Entropy Stable Schemes for the Shear Shallow Water Model Equations

Abstract

The shear shallow water model is an extension of the classical shallow water model to include the effects of vertical shear. It is a system of six non-linear hyperbolic PDE with non-conservative products. We develop a high-order entropy stable finite difference scheme for this model in one dimension and extend it to two dimensions on rectangular grids. The key idea is to rewrite the system so that non-conservative terms do not contribute to the entropy evolution. Then, we first develop an entropy conservative scheme for the conservative part, which is then extended to the complete system using the fact that the non-conservative terms do not contribute to the entropy production. The entropy dissipative scheme, which leads to an entropy inequality, is then obtained by carefully adding dissipative flux terms. The proposed schemes are then tested on several one and two-dimensional problems to demonstrate their stability and accuracy.

Appendices

A A Note on Non-symmetrizability of Shear Shallow Water Model

In this section, we will discuss the symmetrizability of the following SSW model in one dimension, i.e., we consider,

$$\begin{aligned} \frac{\partial \varvec{U}}{\partial t}+\frac{\partial \varvec{F}^x(\varvec{U})}{\partial x}+{\tilde{\varvec{B}^x}}(\varvec{U})\frac{\partial \varvec{U}}{\partial x}=0, \end{aligned}$$

(22)

where $\varvec{U}, \varvec{F}^x$ and ${\tilde{\varvec{B}^x}}$ are defined in Sect. 3. Additionally, this system has the entropy pair $(\eta ,q)$ (2), such that in addition to (1) the following equality holds,

$$\begin{aligned} \frac{\partial \eta }{\partial t}+\frac{\partial q}{\partial x}=0. \end{aligned}$$

for smooth solutions. For detailed proof, refer to Lemma 1. In the standard symmetrization theory [19, 21, 24, 27], one seeks a change of variable $\textbf{U} \rightarrow \mathbf {\varvec{V}}$ applied to (22) so that when transformed

$$\begin{aligned} \frac{\partial \textbf{U}}{\partial \varvec{V}}\frac{\partial \varvec{V}}{\partial t}+\bigg (\frac{\partial \varvec{F}^x}{\partial \textbf{U}}+{\tilde{\varvec{B}^x}}\bigg )\frac{\partial \textbf{U}}{\partial \varvec{V}}\frac{\partial \varvec{V}}{\partial x}=0, \end{aligned}$$

the matrix $\frac{\partial \textbf{U}}{\partial \varvec{V}}$ is symmetric, positive definite and the matrix ${\tilde{A}}_1 = \bigg (\frac{\partial \varvec{F}^x}{\partial \textbf{U}}+{\tilde{\varvec{B}^x}}\bigg )\frac{\partial \textbf{U}}{\partial \varvec{V}}$ is symmetric. For the SSW system (22), we calculate the matrix ${\tilde{A}}_1$ to check it’s symmetry, which yields

$$\begin{aligned} {\tilde{A}}_1 - {\tilde{A}}_1^\top = \begin{pmatrix} 0 &{} -\alpha &{} 0 &{} -\alpha v_1 &{} -\frac{1}{2} \alpha v_2 &{} 0 \\ \alpha &{} 0 &{} \alpha v_2 &{} -\beta _1 &{} \frac{1}{2} \alpha \mathcal {P}_{12} &{} \beta _2 \\ 0 &{} -\alpha v_2 &{} 0 &{} -\alpha v_1 v_2 &{} -\frac{1}{2} \alpha v_2^2 &{} 0 \\ \alpha v_1 &{} \beta _1 &{} \alpha v_1 v_2 &{} 0 &{} a &{} v_1 \beta _2 \\ \frac{1}{2}\alpha v_2 &{} -\frac{1}{2} \alpha \mathcal {P}_{12} &{} \frac{1}{2} \alpha v_2^2 &{} -a &{} 0 &{} \frac{1}{2}v_2 \beta _2\\ 0 &{} -\beta _2 &{} 0 &{} -\beta _2 &{} -\frac{1}{2} v_2 \beta _2 &{} 0 \end{pmatrix}. \end{aligned}$$

where

$$\begin{aligned} \alpha= & {} \frac{gh^2}{2}, \quad \beta _1=\frac{1}{4} g h^2 \left( v_1^2-\mathcal {P}_{11}\right) ,\quad \beta _2=\frac{1}{4} g h^2 \left( v_2^2+\mathcal {P}_{22}\right) \\ a= & {} \frac{1}{8} g h^2 \left( 2 \mathcal {P}_{12} v_1+\left( v_1^2-\mathcal {P}_{11}\right) v_2\right) \end{aligned}$$

