Universit�at KonstanzMultidimensional Contact Problems in Thermoelasticity

Jaime E. Mu~noz RiveraReinhard RackeKonstanzer Schriften in Mathematik und InformatikNr. 21, Dezember 1996ISSN 1430{3558

c Fakult�at f�ur Mathematik und InformatikUniversit�at KonstanzPostfach 5560/D 188, 78434 Konstanz, GermanyEmail: preprints@informatik.uni{konstanz.deWWW: http://www.informatik.uni{konstanz.de/Schriften

Multidimensional contact problems in thermoelasticity�Jaime E. Mu~noz Rivera and Reinhard Racke21/1996Universit�at KonstanzFakult�at f�ur Mathematik und InformatikAbstract: We consider dynamical resp. quasistatic thermoelastic contact problems in Rn modeling theevolution of temperature and displacement in an elastic body that may come into contact with a rigidfoundation. The existence of solutions to these dynamical resp. quasistatic nonlinear problems and theexponential stability are investigated using a penalty method. Interior smoothing e�ects in the quasistaticcase are also disussed.AMS subject classi�cation: 73 B 30, 35 Q 99Keywords and phrases: quasistatic thermoelasticity, asymptotic behavior1 IntroductionWe consider dynamical resp. quasistatic thermoelastic contact problems which model the evolu-tion of temperature and displacement in an elastic body that may come into contact with a rigidfoundation. If � Rn (n � 2) denotes the reference con�guration, we assume that the smoothboundary @ consists of three mutually disjoint parts �D;�N ;�C such that @ = �D [�N [�c,and �D 6= ;. The body is held �xed on �D, tractions are zero on �N , and �c is the part whichmay have contact with a rigid foundation. The temperature is held �xed on @. Then thedynamical initial boundary value problem for the displacement u = u(t; x) and the temperaturedi�erence � = �(t; x), where t � 0 and x 2 , to be considered is the following:%@2t ui � (Cijkluk;l);j + (mij�);j = 0; i = 1; : : : ; n; (1.1)%c@t� � (kij�;i);j +mij@tui;j = 0; (1.2)u(t = 0) = u0; ut(t = 0) = u1; �(t = 0) = �0; (1.3)uj�D = 0; �ij�jj�N = 0; �j@ = 0; (1.4)u� � g; �� � 0; (u� � g)�� = 0; �T = 0 on �c: (1.5)�Supported by a CNPq-KFAJ grant 1

Here % and c denote the density and the heat capacity, respectively, and are assumed in thesequel to be equal to one without loss of generality. Cijkl, mij and kij denote the componentsof the elasticity tensor, those of the thermal expansion tensor and those of the heat conductiontensor, respectively, and will satisfyCijkl 2 L1(); Cijkl = Cjikl = Cklij ; (1.6)9�1 > 0 8�ij = �ji Cijkl�ij�kl � �1j�j2; (1.7)mij 2 W 1;1(); mij = mji � 0; (1.8)kij 2 W 1;1(); kij = kji; (1.9)9�2 > 0 kij�i�j � �2j�j2: (1.10)The comma notation ;j denotes the di�erentiation @j � @=@xj with respect to xj as well asa subindex t will denote the di�erentiation @t � @=@t. The initial values u0; u1 and �0 areprescribed with regularity to be made more precise in the �nal formulation of the system. Thestress tensor is given by � = (�ij) �ij = Cijkluk;l �mij�:Since on the boundary � = 0 we have there�ij = Cijkluk;l:The unit normal vector in x 2 @ is denoted by � = �(x) and the normal component of u byu� : u� = u � �:The normal component �� of the stress tensor is given by�� = �ij�i�jand the tangential part �T is �T = �� � ���:The function g describes the initial gap between the part �c of the reference con�guration andthe rigid foundation and is assumed to satisfyg 2 H1=2(�c); g � 0 a.e. on �c: (1.11)Hence the boundary conditions (1.5) describe Signorini's contact condition on �c for a friction-less (�T = 0) contact.The corresponding quasistatic system arises from (1.1){(1.5) by omitting @2t ui in (1.1) and pre-scribing only �0 in (1.3). 2

These thermoelastic contact problems, being nonlinear because of the contact boundary con-ditions (1.5), arise in applications such as the manufacturing of castings and pistons, see thepaper of Shi & Shillor [13] for more details and references. The mathematical treatment ofthe dynamical problem (1.1){(1.5) in one space dimension was discussed in a paper of Elliott& Tang [5], where more complicated boundary conditions are considered. Existence results inhigher dimensions were announced by Figueiredo & Trabucho in [6]. They considered the caseof constant contact which allows them to use only variational methods to solve the problem;this does not apply to the situation of the contact condition of Signorini's type considered here.We shall investigate the existence in n � 1 space dimensions for radially symmetrical situationsfollowing the approach of Kim [9] who discussed an obstacle problem for a wave equation. Usinga penalty method, we obtain an existence result, and we also prove the exponential stability,which was proved for a special one-dimensional system by Mu~noz Rivera & Lacerda Oliveira[11].As a motivation for the stability results to be expected and as a tool to be used later, we alsodiscuss the linear quasistatic system for classical boundary conditions likeuj@ = 0; �j@ = 0;proving existence, exponential stability and smoothing.For the quastistatic system, Shi & Shillor [13] proved the existence of a solution, while Ames &Payne [2] proved a uniqueness result, see also these papers for references on the one-dimensionalcase, where a series of papers has appeared in the last years.We shall give a new existence proof using a penalty method. This will also allow us to makeconclusions on the exponential stability of the system.Finally, we shall prove that the quasistatic system has a smoothing e�ect in the interior of ,where u and � are shown to be arbitrarily smooth, an e�ect which cannot be expected up to theboundary because of the mixed and contact boundary conditions.We remark that we have assumed exterior forces and exterior heat supply to be zero only forsimplicity.The paper is organized as follows: In section 2 we prove an existence and exponential stabilityresult for the dynamical system(1.1){(1.5) in the case of radial symmetry using a penalty method.Section 3 studies the linear quasistatic system with classical Dirichlet or Neumann type boundaryconditions proving existence, uniqueness and exponential stability. In section 4 the quasistaticcontact problem is investigated, and we obtain an existence result and exponential stabilityusing a penalty method. Finally, section 5 will provide the interior smoothing e�ect for thequasistatic contact problem.Concerning the notation we remark that we use standard notations for Sobolev spaces and alsothe Einstein summation convention for repeated indices. h�; �i denotes the norm in (L2())n; k�kthe corresponding norm.2 Existence for the dynamical contact problemA dynamical thermoelastic contact problem was investigated by Elliott & Tang [5] in one spacedimension following the approach of Kim [9] who used a penalty method for the treatment of3

an obstacle problem for a wave equation. We also adopt this approach. The system characterof our problem arising in multi-dimensional elasticity naturally leads to further problems in theestimates that have to be overcome by additional considerations compared to the two papersabove. We shall discuss the case of radial symmetry.We look for a solution to (1.1){(1.5) in the following sense. LetH1�D := fu 2 (H1()nj uj�D = 0g:De�nition 2.1 (u; �) is a solution to (1.1){(1.5) for given u0 2 H10(); u1 2 L2(); �0 2L2(), if, for any T > 0,u 2 L1((0; T ); H1�D()); ut 2 L1((0; T ); (L2())n);� 2 L1((0; T ); L2())\ L2((0; T ); H10()) (2.1)u(t = 0) = u0; ut(t = 0) = u1; �(t = 0) = �0; (2.2)u� � g on (0; T )� �C (a.e.), (2.3)and for all w 2 L1((0; T ); H1�D()) \W 1;1((0; T ); (L2()n) with w� � g on �c the followinginequality holds, hut(T; �); w(T; �)� u(T; �)i � hu1; w(0; �)� u0i� TZ0 hut; wtidt+ TZ0 fhut; uti � hCijkluk;l; ui;jigdt+ TZ0 hCijkluk;l; wi;jidt� TZ0 hmij�; wi;j � ui;jidt � 0; (2.4)and for all z 2 W 1;2((0; T ); H10()) the following equality holds,� TZ0 h�; ztidt+ h�(T; �); z(T; �)i� h�0; z(0)i+ TZ0 hkij�;i; z;jidt� TZ0 hui;j ; @t(mijz)idt+ hmiju(T; �)i;j; z(T; �)i � hmiju0i;j ; z(0; �)i= 0: (2.5)In the sequel, we shall write L1(H1�D) instead of L1((0; T ); H1�D()), similarly for the otherspaces; moreover we write u(T ) instead of u(T; �), and so on.As a justi�cation for the de�nition of a solution we remark that a smooth classical solution to4

