Convergence to equilibrium of global weak solutions for a Q-tensor problem related to liquid crystals

arXiv:1805.02439v1 [math.AP] 7 May 2018 Convergence to equilibrium of global weak solutions for a Q-tensor problem related to liquid crystals Blanca Climent-Ezquerra and Francisco Guill´en-Gonz´alez∗ Abstract We study a Q-tensor problem modeling the dynamic of nematic liquid crystals in 3D domains. The system consists of the Navier-Stokes equations, with an extra stress tensor depending on the elastic forces of the liquid crystal, coupled with an Allen-Cahn system fo r the Q-tensor variable. This problem has a dissipative in time free-energy which leads, in particular, to prove the existence of global in time weak solutions. We analyze the largetime behavior of the weak solutions. By using a Lojasiewicz-Simon’s result, we prove the convergence as time goes to infinity of the whole trajectory to a single equilibrium. Keywords: Liquid crystals; Allen-Cahn-Navier-Stokes system; Large-time behavior for dissipative systems. 1 Introduction We deal with a generic Q-tensor model, following the theory of Landau-De Gennes, in a smooth and bounded domain Ω ⊂R3, for the unknowns (u, p, Q) : (0, T )×Ω→R3×R×R3×3, satisfying the momentum and incompressibility equations      Dtu−ν∆u+∇p=∇ · τ(Q) + ∇ · σ(H, Q) ∇ · u= 0 (1) and the Q-tensor system: DtQ−S(∇u, Q) = −γ H(Q) (2) in Ω ×(0, T ). In (1) and (2), Dt=∂t+ (u· ∇) denotes the material time derivative, ν > 0 is the viscosity coefficient and γ > 0 is a material-dependent elastic constant. Moreover, S(∇u, Q) = ∇uQt−Qt∇u(3) ∗Departamento de Ecuaciones Diferenciales y An´alisis Num´erico and IMUS. Facultad de Matem´aticas, Universidad de Sevilla, Spain. E-mails:bclimen[email protected], guille[email protected]s. Partially supported by MINECO grant MTM2015-69875-P. 1 is the so-called stretching term. In (1) the tensors τ=τ(Q)∈R3×3and σ=σ(H, Q)∈R3×3are defined by    τij(Q) := −ε(∂jQ:∂iQ) = −ε ∂jQkl ∂iQkl, σ(H, Q) := H Q −Q H, where ε > 0 and the tensor H=H(Q) is related to the variational derivative in L2(Ω) of a free energy functional E(Q), in fact E(Q) := ε 2|∇Q|2+F(Q),E(Q) := ZΩ E(Q)dx, H := δE(Q) δQ .(4) Here, we denote A:B=Aij Bij the scalar product of matrices (using the Einstein summation convention over repeated indices) and the potential function F(Q) is defined by F(Q) := a 2|Q|2−b 3(Q2:Q) + c 4|Q|4,(5) with a,b∈Rand c > 0. We denote by |Q|= (Q:Q)1/2the matrix euclidean norm. Then, from (4) and (5) H=H(Q) = −ε∆Q+f(Q) (6) where f(Q) = ∂F ∂Q(Q) = a Q −b 3Q2+QQt+QtQ+c|Q|2Q. Finally, the system is completed with the following initial and boundary conditions over Γ = ∂Ω: u|t=0 =u0, Q|t=0 =Q0in Ω, (7) u|Γ=0, ∂nQ|Γ= 0 in (0, T ), (8) where ndenotes the normal outwards vector on the boundary Γ. The system (1)-(8) is a simplified version of the following Q-tensor model studied by Paicu & Zarnescu in [12] and Abels et al. in [1]:              Dtu−ν∆u+∇p=∇ · τ(Q) + ∇ · σ(Hpz, Q) in Ω ×(0, T ), ∇ · u= 0 in Ω ×(0, T ), DtQ−(WQ−QW) = −γ Hpz(Q) in Ω ×(0, T ), (9) complemented with the initial and boundary conditions (7)-(8), where Wis the antisymmetric part of ∇u, that is W:= (∇u+ (∇u)t)/2, and Hpz(Q) := −ε∆Q+a Q −bQ2−tr(Q2) 3I+c|Q|2Q. The model (1)-(8) was studied in [9], obtaining also the modifications needed to assure symmetry and traceless of Q. In fact, it suffices to replace ∇uby the antisymmetric part W= (∇u+∇ut)/2 in the stretching term S(∇u, Q) defined in (3) and the H(Q) function given in (6) by H(Q) + α(Q)Iwhere α(Q) is an appropriate scalar function [9]. 