The two unit vectors of the plane are given by
\[ \begin{align} \mathbf{i} &\coloneqq \left(\begin{array}{c} 1\\ 0 \end{array}\right),\\ \mathbf{j} &\coloneqq \left(\begin{array}{c} 0\\ 1 \end{array}\right)\\ \end{align} \]
designated. Now we further define a three-element generating system $\left(\mathbf{i}_1, \mathbf{i}_2 \mathbf{i}_3\right)$ by
\[ \begin{align} \mathbf{i}_1 &\coloneqq \left(\begin{array}{c} 1\\ 0 \end{array}\right) = \mathbf{i},\\ \mathbf{i}_2 &\coloneqq \left(\begin{array}{c} -\sin\left(30^\circ\right)\\ \cos\left(30^\circ\right) \end{array}\right) = \left(\begin{array}{c} -\frac{1}{2}\\ \frac{\sqrt{3}}{2} \end{array}\right) = -\frac{1}{2}\mathbf{i} + \frac{\sqrt{3}}{2}\mathbf{j},\\ \mathbf{i}_3 &\coloneqq \left(\begin{array}{c} -\frac{1}{2}\\ -\frac{\sqrt{3}}{2} \end{array}\right) = -\frac{1}{2}\mathbf{i} - \frac{\sqrt{3}}{2}\mathbf{j}, \end{align} \]
whose elements are each rotated by 120$^\circ$ relative to each other. We further define a generating system $\left(\mathbf{j}_1, \mathbf{j}_2 \mathbf{j}_3\right)$ that is rotated by 90$^\circ$
\[ \begin{align} \mathbf{j}_1 &\coloneqq \mathbf{k}\times\mathbf{i}_1 = \left(\begin{array}{c} 0\\ 0\\ 1 \end{array}\right)\times\left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right) = \left(\begin{array}{c} 0\\ 1 \end{array}\right),\\ \mathbf{j}_2 &\coloneqq \mathbf{k}\times\mathbf{i}_2 = \left(\begin{array}{c} 0\\ 0\\ 1 \end{array}\right)\times\left(\begin{array}{c} -\frac{1}{2}\\ \frac{\sqrt{3}}{2}\\ 0 \end{array}\right) = \left(\begin{array}{c} -\frac{\sqrt{3}}{2}\\ -\frac{1}{2} \end{array}\right) = -\frac{\sqrt{3}}{2}\mathbf{i} - \frac{1}{2}\mathbf{j},\\ \mathbf{j}_3 &\coloneqq \mathbf{k}\times\mathbf{i}_3 = \left(\begin{array}{c} 0\\ 0\\ 1 \end{array}\right)\times\left(\begin{array}{c} -\frac{1}{2}\\ -\frac{\sqrt{3}}{2}\\ 0 \end{array}\right) = \left(\begin{array}{c} \frac{\sqrt{3}}{2}\\ -\frac{1}{2} \end{array}\right) = \frac{\sqrt{3}}{2}\mathbf{i} - \frac{1}{2}\mathbf{j}. \end{align} \]
You observe
\[ \begin{align} \mathbf{i}_1 + \mathbf{i}_2 + \mathbf{i}_3 = \mathbf{0}, & {} & \mathbf{j}_1 + \mathbf{j}_2 + \mathbf{j}_3 = \mathbf{0}. \end{align} \]
For a two-dimensional vector $\mathbf{v}$ one can
\[ \begin{align} \mathbf{v} = u\mathbf{i} + v\mathbf{j} \end{align} \]
write, apply here
\[ \begin{align} u = \mathbf{i}\cdot\mathbf{v}, & {} & v = \mathbf{j}\cdot\mathbf{v}. \end{align} \]
Since the $\mathbf{i}_k$, $\mathbf{j}_l$ are each pairwise linearly independent, one can note down
\[ \begin{align} \mathbf{v} = u_k'\mathbf{i}_k + u_l'\mathbf{i}_l = v_k'\mathbf{j}_k + v_l'\mathbf{j}_l. \end{align} \]
Since the choice of $k$, $l$ is not unique and one could also use all three unit vectors, one can introduce another linear condition. You use this freedom to change the spelling
\[ \begin{align} \mathbf{v} &= \frac{2}{3}\left(u_1\mathbf{i}_1 + u_2\mathbf{i}_2 + u_3\mathbf{i}_3\right) = \frac{2}{3}\left(v_1\mathbf{j}_1 + v_2\mathbf{j}_2 + v_3\mathbf{j}_3\right) \end{align} \]
to demand. This is with
\[ \begin{align} u_k = \mathbf{i}_k\cdot\mathbf{v}, & {} & v_k = \mathbf{j}_k\cdot\mathbf{v} \end{align} \]
fulfilled, because it follows from this
\[ \begin{align} \mathbf{v} &= u\mathbf{i} + v\mathbf{j} = \frac{2}{3}\left(\frac{6}{4}u\mathbf{i} + \frac{6}{4}v\mathbf{j}\right)\nonumber\\ &= \frac{2}{3}\left(u\mathbf{i} + \frac{1}{4}u\mathbf{i} - \frac{\sqrt{3}}{4}u\mathbf{j} - \frac{\sqrt{3}}{4}v\mathbf{i} + \frac{3}{4}v\mathbf{j} + \frac{1}{4}u\mathbf{i} + \frac{\sqrt{3}}{4}u\mathbf{j} + \frac{\sqrt{3}}{4}v\mathbf{i} + \frac{3}{4}v\mathbf{j}\right)\nonumber\\ &= \frac{2}{3}\left(u\mathbf{i} + \left(-\frac{1}{2}u + \frac{\sqrt{3}}{2}v\right)\left(-\frac{1}{2}\mathbf{i} + \frac{\sqrt{3}}{2}\mathbf{j}\right) + \left(-\frac{1}{2}u - \frac{\sqrt{3}}{2}v\right)\left(-\frac{1}{2}\mathbf{i} - \frac{\sqrt{3}}{2}\mathbf{j}\right)\right)\nonumber\\ &= \frac{2}{3}\left(u_1\mathbf{i}_1 + u_2\mathbf{i}_2 + u_3\mathbf{i}_3\right), \end{align} \]
where
\[ \begin{align} u_1 &= u, & {} & u_2 = -\frac{1}{2}u + \frac{\sqrt{3}}{2}v, & {} & u_3 = -\frac{1}{2}u - \frac{\sqrt{3}}{2}v. \end{align} \]
was used. This applies analogously to the $v_k$. From this it follows
\[ \begin{align} 0 = \mathbf{v}\cdot\mathbf{0} = \mathbf{v}\cdot\left(\mathbf{i}_1 + \mathbf{i}_2 + \mathbf{i}_3\right) = u_1 + u_2 + u_3\tag{27.18}\label{eq:linear_condition_trivariate} \end{align} \]
and analogously for the $v_k$. For the gradient $\nabla\alpha$ of a scalar field we now get $\alpha$
\[ \begin{align} \nabla\alpha = \frac{2}{3}\left[\left(\mathbf{i}_1\cdot\nabla\alpha\right)\mathbf{i}_1 + \left(\mathbf{i}_2\cdot\nabla\alpha\right)\mathbf{i}_2 + \left(\mathbf{i}_3\cdot\nabla\alpha\right)\mathbf{i}_3\right]. \end{align} \]
Because of
\[ \begin{align} \mathbf{i}_k\cdot\nabla\alpha = \frac{\partial\alpha}{\partial x_k} \end{align} \]
follows
\[ \begin{align} \nabla\alpha = \frac{2}{3}\left(\frac{\partial\alpha}{\partial x_1}\mathbf{i}_1 + \frac{\partial\alpha}{\partial x_2}\mathbf{i}_2 + \frac{\partial\alpha}{\partial x_3}\mathbf{i}_3\right).\tag{27.21}\label{eq:grad_three_elements} \end{align} \]
This also provides:
\[ \begin{align} \frac{\partial\alpha}{\partial x_1} + \frac{\partial\alpha}{\partial x_2} + \frac{\partial\alpha}{\partial x_3} = 0\tag{27.22}\label{eq:gradient_linear_condition} \end{align} \]
firmly. It applies
\[ \begin{align} \mathbf{i}_1 &= \frac{1}{\sqrt{3}}\left(\mathbf{j}_3 - \mathbf{j}_2\right) \end{align} \]
and cyclically:
\[ \begin{align} \mathbf{i}_2 = \frac{1}{\sqrt{3}}\left(\mathbf{j}_1 - \mathbf{j}_3\right), & {} & \mathbf{i}_3 = \frac{1}{\sqrt{3}}\left(\mathbf{j}_2 - \mathbf{j}_1\right) \end{align} \]
From this it follows
\[ \begin{align} \nabla\alpha = \frac{2}{3\sqrt{3}}\left(\frac{\partial\alpha}{\partial x_1}\left(\mathbf{j}_3 - \mathbf{j}_2\right) + \frac{\partial\alpha}{\partial x_2}\left(\mathbf{j}_1 - \mathbf{j}_3\right) + \frac{\partial\alpha}{\partial x_3}\left(\mathbf{j}_2 - \mathbf{j}_1\right)\right). \end{align} \]
Rearrangement results
\[ \begin{align} \nabla\alpha = \frac{2}{3\sqrt{3}}\left(\left(\frac{\partial\alpha}{\partial x_2} - \frac{\partial\alpha}{\partial x_3}\right)\mathbf{j}_1 + \left(\frac{\partial\alpha}{\partial x_3} - \frac{\partial\alpha}{\partial x_1}\right)\mathbf{j}_2 + \left(\frac{\partial\alpha}{\partial x_1} - \frac{\partial\alpha}{\partial x_2}\right)\mathbf{j}_3\right). \end{align} \]
It applies
\[ \begin{align} \mathbf{v} &= \frac{2}{3}\left(u_1\mathbf{i}_1 + u_2\mathbf{i}_2 + u_3\mathbf{i}_3\right) = \frac{2}{3}\left[u_1\left(\begin{array}{c}1\\0\end{array}\right) + u_2\left(\begin{array}{c}-\frac{1}{2}\\\frac{\sqrt{3}}{2}\end{array}\right) + u_3\left(\begin{array}{c}-\frac{1}{2}\\-\frac{\sqrt{3}}{2}\end{array}\right)\right] = \frac{2}{3}\left(\begin{array}{c}u_1 - \frac{u_2}{2} - \frac{u_3}{2}\\\frac{\sqrt{3}}{2}u_2 - \frac{\sqrt{3}}{2}u_3\end{array}\right)\nonumber\\ &= \left(\begin{array}{c}\frac{2}{3}u_1 - \frac{u_2}{3} - \frac{u_3}{3}\\\frac{1}{\sqrt{3}}u_2 - \frac{1}{\sqrt{3}}u_3\end{array}\right). \end{align} \]
