w2 div vh; t 2 0 f
f 0 Z tZ − div(curl vq (w))div vh; t + kvh (0)k2L2 : 2
f 0
(3.48)
Let us begin by estimating the right-hand side terms. To this end, we rst observe that Z tZ Z Z tZ p div vh; t = pt div vh − p(t) div vh (t) − 0
f
0
+
Z
f
f
f
p(0) div vh (0);
(3.49)
however, using the properties of the special basis, Z Z Z tZ Z tZ 1 t pt div vh = − ∇pt vh − wg; t · n div vh 0
f
f 0 0 ≤ k∇pt kL2 (0; T; L2 ( f )2 ) kvh kL2 (0; T; L2 ( f )2 ) 1 + kwg; t · nkL2 (0; T; L2 ( )) kdiv vh kL2 (0; T; L2 ( )) 1 1 ≤ k∇pt k2L2 (0; T; L2 ( f )2 ) + kwg; t · nk2L2 (0; T; L2 ( 2 2 +2T kvh kL∞ (0; T; L2 ( f )2 ) and
Z
Z −
f
p(t) div vh (t) = Z =
f
f
))
(3.50)
Z ∇p(t)vh (t) −
∇p(t) · n div vh (t) Z Z
∇p(t)vh (t) +
0
t
wg ( ) d · n div vh (t)
≤ k∇p(t)kL∞ (0; T; L2 ( f )2 ) kvh (t)kL∞ (0; T; L2 ( f )2 )
Z t
+ w ( ) d · n kdiv vh (t)kL∞ (0; T; L2 ( g
∞ 2 0
L
(0; T; L ( ))
))
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
577
Z
2
1 1 t
≤ k∇p(t)k2L∞ (0; T; L2 ( f )2 ) + w ( ) d · n g
∞ 0 L (0; T; L2 ( +2kvh (t)k2L∞ (0; T; L2 ( f )2 ) :
))
(3.51)
In addition, as p ∈ H 1 (0; T ; H 1 ( f )) we obtain ∇pt ∈ L2 (0; T ; L2 ( f )2 ) and ∇p(t) ∈ L ( f )2 . Moreover, if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) then wg ∈ W 1; 4 (0; T ; W 2; 4 ( p )). As a result, the inequalities (3.50) and (3.51) become Z tZ pt div vh ≤ C + 2T 2 kvh k2L∞ (0; T; L2 ( f )2 ) ; 2
0
f
Z
−
f
p(t) div vh (t) ≤ C + 2kvh (t)k2L∞ (0; T; L2 ( f )2 ) :
Finally, if p0 ∈ H 2 ( f ) and vh0 ∈ H 1 ( f ) then we obtain the following estimate for Eq. (3.49): Z tZ p div wh; t ≤ C + 2(1 + T 2 )kvh k2L∞ (0; T; L2 ( f )2 ) : (3.52) −
f
0
The estimate of the nonlinear terms is as follows: Z tZ 1 − w2 div vh; t ≤ kw2 k2L2 (0; T ; L2 ( f )) + kdiv vh; t k2L2 (0; T ; L2 ( f )) ;
f 0 −
Z tZ 0
1 div(curl vq (w))div wh; t ≤ kdiv(curl vq (w))k2L2 (0; T ; L2 ( f ))
f +kdiv vh; t k2L2 (0; T ; L2 ( f )) ;
but w ∈ L4 (0; T ; W 2; 4 ( f )2 ) and curl vq ∈ L4 (0; T ; W 1; 4 ( f )), therefore w2 ∈ L2 (0; T ; L2 ( f ));
div(curl vq (w)) ∈ L2 (0; T ; L2 ( f )):
So, the two above inequalities are written as Z tZ w2 div vh; t ≤ C + kdiv vh; t k2L2 (0; T ; L2 ( f )) ; − 0
−
f
Z tZ 0
f
div(curl vq (w))div wh; t ≤ C + kdiv vh; t k2L2 (0; T ; L2 ( f )) :
(3.53) (3.54)
If is small enough, then Eq. (3.48) leads to kdiv vh; t k2L2 (0; T ; L2 ( f )) + kvh k2L∞ (0; T ; L2 ( f )2 ) ≤ C; in addition wg ∈ W 1; 4 (0; T ; W 2; 4 ( f )2 ) hence wh = vh + wg ∈ L∞ (0; T ; L2 ( f )2 ):
(3.55)
578
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
As vq , vh and wg belong to L4 (0; T ; W 2; 4 ( f )2 ), then ∇w2 and ∇ belong to L2 (0; T ; H 1 ( f )2 ). Moreover, using Eq. (3.55), we have div wh; t = div vh; t + div wg; t ∈ L2 (0; T ; L2 ( f )2 ) and so 1 ∇(p − div wh ) = − wh; t − ∇w2 − ∇ ∈ L2 (0; T ; H 1 ( f )2 ): 2 Consequently, (p − div v) is well de ned in L2 (0; T ; H 3=2 ( )). 4. Study of Problem (S) We search the displacement under the following form: u=
n X
ui (t) i ;
i=1 2 2 in which { i }∞ i=1 is an orthonormal basis of L ( p ) and H0 ( p ) associated with the spectral problem 2 = 2 in p ; = ∇ · n = 0 on @ p :
In the following paragraph, we provide the estimate of u necessary to the proof of Theorem 1.1. Lemma 4.1. Under the assumption (p− div v)∈ L2 (0; T ; H 3=2 ( )) and if u1 ∈H 3 ( p ), f ∈ L2 (0; T ; H 2 ( p )), we obtain kut k2L∞ (0; T ; L2 ( p )) + k∇ut k2L∞ (0; T ; L2 ( p )2 ) + k2 uk2L∞ (0; T ; L2 ( p )) ≤ C:
(4.1)
Proof. By applying the ∇ operator to the plate equation, multiplying by ∇ut and integrating on Qp , we obtain kut k2L∞ (0; T ; L2 ( p )) + k∇ut k2L∞ (0; T ; L2 ( p )2 ) + k2 uk2L∞ (0; T ; L2 ( p )) Z Z − ∇utt · nut − 2 u∇ut · n = kut (0)k2p + k∇ut (0)k2p
p
+k2 u(0)k2p + 2
Z 0
p
T
Z
p
Z ∇(p − div wh )∇ut − 2
0
T
Z
p
∇f∇ut :
(4.2)
By using the properties of the basis associated with displacement, the trace terms disappear. In addition, the trace of ∇(p− div v) is de ned in L2 (0; T ; H 1=2 ( )). Hence Z TZ 2 ∇(p − div v)∇ut ≤ C + T 2 k∇ut k2L2 (0; T ; L2 ( p )2 ) : (4.3) 0
p
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
579
As f ∈ L2 (0; T ; H 1 ( p )), we also obtain Z 2
T
Z
∇f∇ut ≤ C + T 2 k∇ut k2L∞ (0; T ; L2 ( p )2 ) :
p
0
(4.4)
By inserting Eqs. (4.3) and (4.4) in Eq. (4.2), and if is small enough, we deduce kut k2L∞ (0; T ; L2 ( p )) + k∇ut k2L∞ (0; T ; L2 ( p )2 ) + k2 uk2L∞ (0; T ; L2 ( p )) ≤ K which achieves the proof. Lemma 4.2. If u0 ∈ H 4 ( p ), p0 ∈ H 1 ( f ), wh0 ∈ H 2 ( f )2 and f ∈ H 1 (0; T ; H 2 ( p )) then we obtain k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) ≤ C(1 + kgk2w1; 4 (0; T ; W 3=2;1 ( p )) ):
(4.5)
Proof. Let us di erentiate the plate equation with respect to time and multiply it by utt . After integrating on Qp , we obtain k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) Z =
Z
T
Z −
2 ut utt + k∇utt (0)k2f + kutt (0)k2f
p
0
T
Z
Z
p
0
ft utt +
T
Z
p
0
(p − div wh )uttt
Z −
Z
p
(p(t) − div wh (t))utt (t) +
p
(p(0) − div wh (0))utt (0):
(4.6)
In the same way, let us multiply it by uttt . We thus obtain, after integrating on Qp k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) Z =−
T
0
Z
p
2 ut utt +
Z
p
2 u(t)utt (t) −
+k∇utt (0)k2f + kutt (0)k2f + Z −
p
Z 0
T
p
p
2 u(0)utt (0)
Z
p
ft utt
Z f(t)utt (t) +
Z
Z f(0)utt (0) −
0
T
Z
p
(p − div wh )uttt :
(4.7)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Hence, by summing (4.6) and (4.7) we are led to 2k∇utt k2L∞ (0; T ; L2 ( p )) + 2kutt k2L∞ (0; T ; L2 ( p )) Z 2 2 = 2k∇utt (0)kf + 2kutt (0)kf + 2 u(t)utt (t) Z − Z +
p
p
2 u(0)utt (0) −
Z
p
p
(p(t) − div wh (t))utt (t) Z
(p(0) − div wh (0))utt (0) −
p
Z f(t)utt (t) +
p
f(0)utt (0): (4.8)
By applying Young’s inequality to this last result, we obtain k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) ≤ C(1 + kp(t)k2L∞ (0; T ; L2 ( p )) + · · · + kdiv wh (t)k2L∞ (0; T ; L2 ( p )) +k2 u(t)k2L∞ (0; T ; L2 ( p )) ) + kutt k2L∞ (0; T ; L2 ( p )) :
(4.9)
But, by using Eqs. (3.25), (3.46) and (4.1) we deduce kp(t)k2L∞ (0; T ; L2 ( p )) ≤ Ckpk2H 1 (0; T ; H 1 ( f )) ≤ C(1 + kgk2W 1; 4 (0; T ; W 3=2; 4 ( p )) ); kdiv wh (t)k2L∞ (0; T ; L2 ( p )) ≤ Ckwh (t)k2L∞ (0; T ; L2 ( f )2 ) ≤ C; k2 u(t)k2L∞ (0; T ; L2 ( p )) ≤ C: Therefore, if is suciently small, we obtain k∇u tt k2L∞ (0; T ; L2 ( p )) + ku tt k2L∞ (0; T ; L2 ( p )) ≤ C(1 + kgk2W 1; 4 (0; T ; W 3=2; 4 ( p )) ) which achieves the proof. 5. Proof of the Theorem We apply here Kakutani’s theorem. To this end we note that W 1; 4 (0; T ; H02 ( p )) ⊂ E = W 1; 4 (0; T ; W 3=2; 4 ( p )) algebraically and topologically, hence (4.5) involves ku tt kL4 (0; T ;W 3=2; 4 ( p )) ≤ C 0 ku tt kL4 (0; T ; H02 ( p )) ≤ TC 00 (1 + kgkW 1; 4 (0; T ; W 3=2; 4 ( p )) ): Thus, we can de ne the following map: : g ∈ E → u t ∈ E: Then, if the value of T is small enough and if kgkE ≤ K with K being well chosen, the map applies the ball B(0; K) of E with center 0 and radius K within itself. That is to say (B(0; K)) ⊂ B(0; K):
