On the Landau–de Gennes Elastic Energy of a Q-Tensor Model for Soft Biaxial Nematics

Abstract

In the Landau–de Gennes theory of liquid crystals, the propensities for alignments of molecules are represented at each point of the fluid by an element $\mathbf{Q}$ of the vector space ${\mathcal {S}}_0$ of $3\times 3$ real symmetric traceless matrices, or $\mathbf{Q}$-tensors. According to Longa and Trebin (1989), a biaxial nematic system is called soft biaxial if the tensor order parameter $\mathbf{Q}$ satisfies the constraint $\mathrm{tr}(\mathbf{Q}^2) = \text {const}$. After the introduction of a $\mathbf{Q}$-tensor model for soft biaxial nematic systems and the description of its geometric structure, we address the question of coercivity for the most common four-elastic-constant form of the Landau–de Gennes elastic free-energy (Iyer et al. 2015) in this model. For a soft biaxial nematic system, the tensor field $\mathbf{Q}$ takes values in a four-dimensional sphere ${{\mathbb {S}}}^4_\rho$ of radius $\rho \le \sqrt{2/3}$ in the five-dimensional space ${\mathcal {S}}_0$ with inner product $\langle \mathbf{Q}, {\mathbf {P}} \rangle = \mathrm{tr}(\mathbf{Q}{\mathbf {P}})$. The rotation group $\textit{SO}(3)$ acts orthogonally on ${\mathcal {S}}_0$ by conjugation and hence induces an action on ${{\mathbb {S}}}^4_\rho \subset {\mathcal {S}}_0$. This action has generic orbits of codimension one that are diffeomorphic to an eightfold quotient ${{\mathbb {S}}}^3/{\mathcal {H}}$ of the unit three-sphere ${{\mathbb {S}}}^3$, where ${{\mathcal {H}}}=\{\pm 1, \pm \mathsf{i}, \pm \mathsf{j}, \pm \mathsf{k}\}$ is the quaternion group, and has two degenerate orbits of codimension two that are diffeomorphic to the projective plane ${\mathbb {R}}P^2$. Each generic orbit can be interpreted as the order parameter space of a constrained biaxial nematic system and each singular orbit as the order parameter space of a constrained uniaxial nematic system. It turns out that ${{\mathbb {S}}}^4_\rho$ is a cohomogeneity one manifold, i.e., a manifold with a group action whose orbit space is one-dimensional. Another important geometric feature of the model is that the set $\Sigma _\rho$ of diagonal $\mathbf{Q}$-tensors of fixed norm $\rho$ is a (geodesic) great circle in ${{\mathbb {S}}}^4_\rho$ which meets every orbit of ${{\mathbb {S}}}^4_\rho$ orthogonally and is then a section for ${{\mathbb {S}}}^4_\rho$ in the sense of the general theory of canonical forms. We compute necessary and sufficient coercivity conditions for the elastic energy by exploiting the $\textit{SO}(3)$-invariance of the elastic energy (frame-indifference), the existence of the section $\Sigma _\rho$ for ${{\mathbb {S}}}^4_\rho$, and the geometry of the model, which allow us to reduce to a suitable invariant problem on (an arc of) $\Sigma _\rho$. Our approach can ultimately be seen as an application of the general method of reduction of variables, or cohomogeneity method.

Appendix: The Longa–Monselesan–Trebin Conditions

In this appendix, using the scalar coordinates corresponding to the representation of $\mathbf{Q}$-tensors described in Sect. 2.1, we compute the Longa–Monselesan–Trebin positivity conditions

$$\begin{aligned} 2L_3>L_2,\quad L_3+L_2>0,\quad 10L_1+L_2+6L_3>0 \end{aligned}$$

(5.1)

for the three-elastic-constant form $L_1I_1+L_2I_2+L_3I_3$ of the elastic energy density $\psi _E$ in the general biaxial case (cf. (1.8)).

These conditions were originally obtained by Longa–Monselesan–Trebin (1987) (cf. also Davis and Gartland 1998) by writing the elastic energy density $L_1I_1+L_2I_2+L_3I_3$ as a linear combination of irreducible $\textit{SO}(3)$-invariants, computed using the representation theory of $\textit{SO}(3)$ on spherical tensors and the Clebsch–Gordan coefficients from the angular momentum theory of quantum mechanics (Messiah 2003).