Hence, ${\tilde{A}}_1$ is not a symmetric matrix unless $g=0$, in which case the non-conservative terms vanish from the SSW model. Furthermore, we recall the following result presented in [18] which gives the necessary and sufficient condition for a non-linear system of conservation laws to admit a strictly convex entropy.

Theorem 3

A necessary and sufficient condition for the conservative system,

$$\begin{aligned} \frac{\partial \varvec{U}}{\partial t} + \frac{\partial \varvec{F}^x}{\partial x}=0, \end{aligned}$$

(23)

to posses a strictly convex entropy $\eta$ is that there exists a change of dependent variables $\varvec{U}=\varvec{U}(\varvec{V})$ that symmetrizes (23).

Analogously, we extend the above result for the case of non-conservative hyperbolic systems of the form (22).

Proposition 2

If $\eta$ is a strictly convex entropy for the non-conservative system of the form (22) and $\eta {'}(\varvec{U}){\tilde{\varvec{B}^x}}(\varvec{U})=0$, then the change of variable $\varvec{U}\rightarrow \varvec{V}$ with $\varvec{V}^\top = \eta {'}(\varvec{U})$ symmetrizes the non-conservative system if and only if ${\tilde{\varvec{B}^x}}(\varvec{U})\varvec{U}{'}(\varvec{V})$ is symmetric.

Proof

Define the conjugate functions

$$\begin{aligned} \eta ^{*}(\varvec{V})=\varvec{V}^\top \varvec{U}(\varvec{V})-\eta (\varvec{U}(\varvec{V})), \qquad q^{*}(\varvec{V})=\varvec{V}^\top \varvec{F}(\varvec{U}(\varvec{V}))-q(\varvec{U}(\varvec{V})). \end{aligned}$$

Differentiating with respect to $\varvec{V}$ gives,

$$\begin{aligned} {\eta ^{*}}^{'}(\varvec{V})=\varvec{U}(\varvec{V})^\top -\varvec{V}^\top \varvec{U}'(\varvec{V})-\eta '(\varvec{U}(\varvec{V}))\varvec{U}'(\varvec{V})=\varvec{U}(\varvec{V})^\top , \end{aligned}$$

and

$$\begin{aligned} {q^{*}}^{'}(\varvec{V})&=\varvec{F}(\varvec{U}(\varvec{V}))^\top +\varvec{V}^\top \varvec{F}'(\varvec{U}(\varvec{V}))\varvec{U}'(\varvec{V})-q'(\varvec{U}(\varvec{V}))\varvec{U}'(\varvec{V})&\\&=\varvec{F}(\varvec{U}(\varvec{V}))^\top +[\varvec{V}^\top \varvec{F}'(\varvec{U}(\varvec{V}))-q'(\varvec{U}(\varvec{V}))]\varvec{U}'(\varvec{V})&\\&=\varvec{F}(\varvec{U}(\varvec{V}))^\top \end{aligned}$$

since,

$$\begin{aligned} \varvec{V}^\top \varvec{F}'(\varvec{U}(\varvec{V}))-q'(\varvec{U}(\varvec{V}))=0. \end{aligned}$$

Hence, the matrices $\varvec{U}'(\varvec{V})={\eta ^{*}}^{''}(\varvec{V})$ and $\varvec{F}'(\varvec{U}(\varvec{V}))\varvec{U}'(\varvec{V})={q^{*}}^{''}(\varvec{V})$ are symmetric. Moreover, the matrix $\varvec{U}'(\varvec{V})=\eta {''}(\varvec{U}(\varvec{V}))^{-1}$ is positive definite.

