Numerical Analysis of a Nonlinear Evolutionary System ... - CiteSeerX
Numerical Analysis of a Nonlinear Evolutionary System with Applications in Viscoplasticity Jiuhua Chen1 , Weimin Han1 and Mircea Sofonea2
Abstract. We consider numerical approximations of a class of abstract nonlinear evolution-
ary systems arising in the study of quasistatic frictional contact problems for elastic-viscoplastic materials. Both spatially semi-discrete and fully discrete schemes are analyzed with the nite element method employed to discretize the spatial domain. Strong convergence of both approximations is established under minimal solution regularity. The results are applied to two particular frictional contact problems for viscoplastic bodies. Under appropriate regularity assumptions on the exact solution, optimal order error estimates are derived.
1 Introduction The aim of this paper is to provide numerical analysis of some problems arising in frictional contact between an elastic-viscoplastic body and a rigid foundation. Situations of frictional contact abound in industry and everyday life. Contacts of the braking pads with the wheel, the tire with the road and the piston with the skirt are just a few simple examples. Because of the importance of the process of frictional contact, a considerable e ort has been made in its modeling and numerical simulations. Indeed, the engineering literature concerning this topic is extensive. Most of it, however, is dedicated to simple geometries, speci c settings, and mostly to numerical simulations. In the applied mathematics literature, the study of general models for dynamic or quasistatic contact process involving elastic-viscoplastic materials is very recent. Rate-type viscoplastic constitutive laws of the form (1.1) _ = E "(u_ ) + G( ; "(u)) are used in the literature to describe mechanical responses of such materials as rubber, various metals, rocks, pastes, etc. In (1.1), denotes the stress tensor, u the displacement eld, "(u) the linearized strain tensor, and E and G are material constitutive functions. The function E is assumed to be linear while G is in general nonlinear. Here and throughout the paper, a dot above a quantity represents its derivative with respect to the time variable t, and double dots denote the second-order derivative. Concrete examples, experimental 1 2
Department of Mathematics, University of Iowa, Iowa City, IA 52242, USA Laboratoire de Theorie des Systemes, University of Perpignan, France
1
background and mechanical interpretations of such models may be found in [4] and references therein. Functional and numerical methods are discussed in [12] for initial and boundary value problems involving (1.1) with the usual displacement and traction boundary conditions. Existence of weak solutions to quasistatic frictional problems for materials modeled by a general rate-type constitutive law of the form (1.1) were established in [1, 2, 16]. In the abstract form, the frictional contact problems are formulated as a nonlinear evolution equation (the abstract version of (1.1)) coupled with an evolutionary variational inequality resulted from the equilibrium equation and the boundary conditions. The variational analysis of this abstract problem was done in [1]. In this paper, we consider the numerical analysis of frictional contact problems for elasticviscoplastic materials of the type (1.1). The literature is abundant on numerical treatment of variational inequality, see for instance the monographs [6, 7, 11, 13]. Of particular relevance to this paper are the works on numerical analysis of variational inequalities arising in plasticity, cf. [8, 9, 10]. The paper is organized as follows. In Section 2 we present the abstract problem, state the assumptions on the data and recall the existence and uniqueness result proved in [1]. In Section 3, both spatially discrete and fully discrete approximations to the abstract problem are analyzed and some error estimates are presented. In deriving the error estimates, we will need to apply Gronwall's inequality, which is recalled here for convenience. Suppose f; g 2 C [a; b] and g is nondecreasing, c0 > 0 is a constant, then (1.2)
f (t) g(t) + c0
Zt
a
f (s) ds; t 2 [a; b] =) f (t) ec0 (t?a) g(t); t 2 [a; b]:
Convergence analysis for both approximations is done in Section 4 under the solution regularity condition established in the proof of well-posedness of the problem. Finally, in Section 5 we apply the results to the numerical approximations of two concrete examples of quasistatic frictional contact problems in rate-type viscoplasticity. Under appropriate solution regularity assumptions, we obtain optimal order error estimates.
2 The abstract problem Let H be a real Hilbert space, V a closed subspace of H . We denote by (; )H the inner product of H and by kkH the associated norm. Let us remark that V itself is a real Hilbert space endowed with the inner product of H ; for this reason we shall use sometimes the notations (u; v)V , kukV instead of (u; v)H ; kukH , if u; v 2 V . Let A : H ! H be a linear 2
operator, B : [0; T ] H H ! H a possibly nonlinear operator, and ' : H ! (?1; +1]. Let [0; T ] be the time interval of interest. We consider an abstract problem
y_ (t) = Ax_ (t) + B (t; x(t); y(t)) a:e: t 2 (0; T )
(2.1)
(2.2) y(t) + @'(x_ (t)) 3 f (t) a:e: t 2 (0; T ) (2.3) x(0) = x0 ; y(0) = y0: Here the unknowns are the functions x : [0; T ] ! V and y : [0; T ] ! H , while x0 2 V , y0 2 H and f : [0; T ] ! V are given data. The symbol @' represents the subdi erential of the function ', and the relation (2.2) is understood in the sense that for a.e. t 2 (0; T ), f (t) ? y(t) is a subgradient of ' at x_ (t). We denote by D' the e ective domain of ' de ned by D' = f x 2 H j '(x) < +1 g: We assume in the sequel D' = V . An equivalent formulation of the problem (2.1){(2.3) is: Problem P. Find functions x : [0; T ] ! V and y : [0; T ] ! H such that
x(0) = x0 ; y(0) = y0;
(2.4) and for a.e. t 2 (0; T ),
y_ (t) = Ax_ (t) + B (t; x(t); y(t)); (y(t); w ? x_ (t))H + '(w) ? '(x_ (t)) (f (t); w ? x_ (t))V 8 w 2 V:
(2.5) (2.6)
In the study of the problem P we make the following assumptions: (2.7)
(2.8)
8 > > < > > :
A : H ! H is a positive de nite; symmetric; linear operator; i:e:; (a) there exists c0 > 0 such that (Ax; x)H c0kxk2H 8 x 2 H ; (b) (Ax; y)H = (x; Ay)H 8 x; y 2 H:
8 > > > > > > > > > >
0 such that kB (t; x1 ; y1) ? B (t; x2 ; y2)kH L (kx1 ? x2 kH + ky1 ? y2kH ) > 8 t 2 [0; T ]; 8 x1 ; x2 ; y1; y2 2 H ; > > > > > > (b) the mapping t 7! B (t; x; y) is measurable for all x; y 2 H ; > > > : (c) the mapping t 7! B (t; 0; 0) 2 L1 (0; T ; H ): 3
(2.9) ' : V ! IR+ is a continuous seminorm. (2.10) f 2 W 1;1(0; T ; V ): (2.11) x0 2 V; y0 2 H: (2.12) y0 + @'(0) 3 f (0): We see from the condition (2.7) that the quantity (Ax; y)H de nes an inner product on H , and the corresponding induced norm q
kxkA = (Ax; x)H ; x 2 H is equivalent to the norm kxkH . We also remark that the assumption (2.9) implies that ' is Lipschitz continuous on V . Everywhere in the paper we use the standard notation for Lp, W m;p, H m and C m spaces, 1 p 1, m 2 IN. Moreover, if X and Y are real Hilbert spaces, we denote in the sequel by X Y the product space endowed with the canonical inner product. The well-posedness of the problem P has been investigated in [1] where the following result can be found. Theorem 2.1. Under the assumptions (2:7){(2:12), the problem
x 2 W 1;1(0; T ; V ), y 2 W 1;1(0; T ; H ).
P has a unique solution
The proof of Theorem 2.1 is carried out in several steps. It is based on time discretization method, standard arguments of elliptic variational inequalities and a xed point property. In the next two sections, we assume (2.7){(2.12) are satis ed.
3 Numerical analysis of the abstract problem In this section, we present and analyze approximation schemes for solving the problem P. We will give some preliminary results that will be applied to some concrete examples later.
3.1 Spatially semi-discrete approximation Let h 2 (0; 1] be an index and fH hg a family of nite dimensional subspaces of H . Set V h = V \ H h, which is nonempty since 0 2 V h. Let PH : H ! H h be the orthogonal h
4
projection de ned through the relation (PH q; qh)H = (q; qh)H 8 q 2 H; qh 2 H h:
(3.1)
h
Obviously, we have (3.2) kPH qkH kqkH 8 q 2 H: This property will be used on various occasions. h
Now a spatially discrete approximation of the problem P is
Problem Ph . Find the functions xh : [0; T ] ! V h and y h : [0; T ] ! H h such that
xh(0) = xh0 ; yh(0) = y0h;
(3.3) and for a.e. t 2 (0; T ), (3.4) (3.5)
y_ h(t) = PH Ax_ h(t) + PH B (t; xh(t); yh(t)); (yh(t); wh ? x_ h (t))H + '(wh) ? '(x_ h(t)) (f (t); wh ? x_ h(t))V 8 wh 2 V h: h
h
Here, xh0 = PV x0 2 V h, y0h = PH y0 2 H h are orthogonal projections of x0 and y0 to V h and H h, respectively. The de nition of PV : V ! V h is similar to that of PH . From the de nition of y0h, we see that the discrete analog of (2.12) is valid: h
h
h
h
(y0h; wh)H + '(wh) (f (0); wh)V 8 wh 2 V h: This relation is needed to verify the existence of a solution of the problem Ph. We observe that the projection operator PH is introduced to ensure that the relation (3.4) is well-de ned on the space H h. h
Using the arguments in [1], it can be shown that the problem Ph has a unique solution xh 2 W 1;1(0; T ; V h) and yh 2 W 1;1(0; T ; H h). Our main purpose here is to derive estimates for the errors x ? xh and y ? yh. To this end, let t 2 [0; T ]. We rst integrate (2.5) and (3.4), and use the initial conditions (2.4) and (3.3) to obtain Zt
B (s; x(s); y(s)) ds + y0 ? Ax0 ;
(3.6)
y(t) = Ax(t) +
(3.7)
yh(t) = PH Axh(t) + PH h
0
Zt
h
0
B (s; xh(s); yh(s)) ds + y0h ? PH Axh0 : h
5
Then we subtract (3.7) from (3.6) to get (3.8)
y(t) ? yh(t) Z t h h h = PH A(x(t) ? x (t)) + PH B (s; x(s); y(s)) ? B (s; x (s); y (s)) ds 0 + y0 ? y0h ? PH A(x0 ? xh0 ) + (IH ? PH )(y(t) ? y0); h
h
h
h
where IH : H ! H is the identity operator. Denote (3.9) e0 = kx0 ? xh0 kV + ky0 ? y0hkH : Then we get the following inequality from (3.8), using the assumptions (2.7), (2.8) and the property (3.2), (3.10) ky(t) ? yh(t)kH c kx(t) ? xh(t)kV + k(IH ? PH )(y(t) ? y0)kH Zt + c (ky(s) ? yh(s)kH + kx(s) ? xh(s)kV ) ds + c e0; h
0
where c is a positive constant which depends on operators A and B . Everywhere in this paper, except in Section 5, the symbol c will represent a strictly positive constant which may change its value from place to place, and may depend on A, B , T and ', but not on the time or the input data. Now plugging (3.6) in (2.6) with w = x_ h (t), we have (3.11)
(Ax(t); x_ (t) ? x_ h(t))H (f (tZ); x_ (t) ? x_ h (t))V + '(x_ h(t)) ? '(x_ (t)) t + ( B (s; x(s); y(s)) ds; x_ h(t) ? x_ (t))H + (y0 ? Ax0 ; x_ h(t) ? x_ (t))H : 0
Meanwhile, we plug (3.7) in (3.5) with any wh = wh(t) 2 V h to obtain (3.12)
? (Axh(t); x_ (t) ? x_ h(t))H (f (tZ); x_ h(t) ? wh(t))V + '(wh(t)) ? '(x_ h(t)) t + ( B (s; xh(s); yh(s)) ds; wh(t) ? x_ h(t))H 0 + (y0h ? Axh0 ; wh(t) ? x_ h (t))H + (Axh(t); wh(t) ? x_ (t))H :
Adding (3.11) and (3.12), we have (A(x(t) ? xh(t)); x_ (t) ? x_ h(t))H (A(x(t) ? xh(t)); x_ (t) ? wh(t))H + R(t; x_ (t); wh(t)) + (D(t); x_ h(t) ? wh(t))H ; 6
where
R(t; x_ (t); wh) = (y(t); wh ? x_ (t))H + '(wh) ? '(x_ (t)) ? (f (t); wh ? x_ (t))V ; Zt D(t) = (B (s; x(s); y(s)) ? B (s; xh(s); yh(s))) ds + y0 ? y0h ? A(x0 ? xh0 ):
(3.13) (3.14)
0
Using the assumption (2.7), we then have 1 d kx(t) ? xh(t)k2 1 kx(t) ? xh(t)k2 + 1 kx_ (t) ? wh(t)k2 + R(t; x_ (t); wh(t)) A A 2 A 2 dt 2 +(D(t); x_ h(t) ? x_ (t))H + (D(t); x_ (t) ? wh(t))H : Integrate the above inequality from 0 to t to obtain
kx(t) ? xh(t)k2AZ t kx0 ? xh0 k2A + c 0 (jR(s; x_ (s); wh(s))j + kD(s)k2H ) ds +c
Zt h 2 h 2 (kx(s) ? x (s)kA + kx_ (s) ? w (s)kA) ds + (D(s); x_ h(s) ? x_ (s))H ds: 0 0
Zt
For the last term above, we perform an integration by parts, Zt
(D(s); x_ h(s) ? x_ (s))H ds 0 = (D(t); xh(t) ? x(t))H ? (D(0); xh0 ? x0 )H Zt ? (B (s; x(s); y(s)) ? B (s; xh(s); yh(s)); xh(s) ? x(s))H ds: 0
Using (2.8), we have the estimate Zt
(D(s); x_ h(s) ? x_ (s))H ds Zt c kD(t)k2H + 21 kx(t) ? xh(t)k2A + c e20 + c 0 (kx(s) ? xh(s)k2A + ky(s) ? yh(s)k2H ) ds: From the de nition (3.14) and the assumptions (2.7) and (2.8), we have 0
Zt
kD(t)kH c 0 (kx(s) ? xh(s)kA + ky(s) ? yh(s)kH ) ds + c e0: Combine the last several relations, (3.15)
kx(t) ? Zxh(t)k2A t c e20 + c 0 (kx_ (s) ? wh(s)k2A + jR(s; x_ (s); wh(s))j) ds Zt
+ c (kx(s) ? xh(s)k2A + ky(s) ? yh(s)k2H ) ds: 0
7
This inequality, together with (3.10), implies (3.16) kx(t) ? xh(t)k2A + ky(t) ? yh(t)k2H Zt 2 2 c e0 + c k(IH ? PH )(y(t) ? y0)kH + c (kx_ (s) ? wh(s)k2A + jR(s; x_ (s); wh(s))j) ds h
0
Zt
+ c (kx(s) ? xh(s)k2A + ky(s) ? yh(s)k2H ) ds: 0
Applying the Gronwall inequality (1.2), we get
kx(t) ? xh(t)k2A + ky(t) ? yh(t)k2H c e20 + c sup k(IH ? PH )(y(t) ? y0)k2H +c
ZT 0
h
t2[0;T ]
(kx_ (t) ? wh(t)k2A + jR(t; x_ (t); wh(t))j) dt:
Summarizing, we have shown the following result. Theorem 3.1. Let (x; y ) 2 W 1;1(0; T ; V
H ) and (xh; yh) 2 W 1;1(0; T ; V h H h) be the
solutions of the problems P and Ph respectively. Then we have the error estimate
(3.17)
kx ? xhkL1(0;T ;V ) + ky ? yhkL1(0;T ;H ) c (kx0 ? xh0 kV + k(y0 ? y0hkH ) + c k(IH ? PH )(y ? y0)kL1(0;T ;H ) ) ZT hk 2 k x _ ? w + ( + c w 2L2inf jR(s; x_ (s); wh(s))j ds)1=2 ; L (0;T ;V ) (0;T ;V ) 0 h
h
h
where R(; ; ) is de ned in (3:13).
