The aim of this chapter is to discretize the momentum diffusion and dissipation on the three-dimensional deformed C-lattice.
First of all, the aim is to discretize the momentum diffusion within the framework of the shallow water equations on the two-dimensional regular hexagonal lattice. In the two-dimensional case, according to Eq. Write (8.50) for the frictional acceleration
\[ \begin{align} \frac{\partial\mathbf{v}_\text{diff}}{\partial t} = \frac{1}{h}\nabla\cdot\left[2K\left(\begin{array}{cc} S_{1, 1} & S_{1, 2}\\ S_{2, 1} & S_{2, 2} \end{array}\right)\right] = \frac{1}{h}\nabla\cdot\left[K\left(\begin{array}{cc} 2\frac{\partial v_1}{\partial x_1} & \frac{\partial v_1}{\partial x_2} + \frac{\partial v_2}{\partial x_1}\\ \frac{\partial v_2}{\partial x_1} + \frac{\partial v_1}{\partial x_2} & 2\frac{\partial v_2}{\partial x_2} \end{array}\right)\right].\tag{29.1}\label{eq:momentum_diff_hex_deriv_0} \end{align} \]
Here, the density $\rho$ was replaced by the layer thickness $h$, $K$ is a general mass-weighted diffusion coefficient. The three-dimensional divergence $\nabla\cdot \mathbf{v}$ disappears in the incompressible case. As an approach for discretization one uses
\[ \begin{align} \frac{\partial\mathbf{v}^T_\text{diff}}{\partial t} = \left(\delta_1, \delta_2, \delta_3\right)\cdot\tau = \left(\delta_1, \delta_2, \delta_3\right)\cdot\left( \begin{array}{ccc} \tau_{1, 1} & \tau_{1, 2} & \tau_{1, 3}\\ \tau_{2, 1} & \tau_{2, 2} & \tau_{2, 3} \\ \tau_{3, 1} & \tau_{3, 2} & \tau_{3, 3} \end{array}\right) \end{align} \]
The aim of this section is to clearly determine the $\tau_{i, j}$. Motivated by Eq. (29.1) requires the symmetry of $\tau$, so
\[ \begin{align} \tau_{i, j} = \tau_{j, i}. \end{align} \]
The tensor $\tau$ is developed in several modifying interactions. You start first
\[ \begin{align} \tau = K\left( \begin{array}{ccc} 2\delta_1u_1 & \delta_2u_1 + \delta_1u_2 & \delta_3u_1 + \delta_1u_3\\ \delta_2u_1 + \delta_1u_2 & 2\delta_2u_2 & \delta_3u_2 + \delta_2u_3 \\ \delta_3u_1 + \delta_1u_3 & \delta_3u_2 + \delta_2u_3 & 2\delta_3u_3 \end{array}\right). \end{align} \]
From this it follows
\[ \begin{align} \frac{\partial u_{1, \text{diff}}}{\partial t} &= K\left[\delta_1\left(2\delta_1u_1\right) + \delta_2\left(\delta_2u_1 + \delta_1u_2\right) + \delta_3\left(\delta_3u_1 + \delta_1u_3\right)\right]\nonumber\\ &= K\left(\delta_{1, 1}u_1 + \delta_{1, 1}u_1 + \delta_{2, 2}u_1 + \delta_{3, 3}u_1 + \delta_{2, 1}u_2 + \delta_{3, 1}u_3\right)\nonumber\\ &= K\left[\delta_{1, 1}u_1 + \delta_{2, 2}u_1 + \delta_{3, 3}u_1 + \delta_1\left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right)\right] = K\left(\frac{3}{2}\Delta u_1 + \frac{3}{2}\delta_1D\right). \end{align} \]
It follows from this
\[ \begin{align} \tau = K\left( \begin{array}{ccc} \frac{4}{3}\delta_1u_1 & \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)\\ \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{4}{3}\delta_2u_2 & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) \\ \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right) & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) & \frac{4}{3}\delta_3u_3 \end{array}\right) \end{align} \]
for frictional acceleration
\[ \begin{align} \frac{\partial u_{1, \text{diff}}}{\partial t} &= K\left(\Delta u_1 + \delta_1D\right). \end{align} \]
