Reproductivity for a nematic liquid crystal model

REPRODUCTIVITY FOR A NEMATIC LIQUID CRYSTAL MODEL Blanca Climent-Ezquerra∗ , Francisco Guill´ en-Gonz´ alez∗ Marko Rojas-Medar† Abstract In this paper we prove existence of weak solution with the reproductivity in time property, for a penalized PDE’s system related to a nematic liquid crystal model. This problem is relatively explicit when time-independent Dirichlet boundary conditions are imposed for the orientation of crystal molecules. Nevertheless, for the time-dependent case, the treatment of the problem is completely different. The verification of a maximum principle for weak reproductive solutions is fundamental in the argument. Finally, the relation between reproductive and periodic in time (regular) solutions will be pointed out, differenting the 2Dand 3Dcases. Basically, in two-dimensional domains every reproductive solution is regular and time periodic, whereas the problem remains open for three-dimensional domains. Mathematical Subject Classification (2000): Primary 35Q35, Secondary 76D03, 35K20, 35A35 Keywords: time reproductive solutions, Navier-Stokes equations, Liquid crystal model, coupled non-linear system, Galerkin method. ∗Dpto. Ecuaciones Diferenciales y An´alisis Num´erico, Universidad de Sevilla, Aptdo. 1160, 41080 Sevilla, Spain. E-mails: [email protected], [email protected] †Dpto. Matem´atica Aplicada, IMECC-UNICAMP, C.P. 6065, 13081-970, Campinas-SP, Brazil. E-mail: mark[email protected] 0 1 Introduction In this work, a nematic liquid crystal model in a simplified Ericksen-Leslie version is considered; see for instance [9] for a formulation of a more complete liquid crystal problem. This model can be seen as a variant of the Navier-Stokes problem (respect to the unknowns velocity-pressure (u, p)) coupled with a convection-diffusion system for a new variable d, a unit vectorial function modelling the orientation of the crystal molecules. On the other hand, it is usual to consider an approximation by Ginzburg-Landau penalization ([1]) for the constraint |d|= 1 (|d|=|d(t, x)|denotes the point-wise euclidean norm). This penalized model (in which the constraint |d|= 1 is relaxed by |d| ≤ 1) was introduced by Lin in [7] and studied (from a mathematical point of view) by Lin and Liu in [8, 9] and by Coutand and Shkoller in [2]. The main difficulties of the model coming from the strongly nonlinear coupling between the orientation vector dand the velocity-pressure (u, p) and from the constraint |d| ≤ 1, jointly with the well known difficulties for the Navier-Stokes problem (a nonlinear parabollic system with the free divergence constraint related to the pressure). In all previous works, initial conditions and time-independent Dirichlet boundary conditions for dare considered. Now, we are interested in time reproductive solutions. In particular, we will see that this study is completely different for timeindependent or time-dependent boundary conditions for d. We assume a (newtonian) fluid confined in an open bounded domain Ω ⊂IRN (N= 2 or 3) with regular boundary ∂Ω. In the penalized model the constraint |d|= 1 is partially conserved to |d| ≤ 1 as consequence of the maximum principle for the Ginzburg-Landau equation considering the penalization function f(d) = 1 ε2(|d|2−1)d where ε > 0 is the penalization parameter. There exists a potential function F(d) = 1 4ε2(|d|2−1)2 such that f(d) = ∇d(F(d)) for each d∈IRN. Then, we consider the following PDE system in (0, T)×Ω:    ∂tu+ (u· ∇)u−ν∆u+∇p=−λ∇ · (∇d ∇d),∇ · u= 0, ∂td+ (u· ∇)d=γ(∆d−f(d)),|d| ≤ 1,(1) The constants ν,λand γare positives, representing respectively, the fluid viscosity, an elasticity constant and a relaxation time. Here, the following tensorial notation 1 is used: (∇d ∇d)ij = N X k=1 ∂xidk∂xjdk,∀i, j = 1, . . . , N. The problem (1) is completed with the (Dirichlet) boundary conditions u(x, t) = 0,d(x, t) = h(x, t) on ∂Ω×(0, T) (2) (assuming as novelty a time-depending boundary data for dgiven by h:∂Ω× (0, T)7→ IRN, in [8, 9] only a time-independent boundary data is considered) and the reproductivity conditions: u(x, 0) = u(x, T),d(x, 0) = d(x, T) in Ω (3) These reproductive conditions jointly with the constraint |d| ≤ 1 are the main difficulties of the problem (1)-(3), while the time-dependent boundary conditions d=h(x, t) produces some additional difficulties. In particular, an adequate lifting of this condition must be done. It is important to remark that reproductive solution with the following boundary data independent of time d(x, t)|∂Ω×(0,T )=d0(x) has the trivial stationary (static) solution: u≡0, dsolution of the elliptic problem: −∆d+f(d) = 0 in Ω,d|∂Ω=d0, psuch that ∇p=−λ∇ · (∇d ∇d). Therefore, in this work will be fundamental assume time-dependent boundary data for d. The concept of reproductive solution in the Navier-Stokes context, appears for the first time in [5], see also [10]. The goal of this paper is to obtain existence of (global in time) weak solution of problem (1)–(3). We start defining a variational formulation, testing the maximum principle for dand the energy inequality of the system. Afterwards, we introduce a Galerkin discretization of the problem, proving existence and uniqueness of approximate solution associated to arbitrary initial conditions. Then, a Leray-Schauder argument (by means of fixed point process about initial and final in time values of the solutions) allows us to obtain a reproductive Galerkin solution, which converges towards a continuous reproductive solution. Finally, some comments about the relation between reproductive and periodic (regular) solutions will be pointed out, being completely different the 2Dand 3Dcases. Basically, in two-dimensional domains every reproductive solution associated to regular boundary data is regular and time periodic, whereas the problem remains open for three-dimensional domains. 2 In our opinion, other interesting open problem related with this work, is the asymptotic behavior of reproductive solutions of (1)–(3) when ε→0. In the case of initial-boundary problem, existence of weak solution of the limit problem is obtained in [3], passing to the limit when ε→0. 2 Variational Formulation For simplicity, we denote L2and H1instead of L2(Ω)Nand H1(Ω)N,L∞(H1) instead of L∞(0, T;H1(Ω)N), etc. Also, the scalar product in L2will be denoted by (·,·), and h·,·i will denote some duality products. Let us consider the following function spaces: H={u∈L2:∇ · u= 0 in Ω,u·n= 0 on ∂Ω} V={u∈H1 0:∇ · u= 0 in Ω} Without less of generality, we fix the constants ν=λ=γ= 1. Obviously, the following compatibility conditions will be imposed in this work for the boundary data function h: |h| ≤ 1 on ∂Ω×(0, T) and h(0) = h(T) on ∂Ω.(4) 2.1 The variational problem for (u,d) Definition 1 We say that (u,d)is a weak reproductive solution of (1)–(3) if u∈L2(V)∩L∞(H) d∈L∞(H1),∆d∈L2(L2),d|∂Ω×(0,T)=h verifying h∂tu,vi+ ((u· ∇)u,v)+(∇u,∇v)+(∇dt∆d,v) = 0 ∀v∈V∩L∞, h∂td,ei+ ((u· ∇)d,e)+(f(d),e)−(∆d,e) = 0 ∀e∈L3, u(0) = u(T),d(0) = d(T)in Ω. tu In order to arrive at the previous variational formulation, the following equalities have been used: ∇ · (∇d ∇d) = ∇ |∇d|2 2!+∇dt∆d and ∇ |∇d|2 2!,v!= 0 ∀v∈V. 3 Notice that the reproductivity conditions u(0) = u(T) and d(0) = d(T) have sense, because uand dare (at least weakly) continuous functions from [0, T] onto some Banach spaces. 