# A second-order direct Eulerian GRP scheme for ten-moment Gaussian closure
equations with source terms

Jiangfu Wang<EMAIL_ADDRESS>Huazhong Tang<EMAIL_ADDRESS>Center for Applied Physics and Technology, HEDPS and LMAM, School of
Mathematical Sciences, Peking University, Beijing, 100871, PR China.  Nanchang
Hangkong University, Nanchang, 330000, Jiangxi Province, PR China.

###### Abstract

This paper proposes a second-order accurate direct Eulerian generalized
Riemann problem (GRP) scheme for the ten-moment Gaussian closure equations
with source terms. The generalized Riemann invariants associated with the
rarefaction waves, the contact discontinuity and the shear waves are given,
and the 1D exact Riemann solver is obtained. After that, the generalized
Riemann invariants and the Rankine-Hugoniot jump conditions are directly used
to resolve the left and right nonlinear waves (rarefaction wave and shock
wave) of the local GRP in Eulerian formulation, and then the 1D direct
Eulerian GRP scheme is derived. They are much more complicated, technical and
nontrivial due to more physical variables and elementary waves. Some 1D and 2D
numerical experiments are presented to check the accuracy and high resolution
of the proposed GRP schemes, where the 2D direct Eulerian GRP scheme is given
by using the Strang splitting method for simplicity. It should be emphasized
that several examples of 2D Riemann problems are constructed for the first
time.

###### keywords:

Ten-moment equations; exact Riemann solver; generalized Riemann problem;
generalized Riemann invariants; Rankine-Hugoniot jump conditions.

##  1 Introduction

It is well-known that the compressible Euler equations of gas dynamics can be
derived from the Boltzmann equation [21] by assuming local thermodynamic
equilibrium. However, for many problems, such as collisionless plasma [13, 33,
18, 47, 11] and the non-equilibrium gas dynamics [9], the local thermodynamic
equilibrium assumption does not hold and anisotropic effects are often
present, so that the Euler equations are less suitable. The ten-moment
Gaussian closure equations [22] provide an alternative for such applications,
where the pressure is described by an anisotropic and symmetric tensor.

Some numerical schemes have been proposed for solving the ten-moment equations
in the past decades. A second-order upwind finite volume scheme was introduced
in [9]. First-order relaxation numerical schemes were employed to approximate
the weak solutions of those equations, see [6, 7]. A Harten–Lax–van Leer-
contact (HLLC) approximate Riemann solver was applied in [37] to solve the
ten-moment equations coupled with magnetic field. In recent years, some high-
order numerical methods were also developed, including high-order positivity-
preserving discontinuous Galerkin (DG) methods [31], finite difference
weighted essentially non-oscillatory (WENO) schemes [32], and high-order
entropy stable finite difference methods [39] as well as DG methods [8].
Besides, a second-order robust monotone upwind scheme was formulated in [28],
and a second-order well-balanced (WB) scheme to handle equilibrium states was
constructed in [29]. Additionally, a robust finite volume scheme for the two-
fluid ten-moment plasma flow equations was proposed in [30]. The high-order
accurate positivity-preserving and well-balanced discontinuous Galerkin
schemes for ten-moment equations were also developed in [46], where a special
modification to the numerical HLLC flux was imposed to enforce the well-
balancedness and the geometric quasilinearization approach [50] was applied to
simplify the positivity analysis.

The generalized Riemann problem (GRP) scheme, which serves as an analytic
second-order extension of the Godunov method, was initially developed for the
study of compressible fluid dynamics [2]. This approach employs a piecewise
linear function to approximate the “initial” data. Subsequently, it resolves a
local GRP analytically at each cell interface to determine the numerical flux.
For a detailed description, the readers are referred to [3]. There are two
versions of the original GRP scheme: the Lagrangian and the Eulerian.
Typically, the Eulerian version is derived through a transformation based on
the Lagrangian framework, which can be particularly intricate, especially when
dealing with sonic and multi-dimensional scenarios. To circumvent those
difficulties, second-order accurate direct Eulerian GRP schemes were
respectively developed for the shallow water equations [23], the Euler
equations [5], and a more general weakly coupled system [4]. Those schemes
directly resolve the local GRPs in the Eulerian formulation by using
generalized Riemann invariants and Rankine-Hugoniot jump conditions.

Up to now, the GRP methodology has been widely implemented in many physically
interesting cases, including the reactive flows [1], the motion of elastic
strings [42], the shallow water equations [23], the radially symmetric
compressible flows [26], the one-dimensional (1D) and two-dimensional (2D)
special relativistic hydrodynamics (RHD) [55, 56], the gas-liquid two-phase
flow in high-temperature and high-pressure transient wells [54], the
spherically symmetric general RHD [51], the radiation hydrodynamical equations
[19], the Baser-Nunziato two-phase model [20], the blood flow model in
arteries [40], the compressible flows of real materials [48], the Kapila model
of compressible multiphase flows [12], the laminar two-phase flow model with
two-velocities [58] and so on. Besides, a comparison between the GRP scheme
and the gas-kinetic scheme revealed the good performance of the GRP solver in
simulating some inviscid flows [25]. By integrating the GRP scheme with the
moving mesh method [43], the adaptive direct Eulerian GRP scheme was
effectively developed in [14], resulting in enhanced resolution and accuracy.
The adaptive GRP scheme was further studied in simulating 2D complex wave
configurations formulated with the 2D Riemann problems of Euler equations [15]
and was also extended to unstructured triangular meshes [27]. There are also
many works on the high-order GRP schemes, such as the third-order GRP schemes
for the Euler equations [53], 1D RHD [52] and the general hyperbolic balance
law [35, 36]. Moreover, a two-stage fourth order time-accurate GRP scheme was
also proposed for hyperbolic conservation laws [24] and for the special RHD
[57]. Arbitrary high-order DG schemes based on the GRP solver were developed
[49], where the reconstruction steps were halved compared with the existing
high-order Runge-Kutta DG (RKDG) schemes. The two-stage fourth-order DG method
based on the GRP solver was also developed in [10], and the computational cost
can be considerably reduced by more than $50\%$ compared with the same order
multi-stage strong-stability-preserving (SSP) RKDG method. The GRP-based high
resolution schemes for the multi-medium flows [17, 16] and the axisymmetric
hydrodynamics [59] were recently developed.

This paper develops the second-order direct Eulerian GRP scheme for the ten-
moment Gaussian closure equations with source terms. The generalized Riemann
invariants and the exact solutions of the 1D Riemann problem are given.
Compared to the Euler equations [5], the shallow water equations [23] and the
blood flow in arteries [40] etc., the ten-moment equations have more physical
variables and more elementary waves (e.g. the left and the right shear waves
which are linearly degenerate and separated by the contact discontinuity).
More physical variables means that more linear algebra equations should be
derived to compute the instantaneous time derivatives. Two more elementary
waves means that four different solution states between the left and right
nonlinear waves should be resolved technically. All these features make the
derivation of the direct Eulerian GRP scheme for the ten-moment equations much
more complicated and nontrivial.

This paper is organized as follows. Section 2 introduces the 1D ten-moment
equations and gives corresponding eigenvalues, eigenvectors and generalized
Riemann invariants. Section 3 introduces the outline of the 1D direct GRP
scheme and its extension to 2D case. Section 4 presents the exact solver for
the classical Riemann problem, and Section 5 resolves the generalized Riemann
problems. Some 1D and 2D numerical experiments are conducted in Section 6 to
demonstrate the accuracy and performance of the proposed GRP scheme. It should
be emphasized that several examples of 2D Riemann problems are constructed for
the first time. Section 7 concludes this paper.

##  2 Preliminaries and notations

This section introduces the 1D ten-moment equations, and corresponding
eigenvalues as well as eigenvectors, discusses the characteristic fields, and
gives generalized Riemann invariants.

###  2.1 1D ten-moment equations

The 1D ten-moment Gaussian closure equations with source term may be written
into the form of balance laws as

$\frac{\partial\mathbf{U}}{\partial t}+\frac{\partial\mathbf{F(U)}}{\partial
x}=\mathbf{S}^{x}(\mathbf{U}),$ (2.1)

where the solution vector
$\mathbf{U}=(\rho,m_{1},m_{2},E_{11},E_{12},E_{22})^{\top}$, $\rho$,
$\mathbf{m}=(m_{1},m_{2})^{\top}$, and $E_{ij}$ denote the density, the
momentum vector, and the component of the symmetric energy tensor
$\mathbf{E}=(E_{ij})$, respectively, $1\leq i,j\leq 2$. The flux
$\mathbf{F(U)}$ and the source term $\mathbf{S}^{x}(\mathbf{U})$ are
respectively given by

$\displaystyle\mathbf{F(U)}=\left(\rho u_{1},\rho u_{1}^{2}+p_{11},\rho
u_{1}u_{2}+p_{12},(E_{11}+p_{11})u_{1},E_{12}u_{1}+\frac{1}{2}(p_{11}u_{2}+p_{12}u_{1}),E_{22}u_{1}+p_{12}u_{2}\right)^{\top},$
(2.2)
$\displaystyle\mathbf{S}^{x}(\mathbf{U})=\left(0,-\frac{1}{2}\rho\partial_{x}W,0,-\frac{1}{2}\rho
u_{1}\partial_{x}W,-\frac{1}{4}\rho u_{2}\partial_{x}W,0\right)^{\top},$

where $\mathbf{u}=(u_{1},u_{2})^{\top}={\bf m}/\rho$ denotes the velocity
vector, and the function $W(x)$ is a given potential, which may denote the
electron quiver energy in laser light (see e.g. [33, 38]). The system (2.1) is
closed by the following “equation of state”

$\mathbf{p}=(p_{ij})_{2\times 2}=2\mathbf{E}-\rho\mathbf{u}\otimes\mathbf{u},$
(2.3)

where the symbol $\otimes$ denotes the tensor product, and its solution
$\mathbf{U}$ should stays in the physically admissible state set

$\mathcal{G}:=\left\\{\mathbf{U}\in\mathbb{R}^{6}:\rho>0,~{}\mathbf{x}^{\top}\mathbf{p}\mathbf{x}>0\quad\forall\mathbf{x}\in\mathbb{R}^{2}\setminus\\{\mathbf{0}\\}\right\\},$
(2.4)

which means that $\rho>0$, $p_{11}>0$ and
$\det(\mathbf{p}):=p_{11}p_{22}-p_{12}^{2}>0$.

