Refined asymptotics for Landau-de Gennes minimizers on planar domains

Abstract

In our previous work Golovaty and Montero (Arch Ration Mech Anal 213(2):447–490, 2014), we studied asymptotic behavior of minimizers of the Landau-de Gennes energy functional on planar domains as the nematic correlation length converges to zero. Here we improve upon those results, in particular by sharpening the description of the limiting map of the minimizers. We also provide an expression for the energy valid for a small, but fixed value of the nematic correlation length.

Appendix

In this appendix we provide a short summary of previous results that are relevant to the present manuscript. We start with the following theorem that was proved in [13].

Theorem 5.1

Let $g : \partial \Omega \rightarrow {\mathcal {P}}$ be a non-contractible curve in ${\mathcal {P}}$ and suppose that $u_\varepsilon \in W^{1,2}(\Omega ; F_1)$ is a minimizer of $E_\varepsilon$ among functions $u \in W^{1,2}(\Omega ; F_1)$ that satisfy the Dirichlet boundary condition $u = g$ on $\partial \Omega .$ First, the minimizers $u_\varepsilon$ take values in the convex envelope of ${\mathcal {P}}$; in particular they are uniformly bounded in $\varepsilon$. Second, there is a single point a in the interior of $\Omega$ such that the $u_\varepsilon$ converge strongly (along a subsequence) to $u_0 \in W^{1,2}(\Omega \setminus B_R\{a\}; {{\mathcal {P}}})$ in $W^{1,2}(\Omega \setminus B_R\{a\}; F_1)$ as $\varepsilon \rightarrow 0$ for any fixed $R >0$. Finally, for any open set $U \subset \subset {\overline{\Omega }}\setminus \{a\}$, $u_0$ minimizes $\int _U \left| {\nabla v}\right| ^2$ among functions $v \in W^{1,2}_{loc}(\Omega \setminus \{a\}; {{\mathcal {P}}})$ satisfying $v = u_0$ on $\partial U$.

To describe the structure of $u_0,$ let $M_a^3({\mathbb {R}}^{})$ be the set of antisymmetric $3\times 3$ matrices and let $[A;B]=AB-BA$ denote the commutator of matrices A and B. It turns out that one can consider a vector field $j(u_0)$ with matrix entries

$$\begin{aligned} j(u_0) = \left( \left[ u_0 , \frac{\partial u_0}{\partial x}\right] , \left[ u_0 , \frac{\partial u_0}{\partial y}\right] \right) , \end{aligned}$$

instead of $u_0$ because $u_0$ can always be recovered from $j(u_0)$ (the reason for this reduces to the following standard fact: if $A:[0,T]\rightarrow M_a^3({\mathbb {R}}^{})$, then the solution of the initial value problem

$$\begin{aligned} \gamma ' = [\gamma , A],\ \ \gamma (0) \in {\mathcal {P}}, \end{aligned}$$

takes values in ${\mathcal {P}}$). In light of this observation, the following theorem [13] gives a rough description of the limiting map $u_0$ described in Theorem 5.1.

Theorem 5.2

Let $u_0$ be as in Theorem 5.1. There is a function

$$\begin{aligned} \psi _0 \in (W^{1,2}\cap L^\infty )(\Omega ; M_a^3({\mathbb {R}}^{})) \end{aligned}$$

and a constant anti-symmetric matrix $\Lambda _0$ such that

$$\begin{aligned} j(u_0) = \frac{1}{2 r} ({\hat{\theta }} \Lambda _0) + \nabla ^\perp \psi _0 \end{aligned}$$

in $\Omega .$ Here $a \in \Omega$ is as defined in Theorem 5.1, r and ${\hat{\theta }}$ are the radial variable and the unit vector in an angular direction for polar coordinates centered at a respectively, and we interpret $({\hat{\theta }} \Lambda _0)$ and $\nabla ^\perp \psi _0$ as matrix-valued vector fields according to (5.2) and (5.1) respectively. Further, $\psi _0$ satisfies