The change of variable yields

$$\begin{aligned} \varvec{U}'(\varvec{V})\varvec{V}_t+[\varvec{F}'(\varvec{U}(\varvec{V})) + {\tilde{\varvec{B}^x}}(\varvec{U}(\varvec{V}))]\varvec{U}'(\varvec{V})\varvec{V}_x=0 \end{aligned}$$

We need $[\varvec{F}'(\varvec{U}(\varvec{V}))+{\tilde{\varvec{B}^x}}(\varvec{U}(\varvec{V}))]\varvec{U}'(\varvec{V})$ to be symmetric, since, $\varvec{F}'(\varvec{U}(\varvec{V}))\varvec{U}'(\varvec{V})$ is symmetric we need ${\tilde{\varvec{B}^x}}(\varvec{U}(\varvec{V}))\varvec{U}'(\varvec{V})$ to be symmetric. $\square$

Remark 3

The matrix ${\tilde{\varvec{B}^x}}(\varvec{U}(\varvec{V}))\varvec{U}'(\varvec{V})$ for system (1) is not symmetric, since,

$$\begin{aligned} {\tilde{\varvec{B}^x}}(\textbf{U}({\textbf {V}}))\textbf{U}'({\textbf {V}}) - \left[ {\tilde{\varvec{B}^x}}(\textbf{U}({\textbf {V}}))\textbf{U}'({\textbf {V}}) \right] ^\top = \begin{pmatrix} 0 &{} -\alpha &{} 0 &{} -\alpha v_1 &{} -\frac{1}{2} \alpha v_2 &{} 0 \\ \alpha &{} 0 &{} \alpha v_2 &{} -\beta _1 &{} \frac{1}{2} \alpha \mathcal {P}_{12} &{} \beta _2 \\ 0 &{} -\alpha v_2 &{} 0 &{} -\alpha v_1 v_2 &{} -\frac{1}{2} \alpha v_2^2 &{} 0 \\ \alpha v_1 &{} \beta _1 &{} \alpha v_1 v_2 &{} 0 &{} a &{} v_1 \beta _2 \\ \frac{1}{2}\alpha v_2 &{} -\frac{1}{2} \alpha \mathcal {P}_{12} &{} \frac{1}{2} \alpha v_2^2 &{} -a &{} 0 &{} \frac{1}{2}v_2 \beta _2\\ 0 &{} -\beta _2 &{} 0 &{} -\beta _2 &{} -\frac{1}{2} v_2 \beta _2 &{} 0 \end{pmatrix}, \end{aligned}$$

where $\alpha =\frac{gh^2}{2},~\beta _1=\frac{1}{4} g h^2 \left( v_1^2-\mathcal {P}_{11}\right) ,~\beta _2=\frac{1}{4} g h^2 \left( v_2^2+\mathcal {P}_{22}\right) ,$

$a=\frac{1}{8} g h^2 \left( 2 \mathcal {P}_{12} v_1+\left( v_1^2-\mathcal {P}_{11}\right) v_2\right)$.

The above matrix is identical to ${\tilde{A}}_1 - {\tilde{A}}_1^\top$ which we derived explicitly and shown above.

From the above discussion, we observe that the existence of entropy pair does not guarantee the symmetrizability of the non-conservative hyperbolic systems. In particular, we have seen that the shear shallow water model has the entropy pair $(\eta ,q)$ but it is not symmetrizable.

B Entropy Scaled Right Eigenvectors for Shear Shallow Water Model

In this section, we will calculate the entropy scaled right eigenvectors for the case of x-direction. Consider the conservative part of the SSW system (1),

$$\begin{aligned} \frac{\partial \varvec{U}}{\partial t} + \frac{\partial \varvec{F}^x}{\partial x} =\frac{\partial \varvec{U}}{\partial t} + A_1\frac{\partial \varvec{U}}{\partial x} = 0, \end{aligned}$$

(24)

where $A_1$ is jacobian matrix of the flux function $\varvec{F}^x$. To derive the eigenvalues and right eigenvectors, it is useful to transform the system (24) in terms of the primitive variables $\varvec{W}$. The eigenvalues of the jacobian matrix $A_1$ [6, 30] are given by,

$$\begin{aligned} v_1-\sqrt{3\mathcal {P}_{11}},\quad v_1-\sqrt{\mathcal {P}_{11}}, \quad v_1,\quad v_1,\quad v_1+\sqrt{\mathcal {P}_{11}}, \quad v_1+\sqrt{3\mathcal {P}_{11}}. \end{aligned}$$

