On multi-dimensional hypocoercive BGK models

Franz Achleitner; Anton Arnold; Eric A. Carlen

doi:10.3934/krm.2018038

Article Contents

2018, Volume 11, Issue 4: 953-1009. Doi: 10.3934/krm.2018038

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On multi-dimensional hypocoercive BGK models

1.
Faculty of Mathematics, University of Vienna, Oskar-Morgenstern-Platz 1, Vienna, A-1090, Austria
2.
Institute of Analysis and Scientific Computing, TU Wien, Wiedner Hauptstrasse 8-10, Vienna, A-1040, Austria
3.
Department of Mathematics, Rutgers University, 110 Frelinghuysen Road, Piscataway, NJ 08854, USA

* Corresponding author: A. Arnold
* Corresponding author: A. Arnold

Received: November 2017

Revised: February 2018

Published: April 2018

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Abstract

We study hypocoercivity for a class of linearized BGK models for continuous phase spaces. We develop methods for constructing entropy functionals that enable us to prove exponential relaxation to equilibrium with explicit and physically meaningful rates. In fact, we not only estimate the exponential rate, but also the second time scale governing the time one must wait before one begins to see the exponential relaxation in the $L^1$ distance. This waiting time phenomenon, with a long plateau before the exponential decay "kicks in" when starting from initial data that is well-concentrated in phase space, is familiar from work of Aldous and Diaconis on Markov chains, but is new in our continuous phase space setting. Our strategies are based on the entropy and spectral methods, and we introduce a new "index of hypocoercivity" that is relevant to models of our type involving jump processes and not only diffusion. At the heart of our method is a decomposition technique that allows us to adapt Lyapunov's direct method to our continuous phase space setting in order to construct our entropy functionals. These are used to obtain precise information on linearized BGK models. Finally, we also prove local asymptotic stability of a nonlinear BGK model.

Keywords:

Kinetic equations,
BGK models,
hypocoercivity,
Lyapunov functionals,
perturbation methods for matrix equations.

Mathematics Subject Classification: Primary: 82C40, 35B40; Secondary: 35F25.

Citation:

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Figure 1. These two functions illustrate the time dependent decay estimate from (15). The values of $C_d,\,\lambda^d$ correspond to the 1D case with $L = 2\pi$, and we chose $\mathcal{E}^d(\tilde f^I) = 15$. We also show the two time scales of the BGK equation: $t_{{\rm init}}$ marks the intersection point of the two blue curves and it corresponds to the generic transport time. $t_2: = t_{{\rm init}}+\frac2\lambda$ marks the intersection point of the exponential curve with the value $2/e$, and $t_2-t_{{\rm init}}$ corresponds to the relaxation time scale. For larger values of $L$, $t_{{\rm init}}$ will be much larger

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Figure 2. For each cell length $L$ the constant $2\mu_*(L)$ obtained from Lemma 3.1 and Remark 9(a) yields a bound for the entropy decay rate in Theorem 1.1

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Table 1. We give a classification of Hermitian matrices ${\bf C}_1$, such that the associated matrix ${\bf C} = i{\bf C}_1 +{\rm diag}(0,0,c_2,\ldots,c_n)$ is hypocoercive. The restrictions on the coefficients of ${\bf C}_1$ are depicted as $0$ if zero, $\bullet$ if non-zero, and $\ast$ if there is no restriction. Furthermore, we give the corresponding two-parameter ansatz for the transformation matrix ${\bf P} = {\bf I}+{\bf A}$. The guideline to construct an admissible Hermitian perturbation matrix ${\bf A}$, is to put the parameters $\lambda_j$ at the positions of the (non-zero) coupling elements of ${\bf C}_1$. In case (2B2) this will be apparent after a suitable transformation, see the proof of Theorem 2.9

