Channel flows and steady variational inequalities of the Navier-Stokes type Текст научной статьи по специальности «Математика»

Вычислительные технологии

Том 7, № 1, 2002

CHANNEL FLOWS AND STEADY VARIATIONAL INEQUALITIES OF THE NAVIER-STOKES TYPE*

St. KraCmar Czech Technical University, Prague e-mail: kracmar@marian.fsik.cvut.cz

Исследуется стационарное течение вязкой несжимаемой жидкости в канале с условиями на выходе, отличными от условий Дирихле. Для того чтобы контролировать кинетическую энергию жидкости в канале, предполагается, что возможные обратные течения на выходе в некотором смысле ограничены. Течения, удовлетворяющие этому условию, заполняют выпуклое подмножество пространства определенных функций. На этом выпуклом множестве формулируется вариационное неравенство типа Навье — Стокса и доказывается существование слабого решения. Предположение, используемое для определения выпуклого подмножества, более ограничительно, чем предположение, из работы [3]. С другой стороны, условие в теореме существования менее строго, чем условие из [3]. Кроме того, изучается вопрос о том, в каком смысле слабое решение удовлетворяет уравнениям Навье — Стокса и смешанным граничным условиям, если это решение гладкое.

Introduction

Let Q be a simply connected bounded domain with a Lipschitz boundary dQ in IR3. Q can be considered as a channel filled up by a moving fluid, r will denote the part of the boundary where the fluid is flowing into the channel or where the channel has fixed boundaries and r2 will denote the part of the boundary where the fluid is supposed to leave the channel. Precisely, we will suppose that r^ r2 are open disjoint subsets of dQ such that dQ = r U r2 and the dQ — r — r2 consists of a finite number of closed simple smooth curves whose each point belongs to dr fl dr2 (dr, respectively dr2, denotes r — ri, respectively r2 — r2). We will also suppose that r2 is a union of a finite number of disjoint simple smooth surfaces S1, ... , Sr. Let us denote their boundaries by C1, ... , Cr .It follows from the previous assumptions that Ci, ..., Cr are closed simple smooth curves.

The motion of a viscous incompressible fluid in Q, generally non-stationary, can be described by the Navier — Stokes equation

d u

— + uVu = f — V p + v Au (1)

and the equation of continuity

div u = 0, (2)

*The research was supported by the Grant Agency of the Czech Republic (grant No. 201/99/0267) and by the research plan of the Ministry of Education of the Czech Republic (MSM98/21000010). The authors are responsible for possible misprints and the quality of translation. © St. Kracmar, 2002.

where u = («1, u2, u3) is the velocity, p is the pressure, f = (/1, /2, /3) is an external body force and v is the kinematic coefficient of viscosity. v is a positive constant.

It is natural to prescribe a Dirichlet boundary condition for the velocity of the fluid on r1. However, since the situation on the output of the channel depends on the behaviour of the fluid inside the channel and it is not known in advance, it is not reasonable to use a boundary condition of the same type on r2. It is a matter of discussion which boundary condition should be used on r2. One of the possibilities is the condition

d u

-p n + v— = F, (3)

d n

where n = (n1, n2,n3) is an outer normal vector on r2 and F = (F1, F2,F3) is a prescribed vector function on r2. It can be shown that a weak problem for equations (1), (2) with no condition on r2 involves condition (3) implicitly. It means that if its solution is "smooth enough" then except for equations (1), (2), it must also satisfy condition (3). This is why condition (3) is often called the "do nothing condition".