Rewrite (2.1) into the following quasi-linear form

$\displaystyle\frac{\partial\rho}{\partial
t}+u_{1}\frac{\partial\rho}{\partial x}+\rho\frac{\partial u_{1}}{\partial
x}=0,$ (2.5a) $\displaystyle\frac{\partial u_{1}}{\partial
t}+u_{1}\frac{\partial u_{1}}{\partial x}+\frac{1}{\rho}\frac{\partial
p_{11}}{\partial x}=-\frac{1}{2}W_{x},$ (2.5b) $\displaystyle\frac{\partial
u_{2}}{\partial t}+u_{1}\frac{\partial u_{2}}{\partial
x}+\frac{1}{\rho}\frac{\partial p_{12}}{\partial x}=0,$ (2.5c)
$\displaystyle\frac{\partial p_{11}}{\partial t}+3p_{11}\frac{\partial
u_{1}}{\partial x}+u_{1}\frac{\partial p_{11}}{\partial x}=0,$ (2.5d)
$\displaystyle\frac{\partial p_{12}}{\partial t}+2p_{12}\frac{\partial
u_{1}}{\partial x}+p_{11}\frac{\partial u_{2}}{\partial x}+u_{1}\frac{\partial
p_{12}}{\partial x}=0,$ (2.5e) $\displaystyle\frac{\partial p_{22}}{\partial
t}+p_{22}\frac{\partial u_{1}}{\partial x}+2p_{12}\frac{\partial
u_{2}}{\partial x}+u_{1}\frac{\partial p_{22}}{\partial x}=0.$ (2.5f)

It is obvious that (2.5d) and (2.5b) imply that

$\displaystyle\frac{\partial u_{1}}{\partial
x}=-\frac{1}{3p_{11}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t},$ (2.6)
$\displaystyle\frac{\partial p_{11}}{\partial
x}=-\rho\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\frac{1}{2}W_{x}\right),$
(2.7)

respectively, where
$\frac{\mathcal{D}}{\mathcal{D}t}:=\frac{\partial}{\partial
t}+u_{1}\frac{\partial}{\partial x}$ is the material derivative. Then it
follows that

$\begin{split}&\frac{\partial u_{1}}{\partial
t}=\frac{u_{1}}{3p_{11}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}+\frac{\mathcal{D}u_{1}}{\mathcal{D}t},\\\
&\frac{\partial p_{11}}{\partial
t}=\frac{\mathcal{D}p_{11}}{\mathcal{D}t}+\rho
u_{1}\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\frac{1}{2}\rho
u_{1}W_{x}.\end{split}$ (2.8)

Besides, (2.5e) and (2.5c) imply that

$\displaystyle\frac{\partial u_{2}}{\partial
x}=-\frac{1}{p_{11}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}-\frac{2p_{12}}{3p_{11}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right),$
(2.9) $\displaystyle\frac{\partial p_{12}}{\partial
x}=-\rho\frac{\mathcal{D}u_{2}}{\mathcal{D}t},$ (2.10)

respectively, where (2.6) has been used in (2.9). It follows that

$\begin{split}&\frac{\partial u_{2}}{\partial
t}=\frac{\mathcal{D}u_{2}}{\mathcal{D}t}+\frac{u_{1}}{p_{11}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}-\frac{2p_{12}}{3p_{11}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right),\\\
&\frac{\partial p_{12}}{\partial
t}=\frac{\mathcal{D}p_{12}}{\mathcal{D}t}+\rho
u_{1}\frac{\mathcal{D}u_{2}}{\mathcal{D}t}.\end{split}$ (2.11)

Moreover, (2.5f) implies that

$\frac{\mathcal{D}p_{22}}{\mathcal{D}t}=\frac{p_{11}p_{22}-4p_{12}^{2}}{3p_{11}^{2}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}+\frac{2p_{12}}{p_{11}}\frac{\mathcal{D}p_{12}}{\mathcal{D}t},$
(2.12)

where (2.6) and (2.9) have been used.

###  2.2 Eigenvalues, eigenvectors and generalized Riemann invariants

The eigenvalues of the system (2.1) or (2.5) are given (see e.g. [9]) by

$\lambda_{1}=u_{1}-c,~{}\lambda_{2}=u_{1}-\frac{c}{\sqrt{3}},~{}\lambda_{3}=\lambda_{4}=u_{1},~{}\lambda_{5}=u_{1}+\frac{c}{\sqrt{3}},~{}\lambda_{6}=u_{1}+c,$

where $c:=\sqrt{{3p_{11}}/{\rho}}$ denotes the sound speed, and the linearly
independent six associated right eigenvectors for (2.5) may be chosen (see
e.g. [39]) as follows

$\displaystyle\mathbf{\tilde{r}}_{1}=\left(\rho
p_{11},-cp_{11},-cp_{12},3p_{11}^{2},3p_{11}p_{12},p_{11}p_{22}+2p_{12}^{2}\right)^{\top},$
$\displaystyle\mathbf{\tilde{r}}_{2}=\left(0,0,-\frac{c}{\sqrt{3}},0,p_{11},2p_{12}\right)^{\top},$
$\displaystyle\mathbf{\tilde{r}}_{3}=(0,0,0,0,0,1)^{\top},\ \
\mathbf{\tilde{r}}_{4}=(1,0,0,0,0,0)^{\top},$
$\displaystyle\mathbf{\tilde{r}}_{5}=\left(0,0,\frac{c}{\sqrt{3}},0,p_{11},2p_{12}\right)^{\top},$
$\displaystyle\mathbf{\tilde{r}}_{6}=\left(\rho
p_{11},cp_{11},cp_{12},3p_{11}^{2},3p_{11}p_{12},p_{11}p_{22}+2p_{12}^{2}\right)^{\top}.$

Denote the primitive variables vector by
$\mathbf{V}:=(\rho,u_{1},u_{2},p_{11},p_{12},p_{22})^{\top}$. The associated
right eigenvector matrix
$\mathbf{R(U)}=\frac{\partial{\mathbf{U}}}{\partial{\mathbf{V}}}\cdot(\mathbf{\tilde{r}}_{1},\cdots,\mathbf{\tilde{r}}_{6})$
for (2.1) is given in A.

By simple calculations, one can find that for any physically admissible
$\mathbf{V}$,
$\nabla_{\mathbf{V}}\lambda_{i}(\mathbf{V})\cdot\mathbf{\tilde{r}}_{i}(\mathbf{V})\neq
0$ for $i=1,6$ and
$\nabla_{\mathbf{V}}\lambda_{i}(\mathbf{V})\cdot\mathbf{\tilde{r}}_{i}(\mathbf{V})=0$
for $i=2,3,4,5$. Hence, only the first and the sixth characteristic fields are
genuinely nonlinear, and the other characteristic fields are linearly
degenerate. Possible wave patterns associated with those characteristic fields
are

$\displaystyle\text{for }\lambda_{1}=u_{1}-c:\quad\text{1-shock or
1-rarefaction wave},$ $\displaystyle\text{for
}\lambda_{2}=u_{1}-\frac{c}{\sqrt{3}}:\quad\text{2-shear wave},$
$\displaystyle\text{for }\lambda_{3}=\lambda_{4}=u_{1}:\quad\text{3, 4-contact
wave},$ $\displaystyle\text{for
}\lambda_{5}=u_{1}+\frac{c}{\sqrt{3}}:\quad\text{5-shear wave},$
$\displaystyle\text{for }\lambda_{6}=u_{1}+c:\quad\text{6-shock or
6-rarefaction wave}.$

Note that the characteristic fields of the 1D ten-moment equations are similar
to those of the 1D shear shallow water model [34]. According to the following
relations