$$\begin{aligned} \Delta \psi _0 = 2\left[ \frac{\partial \psi _0}{\partial x}, \frac{\partial \psi _0}{\partial y}\right] + \frac{1}{\pi r} \left[ \nabla \psi _0 \cdot {\hat{\theta }} , \Lambda _0\right] , \end{aligned}$$

in $\Omega$, where we interpret $\nabla \psi _0 \cdot {\hat{\theta }}$ according to (5.3), subject to boundary conditions

$$\begin{aligned} -\nabla \psi _0 \cdot \nu = \left[ g, \frac{dg}{d \tau }\right] - \frac{{\hat{\theta }}\cdot \tau }{2\pi r} \Lambda _0, \end{aligned}$$

on $\partial \Omega ,$ where $\nu$ and $\tau$ are the outward unit normal and unit tangent vector to $\partial \Omega ,$ respectively. Finally, the function $Z_{u_0}(x) := \frac{1}{2\pi r}(\Lambda _0 - u_0 \Lambda _0 - \Lambda _0 u_0) \in L^2(\Omega ; M_a^3({\mathbb {R}}^{})).$

Notice that in the preceding theorem we deal with matrix-valued functions $u:\Omega \rightarrow M^3({\mathbb {R}}^{})$ and matrix-valued vector fields $F : \Omega \rightarrow (M^3({\mathbb {R}}^{}))^2$, $F = (F_1,F_2)$. For a matrix-valued function u, the gradient and its perpendicular are given by the matrix-valued vector fields

$$\begin{aligned} \nabla u = \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} \right) , \,\,\, \nabla ^\perp u = \left( \frac{\partial u}{\partial y}, -\frac{\partial u}{\partial x} \right) , \end{aligned}$$

(5.1)

respectively. For matrix-valued vector fields, the divergence and curl

$$\begin{aligned} \nabla \cdot F = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y}, \,\,\, \nabla ^\perp \cdot F = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \end{aligned}$$

are matrix-valued functions. When $z :\Omega \rightarrow {\mathbb {R}}^{2}$ and $A:\Omega \rightarrow M^3({\mathbb {R}}^{})$, the matrix-valued vector field zA has the entries

$$\begin{aligned} zA = (z_1A, z_2A). \end{aligned}$$

(5.2)

On the other hand, if F is a matrix-valued vector field and $e=(e_1, e_2)\in {\mathbb {R}}^{2}$, we set

$$\begin{aligned} F\cdot e = e_1F_1+e_2F_2, \end{aligned}$$

(5.3)

which is a matrix-valued function. We emphasize the difference between $F\cdot e$ and zA defined in (5.2).

In the next proposition we summarize the properties of the potential we use.

Proposition 5.3

For $u \in M^3_{s, 1}({\mathbb {R}}^{})$, define

$$\begin{aligned} W_\beta (u) = \frac{1}{4}(1-\left| {u}\right| ^2)^2 - \beta \, {\mathrm{det}}(u). \end{aligned}$$

(5.4)

We have

$$\begin{aligned} W_\beta (u) = \frac{1}{4}(1-\left| {u}\right| ^2)^2 - \frac{\beta }{6}(1 - 3\left| {u}\right| ^2 + 2{\mathrm{tr}}(u^3)). \end{aligned}$$

(5.5)

Next, for $u \in M_{s, 1}^3({\mathbb {R}}^{})$ such that $\langle u, P\rangle \ge 0$ for all $P \in {\mathcal {P}}$, we have

$$\begin{aligned} \frac{(1-\left| {u}\right| ^2)^2}{3} \le \left| {u-u^2}\right| ^2 \le \frac{(1-\left| {u}\right| ^2)^2}{2}. \end{aligned}$$

(5.6)

Lastly, we have

$$\begin{aligned} W_\beta (u) \ge \frac{(3-\beta )}{6}({\mathrm{dist}}(u, {\mathcal P}))^2. \end{aligned}$$

(5.7)

Proof

We start by recalling that Cayley-Hamilton theorem for matrices $u\in M_{s, 1}^3({\mathbb {R}}^{})$ tells us

$$\begin{aligned} u^3 - u^2 + \frac{(1-\left| {u}\right| ^2)}{2}u - {\mathrm{det}}(u)I = 0. \end{aligned}$$

Applying this in (5.4) gives us (5.5).

