Stability of point defects of degree $\pm \frac{1}{2}$ in a two-dimensional nematic liquid crystal model

Abstract

We study k-radially symmetric solutions corresponding to topological defects of charge $\frac{k}{2}$ for integer $k\not = 0$ in the Landau-de Gennes model describing liquid crystals in two-dimensional domains. We show that the solutions whose radial profiles satisfy a natural sign invariance are stable when $|k| = 1$ (unlike the case $|k| > 1$ which we treated before). The proof crucially uses the monotonicity of the suitable components, obtained by making use of the cooperative character of the system. A uniqueness result for the radial profiles is also established.

Appendix

Lemma 5.1

Assume that $a^2 \ge 0, b^2, c^2 > 0$. Let

$$\begin{aligned} f(x,y) = -\frac{a^2}{2}(x^2 + y^2) + \frac{c^2}{4}(x^2 + y^2)^2 - \frac{b^2}{3\sqrt{6}}y(y^2 - 3x^2). \end{aligned}$$

Then

$$\begin{aligned} \min _{{\mathbb {R}}^2} f = -\frac{a^2}{3} s_+^2 - \frac{2b^2}{27}\,s_+^3 + \frac{c^2}{6} s_+^4 \end{aligned}$$

which is attained at (and only at) $\big (0,\frac{2}{\sqrt{6}}s_+\big )$ and $\big (\pm \frac{1}{\sqrt{2}}s_+,- \frac{1}{\sqrt{6}}s_+\big )$. Furthermore, the Hessian of f at all these critical points is positive definite.

Proof

We write $x = r\,\sin \varphi$ and $y = r\cos \varphi$ for some $r \ge 0$ and $\varphi \in [0,2\pi )$. Then

$$\begin{aligned} f(x,y) = -\frac{a^2}{2}r^2 + \frac{c^2}{4}r^4 - \frac{b^2}{3\sqrt{6}}\,r^3 \cos 3\varphi \ge -\frac{a^2}{2}r^2 + \frac{c^2}{4}r^4 - \frac{b^2}{3\sqrt{6}}\,r^3 =: {\tilde{f}}(r). \end{aligned}$$

It is easy to check that ${\tilde{f}}$ has three critical points, $r = 0$ and $r = \frac{2}{\sqrt{6}}s_\pm$ where the first one is a local maximum point and the other two are local minimum points. The global minimum of ${\tilde{f}}$ is then verified to be achieved at $r = \frac{2}{\sqrt{6}}s_+$. We have thus shown that

$$\begin{aligned} f(x,y) \ge {\tilde{f}}\left( \frac{2}{\sqrt{6}}s_+\right) = -\frac{a^2}{3} s_+^2 - \frac{2b^2}{27}\,s_+^3 + \frac{c^2}{6} s_+^4 , \end{aligned}$$

and equality is attained if and only if $r =\frac{2}{\sqrt{6}}s_+$ and $\varphi \in \{0, \frac{2\pi }{3}, \frac{4\pi }{3}\}$. The first assertion follows.

Now a computation using $-a^2 - \frac{b^2}{3} s_+ + \frac{2}{3}c^2\,s_+^2 = 0$ leads to

$$\begin{aligned} D^2 f\left( 0,\frac{2}{\sqrt{6}}s_+\right)&= \left[ \begin{array}{cc} b^2\,s_+ &{} 0\\ 0&{} \frac{1}{3}(3a^2 + b^2\,s_+) \end{array}\right] ,\\ D^2 f\left( \pm \frac{1}{\sqrt{2}}s_+,- \frac{1}{\sqrt{6}}s_+\right)&= \left[ \begin{array}{cc} c^2 s_+^2 &{} \pm \frac{1}{\sqrt{3}}(-c^2\,s_+^2+ b^2\,s_+)\\ \pm \frac{1}{\sqrt{3}}(-c^2\,s_+^2+ b^2\,s_+) &{} \frac{1}{3}(c^2\,s_+^2 + 2b^2\,s_+) \end{array}\right] , \end{aligned}$$

from which the last assertion follows. $\square$