On standing waves and gradient-flow for the Landau–De Gennes model of nematic liquid crystals

The article treats the existence of standing waves and solutions to gradient-flow equation for the Landau–De Gennes models of liquid crystals, a state of matter intermediate between the solid state and the liquid one. The variables of the general problem are the velocity field of the particles and the Q-tensor, a symmetric traceless matrix which measures the anisotropy of the material. In particular, we consider the system without the velocity field and with an energy functional unbounded from below. At the beginning we focus on the stationary problem. We outline two variational approaches to get a critical point for the relative energy functional: by the Mountain Pass Theorem and by proving the existence of a least energy solution. Next we describe a relationship between these solutions. Finally we consider the evolution problem and provide some Strichartz-type estimates for the linear problem. By several applications of these results to our problem, we prove via contraction arguments the existence of local solutions and, moreover, global existence for initial data with small L2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^2$$\end{document}-norm.


Introduction
There are different models describing nematic liquid crystals (we avoid long list of references, but refer to the recent one [25] that gives a very detailed panorama of the state of the art). We are interested in the Q-tensor model, i.e. De Gennes type model (see [1,9,16]) for a molecule in a flux of nematic liquid crystals, which is a state of matter intermediate between the solid state and the liquid one. Concerning the physical modeling of nematic liquid crystals, we refer to [1-4, 23, 25]. The model has been actively studied during the last years both on torus [7,24], bounded domains [13], in R n [17,18], and in exterior domains [8,19,20]. In general the model is described by a system of Navier-Stokes equations and an equation for the Q-tensor: We are looking for the functions where S 0 (3, R) denotes the space of 3 × 3 symmetric matrices with zero trace. The function u represents the velocity field of the molecules, p represents the pressure of the fluid and Q is the De Gennes Q-tensor.
In the model, F denotes the energy of liquid crystals and it is given by where a, b, c ∈ R and |·| is the Frobenius norm. It is easy to prove that ∂ F(Q) = a Q + b(Q 2 ) T + c|Q| 2 Q for all Q ∈ M(3, R).
In the case u = 0 the model reduces to the gradient flow equation Our first goal is to find standing waves, i.e. solutions to the equation for Q ∈ H 1 (R 3 ; S 0 (3, R)).
We shall use the following parametrization for the matrices of S 0 (3, R): It can be proved that every critical point q of the following functional: = |∇q| 2 + ∇q 1 · ∇q 2 dx + a(|q| 2 + q 1 q 2 ) + b 3 tr(Q(q) 3 ) + c(|q| 2 + q 1 q 2 ) 2 dx gives a matrix Q(q) which satisfies (1). The part of our work, related to the stationary problem, was announced in [12]. However, for completeness we present the proofs or key points in the proofs, so that the article is self-contained.

Existence of a least energy solution
We study the case a > 0 and c < 0. Definition 2.1 Let d 3, n 2, p = 2d d−2 , G ∈ C 1 (R n \ {0}) and let q be a solution of the system where α ∈ [0, 1] and e 1 , e 2 are the following vectors of R n : We say that q is a least energy solution of (2) if q ∈ C \ {0} and Following [6], we consider a function G ∈ C 1 (R n \ {0}) which satisfies the following properties for p = 2d d−2 : for all v, w ∈ R n ; • there exists a constant C > 0 such that In [6] the existence of a least energy solution is established for α = 0. We need a small generalization to show it for α ∈ [0, 1].
3, n 2 and G : R n → R satisfy the above hypotheses, then there exists a least energy solution q : R d → R n of (2).