Next, again from Cayley-Hamilton theorem, for $u \in M_{s, 1}^3({\mathbb {R}}^{})$, we obtain

$$\begin{aligned} \left| {u-u^2}\right| ^2 + 2{\mathrm{det}}(u) = \frac{(1-\left| {u}\right| ^2)^2}{2}. \end{aligned}$$

(5.8)

In particular, if $u \in M_{s, 1}^3({\mathbb {R}}^{})$ has $\langle u, P\rangle \ge 0$ for all $P \in {\mathcal {P}}$, then

$$\begin{aligned} \frac{(1-\left| {u}\right| ^2)^2}{3} \le \left| {u-u^2}\right| ^2 \le \frac{(1-\left| {u}\right| ^2)^2}{2}. \end{aligned}$$

This is (5.6)

Under these conditions, if ${\mathrm{dist}}(u, {{\mathcal {P}}}) \le \delta \le \frac{1}{4}$, it is not hard to check that

$$\begin{aligned} \frac{2}{3}{\mathrm{dist}}(u, {{\mathcal {P}}}) \le 2\frac{\left| {u-u^2}\right| }{1-\delta } \le \frac{\left| 1-{|u|}^2\right| }{1-\delta } \end{aligned}$$

Furthermore, again for $u \in M_{s, 1}^3({\mathbb {R}}^{})$ such that $\langle u, P\rangle \ge 0$ for all $P \in {\mathcal {P}}$, a lengthy, but ultimately straight forward minimization shows that

$$\begin{aligned} (1-\left| {u}\right| ^2)^2 \ge 12 \,\, {\mathrm{det}}(u). \end{aligned}$$

Hence, for $1\le \beta < 3$, and $u \in M_{s, 1}^3({\mathbb {R}}^{})$ such that $\langle u, P\rangle \ge 0$ for all $P \in {\mathcal {P}}$, we have

$$\begin{aligned} W_\beta (u)&= \frac{1}{4}(1-\left| {u}\right| ^2)^2 - \beta {\mathrm{det}}(u) = \frac{3-\beta }{12} ( 1 - \left| {u}\right| ^2)^2 + \beta ( \frac{1}{12}(1-\left| {u}\right| ^2)^2 - {\mathrm{det}}(u) )\\&\ge \frac{3-\beta }{12}(1-\left| {u}\right| ^2)^2 \ge \frac{(3-\beta )}{6}({\mathrm{dist}}(u, {{\mathcal {P}}}))^2. \end{aligned}$$

This is (5.7). $\square$

Remark 5.4

The potential $W_{\beta }$ in [13] is written as

$$\begin{aligned} W_\beta (u) = \frac{1}{2}(1-\left| {u}\right| ^2)^2 - \beta \, {\mathrm{det}}(u). \end{aligned}$$

As can be seen from (5.8), in order for this potential to be equal $\left| {u-u^2}\right| ^2$ we need to choose $\beta =2$. In [13] we erroneously stated that our results were valid for $2<\beta <6$ while the correct range should be $2\le \beta < 6$. In this paper, however, we use the expression given in (5.4).

