©2022 Ada Academica https://adac.eeEur. J. Math. Anal. 2 (2022) 5doi: 10.28924/ada/ma.2.5 Unilateral Problem for a Viscoelastic Beam Equation Type p-Laplacian with Strong Damping and Logarithmic Source Ducival C. Pereira1, Geraldo M. de Araújo2, Carlos A. Raposo3,∗ 1Department of Mathematics, State University of Pará, Belém, PA, 66113-200, Brazil ducival@uepa.br 2Department of Mathematics, Federal University of Pará, Belém, PA, 66075-110, Brazil gera@ufpa.br 3Department of Mathematics, Federal University of São João del-Rei, São João del-Rei, 36307-352, Brazil ∗Correspondence: raposo@ufsj.edu.br Abstract. In this manuscript, we investigate the unilateral problem for a viscoelastic beam equationof p-Laplacian type. The competition of the strong damping versus the logarithmic source term isconsidered. We use the potential well theory. Taking into account the initial data is in the stabilityset created by the Nehari surface, we prove the existence and uniqueness of global solutions by usingthe penalization method and Faedo-Galerkin’s approximation. 1. Introduction We denote the p-Laplacian operator by ∆pu = div (|∇u|p−2∇u), which can be extended to amonotone, bounded, hemicontinuos and coercive operator between the spaces W 1,p0 (Ω) and its dualby −∆p : W 1,p 0 (Ω) → W −1,q(Ω), 〈−∆pu,v〉p = ∫ Ω |∇u|p−2∇u ·∇v dx. In [3] the authors establish existence of global solution to the problem utt + ∆ 2u − ∆pu + ∫ t 0 g(t − s)∆u(s)ds − ∆ut + f (u) = 0 in Ω ×R+, (1.1) u = ∆u = 0 on Γ ×R+, (1.2) u(x, 0) = u0, ut(x, 0) = u1 in Ω, (1.3) where Ω is a bounded domain of Rn with smooth boundary Γ = ∂Ω.Equations of the type (1.1) are related to models of elastoplastic microstructure flows. Asconsidered by An and Peirce [1, 2], they are essentially of the form utt + uxxxx −a(u2x )x = 0. Received: 13 Nov 2021. Key words and phrases. Unilateral problem; viscoelastic beam equation type p-Laplacian; logarithmic source.1 https://adac.ee https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 2 A more general equation, utt + ∆ 2u −div(σ(|∇u|2)∇u) − ∆ut + h1(ut) + h2(u) = h3(x), was considered by Yang et al [22–24]. They studied de existence of attractors and their Hausdorffdimensions. Another related equation is utt + ∆ 2u −div(f0(∇u)) + kut = ∆(f1(u)) − f2(u), which was considered by Chueshov and Lasiecka [12] The problem (1.1), with its memory term ∫ t 0 g(t−s)∆u(s)ds, can be regarded as a fourth-orderviscoelastic plate equation with a lower order perturbation of the p-Laplacian type. This kind ofproblem can be also regarded as an elastoplastic flow equation with some kind of memory effect.We observe that for viscoelastic plate equation, it is usual consider a memory of the form∫ t 0 g(t − s)∆2u(s)ds, see for instance [10]. However, because the main dissipation of the system (1.1) is given by strongdamping −∆ut, here we consider a weaker memory, acting only on ∆u. There is a large literatureabout stability in viscoelasticity. We refer the reader to [11, 13].A nonlinear perturbation of problem (1.1) is given by utt + ∆ 2u − ∆pu + ∫ t 0 g(t − s)∆u(s)ds − ∆ut + f (u) ≥ 0. (1.4) Variational inequality theory was introduced by Hartman and Stampacchia (1966) [14] as a toolfor the study of partial differential equations with applications principally in mechanics.In [7] the authors investigated the unilateral problem associated with this perturbation, thatis, a variational inequality given for (1.4) (see [16]). Making use of the penalization method andGalerkin’s approximations, they established existence and the uniqueness of strong solutions.The unilateral problem is very interesting because, in general, dynamic contact problems arecharacterized by nonlinear hyperbolic variational inequalities. Variational inequality theory wasintroduced by