Bifurcation Analysis in a Frustrated Nematic Cell

Abstract

Using Landau-de Gennes theory to describe nematic order, we study a frustrated cell consisting of nematic liquid crystal confined between two parallel plates. We prove the uniqueness of equilibrium states for a small cell width. Letting the cell width grow, we study the behavior of this unique solution. Restricting ourselves to a certain interval of temperature, we prove that this solution becomes unstable at a critical value of the cell width. Moreover, we show that this loss of stability comes with the appearance of two new solutions: there is a symmetric pitchfork bifurcation. This picture agrees with numerical simulations performed by Palffy-Muhoray, Gartland and Kelly, and also by Bisi, Gartland, Rosso, and Virga. Some of the methods that we use in the present paper apply to other situations, and we present the proofs in a general setting. More precisely, the paper contains the proof of a general uniqueness result for a class of perturbed quasilinear elliptic systems, and general considerations about symmetric solutions and their stability, in the spirit of Palais’ Principle of Symmetric Criticality.

Appendices

Appendix 1: Principle of Symmetric Criticality

Proposition 7.1

Let $H$ be a Hilbert space, $G$ a group acting linearly and isometrically on $H$ and $\Sigma =H^G$ the subspace of symmetric elements (that is, $x\in \Sigma$ iff $gx=x$ $\forall g\in G$). Let $f:H\rightarrow \mathbb {R}$ be a $G$-invariant $C^1$ function. It holds:

• (i) If $x\in \Sigma$ is a critical point of $f_{|\Sigma }$, then $x$ is a critical point of $f$.

• (ii) If in addition $f$ is $C^2$, it further holds

  $$\begin{aligned} D^2f(x)\cdot h \cdot k = 0 \quad \text {for }h\in \Sigma ,\, k\in \Sigma ^\perp , \end{aligned}$$

  i.e., the orthogonal decomposition $H=\Sigma \oplus \Sigma ^\perp$ is also orthogonal for the bilinear form $D^2f(x)$.

Item (i) of the above proposition is only a particularly simple case of Palais’ Principle of symmetric criticality (Palais 1979). Item (ii), however, does not seem to be explicitly stated in the literature—as far as we know. Using the same tools as in Sect. 2 of Palais (1979), it is not hard to see that an equivalent of (ii) is actually valid if $H$ is replaced by a Riemannian manifold ${\fancyscript{M}}$ on which the group $G$ acts isometrically. In this case, $\Sigma$ is a submanifold of ${\fancyscript{M}}$ and, at a symmetric critical point $x$, the orthogonal decomposition

$$\begin{aligned} T_x {\fancyscript{M}} = T_x \Sigma \oplus (T_x \Sigma )^\perp \end{aligned}$$

is also orthogonal for the bilinear form $D^2 f(x)$.

Proof of Proposition 7.1

As already pointed out, item (i) is a particular case of (Palais (1979), Sect. 2). We nevertheless present a complete Proof of Proposition 7.1 here, since in the simple framework we consider, the proof of (i) is really straightforward.

The fact that $f$ is $G$-invariant means that it holds

$$\begin{aligned} f(g x)=f(x)\quad \forall g\in G,\, x\in H. \end{aligned}$$

(42)

Since the action of $G$ on $H$ is linear, differentiating (42) we obtain

$$\begin{aligned} Df(gx)\cdot gh = Df(x) \cdot h \quad \forall h\in H. \end{aligned}$$

(43)

Applying (43) for a symmetric $x$, i.e., $x\in \Sigma$, we have

$$\begin{aligned} < \nabla f(x) , gh > = < \nabla f(x) , h > \quad \forall h\in H. \end{aligned}$$

Note that here we distinguish between the differential $Df(x)\in H^*$ and the gradient $\nabla f(x)\in H$. Similarly, below we will distinguish between the second-order differential $D^2f(x)\in \fancyscript{L} (H,H^*)$ and the Hessian $\nabla ^2 f(x)\in \fancyscript{L} (H)$. Since $g$ is a linear isometry, we conclude that

$$\begin{aligned} g^{-1} \nabla f(x) = \nabla f(x) \quad \forall g \in G,\qquad \text {i.e., }\nabla f(x)\in \Sigma . \end{aligned}$$

(44)

Therefore, if we know in addition that $x$ is a critical point of $f_{|\Sigma }$, which means that $\nabla f(x)\in \Sigma ^\perp$, it must hold $\nabla f(x)=0$. This proves (i).