From this it follows that the divergence
\[ \begin{align} D \coloneqq \nabla\cdot\mathbf{v} = \frac{2}{3}\frac{u_1}{\partial x} - \frac{1}{3}\frac{u_2}{\partial x} - \frac{1}{3}\frac{u_3}{\partial x} + \frac{1}{\sqrt{3}}\frac{u_2}{\partial y} - \frac{1}{\sqrt{3}}\frac{u_3}{\partial y}. \end{align} \]
It applies
\[ \begin{align} \frac{\partial}{\partial x} = \frac{\partial}{\partial x_1}, \end{align} \]
because of
\[ \begin{align} \mathbf{j} = \frac{1}{\sqrt{3}}\left(\mathbf{i}_2 - \mathbf{i}_3\right) \end{align} \]
is also
\[ \begin{align} \frac{\partial}{\partial y} = \frac{1}{\sqrt{3}}\left(\frac{\partial}{\partial x_2} - \frac{\partial}{\partial x_3}\right).\tag{27.31}\label{eq:ddy_hex} \end{align} \]
From this it follows
\[ \begin{align} D &= \frac{2}{3}\frac{\partial u_1}{\partial x} - \frac{1}{3}\frac{\partial u_2}{\partial x} - \frac{1}{3}\frac{\partial u_3}{\partial x} + \frac{1}{3}\left(\frac{\partial u_2}{\partial x_2} - \frac{\partial u_2}{\partial x_3}\right) - \frac{1}{3}\left(\frac{\partial u_3}{\partial x_2} - \frac{\partial u_3}{\partial x_3}\right)\nonumber\\ &= \frac{2}{3}\frac{\partial u_1}{\partial x_1} + \frac{1}{3}\frac{u_2}{\partial x_2} + \frac{1}{3}\frac{u_3}{\partial x_3} - \frac{1}{3}\frac{u_2}{\partial x_1} - \frac{1}{3}\frac{\partial u_3}{\partial x_1} - \frac{1}{3}\frac{\partial u_2}{\partial x_3} - \frac{1}{3}\frac{\partial u_3}{\partial x_2}\nonumber\\ &= \frac{2}{3}\left(\frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3}\right) - \frac{1}{3}\left(\frac{\partial u_2}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_2}{\partial x_3} + \frac{\partial u_3}{\partial x_1} + \frac{\partial u_3}{\partial x_2} + \frac{\partial u_3}{\partial x_3}\right). \end{align} \]
With Eq. (27.22) follows
\[ \begin{align} D = \frac{2}{3}\left(\frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3}\right).\tag{27.33}\label{eq:div_3elements2d} \end{align} \]
By combining with Eq. (27.21) is still obtained
\[ \begin{align} \Delta\alpha = \frac{2}{3}\left(\frac{\partial^2\alpha}{\partial x_1^2} + \frac{\partial^2\alpha}{\partial x_2^2} + \frac{\partial^2\alpha}{\partial x_3^2}\right). \end{align} \]
It applies
\[ \begin{align} u_1 &= \mathbf{v}\cdot\mathbf{i}_1 = \frac{2}{3}\left(v_1\mathbf{j}_1\cdot\mathbf{i}_1 + v_2\mathbf{j}_2\cdot\mathbf{i}_1 + v_3\mathbf{j}_3\cdot\mathbf{i}_1\right) = \frac{2}{3}\left(0\cdot v_1 - \frac{\sqrt{3}}{2}v_2 + \frac{\sqrt{3}}{2}v_3\right)\nonumber\\ &= \frac{1}{\sqrt{3}}\left(v_3 - v_2\right). \end{align} \]
By cyclically moving the indices you get
\[ \begin{align} u_2 = \frac{1}{\sqrt{3}}\left(v_1 - v_3\right), & {} & u_3 = \frac{1}{\sqrt{3}}\left(v_2 - v_1\right). \end{align} \]
This gives another representation of the divergence:
\[ \begin{align} D = \frac{2}{3\sqrt{3}}\left[\left(\frac{\partial v_2}{\partial x_3} - \frac{\partial v_3}{\partial x_2}\right) + \left(\frac{\partial v_3}{\partial x_1} - \frac{\partial v_1}{\partial x_3}\right) + \left(\frac{\partial v_1}{\partial x_2} - \frac{\partial v_2}{\partial x_1}\right)\right]\tag{27.37}\label{eq:div_3elements2d_mod} \end{align} \]
It applies
\[ \begin{align} \frac{\partial v_2}{\partial x_3} - \frac{\partial v_3}{\partial x_2} &= \frac{\partial v_2}{\partial x_3} + \frac{\partial v_2}{\partial x_2} - \frac{\partial v_2}{\partial x_2} - \frac{\partial v_3}{\partial x_2} = \left(\frac{\partial}{\partial x_3} + \frac{\partial}{\partial x_2}\right)v_2 - \frac{\partial}{\partial x_2}\left(v_2 + v_3\right)\nonumber\\ &= -\frac{\partial v_2}{\partial x_1} + \frac{\partial v_1}{\partial x_2} = \frac{\partial v_1}{\partial x_2} - \frac{\partial v_2}{\partial x_1}. \end{align} \]
By cyclically exchanging this equation, one can see that all three summands in brackets in Eq. (27.37) are the same. Follow from this
\[ \begin{align} D &= \frac{2}{\sqrt{3}}\left(\frac{\partial v_2}{\partial x_3} - \frac{\partial v_3}{\partial x_2}\right),\\ D &= \frac{2}{\sqrt{3}}\left(\frac{\partial v_3}{\partial x_1} - \frac{\partial v_1}{\partial x_3}\right),\\ D &= \frac{2}{\sqrt{3}}\left(\frac{\partial v_1}{\partial x_2} - \frac{\partial v_2}{\partial x_1}\right). \end{align} \]
It applies
\[ \begin{align} \mathbf{v} &= \frac{2}{3}\left(v_1\mathbf{j}_1 + v_2\mathbf{j}_2 + v_3\mathbf{j}_3\right) = \frac{2}{3}\left[v_1\left(\begin{array}{c}0\\1\end{array}\right) + v_2\left(\begin{array}{c}-\frac{\sqrt{3}}{2}\\-\frac{1}{2}\end{array}\right) + v_3\left(\begin{array}{c}\frac{\sqrt{3}}{2}\\-\frac{1}{2}\end{array}\right)\right]\nonumber\\ &= \left(\begin{array}{c}\frac{1}{\sqrt{3}}v_3 - \frac{1}{\sqrt{3}}v_2\\\frac{2}{3}v_1 - \frac{v_2}{3} - \frac{v_3}{3}\end{array}\right). \end{align} \]
For the vorticity one obtains
\[ \begin{align} \zeta &= \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = \frac{\partial}{\partial x}\left(\frac{2}{3}v_1 - \frac{v_2}{3} - \frac{v_3}{3}\right) - \frac{\partial}{\partial y}\left(\frac{1}{\sqrt{3}}v_3 - \frac{1}{\sqrt{3}}v_2\right)\nonumber\\ &= \frac{\partial}{\partial x_1}\left(\frac{2}{3}v_1 - \frac{v_2}{3} - \frac{v_3}{3}\right) - \frac{1}{\sqrt{3}}\left(\frac{\partial}{\partial x_2} - \frac{\partial}{\partial x_3}\right)\left(\frac{1}{\sqrt{3}}v_3 - \frac{1}{\sqrt{3}}v_2\right)\nonumber\\ &= \frac{2}{3}\frac{\partial v_1}{\partial x_1} + \frac{1}{3}\frac{\partial v_2}{\partial x_2} + \frac{1}{3}\frac{\partial v_3}{\partial x_3} - \frac{1}{3}\frac{\partial v_2}{\partial x_1} - \frac{1}{3}\frac{\partial v_3}{\partial x_1} - \frac{1}{3}\frac{\partial v_3}{\partial x_2} - \frac{1}{3}\frac{\partial v_2}{\partial x_3}\nonumber\\ &= \frac{2}{3}\left(\frac{\partial v_1}{\partial x_1} + \frac{\partial v_2}{\partial x_2} + \frac{\partial v_3}{\partial x_3}\right) - \frac{1}{3}\left(\frac{\partial v_2}{\partial x_1} + \frac{\partial v_2}{\partial x_2} + \frac{\partial v_2}{\partial x_3} + \frac{\partial v_3}{\partial x_1} + \frac{\partial v_3}{\partial x_2} + \frac{\partial v_3}{\partial x_3}\right). \end{align} \]
With Eq. (27.22) follows
\[ \begin{align} \zeta = \frac{2}{3}\left(\frac{\partial v_1}{\partial x_1} + \frac{\partial v_2}{\partial x_2} + \frac{\partial v_3}{\partial x_3}\right). \end{align} \]
It applies
\[ \begin{align} v_1 &= \mathbf{v}\cdot\mathbf{j}_1 = \frac{2}{3}\left(u_1\mathbf{i}_1 + u_2\mathbf{i}_2 + u_3\mathbf{i}_3\right)\cdot\mathbf{j}_1 = \frac{2}{3}\left(u_1\mathbf{i}_1 + u_2\mathbf{i}_2 + u_3\mathbf{i}_3\right)\cdot\mathbf{j}\nonumber\\ &= \frac{2}{3}\left(\frac{\sqrt{3}}{2}u_2 - \frac{\sqrt{3}}{2}u_3\right) = \frac{1}{\sqrt{3}}\left(u_2 - u_3\right).\tag{27.45}\label{eq:turn_3elemets2d} \end{align} \]
By cyclically moving the indices you get
\[ \begin{align} v_2 = \frac{1}{\sqrt{3}}\left(u_3 - u_1\right), & {} & v_3 = \frac{1}{\sqrt{3}}\left(u_1 - u_2\right). \end{align} \]
From this follows another representation of the vorticity:
