On the 2D Ericksen-Leslie equations with anisotropic energy and external forces

In this paper we consider the 2D Ericksen-Leslie equations which describes the hydrodynamics of nematic Liquid crystal with external body forces and anisotropic energy modeling the energy of applied external control such as magnetic or electric field. Under general assumptions on the initial data, the external data and the anisotropic energy, we prove the existence and uniqueness of global weak solutions with finitely many singular times. If the initial data and the external forces are sufficiently small, then we establish that the global weak solution does not have any singular times and is regular as long as the data are regular.


Introduction
We consider a hydrodynamical system modeling the flow of liquid crystal materials with anisotropic energy in a 2D bounded domain. More precisely, let T > 0 and Ω ⊂ R 2 be a bounded domain with This article is part of a project that is currently funded by the European Union's Horizon 2020 research and innovation programme under the Marie Sk lodowska-Curie grant agreement No. 791735 "SELEs". P. Razafimandimby is very grateful to the warm hospitality of the Lehrstuhl für Angewandte Mathematik, Montanuniversität Leoben, Leoben (Austria), where part of the paper was written during the period 16 December 2019 till 30 January 2020. G. Deugoué is thankful to the financial support and the hospitality of the Department of Mathematics at the University of York during his visit in June 2019. He is also very grateful to the financial support from the Marie Sk lodowska-Curie grant (agreement No. 791735) "SELEs". a smooth boundary ∂Ω and let us consider (v(0), d(0)) = (v 0 , d 0 ), in Ω, (1.1f) where v : [0, T )×Ω → R 2 , d : [0, T )×Ω → S 2 , where S 2 is the unit sphere in R 3 , and P : [0, T )×Ω → R represent the velocity field of the flow, the macroscopic molecular orientation of the liquid crystal material and the pressure function, respectively. In the system (1.1), the function φ : R 3 → R + is a given map, f : [0, T ) × Ω → R 2 and g : [0, T ) × Ω → R 3 are given external forces. The symbol ν(x) is a unit outward normal vector at each point x ∈ ∂Ω. The matrix ∇d ⊙ ∇d is defined by Using the identities we can rewrite system (1.1) as follows While we focus our mathematical analysis on the system (1.2) with the Dirichlet and the Neumann boundary conditions (1.2d), our results remain valid for the case of the periodic boundary conditions. That is, our results remain true in the case that Ω is a 2D torus T 2 and (1.2d) is replaced by Ω v(t, x) dx = 0, ∀t ∈ (0, T ]. The model (1.2) is an oversimplification of a Ericksen-Leslie model of nematic liquid crystal with a simplified energy density 1 2 |∇d| 2 + φ(d).
The term 1 2 |∇d| 2 represents the one-constant simplification of the Frank-Oseen energy density and φ(d) represents an anisotropic energy density. One example of such anisotropic energy density is the magnetic energy density φ(n) = (n · H) 2 , when the nematic liquid crystal is subjected to the action of a constant magnetic field H ∈ R 3 . We also give different examples of mathematical models of anisotropic energy density later on. For more details on physical modeling of liquid crystal under the action of external control such as magnetic or electric field we refer to the books [9] and [34] and the papers [10] and [23].
We should note that since (1.2) was obtained by neglecting several terms such as the viscous Leslie stress tensor in the equation for v(see for instance [27,28]), the stretching and rotational effects for d, one does not known whether it is thermodynamically stable or consistent with the laws of thermodynamics. However, this model still retains many mathematical and essential features of the hydrodynamic equations for nematic liquid crystals. In recent years, several liquid crystal models which are thermodynamically consistent and stable have been recently developed and analyzed, see for instance the [11], [12], [13], [14], [15], [16], [31], [29] and references therein.
In the absence of external forcings f , g and the anisotropic energy potential φ(d), the system (1.2) has extensively studied and several important results have been obtained. In addition to the papers we cited above we refer, among others, to [17,18,26,27,28,30,41] for results obtained prior to 2013, and to [7,19,20,21,22,24,25,40,42,43] for results obtained after 2014. For detailed reviews of the literature about the mathematical theory of nematic liquid crystals and other related models, we recommend the review articles [29,15,8] and the recent papers [20,25].
Let us now outline the contributions of our manuscript.
(1) In Section 3, we prove by using Banach fixed point theorem that if (v 0 , d 0 ) ∈ D(A  (2) We exploit this result and the local energy method developed in [36], [26] and [17] to show in Section 4 that there exists universal constants ε 0 > 0 and r 0 such that the following statement hold.
We refer to Proposition 4.3 and its proof for more details.
(3) The two results above are exploited in Section 5 in order to prove the global existence of our problem. This is our main result. It holds under weaker assumptions than those listed in (1) and (2) above, and is presented in Theorem 5.6. It can be summarized as follows.  |u(t, y)| 2 + 1 2 |∇d(t, y)| 2 + φ(d(t, y)) dy ≥ ε 2 1 .
