law. In the literature, for the frictional contact problem with a general normal compliance law, some existence and local uniqueness results are available, usually ...
On the Numerical Approximation of a Frictional Contact Problem with Normal Compliance Weimin Han Department of Mathematics University of Iowa Iowa City, IA 52242 U.S.A.
Abstract. In this paper, we consider a frictional contact problem with normal
compliance. The problem is formulated as a variational inequality, and is shown to possess a unique solution. The nite element approximation is analyzed, with a Cealike inequality proved. A major diculty in actually solving the problem is caused by the presence of a non-di erentiable functional in the formulation of the problem. Here we discuss the regularization method to overcome the diculty. A-posteriori error estimates are derived.
1 Introduction Frictional contact problems with normal compliance are the subject of many recent investigations. An early attempt to study frictional contact problems within the framework of variational inequalities was made in [5]. An excellent reference to the eld of contact problems with or without frictions is [13]. Steady-state as well as time-dependent frictional contact problems with normal compliance and their numerical treatments are discussed in numerous papers, e.g., [1], [4], [14][20] and the references therein. In this paper, we consider a frictional contact problem with a reduced normal compliance law. In the literature, for the frictional contact problem with a general normal compliance law, some existence and local uniqueness results are available, usually under the assumption that 1
certain material coecients are suciently small (cf. [15]). In contrast, for our problem, we will show the existence and uniqueness of a solution without any restriction on the size of the material coecients in the next section. For the nite element approximation of the problem, we will prove an optimal order error estimate, once again without any assumptions on the size of material coecients. This is done in Section 3. A major diculty in solving the problem is caused by the presence of a non-di erentiable functional in the formulation of the problem. One approach used in practice to circumvent the diculty is to apply the regularization method. We discuss the regularization method in the last section, where, besides the usual a-priori error estimate, we also provide some a-posteriori error estimates which are highly desirable in actual implementation of the regularization method.
2 The frictional contact problem Consider the deformation of an elastic body unilaterally supported by a rigid foundation. The body occupies a domain RN , N = 2 or 3 in applications. The boundary of the domain, ? = @ , is assumed to be Lipschitz continuous, and is decomposed into three parts, ? = ?D [ ?F [ ?C . The body is subjected to a body force f in and a surface traction t on ?F . The part ?D of the boundary is xed. The actual surface on which the body comes in contact with the foundation is unknown in advance but is contained in ?C . The initial gap between the body and the foundation is given by g 0. Throughout the paper, we adopt the summation convention over repeated indices. Denote by u = (u1; ; uN )T the displacement vector, = (ij )Ni;j=1 the stress tensor of the body. Let n be the unit outward normal vector on ?. We decompose u into its normal component un and tangential component uT : u = un n + uT ; (2:1) where,
un = u n = uini; uT = u ? unn: Similarly, we decompose into its normal component
n = ij ni nj ; and tangential component T : (T )i = ij nj ? nni; 1 i N: 2
Assume the material response of the body follows the generalized Hooke's law
(u) = E(u) in ; where, (u) is the strain tensor,
(2:2)
!