(1.1){(1.5) obviously satis�es (2.5) and also (2.4), because multiplication of (1.1) by u� w andpartial integration yields that the left-hand side of (2.4) equalsZ�c �jCijkluk;l(wi � ui)d� = Z�c ��(u)(w� � u�)d�= Z�c ��(u)(w� � g)d� � 0:On the other hand, if (u; �) is a solution in the sense of De�nition 2.1, and if (u; �) is smooth,then (1.1){(1.5) follows in the classical sense in the following way:Taking w = u � h; h 2 (C10 ((0; T )� ))n and z 2 C10 ((0; T )� ), we conclude from (2.4),(2.5) that the di�erential equations (1.1), (1.2) are satis�ed in the distributional (and hence inthe classical) sense. (1.3), (1.4) and u�j�c � g;are obvious.Taking w = u+ h; h 2 C1([0; T )� ); h� � 0; h = h� � � on �c, we concludeTZ0 Z�c �jCijkluk;lhid� dt � 0which implies TZ0 Z�c �� � h�d� dt � 0;hence �� � 0:Taking w = u� (u� � g) � ��; j�j � 1, we concludeTZ0 Z�C ��(u� � g)� d� dt = 0;which implies ��(u� � g) = 0:Finally, taking w = u+ h; h� � 0; h � h� � � + hT , we obtainTZ0 Z�c �� � h�d� dt+ TZ0 TZ�c �T � hTd� dt � 0which implies, choosing hT appropriately,TZ0 Z�C �ThTd� dt = 05

hence �T = 0;therefore (1.5) is also satis�ed.The existence of a solution will be proved for the radial symmetrical case by using a penaltymethod. For this purpose we consider the following approximating problem for a parameter" > 0. This will be solved | in general, without restriction to radial symmetry, | and thena priori estimates are proved that will show the convergence, as " # 0, of a subsequence to asolution of (1.1){(1.5). The penalized problem is the following, �rst in classical notation. Forgiven u"0; u"1 2 (H2())n\(H10 ())n, �"0 2 H2()\H10 () �nd a solution (u"; �") to the followinginitial boundary value problem (2.6){(2.10):@2t u"i � (Cijklu"k;l);j + (mij�");j = 0; (2.6)@t�" � (kij�";i);j +mij@tu"i;j = 0; (2.7)u"(t = 0) = u"0; u"t (t = 0) = u"1; �"(t = 0) = �"0; (2.8)u"j�D = 0; �"ij�jj�N = 0; �"j@ = 0; (2.9)�"� = �1" (u"� � g)+ � "@tu"� ; �"T = 0; on �c; (2.10)where �" = �(u") is the stress tensor corresponding to u", and f+ denotes the positive part off . Let fwjg1j=1 � H1�D() be an orthonormal basis in (L2())n and let fzjg1j=1 � H10 () be anorthonormal basis in L2().De�nition 2.2 (u"; �") is a solution to (2.6){(2.10) for u"0; u"1 2 (H2() \ H10())n; �"0 2H2() \H10(), if for any T > 0,u" 2 W 2;1(L2)\W 1;1(H1�D); �" 2 W 1;1(L2) \W 1;2(H10); (2.11)u"(t = 0) = u"0; u"t (t = 0) = u"1; �"(t = 0) = �"0; (2.12)and for all p 2 N and for a.e. t 2 [0; T ]:h@2t u"; wpi+ hCijklu"k;l; wpi;ji � hmij�"; wpi;ji =�1" Z�c (u"� � g)+wp� d� dt� " Z�c @tu"�wp�d�; (2.13)h@t�"; zpi+ hkij�";i; zp;ji+ hmij@tu"i;j ; zpi = 0: (2.14)It is again easy to see that a smooth classical solution to (2.6){(2.10) is a solution (in the senseof De�nition 2.2), and a solution (in the sense of De�nition 2.2), which is smooth, is a classicalsolution, compare the considerations following De�nition 2.1.6

Theorem 2.3 For given u"0; u"1 2 (H2() \H10())n; �"0 2 H2() \H10 (), there is a solution(u"; �") to the penalized problem (2.6){(2.10).Remark: There is no assumption on radial symmetry here.Proof: A Galerkin method will be the appropriate tool as in [9], [5]. De�ning v and byv := u" � u"0 � tu"1; := �" � �"0;then (v; ) should satisfy, for i = 1; : : : ; n,@2t vi � (Cijklvk;l);j + (mij );j= (Cijklu"ok;l + tu"1k;l);j � (mij�"o);j =: fi; (2.15)@t � (kij ";i);j +mij@tvi;j = (kij�"o;i);j �miju"1i;j =: g; (2.16)v(t = 0) = 0; vt(t = 0) = 0; (t = 0) = 0; (2.17)as well as the boundary conditions (2.9), (2.10) with (u"; �") being replaced by (v; ), all to beunderstood in the sense of De�nition 2.2. We make the ansatzvm(t; x) = mXp=1amp(t)wp(x); m 2 N; (2.18) m(t; x) = mXp=1 bmp(t)zp(x); (2.19)where (amp)mp=1, (bmp)mp=1 satisfyh@2t vm; wpi+ hCijklvmk;l; wpi;ji � hmij m; wpi;ji= hf; wpi � 1" Z�c (vm� � g)+wp�d�� " Z�c @tvm� wp�d�; (2.20)for p = 1; : : : ; m, as well ash@t m; zpi+ hkij mi ; zp;ji+ hmij@tvmi;j ; zpi = hg; zpi; (2.21)amp(0) = 0; ddtamp(0) = 0; bmp(0) = 0: (2.22)The equations (2.20), (2.21) are a system of nonlinear ordinary di�erential equation for amp andbmp, p = 1; : : : ; m, with prescribed initial values given in (2.22). Since the (only) nonlinearity in(2.20) is Lipschitz continuous, there is a smooth solution famp; bmpgmp=1 in [0; T ] for any T > 0.7

Now we derive a priori estimates for (vm; m). Multiplication of (2.20) by ddtamp(t) and (2.21)by bmp(t) and summation over p yields12 ddtk@tvmk2 + 12 ddthCijklvmk;l; vmi;ji � hmij m; @tvmi;ji= hf; @tvmi � 1" Z�c (vm� � g)+@tvm� d� � " Z�c j@tvm� j2d�; (2.23)12 ddtk mk2 + hkij m;i ; m;j i+ hmij@tvmi;j ; mi = hg; mi; (2.24)or 12 ddtfk@tvmk2 + hCijklvmk;l; vmi;ji+ k mk2 + 1" Z�c j(vm� � g)+j2d�g+hkij m;j ; m;j i= hf; @tvmi � " Z�C j@tvm� j2d� + hg; mi: (2.25)Integrating (2.25) with respect to t 2 [0; T ], we conclude that there is a positive constantc = c(T; ku"0kH2; ku"1kH2 ; k�"0kH2) (2.26)not depending on m, and depending on " only in the way indicated in (2.26), such that for anyt 2 [0; T ] the following estimates (2.27){(2.31) hold:k@tvm(t)k � c; (2.27)hCijklvmk;l; vmi;ji(t) � c; (2.28)tZ0 hkij m;i ; m;j i(s)ds � c; (2.29)1" Z�c j(vm� � g)+j2d� � c; (2.30)" tZ0 Z�c j@tvm� j2d� ds � c: (2.31)Therefore, using Korn's inequality, we obtain that(vm)m is bounded in W 1;1(L2) \ L1(H1�D); (2.32)8

( m)m is bounded in L1(L2) \ L2(H10): (2.33)Next, the boundedness of @2t vm in L2(L2) will be proved. The estimate will be uniform only inm 2 N, not in " > 0 (cf. (2.26)).Di�erentiation of (2.20), (2.21) with respect to t, then multiplication by d2dt2amp(t) and ddtbmp(t),respectively, and summation leads to12 ddt nk@2t vmk2 + hCijkl@tvmk;l; @tvmi;ji+ k@t mk2o+ hkij@t m;i ; @t m;j i= h@tf; @2t vmi � 1" Z�c (@t(vm� � g)+)@2t vm� d�� " Z�c j@2t vm� j2d�� k@tfk k@2t vmk+ 12"3 Z�c j@t(vm� � g)+j2d�� "2 Z�c j@2t vm� jd�: (2.34)Using (2.31) we conclude, by Gronwall's inequality,k@2t vm(t)k � c1("); (2.35)hCijkl@tvmk;l(t); @tvmi;j(t)i � c1("); (2.36)k@t m(t)k � c1("); (2.37)tZ0 hkij@t m;i ; @t m;j i(s)ds � c1("); (2.38)where the positive constant c1(") is independent of m0 and t 2 [0; T ]; T > 0, but may dependon ". Thus we have that, for �xed " > 0,(@2t vm)m is bounded in L1(L2); (2.39)(@tvm)m is bounded in L1(H1�D); (2.40)(@t m)m is bounded in L1(L2) \ L2(H10): (2.41)It follows from (2.32), (2.33) and (2.39){(2.41) that for �xed " there is a subsequence, againdenoted by ((vm; m))m, and (v; ) such that, as m!1,vm �* v in W 2;1(L2)\W 1;1(H1�D); (2.42) m �* in W 1;1(L2); (2.43) m * in W 1;2(H10): (2.44)With the help of Lemma 1.4 from [9] (essentially Gagliardo-Nirenberg type estimates) the con-vergence (vm� � g)+ ! (v� � g)+ in C0([0; T ]; L2(�c)) (2.45)9

