Convergence to line and surface energies in nematic liquid crystal colloids with external magnetic field

Abstract

We use the Landau-de Gennes energy to describe a particle immersed into nematic liquid crystals with a constant applied magnetic field. We derive a limit energy in a regime where both line and point defects are present, showing quantitatively that the close-to-minimal energy is asymptotically concentrated on lines and surfaces nearby or on the particle. We also discuss regularity of minimizers and optimality conditions for the limit energy.

The complex \(\mathcal {T}\)

In this section, we collect and prove all results in connection to the structure of $\mathcal {T}$ as defined in Sect. 4.3. Recall that

$$\begin{aligned} \mathcal {T}:=\{ Q\in \text {Sym} _{0}\, : \,s>0\, , 0\le r<1\, , n_3 = 0 \}\, . \end{aligned}$$

Our first result is a characterization of $\mathcal {T}$ that provides us with a more accessible parametrization.

Proposition A.1

Every matrix $Q\in \mathcal {T}$ can be written as

$$\begin{aligned} Q = \lambda (\textbf{n}\otimes \textbf{n}- R_\textbf{n}^\top M R_\textbf{n})\, , \end{aligned}$$

where $\lambda >0$, $\textbf{n}=(n_1,n_2,0)\in \mathbb {S}^2$, $R_\textbf{n}$ is the rotation around $\textbf{n}\wedge \textbf{e}_3$, such that $R_\textbf{n}\textbf{n}= \textbf{e}_3$ and

with $M'\in \mathbb {R}^{2\times 2}$ symmetric, $\text {tr}(M')=1$ and $\langle M'v,v\rangle >-1$ for all $v\in \mathbb {S}^1$. The matrix Q is oblate uniaxial if and only if $M'=\frac{1}{2}\text {Id}$.

Proof

A matrix Q of the above form $Q = \lambda (\textbf{n}\otimes \textbf{n}- R_\textbf{n}^\top M R_\textbf{n})$ has $\textbf{n}$ as an eigenvector to the eigenvalue $\lambda$ and $\textbf{n}_3=0$ by definition. Furthermore, since $\min _{v\in \mathbb {S}^1}\langle M'v,v\rangle >-1$ the eigenvalue $\lambda$ is strictly bigger than the other eigenvalues, thus $r<1$ and $Q\in \mathcal {T}$. Conversely, we can write every $Q\in \text {Sym} _{0}$ as

$$\begin{aligned} Q \ = \ \lambda _1 \textbf{n}\otimes \textbf{n}+ \lambda _2 \textbf{m}\otimes \textbf{m}+ \lambda _3\textbf{p}\otimes \textbf{p}\, , \end{aligned}$$

with $\lambda _1\ge \lambda _2\ge \lambda _3$ and $\textbf{n},\textbf{m},\textbf{p}\in \mathbb {S}^2$ pairwise orthogonal eigenvectors of Q to $\lambda _1,\lambda _2,\lambda _3$. By definition of $\mathcal {T}$, $n_3=0$ as required for our parametrization and clearly we can identify $\lambda =\lambda _1$. Setting $M = -R_\textbf{n}(\frac{\lambda _2}{\lambda _1}\textbf{m}\otimes \textbf{m}+ \frac{\lambda _3}{\lambda _1}\textbf{p}\otimes \textbf{p})R_\textbf{n}^\top$, it is obvious that M is of the above form and that $Q\in \mathcal {T}$ can be written as claimed.

If $M'=\frac{1}{2}\text {Id}$ then

$$\begin{aligned} Q \ = \ \lambda (\textbf{n}\otimes \textbf{n}- R_\textbf{n}^\top M R_\textbf{n}) \ = \ \frac{3}{2}\lambda \left( \textbf{n}\otimes \textbf{n}- \frac{1}{3}\text {Id}\right) \, , \end{aligned}$$

i.e. Q is oblate uniaxial. The reverse implication follows similarly, since the matrices $R_\textbf{n}^\top ,R_\textbf{n}$ are invertible. $\square$