$\frac{\text{d}\rho}{\tilde{r}_{i}^{(1)}}=\frac{\text{d}u_{1}}{\tilde{r}_{i}^{(2)}}=\frac{\text{d}u_{2}}{\tilde{r}_{i}^{(3)}}=\frac{\text{d}p_{11}}{\tilde{r}_{i}^{(4)}}=\frac{\text{d}p_{12}}{\tilde{r}_{i}^{(5)}}=\frac{\text{d}p_{22}}{\tilde{r}_{i}^{(6)}},~{}i=1,2,\cdots,6,$

where $\tilde{r}_{i}^{(k)}$ denotes the $k$-th component of
$\mathbf{\tilde{r}}_{i}$, $k=1,2\cdots,6$, one can derive the following
generalized Riemann invariants for six characteristic fields

$\displaystyle\text{for
}\lambda_{1}=u_{1}-c:\quad\frac{p_{11}}{\rho^{3}},\frac{p_{12}}{\rho^{3}},u_{1}+c,u_{2}+\frac{\sqrt{3}p_{12}}{\sqrt{\rho
p_{11}}},\frac{\det(\mathbf{p})}{\rho^{4}},$ $\displaystyle\text{for
}\lambda_{2}=u_{1}-\frac{c}{\sqrt{3}}:\quad\rho,u_{1},p_{11},u_{2}+\frac{p_{12}}{\sqrt{\rho
p_{11}}},\det(\mathbf{p}),$ $\displaystyle\text{for
}\lambda_{3}=u_{1}:\quad\rho,u_{1},u_{2},p_{11},p_{12},$
$\displaystyle\text{for }\lambda_{4}=u_{1}:\quad
u_{1},u_{2},p_{11},p_{12},p_{22},$ $\displaystyle\text{for
}\lambda_{5}=u_{1}+\frac{c}{\sqrt{3}}:\quad\rho,u_{1},p_{11},u_{2}-\frac{p_{12}}{\sqrt{\rho
p_{11}}},\det(\mathbf{p}),$ $\displaystyle\text{for
}\lambda_{6}=u_{1}+c:\quad\frac{p_{11}}{\rho^{3}},\frac{p_{12}}{\rho^{3}},u_{1}-c,u_{2}-\frac{\sqrt{3}p_{12}}{\sqrt{\rho
p_{11}}},\frac{\det(\mathbf{p})}{\rho^{4}},$

which are constant across corresponding linearly degenerate wave or
rarefaction wave.

##  3 Numerical schemes

This section introduces the 1D direct Eulerian GRP scheme and its extension to
the 2D case by the Strang splitting method.

###  3.1 Outline of 1D GRP scheme

For the sake of simplicity, the space domain is divided into a uniform mesh
$\\{x_{j+\frac{1}{2}},j\in\mathbb{Z}\\}$ with $\Delta
x:=x_{j+\frac{1}{2}}-x_{j-\frac{1}{2}}$, denote the $j$th cell by
$I_{j}:=[x_{j-\frac{1}{2}},x_{j+\frac{1}{2}}]$. Assume that $\Delta
t:=t_{n+1}-t_{n}$, and at the time $t=t_{n}$, the approximate solution is
piecewisely linear with the slope $\bm{\sigma}_{j}^{n}$, $n\geq 0$, and
reconstructed as

$\mathbf{V}(x,t_{n})=\mathbf{V}_{j}^{n}+\bm{\sigma}_{j}^{n}(x-x_{j})\quad\forall
x\in I_{j},$ (3.1)

where $\mathbf{V}_{j}^{n}$ approximates the cell average of
$\mathbf{V}(x,t_{n})$ over the cell $I_{j}$. Hence the second-order Godunov-
type scheme to solve (2.1) is

$\mathbf{U}_{j}^{n+1}=\mathbf{U}_{j}^{n}-\frac{\Delta t}{\Delta
x}\left(\mathbf{F}_{j+\frac{1}{2}}^{n+\frac{1}{2}}-\mathbf{F}_{j-\frac{1}{2}}^{n+\frac{1}{2}}\right)+\frac{\Delta
t}{2}\left(\mathbf{S}_{j+\frac{1}{2}}^{x,n+\frac{1}{2}}+\mathbf{S}_{j-\frac{1}{2}}^{x,n+\frac{1}{2}}\right),$
(3.2)

where
$\mathbf{F}_{j+\frac{1}{2}}^{n+\frac{1}{2}}=\mathbf{F}\big{(}\mathbf{U}(\mathbf{V}_{j+\frac{1}{2}}^{n+\frac{1}{2}})\big{)}$
and
$\mathbf{S}_{j+\frac{1}{2}}^{x,n+\frac{1}{2}}=\mathbf{S}^{x}(\mathbf{U}\big{(}\mathbf{V}_{j+\frac{1}{2}}^{n+\frac{1}{2}})\big{)}$,
$\mathbf{V}_{j+\frac{1}{2}}^{n+\frac{1}{2}}$ denotes a second-order accurate
approximation of $\mathbf{V}(x_{j+\frac{1}{2}},t_{n}+\frac{1}{2}\Delta t)$ and
is analytically derived by resolving the local GRP at each point
$(x_{j+\frac{1}{2}},t_{n})$, i.e.,

$\begin{cases}\eqref{ten-moment2},\quad t-t_{n}>0,\\\
\mathbf{V}(x,t_{n})=\begin{cases}\mathbf{V}_{j}^{n}+\bm{\sigma}_{j}^{n}(x-x_{j}),\quad
x<x_{j+\frac{1}{2}},\\\
\mathbf{V}_{j+1}^{n}+\bm{\sigma}_{j+1}^{n}(x-x_{j+1}),\quad
x>x_{j+\frac{1}{2}}.\end{cases}\end{cases}$ (3.3)

More specifically, $\mathbf{V}_{j+\frac{1}{2}}^{n+\frac{1}{2}}$ is usually
calculated by

$\mathbf{V}_{j+\frac{1}{2}}^{n+\frac{1}{2}}=\mathbf{V}_{j+\frac{1}{2}}^{n}+\frac{\Delta
t}{2}\left(\frac{\partial\mathbf{V}}{\partial t}\right)_{j+\frac{1}{2}}^{n},$
(3.4)

where $\left(\frac{\partial\mathbf{V}}{\partial t}\right)_{j+\frac{1}{2}}^{n}$
is obtained by resolving the problem (3.3), according to those theorems in
Section 5, and
$\mathbf{V}_{j+\frac{1}{2}}^{n}:=\bm{\omega}\left(0;\mathbf{V}_{j+\frac{1}{2},L}^{n},\mathbf{V}_{j+\frac{1}{2},R}^{n}\right)$,
here
$\bm{\omega}\left(\frac{x-x_{j}}{t-t_{n}};\mathbf{V}_{j+\frac{1}{2},L}^{n},\mathbf{V}_{j+\frac{1}{2},R}^{n}\right)$
is the exact solution of the associated Riemann problem for (2.1) centered at
$(x_{j+\frac{1}{2}},t_{n})$, i.e., the Cauchy problem

$\begin{cases}\displaystyle\frac{\partial\mathbf{U(V)}}{\partial
t}+\frac{\partial\mathbf{F(U(V))}}{\partial x}=0,\quad t-t_{n}>0\\\
\mathbf{V}(x,t_{n})=\begin{cases}\mathbf{V}_{j+\frac{1}{2},L}^{n},\quad
x<x_{j+\frac{1}{2}},\\\ \mathbf{V}_{j+\frac{1}{2},R}^{n},\quad
x>x_{j+\frac{1}{2}},\end{cases}\end{cases}$ (3.5)

where $\mathbf{V}_{j+\frac{1}{2},L}^{n}=\mathbf{V}_{j}^{n}+\frac{\Delta
x}{2}\bm{\sigma}_{j}^{n}$ and
$\mathbf{V}_{j+\frac{1}{2},R}^{n}=\mathbf{V}_{j+1}^{n}-\frac{\Delta
x}{2}\bm{\sigma}_{j+1}^{n}$ are the left and right limiting values of
$\mathbf{V}(x,t_{n})$ at $x_{j+\frac{1}{2}}$. The exact solution of Riemann
problem (3.5) is given in Section 4, while the approximate slope
$\bm{\sigma}_{j}^{n}$ is evolved by some slope limiter.