as well as @tvm ! @tv in C0([0; T ]; L2(@)) (2.46)follow.Using (2.42){(2.46) and letting m!1 in (2.20), (2.21), we conclude thatu" := v + u"0 + tu"1; �" := + �"0satisfy (2.11){(2.14).Q.e.d.Now we turn to the original problem (1.1){(1.5) and show that a subsequence of (u"; �") where(u"; �") solves a penalized problem for " > 0 according to Theorem 2.3, converges to a solutionof (1.1){(1.5).Theorem 2.4 For given u0 2 H10(); u1 2 (L2())3; �0 2 L2() there is a solution (u; �) to(1.1){(1.5) in the case of radial symmetry.Radial symmetry means that the domain is radially symmetrical, i.e. invariant under trans-formations of the special orthogonal group SO(3) if n = 3 resp. O(2) if n = 2. The typicalexamples are balls or annular domains. Radial symmetry of (u; �) means8A 2 SO(3) resp. O(2) 8 x 2 : u(Ax) = Au(x); �(Ax) = �(x);or, equivalently (see [7]), with r := jxj:There exists a function w : R+0 �! R such that u(x) = xw(r), andthere exists a function : R+0 �! R such that �(x) = (r).Of course, for the existence of radially symmetrical solutions to the penalized, and later for theoriginal, problem, the coe�cients have to satisfy invariance conditions, too. As an example weconsider the homogeneous, isotropic case, where, in particular,Cijkl = ��ij�kl + �(�ik�jl + �jk�il); (2.47)and the di�erential equations (1.1),(1.2) turn intoutt � ��u � (� + �)r div u+ �mr� = 0; (2.48)c�t � �2�� + �m div ut = 0; (2.49)where � and � are the Lam�e moduli satisfying � > 0 and 2� + n� > 0; �; c; � > 0 and m 6= 0are constants.We also notice that in this case ��j@ = �div u+ 2�ui;k�i�k;and it is easy to see that the boundary conditions (2.9),(2.10) allow for radially symmetricalsolutions, i.e. if the initial data are radially symmetrical, then the solution according to Theorem2.3 will have the same symmetry; namely, if (u; �) is the solution, then also (v; �) with v(x) :=10

A�1u(Ax); �(x) := �(Ax) will be a solution to the same initial data and boundary conditionsfor any (special) orthogonal A. Hence, by uniqueness, (u; �) = (v; �).Proof of Theorem 2.4: Let " > 0 and let u"0; u"1 2 (H2()\H10())n; �"0 2 H2()\H10() beradially symmetrical such that, as " # 0,u"0 ! u"1 in (H10())n; (2.50)u"1 ! u1 in (L2())n; (2.51)�"0 ! �0 in L2(): (2.52)Let (u"; �") be the solution to (2.6){(2.10) according to Theorem 2.3, and corresponding to theinitial values (u"0; u"1; �"0). By the known regularity of (u"; �") we can substitute wp by @tu" in(2.13), and zp by �" in (2.14), respectively, and we obtain (cf. (2.25))12 ddtfk@tu"k2 + hCijklu"k;l; u"i;ji+ k�"k2 + 1" Z�c j(u"� � g)+j2 d� + hkij�";i; �";ji= �" Z�c j@tu"� j2 d�: (2.53)Hence, there is a positive constantc2 = c2(T; ku0kH1; ku1kL2 ; k�0kL2) (2.54)which is independent of ", such that for any t 2 [0; T ] we getk@tu"k � c2; (2.55)hCijklu"k;l; u"i;ji � c2; (2.56)tZ0 hkij�";i; �";ji(s)ds � c2; (2.57)1" Z�c j(u"� � g)+j2d� � c2 (2.58)" tZ0 Z�c j@tu"� j2d� ds � c2: (2.59)The estimates (2.55){(2.59) imply the existence of a subsequence, again denoted by (u"; �"), andof (u; �) such that, as " # 0, u" �* u in L1(H1�D)\W 1;1(L2); (2.60)�" �* � in L1(L2); (2.61)�" * � in L2(H10): (2.62)11

Using Lemma 1.4 from [9] again, we concludeu" ! u in C0(L2): (2.63)The equations (2.13), (2.14) imply for (u"; �")TZ0 fh@2t u"; w� u"i+ hCijklu"k;l; wi;j � u"i;ji� hmij�"; wi;j � u"i;jigdt= �1" TZ0 Z�c (u"� � g)+(w� � u"�)d� dt� " TZ0 Z�c @tu"�(w� � u"�)d� dt; (2.64)for any w 2 L1(H1�D)\W 1;1(L2) such that w� � g, andTZ0 nh@t�"; zti+ hkij�";i; z;ji+ hmij@tu"i;j ; ziodt = 0; (2.65)for any z 2 W 1;2(H10).Integration by parts in (2.64) yieldshu"t(T ); w(T )� u"(T )i � hu"1; w(0)� u"0i� TZ0 hu"t ; wtidt+ TZ0 hu"t ; u"ti � hCijklu"k;l; u"i;jidt+ TZ0 hCijklu"k;l; wi;jidt� TZ0 hmij�"; wi;j � u"i;jidt� �" TZ0 Z�c @tu"�w�d� dt+ " TZ0 Z�c @tu"�u"�d� dt= �" TZ0 Z�c @tu"�w� d� dt+ "2 8><>:Z�c ju"�(T )j2d�� Z�c ju"0� j2d�9>=>; (2.66)where we used �1" TZ0 Z�c (u"� � g)+(w� � u"�)d� dt= �1" TZ0 Z�c (u�� � g)+(w� � g)d� dt+ 1" TZ0 Z�c j(u"� � g)+j2d� dt� 0: 12

Integration by parts in (2.65) leads to� TZ0 h�"; ztidt+ h�"(T ); z(T )i� h�"0; z(0)i+ TZ0 hkij�";i; z;jidt� TZ0 hu"i;j ; @t(mijz)idt+ hmiju"(T )i;j; z(T )i � hmiju"oi;j ; z(0)i= 0: (2.67)Using (2.60){(2.62) we conclude from (2.67), as " # 0, that (u; �) satisfy (2.5).It remains to justify (2.3) and (2.4). Notice that u satis�es (2.3) becausek(u� � g)+kL2(�c) = lim"#0 k(u"� � g)+kL2(�c) = 0by (2.63),(2.58).We claim that the following inequality holds:TZ0 hut; uti � hCijkluk;l; ui;jidt � lim"#0 sup TZ0 hu"t ; u"ti � hCijklu"k;l; u"i;jidt: (2.68)The convergence ju"t j2 � Cijklu"k;lu"i;j ! jutj2 � Cijkluk;lui;j (2.69)in D0((0; T ) � ), i.e. in the sense of distributions, follows from the theory of compensatedcompactness, see Corollary 4.3 of [3]; takevi := (@tu"i ;�(Cijklu"k;l)j=1;:::;n); wi := (@tu"i ; (u"i;j)j=1;:::;n)and summation over i = 1; : : : ; n, and observe�(mij�");j 2 L2(L2):The di�erential equations (2.6), (2.7) for (u"; �") are satis�ed in the distributional sense. Since�"t 2 L1(L2) and @tu"i;j 2 L1(L2)we have (kij�";i);j 2 L1(L2);hence �" 2 L1(H2 \H10): (2.70)Moreover, @2t u" 2 L1(L2)and (2.69) imply (Cijklu"k;l);j 2 L1(L2)13

from which we conclude interior regularity in the following sense:8� > 0 : u" 2 L1((0; T ); H2(�)); (2.71)where � := fx 2 j dist (x; @) > �g; 0 < � � �0(�0 �xed).Remark: The H2-regularity up to @ cannot be expected because of the mixed boundaryconditions for u".Let � as above, S� := @�, and let h 2 (C1())n such thath(x) = �(x) = exterior normal in x 2 S�;if dist(x; @) = � � �0=2, h(x) = 0if dist(x; @) � �0.Writing (u; �) instead of (u"; �") for simplicity, we multiply the �rst di�erential equation (2.6)by hkui;k and integrate.TZ0 Z� @2t uihkui;kdx dt = Z� @tui(T )hkui;k(T )dx� Z� u1ihku0i;kdx�12 TZ0 ZS� j@tuj2d� dt+ 12 TZ0 Z� hk;k j@tuj2dx dt: (2.72)With ~�ij := �ij +mij� (2.73)we obtain� TZ0 Z� �ij;jhkui;kdx dt = � TZ0 Z� ~�ij;jhkui;kdx dt� TZ0 Z� (mij�);jhkui;kdx dt= � TZ0 ZS� �j ~�ij�kui;kd� dt+ TZ0 Z� ~�ijhk;jui;kdx dt+ TZ0 Z� ~�ijhk(ui;k);jdx dt� TZ0 Z� (mij�);jhkui;kdx dt: (2.74)Since ~�ijhk(ui;k);j = ~�ijhk(ui;j);k = Crsijui;jhk(ur;s);k= (Cijrsur;s);khkui;j � (Cijrs);kur;shkui;j= (~�ij);khkui;j � (Cijrs);kur;shkui;j ;14

we have ~�ijhk(ui;k);j = 12hk(~�ijui;j);k � 12(Cijrs);kur;shkui;j :Inserting this into (2.74), we conclude� TZ0 Z� �ij;jhkui;kdx dt = � TZ0 ZS� �j ~�ij�kui;kd� dt+ TZ0 Z� ~�ijhk;jui;kdx dt�12 TZ0 Z� ~�ijui;jhk;kdx dt+ 12 TZ0 ZS� ~�ijui;jd� dt�12 TZ0 Z� (Cijrs);kur;shkui;j � TZ0 Z� (mij�);jhkui;kdx dt: (2.75)The equations (2.6),(2.72),(2.75) implyTZ0 ZS� jutj2d� dt� TZ0 ZS� Cijkluk;lui;jd� dt = 2 Z� f@tui(T )hkui;k(T )� u1ihku0i;kgdx+2 TZ0 Z� nhk;k jutj2 + ~�ijhk;jui;j�12(Cijrs);kur;shkui;j � (mij�);jhkui;k�dx dt:�2 TZ0 Z� �j ~�ij�kui;kd� dt= C1� �2 TZ0 Z� �j ~�ij�kui;kd� dt| {z }=:R� ; (2.76)for some C1� which satis�es9M =M(ku0kH1 ; ku1kL2 ; k�0kL2) 80 < � � �0 : jC1� j �M (2.77)according to (2.55){(2.57). In particular, M is independent of ".In order to get control of the boundary integral R�, we multiply (2.6) by hm~�im and obtain (cp.Lemma 2.3 in [7])Z� @2t uihm~�imdx = ddt Z� @tuihm~�imdx� Z� hmCimps@tui@s@tupdx= ddt Z� @tuihm~�imdx� 12 Z� hmCimps@s(@tui@tup)dx+12 Z� hmCimpsh@s@tui@tup � @tui@s@tupidx: (2.78)15