Remark A.2

Given a vector $u\in \mathbb {R}^3$ as axis of rotation and an angle $\theta$, then this rotation is described by the matrix R with

$$\begin{aligned} R = \begin{pmatrix} \cos \theta + u_1^2(1-\cos \theta ) &{} u_1 u_2 (1-\cos \theta ) - u_3\sin \theta &{} u_1 u_3 (1-\cos \theta ) + u_2\sin \theta \\ u_1 u_2 (1-\cos \theta ) + u_3\sin \theta &{} \cos \theta + u_2^2(1-\cos \theta ) &{} u_2 u_3 (1-\cos \theta ) - u_1\sin \theta \\ u_1 u_3 (1-\cos \theta ) - u_2\sin \theta &{} u_2 u_3 (1-\cos \theta ) + u_1\sin \theta &{} \cos \theta + u_3^2(1-\cos \theta ) \end{pmatrix}\,. \end{aligned}$$

Corollary A.3

$\mathcal {T}$ is a four dimensional smooth complex and $\partial \mathcal {T}= \mathcal {C}$.

Proof

From the characterization in Proposition A.1, it is clear that one can use the map $Q\mapsto (\lambda ,\textbf{n},m_{11},m_{12})$ to make $\mathcal {T}$ a four dimensional manifold with a conical singularity in $Q=0$. In particular, $\mathcal {T}$ is a smooth complex.

Proposition A.1 furthermore implies that the boundary of $\mathcal {T}$ consists of matrices of the form $\lambda =0$ (from which follows directly $Q=0$) or $M'$ has the eigenvalue $-1$ (which corresponds to $r=1$). In particular, the matrices with $r=0$ are not included in $\partial \mathcal {T}$ as one may think from the definition in (12). This implies the inclusion $\partial \mathcal {T}\subset \mathcal {C}$. For the inverse inclusion, take $Q\in \mathcal {C}$ with orthogonal eigenvectors $\textbf{m},\textbf{p}\in \mathbb {S}^2$ associated to the largest eigenvalue $\lambda _1=\lambda _2$. So in fact we have a two dimensional subspace of eigenvectors to this eigenvalue spanned by $\textbf{m}$ and $\textbf{p}$. Since the hyperplane defined through $\{n_3=0\}$ is of codimension one, there exists a unit vector $\textbf{n}\in \{n_3=0\}\cap \text {span}\{\textbf{m},\textbf{p}\}$ which we were looking for. The unit eigenvector orthogonal to $\textbf{n}$ in the plane $\text {span}\{\textbf{m},\textbf{p}\}$ requires $M'$ to have the eigenvalue $-1$ or in other words $\min _{v\in \mathbb {S}^1}\langle M'v,v\rangle =-1$, so that $Q\in \partial \mathcal {T}$. $\square$

Lemma A.4

Let $Q\in \mathcal {T}\cap \mathcal {N}$. Then, the normal vector $N_Q$ on $\mathcal {T}$ at Q is given by

$$\begin{aligned} N_Q = \frac{3}{2}\lambda \begin{pmatrix} 0 &{} 0 &{} n_1 \\ 0 &{} 0 &{} n_2 \\ n_1 &{} n_2 &{} 0 \end{pmatrix}\, , \end{aligned}$$

where $\textbf{n}=(n_1,n_2,0)\in \mathbb {S}^2$ is the eigenvector associated to the largest eigenvalue $\lambda _1$.

Proof

We are going to prove a slightly more general result by first considering $Q\in \mathcal {T}$ and calculating the tangent vectors to $\mathcal {T}$ in Q. We use the representation from Proposition A.1 and vary $\lambda ,\textbf{n},m_{11},m_{12}$ one after another.

• First, we can easily take the derivative with respect to $\lambda$ and obtain $\textbf{T}_1 = (\textbf{n}\otimes \textbf{n}- R_\textbf{n}^\top M R_\textbf{n})$.