After obtaining $\mathbf{U}_{j}^{n+1}$ by the scheme (3.2), applying some
slope limiter to get the slope of $\mathbf{U}$, denoted by
$\widehat{\bm{\sigma}}_{j}^{n+1}$, and then the slope $\bm{\sigma}_{j}^{n+1}$
of $\mathbf{V}$ can be obtained by the chain rule. The minmod type and van
Leer limiters are utilized here. With the minmod type limiter,
$\widehat{\bm{\sigma}}_{j}^{n+1}$ is calculated componentwisely by

$\widehat{\bm{\sigma}}_{j}^{n+1}=\mathbf{R}_{j}\cdot\text{minmod}\left(\frac{\theta}{\Delta
x}\mathbf{R}_{j}^{-1}\left(\mathbf{U}_{j}^{n+1}-\mathbf{U}_{j-1}^{n+1}\right),\frac{1}{\Delta
x}\mathbf{R}_{j}^{-1}\left(\mathbf{U}_{j+\frac{1}{2}}^{n+1,-}-\mathbf{U}_{j-\frac{1}{2}}^{n+1,-}\right),\frac{\theta}{\Delta
x}\mathbf{R}_{j}^{-1}\left(\mathbf{U}_{j+1}^{n+1}-\mathbf{U}_{j}^{n+1}\right)\right),$
(3.6)

where $\mathbf{R}_{j}:=\mathbf{R}(\mathbf{U}_{j}^{n+1})$, the parameter
$\theta\in[1,2)$, and
$\mathbf{U}_{j+\frac{1}{2}}^{n+1,-}:=\mathbf{U}(\mathbf{V}_{j+\frac{1}{2}}^{n+1,-})$
with
$\mathbf{V}_{j+\frac{1}{2}}^{n+1,-}:=\mathbf{V}_{j+\frac{1}{2}}^{n}+\Delta
t\left(\frac{\partial\mathbf{V}}{\partial t}\right)_{j+\frac{1}{2}}^{n}$.
$\mathbf{R(U)}$ is the right eigenvector matrix of the Jacobian matrix
$\frac{\partial\mathbf{F(U)}}{\partial\mathbf{U}}$ and given in A. With the
van Leer limiter, $\widehat{\bm{\sigma}}_{j}^{n+1}$ is calculated
componentwise by

$\widehat{\bm{\sigma}}_{j}^{n+1}=\mathbf{R}_{j}\cdot\text{vanLeer}\left(\mathbf{R}_{j}^{-1}\mathbf{U}_{j-1}^{n+1},\mathbf{R}_{j}^{-1}\mathbf{U}_{j}^{n+1},\mathbf{R}_{j}^{-1}\mathbf{U}_{j+1}^{n+1}\right),$
(3.7)

where the van Leer limiter function is defined as [45, 60]

$\text{vanLeer}(a_{L},a_{M},a_{R})=\begin{cases}0,&\text{if}~{}a_{L}=a_{R}~{}\text{or}~{}f\leq
0~{}\text{or}~{}f\geq 1,\\\ 4f(1-f)s,&\text{if}~{}a_{L}\neq
a_{R}~{}\text{and}~{}0<f<1,\end{cases}$

where $f=\frac{\Delta_{ML}}{\Delta_{RL}}$, $s=\frac{\Delta_{RL}}{2\Delta x}$,
$\Delta_{ML}=a_{M}-a_{L}$ and $\Delta_{RL}=a_{R}-a_{L}$.

In summary, the second-order direct Eulerian GRP scheme for the 1D ten-moment
equations (2.1) is implemented in the following four steps:

Step 1: Give the piecewise initial data (3.1), and solve the local Riemann
problem (3.5) to get the solution $\mathbf{V}_{j+\frac{1}{2}}^{n}$ by using
the exact Riemann solver stated in Section 4.

Step 2: Calculate $\left(\frac{\partial\mathbf{V}}{\partial
t}\right)_{j+\frac{1}{2}}^{n}$ by those theorems in Section 5, and
$\mathbf{V}_{j+\frac{1}{2}}^{n+\frac{1}{2}}$ in (3.4), then the numerical flux
$\mathbf{F}_{j+\frac{1}{2}}^{n+\frac{1}{2}}$ and the source term
$\mathbf{S}_{j+\frac{1}{2}}^{n+\frac{1}{2}}$ can be obtained.

Step 3: Evaluate the cell averages $\mathbf{U}_{j}^{n+1}$ by the scheme (3.2).

Step 4: Update the slope $\widehat{\bm{\sigma}}_{j}^{n+1}$ of the conservative
variables by the slope limiter (3.6) or (3.7), and calculate the slope
$\bm{\sigma}_{j}^{n+1}$ of the primitive variables by the chain rule, then go
to Step 1 by replacing $n$ with $n+1$.

###  3.2 2D extension of GRP scheme

For simplicity, this work is limited to a 2D uniform Cartesian grid and uses
the Strang splitting method to extend the above GRP scheme to the 2D case.
Consider the 2D ten-moment equations

$\frac{\partial\mathbf{U}}{\partial t}+\frac{\partial\mathbf{F(U)}}{\partial
x}+\frac{\partial\mathbf{G(U)}}{\partial
y}=\mathbf{S}^{x}(\mathbf{U})+\mathbf{S}^{y}(\mathbf{U}),$ (3.8)

where $\mathbf{U}$, $\mathbf{F(U)}$, and $\mathbf{S}^{x}(\mathbf{U})$ are the
same as those in Section 2.1, and the flux $\mathbf{G(U)}$ and the source term
$\mathbf{S}^{y}(\mathbf{U})$ are defined by

$\displaystyle\mathbf{G(U)}=\left(\rho u_{2},\rho u_{1}u_{2}+p_{12},\rho
u_{2}^{2}+p_{22},E_{11}u_{2}+p_{12}u_{1},E_{12}u_{2}+\frac{1}{2}(p_{12}u_{2}+p_{22}u_{1}),(E_{22}+p_{22})u_{2}\right)^{\top},$
$\displaystyle\mathbf{S}^{y}(\mathbf{U})=\left(0,0,-\frac{1}{2}\rho\partial_{y}W,0,-\frac{1}{4}\rho
u_{1}\partial_{y}W,-\frac{1}{2}\rho u_{2}\partial_{y}W\right)^{\top}.$

The 2D ten-moment equations (3.8) are split into two 1D subsystems

$\frac{\partial\mathbf{U}}{\partial t}+\frac{\partial\mathbf{F(U)}}{\partial
x}=\mathbf{S}^{x}(\mathbf{U})\quad\text{and}\quad\frac{\partial\mathbf{U}}{\partial
t}+\frac{\partial\mathbf{G(U)}}{\partial x}=\mathbf{S}^{y}(\mathbf{U}).$

Hence it suffices to consider the derivation of the GRP scheme for the
subsystem in the $x$-direction in Section 3.1. If denoting the 1D GRP
evolution operators for the above two subsystems by $\mathcal{L}_{x}(\Delta
t)$ and $\mathcal{L}_{y}(\Delta t)$, respectively, for one time step $\Delta
t$, then by using the Strang splitting method, the simple 2D GRP scheme can be
given by

$\mathbf{U}^{n+1}=\mathcal{L}_{x}\left(\frac{\Delta
t}{2}\right)\mathcal{L}_{y}(\Delta t)\mathcal{L}_{x}\left(\frac{\Delta
t}{2}\right)\mathbf{U}^{n}.$

##  4 Exact Riemann solver

This section gives the exact Riemann solver for the associated Riemann problem

$\begin{cases}\displaystyle\frac{\partial\mathbf{U(V)}}{\partial
t}+\frac{\partial\mathbf{F(U(V))}}{\partial x}=0,\quad t>0,\\\
\mathbf{V}(x,0)=\begin{cases}\mathbf{V}_{L},\quad x<0,\\\ \mathbf{V}_{R},\quad
x>0.\end{cases}\end{cases}$ (4.1)

As an example, a special local wave configuration shown in Figure 1 for the
associated Riemann problem is only considered here, where there is a
1-rarefaction wave, a 2-shear wave, a 3,4-contact discontinuity, a 5-shear
wave and a 6-shock wave. Other local wave configurations may be similarly
treated. One can see from Figure 1 that there are five solution states between
the given left and right states $\mathbf{V}_{L}$ and $\mathbf{V}_{R}$ to be
determined, that is, the state inside the 1-rarefaction wave denoted by
$\mathbf{V}_{\text{Lfan}}$ and four solution states between the $i$-wave and
the $(i+1)$-wave ($1\leq i<6$) denoted by

$\begin{cases}\mathbf{V}_{\ast L}=(\rho_{\ast L},u_{1,\ast},u_{2,\ast
L},p_{11,\ast},p_{12,\ast L},p_{22,\ast L})^{\top},\\\ \mathbf{V}_{\ast\ast
L}=(\rho_{\ast
L},u_{1,\ast},u_{2,\ast\ast},p_{11,\ast},p_{12,\ast\ast},p_{22,\ast\ast
L})^{\top},\\\ \mathbf{V}_{\ast\ast R}=(\rho_{\ast
R},u_{1,\ast},u_{2,\ast\ast},p_{11,\ast},p_{12,\ast\ast},p_{22,\ast\ast
R})^{\top},\\\ \mathbf{V}_{\ast R}=(\rho_{\ast R},u_{1,\ast},u_{2,\ast
R},p_{11,\ast},p_{12,\ast R},p_{22,\ast R})^{\top}.\end{cases}$

To determine them, the generalized Riemann invariants will be used for the
rarefaction wave, the shear waves and the contact discontinuity, while the
Rankine-Hugoniot jump conditions will be used for the shock wave. For the sake
of convenience, the main steps of the exact Riemann solver for (4.1) is shown
in Figure 2.

Figure 1: The schematic description of a local wave configuration of the
associated Riemann problem (4.1). Compute $u_{1,\ast}$ and $p_{11,\ast}$, see
Theorem 4.1 and $\mathbf{V}_{\text{Lfan}}$, see (4.3). Compute
$\mathbf{V}_{\ast L}$, see (4.4). Compute $\mathbf{V}_{\ast R}$, see (4.6),
(4.7) and (4.8). Compute $\mathbf{V}_{\ast\ast K}$ ($K=L,R$), see (4.11) and
(4.12). Figure 2: The main steps of the exact Riemann solver in Section 4.