\[ \begin{align} \zeta = \frac{2}{3\sqrt{3}}\left(\frac{\partial u_2}{\partial x_1} - \frac{\partial u_3}{\partial x_1} + \frac{\partial u_3}{\partial x_2} - \frac{\partial u_1}{\partial x_2} + \frac{\partial u_1}{\partial x_3} - \frac{\partial u_2}{\partial x_3}\right) \end{align} \]
By rearranging you get
\[ \begin{align} \zeta = \frac{2}{3\sqrt{3}}\left[\left(\frac{\partial u_3}{\partial x_2} - \frac{\partial u_2}{\partial x_3}\right) + \left(\frac{\partial u_1}{\partial x_3} - \frac{\partial u_3}{\partial x_1}\right) + \left(\frac{\partial u_2}{\partial x_1} - \frac{\partial u_1}{\partial x_2}\right)\right]. \end{align} \]
Analogous to the case of Eq. (27.37) also apply here
\[ \begin{align} \zeta &= \frac{2}{\sqrt{3}}\left(\frac{\partial u_3}{\partial x_2} - \frac{\partial u_2}{\partial x_3}\right),\tag{27.49}\label{eq:hex_curl_symm_0}\\ \zeta &= \frac{2}{\sqrt{3}}\left(\frac{\partial u_1}{\partial x_3} - \frac{\partial u_3}{\partial x_1}\right),\tag{27.50}\label{eq:hex_curl_symm_1}\\ \zeta &= \frac{2}{\sqrt{3}}\left(\frac{\partial u_2}{\partial x_1} - \frac{\partial u_1}{\partial x_2}\right).\tag{27.51}\label{eq:hex_curl_symm_2} \end{align} \]
With the main theorem of vector analysis one can get the vector field $\mathbf{v}$ in the form
\[ \begin{align} \mathbf{v} = \mathbf{k}\times\nabla\psi + \nabla\chi \end{align} \]
note down. This applies
\[ \begin{align} \mathbf{k}\times\nabla\psi &= \frac{2}{3}\left(\frac{\partial\psi}{\partial x_1}\mathbf{j}_1 + \frac{\partial\psi}{\partial x_2}\mathbf{j}_2 + \frac{\partial\psi}{\partial x_3}\mathbf{j}_3\right),\\ \nabla\chi &= \frac{2}{3}\left(\frac{\partial\chi}{\partial x_1}\mathbf{i}_1 + \frac{\partial\chi}{\partial x_2}\mathbf{i}_2 + \frac{\partial\chi}{\partial x_3}\mathbf{i}_3\right). \end{align} \]
Follow from this
\[ \begin{align} u_1 &= \frac{1}{\sqrt{3}}\left(\frac{\partial\psi}{\partial x_3} - \frac{\partial\psi}{\partial x_2}\right) + \frac{\partial\chi}{\partial x_1} = -\frac{1}{\sqrt{3}}\left(\frac{\partial\psi}{\partial x_2} - \frac{\partial\psi}{\partial x_3}\right) + \frac{\partial\chi}{\partial x_1}\tag{27.55}\label{eq:trivariate_helmholtz_eq_0},\\ u_2 &= -\frac{1}{\sqrt{3}}\left(\frac{\partial\psi}{\partial x_3} - \frac{\partial\psi}{\partial x_1}\right) + \frac{\partial\chi}{\partial x_2}\tag{27.56}\label{eq:trivariate_helmholtz_eq_1},\\ u_3 &= -\frac{1}{\sqrt{3}}\left(\frac{\partial\psi}{\partial x_1} - \frac{\partial\psi}{\partial x_2}\right) + \frac{\partial\chi}{\partial x_3}\tag{27.57}\label{eq:trivariate_helmholtz_eq_2}. \end{align} \]
and
\[ \begin{align} v_1 &= \frac{\partial\psi}{\partial x_1} + \frac{1}{\sqrt{3}}\left(\frac{\partial\chi}{\partial x_2} - \frac{\partial\chi}{\partial x_3}\right)\tag{27.58}\label{eq:trivariate_helmholtz_eq_3},\\ v_2 &= \frac{\partial\psi}{\partial x_2} + \frac{1}{\sqrt{3}}\left(\frac{\partial\chi}{\partial x_3} - \frac{\partial\chi}{\partial x_1}\right)\tag{27.59}\label{eq:trivariate_helmholtz_eq_4},\\ v_3 &= \frac{\partial\psi}{\partial x_3} + \frac{1}{\sqrt{3}}\left(\frac{\partial\chi}{\partial x_1} - \frac{\partial\chi}{\partial x_2}\right)\tag{27.60}\label{eq:trivariate_helmholtz_eq_5}. \end{align} \]
Because of
\[ \begin{align} D = \Delta\chi, & {} & \zeta = \Delta\psi \end{align} \]
continue to apply
\[ \begin{align} \Delta u_1 &= -\frac{1}{\sqrt{3}}\left(\frac{\partial\zeta}{\partial x_2} - \frac{\partial\zeta}{\partial x_3}\right) + \frac{\partial D}{\partial x_1},\tag{27.62}\label{eq:hex_laplace_u_1}\\ \Delta u_2 &= -\frac{1}{\sqrt{3}}\left(\frac{\partial\zeta}{\partial x_3} - \frac{\partial\zeta}{\partial x_1}\right) + \frac{\partial D}{\partial x_2},\tag{27.63}\label{eq:hex_laplace_u_2}\\ \Delta u_3 &= -\frac{1}{\sqrt{3}}\left(\frac{\partial\zeta}{\partial x_1} - \frac{\partial\zeta}{\partial x_2}\right) + \frac{\partial D}{\partial x_3}\tag{27.64}\label{eq:hex_laplace_u_3}, \end{align} \]
\[ \begin{align} \Delta v_1 &= \frac{\partial\zeta}{\partial x_1} + \frac{1}{\sqrt{3}}\left(\frac{\partial D}{\partial x_2} - \frac{\partial D}{\partial x_3}\right),\\ \Delta v_2 &= \frac{\partial\zeta}{\partial x_2} + \frac{1}{\sqrt{3}}\left(\frac{\partial D}{\partial x_3} - \frac{\partial D}{\partial x_1}\right),\\ \Delta v_3 &= \frac{\partial\zeta}{\partial x_3} + \frac{1}{\sqrt{3}}\left(\frac{\partial D}{\partial x_1} - \frac{\partial D}{\partial x_2}\right). \end{align} \]
Let $\alpha$ be a field of the form
\[ \begin{align} \alpha\left(\mathbf{r}\right) = \exp\left(i\mathbf{k}\cdot\mathbf{r}\right). \end{align} \]
The distance between the centers of two hexagons is called $d$ (from now on this quantity is called the lattice constant). Now define the central difference quotient in the j direction for $1 \leq j \leq 3$
\[ \begin{align} \delta_j\alpha &= \frac{\alpha\left(\mathbf{r} + \mathbf{i}_j\frac{d}{2}\right) - \alpha\left(\mathbf{r} - \mathbf{i}_j\frac{d}{2}\right)}{d}. \end{align} \]
This applies to this
\[ \begin{align} \delta_j\alpha &= \frac{\exp\left[i\mathbf{k}\cdot\left(\mathbf{r} + \mathbf{i}_j\frac{d}{2}\right)\right] - \exp\left[i\mathbf{k}\cdot\left(\mathbf{r} - \mathbf{i}_j\frac{d}{2}\right)\right]}{d}\nonumber\\ &= \exp\left(i\mathbf{k}\cdot\mathbf{r}\right)\frac{\exp\left(i\mathbf{k}\cdot\mathbf{i}_j\frac{d}{2}\right) - \exp\left(-i\mathbf{k}\cdot\mathbf{i}_j\frac{d}{2}\right)}{d} = \alpha\left(\mathbf{r}\right)\frac{\exp\left(i\mathbf{k}\cdot\mathbf{i}_j\frac{d}{2}\right) - \exp\left(-i\mathbf{k}\cdot\mathbf{i}_j\frac{d}{2}\right)}{d}\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{\exp\left(ik_j\frac{d}{2}\right) - \exp\left(-ik_j\frac{d}{2}\right)}{d} = \alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_jd}{2}\right). \end{align} \]
You define it
\[ \begin{align} s_j \coloneqq \sin\left(\frac{k_jd}{2}\right), \end{align} \]
you can do this in the form
\[ \begin{align} \delta_j\alpha = \frac{2i}{d}s_j\alpha \end{align} \]
note down. We now continue to define a simple averaging operator in the j direction
\[ \begin{align} \newoverline{\alpha}^{(j)} & \coloneqq \frac{\alpha\left(\mathbf{r} + \mathbf{i}_j\frac{d}{2}\right) + \alpha\left(\mathbf{r} - \mathbf{i}_j\frac{d}{2}\right)}{2} = \frac{\exp\left[i\mathbf{k}\cdot\left(\mathbf{r} + \mathbf{i}_j\frac{d}{2}\right)\right] + \exp\left[i\mathbf{k}\cdot\left(\mathbf{r} - \mathbf{i}_j\frac{d}{2}\right)\right]}{2}\nonumber\\ &= \exp\left(i\mathbf{k}\cdot\mathbf{r}\right)\frac{\exp\left(i\mathbf{k}\cdot\mathbf{i}_j\frac{d}{2}\right) + \exp\left(-i\mathbf{k}\cdot\mathbf{i}_j\frac{d}{2}\right)}{2}\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{\exp\left(ik_j\frac{d}{2}\right) + \exp\left(-ik_j\frac{d}{2}\right)}{2} = \alpha\left(\mathbf{r}\right)\cos\left(\frac{k_jd}{2}\right).\tag{27.73}\label{eq:hex_simple_average} \end{align} \]
You define it
\[ \begin{align} c_j \coloneqq \cos\left(\frac{k_jd}{2}\right), \end{align} \]
you can do this in the form
\[ \begin{align} \newoverline{\alpha}^{(j)} = c_j\alpha \end{align} \]
note down.