Because of the presence of the anisotropic energy and the external forces, this result is a generalization of the global existence of weak solution obtained in [17] and [26]. (4) Finally, in Section 6 we prove that the set of singular times is empty when the data (v 0 , d 0 ) ∈ H × H 1 and (f, g) ∈ L 2 (0, T ; H −1 × L 2 ) are sufficiently small. We also show that if the data are sufficiently regular and small, i.e. (v 0 , d 0 ) ∈ D(A 1 2 ) × D(Â) and (f, g) ∈ L 2 (0, T ; H × H 1 ), then the weak solution becomes regular for all time. Moreover, for all t ∈ [0, T ) (u(t), d(t)) lies in a compact set of H × H 1 . We refer the reader to Theorems 6.3 and 6.5 for more detail about these results. We close this introduction with the presentation of the layout of the present paper. In Section 2 we fix the frequently used notation in the manuscript. We also state and prove some auxiliary results which are essential to our analysis. Section 3 is devoted to the existence and uniqueness of of a regular solution to Problem (2.13). In Section 4 we prove that one can find a small number R 0 > 0 and a unique maximal local regular solution ((v, d); T 0 ) such that its energy at any time t ∈ [0, T 0 ) does not exceed twice the supremum of all energies on B(x, 2R 0 ), x ∈ Ω of the initial data. This result will play a pivotal role in the proof of the existence of a maximal local strong solution to Problem (2.13) in Section 5.

Notation and preliminaries
Let Ω ⊂ R 2 be an open and bounded set. We denote by Γ = ∂Ω the boundary of Ω. We assume that the closure Ω of the set Ω is a manifold with C ∞ boundary Γ := ∂Ω which is a 1-dimensional infinitely differentiable manifold being locally on one side of Ω.

2.1.
Notations for the velocity field v. The following is an abridged version of notations and preliminary of the paper [2]. The facts we enumerate here can be found in [2, Section 2] and references therein.
Let D(Ω) (resp. D(Ω)) be the set of all C ∞ class vector fields u : R 2 → R 2 with compact support contained in the set Ω (resp. Ω). Then, let us define The inner products in all L 2 spaces will be denoted by ·, · . The space E(Ω) is a Hilbert space with a scalar product u, v E(Ω) := u, v + div u, div v .
We endow the set H with the inner product ·, · and the norm |·| H induced by L 2 .
The space H can also be characterized in the following way, see [ Because of this the norms on the space V induced by |·| 1 H = |·| L 2 + |∇·| L 2 and by |∇·| L 2 are equivalent. Since the space V is densely and continuously embedded into H, by identifying H with its dual H ′ , we have the following embedding Let us observe here that, in particular, the spaces V, H and V ′ form a Gelfand triple. We will denote by | · | V * and ·, · the norm in V * and the duality pairing between V and V * , respectively. Now, define the bilinear form a : V × V → R by setting It is well-known that this bilinear map is V-continuous and V-coercive, i.e. there exist some Next we define an unbounded linear operator A in H as follows  Under our assumption on Ω, A and D(A) can be characterized as follows It is also well-known, see [2,Section] and references therein, that A is a positive self adjoint operator in H and 2 is the complex interpolation functor of order α 2 . Furthermore, for α ∈ (0, 1 2 ) In particular, V = D(A 1/2 ) and |A 1 2 u| 2 = |∇u| 2 =: u 2 for u ∈ V. The equality (2.7) leads to the following result which was proved in [2, Proposition 2.1].
Let us finally recall that the projection Π extends to a bounded linear projection in the space L q , for any q ∈ (1, ∞).
Indeed, b is a continuous trilinear form such that Let us observe that if v ∈ D(A), then B(u, v) ∈ H and the following identity is a direct consequence of (2.8). (2.9) The restriction of the map B to the space D(A) × D(A) has also the following representation where Π is the Leray-Helmholtz projection operator and u∇v = 2 j=1 u j D j v ∈ L 2 (Ω, R 2 ).

2.2.
The Laplacian for the director field d. Throughout this section we still denote L 2 (Ω; R 3 ) and H k (Ω; R 3 ), k ∈ N, by L 2 and H 1 , respectively. We aim in this subsection to introduce the Laplacian for the director d : Ω → R 3 with the Neumann boundary conditions. We can do this by mimicking the way we define the Stokes operator A. We define the bilinear mapâ : It is clear thatâ is continuous, and hemce, by Riesz representation lemma, there exists a unique bounded linear operatorÂ : H 1 → (H 1 ) * such that (H 1 ) * Â d, n H 1 =â(d, n), for d, n ∈ H 1 . Next, we define an unbounded linear operatorÂ in L 2 as follows Under our assumption on Ω, it is known, see for instance [38,Section2,p. 65] thatÂ and D(Â) can be characterized by where ν = (ν 1 , ν 2 , ν 3 ) is the unit outward normal vector field on ∂Ω and ∂d ∂ν is the directional derivative of d in the direction ν.
Let us recall that the operatorÂ is self-adjoint and nonnegative and D Â 1/2 when endowed with the graph norm coincides with H 1 . Moreover, the operator (I +Â) −1 is compact. Furthermore, if we denoteV := D(Â 1/2 ), the (V, L 2 , V * ) is a Gelfand triple and 3. An abstract formulation of problem (1.1). With the notations we have introduced above, we can now rewrite problem (1.2) as an abstract equations. In fact by projecting the first equation in (2.13) onto H we obtain the following system 3. The existence and the uniqueness of a regular solution to Problem (2.13) Throughout the whole section, we fix a map φ : R 3 → R 3 satisfying the following set of conditions.
Example 3.2. Let H ∈ R 3 be a constant vector. Then the anisotropy energy potential φ due to the action of a magnetic or electric is defined by This potential φ satisfies the Assumption 3.1. In this case H represents a constant magnetic or electric field applied to the nematic liquid crystal. Another mathematical example which satisfies Assumption 3.1 is the potential defined by where ξ ∈ R 3 is a fixed constant vector.
Next, we consider the problem (2.13) on a finite time horizon [0, T ]. Throughout the paper we put For this section, we impose the following set of conditions on the data.