We observe that if $\mathcal {P}_{11}>0$ then all eigenvalues are real. The right eigenvector matrix ${R}^x$ for the matrix $A_1$ is given by the relation

$$\begin{aligned} {R}^x=\dfrac{\partial \varvec{U}}{\partial \varvec{W}}{R}_{\varvec{W}}^x, \end{aligned}$$

where $\dfrac{\partial \varvec{U}}{\partial \varvec{W}}$ is the jacobian matrix for the change of variable, given by

$$\begin{aligned} \dfrac{\partial \varvec{U}}{\partial \varvec{W}}=\begin{pmatrix} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ v_1 &{} h &{} 0 &{} 0 &{} 0 &{} 0 \\ v_2 &{} 0 &{} h &{} 0 &{} 0 &{} 0 \\ \frac{1}{2}(\mathcal {P}_{11}+v_1^2) &{} h v_1 &{} 0 &{} \frac{h}{2} &{} 0 &{} 0 \\ \frac{1}{2}(\mathcal {P}_{12}+v_1 v_2) &{} \frac{h v_2}{2} &{} \frac{h v_1}{2} &{} 0 &{} \frac{h}{2} &{} 0 \\ \frac{1}{2}(\mathcal {P}_{22}+v_2^2) &{} 0 &{} h v_2 &{} 0 &{} 0 &{} \frac{h}{2} \end{pmatrix}, \end{aligned}$$

and the matrix $R^x_{\varvec{W}}$ is given by

$$\begin{aligned} R_{\varvec{W}}^x=\begin{pmatrix} h \mathcal {P}_{11} &{} 0 &{} -h &{} 0 &{} 0 &{} h \mathcal {P}_{11} \\ -\sqrt{3\mathcal {P}_{11}}\mathcal {P}_{11} &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{3\mathcal {P}_{11}}\mathcal {P}_{11} \\ -\sqrt{3\mathcal {P}_{11}}\mathcal {P}_{12} &{} -\sqrt{\mathcal {P}_{11}} &{} 0 &{} 0 &{} \sqrt{\mathcal {P}_{11}} &{} \sqrt{3\mathcal {P}_{11}}\mathcal {P}_{12} \\ 2\mathcal {P}_{11}^2 &{} 0 &{} \mathcal {P}_{11} &{} 0 &{} 0 &{} 2\mathcal {P}_{11}^2 \\ 2\mathcal {P}_{11}\mathcal {P}_{12} &{} \mathcal {P}_{11} &{} \mathcal {P}_{12} &{} 0 &{} \mathcal {P}_{11} &{} 2\mathcal {P}_{11}\mathcal {P}_{12} \\ 2\mathcal {P}_{12}^2 &{} 2\mathcal {P}_{12} &{} 0 &{} 1 &{} 2 \mathcal {P}_{12} &{} 2\mathcal {P}_{12}^2 \end{pmatrix}. \end{aligned}$$

We need to find a scaling matrix $T^x$ such that the scaled right eigenvector matrix $\tilde{R}^x=R^x T^x$ satisfies

$$\begin{aligned} \frac{\partial \varvec{U}}{\partial \varvec{V}} ={{\tilde{R}}^x} {{}{{\tilde{R}}^x}}^\top . \end{aligned}$$

(25)

where $\varvec{V}$ is the entropy variable vector as in Eq. (10). We follow Barth scaling process [2] to scale the right eigenvectors. The scaling matrix $T^x$ is the square root of $Y^x$ where $Y^x$ has the expression

$$\begin{aligned} {Y}^x= \left( \tilde{R}^x_{\varvec{W}} \right) ^{-1} \frac{\partial \varvec{W}}{\partial \varvec{V}} \left( \frac{\partial \varvec{U}}{\partial \varvec{W}}\right) ^{-\top } \left( \tilde{R}^x_{\varvec{W}}\right) ^{-\top }, \end{aligned}$$

which results in

$$\begin{aligned} Y^x=\begin{pmatrix} \frac{1}{12 h \mathcal {P}_{11}^2} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} \frac{\mathcal {P}_{11} \mathcal {P}_{22}-\mathcal {P}_{12}^2}{4 h \mathcal {P}_{11}^2} &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{1}{3h} &{} \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{11}} &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{11}} &{} \frac{3(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)^2+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{11}^2} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} \frac{\mathcal {P}_{11} \mathcal {P}_{22}-\mathcal {P}_{12}^2}{4 h \mathcal {P}_{11}^2} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \frac{1}{12 h \mathcal {P}_{11}^2} \end{pmatrix}. \end{aligned}$$

