Entropy Stable Schemes For Ten-Moment Gaussian Closure Equations

Abstract

In this article, we propose high order, semi-discrete, entropy stable, finite difference schemes for Ten-Moment Gaussian Closure equations. The crucial components of these schemes are an entropy conservative flux and suitable high order entropy dissipative operators to ensure entropy stability. We design two numerical fluxes, one is approximately entropy conservative, and another is entropy conservative flux. For the construction of appropriate entropy dissipative operators, we also derive entropy scaled right eigenvectors. This is used for sign preserving reconstruction of scaled entropy variables, which results in second and third order entropy stable schemes. We also extend these schemes to a plasma flow model with source term. Several numerical results are presented for homogeneous and non-homogeneous cases to demonstrate stability and performance of these schemes.

Appendix: Symmetrization and Entropy Scaled Eigenvectors for Ten-Moment Equations

In this section we will derive the entropy scaled right eigenvectors. For simplicity, let us consider the x-directional homogenous part of the system (1), i.e.

$$\begin{aligned} \frac{\partial \mathbf{u }}{\partial t}+\frac{\partial \,\mathbf f ^x(\mathbf{u })}{\partial x}=0. \end{aligned}$$

(26)

The entropy-entropy flux pair for this system is same as for the complete system. So, the results presented here can be easily extended to the complete system. Let us recall that,

$$\begin{aligned} \mathbf{v }\,=\,\frac{\partial e}{\partial \mathbf{u }}\,=\,\left( \begin{array}{c}4-s-\rho \frac{(p^{xx}v^{y2}\,+\,p^{yy}v^{x2}\,-\,2p^{xy}v^xv^y)}{p^{xx}p^{yy}\,-\,p^{xy2}}\\ \frac{2\rho (p^{yy}v^x\,-\,p^{xy}v^y)}{p^{xx}p^{yy}\,-\,p^{xy2}}\\ \frac{2\rho (p^{xx}v^y\,-\,p^{xy}v^x)}{p^{xx}p^{yy}\,-\,p^{xy2}}\\ \frac{-\rho p^{yy}}{p^{xx}p^{yy}\,-\,p^{xy2}}\\ \frac{2\rho p^{xy}}{p^{xx}p^{yy}\,-\,p^{xy2}}\\ \frac{-\rho p^{xx}}{p^{xx}p^{yy}\,-\,p^{xy2}} \end{array}\right) . \end{aligned}$$

and

$$\begin{aligned} {\psi }^x\,=\,\mathbf{v }^{\top }\cdot \mathbf f ^x\,-\,q^x\,=\,2\rho v^x. \end{aligned}$$

Theorem A.1

(Barth [1]) (Eigenvector scaling)

Let $A\in \mathbb {R}^{n\times n}$ be an arbitrary diagonalizable matrix and S the set of all right symmetrizers:

$$\begin{aligned} S=\{B\in \mathbb {R}^{n\times n}| B\quad SPD,\quad AB\,\, \text{ symmetric }\}. \end{aligned}$$

Further, let $\mathbf{R}\in \mathbb {R}^{n\times n}$ denote the right eigenvector matrix which diagonalizes A, $A=\mathbf{R}\Lambda \mathbf{R}^{-1}$ with r distinct eigenvalues, $\Lambda =Diag(\lambda _1I_{m_1\times m_1}, \lambda _2I_{m_2\times m_2}, \ldots , \lambda _rI_{m_r\times m_r}).$ Then for each $B\in S$ there exists a symmetric block diagonal matrix $T=Diag(T_{m_1\times m_1}, T_{m_2\times m_2},\ldots T_{m_r\times m_r} )$, such that the block scales columns of $\mathbf{R}$, $\tilde{\mathbf{R}}=\mathbf{R}T$, with

$$\begin{aligned} B=\tilde{\mathbf{R}}\tilde{\mathbf{R}}^{\top },\quad A=\tilde{\mathbf{R}}\Lambda \tilde{\mathbf{R}}^{-1} \end{aligned}$$

which imply

$$\begin{aligned} AB=\tilde{\mathbf{R}}\Lambda \tilde{\mathbf{R}}^{\top }. \end{aligned}$$

By change of variable from $\mathbf{u}$ to $\mathbf{v}$ we can rewrite (26) as,

$$\begin{aligned} \frac{\partial \mathbf{u }}{\partial \mathbf{v }}\frac{\partial \mathbf{v }}{\partial t}+\frac{\partial \mathbf f ^x(\mathbf{u })}{\partial \mathbf{u }}\frac{\partial \mathbf{u }}{\partial \mathbf{v }}\frac{\partial \mathbf{v }}{\partial x}=\mathbf 0 . \end{aligned}$$

