Rothe method for parabolic variational–hemivariational inequalities

J. Math. Anal. Appl. 423 (2015) 841–862

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Rothe method for parabolic variational–hemivariational inequalities ✩ Krzysztof Bartosz a , Xiaoliang Cheng b , Piotr Kalita a , Yuanjie Yu b,∗ , Cong Zheng b a b

Faculty of Mathematics and Computer Science, ul. Łojasiewicza 6, 30-048 Krakow, Poland Department of Mathematics, Zhejiang University, Hangzhou 310027, PR China

a r t i c l e

i n f o

Article history: Received 6 June 2014 Available online 18 October 2014 Submitted by H. Frankowska Keywords: Parabolic problem Differential inclusion Rothe method Semidiscrete scheme Finite Element Method Convergence rate

a b s t r a c t The paper deals with the convergence analysis of the semidiscrete Rothe scheme for the parabolic variational–hemivariational inequality with the nonlinear pseudomonotone elliptic operator. The problem involves both a discontinuous and nonmonotone multivalued term as well as a monotone term with potentials which assume infinite values and hence are not locally Lipschitz. We prove the existence of a solution and establish a convergence result of a numerical semidiscrete scheme. The proof can be viewed both as the proof of solution existence as well as the proof of the convergence of a numerical semidiscrete scheme. The numerical simulations to present the rate of convergence with respect to space and time for piecewise linear finite elements are presented as well. © 2014 Elsevier Inc. All rights reserved.

1. Introduction In this paper we deal with the convergence of a semidiscrete (Rothe) scheme for the parabolic first order differential inclusion. The elliptic operator in the problem is assumed to be pseudomonotone. This fairly general class allows to consider the highest order term that is monotone and hemicontinuous (for example the p-Laplacian and operators of Leray–Lions type) and lower order terms that are totally continuous. Multivalued term in the inclusion has the form of the sum of two components: Clarke subdifferential of the functional which is locally Lipschitz and convex subdifferential of the functional that can have infinite values. This general setup of the problem allows to consider unilateral constraints (in the term coming from the convex subdifferential) without requirement of the monotonicity. We prove the convergence of the ✩ This research was supported by the Marie Curie International Research Staff Exchange Scheme Fellowship within the 7th European Community Framework Programme under Grant Agreement No. 295118, the Polish National Science Centre under Grant No. N N201 604640, the international project co-financed by the Ministry of Science and Higher Education of the Republic of Poland under Grant No. W111/7.PR/2012, the Polish National Science Centre under Maestro Advanced Project No. DEC-2012/06/A/ST1/00262, and the project Polonium, Grant Agreement No. 31155YH. * Corresponding author. E-mail addresses: [email protected] (K. Bartosz), [email protected] (X. Cheng), [email protected] (P. Kalita), [email protected] (Y. Yu), [email protected] (C. Zheng).

http://dx.doi.org/10.1016/j.jmaa.2014.09.078 0022-247X/© 2014 Elsevier Inc. All rights reserved.

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semidiscrete scheme and at the same time the existence of solutions for a new class of problems. Indeed, although there is a rather considerable literature on stationary variational–hemivariational inequalities (see for example [3,8,13,16,24]), there are only a few papers on evolution ones (see [4–6]) and in these papers a method of sub- and supersolutions is used, i.e., there is a need to assume the existence of sub- and supersolutions to obtain the existence of solution. Here we provide an alternative proof of existence that does not need this hypothesis. Existence results for hemivariational inequalities were obtained, for example, in [14,17–19]. The present paper can be considered as a generalization of these results to variational–hemivariational case. The proof is based on the approach by a priori estimates that, by reflexivity, allow us to deduce enough convergence to be able to pass to the limit with the approximate problems (see [23] for an application of this method for variational inequalities and [14] for hemivariational ones). The convergence and existence argument is accompanied by a numerical test that uses piecewise linear finite elements for the space discretization and primal–dual active set scheme (see [15]) for the solution of the algebraic inclusion. The tests demonstrate that the fully discrete scheme has the linear rate of convergence. Stationary variational–hemivariational inequalities with nonsmooth functionals whose structure is the sum of a locally Lipschitz functional and a lower semicontinuous, convex, and proper one, were studied in Chapter 3 in [21]. For the recent results on numerical methods for static variational inequalities see for example [25], for hemivariational ones see [2,10] and for variational–hemivariational ones see [3,9,13]. Results for evolution variational inequalities are presented in [12,26], while for hemivariational ones in [13,14,27]. Applications of the problem considered here include the temperature control problem (see [27]), the semipermeability problem (see [17]) as well as quasistatic problems in elasticity (see the classical monograph [12]). The structure of this paper is the following. In Section 2 we present the necessary definitions. Section 3 is devoted to the abstract problem formulation and assumptions. In Section 4 we formulate the semidiscrete scheme, prove the existence and a priori estimates on its solution. Section 5 contains the proof that the solutions of semidiscrete scheme converge and the limit is the solution of the original problem while Section 6 contains the proofs of additional properties of the solution such as the regularity and uniqueness. In the final Section 7 the numerical example is presented. 2. Preliminaries In this section we recall some preliminaries which we will refer to in the sequel. We start with the definitions of Clarke directional derivative and Clarke subdifferential. Let X be a Banach space, X ∗ its dual and let J : X → R be a locally Lipschitz functional. Definition 2.1. Generalized directional derivative in the sense of Clarke at the point x ∈ X in the direction v ∈ X, is defined by: J(y + λv) − J(y) . λ y→x,λ0

J 0 (x, v) = lim sup

(1)

Definition 2.2. Clarke subdifferential of J at x ∈ X is defined by   ∂Cl J(x) = ξ ∈ X ∗ | J 0 (x, v) ≥ ξ, vX ∗ ×X for all v ∈ X . For a convex functional we recall the definition of its subdifferential in the sense of convex analysis. Definition 2.3. Let X be a Banach space and Φ : X → R∪{∞} a convex functional. We define a subdifferential of Φ at x ∈ X by

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  ∂Conv Φ(x) = ξ ∈ X ∗ | Φ(x + v) − Φ(x) ≥ ξ, vX ∗ ×X for all v ∈ X . Moreover we use the notation dom Φ = {x ∈ X | Φ(x) < ∞} and dom ∂Conv Φ = {x ∈ X | ∂Conv Φ(x) = ∅}, note that dom ∂Conv Φ ⊂ dom Φ. Now we pass to the definition of a pseudomonotone operator. Definition 2.4. (See [28], Chapter 27.) Let X be a Banach space. A single valued operator A : X → X ∗ is called pseudomonotone, if for any sequence {vn }∞ n=1 ⊂ X such that vn → v weakly in X and lim supn→∞ Avn , vn − v ≤ 0 we have Av, v − y ≤ lim inf n→∞ Avn , vn − y for every y ∈ X. ∗

Definition 2.5. Let X be a real Banach space. The multivalued operator A : X → 2X is called pseudomonotone if the following conditions hold: 1) A has values which are nonempty, weakly compact and convex, 2) A is usc from every finite dimensional subspace of X into X ∗ furnished with weak topology, 3) if vn → v weakly in X and vn∗ ∈ A(vn ) is such that lim supn→∞ vn∗ , vn − v ≤ 0 then for every y ∈ X there exists u(y) ∈ A(v) such that u(y), v − y ≤ lim inf n→∞ vn∗ , vn − y. The following result can be found for example in [20] (see Proposition 3.58). ∗