The matrix $Y^x$ is a block diagonal matrix which contains the blocks of order 1 and 2. It is straightforward to write the square root of a block matrix of order 1. Consider the $2\times 2$ block sub-matrix of the matrix $Y^x$ and denote it by $Y^{x}_b$,

$$\begin{aligned} Y^{x}_b=\begin{pmatrix} \frac{1}{3h} &{} \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{11}} \\ \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{11}} &{} \frac{3(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)^2+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{11}^2} \end{pmatrix}. \end{aligned}$$

We need to find matrix $T^{x}_b=\sqrt{Y^{x}_b}$. To obtain formula for the matrix $T^x_b$ we first consider the characteristic polynomial of $T^{x}_b$,

$$\begin{aligned} {T^{x}_b}^{2}-{{\,\textrm{trace}\,}}(T^{x}_b)T^{x}_b+\det (T^{x}_b)I=0, \end{aligned}$$

(26)

where $\det (T^{x}_b)=\pm \sqrt{det(Y^{x}_b)}=r_1$, say, with $r_1$ being the positive square root, given by

$$\begin{aligned} r_1&=\sqrt{\frac{3(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)^2+\mathcal {P}_{12}^4}{9 h^2 \mathcal {P}_{11}^2}-\frac{\mathcal {P}_{12}^4}{9 h^2 \mathcal {P}_{11}^2}}&\\&=\sqrt{\frac{3(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)^2}{9 h^2 \mathcal {P}_{11}^2}}&\\&=\frac{\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2}{\sqrt{3}h \mathcal {P}_{11}}, \end{aligned}$$

and $I_{2\times 2}$ is the identity matrix. Observe from Eq. (26) that

$$\begin{aligned} {{\,\textrm{trace}\,}}(T^{x}_b)T^{x}_b={T^{x}_b}^2+r_1I=Y^{x}_b+r_1I, \end{aligned}$$

(27)

and,

$$\begin{aligned} ({{\,\textrm{trace}\,}}(T^{x}_b))^2={{\,\textrm{trace}\,}}({{\,\textrm{trace}\,}}(T^{x}_b)T^{x}_b)={{\,\textrm{trace}\,}}(Y^{x}_b+r_1I)={{\,\textrm{trace}\,}}(Y^{x}_b)+2r_1. \end{aligned}$$

Simultaneously solving Eqs. (26), (27) we obtain

$$\begin{aligned} T^{x}_b=\frac{1}{\sqrt{{{\,\textrm{trace}\,}}(Y^{x}_b)+2r_1}}(Y^{x}_b+r_1I). \end{aligned}$$

(28)

Observe that

$$\begin{aligned} {{{\,\textrm{trace}\,}}(Y^{x}_b)+2r_1}&=\frac{1}{3h}+\frac{3(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)^2+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{11}^2}+\frac{2(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)}{\sqrt{3}h \mathcal {P}_{11}} \\&=\frac{\mathcal {P}_{11}^2+{3(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)^2+\mathcal {P}_{12}^4}+2\sqrt{3}\mathcal {P}_{11}(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)}{3 h \mathcal {P}_{11}^2} \\&=\frac{3(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)^2+2\sqrt{3}\mathcal {P}_{11}(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)+\mathcal {P}_{11}^2+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{11}^2}&\\&=\frac{(\sqrt{3}(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)+\mathcal {P}_{11})^2+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{11}^2}. \end{aligned}$$