• Second, we vary the parameter $\textbf{n}$. So, let’s consider $\textbf{n}=(n_1,n_2,0)\in \mathbb {S}^2$. Without loss of generality we assume that $n_2\ne 0$ and write $\textbf{n}(t) = (n_1+t,n_2-\frac{n_1}{n_2}t)$. Then $|\textbf{n}(t)|^2=1+O(t^2)$ and

  $$\begin{aligned} \textbf{n}(t)\otimes \textbf{n}(t) \ {}&= \ \textbf{n}\otimes \textbf{n}+ t D_{\textbf{n}\otimes \textbf{n}} + O(t^2)\, , \qquad D_{\textbf{n}\otimes \textbf{n}} = \begin{pmatrix} 2n_1 &{} n_2-\frac{n_1^2}{n_2} &{} 0 \\ n_2-\frac{n_1^2}{n_2} &{} -2n_1 &{} 0 \\ 0 &{} 0 &{} 0 \end{pmatrix}\, . \end{aligned}$$

  The derivative of the second term $R_{\textbf{n}(t)}^\top M R_{\textbf{n}(t)}$ can be calculated using Remark A.2 with the axis $\textbf{n}^\perp (t):=\textbf{n}(t)\wedge \textbf{e}_3$. Since $\textbf{n}(t)\perp \textbf{e}_3$ we can write

  $$\begin{aligned} R_{\textbf{n}(t)} \ {}&= \ R_{\textbf{n}} + t D_{R_\textbf{n}} + O(t^2)\, , \qquad D_{R_\textbf{n}} = \frac{1}{n_2} \begin{pmatrix} -2n_1 n_2 &{} -n_2^2+n_1^2 &{} -n_2 \\ -n_2^2+n_1^2 &{} 2n_1 n_2 &{} n_1 \\ n_2 &{} -n_1 &{} 0 \end{pmatrix}\, . \end{aligned}$$

  The second tangent vector $\textbf{T}_2$ is thus given by $\textbf{T}_2 = \lambda (D_{\textbf{n}\otimes \textbf{n}} - D_{R_\textbf{n}}^\top M R_\textbf{n}- R_\textbf{n}^\top M D_{R_\textbf{n}})$.

• Third, we can take the derivative with respect to $m_{11}$. This is straightforward and we obtain

  $$\begin{aligned} \textbf{T}_3 = \lambda R_\textbf{n}^\top \begin{pmatrix} 1 &{} 0 &{} \\ 0 &{} -1 &{} \\ &{} &{} 0 \end{pmatrix} R_\textbf{n}\, . \end{aligned}$$

• Last, varying $m_{12}$ we easily calculate

  $$\begin{aligned} \textbf{T}_4 = \lambda R_\textbf{n}^\top \begin{pmatrix} 0 &{} 1 &{} \\ 1 &{} 0 &{} \\ &{} &{} 0 \end{pmatrix} R_\textbf{n}\, . \end{aligned}$$

Before proceeding, we want to calculate a fifth vector by varying $\textbf{n}_3$. As it will turn out later, this is indeed the normal vector.

• Writing once again $\textbf{n}=(n_1,n_2,0)$ and defining $\textbf{n}(t):=(n_1\sqrt{1-t^2},n_2\sqrt{1-t^2},t)$ we can express

  $$\begin{aligned} \textbf{n}(t)\otimes \textbf{n}(t) \ {}&= \ \textbf{n}\otimes \textbf{n}+ t (\textbf{n}\otimes \textbf{e}_3 + \textbf{e}_3\otimes \textbf{n}) + O(t^2)\, . \end{aligned}$$

  As for the second tangent vector, we use Remark A.2 and the rotation around $\textbf{n}^\perp (t) = \textbf{n}(t)\wedge \textbf{e}_3$. Unlike previously, $\textbf{n}(t)$ is no longer orthogonal to $\textbf{e}_3$ for $t\ne 0$, namely $\theta =\arccos (\langle \textbf{n}(t),\textbf{e}_3\rangle )=t$. Substituting this our expression of the rotation matrix we get

  $$\begin{aligned} R_{\textbf{n}(t)} \ = \ R_{\textbf{n}} + t D_{3} + O(t^2)\, ,\qquad D_{3} = \begin{pmatrix} 1-n_2^2 &{} n_1 n_2 &{} 0 \\ n_1 n_2 &{} 1- n_1^2 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix}\, . \end{aligned}$$

  Adding the two partial results, we get

  $$\begin{aligned} N :=\lambda (\textbf{n}\otimes \textbf{e}_3 + \textbf{e}_3\otimes \textbf{n}- D_{3}^\top M R_\textbf{n}- R_\textbf{n}^\top M D_{3})\, . \end{aligned}$$

It remains to show that $\{\textbf{T}_1,\textbf{T}_2,\textbf{T}_3,\textbf{T}_4, N\}$ are pairwise orthogonal if Q is oblate uniaxial. Indeed, then it follows that N is a normal vector, since it is orthogonal to $T_Q\mathcal {T}$.