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
581
Now, we recall that E is a re exive and separable banach space. Consequently, B(0; K) is metrisable for the weak topology and, nally, compact for this same topology. We are therefore able to work with ordinary sequences. We will prove that if we consider gn as a weak convergent sequence towards g in B(0; K) and such that weakly
un; t = (gn ) −→ t then t = (g) = u t . If (wn ; pn ) is a solution associated to (F) with w(0) = wn (0) and p(0) = pn (0), then (wn ; pn ) satis es the estimates for Lemmas 3.1 – 3.8 and (pn − div wn ) weakly converges towards a limit ! in L2 (0; T ; H 3=2 ( )). As problem (F) allows only one solution, then ! = (p − div w). As a result, for the problem (S) we have (un; tt ; )p + (∇un; tt ; ∇ )p + (un ; )p = −(pn − div wn ; )p + (fn ; )p which converges weakly towards ( n ; )p + (∇ tt ; ∇ )p + ( ; )p = −(p − div w; )p + (f; )p = (H; )p weakly
when gn −→ g in B(0; K). Finally, as problem (S) has only one solution for a given H , then is the unique solution associated to (S) and, hence, t = u t . This last result shows that we have a unique xed point and that w ·n = −u t on p in W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )). We have shown the existence for a small time but this result can be extended for all time thanks to the a priori estimates. 6. Numerical implementation 6.1. Approximate solutions In order to facilitate the numerical implementation, we state p = ]0; 1[ and f = p ×]0; 1[. Due to this simpli ed study domain, resolving the spectral problems de ned in Section 3 provides an analytical basis for V orthonormalized in L2 ( f )2 . To obtain a basis for W˜ we solve the problem ˜ ˜ in Qf ; hl = hl ˜ (H) grad h˜ l · n = 0 on f × ]0; T [; grad h˜ l · n = l on f × ]0; T [; 2 in which { l }∞ l=1 is the basis of H0 ( p ) associated with the movement of the plate. To this end, we use the nite elements method and the MODULEF software. The element utilized is the 2P3D, which is an Hermite type element with three degrees of freedom. This element allows us to obtain, in addition to the solution to prob˜ the numerical values of its primary x and y derivatives. Thanks to this lem (H),
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
property, we can therefore construct a basis for W˜ . Consequently, we search to determine the movement of the plate and the velocity in the uid, respectively, as follows: M X
u=
ul (t) l ;
l=1
w=
N X
N X
Aj (t)Curl qj +
j=1
Bi (t)∇hi +
i=1
M X
Cl (t)∇h˜ l :
l=1
In addition, if Bi (t) =
dbi (t) dt
and
Cl (t) =
dcl (t) ; dt
we recall that we can look for the pressure as p=
N X
i bi (t)hi −
i=1
M X
cl (t)h˜ l
l=1
if bi (0) and cl (0) are chosen such that p(0) =
N X
i bi (0)hi −
M X
i=1
cl (0)h˜ l :
l=1
6.2. Projection of the equations In order to obtain a matrix system, we project the equations onto the bases de ned previously. Projection of the equation of problem (R) onto {h˜ l }M l=1 leads to " # Z Z Z M X d 2 cl ˜ ˜ ˜hl h˜ m + dcl ∇hl ∇hm = gh˜ m ; 1 ≤ m ≤ M: (6.1) dt 2 f dt f
p l=1
In the same way, by successively projecting the conservation equation on ∇hn and Curl qn , we obtain n
n d 2 bn dbn + n2 bn + + n2 dt 2 dt 2 Z +
n
f
curl w (w)∇hn =
d 2 an dan + + n2 2 dt dt
M X l=1
Z
f
w2 hn Z
cl n
f
h˜ l hn ;
1≤n≤N
(6.2)
Z
f
curl w (w) curl qn = 0;
1 ≤ n ≤ N:
(6.3)
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
583
Finally, the projection of the plate equation on the m basis functions leads to " ! # Z Z Z M X dul d 2 ul l m + ∇ l ∇ m + l m dt 2 dt p
p
p l=1
Z =
p
+
f m +
N X
dbi −bi + dt
i=1
M X
cl −
l=1
dct dt
Z
p
h˜ l m ;
Z i
p
hi m
1 ≤ m ≤ M:
(6.4)
Remark. By combining equations (6.1)–(6.4), we can easily return to a second order di erential equations system of the form cl cl cl 2 a d a d i a + nonlinear term [A] 2 + [B] i + [C] i = (F) dt bi dt bi bi ul ul ul which we can solve using a Neumark type scheme coupled to a prediction–correction method. This method is not suitable; however; as the solution must also verify the coupling condition M X
ul; t (t) l = −
l=1
M X
Cl (t) grad h˜ l · n = −
l=1
M X
Cl (t) l
l=1
and it is clear that the system is over-determined. To overcome this diculty and, as was seen in the existence Theorem 1.1, we introduce an algorithm based on xed point method. 6.3. Algorithm We set g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )) and we work on the time-step [tn−1 ; tn ]. Given the solution at tn−1 , we solve (R) using an implicit Euler scheme. Once cl (tn ) are known, we solve the homogeneous (FH) problem using a Neumark type scheme by proceeding with i implicit iterations in such a way as to approach the nonlinear term. By using the same Neumark scheme, we can solve the plate problem (S). The solution to (S) being known, we bring g(tn ) = −ut (tn ) to the problem (R). We reiterate the process k times until convergence of u(tn ) towards a xed point. Finally, if at the kth iteration we note R[ck (tn )] = KG (gk (tn )];
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Fig. 1.
FH (ak (tn ); bk (tn )] = B[ak (tn ); bk (tn ); ck (tn )] + Kc [ck (tn )]; S[uk (tn )] = KT [bk (tn ); ck (tn )] + f(tn ) the systems (6.1)–(6.4) with B[ak (tn ); bk (tn ); ck (tn )] as the nonlinear term, we can represent the algorithm as shown in Fig. 1. 6.4. Validation of the algorithm To validate the algorithm, we introduce the following analytical solution: 2 0 2 3 4 5 t ; u(x; t) = 2 (−x + 4x − 5x + 2x ) cos(y) exp − 0 wg (x; y; t) = ∇hg (x; y; t) = ∇
1 3 1 6 2 4 5 − x + x − x + x cos(y) exp − t ; 3 3 0
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
585
Fig. 2.
in which 0 is the density at rest and the viscosity. This solution satis es, in particular, the coupling condition wg · n = −ut
on p
and the boundary conditions u = ∇u · n = 0
for x = 0
and x = 1:
We suppose that 0 = 1 kg=m3 ; −10
I = 2:8 × 10
= 6 × 10−5 m2 =s;
m = 11:7 kg=m2 ;
D = 65:34 N:m;
3
m
with m, D and I the surfacic mass, the rigidity and the inertia moment of the plate, respectively. We test the algorithm for two points taken in the uid (x = 0:05; y = 0:003285) and on the plate (x = 0:05; y = 0:0). Below are represented some results obtained with the method for these values. Fig. 2 makes the comparison between the analytical and the numerical solutions. Figs. 3 and 4 present the relative and absolute errors for hg (x; y; t) and u(x; t). The decrease in the precision of the results over time is due to the form of the solution which approaches 0 when t increases. All in all, however, this method seems to provide encouraging results despite its numerically demand aspects.