2 These properties of symmetry and traceless are assumed (but not rigorously justified) in [12] and [1] for the model (9). Since the model (9) is a particular case of the general model studied in [9], then any weak solution (u, Q) of (9) satisfies that Q(t) is a traceless and symmetric tensor. By simplicity, in this paper we consider the model (1)-(8), because it retains the essential difficulties of a Q-tensor model like (9). In fact, the results obtained here can be extended to the Q-tensor model (9). The large-time behavior of some models for Nematic liquid crystals with unknown vector director are studied in [15], [8] (without stretching terms) and in [11], [7], [14] (with stretching terms) and in [13] (where different results are deduced depending on considering or not the stretching terms). On the other hand, the large-time behavior is also analyzed for others related models, for example in [6] for a Cahn-Hilliard-Navier-Stokes system in 2Ddomains, in [5] for a chemotaxis model, and in [4] and [3], where a Cahn-Hilliard-Navier-Stokes vesicle model and a smectic-A liquid crystals model are studied respectively. In [10], some results of local in time regularity and uniqueness of the model (1)-(8) are proved. Sections 2 and 3 describe the model and the weak solution concept (more details can be seen in [9]). The novelty of this paper is in the last two sections. In Section 4, two precise energy inequalities are proved via Galerkin Method, a time-integral version for all time t and a time-differential version for almost every time. These inequalities will be essential later and they have neither been proved in [13] nor in [2]. Section 5 is devoted to the study of convergence at infinite time for global weak solutions. In fact, we prove first that the ω-limit set for weak solutions consists of critical points of the free-energy. Finally, by using a Lojasiewicz-Simon’s result, we demonstrate the convergence of the whole trajectory to a single equilibrium as time goes to infinity. Notations The notation can be abridged. We set Lp=Lp(Ω), p≥1, H1=H1(Ω), etc. If X=X(Ω) is a space of functions defined in the open set Ω, we denote by Lp(0, T ;X) the Banach space Lp(0, T ;X(Ω)). Also, boldface letters will be used for vectorial spaces, for instance L2=L2(Ω)N, and the type L2=L2(Ω)N×Nfor the tensors. We set Vthe space formed by all fields u∈C∞ 0(Ω)Nsatisfying ∇u= 0. We denote H (respectively V) the closure of Vin L2(respectively H1). Hand Vare Hilbert spaces for the norms | · |2and k · k1, respectively. Furthermore, H={u∈ L2;∇u= 0,u·n= 0 on ∂Ω},V={u∈H1;∇u= 0,u= 0 on ∂Ω}. From now on, C > 0 will denote different constants, depending only on data of the problem. 