Now assume that $f$ is $C^2$ and differentiate (43) to obtain

$$\begin{aligned} D^2 f(gx) \cdot gh \cdot gk = D^2 f(x) \cdot h \cdot k \quad \forall h,k \in H. \end{aligned}$$

(45)

In particular, if $x$ and $h$ are symmetric (i.e., belong to $\Sigma$), and if we denote by $\nabla ^2 f(x)$ the Hessian of $f$ at $x$, (45) becomes

$$\begin{aligned} {<}g^{-1}(\nabla ^2 f(x)h),k>=<\nabla ^2 f(x) h,k>\quad \forall x\in \Sigma ,\, h\in \Sigma ,\, k\in H, \end{aligned}$$

so that $\nabla ^2 f(x) h$ is symmetric. Hence it is orthogonal to any $k\in \Sigma ^\perp$, which proves (ii). $\square$

Appendix 2: Uniqueness of Critical Points for Small $\lambda$

Let $\Omega \subset \mathbb {R}^N$ be a smooth bounded domain, and $f:\mathbb {R}^d\rightarrow \mathbb {R}$ a $W^{2,\infty }_{loc}$ map. We are interested in critical points of functionals of the form

$$\begin{aligned} E_\lambda (u) =\int _\Omega \frac{1}{2\lambda ^2}|\nabla u|^2 + \int _\Omega f(u), \end{aligned}$$

(46)

i.e., solutions $u\in H^1(\Omega )^d$ of the equation

$$\begin{aligned} \Delta u = \lambda ^2\nabla f(u) \quad \text { in }\fancyscript{D}{^\prime }(\Omega ). \end{aligned}$$

(47)

Note that (47) implies in particular that $\nabla f(u) \in L^1_{loc}$.

We prove the following:

Theorem 8.1

Assume that there exists $C>0$ such that $\nabla f(x)\cdot x \ge 0$ for any $x\in \mathbb R^d$ with $|x|\ge C$.

Let $g\in L^\infty \cap H^{1/2}(\partial \Omega )^d$. There exists $\lambda _0=\lambda _0(\Omega ,f,g)$ such that, for any $\lambda \in (0,\lambda _0)$, $E_\lambda$ admits at most one critical point with ${\mathrm {tr}}\, u=g$ on $\partial \Omega$.

Theorem 8.1 is a direct consequence of Lemmas 8.2 and 8.3 below. Indeed, Lemma 8.2 ensures that, for sufficiently small $\lambda$, $E_\lambda$ admits at most one critical point satisfying a given $L^\infty$ bound (independent of $\lambda$). And in Lemma 8.3 we prove that the assumption on $f$ implies such a bound for critical points of $E_\lambda$.

Lemma 8.2

Let $C>0$. There exists $\lambda _0=\lambda _0(C,f,\Omega )$ such that, for any $\lambda \in (0,\lambda _0)$ and any $g\in H^{1/2}(\partial \Omega )^d$, $E_\lambda$ admits at most one critical point $u$ satisfying $|u|\le C$ a.e. and ${\mathrm {tr}}\, u = g$.

Proof

Let

$$\begin{aligned} X:=\left\{ u\in H^1(\Omega ) : \; |u|\le C \text { a.e.} \right\} \!. \end{aligned}$$

We show that, for $\lambda$ small enough, $E_\lambda$ is strictly convex on $X$.