The linearized shallow water equations (13.173) - (13.174) on the f-plane can be written with the geopotential $\Phi \coloneqq gh$ in the form
\[ \begin{align} \frac{\partial\Phi}{\partial t} + \Phi_0\nabla\cdot\mathbf{v} = 0, & {} & \frac{\partial\mathbf{v}}{\partial t} + f_0\mathbf{k}\times\mathbf{v} - \nabla\Phi = 0 \end{align} \]
note down. Here $\Phi_0 \coloneqq gD$. With the equations (27.33) and (27.45) this leads to the evolution equations for the components regarding the three-element generating system:
\[ \begin{align} \frac{\partial\Phi}{\partial t} + \frac{2}{3}\Phi_0\left(\frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3}\right) &= 0,\\ \frac{\partial u_1}{\partial t} - \frac{f_0}{\sqrt{3}}\left(u_2 - u_3\right) + \frac{\partial\Phi}{\partial x_1} &= 0,\\ \frac{\partial u_2}{\partial t} - \frac{f_0}{\sqrt{3}}\left(u_3 - u_1\right) + \frac{\partial\Phi}{\partial x_2} &= 0,\\ \frac{\partial u_3}{\partial t} - \frac{f_0}{\sqrt{3}}\left(u_1 - u_2\right) + \frac{\partial\Phi}{\partial x_3} &= 0. \end{align} \]
Spatial discretization results
\[ \begin{align} \frac{\partial\Phi}{\partial t} + \frac{2}{3}\Phi_0\left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right) &= 0,\tag{27.81}\label{eq:swe_lin_hex_0}\\ \frac{\partial u_1}{\partial t} - \frac{f_0}{\sqrt{3}}\left(\newoverline{u_2}^{3} - \newoverline{u_3}^{(2)}\right) + \delta_1\Phi &= 0,\tag{27.82}\label{eq:swe_lin_hex_1}\\ \frac{\partial u_2}{\partial t} - \frac{f_0}{\sqrt{3}}\left(\newoverline{u_3}^{(1)} - \newoverline{u_1}^{3}\right) + \delta_2\Phi &= 0,\tag{27.83}\label{eq:swe_lin_hex_2}\\ \frac{\partial u_3}{\partial t} - \frac{f_0}{\sqrt{3}}\left(\newoverline{u_1}^{(2)} - \newoverline{u_2}^{(1)}\right) + \delta_3\Phi &= 0.\tag{27.84}\label{eq:swe_lin_hex_3} \end{align} \]
The arithmetic mean of the two nearest speed components in the desired direction was used to reconstruct the wind components for calculating the Coriolis acceleration. If you set a monochromatic plane wave for all fields, i.e
\[ \begin{align} \psi\left(\mathbf{r}, t\right) = \newhat{\psi}\exp\left(i\mathbf{k}\cdot\mathbf{r} - i\omega t\right) \end{align} \]
for any field $\psi$ with a complex amplitude $\newhat{\psi}$, one obtains
\[ \begin{align} -i\omega\newhat{\Phi} + \frac{4i}{3d}\Phi_0\left(s_1\newhat{u}_1 + s_2\newhat{u}_2 + s_3\newhat{u}_3\right) &= 0,\\ -i\omega\newhat{u}_1 - \frac{f_0}{\sqrt{3}}\left(c_3\newhat{u}_2 - c_2\newhat{u}_3\right) + \frac{2is_1}{d}\newhat{\Phi} &= 0,\\ -i\omega\newhat{u}_2 - \frac{f_0}{\sqrt{3}}\left(c_1\newhat{u}_3 - c_3\newhat{u}_1\right) + \frac{2is_1}{d}\newhat{\Phi} &= 0,\\ -i\omega\newhat{u}_3 - \frac{f_0}{\sqrt{3}}\left(c_2\newhat{u}_1 - c_1\newhat{u}_2\right) + \frac{2is_1}{d}\newhat{\Phi} &= 0. \end{align} \]
It should be noted that there is a phase offset between the equations, which arises from the fact that the individual equations apply at different locations. This was edited out. In matrix form you get
\[ \begin{align} \left(\begin{array}{cccc} -i\omega & \frac{4i}{3d}\Phi_0s_1 & \frac{4i}{3d}\Phi_0s_2 & \frac{4i}{3d}\Phi_0s_3 \\ \frac{2is_1}{d} & -i\omega & -\frac{f_0}{\sqrt{3}}c_3 & \frac{f_0}{\sqrt{3}}c_2\\ \frac{2is_2}{d} & \frac{f_0}{\sqrt{3}}c_3 & -i\omega & -\frac{f_0}{\sqrt{3}}c_1\\ \frac{2is_3}{d} & -\frac{f_0}{\sqrt{3}}c_2 & \frac{f_0}{\sqrt{3}}c_1 & -i\omega \end{array}\right)\left(\begin{array}{c} \newhat{\Phi}\\ \newhat{u}_1\\ \newhat{u}_2\\ \newhat{u}_3 \end{array}\right) = \mathbf{0}. \end{align} \]
Multiplying this by the imaginary unit $i$ gives
\[ \begin{align} \left(\begin{array}{cccc} \omega & -\frac{4}{3d}\Phi_0s_1 & -\frac{4}{3d}\Phi_0s_2 & -\frac{4}{3d}\Phi_0s_3 \\ -\frac{2s_1}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_3 & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & \omega & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \frac{if_0}{\sqrt{3}}c_1 & \omega \end{array}\right)\left(\begin{array}{c} \newhat{\Phi}\\ \newhat{u}_1\\ \newhat{u}_2\\ \newhat{u}_3 \end{array}\right) = \mathbf{0}. \end{align} \]
Nontrivial solutions exist if and only if the determinant of the coefficient matrix of this system of linear equations vanishes:
\[ \begin{align} \left|\begin{array}{cccc} \omega & -\frac{4}{3d}\Phi_0s_1 & -\frac{4}{3d}\Phi_0s_2 & -\frac{4}{3d}\Phi_0s_3 \\ -\frac{2s_1}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_3 & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & \omega & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \frac{if_0}{\sqrt{3}}c_1 & \omega \end{array}\right| &\stackrel{!}{=} 0\nonumber\\ \Leftrightarrow\omega\left|\begin{array}{ccc} \omega & -\frac{if_0}{\sqrt{3}}c_3 & \frac{if_0}{\sqrt{3}}c_2\\ \frac{if_0}{\sqrt{3}}c_3 & \omega & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{if_0}{\sqrt{3}}c_2 & \frac{if_0}{\sqrt{3}}c_1 & \omega \end{array} \right| + \frac{4}{3d}\Phi_0s_1\left|\begin{array}{ccc} -\frac{2s_1}{d} & -\frac{if_0}{\sqrt{3}}c_3 & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & \frac{if_0}{\sqrt{3}}c_1 & \omega \end{array}\right| & \nonumber\\ - \frac{4}{3d}\Phi_0s_2\left|\begin{array}{ccc} -\frac{2s_1}{d} & \omega & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \omega \end{array} \right| + \frac{4}{3d}\Phi_0s_3\left|\begin{array}{ccc} -\frac{2s_1}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_3\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & \omega\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \frac{if_0}{\sqrt{3}}c_1 \end{array} \right| &= 0 \end{align} \] \[ \begin{align} \Leftrightarrow \omega^4 - \frac{\omega^2f_0^2}{3}\left(c_1^2 + c_2^2 + c_3^2\right) + \frac{4}{3d}\Phi_0s_1\left|\begin{array}{ccc} -\frac{2s_1}{d} & -\frac{if_0}{\sqrt{3}}c_3 & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & \frac{if_0}{\sqrt{3}}c_1 & \omega \end{array}\right| & \nonumber\\ -\frac{4}{3d}\Phi_0s_2\left|\begin{array}{ccc} -\frac{2s_1}{d} & \omega & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \omega \end{array}\right| + \frac{4}{3d}\Phi_0s_3\left|\begin{array}{ccc} -\frac{2s_1}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_3\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & \omega\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \frac{if_0}{\sqrt{3}}c_1 \end{array} \right| &= 0 \end{align} \]
It applies
\[ \begin{align} \frac{4}{3d}\Phi_0s_1\left|\begin{array}{ccc} -\frac{2s_1}{d} & -\frac{if_0}{\sqrt{3}}c_3 & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & \frac{if_0}{\sqrt{3}}c_1 & \omega \end{array}\right| &= -\frac{8\omega^2}{3d^2}\Phi_0s_1^2 + \frac{8\Phi_0f_0^2}{9d^2}\left(s_1c_1s_3c_3 + s_1c_1s_2c_2 + s_1c_1s_1c_1\right) + \frac{8if_0}{3\sqrt{3}d}s_1\left(s_3c_2 - c_3s_2\right)\nonumber\\ &= -\frac{8\omega^2}{3d^2}\Phi_0s_1^2 + \frac{8\Phi_0f_0^2}{9d^2}s_1c_1\left(s_1c_1 + s_2c_2 + s_3c_3\right) - \frac{8if_0}{3\sqrt{3}d}s_1\left(s_2c_3 - s_3c_2\right). \end{align} \]
The remaining two partial determinants arise analogously by cyclically exchanging the indices:
\[ \begin{align} -\frac{4}{3d}\Phi_0s_2\left|\begin{array}{ccc} -\frac{2s_1}{d} & \omega & \frac{if_0}{\sqrt{3}}c_2\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & -\frac{if_0}{\sqrt{3}}c_1\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \omega \end{array}\right| &= -\frac{8\omega^2}{3d^2}\Phi_0s_2^2 + \frac{8\Phi_0f_0^2}{9d^2}s_2c_2\left(s_1c_1 + s_2c_2 + s_3c_3\right) - \frac{8if_0}{3\sqrt{3}d}s_2\left(s_3c_1 - s_1c_3\right),\nonumber\\ \frac{4}{3d}\Phi_0s_3\left|\begin{array}{ccc} -\frac{2s_1}{d} & \omega & -\frac{if_0}{\sqrt{3}}c_3\\ -\frac{2s_2}{d} & \frac{if_0}{\sqrt{3}}c_3 & \omega\\ -\frac{2s_3}{d} & -\frac{if_0}{\sqrt{3}}c_2 & \frac{if_0}{\sqrt{3}}c_1 \end{array} \right| &= -\frac{8\omega^2}{3d^2}\Phi_0s_3^2 + \frac{8\Phi_0f_0^2}{9d^2}s_3c_3\left(s_1c_1 + s_2c_2 + s_3c_3\right) - \frac{8if_0}{3\sqrt{3}d}s_3\left(s_1c_2 - s_2c_1\right). \end{align} \]
So the characteristic polynomial is finally:
\[ \begin{align} \omega^4 - \omega^2\left[\frac{f_0^2}{3}\left(c_1^2 + c_2^2 + c_3^2\right) + \frac{8\Phi_0}{3d^2}\left(s_1^2 + s_2^2 + s_3^2\right)\right] + \frac{8\Phi_0f_0^2}{9d^2}\left(s_1c_1 + s_2c_2 + s_3c_3\right)^2 = 0.\tag{27.96}\label{eq:disp_rel_hex_c} \end{align} \]
For the case of well-resolved waves $\left|\mathbf{k}\right| \ll \frac{1}{d}$ apply
\[ \begin{align} c_j \approx 1, & {} & s_j \approx \frac{k_jd}{2}. \end{align} \]
Putting this into Eq. (27.96), you get
\[ \begin{align} \omega^4 - \omega^2\left[f_0^2 + \frac{8\Phi_0}{3d^2}\left(\frac{k_1^2d^2}{4} + \frac{k_2^2d^2}{4} + \frac{k_3^2d^2}{4}\right)\right] + \frac{8\Phi_0f_0^2}{9d^2}\underbrace{\left(\frac{k_1d}{2} + \frac{k_1d}{2} + \frac{k_1d}{2}\right)^2}_{\approx 0} &\approx 0\nonumber\\ \Leftrightarrow\omega^4 - \omega^2\left[f_0^2 + \frac{2\Phi_0}{3}\left(k_1^2 + k_2^2 + k_3^2\right)\right] &\approx 0.\nonumber\\ \Leftrightarrow\omega^4 - \omega^2\left[f_0^2 + \Phi\mathbf{k}^2\right] &\approx 0. \end{align} \]
The last step took advantage of the fact that $k_1^2 + k_2^2 + k_3^2 = \frac{3}{2}\mathbf{k}^2$. This can be made clear by: A. $\mathbf{k} \parallel \mathbf{i}_1$ assumes, because then $k_1 = k$, $k_2 = -k\sin\left(30^\circ\right) = -\frac{k}{2}$ and $k_3 = -\frac{k}{2}$. This leads to the two intertio-gravity modes
\[ \begin{align} \omega^2 \approx f_0^2 + \Phi_0\left|\mathbf{k}\right|^2 \end{align} \]
as well as geostrophic fashion
\[ \begin{align} \omega \approx 0. \end{align} \]
However, for less well-resolved waves, the term $\propto\left(s_1c_1 + s_2c_2 + s_3c_3\right)^2$ causes the geostrophic mode to split into two submodes with $\omega \not= 0$. This is problematic and corresponds to the findings of Sect. 26.9.