Proof
The proof is almost the same as in [6], so we just list the steps: Step 1: Let us consider the problem It can be proved also that, passing to a subsequence, for q ∈ C \ {0}.
Step 2: G(q) dx = 1 and In particular, ∇q j → ∇q in L 2 (R d ; R n ) and therefore q j → q in L p (R d ; R n ).
Step 3: Step 4: If q ∈ L ∞ loc (R d ; R n ) ∩ C is a solution of (2), then there holds Step 5: For all v ∈ L ∞ loc (R d ; R n ) ∩ C non-zero and satisfying (2), we have Step 6: We have the following regularity result: Let q ∈ C be a solution of (2) with g(q) ∈ L 1 loc (R d ; R n ), then it satisfies the following properties: • q tends to 0 as |x| → +∞.
Moreover, if there are C, δ > 0 such that then q decays exponentially as |x| → +∞.
This result corresponds to [6,Theorem 2.3] for α = 0. The proof for α ∈ [0, 1] is analogous. Thanks to this theorem, we gain that q is a least energy solution in C. Let us prove the following easy observation.
then G satisfies the condition (3).
Proof Let us suppose by contradiction that there is γ > 0 such that for any m ∈ N we can find u m , w m ∈ R n which satisfy Divide the above inequality by m[|G(w m )| + |w m | p + 1], we have Let us show that {u m + w m } m is bounded: if {u m + w m } m were unbounded, then passing to a subsequence we can suppose that |u m + w m | → +∞ as m → +∞. Then, for any C > 0, by our hypothesis (5), Let us distinguish two cases: • Suppose |u m | R for some R > 0. This means that |w m | → +∞. By continuity of G, we can find K > 0 such that |G(u m )| K . Therefore which is a contradiction. • Suppose {u m } m is unbounded; then passing to a subsequence, we can suppose |u m | → +∞ as m → +∞. Again by our hypothesis, for any C > 0, |G(u m )| C|u m | p , then which is a contradiction.
Thus, {u m + w m } m is bounded. Now we are ready to conclude: By continuity of G, we have that {G(u m + w m )} m is bounded and therefore which is a contradiction.
In our case n = 5, d = 3, p = 6, α = 1 and Taking into account Lemma 2.4 it is easy to show that g(q) = ∇G(q) satisfies all conditions of Theorem 2.2. Moreover, it is easy to see that it also satisfies (4). Thus we have proved: Corollary 2.5 Let a, b, c ∈ R with a > 0 and c < 0, then there exists a least energy solution q in H 1 (R 3 ; R 5 ) for the system Moreover, q decays exponentially as |x| → +∞.

Least energy solution as a saddle point
Let us recall the functional related to our model: We assumed a > 0 and c < 0. This choice of signs for the parameters gives J a particular structure: since c < 0, the functional is unbounded from below; however, the sign of a implies that J is positive near the zero. Such information suggests existence of a saddle point for the functional.

Definition 2.6
Let X be a Hilbert space and J : X → R be a differentiable function, then we say that We say that J : X → R satisfies the Palais-Smale condition (P.S.) when, for each Palais-Smale sequence, there is a convergent subsequence in X . [22]) Let X be a Hilbert space and J ∈ C 1 (X ) be such that:
is a critical value of J in X .

Remark 2.8
Thanks to the geometric hypotheses of the theorem, it is easy to see that any critical point related to M is not 0.
It is not difficult to check that our J (q) belongs to C 1 (H 1 (R d ; R n )) and that it satisfies the geometric hypotheses of the Mountain Pass Theorem. Conversely, it is difficult to prove the P.S. condition is fulfilled. For this reason we use the Mountain Pass Theorem for J restricted to the space H 1 rad (R d ; R n ): this set is weakly closed in H 1 (R d ; R n ), in particular it is a Hilbert space. Moreover, its elements satisfy the following proposition: Its proof follows from [21, Lemma 1]. (7) with a > 0 and c < 0 has a critical point q = 0.

Proposition 2.10 The function J
The idea is to show that J satisfies the hypotheses of the Mountain Pass Theorem. Firstly, we verify the geometric hypotheses: • Obviously, J (0) = 0.
Then, by Sobolev embeddings, there is K > 0 such that Therefore, we can find ρ sufficiently small and α > 0 such that J (q) α for any q H 1 * = ρ. • Fix q ∈ X different from 0 and let q λ . . = λq for λ > 0.
Now let us show that J satisfies the P.S. condition: let {q k } k ⊆ X be a Palais-Smale sequence, that is If we take v = q k , we get Firstly we consider One can check that Notice that Thanks to this, we get Now, consider Combining (9) and (10), we get We have Passing to a subsequence, we can suppose q k q in X . Now let us show the strong convergence. Denote We notice that I k = 2 q, q k − q H 1 * so, by weak convergence, I k → 0. We are then in the following situation: We need to show that the last two terms tend to 0 as k → +∞. Thanks to Proposition 2.9, we can suppose also that q k → q in L 4 (R 3 ; R 5 ). Then The least energy solution q built in the Sect. 2.1 can be seen as a saddle point of the functional J (q), similarly to the one given by the Mountain Pass Theorem: In fact, with the same techniques as in [14], one can prove Theorem 2.11 Let J : H 1 (R d ; R n ) → R with d 3 and n 2 be defined as where α ∈ [0, 1]; let G satisfy the hypotheses of Theorem 2.2 and condition (4). Let us also suppose that there exists ρ 0 > 0 such that where v 2 let q ∈ C be the least energy solution built in Theorem 2.2, then 1] J (γ (t)).
Proof We give a sketch of the proof: Step 1: Since G satisfies condition (4), thanks to Theorem 2.3 we have that q ∈ with L > 0. It can be seen that, for L sufficiently small, J (γ (1)) < 0, so γ ∈ . On the other hand, thanks to Step 4 in the proof of Theorem 2.2, we have 1] J (γ (t)).
Step 2: Let One can check that is invertible as φ : S → P.
Step 3: It can be seen from the proof of Theorem 2.2 that q = φ(q) with q such that On the other hand, q ∈ H 1 (R d ; R n ), so q ∈ H 1 (R d ; R n ) and we also proved that G(q) dx = 1. Then Thanks to the previous step, Step 4: For any γ ∈ , we have that γ We notice that S(q) = d J (q) − 2 ∇q 2 * . For any γ ∈ we then have It is clear that S ∈ C(H 1 (R d ; R n )), so we can find t 0 ∈ [0, 1] such that S(γ (t 0 )) = 0. We also know by hypothesis that Therefore, γ (t 0 ) > ρ 0 . In particular, γ (t 0 ) = 0, so γ (t 0 ) ∈ P. Thus Just to conclude, we note that