###  4.1 Computations of $u_{1,\ast}$ and $p_{11,\ast}$

Analogous to the derivation for the Euler equations in [44, Proposition 4.1],
one can prove the following results for the 1D ten-moment equations.

######  Theorem 4.1 (Computing $p_{11,\ast}$ and $u_{1,\ast}$).

The pressure component $p_{11,\ast}$ of the Riemann problem (4.1) is given by
the root of the algebraic equation

$f(p_{11},\mathbf{V}_{L},\mathbf{V}_{R}):=f_{L}(p_{11},\mathbf{V}_{L})+f_{R}(p_{11},\mathbf{V}_{R})+u_{1,R}-u_{1,L}=0,$
(4.2)

where the function $f_{L}$ is given by

$f_{L}(p_{11},\mathbf{V}_{L})=\begin{cases}(p_{11}-p_{11,L})\left[\frac{1}{\rho_{L}(2p_{11}+p_{11,L})}\right]^{\frac{1}{2}},&\mbox{if
}p_{11}>p_{11,L}~{}\text{(1-shock)},\\\
c_{L}\left[\left(\frac{p_{11}}{p_{11,L}}\right)^{\frac{1}{3}}-1\right],&\mbox{if
}p_{11}\leq p_{11,L}~{}\text{(1-rarefaction)},\end{cases}$

and the function $f_{R}$ is given by

$f_{R}(p_{11},\mathbf{V}_{R})=\begin{cases}(p_{11}-p_{11,R})\left[\frac{1}{\rho_{R}(2p_{11}+p_{11,R})}\right]^{\frac{1}{2}},&\mbox{if
}p_{11}>p_{11,R}~{}\text{(6-shock)},\\\
c_{R}\left[\left(\frac{p_{11}}{p_{11,R}}\right)^{\frac{1}{3}}-1\right],&\mbox{if
}p_{11}\leq p_{11,R}~{}\text{(6-rarefaction)}.\end{cases}$

Moreover, the velocity $u_{1,\ast}$ may be given by

$u_{1,\ast}=\frac{1}{2}(u_{1,L}+u_{1,R})+\frac{1}{2}[f_{R}(p_{11,\ast},\mathbf{V}_{R})-f_{L}(p_{11,\ast},\mathbf{V}_{L})].$

###### Proof.

The proof is similar to that of the Euler equations, see [44, Subsection 4.2],
with replacing the adiabatic index $\gamma$ with 3. ∎

To find the root of the nonlinear equation (4.2), the Newton-Raphson iterative
method is applied, where the initial guess value may be obtained in a adaptive
way, similar to that for the Euler equations [44, Section 9.5 and Figure 9.4].

###  4.2 Computations of $\mathbf{V}_{\text{Lfan}}$

The 1-rarefaction wave, which is identified by the condition $p_{11,\ast}\leq
p_{11,L}$, is enclosed by the head and the tail, whose characteristic speeds
are given respectively by

$S_{\text{HL}}=u_{1,L}-c_{L},\quad S_{\text{TL}}=u_{1,\ast}-c_{\ast L},$

where $c_{\ast
L}=c_{L}\left(\frac{p_{11,\ast}}{p_{11,L}}\right)^{\frac{1}{3}}$. The solution
$\mathbf{V}_{\text{Lfan}}$ inside the 1-rarefaction fan is easily obtained by
considering the characteristic ray through the origin $(0,0)$ and a general
point $(x,t)$ inside the fan. The slope of the characteristic line is

$\frac{\text{d}x}{\text{d}t}=\frac{x}{t}=u_{1}-c.$

Using the generalized Riemann invariants associated with the 1-rarefaction
wave yields

$\mathbf{V}_{\text{Lfan}}\begin{cases}\rho=\frac{\rho_{L}}{2}\left[1+\frac{1}{c_{L}}\left(u_{1,L}-\frac{x}{t}\right)\right],\\\
u_{1}=\frac{1}{2}\left(c_{L}+u_{1,L}+\frac{x}{t}\right),\\\
u_{2}=u_{2,L}+\frac{\sqrt{3}p_{12,L}}{\sqrt{\rho_{L}p_{11,L}}}-\frac{\sqrt{3}p_{12}}{\sqrt{\rho
p_{11}}},\\\
p_{11}=\frac{p_{11,L}}{8}\left[1+\frac{1}{c_{L}}\left(u_{1,L}-\frac{x}{t}\right)\right]^{3},\\\
p_{12}=p_{12,L}\frac{\rho^{3}}{\rho_{L}^{3}},\\\
p_{22}=\left[\frac{\det(\mathbf{p}_{L})}{\rho_{L}^{4}}\rho^{4}+p_{12}^{2}\right]/p_{11}.\end{cases}$
(4.3)

######  Remark 4.2.

For the 6-rarefaction wave, which is identified by the condition
$p_{11,\ast}\leq p_{11,R}$ and enclosed by the characteristic speeds given
respectively by $S_{\text{HR}}=u_{1,R}+c_{R}$ and
$S_{\text{TR}}=u_{1,\ast}+c_{\ast R}$ with $c_{\ast
R}=c_{R}\left(\frac{p_{11,\ast}}{p_{11,R}}\right)^{\frac{1}{3}}$, the solution
$\mathbf{V}_{\text{Rfan}}$ is given by

$\mathbf{V}_{\text{Rfan}}\begin{cases}\rho=\frac{\rho_{R}}{2}\left[1+\frac{1}{c_{R}}\left(\frac{x}{t}-u_{1,R}\right)\right],\\\
u_{1}=\frac{1}{2}\left(u_{1,R}-c_{R}+\frac{x}{t}\right),\\\
u_{2}=u_{2,R}-\frac{\sqrt{3}p_{12,R}}{\sqrt{\rho_{R}p_{11,R}}}+\frac{\sqrt{3}p_{12}}{\sqrt{\rho
p_{11}}},\\\
p_{11}=\frac{p_{11,R}}{8}\left[1+\frac{1}{c_{R}}\left(\frac{x}{t}-u_{1,R}\right)\right]^{3},\\\
p_{12}=p_{12,R}\frac{\rho^{3}}{\rho_{R}^{3}},\\\
p_{22}=\left[\frac{\det(\mathbf{p}_{R})}{\rho_{R}^{4}}\rho^{4}+p_{12}^{2}\right]/p_{11}.\end{cases}$

###  4.3 Computing $\mathbf{V}_{\ast L}$

For the state $\mathbf{V}_{\ast L}$, which is between the 1-rarefaction wave
and the 2-shear wave, utilizing the generalized Riemann invariants across the
1-rarefaction wave can obtain

$\displaystyle\rho_{\ast
L}=\rho_{L}\left(\frac{p_{11,\ast}}{p_{11,L}}\right)^{\frac{1}{3}},\quad
u_{2,\ast
L}=u_{2,L}+\frac{\sqrt{3}p_{12,L}}{\sqrt{\rho_{L}p_{11,L}}}-\frac{\sqrt{3}p_{12,\ast
L}}{\sqrt{\rho_{\ast L}p_{11,\ast}}},$ $\displaystyle p_{12,\ast
L}=p_{12,L}\frac{\rho_{\ast L}^{3}}{\rho_{L}^{3}},\quad p_{22,\ast
L}=\left[\frac{\det(\mathbf{p}_{L})}{\rho_{L}^{4}}\rho_{\ast L}^{4}+p_{12,\ast
L}^{2}\right]/p_{11,\ast}.$ (4.4)

######  Remark 4.3.

For the state $\mathbf{V}_{\ast R}$, which is between the 6-rarefaction wave
and the 5-shear wave, use of the generalized Riemann invariants across the
6-rarefaction wave may give

$\displaystyle\rho_{\ast
R}=\rho_{R}\left(\frac{p_{11,\ast}}{p_{11,R}}\right)^{\frac{1}{3}},\quad
u_{2,\ast
R}=u_{2,R}-\frac{\sqrt{3}p_{12,R}}{\sqrt{\rho_{R}p_{11,R}}}+\frac{\sqrt{3}p_{12,\ast
R}}{\sqrt{\rho_{\ast R}p_{11,\ast}}},$ $\displaystyle p_{12,\ast
R}=p_{12,R}\frac{\rho_{\ast R}^{3}}{\rho_{R}^{3}},\quad p_{22,\ast
R}=\left[\frac{\det(\mathbf{p}_{R})}{\rho_{R}^{4}}\rho_{\ast R}^{4}+p_{12,\ast
R}^{2}\right]/p_{11,\ast}.$

###  4.4 Computing $\mathbf{V}_{\ast R}$

The solution state $\mathbf{V}_{\ast R}$ is between the 5-shear wave and the
6-shock wave, while the 6-shock wave with the following speed

$\sigma_{R}=u_{1,R}+c_{R}\sqrt{\frac{2p_{11,\ast}}{3p_{11,R}}+\frac{1}{3}}{=u_{1,\ast}+\frac{1}{\rho_{\ast
R}}\sqrt{(2p_{11,\ast}+p_{11,R})\rho_{R}}}$ (4.5)

is identified by the condition $p_{11,\ast}>p_{11,R}$. The derivation of (4.5)
is similar to that of the shock speed for the Euler equations, see Section
3.1.3 in [44]. Moreover, (4.5) implies that $u_{1,\ast}<\sigma_{R}$. By
applying the Rankine-Hugoniot jump conditions for the variables $\rho$,
$m_{1}$ and $E_{11}$, similar to the discussion for the Euler equations in
[44], one can obtain

$\rho_{\ast R}=\rho_{R}\frac{2p_{11,\ast}+p_{11,R}}{p_{11,\ast}+2p_{11,R}}.$
(4.6)

The condition $p_{11,\ast}>p_{11,R}$ implies that $\rho_{\ast
R}\in(\rho_{R},2\rho_{R})$.