The inequality (3.17) is the basis of convergence and error analysis for the spatially-discrete solutions. Concrete order error estimates will be established when Theorem 3.1 is applied in Section 5 to some examples arising in mechanics.
3.2 Fully discrete approximation In addition to the nite dimensional spaces V h and H h introduced in the previous subsection, we divide the time interval [0; T ] into N equal parts and denote the step-size by k = T=N , the nodal points by tn = nk, n = 0; 1; : : : ; N , and the sub-intervals In = [tn?1; tn], n = 1; : : : ; N . Here we assume k 2 (0; 21 ]. The arguments and results of this subsection can be easily extended to the case of non-uniform partition of the time interval. For a continuous function w(t) with values in H or V , we use the notation wn w(tn). For a sequence fwngNn=0, we denote wn = (wn ? wn?1)=k, n = 1; : : : ; N . 8
Then a fully discrete approximation based on a forward Euler scheme which we will analyze is the following. N h hk hk N h Problem Phk . Find xhk = fxhk n gn=0 V and y = fyn gn=0 H , such that hk h h xhk 0 = x 0 ; y 0 = y0 ;
(3.18) and for n = 1; : : : ; N , (3.19) (3.20)
hk hk ynhk = PH Axhk n + PH B (tn?1 ; xn?1 ; yn?1); h hk h hk h h (ynhk; wh ? xhk n )H + '(w ) ? '(xn ) (fn ; w ? xn )V ; 8 w 2 V : h
h
Here again, xh0 = PV x0 2 V h, y0h = PH y0 2 H h are orthogonal projections of x0 and y0 to V h and H h, respectively. h
h
hk We rst inductively show the unique solvability of the problem Phk . Let xhk n?1 and yn?1 be given. We rewrite (3.19) as
(3.21)
hk hk hk ynhk = k PH Axhk n + k PH B (tn?1 ; xn?1 ; yn?1) + yn?1 h
h
and use it in (3.20) to yield h hk h hk k(Axhk n ; w ? xn )H + '(w ) ? '(xn ) hk hk h hk h h (fn ? k B (tn?1; xhk n?1 ; yn?1) ? yn?1; w ? xn )H 8 w 2 V :
This is a discrete elliptic variational inequality of the second kind for the variable xhk n . By hk h a standard result (cf. [6]), this inequality has a unique solution xn 2 V from which we hk hk can determine xhk n . Then yn can be obtained from (3.21). Therefore the problem P has a unique solution. N hk N Our goal here is to derive estimates for the errors fxn ? xhk n gn=1 and fyn ? yn gn=1 . Assume x 2 C 1 ([0; T ]; V ). First, we apply (3.21) recursively to get
(3.22)
ynhk = PH Axhk n + h
n X j =1
hk hk hk kPH B (tj?1; xhk j ?1 ; yj ?1) + y0 ? PH Ax0 : h
h
Then subtracting (3.22) from (3.6) at t = tn, we obtain (3.23)
yn ? ynhk = (IH ? PH )(yn ? y0) + PH A(xn ? xhk n) Z t n X hk ) + PH B (s; x(s); y(s)) ds ? kB (tj?1; xhk ; y j ?1 j ?1 h
h
n
h
+ y0
0
? yhk ? P 0
H
h
A(x0 9
? xhk ): 0
j =1
Denote (3.24) Dn = k
Zt
n
0
B (s; x(s); y(s)) ds ?
n X j =1
kB (tj?1; xj?1; yj?1)kH ; n = 1; : : : ; N;
for the numerical quadrature errors, and hk en = kxn ? xhk n kV + kyn ? yn kH ; n = 0; : : : ; N for the numerical solution errors. As a preliminary result which will be used on various estimates, the following is derived based on the assumptions (2.8),
k
(3.25)
Zt
n
0
n X
n X j =1
hk kB (tj?1; xhk j ?1 ; yj ?1)kH
hk kkB (tj?1; xj?1; yj?1) ? B (tj?1; xhk j ?1; yj ?1)kH + Dn
j =1 n X
c
B (s; x(s); y(s)) ds ?
j =1
hk k(kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1 kH ) + Dn :
Then we obtain the following inequality from (3.23) by the use of (3.25), the assumptions (2.7) and the property (3.2), (3.26)
kyn ? ynhk kH k(IH ? PH )(yn ? y0)kH + c kxn ? xhk n kH + Dn n X hk + c e0 + c k(kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1kH ): h
j =1
Now let us bound rn xn ? xhk n , n = 1; : : : ; N . For this, we plug (3.6) into (2.6) at t = tn hk with w = xn and get (3.27)
(Axn ; x_ n ? xhk )H (fn; x_ n ? xhk )V + '(xhk n n n ) ? '(x_ n ) Zt hk + ( B (s; x(s); y(s)) ds; xhk n ? x_ n )H + (y0 ? Ax0 ; xn ? x_ n )H : n
0
Similarly, plugging (3.22) into (3.20) with an arbitrary wh = wnh 2 V h yields (3.28)
h hk hk h h hk ? (Axhk n ; wn ? xn )H (fn ; xn ? wn )V + '(wn ) ? '(xn ) n X hk h hk hk hk h hk + ( kB (tj?1; xhk j ?1 ; yj ?1); wn ? xn )H + (y0 ? Ax0 ; wn ? xn )H : j =1
Let us consider the quantity An (Arn; rn)H . The following lower bound for An can be obtained by using the assumptions (2.7): An 21k krnk2A ? krn?1k2A : 10
To derive an upper bound for An, we write hk An = (A(xn ? xhk n ); xn ? xn )H hk h hk h hk = (Axn ; xn ? x_ n)H + (Axn ; x_ n ? xhk n )H ? (Axn ; xn ? wn )H ? (Axn ; wn ? xn )H :
We then use (3.27) and (3.28) to bound the second and fourth terms of above respectively,
An (Arn; xn ? wnh)H + Rn (x_ n; wnh) + I1(n) + I2(n) + I3 (n) 21 (Arn; rn)H + 21 (A(xn ? wnh); xn ? wnh)H + Rn(x_ n; wnh) + I1 (n) + I2 (n) + I3(n); where (3.29)
Rn(x_ n; wnh) = (yn; wnh ? x_ n)H + '(wnh) ? '(x_ n) ? (fn; wnh ? x_ n)V ; Zt n X hk hk I1(n) = ( B (s; x(s); y(s)) ds ? kB (tj?1; xhk j ?1 ; yj ?1); xn ? xn )H ; n
0
j =1 hk hk I2(n) = (y0 0 0 ? x0 ); xn ? xn )H ; Zt n X hk I3(n) = ( B (s; x(s); y(s)) ds ? kB (tj?1; xhk j ?1 ; yj ?1) 0 j =1 hk h + y0 ? y0 ? A(x0 ? xhk 0 ); xn ? wn )H :
? yhk ? A(x
n
Combining the lower and upper bounds for An, we have (3.30)
krnk2A ? krn?1k2A k krnk2A + k kxn ? wnhk2A + 2 k (Rn(x_ n ; wnh) + I1(n) + I2 (n) + I3(n)):
In the inequality (3.30), we change the index n to j , and sum over j from 1 to n, (3.31)
krnk2A kr0k2A + +2
n X j =1
n X j =1
k krj k2A +
n X j =1
k kxj ? wjhk2A
k [Rj (x_ j ; wjh) + I1(j ) + I2 (j ) + I3(j )]:
In order to handle the last summation, we rst estimate the quantity Dn de ned by (3.24),
n Zt X
(B (s; x(s); y(s)) ? B (tj?1; xj?1; yj?1)) dskH j =1 t ?1 n Zt X kB (s; x(s); y(s)) ? B (tj?1; xj?1; yj?1)kH ds j =1 t ?1 n Zt X (kx(s) ? xj?1kV + ky(s) ? yj?1kH )ds; c j =1 t ?1
Dn = k
j
j
j
j
j
j
11
where the assumption (2.8) is used. Hence,
Dn c tnk kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) :
(3.32) We now write n X
j =1
kI1 (j ) = =
n Z t X
j
j =1 0 Z t1 0
?
B (s; x(s); y(s)) ds ?
j X i=1
hk k B (ti?1 ; xhk i?1 ; yi?1 ); rj ?1 ? rj
hk B (s; x(s); y(s)) ds ? kB (t0 ; xhk 0 ; y0 ); r0
Z t
n
B (s; x(s); y(s)) ds ?
n X
H
hk kB (ti?1 ; xhk i?1 ; yi?1 ); rn
0 H i=1 nX ?1 Z t +1 B (s; x(s); y(s)) ds ? kB (tj ; xhk ; yjhk); rj : + j H j =1 t j
j
Using the assumption (2.7) and the estimates (3.25) and (3.32), we obtain n nX ?1 X 1 kr k2 kI1(j ) c e20 + c k (krj k2V + kyj ? yjhkk2H ) + 16 (3.33) n A j =1 j =1 + c k2(kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) )2: Similarly, we write
n X
j =1
kI2(j ) = =
n X hk (rj?1 ? rj ))H (y0 0 0 ? x0 ); j =1 (y0 ? y0hk ? A(x0 ? xhk 0 ); r0 ? rn )H ;
? yhk ? A(x
and use the assumption (2.7) to obtain n X
(3.34)
j =1
1 kr k2 : kI2 (j ) c e20 + 16 n A
Finally we use the assumption (2.8) and the estimates (3.25) and (3.32) to nd (3.35)
n X j =1
kI3(j ) c e20 + c
nX ?1 j =1
k (krj k2V + kyj ? yjhkk2H )
+ c k2(kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) )2 + c
n X j =1
k kxj ? wjhk2V :
We now combine the estimates (3.26), (3.31) and (3.33){(3.35) to obtain (3.36)
2 hk 2 kxn ? xhk n kV + kyn ? yn kH c Jn + c
12
nX ?1 j =1
2 hk 2 k(kxj ? xhk j kV + kyj ? yj kH );
H
where
Jn = e20 + k2(kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) )2 + k(IH ? PH )(yn ? y0)k2H n n X X + k kxj ? wjhk2V + k jRj (x_ j ; wjh)j: h
j =1
j =1
An error estimate can be derived based on (3.36). hk N Theorem 3.2. Let (x; y ) 2 C 1 ([0; T ]; V ) W 1;1(0; T ; H ) and f(xhk n ; yn )gn=1 be the solu-
tions of the problem P and Phk , respectively. Then we have the error estimate
(3.37)
hk max kxn ? xhk n kV + kyn ? yn kH 1nN hk c (kx0 ? xhk 0 kV + ky0 ? y0 kH ) + c k (kx_ kL1 (0;T ;V ) + ky_ kL1 (0;T ;H ) ) + c 1max k(I ? PH )(yn ? y0)kH nN H h
+c
X N
k inf kxj ? wjhk2V + jRj (x_ j ; wjh)j
j =1 w 2V h j
1=2
h
:
P Proof. Denote En = nj=1 k e2n , then (3.36) can be rewritten as
e2n c Jn + c En?1:
(3.38) Now
En ? En?1 k e2n c k Jn + c k En?1:
Hence we have
En ? (1 + c k)En?1 c k Jn;
or equivalently,
En ? En?1 c k Jn : (1 + c k)n (1 + c k)n?1 (1 + c k)n By an inductive argument, we get n X
J (ec T ? 1) 1max J: En c k (1 + c k)n?iJi ((1 + c k)n ? 1) 1max in i in i i=1
Using (3.38), we have which implies (3.37).
e2n c 1max J; in i
2
The estimate (3.37) will be used for error analysis of the fully discrete solutions provided the exact solution possesses certain regularity. 13
4 Convergence analysis In this section, we analyze the convergence of the spatially and fully discrete solutions for the problem P under the basic regularity condition (x; y) 2 W 1;1(0; T ; V H ), available from Theorem 2.1. First we make the following additional assumptions on the function spaces H; V and the nite dimensional spaces H h and V h. Assumption (H1). There exist a subspace V0
(h) 0 such that
and
lim (h) = 0;
h!0+
hk = kw ? P wk (h)kwk inf k w ? w V V V0 8 w 2 V0 : V w 2V h
h
h
Assumption (H2). There exist a subspace H0
(h) 0 such that and
V which is dense in V and a function
H which is dense in H and a function
lim (h) = 0;
h!0+
inf kz ? zh kH = kz ? PH zkH (h)kzkH0 8 z 2 H0:
z 2H h
h
h
These two hypotheses will be veri ed for the applicaiton problems discussed in the next section. We will need the following result, which can be found in [17]. Lemma 4.1. Assume that X is a Banach space, X0 X is dense in X . Then H 1 (0; T ; X0)
is dense in H 1 (0; T ; X ).