@ui + @uj ; 1 i; j N: (u)ij = 21 @x j @xi The elasticity tensor E = (Eijkl)Ni;j;k;l=1 is assumed to be uniformly bounded: symmetric:
Eijkl 2 L1 ( ); 1 i; j; k; l N
(2:3)
Eijkl = Ejikl = Eklij ; 1 i; j; k; l N
(2:4)
and pointwise stable:
Eijkl (x)ij kl jj2; 8 2 M N ; 8 x 2 ; (2:5) where, > 0 is a constant, M N denotes the set of symmetric matrices of order N . A general static frictional contact problem with normal compliance is
Problem P1 Find the displacement vector u, such that div (u) + f = 0 in ;
(2:6)
u = 0 on ?D ; (2:7) (u)n = t on ?F ; (2:8) (2:9) n = ?cn(un ? g)m+ n on ?C ; ) jT j < cT (un ? g)m+ T ) uT = 0 (2:10) jT j = cT (un ? g)m+ T ) uT = ? T for some 0 on ?C : Here, cn, mn, cT and mT are material interface parameters, (un ? g)+ describes the penetration approach. (x)+ is the positive part function: (x)+ = x for x 0, and (x)+ = 0 for x < 0. The relation (2.9) and (2.10) de ne the normal compliance power law. We assume
f 2 L2( ); t 2 L2(?F ); g 2 H 1=2(?C ); cn ; cT 2 L1(?); cn ; cT 0; 0 mn; mT < 1 when N = 2; 0 mn; mT 3 when N = 3: 3
To form the weak formulation of the problem P, let us introduce the space of admissible displacements V = fv 2 (H 1( ))N : v = 0 on ?D g; and de ne
a(u; v) =
Z
jn (u; v) = jT (u; v) = F (v) =
Z
Z
E(u) (v) dx;
Z?C
cn(un ? g)m+ n vnds; cT (un ? g)m+ T jvT j ds;
?C
f v dx +
Z
?F
t v ds:
The functionals jn(u; v) and jT (u; v) are well-de ned for u; v 2 V , from the conditions on mn and mT . The weak form of the frictional contact problem P1 is
Problem P2 Find u 2 V , such that a(u; v ? u) + jn(u; v ? u) + jT (u; v) ? jT (u; u) F (v ? u); 8 v 2 V:
(2:11)
By a standard technique, it can be shown that the problems P1 and P2 are formally equivalent (cf. [16]). In [14, 15], it is proved that the problem P2 has a locally unique solution, for 1 mn; mT < 1 if N = 2, and 1 mn; mT < 4 if N = 3 (in the case 3 mn; mT < 4 and N = 3, (2.11) is replaced by a weaker formulation). Error estimates for nite element solutions of the problem P2 are derived in [16]. Here, we will analyze the frictional contact problem with a reduced normal compliance law, (2.9) and ) jT j < cT ) uT = 0 on ?C ; (2:12) jT j = cT ) uT = ? T for some 0 i.e., the normal compliance law (2.9) and (2.10) with mT = 0. Physically, this corresponds to a situation in which sliding of the frictional surface occurs when the frictional stress T exceeds a constant independent of the deformation normal to the contact surface. This can be considered as a \ rst approximation" of the interface shear behavior. For this reduced problem, the functional jT (u; v) de ned before is reduced to
jT (v) = and the reduced problem amounts to nding
Z
?C
cT jvT j ds;
u 2 V : a(u; v ? u) + jn(u; v ? u) + jT (v) ? jT (u) F (v ? u); 8 v 2 V: 4
(2:13)
We keep the same assumptions on the data as before. Further assume meas (?D) 6= 0. Then from the uniform boundedness and pointwise stability of E , and Korn's inequality (cf. [5]), q a(; ) de nes a norm on V , which is equivalent to k kV :
c0kvk2V a(v; v) c1kvk2V ; 8 v 2 V; for some constants c0 and c1. The following simple result will be used several times later.
Lemma 2.1 Assume mn 0. Then
(x)m+ n ? (y)m+ n (x ? y) 0; 8 x; y 2 R:
Consequently, for any c 2 R,
(x ? c)m+ n ? (y ? c)m+ n (x ? y) 0; 8 x; y 2 R:
Proof. The rst inequality is easily proved by considering the four possible cases: x 0 and y 0; x 0 and y < 0; x < 0 and y 0; x < 0 and y < 0. Replacing x and y by x ? c and y ? c in the rst inequality, we obtain the second inequality. 2
Theorem 2.2 Under the assumptions made on the data, the problem (2.13) has a unique solution u2V. Proof. The problem (2.13) is equivalent to the following minimization problem
u 2 V : J (u) = vinf J (v); 2V where,
J (v) =
Z
(2:14)
1 E(v) (v) ? f v dx + Z cn (v ? g)mn+1 + c jv j ds ? Z t v ds: n T T + 2 ?C mn + 1 ?F (2:15)
Consider the functional
Jn(v) =
Z
cn (v ? g)mn +1ds: n + ?C mn + 1
For its derivative, we have
hDJn (u); vi =
Z ?n
cn(un ? g)m+ n vnds:
5
Thus, 8 u; v 2 V ,
hDJn (u) ? DJn (v); u ? vi =
Z
h
?n
i
cn (un ? g)m+ n ? (vn ? g)m+ n (un ? vn) ds 0;
by Lemma 2.1. Hence, Jn(v) is a convex functional. Since J (v) ? Jn(v) is strictly convex, we conclude that J (v) = (J (v) ? Jn (v)) + Jn(v) is a strictly convex functional on V . Furthermore, since J (v) ? Jn(v) is coercive and J (v) J (v) ? Jn(v), we have that J (v) is coercive. Therefore, the problem (2.14), and hence the problem (2.13), has a unique solution u 2 V (cf. Theorem 3.4 in [13]). 2 For later use in error analysis of numerical approximations of the problem, we need a result on the uniform boundedness of the solution.