Existence or uniqueness of solutions of system (1), (2) with condition (3) on the part of dQ is known either locally in time (see e.g. P. Kucera, Z. Skalak [4]) or for "small data" (i. e. "small" initial velocity, "small" external force and "small" function F in condition (3) — see e.g. P. Kucera [5, 7]). The problem with condition (3) is difficult mainly because solutions of (1), (2), (3) need not satisfy an energy inequality. This is due to the fact that boundary condition (3) on r2 does not exclude backward flows on r2, bringing into Q an uncontrollable amount of kinetic energy. The kinetic energy in Q can be estimated by an additional condition on r2 which estimates the backward flows. For example if c0 > 0 then the following condition can be used:

dist(u(x), Ka(x)) dSx < c0, (4)

r2

where Ka(x) denotes the cone of vectors in IR3 whose angle with n(x) is less than or equal to

a, a £ ^0, —^ , and dist(u(x), Ka(x)) means the distance between u(x) and Ka(x). However,

condition (4) has the consequence: if we use it then we are searching for a solution not in a whole function space (which will be exactly specified later) but in its convex subset. This is why we do not use the Navier — Stokes equation (1) and instead of it, we describe the flow by means of a certain variational inequality which arises from equation (1). We have already used this approach in paper [2] where we studied the non-stationary case and we have proved the global in time existence of a weak solution without any restriction on the size of the input data.

In this paper, we study a steady motion of a viscous incompressible fluid in channel Q with the Dirichlet boundary condition

u = u* (5)

on r and with conditions (3) and (4) on r2. Analogously as in [2], we formulate a variational inequality of the Navier — Stokes type, which is now steady, and we prove its weak solvability without any requirement on the size of the input data f, F and u*. We also show that if the solution is smooth enough then it satisfies the Navier — Stokes equation and moreover, if it in some sense finds itself in the interior of the convex set which is defined by means of condition (4) then it also satisfies condition (3) on r2.

1. Formulation of the problem and some properties of its solution

Let c0 be a positive real number and a £ (2, 4). Since 2a/(a — 1) < 4, there exists a continuous

I yn (X ) I

operator of traces from V into L2a/(a-1)(dQ). Let Ka(x) = < y £ IR3 : > cos a \ , a £

|y|

(0, f) x £ r

|| . ||r will denote the Lr-norm on domain Q, || . ||r,r2 the Lr-norm on r2 and || . ||r,s will denote the Wr's-norm on Q. The Wr's-norm on dQ will be denoted by || . ||r;S(dn). In order not to complicate the notation, we will denote traces on dQ of functions which are defined a. e. in Q by the same letters as the functions themselves.

Suppose that function u* in boundary condition (5) is such that it can be extended from r onto the whole boundary dQ so that the extended function (it will be also denoted by u*) belongs to W 1/2'2(öQ)3,

J u* ■ n dS = 0

ÖQ

and on each simple smooth surface Sk, which is a component of r2, u* = 0 in a certain neighbourhood S'k of the boundary Ck and u*(x) = k(x) n (for some k(x) > 0) in all other points x £ Sk.

Lemma 1. There exists a function V £ W 1,2(Q)3 with the following properties:

1. div V = 0 a. e. in Q,

2. V = u* a. e. on T1;

J

r2

dist(V(x), Ka(x))

dSx

4. J (v1Vv2)V dx < (v/2) ||Vv1|2 ||Vv2|2 for all v1, v2 £ W 1,2(Q)3 whose traces are n

zero on r1.

Proof: The proof is in many steps similar to the proof of Lemma VIII.4.2 in [1] or Lemma II. 1.8 in [9]. The main differences between our situation and the situation treated in [1] and in [9] are that we require V to be equal to u* only on the part r1 of dQ, our v1, v2 have the traces equal to zero only on r1 and we require function V to satisfy the equality in item 3 of the lemma. We will show the construction of function V and we will especially pay attention to the equality in item 3. We follow the arguments given in [3]:

There exists a function w1 £ W2'2(Q)3 such that u* = curl w1 in the sense of traces on dQ. Moreover,

|| W11| 2,2 < C1 || u* || 1/2,2(dn), (6)

where c1 = c1(Q). (See Lemma VIII.4.1 in [1].) It follows from the Stokes theorem that the line integral of w1 on each closed simple curve C in S'k (where k £ {1; ...; r}) whose interior is a subset of S'k is equal to zero. (By interior we mean interior on surface Sk — i. e. that one of the two components of the set Sk — C which has a positive distance from the boundary Ck of surface Sk.) Furthermore, the line integral of w1 on each closed simple curve C" in S'k whose

a

0

interior contains Sk — S'k and which is positively oriented when observed from the outer part of Sk is equal to a certain nonnegative number — the flux of u* through surface Sk. Suppose for simplicity that > 0 for all k £ {1; ...; r}.