Under this assumption we will prove in this section that Problem (2.13) has a unique local regular solution. Before stating and proving this result we define what we mean by a maximal local regular solution.
Throughout this paper we will denote by ((v, d); T 0 ) a local regular solution defined on [0, We also need the definition of a maximal local regular solution.
for any other local regular solution ((ṽ,d);T 0 ) we havẽ We state the following important remark.
With the definitions and remark in mind we are now ready to formulate our first result.
then the problem (2.13) has a local regular solution In order to prove this theorem we shall introduce the following spaces 2 )), We also set Let (v, n) ∈ X T and consider the following decoupled linear problem Before continuing further, let us recall the following result.
2 )) and T > 0, then the problem has a unique strong solution (u, d) ∈ X T . Moreover, there exists a constant C > 0, independent of T , such that . Now we state the following lemma whose proof will be given in the appendix.
Lemma 3.9. There exists a constant C 0 > 0, independent of T , such that for all v i ∈ X 1 T , d i ∈ X 2 T , i = 1, 2, the following inequalities hold (3.12) Now, we will give the proof of Theorem 3.7.
Proof of Theorem 3.7. Let Ψ : X T → X T be the map defined as follows. If (v, n) ∈ X T , then Ψ(v, n) = (u, d) iff (u, d) is the unique regular solution to (3.10) with right hand side of the form (3.14) Let us observe that by Lemma 3.9 the term (f, g) defined in (3.14) belongs to L 2 (0, T ; H × D(Â 1 2 )). Hence, by Lemma 3.8 (u, d) ∈ X T , and so the map Ψ is well-defined.
. Then, it is easy to check that (u, d) solves the following problem Hence, by Lemma 3.8 there exists a constant C > 0, independent of T , such that Then, by plugging (3.11), (3.12) and (3.13) in the above inequality and performing elementary calculations imply that there exists a constant C 1 > 0, independent of T , such that for all (v i , n i ) ∈ K R 1 ,R 2 , i = 1, 2, Since g ∈ L 2 (0, T ; H 1 ), for any ε > 0 there exists T 1 ∈ (0, T ) such that Next, we choose a number T 2 such that Hence, by choosing ε = 1 4 and setting T 0 = T 1 ∧ T 2 we infer that for all (v i , n i ) ∈ K R 1 ,R 2 , i = 1, 2, Hence, Ψ has a unique fixed point (u, d) ∈ X T 0 satisfying Thus, in order to prove Theorem 3.7 it remains to prove that For this purpose, let We recall that there exists a constant C > 0 such that for all n ∈ H 3 |n| H 5 2 ≤ C|n| Also, since (u, d) ∈ X T 0 we infer from Lemma 3.9 that Using this and d ∈ X 2 T 0 we easily prove that Now we will claim that z satisfies the weak form of the following problem   2 )) and D(Â) ⊂ L ∞ we easily prove that ϕd ∈ C([0, T 0 ]; H 1 ) ⊂ L 2 (0, T 0 ; H 1 ). Also, since (u, d) ∈ X T 0 we infer from Lemma 3.9 that where we used the fact that ϕd ⊥ R 3 d × g. Since d(t) ∈ D(Â) and ϕd(t) ∈ H 1 for all t ∈ [0, T 0 ], by using [3, Equation (2.6)] we infer that Thus, straightforward calculation yields Hence, recalling the definition of z and using the last identity in (3.24) implies This is exactly the weak form of (3.23). By Proposition B.3 z is the unique solution of (3.23) and satisfies sup 0≤t≤T Since z(0) = 0, we infer that which implies that d(t) ∈ M for all t ∈ [0, T 0 ]. This completes the proof of (3.18). This also completes the proof of Theorem 3.7.
4. The existence and the uniqueness of a maximal local regular solution to (2.13) The aim of this section is to prove that Problem (2.13) has a unique maximal local regular solution when the initial data has small energy. The main result of the section is Proposition 4.3 and it is a generalization of [26,Lemma 5.2]. Before proceeding to a precise statement and a detailed proof of the result let us introduce few notations. For R > 0 and (u, n) ∈ H × H 1 we set and We also recall the following important lemma, see [36, Lemma 3. (4.4) We state and prove the following important result.
There exist a constant ε 0 > 0 and a function which is non-increasing w.r.t. the second variable and nondecreasing w.r.t. the first one, such that the following holds: Then, there exists a unique maximal local regular solution ((v, d); T 0 ) to problem (2.13) satisfying In this theorem, the length T 0 is not the length of the existence interval but the length of the existence interval as long as the condition (4.7) is satisfied.
In order to prove the above proposition we need several results. For n ∈ R 3 We state and prove the following elementary results.
Proof. Let us fix u ∈ V, n ∈ D(Â) and m : [0, T * ) → D(Â) ∩ M satisfying the assumptions of Claim 4.5. Then, since div u = 0 and the fact n(x) ∈ S 2 x-a.e. we get which completes the proof of the claim.
From the Cauchy-Schwarz and the Young inequalities and, the fact d(t) ∈ M, t ∈ [0, T * ), which is part (4) of Definition 3.4, we infer that there exists a constant C > 0 such that Absorbing the first term on the right hand side of the last inequality into the left hand side and integrating over [s, t] ⊂ [0, T * ) completes the proof of the lemma.
For any ε > 0 and R > 0 we define the time We state and prove the following lemma.