Proposition 2.12 If G(q) is continuous and satisfies
then there exists ρ 0 > 0 such that So our functional (7) satisfies also the additional hypothesis of Theorem 2.11.

Evolution equation
Consider the system where G(Q) = bQ 2 + c|Q| p−1 Q with a > 0, b, c ∈ R and p ∈ (1, 5). Let us recall that by Duhamel's formula any solution Q of (11) satisfies the following equality: The proof follows using the well-known estimates on K t and the Young inequality.

Strichartz estimates
Consider the problem with Q 0 and F taken in a suitable space to be chosen later.
Our aim is to obtain certain estimates in the space L q (R + ; L r (R 3 ; S 0 (3, R))) for q, r ∈ (1, +∞). We will use the norm In particular, when q > 1, these spaces are reflexive. Let us consider the following spacial class of exponents: Definition 3.3 Let σ > 0, then (q, r ) is said to be a σ -admissible couple if q, r 2, (q, r , σ ) = (2, ∞, 1) and If the equality holds, the couple is called strictly σ -admissible. Moreover, when σ > 1, the couple 2, 2σ σ −1 is called the endpoint.
Let us obtain an estimate for the term , F ∈ Lq t (I ; Lr x ), (13) where β(q, q, r , r ) = − n 2 Proof Firstly F( · , x) is defined only on I but, preserving the notation we can suppose that it is defined on R + with support on I . If sup J < inf I , then the left-hand side of (13) is equal to zero, since τ / ∈ I . In this case the inequality is obvious. Let us suppose now that sup I < inf J . Then since I ⊆ [0, t] for t ∈ J . If q < +∞ by the Hölder inequality .
By hypothesis, d(I , J ) = λ, t ∈ J and τ ∈ I , then |t − τ | λ so where we have used that |I | = |J | = λ. For q = ∞ it is easier: Theorem 3.6 For any r , r ∈ [2, +∞) and q, q ∈ (1, +∞) such that we have the following inequality: Proof If we denote Let us define the bilinear form To show that it is useful to rewrite the operator B as Then we conclude following the proof of [11,Theorem 1.4].
Before passing to the main theorem of this section, let us remark that Let us also assume the condition of Proposition 3.4 (in the case s = 0) holds: Then the condition on the indices in Theorem 3.6 becomes This means that (q, r ) and ( q, r ) are strictly n 2 -admissible. Theorem 3.7 Let a 0, n 3, r, r , q, q ∈ [2, +∞) be such that (q, r ) and ( q, r ) are strictly n 2 -admissible and such that 1 r + 1 r > n−2 n . If F ∈ Lq t (R + ; Lr x (R n ; S 0 (3, R))) and Q 0 ∈ L 2 (R n ; S 0 (3, R)), then there is a solution Q of (12) such that .
The proof follows from Proposition 3.4, Theorem 3.6 and the above remarks.
For simplicity, in what follows we shall take n = 3. If Q 0 ∈ H s (R 3 ; S 0 (3, R)), then we have more freedom in the choice of couples: If we ignore the condition 1 r + 1 r > 1 3 , the two areas representing the admissible couples are the following: Corollary 3.8 Let a 0, s 0, (q, r ) ∈ D s and ( q, r ) ∈ D s be such that 1 r + 1 r > 1 3 . If F ∈ Lq t (R + ; Lr x (R 3 ; S 0 (3, R))) and Q 0 ∈ H s (R 3 ; S 0 (3, R)), then there is a solution Q of (12) such that .

Remark 3.9
The theorem holds also when one of the couples is an endpoint. .
The constant C does not depend on T .
Another important result is the following smoothing inequality.

