Global Weak Solutions to a General Liquid Crystals System

We prove the global existence of finite energy weak solutions to the general liquid crystals system. The problem is studied in bounded domain of $R^3$ with Dirichlet boundary conditions and the whole space $R^3$.


1.
Introduction. Liquid crystals were discovered in 1888 by F.Reinitzer and O.Lehmann. They are often viewed as intermediate states between the solids and fluids, whose molecular arrangements give rise to preferred directions. As a result, they retain several different features: mechanical, electrical, magnetic and optical properties. According to molecular arrangements, it has been widely supported that liquid crystals can be classified into three types: nematics, cholesterics and smectics. The historical development of liquid crystals confronts two theories, including the swarm theory and the distortion theory. The former theory is well established but only applied to nematics and cholesterics, while the later one, successfully explaining the interactions of nematics with magnetic fields, is not so well-known for us. Based on the predecessors' work, in the 1960's, Ericksen and Leslie established the Hydrodynamic theory of nematic liquid crystals system(see [16], [17], [18], [3], [22] and [31]): ρ t + div(ρu) = 0, (1.1) (ρu) t + div(ρu ⊗ u) = ρF + divσ, (1.2) where ρ ≥ 0, u = (u 1 , u 2 , u 3 ), d = (d 1 , d 2 , d 3 ) are the fluid density, velocity and molecular direction respectively. F denotes the external body force, G the external body force for the direction movement, g the internal body force for the direction movement and ρ 1 dw dt the angular movement per unit time. In low frequency, ρ 1 dw dt is so small that can be neglected. The equations (1.1)-(1.3) represent the conservation where p = aρ γ denotes the pressure, ρH the bulk free energy, and let d i,j represent ∂di ∂xj . In equation (1.6), κ is Lagrange multiplier constraint to |d| = 1. According to Frank's formula [7] (Chapter 3), we can get the bulk free energy of nematic types.
Likewise, from [31] we have Here d ⊗ d denotes a matrix whose (i, j)-th entry is given by d i d j . We use the following notations: here N = (N 1 , N 2 , N 3 ) is used to describe the director movements in satellited coordinates, and w = (w 1 , w 2 , w 3 ) is the material derivatives of d. And λ i , µ i satisfy the following formulas: (1.11) is called Parodi's condition, which is derived from the onsager reciprocal relation, see [23], [3], [31]. For simplicity, we assume k 4 = 0, k 1 = k 2 = k 3 = 1.
The case of bounded domain D: We are interested in the global weak solutions to the system (1.17)-(1.19) in a bounded domain D ⊂ R 3 with initial conditions: (1.21) and the boundary conditions In order to give the definition of the weak solutions, we firstly describe energy inequality where B is a smooth function and Then we have the following result: and the Orlicz space L p 2 (R 3 )(see [24] pp.288) We consider our problem (1.17)- (1.19) in the whole space R 3 with initial conditions: (1.28) One can obtain the energy inequality d dt We define the weak solution in the sense of following  There are a lot of results for the incompressible and compressible liquid crystals systems. For the incompressible case with a constant density, F. Lin and C. Liu systematically studied the existence and partial regularity of weak solutions in their papers [9], [10], [11]. For the density is not constant, X. Liu and Z. Zhang [28] proved existence of weak solutions in a bounded domain of R 3 , also to see F. Jiang and Z. Tan [14] for weakening assumption of [28]. For the compressible case, in 2009, X. Liu and J. Qing [29] firstly proved the existence of weak solutions to liquid crystals system (more simpler than one considered in present paper) in a bounded domain of R 3 . Similar results also see D. Wang, Y. Cheng [4]. We also note that papers by S. Ding, C. Wang and H. Wen [26] for one dimension case and by X. Hu, D.Wang [30] for the Besov space case.
In this paper, we consider a more generic(fitting more physical properties) liquid crystals system (1.17)-(1.19) and study its existence of weak solutions in both bounded domain and the whole space. The main difficulties for solving the problem are dealing with pressure term and the nonlinear terms appeared in the stress tensors. Like Navier-Stokes equations, there are phenomenons of losing compactness caused by the higher nonlinear terms of the equations. Fortunately, we overcome those difficulties by using much more technical methods such as compensated compactness used in P. L. Lions' book [24], also to see E. Feireisl [5], [6].
The remaining part of this paper is organized as follows. Section 2, 3, 4 are devoted to proof Theorem 1.1. Using Theorem 1.1, we prove Theorem 1.2 in Section 5.
2. Approximate solutions. In this section, similar to Eduard Feireisl did on Navier-Stokes equations(see [5], [6]), we firstly construct an approximate problem: complemented by the initial conditions: YUMING CHU, YIHANG HAO AND XIANGAO LIU and the boundary conditions: where n is the out normal vector of ∂D. having the following properties: is the unique classical solution of (2.1), (2.4) and (2.7); Let {φ i } ∞ i=1 be the orthogonal basis of H 1 0 (D), which satisfies: Here a i , (i = 1, 2 · · · ) is the eigenvalue of the operate −∆, and 0 < a 1 ≤ a 2 ≤ · · · a n ≤ · · · , a n → ∞, n → ∞. We consider the finite dimensional space And X n is Hilbert space equipped with norm given by scalar product of L 2 . We shall look for the approximate solution u = u n ∈ C([0, T ]; X n ), satisfying the following integral equation .
where X * n is the dual space of X n . Since ρ has a positive lower bound, then M [ρ] is invertible, and satisfies Obviously from above, we have The equation (2.10) can be rewritten as Here q * 0 ∈ X * n , N ∈ X * n satisfy Next we deduce the energy estimates of (2.1)-(2.3) in finite dimensional space X n . Let (S(u), u, R(u)) be a solution of (2.1)-(2.3) for u ∈ X n . We rewrite (2.10) as follows:

YUMING CHU, YIHANG HAO AND XIANGAO LIU
Integrating above equation in time to obtain where E 0,δ is From the energy law and (2.14), using the standard method in [28], the integral equation (2.10) can be solve in any interval [0, T ].
By energy law again, we have the following lemma describing the information of the approximate solution (ρ n , u n , d n ).
Proof. (2.15)-(2.17) can be directly obtained from (2.14). So we only need to consider (2.18), (2.19). The first term of (2.18) is due to Höld inequality. By elliptic estimates, we have and then integrate to obtain By inequality (2.14) and ǫδβρ β−2 Here we use the interpolation of (2.23) and (2.24).

2.2.
The process of n → ∞. In this section, we let n → ∞ in the sequence {(ρ n , u n , d n )}. The following compactness theorem due to J. Lions(see [19] and [25]).
and then By Sobolev embedding theory, we have H 1 (D) ֒→ L k (D) ֒→ W −1,p (D) for 6 5 ≤ k < 6, and H 1 (D) ֒→ L k (D) compactly. Using Lemma 2.3, we obtain Similarly to ρ, we have Then using |d| 2 ≤ 1, we obtain It is easy to get As d n has good regularities, it is easy to deduce σ n ⇀ σ in D ′ ((0, T ) × D) as well as the weak convergence of other terms except for the following two terms In order to continue, we need the following lemma (see [6] Lemma 7.7.5).
Lemma 2.4. Suppose ρ n is a solution of (2.1) supplement with the boundary conditions (2.7) corresponding to u n . Then there exist r > 1, q > 2, such that ∂ t ρ n , ∆ρ n are bounded in L r ((0, T ) × D), ∇ρ n is bounded in L q (0, T ; L 2 (D)) independently of n. Accordingly, the limit function ρ belongs to the same class and satisfies equation (2.1) a.e. on (0, T ) × D together with the boundary conditions (2.7) in the sense of trace.
Since ρ n u n satisfies (2.12), the relation (2.28) can be strengthened as the following term ρ n u n ⇀ ρu in L ∞ (0, T ; L 2γ γ+1 (D)). (2.29) Observing equation (2.2), we have (2.30) Using Lemma 2.3, the above estimates are enough to show and then Next we consider ∇u∇ρ. By virtue of Lemma 2.4, we have Multiplying above equation with ρ − ρ n , we obtain By choosing a suitable β in equation (2.26), the above equation yields Therefore we obtain Summing up the work of this section, we have the following result.
and Lemma 2.4 holds. Moreover, we have the following energy law: (2.32) 3. Taking limit ǫ → 0. We introduce an operator B = [B 1 , B 2 , B 3 ] corresponding in a certain sense to the inverse of div x .
It can be shown (see [21] or [15] Theorem 10.3.3) that (3.1) admits an operator B : g → v enjoying the following properties: • B is a bounded linear operator from L p (D) into W 1,p 0 (D) for any 1 < p < ∞,