Proof. Let r 0 > 0 be the constant from Lemma 4.1, R ∈ [0, r 0 ] and ε ∈ (0, ε 0 ) where ε 0 is number to be chosen later. We set . Since φ is twice continuously differentiable and the 2-sphere S 2 is compact, we can and will assume throughout that for some constant M > 0

From this observation we infer that for all
The last line of the above inequalities, (4.4), (4.9) and (4.11) imply that which completes the proof of (4.12).
We now proceed to the proof of (4.13). For this we observe that by Lemma 4.1 and (4.12) we infer that for all t In a similar way, we prove that for all t which altogether with (4.14) imply (4.13).
We will need the following estimates which will be proved in the Appendix C.
Claim 4.10. There exists a constant K 1 > 0 such that for all v ∈ D(A) and n ∈ D(Â 3/2 ) we have We will also need the following results.
Claim 4.11. There exists a constant K 2 > 0 such that for all n ∈ D(Â 3/2 ) we have Proof. Using the Cauchy-Schwarz inequality and the fact that there exists a constant M > 0 such that |φ ′ (n)| + |φ ′′ (n)| ≤ M, n ∈ S 2 , we infer that there exists a constant C > 0 such that We now complete the proof using the Young inequality in the last line.
Then, there exist constants K 3 > 0 and Proof. Throughout this proof C > 0 will denote an universal constant which may change from one term to the other. Let (v 0 , d 0 ) ∈ H× D(Â) and ((v, d); T * ) be a local regular solution to the problem (2.13). By part (1) and (4) Hence, By using the Cauchy-Schwarz, the Young inequalities and the Ladyzhenskaya inequality ([37, Using the Hölder and Young inequalities, ,and the Sobolev embedding H 1 ֒→ L 4 we also have Plugging these estimates and the ones in Claims 4.10-4.11 into (4.23) yield (4.24) Hence, Let us put Thus, by the Gronwall Lemma, we obtain which along with the inequality (4.9) the fact θ ≤ e θ , θ ≥ 0 implies that Integrating (4.24), using (4.27) and the fact θ ≤ e θ , θ ≥ 0 yield We easily infer from (4.27) and (4.28) that (4.22) holds. This completes the proof of the lemma.
We have the following consequence of the above lemma.
, r 0 and ε 0 be as in Lemma 4.1 and Lemma 4.9. Let (v, d) ∈ X T * be a maximal local regular solution to the problem (2.13). Let Σ 0 be defined as in (4.21).
then we can apply Lemma 4.12 and infer that (4.22) holds for τ = T (ε , R). Hence, in order to complete the proof of the corollary we need to estimate the exponential term in the right-hand side of (4.22). For this purpose, we use (4.13) and infer that there exists a universal constant K > 0 such that for This completes the proof of the corollary.
and (v, d) ∈ X T * be a maximal local regular solution to the problem (2.13). Let r 0 > 0 and ε 0 > 0 be as in Lemma 4.1 and Lemma 4.9, respectively. Then, for all ε ∈ (0, ε 0 ) and R ∈ (0, r 0 ] we have Proof. We argue by contradiction. Assume that there exists ε ∈ (0, ε 0 ) and R ∈ (0, r 0 ] such that T (ε , R) = T * . Let us put By Corollary 4.13 and (4.9) we infer that there exists a constantK 3 > 0 such that for all t ∈ [0, T * ) It is easily seen that ((v,d); T 1 ) is a local regular solution to (2.13) with initial data (v 0 , d 0 ) and time of existence T * + T 1 > T * . This contradicts the fact that ((v, d); T * ), with T * = T (ε , R), is a maximal smooth solution to (2.13). This completes the proof of the corollary.
We now state and prove a local energy inequality which will play an important role in the proof of Proposition 4.3.
, r 0 and ε 0 be as in Lemma 4.9. Also, let R ∈ (0, r 0 ], ε ∈ (0, ε 0 ) and (v, d) ∈ X T * be a local regular solution to the problem (2.13). Then, there exists a function p : Moreover, there exists a constant K 5 > 0, which may depend on the norms of Proof. Let us fix ε ∈ (0, ε 0 ) and R ∈ (0, r 0 ]. We fix t ∈ [0, T * ). From (4.9) and (4.12) we have ). Hence, one can apply [26,Lemma 4.4] and infer that there exists function p : 3 ) and the identity (4.31) holds. Moreover, . (4.32) From the Hölder inequality, (4.9) and (4.13) we infer that |v · ∇v| In a similar way, . We now continue with some estimates of local energy. Hereafter, in order to save space we Let ϕ ∈ C ∞ c (Ω; R) and put We also set Then, for any Proof. We fix ϕ ∈ C ∞ c (Ω; R) and s ≤ t ∈ [0, T * ). We firstly observe that because of the fact for all t ∈ [0, T * ) and a.e. x ∈ Ω. Multiplying ∂ t d + v · ∇d by −ϕR(d) in L 2 and using the pointwise orthogonality yields (4.37) Using integration by parts and [26, Eq. (4.16)] we obtain For the term B it is easy to show that Note also that for all t ∈ [0, T * ) and a.e. x ∈ Ω.
Hence, using integration by parts and the Cauchy-Schwarz inequality we obtain Plugging (4.38), (4.39) and (4.40) into (4.37) yields We can follow the same calculation in [26] to derive the following local inequality for the velocity v Adding up the last inequalities side by side and using the Cauchy-Schwarz inequality and integrating over [s, t] yield the sought estimate (4.35).
The following lemma is also important for our analysis.
We now deal with term containing the pressure p. First by the Using the Hölder and Poincaré inequalities and the estimates (4.9) we obtain .