The term proportional to $\delta_1D$ does not occur in the shallow water equations, since these describe an incompressible medium, and should therefore be eliminated or at least reduced if possible. The terms that are not on the diagonal must not be modified in order not to destroy the symmetry of the tensor. Therefore one starts with a parameter $0 \leq \alpha \leq 1$
\[ \begin{align} \tau &= K\left( \begin{array}{ccc} \frac{4}{3}\delta_1u_1 - \alpha D & \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)\\ \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{4}{3}\delta_2u_2 - \alpha D & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) \\ \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right) & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) & \frac{4}{3}\delta_3u_3 - \alpha D \end{array}\right).\nonumber\\ &= K\left( \begin{array}{ccc} \frac{2}{3}\left(2 - \alpha\right)\delta_1u_1 - \alpha\frac{2}{3}\left(\delta_2u_2 + \delta_3u_3\right) & \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)\\ \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{2}{3}\left(2 - \alpha\right)\delta_2u_2 - \alpha\frac{2}{3}\left(\delta_1u_1 + \delta_3u_3\right) & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) \\ \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right) & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) & \frac{2}{3}\left(2 - \alpha\right)\delta_3u_3 - \alpha\frac{2}{3}\left(\delta_1u_1 + \delta_2u_2\right) \end{array}\right). \end{align} \]
It now follows from this
\[ \begin{align} \frac{\partial u_{1, \text{diff}}}{\partial t} &= K\left[\Delta u_1 + \left(1 - \alpha\right)\delta_1D\right]. \end{align} \]
The factor $1 - \alpha$ should be as close to zero as possible, so $\alpha$ should be as close to one as possible. According to Eq. (8.66) applies to the dissipation rate
\[ \begin{align} \frac{3}{2}h\epsilon &= \frac{2}{3}\left(2 - \alpha\right)\left(\delta_1u_1\right)^2 - \alpha\frac{2}{3}\left(\delta_1u_1\delta_2u_2 + \delta_1u_1\delta_3u_3\right) + \frac{2}{3}\left(\delta_1u_2\delta_2u_1 + \delta_1u_2\delta_1u_2\right) + \frac{2}{3}\left(\delta_1u_3\delta_3u_1 + \delta_1u_3\delta_1u_3\right)\nonumber\\ &+ \frac{2}{3}\left(\delta_2u_1\delta_2u_1 + \delta_2u_1\delta_1u_2\right) + \frac{2}{3}\left(2 - \alpha\right)\left(\delta_2u_2\right)^2 - \alpha\frac{2}{3}\left(\delta_2u_2\delta_1u_1 + \delta_2u_2\delta_3u_3\right) + \frac{2}{3}\left(\delta_2u_3\delta_3u_2 + \delta_2u_3\delta_2u_3\right)\nonumber\\ &+ \frac{2}{3}\left(\delta_3u_1\delta_3u_1 + \delta_3u_1\delta_1u_3\right) + \frac{2}{3}\left(\delta_3u_2\delta_3u_2 + \delta_3u_2\delta_2u_3\right) + \frac{2}{3}\left(2 - \alpha\right)\left(\delta_3u_3\right)^2 - \alpha\frac{2}{3}\left(\delta_3u_3\delta_1u_1 + \delta_3u_3\delta_2u_2\right)\nonumber\\ &= \frac{2}{3}\left(2 - \alpha\right)\left(\delta_1u_1\right)^2 - \alpha\frac{2}{3}\left(\delta_1u_1\delta_2u_2 + \delta_1u_1\delta_3u_3\right) + \textcolor{red}{\frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right)^2} + \textcolor{red}{\frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)^2}\nonumber\\ &+ \frac{2}{3}\left(2 - \alpha\right)\left(\delta_2u_2\right)^2 - \alpha\frac{2}{3}\left(\delta_2u_2\delta_1u_1 + \delta_2u_2\delta_3u_3\right) + \textcolor{red}{\frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right)^2}\nonumber\\ &+ \frac{2}{3}\left(2 - \alpha\right)\left(\delta_3u_3\right)^2 - \alpha\frac{2}{3}\left(\delta_3u_3\delta_1u_1 + \delta_3u_3\delta_2u_2\right). \end{align} \]