Now we are ready to study the convergence of the spatially discrete solution for the problem P, based on the estimate (3.17). Theorem 4.2. Let (x; y )
2 W 1;1(0; T ; V H ) be the solution of the problem P and
(xh; yh) 2 W 1;1(0; T ; V h H h) the solution of the corresponding spatially-discrete problem Ph. Then under the assumptions (H1) and (H2) we have
(4.1)
kx ? xhkL1(0;T ;V ) + ky ? yhkL1(0;T ;H ) ! 0 as h ! 0:
Proof. From the assumptions (H1) and (H2), we have
(4.2)
kx0 ? xh0 kV ! 0; ky0 ? y0hkH ! 0 as h ! 0: 14
Since ' is Lipschitz continuous on V , using the de nition (3.13), we have
jR(t; x_ (t); wh(t))j (ky(t)kH + kf (t)kV + c) kwh(t) ? x_ (t)kV : Hence, the estimate (3.17) can be rewritten as (4.3)
kxh ? xkL1(0;T ;V ) + kyh ? ykL1(0;T ;H ) h k1=2 c k(IH ? PH )(y ? y0)kL1(0;T ;H ) + c e0 + c w 2L2inf k x _ ? w L2 (0;T ;V ) : (0;T ;V ) h
h
h
Using Lemma 4.1 and Assumption (H1), we know that H 1(0; T ; V0) is dense in H 1(0; T ; V ). So for any " 2 (0; 1), there exists x~ 2 H 1(0; T ; V0) such that
kx ? x~kH 1 (0;T ;V ) ";
(4.4)
which can be combined with (H1) again to yield (4.5)
inf kx_ ? whk1L=22(0;T ;V ) w 2L2 (0;T ;V ) kx~_ ? whk1L=22(0;T ;V ) kx_ ? x~_ k1L=22(0;T ;V ) + w 2L2inf (0;T ;V ) p q (h)kx~k1=2 "+ H 1 (0;T ;V0 ) : h
h
h
h
Similarly, using Lemma 4.1 and the assumption (H2), H 1(0; T ; H0) is dense in H 1(0; T ; H ). So there exists y~ 2 H 1(0; T ; H0) such that
ky ? y0 ? y~kH 1(0;T ;H ) ": Now we are ready to bound the rst term on the right hand side of (4.3):
k(IH ? PH )(y ? y0)kL1(0;T ;H ) c k(IH ? PH )(y ? y0)kH 1 (0;T ;H ) c ky ? y0 ? y~kH 1 (0;T ;H ) + k(IH ? PH )~ykH 1 (0;T ;H ) c " + (h) ky~kH 1(0;T ;H0 ): Above we used the embedding result H 1(0; T ; H ) ,! L1(0; T ; H ) (cf. [17]). The convergence result (4.1) now follows from (4.2) and (4.3){(4.6). 2
(4.6)
h
h
h
We now turn to a convergence analysis of the fully discrete scheme. Notice that we can not use the estimate (3.37), because under the basic regularity condition (x; y) 2 W 1;1(0; T ; V H ), the pointwise values x_ j and y_j are not well-de ned. Here we follow the approach developed in [9] for a convergence analysis. For this purpose we will need another density result, which can also be found in [17]. 15
Lemma 4.3. The space C 1 ([0; T ]; V ) is dense in H 1 (0; T ; V ); that is, given w 2 H 1 (0; T ; V ),
for any " > 0 there exists w 2 C 1([0; T ]; V ) such that kw ? wkH 1(0;T ;V ) ":
Let us consider the quantity An = (Arn; rn)H . As in Section 3, a lower bound of An is An 21k (krnk2A ? krn?1k2A): (4.7) To obtain an upper bound we begin with hk An = (A(xn ? xhk n ); xn ? xn )H hk h hk h hk = (Axn; xn ? xhk n )H ? (Axn ; xn ? wn )H ? (Axn ; wn ? xn )H ; where wnh 2 V h is arbitrary. Using (3.28) to bound the last term, we obtain hk h (4.8) An (Axn; xn ? xhk n )H ? (Axn ; xn ? wn )H h h hk + (fn; xhk n ? wn )V + '(wn ) ? '(xn ) n X hk h hk hk hk h hk + ( kB (tj?1; xhk j ?1 ; yj ?1); wn ? xn )H + (y0 ? Ax0 ; wn ? xn )H : j =1
Now integrate (2.6) with w = xhk n from t = tn?1 to tn and use (3.6) to obtain Z Z 1 1 hk hk (4.9) 0 k (Ax(t); xn ? x_ (t))H dt + '(xn ) ? k '(x_ (t))dt I I Z Zt 1 hk + k ( B (s; x(s); y(s))ds; xhk n ? x_ (t))H dt + (y0 ? Ax0 ; xn ? xn )H I 0 Z + 1 (fn; x_ (t) ? xhk n )V dt: k I Then we add (4.9) and (4.8) to obtain (4.10) A n R1 + R2 + R3 + R 4 + R 5 ; where Z hk ; x ? wh) + 1 (Ax(t); xhk ? x_ (t)) dt; R1 = (Axn; xn ? xhk ) ? ( Ax H n H H n n n n k I Z R2 = '(wnh) ? k1 '(x_ (t))dt; I Z 1 hk h R3 = (fn; xn ? wn)V + k (fn; x_ (t) ? xhk n )V dt; I h hk hk R4 = (y0hk ? Axhk 0 ; wn ? xn )H + (y0 ? Ax0 ; xn ? xn )H ; n X hk h hk R5 = ( kB (tj?1; xhk j ?1 ; yj ?1); wn ? xn )H n
n
n
n
n
n
n
j =1
+ k1
Z
Zt
I
n
( B (s; x(s); y(s))ds; xhk n ? x_ (t))H dt: 0
16
We need to nd appropriate bounds for Ri, 1 i 5. First let us estimate R1 . Using the assumption (2.7) and the properties of inner product, we have Z Z 1 1 1 (4.11) R1 = k (Axn ? k Ax(t) dt; rn ? rn?1)H + k (Ax(t); xn ? x_ (t))H dt I I h h + (Aen; xn ? wn)H ? (Axn ; xn ? wn )H Z Z k12 (A I (t ? tn?1 )x_ (t) dt; rn ? rn?1)H + k1 I (Ax(t); xn ? x_ (t))H dt + 1 krnk2A + 1 kxn ? wnhk2A ? (Axn; xn ? wnh)H 2 2 Z c kx_ kL1(I ;V ) (krnkV + krn?1kV ) + kc kxkL1(I ;V ) I kxn ? x_ (t)kV dt + 1 krnk2A + c kxn ? wnhk2V + c kxnkV kxn ? wnhkV : 2 Using the Lipschitz continuity of ' on V , we nd a bound for R2 , Z Z c 1 h (4.12) jR2j = k j I ('(wn) ? '(x_ (t)) dtj k I kwnh ? x_ (t)kV dt: For R3, we have Z 1 R3 = k2 ( (t ? tn?1 )f_(t)dt; rn?1 ? rn)H + (fn; xn ? wnh)H (4.13) I Z 1 ? k I (f (t); xn ? x_ (t))H dt kf_kL1(I ;V ) (krnkV + krn?1kV ) + kfnkV kxn ? wnhkV Z + k1 kf kL1(I ;V ) kxn ? x_ (t)kV dt: n
n
n
n
n
n
n
n
n
n
n
n
n
I
n
For R4, we have
h R4 = k1 (y0 ? y0hk ? A(x0 ? xhk 0 ); rn?1 ? rn )H ? (y0 ? Ax0 ; xn ? wn )H h + (y0 ? y0hk ? A(x0 ? xhk 0 ); xn ? wn )H k1 (y0 ? y0hk ? A(x0 ? xhk 0 ); rn?1 ? rn )H + c (e0 + ky0kH + kx0 kV )kxn ? wnhkV : Express R5 as
(4.14)
(4.15)
R5 =
Zt hk hk ( kB (tj?1; xj?1; yj?1) ? B (s; x(s); y(s)) ds; wnh ? xhk n )H 0 j =1 Zt + ( B (s; x(s); y(s)) ds; wnh ? xhk n )H 0 Z Zt + k1 ( B (s; x(s); y(s)) ds; (xhk n ? xn ) + (xn ? x_ (t)))H dt: I 0
n X
n
n
n
17
Integrating by parts, we obtain Z Zt Z (4.16) k1 ( B (s; x(s); y(s))ds; xn ? x_ (t))H dt = k1 (B (t; x(t); y(t)); x(t))H dt; I 0 I where x(t) = x(t) ? t ?ktn?1 xn ? tn k? t xn?1 ; tn?1 t tn : (4.17) By an elementary manipulation, we have 1 Z (Z t B (s; x(s); y(s)) ds; xhk ? x ) dt (4.18) n H n kZ I 0 t = ( B (s; x(s); y(s)) ds; xhk n ? xn )H 0 Z ? k1 ( I (s ? tn?1)B (s; x(s); y(s)) ds; xhk n ? xn )H : Using (4.16) and (4.18), we can rewrite R5 as n
n
n
n
n
R5 =
Zt hk hk ( kB (tj?1; xj?1; yj?1) ? B (s; x(s); y(s)) ds; wh ? xn)H 0 j =1 Z n X hk ) ? t B (s; x(s); y (s)) ds; r ? r ) ; y + k1 ( k B (tj?1; xhk n n?1 H j ?1 j ?1 0 j =1 Z ? k12 ( I (s ? tn?1 )B (s; x(s); y(s)) ds; rn?1 ? rn)H Z Zt + k1 (B (t; x(t); y(t)); x(t))H dt + ( B (s; x(s); y(s)) ds; wnh ? xn)H : I 0
n X
n
n
n
n
n
We then apply the estimates (3.25) and (3.32), (4.19)
R5 c
n X j =1
2 hk 2 k(kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1kV )
+ c k2(kx_ k2L1(0;t ;V ) + ky_ k2L1(0;t ;H )) Z 1 1 h 2 1 + 2 kwn ? xn kV + k kB (; x; y)kL (I ;H ) kx(t)kH dt I Zt n X 1 hk + k ( kB (tj?1; xhk B (s; x(s); y(s))ds; rn ? rn?1)H j ?1 ; yj ?1) ? n
n
n
n
n
j =1
+ kB (; x; y)kL1(I
n
0
krnkV + krn?1kV + tn kwnh ? xn kV ):
;H ) (
Combining the relations (4.7), (4.10){(4.14) and (4.19), we nd that (4.20) krnk2A ? krn?1k2A k krnk2A + c k (kx_ kL1(I ;V ) + kf_kL1(I n
n
;V )
+ kB (; x; y)kL1(I 18
n
krnkV + krn?1kV )
;V ) ) (
Z
+ c (kxkL1(I ;V ) + kf kL1(I ;V )) kxn ? x_ (t)kV dt + ckkxn ? wnhk2V I + c (ZkxnkV + kfnkV + e0 + ky0kH + kx0 kV + kB (; x; y)kL1(I ;V ) )kxn ? wnhkV + c kwnh ? x_ (t)kV dt + 2 (y0 ? y0hk ? A(x0 ? xhk 0 ); rn?1 ? rn )H n
n
n
n
I n X n
+2( +c
hk kB (tj?1; xhk j ?1 ; yj ?1) ?
j =1 n X 2
j =1
Zt
n
0
B (s; x(s); y(s)) ds; rn ? rn?1)H
2 hk 2 k (kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1kV )
+ c kB (; x; y)k
L1(I ;H ) n
Z
I
n
kx(t)kH dt + c k3(kx_ k2L1(0;t ;V ) + ky_ k2L1(0;t ;V ) ): n
n
By a series of manipulations similar to those leading to (3.36), we obtain from (4.20) that (4.21)
kenk2V + kyn ? ynhk k2H nX ?1 2 hk 2 2 c k(kxj ? xhk j kV + kyj ? yj kV ) + c e0 j =1
+ c k (kxk2W 1 1(0;T ;V ) + kf k2W 1 1(0;T ;V ) + ky0k2H + kx0 k2V + kB (; x; y)k2L1(0;T ;V )) ;
;
+ c (kxkL1(0;T ;V ) + kf kL1(0;T ;V ) ) +c
n Z X
j =1 I
j
kxj ? x_ (t)kV dt + c
kwjh ? x_ (t)kV dt + c kB (; x; y)kL1(0;t ;H )
j =1 I + k(IH ? PH j
n Z X
n
h
)(yn ? y0)k2H :
n Z X j =1 I
j
n X
j =1
k kxj ? wjhk2V
kx(t)kH dt
We can then apply the technique of the proof of Theorem 3.2 to nd the estimate (4.22)
2 hk 2 max (kxn ? xhk n kV + kyn ? yn kV )
1nN
c e20 + 1max k(I ? PH )(yn ? y0)k2H nN H + k (kxk2W 1 1(0;T ;V ) + kf k2W 1 1(0;T ;V ) + ky0k2H + kx0 k2V + kB (; x; y)k2L1(0;T ;V ) ) h
;
;
+ (kxkL1(0;T ;V ) + kf kL1(0;T ;V ) ) N Z X + j =1 I
j
N Z X
j =1 I
j
kxj ? x_ (t)kV dt +
kwjh ? x_ (t)kV dt + kB (; x; y)kL1(0;T ;H )
N Z X j =1 I
j
N X
j =1
k kxj ? wjhk2V
kx(t)kH dt :
We now derive a convergence result from (4.22). Theorem 4.4. Let (x; y )
2 W 1;1(0; T ; V H ) be the solution of the problem P and 19
hk N hk f(xhk n ; yn )gn=1 be the solution of the corresponding fully discrete problem P . Then un-
der the assumptions (H1) and (H2) we have
2 hk 2 max (kxn ? xhk n kV + kyn ? yn kV ) ! 0 as h; k ! 0:
(4.23)
1nN
Proof. We still have (4.2). By Lemma 4.3, for any " > 0, there exists x 2 C 1([0; T ]; V )
such that
kx ? xkH 1 (0;T ;V ) ":
We now estimate the term PNj=1 Z
I Z
j
=
I
R
I
j
kxj ? x_ (t)kV dt. Since
kxj ? x_ (t)kV dt Z
k k1 I (x_ (s) ? x_ (t))dskV dt
Z Z k1 I I [kx_ (s) ? x_ (s)kV + kx_ (t) ? x_ (t)kV + kx_ (s) ? x_ (t)kV ] ds dt Z Z 2 kx_ (t) ? x_ (t)kV dt + k kx(t)kV dt; j
j
j
j
I
I
j
we have (4.24)
N Z X j =1 I
j
Next, from
It suces
kxj ? x_ (t)kV dt c " + k kx(t)kL1 (0;T ;V ) :
kwjh ? x_ (t)kV kwjh ? xj kV + kxj ? x_ (t)kV ;
we see that (4.25)
j
N Z N X X h h kwj ? x_ (t)kV dt k kwj ? xj kV + I kxj ? x_ (t)kV dt: j =1 j =1 j =1 I to estimate the term k PNj=1 kwjh ? xj kV . Noticing that Z wjh ? xj = wjh ? x~j + k1 (x~_ (t) ? x_ (t))dt; I N Z X
j
j
j
we have (4.26)
k
N X j =1
kwjh ? xj kV
k k
N X j =1 N X j =1
kwjh ? x~j kV + kx_ ? x~_ kL1 (0;T ;V ) kwjh ? x~j kV + c ": 20
Here x~ is the function used in the proof of Theorem 4.2 such that (4.4) holds. It remains to bound the last term in (4.22). From the de nition (4.17), we have Z t t ?t Z t t?t n n?1 x(t) = x _ (s)ds ? k x_ (s)ds; tn?1 t tn : t ?1 k t Therefore, N Z N Z X X (4.27) kx(t)kH dt k kx_ (s)kV ds = k kx_ kL1(0;T ;V ): n
n
j =1 I
j
j =1
I
j
The bounds (4.24){(4.27) are now used in the estimate (4.22). Noticing the arbitrariness of wjh 2 V h, 1 j N , and recalling the estimate (4.6), we get (4.28)
2 hk 2 2 max (kxn ? xhk x)); n kV + kyn ? yn kV ) c (e0 + " + (h) + k + Dhk (~
1nN
where
Dhk (~x) = k
N X
inf kx~j ? wjhkV
j =1 w 2V h j
h
and the constant c depends on x; y; y~; f; B (t; x(t); y(t)) but is independent of "; h; k. By the assumption (H1), we see that Z ( h ) h _ inf kx~j ? wj kV (h)kx~j kV0 k I kx~(t)kV0 dt; w 2V h j
h
j
and thus (4.29) Dhk (~x) (h) kx~_ kL1(0;T ;V0 ): Then the convergence result (4.23) follows directly from (4.2), (4.28) and (4.29).
2
We remark here that the estimate (3.37) is used for deriving optimal order error estimates provided the exact solution has higher degree regularity, while the estimate (4.22) is suitable for convergence analysis under the basic solution regularity condition.