Now we shall use | for the �rst time | the radial symmetry, i.e.ui(t; x) = xiw(t; r); r = jxj;for some function w, in particular, rotu = 0;which implies h@s@tui@tup � @tui@s@tupi = (@tup@i � @tui@p) @tus= @tw (xp@i � xi@p)xs@tw� G; (2.79)and G contains at most �rst-order derivatives (wt) of u because(xp@i � xi@p) @tw = �xpxir � xixpr � (@=@r)@tw = 0:Combining (2.78) and (2.79) we getZ� @2t uihm~�imdx = ddt Z� @tuihm~�imdx� 12 ZS� Cimps�m@tui�s@tupd�+12 Z� (hmCimps);s@tui@tupdx+ 12 Z� hmCimpsGdx: (2.80)Moreover,� Z� ~�ij;jhm~�imdx = � ZS� �j ~�ij�m~�imd� + Z� ~�ijhm;j ~�imdx+ Z� ~�ijhm~�im;jdx= � ZS� j~� � �j2d� + Z� ~�ijhm;j ~�imdx+ Z� ~�ijhj ~�im;mdx+ Z� ~�ij fhm@j � hj@mg ~�im;mdx:Observing that for the radial symmetrical casehk = xk=r;the last integral vanishes, and we obtain� Z� ~�ij;jhm~�imdx = �12 ZS� j~� � �j2d� + 12 Z� ~�ijhm;j ~�imdx: (2.81)Combining (2.6),(2.80),(2.81), and integrating with respect to t yieldsTZ0 ZS� j~� � �j2d� dt+ TZ0 ZS� Cimps�m@tui�s@tup| {z }�0 by (1:7) d� dt = C2� ; (2.82)16

where C2� satis�es the same estimate as C1� in (2.77).The boundary integral R� can now be estimated byjR�j � ��1 TZ0 ZS� j~� � �j2d� dt+ � TZ0 ZS� jruj2d� dt� ��1jC2� j+ � TZ0 ZS� jruj2d� dt; (2.83)where � > 0 is still arbitrary.For � > 0 let %� 2 C1(R); %� increasing, %�(s) = 0 if s � �=2; %�(s) = 1 if s � �, and let �(x) := %�(dist (x; @)):Then � is constant on S� for each �. Integrating (2.76) with respect to � 2 [0; ��]; 0 < �� ��0=2, after multiplication with (1� �(x)), and using (2.83), we obtain������� TZ0 ZU�� (1� �) �ju"t j2 � Cijklu"k;lu"i;j�dx dt������� � ��M + ��1��M + �M;where U�� := fx 2 j dist (x; @) < ��g = n �� :As a consequence, we have for any � > 0 the existence of a �� = ��(�) (independent of ") suchthat ������� TZ0 ZU�� (1� �) �ju"t j2 � Cijklu"k;lu"i;j�dx dt������� � �; (2.84)if �� � ��(�). (Choose � = �(�) := �=(3M), �� := �=(3M)min f1; �g = �=(3M)minf1; �=(3M)g.)Writing 1 = � + (1� �) we conclude by (2.69) and (2.84) the claimed inequality (2.68).Now, the desired inequality (2.4) follows as " # 0 from (2.66), using (2.60){(2.63) and (2.68).Q.e.d.Finally, we shall prove the exponential decay for the radially symmetrical, homogeneous case.We consider the system (2.48), (2.49), now de�ned over the set: := B(r1; 0) nB(r0; 0);where we are denoting by B(r; 0) the ball of radius r with center in 0, i.e. is an annular region.First we treat the penalized problem, and we assume the following boundary conditions:u = 0; @�@� = 0; on @B(r0; 0) = �D; (2.85)and the contact conditions on @B(r1; 0) = �c which in our case can be written as:2�@u@� � � + �divu = �1" (u� � g)+ on �c; (2.86)� = 0 on �c: (2.87)17

Observe that the boundary condition for � in (2.85) is di�erent from the Dirichlet boundarycondition studied before, but the corresponding existence theorems, both for the penalized andfor the original contact problem, hold as well, the proof carries over almost literally, noticing,for example, that the Dirichlet boundary condition is still given on �c.For simplicity in our notations we will drop the upper index ". Since u; � are radially symmetricalsolutions we have ui(t; x) = xiw(t; r); r = jxj;�(t; x) = (t; r);for some w and . Then the system (2.48), (2.49) is equivalent towtt � (2�+ �)wrr � (2�+ �)n+ 1r wr + �mr �r = 0; (2.88)c�t � �2�rr � �2(n � 1)�rr + �mnwt + �mrwrt = 0; (2.89)where w and � satisfy w(r0) = 0; �r(r0) = 0;among other boundary conditions. Note that there exist positive constants C0; C1 (which donot depend on ") satisfyingC0 Z jruj2 dx � Z 1r0 jwrj2 dr � C1 Z jruj2 dx;C0 Z jutj2 dx � Z 1r0 jwtj2 dr � C1 Z jutj2 dx:To show the exponential decay we shall exploit the following two Lemmas.Lemma 2.5 Let q be any C2([r0; 1])-function such that q(r0) = 0 and q(1) = 1, let � > 0 andlet f be a function in W 1;2(L2). Then for any solution ' 2 L2(H1)\W 1;2(H1) \W 2;2(L2) of'tt � �'rr = f; (2.90)we have that� ddt �Z 1r0 q(x)'t'r dr� = �12 nj't(1; t)j2+ �j'r(1; t)j2o+12 Z 1r0 q0(r)nj'tj2 + �j'rj2o dr � Z 1r0 q(r)'rf dr:Proof: To get the above equality, multiply equation (2.90) by q'r and make an integration byparts.Q.e.d.Let us introduce the function by (t; r) := Z rr0 �(t; �) d�:Then satis�es the following equationc t � �2 rr � �2(n� 1) Z rr0 �rr d�+ �mn Z rr0 wt d�� �m Z rr0 wt d�+ �mrwt = 0: (2.91)With these notations we have 18

Lemma 2.6 Let q be any C2([r0; 1])-function such that q(r0) = 0 and q(1) = 1. Then, for � > 0small enough, we haveddt �Z 1r0 wt dr � � Z 1r0 q(r)wtwr dr� � � �m4 Z 1r0 jwtj2 dr��4 njwt(1; t)j2+ (2�+ �)jwr(1; t)j2o+� Z 1r0 jwrj2 dr+ C Z 1r0 j�xj2 dr:Proof: Multiplying equation (2.91) by wt we getddt Z 1r0 c wt dr = Z 1r0 c twt dr+ Z 1r0 c wtt dr= �2 Z 1r0 rrwt dr + (n� 1)�2 Z 1r0 �Z rr0 �rr d��wt dr� �m(n� 1) Z 1r0 �Z rr0 wt d��wt dr � �m Z 1r0 rjwtj2 dr + Z 1r0 c wtt dr= �2 Z 1r0 �rwt dr+ (n� 1)�2 Z 1r0 Z rr0 �rr d�wt dr� �m(n� 1)2 �Z 1r0 wt d��2 � �m Z 1r0 rjwtj2 dr � (2�+ �)c Z 1r0 �wr dr+(2�+ �)(n+ 1) Z 1r0 wrr d�� �mc Z 1r0 �rr dr + c(2�+ �) (1; t)wr(1; t):Using Sobolev's embedding theorem, we conclude from the above identityddt Z 1r0 wt dr � � �m2 Z 1r0 rjwtj2 d�+ � Z 1r0 jwrj2 dr + C� Z 1r0 j�rj2 dr + �4(2�+ �)jwr(1; t)j2:Since w 2 L2(H2)\W 1;2(H1), � 2 W 1;2(H1) we can use Lemma 2.5 for ' = w, � = 2�+� andf = �(2�+ �)nwrr + �m �rr 2 W 1;2(L2), which proves our conclusion.Q.e.d.Now we are able to prove the exponential decay for the penalized problem associated to thedynamical contact case.Theorem 2.7 Under the above conditions, and with u0 such that u0 � � � g, the energy E" =E"(t) to the penalized system (2.48), (2.49) de�ned byE"(t) := 12 Z ju"t j2 + Cijklu"k;lu"i;j + j�"j2 dx+ 12" Z�c j(u"� � g)+j2 d�cdecays exponentially as time goes to in�nity, i.e.9C > 0; > 0; 8 t � 0 : E"(t) � C e� t; (2.92)where and C are independent of ". 19

