Orientability and asymptotic convergence of Q-tensor flow of biaxial nematic liquid crystals

Abstract

In recent paper, we will consider two contents on the maximal biaxial nematic liquid crystals. In the first part, we get an orientability issue, that is if $Q\in W^{1,p}(\Omega ,\mathcal {N}),$ $p\ge 2,$ then there is $(n,m)\in \mathcal {M}$ with $n,m\in W^{1,p}(\Omega ),$ such that $Q=r(n\otimes n-m\otimes m).$ Unlike $\mathbb {S}^2,$ the set $\mathcal {M}$ is not simple connect. Our orientability result extends to maximal biaxial nematics from earlier conclusions corresponding to uniaxial nematics in Ball and Zarnescu (Arch Rational Mech Anal 202:493–535, 2011). In the second part, we study an asymptotic convergence of approximate solutions $Q_\epsilon$ of the Q-tensor flow in $\mathbb {R}^3$ as the parameter $\epsilon$ goes to zero. The limiting direction map (n, m) satisfies a gradient flow, which is different from the heat flow of harmonic map that takes value into $\mathbb {S}^2$ or $\mathcal {M}.$ A partial regularity of this gradient flow is also derived. We extend the works in Ball and Zarnescu (Arch Rational Mech Anal 202:493–535, 2011) and Wang et al. (Arch Rational Mech Anal 225:663–683, 2017) to the maximal biaxial nematic liquid crystals.

Appendix A. Derivation of system (1.10)

Our goal of this section is to derive (1.10).

In fact, let $\lambda _1$ and $\lambda _2$ be two Lagrange multipliers corresponding to the constraint $|n|=1$ and $|m|=1$ respectively and $\mu$ be the Lagrange multiplier corresponding to the constraint $n\cdot m=0.$ For any $\varphi \in C^{\infty }_c(\mathbb {R}^3,\mathbb {R}^3),$ consider

$$\begin{aligned}&\frac{d}{ds}|_{s=0}\int _{\Omega }[\frac{1}{2}|\nabla (n+s\varphi )|^2+\frac{1}{2}|\nabla m|^2+|(n+s\varphi )\cdot \nabla m|^2+\lambda _1(1-|n+s\varphi |^2)\\&\quad +\lambda _2(1-|m|^2)+\mu (n+s\varphi )\cdot m]dx=0 \end{aligned}$$

and

$$\begin{aligned}&\frac{d}{ds}|_{s=0}\int _{\Omega }[\frac{1}{2}|\nabla (m+s\varphi )|^2+\frac{1}{2}|\nabla n|^2+|n\cdot \nabla (m+s\varphi )|^2+\lambda _1(1-|n|^2)\\&\quad +\lambda _2(1-|m+s\varphi |^2)+\mu n\cdot (m+s\varphi )]dx=0. \end{aligned}$$

Then we have

$$\begin{aligned}&\int _{\Omega }\nabla n\cdot \nabla \varphi dx+2\int _{\Omega }(n\cdot \nabla m)\nabla m\cdot \varphi dx=2\int _{\Omega }\lambda _1n\cdot \varphi dx-\int _{\Omega }\mu m\cdot \varphi dx, \end{aligned}$$

(A.1)

$$\begin{aligned}&\int _{\Omega }\nabla m\cdot \nabla \varphi dx+2\int _{\Omega }(n\cdot \nabla m)(n\cdot \nabla \varphi )dx=2\int _{\Omega }\lambda _2m\cdot \varphi dx-\int _{\Omega }\mu n\cdot \varphi dx. \end{aligned}$$

(A.2)

By $n\cdot m=0$ and $|n|=|m|=1,$ taking $\varphi =\eta n$ in (A.1) and (A.2), where $\eta$ is a smooth cut-off function, we get that

$$\begin{aligned} \lambda _1=\frac{1}{2}|\nabla n|^2+|n\cdot \nabla m|^2 \end{aligned}$$

and

$$\begin{aligned} \int _{\Omega }\mu \eta dx= & {} -\int _{\Omega }\nabla m\cdot \nabla (\eta n)dx-2\int _{\Omega }(n\cdot \nabla m)\nabla \eta dx\nonumber \\= & {} -\int _{\Omega }\nabla m\cdot \nabla n\eta dx-3\int _{\Omega }\nabla m\cdot \nabla \eta n dx, \end{aligned}$$

(A.3)

where we have used $\nabla m\cdot n=-n\cdot \nabla m$ and $\nabla n\cdot n=0.$

Taking $\varphi =\eta m$ in (A.1) and (A.2), we have

$$\begin{aligned} \lambda _2=\frac{1}{2}|\nabla m|^2+|n\cdot \nabla m|^2 \end{aligned}$$

and

$$\begin{aligned} \ \int _{\Omega }\mu \eta dx=-\int _{\Omega }\nabla n\cdot \nabla m \eta dx-3\int _{\Omega }\nabla n\cdot \nabla \eta m dx. \end{aligned}$$

(A.4)

From (A.3) and (A.4), we have

$$\begin{aligned} \int _{\Omega }\mu \eta dx=-\int _{\Omega }\nabla n\cdot \nabla m \eta dx. \end{aligned}$$

Then we get $\mu =-\nabla n\cdot \nabla m.$ Hence from (A.3) we also have $\int _{\Omega }\nabla m\cdot n\nabla \eta dx=0,$ which implies that $\triangle m\cdot n+\nabla n\cdot \nabla m=0$ in the sense of distribution.

Therefore we have the Euler-Lagrange equation associated with the simple functional $\int _{\Omega }E(n,m)dx,$

$$\begin{aligned} {\left\{ \begin{array}{ll}-\triangle n-2(\nabla n\cdot m)\nabla m=|\nabla n|^2n+2|n\cdot \nabla m|^2n+(\nabla n\cdot \nabla m)m \ \ \text{ in }\ \ \Omega ,\\ -\triangle m-2(\nabla m\cdot n)\nabla n=|\nabla m|^2m+2|n\cdot \nabla m|^2m+(\nabla n\cdot \nabla m)n \ \ \text{ in }\ \ \Omega ,\end{array}\right. } \end{aligned}$$

here we have used $\triangle m\cdot n=-\nabla n\cdot \nabla m$ and then $\nabla \cdot [(n\cdot \nabla m)n]=(n\cdot \nabla m)\nabla n.$ Then we obtain the derivation of system (1.10).