Hartman and Stampacchia (1966) [14] as a tool for the study of partial differentialequations with applications principally in mechanics. Bensoussan and Lions (1982) [9] used vari-ational inequalities initially in the study of stochastic control. In [5] was obtained a variationalinequality for the Navier-Stokes operator with variable viscosity. In [6] was studied the contactproblem on the Oldroyd model of viscoelastic fluids. By using results from the theory of monotoneoperators, was established the existence of weak solutions. In [8] was studied the problem forparabolic variational inequalities with Volterra type operators. The authors proved the existenceand the uniqueness of the solution. For contact problems on elasticity and finite element method,see Kikuchi-Oden [15] and reference therein. In [18] was studied the unilateral problem for the https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 3 Klein-Gordon operator with the nonlinearity of Kirchhoff-Carrier type. By using an appropriate pe-nalization was shown the existence and uniqueness of solutions for the perturbed equation. In [19]was considered the unilateral problem for a nonlinear wave equation with p-Laplacian operatorand source term. By using an appropriate penalization, authors obtained a variational inequalityfor the equation perturbed and then the existence of solutions was proved.In this work, we propose to investigate the existence and uniqueness of solutions for the vari-ational inequality associated with the problem (1.4) with the source term f (u) = −|u|r−2u ln |u|.More precisely, we investigate the existence and uniqueness of solutions for the unilateral problem utt + ∆ 2u − ∆pu + ∫ t 0 g(t − s)∆u(s)ds − ∆ut ≥ |u|r−2u ln |u| in Ω×R+, (1.5) u(x, 0) = u0(x),ut(x, 0) = u1(x) in Ω, (1.6) u(x,t) = ∆u(x,t) = 0 on Γ ×R+. (1.7) This work is organized as follows: In section 2 we introduce the notation and some well-knownresults. In section 3 we introduce the potential theory suitable for our problem. In section 4 definestrong solution to the boundary value problem (1.5)-(1.7) and present the theorem of existence ofstrong solution. In section 5 we apply the penalization method. The existence of global solutionsis given by using Faedo-Galerkin approximation. Finally, in Section 6 we prove the result ofuniqueness. 2. Preliminaries Let Ω be a bounded domain in Rn with the boundary Γ of class C2. For T > 0, we denote by Qthe cylinder Ω×(0,T ), with lateral boundary Σ = Γ×(0,T ). By 〈·, ·〉 we will represent the dualitypairing between a Banach space X and X′, X′ being the topological dual of the space X, and by C we denote various positive constants. The inner product in H10 (Ω) and L2(Ω) , respectively, willbe denoted by (∇·,∇·), (·, ·). The norm in Lp(Ω) will be denoted by | · |p.The inequality (1.5) must be satisfied in the following sense. Let K = {v ∈ H10 (Ω); v ≥ 0 a.e. in Ω} be a closed and convex subset of H10 (Ω), the unilateral problem consists to find a solution u(x,t)satisfying∫ Q (utt + ∆ 2u − ∆pu + ∫ t 0 g(t − s)∆u(s)ds − ∆ut −|u|r−2u ln |u|)(v −ut) ≥ 0, (2.1) for all v ∈ K with ut(x,t) ∈ K a.e. on [0,T ] and the initial and boundary data u = ∆u = 0 in Γ × (0,T ), (2.2) u(x, 0) = u0, ut(x, 0) = u1 in Ω. (2.3) https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 4 To study the existence and uniqueness of the problem (1.5)-(1.7), let us consider the followinghypotheses: H1. Suppose that  2 ≤ p , if n = 1, 2 2 ≤ p ≤ 2n− 2 n− 2 , if n ≥ 3. H2. With respect to the power r , let us suppose that 2 < r < +∞ , if n = 1, 2 2 < r < 2n n− 2 , if n ≥ 3. H3. With respect to the function g : [0, +∞) →R, we will assume that g ∈ C1[0,T ] and g(0) > 0, I = 1 −µ ∫ ∞ 0 g(s)ds > 0, where µ > 0 is the embedding constant for |∇u| ≤√µ |∆u|, for all u ∈ H10 (Ω) ∩H2(Ω). H4. There exists a constant k1 > 0 such that g′(t) ≤−k1g(t), ∀t ≥ 0. By H1 we have H10 (Ω) ∩H 2(Ω) ↪→ W 1,2(p−1)0 (Ω) ↪→ H 1 0 (Ω) ↪→ L 2(Ω). The Lemmas below will be a important role in this manuscript. Lemma 2.1. (Sobolev Poincaré inequality) Let p be a number with 2 < p < ∞ if n = 1, 2 or 2 ≤ p ≤ 2n n− 2 if n ≥ 3, then there exists a constant C > 0 such that |u|p ≤ C|∇u|,∀u ∈ H10 (Ω) Lemma 2.2. (Technical lemma) For v ∈ C1(0,T ; H10 (Ω)), we have∫ Ω ∫ t 0 g(t − s)∇v ·∇vtdsdx = 1 2 (g′ �∇v)(t) − 1 2 g(t)|∇v(t)|2 − 1 2 d dt [ (g �∇v)(t) − (∫ t 0 g(s)ds ) |∇v(t)|2 ] , where (g �∇u)(t) = ∫ t 0 g(t − s)|∇u(s) −∇u(t)|2ds. Proof. Differentiating the term (g �∇u)(t) we arrive to the above inequality. � https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 5 3. Potential well In this section, we use the potential well theory, a power full tool in the study of the globalexistence of solution in partial differential equation. See Payne-Sattinger [17]. It is well-knownthat the energy of a PDE system, in some sense, splits into kinetic and potential energy. The sourceterm induces potential energy in the system that acts in opposition to the effect of the stabilizingmechanism. In this sense, it is possible that the energy from the source term destabilizes all thesystem and produces a blow-up in a finite time. To provide a global solution, we are able toconstruct a stability set corresponding to the source term created from the Nehari Manifold, see Y.Ye [20]. In the stability set, there exists a valley or a well of the depth d created in the potentialenergy. If d is strictly positive, then we find that, for solutions with the initial data in the goodpart of the potential well, the potential energy of the solution can never escape the potential well.In general, the energy from the source term causes the blow-up in a finite time. However, thegood part of the potential well is an invariant set where it remains bounded. As a result, the totalenergy of the solution remains finite for any time interval [0,T ], providing the global existence ofthe solution.For the model considered here, the total energy is given by E(t) = 1 2 [ |ut(t)|2 + |∆u(t)|2 + 2 p |∇u(t)|pp + (g �∇u)(t) − (∫ t 0 g(s)ds ) |∇u(t)|2 + 2 r2 |u(t)|rr − 2 r ∫ Ω |u(t)|r ln |u(t)|dx ] (3.1) and satisfies d dt E(t) ≤−|∇ut(t)|2. (3.2) From (H3) we get I(t) = 1 2 [ |ut(t)|2 + ( 1 −µ ∫ t 0 g(s)ds ) |∆u(t)|2 +(g �∇u(t)+ 2 p |∇u(t)|pp + 2 r2 |u(t)|rr − 2 r ∫ Ω |u(t)|r ln |u(t)|dx ] . (3.3) Then, we introduce the functional J : H10 (Ω) ∩H2(Ω) →R defined by J(u) = 1 2 [( 1 −µ ∫ t 0 g(s)ds ) |∆u|2 + 2 p |∇u(t)|pp + (g �∇u)(t) + 2 r2 |u(t)|rr − 2 r ∫ Ω |u(t)|r ln |u(t)|dx ] . (3.4) https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 6 For u ∈ H10 (Ω) ∩H2(Ω), we have J(λu) = λ2 2 ( 1 −µ ∫ t 0 g(s)ds ) |∆u|2 + λp p |∇u(t)|pp + λ 2 (g �∇u)(t) + λr r2 |u(t)|rr − λr r ∫ Ω |u(t)|r ln |u(t)|dx. (3.5) Associated with J, we have the well-known Nehari Manifold given by N def= { u ∈ H10 (Ω) ∩H 2(Ω) \{0}; [ d dλ J(λu) ] λ=1 = 0 } (3.6) or equivalently, N = { u ∈ H10 (Ω) ∩H 2(Ω)) \{0}; ( 1 −µ ∫ t 0 g(s)ds ) |∆u|2 + |∇u(t)|pp + 1 2 (g �∇u)(t) = ∫ Ω |u(t)|r ln |u(t)|dx } . (3.7) We define as in the Mountain Pass theorem due to Ambrosetti and Rabinowitz [4] d def = inf u∈(H10 (Ω)∩H2(Ω)\{0} sup λ≥0 J(λu). similar to the result in [21] one has 0 < d = inf u∈N J(u).Now, we introduce W = {u ∈ H10 (Ω) ∩H 2(Ω); J(u) < d}∪{0} and partition it into two sets W = W1 ∪W2 as follows W1 = { u ∈ W ; ( 1 −µ ∫ t 0 g(s)ds ) |∆u|2 + 2 p |∇u(t)|pp + (g �∇u)(t) + 2 r2 |u(t)|rr > 2 r ∫ Ω |u(t)|r ln |u(t)|dx } ∪{0} (3.8) and W2 = { u ∈ W ; ( 1 −µ ∫ t 0 g(s)ds ) |∆u|2 + 2 p |∇u(t)|pp + (g �∇u)(t) + 2 r2 |u(t)|rr < 2 r ∫ Ω |u(t)|r ln |u(t)|dx } . (3.9) So, we define by W1 the set of stability