Proof: Multiplying as usual equation (2.48) by ut and equation (2.49) by � we get, dropping "for simplicity, ddtE(t) = � Z jr�j2 dx:Multiplying equation (2.48) by u and integrating by parts over we getddt �Z uut dx� = Z jutj2 dx� Z �jruj2 + (�+ �)jdivuj2 dx+ �m Zr�u dx� 12" Z�c j(u� � g)+j2 d�c;from where it follows thatddt �Z uut dx� � Z jutj2 dx� 12 Z Cijkluk;lui;j dx� 12" Z�c j(u� � g)+j2 d�c + C1 Z jr�j2 dx;where C1, and C2; C3 below, are positive constants. Using Lemma 2.6 and choosing � and �small enough we get ddt �Z 1r0 wt dr� � Z 1r0 q(x)wtwr dr + � Z uut dx� ���0�Z jutj2 + jruj2 + 12" Z�c j(u� � g)+j2 d�c�+ C Z jr�j2 dx:So taking N large enough we obtainddt �NE(t) + Z 1r0 wt dr � � Z 1r0 q(x)wtwr dr + � Z uut dx� ���0 �Z jutj2 + jruj2 + jr�j2 dx+ 12" Z�c j(u� � g)+j2 d�c�for some �0 > 0 being independent of ". Since we also have the relationj Z 1r0 wt dr � � Z 1r0 q(x)wtwr dr + � Z uut dxj � C2E(t);we conclude, taking N large enough, thatL(t) := NE(t) + Z 1r0 wt dr� � Z 1r0 q(x)wtwr dr+ � Z uut dxsatis�es ddtL(t) � � L(t)for some positive constant which is independent of ". So we concludeL(t) � L(0)e� tand hence E(t) � C3E(0)e� t:20

By our choice of the initial data, the right hand side of the above inequality is bounded, whichproves the Theorem.Q.e.d.Using the lower semicontinuity of norms, when "! 0 we �nally obtain that the energy associatedto the original contact problem also decays exponentially, that is we have provedTheorem 2.8 The energy E = E(t) to the system (1.1), (1.2), (1.3), (1.5), (2.85), (2.86),E(t) := 12 Z jutj2 + Cijkluk;lui;j + j�j2 dxdecays exponentially as time goes to in�nity, i.e.9C > 0; > 0; 8 t � 0 : E(t) � CE(0) e� t: (2.93)3 Linear quasistatic thermoelasticityThe linear quasistatic problem arises from the full dynamical problem (1.1){(1.5) by neglectingthe acceleration term �@2t ui in (1.1), and, consequently, prescribing only �0 in (1.3), and replacingthe boundary conditions for u by Dirichlet type boundary conditions (see the remarks below forother linear boundary conditions). That is, we consider the following linear elliptic-parabolicinitial boundary value problem:� (Cijkluk;l);j + (mij�);j = 0; i = 1; : : : ; n; (3.1)@t� � (kij�;i);j +mij@tui;j = 0; (3.2)�(t = 0) = �0; (3.3)u=@ = 0; �=@ = 0: (3.4)According to the paper of Shi & Xu [14], the known results on well-posedness are restrictedto one-dimensional situations, see Day [4], or to homogeneous and isotropic media for beingequal to the unit disk on R2; see [14], where methods from the theory of complex functions arebasic ingredients of the proofs.Here, we discuss the general case of a bounded C2-reference con�guration in any spacedimension, allowing the medium to be anisotropic and non-homogeneous (see below for weak-ening the C2-assumption on the boundary). We consider Dirichlet boundary conditions for thedisplacement vector and for the temperature, but it will be pointed out at the end of this sectionhow other boundary conditions can be dealt with. In passing, we note that the questions in[14] concerning well-posedness are answered. The analyticity of solutions is not touched in thegeneral situation.The well-posedness of the problem will be obtained, i.e. the unique existence of a solutionand the continuous dependence on the data is proved. Moreover, the exponential decay toequilibrium is obtained as a by-product.All coe�cients are assumed to be smooth functions, e.g. C1(�) should be an appropriateregularity class for Theorem 3.1.(i)). In the homogeneous case all coe�cients would be inde-pendent of x, in the speci�c homogeneous and isotropic case the equations (3.1), (3.2) reduce21

to � (�+ �)rdiv u� ��u+ �mr� = 0; (3.5)@t� � �2�� + �mdiv @tu = 0: (3.6)Here � and � are the Lam�e constants, �m > 0 is the coupling (interaction) constant and �2 isa positive constant representing the heat conductivity constant. The equations (3.5), (3.6) areessentially those considered for the unit disk in R2 in [14].We look for solutions u 2 C0([0;1); (H3()\H10())3); (3.7)� 2 C1([0;1); L2())\ C0([0;1); H2() \H10()) (3.8)to the elliptic-parabolic problem (3.1){(3.4). We will obtain such solutions via the naturaldecoupling arising from (3.1),(3.2).Theorem 3.1 (i) Let �0 2 H2()\H10(). Then there exists a unique solution (u; �) of (3.1){(3.4) with u 2 C0([0;1); (H3()\H10())3);� 2 C1([0;1); L2())\ C0([0;1); H2()\H10()):� and u decay to zero exponentially as t !1, i.e.9d1; d2; d3 > 0 8t � 0 : k�(t; �)k � d1e�d2tk�0k;ku(t)kH1 � d3e�d2tk�0k:(ii) If �m := k(mij)ijkL1 is su�ciently small, and if @ is smooth, and additionally �0 2H2(m�1)(), m � 2, satis�es the usual compatibility conditions, thenu 2 m\j=0Cj([0;1); (H2(m�j)+1())3);� 2 m\j=0Cj([0;1); H2(m�j)()):The higher derivatives decay to zero exponentially, i.e.9d4; d5 > 0 8t � 0 : k�(t; �)kH2(m�1) � d4e�d2tk�0kH2(m�1) ;ku(t)kH2m�1 � d5e�d2tk�0kH2(m�1) :We distinguish between d1 and d3 because the dependence on the coe�cients is di�erent, seethe remarks below. We note that a posteriori the acceleration utt goes to zero as t ! 1 andhence remains small for all times.Proof of Theorem 3.1:The natural decoupling arising from (3.1), (3.2) is obtained using the elasticity operatorE = � ((@jCijkl@l)ik ;22

formally �rst, which is well-de�ned byE : D(E) � (L2())n ! (L2())n;D(E) := fu 2 (H10))njEu 2 (L2())ng= (H2()\H10 ())n:The latter equality follows from elliptic regularity theory. E is an elliptic, self-adjoint, positiveoperator, cf. [10], and E�1 : (L2())n ! D(E)is continuous as well as E�1 : H�1()! (H10())n;where H�1() is the dual space to (H10())n.These mapping properties are known, see [10] for the former; the latter follows from the repre-sentation theorem of Lax&Milgram for coercive forms on (H10())n).The system (3.1), (3.2) is then equivalent tou = �E�1(r0M 0�); (3.9)@t� � (kij�;i);j �MrE�1r0M 0@t� = 0; (3.10)where M = (mij)ij :Remark: This natural ansatz is also discussed in [1], where an elliptic-parabolic problem for ascalar u is studied in a general framework for nonlinear equations. The corresponding operatorE is the Laplace operator ��, the solutions live in Lp-spaces, with p > 2.An appropriate solution � of the decoupled equation (3.10), together with initial and bound-ary conditions given in (3.3), (3.4), will solve the problem.Theorem 3.2 Let �0 2 H2()\H10(). Then there exists a unique solution� 2 C1(([0;1); L2())\ C0([0;1); H2() \H10())to (3.10), (3.3), (3.4). � decays exponentially to zero as t!1, i.e.9d1; d2 > 0 8t � 0 : k�(t; �)k � d1e�d2tk�0k:If �m (= k(mij)ij jL1) is small with respect to the norm of E�1, and if @ is smooth, andadditionally �0 2 H2(m�1)(); m � 2, then� 2 m\j=0Cj([0;1); H2(m�j)())and 9d4 > 0 8t � 0 : k�(t; �)kH2(m�1) � d4e�d2tk�0kH2(m�1) :23

By (3.9) we obtain the complete unique solution (u; �) of the original problem in the classdescribed in (3.7), (3.8), yielding the decay of u too, which will prove Theorem 3.1.Proof of Theorem 3.2:The equation (3.10) is equivalent to G(�t)� (kij�;i);j = 0; (3.11)where G : L2()! L2() is given byGv := v �MrE�1r0M 0v:Since for v; w 2 L2() we havehGv;wi = hv; wi+ hhE�1r0M 0v;r0M 0wii= hv;Gwi;where hhg; hii denotes the dual pairing for g 2 H10(); h 2 H�1() (= dual space of H10()here), we conclude that G is a positive, self-adjoint operator,hGv; viL2 � kvk2L2: (3.12)Thus G�1 : L2()! L2() is a well-de�ned continuous operator, and (3.11) is equivalent to�t �G�1(kij�;i);j = 0: (3.13)This evolution equation can be solved uniquely for an initial value given in (3.3) in the followingHilbert space. Let H := L2() equipped with the inner product h�; �iH de�ned byhv; wiH := hGv;wi:Let A : D(A) � H ! H, D(A) := H2() \H10(); Av := �G�1(kijv;i);j :A is densely de�ned and symmetric, hAv;wiH = h�r0Lrv; wi = hv;�r0Lrwi = hv; AwiH,and w 2 D(A�); A� denoting the adjoint operator, implies9f 2 H 8v 2 D(A) : h�(kijv;i);j ; wi = hv;Gfi;hence, by the self-adjointness of the realization of @jkij@i on D(A) in L2(), we conclude w 2D(A). Therefore A is self-adjoint, and our problem reads as�t + A� = 0; (3.14)with initial condition given by (3.3). We apply the spectral theorem for self-adjoint operators(or semigroup theory) and obtain a unique solution � satisfying� 2 C1([0;1);H)\ C0([0;1); D(A));provided �0 2 D(A) = H2()\H10(). This proves the unique existence of a solution (u; �) inthe desired class described in (i).Since hAv; viH = hkijv;i; v;ji � �2krvk2L2 � l1cpkvk2 � �2cp=kkvk2H; (3.15)24