$ \underline{{\rm Case~ 2A:}}$

${\bf C}_1 = \left(\begin{array}{c|c} \begin{matrix}\ast&\ast \\ \ast&\ast \end{matrix} &\begin{matrix}\ast&\bullet&\ast&\cdots&\ast \\ \bullet&\ast&\ast&\cdots&\ast \end{matrix}\\ \hline \begin{matrix}\ast&\bullet \\ \bullet&\ast \\ \ast&\ast \\ \vdots&\vdots \\ \ast&\ast \end{matrix} &\bf{*} \end{array}\right)\!,~~{\bf P}={\bf I} + \left(\begin{array}{c|c} \begin{matrix}0&0&0&\lambda_1 \\ 0&0&\lambda_2&0 \\ 0&\overline{\lambda_2}&0&0 \\ \overline{\lambda_1}&0&0&0 \end{matrix} &{\bf 0}\\ \hline {\bf 0}& {\bf 0} \end{array}\right)\!,~~ (40) $

where the upper right submatrix ${\bf C}_1^{ur}\in\mathbb{C}^{2\times(n-2)}$ has rank 2. Here, we assume w.l.o.g. that $|c_{1,4}c_{2,3}|\geq |c_{1,3}c_{2,4}|$ and $c_{1,4}\,c_{2,3} \ne c_{1,3}\,c_{2,4}$, such that $c_{2,3}\ne 0$ and $c_{1,4}\ne 0$.

$ \underline{{\rm Case~ 2B:}}$

${\bf C}_1 = \left(\begin{array}{c|c} \begin{matrix}\ast&\ast \\ \ast&\ast \end{matrix}&\begin{matrix}\ast&\ast&\cdots&\ast \\ \bullet&\ast&\cdots&\ast \end{matrix}\\ \hline \begin{matrix}\ast&\bullet \\ \ast&\ast \\ \vdots&\vdots \\ \ast&\ast \end{matrix} &\bf * \end{array}\right)\!, ~~~~~ ~~ {\bf P}={\bf I} +{\bf U} \left(\begin{array}{c|c}\begin{matrix}0&\lambda_1&0 \\ \overline{\lambda_1}&0&\lambda_2 \\ 0&\overline{\lambda_2}&0 \end{matrix}&{\bf 0}\\ \hline {\bf 0}& {\bf 0} \end{array}\right) {\bf U}^* \!,~ (41) $

where the upper right submatrix ${\bf C}_1^{ur}\in\mathbb{C}^{2\times(n-2)}$ has rank 1. Again, we assume w.l.o.g. that $c_{2,3}\ne 0$. The right choice for the unitary matrix ${\bf U}$ depends on the structure of ${\bf C}_1$:

$ {\rm (2B1)}~~~~ {\bf C}_1 = \left(\begin{array}{c|c} \begin{matrix}\ast&\bullet \\ \bullet&\ast \end{matrix}& \begin{matrix}0&0&\cdots&0 \\ \bullet&\ast&\cdots&\ast \end{matrix} \\ \hline \begin{matrix}0&\bullet \\ 0&\ast \\ \vdots&\vdots \\ 0&\ast \end{matrix} &\bf * \end{array}\right)\!, ~~~~~ {\bf U} ={\bf I}\,, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (42)$

${\rm (2B2)} ~~~~~ {\bf C}_1 = \left(\begin{array}{c|c} \begin{matrix}\ast&\ast \\ \ast&\ast \end{matrix}& \begin{matrix}\bullet&\ast&\cdots&\ast \\ \bullet&\ast&\cdots&\ast \end{matrix} \\ \hline \begin{matrix}\bullet&\bullet \\ \ast&\ast \\ \vdots&\vdots \\ \ast&\ast \end{matrix} &\bf * \end{array}\right)\!,~~~~~{\bf U} = \left(\begin{array}{c|c} {\bf U}^{ul}&{\bf 0}\\ \hline {\bf 0}& {\bf I} \end{array}\right) \!,~~~~~~~~~ (43) $

with upper left submatrix ${\bf U}^{ul}=\tfrac1{\sqrt{|c_{1,3}|^2 +|c_{2,3}|^2}}\begin{pmatrix}\overline{c_{2,3}}&c_{1,3} \\ -\overline{c_{1,3}}&c_{2,3} \end{pmatrix}$.