Proposition 3.11
Let s 0 and a, T > 0 be constants, let Q 0 ∈ H s (R 3 ; S 0 (3, R)) and F ∈ L 2 ([0, T ]; H s−1 (R 3 ; S 0 (3, R))), let Q be a solution of (12), then Q satisfies Proof Firstly we prove the case s = 0: let us multiply (12) by Q(t, x): Since a > 0, it is clear that x for any t 0. Then we have By the Plancherel identity, we have If we take ε < 1/C 2 , then and taking the integral from 0 to T we achieve the result. For the case s > 0 we only need to apply this result to (Id − ) s/2 Q.

Local existence of the solutions
We consider the following spaces: In the following, for simplicity, we omit the index s. With respect to the previous theorems, we assume r , r < 6 so that the condition 1/r + 1/ r > 1/3 is automatically satisfied. In this case, thanks to Proposition 3.4 and Theorem 3.6 we obtain Moreover, Remark 3.10 tells us that this inequality holds for any T > 0.
The proof is technical and follows from several uses of interpolation and Hölder inequalities.

Remark 3.13
If we want to capture all the powers p ∈ 5 3 , 5 , it is necessary to assume that s 4 5 . We are finally ready to prove the existence and uniqueness of the solution in S T . Theorem 3.14 Let s 0, p ∈ 5 3 , 5 be such that p < 7 3−2s , let a 0 and b, c ∈ R.
Proof The idea of the proof is to apply the Banach Fixed Point Theorem to the operator for T and R sufficiently small (the couples (q i , k i ) are the ones of Lemma 3.12 for both the powers in G(Q)). In particular, for any Q 1 , Q 2 ∈ B T , we have the inequality Therefore On the other hand, So, exactly as before, it can be proved that there exists a unique Q ∈ B(0, R) ⊆ BT fixed point for K . Moreover the choice of T is the same as before: it depends only on some global constants (thanks to Remark 3.10) and on R, which can be taken as before.
In conclusion, we can choose T = T . On the other hand, B T ⊆ B T , in particular the R-balls of the two spaces share the same inclusion. So, by uniqueness of the fixed point in the greater ball, Q = Q and therefore Q ∈ L q t ([0, T ]; L r x ). Certainly, this argument can be repeated for all (q, r ) ∈ S, so Q ∈ B(0, R) ⊆ S T .
Before finishing this section, let us present some regularity results.
Proof Thanks to Proposition 3.11, we have Therefore, for a fixed p ∈ 5 3 , 5 , we only need to see that (2 p, 2 p) ∈ D 1 :
Proof Let us take s, t 0.
Let us see what happens to the first term when s → t: thanks to Proposition 3.1, it is easy to see that for any p ∈ 5 3 , 5 and r Let us evaluate the part related with the L 2 -norm of the gradient: Thanks to Proposition 3.15, we know that Q ∈ L ∞ ([0, T ]; L q (R 3 ; S 0 (3, R))) and that ∇ Q ∈ L 2 ([0, T ]; L q (R 3 ; S 0 (3, R))) for any q ∈ [2,6]. So, if we take r = 6 p , α = 6 r ( p−1) and β = (t − τ ) As for the second term, by similar calculations we have with r = 6 p . In particular, this term is bounded for s → t. On the other hand, it is easy to see that e ( −a)ε → Id as ε → 0+ as a map from H 1 (R 3 ; S 0 (3, R)) to itself.

Global existence and decay for t → +∞
Let us show that any solution of (11) does not blow up in a finite time when the data are "small".
It can be seen that, for Q 0 H 1 ε sufficiently small, the local solution exists globally. However we can do better: we only need to require that Q 0 L 2 is small.
Proof We start from equality (16), which can be rewritten as x + e 2at Q(t) 3 2 L 2 x ∇ Q(t) 3 2 L 2 x + e 2at Q(t) If we denote f (t) . . = e 2at |Q(t, x)| 2 dx and g(t) . . = 2e 2at |∇ Q(t, x)| 2 dx, then we have for all δ > 0, where we have used that 3( p−1) 4 < 1 for p < 7 3 . Therefore, if we take δ 4 + δ Let us show that there exists ε sufficiently small such that y(t) is uniformly bounded on t ∈ R + . Let us start with the case of just one non-linearity: With A > 1 and B, C > 0, it is easy to prove that If we take Q 0 L 2 ε sufficiently small such that γ < 0, then y(t) − 1 (A−1)γ for all t 0.
Let us return to the case of two non-linearities. If, by contradiction, y(t) were unbounded, then, for any ε > 0, there would exist T ε > 0 such that y(t) 1 for all t < T ε , y(T ε ) = 1.
If we multiply this equation by ∂ t V , we get By Proposition 3.17 we have that V L 2 Finally, we apply the Grönwall inequality and obtain ∇V (τ ) Ce t 0 e −aτ +e −a( p−1)τ dτ 1 for all t 0.
We are done, since ∇ Q(t) L 2 x = e −at ∇V (t) L 2 x .