Proposition 2.1. Let X be a real reflexive Banach space, and assume that A : X → 2X satisfies the following conditions 1) for each v ∈ X we have that A(v) is a nonempty, closed and convex subset of X ∗ , 2) A is bounded, i.e., it maps bounded sets into bounded ones, 3) if vn → v weakly in X and vn∗ → v ∗ weakly in X ∗ with vn∗ ∈ A(vn ) and if lim supn→∞ vn∗ , vn − v ≤ 0, then v ∗ ∈ A(v) and vn∗ , vn  → v ∗ , v. Then the operator A is pseudomonotone. Let X be a Banach space and I = (0, T ) be a time interval. We introduce the space BV (I; X) of functions of bounded total variation on I. Let π denote any finite partition of I by a family of disjoint subintervals n {σi = (ai , bi )} such that I¯ = i=1 σ ¯i . Let F denote the family of all such partitions. Then for a function x : I → X we define its total variation by

x BV (I;X) = sup



π∈F

 . X

  x(bi ) − x(ai )

σi ∈π

As a generalization of the above definition, for 1 ≤ q < ∞, we define a seminorm

x qBV q (I;X) = sup

π∈F

 σi ∈π

  x(bi ) − x(ai )q



X

and the space   BV q (I; X) = x : I → X; x BV q (I;X) < ∞ . For 1 ≤ p ≤ ∞, 1 ≤ q < ∞ and Banach spaces X, Z such that X ⊂ Z we introduce a vector space

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M p,q (I; X, Z) = Lp (I; X) ∩ BV q (I; Z). Then M p,q (I; X, Z) is also a Banach space with the norm given by · Lp (I;X) + · BV q (I;Z) . 3. Problem formulation Let V be a strictly convex, reflexive and separable Banach space and let V ∗ denote its dual. Let H be a separable Hilbert space and let V ⊂ H ⊂ V ∗ be an evolution triple with all embeddings being continuous, dense and compact. We denote by i the embedding between V and H. We denote by ·, · and · the duality pairing between V and V ∗ and a norm in V respectively. We also denote by (·, ·) and | · |H the scalar product and the norm in H. Moreover we denote by | · | the norm in Rn . Let U be another reflexive Banach space and let ·, ·U ∗ ×U and · U denote the duality pairing and the norm in U respectively. For T > 0 we define the time dependent spaces V = L2 (0, T ; V ), H = L2 (0, T ; H), U = L2 (0, T ; U ) and W = {u ∈ V | u ∈ V ∗ }, where u is the time derivative of u understood in the sense of distributions. Now we consider an abstract variational–hemivariational inequality of parabolic type. Problem (V). Find u ∈ W such that u(0) = u0 and for all v ∈ V, we have T

 



u (t) + Au(t) + ι∗ ξ(t) − f (t), v(t) − u(t) + Φ v(t) − Φ u(t) dt ≥ 0,

(2)

0

where ξ ∈ U ∗ is such that ξ(t) ∈ ∂Cl J(ιu(t)) for a.e. t ∈ (0, T ). We impose the following assumptions on the data of the problem. H(A): The operator A : V → V ∗ satisfies (i) (ii) (iii) (iv)

A is pseudomonotone, A satisfies the growth condition Av V ∗ ≤ a + b v for all v ∈ V with a ≥ 0, b > 0, A is coercive Av, v ≥ α v 2 − β v 2H for all v ∈ V with α > 0 and β ≥ 0, Au − Av, u − v ≥ m1 u − v 2 − m2 u − v 2H for all u, v ∈ V with m1 ≥ 0 and m2 ≥ 0.

H(J): The functional J : U → R satisfies (i) J is locally Lipschitz, (ii) ∂Cl J satisfies the growth condition ξ U ∗ ≤ c(1 + u U ) for all u ∈ U and ξ ∈ ∂Cl J(u) with c ≥ 0, (iii) −∞ < m(∂Cl J) :=

inf

u,v∈U,u=v ξ∈∂Cl J(u),η∈∂Cl J(v)

ξ − η, u − vU ∗ ×U .

u − v 2U

H(Φ): The functional Φ : V → R ∪ {+∞} is convex, proper and lower semicontinuous. H(U ): There exists a linear, continuous and compact mapping ι : V → U such that (i) the associated Nemytskii mapping ¯ι : M 2,2 (0, T ; V, V ∗ ) → U defined by (¯ιv)(t) = ι(v(t)) is also compact, (ii) for every > 0 there exists a constant C( ) > 0 such that for all v ∈ V we have

ιv U ≤ v + C( ) v H .

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H0 : f ∈ H 1 (0.T ; V ∗ ) and u0 ∈ dom ∂Conv Φ is such that the following compatibility condition holds: there exists ξ0 ∈ ∂Cl J(ιu0 ) and η0 ∈ ∂Conv Φ(u0 ) such that Au0 + ι∗ ξ0 + η0 − f (0) ∈ H. Remark 3.1. For a typical setup, where V is a closed subspace of H 1 (Ω), H = L2 (Ω) and U = L2 (Γ ), Γ ⊂ ∂Ω is a set of positive boundary measure, the assumption H(U )(ii) follows from the Ehrling lemma (cf. [14]). 4. The Rothe problem In this section we consider an approximate problem based on time discretization of Problem (V). It is known as the Rothe problem.  kτ T Let us fix N ∈ N and take τ = N . We define fτk = τ1 (k−1)τ f (t) dt for k = 1, . . . , N . We define the following Rothe problem. k ∞ ∗ 0 Problem (Vτ ). Find {ukτ }N k=0 ⊂ V and {ξτ }k=1 ⊂ U such that uτ = u0 and









 ukτ − uk−1 τ , v − ukτ + Aukτ , v − ukτ + ξτk , ι v − ukτ U ∗ ×U + Φ(v) − Φ ukτ τ H 

k for all v ∈ V, ≥ fτ , v − ukτ

(3)

with ξτk ∈ ∂Cl J(ιukτ ) and k = 1, . . . , N . Lemma 4.1. Under assumptions H(A), H(J), H(Φ), H(U ) and H0 , there exists τ0 > 0 such that Problem (Vτ ) has at least one solution for τ ∈ (0, τ0 ). Proof. The proof will be done in two steps. First, we formulate an auxiliary problem where we use the Yosida approximation of the functional Φ. Then, in Step 2, we pass to the limit in the Yosida approximation and recover the solution of original problem. Step 1. Given ε > 0 let Φε : V → R be the Yosida approximation of Φ defined by  Φε (u) = inf

v∈V



u − v 2 + Φ(v) 2ε

for u ∈ V.

We recall the following lemma (cf. e.g. [23], Lemma 5.15). Lemma 4.2. If H(Φ) holds then Φε is convex, lower semicontinuous and Gâteaux differentiable functional with a monotone, bounded and demicontinuous differential Φε : V → V ∗ . Moreover, we have lim sup Φε (v) ≤ Φ(v) ε→0

vε → v weakly in V

=⇒

for all v ∈ V, lim inf Φε (vε ) ≥ Φ(v). ε→0

(4) (5)

For a given ε > 0 we formulate an auxiliary problem corresponding to Problem (Vτ ). k ∞ ∗ 0 Problem (Vτ ε ). Find {ukτε }N k=0 ⊂ V and {ξτ ε }k=1 ⊂ U such that uτ ε = u0 and



ukτε − uk−1 τε ,v τ

 H







 + Aukτε , v + ξτkε , ιv U ∗ ×U + Φε ukτε , v = fτk , v



for all v ∈ V with ξτkε ∈ ∂Cl J ιukτε and k = 1, . . . , N.