If x £ Q then we define $(x) as the infimum of lengths of all possible curves in Q whose initial point is x and their terminal point is on r. (I. e. $(x) is a distance of x from r, measured only on trajectories never leaving Q.)

Using the same approach as in the proof of Lemma III.6.2 and in the proof of Lemma III.6.3 in [1], it can be shown that there exists c3 > 0 such that

||u/tf||2 < C3 ||u||i, 2 (7)

for all u £ W1 ' 2(Q)3 whose trace on r1 is zero.

Let e be a positive parameter. We define the function

( 1 if A < e-2/e,

fe(A) = < —1 — e ln A if e-2/e < A < e-1/e, [o if e-1/e < A.

Let Re be the mollifier with the kernel whose support has the diameter ^ e-2/e. We define

^(x) = Re£e(£(x)).

Then _

i) |^e(x)| < 1 for all x £ Q,

ii) ^e(x) = 1 if ¿(x) < 1 e-2/e,

iii) 0 < ^e(x) < 1 if 1 e-2/e < ¿(x) < e-1/e + 1 e-2/e, 212

iv) ^e(x) = 0 if 5(x) > e-1/e ^ e

_ e-2/e 2

1

_ e-2/e

v) |V^e(x)| < e/5(x) if 5(x) < e-1/e + — e

It can be shown that if e is sufficiently small (what we will further assume) then the following assertions hold for each k £ {1; ...; r}:

— The set S'k€ = {x £ Sk; V^e(x) = 0} is a subset of S'k and S'k€ can be expressed as a union of mutually disjoint closed simple smooth curves Cy e (for y £ (0,1)), each of whose is an equipotential line of function Cy e = {x £ Sk; ^e(x) = y}.

— Each of the curves Cy e contains the set Sk — S'k in its interior.

— The vector n x is tangent to each of the curves Cy e. Let us further assume that this vector defines the orientation of Cy e.

— The system S of curves in Sk' e whose tangent vector is — (V^en) n (the tangent to Sk component of V^e) is perpendicular to the system of curves Cy e and these curves also cover the whole set Sk' e.

Suppose further that P(t); t £ [0,^k] is a parametrization of one of the curves Cy e — let it

be e.g. the curve Ck /£2. Due to the smoothness of the curve Ck /£2, the parametrization can be

chosen so that P+(0) = P-(,5k). Let us denote by e (for t £ [0,^k)) that one of the curves

from system S defined above, whose intersection with curve C^2 is the point P(t).

Let us now define a real function on Sk e in this way: = t in all points x G ^Cjt,

Put w2 = V01 in Ske- In fact, since function has a discontinuity on curve e, V01 is not defined on e. However, it follows from the introduction of function that V01 can be continuously extended to ^CjO e. Thus, we understand by V01 the value of this extension on ^C0 e. Function w2 has obviously these properties:

— The line integral of w2 on each closed simple curve C in Sk whose interior is a subset of Sk is equal to zero and the line integral of w2 on each closed simple curve C" in Sk whose interior contains Sk — Sk and which is positively oriented when observed from the outer part of Sk is equal to .