for the problem (1.5)-(1.7), and before starting the sectionof existence and uniqueness of solution, we will prove that W1 is invariant set for sub-critical initialenergy. Proposition 1. Let u0 ∈ W1 and u1 ∈ H10 (Ω). If E(0) < d then u(t) ∈W1. https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 7 Proof. Let T > 0 be the maximum existence time. From (3.2) we get E(t) ≤ E(0) < d, for all t ∈ [0,T ). and then, 1 2 ∫ Ω |ut(t)|2 dx + J(u(t)) < d, for all t ∈ [0,T ). (3.10) Arguing by contradiction, we suppose that there exists a first t0 ∈ (0,T ) such that I(u(t0)) = 0and I(u(t)) > 0 for all 0 ≤ t < t0, that is,( 1 −µ ∫ t 0 g(s)ds ) |∆u(t0)|2 + 2 p |∇u(t0)|pp + (g �∇u)(t0) + 2 r2 |u(t0)|rr = 2 r ∫ Ω |u(t0)|r ln |u(t0)|dx From the definition of N , we have that u(t0) ∈N , which leads to J(u(t0)) ≥ inf u(t)∈N J(u(t)) = d. We deduce 1 2 ∫ Ω |ut(t0)|2 dx + J(u(t0)) ≥ d, which contradicts with (3.10). Then u(t) ∈W1 for all t ∈ [0,T ). � 4. Existence of strong solutions Next, we shall state the main results of this paper. Theorem 4.1. Consider the space H3Γ(Ω) = {u ∈ H 3(Ω)|u = ∆u = 0 on Γ}. If u0 ∈ W1 ∩H3Γ(Ω),J(u0) < d,u1 ∈ H 1 0 (Ω) and the hypothesis (H1)-(H4) holds, then there exists a function u : Ω × (0,T ) →R such that u ∈ L∞(0,T ; (H10 (Ω) ∩H 2(Ω))) ∩L∞(0,T ; H3Γ(Ω)), (4.1) ut ∈ L∞(0,T ; L2(Ω)) ∩L2(0,T ; H10 (Ω) ∩H 2(Ω)), (4.2) utt ∈ L∞(0,T ; H−1(Ω)), (4.3) ut(t) ∈ K a.e. in [0,T ], (4.4) ∫ T 0 [ 〈utt,v −ut〉 + (∆2u,v −ut) − (∆pu,v −ut) + (∫ t 0 g(t − s)∆u(s)ds,v −ut ) − (∆ut,v −ut) − (|u|r−2u ln |u|,v −ut) ] ≥ 0, (4.5) https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 8 for all v ∈ L2(0,T ; H10 (Ω)), v(t) ∈ K a.e. in t and initial data u(0) = u0, ut(0) = u1. The proof of Theorem 4.1 is given in Section 5 by the penalization method. It consists in con-sidering a perturbation of the problem (1.5) adding a singular term called penalization, dependingon a parameter � > 0. We solve the mixed problem in Q for the penalization operator and theestimates obtained for the local solution of the penalized equation, allow to pass to limits, when �goes to zero, in order to obtain a function u which is the solution of our problem. First of all, letus consider the penalization operator β : H10 (Ω) −→ H −1(Ω) associated to the closed convex set K, cf. Lions [16], p. 370. The operator β is monotonous,hemicontinuous, takes bounded sets of H10 (Ω) into bounded sets of H−1(Ω), its kernel is K and β : L2(0,T ; H10 (Ω)) −→ L 2(0,T ; (H−1(Ω)) is monotone and hemicontinous. The penalized problem associated with the variational inequality(1.5)-(1.7), consists in given 0 < � < 1, find u� satisfying u�tt + ∆ 2u� − ∆pu� + ∫ t 0 g(t − s)∆u�(s)ds − ∆u�t + 1 � (β(u�t )) −|u �|r−2u� ln |u| = 0, in Q (4.6) and u�(x, 0) = u�0(x),u � t (x, 0) = u�1(x) in Ω. u�(x.t) = ∆u�(x,t) = 0 on ∂Ω ×R+. (4.7) Definition 4.2. Suppose that u�0 ∈ W1, J(u�0) < d, u�1 ∈ H10 (Ω) and hypothesis (H1) − (H4) holds. A strong solution to the boundary value problem (4.6)-(4.7) is a function u� such that u� ∈ L∞(0,T ; H10 (Ω) ∩H 2(Ω)), u�t ∈ L ∞(0,T ; L2(Ω)) ∩L2(0,T ; H10 (Ω)), u�tt ∈ L 2(0,T ; (H10 (Ω) ∩H 2(Ω))′) satisfying for all w ∈ H10 (Ω) ∩H 2(Ω) d dt (u�t (t),w) + (∆u �(t), ∆w) + (−∆pu�(t),w) + ∫ t 0 g(t − s)(∆u�(s),w)ds +(∇u�t (t),∇w) + 1 � (β(u�t (t)),w)−(|u �(t)|r−2u�(t) ln |u�(t)|),w) = 0 https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 9 and initial data u�(0) = u�0, u � t (0) = u�1. The solution of problem (4.6)-(4.7) is given by the following theorem: Theorem 4.3. Assume that hypotheses (H1) − (H4) holds, u�0 ∈ W1, J(u�0) < d and u�1 ∈ H10 (Ω), (4.8) then, for each 0 < � < 1, there exists a function u� strong solution of (4.6)-(4.7). 