(6)

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First we will show that the above problem has a solution. To this end, it is enough to prove that for a k given uk−1 τ ε there exists uτ ε ∈ V which satisfies (6). We will do this by showing that the multivalued operator V∗ L : V → 2 defined by L=

i∗ i + A + ι∗ ∂Cl J(ι·) + Φε τ

is surjective. We will use Theorem 1.3.70 in [11]. We show that L is pseudomonotone and coercive, i.e., inf v∗ ∈Lv v ∗ ,v

→ ∞ as v → ∞. v First, we show the coercivity of L. Let v ∈ V and v ∗ ∈ L(v). For some ξ ∈ ∂Cl J(ιv), we have

∗





v ,v =



i∗ i v, v τ

 + Av, v + ξ, ιvU ∗ ×U + Φε (v), v

H

 1

v 2H + α v 2 − β v 2H + ξ, ιvU ∗ ×U + Φε (v) − Φε (0) + Φε (0), v − 0 τ  

 1 ≥ − β v 2H + α v 2 + ξ, ιvU ∗ ×U + Φε (0), v τ     1 − β v 2H + α v 2 + ξ, ιvU ∗ ×U − Φε (0)V ∗ v . ≥ τ ≥

Using H(J)(ii) and H(U )(ii) we get ξ, ιvU ∗ ×U

 

1 c2

ιv 2U ≥ − ξ U ∗ ιv U ≥ −c 1 + ιv U ιv U ≥ − − c + 2 2  

1 c2 v 2 + C( ) v 2H ≥− − c+ 2 2

(7)

and  v∗ , v ≥





      1 1 c2  1 − β − C( ) c +

v 2H + α − c +

v 2 − − Φε (0)V ∗ v . τ 2 2 2

α We choose and τ such that = 2c+1 and τ1 ≥ β + C( )(c + 12 ). Then the operator L is coercive. For the proof of pseudomonotonicity of L, observe that the sum of two pseudomonotone operators is ∗ pseudomonotone (see Proposition 6.3.63 in [11]). Pseudomonotonicity of iτ i + A + ι∗ ∂Cl J(ι·) is proved in Lemma 2 in [14]. Now Φε is single valued and bounded, so conditions 1) and 2) of Proposition 2.1 hold. Since Φε is bounded, demicontinuous and monotone, then it is also pseudomonotone single valued operator (see Lemma 2.9 in [23]), and hence, by standard calculation it follows that condition 3) of Proposition 2.1 holds and hence Φε is a pseudomonotone multivalued operator.

Step 2. Passing to the limit. Let {εn } be a sequence of positive number such that εn → 0 as n → ∞. We will show, that for a subsequence of {εn } (which we denoted for simplicity by ε) we have ukτε → ukτ strongly in H for all k ∈ {0, . . . , N } and weakly in V for all k ∈ {1, . . . , N } as ε → 0 where {ukτ }N k=0 solves (3). We k−1 ∗ k proceed by induction. Assume that uk−1 → u strongly in H. Then, for some v ∈ Lu τε τ τ ε , we have

∗



v ,v =

Let w ∈ V and v = w − ukτε . Then



fτk , v



 +

uk−1 τε ,v τ

 for all v ∈ V. H

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862





 v ∗ , w − ukτε = fτk , w − ukτε +



uk−1 τε , w − ukτε τ

847

 . H

Moreover, for some ξ ∈ ∂Cl J(ιukτε ), we have





 1 k uτ ε , w − ukτε H + Aukτε , w − ukτε + ξ, ι w − ukτε U ∗ ×U + Φε ukτε , w − ukτε τ   k−1

 uτ ε , w − ukτε = fτk , w − ukτε + . τ H Hence, by convexity of Φε , we deduce



 1 k u , w − ukτε H + Aukτε , w − ukτε + ξ, ι w − ukτε U ∗ ×U + Φε (w) − Φε ukτε τ τε   k−1

k  uτ ε k k , w − uτ ε ≥ fτ , w − uτ ε + , τ H and we rewrite the last inequality as follows LHS :=





 1 k

1 ukτε 2 + Aukτε , ukτε + ξ, ιukτε ∗ uτ ε , w H + Φε ukτε ≤ H U ×U τ τ  k−1 

k 

k k  uτ ε , ukτε − w + Auτ ε , w + ξ, ιwU ∗ ×U + Φε (w) + fτ , uτ ε − w + := RHS. τ H

Using (4), Lemma 2.5 in [22], by the estimate analogous to the proof of coercivity, we get 

     2    k 2  k 2 1 1 1   uτ ε  − k1 ukτε  − k2 − c − β − C( ) c + uτ ε H + α − c + τ 2 2 2      1 uk 2 + 1 w 2H + a + bukτε  w + c 1 + ιukτε  ιw U ≤ LHS ≤ RHS ≤ U 2τ τ ε H 2τ  k   k 

1  k−1   k 

+ Φ(w) + fτ V ∗ uτ ε  + w + uτ ε H uτ ε H + w H . τ

Applying the Cauchy inequality with , we obtain RHS ≤

2 2 2 2    1 2 ukτε 2 + 1 w 2H + a w + ukτε 2 + b w 2 + c ιw U + ukτε 2 + c ι 4 L(V ;U ) w 2 H 2τ 2τ 2 2 2 2 2 2     2  2 1  2 1  w H + ukτε 2 + ukτε  + 2 fτk V ∗ + Φ(w) + w fτk V ∗ + uk−1 τε H 2 2 τ 2   1 2  . + 2 2 i 2L(V ;H) uk−1 τε H 2τ

Thus    1 ukτε 2 + 1 w 2H + 2 2 ukτε 2 + a w + b w 2 + c ιw U H 2τ 2τ 2 2      c2 1  2 1  w + 1 uk−1  . + 2 w 2 ι 4L(V ;U ) + 2 fτk V ∗ + w fτk V ∗ + uk−1 τε τε 2 H H 2 2 τ 2τ

RHS ≤ Φ(w) +

Let w ∈ dom Φ (this set is nonempty). Now, since uk−1 τ ε H is bounded, it is easily seen that 

          1 1  1 c2 1 k 2 2  k 2  uτ ε − − β − C( ) c + − uτ ε H + α − c + − 2 ≤D τ 2 2τ 2 2

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where the constant D does not depend on . By the appropriate choice of , for τ small enough (i.e., 1 1 k 2τ ≥ β + C( )(c + 2 )), we have that uτ ε is uniformly bounded with respect to τ . Hence for a subsequence k k we may suppose that uτ ε → uτ weakly in V , ukτε → ukτ strongly in H, and ιukτε → ιukτ strongly in U . It remains to show that ukτ satisfies (3). Let w ∈ V . Taking v = w − ukτε in (6), by the convexity of Φε , we have 







 ukτε − uk−1 τε , w − ukτε + Aukτε , w − ukτε + ξτkε , ι w − ukτε U ∗ ×U + Φε (w) τ H  k

k k ≥ fτ , w − uτ ε + Φε uτ ε ,

(8)

where ξτkε ∈ ∂Cl J(ιukτε ). We pass to the limit in (8). Since ιukτε is bounded in U , then by H(J)(ii) we have that ξτkε is bounded in U ∗ . Hence by reflexivity of U ∗ , we may assume that for a subsequence we have ξτkε → ξτk weakly in U ∗ as ε → 0. Since the graph of ∂Cl J is closed in (s − U ) × (w − U ∗ ) topology, we have ξτk ∈ ∂Cl J(ιukτ ). From (8) we obtain 







 ukτ − uk−1 τ , w − ukτ + lim sup Aukτε , w − ukτε + ξτk , ι w − ukτ U ∗ ×U + Φ(w) τ ε→0 H  k

k k ≥ fτ , w − uτ + Φ uτ .