In a planar triangular or hexagonal C-grid discretization, the velocity components are rotated by 120 or 60$^\circ$ with respect to each other, but the components are not defined at the same location. Therefore, the condition Eq. (27.18) not simply for the vector components, but only after interpolation to reference locations, for which the centers of the cells are used here. This interpolation is done using operators
\[ \begin{align} \newtilde{u}^{(i)} \end{align} \]
with $1 \leq i \leq 3$, where $u$ is any vector component. A separate averaging operator is permitted here for each spatial direction $i$. The conditions Eq. (27.18) can be expressed here in the form
\[ \begin{align} \newtilde{u}_1^{(1)} + \newtilde{u}_2^{(2)} + \newtilde{u}_3^{(3)} \hastobe 0, & {} & \newtilde{v}_1^{(1)} + \newtilde{v}_2^{(2)} + \newtilde{v}_3^{3} \hastobe 0\tag{27.102}\label{eq:linear_condition_trivariate_discrete} \end{align} \]
note down. These equations are to be understood as requirements for vector fields $\mathbf{v}$. If gradient fields $\nabla\alpha$ satisfy the condition Eq. (27.102) is not, so you have a problem, since the calculation of gradient fields is fixed on the C-grid. Therefore, the averaging of Eq. is used as a requirement for the averaging operator to be found. (27.22):
\[ \begin{align} \newtilde{\delta_1\alpha}^{(1)} + \newtilde{\delta_2\alpha}^{(2)} + \newtilde{\delta_3\alpha}^{3} \stackrel{!}{=} 0\tag{27.103}\label{eq:linear_condition_trivariate_discrete_gradient} \end{align} \]
For the Helmholtz distribution Equations (27.55) - (27.60) follows by averaging in the relevant spatial directions
\[ \begin{align} \newtilde{u}_1^{(1)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_3\psi}^{(1)} - \newtilde{\delta_2\psi}^{(1)}\right) + \newtilde{\delta_1\chi}^{(1)},\\ \newtilde{u}_2^{(2)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_1\psi}^{(2)} - \newtilde{\delta_3\psi}^{(2)}\right) + \newtilde{\delta_2\chi}^{(2)},\\ \newtilde{u}_3^{3} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_2\psi}^{3} - \newtilde{\delta_1\psi}^{3}\right) + \newtilde{\delta_3\chi}^{3},\\ \newtilde{v}_1^{(1)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_2\chi}^{(1)} - \newtilde{\delta_3\chi}^{(1)}\right) + \newtilde{\delta_1\psi}^{(1)},\\ \newtilde{v}_2^{(2)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_3\chi}^{(2)} - \newtilde{\delta_1\chi}^{(2)}\right) + \newtilde{\delta_2\psi}^{(2)},\\ \newtilde{v}_3^{3} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_1\chi}^{3} - \newtilde{\delta_2\chi}^{3}\right) + \newtilde{\delta_3\psi}^{3}. \end{align} \]
This does not yet fulfill the equations (27.102) - (27.102), rather a preparatory averaging of the scalar fields is required:
\[ \begin{align} \newtilde{u}_1^{(1)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_3\newtilde{\psi}^{(2)}}^{(1)} - \newtilde{\delta_2\newtilde{\psi}^{3}}^{(1)}\right) + \newtilde{\delta_1\chi}^{(1)}\\ \newtilde{u}_2^{(2)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_1\newtilde{\psi}^{3}}^{(2)} - \newtilde{\delta_3\newtilde{\psi}^{(1)}}^{(2)}\right) + \newtilde{\delta_2\chi}^{(2)}\\ \newtilde{u}_3^{3} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_2\newtilde{\psi}^{(1)}}^{3} - \newtilde{\delta_1\newtilde{\psi}^{(2)}}^{3}\right) + \newtilde{\delta_3\chi}^{3}\\ \newtilde{v}_1^{(1)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_2\newtilde{\chi}^{3}}^{(1)} - \newtilde{\delta_3\newtilde{\chi}^{(2)}}^{(1)}\right) + \newtilde{\delta_1\psi}^{(1)}\\ \newtilde{v}_2^{(2)} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_3\newtilde{\chi}^{(1)}}^{(2)} - \newtilde{\delta_1\newtilde{\chi}^{3}}^{(2)}\right) + \newtilde{\delta_2\psi}^{(2)}\\ \newtilde{v}_3^{3} &= \frac{1}{\sqrt{3}}\left(\newtilde{\delta_1\newtilde{\chi}^{(2)}}^{3} - \newtilde{\delta_2\newtilde{\chi}^{(1)}}^{3}\right) + \newtilde{\delta_3\psi}^{3} \end{align} \]
This implies
\[ \begin{align} u_1 &= \frac{1}{\sqrt{3}}\left(\delta_3\newtilde{\psi}^{(2)} - \delta_2\newtilde{\psi}^{3}\right) + \delta_1\chi\tag{27.116}\label{eq:hex:helmholtz_0},\\ u_2 &= \frac{1}{\sqrt{3}}\left(\delta_1\newtilde{\psi}^{3} - \delta_3\newtilde{\psi}^{(1)}\right) + \delta_2\chi\tag{27.117}\label{eq:hex:helmholtz_1},\\ u_3 &= \frac{1}{\sqrt{3}}\left(\delta_2\newtilde{\psi}^{(1)} - \delta_1\newtilde{\psi}^{(2)}\right) + \delta_3\chi\tag{27.118}\label{eq:hex:helmholtz_2},\\ v_1 &= \frac{1}{\sqrt{3}}\left(\delta_2\newtilde{\chi}^{3} - \delta_3\newtilde{\chi}^{(2)}\right) + \delta_1\psi,\\ v_2 &= \frac{1}{\sqrt{3}}\left(\delta_3\newtilde{\chi}^{(1)} - \delta_1\newtilde{\chi}^{3}\right) + \delta_2\psi,\\ v_3 &= \frac{1}{\sqrt{3}}\left(\delta_1\newtilde{\chi}^{(2)} - \delta_2\newtilde{\chi}^{(1)}\right) + \delta_3\psi. \end{align} \]
Now the question arises as to what the averaging operator $\newtilde{\psi}^{(j)}$ should actually look like. For the one in Eq. (27.73) operator applies
\[ \begin{align} &\newoverline{\delta_1\psi}^{(1)} + \newoverline{\delta_1\psi}^{(2)} + \newoverline{\delta_1\psi}^{3}\nonumber\\ &= \psi\left(\mathbf{r}\right)\frac{2i}{d}\left[\sin\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_1d}{2}\right) + \sin\left(\frac{k_2d}{2}\right)\cos\left(\frac{k_2d}{2}\right) + \sin\left(\frac{k_3d}{2}\right)\cos\left(\frac{k_3d}{2}\right)\right]\nonumber\\ &= \psi\left(\mathbf{r}\right)\frac{i}{d}\left[2\sin\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_1d}{2}\right) + 2\sin\left(\frac{k_2d}{2}\right)\cos\left(\frac{k_2d}{2}\right) + 2\sin\left(\frac{k_3d}{2}\right)\cos\left(\frac{k_3d}{2}\right)\right]\nonumber\\ &= \psi\left(\mathbf{r}\right)\frac{i}{d}\left[\sin\left(k_1d\right) + \sin\left(k_2d\right) + \sin\left(k_3d\right)\right] \stackrel{\text{i. A.}}{\not=} 0. \end{align} \]
This operator satisfies the condition Eq. So (27.103) not. Therefore, a second averaging operator is defined
\[ \begin{align} \newoverline{\psi}^{((1))} & \coloneqq \newoverline{\psi}^{(2, 3)} = \newoverline{\newoverline{\psi}^{3}}^{(2)} = \newoverline{\newoverline{\psi}^{(2)}}^{3} = \psi\left(\mathbf{r}\right)\cos\left(\frac{k_2d}{2}\right)\cos\left(\frac{k_3d}{2}\right).\text{ (und zyklisch)} \end{align} \]
This applies to this one
\[ \begin{align} \newoverline{\delta_1\alpha}^{((1))} + \newoverline{\delta_2\alpha}^{((2))} + \newoverline{\delta_3\alpha}^{(3)} &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\Big[\sin\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right)\cos\left(\frac{k_3d}{2}\right) + \cos\left(\frac{k_1d}{2}\right)\sin\left(\frac{k_2d}{2}\right)\cos\left(\frac{k_3d}{2}\right)\nonumber\\ & + \cos\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right)\sin\left(\frac{k_3d}{2}\right)\Big]\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\Big[\sin\left(\frac{k_1d + k_2d}{2}\right)\cos\left(\frac{k_3d}{2}\right) + \cos\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right)\sin\left(\frac{k_3d}{2}\right)\Big]\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\Big[\sin\left(\frac{k_1 + k_2 + k_3}{2}d\right) - \cos\left(\frac{k_1 + k_2}{2}d\right)\sin\left(\frac{k_3}{2}d\right)\nonumber\\ &+ \cos\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right)\sin\left(\frac{k_3d}{2}\right)\Big]. \end{align} \]