From the last line and Lemma 4.15 we infer that
.
We now deal with the terms containing |ϕ|. Applying the Hölder inequality and (4.13) yields 1 We also have t 0 Ω |f ||v||ϕ| dxdr ≤ Ct . We are now ready to give the proof of Proposition 4.3.
Proof of Proposition 4.3. We recall that under the assumption of Proposition 4.3 there exists a unique solution (v, d) ∈ X T * to the problem (2.13), see Theorem 3.7. We can assume that T * > 0 is the maximal time of existence of (v, d). Let r 0 > 0 and ε 0 > 0 be as in Lemma 4.1 and Lemma 4.9, respectively. Let R 0 ∈ (0, r 0 ) be chosen such that dx is absolutely continuous, then it is possible to choose such R 0 . We also observe that ε 2 1 < E 0 . Now, let .
• If T 0 > R 2 0 , then because θ 0 (ε 1 , E 0 ) ∈ (0, 3 4 ], , then by Corollary 4.14, the definition of T 0 and the continuity of (v, d) at t = T 0 we infer that Hence, by using the inequality (4.42) with t = T 0 and the fact ε 2 1 < E 0 , we infer the existence of a universal constant K 7 > 0 such that where Ψ is defined in (4.19). Since R 0 ≤ r 0 we have from which we deduce that By the definition of T 0 = T (ε 1 , R 0 ), see (4.10), and the fact T 0 < T * we automatically obtain (4.4). Thus, the proof of Proposition 4.3 is complete.

The existence and the uniqueness of a global weak solution
In this section we will prove global existence of a weak solutions to problem (2.13). Before we state and prove this result let us define the concept of a weak solution.
We also introduce the notion of local strong solution which will be needed to prove the existence of a global weak solution to our problem.  Similarly to Definition 3.5, one we can also define the notion of a maximal local strong solution.
We state the following important remark.
Let us now state the standing assumptions of this section.
Assumption 5.4. Let T > 0 and assume that (f, g) ∈ L 2 (0, T ; H −1 × L 2 ). We also assume that The first main result of this section is the following uniqueness result.
2 ), i = 1, 2, be two strong solutions to (2.13) defined on [0, T ]. Then, Proof. In order to prove this result we closely follow the approach of [24].
Let (v i , d i ) be a two strong solutions to (2.13), By parts (1) and (4)  and using the facts that A is self-adjoint and div w = 0 we infer that We also used the integration by parts to obtain the second line. Let us now estimate the terms on the right hand side of the last line of the chain of identities above.
Firstly, by using the Hölder, the Young inequalities and the Ladyzhenskaya inequality ([37, Lemma III.3.3]) we infer that Observe that |v| 2 In a similar way, we can prove that

Secondly, making use of the Ladyzhenskaya inequality ([37, Lemma III.3.3]), the Hölder and the Young inequalities we obtain
Collecting all these inequalities we obtain

Let us turn our attention to the function
Since d 1 , d 2 ∈ L 2 (0, T ; D(Â)), and ∂ t d 1 , ∂ t d 2 ∈ L 2 (0, T ; L 2 ), we infer by applying Lions-Magenes Lemma ([37, Lemma III.1.2]), and the facts v 2 · ∇d, d = 0 and d × g, d = 0 (because d × g ⊥ R 3 d) that Let us estimate the terms in the right hand side of (5.2). First, by using the Hölder, the Young inequalities and the Gagliardo-Nirenberg inequality ([1, Section 9.8, Example C.3]) we show that With the same idea, we prove that In the last line we used the fact that |d 1 | L ∞ ≤ 1.
Utilizing the Hölder, the Young inequalities and the Gagliardo-Nirenberg inequality ([1, Section 9.8, Example C.3]) we obtain Using the definition of α(d 2 ) = (φ ′ (d 2 ) · d 2 ), the fact |d 2 | = 1 and (3.1) we have Using again the definition of α(d 1 ) and α(d 2 ) we obtain Since φ ′ is Lipschitz, d i (t) ∈ M for all t ∈ [0, T ], we show with the same ideas as used in (5.3) and Hence, collecting all these inequalities related to the terms in the right hand side of the equation (5.2) we obtain Thus, summing (5.1) and (5.5) up, we have 1 2 . Then, we see from (5.6) that y satisfieṡ Thus, one can apply the Gronwall inequality to (5.7) and deduce that This completes the proof of the Proposition 5.5.
The second main result of this paper is the following theorem.
for all R ∈ (0, ̺ 0 ]. (4) At each T i there is a loss of energy at least ε 2 1 ∈ (0, ε 2 0 ), i.e., The proof of this theorem is established in several steps. The first of such steps is the proof of the existence of a maximal local strong solution for the Ericksen-Leslie system (2.13) with data satisfying Assumption 5.4.  For the proof of this theorem, we first prove the existence of a local strong solution which is given by the following proposition. which is non-increasing w.r.t. the second variable and nondecreasing w.r.t. the first one such that the following holds. Let r 0 > 0 be the constant from Lemma 4.1 and assume that the initial data (v 0 , d 0 ) and the forcing (f, g) satisfies Assumption 5.4. Then, there exists ̺ 0 ∈ (0, r 0 ] such that Moreover, there exists a local strong solution ((v, d); T 0 ) satisfying see [32,Section 4]. By definition, v 0 ∈ H can be approximated by a sequence {v k Since embedding L 2 ֒→ H −1 and H 1 ⊂ L 2 are dense, then one can also approximate (f, g) ∈ L 2 (0, T ; H −1 × L 2 ) by a sequence ((f k , g k )) k∈N ⊂ L 2 (0, T ; L 2 × H 1 ) in the following sense Let ε 0 > 0 and R 0 ∈ (0, r 0 ] be the constants from Proposition 4.3. Since E(v 0 , d 0 ) < ∞ and µ(A) = Ω∩A |v 0 | 2 + |∇d 0 | 2 + φ(d 0 ) dx is absolutely continuous, then there existsR 0 > 0 such that Choosing ̺ 0 =R 0 ∧ R 0 yields (5.9). Let ε > 0 be an arbitrary real number.