It is easy to see that since the trace is invariant by change of basis and since $R_\textbf{n}^\top =R_\textbf{n}^{-1}$

$$\begin{aligned} \langle \textbf{T}_3,\textbf{T}_4\rangle \ {}&= \ \lambda ^2\text {tr}\left( \begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix}\begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix} \right) \ = \ \lambda ^2 \text {tr}\left( \begin{pmatrix} 0 &{} 1 \\ -1 &{} 0 \end{pmatrix} \right) \ = \ 0\, . \end{aligned}$$

Noting that $\textbf{n}\otimes \textbf{n}R_\textbf{n}^\top M R_\textbf{n}=0$ for $M\in \text {Sym} _{0}$ with $m_{ij}=0$ if $i=3$ or $j=3$, we get

$$\begin{aligned} \langle \textbf{T}_1,\textbf{T}_3\rangle \ {}&= \ \lambda \text {tr}\left( (\textbf{n}\otimes \textbf{n}- R_\textbf{n}^\top M R_\textbf{n})\left( R_\textbf{n}^\top \begin{pmatrix} 1 &{} 0 &{} \\ 0 &{} -1 &{} \\ &{} &{} 0 \end{pmatrix} R_\textbf{n}\right) \right) \\ \ {}&= \ \lambda \text {tr}\left( M\begin{pmatrix} 1 &{} 0 &{} \\ 0 &{} -1 &{} \\ &{} &{} 0 \end{pmatrix}\right) \ = \ \lambda \text {tr}\left( \begin{pmatrix} m_{11} &{} -m_{12} &{} \\ m_{12} &{} -m_{22} &{} \\ &{} &{} 0 \end{pmatrix}\right) \ = \ \lambda (2 m_{11}-1) \, . \end{aligned}$$

With the same argument we also find

$$\begin{aligned} \langle \textbf{T}_1,\textbf{T}_4\rangle \ {}&= \ \lambda \text {tr}\left( (\textbf{n}\otimes \textbf{n}- R_\textbf{n}^\top M R_\textbf{n})\left( R_\textbf{n}^\top \begin{pmatrix} 0 &{} 1 &{} \\ 1 &{} 0 &{} \\ &{} &{} 0 \end{pmatrix} R_\textbf{n}\right) \right) \\ \ {}&= \ \lambda \text {tr}\left( M \begin{pmatrix} 0 &{} 1 &{} \\ 1 &{} 0 &{} \\ &{} &{} 0 \end{pmatrix}\right) \ = \ \lambda \text {tr}\left( \begin{pmatrix} m_{12} &{} m_{11} &{} \\ m_{22} &{} m_{12} &{} \\ &{} &{} 0 \end{pmatrix}\right) \ = \ 2\lambda m_{12} \, . \end{aligned}$$

Furthermore, we claim that

$$\begin{aligned} \langle \textbf{T}_1,\textbf{T}_2\rangle \ {}&= \ \lambda \, \text {tr}\Bigg ((\textbf{n}\otimes \textbf{n}- R_\textbf{n}^\top M R_\textbf{n})(D_{\textbf{n}\otimes \textbf{n}} - D_{R_\textbf{n}}^\top M R_\textbf{n}- R_\textbf{n}^\top M D_{R_\textbf{n}})\Bigg ) \ = \ 0\, . \end{aligned}$$

Indeed, one can check that

$$\begin{aligned} \text {tr}(\textbf{n}\otimes \textbf{n}D_{\textbf{n}\otimes \textbf{n}}) \ {}&= \ 0 \ = \ \text {tr}(\textbf{n}\otimes \textbf{n}D_{R_\textbf{n}}^\top M R_\textbf{n}) \, , \\ \text {tr}(\textbf{n}\otimes \textbf{n}R_\textbf{n}^\top M D_{R_\textbf{n}}) \ {}&= \ 0 \ = \ \text {tr}(R_\textbf{n}^\top M R_\textbf{n}D_{\textbf{n}\otimes \textbf{n}}) \, , \\ \text {tr}(R_\textbf{n}^\top M R_\textbf{n}D_{R_\textbf{n}}^\top M R_\textbf{n}) \ {}&= \ 0 \ = \ \text {tr}(R_\textbf{n}^\top M R_\textbf{n}R_\textbf{n}^\top M D_{R_\textbf{n}}) \, . \end{aligned}$$