where cp ist the Poincar�e constant and k is the norm of G in L2(), which implies that A ispositive. The inverse A�1 is compact, because A�1 = �(@jkij@i)�1G with compact (@jkij@i)�1and continuous G. Therefore, the spectrum of A is its discrete spectrum, a subset of (0;1) withonly accumulation point 1 and smallest eigenvalue denoted by d2. The spectral theorem forself-adjoint operators implies �(t) = 1Zd2 e��tdP��0;(P�)�2Rbeing the spectral resolution for A.If d1 = d1(E�1; (mij)) denotes the square root of the norm of G in L2(), we concludek�(t; �)k � k�(t; �)kH � e�d2tk�0kH � d1e�d2tk�0k:This completes the proof of Theorem 3.2.(i), and thus of Theorem 3.1.(i) too.The proof of (ii) immediately follows from the following Lemma 3.3 by di�erentiating the equa-tion for � with respect to t. Q.e.d.Lemma 3.3 Let s 2 N0.(i) G : Hs() �! Hs() is continuous.(ii) A has the usual elliptic regularity property, i.e. v 2 H2()\H10() and Av 2 Hs() implyv 2 Hs+2() and kvkHs+2 � cskAvkHs.(iii) If �m (= k(mij)ijkL1) is small with respect to the norm of E�1, then G is a homeomorphismof Hs().Proof: (i): Using the boundedness of E�1 : Hs�1() �! Hs+1(), cf. [8], the assertion isobvious.(ii) is a consequence of (i) and the elliptic regularity of @jkij@i.(iii): G = (Id + B) with the bounded operator B := MrE�1r0M 0, and the inverse of G isgiven by a Neumann series if kBkHs!Hs < 1; this is ful�lled, if �m2 < cs(E�1), where cs(E�1) isessentially the norm of E�1 : Hs�1() �! Hs+1(). Q.e.d.Remarks:1. The smallness assumption on �m is satis�ed e.g. for homogeneous, isotropic media, cf. [10].2. A generates an analytic semigroup.3. In the homogeneous, isotropic case we have d1 � 1 + ~d1(E�1) �m2.4. A lower bound for d2 is given by �2cp=k according to (3.15). The constant d1 depends on�m. The decay rate d2 for u in ku(t)k � d3e�d2tk�0kis bounded from below as �m vanishes, while d3 !1 as �m vanishes.25

5. The C2-condition on in Theorem 3.1.(i) and Theorem 3.2.(i) can be dropped if we replace(H2()\H10())n by D(E) = fv 2 (H10())njEV 2 (L2())n)gand D(A) = H2() \H10() by D(�@jkij@i) := fv 2 H10()j(kijvi)j 2 L2()gthroughout the text, i.e. in this case the boundary of may be arbitrary.The Dirichlet (u)-Dirichlet (�) boundary conditions (3.4) have been chosen for simplicity ofthe presentation, other boundary conditions such as Neumann-Neumann, Neumann-Dirichlet orDirichlet-Neumann can be dealt with similarly, for example Neumann-Neumann,�ij�j =@ = 0; �jkij�i =@ = 0;There are nontrivial solutions in the null space of the corresponding elasticity operator E (cf.[10], [12]), as well as in the null space of the corresponding realization of @jkij@i (consisting ofthe constant functions). In order to carry over the arguments from the proof of Theorem 3.1,one concentrates on initial values �0 from the orthogonal complement of the constant functions,i.e. on initial values with mean value R �0 zero. Observe that Mr� lies in the orthogonalcomplement of the null space of E, and that a Poincar�e inequality holds. We do not go intofurther details and end up with the remark that has to satisfy the restricted cone property(cf. [12], [10]).As the proof of Theorem 3.1 shows, it would be possible to deal with exterior forces andexterior heat supply, i.e. with non-zero right-hand sides f and g in (3.1) and (3.2), respectively.4 Quasistatic contact problems | existence and stabilityThe quasistatic contact problem arises from the full dynamical problem (1.1){(1.5) by neglectingthe acceleration term %@2t ui in (1.1), and, consequently, prescribing only �0 in (1.3). Therefore,we consider the following problem� (Cijkluk;l);j + (mij�);j = 0; i = 1; : : : ; n; (4.1)@t� � (kij�;i);j +mij@tui;j = 0; (4.2)�(t = 0) = �0; (4.3)uj�D = 0; �ij�jj�N = 0; �j@ = 0; (4.4)u� � g; �� � 0; (u� � g)�� = 0; �T = 0; on �c: (4.5)Shi & Shillor [13] gave an existence proof, providedm := supx;i;j jmij(x)j is small enough. (4.6)Ames & Payne [2] proved a uniqueness and continuous dependence result.We shall prove an existence result using a penalty method, which will also allow us to prove theexponential stability. The condition (4.6) will also be required.26

De�nition 4.1 (u; �) is a solution to (4.1){(4.5) for given �0 2 H10 (), if, for any T > 0,u 2 L2(H1�D); ((Cijkluk;l);j)i=1;:::;n 2 L2(L2); ut 2 L2(H1); (4.7)� 2 L2(H2 \H10) \ C0(H10); �t 2 L2(L2); (4.8)�(t = 0) = �0; (4.9)u� � g on (0; T )� �c (a.e.); (4.10)and for all w 2 L1(H1�D) with w� � g on �c the following inequality holds,TZ0 hCijkluk;l; wi;jidt� TZ0 hCijkluk;l; ui;jidt� TZ0 hmij�; wi;j � ui;jidt � 0; (4.11)and for all z 2 L2(H10) the following equality holds,TZ0 h�t; zi+ hkij i; z;ji+ hmij@tui;j ; zidt = 0: (4.12)A solution will be obtained by studying an associated penalized problem for the parameter " > 0,then proving a priori estimates and �nally letting " tend to zero. The advantage will be that�nally the exponential stability can be proved. The penalized problem to be solved �rst is givenas follows: � (Cijklu"k;l);j + (mij�");j = 0; i = 1; : : : ; n; (4.13)@t�" � (kij�";i);j +mij@tu"i;j = 0; (4.14)�"(t = 0) = �0 2 H10(); (4.15)u" 2 L2(H1�D); ((Cijklu"k;l);j)i=1;:::;n 2 L2(L2); u"t 2 L2(H1); (4.16)�" 2 L2(H2 \H10); �"t 2 L2(L2); (4.17)u"j�D = 0; �jCijklu"k;lj�N = 0; �Tj�c = 0; (4.18)��(u")j�c = �1" (u"� � g)+j�c : (4.19)A solution of this penalized problem, for �xed " > 0, will be obtained by a �xed point argument.For this purpose let � > 0 andW� := fu 2 L2(H1) j ((Cijkluk;l);j)i=1;:::;n 2 L2(L2); ut 2 L2(H1);kutkL2(H1) � �; k((Cijkluk;l);j)i=1;:::;nkL2(L2) � �g:27

Lemma 4.2 W� is a Banach space with norm kukW� := kukL2(H1).The Proof follows from the completeness of L2(H1) with respect to the norm k � kW� and theweak compactness of balls in L2(L2) resp. L2(H1).Q.e.d.For v 2 W� we de�ne u = Sv;and hence the map S as follows. Let � be the solution to the parabolic equation�t � (kij�;i);j = �mij@tvi;j ; (4.20)with initial value �(t = 0) = �0: (4.21)Then there exists a solution � with the regularity� 2 L2(H2 \H10) \ C0(H10); �t 2 L2(L2): (4.22)Now let u be the solution to the penalized problem� (Cijkluk;l);j + (mij�);j = 0; i = 1; : : : ; n; (4.23)uj�D = 0; �ij�jj�N = 0; (4.24)��(u)j�c = �1" (u�j�c � g)+; �T = 0; on �c: (4.25)A solution u to (4.23){(4.25) is obtained by minimizingJ(u) := hCijkluk;l; ui;ji � 2h(mij�);j ; uii+ 1" Z�C j(u� � g)+j2d�on fw 2 (H1())nj w = 0 on �Dg;where t is regarded as a parameter.Theorem 4.3 If � = �(�0) is chosen large enough, and if �m is small enough, where the requiredsmallness is independent of ", then S : W� ! W�and S is a contraction mapping.Proof: Let v 2 W� and u := Sv. By the de�nition of u we immediately obtainkukL2(H1) � c1 �mk�kL2(L2); (4.26)k((Cijkluk;l);j)i=1;:::;nk2L2(L2) � c22 �m2k�kL2(H1)� c22 �m2k�0k2 + c23 �m4kvtk2L2(L2)� c22 �m2k�0k2 + c23 �m4�2: (4.27)28