The red marked terms arise from the non-diagonal elements and are all square and therefore non-negative. applies to the remaining terms
\[ \begin{align} &\frac{2}{3}\left(2 - \alpha\right)\left(\delta_1u_1\right)^2 - \alpha\frac{2}{3}\left(\delta_1u_1\delta_2u_2 + \delta_1u_1\delta_3u_3\right) + \frac{2}{3}\left(2 - \alpha\right)\left(\delta_2u_2\right)^2 - \alpha\frac{2}{3}\left(\delta_2u_2\delta_1u_1 + \delta_2u_2\delta_3u_3\right)\nonumber\\ &+ \frac{2}{3}\left(2 - \alpha\right)\left(\delta_3u_3\right)^2 - \alpha\frac{2}{3}\left(\delta_3u_3\delta_1u_1 + \delta_3u_3\delta_2u_2\right)\nonumber\\ &= \frac{2}{3}\left(2 - \alpha\right)\left(\delta_1u_1\right)^2 + \frac{2}{3}\left(2 - \alpha\right)\left(\delta_2u_2\right)^2 + \frac{2}{3}\left(2 - \alpha\right)\left(\delta_3u_3\right)^2 - \alpha\frac{4}{3}\left(\delta_1u_1\delta_2u_2 + \delta_1u_1\delta_3u_3 + \delta_2u_2\delta_3u_3\right)\nonumber\\ &\hastobe \alpha\frac{2}{3}\left(\delta_1u_1 - \delta_2u_2\right)^2 + \alpha\frac{2}{3}\left(\delta_1u_1 - \delta_3u_3\right)^2 + \alpha\frac{2}{3}\left(\delta_2u_2 - \delta_3u_3\right)^2. \end{align} \]
The requirement in the last step is based on wanting to obtain a non-negative dissipation rate. It turns out
\[ \begin{align} \frac{2}{3}\left(2 - \alpha\right) = \frac{4}{3}\alpha \Leftrightarrow \frac{4}{3} - \alpha\frac{2}{3} = \frac{4}{3}\alpha \Rightarrow \frac{4}{3} = \frac{6}{3}\alpha \Leftrightarrow\alpha = \frac{2}{3}. \end{align} \]
With this choice of $\alpha$, the terms on the main diagonal also contribute non-negatively to the dissipation rate and it holds
\[ \begin{align} \epsilon \geq 0. \end{align} \]
The result for the friction tensor is therefore:
\[ \begin{align} \tau &= K\left( \begin{array}{ccc} \frac{4}{9}\left[2\delta_1u_1 - \left(\delta_2u_2 + \delta_3u_3\right)\right] & \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)\\ \frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & \frac{4}{9}\left[2\delta_2u_2 - \left(\delta_1u_1 + \delta_3u_3\right)\right] & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right)\\ \frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right) & \frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) & \frac{4}{9}\left[2\delta_3u_3 - \left(\delta_1u_1 + \delta_2u_2\right)\right] \end{array}\right). \end{align} \]
The generalization
\[ \begin{align} \tau &= \left( \begin{array}{ccc} K_c\frac{4}{9}\left[2\delta_1u_1 - \left(\delta_2u_2 + \delta_3u_3\right)\right] & K_1\frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & K_2\frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)\\ K_1\frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right) & K_c\frac{4}{9}\left[2\delta_2u_2 - \left(\delta_1u_1 + \delta_3u_3\right)\right] & K_3\frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right)\\ K_2\frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right) & K_3\frac{2}{3}\left(\delta_3u_2 + \delta_2u_3\right) & K_c\frac{4}{9}\left[2\delta_3u_3 - \left(\delta_1u_1 + \delta_2u_2\right)\right] \end{array}\right)\tag{29.15}\label{eq:stress_tensor_hex_regular} \end{align} \]
does not destroy any of the verified statements.