Theorem 2.3 Let u 2 V be the solution of the problem (2.13). Then
kukH ( ) c kf kL ( ) + ktkL (?F ) 1
2
(2:16)
2
for some constant c depending only on E and . Proof. We take v = 0 in (2.13) to obtain
a(u; u) + jn (u; u) + jT (u) F (u); i.e.,
Z
E(u) (u) dx +
Z
h ?C
cn(un
? g ) mn u +
n + cT juT j
i
ds
Z
f u dx +
Z ?F
t u ds:
We observe that (un ? g)m+ n un 0 is always true, since g 0 by assumption. Thus, Z
E(u) (u) dx
Z
f u dx +
Z
?F
t u ds;
from which the uniform bound on the solution follows in a standard way.
2
3 Finite element approximation Let V h V be a nite element subspace. The nite element approximation of the problem (2.13) is
uh 2 V h : a(uh; vh ? uh) + jn(uh ; vh ? uh ) + jT (vh) ? jT (uh) F (vh ? uh); 8 vh 2 V h: (3:1) 6
Applying Theorem 2.2 for the case when V is replaced by V h, we nd that the problem (3.1) has a unique solution uh 2 V h. By Theorem 2.3, we further have a uniform bound result
kuh kH ( ) c kf kL ( ) + ktkL (?F ) 1
2
(3:2)
2
where the constant c depends only on E and .
Theorem 3.1 Assume 1 mn 3, and for some s > 3=2, T (u) 2 (H s?3=2(?C ))N ; cT 2 H s?3=2(?C ): Then, for the nite element solution uh de ned by (3.2), we have the following error estimate:
1=2 h ? uk h kuh ? ukH ( ) c vhinf k v H ( ) + kv ? ukH ?s h 2V 1
1
+2
( )
:
(3:3)
Proof. For any v h 2 V h , we write
a(u ? uh; u ? uh) = a(u ? uh; u ? vh) + a(u; vh ? u) ? a(u; uh ? u) ? a(uh; vh ? uh): Taking v = uh in (2.13), we obtain
?a(u; uh ? u) jn(u; uh ? u) + jT (uh ) ? jT (u) ? F (uh ? u): From (3.1), we have
?a(uh; vh ? uh) jn(uh; vh ? uh) + jT (vh) ? jT (uh) ? F (vh ? uh): Thus,
a(u ? uh; u ? uh ) a(u ? uh; u ? vh) + a(u; vh ? u) ? F (vh ? u) + jn(u; uh ? u) + jn(uh; vh ? uh) + jT (vh) ? jT (u): Using Green's formula, and (2.6){(2.9), we get aZ (u; vh ? u) = (u) (vh ? u) dx
=
= =
Z
Z h h ? u) dx ( u ) n ( v ? u ) ds ? div ( u ) ( v Z? Z h
h t (v ? u) ds + ? ?cn(un ? g)m+ n (vnh ? un ) + T (u) (vTh ?F C Z h + f (v ? u) dx
Z h i h n(u)(vnh ? un) + T (u) (vTh ? uT ) ds: F (v ? u) + ? C
7
i
? uT ) ds
(3:4)
So, from (3.4), Z h i a(u ? uh; u ? uh) a(u ? uh; u ? vh) + ? cn (uhn ? g)m+ n ? (un ? g)m+ n (vnh ? uhn ) ds C Z h i (3:5) + T (u) (vTh ? uT ) + cT (jvTh j ? juT j) ds: ?C
Consider the second term on the right side of (3.5). We use Lemma 2.1 and proceed as in the proof of Lemma 5 in [16], Z h i cn (uhn ? g)m+ n ? (un ? g)m+ n (vnh ? uhn) ds =
Z?C
?C
+
h
i
cn (uhn ? g)m+ n ? (vnh ? g)m+ n (vnh ? uhn) ds
Z
?C
Z
h
i
cn (vnh ? g)m+ n ? (un ? g)m+ n (vnh ? uhn) ds i
h
? cn (vnh ? g)m+ n ? (un ? g)m+ n (vnh ? uhn) ds C c k(vnh ? g)m+ n ? (un ? g)m+ n kL = (?C )kvnh ? uhnkL (?C ) c kvnh ? unkL (?C )k(vnh ? g)m+ n ?1 + (un ? g)+mn?1 kL (?C )kvnh ? uhnkL (?C ) c k(vnh ? g)+mn?1kL (?C ) + k(un ? g)m+ n ?1kL (?C ) kvh ? ukH ( )kvh ? uh kH ( ) = c k(vnh ? g)+kmL nm?n1? (?C ) + k(un ? g)+kmL nm?n1? (?C ) kvh ? ukH ( )kvh ? uhkH ( ): 4