In general, by (2.7) we have $I_3=|\nabla \mathbf{u}|^2$. Also, by (3.7), in terms of $\mathbf{u}$ we have

$$\begin{aligned} 2I_1= & {} \left( {\frac{1}{\sqrt{3}}}\partial _1u^1+\partial _1u^2+\partial _2u^3+\partial _3u_4\right) ^2+ \left( {\frac{1}{\sqrt{3}}}\partial _2u^1-\partial _2u^2+ \partial _3u^5+\partial _1u^3\right) ^2\\&+\left( -{\frac{2}{\sqrt{3}}}\partial _3u^1 +\partial _1u^4+\partial _2u^5\right) ^2. \end{aligned}$$

Similarly, by (3.8) we find the formula

$$\begin{aligned} 2I_2= & {} \displaystyle \left( {\frac{1}{\sqrt{3}}}\partial _1u^1+\partial _1u^2\right) ^2+(\partial _2u^3)^2+(\partial _3u^4)^2\\&+\,2\Bigl (-{\frac{2}{\sqrt{3}}}\partial _1u^1 \partial _3u^4+{\frac{1}{\sqrt{3}}}\partial _1u^1\partial _2u^3 - \partial _1u^2\partial _2u^3\Bigr )\\&+\,\displaystyle \left( {\frac{1}{\sqrt{3}}}\partial _2u^1-\partial _2u^2\right) ^2+(\partial _3u^5)^2+(\partial _1u^3)^2\\&+\,2\left( -{\frac{2}{\sqrt{3}}}\partial _2u^1\partial _3u^5+{\frac{1}{\sqrt{3}}}\partial _2u^1\partial _1u^3 - \partial _2u^2\partial _1u^3\right) \\&+\,\displaystyle \Bigl (-{\frac{2}{\sqrt{3}}}\partial _3u^1\Bigr )^2+(\partial _1u^4)^2+(\partial _2u^5)^2\\&+\,2\left( {\frac{1}{\sqrt{3}}}\partial _3u^1\partial _1u^4 + \partial _3u^2\partial _1u^4+ {\frac{1}{\sqrt{3}}}\partial _3u^1\partial _2u^5 - \partial _3u^2\partial _2u^5\right) \\&+\,2\,\bigl (\partial _3u^3\partial _2u^4+\partial _3u^3\partial _1u^5+\partial _2u^4\partial _1u^5\bigr ) . \end{aligned}$$

For any choice of the real coefficients $L_i$ we thus may decompose into four quadratic forms:

$$\begin{aligned} L_1I_1+L_2I_2+L_3I_3={{\mathcal {F}}}_1+{{\mathcal {F}}}_2+{{\mathcal {F}}}_3+{{\mathcal {F}}}_4. \end{aligned}$$

The quadratic form ${{\mathcal {F}}}_1$ depends on the four variables $\partial _1u^1,\,\partial _1u^2,\,\partial _2u^3,\,\partial _3u_4$, and its related matrix is:

$$\begin{aligned} M_1:=\begin{pmatrix} \displaystyle L_3+{\frac{L_1+L_2}{6}} &{}\quad \displaystyle {\frac{1}{2\sqrt{3}}}(L_1+L_2) &{}\quad \displaystyle {\frac{1}{2\sqrt{3}}}(L_1+L_2) &{}\quad \displaystyle {\frac{1}{2\sqrt{3}}}(L_1-2L_2) \\ \displaystyle {\frac{1}{2\sqrt{3}}}(L_1+L_2) &{}\quad \displaystyle {\frac{K}{2}} &{}\quad \displaystyle {\frac{1}{2}}(L_1-L_2) &{}\quad \displaystyle {\frac{L_1}{2}} \\ \displaystyle {\frac{1}{2\sqrt{3}}}(L_1+L_2) &{}\quad \displaystyle {\frac{1}{2}}(L_1-L_2) &{}\quad \displaystyle {\frac{K}{2}} &{}\quad \displaystyle {\frac{L_1}{2}} \\ \displaystyle {\frac{1}{2\sqrt{3}}}(L_1-2L_2) &{}\quad \displaystyle {\frac{L_1}{2}} &{}\quad \displaystyle {\frac{L_1}{2}} &{}\quad \displaystyle {\frac{K}{2}} \end{pmatrix} \end{aligned}$$

where we have denoted $K:=L_1+L_2+2L_3$. The second one, ${{\mathcal {F}}}_2$, depends on the four variables $\partial _2u^1,$ $\partial _2u^2,$ $\partial _3u^5,$ $\partial _1u^3$, and its corresponding matrix is:

$$\begin{aligned} M_2:=\begin{pmatrix} \displaystyle L_3+{\frac{L_1+L_2}{6}} &{}\quad \displaystyle -{\frac{1}{2\sqrt{3}}}(L_1+L_2) &{}\quad \displaystyle {\frac{1}{2\sqrt{3}}}(L_1-2L_2) &{}\quad \displaystyle {\frac{1}{2\sqrt{3}}}(L_1+L_2) \\ \displaystyle -{\frac{1}{2\sqrt{3}}}(L_1+L_2) &{}\quad \displaystyle {\frac{K}{2}} &{}\quad \displaystyle -{\frac{L_1}{2}} &{}\quad -\displaystyle {\frac{1}{2}}(L_1-L_2) \\ \displaystyle {\frac{1}{2\sqrt{3}}}(L_1-2L_2) &{}\quad -\displaystyle {\frac{L_1}{2}} &{}\quad \displaystyle {\frac{K}{2}} &{}\quad \displaystyle {\frac{L_1}{2}} \\ \displaystyle {\frac{1}{2\sqrt{3}}}(L_1+L_2) &{}\quad \displaystyle -{\frac{1}{2}}(L_1-L_2) &{}\quad \displaystyle {\frac{L_1}{2}} &{}\quad \displaystyle {\frac{K}{2}} \end{pmatrix}. \end{aligned}$$