In a similar way one can prove that, Observe that Since |d k 0 | = |d 0 | = 1, sup Hence, there exists a constant k 0 ∈ N such that for all k ≥ k 0 Without loss of generality, we will assume that for all k ≥ 1 3 4 ] satisfying the properties stated in Proposition 5.8, a sequence of time T k 0 satisfying (5.10) and a sequence of regular solutions We recall that for all k ≥ 1 and t ∈ [0, T k 0 ), we have Hereafter, we put We also recall that there exists C > 0 such that for all k ∈ N and R ∈ (0, ̺ 0 ] We now estimate the time derivatives. Let us put Then, we deduce from Claim 4.5 that Now we estimate the integral on the right hand side of the last line as follow: By employing (5.16) in the last line we get Summing up these discussion, we get Thus, we obtain the following uniform estimate of ∂ t d We now estimate the time derivative ∂ t v k . Let ϕ ∈ V. We have Hence This altogether with (5.16) and (5.17) imply that there exists C > 0 such that for all k ∈ N Now, let us set Since for all k ≥ 1 then by the definition of T 0 , we get It then follows from the previous analysis that the sequence . Hence by Aubin-Lions compactness lemma and Banach-Alaoglu theorem, one can extract a subsequence Let t ∈ [0, T 0 ]. Then, the sequence (v k (t), d k (t)) k∈N is bounded in H × H 1 . Hence, thanks to the compact embedding H 1 ֒→ L 2 we can and we will assume that the subsequence (v k j , d k j ) j∈N satisfies, for all t ∈ [0, T 0 ] d k j (t) → d(t) strongly in L 2 . This and the fact d k j (t) ∈ M for all j ∈ N, t ∈ [0, T 0 ] implies that there exists a constant C > 0 such that for all j ∈ N, t ∈ [0, Passing to the limit as j → ∞ yields (5.21) Our next step is to show that the limit (v, d) satisfies the system (2.13). Hence, we need to pass to the limit in the nonlinear terms. In order to do this, we firstly observe that the convergences are now well-known, see, for instance, [37] or [4], hence we omit their proof.
Secondly, the most difficult point is the convergence lim k→∞ T 0 0 ||∇d k | 2 d k − |∇d| 2 d| L 2 dt = 0, (5.22) and hence we prove it here. For this quest we notice that there exists a constant C > 0 such that for all k ≥ 1 By the Hölder inequality, we have By the Ladyzhenskaya inequality ([37, Lemma III.3.3]) and the Sobolev embedding H 1+θ ֒→ L ∞ (θ ∈ (0, 1)), we arrive at where we have also used the fact that The strong convergence (5.20) and the fact that completes the proof of (5.22). We now study the convergence of the term φ ′ (d k ). Since d k j → d strongly in L 2 (0, T 0 ; H 1 ), we can assume that d k j → d a.e. (t, x) ∈ [0, T 0 ] × Ω. Thus, by the continuity of φ ′ (.), we get Since d k j (t) ∈ M, t ∈ [0, T 0 ) and, by assumption, |φ ′ (d k j )| ≤ M , the Lebesgue dominated convergence theorem implies that and d k j → d strongly in L 2 (0, T 0 ; L ∞ ), we get We will now prove that (v, d) satisfies the initial conditions and that (v, d) ∈ C([0, T 0 ]; H × H 1 ). Towards these goals, we first observe that since (v, d) ∈ L ∞ (0, T 0 ; H × H 1 ) and then by the Strauss theorem, see [37, Lemma III.1.2], we get From all these passages to the limits we see that the limit v and d satisfy the equations (3.4) and (3.4) in V * and L 2 , respectively. The estimates (5.11) and (5.12) are established by passing to the limit and using the weak lower semicontinuity of the norms in the estimates (5.18), (5.17) and (5.15).
What remains to prove is the continuity of (v, d) 2 ). For this we will firstly establish that (∂ t v, ∂d) ∈ L 2 (0, T ; V * × L 2 ).
Proof of Theorem 5.7. In order to prove the theorem, let us denote by Σ the set of local solutions to problem (2.13). By Proposition 5.8, the set Σ is non empty and we can and will assume that the time of existence T 0 of any local solution (v, d) ∈ Σ satisfies the property (5.10) stated in Proposition 5.8. Let us define the relation on Σ by (y 1 ; σ 1 ) (y 2 ; σ 2 ) if σ 1 ≤ σ 2 and y 2 = y 1 on [0, σ 1 ], for all (y i ; σ i ) := ((u i , d i ); σ i ) ∈ Σ, i = 1, 2.