where, in this proof, c1; c2; : : : will denote constants that do neither depend on �m nor on ". Inorder to see that u 2 W�, we have to estimate ut. Let h 6= 0 andvh(t; �) := (v(t+h; �)�v(t; �))=h; uh(t; �) := (u(t+h; �)�u(t; �))=h; �h(t; �) := (�(t+h; �)��(t; �))=h:From the obvious di�erential equations for uh and �h we obtainZ Cijkluhk;luhi;jdx� Z�c �jCijkluhk;luhi d� = Z �hmijuhi;jdx: (4.28)For the boundary term, we obtain, using the notationu1(t; �) := u(t+ h; �); u2(t; �) := u(t; �);Z�c �jCijkluhk;luhi d� = Z�c �1�(u1� � g) + �2�(u2� � g)d�� Z�c �1�(u2� � g) + �2�(u1� � g)d�= �1" Z�c j(u1� � g)+j2 + j(u2� � g)+j2d� + 1" Z�c (u1� � g)+(u2� � g) + (u2� � g)+(u1� � g)d�= Z�c\�+: : : d� + Z�cn�+: : : d�; (4.29)where �+ := fx 2 �cj u1� > g and u2� > gg:Z�c\�+: : : d� = �1" Z�+ (u1� � u2�)2d� � 0; (4.30)Z�cn�+: : : d� = �1" Z�cn�+ j(u1� � g)+j2 + j(u2� � g)+j2d�+ 1" Z�cn�+(u1� � g)+(u2� � g) + (u2� � g)+(u1� � g)d�� �1" Z�cn�+j(u1� � g)+j2 + j(u2� � g)+j2d�� 0: (4.31)The estimates (4.29){(4.31) imply Z�c �jCijkluhk;luhi d� � 0: (4.32)Combining (4.28) and (4.32) we get the estimatekuhk2L2(H1) � c24 �m2k�hk2L2(L2)� c24 �m2k�tk2L2(L2)� c24 �m2k�0k2 + c25 �m4kvtk2L2(H1); (4.33)29

which implies ut 2 L2(H1)and kutk2L2(H1) � c24 �m2k�0k2 + c25 �m4kvtk2L2(H1)� c24 �m2k�0k2 + c25 �m4�2: (4.34)Choosing �m and � such that �m2(c23 + c25) � 12 ; (4.35)�2 = �2(�0) = 2k�0k2max fc22; c24g; (4.36)we conclude from (4.27) and (4.34) that S maps W� into itself.Now we prove the contraction property. For this purpose letvj 2 W�; uj := Svj ; j = 1; 2:Let �j denote the solution to (4.20), (4.21) with respect to vj. Replacing uh by u12 := u1 � u2and so on, we can derive analogous estimates to those given in (4.28){(4.33), and concludeku12k2L2(H1) � c24 �m2k�12k2L2(L2)� c25 �m4kv12k2L2(H1); (4.37)which proves that S is a contraction, if c5 �m2 < 1: (4.38)Q.e.d.The unique �xed point u of S together with the associated � is then the desired solution to thepenalized problem, i.e. we have provedTheorem 4.4 For given " > 0 and �0 2 H10 () there is a unique solution (u"; �") to thepenalized problem (4.13){(4.19), provided �m is small enough.From the estimates in the proof of Theorem 4.3, we conclude the validity of the following lemma.Lemma 4.5 (�")" is bounded in W 1;2(L2) \ L2(H2 \ H10) \ C0(H10), (u")" is bounded inW 1;2(H1), and ((Cijklu"k;l);j)i=1;:::;n is bounded in L2(L2).Now we can prove the existence of a solution to the quasistatic contact problem.Theorem 4.6 For given �0 2 H10 () there exists a solution (u; �) to (4.1){(4.5) in the sense ofDe�nition 4.1, provided �m is small enough.Remark: The solution is unique, which follows from the result of Ames & Payne [2].Proof: Let (u"; �") be the solution to the penalized problem according to Theorem 4.4. From30

Lemma 4.5 we conclude that there exists a subsequence, again denoted by (u"; �"), and (u; �)such that �" �* � in L1(H10); (4.39)�" * � in W 1;2(L2) \ L2(H2 \H10 ); (4.40)u" * u in W 1;2(H1); (4.41)((Cijklu"k;l);j)i=1;:::;n * ((Cijkluk;l);j)i=1;:::;n in L2(L2): (4.42)It shall be proved that (u; �) is a solution to (4.1){(4.5). It only remains to justify the relations(4.11) and (4.12). Since (u"; �") sati�es (4.12), we can use (4.39){(4.42) to see that also (u; �)satis�es (4.12).For w 2 L1(H1�D), w� � g on �c, we haveTZ0 hCijklu"k;l; wi;j � u"i;jidt� TZ0 hmij "t ; wi;j � u"i;jidt= TZ0 Z�c �"�(w� � u"�)d� dt= �1" TZ0 Z�c (u"� � g)+(w� � u"�)d� dt= �1" TZ0 Z�c (u"� � g)+(w� � g)d� dt+ 1" TZ0 Z�c (u"� � g)+(u"� � g)d� dt� 0: (4.43)Using the lower semi-continuity of norms and (4.39){(4.42), we obtainTZ0 h�Cijkluk;l +mij�; ui;jidt � lim"#0 sup TZ0 h�Cijklu"k;l +mij�"; u"i;jidt: (4.44)From (4.43) and (4.44), we conclude the validity of (4.11).Q.e.d.Finally, we want to describe the exponential stability.Let u"; �" denote the solution to the penalized problem as given in Theorem 4.4. Similarly as in[2] we change variables from �" to ", where "(t; x) := tZ0 �"(s; x)ds+ "0(x): (4.45)Here, "0 2 H2()\H10 () is de�ned as solution to(kij 0;i);j = �0 +miju"i (t = 0);j; (4.46)31

where u"i (t = 0) is de�ned by �0 via (4.13). Then (u"; ") satis�es "t � (kij ;i);j +miju"i;j = 0: (4.47)Multiplying (4.47) by t we obtain12 ddt Z kij ";i ";jdx = � Z j "t j2dx� Z miju"i;j "tdx: (4.48)From (4.13) we obtain� Z miju"i;j "tdx = � Z Cijklu"k;lu"i;jdx+ Z�c �jCijklu"k;lu"id�= � Z Cijklu"k;lu"i;jdx� 1" Z�c j(u"� � g)+j2d�: (4.49)From (4.48), (4.49) we have12 ddt Z kij ";i ";jdx � � Z j "t j2dx� Z Cijklu"k;lu"i;jdx� 1" Z�c j(u"� � g)+j2d�: (4.50)On the other hand we obtain from (4.13), (4.14)� Z (Cijklu"k;l);j@tu"idx+ Z �"t�"dx� Z (kij�";i);j�"dx = 0;which implies 12 ddt 8<:Z Cijklu"k;lu"i;j + j�"j2dx9=; = � Z k;j�";i�";jdx+ Z�c �"�@t(u"� � g)d�= � Z kij�";i�";jdx� 1" Z�c (u� � g)+@t(u� � g)d�;where 12 ddt 8><>:Z Cijklu"k;lu"i;j + j�"j2dx+ 1" Z�c j(u"� � g)+j2d�9>=>; � � Z kij�";i�";jdx: (4.51)Combining (4.50) and (4.51) we obtainddt 12 8<:Z Cijklu"k;lu"i;j + j�"j2 + kij ;i ";j9=;dx+ 1" Z�c ju"� � g)+j2d�g� �8<:Z Cijklu"k;lu"i;j + j�"j2 + ki;j�";i�";jdx+ 1" Z�c j(u� � g)+j2d�9>=>; : (4.52)32

Multiplying (4.14) by ", we obtain12 ddt Z j "j2dx = � Z kij ";i ";jdx� Z miju"i;j "dx� �12 Z kij ";i ";jdx+ c1 Z Cijklu"k;lu"i;jdx; (4.53)where the constant c1 > 0 is independent of ".Let E"(t) := 12 8<:Z Cijklu"k;lu"i;j + j�"j2 + kij ";i ";jdx+ 1" Z�c j(u"� � g)+j2d� + 12c1 Z j "j2dx9>=>; (4.54)Then the estimates (4.52) and (4.53), together with Poincar�e's estimate applied to ", implyddtE"(t) � ��E"(t); t � 0; (4.55)for some constant � > 0 which is independent of ". Hence we have proved:Theorem 4.7 For �xed " > 0 the "energy" E", de�ned for solutions (u"; ") to the penalizedproblem (4.13){(4.19) in (4.54), decays to zero exponentially, i.e.9� > 0 8t � 0 E"(t) � E"(0)e��t; (4.56)and � is independent of ".If E(u"; ")(t) := E"(t)� 12" Z�C j(u"� � g)+j2d�;then obviously E(u"; ")(t) � E"(0)e��t:Lemma 4.8(i) 9�1 > 0 8" > 0 : E(u"; ")(0) � �1k�0k2; (4.57)(ii) 9�2 > 0 8" > 0 8t � 0 : 1" Z�c j(u"� � g)+j2d� � �2k�0k2: (4.58)33