Proposition 5.5

Let $\Omega \subset {\mathbb {R}}^{2}$ be a smooth, bounded, simply-connected open set, and $u_\varepsilon : \Omega \rightarrow M_{s, 1}^3({\mathbb {R}}^{})$ be a minimizer of the LdG energy with non-contractible boundary data in ${\mathcal {P}}$. Let $a\in \Omega$ be the distinguished point in $\Omega$ that Theorem 5.1 shows exist. For $r>0$ such that $B_{2r}(a)\subset \Omega$, there is $\varepsilon _0 > 0$ and a constant $C>0$ such that

$$\begin{aligned} \left| {(\nabla u_\varepsilon )(x)}\right| + \frac{(1-\left| {u_\varepsilon }\right| ^2)}{\varepsilon ^2} \le C \end{aligned}$$

for all $x\in \Omega \setminus B_r(a)$, and all $0 < \varepsilon \le \varepsilon _0$.

Proof

The proof follows [3]. Let us observe that the end of the proof of Lemma 8 of [13] shows that minimizers $u_\varepsilon$ satisfy

$$\begin{aligned} \limsup _{\varepsilon \rightarrow 0} \int _{\Omega \setminus B_r(a)} \frac{W(u_\varepsilon )}{\varepsilon ^2} = 0. \end{aligned}$$

We now appeal to Steps A.2 and B.2 of the proof of Theorem 1 of [2], to conclude that $W(u_\varepsilon ) \rightarrow 0$ uniformly in $\Omega \setminus B_r(a)$. In particular, for $\delta > 0$ we can choose $\varepsilon _0 > 0$ such that

$$\begin{aligned} 0 \le 1-\left| {u_\varepsilon ^2}\right| (x) \le \delta \end{aligned}$$

for all $x\in \Omega \setminus B_r(a)$ and all $0 < \varepsilon \le \varepsilon _0$.

We next recall from the appendix of [13] that

$$\begin{aligned} \frac{4-\beta }{\varepsilon ^2}(1-\left| {u}\right| ^2) - 4W_{\frac{3\beta }{4}} = -\Delta \frac{\left| {u}\right| ^2}{2} + \left| {\nabla u}\right| ^2 = -\langle u , \Delta u\rangle . \end{aligned}$$

We know from [13] that $\langle u_\varepsilon , P \rangle \ge 0$ for all $P \in {\mathcal {P}}$, so we deduce

$$\begin{aligned} 4W_{\frac{3\beta }{4}}(u_\varepsilon ) \le (1-\left| {u}\right| ^2)^2. \end{aligned}$$

Hence, we can choose $\varepsilon _0 > 0$ small enough for

$$\begin{aligned} \frac{4-\beta }{\varepsilon ^2}(1-\left| {u_\varepsilon }\right| ^2) - 4W_{\frac{3\beta }{4}}(u_\varepsilon ) \ge \delta (1-\left| {u_\varepsilon }\right| ^2) \end{aligned}$$

in $\Omega \setminus B_r(a)$, for all $0 <\varepsilon \le \varepsilon _0$. Since $\left| {u_\varepsilon }\right| \le 1$, we conclude that

$$\begin{aligned} \left| {\Delta u_\varepsilon }\right| \ge \delta (1-\left| {u_\varepsilon }\right| ^2) \end{aligned}$$

in $\Omega \setminus B_r(a)$, for all $0 <\varepsilon \le \varepsilon _0$.

From the Euler-Lagrange equation for $u_\varepsilon$, we obtain

$$\begin{aligned} -\Delta \frac{\partial u_\varepsilon }{\partial x_k} + \frac{1}{\varepsilon ^2} (D^2_u W)(u)\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) = \frac{\partial \lambda _\varepsilon }{\partial x_k}I, \end{aligned}$$

and then

$$\begin{aligned} \Delta \left| {\nabla u_\varepsilon }\right| ^2 = 2\left| {D^2_x u}\right| ^2 + \frac{2}{\varepsilon ^2}\sum _{k=1}^2 \langle (D^2_u W)(u)\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) , \frac{\partial u_\varepsilon }{\partial x_k} \rangle . \end{aligned}$$