The following result gives an important property of the energy of the maximal solution ((v, d); T * ) near the point T * . Proposition 5.9. Let ε 0 > 0, ̺ 0 > 0 and θ 0 be the constants and the function from Proposition 5.8 and d); T * ) be the maximal solution defined in Theorem 5.7. Then, Proof. We prove the proposition by contradiction. Suppose that there exists R > 0 such that Thus, there exists an increasing sequence (t n ) n∈N such that t n ր T * as n → ∞, |v(t n )| 2 + |∇d(t n )| 2 + φ(d(t n )) dy < ε 2 1 .
By the global energy inequality (4.9), we get Hence, by Proposition 5.8, there exists a solution (ṽ,d) defined on [t m , τ + t m ] with τ ≥ θ 0 (ε 2 1 , E(v(t m ), d(t m ))R 2 0 . But observe that θ 0 (ε 2 1 , E 0 ) is a non-increasing function of the initial energy E 0 . Hence, by (5.33) By doing elementary calculation we obtain Hence, we get the existence of a local solution (ṽ,d) . This contradicts the fact that ((v, d); T * ) is a maximal solution.
We now give the promised proof of Theorem 5.6.
Proof of Theorem 5.6. Let ((v, d), T * ) be the maximal local strong solution to (2.13) constructed from Theorem 5.7. Firstly, we set T * = T 1 and prove the following result.
We can now continue with the proof of Theorem 5.7. From Lemma 5.10, we can define Since ∇d ∈ L ∞ (0, T 1 ; L 2 (Ω)), then ∇d(t) ⇀ ∇d(T 1 ) weakly in L 2 (Ω). This and Theorem D.1 implies that v(T 1 ) ∈ H and d(T 1 ) ∈ H 1 and (v, d) ∈ C w ([0, T 1 ]; H × H 1 ). Moreover, thanks to the strong convergence d(t) → d(T 1 ) in L 2 (Ω), we show using the same idea as in the proof of (5.21) that Also, it is not difficult to prove that We also prove that In fact, the inequality (5.45) implies that there exists a sequence t n ր T 1 and x 0 ∈ Ω such that lim sup Therefore, This completes the proof of (5.39). With (5.38) and (5.37) at hand, one can apply Proposition 5.8 to construct a local strong solution (ṽ,d) : Furthermore, Hence, we can construct a sequence of maximal local strong solutions ( In order to construct the global solution we consider a function (v, d) defined and prove that L < ∞. Towards this aim, we first deduce from (5.43) that In order to complete the proof we need to check that (v, d) is indeed a global weak solution. But this follows from the definition (5.44), the fact that each ((v i , d i ); T i ) are maximal local strong solution defined on [T i−1 , T i ) and satisfying 6. On the regularity and the set of singular times of the solutions when the data is small In this section we prove that the set of singular time reduces to the final time horizon T when the data are small enough. Let us start with the following conditional regularity of a strong solution ((v, d); T * ). Proof of Proposition 6.1. We start the proof with the following claim. 1 In fact,ε 1 = min{ε i; 1 ≤ i ≤ L} Claim 6.2. There exist constants K > 0 and K 7 > 0 depending only on the norms of (v 0 , d 0 ) ∈ V × D(Â 1 2 ) and the norms of (f, g) ∈ L 2 ([0, T ]; H × H 1 ) such that the following holds. then, This and the assumption (6.1) implies the desired inequality (6.2). This proves the claim.
Now we give the proof of Proposition. ). Thanks to the claim again, we can repeat the above procedure finitely many times, say, on [T 0 , T 1 ], . . . [T N , T * ] to assert that With this we conclude the proof of the proposition. Theorem 6.3. Let ε 0 > 0 andε 1 ∈ (0, ε 0 ) be the constants from Proposition 5.8 and Theorem 5.6, respectively. If the data For this purpose, we argue by contradiction. Assume that T 1 < T and that T 2 = T . Then, by part (2) of Theorem 5.6 which altogether with the assumption (6.4) yields lim sup This clearly contradicts the fact that lim sup |v(t, y)| 2 + |∇d(t, y)| 2 + φ(d(t, y)) dy ≥ε 2 1 > 0, for all R ∈ (0, ̺ 0 ]. This completes the proof of the first part of the theorem. The second part of the theorem follows easily from the Proposition 6.1. Hence, the proof of the theorem is completed. The last, but not the least, result of this section is about the precompactness of the orbit (v(t), d(t)), t ∈ [0, T ), in H × D(Â). This result will require the following set of conditions on the map φ.
It is clear that if φ satisfies this assumption, the it also satisfies Assumption 3.1.
then any regular solution (v, d) to (2.13) defined on [0, T ) satisfies . Let ε 0 > 0 and ε 1 ∈ (0, ε 0 ) be the constants from Proposition 5.8 and Theorem 5.6, respectively. Let κ 0 ∈ (0, ε 1 ] be a number to be chosen later such that (6.7) holds. Let (v, d) be a regular solution to (2.13) defined on [0, T ). Then, multiplying the velocity equation by v and using the Cauchy-Schwarz and Young inequalities imply 1 2 We multiply the optical director equation by −∆d, then use Cauchy-Schwarz and Young inequalities and the constraint d(t) ∈ M, t ∈ [0, T ], and obtain 1 2 Using the fact φ ′ (d) = d − ξ and the integration-by-parts on the torus we find 1 2 Before proceeding further, let us estimate the last term of the above inequality. Toward this end we divide the task into two parts. Firstly, we use fact that ∆d · d = −|∇d| 2 , the Hölder, the Gagliardo-Nirenberg ([1, Section 9.8, Example C.3]) , the Young inequalities and [33,Theorem 3.4] to get the following chain of inequalities Plugging (6.14) and (6.15) into the inequality (6.13) yields (6.16) Now adding the inequalities (6.12) and (6.16) side by side, and using (4.8) imply Thanks to the energy inequality (4.9) we obtain We now easily conclude the proof of (6.10) in the proposition by integrating (6.17), taking the supremum over t ∈ [0, T 0 ] and choosing In order to prove the second estimate, we use the Gagliardo-Nirenberg inequality (see [1, Section 9.8, Example C.3]) and (6.10) to obtain This completes the proof of the proposition.