YUMING CHU, YIHANG HAO AND XIANGAO LIU
Then we have We estimate each I i as follows: T 0 |ψ t |dt, and Summing up above estimates, we have the following lemma: Due to the Proposition 2.2, we have As we have u ǫ ⇀ u in L 2 (0, T ; H 1 0 (D)), the following term holds weak (D)). Then it is natural to obtain
Passing to the limit for m → ∞, we have So we conclude We introduce Riesz integral operator R and singular integral operatorA.
, and

GLOBAL WEAK SOLUTIONS TO A GENERAL LIQUID CRYSTALS SYSTEM 15
The next lemma is from [6](Corollary 6.6.1).
Lemma 3.5. Let {v n }, {w n } be the two sequences of vector functions, T ) and B is a bounded measurable function satisfying

Proof. Form the definition of A, we have
Taking above relations into (3.5), we conclude this lemma.

YUMING CHU, YIHANG HAO AND XIANGAO LIU
Before proving the lemma, we prolong ρ ǫ to be zero outside D Since ρ, u vanish outsider D, we have div(ρu) = 0, x ∈ R 3 \ D and Proof. Here ρ ǫ , ρ are extended to zero outside of D. Taking ϕ = ψηA[ρ ǫ ] as a test function and using Lemma 3.6, we get Similarly, taking ϕ = ψηA[ρ] as a test function of (3.5), we have
We need the following lemma about renormalized solution(see [6]). For the sake of comleteness, we rewrite the proof. Then (ρ, u) is a renormalized solution of (3.4).
Proof. Taking the mollified operator on both side of (3.4), we have (3.20) and the second term of the right side of (3.19) is bounded, we have Passing to the limit for ǫ → 0, we conclude The term ∇ρ ǫ · n = 0 leads to D ∆ (B(ρ ǫ )) dx = 0. Then we have , satisfying ψ m → 1 and η m → 1. By virtue of Lemma 3.7, we have lim sup The last inequality is due to the fact: B(z) is convex and globally lipschitz on R + . Thus we have proved Setting p(z) = az γ + δz β , it holds Choosing v = ρ + ζϕ, for any ϕ, then ζ → 0 yields p = aρ γ + δρ β . (3.23)

YUMING CHU, YIHANG HAO AND XIANGAO LIU
Thus we have proved the following lemma:

) be a solution to the problem (3.24)-(3.26), then there exists a constant
Passing to the limit for δ → 0, (ρ, u, d) satisfies Let Υ denote the weak convergence limit of Υ in this paper. We have the following lemma.
Proof. By virtue of Lemma 3.2, (3.24) still holds. Taking as a test function for (3.25), with a straightforward computation, we get Using ϕ = ψηA[T k (ρ)] as a test function of (4.5), we have One easily observes where θ is defined in Lemma 4.1. Noting T k (z) is bounded, we can get I δ i → I i term by term as in the proof of Lemma 3.7.
Next we define oscillations defect measure where T k is defined by (4.7). Proof. Using (4.8), we have Moreover, one easily show, (4.10) Here we use p(z) is convex and T k (z) is concave. For p(z) = az γ , we have Considering the definition of T k , the inequality (4.11) yields Thus we have Using (4.10), we get And we easily obtain (4.14) Then we complete the proof.
Proof. Using (ρ δ , u δ ) is renormalized solution of (3.24), we have Passing to the limit for δ → 0, we obtain Similarly to Lemma 3.8, we get the following equation, where B(z) satisfies Utilizing the weak lower semi-continuity of function's norm, we deduce And passing to the limit for k → 0, we have B(T k (ρ)) → B(ρ), B ′ (T k (ρ)) → B ′ (ρ) in the corresponding spaces.
In order to complete the proof, we should show the right side of (4.17) tends to zero as k → ∞. One easily show By interpolation, we get .

(4.19)
Observing the definition of T k , we have In the last step, we prove ρ δ → ρ in L 1 ((0, T ) × D). Let's define L k : Observing (ρ δ , u δ ) is a renormalized solution of (3.24) and (ρ, u) is a renormalized solution of (4.4), we have For any η n ∈ C ∞ 0 (D), It holds From the definition of L k , we can get L k (z) = 2kz − 2k, z ≥ 3k. Let {η n } be sequence such that η n → 1 in D. Passing to the limit for n → ∞ in (4.29), we have  Then we have ρ log(ρ)(t) ≤ ρ log(ρ)(t).