With $k_1 + k_2 + k_3 = 0$ follows
\[ \begin{align} \newoverline{\delta_1\alpha}^{((1))} + \newoverline{\delta_2\alpha}^{((2))} + \newoverline{\delta_3\alpha}^{(3)} &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\Big[-\cos\left(\frac{k_1 + k_2}{2}d\right)\sin\left(\frac{k_3}{2}d\right) + \cos\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right)\sin\left(\frac{k_3d}{2}\right)\Big]\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_3}{2}d\right)\Big[\cos\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right) - \cos\left(\frac{k_1 + k_2}{2}d\right)\Big]\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_3}{2}d\right)\Big[\cos\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right) - \cos\left(\frac{k_1d}{2}\right)\cos\left(\frac{k_2d}{2}\right)\nonumber\\ &+ \sin\left(\frac{k_1d}{2}\right)\sin\left(\frac{k_2d}{2}\right)\Big]\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2d}{2}\right)\sin\left(\frac{k_3d}{2}\right) \stackrel{\text{i. A.}}{\not=} 0.\tag{27.125}\label{eq:thuburn_op_deriv_0} \end{align} \]
This operator also satisfies the condition Eq. So (27.103) not. However, one continues the calculation with $k_1 + k_2 + k_3 = 0$:
\[ \begin{align} & \newoverline{\delta_1\alpha}^{((1))} + \newoverline{\delta_2\alpha}^{((2))} + \newoverline{\delta_3\alpha}^{(3)} = \alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2d}{2}\right)\sin\left(\frac{k_3d}{2}\right)\nonumber\\ &= \alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2d}{2}\right)\sin\left(-\frac{k_1 + k_2}{2}d\right) = -\alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2d}{2}\right)\sin\left(\frac{k_1 + k_2}{2}d\right)\nonumber\\ &= -\alpha\left(\mathbf{r}\right)\frac{2i}{d}\sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2d}{2}\right)\left[\sin\left(\frac{k_1}{2}d\right)\cos\left(\frac{k_2}{2}d\right) + \cos\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)\right]\nonumber\\ &= -\alpha\left(\mathbf{r}\right)\frac{2i}{d}\left[\sin\left(\frac{k_2}{2}d\right)\sin\left(\frac{k_1}{2}d\right)^2\cos\left(\frac{k_2}{2}d\right) + \sin\left(\frac{k_1}{2}d\right)\cos\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)^2\right]\nonumber\\ &= -\alpha\left(\mathbf{r}\right)\frac{i}{d}\left[\sin\left(k_2d\right)\sin\left(\frac{k_1}{2}d\right)^2 + \sin\left(k_1d\right)\sin\left(\frac{k_2}{2}d\right)^2\right]\nonumber\\ &= -\alpha\left(\mathbf{r}\right)\frac{i}{d}\left[\sin\left(k_2d\right) + \sin\left(k_1d\right) - \sin\left(k_2d\right)\cos\left(\frac{k_1}{2}d\right)^2 - \sin\left(k_1d\right)\cos\left(\frac{k_2}{2}d\right)^2\right]\nonumber \end{align} \] \[ \begin{align} &= -\alpha\left(\mathbf{r}\right)\frac{i}{d}\Big[\sin\left(k_2d\right) + \sin\left(k_1d\right) - \sin\left(k_2d\right)\cos\left(k_1\right) - \sin\left(k_1d\right)\cos\left(k_2d\right) + \sin\left(k_2d\right)\sin\left(\frac{k_1}{2}d\right)^2\nonumber\\ & + \sin\left(k_1d\right)\sin\left(\frac{k_2}{2}d\right)^2\Big]\nonumber\\ &= -\alpha\left(\mathbf{r}\right)\frac{i}{d}\Big[\sin\left(k_2d\right) + \sin\left(k_1d\right) - \sin\left[\left(k_1 + k_2\right)d\right] + \sin\left(k_2d\right)\sin\left(\frac{k_1}{2}d\right)^2 + \sin\left(k_1d\right)\sin\left(\frac{k_2}{2}d\right)^2\Big]\nonumber\\ &= -\alpha\left(\mathbf{r}\right)\frac{i}{d}\Big[\sin\left(k_2d\right) + \sin\left(k_1d\right) - \sin\left(-k_3d\right) + \sin\left(k_2d\right)\sin\left(\frac{k_1}{2}d\right)^2 + \sin\left(k_1d\right)\sin\left(\frac{k_2}{2}d\right)^2\Big]\nonumber\\ &= -\alpha\left(\mathbf{r}\right)\frac{i}{d}\Big[\sin\left(k_2d\right) + \sin\left(k_1d\right) + \sin\left(k_3d\right) + \sin\left(k_2d\right)\sin\left(\frac{k_1}{2}d\right)^2 + \sin\left(k_1d\right)\sin\left(\frac{k_2}{2}d\right)^2\Big]\nonumber\\ &= -\left(\newoverline{\delta_1\alpha}^{(1)} + \newoverline{\delta_1\alpha}^{(2)} + \newoverline{\delta_1\alpha}^{3}\right) - \alpha\left(\mathbf{r}\right)\frac{i}{d}\Big[\sin\left(k_2d\right)\sin\left(\frac{k_1}{2}d\right)^2 + \sin\left(k_1d\right)\sin\left(\frac{k_2}{2}d\right)^2\Big] \end{align} \]
Now you do the math
\[ \begin{align} &\sin\left(k_2d\right)\sin\left(\frac{k_1}{2}d\right)^2 + \sin\left(k_1d\right)\sin\left(\frac{k_2}{2}d\right)^2 = \sin\left(\frac{k_2}{2}d + \frac{k_2}{2}d\right)\sin\left(\frac{k_1}{2}d\right)^2 + \sin\left(\frac{k_1}{2}d + \frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)^2\nonumber\\ &= 2\sin\left(\frac{k_1}{2}d\right)^2\sin\left(\frac{k_2}{2}d\right)\cos\left(\frac{k_2}{2}d\right) + 2\sin\left(\frac{k_2}{2}d\right)^2\sin\left(\frac{k_1}{2}d\right)\cos\left(\frac{k_1}{2}d\right)\nonumber\\ &= \sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)\left[2\sin\left(\frac{k_1}{2}d\right)\cos\left(\frac{k_2}{2}d\right) + 2\cos\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)\right]\nonumber\\ &= \sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)\sin\left(\frac{k_1 + k_2}{2}d\right) = \sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)\sin\left(-\frac{k_3}{2}d\right)\nonumber\\ &= -\sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)\sin\left(\frac{k_3}{2}d\right). \end{align} \]
Thus follows
\[ \begin{align} & \newoverline{\delta_1\alpha}^{((1))} + \newoverline{\delta_2\alpha}^{((2))} + \newoverline{\delta_3\alpha}^{(3)} = -\left(\newoverline{\delta_1\alpha}^{(1)} + \newoverline{\delta_1\alpha}^{(2)} + \newoverline{\delta_1\alpha}^{3}\right) + \alpha\left(\mathbf{r}\right)\frac{i}{d}\sin\left(\frac{k_1}{2}d\right)\sin\left(\frac{k_2}{2}d\right)\sin\left(\frac{k_3}{2}d\right)\nonumber\\ &\stackrel{\href{#eq:thuburn_op_deriv_0}{\text{Glg. (27.125)}}}{=} -\left(\newoverline{\delta_1\alpha}^{(1)} + \newoverline{\delta_1\alpha}^{(2)} + \newoverline{\delta_1\alpha}^{3}\right) + \frac{1}{2}\left(\newoverline{\delta_1\alpha}^{((1))} + \newoverline{\delta_2\alpha}^{((2))} + \newoverline{\delta_3\alpha}^{(3)}\right)\nonumber\\ &\Rightarrow\newoverline{\delta_1\alpha}^{((1))} + \newoverline{\delta_2\alpha}^{((2))} + \newoverline{\delta_3\alpha}^{(3)} = -\frac{1}{2}\left(\newoverline{\delta_1\alpha}^{(1)} + \newoverline{\delta_1\alpha}^{(2)} + \newoverline{\delta_1\alpha}^{3}\right).\tag{27.128}\label{eq:thuburn_op_deriv_1} \end{align} \]
You now make the approach
\[ \begin{align} \newtilde{\alpha}^{(j)} = \beta\newoverline{\alpha}^{(j)} + \left(1 - \beta\right)\newoverline{\alpha}^{((j))} \end{align} \]
with $0 \leq \beta \leq 0$. If you put Eq. (27.128), you get
\[ \begin{align} \newtilde{\delta_1\alpha}^{(1)} + \newtilde{\delta_2\alpha}^{(2)} + \newtilde{\delta_3\alpha}^{3} = \left(\beta - \frac{1}{2}\left(1 - \beta\right)\right)\left(\newoverline{\delta_1\alpha}^{(1)} + \newoverline{\delta_1\alpha}^{(2)} + \newoverline{\delta_1\alpha}^{3}\right) \stackrel{!}{=} 0. \end{align} \]
This is fulfilled for
\[ \begin{align} \beta - \frac{1}{2}\left(1 - \beta\right) &= 0 \Leftrightarrow \beta - \frac{1}{2} + \frac{\beta}{2} = \frac{3\beta}{2} - \frac{1}{2} = 0\nonumber\\ \Leftrightarrow \beta = \frac{1}{3}. \end{align} \]
Therefore applies
\[ \begin{align} \newtilde{\alpha}^{(j)} = \frac{1}{3}\newoverline{\alpha}^{(j)} + \frac{2}{3}\newoverline{\alpha}^{((j))}.\tag{27.132}\label{eq:thuburn_op} \end{align} \]
This operator is called the Thuburn operator because it was first used by Thuburn in [38].