Let $u,v\in X$. Then $u-v\in H^1_0(\Omega )^d$. Using Poincaré’s inequality, we obtain

$$\begin{aligned} E_\lambda \left( \frac{u+v}{2}\right)&= \frac{1}{8\lambda ^2}\int |\nabla u + \nabla v|^2 + \int f\left( \frac{u+v}{2}\right) \nonumber \\&\!=\! \frac{1}{4\lambda ^2}\int |\nabla u|^2 \!+\! \frac{1}{4\lambda ^2}\int |\nabla v|^2 \!-\!\frac{1}{8\lambda ^2}\int |\nabla (u\!-\!v)|^2 \!+\! \int f\left( \frac{u+v}{2}\right) \nonumber \\&= \frac{1}{2} E_\lambda (u) + \frac{1}{2} E_\lambda (v) -\frac{1}{8\lambda ^2}\int |\nabla (u-v)|^2 \nonumber \\&\quad + \int \left[ f\left( \frac{u+v}{2}\right) -\frac{1}{2}f(u)-\frac{1}{2}f(v)\right] \nonumber \\&\le \frac{1}{2} E_\lambda (u) + \frac{1}{2} E_\lambda (v) - \frac{c_1(\Omega )}{\lambda ^2}\Vert u-v\Vert _{L^2}^2 \nonumber \\&\quad + \int \left[ f\left( \frac{u+v}{2}\right) -\frac{1}{2}f(u)-\frac{1}{2}f(v)\right] . \end{aligned}$$

(48)

On the other hand, for any $x,y\in \mathbb {R}^d$ satisfying $|x|,|y|\le C$, it holds

$$\begin{aligned} f\left( \frac{x+y}{2} \right) -\frac{1}{2}f(x)-\frac{1}{2}f(y) \le \Vert f\Vert _{W^{2,\infty }(B_C)}|x-y|^2. \end{aligned}$$

(49)

Plugging (49) into (48) we obtain, for some $c_2=c_2(\Omega ,f,C)>0$,

$$\begin{aligned} E_\lambda \left( \frac{u+v}{2}\right)&\le \frac{1}{2} E_\lambda (u) + \frac{1}{2} E_\lambda (v) - \frac{c_1}{\lambda ^2}\Vert u-v\Vert _{L^2}^2 + c_2 \Vert u-v\Vert _{L^2}^2 \\&= \frac{1}{2} E_\lambda (u) + \frac{1}{2} E_\lambda (v) - \frac{c_1}{2\lambda ^2}\Vert u-v\Vert _{L^2}^2 -c_2\left( \frac{c_1}{2c_2\lambda ^2}-1\right) \Vert u-v\Vert _{L^2}^2. \end{aligned}$$

Hence, for $\lambda \le \lambda _0 := \sqrt{c_1/(2c_2)}$, it holds

$$\begin{aligned} E_\lambda \left( \frac{u+v}{2}\right) < \frac{1}{2} E_\lambda (u) + \frac{1}{2} E_\lambda (v) \quad \forall u,v\in X,\; u\ne v. \end{aligned}$$

Thus, $E_\lambda$ is strictly convex on $X$.

To conclude the proof, assume that for a $\lambda \in (0,\lambda _0)$, there exist two solutions $u_1$ and $u_2$ of (47), belonging to $X$. Then one easily shows that $[0,1]\ni t\mapsto E_\lambda (tu_1 + (1-t)u_2)$ is $C^1$ and that its derivative vanishes at 0 and 1, which is incompatible with the strict convexity of $E_\lambda$. $\square$

Lemma 8.3

Assume that there exists $C>0$ such that

$$\begin{aligned} |x|\ge C \quad \Rightarrow \quad \nabla f(x)\cdot x \ge 0. \end{aligned}$$

Let $g\in L^\infty \cap H^{1/2}(\partial \Omega )^d$. If $u\in H^1_g(\Omega )^d$ is a critical point of $E_\lambda$, then it holds

$$\begin{aligned} |u|\le \max (C,\Vert g\Vert _\infty ) \quad \text {a.e.} \end{aligned}$$

Proof

We may assume $C = \max ( C, \Vert g\Vert _\infty ) > 0$.

Let $\varphi \in C^\infty (\mathbb R)$ be such that:

$$\begin{aligned} \left\{ \begin{array}{l} \varphi \ge 0,\\ \varphi {^\prime } \ge 0,\\ \varphi (t) = 0\qquad \text {for }t\le C^2,\\ \varphi (t) = 1\qquad \text {for }t\ge T,\text { for some }T> C^2. \end{array} \right. \end{aligned}$$

(50)

Let $w=\varphi (|u|^2)$. The assumptions on $\varphi$ ensure that $w\ge 0$, and $w=0$ in $\lbrace |u|\le C\rbrace$.