The third one, ${{\mathcal {F}}}_3$, depends on the four variables $\partial _3u^1,\,\partial _3u^2,\,\partial _1u^4,\,\partial _2u^5$, and its corresponding matrix is:

$$\begin{aligned} M_3:=\begin{pmatrix} \displaystyle L_3+{\frac{2}{3}}(L_1+L_2) &{}\quad \displaystyle 0 &{}\quad \displaystyle {\frac{1}{2\sqrt{3}}}(-2L_1+L_2) &{} \displaystyle {\frac{1}{2\sqrt{3}}}(-2L_1+L_2) \\ \displaystyle 0 &{}\quad \displaystyle L_3 &{}\quad \displaystyle {\frac{L_2}{2}} &{}\quad -\displaystyle {\frac{L_2}{2}} \\ \displaystyle {\frac{1}{2\sqrt{3}}}(-2L_1+L_2) &{}\quad \displaystyle {\frac{L_2}{2}}&{}\quad \displaystyle {\frac{K}{2}} &{} \quad \displaystyle {\frac{L_1}{2}} \\ \displaystyle {\frac{1}{2\sqrt{3}}}(-2L_1+L_2) &{}\quad \displaystyle -{\frac{L_2}{2}} &{}\quad \displaystyle {\frac{L_1}{2}} &{}\quad \displaystyle {\frac{K}{2}} \end{pmatrix}. \end{aligned}$$

Finally, the fourth one, ${{\mathcal {F}}}_4$, depends on the three remaining variables $\partial _3u^3,\,\partial _2u^4,\,\partial _1u^5$, and its corresponding matrix is:

$$\begin{aligned} M_4:=\begin{pmatrix} \displaystyle L_3 &{} \displaystyle {\frac{L_2}{2}} &{}\quad \displaystyle {\frac{L_2}{2}} \\ \displaystyle \displaystyle {\frac{L_2}{2}} &{}\quad \displaystyle L_3 &{}\quad \displaystyle {\frac{L_2}{2}} \\ \displaystyle {\frac{L_2}{2}}&{}\quad \displaystyle {\frac{L_2}{2}} &{}\quad L_3 \end{pmatrix}. \end{aligned}$$

The matrix $M_4$ has determinant $(L_3-L_2/2)^2(L_3+L_2)$, whence ${{\mathcal {F}}}_4$ is positive definite if and only if the conditions $L_3>0$, $2L_3>|L_2|$, $L_3+L_2>0$ hold, which clearly reduce to the system

$$\begin{aligned} L_3>0,\quad 2L_3>L_2,\quad L_3+L_2>0. \end{aligned}$$

(5.2)

Dealing with $M_3$, and starting from the right-bottom corner, we obtain the first two conditions $K>0$ and $K>|L_1|$, which reduce to

$$\begin{aligned} L_2+2L_3>0,\quad 2L_1+L_2+2L_3>0, \end{aligned}$$

(5.3)

where the first inequality follows from (5.2). The determinant of the $3\times 3$ right-bottom minor is

$$\begin{aligned}&{\frac{1}{4}}\,(2L_1+L_2+2L_3)(2L_3^2+L_2L_3-L_2^2)\\&\quad ={\frac{1}{4}}\,(2L_1+L_2+2L_3)(2L_3-L_2)(L_3+L_2) \end{aligned}$$

and hence it is positive under the assumptions (5.2) and (5.3). Finally, we compute $\mathrm{det}\,M_3$ by applying Laplace’s formula w.r.t. the first column, and we write

$$\begin{aligned} \mathrm{det}\,M_3=A_3^1-A_3^2+A_3^3-A_3^4 \end{aligned}$$

where $A_3^2=0$ and we, respectively, compute:

$$\begin{aligned} A_3^1= & {} \left( L_3+{\frac{2}{3}}\,(L_1+L_2) \right) \cdot {\frac{1}{4}}\,(2L_1+L_2+2L_3)(2L_3-L_2)(L_3+L_2),\\ A_3^3-A_3^4= & {} -{\frac{1}{12}}\,(-2L_1+L_2)^2\,(2L_3-L_2)(L_3+L_2) \end{aligned}$$

and hence

$$\begin{aligned} \begin{array}{rl} 12\mathrm{det}\,M_3= &{} (2L_3-L_2)(L_3+L_2)\,[(3L_3+2L_1+2L_2)(2L_1+L_2+2L_3)\\ &{}\quad -(-2L_1+L_2)^2] \\ =&{} \displaystyle (2L_3-L_2)(L_3+L_2)^2(6L_3+L_1+10L_2). \end{array} \end{aligned}$$