(9)

Next we show that



 lim sup Aukτε , w − ukτε ≤ Aukτ , w − ukτ . ε→0

By the pseudomonotonicity of A, it is enough to prove that lim inf ε→0 Aukτε , ukτ − ukτε  ≥ 0. To this end, we take w = ukτ in (8) and obtain  ukτε − uk−1 τε , ukτ − ukτε τ H k k k k

 k + Φε uτ ε − Φε uτ − ξτ ε , ι uτ − uτ ε U ∗ ×U .





 Aukτε , ukτ − ukτε ≥ fτk , ukτ − ukτε −



Passing to the limit with ε → 0, we get

 lim inf Aukτε , ukτ − ukτε ≥ lim inf Φε ukτε − lim sup Φε ukτ ≥ Φ ukτ − Φ ukτ = 0, ε→0

ε→0

ε→0

and the proof is complete. 2 Now we derive an appropriate estimates for the solution of the Rothe problem. Lemma 4.3. If assumptions H(A), H(J), H(Φ), H(U ) and H0 hold, then there exists τ0 > 0 such that for all τ ∈ (0, τ0 ) the solution of Problem (Vτ ) satisfies   max ukτ H ≤ const,

k=1,...,N

N   k 2 uτ − uk−1  ≤ const, τ H

(10) (11)

k=1

τ

N   k 2 uτ  ≤ const, k=1

(12)

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

  k   uτ − uk−1 τ   max   k=1,...,N τ

849

≤ const,

(13)

H

2 N  k    uτ − uk−1 τ  ≤ const,  τ   τ

(14)

k=1

with the constant independent of τ . Proof. Taking v0 ∈ dom Φ as a test function in (3), we have 



 ukτ − uk−1 τ , ukτ + Aukτ , ukτ + ξτk , ιukτ U ∗ ×U + Φ ukτ τ H   k k−1 

 



u τ − uτ , v0 + Aukτ , v0 + ξτk , ιv0 U ∗ ×U + Φ(v0 ) + fτk , ukτ − v0 := RHS. ≤ τ H 

LHS :=

Using Lemma 2.5 in [22] as well as the identity (a −b, a)H = 12 a 2H − 12 b 2H + 12 a −b 2H , for a, b ∈ H and assumptions H(A)(iii), H(J)(ii) and H(U )(ii) by an argument analogous to the one used in the estimate (7), we obtain that LHS ≥

2        2 2 1  ukτ 2 − uk−1  + ukτ − uk−1  + αukτ 2 − β ukτ 2 − c τ τ H H H H 2τ 2       

1  2 2 − c+ ukτ  + C( )ukτ H − k1 ukτ  − k2 , 2

where k1 , k2 are positive constants. Subsequently, it follows that  RHS ≤ Φ(v0 ) +

ukτ − uk−1 τ , v0 τ



    + v0 a + bukτ  + c ι L(V ;U ) v0 1 + ι L(V ;U ) ukτ  H

      + fτk V ∗ v0 + fτk V ∗ ukτ  ≤





1  2 + Φ(v0 ) + a v0 + c ι L(V ;U ) v0 + fτk V ∗ 2 H     1 2 2 + v0 2 + ukτ  + C( ) b2 v0 2 + c2 ι 4L(V ;U ) v0 2 + fτk V ∗ . 2 ukτ − uk−1 τ , v0 τ

From the last two inequalities, we get         2  2 1 1 1 −β− c+ C( ) ukτ H + α − c + − 2 ukτ  2τ 2 2   k k−1 2 1  + C2 ( ) v0 2 + uτ − uτ , v0 uk−1 + + Φ(v0 ) + C3 ( ), τ H 2τ τ H

2 1  + ukτ − uk−1 τ H 2τ  2 ≤ C1 ( )fτk V ∗



where C( ), C1 ( ), C2 ( ), C3 ( ) are positive constants dependent only on and the constants that appear in the assumptions of the problem. Choosing > 0 small enough such that α > (c + 12 ) + 2 and multiplying by 2τ , we arrive at         k 2 2

uτ − uk−1  + (1 − C4 τ )ukτ 2 + C5 τ ukτ 2 ≤ C6 τ fτk 2 ∗ + uk−1  + 2 ukτ − uk−1 , v0 H + C7 τ, τ τ τ H H V H where Ci > 0, for i = 4, 5, 6, 7, are positive constants. Fixing n ∈ {1, . . . , N } and summing up the above inequalities from k = 1 to n, we obtain n n N n      k 2   k 2  k 2  k 2 1 n 2 uτ − uk−1      uτ  + C8 ,  u f u + + C τ ≤ C τ + C τ 5 6 4 τ τ H τ τ V∗ H H 2

k=1

k=1

k=1

k=1

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where C8 > 0. By a straightforward computation, we have τ

N k=1

fτk 2V ∗ ≤ f 2V ∗ which entails that

n n n     k 2   k 2  k 2 1 n 2 uτ − uk−1    uτ  + C9 ,  u u + + C τ ≤ C τ 5 4 τ τ H τ H H 2

k=1

k=1

k=1

with C9 > 0. Choosing τ < 4C1 4 in the last inequality, by the discrete Gronwall inequality (see (1.68)–(1.69) in [23]), we obtain the estimates (10)–(12). ∗ We pass to the proof of (13) and (14). Let us define u−1 τ = u0 +τ (Au0 +ι ξ0 +η0 −f (0)), η0 ∈ ∂Conv Φ(u0 ), and ξ0 ∈ ∂Cl J(u0 ). We rewrite the last inequality as, u0τ − u−1 τ + Au0 + ι∗ ξ0 + η0 = f (0). τ

(15)

From (3), we have for k = 1, . . . , N 





 ukτ − uk−1 τ k−1 k , uτ − uτ + Aukτ , uk−1 − ukτ + ξτk , ι uk−1 − ukτ U ∗ ×U τ τ τ H

k k k−1  + Φ uk−1 − Φ u ≥ fτ , uτ − ukτ with ξτk ∈ ∂Cl J ιukτ τ τ

and, defining fτ0 = f (0), 



 k−1 k

 − uk−2 uk−1 τ τ k k−1 , uτ − uτ + Auk−1 , ukτ − uk−1 + ξτ , ι uτ − uk−1 τ τ τ U ∗ ×U τ H

k−1 k 

+ Φ ukτ − Φ uk−1 ≥ fτ , uτ − uk−1 with ξτk−1 ∈ ∂Cl J ιuk−1 , τ τ τ

which, for k = 1 follows from (15), and for k = 2, . . . , N from (3). Adding the two inequalities, we obtain 



 −uk−1 + uk−2 + ukτ − uk−1 τ τ τ k k−1 , uτ − uτ + Aukτ − Auk−1 , ukτ − uk−1 τ τ τ H

k k



k  k−1 k−1 + ξτ − ξτ , ι uτ − uτ + ≤ fτ − fτk−1 , ukτ − uk−1 . τ U ∗ ×U

Next from H(J)(iii), we have m(∂Cl J) ≤

ξ − η, u − vU ∗ ×U

u − v 2U

for all u, v ∈ U, u = v, ξ ∈ ∂Cl J(u), η ∈ ∂Cl J(v).