4 3
4
2
2
2
2(
Now
4
1)
2(
1
1
1)
k(un ? g)+ kL mn ? (?C ) kun kL mn? (?C ) + kgkL mn ? c kukH ( ) + kgkL mn ? (?C ) 2(
2(
1)
2(
2(
1)
(?C )
1)
is bounded by Theorem 2.3. Also, k(vnh ? g)+kL mn ? (?C ) c kvhkH ( ) + kgkL mn? (?C ) c kvh ? ukH ( ) + kukH ( ) + kgkL Hence, Z h i cn (uhn ? g)m+ n ? (un ? g)m+ n (vnh ? uhn) ds 1
1)
2(
1)
1
1
?C
c kvh ? ukH ( ) + kukH ( ) + kgkL 1
1
1
1)
1
2(
1
mn ?1) (?C )
2(
mn ?1
mn ?1) (?C ) :
2(
kvh ? ukH ( )kvh ? uh kH ( ): 1
1
For the third term on the right side of (3.5), Z h i T (u) (vTh ? uT ) + cT (jvTh j ? juT j) ds ?C
kT (u)kH s? = (?C ) + kcT kH s? = (?C ) kvTh ? uT kH ?s c kT (u)kH s? = (?C ) + kcT kH s? = (?C ) kvh ? ukH ?s 3 2
3 2
3 2
3 2
8
= (?C )
+3 2 +2
( )
:
Thus, from (3.5), we have, 8 vh 2 V h,
a(u ? uh; u ? uh ) a(u ? uh; u ? vh) m ?1 + c kvh ? ukH ( ) + kukH ( ) + kgkL mn? (?C ) n kvh ? ukH ( )kvh ? uhkH ( ) + c kT (u)kH s? = (?C ) + kcT kH s? = (?C ) kvh ? ukH ?s ( ): 1
1
2(
3 2
1
1)
1
+2
3 2
q
Using the continuity of a and the equivalence of a(; ) with k kV , we then have
kuh ? uk2H ( ) c kuh ? ukH ( )kvh ? ukH ( ) ?1 ) kv h ? uk h h + c (1 + kvh ? ukHmn( ) H ( ) kv ? ukH ( ) + ku ? ukH ( ) + ckvh ? ukH ?s ( ) +1 + kv h ? uk2mn 21 kuh ? uk2H ( ) + c kvh ? uk2H ( ) + kvh ? ukHmn( ) H ( ) h + ckv ? ukH ?s ( ): 1
1
1
1
1
1
1
+2
1
1
1
1
+2
So,
+1 + kv h ? uk2mn + kv h ? uk ?s kuh ? uk2H ( ) c kvh ? uk2H ( ) + kvh ? ukHmn( ) H H ( ) 1
1
1
+2
1
( )
:
Since mn 1, we have mn + 1 2 and 2mn 2. So +1 c inf kv h ? uk2 inf kvh ? ukHmn( ) H ( ) ; vh 2V h
vh 2V h
and Therefore,
1
1
n c inf kv h ? uk2 inf kvh ? ukH2m( ) H ( ) : vh 2V h
vh 2V h
1
1
kvh ? uk2H ( ) + kvh ? ukH ?s kuh ? uk2H ( ) c vhinf 2V h
+2
1
1
( )
2
from which, we get the Cea-like inequality (3.3). by
;
Alternatively, if we only assume T (u) 2 (L2(?C ))N , then the estimate (3.3) can be replaced
kvh ? ukH ( ) + kvh ? uk1L=2(?C ) : kuh ? ukH ( ) c vhinf 2V h 1
1
2
(3:6)
It is easy to see that the results (3.3) and (3.6) are valid when mn = 0. It would be interesting to nd the optimal error estimate when mn 2 (0; 1). The inequalities (3.3) and (3.6) are basis for various convergence order estimates for di erent nite elements under di erent assumptions on the regularity of the solution, cf. e.g., [2, 3]. 9
We notice in similar error estimates in [16], the material coecients cn and cT are assumed to be suciently small. In the proof of the error estimate (3.3) here, we used the inequality i h (uhn ? g)m+ n ? (vnh ? g)m+ n (vnh ? uhn) 0; and thus an assumption on the smallness of the material coecient cn is not needed. If we apply the same technique in the proof of Theorem 6 and Theorem 8 of [16], we observe that the assumptions there on the smallness of the material coecient cn can be removed.
