On the Equilibria of the Extended Nematic Polymers under Elongational Flow

We classify the equilibrium solutions of the Smoluchowski equation for dipolar (extended) rigid nematic polymers under imposed elongational flow. The Smoluchowski equation couples the Maier-Saupe short-range interaction, dipole-dipole interaction, and an external elongational flow. We show that all stable equilibria of rigid, dipolar rod dispersions under imposed uniaxial elongational flow field are axisymmetric. This finding of axisymmetry significantly simplifies any procedure of obtaining experimentally observable equilibria.


Introduction
Nematic liquid crystal polymers are viscoelastic anisotropic materials that have many important applications [1].The dynamic behavior of rigid rod nematic liquid crystal polymers is modeled by the Smoluchowski equation [2][3][4].Analytical results on pure nematic equilibria have been obtained in a series of papers [5][6][7][8][9][10][11].In [12] a 2D Smoluchowski equation under weak shear is analyzed.In [13] a coplanar magnetic field is coupled in the Smoluchowski equation to investigate the monodomain dynamics for rigid rod and platelet suspensions.However, dipole-dipole interaction is not included in [13].Recently, the authors [14,15] studied kinetic equilibria of rigid, dipolar rod ensembles for coupled dipole-dipole and Maier-Saupe short-range potentials.This work is a natural extension of [14,15] to include an external elongational flow.For completeness and for reader's convenience, we include all the lemmas and theorems necessary for reaching the conclusions even though some of the lemmas and theorems have appeared in previous works.
This paper is organized as follows.We first briefly give the mathematical formulation of the Smoluchowski equation for extended (polar) nematics in Section 2. Our main theoretical results of the equilibrium solutions of the Smoluchowski equation are presented in Sections 3, 4, and 5.More specifically, we first show that the first moment of an equilibrium solution must be aligned with one of the principal axes of the second moment.Then in Section 4 we exploit free energy to show that an equilibrium solution whose first moment is not parallel to the imposed external elongational flow field is unstable.Finally in Section 5 we prove the most important result of this paper: all stable equilibrium solutions are axisymmetric.We give concluding remarks in Section 6.

The Smoluchowski equation for extended (polar) nematics
In this paper, we study equilibrium solutions of the Smoluchowski equation for rigid extended (polar) nematics under elongational flow.Here the nematic molecules are magnetically polar.For these extended nematics, the molecular interaction includes both the dipole-dipole interaction and the short-range Maier-Saupe interaction.We assume the system has an imposed elongational flow field.The potential due to the external elongational field is given by [16], where E is the direction of the elongation, α 0 > 0 corresponds to the stretching in the E direction (uniaxial elongation) and α 0 < 0 corresponds to compressing, k is the Boltzmann constant, and T is the absolute temperature.Now let ρ(m,t) denote the probability density function (pdf) for dipolar rod-like nematic molecules in unit direction m at time t.The dynamic evolution of the orientational pdf for the ensembles of rigid rods with inherent dipoles is governed by the Smoluchowski equation [3]: Here ∂/∂m represents the orientational gradient operator [17], and the total potential is given by where α is the strength of the dipole-dipole interaction, b denotes the strength of the Maier-Saupe short-range interaction, and is the ensemble average with respect to the pdf ρ, which is a solution of the Smoluchowski equation (2.2).For simplicity, from now on we assume kT = 1, or equivalently we assume that all energies are normalized by kT.The total potential (2.3) can be split into the sum of the internal potential representing the interaction among polymer rods with themselves and the external potential: Hong Zhou et al. 3 where the internal and external potentials are given by In this paper, we restrict our study to the case of α > 0, α 0 > 0, and b > 0. In particular, α 0 > 0 (uniaxial elongation) will help us eliminate many unstable equilibrium solutions.