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Table 2. Let $\delta_j(\kappa,\alpha,\beta,\gamma,\omega)$ denote the determinant of the upper left $j\times j$ submatrix of ${\bf D}_{\kappa,\alpha,\beta,\gamma,\omega}$ for integers $j = 1,2,\ldots,11$. For our choice $\beta = 2\alpha$, $\gamma = \alpha$ and $\omega = \sqrt{6}\alpha$, the minors $\delta_j(\kappa,\alpha) = \delta_j(\kappa,\alpha,2\alpha,\alpha,\sqrt{6}\alpha)$ are given in this table

$\delta_1(\kappa,\alpha)$ = $2\ell\alpha$

$\delta_2(\kappa,\alpha)$ = $4\ell^2\alpha^2$

$\delta_3(\kappa,\alpha)$ = $8\ell^3\alpha^3$

$\delta_4(\kappa,\alpha)$ = $44\ell^4 \alpha^4$

$\delta_5(\kappa,\alpha)$ = $22\ell^3 \alpha^4 (4\ell -4\ell^2 \alpha -{\alpha}/{\kappa^2})$

$\delta_6(\kappa,\alpha)$ = $\delta_5(\kappa,\alpha) p_6(\kappa,\alpha)/\ell$
with $p_6(\kappa,\alpha) :=-\tfrac{54}{11} \ell^2 \alpha +2\ell -2\alpha /{\kappa^2}$.

$\delta_7(\kappa,\alpha)$ = $\frac{2}{11\ell^2} \delta_5(\kappa,\alpha) p_7(\kappa,\alpha)$
    with $p_{7}(\kappa,\alpha) = \big(p_{7,0}(\alpha) +p_{7,1}(\alpha) \frac1{\kappa^2}\big) \frac1{\kappa^2} + p_{7,2}(\alpha)$,
    $p_{7,0}(\alpha) = 93\ell^2 \alpha^2 -34\ell\alpha$, $p_{7,1}(\alpha) = 12\alpha^2$,
    $p_{7,2}(\alpha) = 162\ell^4 \alpha^2 -120\ell^3 \alpha +22\ell^2$.

$\delta_8(\kappa,\alpha)$ = $44 \ell^3 \alpha^4 \frac{\delta_7(\kappa,\alpha)}{\delta_5(\kappa,\alpha)} p_8(\kappa,\alpha)$
with $p_8(\kappa,\alpha) = 2\ell^3 \alpha^2 -6\ell^2\alpha +4\ell -{\alpha}/{\kappa^2}$.

$\delta_9(\kappa,\alpha)$ = $8 \ell \alpha^4 p_8(\kappa,\alpha) p_9(\kappa,\alpha)$
    with $p_{9}(\kappa,\alpha) = \big(p_{9,0}(\alpha) +p_{9,1}(\alpha) \frac1{\kappa^2}\big) \frac1{\kappa^2} + p_{9,2}(\alpha)$,
    $p_{9,0}(\alpha) = -12\ell^3 \alpha^3 +198\ell^2 \alpha^2 -68\ell \alpha$, $p_{9,1}(\alpha) = 24\alpha^2$,
    $p_{9,2}(\alpha) = -81\ell^5 \alpha^3 +411\ell^4 \alpha^2 -262\ell^3 \alpha +44\ell^2$.

$\delta_{10}(\kappa,\alpha)$ = $2 \delta_9(\kappa,\alpha)$,

$\delta_{11}(\kappa,\alpha)$ = $64 \ell \alpha^4 p_8(\kappa,\alpha) p_{11}(\kappa,\alpha)$
    with $p_{11}(\kappa,\alpha) = \big(p_{11,0}(\alpha) +p_{11,1}(\alpha) \frac1{\kappa^2}\big) \frac1{\kappa^2} + p_{11,2}(\alpha)$,
    $p_{11,0}(\alpha) = -72\ell^4 \alpha^4 -300\ell^3 \alpha^3 +294\ell^2 \alpha^2 -68\ell\alpha$, $p_{11,1}(\alpha) = 24\alpha^2$,
    $p_{11,2}(\alpha) = 162\ell^6 \alpha^4 -909\ell^5 \alpha^3 +963\ell^4 \alpha^2 -358\ell^3 \alpha +44\ell^2$.