Using the Rankine-Hugoniot jump conditions for the variables $m_{2}$ and
$E_{12}$ yields

$\mathbf{A}^{R}\left(\begin{array}[]{c}u_{2,\ast R}\\\ p_{12,\ast
R}\end{array}\right)=\left(\begin{array}[]{c}a_{1}^{R}\\\
a_{2}^{R}\end{array}\right),$ (4.7)

where

$\displaystyle\mathbf{A}^{R}:=\begin{pmatrix}\rho_{\ast
R}(u_{1,\ast}-\sigma_{R})&1\\\ E_{11,\ast R}-\frac{1}{2}\rho_{\ast
R}u_{1,\ast}\sigma_{R}&u_{1,\ast}-\frac{1}{2}\sigma_{R}\end{pmatrix},$
$\displaystyle a_{1}^{R}:=\rho_{R}(u_{1,R}-\sigma_{R})u_{2,R}+p_{12,R},$
$\displaystyle
a_{2}^{R}:=E_{12,R}(u_{1,R}-\sigma_{R})+\frac{1}{2}(p_{11,R}u_{2,R}+p_{12,R}u_{1,R}).$

Due to

$\det(\mathbf{A}^{R})=\frac{p_{11,\ast}(2\rho_{R}-\rho_{\ast
R})+p_{11,R}\rho_{R}}{2\rho_{\ast R}}>0,$

the solution of (4.7) exists and is unique.

By the Rankine-Hugoniot jump conditions for the variable $E_{22}$, one has

$E_{22,\ast R}u_{1,\ast}+p_{12,\ast R}u_{2,\ast
R}-(E_{22,R}u_{1,R}+p_{12,R}u_{2,R})=\sigma_{R}(E_{22,\ast R}-E_{22,R}),$

which implies that

$E_{22,\ast
R}=\frac{E_{22,R}u_{1,R}+p_{12,R}u_{2,R}-\sigma_{R}E_{22,R}-p_{12,\ast
R}u_{2,\ast R}}{u_{1,\ast}-\sigma_{R}},$

and then using the “equation of state” yields

$p_{22,\ast R}=2E_{22,\ast R}-\rho_{\ast R}u_{2,\ast R}^{2}.$ (4.8)

######  Remark 4.4.

For the 1-shock wave with the speed

$\sigma_{L}=u_{1,L}-c_{L}\sqrt{\frac{2p_{11,\ast}}{3p_{11,L}}+\frac{1}{3}}={u_{1,\ast}-\frac{1}{\rho_{\ast
L}}\sqrt{(2p_{11,\ast}+p_{11,L})\rho_{L}}},$

which is identified by the condition $p_{11,\ast}>p_{11,L}$, one can obtain

$\rho_{\ast L}=\rho_{L}\frac{2p_{11,\ast}+p_{11,L}}{p_{11,\ast}+2p_{11,L}}.$

The variables $u_{2,\ast L}$ and $p_{12,\ast L}$ are obtained by solving the
following system

$\mathbf{A}^{L}\left(\begin{array}[]{c}u_{2,\ast L}\\\ p_{12,\ast
L}\end{array}\right)=\left(\begin{array}[]{c}a_{1}^{L}\\\
a_{2}^{L}\end{array}\right),$

where

$\displaystyle\mathbf{A}^{L}:=\begin{pmatrix}\rho_{\ast
L}(u_{1,\ast}-\sigma_{L})&1\\\ E_{11,\ast L}-\frac{1}{2}\rho_{\ast
L}u_{1,\ast}\sigma_{L}&u_{1,\ast}-\frac{1}{2}\sigma_{L}\end{pmatrix},$
$\displaystyle a_{1}^{L}:=\rho_{L}(u_{1,L}-\sigma_{L})u_{2,L}+p_{12,L},$
$\displaystyle
a_{2}^{L}:=E_{12,L}(u_{1,L}-\sigma_{L})+\frac{1}{2}(p_{11,L}u_{2,L}+p_{12,L}u_{1,L}).$

The variable $p_{22,\ast L}$ is given by

$p_{22,\ast L}=2E_{22,\ast L}-\rho_{\ast L}u_{2,\ast L}^{2},$

where

$E_{22,\ast
L}=\frac{E_{22,L}u_{1,L}+p_{12,L}u_{2,L}-\sigma_{L}E_{22,L}-p_{12,\ast
L}u_{2,\ast L}}{u_{1,\ast}-\sigma_{L}}.$

###  4.5 Computing $\mathbf{V}_{\ast\ast L}$ and $\mathbf{V}_{\ast\ast R}$

The solution state $\mathbf{V}_{\ast\ast L}$ is between the 2-shear wave and
the contact discontinuity, while $\mathbf{V}_{\ast\ast R}$ is between the
5-shear wave and the contact discontinuity. Across the 2-shear wave, utilizing
the generalized Riemann invariant $u_{2}+\frac{p_{12}}{\sqrt{\rho p_{11}}}$
gives

$u_{2,\ast\ast}+\frac{p_{12,\ast\ast}}{\sqrt{\rho_{\ast
L}p_{11,\ast}}}=a_{3}^{L},$ (4.9)

with $a_{3}^{L}:=u_{2,\ast L}+\frac{p_{12,\ast L}}{\sqrt{\rho_{\ast
L}p_{11,\ast}}}$. For the 5-shear wave, using the generalized Riemann
invariant $u_{2}-\frac{p_{12}}{\sqrt{\rho p_{11}}}$ yields

$u_{2,\ast\ast}-\frac{p_{12,\ast\ast}}{\sqrt{\rho_{\ast
R}p_{11,\ast}}}=a_{3}^{R},$ (4.10)

with $a_{3}^{R}:=u_{2,\ast R}-\frac{p_{12,\ast R}}{\sqrt{\rho_{\ast
R}p_{11,\ast}}}$. Solving (4.9) and (4.10) gives

$\begin{split}&u_{2,\ast\ast}=\frac{a_{3}^{L}\sqrt{\rho_{\ast
L}}+a_{3}^{R}\sqrt{\rho_{\ast R}}}{\sqrt{\rho_{\ast L}}+\sqrt{\rho_{\ast
R}}},\\\ &p_{12,\ast\ast}=\frac{(a_{3}^{L}-a_{3}^{R})\sqrt{\rho_{\ast
L}\rho_{\ast R}p_{11,\ast}}}{\sqrt{\rho_{\ast L}}+\sqrt{\rho_{\ast
R}}}.\end{split}$ (4.11)

Moreover, the fact that $\det(\mathbf{p})$ is the generalized Riemann
invariant of both the 2-shear wave and the 5-shear wave implies that

$p_{22,\ast\ast K}=\frac{\det(\mathbf{p}_{\ast
K})+p_{12,\ast\ast}^{2}}{p_{11,\ast}},\quad K=L,R.$ (4.12)

##  5 Resolution of the GRP

This section resolves the following GRP problem

$\begin{cases}\eqref{ten-moment2},\quad t>0,\\\
\mathbf{V}(x,0)=\begin{cases}\mathbf{V}_{L}+x\mathbf{V}_{L}^{\prime},\quad
x<0,\\\ \mathbf{V}_{R}+x\mathbf{V}_{R}^{\prime},\quad
x>0,\end{cases}\end{cases}$ (5.1)

to derive the limiting value of $\frac{\partial\mathbf{V}}{\partial t}$ at
$x=0$, as $t\rightarrow 0+$, where
$\mathbf{V}_{K}^{\prime}:=(\rho_{K}^{\prime},u_{1,K}^{\prime},u_{2,K}^{\prime},p_{11,K}^{\prime},p_{12,K}^{\prime},p_{22,K}^{\prime})^{\top}$,
$K=L,R$, are the constant slope vectors. The initial structure of the solution
$\mathbf{V}^{{\rm GRP}}(x,t)$ of (5.1) is determined by the exact solution
$\bm{\omega}\left(\frac{x}{t};\mathbf{V}_{L},\mathbf{V}_{R}\right)$ of the
Riemann problem (4.1) and

$\lim_{t\rightarrow 0+}\mathbf{V}^{{\rm GRP}}(\lambda
t,t)=\bm{\omega}\left(0;\mathbf{V}_{L},\mathbf{V}_{R}\right)=:\mathbf{V}^{{\rm
RP}},\quad x=\lambda t.$

The local wave pattern around the singularity point $(x,t)=(0,0)$ of the GRP
(5.1) typically exhibits piecewise smoothness and comprises elementary wave
types, including the rarefaction wave, the shock wave, the shear wave, and the
contact discontinuity, as depicted schematically in Figure 3.

Figure 3: The schematic description of a local wave configuration of the GRP
(5.1) with $0\leq t\ll 1$.