Therefore, taking the scalar product of (47) with $wu$ and using the assumption that $\nabla f(u)\cdot u \ge 0$ outside of $\lbrace |u|\le C \rbrace$, we obtain

$$\begin{aligned} \frac{1}{\lambda ^2} w u\cdot \Delta u = w \nabla f(u)\cdot u \ge 0\quad \text {a.e.} \end{aligned}$$

(51)

Since $wu\in H_0^1(\Omega )^d$, we may apply Lemma 8.4 below, to deduce

$$\begin{aligned} \int _\Omega \nabla u \cdot \nabla (wu) \le 0. \end{aligned}$$

(52)

On the other hand, it holds

$$\begin{aligned} \int _\Omega \nabla u \cdot \nabla (wu) = \int _\Omega w |\nabla u|^2 + \int _\Omega 2 \sum _k(u\cdot \partial _k u)^2 \varphi {^\prime }(|u|^2), \end{aligned}$$

so that we have in fact

$$\begin{aligned} \int _\Omega w |\nabla u|^2\le 0. \end{aligned}$$

(53)

Finally we may choose an increasing sequence $\varphi _k$ of smooth maps satisfying (50) and converging to $\mathbf {1}_{t > C^2}$. Then, $w_k=\varphi _k(|u|^2)$ is increasing and converges a.e. to $\mathbf {1}_{|u| > C}$, and we conclude that

$$\begin{aligned} \int _{|u| > C} |\nabla u|^2 = 0, \end{aligned}$$

so that $|u|\le C$ a.e. $\square$

The following result, which we used in the proof of Lemma 8.3, is due to Pierre Bousquet.

Lemma 8.4

Let $u\in H^1(\Omega )^d$ and assume that $\Delta u = g \in L^1_{loc}(\Omega )^d$. Then, for any $\zeta \in H_0^1(\Omega )^d$,

$$\begin{aligned} \zeta \cdot g \ge 0 \; \text {a.e. } \Longrightarrow \; \int \nabla \zeta \cdot \nabla u \le 0. \end{aligned}$$

(54)

Proof

We proceed in three steps: first we show that (54) is valid for $\zeta \in H^1\cap L^\infty (\Omega )^d$ with compact support in $\Omega$, then for $\zeta \in H^1_0 \cap L^\infty (\Omega )^d$, and eventually for $\zeta \in H^1_0(\Omega )^d$.

Step 1: $\zeta \in H^1_c\cap L^\infty$.

Since $\zeta$ is bounded and compactly supported, there exists a sequence $\zeta _k$ of $C_c^\infty$ functions, a constant $C>0$, and a compact $K\subset \Omega$, such that

$$\begin{aligned} {\mathrm {supp}}\,\zeta _k \subset K,\quad \Vert \zeta _k \Vert _\infty \le C,\;\text {and }\zeta _k \longrightarrow \zeta \text { in }H^1\text { and a.e.} \end{aligned}$$

Since $\zeta _k\in C^\infty _c(\Omega )^d$, it holds, by definition of the weak laplacian,

$$\begin{aligned} \int \zeta _k\cdot g = - \int \nabla \zeta _k\cdot \nabla u, \end{aligned}$$

and we may pass to the limit (using dominated convergence on the compact $K$ for the left-hand side) to obtain

$$\begin{aligned} \int _\Omega \zeta \cdot g = - \int \nabla \zeta \cdot \nabla u, \end{aligned}$$

which implies (54).

Step 2: $\zeta \in H^1_0\cap L^\infty$.

Let $\theta _k\in C_c^\infty (\Omega )$ be such that

$$\begin{aligned} 0\le \theta _k \le 1,\; \theta _k(x)=1\text { if }d(x,\partial \Omega )>\frac{1}{k},\;\text {and }|\nabla \theta _k (x) | \le \frac{c}{d(x,\partial \Omega )}, \end{aligned}$$

and define $\zeta _k = \theta _k \zeta \in H^1_c \cap L^\infty (\Omega )^d$.