5 Applications to contact problems in rate-type viscoplasticity The aim of this section is to apply the results obtained in Sections 3 and 4 to the numerical analysis of two nonlinear quasistatic frictional contact problems for viscoplastic materials. In this section, the symbol c may depend on the exact solution, but it is independent of the discretization parameters h and k. 21
Let us consider a viscoplastic body whose material particles occupy a bounded domain
IRd (d = 2; 3 in applications). The boundary of , being assumed Lipschitz continuous, is partitioned into three disjoint measurable parts ?1; ?2 and ?3, with meas (?1) > 0. Displacement, surface traction and contact conditions will be speci ed on ?1 ; ?2 and ?3 , respectively. Since the boundary is Lipschitz continuous, the unit outward normal vector exists a.e. on the boundary. Let [0; T ] be a time interval of interest. We assume that the body is xed on ?1 , a body force of density b acts in , and a surface traction of density F acts on ?2 . Both b and F can be time dependent but we assume a slowly variation of these functions in time so that the inertia terms in the equations of motion may be neglected. We choose (1.1) as the constitutive relation for the viscoplastic material, in which E is a fourth order tensor and G is a given constitutive function, possibly nonlinear. The boundary value problem is then (5.1) (5.2) (5.3) (5.4) (5.5)
_ = E "(u_ ) + G( ; "(u)) in (0; T ); Div + b = 0 in (0; T ); u = 0 on ?1 (0; T ); = F on ?2 (0; T ); u(0) = u0 ; (0) = 0 on :
The unknowns of the problem are the displacement eld u : [0; T ] ! IRd and the stress eld : [0; T ] ! Sd , Sd being the set of second order symmetric tensors on IRd . Here u0 and 0 are the given initial data. We de ne the inner product and the corresponding norms on IRd and Sd by u v = ui vi ; jvj = (v v )1=2 ; = ij ij ; jj = ( )1=2 ;
8 u; v 2 IRd; 8 ; 2 Sd :
For every vector eld v, we denote by v and v the normal and the tangential components of v on the boundary given by
v = v ; v = v ? v and let "(v) denote the tensor eld de ned by 1
"(v ) = ("ij (v )); "ij (v ) = (vi;j + vj;i ); 2
where the index that follows a comma indicates a partial derivative with respect to the corresponding component of the independent variable. 22
We denote in the sequel by H the real Hilbert space de ned by
H = f = ( ij ) j ij = ji 2 L2 ( ); 1 i; j dg with the inner product
(; )H =
Z
ij ij dx; ; 2 H:
We assume in the sequel that E : Sd ! Sd and G : Sd Sd ! Sd satisfy the following assumptions. 8 > > > >
(c) There exists an 0 > 0 such that > > > : E 0 j j2 8 2 Sd; a:e: in : 8 > (a) There exists an L > 0 such that 8 1; 2; "1; "2 2 Sd; a:e: in ; > > > < jG(x; 1 ; "1 ) ? G(x; 2 ; "2 )j L (j1 ? 2 j + j"1 ? "2j): > (b) For any ; " 2 Sd ; x 7! G(x; ; ") is measurable: > > > : (c) The mapping x 7! G(x; 0; 0) 2 H: These assumptions will be used to verify (2.7) and (2.8) in the context of mechanical applications later. For the input data, we assume that b 2 W 1;1(0; T ; (L2( ))d), F 2 W 1;1(0; T ; (L2(?2))d ). Finally, we suppose that the viscoplastic body is in contact with a rigid foundation on ?3 (0; T ). This contact involves friction. In the sequel we consider two di erent friction laws which lead us to the following examples.
5.1 Bilateral contact with Tresca's friction law We assume a bilateral contact modeled by Tresca's friction law (see e.g. [2, 5]), i.e. (5.6)
8 > > < u = 0; j j g; j j < g ) u_ = ; > > : j j = g ) there exists 0
0
such that = ?u_ ;
on ?3 (0; T );
where u represents the normal displacement, u_ denotes the tangential velocity, is the tangential force on the contact boundary and g 0 is the friction yield limit. 23
Let
U = fv 2 (H 1( ))d j vZ= 0 on ?1; v = 0 on ?3 g; j : U ! IR+; j (v) = g jv j ds; ?3
L : [0; T ] U ! IR; L(t; v) =
Z
b(t) v dx +
and let u0 2 U , 0 2 H be given initial such that
Z
?2
F (t) v ds
(0 ; "(v))H + j (v ) L(0; v) 8 v 2 U: In [2] the following weak formulation of the mechanical problem (5.1){(5.6) was derived. Problem P1 . Find the displacement eld u : [0; T ] ! U and the stress eld : [0; T ] ! H
such that (5.7) and for a.e. t 2 (0; T ),
(5.8) (5.9)
u(0) = u0 ; (0) = 0 ;
_ (t) = E "(u_ (t)) + G( (t); "(u(t))); ((t); "(v ? u(t)))H + j (v) ? j (u(t)) L(t; v ? u(t))
8 v 2 U:
Let V be the subspace of H given by
V = "(U ) = f"(v) j v 2 U g: Since meas (?1) > 0, Korn's inequality holds (see [14]):
kvkU c k"(v)kH 8 v 2 U: It follows that V is a closed subspace of H and that the deformation operator " : U ! V is a linear continuous invertible operator. We denote the inverse of " : U ! V by "?1 : V ! U , which is a linear, continuous operator. Now the variational problem P1 can be viewed as a special case of the abstract problem P, after we make the following identi cations (5.10)
(
x $ "(u); y $ ; x0 $ "(u0); y0 $ 0 ; A $ E ; B $ G; '(w) $ j ("?1(w)); (f (t); w)V $ L(t; "?1(w)) for w 2 V:
The conditions (2.7){(2.12) can then be veri ed by using the assumptions made on the constitutive functions E and G as well as on the data b, F , u0 , 0 and g. Therefore, 24
using Theorem 2.1 is follows that problem P1 has a unique solution having the regularity (u; ) 2 W 1;1(0; T ; U H ). We now brie y specify how to construct the nite dimensional spaces V h and H h via the nite element method. Details can be found in [3]. For simplicity, we assume that is polygonal. Let T h be a regular nite element partition of in such a way that if a side of an element lies on the boundary, the side belongs entirely to one of the subsets ?1 , ?2 and ?3. Let U h U consist of linear elements, let us use piecewise constants for H h and recall that V h = H h \ V . It could be shown that V h = "(U h). With the above speci cations, let us show that the assumptions (H1) and (H2) are satis ed. Let
U0 = U \ (H 2( ))d; V0 = "(U0 ); H0 = H \ (H 1( ))dd : The spaces V0 and H0 are dense in V and H , respectively. For any w 2 V0 ; there exists 2 U0 such that w = "(w ). Let hw 2 U h be the piecewise linear interpolant of w . Then w from the standard nite element interpolation theory (cf. [3]), we have
kw ? hw k(H 1 ( )) c h kw k(H 2( )) : d
d
By Korn's inequality, we get h kw ? z hk(H 1 ( )) winf2V kw ? w kV c z inf 2U c kw ? hw k(H 1 ( )) c h kw k(H 2 ( )) c h kwk(H 1 ( )) : h
h
h
d
h
d
d
d
So the assumption (H1) is satis ed and we may take (h) = c h. The assumption (H2) can be veri ed similarly. Therefore by Theorems 4.2 and 4.4, both the spatially discrete and fully discrete solutions corresponding to Problem P1 converge to the solution of Problem P1 as h, and h and k go to zero. In order to derive error estimates via (3.17) and (3.37), we need to make assumptions on the regularity of the exact solution. First let us specialize the estimate (3.17) for the spatial discretization of the problem P1 and obtain (5.11)
kuh ? ukL1(0;T ;U ) + kh ? kL1(0;T ;H ) c k(IH ? PH )( (? 0)kL1(0;T ;H ) + c (kuh0 ? u0 kU + kh0 ? 0 kH ) ) ZT ku_ ? z hkL2(0;T ;U ) + ( 0 jR(t; "(u_ (t)); "(z h(t)))j dt)1=2 : + c z 2L2inf (0;T ;U ) h
h
h
25
The error analysis of the spatially-discrete solution is given by the following result. Theorem 5.1. Let (u; ) 2 W 1;1(0; T ; U H ) be the solution of the problem P1 and (uh; h) 2 W 1;1(0; T ; U h H h) be the corresponding spatially-discrete solution. For the initial values, we choose uh0 2 U h to be the orthogonal projection of u0 into U h with respect to the inner product ("(u); "(v))V , and h0 2 H h the orthogonal projection of 0 into H h
with respect to the inner product of H . Assume
2 W 1;1(0; T ; H0); u 2 W 1;1(0; T ; U ) \ H 1 (0; T ; U0 ); u
2 H 1(0; T ; H 2(?3 )):
Then we have the optimal order error estimates
kuh ? ukL1(0;T ;U ) + kh ? kL1(0;T ;H ) c h:
(5.12)
Proof. First from the choice of the initial values uh0 and h0 , we have
ku0 ? uh0 kU c h; k0 ? h0 kH c h:
(5.13)
Let hu_ (t) 2 U h be the piecewise linear interpolant of u_ (t) for a.e. t 2 (0; T ). Then we have the estimates (cf. [3])
ku_ (t) ? hu_ (t)kU c h ku_ (t)k(H 2 ( )) ; ku_ (t) ? (hu_ (t)) kL2(?3 ) c h2ku_ (t)kH 2 (?3 ): d
Keep in mind the de nition (3.13) and the transformation (5.10). We perform an integration by parts to obtain
R (t; "(u_ (t)); "(hu_ (t))) Z Z h = (t)(( u_ (t)) ? u_ (t)) ds + g (ju_ (t)j ? j(h u_ (t)) j) ds: ?3
?3
Hence, (5.14)
jR(t; "(u_ (t)); "(hu_ (t)))j (k (t)kL2(?3 ) + c g) ku_ (t) ? (hu_ (t)) kL2 (?3 ) c (k(t)kH0 + g)h2ku_ (t)kH 2(?3 ) :
From the property of the projection, we have (5.15)
k(IH ? PH )((t) ? 0 )kH c h k(t) ? 0kH0 : h
Then the error estimate (5.12) follows from (5.11), (5.13), (5.14) and (5.15). 26
2
Now we turn to error analysis of the fully discrete solution. Theorem 5.2. Let (u; ) 2 W 1;1(0; T ; U H ) be the solution of the problem P1 and hk N f(uhk n ; n )gn=1 be the corresponding fully-discrete solution. For the initial values, we choose uh0 2 U h to be the orthogonal projection of u0 into U h with respect to the inner product ("(u); "(v))V , and h0 2 H h the orthogonal projection of 0 into H h with respect to the
inner product of H . Assume 2 W 1;1(0; T ; H0); u 2 H 2 (0; T ; U ) \ C 1 ([0; T ]; U0 ); u 2 C 1 ([0; T ]; H 2(?3 )): Then we have the optimal order error estimates hk ? u k + k hk ? k c (h + k): k u (5.16) max n U n H n n 1nN
Proof. We rst specialize the estimate (3.37) for the case of the full discretization of the
problem P1: hk k + k ? hk k (5.17) max k u ? u n U n H n n 1nN c k (ku_ kL1(0;T ;U ) + k_ kL1(0;T ;H )) +c +c
1=2 h + jRj ("(u_ j ); "(z ))j k inf j =1 z 2U hk k + k ? hk k : max k ( I ? P )( ? ) k + c k u ? u H n 0 H 0 U 0 H H 0 0 1nN
X N
h
h
kuj ? z hk2U h
Again, we have the estimate (5.13) for the approximation of the initial values. For each j , let hu_ j be the piecewise linear interpolant of u_ j . Then kuj ? hu_ j kU c (kZuj ? u_ j kU + ku_ j ? hu_ j kU ) c ( k1 I ku_ (t) ? u_ j kU dt + h ku_ j kU0 ) Z c ku (t)kU dt + c h ku_ j kU0 : j
So
I
j
N X j =1
kkuj ? hu_ j k2U
c
N X j =1
k
!2
Z
I
j
ku (t)kU dt + c h2 1max ku_ k2 j N j U0
c k2ku kL2 (0;T ;U ) + c h2ku_ k2L1(0;T ;U0):
Similar to (5.14) we have jRj ("(u_ j ); "(zh))j c h2(kkL1(0;T ;H0) + 1) ku_ kL1(0;T ;H 2 (?3 )) : The error estimate (5.16) then follows from (5.17) together with the bounds for the various terms on the right hand side of (5.17). 2 27
5.2 Unilateral contact with simpli ed Coulomb's friction law Now we consider an unilateral contact problem modeled by a simpli ed version of Coulomb's law of dry friction (see e.g. [5, 15]), i.e. (5.18)
8 > > < = S; j j g j j; j j < g j j ) u_ = ; > > : j j = g j j ) there exists 0 such
0
that = ?u_ ;
on ?3 (0; T ):
Here denotes the normal force on the contact boundary, S 2 L1(?3) is a given function and g 0 is the coecient of friction. Let
U = fv 2 (H 1( ))d j vZ= 0 on ?1 g; j : U ! IR+; j (v) = g jS j jv j ds; ?3
L : [0; T ] U ! IR; L(t; v ) =
Z
b(t) v dx +
Z
and let u0 2 U , 0 2 H denote initial data such that
?2
F (t) v ds +
Z ?3
S v ds;
(0 ; "(v))H + j (v ) L(0; v) 8 v 2 U: The weak formulation of the mechanical problem (5.1){(5.5), (5.18) is (see e.g. [1]): Problem P2 . Find the displacement eld u : [0; T ] ! U and the stress eld : [0; T ] ! H
such that (5.19) and for a.e. t 2 (0; T ),
(5.20) (5.21)
u(0) = u0 ; (0) = 0 ;
_ (t) = E "(u_ (t)) + G( (t); "(u(t))); ((t); "(v ? u(t)))H + j (v) ? j (u(t)) L(t; v ? u(t))
8 v 2 U:
Let V be the subspace of H given by
V = "(U ) = f"(v) j v 2 U g: Again with the identi cations (
x $ "(u); y $ ; x0 $ "(u0); y0 $ 0 ; A $ E ; B $ G; '(w) $ j ("?1(w)); (f (t); w)V $ L(t; "?1(w)) for w 2 V; 28
the problem P2 can be rewritten in the form of Problem P. Moreover, from the assumptions made on the constitutive functions and data, it follows that conditions (2.7){(2.12) are satis ed in this case. Therefore, we can readily extend all the discussions in Subsection 5.1 to obtain similar results on convergence and error estimates for spatially and fully discrete approximations of the problem P2.
References [1] A. Amassad and M. Sofonea, Analysis of some nonlinear evolution systems arising in rate-type viscoplasticity, in Dynamical System and Di erential Equations, Eds. W. Chen and S. Hu, An added volume to Discrete and Continuous Dynamical Systems , 1998, pp. 58{71. [2] A. Amassad and M. Sofonea, Analysis of a quasistatic viscoplastic problem involving Tresca friction law, Discrete and Continuous Dynamical Systems 4 (1998), pp. 55{72. [3] P.G. Ciarlet, The Finite Element Method for Elliptic Problems , North Holland, Amsterdam, 1978. [4] N. Cristescu and I. Suliciu, Viscoplasticity , Martius Nijho , Editura Tehnica, Bucharest, 1982. [5] G. Duvaut and J.L. Lions, Inequalities in Mechanics and Physics, Springer-Verlag, Berlin, 1976. [6] R. Glowinski, Numerical Methods for Nonlinear Variational Problems, Springer-Verlag, New York, 1984. [7] R. Glowinski, J.-L. Lions and R. Tremolieres, Numerical Analysis of Variational Inequalities , North-Holland, Amsterdam, 1981. [8] W. Han and B.D. Reddy, Computational plasticity: the variational basis and numerical analysis, Computational Mechanics Advances 2 (1995), pp. 283{400. [9] W. Han and B.D. Reddy, Convergence of approximations to the primal problem in plasticity under conditions of minimal regularity, submitted. [10] W. Han and B.D. Reddy, Plasticity: Mathematical Theory and Numerical Analysis , Springer-Verlag, New York, 1999. 29
[11] I. Hlavacek, J. Haslinger, J. Necas and J. Lovsek, Solution of Variational Inequalities in Mechanics, Springer-Verlag, New York, 1988. [12] I. Ionescu and M. Sofonea, Functional and Numerical Methods in Viscoplasticity, Oxford University Press, Oxford 1993. [13] N. Kikuchi and J.T. Oden, Contact Problems in Elasticity: A Study of Variational Inequalities and Finite Element Methods , SIAM, Philadelphia, 1988. [14] J. Necas and I. Hlavacek, Mathematical Theory of Elastic and Elastoplastic Bodies: An Introduction, Elsevier, Amsterdam, 1981. [15] P. D. Panagiotopoulos, Inequality Problems in Mechanics and Applications, Birkhauser, Basel, 1985. [16] M. Sofonea and M. Shillor, Variational analysis of quasistatic viscoplastic contact problems with friction, to appear in Communications in Applied Analysis . [17] E. Zeidler, Nonlinear Functional Analysis and its Applications, Volume II/A: Linear Monotone Operators , Springer-Verlag, New York, 1985.