2.2 Weak Maximum Principle for d An essential characteristic of the problem for d(given u) is the following weak maximum principle: Lemma 2.1 Assume (4). Then, given u∈L2(0, T;V)∩L∞(0, T;H), any weak solution for the d-problem, i.e. d∈L2(0, T;H1)∩L∞(0, T;L2)such that h∂td,ei+ ((u· ∇)d,e)+(f(d),e)+(∇d,∇e) = 0,∀e∈H1 0,(5) d|∂Ω×(0,T)=hand d(0) = d(T),(6) verifies |d(x, t)| ≤ 1a.e. in Ω×(0, T). tu Proof. Let us define the function: ϕ(x, t) = (|d(x, t)|2−1)+ where z+=max(z, 0) for each z∈IR. Hypothesis |h| ≤ 1 on ∂Ω×(0, T) implies that also ϕ= 0 on ∂Ω×(0, T). Taking e=ϕdas test function (5), one has: 1 2ZΩ∂t(|d|2)ϕ+ZΩ(u· ∇)|d|2ϕ+ZΩ∇d:∇(ϕd) + ZΩϕ f(d)·d= 0 (7) The first three terms of (7) can be written as follows: 1 2ZΩ∂t(|d|2)ϕ=1 2ZΩ∂t(|d|2−1)ϕ=1 4 d dtkϕk2 L2(Ω), ZΩ(u· ∇)|d|2ϕ=ZΩ(u· ∇)(|d|2−1)ϕ=ZΩ(u· ∇)ϕ·ϕ= 0, ZΩ∇d:∇(ϕd) = 1 2ZΩ∇(|d|2)· ∇ϕ+ZΩ|∇d|2ϕ ≥1 2ZΩ∇(|d|2−1)∇ϕ=1 2k∇ϕk2 L2(Ω). Taking into account in the last term of (7) that f(d)·d>0 = 1 ε2(|d|2−1)|d|2as |d|>1 (and f(d)·d= 0 as |d| ≤ 1), we arrive at the differential inequality: d dtkϕk2 L2+ 2k∇ϕk2 L2≤0. 4 Integrating in [0, T] one has kϕ(T)k2 L2+ 2 ZT 0k∇ϕk2 L2≤ kϕ(0)k2 L2 Since ϕ(T) = ϕ(0), the previous inequality implies ∇ϕ= 0 a.e. in Ω ×(0, T). Therefore ϕ(·, t)≡constant in Ω. But, since ϕ(·, t) = 0 on ∂Ω (thanks to |d|= |h| ≤ 1 on ∂Ω), we conclude ϕ(x, t) = 0 a.e. in (x, t)∈Ω×(0, T), i.e. |d(x, t)| ≤ 1 a.e. in Ω ×(0, T). tu Notice that, although the constraint |d| ≤ 1 has been explicitly included in the problem (1)–(3), the previous Lemma say us that it is not necessary because this constraint can be obtained a posteriori. In the sequel, we will consider the penalized function fas a bounded function, since from maximum principle one has |d| ≤ 1, hence one can consider the problem (1) changing fby e fthe auxiliary function e f(d) =    f(d) if |d| ≤ 1 0 if |d|>1 Indeed, if (u, p, d) is a solution of (1) with e f, in particular |d| ≤ 1 (because the maximum principle is also verified, since e f(d)·d>0 as |d|>1), then (u, p, d) is also a solution of (1) with f. The inverse implication is easy to verify . Notice that, |e f(d)| ≤ 1/ε2for each d∈IRN, hence all estimates below will be dependent on ε(recall that the asymptotic behavior as ε→0 for reproductive solutions remains as open problem). 2.3 Variational Formulacion in uand b d Since a time-dependent boundary data h(x, t) on ∂Ω×(0, T) has been considered, an adequate lifting is necessary. Assuming h∈H1(0, T;H1/2(∂Ω)N), if we define e d(t) as the weak solution of the Laplace-Dirichlet problem    −∆e d= 0 in Ω, e d|∂Ω=h(t) on ∂Ω, then e d∈H1(0, T;H1(Ω)N) (notice that only weak regularity of this Laplace-Dirichlet problem is used, therefore it is suffices Lipschitz regularity for the domain Ω). Moreover, since h(0) = h(T) on ∂Ω, then e d(0) = e d(T) in Ω. Therefore, if we define b d(t) = d(t)−e d(t), then b d(t)∈H1 0(Ω)N, ∆b d= ∆din Ω×(0, T) and d(0) = d(T) if and only if b d(0) = b d(T). Then, we can rewrite the problem in the variables (u,b d) as follows: 5 Notice that (u,d) is a weak reproductive solution of (1)–(3) if and only if (u,b d) with d=b d+e d(e dpreviously defined) verifies u∈L2(V)∩L∞(H), b d∈L∞(H1 0),∆b d∈L2(L2), and h∂tu,vi+ ((u· ∇)u,v)+(∇u,∇v)+(∇dt∆b d,v)=0 ∀v∈V∩L∞, h∂tb d,ei+ ((u· ∇)d,e)+(f(d),e)−(∆b d,e) = −(∂te d,e)∀e∈L3, u(0) = u(T),b d(0) = b d(T) in Ω. 2.4 Energy Inequality Taking v=uand e=−∆b das test functions in previous formulation, one has    (∂tu,u) + ((u· ∇)u,u)+(∇u,∇u)+(∇dt∆b d,u) = 0 −(∂tb d,∆b d)−((u· ∇)d,∆b d)−(f(d),∆b d) + (∆b d,∆b d) = (∂te d,∆b d). Adding up, taking into account that ((u· ∇)u,u) = 0 and (∇dt∆b d,u)−((u· ∇)d,∆b d)=0, one arrives (at least formally) at the following energy equality: 1 2 d dt kuk2 L2+k∇b dk2 L2+k∇uk2 L2+k∆b dk2 L2= (f(d),∆b d)+(∂te d,∆b d). Consequently, one has the energy inequality: d dt kuk2 L2+k∇b dk2 L2+ 2k∇uk2 L2+k∆b dk2 L2≤2kf(d)k2 L2+k∂te dk2 L2,(8) where the right hand side is bounded in L1(0, T) since ∂te d∈L2(L2). 