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Table 3. Matrix ${\bf D}_{\kappa,\alpha,\beta,\gamma,\omega,\eta}$ with $\ell: = 2\pi/L>0$ and $A : = \tfrac{\ell}{\sqrt3} (\sqrt2 \alpha -\sqrt3 \beta)$, $B : = \tfrac2{\sqrt3} \ell (\sqrt2 \beta -\sqrt3 \alpha)$, $C : = \tfrac{\ell}{\sqrt3} (\sqrt2 \eta -\sqrt3 \beta)$, $D : = \tfrac{\ell}{\sqrt3} (\sqrt3 \eta -\sqrt2 \beta)$, $E : = 2-2\ell\omega$, and $F : = 2-2\tfrac{\sqrt2}{\sqrt3} \ell\beta$

$ \begin{pmatrix} 2\ell\alpha&0&0&0&\tfrac{\sqrt2}{\sqrt3} \ell\alpha&0&0&A&0&\tfrac{\sqrt2}{\sqrt3} \ell\alpha&0&0&0&0&0&0&0&0&0&0&0 \\ 0&B&0&0&0&0&0&- \frac{i}{\kappa} \beta&0&0&-C&0&0&\tfrac{3+\sqrt3}{6} \ell\beta&0&\tfrac{3-\sqrt3}{6} \ell\beta&0&0&0&0&0 \\ 0&0&2\ell \gamma&0&0&-\frac{i}{\kappa} \gamma&0&0&0&0&0&\sqrt2 \ell \gamma&0&0&0&0&0&0&0&0&0 \\ 0&0&0&2\ell\omega&0&0&-\frac{i}{\kappa} \omega&0&0&0&0&0&\sqrt2 \ell \omega&0&0&0&0&0&0&0&0 \\ \tfrac{\sqrt2}{\sqrt3} \ell\alpha&0&0&0&2\ell\eta&0&0&D&0&\ell\eta&-\frac{i}{\kappa} \eta&0&0&0&0&0&0&0&0&0&2\ell\eta \\ 0&0&\frac{i}{\kappa} \gamma&0&0&2-2\ell\gamma&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \\ 0&0&0&\frac{i}{\kappa} \omega&0&0&E&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \\ A&\frac{i}{\kappa} \beta&0&0&D&0&0&F&0&-\tfrac{\sqrt2}{\sqrt3} \ell\beta&0&0&0&0&0&0&0&0&0&0&0 \\ 0&0&0&0&0&0&0&0&2&0&0&0&0&0&0&0&0&0&0&0&0 \\ \tfrac{\sqrt2}{\sqrt3} \ell\alpha&0&0&0&\ell\eta&0&0&-\tfrac{\sqrt2}{\sqrt3} \ell\beta&0&2&0&0&0&0&0&0&0&0&0&0&0 \\ 0&-C&0&0&\frac{i}{\kappa} \eta&0&0&0&0&0&2-2\ell\eta&0&0&-\tfrac{1}{\sqrt{3}} \ell\eta&0&-\tfrac{1}{\sqrt{3}} \ell\eta&0&0&0&0&0 \\ 0&0&\sqrt2 \ell\gamma&0&0&0&0&0&0&0&0&2&0&0&0&0&0&0&0&0&0 \\ 0&0&0&\sqrt2 \ell\omega&0&0&0&0&0&0&0&0&2&0&0&0&0&0&0&0&0 \\ 0&\tfrac{3+\sqrt3}{6} \ell\beta&0&0&0&0&0&0&0&0&-\tfrac{1}{\sqrt{3}} \ell\eta&0&0&2&0&0&0&0&0&0&0 \\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&2&0&0&0&0&0&0 \\ 0&\tfrac{3-\sqrt3}{6} \ell\beta&0&0&0&0&0&0&0&0&-\tfrac{1}{\sqrt{3}} \ell\eta&0&0&0&0&2&0&0&0&0&0 \\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&2&0&0&0&0 \\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&2&0&0&0 \\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&2&0&0 \\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&2&0 \\ 0&0&0&0&2\ell\eta&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&2 \\ \end{pmatrix}$

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Table 4. Matrix ${\bf D}_{\kappa,\alpha,\sqrt3 \alpha,\alpha,\alpha,\alpha}$ with $\ell: = 2\pi/L>0$ and $B: = 2 (\sqrt2 -1) \ell\alpha$, $E: = 2-2\ell\alpha$, $F: = 2-2\sqrt2 \ell \alpha$