5. Penalization method In order to prove Theorem 4.1, we first prove the penalized Theorem 4.3. The existence ofglobal solutions will be given by using Faedo-Galerkin method. First we consider the approximateproblem. Then we obtain the a priori estimates needed to passage to the limit in the approximatesolutions. 5.1. Approximate problem. Let {wj} be the Galerkin basis given by eigenfunctions of ∆2 withboundary condition u = ∆u = 0 on Γ ×R+ and let Vm ⊂ N be the subspace spanned by thevectors w1,w2, ...,wm.. Consider u�m(t) = m∑ j=1 g�jm(t)wj solution of approximate problem (u�mtt (t),w)+(∆u �m(t), ∆w)+(−∆pu�m(t),w)+ ∫ t 0 g(t − s)(∆u�m(t),w)ds −(|u�m(t)|r−2u�m(t) ln |u�m|,w) + (∇u�m(t),∇w) + 1 � (β(u�mt )(t),w) = 0 (5.1) with initial conditions u�m(0) = u�0m → u�0 strongly in H2(Ω) ∩H10 (Ω), (5.2) u�mt (0) = u�1m → u�1 strongly in L2(Ω). (5.3) The system of ordinary differential equation (5.1) in the variable t has a local solution u�m(t)defined in [0,tm[, 0 < tm ≤ T . In the next step obtain priori estimates for the solution u�m(t) thatpermits us to extend this solution to the whole interval [0,T ]. https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 10 5.2. First estimate. We consider w = u�mt in (5.1) to obtain d dt [ 1 2 |u�mt (t)| 2 + 1 2 |∆u�m(t)|2 + 1 p |∇u�m(t)|pp + 1 r2 |u�m(t)|pp − 1 r ∫ Ω |u�m(t)|r ln |u�m(t)|dx ] + |∇u�mt (t)|2 + 1 � (β(u�mt (t)),u �m t (t)) = ∫ t 0 g(t − s)(∇u�m(s),∇u�mt (t))ds. (5.4) We have (β(u�mt (t)),u�mt (t)) ≥ 0. Then from Lemma 2.2 and (H4) 1 2 d dt [ |u�mt (t)| 2 + |∆u�m(t)|2 + 2 p |∇u�m(t)|pp + (g �∇u �m)(t) − (∫ t 0 g(s)ds ) |∇u�m(t)|2 + 2 r2 |u�m(t)|rr − 2 r ∫ Ω |u�m(t)|r ln |u�m(t)|dx ] + |∇u�mt (t)|2+ ≤ 1 2 (g′ �∇u�m)(t) − 1 2 g(t)|∇u�m(t)|2 ≤ 0. (5.5) Let E�m(t) = 1 2 [ |u�mt (t)| 2 + |∆u�m(t)|2 + 2 p |∇u�m(t)|pp + (g �∇u �m)(t) − (∫ t 0 g(s)ds ) |∇u�m(t)|2 + 2 r2 |u�m(t)|rr − 2 r ∫ Ω |u�m(t)|r ln |u�m(t)|dx ] . (5.6) So, by (5.5) and (5.8), we have d dt E�m(t) ≤−|∇u�mt (t)| 2. Integrating from 0 to t, t ≤ tm, we obtain E�m(t) + ∫ t 0 |∇u�mt (t)| 2 ≤ E�m(0). (5.7) By (H3), it follows 1 2 [ |u�mt (t)| 2 + ( 1 −µ ∫ t 0 g(s)ds ) |∆u�m(t)|2 +(g �∇u�m)(t)+ 2 p |∇u�m(t)|pp + 2 r2 |u�m(t)|rr − 2 r ∫ Ω |u�m(t)|r ln |u�m(t)|dx ] + ∫ t 0 |∇u�mt | 2ds ≤ E�m(t) ≤ E�m(0) = 1 2 |u�1m|2 + C1J(u�0m), (5.8) where C1 > 0 is a positive constant, independent of m and t. https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 11 We have J(u0�m) < d and by (5.3), there exists a constant C2 > 0 such that |u�mt (t)| 2 + ( 1 −µ ∫ t 0 g(s)ds ) |∆u�m(t)|2 +(g �∇u�m)(t)+ 2 p |∇u�m(t)|pp + 2 r2 |u�m(t)|rr − 2 r ∫ Ω |u�m(t)|r ln |u�m(t)|dx + ∫ t 0 |∇u�mt | 2ds ≤ C2. (5.9) From (3.7) and (5.9) we get ∆u�m ⇀ ∆u� in L∞(0,T ; L2(Ω)), (5.10) u�m ⇀ u� in L∞(0,T ; H10 (Ω) ∩H2(Ω)), (5.11) −∆pu�m ⇀ χ in L2(0,T ; H−1(Ω)), (5.12) u�mt ⇀ u � t in L∞(0,T ; L2(Ω)) ∩L2(0,T ; H10 (Ω)), (5.13) β(u�mt ) ⇀ ψ in L2(0,T ; H−1(Ω)). (5.14) Follows from(5.11), (5.13) and Aubin-Lions Theorem, for any T > 0, u�m → u� in L2(0,T ; H10 (Ω)), strong and a.e. in Q. (5.15) Now, we prove that χ(t) = −∆pu�(t). We consider x,y ∈ R,p ≥ 2. Then the elementaryinequality ∣∣|x|p−2x −|y|p−2y∣∣ ≤ C (|x|p−2 + |y|p−2) |x −y| (5.16) is a consequence of the mean value theorem. Using (5.16) and Hölder generalized inequality with p− 2 2(p− 1) + 1 2 + 1 2(p− 1) = 1, we deduce for θ ∈D(0,T ) and v ∈ Vm,∣∣∣∣∫ T 0 〈(−∆u�mp (t)) − (−∆u � p(t)),v〉pθ(t)dt ∣∣∣∣ = ∣∣∣∣∫ T 0 ∫ Ω ( |∇u�m(t)|p−2∇u�m(t) −|∇u�(t)|p−2∇u�(t) ) ∇v dx θ(t) dt ∣∣∣∣ ≤ C|θ|∞ ∫ T 0 ∫ Ω ( |∇u�m(t)|p−2 + |∇u�(t)|p−2 ) |∇u�m(t) −∇u�(t)||∇v|dx dt ≤ C1 ∫ T 0 ( |∇u�m(t)|p−2 2(p−1) + |∇u �(t)|p−2 2(p−1) ) |∇u�m(t) −∇u�(t)||∇v|2(p−1) dt ≤ C2 ∫ T 0 |∇u�m(t) −∇u�(t)|dt (5.17) where C1 and C2 are positive constants independent of m and t. https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 12 Now, from estimate (5.10) and (5.11), we have d dt |∇u�m(t) −∇u�(t)|2 ≤ 