4 Regularization and a-posteriori error estimate A major diculty in solving the problem (3.1) is caused by the presence of the non-di erentiable functional jT . Several approaches can be used to circumvent the diculty. One approach is to introduce a Lagrange multiplier for the non-di erentiable term, and the problem (3.1) is solved by an iterative procedure, for detail, see, e.g., [7], [10]. Another approach is through the use of the regularization method. The idea of the regularization method is to approximate the nondi erentiable term by a sequence of di erentiable ones. The regularization method has been widely used in applications (cf. [7, 8, 13, 21]). An approximating di erentiable sequence in the regularization method depends on a small parameter " > 0. The convergence is obtained when " goes to 0. In this paper, we will discuss the regularization method only on the continuous problem level, i.e., the regularization method for the problem (2.13). Similar arguments below can be used for an a-posteriori error analysis for solving the discrete system (3.1) by the regularization method. In [16], the following functional is used to regularize jT : Z jT" (v) = cT "(vT ) ds; (4:1) where,
?C
8 > > < " (v ) = T > > :
jvT j ? 21 " if jvT j "; 1 jv j2 if jv j < ": T 2" T
The functional jT" is di erentiable, and we have Z hDjT" (u); vi = cT "(uT ) vT ds; with
?C
8 > > < " (uT ) = > > :
1 v if jv j "; T jvT j T 1 v if jv j < ": T " T 10
(4:2) (4:3) (4:4)
Thus, the solution u of (2.13) is approximated by a sequence of approximate solutions fu" g, where,
u" 2 V : a(u"; v ? u") + jn(u" ; v ? u") + jT" (v) ? jT" (u" ) F (v ? u" ); 8 v 2 V:
(4:5)
Since jT" is di erentiable, (4.3) is equivalent to a boundary value equation problem,
u" 2 V : a(u"; v) + jn (u"; v) + hDjT" (u"); vi = F (v); 8 v 2 V:
(4:6)
Following the proof of Theorem 2.2, one can show that the regularized problem has a unique solution u" 2 V . For the convergence and an a-priori error estimate for the regularization method, we have
Theorem 4.1 There exists a constant c independent of u and ", such that ku ? u"kH ( ) c p": Thus, u" ! u in H 1 ( ) as " ! 0. 1
(4:7)
Proof. From Lemma 9 in [16], we have
jjT (v) ? jT" (v)j c "; 8 v 2 V:
(4:8)
Using (2.13) and (4.6), we have
a(u ? u"; u ? u") = a(u; u ? u") ? a(u"; u ? u") jn (u"; u ? u") ? jn (u; u ? u") + jT (u") ? jT (u) + hDjT" (u"); u ? u"i: Notice that
jn
(u"; u ? u") ? j
n
(u; u ? u" ) =
Z ?C
h
i
cn (u"n ? g)m+ n ? (un ? g)m+ n (un ? u"n ) ds 0;
by Lemma 2.1. From the convexity of jT" , we have
hDjT" (u"); u ? u"i jT" (u) ? jT" (u"): Hence,
jT (u") ? jT (u) + hDjT" (u"); u ? u"i [jT (u") ? jT" (u")] + [jT" (u) ? jT (u)] c "; by using (4.8). Thus,
a(u ? u"; u ? u") c "; 11
2
which completes the proof of (4.7).
