Mean-Field Limit and Phase Transitions for Nematic Liquid Crystals in the Continuum

Abstract

We discuss thermotropic nematic liquid crystals in the mean-field regime. In the first part of this article, we rigorously carry out the mean-field limit of a system of N rod-like particles as $N\rightarrow \infty$, which yields an effective ‘one-body’ free energy functional. In the second part, we focus on spatially homogeneous systems, for which we study the associated Euler–Lagrange equation, with a focus on phase transitions for general axisymmetric potentials. We prove that the system is isotropic at high temperature, while anisotropic distributions appear through a transcritical bifurcation as the temperature is lowered. Finally, as the temperature goes to zero we also prove, in the concrete case of the Maier–Saupe potential, that the system converges to perfect nematic order.

Appendix A: Laplace’s Method

Lemma 15

Let $f\in C^3 ([0,\pi /2];\mathbb {R})$ be such that 0 is its unique global minimum. Assume that $f''(0)\ne 0$. Then the following holds.

• (a) As $\beta \rightarrow \infty$,

  $$\begin{aligned} \int _{0}^{\pi /2}\mathrm {e}^{-\beta f(\theta ) }\sin \theta \,\mathrm {d}{\theta } =\mathrm {e}^{-\beta f(0) }\left[ \frac{1}{f''(0)}\beta ^{-1}-\frac{1 }{2}\sqrt{\frac{\pi }{2}}\frac{f'''(0)}{f''(0)^{5/2}}\beta ^{-3/2}+o(\beta ^{-3/2})\right] . \end{aligned}$$

  (35)

• (b) Then,

  $$\begin{aligned} \lim _{\beta \rightarrow \infty }\langle g \rangle _\beta ^f=g(0). \end{aligned}$$

  (36)

• (c) If $g\in C^2 ([0,\pi /2];\mathbb {R})$ we have, as $\beta \rightarrow \infty$,

  $$\begin{aligned} \langle g \rangle _\beta ^f=g(0)+\sqrt{\frac{\pi }{2}} \frac{g'(0)}{f''(0)^{1/2}}\beta ^{-1/2} +\left[ \frac{g''(0)}{f''(0)}+\frac{3\pi -16}{12}\frac{g'(0)f'''(0)}{f''(0)^2} \right] \beta ^{-1}+o(\beta ^{-1}). \end{aligned}$$

  (37)

Proof

• (a) We first write

  $$\begin{aligned} \int _{0}^{\pi /2}\mathrm {e}^{-\beta f(\theta ) }\sin \theta \,\mathrm {d}{\theta } = \mathrm {e}^{-\beta f(0) } \int _{0}^{\pi /2}\mathrm {e}^{-\beta (f(\theta ) -f(0))}\sin \theta \,\mathrm {d}{\theta }. \end{aligned}$$

  By assumption, there is $\epsilon >0$ such that, for all $0<\theta <\epsilon$, $f(\theta )- f(0) \ge \frac{1}{4}f''(0)\epsilon ^2$. Hence,

  $$\begin{aligned}&\int _\epsilon ^{\pi /2}\mathrm {e}^{-\beta (f(\theta ) - f(0))}\sin \theta \,\mathrm {d}\theta \le \bigg (\sup _{\theta \in [\epsilon ,\pi /2]}\mathrm {e}^{-\frac{\beta }{2} (f(\theta ) - f(0))}\bigg )\\&\times \int _\epsilon ^{\pi /2}\mathrm {e}^{-\frac{\beta }{2} (f(\theta ) - f(0))}\sin \theta \,\mathrm {d}\theta \le C \mathrm {e}^{-\frac{\beta }{8} f''(0)\epsilon ^2}, \end{aligned}$$

  where C is independent of $\beta ,\epsilon$. Let now $\epsilon = \beta ^{-1/4}$. Then $\beta \epsilon ^2 = \beta ^{1/2}\rightarrow \infty$ and the above integral vanishes exponentially as $\beta \rightarrow \infty$. In a neighbourhood of the minimum, Taylor expansions yield

  $$\begin{aligned}&\int _0^\epsilon \mathrm {e}^{-\beta (f(\theta ) - f(0))}\sin \theta \,\mathrm {d}\theta \!= \!\int _0^\epsilon \mathrm {e}^{-\frac{\beta }{2}f''(0)\theta ^2}\left( 1 \!-\! \frac{\beta }{6}f'''(0) \theta ^3 \!+\! \beta o(\theta ^3)\right) \left( \theta \!+\! o(\theta ^2)\right) \,\mathrm {d}\theta \\&\quad = \beta ^{-1/2}\int _0^{\beta ^{1/2}\epsilon } \mathrm {e}^{-\frac{1}{2}f''(0)\zeta ^2}\left( \beta ^{-1/2}\zeta - \beta ^{-1} \frac{f'''(0)}{6} \zeta ^4 + o(\beta ^{-1})\right) \,\mathrm {d}\zeta , \end{aligned}$$

  where we let $\theta = \beta ^{-1/2}\zeta$. Since $\beta ^{1/2}\epsilon \rightarrow \infty$, the error in replacing the upper bound of integration by $\infty$ is exponentially small indeed. The integrals can finally be carried out explicitly to yield (35).

• (b) Here again, we start by writing

  $$\begin{aligned} \langle g\rangle _\beta ^f = g(0) + \langle g - g(0)\rangle _\beta ^f = g(0) + \frac{\int _0^{\pi /2}\mathrm {e}^{-\beta (f(\theta )-f(0))}(g(\theta ) - g(0))\sin \theta \,\mathrm {d}\theta }{\int _0^{\pi /2}\mathrm {e}^{-\beta (f(\theta )-f(0))}\sin \theta \,\mathrm {d}\theta }. \end{aligned}$$

  (38)

  Proceeding as in (a), we can restrict our attention to $[0,\beta ^{-1/4})$ in the integrals. By continuity of g, for any $\tilde{\epsilon }>0$, there is $\beta <\infty$ such that $\theta <\beta ^{-1/4}$ implies $\vert g(\theta ) - g(0)\vert < \tilde{\epsilon }$. Hence,

  $$\begin{aligned} \big \vert \langle g\rangle _\beta ^f - g(0) \big \vert = \big \vert \langle g - g(0)\rangle _\beta ^f\big \vert \le \tilde{\epsilon }. \end{aligned}$$

• (c) We consider again (38). Proceeding as in (a), we expand the relevant part of the numerator as

  $$\begin{aligned}&\int _0^\epsilon \mathrm {e}^{-\beta (f(\theta )-f(0))}(g(\theta ) - g(0))\sin \theta \,\mathrm {d}\theta = \beta ^{-1/2}\int _0^{\beta ^{1/2}\epsilon } \mathrm {e}^{-\frac{1}{2}f''(0)\zeta ^2}\\&\qquad \times \left( g'(0)\zeta ^2 \beta ^{-1} + \frac{1}{2}g''(0)\zeta ^3\beta ^{-3/2} - \frac{1}{6}f'''(0)g'(0)\beta ^{-3/2}\zeta ^5 + o(\beta ^{-3/2})\right) \,\mathrm {d}\zeta \\&\quad = \sqrt{\frac{\pi }{2}}\frac{g'(0)}{f''(0)^{3/2}}\beta ^{-3/2} + \left[ \frac{g''(0)}{f''(0)^{2}} - \frac{4}{3}\frac{g'(0)f'''(0)}{f''(0)^{3}}\right] \beta ^{-2} + o(\beta ^{-2}). \end{aligned}$$

  Combining this with the expansion (35), we obtain the claim. $\square$