This implies that

$$\begin{aligned} \langle N,\textbf{T}_3\rangle \ {}&= \ \lambda ^2 \text {tr}\left( (\textbf{n}\otimes \textbf{e}_3 + \textbf{e}_3\otimes \textbf{n}- D_{3}^\top M R_\textbf{n}- R_\textbf{n}^\top M D_{3})\left( R_\textbf{n}^\top \begin{pmatrix} 1 &{} 0 &{} \\ 0 &{} -1 &{} \\ &{} &{} 0 \end{pmatrix} R_\textbf{n}\right) \right) \ = \ 0\, , \end{aligned}$$

since again the traces of all cross terms vanish. Similarly,

$$\begin{aligned} \langle N,\textbf{T}_4\rangle \ {}&= \ 0 \, . \end{aligned}$$

Next, we have the equality

$$\begin{aligned} \langle \textbf{T}_2,\textbf{T}_3\rangle \ {}&= \ -4 \lambda ^2 \frac{m_{12}}{n_2} \, . \end{aligned}$$

This follows since $\text {tr}(D_{\textbf{n}\otimes \textbf{n}} \textbf{T}_3) = 0$ and $\text {tr}(D_{3} M R_\textbf{n}\textbf{T}_3) = \frac{2m_{12}}{n_2}$. The latter fact is evident if one calculates $M \begin{pmatrix} 1 &{} 0 &{} \\ 0 &{} -1 &{} \\ &{} &{} 0 \end{pmatrix} \ = \ \begin{pmatrix} m_{11} &{} -m_{12} &{} \\ m_{12} &{} -m_{22} &{} \\ &{} &{} 0 \end{pmatrix}$ and $R_\textbf{n}D_3^\top = \begin{pmatrix} 0 &{} -1/n_2 &{} 1 \\ 1/n_2 &{} 0 &{} -n_1/n_2 \\ -1 &{} n_1/n_2 &{} 0 \end{pmatrix}$. This also permits us to derive

$$\begin{aligned} \langle \textbf{T}_2,\textbf{T}_4\rangle \ {}&= \ 2\lambda ^2\frac{2m_{11}-1}{n_2} \, . \end{aligned}$$

Again, we simply calculate the traces of all cross terms. For example

$$\begin{aligned} \text {tr}(\textbf{n}\otimes \textbf{e}_3 D_{\textbf{n}\otimes \textbf{n}}) \ {}&= \ 0 \, , \\ \text {tr}(\textbf{n}\otimes \textbf{e}_3 R_{\textbf{n}}^\top M D_{R_{\textbf{n}}}) \ {}&= \ 0 \, , \\ \text {tr}(\textbf{n}\otimes \textbf{e}_3 D_{R_\textbf{n}}^\top MR_\textbf{n}) \ {}&= \ \frac{m_{12}}{n_2}(n_1^2-n_2^2)-n_1(2m_{11}-1) \, , \\ \text {tr}(D_{R_\textbf{n}}^\top M R_\textbf{n}D_{\textbf{n}\otimes \textbf{n}}) \ {}&= \ 2\frac{m_{11}n_1}{n_2} + \frac{1}{n_2^2}(n_1^2(2m_{11}-1)+m_{11}) \, , \\ \text {tr}(D_{R_\textbf{n}}^\top M R_\textbf{n}R_{\textbf{n}}^\top M D_{R_{\textbf{n}}}) \ {}&= \ -2\frac{n_1 m_{12}}{n_2} + \frac{1}{n_2^2}\left( 3(m_{11}^2 + m_{12}^2) - (1 + n_1^2)(2m_{11}-1) \right) \, , \\ \text {tr}(D_{R_\textbf{n}}^\top M R_\textbf{n}D_{R_\textbf{n}}^\top MR_\textbf{n}) \ {}&= \ 2\frac{m_{11}m_{22}+m_{12}^2}{n_2^2}\, , \end{aligned}$$