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Fig. 3.
Fig. 4.
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
587
References [1] J. Boujot, Mathematical formulation of uid structure interaction problems, Math. Mod. Numer. Anal. 21 (2) (1987) 239 – 260. [2] S. Cerneau, E. Sanchez-Palencia, Sur les vibrations des corps e lastiques plonges dans des uides, J. de Mecanique 15 (3) (1976). [3] F.J. Chatelon, P. Orenga, On a Nonhomogeneous Shallow Water Problem, M2 AN, 31 (1997) 27–55. [4] P.G. Ciarlet, Plates and junctions in elastic multi-structures, Collection Recherches en Mathematiques appliquees, RMA 14, 1990. ements Finis, Maloine Editeur, 1984. [5] G. Dhatt, G. Touzot, Une Presentation de la Methode des El [6] M.J. Ghaleb, Vibrations d’une plaque e lastique in nie couplee avec un uide sous l’e et d’un impact ponctuel, These de troisieme cycle, Institut National Polytechnique de Grenoble, 1987. [7] O.A. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Flow, Gordon and Breach, New York, 1963. [8] J.L. Lions, Quelques methodes de resolution des problemes aux limites non lineaires, Dunod, Paris, 1969. [9] J.L. Lions, E. Magenes, Problemes aux Limites Nonhomogenes et Applications, Dunod, Paris, 1972. [10] C. Marchioro, M. Pulvirenti, Mathematical Theory of Incompressible Nonviscous Fluids, Applied Mathematical Sciences, vol. 96, Springer, Berlin. [11] H. Morand, R. Ohayon, Interactions Fluides–Structures. Collection recherches en Mathematiques appliquees, RMA 23, 1992. [12] P. Orenga, Analyse de quelques problemes en oceanographie physique, These d’habilitation, Universite de Corse, 1991. [13] P. Orenga, Un theoreme d’existence de solution d’un probleme de Shallow Water, Archive Rational Mech. Anal., Universite de Corse, 1994. [14] R. Papi, P. Orenga, Analyse numerique d’un probleme de couplage uide structure, C.R.A.S. Paris 317, serie I (1993) 1087–1090. [15] Flori-Orenga, On a uid-structure interaction problem in velocity-displacement formulation, M3AS, to appear.
Analysis of a nonlinear uid–structure interaction problem in velocity–displacement formulation Fabien Flori∗ , Pierre Orenga Centre de Mathematique et de Calcul Scienti que, URA 2053, Universite de Corse, Quartier Grossetti, 20250 Corte, France Received 15 March 1997; accepted 20 April 1997
Keywords: Compressible Navier–Stokes equations; Fluid–structure interaction; Problem velocity–displacement formulation
in
0. Introduction In the present study, we propose an existence and uniqueness theorem for the solution of a bidimensional nonlinear problem of uid–structure coupling as well a method of numerical resolution. If u; w and p represent the displacement within the structure, the velocity and the pressure in the uid, respectively, the problem (P) to be studied is the following: Fluid equations:
(F)
∗
wt + 12 ∇w2 + curl w ∧ w − w + ∇p = 0 pt + div w = 0 in Qf curl w = 0 on ×]0; T [ w · n = 0 on f ×]0; T [ w(t = 0) = w0 in f p(t = 0) = p0 in f
Corresponding author.
0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 7 1 7 - 7
in Qf
(I)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Plate equations: 2 utt − utt + u + p − div w = f (S) u = ∇u · n = 0 on p ×]0; T [ u(t = 0) = u0 ; ut (t = 0) = u1 in p Coupling condition: @u (C) − =w · n @t
in Qp (II)
(III)
in Qp
with Qp = p ×]0; T [; Qf = f ×]0; T [ where p is an open domain of R and f is an open domain of R2 simply connected with boundary = f ∪ p of class C 1; 1 . We classically denote by v the viscosity of the uid. In a previous study we determined the existence conditions of the linearized problem (P). This last study was carried out in order to point out the problems inherent to the coupling which arise in particular from the meaning given to the trace of p − div w in the plate equation. Here, the nonlinearity of the momentum equation accentuates these diculties and requires the introduction of smoothness results. We show that problem (P) admits a xed point such that w · n=−
@u ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )): @t
Sections 1–5 are entirely devoted to the proof of the theorem and in Section 6 we present the implementation of the numerical method. 1. An existence and uniqueness result Let us begin by providing several elements necessary to state the theorem. We consider the following functional space W = {w ∈ H 1 ( f )2 ; w · n = 0 on
f}
which we can decompose in the following manner [15]: W = V + W˜ with V = {v = grad h; h ∈ H 2 ( f ); grad h · n = 0 on
} ⊕ Rot(H 2 ( f ) ∩ H01 ( f ))
and ˜ grad h˜ · n = 0 on ˜ h˜ ∈ H 2 ( f ); h˜ = h; W˜ = {w˜ = grad h;
f }:
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
563
Thus, all functions of W are represented in the following unique manner: ˜ w = v + w˜ = vq + vh + w˜ = Curl q + grad h + grad h; where q; h are the solutions to the following scalar problems: q = curl vq in f ; (Q) q=0 on and
( (H)
h = div vh grad h · n = 0
in f ; on
:
In order to mute the problem (F) homogeneous, we then introduce the function hg such that hg; t − hg = 0 in Qf ; grad hg · n = 0 on f ×]0; T [; (R) grad hg · n = g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) on Qp ; hg (0) = 0 in f : We easily establish that wg = grad hg ∈ W 1; 4 (0; T ; W 2; 4 ( f )2 ); div wg ∈ L4 (0; T ; L∞ ( f )); wg · n = g
on Qp ;
curl wg = 0
in Qf :
We note by CS , the imbedding constant of H 1 in L2 , and by CG , Gagliardo– Nirenberg’s optimal constant of inequality kvk2L4 ≤ CG kvkL2 kvkH 1 . Thus, if we consider the real numbers ∈]0; 1[; ¿ 12 CG CS 1 ¿0, 2 ¡ 12 , and the functions f ∈ H 1 (0; T ; H 1 ( p ));
(1.1)
g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ));
(1.2)
w0 ∈ H 2 ( f )2 ;
(1.3)
p0 ∈ H 2 ( f );
(1.4)
u0 ∈ H 4 ∩ H02 ( p );
(1.5)
u1 ∈ H 3 ∩ H02 ( p );
(1.6)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
we assume kv0 k2f + kp0 k2f +
CG CS 2 2 A2 kwg k2L4 (0; T ; H 1 ( f )3 ) + khg; t k2L1 (0; t1 ; L2 ( p )) ¡ 4 2 1 2 CG
(1.7)
with A= −
CG CS 1 − 2CS CG kwg (t)kL∞ (0; T ; L2 ( f )2 ) ¿0; 2
(1.8)
we obtain the existence and uniqueness result shown below. Theorem 1.1. Under the conditions (1.1)–(1.8), the trace of p − div w is de ned in L2 (0; T ; H 3=2 ( )) and the problem (P) admits a unique xed point (w; u) such that w · n = −ut ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )):
(1.9)
The proof of this theorem is made possible using Kakutani’s xed point theorem. To this end, we prescribe w · n = g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) and we solve problems (F) and (S). The main diculty is to give a meaning to the trace of p − div w on p . We will see that problem (P) admits a xed point such that w · n = −ut ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )) and with this regularity the solution (w; p) of (F) is such that
(p − div w) ∈ L2 (0; T ; H 3=2 ( )):
2. Study of problem (F) By using the orthogonal decomposition introduced in the previous section, Eq. (I) are transformed in order to obtain a homogeneous problem denoted as (Fh ). By setting w = v + wg , we obtain the following problem (Fh ) vt + 12 ∇w 2 + curl v (v + wg ) − v + ∇p = 0 pt + div v = −div wg in Qf ; v · n = 0 on ; curl v = 0 on ; v(t = 0) = w0 − wg (t = 0) = v0 in f ; p(t = 0) = p0 in f :
in Qf ;
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
565
The problem (Fh ) is then solved using the following weak formulation (VFh ): Find (v; p) ∈ L∞ (0; T ; L 2 ( f ) 2 ) ∩ L 2 (0; T ; V ) × L∞ (0; T ; L 2 ( f )) such that (vt ; ’)f − 12 (v 2 ; div ’)f − (vwg ; div ’)f + (curl v (v); ’)f +(rot v (wg ); ’)f + a(v; ’) − (p; div ’)f = 12 (wg2 ; div ’)f v(t = 0) = v0 ; ( (pt ; )f + (div v; )f = −(div wg ; )f ∀ ∈ L2 ( f ) p(t = 0) = p ; 0
∀’ ∈ V
where a(v; ’) is the bilinear form Z
Z a(v; ’) =
f
div v div ’ +
f
curl v curl ’:
If v0 ∈ L2 ( f ) 2 and p0 ∈ L 2 ( f ) then we obtain Theorem 2.1. If Eqs. (1.7) and (1.8) hold and if g ∈ H 1 (0; T ; H 1=2 ( p )); then the problem (VFh ) admits a unique solution such that (v; p) ∈ L2 (0; T ; V ) ∩ L 2 (0; T ; L∞ ( f ) 2 ) × L∞ (0; T ; L∞ ( f )); in addition; the solution is such that kv(t)kL∞ (0; T ; L 2 ( f ) 2 ) ≤ 2 kvkL22 (0; T ; V ) ≤ 2
A ; CG
2 A ; 1 − CG2
(2.1) kp(t)kL2∞ (0; T ; L 2 ( f ) 3 ) ≤
4 2 A : − 22 )
CG2 (1
(2.2)
For the proof of this theorem, we refer to the results developed in [12]. It should be noted that the main diculty in obtaining an estimate is in the estimation of the following term 1 2 (v ; div v)f ; 2 which is not necessarily bounded. To overcome this diculty we must build a stability space. These results show that w = v + wg ∈ L 2 (0; T ; W ) and that p ∈ L∞ (0; T ; L 2 ( f )) and with this regularity we are unable to give a meaning to the trace of p − div w in the plate equation. We therefore introduce a series of smoothness results in Section 3 which will allow us, based on several additional assumptions, to prove that (p − div w) has a meaning in L 2 (0; T ; H 3=2 ( )). This result is sucient to solve the plate equation.