— w2 is tangent to the curves Cy e and so it has the same direction as n x V^e in all points of Sk,e-

Thus, the line integral of w1 — w2 does not depend on the path in Sk e- Hence we can define a scalar function 02 on Sk' e so that we choose a fixed point x0 G Sk' e and we put 02(x) equal to the value of the line integral of w1 — w2 on any curve in Sk' e which starts in xo and terminates in x. Function 02 can be extended from the union of all Sk' e (k G {1; ... ; r}) to a smooth function in Q such that its gradient is in W2 '2(Q)3 and V02n = 0 on Sk' e (k G {1; ... ; r}). Then the function w = w1 — V02 coincides with w2 on Sk' e and curl w is identical with curl w1. It also follows from (6) that for some c2 > 0

w

12,2 < c1 ||u*|i/2)2(ön) + C2. (8)

It follows from the smoothness of curves Ck that c2 can be chosen so that it is independent of e.

We put Ve = curl w). Then Ve = ^ curl w + x w. It follows from our choice of e that ^e(x) = 0 in all points x £ Sk where curl w(x) = 0. Thus, Ven = (V^e x w)n = (n x V^e)w. This is obviously positive in all points of Sk e and equal to zero in all points of Sk — Sk e (for each k £ {1; ...; r}). Moreover, it follows from the smoothness of Sk that fo sufficiently small e the angle between and n(x) on Sk' e is from [n/2 — a/2,n/2 + a/2]. Hence, the angle between Ve = x w and n is from [0,a/2], and function Ve satisfies condition in item 3 of the lemma.

It remains to verify that if e is small enough and we put V = Ve then the inequality in item 4 is satisfied. However, the proof can be done in the same way as the proof of estimate (4.38) in [1], p. 32, and so we do not show it here. ■

We will further search for the velocity u in the form u = V+v where V is the function given by Lemma 1 and v £ W1 ' 2(Q)3 is a new unknown function such that v|ri = 0. Let V be a set of infinitely differentiable divergence-free vector functions in Q which have a compact support in Q U r2. Let K be a subset of V which contains only functions h satisfying the condition

a

J |dist(V(x) + h(x),Ka(x))J dSx < co. (9)

r2

Denote by H (respectively V) the closure of V in L2(Q)3 (respectively in the norm || V. ||2) and denote by K the closure of K in V. The dual space to V will be denoted by V' and the duality between elements of V' and V will be denoted by (. , .).

It follows from the existence of a continuous operator of traces from V into L"(r2)3 that

K = < h G V : J

r2

dist(V(x) + h(x),Ka(x))

dSx < co

a

If we also use item 3 of Lemma 1, we can show that there exists e1 > 0 so that K contains the ei-neighbourhood of zero in V. Moreover, K is a convex set in V; This can be proved by means of the Minkowski inequality.

Let us formally derive the Navier — Stokes variational inequality now. (See e.g. [2] or [8] for similar approaches.) We use the steady Navier — Stokes equation in the form

uVu - f + Vp — v Au = 0, (10)

we multiply the left hand side by q — v (where q is a test function from K), we integrate over Q and we require the result to be greater or equal to zero. We obtain:

J[ uVu - f + Vp - v Au ](q - v) dx > 0.

If we express u in the form V+v , integrate by parts, use boundary condition (3) and write the duality (f , q — v) instead of the scalar product of f and q — v in H, we obtain the inequality

J[(V + v)V(V + v)](q — v) dx + v J V(V + v)V(q — v) dx — n n

— (f , q — v) — y F(q — v) dS > 0. (11)

r2

We will solve the following problem:

Problem 1. Let f G V' and F G L2(r2)3. We are looking for function v G K such that inequality (11) holds for all functions q G K.

The next two theorems show that if a solution of Problem 1 is aposteriori smooth enough then it satisfies the Navier — Stokes equation (10) and the inequality holds only on r2 (Theorem 1) or the solution satisfies boundary condition (3) on r2 (Theorem 2).

Theorem 1. Let f G L2(Q)3, let function u* can be extended from r1 to dQ so that except for already mentioned properties, the extended function belongs to W3/2,2(dQ)3 and let v be a solution of Problem 1 such that v G W2'2(Q)3. Then there exists p G W 1,2(Q) such that u = V + v and p satisfy the steady Navier — Stokes equation (10) in Q in a strong sense and

d u

v — — p n — F d n

(q' - v) dS > 0 (12)

r2

for all q' G K.