Equilibrium solutions for extended nematics
An equilibrium solution of (2.2) is given by the Boltzmann distribution [3] The nonlinear integral equation for the first moment vector m is whereas the nonlinear integral equation for the second moment tensor mm is We establish the coordinate system as follows.We select the uniaxial elongation field E as the z-axis.We select the x-axis and the y-axis to be perpendicular to the z-axis but otherwise arbitrary.In this Cartesian coordinate system, we have (3.4) The nonlinear equations for the first and second moments become S m i m j ρ eq (m)dm. (3.5) We select the spherical coordinate system (φ,θ) using the y-axis as the pole.In this spherical coordinate system, we have m 1 ,m 2 ,m 3 = (sinφ sinθ,cosφ,sinφ cos θ).
(3.6) Theorem 3.1.For an equilibrium probability density, U Ext satisfies Proof.We first prove that This is not surprising.Physically, this quantity is the (negative) total torque on the system about the y-axis.Since the system is in equilibrium, the total torque should be zero.Mathematically, we have The second part is to show that Again, this is not surprising at all.Physically, this quantity is the torque on the system from the mutual interaction.If there is a torque on polymer rod A from other polymer rods, then by Newton's third law, rod A exerts a torque of the opposite sign on other rods.Thus, the sum of the torques on all rods from the mutual interaction is zero.Mathematically, we prove it as follows.We notice that Using this fact, we immediately obtain (3.12) Hong Zhou et al. 5 Here we have adopted the more concise and more convenient way of writing the spherical integrals in terms of m instead of (φ,θ).For the second part of the internal potential, we have Theorem 3.2.For an equilibrium probability density, the z-axis is an eigenvector of the second moment tensor mm .
Proof.Differentiating the external potential, we have Applying Theorem 3.1, we obtain Since the x-axis and the y-axis are selected arbitrarily, exchanging the roles of the x-axis and the y-axis yields immediately Thus, the second moment tensor mm has the form It is obvious that the z-axis is an eigenvector of the the second moment tensor.
To facilitate the analysis, let us select the x-axis and the y-axis properly in the xysubspace to diagonalize the matrix mm .If m is not parallel to the z-axis, then we can select the positive directions of x-axis and y-axis such that q 1 > 0 and q 2 ≥ 0. In the new Cartesian coordinate system, we have where Our next move is to prove two theorems which establish the relationship between the first moment vector and the second moment tensor.Theorem 3.3.If an equilibrium solution satisfies q 1 > 0, then q 3 = 0.
Proof.We prove the theorem by contradiction.Suppose there is a solution satisfying q 1 > 0 and q 3 = 0. We are going to show that q 1 > 0 and q 3 = 0 lead to (1/q 3 ) m 1 m 3 > 0, which contradicts with the result of Theorem 3.2: (3.20) In the above, the second factor of the integrand is Notice that both U ext (m) and U 2 (m) are even functions of m 1 , m 2 , and m 3 .Substituting into (3.20),we get Since the integrand satisfies we have (1/q 3 ) m 1 m 3 > 0, which contradicts with the result of Theorem 3.2.
Remark 3.4.Theorem 3.3 tells us that if the first moment vector m is not parallel to the z-axis, then it must be perpendicular to the z-axis.
Proof.The proof of this theorem is very similar to the proof of Theorem 3.3.In the proof of Theorem 3.3, we did not use that the z-axis is the direction of E. We only used the fact that both U ext (m) and U 2 (m) are even functions of m 1 , m 2 , and m 3 .So by exchanging the roles of m 2 and m 3 , we can extend the proof of Theorem 3.3 to Theorem 3.5.
Hong Zhou et al. 7 Remark 3.6.Theorem 3.5 indicates that if the first moment m is not parallel to the z-axis, then it must be parallel to either the x-axis or y-axis.
Combining the results of Theorems 3.3 and 3.5, we conclude that the first moment m of the equilibrium solution of (2.2) must be aligned with one of the major axes of the second moment tensor mm .