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3. Smoothness of the solutions of (F) 3.1. Preliminary results The results presented in this paragraph are demonstrated in detail in [3]. We search w = vh + vq + wg under the approximated forms [15] w=
N X
Aj (t) Curl qj +
j=1
N X
Bi (t)∇hi +
i=1
M X
Cl (t)∇h˜ l
l=1
in which hi and qj are solutions of the following spectral problems: ( ( −qj = j qj in f −hi = i hi in f and ∇hi ·n = 0 on qj = 0 on and where ˜ (H)
˜ ˜ hl = hl in Qf grad h˜ l · n = 0 on f ×]0; T [ grad h˜ l · n = l on f ×]0; T [
2 with { l }∞ l=1 a basis of H0 ( p ) associated to the motion of the plate. In addition, as the continuity equation must be veri ed, if we set
Bi (t) =
dbi (t) dt
and
Cl (t) =
dcl (t) ; dt
then we look for the pressure as p=
N X
i bi (t)hi −
i=1
M X
cl (t)h˜ l
l=1
if bi (0) and cl (0) are chosen such that p(0) =
N X i=1
i bi (0)hi −
M X
cl (0)h˜ l :
l=1
The projection for the scalar product of L 2 ( f ) 2 in the momentum equation on the eld of rotational vectors allows us to write vq; t − Rot rot vq + P(rot vq (w)) = 0; where P is the projection operator on the subspace of the rotational eld vectors. In the same way, the projection onto the eld of gradient vectors leads to ! vq2 (h + hg ) 2 ∇ ht − h + + vq ∇(h + hg ) + + + p = 0; 2 2
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
567
that is to say, if we set ˜ = (h + hg ), ht − h +
vq2 ˜2 + vq ∇˜ + + + p= 2 2
(3.1)
with such that ∇ represents the projection, for the scalar product L 2 ( f ) 2 , of curl vq (w) on the eld gradient vectors and a function which is only dependent on time. We will now give some results dealing with the regularity of vq and . Lemma 3.1. If we assume that w0 ∈ H 1 ( f ) 2 ; then vq ∈ L 4 (0; T ; W 1; 4 ( f ) 2 ) ∩ L∞ (0; T ; H 1 ( f ) 2 );
(3.2)
vq; t ∈ L 2 (Qf ):
(3.3)
If; in addition w0 ∈ H 2 ( f ) 2 and wh; t = ∇˜t ∈ L 2 (Qf ); we obtain vq; t ∈ L∞ (0; T ; L 2 ( f ) 2 );
(3.4)
curl vq; t ∈ L 2 (Qf );
(3.5)
vq ∈ W 1; 4 (Qf ):
(3.6)
And nally; if wh = L 2 (0; T ; W 1; 4 ( f ) 2 ); then vq ∈ L 4 (0; T ; W 2; 4 ( f ) 2 );
(3.7)
curl vq ∈ L 4 (0; T ; L∞ ( f ) 2 ):
(3.8)
Lemma 3.2. Given that represents the function de ned above; then veri es ∈ L 4 (Qf );
(3.9)
k∇ kL 2 ( f ) ≤ C(1 + kvq; t kL 2 ( f ) 2 )
a.e. on [0; T ]:
(3.10)
3.2. Smoothness of w and P The diculty in the following estimates is determining the proper smoothness for g. Indeed, this smoothness must permit the obtention of estimates which allow a meaning to be given to the trace of p − div w. We present here only the obtention of a priori estimates as the passage to the limit is of no great diculty. Let us introduce a rst lemma on the smoothness of . Lemma 3.3. The function de ned above satis es ∈ L∞ (0; T ) t mes( f ) =
Z
f
wwt −
1
(3.11)
Z
f
hg; t :
(3.12)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Proof. By integrating Eq. (3.1) with respect to f , we obtain Z Z Z Z Z @ 1 h− h + w2 + + p: (t)· mes( f ) = @t f 2 f
f
f
f First of all, it should be noted that Z h = 0:
f
In addition, as h and can be expressed as the sum of a scalar function and the sum of a constant, we can always choose a constant such that Z Z h= = 0:
f
f
This yields mes ( f ) =
1 2
Z
f
w2 +
Z p:
f
(3.13)
In this way, as p∈L∞ (0; T ; L2 ( f )) and w2 ∈L∞ (0; T ; L1 ( f )), we obtain Eq. (3.11). The result (3.12) is then obtained by di erentiating Eq. (3.13) with respect to time and by observing that Z Z Z 1 Pt = − hg = − hg; t : v f
f
f Lemma 3.4. If p0 ∈ H 1 ( f ); w0 ∈ H 2 ( f )2 then ˜t ∈ L4 (Qf );
(3.14)
p ∈ L4 (Qf );
(3.15)
and if; in addition; we consider that g ∈ H 1 (0; T ; W 1=2; 4 ( p )) we obtain div wh = ˜ ∈ L4 (Qf ):
(3.16)
Proof. Let us begin by proving Eq. (3.14). Di erentiating Eq. (3.1) with respect to time, multiplying it by 2ht and integrating it with respect to Qf , we obtain kht (t)k2L∞ (0; T ; L2 ( f )) + 2vk∇ht k2L2 (0; T ; L2 ( f )2 ) +
Z tZ 0
f
wwt ht + 2
Z tZ 0
f
t ht + 2
Z tZ 0
f
pt ht = kht (0)k2f + 2
Z tZ 0
f
t ht : (3.17)
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
569
The di erent terms are estimated in the following manner: Z tZ Z tZ wwt ht = w(∇ht + ∇hg; t + vq; t )ht
f
f 0 0 Z Z 3 t ≤ C + k∇ht k2L2 (0; T ; L2 ( f )2 ) + |w|2 |ht |2 ; 0 f Z tZ
2
0
2
Z t ht ≤ 2
f
t
0
k∇ t kL1 kht kf ≤ C + kht k2L∞ (0; T ; L2 ( f )) ;
Z tZ
1 pt ht = kdiv wk2L1 (0; T ; L2 ( f )) + kht k2L∞ (0; T ; L2 ( f ))
f
0
≤ C + kht k2L∞ (0; T ; L2 ( f )) : The last term is estimated to be ! Z # Z t" Z Z Z tZ 2 t ht ≤ |wwt | + |hg; t | · |ht | ; 2 mes( f ) 0
f
f
f
f 0 however Z t Z 0
f
Z ≤
t
0
!