2|∆(u�m(t) −u�(t))||∇(u�mt (t) −u � t (t))| ≤ C3, where C3 is a constant independent of m and t. So, |u�m(t) −u�(t)|H10 (Ω) ∈ H 1[0,T ] ↪→ C[0,T ], whence ∇u�m(t) →∇u�(t) a. e. in [0,T ]. Therefore, χ = −∆pu�. Now, we observe that Sobolev inequality∫ Ω ||u�m(t)|r−2u�m(t) ln |u�m(t)||2dx ≤ |u�m(t)|2r2r ≤ C 2r|∇u�m(t)|2r ≤ µrC2r|∆u�m(t)|r ≤ C4, where C4 is a constant independent of m and t.Then (|u�m|r−2u�m ln |u�m|) is bounded in L2(0,T ; L2(Ω)) = L2(Q). (5.18) Using continuity of function s →|s|r−2s ln |s| and (5.15) we have |u�m|r−2u�m ln |u�m|→ |u�|r−2u� ln |u�| a.e. in Q. (5.19) By (5.18), (5.19) and applying Lions Lemma (Lemma 1.3, page 12, [16]), we get |u�m|r−2u�m ln |u�m| ⇀ |u�|r−2u� ln |u�| weakly in L2(0,T ; L2(Ω)). (5.20) 5.3. Second estimate. Let us consider the initial data u�0 ∈ H3Γ(Ω),u�1 ∈ H 1 0 (Ω) and u�m0 = ∆u�m0 = 0 on Γ. (5.21) We consider w = −∆u�mt in approximate equation (5.1).Then we have d dt { 1 2 |∇u�mt (t)| 2 + 1 2 |∇∆u�m(t)|2 } + 〈∆pu�mt (t), ∆u �m t (t)〉 +|∆u�mt (t)|2 + 1 � (β(u�mt (t),−∆u �m t (t)) = (|u�m(t)|r−2u�m ln |u�m(t)|,−∆u�mt (t)) + ∫ t 0 g(t − s)(∆u�m(s), ∆u�mt (t))ds. Now, 〈∆pu�m(t), ∆u�mt (t)〉 = d dt 〈∆pu�m(t), ∆u�m(t)〉−J1, https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 13 where J1 = ∫ Ω { (p− 2)|∇u�m(t)|p−4(∇u�m(t) ·∇u�mt (t))∇u �m(t) +|∇u�m(t)|p−2∇u�mt (t) } ·∇∆u�m(t)dx. Then d dt { 1 2 |∇u�mt (t)| 2 + 1 2 |∇∆u�m(t)|2 + 〈∆pu�m(t), ∆u�m(t)〉 } +|∆u�mt (t)|2 + 1 � (β(u�mt (t)),−∆u �m t (t)) = J1 + J2 + J3. (5.22) where J2 = ∫ Ω |u�m(t)|r−2u�m ln |u�m(t)|∆u�mt (t) and J3 = ∫ t 0 g(t − s)(∆u�m(s), ∆u�mt (t))ds. Let us the right hand side of (5.22). We denote by C a generic positive constant not dependingon m,t. By estimate (5.9) and p− 2 2(p− 1) + 1 2(p− 1) + 1 2 = 1, |J1| ≤ (p− 1) ∫ Ω |∇u�m(t)|p−2|∇u�mt (t)||∇∆u �m(t)|dx ≤ (p− 1)|∇u�m(t)|p−2 2(p−1)|∇u �m t (t)|2(p−1)|∇∆u �m(t)| ≤ C|∇u�mt (t)|2(p−1)|∇∆u �m(t)|. How H10 (Ω) ∩H2(Ω) ↪→ W 1,20 (Ω), we have |∇u�mt (t)| 2 2(p−1) ≤ µ2|∆u �m t (t)| 2, where µ2 > 0 is the corresponding embedding constant. Then |J1| ≤ 1 2 |∆u�mt (t)| 2 + C|∇∆u�m(t)|2. (5.23) https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 14 Let Ω1 = {x ∈ Ω : |u�m(t)| < 1} and Ω2 = {x ∈ Ω : |u�m(t)| ≥ 1}. By (5.9) and Sobolevinequality |J2| ≤ ∫ Ω1 ||u�m(t)|r−2u�m ln |u�m(t)|∆u�mt (t)|dx + ∫ Ω2 ||u�m(t)|r−2u�m ln |u�m(t)|∆u�mt (t)|dx ≤ (e(r − 1))−1 ∫ Ω |∆u�mt (t)|dx + (e(r − 1)) −1 ∫ Ω |u�m(t)|r−1|∆u�mt (t)|dx ≤ 2(e(r − 1))−2 + 1 8 |∆u�mt (t)| 2 + 2(e(r − 1))−2|u�m(t)|2(r−1) 2(r−1) + 1 8 |∆u�mt (t)| 2 ≤ 2(e(r − 1))−2 + 1 4 |∆u�mt (t)| 2 + 2C(e(r − 1))−2|∇u�m(t)|2(r−1) ≤ C + 1 4 |∆u�mt (t)| 2 (5.24) where we have used |xr−1 ln x| ≤ (e(r − 1))−1 for 0 < x < 1 and ln x ≤ (e(r − 1))−1xr−1, if x ≥ 1. Remark 5.1. We note from the Cauchy-Schwarz inequality and Fubini’s Theorem follows ‖g �∇u‖L2(Q) ≤‖g‖L1(0,∞)‖∇u‖L2(Q) Again from estimate (5.9) and remark 5.1 |J3| ≤ (∫ t 0 g(t − s)|∆u�m(t)|ds ) |∆u�mt (t)| (5.25) ≤ C‖g‖L1(R+)|∆u �m t (t)| ≤ C + 1 4 |∆u�mt (t)| 2. Follows from (5.22)-(5.25) that d dt [ 1 2 |∇u�mt (t)| 2 + 1 2 |∇∆u�m(t)|2 + 〈∆pu�m(t), ∆u�m(t)〉 ] + 1 2 |∆u�mt (t)| 2 + 1 � (β(u�mt (t)),−∆u �m t (t)) ≤ C + C|∇∆u�m(t)|2. (5.26) Now, observe that |〈∆pu�m(t), ∆u�m(t)〉| ≤ ∫ Ω |∇u�m(t)|p−1|∇∆u�m(t)|dx ≤ |∆u�m(t)|p−1 2(p−1)|∇∆u �m(t)| (5.27) ≤ C + |∇∆u�m(t)|2, https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 15 and then C + |∇∆u�m(t)|2 + 〈∆pu�m(t), ∆u�m(t)〉≥ 0. Therefore, there exists C0 > 0 such that d dt [ 1 2 |∇u�mt (t)| 2 + 1 2 |∇∆u�m(t)|2 + 〈∆pu�m(t), ∆u�m(t)〉 ] + 1 2 |∆u�mt (t)| 2 + 1 � (β(u�mt (t)),−∆u �m t (t)) ≤ C0 + C0|∇∆u�m(t)|2 + 〈∆pu�m(t), ∆u�m(t)〉. (5.28) Taking into account that (β(u�mt (t),−∆u�mt (t)) ≥ 0, (5.21), integrating from 0 to t and applyingGronwall inequality, we obtain |∇u�mt (t)| 2 + |∇∆u�m(t)|2 + ∫ t 0 |∆u�mt (t)| 2 ≤ C, (5.29) then u�m ⇀ u� in L∞(0,T ; H3Γ(Ω)), weakly star. (5.30) u�mt ⇀ u � t in L2(0,T ; H10 (Ω) ∩H2(Ω)), weakly (5.31) ∆u�m ⇀ ∆u� in L∞(0,T ; H10 (Ω)), weakly star. (5.32) 5.4. Third estimate. Let Pm be the ortogonal projection Pm : L2(Ω) → Vm, that is Pmφ = m∑ n=1 (φ,wj)wj, φ ∈ L2(Ω). Remark 5.2. By remark 5.1, we observe that if ψ ∈ L2(0,T ; H10 (Ω)) then ∫ t 0 g(t − s)ψ(s)ds ∈ L2(0,T ; H−1(Ω)) and by (5.12) −∆pu�m ∈ L2(0,T ; (H−1(Ω)). We obtain using the notation and ideas of Lions [16], pages 75-76, remark 5.2 and estimatesabove that u�mtt ⇀ u � tt in L2(0,T ; (H−1(Ω)), weakly. (5.33) (5.31), (5.33) and Aubin-Lions compactness Theorem imply that there exists a subsequence from (u�mt ), still denoted by (u�mt ), such that u�mt → u � t strongly in L2(0,T ; H10 (Ω)) and a.e. in Q. (5.34) Now, we are in position to prove Theorem 4.1. https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 16 5.5. Strong solution. Let v ∈ L2(0,T ; H10 (Ω)) be v(t) ∈ K a. e. for t ∈ (0,T ). From (4.6)1follows that ∫ T 0 (u�tt,v −u � t )dt + ∫ T 0 (∆2u�,v −u�t )dt + ∫ T 0 (−∆pu�,v −u�t )dt + ∫ T 0 (∫ t 0 g(t − s)∆u�(s)ds,v −u�t ) dt + ∫ T 0 (−∆u�t,v −u � t )dt − ∫ T 0 (|u�|r−2u� ln |u�|,v −u�t )dt = 1 � ∫ T 0 (β(u�t ),u � t −v) dt = 1 � ∫ T 0 (β(u�t ) −βv,u � t −v) dt ≥ 0, (5.35) because v ∈ K (β(v) = 0) and β is monotone.From (5.11), (5.12), (5.15), (5.20), (5.30), (5.31), (5.33), (5.34) and the Bannach-Steinhauss The-orem, it follows that there exists a subsequence (u�)0<�<1, such that it converge to u as � → 0,that is u� ⇀ u in L∞(R+; H10 (Ω) ∩H2(Ω)), (5.36) −∆pu� ⇀ −∆pu in L2(0,T ; H−1(Ω), (5.37) u� → u in L2(0,T ; H10 (Ω))and a.e. in Q, (5.38) u� ⇀ u in L∞(0,T ; H3Γ(Ω)), (5.39) u�t ⇀ ut in L2(0,T ; H10 (Ω) ∩H2(Ω)), (5.40) u�tt ⇀ utt in L2(0,T ; H−1(Ω)), (5.41) |u�|r−2u� ln |u�| ⇀ |u|r−2u ln |u| in L2(0,T ; L2(Ω)), (5.42) u�t → ut in L2(0,T ; H10 (Ω)) and a.e. in Q. (5.43) The convergences above are sufficient to pass to the limit in (5.35) with � > 0 to conclude that(4.5) is valid. To complete the proof of Theorem 4.1, it remains to show that ut(t) ∈ K a.e.In the position, we observe that using convergences (5.10)-(5.16) and (5.30)-(5.32), making m → ∞ in (5.1), we can find u� such that u�tt + ∆ 2u� − ∆pu� + ∫ t 0 g(t − s)∆u�(s)ds − ∆u�t −|u �|r−2u� ln |u�| + 1 � β(u�t ) = 0 in L2(0,T ; H−1(Ω). (5.44) Then, β(u�t ) = �[−u � tt − ∆ 2u� + ∆pu � − ∫ t 0 g(t − s)∆u�(s)ds + ∆u�t + |u �|r−2u� ln |u�|]. (5.45) So, β(u�t ) → 0 in D′(0,T ; H−1Ω). https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 17 From (5.45) it follows that β(u�t ) is bounded in L2(0,T ; H−1(Ω)), therefore β(u�t ) ⇀ 0 weak in L2(0,T ; H−1Ω). (5.46) On the other hand we deduce from (5.45) that 0 ≤ ∫ T 0 (β(u�t ),u � t ) dt ≤ � C. (5.47) Thus ∫ T 0 (β(u�t ),u � t )dt −→ 0. (5.48) We have that ∫ T 0 (β(u�t ) −β(ϕ),u � t −ϕ) dt ≥ 0, ∀ϕ in L2(0,T ; H10 (Ω)), because β is a monotonous operator. Thus,∫ T 0 (β(u�t ),u � t ) dt − ∫ T 0 (β(u�t ),ϕ) dt − ∫ T 0 (β(ϕ),u�t −ϕ) dt ≥ 0. (5.49) From (5.40), (5.46) and (5.48) we obtain∫ T 0 (β(ϕ),ut(t) −ϕ) dt ≤ 0. (5.50) Taking ϕ = ut −λv , with v ∈ L2(0,T ; H10 (Ω)) and λ > 0, we deduce using the hemicontinuityof β that β(ut(t)) = 0, (5.51) and this implies that ut(t) ∈ K a. e. 6. Uniqueness Let u1,u2 two solutions of (4.5) , w = u2 −u1 and t ∈ (0,T ). Because ut ∈ L2(0,T ; H10 (Ω), wecan talking u1t (resp. u2t ) in the inequality (4.5) relative to v2 (resp. v1) and adding up the resultswe obtain − ∫ t 0 (wtt,wt)ds − ∫ t 0 (∆2w,wt)ds + ∫ t 0 (∆pu 1,wt)ds − ∫ t 0 (∆pu 2,wt)ds + ∫ t 0 (∫ t 0 g(t − s)∆w(s)ds,wt ) ds + ∫ t 0 (∆wt,wt)ds − ∫ t 0 (|u1|r−2u1 ln |u1|,wt)ds + ∫ t 0 (|u2|r−2u2 ln |u2|,wt)ds ≥ 0, https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 18 thus, we have 1 2 ∫ t 0 d dt ( |wt(t)|2 + |∆w(t)|2 ) ds + ∫ t 