3 2 The Landau-De Gennes theory Liquid crystals can be seen as an intermediate phase of matter between crystalline solids and isotropic fluids. Nematic liquid crystals consist of molecules with, for instance, rod-like shape whose center of mass is isotropically distributed and whose direction is anisotropic, almost constant on average over small regions. In the Landau-De Gennes theory, the symmetric and traceless matrix Q∈R3×3, known as the Q-tensor order parameter, measures the deviation of the second moment tensor from its isotropic value. A nematic liquid crystal is said to be isotropic when Q= 0, uniaxial when the Q-tensor has two equal non-zero eigenvalues and can be written in the special form: Q=sn⊗n−1 3Iwith s∈R\{0},n∈S2 and biaxial when Qhas three different eigenvalues and can be represented as follows: Q=sn⊗n−1 3I+rm⊗m−1 3I where s, r ∈R;n,m∈S2. The definition of the Q-tensor is related to the second moment of a probability measure µ(x,·) : L(S2)→[0,1] for each x∈Ω, being L(S2) the family of Lebesgue measurable sets on the unit sphere. For any A⊂S2,µ(x, A) is the probability that the molecules with centre of mass in a very small neighborhood of the point x∈Ω are pointing in a direction contained in A. This probability must satisfy µ(x, A) = µ(x,−A) in order to reproduce the so-called “head-to-tail” symmetry. As a consequence, the first moment of the probability measure vanishes, that is hpi(x) = ZS2 pidµ(x, p) = 0. Then, the main information on µcomes from the second moment tensor M(µ)ij =ZS2 pipjdµ(p), i, j = 1,2,3. It is easy to see that M(µ) = M(µ)tand tr(M) = 1. If the orientation of the molecules is equally distributed, then the distribution is isotropic and µ=µ0,dµ0(p) = 1 4πdA and M(µ0) = 1 3I. The deviation of the second moment tensor from its isotropic value is therefore measured as: Q=M(µ)−M(µ0) = ZS2p⊗p−1 3Idµ(p). From this equality, Qis symmetric and traceless. 3 Weak solutions Definition 1 (Weak solution) It will be said that (u, Q)is a weak solution in (0,+∞)of problem (1)-(8) if:    u∈L∞(0,+∞;H)∩L2(0,+∞;V), Q∈L∞(0,+∞;H1(Ω)) ∩L2(0; T;H2(Ω)) ∀T > 0, (10) 4 and satisfies the variational formulation (11) and (12) (defined below), the initial conditions (7) and the boundary conditions (8). Note that the regularity imposed in (10) is satisfied up to infinite time excepting the H2(Ω)-regularity for Q. In [9] the following result is proved by means of a Galerkin approximation. Theorem 2 (Existence of weak solutions) If (u0, Q0)∈H×H1(Ω), there exists a weak solution (u, Q)of system (1)-(8) in (0,+∞). Variational formulation Taking into account that ∂iF(Q) = F′(Q) : ∂iQ=f(Q) : ∂iQ, the term of the symmetric tensor τ(Q) can be rewritten as: (∇ · τ(Q))i=H(Q) : ∂iQ−∂iF(Q) + ε 2|∇Q|2, where |∇Q|2=∂jQ:∂jQ. Then, testing (1) by any e u: Ω →R3with e u|∂Ω=0and ∇· e u= 0 in Ω, we arrive at the following variational formulation of (1): (Dtu,e u) + ν(∇u,∇e u)−((e u· ∇)Q, H) + (σ(H, Q),∇e u) = 0.(11) On the other hand, testing (2) by any e Hand the system −ε∆Q+f(Q) = Hby any e Q, we arrive at the variational formulation:    (∂tQ, e H) + ((u· ∇)Q, e H)−(S(∇u, Q),e H) + γ(H, e H) = 0, ε(∇Q, ∇e Q) + (f(Q),e Q)−(H, e Q) = 0, (12) for any e H,e Q: Ω →R3×3. From (12), one has in particular: (∂tQ, e Q) + ((u· ∇)Q, e Q)−(S(∇u, Q),e H)−ε γ (∆Q, e Q) + γ(f(Q),e Q) = 0.