30
Abstract. We consider numerical approximations of a class of abstract nonlinear evolution-
ary systems arising in the study of quasistatic frictional contact problems for elastic-viscoplastic materials. Both spatially semi-discrete and fully discrete schemes are analyzed with the nite element method employed to discretize the spatial domain. Strong convergence of both approximations is established under minimal solution regularity. The results are applied to two particular frictional contact problems for viscoplastic bodies. Under appropriate regularity assumptions on the exact solution, optimal order error estimates are derived.
1 Introduction The aim of this paper is to provide numerical analysis of some problems arising in frictional contact between an elastic-viscoplastic body and a rigid foundation. Situations of frictional contact abound in industry and everyday life. Contacts of the braking pads with the wheel, the tire with the road and the piston with the skirt are just a few simple examples. Because of the importance of the process of frictional contact, a considerable e ort has been made in its modeling and numerical simulations. Indeed, the engineering literature concerning this topic is extensive. Most of it, however, is dedicated to simple geometries, speci c settings, and mostly to numerical simulations. In the applied mathematics literature, the study of general models for dynamic or quasistatic contact process involving elastic-viscoplastic materials is very recent. Rate-type viscoplastic constitutive laws of the form (1.1) _ = E "(u_ ) + G( ; "(u)) are used in the literature to describe mechanical responses of such materials as rubber, various metals, rocks, pastes, etc. In (1.1), denotes the stress tensor, u the displacement eld, "(u) the linearized strain tensor, and E and G are material constitutive functions. The function E is assumed to be linear while G is in general nonlinear. Here and throughout the paper, a dot above a quantity represents its derivative with respect to the time variable t, and double dots denote the second-order derivative. Concrete examples, experimental 1 2
Department of Mathematics, University of Iowa, Iowa City, IA 52242, USA Laboratoire de Theorie des Systemes, University of Perpignan, France
1
background and mechanical interpretations of such models may be found in [4] and references therein. Functional and numerical methods are discussed in [12] for initial and boundary value problems involving (1.1) with the usual displacement and traction boundary conditions. Existence of weak solutions to quasistatic frictional problems for materials modeled by a general rate-type constitutive law of the form (1.1) were established in [1, 2, 16]. In the abstract form, the frictional contact problems are formulated as a nonlinear evolution equation (the abstract version of (1.1)) coupled with an evolutionary variational inequality resulted from the equilibrium equation and the boundary conditions. The variational analysis of this abstract problem was done in [1]. In this paper, we consider the numerical analysis of frictional contact problems for elasticviscoplastic materials of the type (1.1). The literature is abundant on numerical treatment of variational inequality, see for instance the monographs [6, 7, 11, 13]. Of particular relevance to this paper are the works on numerical analysis of variational inequalities arising in plasticity, cf. [8, 9, 10]. The paper is organized as follows. In Section 2 we present the abstract problem, state the assumptions on the data and recall the existence and uniqueness result proved in [1]. In Section 3, both spatially discrete and fully discrete approximations to the abstract problem are analyzed and some error estimates are presented. In deriving the error estimates, we will need to apply Gronwall's inequality, which is recalled here for convenience. Suppose f; g 2 C [a; b] and g is nondecreasing, c0 > 0 is a constant, then (1.2)
f (t) g(t) + c0
Zt
a
f (s) ds; t 2 [a; b] =) f (t) ec0 (t?a) g(t); t 2 [a; b]:
Convergence analysis for both approximations is done in Section 4 under the solution regularity condition established in the proof of well-posedness of the problem. Finally, in Section 5 we apply the results to the numerical approximations of two concrete examples of quasistatic frictional contact problems in rate-type viscoplasticity. Under appropriate solution regularity assumptions, we obtain optimal order error estimates.
2 The abstract problem Let H be a real Hilbert space, V a closed subspace of H . We denote by (; )H the inner product of H and by kkH the associated norm. Let us remark that V itself is a real Hilbert space endowed with the inner product of H ; for this reason we shall use sometimes the notations (u; v)V , kukV instead of (u; v)H ; kukH , if u; v 2 V . Let A : H ! H be a linear 2
operator, B : [0; T ] H H ! H a possibly nonlinear operator, and ' : H ! (?1; +1]. Let [0; T ] be the time interval of interest. We consider an abstract problem
y_ (t) = Ax_ (t) + B (t; x(t); y(t)) a:e: t 2 (0; T )
(2.1)
(2.2) y(t) + @'(x_ (t)) 3 f (t) a:e: t 2 (0; T ) (2.3) x(0) = x0 ; y(0) = y0: Here the unknowns are the functions x : [0; T ] ! V and y : [0; T ] ! H , while x0 2 V , y0 2 H and f : [0; T ] ! V are given data. The symbol @' represents the subdi erential of the function ', and the relation (2.2) is understood in the sense that for a.e. t 2 (0; T ), f (t) ? y(t) is a subgradient of ' at x_ (t). We denote by D' the e ective domain of ' de ned by D' = f x 2 H j '(x) < +1 g: We assume in the sequel D' = V . An equivalent formulation of the problem (2.1){(2.3) is: Problem P. Find functions x : [0; T ] ! V and y : [0; T ] ! H such that
x(0) = x0 ; y(0) = y0;
(2.4) and for a.e. t 2 (0; T ),
y_ (t) = Ax_ (t) + B (t; x(t); y(t)); (y(t); w ? x_ (t))H + '(w) ? '(x_ (t)) (f (t); w ? x_ (t))V 8 w 2 V:
(2.5) (2.6)
In the study of the problem P we make the following assumptions: (2.7)
(2.8)
8 > > < > > :
A : H ! H is a positive de nite; symmetric; linear operator; i:e:; (a) there exists c0 > 0 such that (Ax; x)H c0kxk2H 8 x 2 H ; (b) (Ax; y)H = (x; Ay)H 8 x; y 2 H:
8 > > > > > > > > > >
0 such that kB (t; x1 ; y1) ? B (t; x2 ; y2)kH L (kx1 ? x2 kH + ky1 ? y2kH ) > 8 t 2 [0; T ]; 8 x1 ; x2 ; y1; y2 2 H ; > > > > > > (b) the mapping t 7! B (t; x; y) is measurable for all x; y 2 H ; > > > : (c) the mapping t 7! B (t; 0; 0) 2 L1 (0; T ; H ): 3
(2.9) ' : V ! IR+ is a continuous seminorm. (2.10) f 2 W 1;1(0; T ; V ): (2.11) x0 2 V; y0 2 H: (2.12) y0 + @'(0) 3 f (0): We see from the condition (2.7) that the quantity (Ax; y)H de nes an inner product on H , and the corresponding induced norm q
kxkA = (Ax; x)H ; x 2 H is equivalent to the norm kxkH . We also remark that the assumption (2.9) implies that ' is Lipschitz continuous on V . Everywhere in the paper we use the standard notation for Lp, W m;p, H m and C m spaces, 1 p 1, m 2 IN. Moreover, if X and Y are real Hilbert spaces, we denote in the sequel by X Y the product space endowed with the canonical inner product. The well-posedness of the problem P has been investigated in [1] where the following result can be found. Theorem 2.1. Under the assumptions (2:7){(2:12), the problem
x 2 W 1;1(0; T ; V ), y 2 W 1;1(0; T ; H ).
P has a unique solution
The proof of Theorem 2.1 is carried out in several steps. It is based on time discretization method, standard arguments of elliptic variational inequalities and a xed point property. In the next two sections, we assume (2.7){(2.12) are satis ed.
3 Numerical analysis of the abstract problem In this section, we present and analyze approximation schemes for solving the problem P. We will give some preliminary results that will be applied to some concrete examples later.
3.1 Spatially semi-discrete approximation Let h 2 (0; 1] be an index and fH hg a family of nite dimensional subspaces of H . Set V h = V \ H h, which is nonempty since 0 2 V h. Let PH : H ! H h be the orthogonal h
4
projection de ned through the relation (PH q; qh)H = (q; qh)H 8 q 2 H; qh 2 H h:
(3.1)
h
Obviously, we have (3.2) kPH qkH kqkH 8 q 2 H: This property will be used on various occasions. h
Now a spatially discrete approximation of the problem P is
Problem Ph . Find the functions xh : [0; T ] ! V h and y h : [0; T ] ! H h such that
xh(0) = xh0 ; yh(0) = y0h;
(3.3) and for a.e. t 2 (0; T ), (3.4) (3.5)
y_ h(t) = PH Ax_ h(t) + PH B (t; xh(t); yh(t)); (yh(t); wh ? x_ h (t))H + '(wh) ? '(x_ h(t)) (f (t); wh ? x_ h(t))V 8 wh 2 V h: h
h
Here, xh0 = PV x0 2 V h, y0h = PH y0 2 H h are orthogonal projections of x0 and y0 to V h and H h, respectively. The de nition of PV : V ! V h is similar to that of PH . From the de nition of y0h, we see that the discrete analog of (2.12) is valid: h
h
h
h
(y0h; wh)H + '(wh) (f (0); wh)V 8 wh 2 V h: This relation is needed to verify the existence of a solution of the problem Ph. We observe that the projection operator PH is introduced to ensure that the relation (3.4) is well-de ned on the space H h. h
Using the arguments in [1], it can be shown that the problem Ph has a unique solution xh 2 W 1;1(0; T ; V h) and yh 2 W 1;1(0; T ; H h). Our main purpose here is to derive estimates for the errors x ? xh and y ? yh. To this end, let t 2 [0; T ]. We rst integrate (2.5) and (3.4), and use the initial conditions (2.4) and (3.3) to obtain Zt
B (s; x(s); y(s)) ds + y0 ? Ax0 ;
(3.6)
y(t) = Ax(t) +
(3.7)
yh(t) = PH Axh(t) + PH h
0
Zt
h
0
B (s; xh(s); yh(s)) ds + y0h ? PH Axh0 : h
5
Then we subtract (3.7) from (3.6) to get (3.8)
y(t) ? yh(t) Z t h h h = PH A(x(t) ? x (t)) + PH B (s; x(s); y(s)) ? B (s; x (s); y (s)) ds 0 + y0 ? y0h ? PH A(x0 ? xh0 ) + (IH ? PH )(y(t) ? y0); h
h
h
h
where IH : H ! H is the identity operator. Denote (3.9) e0 = kx0 ? xh0 kV + ky0 ? y0hkH : Then we get the following inequality from (3.8), using the assumptions (2.7), (2.8) and the property (3.2), (3.10) ky(t) ? yh(t)kH c kx(t) ? xh(t)kV + k(IH ? PH )(y(t) ? y0)kH Zt + c (ky(s) ? yh(s)kH + kx(s) ? xh(s)kV ) ds + c e0; h
0
where c is a positive constant which depends on operators A and B . Everywhere in this paper, except in Section 5, the symbol c will represent a strictly positive constant which may change its value from place to place, and may depend on A, B , T and ', but not on the time or the input data. Now plugging (3.6) in (2.6) with w = x_ h (t), we have (3.11)
(Ax(t); x_ (t) ? x_ h(t))H (f (tZ); x_ (t) ? x_ h (t))V + '(x_ h(t)) ? '(x_ (t)) t + ( B (s; x(s); y(s)) ds; x_ h(t) ? x_ (t))H + (y0 ? Ax0 ; x_ h(t) ? x_ (t))H : 0
Meanwhile, we plug (3.7) in (3.5) with any wh = wh(t) 2 V h to obtain (3.12)
? (Axh(t); x_ (t) ? x_ h(t))H (f (tZ); x_ h(t) ? wh(t))V + '(wh(t)) ? '(x_ h(t)) t + ( B (s; xh(s); yh(s)) ds; wh(t) ? x_ h(t))H 0 + (y0h ? Axh0 ; wh(t) ? x_ h (t))H + (Axh(t); wh(t) ? x_ (t))H :
Adding (3.11) and (3.12), we have (A(x(t) ? xh(t)); x_ (t) ? x_ h(t))H (A(x(t) ? xh(t)); x_ (t) ? wh(t))H + R(t; x_ (t); wh(t)) + (D(t); x_ h(t) ? wh(t))H ; 6
where
R(t; x_ (t); wh) = (y(t); wh ? x_ (t))H + '(wh) ? '(x_ (t)) ? (f (t); wh ? x_ (t))V ; Zt D(t) = (B (s; x(s); y(s)) ? B (s; xh(s); yh(s))) ds + y0 ? y0h ? A(x0 ? xh0 ):
(3.13) (3.14)
0
Using the assumption (2.7), we then have 1 d kx(t) ? xh(t)k2 1 kx(t) ? xh(t)k2 + 1 kx_ (t) ? wh(t)k2 + R(t; x_ (t); wh(t)) A A 2 A 2 dt 2 +(D(t); x_ h(t) ? x_ (t))H + (D(t); x_ (t) ? wh(t))H : Integrate the above inequality from 0 to t to obtain
kx(t) ? xh(t)k2AZ t kx0 ? xh0 k2A + c 0 (jR(s; x_ (s); wh(s))j + kD(s)k2H ) ds +c
Zt h 2 h 2 (kx(s) ? x (s)kA + kx_ (s) ? w (s)kA) ds + (D(s); x_ h(s) ? x_ (s))H ds: 0 0
Zt
For the last term above, we perform an integration by parts, Zt
(D(s); x_ h(s) ? x_ (s))H ds 0 = (D(t); xh(t) ? x(t))H ? (D(0); xh0 ? x0 )H Zt ? (B (s; x(s); y(s)) ? B (s; xh(s); yh(s)); xh(s) ? x(s))H ds: 0
Using (2.8), we have the estimate Zt
(D(s); x_ h(s) ? x_ (s))H ds Zt c kD(t)k2H + 21 kx(t) ? xh(t)k2A + c e20 + c 0 (kx(s) ? xh(s)k2A + ky(s) ? yh(s)k2H ) ds: From the de nition (3.14) and the assumptions (2.7) and (2.8), we have 0
Zt
kD(t)kH c 0 (kx(s) ? xh(s)kA + ky(s) ? yh(s)kH ) ds + c e0: Combine the last several relations, (3.15)
kx(t) ? Zxh(t)k2A t c e20 + c 0 (kx_ (s) ? wh(s)k2A + jR(s; x_ (s); wh(s))j) ds Zt
+ c (kx(s) ? xh(s)k2A + ky(s) ? yh(s)k2H ) ds: 0
7
This inequality, together with (3.10), implies (3.16) kx(t) ? xh(t)k2A + ky(t) ? yh(t)k2H Zt 2 2 c e0 + c k(IH ? PH )(y(t) ? y0)kH + c (kx_ (s) ? wh(s)k2A + jR(s; x_ (s); wh(s))j) ds h
0
Zt
+ c (kx(s) ? xh(s)k2A + ky(s) ? yh(s)k2H ) ds: 0
Applying the Gronwall inequality (1.2), we get
kx(t) ? xh(t)k2A + ky(t) ? yh(t)k2H c e20 + c sup k(IH ? PH )(y(t) ? y0)k2H +c
ZT 0
h
t2[0;T ]
(kx_ (t) ? wh(t)k2A + jR(t; x_ (t); wh(t))j) dt:
Summarizing, we have shown the following result. Theorem 3.1. Let (x; y ) 2 W 1;1(0; T ; V
H ) and (xh; yh) 2 W 1;1(0; T ; V h H h) be the
solutions of the problems P and Ph respectively. Then we have the error estimate
(3.17)
kx ? xhkL1(0;T ;V ) + ky ? yhkL1(0;T ;H ) c (kx0 ? xh0 kV + k(y0 ? y0hkH ) + c k(IH ? PH )(y ? y0)kL1(0;T ;H ) ) ZT hk 2 k x _ ? w + ( + c w 2L2inf jR(s; x_ (s); wh(s))j ds)1=2 ; L (0;T ;V ) (0;T ;V ) 0 h
h
h
where R(; ; ) is de ned in (3:13).