The divergence of the tensor Eq. (29.15) does not yet satisfy the Thuburn condition. Motivated by section 27.6.1, one therefore makes the approach for the frictional acceleration on the edge $e$
\[ \begin{align} h_e\frac{\partial u_{e, \text{diff}}}{\partial t} &= \left[\nabla\left(K_cE_{c(e)}\right)\right]_e + \frac{1}{l_e}\left(\frac{1}{3}\sum_{r\in t_1}K_rF_{r(e)} - \frac{1}{3}\sum_{r\in t_2}K_rF_{r(e)}\right)\nonumber\\ &= \left[\nabla\left(K_cE_{c(e)}\right)\right]_e + \frac{1}{3l_e}\left(\sum_{r\in t_1}K_rF_{r(e)} - \sum_{r\in t_2}K_rF_{r(e)}\right). \end{align} \]
Here $E_c$ and $F_r$ are deformations, which are localized in the cell centers and at the edges, respectively. The $E_c$ are located on the diagonal of the stress tensor, they are referred to as shear deformations. The $F_r$ are not located on the diagonal of the stress tensor, they are called shear deformation. Here you allow that both can depend not only on their position but also on the edge on which you want to calculate the acceleration, i.e. $E_{c(e)}$ or $F_{r(e)}$. The triangle $t_1$ is located with respect to the edge $e$ in the direction $\mathbf{k}\times\mathbf{n}_e$, the edge $t_2$ in the direction $-\mathbf{k}\times\mathbf{n}_e$. $K_c$ and $K_r$ are nonnegative diffusion coefficients, which are left open in this chapter. O.B.d. A. assume that $\mathbf{n}_e$ is aligned parallel to $\mathbf{i}_1$.
On the regular grid one obtains for the shear deformation
\[ \begin{align} E_{c(e)} &= \frac{4}{9}\left[2\delta_1u_1 - \left(\delta_2u_2 + \delta_3u_3\right)\right] = \frac{4}{9}\left[3\delta_1u_1 - \left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right)\right] = \frac{4}{3}\delta_1u_1 - \frac{4}{9}\left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right)\nonumber\\ &= -\frac{4}{9}\left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right) + \frac{4}{3}\delta_1u_1 = -\frac{2}{3}\frac{2}{3}\left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right) + \frac{4}{3}\delta_1u_1\nonumber\\ &\stackrel{\href{ch-26-problems-of-a-three-element-generating-s.html#eq:div_3elements2d}{\text{Glg. (27.33)}}}{=} -\frac{2}{3}D + \frac{4}{3}\delta_1u_1 \end{align} \]
For the shear deformations, according to Eq. (29.15)
\[ \begin{align} &\delta_2\left[\frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right)\right] + \delta_3\left[\frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)\right] \hastobe \frac{1}{3}\delta_y\left(F_{1(e)} + F_{2(e)} + F_{3(e)}\right)\nonumber\\ &\stackrel{\href{ch-26-problems-of-a-three-element-generating-s.html#eq:ddy_hex}{\text{Glg. (27.31)}}}{=} \frac{1}{3\sqrt{3}}\left(\delta_2 - \delta_3\right)\left(F_{1(e)} + F_{2(e)} + F_{3(e)}\right)\tag{29.18}\label{eq:deriv_hex_diff_deformed_0} \end{align} \]