Proof: Since function u* has a higher regularity than it was assumed in Lemma 1, function V given by Lemma 1 can also be found so that it has a higher regularity, namely that it belongs to W2'2(H)3. Let q' G W2'2(H)3 n K at first and 0 G (0,1). There exists a sequence {qn} in K such that qn ^ 0q' + (1 - 0)v in W2'2(^)3. If we use q = qn in inequality (11) and assume that n ^ we obtain

0 J[(V + v)V(V + v) ](q' - v) dx + v0 j V(V + v)V(q' - v) dx -n n

- 0 y f (q' - v) dx - 0 y F(q' - v) dS > 0.

n r2

n

If we divide this inequality by 0, assume that 0 ^ 0+ and write u instead of w + v, we get J[ uVu - f ](q' - v) dx + v j VuV(q' - v) dx - J F(q' - v) dS > 0,

r2

J [ uVu - v Au - f ](q' - v) dx + J

n r2

d u ^

"IT - F

d n

(q' - v) dS > 0.

(13)

Let 0 be any function from W2'2(H)3 fl W0' (H)3 such that div0 = 0. If we successively use q' = v + 0 and q' = v - 0 in (13), we obtain

[ uVu - v Au - f ]0 dx = 0.

This implies the existence of p G W 1 (H) such that ||p||2 < c3(H)||Vp||2 and

n

n

n

uVu - v Au - f = -Vp (14)

a. e. in H (see [1]). Let us now return to inequality (13). If we use (14), apply the integration by parts and use the fact that functions q' and v are divergence-free, we obtain (12). The validity of (12) for all q' G K follows from the density of W2>2(H)3 f K in K. ■

Theorem 2. Let the assumptions of Theorem 1 be fulfilled and moreover, let there exist a neighbourhood U of zero in V so that v + h G K for all h G U. Then u = V + v and p (given by Theorem 1) satisfy boundary condition (3) a. e. on r2.

Proof: Let h G U. The functions q1 = v + h and q2 = v - h belong to K. If we successively use them in (12) instead of q', we obtain:

du _

v — - p n - F dn

h dS = 0.

r2

Since this holds for all h G U, u and p satisfy condition (3) a. e. on r2.

2. Approximations and their estimates

We will prove that if c0 is an arbitrary positive constant then Problem 1 has at least one solution in Section 4. In Section 3, we will construct a sequence of approximations and derive some estimates. We will use the Galerkin method combined with the method of penalisation.

Let P be the projector of V onto K which assigns to each element of V the nearest element in K and put ^(h) = h - P h for h G V .It follows from the convexity of K that ^ is a monotone operator in V. It will be used as a penalisation in the following. Let us prove that

(tf(h), h )v >||tf(h)||V, and (tf(h), h )v > ei||tf(h)||v (15)

(for all h G V) at first. (We remind that K contains the e1-neighbourhood of zero in V.) The first inequality in (15) is obvious:

(tf(h), h)v = (h - Ph, h)v = (h - Ph, h - Ph)v + (h - Ph, Ph - 0)v

and since 0 £ K and K is convex, (h — Ph, Ph — 0)V > 0 and so we obtain the desired inequality. The second inequality is clearly satisfied if h £ K. Thus, let h £ K. Put p = e1(h — Ph)/||h — Ph||V. Then p £ K and

(tf(h), h)v = (h — Ph, h)v = (h — Ph, h — p)y + (h — Ph, p)y = = (h — Ph, h — Ph)V + (h — Ph, Ph — p)V + e1(h — Ph, h — Ph)V/||h — Ph||V >

> ejh — Ph|y = e1 ||*(h)||v .