(27)

As we have $(e,q_{x})$ as entropy pair for the system (26), the above change of variable should symmetrize the system (see [1]). Furthermore, the matrix $\frac{\partial \mathbf{u }}{\partial \mathbf{v }}$ is symmetric positive definite. We now want to find scaled entropy right eigenvectors $\tilde{\mathbf{R}}$ such that,

$$\begin{aligned} \frac{\partial \mathbf{u }}{\partial \mathbf{v }}=\tilde{\mathbf{R}}\tilde{\mathbf{R}}^{\top }. \end{aligned}$$

(28)

To simplify the calculation of the entropy scaling, let us introduce primitive variables $\mathbf{w}=\{\rho , \vec {v},\mathbf{p}\}^{\top }.$ Rewriting the system (26) in primitive variables, we get eigenvalues as,

$$\begin{aligned} v^x-\sqrt{\frac{3p^{xx}}{\rho }},\,v^x-\sqrt{\frac{p^{xx}}{\rho }},\,v^x,\,v^x,\,v^x+\sqrt{\frac{p^{xx}}{\rho }},\,v^x+\sqrt{\frac{3p^{xx}}{\rho }}, \end{aligned}$$

and corresponding right eigenvector matrix as,

$$\begin{aligned} \mathbf{R}_{\mathbf{w}}^x= \left( \begin{array}{c c c c c c} \rho p^{xx} &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \rho p^{xx}\\ -\sqrt{\frac{3p^{xx}}{\rho }}p^{xx} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \sqrt{\frac{3p^{xx}}{\rho }}p^{xx}\\ -\sqrt{\frac{3p^{xx}}{\rho }}p^{xy} &{}\quad -\,\sqrt{\frac{p^{xx}}{\rho }} &{}\quad 0 &{}\quad 0 &{}\quad \sqrt{\frac{p^{xx}}{\rho }} &{} \sqrt{\frac{3p^{xx}}{\rho }}p^{xy} \\ 3p^{xx2} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 3p^{xx2}\\ 3p^{xx}p^{xy} &{}\quad p^{xx} &{}\quad 0 &{}\quad 0 &{}\quad p^{xx} &{}\quad 3p^{xx}p^{xy}\\ 2p^{xy2}+p^{xx}p^{yy} &{}\quad 2p^{xy} &{}\quad 0 &{}\quad 1 &{}\quad 2p^{xy} &{}\quad 2p^{xy2}+p^{xx}p^{yy} \end{array}\right) . \end{aligned}$$

Once the entropy scaled right eigenvector for primitive variable system, $\tilde{\mathbf{R}}_{\mathbf{w}}^x$ is found, entropy scaled right eigenvectors for the conservative system can be easily calculated using,

$$\begin{aligned} \tilde{\mathbf{R}}^x=\frac{\partial \mathbf{u }}{\partial \mathbf{w }}\,\tilde{\mathbf{R}}_{\mathbf{w}}^x. \end{aligned}$$

(29)

Here, the jacobian matrix for the change of variable is,

$$\begin{aligned} \frac{\partial \mathbf{u }}{\partial \mathbf{w }}\,=\,\left( \begin{array}{c c c c c c} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ v^x &{}\quad \rho &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ v^y &{}\quad 0 &{}\quad \rho &{}\quad 0 &{}\quad 0 &{}\quad 0\\ v^{x2} &{}\quad 2\rho v^x &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0\\ v^xv^y &{}\quad \rho v^y &{}\quad \rho v^x &{}\quad 0 &{}\quad 1 &{}\quad 0\\ v^{y2} &{}\quad 0 &{}\quad 2\rho v^y &{}\quad 0 &{}\quad 0 &{}\quad 1 \end{array}\right) . \end{aligned}$$

Using Eqs.(28) and (29), we note that,

$$\begin{aligned} \frac{\partial \mathbf{u}}{\partial \mathbf{v}}= \frac{\partial \mathbf{u}}{\partial \mathbf{w}}\frac{\partial \mathbf{w}}{\partial \mathbf{v}}=\tilde{\mathbf{R}}^x \tilde{\mathbf{R}}^{x,\top }= \frac{\partial \mathbf{u }}{\partial \mathbf{w }}\,\tilde{\mathbf{R}}_{\mathbf{w}}^x \tilde{\mathbf{R}}_{\mathbf{w}}^{x,\top }{\frac{\partial \mathbf{u }}{\partial \mathbf{w }}}^{\top } \end{aligned}$$

which implies,

$$\begin{aligned} \tilde{\mathbf{R}}_{\mathbf{w}}^x \tilde{\mathbf{R}}_{\mathbf{w}}^{x,\top }= \frac{\partial \mathbf{w }}{\partial \mathbf{v }}\frac{\partial \mathbf{u }}{\partial \mathbf{w }}^{(-\top )}. \end{aligned}$$