Hence we obtain  k   k−1  k−1 2 k−2 2    2  2

2 τ  uτ − uτ  − τ  uτ − uτ  + m1 ukτ − uk−1   − m2 ukτ − uk−1  + m(∂Cl J)ι ukτ − uk−1 τ τ τ    H 2 τ 2 τ H H  2    + 1 fτk − fτk−1 2 ∗ . ≤ ukτ − uk−1 τ V 4 Using H(U )(ii), we deduce  k   k−1  k−1 2 k−2 2  2  τ  uτ − uτ  − τ  uτ − uτ  + (m1 − )ukτ − uk−1  τ     2 τ 2 τ H H     2  2

2

2 1  + C( ) ukτ − uk−1  ,  + m(∂Cl J)  ukτ − uk−1 ≤ fτk − fτk−1 V ∗ + m2 ukτ − uk−1 τ τ τ H H 4

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and  k   k−1  k−1 2 k−2 2     2 τ  uτ − uτ  − τ  uτ − uτ  + m1 − − m(∂Cl J) ukτ − uk−1  τ    2 τ 2 τ H H   2 2

 1  . ≤ fτk − fτk−1 V ∗ + m2 + m(∂Cl J)C( ) ukτ − uk−1 τ H 4 We choose such that 0 < <

m1 1+|m(∂Cl J)| .

(16)

Then m1 − − |m(∂Cl J)| > 0. Then (16) becomes

 k   k−1  k−1 2 k−2 2   2    2 τ  uτ − uτ  − τ  uτ − uτ  + D1 ukτ − uk−1  ≤ D2 fτk − fτk−1 2 ∗ + D3 ukτ − uk−1  . τ τ     V H 2 τ 2 τ H H Summing up, from k = 1 to n, we get 2 2  n n  k     uτ − un−1  uτ − uk−1   τ τ  + 2D1 τ  ≤ Au0 + ι∗ ξ0 + η0 − f (0)2       H τ τ H

+ 2D2 τ

k=1

n  

 fτk  

k=1

2 fτk−1  

− τ



V∗

2 n  k    uτ − uk−1 τ  .  + 2D3 τ   τ

(17)

H

k=1

Taking f (t) = f (0) for t ∈ (−τ, 0), we extend f to the interval (−τ, T ) in a way such that f ∈ H 1 (−τ, T ; V ∗ ). By the definition of fτk , we have  kτ 2 (k−1)τ   2 1 1   k k−1 fτ − fτ  ∗ = f (t)dt − f (t)dt   V  ∗ τ2 τ4  (k−1)τ

V

(k−2)τ

2  kτ  1 1   = 4 f (t) − f (t − τ )dt ≤ 2   τ τ ∗ (k−1)τ

V

kτ τ

   f (t − s)2 ∗ dsdt. V

(k−1)τ 0

So  kτ τ τ n kτ n  k n   fτ − fτk−1 2     2  1 1     f (t − s)2 ∗ dtds  f (t − s) V ∗ dsdt = τ ≤   V τ τ τ V∗ k=1

k=1(k−1)τ 0

1 ≤ τ

τ T 0

   f (t − s)2 ∗ dtds ≤ 1 V τ

0

0 k=1(k−1)τ

τ T −s   2   f (t) ∗ dtds = f  2 ∗ . V

V

0 (−s)

Since f  (t) = 0 for a.e. t ∈ (−τ, 0), from (17), we get 2 2  n n  k     uτ − un−1  uτ − uk−1   τ τ  + 2D1 τ  ≤ Au0 + ι∗ ξ0 + η0 − f (0)2 + 2D2 f  2V ∗       H τ τ H k=1 2 n  k    uτ − uk−1 τ  ,  + 2D3 τ   τ H k=1

and finally

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2 2  n n  k     uτ − un−1  uτ − uk−1 τ τ  ≤ D4 + 2D3 τ  .       τ τ H

k=1

H

Consequently, by the Gronwall inequality (see (1.68)–(1.69) in [23]), it follows that there exists τ0 > 0 1 (it suffices to take 0 < τ0 < 2D ) such that for all τ ∈ (0, τ0 ), the estimations (13) and (14) are satisfied, 3 which completes the proof of the lemma. 2 5. Convergence of the Rothe method In this section we construct sequences of time dependent piecewise constant and piecewise linear interpolants of the solution of Problem (V). Next we study their convergence to the exact solution. Let the functions uτ : [0, T ] → V and u ¯τ : [0, T ] → V be defined by the formulae  u ¯τ (t) =

ukτ , t ∈ ((k − 1)τ, kτ ] u0τ , t = 0

and  uτ (t) =

ukτ

+



t − k ukτ − uk−1 τ τ

 for t ∈ (k − 1)τ, kτ .

The piecewise constant interpolant ξ¯τ : (0, T ] → U ∗ is given by ξ¯τ (t) = ξτk

 for t ∈ (k − 1)τ, kτ .

Moreover, we define fτ : (0, T ] → V ∗ as follows fτ (t) = fτk

 for t ∈ (k − 1)τ, kτ .

By Lemma 3.3 from [7], we know that fτ → f strongly in V ∗ as τ → 0. We observe that (3) can be written equivalently as







 uτ (t), v − u ¯τ (t) H + A¯ uτ (t), v − u ¯τ (t) + ξ¯τ (t), ¯ι v − u ¯τ (t) U ∗ ×U



 + Φ(v) − Φ u ¯τ (t) ≥ fτ (t), v − u ¯τ (t) a.e. t ∈ (0, T ) for all v ∈ V.

(18)

Defining the Nemytskii operator A : V → V ∗ by (Av)(t) = A(v(t)), we have

uτ , v − u ¯τ T +

H

 + A¯ uτ − fτ , v − u ¯τ V ∗ ×V + ξ¯τ , ¯ι(v − u ¯τ ) U ∗ ×U





Φ v(t) − Φ u ¯τ (t) dt ≥ 0 for all v ∈ V.

(19)

0

Lemma 5.1. Under the assumptions H(A), H(J), H(Φ), H(U ) and H0 , there exists τ0 > 0 such that for all τ ∈ (0, τ0 ), we have

¯ uτ L∞ (0,T ;H) + ¯ uτ M 2,2 (0,T ;V,V ∗ ) ≤ const,  

uτ C(0,T ;H) + uτ V + uτ V ≤ const,

(20)

ξ¯τ U ∗ ≤ const,

(22)

with the constants independent of τ .