Z |wwt |
f
|ht | Z
kwkf kvq; t kf kht kL1 +
t
0
Z kwkf k∇ht kf kht kL1 +
t
0
kwkf k∇hg; t kf kht kL1
Z t 3 2 2 2 ≤ C + mes( f ) kwkf kht kf + k∇ht kL2 (0; T ; L2 ( f )2 ) 0 and Z
f
0
hence 2
Z tZ 0
f
!
Z
Z
t
|hg; t |
f
|ht |
≤ C + mes( f )kht k2L∞ (0; T ; L2 ( f ))
t ht ≤ C + 2kht (t)k2L∞ (0; T ; L2 ( f ))
+2kht k2L2 (0; T ; H 1 ( f )) +
6
Z 0
t
kwk2f kht k2f :
Finally, if the value of is chosen small enough, then Eq. (3.17) becomes Z t kwk2f kht k2f ; kht (t)k2L∞ (0; T ; L2 ( f )) + k∇ht k2L2 (0; T ; L2 ( f )2 ) ≤ C1 + C2 0
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
w, however, is bounded into L2 (0; T ; L2 ( f )2 ), therefore, by applying Gronwall’s lemma to kht (t)kL∞ (0; T ; L2 ( f )) we obtain kht (t)kL∞ (0; T ; L2 ( f )) ≤ C and, subsequently k∇ht k2L2 (0; T ; L2 ( f )2 ) ≤ C: Therefore ht ∈ L4 (Qf ) and as hg; t ∈ L4 (Qf ) then ˜t ∈ L4 (Qf ). By multiplying the continuity equation by 4|p|2 p we deduce dp4 + 4˜ · |p|2 p = 0: dt As a result, by expressing ˜ using Eq. (3.1), we obtain dp4 + 4p4 = −4˜t · |p|2 p − 2w2 · |p|2 p − 4 · |p|2 p + 4 · |p|2 p: (3.18) dt Then, by integrating Eq. (3.18) with respect to Qf , we get the following estimate:
kp(t)k4L∞ (0; T ; L4 ( f )) + 4kpk4L4 (Qf ) Z t Z t ≤ kp0 k4L4 + 4 k˜t kL4 kp3 kL4=3 + 4 k kL4 kp3 kL4=3 Z +4
0
0
t
kkL4 kp3 kL4=3 + 4
Z 0
0
t
kw2 kL4 kp3 kL4=3 :
The right-hand side terms are estimated as follows: Z t 3 4 k˜t kL4 kp3 kL4=3 ≤ k˜t k4L4 (Qf ) + 3kpk4L4 (Qf ) ; 0 Z t 3 4 k kL4 kp3 kL4=3 ≤ k k4L4 (Qf ) + 3kpk4L4 (Qf ) ; 0 Z t 3 kkL4 kp3 kL4=3 ≤ kk4L4 (Qf ) + 3kpk4L4 (Qf ) ; 4 0 Z t Z t 2 3 4 kw kL4 kp kL4=3 ≤ 4 kw2 kL4 kpk3L4 ≤ 4kw2 kL1 (0; T ; L4 ( f )) kPk3L∞ (0; T ; L4 ( f )) 0
0
1 ≤ kw2 k4L1 (0; T ; L4 ( f )) + kPk4L∞ (0; T ; L4 ( f )) : As a result, if is small enough and if the fact that p0 ∈ H 1 ( f ), , , and ˜t ∈ L4 (Qf ) is taken into consideration, then kp(t)k4L∞ (0; T ; L4 ( f )) + kpk4L4 (Qf ) ≤ C(1 + kw2 k4L1 (0; T ; L4 ( f )) ): Moreover, w ∈ L2 (0; T ; H 1 ( f )2 )⊂L2 (0; T ; L8 ( f )2 ), therefore w2 ∈L1 (0; T ; L2 ( f )) and so p ∈ L4 (Qf ).
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
571
Eq. (3.16) remains to be proven. It has been shown in [3] that h ∈ L4 (Qf ) if ht ∈ L4 (Qf ). In addition, if g ∈ H 1 (0; T ; W 1=2; 4 ( p )) then wg ∈ H 1 (0; T ; H 2 ( f )) and as the embedding of H 2 ( f ) is compact in W 1; 4 ( f ) in dimension 2, we obtain div wg = hg ∈ H 1 (0; T ; L4 ( f )). So ˜ = (h + hg ) ∈ L4 (Qf ). As a continuation of the proof, we will now present a regularity and fundamental trace result. Lemma 3.5. Under the assumptions of Lemma 3.4, we have p ∈ H 1 (0; T ; H 1 ( f )):
(3.20)
Proof. By multiplying the momentum equation by ∇div vh and by integrating with respect to Qf , we obtain 1 1 kdiv vh (t)k2L∞ (0; T ; L2 ( f )) + k∇vh k2L2 (0; T ; L2 ( f )2 ) + k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) 2 2 Z Z Z tZ Z tZ 1 t + ∇p∇div wg − ∇w2 vh − curl vq (w)vh 2 0 f
f
f 0 0 =
1 1 kdiv vh (0)k2f + k∇p(0)k2f : 2 2
(3.21)
To estimate the non-linear terms we proceed as follows: Z tZ
curl vq (w)vh ≤
f
0
3 kcurl vq (w)k2L2 (Qf ) + kvh k2L2 (0; T ; L2 ( f )2 ) : 3
Let us show that curl vq (w) is bounded into L2 (Qf ). To this end we observe that vq ∈ L4 (0; T ; W 1; 4 ( f )2 ) ⇒ rot vq ∈ L4 (Qf ) and that vq ∈ L4 (0; T ; W 1; 4 ( f )3 ) wh = ∇˜ ∈ L4 (0; T ; W 1; 4 ( f )3 )
) ⇒ w ∈ L4 (0; T ; W 1; 4 ( f )2 ):
As a result kcurl vq (w)k2L2 (Qf ) ≤ kcurl vq k2L4 (Qf ) k (w)k2L4 (Qf ) ≤ C: In the same way, as w ∈ L4 (0; T ; w1; 4 ( f )2 ) we prove that ∇w2 is bounded into L (Qf ) and, consequently, that 2
1 2
Z tZ 0
∇w2 vh ≤ C + kvh k2L2 (0; T ; L2 ( f )2 ) : 3
f
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Finally, since the solution of problem (R) satis es kdiv wg (t)k2L∞ (0; T ; L2 ( f )) + kwg k2L2 (0; T ; L2 ( f )2 ) ≤ C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ) (3.22) we obtain Z tZ 0
f
∇p∇ div wg ≤ k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) + C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ):
(3.23)
Consequently, if is small enough, then Eq. (3.21) becomes kdiv vh (t)k2L∞ (0; T ; L2 ( f )) + kvh k2L2 (0; T ; L2 ( f )2 ) +k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) ≤ C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ):
(3.24)
However, by adding the inequalities (3.22), (3.24) and if we recall that pt = −div vh − div wg , then we easily prove that kpt (t)k2L∞ (0; T ; L2 ( f )) + k∇pt k2L2 (0; T ; L2 ( f )2 ) +k∇p(t)k2L∞ (0; T ; L2 ( f )2 ) ≤ C(1 + kgt k2L2 (0; T ; H 1=2 ( p )) ):
(3.25)
In addition, thanks to Theorem 2.1 we know that p ∈ L∞ (0; T ; L2 ( f )). As a result, the inequality (3.25) shows that p ∈ L∞ (0; T ; H 1 ( f )) and that pt ∈ L2 (0; T ; H 1 ( f )) which leads to Eq. (3.20). Also, by using the continuity equation, this result demonstrates that div wh ∈ L2 (0; T ; H 1 ( f )) and that the trace of p − div wh has a meaning in L2 (0; T ; H 1=2 ( )). Lemma 3.6. If w0 ∈ H 2 ( f ) 2 and if g ∈ H 1 (0; T ; H 1=2 ( p )); then we obtain vh; t ∈ L4 (Qf ):