(30)

To calculate the matrix $\frac{\partial \mathbf{w }}{\partial \mathbf{v }}$ we need to write primitive variables in terms of entropy variables $\mathbf{v}$. Let us define the notations,

Plugging in the expression for $v_{4},v_{5}$ and $v_{6}$ into the equation for $v_{2}$ and $v_{3}$ we get,

$$\begin{aligned} v_2= & {} -2v_4v^x-v_5v^y,\\ v_3= & {} -2v_6v^y-v_5v^x. \end{aligned}$$

On solving for $v^{x}$ and $v^{y}$, we get,

$$\begin{aligned} v^x= & {} \frac{v_5v_3-2v_2v_6}{4v_4v_6-v^2_5},\end{aligned}$$

(31)

$$\begin{aligned} v^y= & {} \frac{v_2v_5-2v_3v_4}{4v_4v_6-v^2_5}. \end{aligned}$$

(32)

We now rewrite $v_{1}$ as,

$$\begin{aligned} v_1= & {} 4-s+(v_6v^{y2}+v_4v^{x2}+v_5v^xv^y)\\= & {} 4-\log (4)+\log (4v_4v_6-v^2_5)+2\log (\rho )+(v_6v^{y2}+v_4v^{x2}+v_5v^xv^y), \end{aligned}$$

where we have used

$$\begin{aligned} \log (p^{xx}p^{yy}-p^{xy2})= & {} \log (\rho ^2)+\log (4)-\log (4v_4v_6-v^2_5). \end{aligned}$$

This results in,

$$\begin{aligned} \rho =e^{(v_1-4+\log (4)-\log (4v_4v_6-v^2_5)-(v_6v^{y2}+v_4v^{x2}+v_5v^xv^y))/2}. \end{aligned}$$

(33)

Now using expression for $v_{4}$,

$$\begin{aligned} p^{yy}=-\frac{4v_4\rho }{4v_4v_6-v^2_5},\nonumber= & {} -\frac{4v_4e^{(v_1-4+\log (4)-\log (4v_4v_6-v^2_5)-(v_6v^{y2}+v_4v^{x2}+v_5v^xv^y))/2}}{4v_4v_6-v^2_5}.\\ \end{aligned}$$

(34)

Similarly,

$$\begin{aligned} p^{xy}= & {} \frac{2v_5e^{(v_1-4+\log (4)-\log (4v_4v_6-v^2_5)-(v_6v^{y2}+v_4v^{x2}+v_5v^xv^y))/2}}{4v_4v_6-v^2_5},\end{aligned}$$

(35)

$$\begin{aligned} p^{xx}= & {} -\frac{4v_6e^{(v_1-4+\log (4)-\log (4v_4v_6-v^2_5)-(v_6v^{y2}+v_4v^{x2}+v_5v^xv^y))/2}}{4v_4v_6-v^2_5}. \end{aligned}$$

(36)

The matrix $\frac{\partial \mathbf{w }}{\partial \mathbf{v }}$ can now be easily calculated and is found to be,