$ \begin{pmatrix} 2\ell\alpha & 0 & 0 & 0 & \tfrac{\sqrt2}{\sqrt3} \ell\alpha & 0 & 0 & \tfrac{\sqrt2 -3}{\sqrt3} \ell \alpha & 0 & \tfrac{\sqrt2}{\sqrt3} \ell\alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & B & 0 & 0 & 0 & 0 & 0 & - \frac{i}{\kappa} \sqrt{3}\alpha & 0 & 0 & \tfrac{3-\sqrt2}{\sqrt3} \ell\alpha & 0 & 0 & \tfrac{\sqrt3+1}{2} \ell\alpha & 0 & \tfrac{\sqrt3-1}{2} \ell\alpha & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2\ell\alpha & 0 & 0 & -\frac{i}{\kappa} \alpha & 0 & 0 & 0 & 0 & 0 & \sqrt2 \ell \alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2\ell\alpha & 0 & 0 & -\frac{i}{\kappa} \alpha & 0 & 0 & 0 & 0 & 0 & \sqrt2 \ell \alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \tfrac{\sqrt2}{\sqrt3} \ell\alpha & 0 & 0 & 0 & 2\ell\alpha & 0 & 0 & -\frac{B}{2} & 0 & \ell\alpha & -\frac{i}{\kappa} \alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2\ell\alpha \\ 0 & 0 & \frac{i}{\kappa} \alpha & 0 & 0 & E & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{i}{\kappa} \alpha & 0 & 0 & E & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \tfrac{\sqrt2 -3}{\sqrt3} \ell\alpha & \frac{i}{\kappa} \sqrt{3}\alpha & 0 & 0 & -\frac{B}{2} & 0 & 0 & F & 0 & -\sqrt2 \ell \alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \tfrac{\sqrt2}{\sqrt3} \ell\alpha & 0 & 0 & 0 & \ell\alpha & 0 & 0 & -\sqrt2 \ell \alpha & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \tfrac{3-\sqrt2}{\sqrt3} \ell\alpha & 0 & 0 & \frac{i}{\kappa} \alpha & 0 & 0 & 0 & 0 & 0 & E & 0 & 0 & -\tfrac{1}{\sqrt{3}} \ell\alpha & 0 & -\tfrac{1}{\sqrt{3}} \ell\alpha & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt2 \ell\alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \sqrt2 \ell\alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \tfrac{\sqrt3+1}{2} \ell \alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -\tfrac{1}{\sqrt{3}} \ell\alpha & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \tfrac{\sqrt3-1}{2} \ell \alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -\tfrac{1}{\sqrt{3}} \ell\alpha & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\ell\alpha & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{pmatrix} ~~~~(114)$

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Table 5. Let $\delta_j(\kappa,\alpha,\beta,\gamma,\omega,\eta)$ denote the determinant of the upper left $j\times j$ submatrix of ${\bf D}_{\kappa,\alpha,\beta,\gamma,\omega,\eta}$ for integers $j = 1,2,\ldots,21$. For our choice $\beta = \sqrt3 \alpha$, $\gamma = \alpha$, $\omega = \alpha$ and $\eta = \alpha$, the minors $\delta_j(\kappa,\alpha) = \delta_j(\kappa,\alpha,\sqrt3 \alpha,\alpha,\alpha,\alpha)$ for integers $j = 1,2,\ldots,13$, are given in this table

$\delta_1(\kappa,\alpha)$ = $2\ell\alpha$

$\delta_2(\kappa,\alpha)$ = $4(\sqrt2 -1)\ell^2\alpha^2$

$\delta_3(\kappa,\alpha)$ = $8(\sqrt2 -1)\ell^3\alpha^3$

$\delta_4(\kappa,\alpha)$ = $16(\sqrt2 -1)\ell^4 \alpha^4$

$\delta_5(\kappa,\alpha)$ = $\tfrac{80}{3} (\sqrt2 -1)\ell^5 \alpha^5$

$\delta_6(\kappa,\alpha)$ = $\tfrac{40}{3} (\sqrt2 -1)\ell^4 \alpha^5 p_6(\kappa,\alpha)$
with $p_6(\kappa,\alpha) = -4\ell^2 \alpha -\frac{\alpha}{\kappa^2} +4\ell$.

$\delta_7(\kappa,\alpha)$ = $\tfrac{20}{3} (\sqrt2 -1)\ell^3 \alpha^5 p_6(\kappa,\alpha)^2$

$\delta_8(\kappa,\alpha)$ = $12 \ell^2\ \alpha^5\ p_6(\kappa,\alpha)^2\ p_8(\kappa,\alpha)$
with $p_8(\kappa,\alpha) = \tfrac{2-3\sqrt2}{3} \ell^2 \alpha -\tfrac56 \frac{\alpha}{\kappa^2} +\tfrac{10}{9}(\sqrt2 -1)\ell$.