(13) On the other hand, by applying regularity (10) to the systems (11) and (13), one has ∂tu∈L4/3 loc ([0,+∞); V′) and ∂tQ∈L4/3 loc ([0,+∞); L2(Ω)), hence, the following time-continuity can be deduced: u∈C([0,+∞); V′)∩Cw([0,+∞); H), Q ∈C([0,+∞); L2(Ω)) ∩Cw([0,+∞); H1). In particular, the initial conditions (7) have sense. Dissipative energy law and global in time a priori estimates Now, we argue in a formal manner, assuming a regular enough solution (u, p, Q) of (1)-(8). By taking e u=uin (11) and ( e H, e Q) = (H, ∂tQ) in (12) then the stretching term cancels with the term dependent on the tensor σ(H, Q), the term ((u· ∇)Q, H) appearing in both 5 (11) and (12) also cancel and the convection term ((u· ∇)u,u) vanishes, hence the following “energy equality” holds: d dt 1 2kuk2 L2(Ω) +ZΩ E(Q)dx+νk∇uk2 L2+γkHk2 L2= 0.(14) For the moment, bounds for (u, Q) are not guaranteed from (16) because ZΩ E(Q)dxis not a positive term due to F(Q). However, it is possible to find a large enough constant µ > 0 depending on parameters a,band cgiven in the definition of F(Q) in (5), such that Fµ(Q) := F(Q) + µ≥c 8|Q|4.(15) By replacing E(Q) in (14) by Eµ(Q) := 1 2|∇Q|2+Fµ(Q)≥0, and denoting the kinetic and phase energies as Ek(u(t)) := 1 2kuk2 L2and Eµ(Q) := ZΩ Eµ(Q)dx and the total energy as E(u, Q) := Ek(u) + Eµ(Q), then (14) implies d dtE(u(t), Q(t)) + νk∇uk2 L2+γkHk2 L2= 0.(16) This energy equality shows the dissipative character of the model with respect to the total free-energy E(u(t), Q(t)). In fact, assuming finite total energy of initial data, i.e. ZΩ Eµ(Q0)dx+1 2ku0k2 L2(Ω) <+∞, then the following estimates hold: u∈L∞(0,+∞; L2(Ω)) ∩L2(0,+∞;H1(Ω)), ∇Q∈L∞(0,+∞;L2(Ω)), Fµ(Q)∈L∞(0,+∞;L1(Ω)), H∈L2(0,+∞;L2(Ω)). (17) In particular, from (15) and (17), we deduce the regularity: Q∈L∞(0,+∞;L4(Ω)) and Q∈L∞(0,+∞;H1(Ω)), hence, in particular Q∈L∞(0,+∞;L6(Ω)).(18) Since f(Q) is a third order polynomial function, |f(Q)| ≤ C(a, b, c)|Q|+|Q|2+|Q|3 6 which, together with (18), gives f(Q)∈L∞(0,+∞;L2(Ω)). Then, using that H(Q) = −ε∆Q+f(Q), we obtain: ∆Q∈L∞(0 + ∞;L2(Ω)) + L2(0,+∞;L2(Ω)) hence ∆Q∈L2(0, T ;L2(Ω)) ∀T > 0. Finally, by using the H2-regularity of the Poisson problem:    −ε∆Q+Q=f(Q) + Qin Ω, ∂nQ|Γ= 0 we deduce that: Q∈L2(0, T ;H2(Ω)) ∀T > 0. 4 Two improved energy inequalities Now, we are in order to prove the following technical lemma. Lemma 3 Let (u, Q)be a weak solution in (0,+∞)of problem (1)-(8) furnished by a Galerkin approximation. Then, (u, Q)satisfies the following energy inequality a.e. t1, t0: t1≥t0≥0: E(u(t1), Q(t1)) − E(u(t0), Q(t0)) + Zt1 t0 (νk∇u(s)k2 L2+γkH(s)k2 L2)ds ≤0.(19) Moreover, there exists a special function e E=e E(t)∈Rdefined for all t≥0, which satisfies the following integral inequality for all t1, t0:t1≥t0≥0: e E(t1)−e E(t0) + Zt1 t0 (νk∇u(s)k2 L2+γkH(s)k2 L2)ds ≤0,(20) and the following differential version a.e. t≥0: d dt e E(t) + νk∇u(t)k2 L2+γkH(t)k2 L2≤0.(21) Proof: To prove (20) we start from the following energy equality satisfied by the Galerkin approximate solutions (see [9]) for all t, t0with t≥t0≥0: E(um(t), Qm(t)) − E(um(t0), Qm(t0)) + Zt t0 (νk∇um(s)k2 L2+γkHm(s)k2 L2)ds ≤0.(22) Moreover, um(t) and Qm(t) have sufficient estimates to obtain E(um(t), Qm(t)) → E(u(t), Q(t)) in L1(0, T ), and in particular a.e. t≥0.