$$\begin{aligned} \left( \begin{array}{c c c c c c} \frac{\rho }{2} &{}\quad \frac{\rho v^x}{2} &{}\quad \frac{\rho v^y}{2} &{}\quad \frac{e^{xx}}{2} &{}\quad \frac{e^{xy}}{2} &{} \frac{e^{yy}}{2}\\ &{}\quad &{}\quad &{} \quad &{}\quad &{}\quad \\ 0 &{}\quad \frac{p^{xx}}{2\rho } &{}\quad \frac{p^{xy}}{2\rho } &{}\quad \frac{p^{xx}v^x}{\rho } &{}\quad \frac{p^{xy}v^x+p^{xx}v^y}{2\rho } &{}\quad \frac{p^{xy}v^y}{\rho }\\ &{}\quad &{}\quad &{}\quad &{}\quad &{}\\ 0 &{}\quad \frac{p^{xy}}{2\rho } &{}\quad \frac{p^{yy}}{2\rho } &{}\quad \frac{p^{xy}v^x}{\rho } &{}\quad \frac{p^{yy}v^x+p^{xy}v^y}{2\rho } &{}\quad \frac{p^{yy}v^y}{\rho }\\ &{}\quad &{}\quad &{}\quad &{}\quad &{}\\ \frac{p^{xx}}{2} &{}\quad \frac{p^{xx}v^x}{2} &{}\quad \frac{p^{xx}v^y}{2} &{}\quad \frac{3 p^{xx2}}{2\rho } +\frac{p^{xx}v^{x2}}{2} &{}\quad \frac{3 p^{xx}p^{xy}}{2\rho } +\frac{p^{xx}v^{x}v^y}{2} &{}\quad \frac{p^{xy2}}{\rho }+\frac{p^{xx}p^{yy}}{2\rho }+\frac{p^{xx}v^{y2}}{2}\\ &{}\quad &{}\quad &{}\quad &{}\quad &{}\\ \frac{p^{xy}}{2} &{}\quad \frac{p^{xy}v^x}{2} &{}\quad \frac{p^{xy}v^y}{2} &{}\quad \frac{3 p^{xx}p^{xy}}{2\rho } +\frac{p^{xy}v^{x2}}{2} &{}\quad \frac{p^{xy2}}{\rho }+\frac{p^{xx}p^{yy}}{2\rho }+\frac{p^{xy}v^{x}v^y}{2} &{} \frac{3 p^{xy}p^{yy}}{2\rho } +\frac{p^{xy}v^{y2}}{2} \\ &{}\quad &{}\quad &{}\quad &{}\quad &{}\\ \frac{p^{yy}}{2} &{}\quad \frac{p^{yy}v^x}{2} &{}\quad \frac{p^{yy}v^y}{2} &{}\quad \frac{p^{xy2}}{\rho }+\frac{p^{xx}p^{yy}}{2\rho }+\frac{p^{yy}v^{x2}}{2} &{}\quad \frac{3 p^{xy}p^{yy}}{2\rho } +\frac{p^{yy}v^xv^y}{2} &{}\quad \frac{3 p^{yy2}}{2\rho } +\frac{p^{yy}v^{y2}}{2}\\ \end{array}\right) . \end{aligned}$$

Furthermore, the matrix $\frac{\partial \mathbf{w }}{\partial \mathbf{v }}\frac{\partial \mathbf{u }}{\partial \mathbf{w }}^{-\top }$ is,

$$\begin{aligned} \left( \begin{array}{c c c c c c} \frac{\rho }{2} &{} 0 &{} 0 &{} \frac{p^{xx}}{2} &{} \frac{p^{xy}}{2} &{} \frac{p^{yy}}{2}\\ &{} &{} &{} &{} &{}\\ 0 &{} \frac{p^{xx}}{2\rho ^2} &{} \frac{p^{xy}}{2\rho ^2} &{} 0 &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} \frac{p^{xy}}{2\rho ^2} &{} \frac{p^{yy}}{2\rho ^2} &{} 0 &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ \frac{p^{xx}}{2} &{} 0 &{} 0 &{} \frac{3 p^{xx2}}{2\rho } &{} \frac{3 p^{xx}p^{xy}}{2\rho } &{} \frac{p^{xy2}}{\rho }+\frac{p^{xx}p^{yy}}{2\rho }\\ &{} &{} &{} &{} &{}\\ \frac{p^{xy}}{2} &{} 0 &{} 0 &{} \frac{3 p^{xx}p^{xy}}{2\rho } &{} \frac{p^{xy2}}{\rho }+\frac{p^{xx}p^{yy}}{2\rho } &{} \frac{3 p^{xy}p^{yy}}{2\rho } \\ &{} &{} &{} &{} &{}\\ \frac{p^{yy}}{2} &{} 0 &{} 0 &{} \frac{p^{xx}p^{yy}}{2\rho }+\frac{p^{xy2}}{\rho } &{} \frac{3 p^{xy}p^{yy}}{2\rho } &{} \frac{3 p^{yy2}}{2\rho } \\ \end{array}\right) . \end{aligned}$$