3 The Main Result Theorem 3.1 Let T > 0and Ω⊂IRN(N= 2 or 3) an open bounded domain with Lipschitz boundary. Assume h∈H1(0, T;H1/2(∂Ω)) such that (4) hold. Then there exists a weak reproductive solution (u,d)of problem (1)–(3) tu 6 Remark. Hypothesis of regularity for boundary data hcan be relaxed by h∈L∞(0, T;H1/2(∂Ω)) with ∂th∈L2(0, T;L2(∂Ω)). Indeed, this regularity implies the following regularity for the lifting function e d∈L∞(0, T;H1(Ω)) with ∂te d∈L2(0, T;L2(Ω)), which will be sufficient in the sequel. Notice that regularity for time derivative can be proved using the “transposition solution” of the Laplace-Dirichlet problem with boundary data equal to ∂th(due to ∂thhas not a trace sense). By the contrary, further regularity for the domain Ω is necessary (which implies to have H2-regularity for the adjoint problem), in order to define the transposition solution. In the proof of this theorem, the Galerkin method will be used. Firstly, we consider the initial-boundary problem associated to arbitrary initial data. Afterwards, the key is to find certain initial data at t= 0 that are “reproduced” at final time t=T. 3.1 The Galerkin Initial-Boundary Problem Let {φi}n≥1 and {ϕi}n≥1 “special” basis of Vand H1 0(Ω), respectively, formed by eigenfunctions of the Stokes problem (∇φi,∇v) = λi(φi,v)∀v∈V, φi∈V,con kφikL2= 1, λi%+∞ and of the Poisson problem (∇ϕi,∇w) = µi(ϕi,w)∀w∈H1 0, ϕi∈H1,con kϕikL2= 1, µi%+∞ Let Vmand Wmbe the finite-dimensional subspaces spanned by {φ1, φ2, . . . , φn} and {ϕ1, ϕ2, . . . , ϕn}respectively. Given u0∈Hand d0∈H1(verifying the compatibility condition d0|∂Ω=h(0)), for each m≥1, we seek an approximate solution (um,dm), with um: [0, T]7→ Vm and dm=b dm+e d, with b dm: [0, T]7→ Wm, verifying the following variational formulation a.e. in t:                        (∂tum(t),vm) + ((um(t)· ∇)um(t),vm)+(∇um(t),∇vm) +(∇dt m(t)∆b dm(t),vm) = 0 ∀vm∈Vm (∂tb dm(t),em) + ((um(t)· ∇)dm(t),em)+(f(dm(t)),em) +(∇b dm(t),∇em) = −(∂te d(t),em)∀em∈Wm um(0) = u0m=Pm(u0),dm(0) = d0m=Qm(d0), (9) 7 Here, Pm:H7→ Vmdenotes the usual orthogonal projector of Honto Vm, and Qm:L27→ Wmthe orthogonal projector of L2onto Wm. In particular, u0m→u0 in L2and d0m→d0in H1(as m→0). If we put um(t) = m X j=1 ξi,m(t)φiand b dm(t) = m X j=1 ζi,m(t)ϕi, then (9) can be rewritten as a first order ordinary differential system (in normal form) associated to the unknowns (ξi,m(t), ζi,m(t)). Then, one has existence of a maximal solution (defined in some interval [0, τm)⊂[0, T]) of the related Cauchy problem. Moreover, from a priori estimates (independent on m) which will be obtained below, in particular one has that τm=T. Finally, using regularity of the chosen spectral basis, uniqueness of approximate solution will be proved in section 3.3. Remark. Since a discretization in space has been done in definition of approximate solution, the maximum principle is not always verified, therefore the constraint |dm| ≤ 1 is not true in general. 3.2 “A priori” estimates Taking um(t)∈Vmas test function in the u-system of (9) and −∆b dm(t)∈Wmin the d-system (latter is possible due to consider the special eigenfunction basis), and following the argument that yields to energy inequality (Section 2.4), one has d dt kumk2 L2+k∇b dmk2 L2+ 2k∇umk2 L2+k∆b dmk2 L2≤2kf(dm)k2 L2+k∂tk2 L2,(10) hence, using the initial estimates kum(0)k2 L2≤Cand k∇b dm(0)k2 L2≤C, Gronwall’s lemma implies (um) is uniformly bounded in L∞(H)∩L2(V) and (b dm) is uniformly bounded in L∞(H1 0)∩L2(H2) Therefore, since dm=b dm+e dwith e d∈L∞(H1) and ∆dm= ∆b dm, one has (dm) is uniformly bounded in L∞(H1) and (∆dm) is uniformly bounded in L2(L2). 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