The rarefaction waves in the solution of the Riemann problem (4.1) exhibit
isentropic flow properties, making the generalized Riemann invariants constant
with vanishing derivatives within $i$th-rarefaction wave ($i=1,6$), but
unfortunately, those properties do not generally hold for the generalized
Riemann problem (5.1) due to the need to consider the curved rarefaction
waves. Nonetheless, for a short time period $0\leq t\ll 1$, the solution of
(5.1) may be considered as a perturbation of the solution of (4.1), allowing
us to expect that the generalized Riemann invariants remain regular within the
$i$th-rarefaction waves ($i=1,6$) of the solution $\mathbf{V}^{{\rm
GRP}}(x,t)$ near the singularity point $(x,t)=(0,0)$. Consequently, the
generalized Riemann invariants are still utilized to resolve the rarefaction
waves around this singularity point.

From this, as an example, we will continue to focus our attention on the
specific local wave configuration depicted in Figure 3, corresponding to
Figure 1. In this configuration, a rarefaction wave propagates to the left,
while a shock wave moves to the right. The intermediate region between them is
separated by two shear waves and a contact discontinuity. It is worth noting
that other local wave configurations can be analyzed in a similar manner. In
the subsequent subsections, the nonsonic case (see Subsection 5.1), the sonic
case (see Subsection 5.2), and the acoustic case (see Subsection 5.3) will be
separately discussed in detail to compute the limiting value of
$\frac{\partial\mathbf{V}}{\partial t}(0,t)$ as $t\rightarrow 0+$.

###  5.1 Nonsonic case

If the $t$-axis is located between the 1-wave and the 6-wave, the nonsonic
case happens. Denote the limiting values of
$\frac{\partial\mathbf{V}}{\partial t}(0,t)$ when $t\rightarrow 0+$ in the
four middle domains by $\left(\frac{\partial\mathbf{V}}{\partial
t}\right)_{\ast L}$, $\left(\frac{\partial\mathbf{V}}{\partial
t}\right)_{\ast\ast L}$, $\left(\frac{\partial\mathbf{V}}{\partial
t}\right)_{\ast\ast R}$ and $\left(\frac{\partial\mathbf{V}}{\partial
t}\right)_{\ast R}$, respectively. The derivations of those limiting values
are very long-winded and tedious, and for the sake of convenience, the main
steps in this subsection are outlined in Figure 4.

Compute $(\frac{\partial u_{1}}{\partial t})_{\ast K}$ (=$(\frac{\partial
u_{1}}{\partial t})_{\ast\ast K}$), $(\frac{\partial p_{11}}{\partial
t})_{\ast K}$ (=$(\frac{\partial p_{11}}{\partial t})_{\ast\ast K}$) and
$(\frac{\partial\rho}{\partial t})_{\ast K}$ (=$(\frac{\partial\rho}{\partial
t})_{\ast\ast K}$) ($K=L,R$), see Theorem 5.6. Compute $(\frac{\partial
u_{2}}{\partial t})_{\ast L}$, $(\frac{\partial p_{12}}{\partial t})_{\ast L}$
and $(\frac{\partial p_{22}}{\partial t})_{\ast L}$, see Theorem 5.10. Compute
$(\frac{\partial u_{2}}{\partial t})_{\ast R}$, $(\frac{\partial
p_{12}}{\partial t})_{\ast R}$ and $(\frac{\partial p_{22}}{\partial t})_{\ast
R}$, see Theorem 5.14. Compute $\left(\frac{\partial u_{2}}{\partial
t}\right)_{\ast\ast K}$, $\left(\frac{\partial p_{12}}{\partial
t}\right)_{\ast\ast K}$ and $\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast\ast K}$ $(K=L,R)$, see Theorem 5.20 and Theorem 5.21. Figure
4: The main steps for the nonsonic case in Subsection 5.1.

####  5.1.1 Computing $(\frac{\partial u_{1}}{\partial t})_{\ast K}$,
$(\frac{\partial p_{11}}{\partial t})_{\ast K}$ and
$(\frac{\partial\rho}{\partial t})_{\ast K}$ ($K=L,R$)

Similar to the second-order accurate GRP scheme in [5, 55, 56, 19], the region
of the 1-rarefaction wave associated with $\lambda_{1}=u_{1}-c$ can be
described by the characteristic coordinates $(\alpha,\beta)$, where
$\alpha\in[-\infty,0]$, $\beta\in[\beta_{L},\beta_{\ast L}]$ with
$\beta_{L}=\lambda_{1}(\mathbf{V}_{L})$ and $\beta_{\ast
L}=\lambda_{1}(\mathbf{V}_{\ast L})$, see Figure 3, and $\beta=\beta(x,t)$ and
$\alpha=\alpha(x,t)$ are the integral curves of the following two equations

$\frac{\text{d}x}{\text{d}t}=u_{1}-c,\quad\frac{\text{d}x}{\text{d}t}=u_{1}+c,$
(5.2)

respectively. Here, $\beta$ represents the initial value of the slope
$\lambda_{1}=u_{1}-c$ at the singularity point (0,0), and $\alpha$ denotes the
$x$-coordinate of the intersection point between the transversal
characteristic curve and the leading $\beta$-curve, as illustrated in Figure
3. Two equations in (5.2) may give

$\frac{\partial x}{\partial\alpha}=(u_{1}-c)\frac{\partial
t}{\partial\alpha},\quad\frac{\partial
x}{\partial\beta}=(u_{1}+c)\frac{\partial t}{\partial\beta},$ (5.3)

which further imply that

$\frac{\partial^{2}t}{\partial\alpha\partial\beta}=\frac{1}{2c}\left(\frac{\partial(u_{1}-c)}{\partial\beta}\frac{\partial
t}{\partial\alpha}-\frac{\partial(u_{1}+c)}{\partial\alpha}\frac{\partial
t}{\partial\beta}\right).$ (5.4)

Similarly, for the associated Riemann problem, there exists a local coordinate
transformation $x_{\text{ass}}=x(\alpha,\beta)$,
$t_{\text{ass}}=t(\alpha,\beta)$. Both local coordinate transformations
satisfy the properties [5]

$\frac{\partial\lambda_{1}}{\partial t}(0,\beta)=1,~{}\frac{\partial
t}{\partial\alpha}(0,\beta)=\frac{\partial
t_{\text{ass}}}{\partial\alpha}(0,\beta),~{}\frac{\partial
t}{\partial\beta}(0,\beta)=0,~{}\beta_{L}\leq\beta\leq\beta_{\ast L}.$ (5.5)

It follows from (5.4) that

$\frac{\partial^{2}t}{\partial\alpha\partial\beta}(0,\beta)=\frac{1}{2c(0,\beta)}\frac{\partial
t}{\partial\alpha}(0,\beta).$ (5.6)

Let $\psi_{1}:=u_{1}+c$ and $S_{1}:=\frac{p_{11}}{\rho^{3}}$. Following the
same derivation in [5], one has the following explicit expressions of the
local coordinate transformation in the 1-rarefaction wave of the associated
Riemann problem

$t_{\text{ass}}(\alpha,\beta)=\frac{\alpha}{\psi_{1,L}-\beta},\ \
x_{\text{ass}}(\alpha,\beta)=\frac{\alpha\beta}{\psi_{1,L}-\beta}.$ (5.7)

Because $\psi_{1}=u_{1}+c$, $\beta=u_{1}(0,\beta)-c(0,\beta)$ and $\psi_{1}$
is a generalized Riemann invariant across the 1-rarefaction wave, one has

$c(0,\beta)=\frac{1}{2}(\psi_{1,L}-\beta),~{}u_{1}(0,\beta)=\frac{1}{2}(\psi_{1,L}+\beta).$
(5.8)

Combining (5.5), (5.7) with (5.8) gives

$\frac{\partial t}{\partial\alpha}(0,\beta)=\frac{1}{2c(0,\beta)}.$ (5.9)

Due to
$\text{d}\psi_{1}=\text{d}u_{1}-\frac{c}{2\rho}\text{d}\rho+\frac{3}{2\rho
c}\text{d}p_{11}$ and
$\text{d}S_{1}=\frac{-3p_{11}}{\rho^{4}}\text{d}\rho+\frac{1}{\rho^{3}}\text{d}p_{11}$,
using (2.5a), (2.5b) and (2.5d) yields

$\displaystyle\frac{\partial\psi_{1}}{\partial
t}+(u_{1}+c)\frac{\partial\psi_{1}}{\partial x}=\Pi_{1},$ (5.10)
$\displaystyle\frac{\partial S_{1}}{\partial t}+u_{1}\frac{\partial
S_{1}}{\partial x}=0,$ (5.11)

where $\Pi_{1}:=-\frac{c^{2}}{2\rho}\frac{\partial\rho}{\partial
x}+\frac{1}{2\rho}\frac{\partial p_{11}}{\partial x}-\frac{1}{2}W_{x}$.
Besides, one has

$\text{d}p_{11}=c^{2}\text{d}\rho+\rho^{3}\text{d}S_{1}.$ (5.12)

It follows that

$\Pi_{1}=\frac{1}{2\rho}\left(\frac{\partial p_{11}}{\partial
x}-c^{2}\frac{\partial\rho}{\partial
x}\right)-\frac{1}{2}W_{x}=\frac{1}{2}\rho^{2}\frac{\partial S_{1}}{\partial
x}-\frac{1}{2}W_{x}.$ (5.13)

For $\psi_{1}$, by the total differentials of $\psi_{1}$ and $S_{1}$, one has

$\text{d}\psi_{1}=\text{d}u_{1}+\frac{1}{\rho
c}\text{d}p_{11}+\frac{\rho^{2}}{2c}\text{d}S_{1}.$ (5.14)

The following lemma gives the expressions of $\left(\rho^{2}\frac{\partial
S_{1}}{\partial t}\right)(0,\beta)$ and $\frac{\partial\psi_{1}}{\partial
t}(0,\beta)$.