Following the Theorem (A.1) in [1], the scaling matrix $\mathbf{T}^x$ is square root of matrix $\mathbf{Y} =\mathbf{R}_{\mathbf{w}}^{-1x} \frac{\partial \mathbf{w }}{\partial \mathbf{v }}\frac{\partial \mathbf{u }}{\partial \mathbf{w }}^{-\top } \mathbf{R}_{\mathbf{w}}^{-\top x}$, which results in,

$$\begin{aligned} \mathbf{T}^x=\left( \begin{array}{c c c c c c} \frac{1}{\sqrt{12p^{xx2}}\rho } &{} 0 &{} 0 &{} 0 &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} \sqrt{\frac{p^{xx}p^{yy}-p^{xy2}}{4\rho p^{xx2}}} &{} 0 &{} 0 &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} \frac{\frac{\rho }{3}+s^x}{t^x} &{} \frac{p^{xx}p^{yy}-p^{xy2}}{3p^{xx}t^x} &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} \frac{p^{xx}p^{yy}-p^{xy2}}{3p^{xx}t^x} &{} \frac{\frac{4(p^{xx}p^{yy}-p^{xy2})^2}{3p^{xx2}\rho }+s^x}{t^x} &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{p^{xx}p^{yy}-p^{xy2}}{4\rho p^{xx2}}} &{} 0 \\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{1}{12p^{xx2}\rho }} \\ \end{array}\right) , \end{aligned}$$

where $s^x=\frac{p^{xx}p^{yy}-p^{xy2}}{\sqrt{3}p^{xx}}$ and $t^{x2}=\frac{\rho }{3}+\frac{4}{3}\frac{(p^{xx}p^{yy}-p^{xy2})^2}{3p^{xx2}\rho }+\frac{2}{\sqrt{3}}\frac{p^{xx}p^{yy}-p^{xy2}}{p^{xx}}$. So, the entropy scaled right eigenvector matrix in primitive variables is,

$$\begin{aligned} \tilde{\mathbf{R}}_{\mathbf{w}}^x= & {} \,\mathbf{R}_{\mathbf{w}}^x\mathbf{T}^x\\= & {} \left( \begin{array}{c c c c c c} A^x\rho p^{xx} &{} 0 &{} \frac{\frac{\rho }{3}+s^x}{t^x} &{} \frac{1}{3t^x\beta _1} &{} 0 &{} A^x\rho p^{xx}\\ -A^x\sqrt{\frac{3p^{xx}}{\rho }}p^{xx} &{} 0 &{} 0 &{} 0 &{} 0 &{} A^x\sqrt{\frac{3p^{xx}}{\rho }}p^{xx}\\ -A^x\sqrt{\frac{3p^{xx}}{\rho }}p^{xy} &{} -B^x\sqrt{\frac{p^{xx}}{\rho }} &{} 0 &{} 0 &{} B^x\sqrt{\frac{p^{xx}}{\rho }} &{} A^x\sqrt{\frac{3p^{xx}}{\rho }}p^{xy} \\ A^x(3p^{xx2}) &{} 0 &{} 0 &{} 0 &{} 0 &{} A^x(3p^{xx2})\\ A^x(3p^{xx}p^{xy}) &{} B^xp^{xx} &{} 0 &{} 0 &{} B^xp^{xx} &{} A^x(3p^{xx}p^{xy})\\ A^x(2p^{xy2}+p^{xx}p^{yy}) &{} 2p^{xy}B^x &{} \frac{1}{3t^x\beta _1} &{} \frac{\frac{4}{3\rho \beta _1^2}+s^x}{t^x} &{} 2p^{xy}B^x &{} A^x(2p^{xy2}+p^{xx}p^{yy}) \end{array}\right) , \end{aligned}$$

where $A^x=\frac{1}{2\sqrt{3}p^{xx}\sqrt{\rho }}$, $B^x=\frac{\sqrt{p^{xx}p^{yy}-p^{xy2}}}{2p^{xx}\sqrt{\rho }}$, $\beta _1=\frac{p^{xx}}{(p^{xx}p^{yy}-p^{xy2})}$. Using (29), we can now calculate entropy scaled right eigenvectors in conservative variables which satisfy (28).

Remark A.1

Proceeding similarly in y-direction, with right eigenvector matrix for system in primitive variable as,