(21)

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

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Proof. From (10), it follows that ¯ uτ L∞ (0,T ;H) ≤ const. Let us assume that the semi-norm BV 2 (0, T ; V ∗ ) of piecewise constant function u ¯τ is realized by some division 0 = a0 < a1 < . . . < an = T , and each ai i is in different interval ((mi − 1)τ, mi τ ] such that u ¯τ (ai ) = um with m0 = 0, mn = N and mi+1 > mi for τ i = 1, . . . , N − 1. Thus by (14) we obtain

¯ uτ 2BV 2 (0,T ;V ∗ )

 n n    m 2 m i i−1 uτ − uτ  ∗≤ = (mi − mi−1 ) V i=1

i=1

 n  n   ≤ (mi − mi−1 ) i=1

= Tτ

mi 

 k 2 uτ − uk−1  ∗ τ



V

k=mi−1 +1 mi 

 k 2 uτ − uk−1  ∗ τ V

 =N

i=1 k=mi−1 +1

N   k 2 uτ − uk−1  ∗ τ V

k=1

2 2 N  k N  k      uτ − uk−1  uτ − uk−1 τ τ  ≤ CT τ  ≤ const,       τ τ ∗ V

k=1

(23)

k=1

and by (12) we have

¯ uτ 2V

kτ N N    k 2  k 2   uτ  ≤ const. uτ dt = τ = k=1 (k−1)τ

(24)

k=1

Consequently, u ¯τ is bounded in M 2,2 (0, T ; V, V ∗ ) which together with (10), gives us (20). For the proof of (21), we have

uτ 2V

   2   kτ  kτ  N N    k  k   k 2 k

2 t t k−1  k−1 2     u τ − uτ dt uτ + −k =  dt ≤ 2 uτ + τ − k uτ − uτ τ k=1 (k−1)τ

k=1 (k−1)τ

 2  kτ  N N    k  k 2 2 t k−1 uτ − uτ  uτ  + 2 ≤ 2τ − k dt τ k=1

≤ 2τ

k=1

(k−1)τ

N N   k 2 2τ   k 2 uτ  + uτ − uk−1  ≤ const, τ 3

k=1

k=1

and the assertion follows from (10), (12), (14), and the fact that 2 N  k    uτ − uk−1   τ  .  uτ  = τ   V τ k=1

Finally, for the proof of (22) observe that

ξ¯τ 2U ∗ =

N N N     2   2  k 2 uτ  ≤ const, τ ξτk U ∗ ≤ τ c2 1 + ιukτ U ≤ 2T c2 + 2c2 τ ι 2 k=1

k=1

k=1

which completes the proof of the lemma. 2 Theorem 5.1. Let {τn } be a sequence such that τn → 0. For a subsequence, still denoted by τ we have

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

854

u ¯τ → u u ¯τ → u

weakly in V,

(25)

strongly in H,

(26)

strongly in U,

(27)

¯ιu ¯τ → ¯ιu

uτ

∗

ξ¯τ → ξ

weakly in U ,

(28)

uτ → u

weakly in V,

(29)

weakly in V and H,

(30)

→u



where u and ξ satisfy Problem (V). Proof. The convergence (25) follows from the estimate (20) and the reflexivity of V. To get (26) we use the bound (20) in M 2,2 (0, T ; V, V ∗ ) and Proposition 2 in [14]. The same bound (20) in M 2,2 (0, T ; V, V ∗ ) together with H(U )(i) gives the convergence (27). The convergence (28) follows from the estimate (22). Moreover, since ξ¯τ (t) ∈ ∂Cl J(¯ιu ¯τ (t)) for a.e. t ∈ (0, T ) from (27), (28), and the convergence theorem of Aubin and Cellina (see [1]), we have

ξ(t) ∈ ∂Cl J ¯ιu(t)

for a.e. t ∈ (0, T ).

Since, by (21) uτ is bounded in V, for a subsequence, we may assume that uτ → u1 weakly in V. From (25), it follows that u ¯τ − uτ → u − u1 weakly in V. We observe that

¯ uτ −

uτ 2V

=

kτ N 



 k 2  uτ

(kτ − t) 

k=1 (k−1)τ

2 2   − uk−1 τ  dt = τ uτ 2 ,  V τ 3 V

and from the bound on uτ in (21) we get u = u1 and (29) follows. The same bound on uτ in (21) implies (30). From (29) and (30) it follows that uτ (0) → u(0) weakly in H. Hence, since uτ (0) = u0 for all τ we get u(0) = u0 . Now we pass to the limit with τ → 0 in (19). Let v ∈ V. From (26) and (30), we have

¯τ uτ , v − u

H



→ u , v − u H .

(31)

Since fτ → f strongly in V ∗ , by (25), we have fτ , w − u ¯τ V ∗ ×V → f, w − uV ∗ ×V .

(32)

From (27) and (28), we get



ξ¯τ , ¯ι(v − u ¯τ )

U ∗ ×U

 → η, ¯ι(v − u) U ∗ ×U .

(33)

T We define Ψ : V → R ∪ {+∞} by Ψ (v) = 0 Φ(v(t))dt. Since Φ is convex, then so is Ψ . We show that Ψ is lower semicontinuous. To this end let un → u strongly in V. Since Φ is convex there exist k1 , k2 ∈ R such that for v ∈ V we have Φ(v) ≥ k1 v V + k2 . It follows that T 0



Φ un (t) dt ≥ k2 T + k1

T

√   un (t) dt ≥ k2 T − |k1 | T un V ≥ const. V

(34)

0

Let us take any convergent subsequence of Ψ (un ), i.e., Ψ (un ) → M as n → ∞. For another subsequence, we have unk (t) → u(t) strongly in V for a.e. t ∈ (0, T ). From lower semicontinuity of Φ, we have

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862





Φ u(t) ≤ lim inf Φ unk (t)

855

a.e. t ∈ (0, T ).

n→∞

Consequently, by (34), we can use the Fatou lemma, and get T



Φ u(t) dt ≤

0

T



lim inf Φ unk (t) dt ≤ lim inf nk →∞

T

nk →∞

0



Φ unk (t) dt = M,

0

hence Ψ is lower semicontinuous. Since it is also convex, then it is weakly sequentially lower semicontinuous. From (25), we have Ψ (u) ≤ lim inf τ →0 Ψ (¯ uτ ), and T lim sup τ →0





Φ v(t) − Φ u ¯τ (t) dt ≤

0

T





Φ v(t) − Φ u(t) dt

(35)

0

which proves that Ψ is weakly lower semicontinuous. Next, we will prove that for all v ∈ V lim supA¯ uτ , v − u ¯τ V×V ∗ ≤ Au, v − uV×V ∗

(36)

lim inf A¯ uτ , u ¯τ − vV×V ∗ ≥ Au, u − vV×V ∗ .

(37)

lim supA¯ uτ , u ¯τ − uV×V ∗ ≤ 0.