Proof: �Cijklu"k;l(0) = �(m;j�0);j ; i = 1; : : : ; n:This impliesZ Cijklu"k;l(0)u"i;j(0)dx+ 1" Z�c j(u"�(0)� g)+j2d� � � Z mij�0ui;j(0)� c1k�0k2 + 12 Z Cijklu"k;l(0)u"i;j(0)dx;where c1 > 0 is a constant. ThereforeZ Cijklu"k;l(0)u"i;j(0)dx+ 1" Z�c j(u"�(0)� g)+j2 � 2c1k�0k2: (4.59)Z j "(0)j2dx+ Z kij ";i(0) ";j(0)dx � c1k "0k2H1� c1(k�0k2 + kmijui;j(0)k2 (by (4.46))� c1k�0k2 (by (4.59)): (4.60)We use c1 to denote various constants being independent of ".The estimates (4.59), (4.60) prove (4.57). From (4.51) and (4.59) we conclude for t � 01" Z�c j(u"�(t)� g)j2d� � �2k�0k2; (4.61)for some �2 > 0, hence (4.58) is proved.Q.e.d.Corollary 4.9 9�1; �2 > 0 8t � 0 8" > 0 : E(u"; ")(t) � �1k�0k2e�2t:Now let (u; ) be the solution to the quasistatic problem obtained from (u"; ")" as " # 0(Theorem 4.5). Let the "energy" of (u; ) be de�ned byE(t) := Z Cijkluk;lui;j + j tj2 + kij ;i ;jdx: (4.62)Then the exponential stability is expressed in the next theorem.Theorem 4.10 8t � 0 : E(t) � �1k�0k2e��2t; (4.63)(�1; �2 from Corollary 4.8).Proof: Corollary 4.8 and the weak lower semi-continuity of the norm yield the proof.Q.e.d. 34

5 Smoothing in the interior for the quasistatic contact problemThe linear, quasistatic problem with Dirichlet boundary conditions has smoothing propertieslike those known for the solutions to heat equations, in particular, the solution (u; �)(t; x) isin�nitely smooth in t and x as soon as t > 0, no matter how smooth the initial data is (cf. theremarks in section 3). This behavior cannot be expected in general for the quasistatic contactproblem because of the mixed boundary conditions for u. We shall prove that the solution (u; �)to the quasistatic thermoelastic contact problem (4.1){(4.5) is in�nitely smooth with respect tox in the interior of , if t > 0.Naturally, we assume in this section that all coe�cients are C1-smooth. We also assume that(u; �) is a solution to (1.1){(1.5) as given in the previous section. Let ' 2 C10 () such thatsupp' � �0=2 for some �0 > 0, with �0=2 open such that�0=2 = fx 2 jdist (x; @) > �0=2g:Also let ' = 1 in 3�0=4:Let (~u; ~�) := ('u; '�):Then (~u; ~�) satisfy� (Cijkl~uk;l);j + (mij ~�);j = Ri(u;ru; �); i = 1; : : : ; n; (5.1)~�t � (kij ~�;i);j +mij@t~ui;j = Q(ut; �;r�); (5.2)~�(t = 0) = ~�0 := '�0; (5.3)supp ~u; supp ~� �� : (5.4)Here Ri and Q are given byRi = �';l;jCijkluk � ';l(Cijkluk);j � ';jCijkluk;l + ';jmij�; (5.5)Q = �';i;jkij� � ';i(kij�);j � ';j�;i + ';jmij@tui: (5.6)In particular, suppRi [ suppQ � suppr' �� :Let r1 > 0 such that �� B := B(0; r1).Let (w; #) solve in (0;1)� B:� (Cijklwk;l);j + (mij#);j = 0; i = 1; : : : ; n; (5.7)#t � (kij#;i);j +mij@twi;j = 0; (5.8)35

#(t = 0) = ~�0; (5.9)wj@B = 0; #j@B = 0: (5.10)The linear, quasistatic problem (5.7){(5.10) can be solved as shown in Section 2 assuming�0 2 H2()\H10(); (5.11)and (w; #) is smooth in B as t > 0.Let (v; %) be the solution in (0;1)�B to� (Cijklvk;l);j + (mij%);j = Ri; i = 1; : : : ; n; (5.12)%t � (kij%;i);j +mij@tvi;j = Q; (5.13)%(t = 0) = 0; (5.14)vj@B = 0; %j@B = 0: (5.15)The right-hand sides Ri; Q satisfyR = (R1; : : : ; Rn) 2 W 1;2(L2); Q 2 L2(H1): (5.16)The system (5.12){(5.15) can be transformed into a parabolic equation for %, essentially, as inSection 2: v = �E�1r0M� +E�1R; vj@B = 0; (5.17)where M = (mij)ij ; E = (�@jCijkl@l)i;hence %t + A% = F; (5.18)%(t = 0) = 0; (5.19)%j@B = 0; (5.20)where A% = �(Id �MrE�1r0M 0)�1(kij%;i);j ; (5.21)F = Q�Mr@tE�1R: (5.22)36

The regularity of R and Q given in (5.16) as well as the mapping properties of the operator E(cf. Section 2) imply F 2 L2(H1): (5.23)This regularity of F implies for the (existing) solution % to (5.18){(5.20):% 2 L2(H3(�0=2); %t 2 L2(H1(�0=2)) (5.24)by the usual regularity for parabolic equations, see for example p.11 in [15], applied to @j%,smoothly cut o� in �0=2.By (5.17) and (5.24) we conclude v 2 W 1;2(H2(�0=2)): (5.25)By uniqueness we have in (0;1)� B~u = w + v; ~� = #+ %:Since (w; #) is smooth for t > 0, we conclude from (5.24), (5.25) and the fact thatu = ~u; � = ~� in 3�0=4 (5.26)that, for any � > 0,u 2 W 1;2((�;1); H2(3�0=4); � 2 L2((�;1); H3(3�0=4)); �t 2 L2((�;1); H1(3�0=4));(5.27)i.e. more regularity for (u; �).This implies (cp. (5.16))R 2 W 1;2((�;1); H1(3�0=4); Q 2 L2((0;1); H2(3�0=4)); (5.28)i.e. higher regularity for F in (5.18).Proceeding by induction (�0 > 0 chosen appropriately) we have provedTheorem 5.1 Let �0 2 H10(). Then the solution (u; �) to (4.1){(4.5) satis�es8� > 0 80 �� 8m 2 N : u 2 W 1;2((�;1); Hm(0)); � 2W 1;2((�;1); Hm(0)): (5.29)This interior smoothing e�ect in x carries over to smoothing in t, which can be seen by di�er-entiating the equation for (~u; ~�) with respect to t and applying the same arguments again. Asa corollary we haveCorollary 5.2 Let �0; (u; �) as in Theorem 5.1. Then8� > 0 80 �� : u 2 C1((�;1)� 0); � 2 C1((�;1)� )): (5.30)37

References[1] H. Amann, Nonhomogeneous linear and quasilinear elliptic and parabolic boundary valueproblems, Manuscript, 1993.[2] K.A. Ames, L.E. Payne, Uniqueness and continuous dependence of solutions to a mul-tidimensional thermoelastic contact problem, J. Elasticity, 34 (1994), pp. 139{148.[3] B. Dacorogna,Weak continuity and weak lower semicontinuity of non-linear functionals,Lec. Notes Math., 922, 1982.[4] W.A. Day, Heat conduction within linear thermoelasticity, Springer Tracts Nat. Philos.30, Springer-Verlag, New York et al., 1985.[5] C.M. Elliott, Q. Tang, A dynamic contact problem in thermoelasticity, NonlinearAnal., T.M.A., 23 (1994), pp. 883{898.[6] I. Figueiredo, L. Trabucho, Some existence results for contact and friction problemsin thermoelasticity and in thermoviscoelasticity, In: "Asymptotic Methods For ElasticStructures", Proceedings of the International Conf., Lisbon (1993), P.G. Ciarlet et al.(eds.), de Gruyter (1995), pp. 223{235.[7] S. Jiang, J.E. Mu~noz Rivera, R. Racke, Asymptotic stability and global existence inthermoelasticity with symmetry, Quart. Appl. Math., to appear.[8] S. Jiang, R. Racke, On some quasilinear hyperbolic-parabolic initial boundary valueproblems, Math. Meth. Appl. Sci., 12 (1990), pp. 315{339.[9] J.U. Kim, A boundary thin obstacle problem for a wave equation, Comm. PDE, 14 (1989),pp. 1011{1026.[10] R. Leis, Initial boundary value problems in mathematical physics. B.G. Teubner-Verlag,Stuttgart; John Wiley & Sons, Chichester et al., 1986.[11] J.E. Mu~noz Rivera, M. de Lacerda Oliveira, Exponential stability for a contactproblem in thermoelasticity, IMA J. Appl. Math., to appear.[12] R. Racke On the time-asymptotic behaviour of solutions in thermoelasticity, Proc. Roy.Soc. Edinburgh, 107A (1987), pp. 289{298.[13] P. Shi, M. Shillor, Existence of a solution to the n dimensional problem of thermoelasticcontact, Comm. PDE 17 (1992), pp. 1597{1618.[14] P. Shi, Y. Xu, Decoupling of the quasistatic system of thermoelasticity on the unit disk,J. Elasticity, 31 (1993), pp. 209-218.[15] S. Zheng Nonlinear parabolic equations and hyperbolic-parabolic coupled systems, PitmanMonographs Surv. Pure Appl. Math. 76, Longman; John Wiley & Sons, New York, 1995.Jaime E.Mu~noz Rivera, Department of Research and Development, National Laboratory for Scienti�cComputation, Rua Lauro M�uller 455, 22290 Rio de Janeiro, RJ, Brasile-mail: rivera@server01.lncc.brReinhard Racke, Faculty of Mathematics and Computer Science, University of Constance, P.O. Box5560, 78434 Konstanz, Germanye-mail: racke@dirichlet.mathe.uni-konstanz.de 38