$\delta_9(\kappa,\alpha)$ = $2\ \delta_8(\kappa,\alpha)$

$\delta_{10}(\kappa,\alpha)$ = $\tfrac43 \ell^2\ \alpha^5\ p_6(\kappa,\alpha)^2\ p_{10}(\kappa,\alpha)$
with $p_{10}(\kappa,\alpha) = 9((\sqrt2 -1)\ell^2 +\frac1{\kappa^2})\ell \alpha^2$
$-6((8\sqrt2 -6)\ell^2 +\frac5{\kappa^2})\alpha +40(\sqrt2 -1)\ell$.

$\delta_{11}(\kappa,\alpha)$ = $\tfrac29 \ell\ \alpha^5\ p_6(\kappa,\alpha)^2\ p_{11}(\kappa,\alpha)$
     with $p_{11}(\kappa,\alpha) = \big(p_{11,0}(\alpha) +p_{11,1}(\alpha) \frac1{\kappa^2}\big) \frac1{\kappa^2} + p_{11,2}(\alpha) \ell^2$,
     $p_{11,0}(\alpha) = (54\sqrt2 -144)\ell^3 \alpha^3 +(672 -72\sqrt2)\ell^2 \alpha^2 -(216 +144\sqrt2)\ell\alpha$,
     $p_{11,1}(\alpha) =18(6 -\ell\alpha) \alpha^2$,

$p_{11,2}(\alpha) =(9 -54\sqrt2)\ell^3 \alpha^3 +(456\sqrt2 -24) \ell^2 \alpha^2$

$+(472 -816\sqrt2) \ell \alpha + 480(\sqrt2 -1)$.

$\delta_{12}(\kappa,\alpha)$ = $\delta_{11}(\kappa,\alpha) \frac{p_{12}(\kappa,\alpha)}{p_6(\kappa,\alpha)}$ = $\tfrac29 \ell\ \alpha^5\ p_6(\kappa,\alpha)\ p_{11}(\kappa,\alpha)\ p_{12}(\kappa,\alpha)$
with $p_{12}(\kappa,\alpha) = 4\ell^3 \alpha^2 -12\ell^2 \alpha +8\ell -\frac{2 \alpha}{\kappa^2}$.

$\delta_{13}(\kappa,\alpha)$ = $\delta_{12}(\kappa,\alpha) \frac{p_{12}(\kappa,\alpha)}{p_6(\kappa,\alpha)} = \delta_{11}(\kappa,\alpha) \Big(\frac{p_{12}(\kappa,\alpha)}{p_6(\kappa,\alpha)}\Big)^2 = \tfrac29 \ell\ \alpha^5\ p_{11}(\kappa,\alpha) p_{12}(\kappa,\alpha)^2$

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Table 6. Let $\delta_j(\kappa,\alpha,\beta,\gamma,\omega,\eta)$ denote the determinant of the upper left $j\times j$ submatrix of ${\bf D}_{\kappa,\alpha,\beta,\gamma,\omega,\eta}$ for integers $j = 14,\ldots,21$. For our choice $\beta = \sqrt3 \alpha$, $\gamma = \alpha$, $\omega = \alpha$ and $\eta = \alpha$, the minors $\delta_j(\kappa,\alpha) = \delta_j(\kappa,\alpha,\sqrt3 \alpha,\alpha,\alpha,\alpha)$ are given in this table

$\delta_{14}(\kappa,\alpha)$ = $\tfrac1{9\ (1+\sqrt3)^2} \ell\ \alpha^5\ p_{12}(\kappa,\alpha)^2\ p_{14}(\kappa,\alpha)$
with $p_{14}(\kappa,\alpha) = \big(p_{14,0}(\alpha) +p_{14,1}(\alpha) \frac1{\kappa^2}\big) \frac1{\kappa^2} +\ell^2 p_{14,2}(\alpha)$,

$p_{14,0}(\alpha) = (-108\sqrt6 -72\sqrt3 -180\sqrt2 -144) \ell^4 \alpha^4$

$+(360\sqrt6 -1824\sqrt3 +720\sqrt2 -3396) \ell^3 \alpha^3$

$+(-576\sqrt6 +5952\sqrt3 -1152\sqrt2 +11760) \ell^2 \alpha^2$

       $+(-1152\sqrt6 -1728\sqrt3 -2304\sqrt2 -3456) \ell \alpha$,
     $p_{14,1}(\alpha) = 144\ (\sqrt3 +2)\ (6 -\ell \alpha)\ \alpha^2$,