(38)

τ →0

or equivalently τ →0

We will prove that

n→∞

Indeed, taking v = u in (19), we get

 A¯ uτ , u ¯τ − uV ∗ ×V ≤ uτ , u − u ¯τ H + ξ¯τ , ¯ι(u − u ¯τ ) U ∗ ×U + Ψ (u) − Ψ (¯ uτ ) + fτ , u ¯τ − uV ∗ ×V . From (31)–(33) by the sequential weak lower semicontinuity of Ψ , it follows that lim supA¯ uτ , u ¯τ − uV ∗ ×V ≤ Ψ (u) − lim inf Ψ (¯ uτ )dt ≤ 0, τ →0

τ →0

(39)

and we obtain (38). From (38), H(A) and the fact that {¯ uτ } remains in a bounded subset of M 2,2 (0, T ; V, V ∗ ) using an argument of Lemma 1 in [14] we deduce (37). Finally, from (31)–(33), (35) and (36), we pass to the limit in (19) and conclude that u and ξ solve Problem (V), which completes the proof. 2 6. Uniqueness and regularity of solutions In this section we formulate and prove two theorems providing both a uniqueness and an appropriate regularity of the solution of Problem (V). Theorem 6.1. Under assumptions H(A), H(J), H(Φ), H(U ) and H0 , the solution of Problem (V) is unique. Proof. Suppose u1 and u2 are two solutions of Problem (V). Taking u1 as v in (2) written for u2 and u2 as v in (2) written for u1 , and adding the two resulting inequalities, we get

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

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u1 − u2 , u2 − u1

H

 + Au1 − Au2 , u2 − u1 V ∗ ×V + ξ1 − ξ2 , ¯ι(u2 − u1 ) U ∗ ×U ≥ 0,

(40)

where ξ1 (t) ∈ ∂J(ιu1 (t)) and ξ2 (t) ∈ ∂J(ιu2 (t)) for a.e. t ∈ (0, T ). By H(A)(iv) and H(J)(iii), we obtain  1 u1 (T ) − u2 (T )2 + H 2

T



2 2   m1 u1 (t) − u2 (t) − m2 u1 (t) − u2 (t)H dt

0

T −

   m(∂Cl J)ιu1 (t) − ιu2 (t)2 dt ≤ 0. U

0

By H(U )(ii), we have  1 u1 (T ) − u2 (T )2 + H 2

T

2  m1 u1 (t) − u2 (t) dt ≤

0

  + m(∂Cl J)

T

T

2  m2 u1 (t) − u2 (t)H dt

0

2  2  u1 (t) − u2 (t) + C( )u1 (t) − u2 (t)H dt,

0

with C( ) > 0. Choosing > 0 such that |m(∂Cl J)| = m1 , we get   u1 (T ) − u2 (T )2 ≤ D H

T

  u1 (t) − u2 (t)2 dt, H

0

where D > 0 is a constant. Repeating this argument for s ∈ [0, T ] in place of T , we get   u1 (s) − u2 (s)2 ≤ D H

s

  u1 (t) − u2 (t)2 dt. H

0

Consequently, from the Gronwall lemma, we obtain u1 = u2 , which completes the proof. 2 Theorem 6.2. Under assumptions H(A), H(J), H(Φ), H(U ) and H0 , the unique solution u of Problem (V) satisfies u ∈ V and u : [0, T ] → V is Hölder continuous with the exponent 1/2. Proof. From (30) we have u ∈ V. Now taking t1 , t2 ∈ [0, T ] it follows that   u(t1 ) − u(t2 ) ≤

 T  12 t2      2 u (s) ds ≤ |t1 − t2 | 12 u (s) ds . t1

(41)

0

The proof is complete. 2 7. Numerical simulations In this section we present the results of a numerical experiment for the proposed problem. More specifically, we take Ω = (0, 1) × (0, 1), ΓS = {0} × (0, 1) and ΓD = ∂Ω \ ΓS . The problem under consideration is as follows

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Fig. 1. Graph of the nonmonotone and multivalued boundary condition.

∂u(x, t) − Δu(x, t) = f (x, t) ∂t u(x, t) = 0





∂u(x, t) − ∈ M u(x, t) = ∂Cl j u(x, t) + ∂Conv φ u(x, t) ∂n u(x, 0) = 0

in Ω × [0, T ],

(42)

on ΓD × [0, T ],

(43)

on ΓS × [0, T ],

(44)

in Ω.

(45)

The problem constitutes a heat equation with a multivalued feedback control law on the boundary ΓS . Indeed the heat flow through this boundary changes with temperature u as given by the multifunction M . The discontinuities (represented by vertical lines) in the graph of M correspond to changes of the controller behavior once the temperature has a certain threshold value (see [27]). The convex component represents the impenetrable lower or upper bound (or both) that the temperature cannot exceed. In the following numerical example, we take T = 0.5 and f = −30. The example of multivalued function M : R → 2R is given by ⎧ ∅ ⎪ ⎪ ⎪ ⎪ (∞, 0] ⎨ M (r) = {−2 − r} ⎪ ⎪ ⎪ [−1, 0] ⎪ ⎩ {0}

for for for for for

r r r r r

∈ (−∞, −2), = −2, ∈ (−2, −1), = −1, ∈ (−1, +∞),

and the graph of M is presented in Fig. 1. We decompose M into the following two multifunctions ⎧ for r ∈ (−∞, −2), ⎨∅ ∂Conv φ(r) = (∞, 0] for r = −2, ⎩ {0} for r > −2,

⎧ {0} for r ∈ (−∞, −2), ⎪ ⎪ ⎨ {−2 − r} for r ∈ (−2, −1), ∂Cl j(r) = ⎪ [−1, 0] for r = −1, ⎪ ⎩ {0} for r ∈ (−1, +∞).

Now we have  φ(r) =

+∞ for r ∈ (−∞, −2), 0 for r ∈ [−2, +∞),

⎧ for r ∈ (−∞, −2), ⎨0 j(r) = − 12 r2 − 2r − 2 for r ∈ [−2, −1), ⎩ 1 for r ∈ [−1, +∞). −2

Note that the defined problem is consistent with the abstract scheme defined previously. Indeed let V = {v ∈ H 1 (Ω) | v = 0 on ΓD }, H = L2 (Ω), U = L2 (ΓS ), Au, v = (∇u, ∇v)L2 (Ω;R2 ) and ι : V → U be

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

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Fig. 2. Domain Ω with ΓS and ΓD , and the cross mesh triangulation.

 the trace operator. Moreover, Φ, J : U → R ∪ {+∞} are defined by Φ(u) = ΓS φ(u(x)) dΓ (x), J(u) =  j(u(x)) dΓ (x), respectively. Then the assumptions H(A), H(J), H(Φ), H(U ) and H0 hold. ΓS We will use the implicit Euler scheme for the time discretization and first order finite elements for the space discretization. Let Th be a family of regular triangulations of the domain Ω with simplicial elements, and Sh the triangulation of ΓS , which is compliant with Th , i.e., elements of Sh are the edges of triangles from Th . Now Vh will be standard piecewise linear finite element space over the triangulation Th defined by   Vh = v ∈ C(Ω) | ∀K ∈ Th , v|K ∈ P1 (K), v|ΓS = 0 . Moreover, Wh is the space of the piecewise constant functions defined on the boundary ΓS , i.e.   Wh = v : ΓS → R | ∀K ∈ Sh , v|K ∈ P0 (K) . Finally, Πh : Vh → Wh denotes the 0-th order quasi-interpolation operator on the boundary ΓS defined by Πh (v)(x) =

 K∈Sh

χK (x)

1 |K|

v(x) dΓ (x) for v ∈ Vh . K

In our implementation, we use the cross mesh triangulation as shown in Fig. 2 and time mesh of equidistant intervals. Now the fully discrete problem to be solved in each time step is given by n ∈ Wh such that for all vh ∈ Vh Problem (Vhτ ). Find unhτ ∈ Vh and ξhτ



unhτ − un−1 hτ , vh τ

 L2 (Ω)

n ξhτ



n

+ ∇unhτ , ∇vh L2 (Ω;R2 ) + ξhτ , Πh (vh ) L2 (Γ u0hτ = 0,



∈ ∂Cl J Πh unhτ + ∂Conv Φ Πh unhτ .