(3.26)
Proof. Let us di erentiate Eq. (3.1) with respect to time, multiply it by vh; t and integrate it with respect to Qf . We obtain 1 kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) + kdiv vh; t k2L2 (0;T ; L2 ( f )) 2 Z tZ Z tZ Z tZ 1 = kvh; t (0)k2f + wwt div vh; t − ∇ t vh; t + pt div vh; t : 2
f
f
f 0 0 0 Thanks to the properties of the special basis, the trace terms are cancelled. In addition, we estimate the right-hand side terms in the following manner: Z tZ 0
f
wwt div vh; t ≤ kwkL∞ (0;T ; L4 ( f ) 2 ) kwt kL2 (0;T ; L4 ( f ) 2 ) kdiv vh; t kL2 (0;T ; L2 ( f ))
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
573
≤ CG kwkL∞ (0;T ; L4 ( f ) 2 ) kwt k1=2 L2 (0;T ; L2 ( f ) 2 ) × kcurlvq; t + div wg; t k1=2 kdiv vh; t kL2 (0;T ; L2 ( f )) L2 (0;T ; L2 ( f ) 2 ) + CG kwkL∞ (0;T ; L4 ( f ) 2 ) kwt kt=2 kdiv vh; t k3=2 L2 (0;T ; L2 ( f ) 2 ) L2 (0;T ; L2 ( f ) 2 ) ≤
CG kwk2L∞ (0;T ; L4 ( f ) 2 ) kvq; t + vh; t + wg; t kL2 (0;T ; L2 ( f ) 2 ) Z CG t × kcurlvq; t + div wg; t kL2 (0;T ; L2 ( f ) 2 ) + kwk4L4 kwt k2L2 0 + 2 CG kdiv vh; t k2L2 (0;T ; L2 ( f ))
≤
with
Z 0
t
CG kwk2L∞ (0;T ; L4 ( f ) 2 ) kvq; t + wg; t kL2 (0;T ; L2 ( f ) 2 ) CG × kcurlvq; t + div wg; t kL2 (0;T ; L2 ( f ) 2 ) + 3 kwk4L∞ (0;T ; L4 ( f ) 2 ) 2 × kcurlvq; t + div wg; t kL2 (Qf ) + CG kvh; t k2L2 (0;T ; L2 ( f ) 2 ) Z CG t + kwk4L4 kwt k2L2 + 2 CG kdiv vh; t k2L2 (0;T ; L2 ( f )) (3.27) 0
kwk4L4 kwt k2L2 ≤ kwk4L∞ (0;T ; L4 ( f ) 2 ) kvq; t + wg; t k2L2 (Qf ) 2 kwk8L∞ (0;T ; L4 ( f ) 2 ) kvq; t + wg; t k2L2 (Qf ) 2 Z t + 22 T 2 kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) + kwk4L4 kvh; t k2L2 : +
0
(3.28)
Moreover, pt = −(div vh + div wg ) ∈ L∞ (0; T ; L2 ( f )) and wg ∈ H 1 (0; T ; H 1 ( f ) 2 );
(3.29)
hence vh ∈ L∞ (0; T ; H 1 ( f ) 2 ):
(3.30)
In addition, as w0 ∈ H 2 ( f ) 2 , Lemma 3.1 and the results (3.29) and (3.30) show that w = vq + vh + wg ∈ L∞ (0; T ; L4 ( f ) 2 );
(3.31)
2
vq; t + wg; t ∈ L (Qf );
(3.32)
curl vq; t + div wg; t ∈ L2 (Qf ):
(3.33)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
and, nally, (3.27) and (3.28) lead to Z tZ wwt div vh; t ≤ C + 2CG (T 2 kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) + kdiv vh; t k2L2 (Qf ) ) 0
f
CG +
Z 0
t
kwk4L4 |vh; t |2L2 :
The remaining terms are estimated as follows: Z tZ 1 pt div vh; t ≤ kpt k2L2 (0;T ; L2 ( f )) + kdiv vh; t k2L2 (0;T ; L2 ( f ))
f 0 ≤ C + kdiv vh; t k2L2 (0;T ; L2 ( f )) ; Z tZ 0
f
t div vh; t ≤
1
≤
C
Z
t
k t k2L2 + kdiv vh; t k2L2 (0;T ; L2 ( f ))
0
Z
t
0
k∇ t k2L1 + kdiv vh; t k2L2 (0;T ; L2 ( f ))
≤ C + kdiv vh; t k2L2 (0;T ; L2 ( f )) : Therefore, if is small enough, we obtain the following estimate: Z t kwk4L4 kvh; t k2L2 ; kvh; t (t)k2L2 ( f ) 2 + kdiv vh; t k2L2 (0;T ; L2 ( f )) ≤ C1 + C2 0
(3.35)
as w ∈ L∞ (0; T ; L4 ( f ) 2 ) we can apply Gronwall’s lemma to kvh; t (t)k2L2 ( f ) 2 . In this way, we obtain kvh; t k2L∞ (0;T ; L2 ( f ) 2 ) ≤ C and consequently, kdiv vh; t k2L2 (0;T ; L2 ( f ) 2 ) ≤ C which shows that vh; t ∈ L4 (Qf ). Lemma 3.7. If p0 ∈ W 1; 4 ( f ) and if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) then vh ∈ L4 (0; T ; W 2; 4 ( f )):
(3.36)
Proof. Let us apply the operator ∇ to the mass equation and multiply it by 4∇p|∇p|2 . By integrating with respect to f , we obtain 1 1 d k∇pk4L4 ≤ kvh |4L4 + kwg k4L4 + 2k∇pk4L4 : dt
(3.37)
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
575
It should be noted that if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) then wg ∈ W 1; 4 (0; T ; W 2; 4 ( f ) 2 ). As a result Eq. (3.37) becomes d 1 k∇pk4L4 ≤ C + kvh k4L4 + 2k∇pk4L4 : dt
(3.38)
Let us provide an estimate of vh . To this end, the momentum equation is multiplied by 4vh |vh |2 and integrated on f . 1 kvh k4L4 ≤ (kvh; t k4L4 + k∇w2 k4L4 + k∇ k4L4 + k∇pk4L4 ) + 4kvh k4L4 :
(3.39)
First, we recall that div vh and vh; t ∈ L4 (Qf ); thus vh ∈ W 1; 4 (Qf ):
(3.40)
Moreover, we have wg ∈ W 1; 4 (0; T ; W 2; 4 ( f ) 2 );
vq ∈ W 1; 4 (Qf );
which imply that w = vq + wh = vq + vh + wg ∈ W 1; 4 (Qf ): Using Sobolev’s inequality we obtain ∇w2 ∈ W 1; 4 (Qf ) and then ∇w2 ∈ L4 (Qf ):
(3.41)
In addition, using inequality (3.35) we have vh ∈ H 1 (0; T ; H 1 ( f ) 2 ); moreover vq ∈ L∞ (0; T ; H 1 ( f ) 2 ) and wg ∈ H 1 (0; T ; H 1 ( f ) 2 ), thus w ∈ L∞ (0; T ; H 1 ( f ) 2 ): Finally, as curl vq ∈ L4 (0; T ; W 1; 4 ( f ) 2 ) we deduce that ∇ ∈ L4 (0; T ; W 1; 4=3 ( f ) 2 ) ⊂ L4 (Qf ):
(3.42)
Then, if in (3.39) is small enough, these estimates show that kvh k4L4 ≤ C(1 + k∇pk4L4 ):
(3.43)
By combining Eqs. (3.38) and (3.43) we obtain d k∇pk4L4 ≤ C1 + C2 k∇pk4L4 dt
(3.44)
and as p0 ∈ W 1; 4 ( f ); we arrive at k∇pk4L∞ (0;T ; L4 ( f ) 2 ) ≤ C;
(3.45)
so by Eq. (3.43), vh ∈ L4 (Qf ). Finally, we have vh ∈ L4 (Qf ); vh ∈ L4 (Qf ) and vh · n = 0 which implies that vh ∈ L4 (0; T ; W 2; 4 ( f ) 2 ).