The application of the Thuburn operator in the linearized shallow water equations on the hexagonal lattice equations (27.81) - (27.84) shows
\[ \begin{align} \frac{\partial\Phi}{\partial t} + \frac{2}{3}\Phi_0\left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right) &= 0,\\ \frac{\partial u_1}{\partial t} - \frac{f_0}{\sqrt{3}}\left(\newtilde{u}_2^{(3)} - \newtilde{u}_3^{(2)}\right) + \delta_1\Phi &= 0,\tag{27.134}\label{eq:x_1_momentum_thuburn_c-grid}\\ \frac{\partial u_2}{\partial t} - \frac{f_0}{\sqrt{3}}\left(\newtilde{u}_3^{(1)} - \newtilde{u}_1^{(3)}\right) + \delta_2\Phi &= 0,\\ \frac{\partial u_3}{\partial t} - \frac{f_0}{\sqrt{3}}\left(\newtilde{u}_1^{(2)} - \newtilde{u}_2^{(1)}\right) + \delta_3\Phi &= 0. \end{align} \]
From this it follows immediately
\[ \begin{align} \frac{\partial}{\partial t}\left(\newtilde{u}_1^{(1)} + \newtilde{u}_2^{(2)} + \newtilde{u}_3^{(3)}\right) = 0.\tag{27.137}\label{eq:thuburn_con_swe_c-grid} \end{align} \]
The dispersion relation of this system of equations can be calculated analogously to the derivation of Eq. (27.96) by doing the substitution
\[ \begin{align} c_1 \to a_1 \coloneqq\frac{c_1 + 2c_2c_3}{3}\text{ (und zyklisch)} \end{align} \]
carries out. This leads to
\[ \begin{align} \omega^4 - \omega^2\left[\frac{f_0^2}{3}\left(a_1^2 + a_2^2 + a_3^2\right) + \frac{8\Phi_0}{3d^2}\left(s_1^2 + s_2^2 + s_3^2\right)\right] + \frac{8\Phi_0f_0^2}{9d^2}\left(s_1a_1 + s_2a_2 + s_3a_3\right)^2 = 0.\tag{27.139}\label{eq:disp_rel_hex_c_mod_0} \end{align} \]
If one sets $\alpha = \exp\left(i\mathbf{k}\cdot\mathbf{r}\right)$ in Eq. (27.103), you get
\[ \begin{align} \frac{2i}{d}\left(a_1s_1 + a_2s_2 + a_3s_3\right)\alpha = 0 \Leftrightarrow a_1s_1 + a_2s_2 + a_3s_3 = 0. \end{align} \]
Thus, Eq. (27.139) to
\[ \begin{align} \omega^4 - \omega^2\left[\frac{f_0^2}{3}\left(a_1^2 + a_2^2 + a_3^2\right) + \frac{8\Phi_0}{3d^2}\left(s_1^2 + s_2^2 + s_3^2\right)\right] = 0.\tag{27.141}\label{eq:thuburn_leads_to_geostrophic_mode} \end{align} \]
If you use Eq. (27.132) to reconstruct the tangential velocity components, thus creating a stationary geostrophic mode.
On a regular hexagonal lattice with lattice constant $d$, a triangle has area
\[ \begin{align} A_v = \frac{d}{2}l. \end{align} \]
This applies
\[ \begin{align} l = \cos\left(30^\circ\right)d = \frac{\sqrt{3}}{2}d, \end{align} \]
from what
\[ \begin{align} A_v = \frac{d^2\sqrt{3}}{4}\tag{27.144}\label{eq:hex_face_triangle} \end{align} \]
follows. Thus, on a triangle (superscript $t$) that points upwards (index $u$), one can find the divergence according to Gauss's theorem in the form
\[ \begin{align} D^{(t)}_u = -\frac{dv_1 + dv_2 + dv_3}{A_v} = -\frac{4}{\sqrt{3}d}\left(v_1 + v_2 + v_3\right) \end{align} \]
calculate. The minus sign comes from the fact that in such a triangle the coordinate axes of the j-system point inwards. On a triangle that points downwards (index $l$), this is the other way around, it applies
\[ \begin{align} D^{(t)}_l = \frac{dv_1 + dv_2 + dv_3}{A_v} = \frac{4}{\sqrt{3}d}\left(v_1 + v_2 + v_3\right). \end{align} \]
This can be done in the form
\[ \begin{align} D^{(t)}_{u, l} = \mp\frac{dv_1 + dv_2 + dv_3}{A_v} = \mp\frac{4}{\sqrt{3}d}\left(v_1 + v_2 + v_3\right) = \mp\frac{4}{\sqrt{3}d}\sum_{e \in t}v_e \end{align} \]
summarize, where $e$ denotes the edges of the triangle. Analogously, for the relative vorticity according to Stokes' theorem applies
\[ \begin{align} \zeta^{(t)}_{u, l} = \pm\frac{du_1 + du_2 + du_3}{A_v} = \pm\frac{4}{\sqrt{3}d}\sum_{e \in t}u_e. \end{align} \]
Let $h$ now denote a hexagon. Then $u \in h$ denote the three upward-pointing triangles that overlap with the hexagon $h$ under consideration and similarly $l \in h$ denote the three downward-pointing triangles that overlap with $h$. With these names you can
\[ \begin{align} \frac{\sqrt{3}d}{4}\left(\sum_{l \in h}D_l^{(t)} - \sum_{u \in h}D_u^{(t)}\right) &= \sum_{l \in h}\sum_{e \in l}v_e + \sum_{u \in h}\sum_{e \in u}v_e = \sum_{l \in h}\left(\sum_{e \in l, e \in h}v_e + \sum_{e \in l, e \not\in h}v_e\right) + \sum_{u \in h}\left(\sum_{e \in u, e \in h}v_e + \sum_{e \in u, e \not\in h}v_e\right) \end{align} \]
note down. During the last transformation step, the edges of the triangles were divided into those that are also edges of the hexagon under consideration ($e \in h$) and those that are not ($e \not\in h$). Excluding results
\[ \begin{align} \frac{\sqrt{3}d}{4}\left(\sum_{l \in h}D_l^{(t)} - \sum_{u \in h}D_u^{(t)}\right) &= \sum_{l \in h}\sum_{e \in l, e \in h}v_e + \sum_{l \in h}\sum_{e \in l, e \not\in h}v_e + \sum_{u \in h}\sum_{e \in u, e \in h}v_e + \sum_{u \in h}\sum_{e \in u, e \not\in h}v_e. \end{align} \]
As can be seen, all velocity components with a positive sign contribute to $\sum_{l \in h}D_l^{(t)} - \sum_{u \in h}D_u^{(t)}$. In the sums $\sum _{l \in h}\sum _{e \in l, e \in h}$ and $\sum _{u \in h}\sum _{e \in u, e \in h}$, each edge of the hexagon $h$ occurs once. Therefore applies
\[ \begin{align} \frac{\sqrt{3}d}{4}\left(\sum_{l \in h}D_l^{(t)} - \sum_{u \in h}D_u^{(t)}\right) &= 2\sum_{e \in h}v_e + \sum_{l \in h}\sum_{e \in l, e \not\in h}v_e + \sum_{u \in h}\sum_{e \in u, e \not\in h}v_e. \end{align} \]
Those edges that do not border the hexagon under consideration each contribute once to $\sum _{l \in h}D_l^{(t)} - \sum _{u \in h}D_u^{(t)}$:
\[ \begin{align} \frac{\sqrt{3}d}{4}\left(\sum_{l \in h}D_l^{(t)} - \sum_{u \in h}D_u^{(t)}\right) &= 2\sum_{e \in h}v_e + \sum_{t \in h}\sum_{e \in t, e \not\in h}v_e \end{align} \]
At this point, consider the size $\newtilde{v}_1^{(1)}$, which is defined in the center of the hexagon under consideration. This is a superposition of four velocity components, two of which lie on the i-1 axis and two perpendicular to it. The former contribute with the weight $1/3$ to $\newtilde{v}_1^{(1)}$, the latter with the weight $1/6.$ This can also be transferred cyclically to the other spatial directions. Therefore applies
\[ \begin{align} \frac{\sqrt{3}d}{4}\left(\sum_{l \in h}D_l^{(t)} - \sum_{u \in h}D_u^{(t)}\right) &= 6\left(\newtilde{v}_1^{(1)} + \newtilde{v}_2^{(2)} + \newtilde{v}_3^{3}\right).\tag{27.153}\label{eq:checkerboard_triangular} \end{align} \]
If the condition $\newtilde{v}_1^{(1)} + \newtilde{v}_2^{(2)} + \newtilde{v}_3^{3} = 0$ is not met, then the mean of the divergence of the three upward-pointing triangles is i. A. not equal to the mean of the divergence of the downward triangles. This leads to the creation of a so-called Checkerboard pattern in the divergence. This noise propagates from there to other fields. It is therefore important to always adhere to the condition $\newtilde{v}_1^{(1)} + \newtilde{v}_2^{(2)} + \newtilde{v}_3^{3} = 0$.
On the hexagonal grid, the train of thought can be transferred completely analogously to the relative vorticity, which is based on
\[ \begin{align} \frac{\sqrt{3}d}{4}\left(\sum_{u \in h}\zeta_u^{(t)} - \sum_{l \in h}\zeta_l^{(t)}\right) &= 6\left(\newtilde{u}_1^{(1)} + \newtilde{u}_2^{(2)} + \newtilde{u}_3^{3}\right) \end{align} \]
leads. Failure to comply with the condition $\newtilde{u}_1^{(1)} + \newtilde{u}_2^{(2)} + \newtilde{u}_3^{3} = 0$ also leads to a checkerboard pattern on the hexagonal grid (but here in the vorticity) and must therefore also be taken into account here.
In the previous section it was shown that a checkerboard pattern exists on the hexagonal grid in the vorticity calculated on triangles if the condition $\newtilde{u}_1^{(1)} + \newtilde{u}_2^{(2)} + \newtilde{u}_3^{3} = 0$ for the wind field $\mathbf{v}$ is not met. However, in order for this condition for the wind field to be fulfilled for all time steps, this must apply to the wind field in the initial state as well as to all forcings $\mathbf{w}$. However, simulations are usually carried out with momentum diffusion to avoid a build-up of kinetic energy on the grid scale. These diffusive terms break down gradients on the grid scale and can therefore also break down an existing checkerboard pattern as long as it is not too strong. Accordingly, the requirement $\newtilde{u}_1^{(1)} + \newtilde{u}_2^{(2)} + \newtilde{u}_3^{3} = 0$ for the initial wind field $\mathbf{v}$ can be dispensed with. However, the following is important:
For all velocity tendencies $\mathbf{w}$ $\newtilde{w}_1^{(1)} + \newtilde{w}_2^{(2)} + \newtilde{w}_3^{3} = 0$ must hold.
The analogous facts apply on the triangular grid.