$p_{14,2}(\alpha) = (1440 -180\sqrt6 +828\sqrt3 -324\sqrt2) \ell^4 \alpha^4$

$-(9348 -336\sqrt6 -5400\sqrt3 -624\sqrt2) \ell^3 \alpha^3$

$+(11056 +3424\sqrt6 +6368\sqrt3 +6864\sqrt2) \ell^2 \alpha^2$

$+(4192 -6528\sqrt6 +1856\sqrt3 -13056\sqrt2) \ell \alpha$

$+(3840\sqrt6 -3840\sqrt3 +7680\sqrt2 -7680)$.

$\delta_{15}(\kappa,\alpha)$ = $2\ \delta_{14}(\kappa,\alpha)$

$\delta_{16}(\kappa,\alpha)$ = $\tfrac89 \tfrac{2+\sqrt3}{(1+\sqrt3)^2} \ell\ \alpha^5\ p_{12}(\kappa,\alpha)^2\ p_{16}(\kappa,\alpha)$
with $p_{16}(\kappa,\alpha) = \big(p_{16,0}(\alpha) +p_{16,1}(\alpha) \frac1{\kappa^2}\big) \frac1{\kappa^2} +\ell^2 p_{16,2}(\alpha)$,

$p_{16,0}(\alpha) = -36 (\sqrt2 +2) \ell^4 \alpha^4 +(144 \sqrt2 -744) \ell^3 \alpha^3$

       $+(-288 \sqrt2 +2976) \ell^2 \alpha^2 +(-576 \sqrt2 -864) \ell \alpha$,
     $p_{16,1}(\alpha) = 72 (6 -\alpha \ell) \alpha^2$,

$p_{16,2}(\alpha) = 27 \ell^5 \alpha^5 +(-144 \sqrt2 +216) \ell^4 \alpha^4 +(-24 \sqrt2 -2412) \ell^3 \alpha^3$

$+(1632 \sqrt2 +3104) \ell^2 \alpha^2 +(-3264 \sqrt2 +928) \ell \alpha +1920 (\sqrt2 -1)$.

$\delta_{17}(\kappa,\alpha)$ = $2\ \delta_{16}(\kappa,\alpha)$

$\delta_{18}(\kappa,\alpha)$ = $2^2\ \delta_{16}(\kappa,\alpha)$

$\delta_{19}(\kappa,\alpha)$ = $2^3\ \delta_{16}(\kappa,\alpha)$

$\delta_{20}(\kappa,\alpha)$ = $2^4\ \delta_{16}(\kappa,\alpha)$

$\delta_{21}(\kappa,\alpha)$ = $\tfrac{256 (\sqrt3 +2) (24\sqrt2 +61)}{23121 (\sqrt3 +1)^2} \ell\ \alpha^5\ p_{12}(\kappa,\alpha)^2\ p_{21}(\kappa,\alpha)$
with $p_{21}(\kappa,\alpha) = \Big(p_{21,0}(\alpha) +p_{21,1}(\alpha) \frac1{\kappa^2}\Big) \frac1{\kappa^2} +\ell^2 p_{21,2}(\alpha)$,

$p_{21,0}(\alpha) = (-1152 \sqrt2 +2928) \ell^5 \alpha^5 +(-468 \sqrt2 -2664) \ell^4 \alpha^4$

$+(75024 \sqrt2 -175272) \ell^3 \alpha^3 +(-130464 \sqrt2 +300768) \ell^2 \alpha^2$

       $+(-14400 \sqrt2 -25056) \ell \alpha$,
     $p_{21,1}(\alpha) = (-1728 \sqrt2 +4392) (6 -\ell \alpha) \alpha^2$,

$p_{21,2}(\alpha) = 7707 \ell^5 \alpha^5 +(-25248 \sqrt2 +95000) \ell^4 \alpha^4$

$+(89448 \sqrt2 -353228) \ell^3 \alpha^3 +(158880 \sqrt2 +38048) \ell^2 \alpha^2$

$+(-417216 \sqrt2 +464416) \ell \alpha +1920 (85\sqrt2 -109)$.

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