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Lemma 3.8. If vh0 ∈ H 2 ( f ) 2 and if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )); then wh ∈ L∞ (0; T ; L2 ( f ) 2 );
(3.46)
(p − div w) ∈ L2 (0; T ; H 3=2 ( p )):
(3.47)
Proof. Let us apply the divergence operator to the conservation equation and multiply by div vh; t . After integrating on Qf , we deduce kdiv vh; t k2L2 (0; T ; L2 ( f )) + kvh k2L∞ (0; T ; L2 ( f )2 ) 2 Z Z Z tZ 1 t =− p div vh; t − w2 div vh; t 2 0 f
f 0 Z tZ − div(curl vq (w))div vh; t + kvh (0)k2L2 : 2
f 0
(3.48)
Let us begin by estimating the right-hand side terms. To this end, we rst observe that Z tZ Z Z tZ p div vh; t = pt div vh − p(t) div vh (t) − 0
f
0
+
Z
f
f
f
p(0) div vh (0);
(3.49)
however, using the properties of the special basis, Z Z Z tZ Z tZ 1 t pt div vh = − ∇pt vh − wg; t · n div vh 0
f
f 0 0 ≤ k∇pt kL2 (0; T; L2 ( f )2 ) kvh kL2 (0; T; L2 ( f )2 ) 1 + kwg; t · nkL2 (0; T; L2 ( )) kdiv vh kL2 (0; T; L2 ( )) 1 1 ≤ k∇pt k2L2 (0; T; L2 ( f )2 ) + kwg; t · nk2L2 (0; T; L2 ( 2 2 +2T kvh kL∞ (0; T; L2 ( f )2 ) and
Z
Z −
f
p(t) div vh (t) = Z =
f
f
))
(3.50)
Z ∇p(t)vh (t) −
∇p(t) · n div vh (t) Z Z
∇p(t)vh (t) +
0
t
wg ( ) d · n div vh (t)
≤ k∇p(t)kL∞ (0; T; L2 ( f )2 ) kvh (t)kL∞ (0; T; L2 ( f )2 )
Z t
+ w ( ) d · n kdiv vh (t)kL∞ (0; T; L2 ( g
∞ 2 0
L
(0; T; L ( ))
))
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
577
Z
2
1 1 t
≤ k∇p(t)k2L∞ (0; T; L2 ( f )2 ) + w ( ) d · n g
∞ 0 L (0; T; L2 ( +2kvh (t)k2L∞ (0; T; L2 ( f )2 ) :
))
(3.51)
In addition, as p ∈ H 1 (0; T ; H 1 ( f )) we obtain ∇pt ∈ L2 (0; T ; L2 ( f )2 ) and ∇p(t) ∈ L ( f )2 . Moreover, if g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p )) then wg ∈ W 1; 4 (0; T ; W 2; 4 ( p )). As a result, the inequalities (3.50) and (3.51) become Z tZ pt div vh ≤ C + 2T 2 kvh k2L∞ (0; T; L2 ( f )2 ) ; 2
0
f
Z
−
f
p(t) div vh (t) ≤ C + 2kvh (t)k2L∞ (0; T; L2 ( f )2 ) :
Finally, if p0 ∈ H 2 ( f ) and vh0 ∈ H 1 ( f ) then we obtain the following estimate for Eq. (3.49): Z tZ p div wh; t ≤ C + 2(1 + T 2 )kvh k2L∞ (0; T; L2 ( f )2 ) : (3.52) −
f
0
The estimate of the nonlinear terms is as follows: Z tZ 1 − w2 div vh; t ≤ kw2 k2L2 (0; T ; L2 ( f )) + kdiv vh; t k2L2 (0; T ; L2 ( f )) ;
f 0 −
Z tZ 0
1 div(curl vq (w))div wh; t ≤ kdiv(curl vq (w))k2L2 (0; T ; L2 ( f ))
f +kdiv vh; t k2L2 (0; T ; L2 ( f )) ;
but w ∈ L4 (0; T ; W 2; 4 ( f )2 ) and curl vq ∈ L4 (0; T ; W 1; 4 ( f )), therefore w2 ∈ L2 (0; T ; L2 ( f ));
div(curl vq (w)) ∈ L2 (0; T ; L2 ( f )):
So, the two above inequalities are written as Z tZ w2 div vh; t ≤ C + kdiv vh; t k2L2 (0; T ; L2 ( f )) ; − 0
−
f
Z tZ 0
f
div(curl vq (w))div wh; t ≤ C + kdiv vh; t k2L2 (0; T ; L2 ( f )) :
(3.53) (3.54)
If is small enough, then Eq. (3.48) leads to kdiv vh; t k2L2 (0; T ; L2 ( f )) + kvh k2L∞ (0; T ; L2 ( f )2 ) ≤ C; in addition wg ∈ W 1; 4 (0; T ; W 2; 4 ( f )2 ) hence wh = vh + wg ∈ L∞ (0; T ; L2 ( f )2 ):
(3.55)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
As vq , vh and wg belong to L4 (0; T ; W 2; 4 ( f )2 ), then ∇w2 and ∇ belong to L2 (0; T ; H 1 ( f )2 ). Moreover, using Eq. (3.55), we have div wh; t = div vh; t + div wg; t ∈ L2 (0; T ; L2 ( f )2 ) and so 1 ∇(p − div wh ) = − wh; t − ∇w2 − ∇ ∈ L2 (0; T ; H 1 ( f )2 ): 2 Consequently, (p − div v) is well de ned in L2 (0; T ; H 3=2 ( )). 4. Study of Problem (S) We search the displacement under the following form: u=
n X
ui (t) i ;
i=1 2 2 in which { i }∞ i=1 is an orthonormal basis of L ( p ) and H0 ( p ) associated with the spectral problem 2 = 2 in p ; = ∇ · n = 0 on @ p :
In the following paragraph, we provide the estimate of u necessary to the proof of Theorem 1.1. Lemma 4.1. Under the assumption (p− div v)∈ L2 (0; T ; H 3=2 ( )) and if u1 ∈H 3 ( p ), f ∈ L2 (0; T ; H 2 ( p )), we obtain kut k2L∞ (0; T ; L2 ( p )) + k∇ut k2L∞ (0; T ; L2 ( p )2 ) + k2 uk2L∞ (0; T ; L2 ( p )) ≤ C:
(4.1)
Proof. By applying the ∇ operator to the plate equation, multiplying by ∇ut and integrating on Qp , we obtain kut k2L∞ (0; T ; L2 ( p )) + k∇ut k2L∞ (0; T ; L2 ( p )2 ) + k2 uk2L∞ (0; T ; L2 ( p )) Z Z − ∇utt · nut − 2 u∇ut · n = kut (0)k2p + k∇ut (0)k2p
p
+k2 u(0)k2p + 2
Z 0
p
T
Z
p
Z ∇(p − div wh )∇ut − 2
0
T
Z
p
∇f∇ut :
(4.2)
By using the properties of the basis associated with displacement, the trace terms disappear. In addition, the trace of ∇(p− div v) is de ned in L2 (0; T ; H 1=2 ( )). Hence Z TZ 2 ∇(p − div v)∇ut ≤ C + T 2 k∇ut k2L2 (0; T ; L2 ( p )2 ) : (4.3) 0
p
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
579
As f ∈ L2 (0; T ; H 1 ( p )), we also obtain Z 2
T
Z
∇f∇ut ≤ C + T 2 k∇ut k2L∞ (0; T ; L2 ( p )2 ) :
p
0
(4.4)
By inserting Eqs. (4.3) and (4.4) in Eq. (4.2), and if is small enough, we deduce kut k2L∞ (0; T ; L2 ( p )) + k∇ut k2L∞ (0; T ; L2 ( p )2 ) + k2 uk2L∞ (0; T ; L2 ( p )) ≤ K which achieves the proof. Lemma 4.2. If u0 ∈ H 4 ( p ), p0 ∈ H 1 ( f ), wh0 ∈ H 2 ( f )2 and f ∈ H 1 (0; T ; H 2 ( p )) then we obtain k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) ≤ C(1 + kgk2w1; 4 (0; T ; W 3=2;1 ( p )) ):
(4.5)
Proof. Let us di erentiate the plate equation with respect to time and multiply it by utt . After integrating on Qp , we obtain k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) Z =
Z
T
Z −
2 ut utt + k∇utt (0)k2f + kutt (0)k2f
p
0
T
Z
Z
p
0
ft utt +
T
Z
p
0
(p − div wh )uttt
Z −
Z
p
(p(t) − div wh (t))utt (t) +
p
(p(0) − div wh (0))utt (0):
(4.6)
In the same way, let us multiply it by uttt . We thus obtain, after integrating on Qp k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) Z =−
T
0
Z
p
2 ut utt +
Z
p
2 u(t)utt (t) −
+k∇utt (0)k2f + kutt (0)k2f + Z −
p
Z 0
T
p
p
2 u(0)utt (0)
Z
p
ft utt
Z f(t)utt (t) +
Z
Z f(0)utt (0) −
0
T
Z
p
(p − div wh )uttt :
(4.7)
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Hence, by summing (4.6) and (4.7) we are led to 2k∇utt k2L∞ (0; T ; L2 ( p )) + 2kutt k2L∞ (0; T ; L2 ( p )) Z 2 2 = 2k∇utt (0)kf + 2kutt (0)kf + 2 u(t)utt (t) Z − Z +
p
p
2 u(0)utt (0) −
Z
p
p
(p(t) − div wh (t))utt (t) Z
(p(0) − div wh (0))utt (0) −
p
Z f(t)utt (t) +
p
f(0)utt (0): (4.8)
By applying Young’s inequality to this last result, we obtain k∇utt k2L∞ (0; T ; L2 ( p )) + kutt k2L∞ (0; T ; L2 ( p )) ≤ C(1 + kp(t)k2L∞ (0; T ; L2 ( p )) + · · · + kdiv wh (t)k2L∞ (0; T ; L2 ( p )) +k2 u(t)k2L∞ (0; T ; L2 ( p )) ) + kutt k2L∞ (0; T ; L2 ( p )) :
(4.9)
But, by using Eqs. (3.25), (3.46) and (4.1) we deduce kp(t)k2L∞ (0; T ; L2 ( p )) ≤ Ckpk2H 1 (0; T ; H 1 ( f )) ≤ C(1 + kgk2W 1; 4 (0; T ; W 3=2; 4 ( p )) ); kdiv wh (t)k2L∞ (0; T ; L2 ( p )) ≤ Ckwh (t)k2L∞ (0; T ; L2 ( f )2 ) ≤ C; k2 u(t)k2L∞ (0; T ; L2 ( p )) ≤ C: Therefore, if is suciently small, we obtain k∇u tt k2L∞ (0; T ; L2 ( p )) + ku tt k2L∞ (0; T ; L2 ( p )) ≤ C(1 + kgk2W 1; 4 (0; T ; W 3=2; 4 ( p )) ) which achieves the proof. 5. Proof of the Theorem We apply here Kakutani’s theorem. To this end we note that W 1; 4 (0; T ; H02 ( p )) ⊂ E = W 1; 4 (0; T ; W 3=2; 4 ( p )) algebraically and topologically, hence (4.5) involves ku tt kL4 (0; T ;W 3=2; 4 ( p )) ≤ C 0 ku tt kL4 (0; T ; H02 ( p )) ≤ TC 00 (1 + kgkW 1; 4 (0; T ; W 3=2; 4 ( p )) ): Thus, we can de ne the following map: : g ∈ E → u t ∈ E: Then, if the value of T is small enough and if kgkE ≤ K with K being well chosen, the map applies the ball B(0; K) of E with center 0 and radius K within itself. That is to say (B(0; K)) ⊂ B(0; K):