0 |∇wt(t)|2ds ≤ ∫ t 0 〈∆pu1(t) − ∆pu2(t),wt(t)〉ds + ∫ t 0 ∫ t 0 g(t − s)(∇w(s),∇wt(t))dsdσ∫ t 0 ( |u1(t)|r−2u1(t) ln |u1(t)|− |u2(t)|r−2u2(t) ln |u2(t)|,wt(t) ) ds. By Lemma 2.2, we derive 1 2 ∫ t 0 d dt { |wt(t)|2 + |∆w(t)|2 − (∫ t 0 g(s)ds ) |∇w(t)|2 + (g �∇w)(t) } ds + ∫ t 0 |∇wt(t)|2ds ≤ ∫ t 0 |〈∆pu1(t) − ∆pu2(t),wt(t)〉|ds + ∫ t 0 ∫ Ω ( |u1(t)|r−2u1(t) ln |u1(t)|− |u2(t)|r−2u2(t) ln |u2(t)|,wt(t) ) dxds. (6.1) From Mean Value Theorem, |〈∆pu1(t) − ∆pu2(t),wt(t)〉| ≤ C ( |∇u1(t)|p−2 2(p−1) + |∇u2(t)| p−2 2(p−1) ) |∇w(t)|2(p−1)|∇wt(t)| ≤ C|∆w(t)|2 + 1 4 |∇wt(t)|2, (6.2) for some constant C > 0, and∫ t 0 ∫ Ω ( |u1(t)|r−2u1(t) ln |u1(t)|− |u2(t)|r−2u2(t) ln |u2(t)|,wt(t) ) dxds ≤ ∫ t 0 ∫ Ω |θu1(t) + (1 −θ)u2(t))|r−2|w(t)||wt(t)|dxds +(r − 1) ∫ t 0 ∫ Ω |θu1(t) + (1 −θ)u2|r−2 ln |θu1(t) +(1 −θ)u2(t)||w(t)|wt(t)|dxds = I1 + I2, 0 < θ < 1. (6.3) Hence, from the Hölder inequality and Sobolev inequality, we have∫ Ω |θu1(t) + (1 −θ)u2(t))|r−2|w(t)||wt(t)|dx ≤ |θu1(t) + (1 −θ)u2(t)|r−2 n(r−2)|w(t)| 2nn−2 |wt(t)| ≤ Cr−21 C2C3|∆w(t)||∇wt(t)| ≤ C|∆w(t)| 2 + 1 4 |∇wt(t)|2, (6.4) where C1 ,C2 and C3 are constants satisfying |θu1(t) + (1 −θ)u2(t)|r−2 n(r−2)| ≤ C1|θu 1(t) + (1 −θ)u2(t)|, |w(t)| 2n n−2 ≤ C|w(t)| ≤ C2|∆w(t)| and |w(t)| ≤ C3|∇w(t)|. https://doi.org/10.28924/ada/ma.2.5 Eur. J. Math. Anal. 10.28924/ada/ma.2.5 19 Also we used the condition n(p− 2) < 2n n− 2 . Now, using the calculation similar to (5.24), it follows that∫ Ω |θu1(t) + (1 −θ)u2|r−2 ln |θu1(t) + (1 −θ)u2(t)|ndx ≤ (e(r − 2)−n)|Ω| + (e(r − 2))−n|θu1(t) + (1 −θ)u2(t)|n(r−2) n(r−2) ≤ (e(r − 2)−n)|Ω| + (e(r − 2))−1C4|θu1(t) + (1 −θ)u2(t)|n(r−2) ≤ C. (6.5) Inserting (6.5) into I2, we have I2 = (r − 1) ∫ t 0 ∫ Ω |θu1(t) + (1 −θ)u2|r−2 ln |θu1(t) +(1 −θ)u2(t)||w(t)|wt(t)|dxds ≤ (r − 1) ∫ t 0 (∫ Ω ||θu1(t) + (1 −θ)u2|r−2 ln |θu1(t) + θu2(t)||ndx )1 n ×|wt(t)||w(t)| 2n n−2 ds ≤ C|∆w(t)|2 + 1 4 |∇wt(t)|2. (6.6) By (6.1), (6.2), (6.4) and (6.6) we get∫ t 0 d dt { |wt(t)|2 + |∆w(t)|2 − (∫ t 0 g(s)ds ) |∇w(t)|2 + (g �∇w)(t) } ds + ∫ t 0 |∇wt(t)|2ds ≤ C ∫ t 0 (|∆w(t)|2 + |∇wt(t)|2)ds. (6.7) Putting, Φ(t) = |wt(t)|2 + |∆w(t)|2 − (∫ t 0 g(s)ds ) |∇w(t)|2 + (g �∇w)(t) and using (H3), we have |∆w(t)|2 − (∫ t 0 g(s)ds ) |∇w(t)|2 ≥ I|∆w(t)|2 ≥ 0. As (g �∇w)(t) ≥ 0, we have from (6.7) that ∫ t 0 d dt Φ(t) ≤ CΦ(t) and because Φ(0) = 0, followsfrom the Gronwall lemma that |wt(t)|2 + I|∆w(t)|2 ≤ Φ(t) ≤ 0, which proves that w = 0 in H10 (Ω) ∩H2(Ω). References [1] L. An, A. Pierce, The effect of microstructure on elastic-plastic models, SIAM J. Appl. Math. 54(3) (1994) 708-730. https://doi.org/10.1137/S0036139992238498[2] L. An, A. 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Phys. 50(3) (2009) 032701. https: //doi.org/10.1063/1.3085951 https://doi.org/10.28924/ada/ma.2.5 https://doi.org/10.1016/0022-1236(73)90051-7 https://doi.org/10.3934/cpaa.2006.5.583 http://ejde.math.txstate.edu https://doi.org/10.1080/00036811.2020.1766028 https://doi.org/10.13189/ms.2019.070504 https://doi.org/10.1137/S0363012902408010 https://doi.org/10.3934/dcds.2006.15.777 https://doi.org/10.3934/dcds.2006.15.777 https://doi.org/10.1007/BF00251609 https://doi.org/10.1007/BF00251609 https://doi.org/10.1007/BF02392210 https://doi.org/10.1007/BF02761595 https://ejde.math.txstate.edu/Volumes/2015/137/abstr.html https://ejde.math.txstate.edu/Volumes/2015/137/abstr.html https://doi.org/10.11948/20200147 https://doi.org/10.1155/2007/19685 https://doi.org/10.1002/MMA.1080 https://doi.org/10.1063/1.3303633 https://doi.org/10.1063/1.3085951 https://doi.org/10.1063/1.3085951 1. Introduction 2. Preliminaries 3. Potential well 4. Existence of strong solutions 5. Penalization method 5.1. Approximate problem 5.2. First estimate 5.3. Second estimate 5.4. Third estimate 5.5. Strong solution 6. Uniqueness References