The inequality (3.17) is the basis of convergence and error analysis for the spatially-discrete solutions. Concrete order error estimates will be established when Theorem 3.1 is applied in Section 5 to some examples arising in mechanics.
3.2 Fully discrete approximation In addition to the nite dimensional spaces V h and H h introduced in the previous subsection, we divide the time interval [0; T ] into N equal parts and denote the step-size by k = T=N , the nodal points by tn = nk, n = 0; 1; : : : ; N , and the sub-intervals In = [tn?1; tn], n = 1; : : : ; N . Here we assume k 2 (0; 21 ]. The arguments and results of this subsection can be easily extended to the case of non-uniform partition of the time interval. For a continuous function w(t) with values in H or V , we use the notation wn w(tn). For a sequence fwngNn=0, we denote wn = (wn ? wn?1)=k, n = 1; : : : ; N . 8
Then a fully discrete approximation based on a forward Euler scheme which we will analyze is the following. N h hk hk N h Problem Phk . Find xhk = fxhk n gn=0 V and y = fyn gn=0 H , such that hk h h xhk 0 = x 0 ; y 0 = y0 ;
(3.18) and for n = 1; : : : ; N , (3.19) (3.20)
hk hk ynhk = PH Axhk n + PH B (tn?1 ; xn?1 ; yn?1); h hk h hk h h (ynhk; wh ? xhk n )H + '(w ) ? '(xn ) (fn ; w ? xn )V ; 8 w 2 V : h
h
Here again, xh0 = PV x0 2 V h, y0h = PH y0 2 H h are orthogonal projections of x0 and y0 to V h and H h, respectively. h
h
hk We rst inductively show the unique solvability of the problem Phk . Let xhk n?1 and yn?1 be given. We rewrite (3.19) as
(3.21)
hk hk hk ynhk = k PH Axhk n + k PH B (tn?1 ; xn?1 ; yn?1) + yn?1 h
h
and use it in (3.20) to yield h hk h hk k(Axhk n ; w ? xn )H + '(w ) ? '(xn ) hk hk h hk h h (fn ? k B (tn?1; xhk n?1 ; yn?1) ? yn?1; w ? xn )H 8 w 2 V :
This is a discrete elliptic variational inequality of the second kind for the variable xhk n . By hk h a standard result (cf. [6]), this inequality has a unique solution xn 2 V from which we hk hk can determine xhk n . Then yn can be obtained from (3.21). Therefore the problem P has a unique solution. N hk N Our goal here is to derive estimates for the errors fxn ? xhk n gn=1 and fyn ? yn gn=1 . Assume x 2 C 1 ([0; T ]; V ). First, we apply (3.21) recursively to get
(3.22)
ynhk = PH Axhk n + h
n X j =1
hk hk hk kPH B (tj?1; xhk j ?1 ; yj ?1) + y0 ? PH Ax0 : h
h
Then subtracting (3.22) from (3.6) at t = tn, we obtain (3.23)
yn ? ynhk = (IH ? PH )(yn ? y0) + PH A(xn ? xhk n) Z t n X hk ) + PH B (s; x(s); y(s)) ds ? kB (tj?1; xhk ; y j ?1 j ?1 h
h
n
h
+ y0
0
? yhk ? P 0
H
h
A(x0 9
? xhk ): 0
j =1
Denote (3.24) Dn = k
Zt
n
0
B (s; x(s); y(s)) ds ?
n X j =1
kB (tj?1; xj?1; yj?1)kH ; n = 1; : : : ; N;
for the numerical quadrature errors, and hk en = kxn ? xhk n kV + kyn ? yn kH ; n = 0; : : : ; N for the numerical solution errors. As a preliminary result which will be used on various estimates, the following is derived based on the assumptions (2.8),
k
(3.25)
Zt
n
0
n X
n X j =1
hk kB (tj?1; xhk j ?1 ; yj ?1)kH
hk kkB (tj?1; xj?1; yj?1) ? B (tj?1; xhk j ?1; yj ?1)kH + Dn
j =1 n X
c
B (s; x(s); y(s)) ds ?
j =1
hk k(kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1 kH ) + Dn :
Then we obtain the following inequality from (3.23) by the use of (3.25), the assumptions (2.7) and the property (3.2), (3.26)
kyn ? ynhk kH k(IH ? PH )(yn ? y0)kH + c kxn ? xhk n kH + Dn n X hk + c e0 + c k(kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1kH ): h
j =1
Now let us bound rn xn ? xhk n , n = 1; : : : ; N . For this, we plug (3.6) into (2.6) at t = tn hk with w = xn and get (3.27)
(Axn ; x_ n ? xhk )H (fn; x_ n ? xhk )V + '(xhk n n n ) ? '(x_ n ) Zt hk + ( B (s; x(s); y(s)) ds; xhk n ? x_ n )H + (y0 ? Ax0 ; xn ? x_ n )H : n
0
Similarly, plugging (3.22) into (3.20) with an arbitrary wh = wnh 2 V h yields (3.28)
h hk hk h h hk ? (Axhk n ; wn ? xn )H (fn ; xn ? wn )V + '(wn ) ? '(xn ) n X hk h hk hk hk h hk + ( kB (tj?1; xhk j ?1 ; yj ?1); wn ? xn )H + (y0 ? Ax0 ; wn ? xn )H : j =1
Let us consider the quantity An (Arn; rn)H . The following lower bound for An can be obtained by using the assumptions (2.7): An 21k krnk2A ? krn?1k2A : 10
To derive an upper bound for An, we write hk An = (A(xn ? xhk n ); xn ? xn )H hk h hk h hk = (Axn ; xn ? x_ n)H + (Axn ; x_ n ? xhk n )H ? (Axn ; xn ? wn )H ? (Axn ; wn ? xn )H :
We then use (3.27) and (3.28) to bound the second and fourth terms of above respectively,
An (Arn; xn ? wnh)H + Rn (x_ n; wnh) + I1(n) + I2(n) + I3 (n) 21 (Arn; rn)H + 21 (A(xn ? wnh); xn ? wnh)H + Rn(x_ n; wnh) + I1 (n) + I2 (n) + I3(n); where (3.29)
Rn(x_ n; wnh) = (yn; wnh ? x_ n)H + '(wnh) ? '(x_ n) ? (fn; wnh ? x_ n)V ; Zt n X hk hk I1(n) = ( B (s; x(s); y(s)) ds ? kB (tj?1; xhk j ?1 ; yj ?1); xn ? xn )H ; n
0
j =1 hk hk I2(n) = (y0 0 0 ? x0 ); xn ? xn )H ; Zt n X hk I3(n) = ( B (s; x(s); y(s)) ds ? kB (tj?1; xhk j ?1 ; yj ?1) 0 j =1 hk h + y0 ? y0 ? A(x0 ? xhk 0 ); xn ? wn )H :
? yhk ? A(x
n
Combining the lower and upper bounds for An, we have (3.30)
krnk2A ? krn?1k2A k krnk2A + k kxn ? wnhk2A + 2 k (Rn(x_ n ; wnh) + I1(n) + I2 (n) + I3(n)):
In the inequality (3.30), we change the index n to j , and sum over j from 1 to n, (3.31)
krnk2A kr0k2A + +2
n X j =1
n X j =1
k krj k2A +
n X j =1
k kxj ? wjhk2A
k [Rj (x_ j ; wjh) + I1(j ) + I2 (j ) + I3(j )]:
In order to handle the last summation, we rst estimate the quantity Dn de ned by (3.24),
n Zt X
(B (s; x(s); y(s)) ? B (tj?1; xj?1; yj?1)) dskH j =1 t ?1 n Zt X kB (s; x(s); y(s)) ? B (tj?1; xj?1; yj?1)kH ds j =1 t ?1 n Zt X (kx(s) ? xj?1kV + ky(s) ? yj?1kH )ds; c j =1 t ?1
Dn = k
j
j
j
j
j
j
11
where the assumption (2.8) is used. Hence,
Dn c tnk kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) :
(3.32) We now write n X
j =1
kI1 (j ) = =
n Z t X
j
j =1 0 Z t1 0
?
B (s; x(s); y(s)) ds ?
j X i=1
hk k B (ti?1 ; xhk i?1 ; yi?1 ); rj ?1 ? rj
hk B (s; x(s); y(s)) ds ? kB (t0 ; xhk 0 ; y0 ); r0
Z t
n
B (s; x(s); y(s)) ds ?
n X
H
hk kB (ti?1 ; xhk i?1 ; yi?1 ); rn
0 H i=1 nX ?1 Z t +1 B (s; x(s); y(s)) ds ? kB (tj ; xhk ; yjhk); rj : + j H j =1 t j
j
Using the assumption (2.7) and the estimates (3.25) and (3.32), we obtain n nX ?1 X 1 kr k2 kI1(j ) c e20 + c k (krj k2V + kyj ? yjhkk2H ) + 16 (3.33) n A j =1 j =1 + c k2(kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) )2: Similarly, we write
n X
j =1
kI2(j ) = =
n X hk (rj?1 ? rj ))H (y0 0 0 ? x0 ); j =1 (y0 ? y0hk ? A(x0 ? xhk 0 ); r0 ? rn )H ;
? yhk ? A(x
and use the assumption (2.7) to obtain n X
(3.34)
j =1
1 kr k2 : kI2 (j ) c e20 + 16 n A
Finally we use the assumption (2.8) and the estimates (3.25) and (3.32) to nd (3.35)
n X j =1
kI3(j ) c e20 + c
nX ?1 j =1
k (krj k2V + kyj ? yjhkk2H )
+ c k2(kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) )2 + c
n X j =1
k kxj ? wjhk2V :
We now combine the estimates (3.26), (3.31) and (3.33){(3.35) to obtain (3.36)
2 hk 2 kxn ? xhk n kV + kyn ? yn kH c Jn + c
12
nX ?1 j =1
2 hk 2 k(kxj ? xhk j kV + kyj ? yj kH );
H
where
Jn = e20 + k2(kx_ kL1(0;T ;V ) + ky_ kL1(0;T ;H ) )2 + k(IH ? PH )(yn ? y0)k2H n n X X + k kxj ? wjhk2V + k jRj (x_ j ; wjh)j: h
j =1
j =1
An error estimate can be derived based on (3.36). hk N Theorem 3.2. Let (x; y ) 2 C 1 ([0; T ]; V ) W 1;1(0; T ; H ) and f(xhk n ; yn )gn=1 be the solu-
tions of the problem P and Phk , respectively. Then we have the error estimate
(3.37)
hk max kxn ? xhk n kV + kyn ? yn kH 1nN hk c (kx0 ? xhk 0 kV + ky0 ? y0 kH ) + c k (kx_ kL1 (0;T ;V ) + ky_ kL1 (0;T ;H ) ) + c 1max k(I ? PH )(yn ? y0)kH nN H h
+c
X N
k inf kxj ? wjhk2V + jRj (x_ j ; wjh)j
j =1 w 2V h j
1=2
h
:
P Proof. Denote En = nj=1 k e2n , then (3.36) can be rewritten as
e2n c Jn + c En?1:
(3.38) Now
En ? En?1 k e2n c k Jn + c k En?1:
Hence we have
En ? (1 + c k)En?1 c k Jn;
or equivalently,
En ? En?1 c k Jn : (1 + c k)n (1 + c k)n?1 (1 + c k)n By an inductive argument, we get n X
J (ec T ? 1) 1max J: En c k (1 + c k)n?iJi ((1 + c k)n ? 1) 1max in i in i i=1
Using (3.38), we have which implies (3.37).
e2n c 1max J; in i
2
The estimate (3.37) will be used for error analysis of the fully discrete solutions provided the exact solution possesses certain regularity. 13
4 Convergence analysis In this section, we analyze the convergence of the spatially and fully discrete solutions for the problem P under the basic regularity condition (x; y) 2 W 1;1(0; T ; V H ), available from Theorem 2.1. First we make the following additional assumptions on the function spaces H; V and the nite dimensional spaces H h and V h. Assumption (H1). There exist a subspace V0
(h) 0 such that
and
lim (h) = 0;
h!0+
hk = kw ? P wk (h)kwk inf k w ? w V V V0 8 w 2 V0 : V w 2V h
h
h
Assumption (H2). There exist a subspace H0
(h) 0 such that and
V which is dense in V and a function
H which is dense in H and a function
lim (h) = 0;
h!0+
inf kz ? zh kH = kz ? PH zkH (h)kzkH0 8 z 2 H0:
z 2H h
h
h
These two hypotheses will be veri ed for the applicaiton problems discussed in the next section. We will need the following result, which can be found in [17]. Lemma 4.1. Assume that X is a Banach space, X0 X is dense in X . Then H 1 (0; T ; X0)
is dense in H 1 (0; T ; X ).