Put V2 = V fl W2 ' 2(Q)3. Then V2 is a Hilbert space with the same scalar product as W2 ' 2(Q)3. Let functions ek (k = 1, 2,...) form a basis of V2 which is orthonormal in H. It follows from the density of V2 in V and in H and from the continuity of imbeddings of V2 into V and into H that { e1, e2,... } is also a basis in V and in H. Functions ek (k = 1, 2,...) can be chosen so that they all belong to V. Let n £ IN be given. We are looking for 0J! £ IR (k =1, ... , n) so that the function

v

ek

k=1

(16)

satisfies for k = 1, . . . , n the equations

(Pvn + V)V(vra + V) ek dx + v J V(vra + V)V ek dx —

n

— (f , ek ) — F ek dS + n (1 + ||vn||2r2) (^(vn) , ek)

V

(17)

r2

where q £ (1, 4) will be chosen later. Substituting here from (16), we get a system of algebraic equations for unknowns

n

1=1

m=1

e1 + VJ em + V ekdx + v V (J^ ^ + VJ Vekdx — (f, ek) —

1=1

— J F ek dS + n r2

1 +

V

¿=1

in_Z

q \ q

dS

q \

*(£ ^ em) , ek

/

m=1

V

(18)

for k = 1, ... , n. Let us denote by 0n the n-tuple [0n, ... , ^L by Gk(0n) the left-hand side of equation (18) and by G(0n) the n-tuple [G1 (0n), ... , Gn(0n)]. Then system (18) is equivalent to the equation

G (0n) = 0n, (19)

where 0n is the zero element in IRn. G is a continuous mapping from IRn to IRn. We will show that there exists R > 0 (independent of n) such that

G(r)r > 0

(20)

for all r £ IRn such that |0n | = R:

n

n

0

n

n

n

n

n

n

0

2

Multiplying (0n) by 0J!, summing over k from 1 to n, using (16), and using item 4 of Lemma 1 we obtain

(Pvn + V)V(vn + V) vn dx + v V(vn + V)Vvn dx -

fc=i

-(f,vn) -J FvndS + n r2

v

№

vn), vn)

V

(Pvn + V)V(vn + V) J (vn + V) dx - J [ (Pvn + V)V(vn + V) n n

+ v J V(vn + V) Vvn dx - (f , vn ) - J Fvn dS + n|i + II V" n r2

Iq,r2

V dx + vn), vra^

(Pvra + V)n 1 |vn + V|2 dS + f (Vn) 2 |V|2 dS -

P vnVvn

V dx -

r2

ri

P vnVV

V dx -

VV(vn + V)

V dx +

n n

+ v ||Vvn||2 + v J VVVv" dx - (f , vn ) - J Fvn dS + n n r2

v

Iq,r2

vn), vn >

V

>

(Pvn + V)n) |vn + V|2 dS + I (u*n) 2 |u*|2 dS - 1 v ||VPvn||2 ||Vvn||2 -

ri

- C4 || Vvn 12 - C5 + v ||Vvn|2 + n (l+ KH^) (^OO, vj

V

where r2 is the part of r2 with (Pvn + V)n < 0.

1 f 2 The term ^ [(Pvn + V) n] |vra + V| dS can be estimated by using Holder's inequality.

There exists e G (0, 2/3), r G [1, a], and q G (1,4) ( r = r(e) > 4ae/(2a - 4 + ae)), such that

2 i [(Pvn + V) n] |vn + V|2 dS >

>-

i

V

|(Pvn + V) n|° dS

V-

/

( \

I |vn + V|q dS

V2- )

(2-e)

i

\1

|vn + V|r dS

v2

/

>

>-Coac6 ||vn + V^"- K + V||«r- >

>-c0C6 ||vn + V|q,r-

12-e

vn - pv"l

+ ||P vn + V||,

>

> -CoaC6 ||vn + V|2"r- ( ||vn - Pvn||fr- + ||Pvn + V||« ) >

n

n

2

n

2

r

2

r

2

i

q

2

e

i

> — cQce ||vn + V

|2 —£

l^(vn +

( (

c7 cos a

\

dist (Pvn + V,K«)a dS

\ V2-

(because, if wn < 0 then |w| < (1/cos a) dist (w, Ka))