In actual implementation of the regularization method, it is highly desirable to have aposteriori error estimates which can give us computable error bounds once we have solutions of regularized problems. We will use the duality theory from the convex analysis (cf. [6]) to derive the a-posteriori error estimates. The idea was rst used in analyzing a regularization method for a simpli ed friction problem ([9]) and a holonomic elastic-plastic problem ([10]). In [12], a detailed discussion is given on the regularization method with various choices of the regularization sequence for an obstacle problem. In [11], an a-posteriori error analysis is given for solving some mixed variational inequalities arising in elastoplasticity by the regularization method. We will use a result from convex analysis, for detail, see [6]. Let V and Q be two normed spaces, V and Q their dual spaces. Assume there exists a linear continuous operator 2 L(V; Q), with transpose 2 L(Q; V ). Let J be a function mapping V Q into R|the extended real line. Consider the minimization problem: inf J (v; v)
(4:9)
J (v; q) = sup [hv; vi + hq; qi ? J (v; q)]
(4:10)
De ne the conjugate function of J by:
v2V
v2V;q2Q
Theorem 4.2 Assume
(1) V is a re exive Banach space, Q a normed space. (2) J : V Q ! R is a proper, l.s.c., strictly convex function. (3) 9 u0 2 V , such that J (u0; u0) < 1 and q 7! J (u0; q) is continuous at u0. (4) J (v; v) ! +1, as kvk ! 1; v 2 V . Then problem (4.9) has a unique solution u 2 V , and
? J (u; u) J (q; ?q); 8 q 2 Q For the frictional contact problem (2.13) considered in this paper, we take
V = fv 2 (H 1( ))N : v = 0 on ?D g; Q = Q 1 Q 2 Q 3 Q4 ; Q1 = fq1 = (q1;ij ) 2 (L2( ))N N : q1;ij = q1;ji a:e: in ; 1 i; j N g; Q2 = Lmn +1(?C ); 12
(4:11)
Q3 = (L2(?C ))3; Q4 = (L2(?F ))3; Q = Q1 L1+1=mn (?C ) Q3 Q4; v = ((v); vnj?C ; vT j?C ; vj?F ) ; Z 1 J (v; q) = Eq1 q1 ? f v dx
Z 2 cn (q ? g)mn+1 + c jq j ds ? Z t q ds: + 2 T 3 4 + ?F ?C mn + 1 Here and in what follows, we use the notation q = (q1; q2; q3; q4) for an element in Q, with q1 2 (L2( ))N N , q2 2 Lmn +1(?C ), q3 2 (L2(?C ))3 and q4 2 (L2(?F ))3 being the auxiliary variables for (v), vnj?C , vT j?C and vj?F , respectively. We will use q = (q1; q2; q3; q4) 2 Q to denote the dual variable of q 2 Q. In this section, we assume 0 < mn < 1; inf c (x) > 0: x2?C n Let us rst compute J (q; ?q). By the de nition (4.10), J (q; ?q) = sup [hv; qi + hq; ?qi ? J (v; q)] v2V;q2Q = sup [hv; qi ? hq; qi ? J (v; q)] v2V;q2Q = I 1 + I2 + I 3 + I4 + I5 ; where, Z Z Z I1 = sup ((v) q1 ? f v) dx + (vnq2 + vT q3) ds + v q4 ds ; ?C ?F v2V
Z I2 = sup ? 21 Eq1 q1 + q1 q1 dx ;
q 2Q Z c n m +1 n ds ; q2q2 + m + 1 (q2 ? g)+ I3 = sup ? 1
1
q2 2Q2
Z
I4 = sup ? q3 2Q3
I5 = sup
Z
q4 2Q4 ?F
?C
?C
n
(q3 q3 + cT jq3j) ds ;
(t ? q4) q4 ds:
Let us compute each of Ii, i = 1; ; 5. For v 2 V , we have Z (v) q1dx = =
Z
Z?