Now we are ready to study the convergence of the spatially discrete solution for the problem P, based on the estimate (3.17). Theorem 4.2. Let (x; y )
2 W 1;1(0; T ; V H ) be the solution of the problem P and
(xh; yh) 2 W 1;1(0; T ; V h H h) the solution of the corresponding spatially-discrete problem Ph. Then under the assumptions (H1) and (H2) we have
(4.1)
kx ? xhkL1(0;T ;V ) + ky ? yhkL1(0;T ;H ) ! 0 as h ! 0:
Proof. From the assumptions (H1) and (H2), we have
(4.2)
kx0 ? xh0 kV ! 0; ky0 ? y0hkH ! 0 as h ! 0: 14
Since ' is Lipschitz continuous on V , using the de nition (3.13), we have
jR(t; x_ (t); wh(t))j (ky(t)kH + kf (t)kV + c) kwh(t) ? x_ (t)kV : Hence, the estimate (3.17) can be rewritten as (4.3)
kxh ? xkL1(0;T ;V ) + kyh ? ykL1(0;T ;H ) h k1=2 c k(IH ? PH )(y ? y0)kL1(0;T ;H ) + c e0 + c w 2L2inf k x _ ? w L2 (0;T ;V ) : (0;T ;V ) h
h
h
Using Lemma 4.1 and Assumption (H1), we know that H 1(0; T ; V0) is dense in H 1(0; T ; V ). So for any " 2 (0; 1), there exists x~ 2 H 1(0; T ; V0) such that
kx ? x~kH 1 (0;T ;V ) ";
(4.4)
which can be combined with (H1) again to yield (4.5)
inf kx_ ? whk1L=22(0;T ;V ) w 2L2 (0;T ;V ) kx~_ ? whk1L=22(0;T ;V ) kx_ ? x~_ k1L=22(0;T ;V ) + w 2L2inf (0;T ;V ) p q (h)kx~k1=2 "+ H 1 (0;T ;V0 ) : h
h
h
h
Similarly, using Lemma 4.1 and the assumption (H2), H 1(0; T ; H0) is dense in H 1(0; T ; H ). So there exists y~ 2 H 1(0; T ; H0) such that
ky ? y0 ? y~kH 1(0;T ;H ) ": Now we are ready to bound the rst term on the right hand side of (4.3):
k(IH ? PH )(y ? y0)kL1(0;T ;H ) c k(IH ? PH )(y ? y0)kH 1 (0;T ;H ) c ky ? y0 ? y~kH 1 (0;T ;H ) + k(IH ? PH )~ykH 1 (0;T ;H ) c " + (h) ky~kH 1(0;T ;H0 ): Above we used the embedding result H 1(0; T ; H ) ,! L1(0; T ; H ) (cf. [17]). The convergence result (4.1) now follows from (4.2) and (4.3){(4.6). 2
(4.6)
h
h
h
We now turn to a convergence analysis of the fully discrete scheme. Notice that we can not use the estimate (3.37), because under the basic regularity condition (x; y) 2 W 1;1(0; T ; V H ), the pointwise values x_ j and y_j are not well-de ned. Here we follow the approach developed in [9] for a convergence analysis. For this purpose we will need another density result, which can also be found in [17]. 15
Lemma 4.3. The space C 1 ([0; T ]; V ) is dense in H 1 (0; T ; V ); that is, given w 2 H 1 (0; T ; V ),
for any " > 0 there exists w 2 C 1([0; T ]; V ) such that kw ? wkH 1(0;T ;V ) ":
Let us consider the quantity An = (Arn; rn)H . As in Section 3, a lower bound of An is An 21k (krnk2A ? krn?1k2A): (4.7) To obtain an upper bound we begin with hk An = (A(xn ? xhk n ); xn ? xn )H hk h hk h hk = (Axn; xn ? xhk n )H ? (Axn ; xn ? wn )H ? (Axn ; wn ? xn )H ; where wnh 2 V h is arbitrary. Using (3.28) to bound the last term, we obtain hk h (4.8) An (Axn; xn ? xhk n )H ? (Axn ; xn ? wn )H h h hk + (fn; xhk n ? wn )V + '(wn ) ? '(xn ) n X hk h hk hk hk h hk + ( kB (tj?1; xhk j ?1 ; yj ?1); wn ? xn )H + (y0 ? Ax0 ; wn ? xn )H : j =1
Now integrate (2.6) with w = xhk n from t = tn?1 to tn and use (3.6) to obtain Z Z 1 1 hk hk (4.9) 0 k (Ax(t); xn ? x_ (t))H dt + '(xn ) ? k '(x_ (t))dt I I Z Zt 1 hk + k ( B (s; x(s); y(s))ds; xhk n ? x_ (t))H dt + (y0 ? Ax0 ; xn ? xn )H I 0 Z + 1 (fn; x_ (t) ? xhk n )V dt: k I Then we add (4.9) and (4.8) to obtain (4.10) A n R1 + R2 + R3 + R 4 + R 5 ; where Z hk ; x ? wh) + 1 (Ax(t); xhk ? x_ (t)) dt; R1 = (Axn; xn ? xhk ) ? ( Ax H n H H n n n n k I Z R2 = '(wnh) ? k1 '(x_ (t))dt; I Z 1 hk h R3 = (fn; xn ? wn)V + k (fn; x_ (t) ? xhk n )V dt; I h hk hk R4 = (y0hk ? Axhk 0 ; wn ? xn )H + (y0 ? Ax0 ; xn ? xn )H ; n X hk h hk R5 = ( kB (tj?1; xhk j ?1 ; yj ?1); wn ? xn )H n
n
n
n
n
n
n
j =1
+ k1
Z
Zt
I
n
( B (s; x(s); y(s))ds; xhk n ? x_ (t))H dt: 0
16
We need to nd appropriate bounds for Ri, 1 i 5. First let us estimate R1 . Using the assumption (2.7) and the properties of inner product, we have Z Z 1 1 1 (4.11) R1 = k (Axn ? k Ax(t) dt; rn ? rn?1)H + k (Ax(t); xn ? x_ (t))H dt I I h h + (Aen; xn ? wn)H ? (Axn ; xn ? wn )H Z Z k12 (A I (t ? tn?1 )x_ (t) dt; rn ? rn?1)H + k1 I (Ax(t); xn ? x_ (t))H dt + 1 krnk2A + 1 kxn ? wnhk2A ? (Axn; xn ? wnh)H 2 2 Z c kx_ kL1(I ;V ) (krnkV + krn?1kV ) + kc kxkL1(I ;V ) I kxn ? x_ (t)kV dt + 1 krnk2A + c kxn ? wnhk2V + c kxnkV kxn ? wnhkV : 2 Using the Lipschitz continuity of ' on V , we nd a bound for R2 , Z Z c 1 h (4.12) jR2j = k j I ('(wn) ? '(x_ (t)) dtj k I kwnh ? x_ (t)kV dt: For R3, we have Z 1 R3 = k2 ( (t ? tn?1 )f_(t)dt; rn?1 ? rn)H + (fn; xn ? wnh)H (4.13) I Z 1 ? k I (f (t); xn ? x_ (t))H dt kf_kL1(I ;V ) (krnkV + krn?1kV ) + kfnkV kxn ? wnhkV Z + k1 kf kL1(I ;V ) kxn ? x_ (t)kV dt: n
n
n
n
n
n
n
n
n
n
n
n
n
I
n
For R4, we have
h R4 = k1 (y0 ? y0hk ? A(x0 ? xhk 0 ); rn?1 ? rn )H ? (y0 ? Ax0 ; xn ? wn )H h + (y0 ? y0hk ? A(x0 ? xhk 0 ); xn ? wn )H k1 (y0 ? y0hk ? A(x0 ? xhk 0 ); rn?1 ? rn )H + c (e0 + ky0kH + kx0 kV )kxn ? wnhkV : Express R5 as
(4.14)
(4.15)
R5 =
Zt hk hk ( kB (tj?1; xj?1; yj?1) ? B (s; x(s); y(s)) ds; wnh ? xhk n )H 0 j =1 Zt + ( B (s; x(s); y(s)) ds; wnh ? xhk n )H 0 Z Zt + k1 ( B (s; x(s); y(s)) ds; (xhk n ? xn ) + (xn ? x_ (t)))H dt: I 0
n X
n
n
n
17
Integrating by parts, we obtain Z Zt Z (4.16) k1 ( B (s; x(s); y(s))ds; xn ? x_ (t))H dt = k1 (B (t; x(t); y(t)); x(t))H dt; I 0 I where x(t) = x(t) ? t ?ktn?1 xn ? tn k? t xn?1 ; tn?1 t tn : (4.17) By an elementary manipulation, we have 1 Z (Z t B (s; x(s); y(s)) ds; xhk ? x ) dt (4.18) n H n kZ I 0 t = ( B (s; x(s); y(s)) ds; xhk n ? xn )H 0 Z ? k1 ( I (s ? tn?1)B (s; x(s); y(s)) ds; xhk n ? xn )H : Using (4.16) and (4.18), we can rewrite R5 as n
n
n
n
n
R5 =
Zt hk hk ( kB (tj?1; xj?1; yj?1) ? B (s; x(s); y(s)) ds; wh ? xn)H 0 j =1 Z n X hk ) ? t B (s; x(s); y (s)) ds; r ? r ) ; y + k1 ( k B (tj?1; xhk n n?1 H j ?1 j ?1 0 j =1 Z ? k12 ( I (s ? tn?1 )B (s; x(s); y(s)) ds; rn?1 ? rn)H Z Zt + k1 (B (t; x(t); y(t)); x(t))H dt + ( B (s; x(s); y(s)) ds; wnh ? xn)H : I 0
n X
n
n
n
n
n
We then apply the estimates (3.25) and (3.32), (4.19)
R5 c
n X j =1
2 hk 2 k(kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1kV )
+ c k2(kx_ k2L1(0;t ;V ) + ky_ k2L1(0;t ;H )) Z 1 1 h 2 1 + 2 kwn ? xn kV + k kB (; x; y)kL (I ;H ) kx(t)kH dt I Zt n X 1 hk + k ( kB (tj?1; xhk B (s; x(s); y(s))ds; rn ? rn?1)H j ?1 ; yj ?1) ? n
n
n
n
n
j =1
+ kB (; x; y)kL1(I
n
0
krnkV + krn?1kV + tn kwnh ? xn kV ):
;H ) (
Combining the relations (4.7), (4.10){(4.14) and (4.19), we nd that (4.20) krnk2A ? krn?1k2A k krnk2A + c k (kx_ kL1(I ;V ) + kf_kL1(I n
n
;V )
+ kB (; x; y)kL1(I 18
n
krnkV + krn?1kV )
;V ) ) (
Z
+ c (kxkL1(I ;V ) + kf kL1(I ;V )) kxn ? x_ (t)kV dt + ckkxn ? wnhk2V I + c (ZkxnkV + kfnkV + e0 + ky0kH + kx0 kV + kB (; x; y)kL1(I ;V ) )kxn ? wnhkV + c kwnh ? x_ (t)kV dt + 2 (y0 ? y0hk ? A(x0 ? xhk 0 ); rn?1 ? rn )H n
n
n
n
I n X n
+2( +c
hk kB (tj?1; xhk j ?1 ; yj ?1) ?
j =1 n X 2
j =1
Zt
n
0
B (s; x(s); y(s)) ds; rn ? rn?1)H
2 hk 2 k (kxj?1 ? xhk j ?1 kV + kyj ?1 ? yj ?1kV )
+ c kB (; x; y)k
L1(I ;H ) n
Z
I
n
kx(t)kH dt + c k3(kx_ k2L1(0;t ;V ) + ky_ k2L1(0;t ;V ) ): n
n
By a series of manipulations similar to those leading to (3.36), we obtain from (4.20) that (4.21)
kenk2V + kyn ? ynhk k2H nX ?1 2 hk 2 2 c k(kxj ? xhk j kV + kyj ? yj kV ) + c e0 j =1
+ c k (kxk2W 1 1(0;T ;V ) + kf k2W 1 1(0;T ;V ) + ky0k2H + kx0 k2V + kB (; x; y)k2L1(0;T ;V )) ;
;
+ c (kxkL1(0;T ;V ) + kf kL1(0;T ;V ) ) +c
n Z X
j =1 I
j
kxj ? x_ (t)kV dt + c
kwjh ? x_ (t)kV dt + c kB (; x; y)kL1(0;t ;H )
j =1 I + k(IH ? PH j
n Z X
n
h
)(yn ? y0)k2H :
n Z X j =1 I
j
n X
j =1
k kxj ? wjhk2V
kx(t)kH dt
We can then apply the technique of the proof of Theorem 3.2 to nd the estimate (4.22)
2 hk 2 max (kxn ? xhk n kV + kyn ? yn kV )
1nN
c e20 + 1max k(I ? PH )(yn ? y0)k2H nN H + k (kxk2W 1 1(0;T ;V ) + kf k2W 1 1(0;T ;V ) + ky0k2H + kx0 k2V + kB (; x; y)k2L1(0;T ;V ) ) h
;
;
+ (kxkL1(0;T ;V ) + kf kL1(0;T ;V ) ) N Z X + j =1 I
j
N Z X
j =1 I
j
kxj ? x_ (t)kV dt +
kwjh ? x_ (t)kV dt + kB (; x; y)kL1(0;T ;H )
N Z X j =1 I
j
N X
j =1
k kxj ? wjhk2V
kx(t)kH dt :
We now derive a convergence result from (4.22). Theorem 4.4. Let (x; y )
2 W 1;1(0; T ; V H ) be the solution of the problem P and 19
hk N hk f(xhk n ; yn )gn=1 be the solution of the corresponding fully discrete problem P . Then un-
der the assumptions (H1) and (H2) we have
2 hk 2 max (kxn ? xhk n kV + kyn ? yn kV ) ! 0 as h; k ! 0:
(4.23)
1nN
Proof. We still have (4.2). By Lemma 4.3, for any " > 0, there exists x 2 C 1([0; T ]; V )
such that
kx ? xkH 1 (0;T ;V ) ":
We now estimate the term PNj=1 Z
I Z
j
=
I
R
I
j
kxj ? x_ (t)kV dt. Since
kxj ? x_ (t)kV dt Z
k k1 I (x_ (s) ? x_ (t))dskV dt
Z Z k1 I I [kx_ (s) ? x_ (s)kV + kx_ (t) ? x_ (t)kV + kx_ (s) ? x_ (t)kV ] ds dt Z Z 2 kx_ (t) ? x_ (t)kV dt + k kx(t)kV dt; j
j
j
j
I
I
j
we have (4.24)
N Z X j =1 I
j
Next, from
It suces
kxj ? x_ (t)kV dt c " + k kx(t)kL1 (0;T ;V ) :
kwjh ? x_ (t)kV kwjh ? xj kV + kxj ? x_ (t)kV ;
we see that (4.25)
j
N Z N X X h h kwj ? x_ (t)kV dt k kwj ? xj kV + I kxj ? x_ (t)kV dt: j =1 j =1 j =1 I to estimate the term k PNj=1 kwjh ? xj kV . Noticing that Z wjh ? xj = wjh ? x~j + k1 (x~_ (t) ? x_ (t))dt; I N Z X
j
j
j
we have (4.26)
k
N X j =1
kwjh ? xj kV
k k
N X j =1 N X j =1
kwjh ? x~j kV + kx_ ? x~_ kL1 (0;T ;V ) kwjh ? x~j kV + c ": 20
Here x~ is the function used in the proof of Theorem 4.2 such that (4.4) holds. It remains to bound the last term in (4.22). From the de nition (4.17), we have Z t t ?t Z t t?t n n?1 x(t) = x _ (s)ds ? k x_ (s)ds; tn?1 t tn : t ?1 k t Therefore, N Z N Z X X (4.27) kx(t)kH dt k kx_ (s)kV ds = k kx_ kL1(0;T ;V ): n
n
j =1 I
j
j =1
I
j
The bounds (4.24){(4.27) are now used in the estimate (4.22). Noticing the arbitrariness of wjh 2 V h, 1 j N , and recalling the estimate (4.6), we get (4.28)
2 hk 2 2 max (kxn ? xhk x)); n kV + kyn ? yn kV ) c (e0 + " + (h) + k + Dhk (~
1nN
where
Dhk (~x) = k
N X
inf kx~j ? wjhkV
j =1 w 2V h j
h
and the constant c depends on x; y; y~; f; B (t; x(t); y(t)) but is independent of "; h; k. By the assumption (H1), we see that Z ( h ) h _ inf kx~j ? wj kV (h)kx~j kV0 k I kx~(t)kV0 dt; w 2V h j
h
j
and thus (4.29) Dhk (~x) (h) kx~_ kL1(0;T ;V0 ): Then the convergence result (4.23) follows directly from (4.2), (4.28) and (4.29).
2
We remark here that the estimate (3.37) is used for deriving optimal order error estimates provided the exact solution has higher degree regularity, while the estimate (4.22) is suitable for convergence analysis under the basic solution regularity condition.