Free energy and stability
In this section we consider the stability of an equilibrium solution.The study of stability allows us to narrow down the possible solutions that can be observed experimentally.To investigate the stability of an equilibrium solution, it is useful to exploit the free energy of the system.
To do so, consider an arbitrary probability density ρ(m), which is not necessarily an equilibrium probability density.The free energy of the probability density ρ(m) can be written as [8] In the above expression G ent [ρ] corresponds to the entropic part of the free energy, G 1 [ρ] and G 2 [ρ] are free energy parts associated with the two mutual interactions, and G ext [ρ] is the free energy part associated with the external field.Here • represents the mean taken with respect to the probability density ρ(m).For the clarity of analysis below, we introduce two different notations for means taken with respect to two different probability densities.
(i) • eq denotes the mean taken with respect to the equilirium probability density, whereas • represents the mean taken with respect to a general probability density (usually a perturbed probability density near the equilirium probability density).Recall from Theorem 3.5 that for an equilibrium solution of the Smoluchowski equation (2.2), if the first moment m is not parallel to the z-axis (which is the direction of the external elongational flow), then it must be parallel to either the x-axis or y-axis.We call these solutions with m not parallel to the z-axis nonparallel solutions.Now we show that nonparallel equilibrium solutions are actually unstable.Therefore, we can exclude them from our study and focus on the parallel solutions.
Theorem 4.1.If an equilibrium solution satisfies s 3 < s 1 or s 3 < s 2 , then it is unstable.
Proof.We present the proof for the case of s 3 < s 1 .The proof for the case of s 3 < s 2 is similar.
In order to prove that an equilibrium solution with s 3 < s 1 is unstable, we show that the free energy G[ρ] does not reach a local minimum at ρ eq (m).In particular, we construct a perturbed probability density ρ(m) arbitrarily close to the equilibrium probability density ρ eq (m) such that G ρ(m) < G ρ eq (m) . (4.2) We rotate the equilibrium probability density ρ eq (m) about the y-axis by ε (a small angle) and use the result as the perturbed probability density ρ(m).Mathematically this is equivalent to keeping the probability density fixed but rotating the external elongational flow field about the y-axis by −ε.After the rotation, the external elongational flow field is We have If s 3 < s 1 , then for ε sufficiently small we have and the equilibrium solution ρ eq is unstable.
Theorem 4.2.If the first moment of an equilibrium solution satisfies q 1 > 0, then the dipoledipole interaction strength α is related to the second moment through the inequality αs 1 > 1.
The parameter q 1 has a fixed value in the equilibrium solution ρ eq (m).Here we rename it v and treat it as an independent variable.We consider the probability density For clarity, let • v denote the mean taken with respect to the probability density ρ(m,v).Note that ρ(m,v)| v=q1 = ρ eq (m).Consider the function Hong Zhou et al. 9 which satisfies F(0) = 0 and F(q 1 ) = 0. We are going to show that αs 1 ≤ 1 implies that F (v) > 0 for v ∈ (0, q 1 ), which contradicts F(0) = F(q 1 ) = 0. Hence αs 1 > 1.We do it in several steps.
Proof.This theorem does not specify any condition on q 1 .Recall the method we used in selecting the axes: if the equilibrium solution is not parallel to E, then we select the axes to make q 1 > 0 and q 2 ≥ 0. Theorem 3.5 tells us that if q 1 > 0, then q 2 = 0 and q 3 = 0.So we always have q 2 = 0.
We only need to show that the free energy G[ρ] does not attain a local minimum at ρ eq (m).More precisely, we show that there exists a perturbed probability density ρ(m) arbitrarily close to the equilibrium probability density ρ eq (m) such that G ρ(m) < G ρ eq (m) . (4.11) We consider ρ(m) = 1 + εm 3 ρ eq (m).(4.12) Since m 3 eq = q 3 = 0, we have S ρ(m)dm = 1, which means ρ(m) is a probability density.We calculate the four parts of the free energy of the perturbed probability density ρ(m).
Because q 2 = 0 and q 3 = 0, the equilibrium probability density ρ eq (m) is symmetric with respect to m 2 and m 3 .Using the Taylor expansion we have The symmetry of ρ eq (m) gives us (4.17) If αs 3 > 1, then for ε sufficiently small we have G[ ρ ] < G[ρ eq ], which means the equilibrium solution ρ eq is unstable.
Hong Zhou et al. 11 Proof.Theorem 3.5 tells us that q 1 > 0 implies q 2 = 0 and q 3 = 0. We discuss two cases.(i) Case 1: s 3 < s 1 .Theorem 4.1 tells us that the equilibrium solution is unstable.
(ii) Case 2: s 3 ≥ s 1 .Using Theorem 4.2, we have αs 1 > 1.It follows that αs 3 > 1.Using Theorem 4.3, we conclude that the equilibrium solution is unstable.Therefore, all nonparallel solutions are unstable.In this way, we have excluded all nonparallel solutions from our study.

All stable equilibria are axisymmetric
In the previous section we have concluded that all nonparallel solutions are unstable.So from now on we only consider parallel solutions.It is well known that for pure (nondipolar) nematic rod ensembles where m = 0 [8,11], the stable equilibria satisfy either s 1 = s 2 = s 3 (isotropic phase) or s 3 > s 1 = s 2 (prolate uniaxial phase) in the selected coordinate system.In [15] we showed that this is also the case for extended (dipolar) nematics.In this section we are going to show that in the presence of imposed uniaxial elongational flow the system still retains this axisymmetry.
Below for the case of extended nematic equilibria in which m may be nonzero, we will show that if m = 0 (i.e., q 1 = q 2 = 0 and q 3 > 0 by the selection of the coordinate system and by the result of Theorem 3.1), then a stable equilibrium solution must be uniaxial.Furthermore, the axis of symmetry must be the major director (i.e., the eigenvector of the second moment corresponding to the largest eigenvalue).That is, m 2 1 = m 2 2 < m 2 3 .It should be point out that in [15] we proved five lemmas that paved the roads to reach axisymmetry.In order to prove axisymmetry for the extended nematics in the presence of an external elongational flow, we will take full advantage of the well-established results in [15].
First, we recall that in [15] there is no external flow field and the probability density is given by where the components of the second moment and the first moment are ) (5.3)Under the constraints s 1 < s 3 and s 2 < s 3 we proved in [15] that there is no equilibrium solution such that s 1 < s 2 .This holds for all α ≥ 0 and b ≥ 0. Since the nonexistence of s 1 < s 2 (where s 1 < s 3 and s 2 < s 3 ) is true for all α ≥ 0 and b ≥ 0, we introduce λ = αq 3 as a parameter and treat α as unknown.Equation (5.3) becomes (5.4)