If you apply the Laplace operator to equations (27.116) - (27.118), you get
\[ \begin{align} \Delta u_1 &= \frac{1}{\sqrt{3}}\left(\delta_3\Delta\newtilde{\psi}^{(2)} - \delta_2\Delta \newtilde{\psi}^{3}\right) + \delta_1\Delta\chi,\tag{27.155}\label{eq:rhombus_curl_deriv_0}\\ \Delta u_2 &= \frac{1}{\sqrt{3}}\left(\delta_1\Delta\newtilde{\psi}^{3} - \delta_3\Delta \newtilde{\psi}^{(1)}\right) + \delta_2\Delta\chi,\tag{27.156}\label{eq:rhombus_curl_deriv_1}\\ \Delta u_3 &= \frac{1}{\sqrt{3}}\left(\delta_2\Delta\newtilde{\psi}^{(1)} - \delta_1\Delta \newtilde{\psi}^{(2)}\right) + \delta_3\Delta\chi.\tag{27.157}\label{eq:rhombus_curl_deriv_2} \end{align} \]
Since discretized partial derivatives and averaging operators swap, it follows immediately
\[ \begin{align} \newtilde{\Delta u_1}^{(1)} + \newtilde{\Delta u_2}^{(2)} + \newtilde{\Delta u_3}^{(3)} = 0. \end{align} \]
If you discretize the equations (27.62) - (27.64), you get
\[ \begin{align} \Delta u_1 &= \frac{1}{\sqrt{3}}\left(\delta_3\zeta - \delta_2\zeta\right) + \delta_1 D,\tag{27.159}\label{eq:rhombus_curl_deriv_3}\\ \Delta u_2 &= \frac{1}{\sqrt{3}}\left(\delta_1\zeta - \delta_3\zeta\right) + \delta_2 D,\tag{27.160}\label{eq:rhombus_curl_deriv_4}\\ \Delta u_3 &= \frac{1}{\sqrt{3}}\left(\delta_2\zeta - \delta_1\zeta\right) + \delta_3 D.\tag{27.161}\label{eq:rhombus_curl_deriv_5} \end{align} \]
By comparison, one finds that there are three types of vorticities in the $\mathbf{i}$ system:
\[ \begin{align} \zeta_1 \coloneqq \Delta\newtilde{\psi}^{(1)}, & {} & \zeta_2 \coloneqq \Delta\newtilde{\psi}^{(2)}, & {} & \zeta_3 \coloneqq \Delta\newtilde{\psi}^{3} \end{align} \]
It therefore applies
\[ \begin{align} \Delta u_1 &= \frac{1}{\sqrt{3}}\left(\delta_3\zeta_2 - \delta_2\zeta_3\right) + \delta_1 D,\tag{27.163}\label{eq:rhombus_curl_deriv_6}\\ \Delta u_2 &= \frac{1}{\sqrt{3}}\left(\delta_1\zeta_3 - \delta_3\zeta_1\right) + \delta_2 D,\tag{27.164}\label{eq:rhombus_curl_deriv_7}\\ \Delta u_3 &= \frac{1}{\sqrt{3}}\left(\delta_2\zeta_1 - \delta_1\zeta_2\right) + \delta_3 D.\tag{27.165}\label{eq:rhombus_curl_deriv_8} \end{align} \]
For the divergence, according to Eq. (27.33)
\[ \begin{align} D &= \frac{2}{3}\left(\delta_1u_1 + \delta_2u_2 + \delta_3u_3\right). \end{align} \]
If you insert this into the equations (27.163) - (27.165), you get
\[ \begin{align} \Delta u_1 &= \frac{1}{\sqrt{3}}\left(\delta_3\zeta_2 - \delta_2\zeta_3\right) + \frac{2}{3}\left(\delta_1\delta_1u_1 + \delta_1\delta_2u_2 + \delta_1\delta_3u_3\right),\tag{27.167}\label{eq:hex_laplace_vec_0}\\ \Delta u_2 &= \frac{1}{\sqrt{3}}\left(\delta_1\zeta_3 - \delta_3\zeta_1\right) + \frac{2}{3}\left(\delta_2\delta_1u_1 + \delta_2\delta_2u_2 + \delta_2\delta_3u_3\right),\tag{27.168}\label{eq:hex_laplace_vec_1}\\ \Delta u_3 &= \frac{1}{\sqrt{3}}\left(\delta_2\zeta_1 - \delta_1\zeta_2\right) + \frac{2}{3}\left(\delta_3\delta_1u_1 + \delta_3\delta_2u_2 + \delta_3\delta_3u_3\right).\tag{27.169}\label{eq:hex_laplace_vec_2} \end{align} \]
For the vorticities $\zeta_i$, the approaches are made, motivated by the equations (27.49) - (27.51)
\[ \begin{align} \zeta_1 &= \frac{2}{\sqrt{3}}\left(\delta_2u_3 - \delta_3u_2\right),\tag{27.170}\label{eq:hex_curl_rhombus_0}\\ \zeta_2 &= \frac{2}{\sqrt{3}}\left(\delta_3u_1 - \delta_1u_3\right),\tag{27.171}\label{eq:hex_curl_rhombus_1}\\ \zeta_3 &= \frac{2}{\sqrt{3}}\left(\delta_1u_2 - \delta_2u_1\right).\tag{27.172}\label{eq:hex_curl_rhombus_2} \end{align} \]
If you put Eq. (27.170) in Eq. (27.167), you get
\[ \begin{align} \Delta u_1 &= \frac{2}{3}\left(\delta_3\delta_3u_1 - \delta_3\delta_1u_3 - \delta_2\delta_1u_2 + \delta_2\delta_2u_1\right) + \frac{2}{3}\left(\delta_1\delta_1u_1 + \delta_1\delta_2u_2 + \delta_1\delta_3u_3\right) = \frac{2}{3}\left(\delta_1\delta_1u_1 + \delta_2\delta_2u_1 + \delta_3\delta_3u_1\right). \end{align} \]
By cyclically swapping, the analogous statement is also obtained for the other components. The approaches of the equations (27.170) - (27.172) are therefore justified. Two neighboring triangles together form a rhombus (a parallelogram), according to Eq. (27.144) the area
\[ \begin{align} A_r = 2A_v = 2\frac{d^2\sqrt{3}}{4} = \frac{d^2\sqrt{3}}{2}. \end{align} \]
Thus, $\zeta_i$ corresponds to the vorticity calculated using Stokes' theorem over a rhombus whose short axis of symmetry is parallel to the i-axis. On the hexagonal grid, the rhombuses form the natural sets on which Stokes' theorem must be evaluated. This is a deviation from the actually intuitive assumption that the vorticity must be calculated via the dual grid (the triangular grid).
As usual, divergence and vorticity appear in reversed roles on the triangular grid. The vorticity has the clear value there
\[ \begin{align} \zeta = \frac{2}{3}\left(\delta_1v_1 + \delta_2v_2 + \delta_3v_3\right) \end{align} \]
However, three types of divergences occur there, each of which is calculated using Gauss's theorem over the respective rhombuses:
\[ \begin{align} D_1 &= \frac{2}{\sqrt{3}}\left(\delta_3v_2 - \delta_2v_3\right)\\ D_2 &= \frac{2}{\sqrt{3}}\left(\delta_1v_3 - \delta_3v_1\right)\\ D_3 &= \frac{2}{\sqrt{3}}\left(\delta_2v_1 - \delta_1v_2\right) \end{align} \]
This is also achieved by the discretization of the vector Laplace operator according to Eq. (B.54) underlined which
\[ \begin{align} \Delta\mathbf{v} = \nabla\left(\nabla\cdot\mathbf{v}\right) - \nabla\times\left(\nabla\times\mathbf{v}\right) \end{align} \]
is. Applying this to the vector component $u_1$ gives
\[ \begin{align} \Delta u_1 = \delta_1D_h - \delta_{i, \perp}\zeta_t. \end{align} \]
Here $D_h$ is the divergence on hexagons and $\zeta_t$ is the vorticity on triangles. $\delta_{i, \perp}$ denotes the central difference quotient in the $\mathbf{j}_1-$direction. This applies to this one
\[ \begin{align} -\delta_{i, \perp}\zeta_t &= \frac{\sqrt{3}}{d}\left(\zeta_{t, l} - \zeta_{t, u}\right). \end{align} \]
From Eq. (27.163) follows
\[ \begin{align} \Delta u_1 &= \delta_1D_h + \frac{\sqrt{3}}{d}\left(\zeta_{t, l} - \zeta_{t, u}\right) = \delta_1 D_h + \frac{1}{\sqrt{3}}\left(\delta_3\zeta_2 - \delta_2\zeta_3\right)\nonumber\\ & \Leftrightarrow \sqrt{3}\frac{\zeta_{t, l} - \zeta_{t, u}}{d} = \frac{\sqrt{3}}{3}\left(\delta_3\zeta_2 - \delta_2\zeta_3\right) \Leftrightarrow \sqrt{3}\frac{\zeta_{t, l} - \zeta_{t, u}}{d} = \frac{\sqrt{3}}{3}\frac{\zeta_{2, l} - \zeta_{2, u} - \zeta_{3, u} + \zeta_{3, l}}{d}\nonumber\\ & \Leftrightarrow \sqrt{3}\frac{\zeta_{t, l} - \zeta_{t, u}}{d} = \frac{\sqrt{3}}{3}\frac{\zeta_{2, l} + \zeta_{3, l} - \left(\zeta_{2, u} - \zeta_{3, u}\right)}{d}\nonumber\\ & \Leftrightarrow \sqrt{3}\frac{\zeta_{t, l} - \zeta_{t, u}}{d} = \frac{\sqrt{3}}{3}\frac{\zeta_{1, l} + \zeta_{2, l} + \zeta_{3, l} - \left(\zeta_{1, u} + \zeta_{2, u} - \zeta_{3, u}\right)}{d}. \end{align} \]
The final equivalent transformation follows from the fact $\zeta _{1, l} = \zeta _{1, u}$. This is with the definitions
\[ \begin{align} \zeta_{t, u} &\coloneqq \frac{\zeta_{1, u} + \zeta_{2, u} + \zeta_{3, u}}{3},\\ \zeta_{t, l} &\coloneqq \frac{\zeta_{1, l} + \zeta_{2, l} + \zeta_{3, l}}{3} \end{align} \]
fulfilled. The rotation on a triangle can therefore be calculated as the average of the three rhombuses that overlap with this triangle.
On the hexagonal C-lattice, gradient fields $\mathbf{u} = \nabla\psi$ satisfy the condition
\[ \begin{align} \newtilde{u}_1^{(1)} + \newtilde{u}_2^{(2)} + \newtilde{u}_3^{3} = 0. \end{align} \]
On the triangular C-lattice, a gradient field $\mathbf{v} = \nabla\psi$ naturally has components parallel to the basis elements $\mathbf{j}_i$. This fulfills
\[ \begin{align} \newtilde{v}_1^{(1)} + \newtilde{v}_2^{(2)} + \newtilde{v}_3^{3} = 0 \end{align} \]
not, as one can easily see. Therefore, gradient fields on the triangular C-lattice produce according to Eq. (27.153) a checkerboard pattern in divergence, which would require unphysical numerical stabilizers such as divergence damping. Therefore, this grid was rightly excluded in Section 26.9.1. This finding was first put forward by Gassmann in [24].