/ / J

>

> — cQce ||vn + V

|2 —£ q,r2

n\ 11 £

+ I — cq^X 2 cos a

>

>—c (KCr- + 1

I^(vn)|£,r2 + C

Constants C1,C2 depend on Q, r1, r2, u*, V, a, a, cq, e, and q. Using now the estimate of the term 2 f [(Pvn + V) n] |vn + V|2 dS for evaluation of (G0n0n) we get

Vvn

Vvn

I 1 —£

v

G (0n) 0n > Vvn 12 — C1c9 (1 + C2)) — c4

— ca +

>< +n (|vn|2,r2 + 1) (vn) , vn)V , (v

'n)|q,r2 < 1,

Vvn

12^2 ||Vvn|2 — c4 )—'

)— c8 + (n — n,)(||vnn2>ra + 1)(^(vn) , vn)v , ||^(vn)Hq>r2 > 1

where nQ is such that nQe1 > C1(1 + C2)c9, with e1 from relation (15). c4, ... , c9 are appropriate constants which depend only on Q, r1, r2, u*, V, a, a, cq, e, and q. Let us define constant RQ with the following property:

|| Vvn||2 > Rq ^ G (0n) > 0. It follows from (21) that RQ can be taken e. g.

Rq = max

2

v

C1 (C2 + 1) c9 + 1

2

(c4 + 1) 1—£ ,c8 + 1,- (c4 + 1)

v

(21)

Suppose that

|0n| = ||vn||2 = R>Rq c—q1, (22)

where c1Q is the constant from the inequality ||Vvn||2 > c1Q ||vn||2. Then

Rq < c1qR = c1q || vn || 2 < ||Vv||2,

and so from (21) we have

G (0n) r > 0.

Hence equation (19) has a solution 0n in the ball BR(0n) in IRn (see [9], Lemma II.1.4). The radius R of this ball does not depend on n.

The fact that solution 0n of equation (19) satisfies |0n| < R means that function vn given by (16) satisfies the estimate ||vn||2 < R. However, we will also need estimates of ||Vvn||2

i

r

2

2

2

2

and the penalisation term. If we assume a solution 0n of equation (19) and its corresponding vn we get from (21) ||Vvn|| < R0 because G(0n)0n = 0.

Taking into account these two facts together with estimate (21) we get an estimate of the penalisation term. So we have

|| Vvn|| < Ro, (n - no) (|vn|2,r2 + ^ (* (vn), vn)v < cii (23)

for n > n0 and a positive constant c11 which does not depend on n.

3. The limit procedure for n —►

It follows from (23) that there exists a function v G V and a subsequence of {vn} (which will be also denoted by {vn} in order to keep a simple notation) such that

vn ^ v weakly in V. (24)

The operator of traces from V into Lq(dH)3 is compact for q G (1, 4). This implies that

vn ^ v strongly in Lq(dH)3 (25)

for q G (1, 4). It follows from (15) and (23) that

^(vn) ^ 0 strongly in V. (26)

Due to the monotonicity of operator we also have

^(vn) - tf(z) , vn - z)v > 0 (27)

for all n G IN and z G V. Thus, using the boundedness of the sequence {vn - z} in V and

(26), we get lim f^(vn), vn - z) = 0. Condition (24) implies that lim f^(z), vn) =

«,^+TO V /v «,^+TO V /v

(^(z), v)v • If we pass to the limit for n ^ in (27), we obtain - ^^(z), v - z^ > 0. Put z = v - e ^(v) where e > 0. Dividing the inequality by e and passing to the limit for e ^ 0+, we get: - ^^(v), ^(v)^ > 0 which means that ^(v) = 0. Hence v G K and Pv = v.

Let M be a set of functions from K which are linear combinations of a finite number of functions e1, e2, ...