We end up with

$$\begin{aligned} \langle N,\textbf{T}_2\rangle \ {}&= \ \frac{6\lambda ^2 m_{12} (n_1^2 - n_2^2)}{n_2} - 6\lambda ^2 n_1(2 m_{11} -1) \, . \end{aligned}$$

Another straightforward calculation shows that

$$\begin{aligned} \langle N,\textbf{T}_1\rangle \ {}&= \ \lambda 2 n_1 m_{12}(n_1^5 m_{12} - 2 n_{1}^4 n_{2} m_{11} - 2 n_{1}^3 m_{12} - 2n_1^2 n_2^3 m_{11} + 3 n_1^2 n_2 m_{11} \\&-2n_1^2 n_2- n_1 n_2^4m_{12} + n_1 m_{12} + n_2^3 m_{11} - n_2 m_{11} - 2 n_2^3 + 2n_2 ) \, . \end{aligned}$$

After these calculations, it is apparent that for prolate uniaxial $Q\in \text {Sym} _{0}$ (and in particular $Q\in \mathcal {N}$), i.e. $M'=\frac{1}{2}\text {Id}$ all inner products vanish. In order to form a basis, we must prove that the vectors themselves never vanish. We find

$$\begin{aligned} \Vert \textbf{T}_1\Vert ^2 \ {}&= \ 2(m_{11}^2 - m_{11} + m_{12} + 1) \, , \\ \Vert \textbf{T}_2\Vert ^2 \ {}&= \ \frac{2}{n_2^2}(6n_1^2(1-2m_{11}) - 6m_{12}n_1 n_2 + 5m_{11}^2 - 2m_{11} + 5m_{12} + 2) \, , \\ \Vert \textbf{T}_3\Vert ^2 \ {}&= \ 2\lambda ^2 \, , \\ \Vert \textbf{T}_4\Vert ^2 \ {}&= \ 2\lambda ^2 \, , \\ \Vert N \Vert ^2 \ {}&= \ \lambda ^2(12m_{11}n_1^2 -6n_1^2 + 12 m_{12}n_1n_2 + 2m_{11}^2 - 8m_{11} + 2m_{12}^2 + 8) \, , \end{aligned}$$

and thus for $M'=\frac{1}{2}\text {Id}$ it holds that $\Vert \textbf{T}_1\Vert ^2 = \frac{6}{4}$, $\Vert \textbf{T}_2\Vert ^2=\frac{9}{2}\lambda ^2 n_2^{-2}$ and $\Vert N\Vert ^2 = \frac{9}{2}\lambda ^2$.

This concludes the proof that $\{\textbf{T}_1,\textbf{T}_2,\textbf{T}_3,\textbf{T}_4\}$ form indeed a basis of $T_Q\mathcal {T}$, and since N is orthogonal to $T_Q\mathcal {T}$, the result follows. $\square$

Proposition A.5

There exists $C,\alpha _0>0$ such that for all $\alpha \in (0,\alpha _0)$ and $Q\in \mathcal {N}$ it holds

$$\begin{aligned} \mathcal {H}^4(B_\alpha (Q)\cap \mathcal {T}) \ \le \ C \alpha ^4\,. \end{aligned}$$

Proof

As seen before, $\mathcal {T}$ has the structure of a smooth manifold around $\mathcal {N}$. By invariance of $\mathcal {N}$ under rotations, it is enough to show that the claim holds around one $Q\in \mathcal {N}$. The Ricci curvature $\kappa$ of $\mathcal {N}$ is bounded so that we can choose $\alpha _0>0$ small enough such that $B_\alpha (Q)\cap \mathcal {T}$ is contained in the geodesic ball in $\mathcal {T}$ of size $2\alpha$ around Q for all $\alpha \in (0,\alpha _0)$. Furthermore, if needed, we can choose $\alpha _0>0$ even smaller such that $1-\frac{\kappa }{36\alpha _0^2}\le 2$. Theorem 3.1 in [38] then implies that

$$\begin{aligned} \mathcal {H}^4(B_\alpha (Q)\cap \mathcal {T}) \ {}&\le \ \text {vol}_\mathcal {T}(B_{2\alpha }(Q)) \ \le \ 16\pi ^2\alpha ^4\, . \end{aligned}$$

$\square$