S)

= (f, vh )L2 (Ω) ,

(46) (47) (48)

The following iterative scheme based on the primal–dual active set approach (see [15]) is used to find (m) (m) (m) solutions of Problem (Vhτ ) in every time step. All edges of Sh are divided into four sets A1 , A2 , A3 (m) and A4 , where m is the number of the inner iteration step. Initially, m = 0 and we assume that the sets (0) are the same as in the previous time step. In the first time step, we assume that A1 = Sh while all other (m) n sets are empty. On the edges of A1 , we assume that ξhτ = 0 (which corresponds to horizontal line in

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

859

Fig. 3. Evolution of u with respect to time for the finest mesh. Plots (A)–(G) present the values of u after 4, 12, 18, 24, 30, 42 and 64 time steps respectively. Plot (H) presents the evolution of the solution on boundary ΓS where the four shades of gray represent four lines in the graph of the multifunction M .

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

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Fig. 4. Dependence of the L∞ norm error on the length of the space step h and the time step τ .

Fig. 5. Dependence of the energy norm error on the length of the space step h and the time step τ . (m)

the graph in Fig. 2), on the edges of A2 , we assume that Πh unhτ = −1 (which corresponds to vertical (m) n line at u = −1 in the graph in Fig. 2), on the edges of A3 , we assume that ξhτ = −2 − Πh unhτ (which (m) corresponds to oblique line at u = −1 in the graph in Fig. 2), and finally on the edges of A4 , we assume that Πh unhτ = −2 (which corresponds to the vertical line at u = −2 in the graph in Fig. 2). This gives the dim Wh conditions, which together with (46) constitute a linear system of dim Wh + dim Vh linear equations n(m) n(m) (m+1) (m) that to be solved for uhτ and ξhτ . We assume that Ai = Ai and we modify the sets according to the following scheme: (m)

n(m)

(m+1)

(m+1)

(1) if K ∈ A1 and Πh uhτ |K < −1, then K is removed from A1 and added to A2 (m) n(m) (m+1) (m+1) (2) if K ∈ A2 and ξhτ |K > 0, then K is removed from A2 and added to A1 , (m) n(m) (m+1) (m+1) (3) if K ∈ A2 and ξhτ |K < −1, then K is removed from A2 and added to A3 ,

,

K. Bartosz et al. / J. Math. Anal. Appl. 423 (2015) 841–862

(m)

n(m)

(m+1)

861

(m+1)

(4) if K ∈ A3 and Πh uhτ |K > −1, then K is removed from K ∈ A3 and added to A2 , (m) n(m) (m+1) (m+1) (5) if K ∈ A3 and Πh uhτ |K < −2, then K is removed from K ∈ A3 and added to A4 , (m) n(m) (m+1) (m+1) (6) if K ∈ A4 and ξhτ |K > 0, then K is removed from K ∈ A4 and added to A3 . Finally, we increase m. We stop the iteration if no edges are transferred from one set to another. Note that the scheme, if it converges, achieves a fixed point in a finite number of iterations, and a fixed point is guaranteed to be a solution of Problem (Vhτ ). The following simulations were run: 1 1 (1) The test for the finest mesh. We took τ = 128 and h = 64 . The results are shown in Fig. 3. 1 1 1 (2) Convergence rate with respect to a space step. We took h = 14 , 18 , 16 , 32 and τ = 128 and compared the 1 − solution with the one for the finest mesh obtained in the simulation (1). We denoted by eh the error uh, 128 ∞ 1 1 , where the norm is either L 2 2 u 64 (Ω × (0, T )) or the energy norm given as max

∇u . t∈[0,T ] L (Ω;R ) We , 128 ∞ list the results in the following table (also see Fig. 4 for the errors in L norm and Fig. 5 for the errors in energy norm).

h ∞

eh (L norm) eh (energy norm)

1 4

1 8

1 16

1 32

0.2800 1.4702

0.0802 0.7665

0.0357 0.3770

0.0173 0.1682

1 1 1 1 (3) Convergence rate with respect to time step. We took h = 64 and τ = 18 , 16 , 32 , 64 . As in the previous case we compared the solution with the one for the finest mesh obtained in the simulation (1). Similarly, 1 , and consider the same norms as for the simulation (2). 1 1 we denote the error eτ by eτ = u 64 ,τ − u 64 , 128 The results are presented in the following table (also see Fig. 4 for the errors in L∞ norm and Fig. 5 for the errors in energy norm).

τ ∞

eτ (L norm) eτ (energy norm)

1 8

1 16

1 32

1 64

0.5073 1.1811

0.3027 0.7041

0.1433 0.3281

0.0532 0.1269

The simulations indicate that the convergence rate is linear both with respect to time and space step length. Acknowledgment We thank the referee for valuable comments and remarks. References [1] J.P. Aubin, H. Frankowska, Set-Valued Analysis, Birkhäuser, Boston, Basel, Berlin, 1990. [2] M. Barboteu, K. Bartosz, P. Kalita, An analytical and numerical approach to a bilateral contact problem with nonmonotone friction, Int. J. Appl. Math. Comput. Sci. 23 (2013) 263–276. [3] M. Barboteu, K. Bartosz, P. Kalita, A. Ramadan, Analysis of a contact problem with normal compliance, finite penetration and nonmonotone slip dependent friction, Commun. Contemp. Math. 16 (2014) 1350016 (29 pages). [4] S. Carl, Existence and extremal solutions of parabolic variational hemivariational inequalities, Monatsh. Math. 172 (2013) 29–54. [5] S. Carl, V.K. Le, Quasilinear parabolic variational inequalities with multi-valued lower order terms, Z. Angew. Math. Phys. 65 (2014) 845–864. [6] S. Carl, D. Motreanu, Directedness of solution set for some quasilinear multi-valued parabolic problems, Appl. Anal. 89 (2010) 161–174. [7] C. Carstensen, J. Gwinner, A theory of discretisation for nonlinear evolution inequalities applied to parabolic Signorini problems, Ann. Mat. Pura Appl. (4) 177 (1999) 363–394. [8] N. Costea, C. Lupu, On a class of variational–hemivariational inequalities involving set valued mappings, Adv. Pure Appl. Math. 1 (2010) 233–246. [9] J. Czepiel, P. Kalita, Numerical solution of a variational hemivariational inequality modelling simplified adhesion of an elastic body, IMA J. Numer. Anal. (2013), http://dx.doi.org/10.1093/imanum/drt058, in press. [10] M.N. Dao, J. Gwinner, D. Noll, N. Ovcharova, Nonconvex bundle method with application to a delamination problem, arXiv:1401.6807, 2014.

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