$$\begin{aligned} \mathbf{R}_{\mathbf{w}}^y= \left( \begin{array}{c c c c c c} \rho p^{yy} &{} 0 &{} 1 &{} 0 &{} 0 &{} \rho p^{yy}\\ -\sqrt{\frac{3p^{yy}}{\rho }}p^{xy} &{} -\sqrt{\frac{p^{yy}}{\rho }} &{} 0 &{} 0 &{} \sqrt{\frac{p^{yy}}{\rho }} &{} \sqrt{\frac{3p^{yy}}{\rho }}p^{xy}\\ -\sqrt{\frac{3p^{yy}}{\rho }}p^{yy} &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{3p^{yy}}{\rho }}p^{yy}\\ 2p^{xy2}+p^{xx}p^{yy} &{} 2p^{xy} &{} 0 &{} 1 &{} 2p^{xy} &{} 2p^{xy2}+p^{xx}p^{yy} \\ 3p^{xy}p^{yy} &{} p^{yy} &{} 0 &{} 0 &{} p^{yy} &{} 3p^{xy}p^{yy}\\ 3p^{yy2} &{} 0 &{} 0 &{} 0 &{} 0 &{} 3p^{yy2} \end{array}\right) , \end{aligned}$$

we obtained scaling matrix $\mathbf{T}^y$ as,

$$\begin{aligned} \mathbf{T}^{y}=\left( \begin{array}{c c c c c c} \frac{1}{\sqrt{12p^{yy2}}\rho } &{} 0 &{} 0 &{} 0 &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} \sqrt{\frac{p^{xx}p^{yy}-p^{xy2}}{4\rho p^{yy2}}} &{} 0 &{} 0 &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} \frac{\frac{\rho }{3}+s^y}{t^y} &{} \frac{p^{xx}p^{yy}-p^{xy2}}{3p^{yy}t^y} &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} \frac{p^{xx}p^{yy}-p^{xy2}}{3p^{yy}t^y} &{} \frac{\frac{4(p^{xx}p^{yy}-p^{xy2})^2}{3p^{yy2}\rho }+s^y}{t^y} &{} 0 &{} 0\\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{p^{xx}p^{yy}-p^{xy2}}{4\rho p^{yy2}}} &{} 0 \\ &{} &{} &{} &{} &{}\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{\frac{1}{12p^{yy2}\rho }} \\ \end{array}\right) , \end{aligned}$$

where $s^y=\frac{p^{xx}p^{yy}-p^{xy2}}{\sqrt{3}p^{yy}}$ and $t^{y2}=\frac{\rho }{3}+\frac{4}{3}\frac{(p^{xx}p^{yy}-p^{xy2})^2}{3p^{yy2}\rho }+\frac{2}{\sqrt{3}}\frac{p^{xx}p^{yy}-p^{xy2}}{p^{yy}}$. Using this, the entropy scaled right eigenvectors in primitive variable in y-direction are,

$$\begin{aligned} \tilde{\mathbf{R}}^y_{\mathbf{w}}= & {} \,\mathbf{R}^y_{\mathbf{w}}{} \mathbf{T}^y\\= & {} \, \left( \begin{array}{c c c c c c} A^y\rho p^{yy} &{} 0 &{} \frac{\frac{\rho }{3}+s^y}{t^y} &{} \frac{1}{3t^y\beta _2} &{} 0 &{} A^y\rho p^{yy}\\ -A^y\sqrt{\frac{3p^{yy}}{\rho }}p^{xy} &{} -B^y\sqrt{\frac{p^{yy}}{\rho }} &{} 0 &{} 0 &{} B^y\sqrt{\frac{p^{yy}}{\rho }} &{} A^y\sqrt{\frac{3p^{yy}}{\rho }}p^{xy}\\ -A^y\sqrt{\frac{3p^{yy}}{\rho }}p^{yy} &{} 0 &{} 0 &{} 0 &{} 0 &{} A^y\sqrt{\frac{3p^{yy}}{\rho }}p^{yy}\\ A^y(2p^{xy2}+p^{xx}p^{yy}) &{} 2p^{xy}B^y &{} \frac{1}{3t^y\beta _2} &{} \frac{\frac{4}{3\rho \beta _2^2}+s^y}{t^y} &{} 2p^{xy}B^y &{} A^y(2p^{xy2}+p^{xx}p^{yy}) \\ A^y(3p^{xy}p^{yy}) &{} B^yp^{yy} &{} 0 &{} 0 &{} B^yp^{yy} &{} A^y(3p^{xy}p^{yy})\\ A^y(3p^{yy2}) &{} 0 &{} 0 &{} 0 &{} 0 &{} A^y(3p^{yy2}) \end{array}\right) \end{aligned}$$

with $A^y=\frac{1}{2\sqrt{3}p^{yy}\sqrt{\rho }}$, $B^y=\frac{\sqrt{p^{xx}p^{yy}-p^{xy2}}}{2p^{yy}\sqrt{\rho }}$, $\beta _2=\frac{p^{yy}}{(p^{xx}p^{yy}-p^{xy2})}$.