Now, writing $v_\varepsilon$ for the nearest element of ${\mathcal {P}}$ to $u_\varepsilon$, we have

$$\begin{aligned} \langle (D^2_u W_\beta )(u_\varepsilon )\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) , \frac{\partial u_\varepsilon }{\partial x_k} \rangle&= \langle (D^2_u W)(v_\varepsilon )\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) , \frac{\partial u_\varepsilon }{\partial x_k} \rangle \\&\quad + \langle (D^2_u W_\beta )(u_\varepsilon )\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) - (D^2_u W_\beta )(v_\varepsilon )\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) , \frac{\partial u_\varepsilon }{\partial x_k} \rangle . \end{aligned}$$

Now $v_\varepsilon \in {\mathcal {P}}$, which is the set of minimizers of $W_\beta$, and it is easy to check that ${\mathrm{tr}}(\frac{\partial u_\varepsilon }{\partial x_k}) = 0$. Hence

$$\begin{aligned} \langle (D^2_u W)(v_\varepsilon )\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) , \frac{\partial u_\varepsilon }{\partial x_k} \rangle \ge 0. \end{aligned}$$

We deduce that

$$\begin{aligned} \langle (D^2_u W_\beta )(u_\varepsilon )\left( \frac{\partial u_\varepsilon }{\partial x_k}\right) , \frac{\partial u_\varepsilon }{\partial x_k} \rangle \ge -C\left| {u_\varepsilon - v_\varepsilon }\right| \left| {\frac{\partial u_\varepsilon }{\partial x_k}}\right| ^2 \ge -C(1-\left| {u_\varepsilon }\right| ^2)\left| {\frac{\partial u_\varepsilon }{\partial x_k}}\right| ^2, \end{aligned}$$

where the last inequality holds because from the comments before the proposition we have

$$\begin{aligned} \left| {u_\varepsilon -v_\varepsilon }\right| = {\mathrm{dist}}(u_\varepsilon , {{\mathcal {P}}}) \le C(1-\left| {u}\right| ^2). \end{aligned}$$

We conclude that

$$\begin{aligned} \Delta \left| {\nabla u_\varepsilon }\right| ^2 \ge 2\left| {D^2_x u}\right| ^2 - C\frac{(1-\left| {u_\varepsilon }\right| ^2)}{\varepsilon ^2}\left| {\nabla u}\right| ^2 \ge 2\left| {D^2_x u}\right| ^2 - C\left| {\Delta u_\varepsilon }\right| \left| {\nabla u}\right| ^2. \end{aligned}$$

Since this implies

$$\begin{aligned} \Delta \left| {\nabla u_\varepsilon }\right| ^2 \ge \left| {D^2_x u_\varepsilon }\right| ^2 - C\left| {\nabla u}\right| ^4 \end{aligned}$$

in $\Omega \setminus B_r(a)$, we can apply Steps A.4 and B.3 of the proof of Theorem 1 of [2] to conclude that

$$\begin{aligned} \left| {\nabla u_\varepsilon }\right| \le C \end{aligned}$$

in $\Omega \setminus B_r(a)$, for some constant independent of $\varepsilon \in ]0, \varepsilon _0]$.

Finally, we recall from [13] that

$$\begin{aligned} \Delta \frac{\left| {u_\varepsilon }\right| ^2}{2} = \frac{4}{\varepsilon ^2} W_{\frac{3\beta }{4}}(u_\varepsilon ) -\frac{4-\beta }{\varepsilon ^2}(1-\left| {u_\varepsilon }\right| ^2) + \left| {\nabla u_\varepsilon }\right| ^2. \end{aligned}$$

From here, $\zeta = 1-\left| {u_\varepsilon }\right| ^2$ satisfies

$$\begin{aligned} -\Delta \zeta + \frac{4-\beta }{\varepsilon ^2}\zeta = \left| {\nabla u_\varepsilon }\right| ^2. \end{aligned}$$

Steps A.5 and B.4 of Theorem 1 of [2] give us the last conclusion of the proposition. $\square$