Z
q1n v ds ? div q1 v dx
Z
Z
(q1;nvn + q1;T vT ) ds + q1n v ds ? div q1 v dx:
?F ?C 13
Hence,
8 > > < I1 = > > :
0; if ? div q1 = f in ; q1;n + q2 = 0 and q1;T + q3 = 0 on ?C ; q1n + q4 = 0 on ?F ; 1; otherwise:
It is easy to nd that For I3, we write
I2 = Z
Z
1 E ?1q qdx: 1 1
2 Z
q2q2 + m cn+ 1 (q2)+mn+1 ds ?C ?C q 2Q n Z Z c n m +1 n = ? gq2 ds + sup ?q2q2 ? m + 1 (q2)+ ds: ?C ?C q 2R n When q2 > 0, we have c n m +1 n = sup jq2q2j = 1: sup ?q2q2 ? m + 1 (q2)+ q < ?gq2 + m + 1 cn jq2 j ds; if q2 0 a:e: on ?C ; I3 = > ?C n : 1; otherwise: I3 = ?
gq2ds + sup ? 2
2
2
2
2
2
2
Easily, we have
8 < I4 = sup ? (jq3j ? cT ) jq3j ds = : q3 2Q3 C Z
and
8 < I5 = :
0; if jq3j cT on ?C ; 1; otherwise;
0; if q4 = t on ?F ; 1; otherwise:
In conclusion, if q 2 Q satis es the following relations
?div q1 = f in ;
q1;n + q2 = 0; q1;T + q3 = 0; q2 0 and jq3j cT on ?C ; q1n + q4 = 0 and q4 = t on ?F ; 14
(4.12) (4.13) (4.14)
then,
1 E ?1q qdx + Z ?gq + mn c?1=mn jqj1+1=mn ds: (4:15) 2 1 1 2 mn + 1 n ?C
2 We will call a q 2 Q satisfying the relations (4.12){(4.14) an admissible dual variable, which plays an important role in the a-posteriori error estimate below.
J (q; ?q) =
Z
Now we try to de ne an admissible dual variable from the solution of the regularized problem. From (4.6), we have Z
E(u") (v) dx +
Z
h ?C
cn(u"n ? g)m+ n vn + cT "(u"T ) vT
i
ds =
Z
f v dx +
Z
?F
t v ds; 8 v 2 V:
Applying Green's formula to the rst term on the left side, we get the relation
Z i " m " " " n n (u")n v ds (n n (un ? g )+ )vn + (T (u ) + cT (uT )) vT ds + ?F ?C Z Z Z " ? div (u ) v dx = f v dx + t v ds; 8 v 2 V:
Z
h
(u") + c
?F
Thus,
?div (u") = f in ; n(u" ) + cn(u"n ? g)m+ n = 0; T (u") + cT "(u"T ) = 0 on ?C ; n(u" )n = t on ?F :
Comparing these relations with (4.12){(4.14), we see that the following is an admissible dual variable: 8 > q1 = (u") in ; > > > > < q = cn (u" ? g )mn on ?C ; n 2 + (4:16) = c " (u" ) on ? ; > > q T C > 3 T > > : q = t on ? : F 4 Here, the property jq3j cT on ?C follows from (4.4). In order to derive an a-posteriori error estimate, we consider
D(u; v) = J (v; v) ? J (u; u): For v 2 V , a lower bound for D(u; v) is derived as following: D(u; v) = 21 a(v; v) + Jn(v) + jT (v) ? F (v) ? 21 a(u; u) ? Jn(u) ? jT (u) + F (u) 21 a(u ? v; u ? v) + Jn (v) ? Jn(u) ? jn (u; v ? u); 15
(4:17)
where, we used (2.13). Now since h(t) = m 1+ 1 t+mn+1 is a convex function,
Jn (v) ? Jn (u) ? jn (u; v ? u) =
Therefore, In particular,
Z
?C
n
cn [h(vn) ? h(un ) ? h0(un )(vn ? un )] ds 0:
D(u; v) 21 a(u ? v; u ? v); 8 v 2 V: D(u; u" ) 21 a(u ? u"; u ? u"):
(4:18)
Now we use Theorem 4.2 to derive an upper bound for D(u; u"). Let q 2 Q be the admissible dual variable de ned in (4.16). We have D(u; u" ) = J (u"; u") ? J (u; u) J (u"; u") + J (qZ; ?hq) i cn (u"n ? g)+mn+1 + cn(u"n ? g)m+ n g + cT ju"T j ds = a(u"; u") ? F (u") + ?C (using (4:6) with v = u") Z " " " " = c T [juT j ? (uT ) uT ] ds ?C Z " j! j u = cT 1 ? "T ju"T j ds: " ?C \fjuT j