The next result is crucial for the proof of Theorem 6.5.
Proposition 6.7. Let ε 0 be as in Proposition 5.8 and (v 0 , d 0 ) ∈ V × D(Â), f ∈ L 2 (0, T ; L 2 ), g ∈ L 2 (0, T ; H 1 ). Assume that where κ 1 ∈ (0, ε 0 ) is defined in Proposition 6.6. Then, there exist a constant K 2 > 0, which depends only on the norms of (v 0 , d 0 ) ∈ V × D(Â) and (f, g) ∈ L 2 (0, T ; H × H 1 ), such that for a regular solution (v, d) to (2.13) we have Proof. The proof of the estimate (6.19) is very similar to the proof of (4.22), hence we only sketch the proof. We start with the same idea as in the proof of (4.22), i.e., we multiply the velocity and optical director equations by Av and A 2 d, respectively. This procedure implies the equation (4.23).
Appendix A. Proof of Lemma 3.9 In this section we will prove Lemma 3.9.
Proof of Lemma 3.9. Firstly, we infer from [5, Lemma 2.5] 2 that there exists a constant C > 0 such that for all This clearly implies that there exists a constant C > 0 such that for all (v i , d i ) ∈ X T , i = 1, 2, Hence, we infer that there exists a constant C > 0 such that for all (v i , d i ) ∈ X T , i = 1, 2, from which we easily deduce the inequality (3.11).
We will now proceed to the proof of (3.12) which will be divided into four steps. Firstly, because of the assumption (3.3) the map φ ′ : R 3 → R 3 is Lipschitz continuous. Hence, for all n i ∈ H 2 , i = 1, 2, Using (3.2), (3.3) Thus, the proof of Claim A.1 is complete.
In a similar way, one can show that there exists a constant C > 0 such that for all n i ∈ X 2 T , i = 1, 2, We easily complete the proof of (A.6) by using (A.7) and (A.8) in the following inequality In order to complete the proof of Lemma 3.9 we need to establish the inequality (3.13). For this purpose, we firstly observe that it is not difficult to prove that there exists a constant C > 0 such that for all n i , i = 1, 2, Hence, there exists a constant C > 0 such that for all n i ∈ X 2 T , i = 1, 2, we have which is the inequality (3.13). Hence, the proof of Lemma 3.9 is complete.
Appendix B. Weak solution of a modified viscous transport equation Proof. Throughout this proof c > 0 will denote an universal constant which depends only on Ω and may change from one term to the other. For the sake of simplicity we will omit the dependence on the space variable inside any integral over Ω.
Proof of inequality (4.16). Let us choose and fix v ∈ D(A) and n ∈ D(Â 3 2 ). Since D(Â This complete the proof of the inequality (4.16).

Appendix D. A weak continuity of Banach space valued functions
In this section we will state and prove of continuity theorem for Banach-valued map similar to [35]. This theorem was initially proven in the work in progress [6], but for the sake of completeness we repeat the proof here.
Theorem D.1. Let X and Y be two Banach spaces such that X is reflexive, X ⊂ Y and the canonical injection i : X → Y is dense and continuous. Let T > 0 be fixed and u ∈ L ∞ ([0, T ); X). is weakly continuous.
Proof. Let X, Y , b ∈ Y , T > 0, u ∈ L ∞ ([0, T ); X), v : [0, T ] → Y be as in the statement of the theorem. Let also (t n ) n∈N ⊂ [0, T ) be a sequence such that t n ր T . Let us prove the first part of the theorem, i.e., that b ∈ X. For this aim we observe that by assumption there exists M > 0 such that |u(t)| X ≤ M for all t ∈ [0, T ).
Hence, by the Banach-Alaoglu theorem we can extract from (t n ) n∈N a subsequence, which is still denoted by (t n ) n∈N , such that t n ր T and u(t n ) → x weakly in X.
Since, by assumption, X ⊂ Y we infer that v(t) = i(u(t)) = u(t) for all t ∈ [0, T ]. Hence, by the weak continuity of v we infer that v(t n ) = u(t n ) → b weakly in Y.
By the uniqueness of weak limit we infer that x = b ∈ X.
It remains to prove the second part of the theorem, i.e., we shall show that the mapũ : [0, T ] → X defined in (D.1) is weakly continuous. For this purpose, we will closely follow the proof of [37,Lemma III.1.4 ]. We will divide the task into two cases.
Case 1 Let t, t 0 ∈ [0, T ). Let X * and Y * be the dual spaces of X and Y , respectively. Recall that since the embedding X ⊂ Y is dense and continuous, the embedding Y * ⊂ X * is dense and continuous, too. Let us choose and fix ε > 0 and η ∈ X * . Then, there exists η ε ∈ Y * such that |η − η ε | X * < ε .
Thus, using the boundedness of u =ũ [0, T ) we infer that By the weak continuity of v [0,T ) = u =ũ we have Since ε > 0 is arbitrary we infer that is weakly continuous.
Thus,ũ is weakly continuous at t = T . This completes the proof of Case 2, the second part of the theorem and hence the whole theorem.