Since $h>0$, we have ${{\,\textrm{trace}\,}}(Y^{x}_b)+2r_1>0$. We use notation $\alpha _1 = \sqrt{{{\,\textrm{trace}\,}}(Y^{x}_b)+2r_1}$. A long simplification using the block matrix $T^x_b$ as in Eq. (28) results in

$$\begin{aligned} T^x=\begin{pmatrix} \sqrt{\frac{1}{12 h \mathcal {P}_{11}^2}} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} \sqrt{\frac{\mathcal {P}_{11} \mathcal {P}_{22}-\mathcal {P}_{12}^2}{4 h \mathcal {P}_{11}^2}} &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{\frac{1}{3h}+r_1}{\alpha _1} &{} \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{11}\alpha _1} &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{11}\alpha _1} &{} \frac{\beta _1(\beta _1+\mathcal {P}_{11})+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{11}^2\alpha _1} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{\mathcal {P}_{11} \mathcal {P}_{22}-\mathcal {P}_{12}^2}{4 h \mathcal {P}_{11}^2}} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{1}{12 h \mathcal {P}_{11}^2}} \end{pmatrix}. \end{aligned}$$

Remark 4

We proceed similarly in the y-direction, the eigenvalues for the jacobian matrix $A_2=\frac{\partial \varvec{F}_2}{\partial \varvec{U}}$ are given by

$$\begin{aligned} v_2-\sqrt{3\mathcal {P}_{22}},\quad v_2-\sqrt{\mathcal {P}_{22}},\quad v_2,\quad v_2,\quad v_2+\sqrt{\mathcal {P}_{22}},\quad v_2+\sqrt{3\mathcal {P}_{22}}. \end{aligned}$$

and right eigenvector matrix is given by the relation

$$\begin{aligned} {R}^y=\dfrac{\partial \varvec{U}}{\partial \varvec{W}}{R}_{\varvec{W}}^y, \end{aligned}$$

where

$$\begin{aligned} R_{\varvec{W}}^y=\begin{pmatrix} h \mathcal {P}_{22} &{} 0 &{} -h &{} 0 &{} 0 &{} h \mathcal {P}_{22} \\ -\sqrt{3\mathcal {P}_{22}}\mathcal {P}_{12} &{} -\sqrt{\mathcal {P}_{22}} &{} 0 &{} 0 &{} \sqrt{\mathcal {P}_{22}} &{} \sqrt{3\mathcal {P}_{22}}\mathcal {P}_{12} \\ -\sqrt{3\mathcal {P}_{22}}\mathcal {P}_{22} &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{3\mathcal {P}_{22}}\mathcal {P}_{22} \\ 2\mathcal {P}_{12}^2 &{} 2 \mathcal {P}_{12} &{} 0 &{} 1 &{} 2\mathcal {P}_{12} &{} 2\mathcal {P}_{12}^2 \\ 2\mathcal {P}_{22}\mathcal {P}_{12} &{} \mathcal {P}_{22} &{} \mathcal {P}_{12} &{} 0 &{} \mathcal {P}_{22} &{} 2\mathcal {P}_{22}\mathcal {P}_{12} \\ 2\mathcal {P}_{22}^2 &{} 0 &{} \mathcal {P}_{22} &{} 0 &{} 0 &{} 2\mathcal {P}_{22}^2 \end{pmatrix}. \end{aligned}$$

Accordingly, we obtain the scaling matrix $T^y$ as,

$$\begin{aligned} T^y=\begin{pmatrix} \sqrt{\frac{1}{12 h \mathcal {P}_{22}^2}} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} \sqrt{\frac{\mathcal {P}_{11} \mathcal {P}_{22}-\mathcal {P}_{12}^2}{4 h \mathcal {P}_{22}^2}} &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{\frac{1}{3h}+r_2}{\alpha _2} &{} \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{22}\alpha _2} &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{\mathcal {P}_{12}^2}{3 h \mathcal {P}_{22}\alpha _2} &{} \frac{\beta _1(\beta _1+\mathcal {P}_{22})+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{22}^2\alpha _2} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{\mathcal {P}_{11} \mathcal {P}_{22}-\mathcal {P}_{12}^2}{4 h \mathcal {P}_{22}^2}} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{1}{12 h \mathcal {P}_{22}^2}} \end{pmatrix}, \end{aligned}$$

where $r_2=\frac{\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2}{\sqrt{3}h \mathcal {P}_{22}}$ and $\alpha _2=\sqrt{\frac{(\sqrt{3}(\mathcal {P}_{11}\mathcal {P}_{22}-\mathcal {P}_{12}^2)+\mathcal {P}_{22})^2+\mathcal {P}_{12}^4}{3 h \mathcal {P}_{22}^2}}$.