apply. The choice of $F_{1(e)}$ is arbitrary because this term is first added and then subtracted again. It is therefore obvious to choose
\[ \begin{align} F_{1(e)} &= \zeta_1. \end{align} \]
First, one treats Eq. (29.18) in the form
\[ \begin{align} &\delta_2\left[\frac{2}{3}\left(\delta_2u_1 + \delta_1u_2\right)\right] + \delta_3\left[\frac{2}{3}\left(\delta_3u_1 + \delta_1u_3\right)\right] = \frac{1}{\sqrt{3}}\left(\delta_2F_{3(e)} - \delta_3F_{2(e)}\right). \end{align} \]
This gives you
\[ \begin{align} F_{2(e)} &= -\frac{2}{\sqrt{3}}\left(\delta_3u_1 + \delta_1u_3\right) = \frac{2}{\sqrt{3}}\left(-\delta_3u_1 - \delta_1u_3\right) = \frac{2}{\sqrt{3}}\left[\left(\delta_3u_1 - \delta_1u_3\right) - 2\delta_3u_1\right] \stackrel{\href{ch-26-problems-of-a-three-element-generating-s.html#eq:hex_curl_rhombus_1}{\text{Glg. (27.171)}}}{=} \zeta_2 - \frac{4}{\sqrt{3}}\delta_3u_1,\\ F_{3(e)} &= \frac{2}{\sqrt{3}}\left(\delta_2u_1 + \delta_1u_2\right) = \frac{2}{\sqrt{3}}\left[\left(\delta_1u_2 - \delta_2u_1\right) + 2\delta_2u_1\right] \stackrel{\href{ch-26-problems-of-a-three-element-generating-s.html#eq:hex_curl_rhombus_2}{\text{Glg. (27.172)}}}{=} \zeta_3 + \frac{4}{\sqrt{3}}\delta_2u_1. \end{align} \]
Analogous to section 27.6.1, the averaged shear deformations are now used, which is based on Eq. (29.18) leads. The procedure for calculating the gradient of the component $u_1$ is presented in Section 29.2.0.1 in the context of the deformed grid.
In this section the results of the previous section are generalized to the deformed hexagonal C-lattice. Here too one remains within the framework of the shallow water equations. Essentially you can use the same formulas as on the regular grid, but the velocity gradients can be noted slightly differently. Because of
\[ \begin{align} F_{2(e)} + F_{3(e)} &= \zeta_2 - \frac{4}{\sqrt{3}}\delta_3u_1 + \zeta_3 + \frac{4}{\sqrt{3}}\delta_2u_1 = \zeta_2 + \zeta_3 + 4\frac{1}{\sqrt{3}}\left(\delta_2 - \delta_3\right)u_1 = \zeta_2 + \zeta_3 + 4\delta_yu_1\nonumber\\ &= \zeta_2 + 2\delta_yu_1 + \zeta_3 + 2\delta_yu_1 \end{align} \]
you can also see the deformations in the form
\[ \begin{align} E_{c(e)} &= -\frac{2}{3}D + \frac{4}{3}\delta_xu_1\vline_c,\\ F_{1(e)} &= \zeta_1,\\ F_{2(e)} &= \zeta_2 + 2\delta_yu_1\vline_2,\\ F_{3(e)} &= \zeta_3 + 2\delta_yu_1\vline_3 \end{align} \]
write down. The speed gradients are calculated via cells or rhombuses.
For the dissipation rate $\epsilon_\text{hex}$ can simply
\[ \begin{align} \epsilon_\text{hex} = -\rho\frac{\partial\mathbf{v}_\text{diff}}{\partial t}\cdot\mathbf{v} \end{align} \]
be set. Although this does not correspond exactly to the molecular dissipation rate Eq. (8.66), but the global integrals over both approaches are according to Eq. (8.96) identical. This creates energetic self-consistency.