(23) Since um→uweakly in L2(0, T ;H1) and Hm→Hweakly in L2(0, T ;L2), lim inf m→+∞Zt1 t0 (νk∇um(s)k2 L2+γkHm(s)k2 L2)ds ≥Zt1 t0 (νk∇u(s)k2 L2+γkH(s)k2 L2)ds (24) 7 for all t1, t0:t1≥t0≥0. By taking lim infm→+∞in (22), we obtain that for all t1≥t0≥0, lim inf m→+∞E(um(t), Qm(t)) + lim inf m→+∞Zt1 t0 (νk∇um(s)k2 L2+γkHm(s)k2 L2)ds ≤lim sup m→+∞ E(um(t0), Qm(t0)). (25) By using (23) and (24) in (25), we obtain (19). On the other hand, since the inequality (19) is satisfied for all t0, t1∈[0,+∞)\N, where Nis a set of null Lebesgue measure, then the map t∈[0,+∞)\N→ E(u(t), Q(t)) ∈Ris a real decreasing (and bounded) function. The, we can define a special function e E(t) for all t∈[0,+∞) as: e E(0) := E(u0, Q0),e E(t) := lim s→t− s∈[0,+∞)\N E(u(s), Q(s)). This function e Eis “continuous from the left” and decreasing for all t≥0. Indeed, for any t1, t2∈[0,+∞), for instance t1< t2, we can choose sequences {s1 n},{s2 n} ⊂ [0,+∞)\Nsuch that s1 n→t− 1,s2 n→t− 2and, s1 n≤s2 nfor all n≥n0. Since s1 nand s2 nare not in N, we know that E(u(s1 n), Q(s1 n)) ≥ E(u(s2 n), Q(s2 n)). By taking limit as s1 n→t− 1and s2 n→t− 2, we obtain that e E(t1)≥e E(t2). Since e E(t) is decreasing for all t∈[0,+∞), it is derivable (and absolutely continuous) almost everywhere t∈(0,+∞). Since the inequality (19) is satisfied for all t0, t1∈[0,+∞)\Nwhere the measure of N is zero, given any t0< t1, we can take δn>0 and ηn>0 such that t0−δn, t1−ηn6∈ Nand δn, ηn→0, hence e E(t1−ηn)−e E(t0−δn) + Zt1−ηn t0−δn (νk∇u(s)k2 L2+γk∇H(s)k2 L2)ds ≤0. By taking δn→0 and ηn→0, we obtain (20). In particular, by choosing t0=tand t1=t+hin (20), we obtain e E(t+h)−e E(t) h+1 hZt+h t (νk∇u(s)k2 L2+γk∇H(s)k2 L2)ds ≤0,∀t, h ≥0.(26) Observe that lim h→0 1 hZt+h t (νk∇u(s)k2 L2+γk∇H(s)k2 L2)ds =νk∇u(t)k2 L2+γk∇H(t)k2 L2, a.e. t≥0 because the map, s∈[0,+∞)→νk∇u(s)k2 L2+γk∇H(s)k2 L2∈R, belongs to L1(0,+∞). Accordingly, by taking h→0 in (26), we obtain (21) a.e. t≥0.  5 Convergence at infinite time. Let (u, Q) be a weak solution of (1)-(8) in (0,+∞) associated to an initial data (u0, Q0)∈H× H1(Ω) (see Definition 1) satisfying Lemma 3. From the energy inequality (19), there exists 8 a real number E∞≥0 such that the total energy evaluated in the trajectory (u(t), Q(t)) satisfies E(u(t), Q(t)) ցE∞in Ras t↑+∞.(27) Let us define the ω-limit set of this global weak solution (u, Q) as follows: ω(u, Q) = {(u∞, Q∞)∈H×H1:∃{tn} ↑ +∞s.t. (u(tn), Q(tn)) →(u∞, Q∞) weakly in L2×H1}. Let Sbe the set of critical points of the energy E(Q) defined in (4), that is S={Q∈H2:−ε∆Q+f(Q) = 0 in Ω, ∂nQ|Γ= 0}. Theorem 4 Assume that (u0, Q0)∈H×H1. Fixed (u, Q)a weak solution of (1)-(8) in (0,+∞)satisfying Lemma 3, then ω(u, Q)is nonempty and ω(u, Q)⊂ {0} × S. Moreover, for any Q∞∈ S such that (0, Q∞)∈ω(u, Q), it holds Eµ(Q∞) = E∞. In particular, u(t)→0weakly in L2and Eµ(Q(t)) → Eµ(Q∞)in Ras t↑+∞. Proof: Observe that since (u, Q)∈L∞(0,+∞;H×H1), for any sequence {tn} ↑ +∞there exists a subsequence (equally denoted) and suitable limit functions (u∞, Q∞)∈H×H1, such that u(tn)→u∞weakly in H, Q(tn)→Q∞weakly in H1.(28) We consider the initial and boundary-value problem associated to (1)-(8) restricted on the time interval [tn, tn+ 1] with initial values u(tn) and Q(tn). If we define un(s) := u(s+tn), Qn(s) := Q(s+tn), Hn(s) := H(s+tn) for a.e. s∈[0,1], then, (un, Qn) is a weak solution to the problem (1)-(8) in the time interval [0,1]. From the energy inequality (19), we have that Z1 0 (νk∇un(s)k2 L2+γkHn(s)k2 L2)ds =Ztn+1 tn (ν|k∇u(t)k2 L2+γkH(t)k2 L2)dt ≤ Eµ(Q(tn)) − Eµ(Q(tn+ 1)) −→ 0 as n→ ∞, hence, ∇un→0 strongly in L2(0,1; L2) and Hn→0 strongly in L2(0,1; L2). 9