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
581
Now, we recall that E is a re exive and separable banach space. Consequently, B(0; K) is metrisable for the weak topology and, nally, compact for this same topology. We are therefore able to work with ordinary sequences. We will prove that if we consider gn as a weak convergent sequence towards g in B(0; K) and such that weakly
un; t = (gn ) −→ t then t = (g) = u t . If (wn ; pn ) is a solution associated to (F) with w(0) = wn (0) and p(0) = pn (0), then (wn ; pn ) satis es the estimates for Lemmas 3.1 – 3.8 and (pn − div wn ) weakly converges towards a limit ! in L2 (0; T ; H 3=2 ( )). As problem (F) allows only one solution, then ! = (p − div w). As a result, for the problem (S) we have (un; tt ; )p + (∇un; tt ; ∇ )p + (un ; )p = −(pn − div wn ; )p + (fn ; )p which converges weakly towards ( n ; )p + (∇ tt ; ∇ )p + ( ; )p = −(p − div w; )p + (f; )p = (H; )p weakly
when gn −→ g in B(0; K). Finally, as problem (S) has only one solution for a given H , then is the unique solution associated to (S) and, hence, t = u t . This last result shows that we have a unique xed point and that w ·n = −u t on p in W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )). We have shown the existence for a small time but this result can be extended for all time thanks to the a priori estimates. 6. Numerical implementation 6.1. Approximate solutions In order to facilitate the numerical implementation, we state p = ]0; 1[ and f = p ×]0; 1[. Due to this simpli ed study domain, resolving the spectral problems de ned in Section 3 provides an analytical basis for V orthonormalized in L2 ( f )2 . To obtain a basis for W˜ we solve the problem ˜ ˜ in Qf ; hl = hl ˜ (H) grad h˜ l · n = 0 on f × ]0; T [; grad h˜ l · n = l on f × ]0; T [; 2 in which { l }∞ l=1 is the basis of H0 ( p ) associated with the movement of the plate. To this end, we use the nite elements method and the MODULEF software. The element utilized is the 2P3D, which is an Hermite type element with three degrees of freedom. This element allows us to obtain, in addition to the solution to prob˜ the numerical values of its primary x and y derivatives. Thanks to this lem (H),
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F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
property, we can therefore construct a basis for W˜ . Consequently, we search to determine the movement of the plate and the velocity in the uid, respectively, as follows: M X
u=
ul (t) l ;
l=1
w=
N X
N X
Aj (t)Curl qj +
j=1
Bi (t)∇hi +
i=1
M X
Cl (t)∇h˜ l :
l=1
In addition, if Bi (t) =
dbi (t) dt
and
Cl (t) =
dcl (t) ; dt
we recall that we can look for the pressure as p=
N X
i bi (t)hi −
i=1
M X
cl (t)h˜ l
l=1
if bi (0) and cl (0) are chosen such that p(0) =
N X
i bi (0)hi −
M X
i=1
cl (0)h˜ l :
l=1
6.2. Projection of the equations In order to obtain a matrix system, we project the equations onto the bases de ned previously. Projection of the equation of problem (R) onto {h˜ l }M l=1 leads to " # Z Z Z M X d 2 cl ˜ ˜ ˜hl h˜ m + dcl ∇hl ∇hm = gh˜ m ; 1 ≤ m ≤ M: (6.1) dt 2 f dt f
p l=1
In the same way, by successively projecting the conservation equation on ∇hn and Curl qn , we obtain n
n d 2 bn dbn + n2 bn + + n2 dt 2 dt 2 Z +
n
f
curl w (w)∇hn =
d 2 an dan + + n2 2 dt dt
M X l=1
Z
f
w2 hn Z
cl n
f
h˜ l hn ;
1≤n≤N
(6.2)
Z
f
curl w (w) curl qn = 0;
1 ≤ n ≤ N:
(6.3)
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
583
Finally, the projection of the plate equation on the m basis functions leads to " ! # Z Z Z M X dul d 2 ul l m + ∇ l ∇ m + l m dt 2 dt p
p
p l=1
Z =
p
+
f m +
N X
dbi −bi + dt
i=1
M X
cl −
l=1
dct dt
Z
p
h˜ l m ;
Z i
p
hi m
1 ≤ m ≤ M:
(6.4)
Remark. By combining equations (6.1)–(6.4), we can easily return to a second order di erential equations system of the form cl cl cl 2 a d a d i a + nonlinear term [A] 2 + [B] i + [C] i = (F) dt bi dt bi bi ul ul ul which we can solve using a Neumark type scheme coupled to a prediction–correction method. This method is not suitable; however; as the solution must also verify the coupling condition M X
ul; t (t) l = −
l=1
M X
Cl (t) grad h˜ l · n = −
l=1
M X
Cl (t) l
l=1
and it is clear that the system is over-determined. To overcome this diculty and, as was seen in the existence Theorem 1.1, we introduce an algorithm based on xed point method. 6.3. Algorithm We set g ∈ W 1; 4 (0; T ; W 3=2; 4 ( p ) ∩ H02 ( p )) and we work on the time-step [tn−1 ; tn ]. Given the solution at tn−1 , we solve (R) using an implicit Euler scheme. Once cl (tn ) are known, we solve the homogeneous (FH) problem using a Neumark type scheme by proceeding with i implicit iterations in such a way as to approach the nonlinear term. By using the same Neumark scheme, we can solve the plate problem (S). The solution to (S) being known, we bring g(tn ) = −ut (tn ) to the problem (R). We reiterate the process k times until convergence of u(tn ) towards a xed point. Finally, if at the kth iteration we note R[ck (tn )] = KG (gk (tn )];
584
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Fig. 1.
FH (ak (tn ); bk (tn )] = B[ak (tn ); bk (tn ); ck (tn )] + Kc [ck (tn )]; S[uk (tn )] = KT [bk (tn ); ck (tn )] + f(tn ) the systems (6.1)–(6.4) with B[ak (tn ); bk (tn ); ck (tn )] as the nonlinear term, we can represent the algorithm as shown in Fig. 1. 6.4. Validation of the algorithm To validate the algorithm, we introduce the following analytical solution: 2 0 2 3 4 5 t ; u(x; t) = 2 (−x + 4x − 5x + 2x ) cos(y) exp − 0 wg (x; y; t) = ∇hg (x; y; t) = ∇
1 3 1 6 2 4 5 − x + x − x + x cos(y) exp − t ; 3 3 0
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
585
Fig. 2.
in which 0 is the density at rest and the viscosity. This solution satis es, in particular, the coupling condition wg · n = −ut
on p
and the boundary conditions u = ∇u · n = 0
for x = 0
and x = 1:
We suppose that 0 = 1 kg=m3 ; −10
I = 2:8 × 10
= 6 × 10−5 m2 =s;
m = 11:7 kg=m2 ;
D = 65:34 N:m;
3
m
with m, D and I the surfacic mass, the rigidity and the inertia moment of the plate, respectively. We test the algorithm for two points taken in the uid (x = 0:05; y = 0:003285) and on the plate (x = 0:05; y = 0:0). Below are represented some results obtained with the method for these values. Fig. 2 makes the comparison between the analytical and the numerical solutions. Figs. 3 and 4 present the relative and absolute errors for hg (x; y; t) and u(x; t). The decrease in the precision of the results over time is due to the form of the solution which approaches 0 when t increases. All in all, however, this method seems to provide encouraging results despite its numerically demand aspects.
586
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
Fig. 3.
Fig. 4.
F. Flori, P. Orenga / Nonlinear Analysis 35 (1999) 561 – 587
587
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