Let us assume that q G M at first. Then there exists m G IN (depending on q) and real

m

numbers A (k = 1, ... , m) such that q = ek. Let us choose n G IN so that n > m.

fc=1

Let us multiply (17) by (-0J? + ) if k < m and by (-0J?) if m < k < n. Let us sum for k = 1, ... , n the obtained equations. We get

v J V(vn + V)V(q - vn) dx -J F(q - vn) dS + n (j|vn ||2r2 + (^(vn), q - vra)v = 0. (28) n r2

It follows from the monotonicity of operator ^ and from the fact that q G K (which means that ^(q) = 0) that the last term on the left hand side of (28) is nonpositive. If we omit this

term, we get

(Pvn + V)V(vn + V) (q - vn) dx - (f , q - vn) +

+ v V(vn + V) V(q - vn) dx - / F(q - vn) dS > 0.

(29)

r2

We are going to pass to the limit for n — in (29) now. We will use all convergences (24)-(26). Since some steps are standard, we do not show all the details here. Let us deal for example with the two nonlinear terms:

a)

b)

lim inf I —v I VvnVvn dx | < -v I VvVv dx,

«,^+TO

Q / „ Q

(P(vn)+ V)V(vn + V) (q - vn) dx

(Pvn + V)V(vn + V) (V + q) dx -

1

1

2J (Vn) |V|2 dS - 2 ri

(Pvn + V)n |vn + V|2 dS.

r2

Due to (25), we have

(Pvn + V)nl |vn + V|2 dS

(v + V)n |v + V|2 dS.

r2

r2

If we use (26), the strong convergence of vn to v in L4(Q)3 (following from (24)) and the decomposition Pvn — v = (Pvn — vn) + (vn — v) = —^(vn) + (vn — v), we obtain that Pvn ^ v in L4(Q)3). Hence

lim

ra^+TO

Q

Thus, we have lim

ra^+TO I

(P(vn)+ V)V(vn + V) (V + q) dx

(P(vn)+ V)V(vn + V) (q - vn) dx

(v + V)V(v + V) (V + q) dx.

(v + V)V(v + V) (q - v) dx.

Using this all in (29), we get (11).

We need to show that (11) is satisfied not only for all q £ M but for all q £ K now. In order to do that, it is sufficient to show that M is dense in K in the norm of V. Let e > 0 and q £ K be such that

dist(V(x) + q(x), Ka(x))

dSx < Co — c

r2

at first. There exists £ > 0 so that if q1 £ V, ||q1 — q||V < £ then q1 £ K. Let qm be the orthogonal (in V) projection of q onto the subspace of V which is generated by the functions e1, ... , em. Then ||qm — q||V ^ 0 if m ^ This means that qm £ K for m large enough. Thus, qm £ M for m large enough and hence q can be approximated with an arbitrary accuracy (in the norm of V) by a function from M. We can get the same result for all q £ K if we let e ^ 0+.

We have proved the following theorem: Theorem 3. There exists a solution of Problem 1.

Q

Q

Q

Q

—>

Q

Q

Q

References

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[2] Kracmar S., NeüSTüPA J. Global existence of weak solutions of a nonsteady variational inequality of the Navier — Stokes type with mixed boundary conditions // Proc. Intern. Symp. on Numer. Analysis. Charles Univ. Prague, 1993. P. 156-177.

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[6] KüCERA P. A structure of the set of critical points to the Navier — Stokes equations with mixed boundary conditions // Proc. Conf. "Navier —Stokes Equations: Theory and Numer. Methods" / Ed. R. Salvi. Pitman Research Notes in Math. Ser. 388. L.: Addison Wesley, 1998. P. 201-205.

[7] KüCERA P. Some properties of solutions to the Navier — Stokes equations with mixed boundary conditions on an infinite time interval // Proc. 3rd Conf. "Numerical Modelling in Continuum Mechanics" / Eds M. Feistauer, K. Kozel & R. Rannacher. Prague: Charles Univ., 1997. P. 375-383.

[8] LiüNS J.L. Quelques methodes de resolution des problèmes aux limites non lineaires. Paris: Dunod — Gauthier — Villars, 1969.

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Received for publication October 17, 2001