Therefore, under the conditions (5.2) and (5.3), the determinant of $M_3$ is positive if and only if we also have

$$\begin{aligned} 6L_3+L_1+10L_2>0 . \end{aligned}$$

(5.4)

We now consider the matrix $M_2$. Starting from the right-bottom corner, we again obtain the first two conditions $K>0$ and $K>|L_1|$. This time, the determinant D of the $3\times 3$ right-bottom minor is

$$\begin{aligned} D= & {} {\frac{1}{8}}\,\bigl [K(K^2-2L_1^2-(L_2-L_1)^2)-2L_1^2(L_2-L_1) \bigr ]\nonumber \\= & {} {\frac{1}{2}}\,(L_3+L_2)(2L_3^2+(3L_1+L_2)L_3+L_1L_2) \end{aligned}$$

(5.5)

and hence it is positive under the additional assumption

$$\begin{aligned} 2L_3^2+(3L_1+L_2)L_3+L_1L_2>0. \end{aligned}$$

(5.6)

As before, we now compute $\mathrm{det}\,M_2$ by applying Laplace’s formula w.r.t. the first column, and we write

$$\begin{aligned} \mathrm{det}\,M_2=A_2^1-A_2^2+A_2^3-A_2^4. \end{aligned}$$

We have:

$$\begin{aligned} A_2^1= & {} {\frac{6L_3+L_1+L_2}{6}}\cdot {\frac{1}{2}}\,(L_3+L_2)(2L_3^2+(3L_1+L_2)L_3+L_1L_2),\\ -A_2^2= & {} -{\frac{1}{24}}\,(L_3+L_2)(L_1+L_2)(4L_1L_2+L_2^2+2L_1L_3+2L_2L_3),\\ A_2^3= & {} {\frac{1}{24}}\,(L_3+L_2)(L_1-2L_2)(3L_1L_2-L_1L_3+2L_2L_3),\\ -A_2^4= & {} -{\frac{1}{24}}\,(L_3+L_2)(L_1+L_2)(4L_1L_2+L_2^2+2L_1L_3+2L_2L_3)=-A_2^2, \end{aligned}$$

and hence we obtain:

$$\begin{aligned} \begin{array}{rl} 12\,\mathrm{det}\,M_2= (L_2+L_3)\,\bigl [ &{} (6L_3+L_1+L_2)(2L_3^2+(3L_1+L_2)L_3+L_1L_2) \\ &{}-(L_1+L_2)(4L_1L_2+L_2^2+2L_1L_3+2L_2L_3) \\ &{} + (L_1-2L_2)(3L_1L_2-L_1L_3+2L_2L_3) \,\,\bigr ]\\ = \,\, (L_2+L_3)\,\bigl [ &{} 12L_3^3+4(5L_1+2L_2)L_3^2+(10L_1L_2-5L_2^2)L_3\\ &{}-(10L_1L_2^2+L_2^3) \,\,\bigr ] \\ = \,\, (L_2+L_3)\,\bigl [ &{} (L_2+L_2)(2L_3-L_2)(6L_3+10L_1+L_2)\,\,\bigr ]. \end{array} \end{aligned}$$

Therefore, condition $\mathrm{det}\,M_3>0$ follows from (5.2), (5.3), and (5.4).

We finally deal with the matrix $M_1$, obtaining exactly the same positivity conditions as for $M_2$. In fact, starting from the right-bottom corner, the first two conditions are again $K>0$ and $K>|L_1|$, whereas the determinant D of the $3\times 3$ right-bottom minor is equal to the expression from (5.5). Computing as before the determinant of $M_1$ by means of Laplace’s formula w.r.t. the first column, and writing

$$\begin{aligned} \mathrm{det}\,M_1=A_1^1-A_1^2+A_1^3-A_1^4, \end{aligned}$$

it is not difficult to check that we have:

$$\begin{aligned} A_1^1=A_1^2,\quad -A_1^2=-A_2^2,\quad A_1^3=-A_2^4,\quad -A_1^4=A_2^3 \end{aligned}$$

and hence we get $\mathrm{det}\,M_1=\mathrm{det}\,M_2$.

In conclusion, arguing as in Remark 3.4 we deduce that the system (5.2), (5.3), (5.4), and (5.6) is equivalent to the one in (5.1).