5 Applications to contact problems in rate-type viscoplasticity The aim of this section is to apply the results obtained in Sections 3 and 4 to the numerical analysis of two nonlinear quasistatic frictional contact problems for viscoplastic materials. In this section, the symbol c may depend on the exact solution, but it is independent of the discretization parameters h and k. 21
Let us consider a viscoplastic body whose material particles occupy a bounded domain
IRd (d = 2; 3 in applications). The boundary of , being assumed Lipschitz continuous, is partitioned into three disjoint measurable parts ?1; ?2 and ?3, with meas (?1) > 0. Displacement, surface traction and contact conditions will be speci ed on ?1 ; ?2 and ?3 , respectively. Since the boundary is Lipschitz continuous, the unit outward normal vector exists a.e. on the boundary. Let [0; T ] be a time interval of interest. We assume that the body is xed on ?1 , a body force of density b acts in , and a surface traction of density F acts on ?2 . Both b and F can be time dependent but we assume a slowly variation of these functions in time so that the inertia terms in the equations of motion may be neglected. We choose (1.1) as the constitutive relation for the viscoplastic material, in which E is a fourth order tensor and G is a given constitutive function, possibly nonlinear. The boundary value problem is then (5.1) (5.2) (5.3) (5.4) (5.5)
_ = E "(u_ ) + G( ; "(u)) in (0; T ); Div + b = 0 in (0; T ); u = 0 on ?1 (0; T ); = F on ?2 (0; T ); u(0) = u0 ; (0) = 0 on :
The unknowns of the problem are the displacement eld u : [0; T ] ! IRd and the stress eld : [0; T ] ! Sd , Sd being the set of second order symmetric tensors on IRd . Here u0 and 0 are the given initial data. We de ne the inner product and the corresponding norms on IRd and Sd by u v = ui vi ; jvj = (v v )1=2 ; = ij ij ; jj = ( )1=2 ;
8 u; v 2 IRd; 8 ; 2 Sd :
For every vector eld v, we denote by v and v the normal and the tangential components of v on the boundary given by
v = v ; v = v ? v and let "(v) denote the tensor eld de ned by 1
"(v ) = ("ij (v )); "ij (v ) = (vi;j + vj;i ); 2
where the index that follows a comma indicates a partial derivative with respect to the corresponding component of the independent variable. 22
We denote in the sequel by H the real Hilbert space de ned by
H = f = ( ij ) j ij = ji 2 L2 ( ); 1 i; j dg with the inner product
(; )H =
Z
ij ij dx; ; 2 H:
We assume in the sequel that E : Sd ! Sd and G : Sd Sd ! Sd satisfy the following assumptions. 8 > > > >
(c) There exists an 0 > 0 such that > > > : E 0 j j2 8 2 Sd; a:e: in : 8 > (a) There exists an L > 0 such that 8 1; 2; "1; "2 2 Sd; a:e: in ; > > > < jG(x; 1 ; "1 ) ? G(x; 2 ; "2 )j L (j1 ? 2 j + j"1 ? "2j): > (b) For any ; " 2 Sd ; x 7! G(x; ; ") is measurable: > > > : (c) The mapping x 7! G(x; 0; 0) 2 H: These assumptions will be used to verify (2.7) and (2.8) in the context of mechanical applications later. For the input data, we assume that b 2 W 1;1(0; T ; (L2( ))d), F 2 W 1;1(0; T ; (L2(?2))d ). Finally, we suppose that the viscoplastic body is in contact with a rigid foundation on ?3 (0; T ). This contact involves friction. In the sequel we consider two di erent friction laws which lead us to the following examples.
5.1 Bilateral contact with Tresca's friction law We assume a bilateral contact modeled by Tresca's friction law (see e.g. [2, 5]), i.e. (5.6)
8 > > < u = 0; j j g; j j < g ) u_ = ; > > : j j = g ) there exists 0
0
such that = ?u_ ;
on ?3 (0; T );
where u represents the normal displacement, u_ denotes the tangential velocity, is the tangential force on the contact boundary and g 0 is the friction yield limit. 23
Let
U = fv 2 (H 1( ))d j vZ= 0 on ?1; v = 0 on ?3 g; j : U ! IR+; j (v) = g jv j ds; ?3
L : [0; T ] U ! IR; L(t; v) =
Z
b(t) v dx +
and let u0 2 U , 0 2 H be given initial such that
Z
?2
F (t) v ds
(0 ; "(v))H + j (v ) L(0; v) 8 v 2 U: In [2] the following weak formulation of the mechanical problem (5.1){(5.6) was derived. Problem P1 . Find the displacement eld u : [0; T ] ! U and the stress eld : [0; T ] ! H
such that (5.7) and for a.e. t 2 (0; T ),
(5.8) (5.9)
u(0) = u0 ; (0) = 0 ;
_ (t) = E "(u_ (t)) + G( (t); "(u(t))); ((t); "(v ? u(t)))H + j (v) ? j (u(t)) L(t; v ? u(t))
8 v 2 U:
Let V be the subspace of H given by
V = "(U ) = f"(v) j v 2 U g: Since meas (?1) > 0, Korn's inequality holds (see [14]):
kvkU c k"(v)kH 8 v 2 U: It follows that V is a closed subspace of H and that the deformation operator " : U ! V is a linear continuous invertible operator. We denote the inverse of " : U ! V by "?1 : V ! U , which is a linear, continuous operator. Now the variational problem P1 can be viewed as a special case of the abstract problem P, after we make the following identi cations (5.10)
(
x $ "(u); y $ ; x0 $ "(u0); y0 $ 0 ; A $ E ; B $ G; '(w) $ j ("?1(w)); (f (t); w)V $ L(t; "?1(w)) for w 2 V:
The conditions (2.7){(2.12) can then be veri ed by using the assumptions made on the constitutive functions E and G as well as on the data b, F , u0 , 0 and g. Therefore, 24
using Theorem 2.1 is follows that problem P1 has a unique solution having the regularity (u; ) 2 W 1;1(0; T ; U H ). We now brie y specify how to construct the nite dimensional spaces V h and H h via the nite element method. Details can be found in [3]. For simplicity, we assume that is polygonal. Let T h be a regular nite element partition of in such a way that if a side of an element lies on the boundary, the side belongs entirely to one of the subsets ?1 , ?2 and ?3. Let U h U consist of linear elements, let us use piecewise constants for H h and recall that V h = H h \ V . It could be shown that V h = "(U h). With the above speci cations, let us show that the assumptions (H1) and (H2) are satis ed. Let
U0 = U \ (H 2( ))d; V0 = "(U0 ); H0 = H \ (H 1( ))dd : The spaces V0 and H0 are dense in V and H , respectively. For any w 2 V0 ; there exists 2 U0 such that w = "(w ). Let hw 2 U h be the piecewise linear interpolant of w . Then w from the standard nite element interpolation theory (cf. [3]), we have
kw ? hw k(H 1 ( )) c h kw k(H 2( )) : d
d
By Korn's inequality, we get h kw ? z hk(H 1 ( )) winf2V kw ? w kV c z inf 2U c kw ? hw k(H 1 ( )) c h kw k(H 2 ( )) c h kwk(H 1 ( )) : h
h
h
d
h
d
d
d
So the assumption (H1) is satis ed and we may take (h) = c h. The assumption (H2) can be veri ed similarly. Therefore by Theorems 4.2 and 4.4, both the spatially discrete and fully discrete solutions corresponding to Problem P1 converge to the solution of Problem P1 as h, and h and k go to zero. In order to derive error estimates via (3.17) and (3.37), we need to make assumptions on the regularity of the exact solution. First let us specialize the estimate (3.17) for the spatial discretization of the problem P1 and obtain (5.11)
kuh ? ukL1(0;T ;U ) + kh ? kL1(0;T ;H ) c k(IH ? PH )( (? 0)kL1(0;T ;H ) + c (kuh0 ? u0 kU + kh0 ? 0 kH ) ) ZT ku_ ? z hkL2(0;T ;U ) + ( 0 jR(t; "(u_ (t)); "(z h(t)))j dt)1=2 : + c z 2L2inf (0;T ;U ) h
h
h
25
The error analysis of the spatially-discrete solution is given by the following result. Theorem 5.1. Let (u; ) 2 W 1;1(0; T ; U H ) be the solution of the problem P1 and (uh; h) 2 W 1;1(0; T ; U h H h) be the corresponding spatially-discrete solution. For the initial values, we choose uh0 2 U h to be the orthogonal projection of u0 into U h with respect to the inner product ("(u); "(v))V , and h0 2 H h the orthogonal projection of 0 into H h
with respect to the inner product of H . Assume
2 W 1;1(0; T ; H0); u 2 W 1;1(0; T ; U ) \ H 1 (0; T ; U0 ); u
2 H 1(0; T ; H 2(?3 )):
Then we have the optimal order error estimates
kuh ? ukL1(0;T ;U ) + kh ? kL1(0;T ;H ) c h:
(5.12)
Proof. First from the choice of the initial values uh0 and h0 , we have
ku0 ? uh0 kU c h; k0 ? h0 kH c h:
(5.13)
Let hu_ (t) 2 U h be the piecewise linear interpolant of u_ (t) for a.e. t 2 (0; T ). Then we have the estimates (cf. [3])
ku_ (t) ? hu_ (t)kU c h ku_ (t)k(H 2 ( )) ; ku_ (t) ? (hu_ (t)) kL2(?3 ) c h2ku_ (t)kH 2 (?3 ): d
Keep in mind the de nition (3.13) and the transformation (5.10). We perform an integration by parts to obtain
R (t; "(u_ (t)); "(hu_ (t))) Z Z h = (t)(( u_ (t)) ? u_ (t)) ds + g (ju_ (t)j ? j(h u_ (t)) j) ds: ?3
?3
Hence, (5.14)
jR(t; "(u_ (t)); "(hu_ (t)))j (k (t)kL2(?3 ) + c g) ku_ (t) ? (hu_ (t)) kL2 (?3 ) c (k(t)kH0 + g)h2ku_ (t)kH 2(?3 ) :
From the property of the projection, we have (5.15)
k(IH ? PH )((t) ? 0 )kH c h k(t) ? 0kH0 : h
Then the error estimate (5.12) follows from (5.11), (5.13), (5.14) and (5.15). 26
2
Now we turn to error analysis of the fully discrete solution. Theorem 5.2. Let (u; ) 2 W 1;1(0; T ; U H ) be the solution of the problem P1 and hk N f(uhk n ; n )gn=1 be the corresponding fully-discrete solution. For the initial values, we choose uh0 2 U h to be the orthogonal projection of u0 into U h with respect to the inner product ("(u); "(v))V , and h0 2 H h the orthogonal projection of 0 into H h with respect to the
inner product of H . Assume 2 W 1;1(0; T ; H0); u 2 H 2 (0; T ; U ) \ C 1 ([0; T ]; U0 ); u 2 C 1 ([0; T ]; H 2(?3 )): Then we have the optimal order error estimates hk ? u k + k hk ? k c (h + k): k u (5.16) max n U n H n n 1nN
Proof. We rst specialize the estimate (3.37) for the case of the full discretization of the
problem P1: hk k + k ? hk k (5.17) max k u ? u n U n H n n 1nN c k (ku_ kL1(0;T ;U ) + k_ kL1(0;T ;H )) +c +c
1=2 h + jRj ("(u_ j ); "(z ))j k inf j =1 z 2U hk k + k ? hk k : max k ( I ? P )( ? ) k + c k u ? u H n 0 H 0 U 0 H H 0 0 1nN
X N
h
h
kuj ? z hk2U h
Again, we have the estimate (5.13) for the approximation of the initial values. For each j , let hu_ j be the piecewise linear interpolant of u_ j . Then kuj ? hu_ j kU c (kZuj ? u_ j kU + ku_ j ? hu_ j kU ) c ( k1 I ku_ (t) ? u_ j kU dt + h ku_ j kU0 ) Z c ku (t)kU dt + c h ku_ j kU0 : j
So
I
j
N X j =1
kkuj ? hu_ j k2U
c
N X j =1
k
!2
Z
I
j
ku (t)kU dt + c h2 1max ku_ k2 j N j U0
c k2ku kL2 (0;T ;U ) + c h2ku_ k2L1(0;T ;U0):
Similar to (5.14) we have jRj ("(u_ j ); "(zh))j c h2(kkL1(0;T ;H0) + 1) ku_ kL1(0;T ;H 2 (?3 )) : The error estimate (5.16) then follows from (5.17) together with the bounds for the various terms on the right hand side of (5.17). 2 27
5.2 Unilateral contact with simpli ed Coulomb's friction law Now we consider an unilateral contact problem modeled by a simpli ed version of Coulomb's law of dry friction (see e.g. [5, 15]), i.e. (5.18)
8 > > < = S; j j g j j; j j < g j j ) u_ = ; > > : j j = g j j ) there exists 0 such
0
that = ?u_ ;
on ?3 (0; T ):
Here denotes the normal force on the contact boundary, S 2 L1(?3) is a given function and g 0 is the coecient of friction. Let
U = fv 2 (H 1( ))d j vZ= 0 on ?1 g; j : U ! IR+; j (v) = g jS j jv j ds; ?3
L : [0; T ] U ! IR; L(t; v ) =
Z
b(t) v dx +
Z
and let u0 2 U , 0 2 H denote initial data such that
?2
F (t) v ds +
Z ?3
S v ds;
(0 ; "(v))H + j (v ) L(0; v) 8 v 2 U: The weak formulation of the mechanical problem (5.1){(5.5), (5.18) is (see e.g. [1]): Problem P2 . Find the displacement eld u : [0; T ] ! U and the stress eld : [0; T ] ! H
such that (5.19) and for a.e. t 2 (0; T ),
(5.20) (5.21)
u(0) = u0 ; (0) = 0 ;
_ (t) = E "(u_ (t)) + G( (t); "(u(t))); ((t); "(v ? u(t)))H + j (v) ? j (u(t)) L(t; v ? u(t))
8 v 2 U:
Let V be the subspace of H given by
V = "(U ) = f"(v) j v 2 U g: Again with the identi cations (
x $ "(u); y $ ; x0 $ "(u0); y0 $ 0 ; A $ E ; B $ G; '(w) $ j ("?1(w)); (f (t); w)V $ L(t; "?1(w)) for w 2 V; 28
the problem P2 can be rewritten in the form of Problem P. Moreover, from the assumptions made on the constitutive functions and data, it follows that conditions (2.7){(2.12) are satis ed in this case. Therefore, we can readily extend all the discussions in Subsection 5.1 to obtain similar results on convergence and error estimates for spatially and fully discrete approximations of the problem P2.
References [1] A. Amassad and M. Sofonea, Analysis of some nonlinear evolution systems arising in rate-type viscoplasticity, in Dynamical System and Di erential Equations, Eds. W. Chen and S. Hu, An added volume to Discrete and Continuous Dynamical Systems , 1998, pp. 58{71. [2] A. Amassad and M. Sofonea, Analysis of a quasistatic viscoplastic problem involving Tresca friction law, Discrete and Continuous Dynamical Systems 4 (1998), pp. 55{72. [3] P.G. Ciarlet, The Finite Element Method for Elliptic Problems , North Holland, Amsterdam, 1978. [4] N. Cristescu and I. Suliciu, Viscoplasticity , Martius Nijho , Editura Tehnica, Bucharest, 1982. [5] G. Duvaut and J.L. Lions, Inequalities in Mechanics and Physics, Springer-Verlag, Berlin, 1976. [6] R. Glowinski, Numerical Methods for Nonlinear Variational Problems, Springer-Verlag, New York, 1984. [7] R. Glowinski, J.-L. Lions and R. Tremolieres, Numerical Analysis of Variational Inequalities , North-Holland, Amsterdam, 1981. [8] W. Han and B.D. Reddy, Computational plasticity: the variational basis and numerical analysis, Computational Mechanics Advances 2 (1995), pp. 283{400. [9] W. Han and B.D. Reddy, Convergence of approximations to the primal problem in plasticity under conditions of minimal regularity, submitted. [10] W. Han and B.D. Reddy, Plasticity: Mathematical Theory and Numerical Analysis , Springer-Verlag, New York, 1999. 29
[11] I. Hlavacek, J. Haslinger, J. Necas and J. Lovsek, Solution of Variational Inequalities in Mechanics, Springer-Verlag, New York, 1988. [12] I. Ionescu and M. Sofonea, Functional and Numerical Methods in Viscoplasticity, Oxford University Press, Oxford 1993. [13] N. Kikuchi and J.T. Oden, Contact Problems in Elasticity: A Study of Variational Inequalities and Finite Element Methods , SIAM, Philadelphia, 1988. [14] J. Necas and I. Hlavacek, Mathematical Theory of Elastic and Elastoplastic Bodies: An Introduction, Elsevier, Amsterdam, 1981. [15] P. D. Panagiotopoulos, Inequality Problems in Mechanics and Applications, Birkhauser, Basel, 1985. [16] M. Sofonea and M. Shillor, Variational analysis of quasistatic viscoplastic contact problems with friction, to appear in Communications in Applied Analysis . [17] E. Zeidler, Nonlinear Functional Analysis and its Applications, Volume II/A: Linear Monotone Operators , Springer-Verlag, New York, 1985.
30