Assuming that $\zeta \cdot g \ge 0$ a.e., we deduce that $\zeta _k\cdot g \ge 0$ a.e., and thus we may apply Step 1 to $\zeta _k$: it holds

$$\begin{aligned} 0\ge \int \nabla \zeta _k \cdot \nabla u = \int \theta _k \nabla \zeta \cdot \nabla u + \int \nabla \theta _k \cdot \nabla u \cdot \zeta . \end{aligned}$$

(55)

The first term in the right-hand side of (55) converges to $\int \nabla \zeta \cdot \nabla u$, by dominated convergence. Therefore we only need to prove that the second term in the right-hand side of (55) converges to zero. To this end we use the following Hardy-type inequality:

$$\begin{aligned} \int \frac{|\zeta |^2}{d(x,\partial \Omega )^2} \le C \int |\nabla \zeta |^2,\qquad \forall \zeta \in H_0^1(\Omega ). \end{aligned}$$

(56)

Using (56) and the Hölder inequality, we obtain

$$\begin{aligned} \left| \int \nabla \theta _k \cdot \nabla u \cdot \zeta \right| ^2 \le C \Vert \nabla \zeta \Vert ^2_{L^2} \int _{d(x,\partial \Omega )>1/k}\!\! |\nabla u|^2 \longrightarrow 0, \end{aligned}$$

which concludes the proof of Step 2.

Step 3: $\zeta \in H^1_0$.

We define $\zeta _k = P_k(\zeta )$, where $P_k:\mathbb R^d \rightarrow \mathbb R^d$ is given by

$$\begin{aligned} P_k(x)={\left\{ \begin{array}{ll} x &{} \quad \text {if }|x|\le k, \\ \frac{k}{|x|}x &{} \quad \text {if }|x|>k. \end{array}\right. } \end{aligned}$$

Then $\zeta _k\in H_0^1\cap L^\infty (\Omega )$ and $\zeta _k \rightarrow \zeta$ in $H^1$.

If $\zeta \cdot g\ge 0$, then it obviously holds $\zeta _k\cdot g \ge 0$, so that we may apply Step 2 to $\zeta _k$ and obtain

$$\begin{aligned} \int \nabla \zeta _k \cdot \nabla u \le 0. \end{aligned}$$

Letting $k$ go to $\infty$ in this last inequality provides the desired conclusion. $\square$

Appendix 3: Second Variation of the Energy

At a map $Q\in H^1(-1,1)^3$, the second variation of the energy reads

$$\begin{aligned} D^2 E(Q) [H] = \int \left( \frac{1}{\lambda ^2} (H{^\prime })^2 + D^2f(Q)[H]\right) \mathrm{d}x, \end{aligned}$$

where

$$\begin{aligned} D^2f(Q)[H] = \frac{\theta }{3}|H|^2 - 4 Q \cdot H^2 + (Q\cdot H)^2 + \frac{1}{2} |Q|^2 |H|^2. \end{aligned}$$

If we take $Q=\chi =(q_1,q_2,0)$, and consider separately perturbations $H_{sp}=(h_1,h_2,0)$ and $H_\mathrm{sb}=(0,0,h_3)$, we have

$$\begin{aligned} |H_\mathrm{sp}|^2&= 6h_1^2+2h_2^2&|H_\mathrm{sb}|^2&= 2h_3^2 \\ \chi \cdot H_\mathrm{sp}^2&= 2q_1(h_2^2-3h_1^2)+4q_2 h_1 h_2&\chi \cdot H_\mathrm{sb}^2&=2q_1h_3^2 \\ \chi \cdot H_\mathrm{sp}&= 6q_1h_1+2q_2h_2&\chi \cdot H_\mathrm{sb}&=0, \end{aligned}$$

so that we can compute

$$\begin{aligned} D^2f(\chi )[H_\mathrm{sp}]&= 6\left( \frac{\theta }{3}+2q_1+9q_1^2+q_2^2\right) h_1^2 \\&\quad + 2\left( \frac{\theta }{3}-4q_1 + 3q_1^2+3q_2^2\right) h_2^2 \\&\quad + 8q_2(3q_1-2)h_1h_2\\ D^2f(\chi )[H_\mathrm{sb}]&= 2\left( \frac{\theta }{3}-4q_1+3q_1^2+q_2^2\right) h_3^2. \end{aligned}$$