######  Lemma 5.1.

If assuming that the 1-rarefaction wave associated with $u_{1}-c$ moves to the
left, and considering the generalized Riemann invariants $\psi_{1},S_{1}$ and
their time derivatives $\frac{\partial\psi_{1}}{\partial t}$, $\frac{\partial
S_{1}}{\partial t}$ as continuous functions of $\alpha,\beta$ with
$-\alpha_{0}\leq\alpha\leq 0$ and $\beta_{L}\leq\beta\leq\beta_{\ast L}$, then
one has

$\displaystyle\left(\rho^{2}\frac{\partial S_{1}}{\partial
t}\right)(0,\beta)=-\frac{\rho_{L}^{2}S_{1,L}^{\prime}}{c_{L}^{3}}[\beta+c(0,\beta)]\cdot
c^{3}(0,\beta),$ (5.15) $\displaystyle\frac{\partial\psi_{1}}{\partial
t}(0,\beta)=\frac{\rho_{L}^{2}S_{1,L}^{\prime}}{8c_{L}^{3}}[\psi_{1,L}(3c_{L}^{2}-c^{2})-2\beta
c^{2}](0,\beta)-\psi_{1,L}^{\prime}\psi_{1}(0,\beta)-\frac{1}{2}W_{x}(0),$
(5.16)

where

$\displaystyle
S_{1,L}^{\prime}=\frac{1}{\rho_{L}^{3}}(p_{11,L}^{\prime}-c_{L}^{2}\rho_{L}^{\prime}),\quad\psi_{1,L}^{\prime}=u_{1,L}^{\prime}+\frac{1}{\rho_{L}c_{L}}p_{11,L}^{\prime}+\frac{\rho_{L}^{2}}{2c_{L}}S_{1,L}^{\prime},$
(5.17)

obtained by (5.12) and (5.14).

###### Proof.

(i) Computing $\left(\rho^{2}\frac{\partial S_{1}}{\partial
t}\right)(0,\beta)$. Using (5.3) and (5.11) gives

$\displaystyle\frac{\partial S_{1}}{\partial\beta}=\frac{\partial
t}{\partial\beta}\left[\frac{\partial S_{1}}{\partial
t}+(u_{1}+c)\frac{\partial S_{1}}{\partial x}\right]=\frac{\partial
t}{\partial\beta}\cdot c\frac{\partial S_{1}}{\partial x},$ (5.18)
$\displaystyle\frac{\partial S_{1}}{\partial\alpha}=\frac{\partial
t}{\partial\alpha}\left[\frac{\partial S_{1}}{\partial
t}+(u_{1}-c)\frac{\partial S_{1}}{\partial x}\right]=-\frac{\partial
t}{\partial\alpha}\cdot c\frac{\partial S_{1}}{\partial x}.$ (5.19)

Differentiating (5.18) with respect to $\alpha$ and noting that
$\frac{\partial t}{\partial\beta}(0,\beta)=0$, one gets

$\frac{\partial}{\partial\beta}\left(\frac{\partial
S_{1}}{\partial\alpha}(0,\beta)\right)=\frac{1}{2}\frac{\partial
t}{\partial\alpha}(0,\beta)\frac{\partial S_{1}}{\partial
x}(0,\beta)=-\frac{1}{2c(0,\beta)}\frac{\partial
S_{1}}{\partial\alpha}(0,\beta),$ (5.20)

where (5.6) and (5.19) have been used in the first and second equality,
respectively. Integrating (5.20) from $\beta_{L}$ to $\beta$ yields

$\frac{\partial S_{1}}{\partial\alpha}(0,\beta)=\frac{\partial
S_{1}}{\partial\alpha}(0,\beta_{L})\cdot\exp\left(-\int_{\beta_{L}}^{\beta}\frac{1}{2c(0,\eta)}\text{d}\eta\right)=\frac{\partial
S_{1}}{\partial\alpha}(0,\beta_{L})\cdot\frac{c(0,\beta)}{c_{L}}.$ (5.21)

Similarly, from (5.6), one has

$\frac{\partial t}{\partial\alpha}(0,\beta)=\frac{\partial
t}{\partial\alpha}(0,\beta_{L})\cdot\frac{c_{L}}{c(0,\beta)}.$ (5.22)

By (5.19), (5.21) and (5.22), one has

$\displaystyle\frac{\partial S_{1}}{\partial x}(0,\beta)$
$\displaystyle=-\frac{\partial
S_{1}}{\partial\alpha}(0,\beta)\cdot\left(c\frac{\partial
t}{\partial\alpha}\right)^{-1}(0,\beta)$ $\displaystyle=-\frac{\partial
S_{1}}{\partial\alpha}(0,\beta_{L})\cdot\frac{c(0,\beta)}{c_{L}}\frac{1}{c(0,\beta)}\left(\frac{\partial
t}{\partial\alpha}\right)^{-1}(0,\beta_{L})\cdot\frac{c(0,\beta)}{c_{L}}$
$\displaystyle=-\frac{\partial
S_{1}}{\partial\alpha}(0,\beta_{L})\cdot\left(\frac{\partial
t}{\partial\alpha}\right)^{-1}(0,\beta_{L})\cdot\frac{c(0,\beta)}{c_{L}^{2}}$
$\displaystyle=c_{L}\frac{\partial S_{1}}{\partial
x}(0,\beta_{L})\cdot\frac{c(0,\beta)}{c_{L}^{2}}=\frac{S_{1,L}^{\prime}}{c_{L}}c(0,\beta),$
(5.23)

It follows that

$\frac{\partial S_{1}}{\partial t}(0,\beta)=-u_{1}(0,\beta)\cdot\frac{\partial
S_{1}}{\partial
x}(0,\beta)=-\frac{S_{1,L}^{\prime}}{c_{L}}[\beta+c(0,\beta)]\cdot
c(0,\beta).$ (5.24)

Because $S_{1}=p_{11}/\rho^{3}$ is a generalized Riemann invariant for the
1-rarefaction wave, it holds that

$\frac{\rho^{2}(0,\beta)}{\rho_{L}^{2}}=\frac{(3p_{11}/\rho)(0,\beta)}{3p_{11,L}/\rho_{L}}=\frac{c^{2}(0,\beta)}{c_{L}^{2}},$
(5.25)

and then combining (5.24) and (5.25) gives (5.15).

(ii) Computing $\frac{\partial\psi_{1}}{\partial t}(0,\beta)$. Using (5.10)
gives

$\displaystyle\frac{\partial\psi_{1}}{\partial\beta}=\frac{\partial
t}{\partial\beta}\Pi_{1}(\alpha,\beta),$ (5.26)
$\displaystyle\frac{\partial\psi_{1}}{\partial\alpha}=\frac{\partial
t}{\partial\alpha}\left[\frac{\partial\psi_{1}}{\partial
t}+(u_{1}-c)\frac{\partial\psi_{1}}{\partial x}\right]=\frac{\partial
t}{\partial\alpha}\left[\Pi_{1}(\alpha,\beta)-2c\frac{\partial\psi_{1}}{\partial
x}\right].$ (5.27)

Differentiating (5.26) with respect to $\alpha$ and using (5.5), (5.6) gets

$\frac{\partial}{\partial\beta}\left(\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta)\right)=\frac{1}{2c(0,\beta)}\frac{\partial
t}{\partial\alpha}(0,\beta)\cdot\Pi_{1}(0,\beta).$

Integrating the above equation from $\beta_{L}$ to $\beta$ yields

$\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta)=\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta_{L})+\int_{\beta_{L}}^{\beta}\frac{1}{2c(0,\eta)}\frac{\partial
t}{\partial\alpha}(0,\eta)\cdot\Pi_{1}(0,\eta)\text{d}\eta.$ (5.28)

Due to (5.13) and (5.27), the initial value is computed as

$\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta_{L})=\frac{\partial
t}{\partial\alpha}(0,\beta_{L})\cdot\left[\frac{1}{2}\rho_{L}^{2}S_{1,L}^{\prime}-\frac{1}{2}W_{x}(0)-2c_{L}\psi_{1,L}^{\prime}\right].$
(5.29)

Moreover, by (5.8), (5.13), (5.23) and (5.25), one has

$\Pi_{1}(0,\beta)=\frac{\rho_{L}^{2}S_{1,L}^{\prime}}{16c_{L}^{3}}(\psi_{1,L}-\beta)^{3}-\frac{1}{2}W_{x}(0).$
(5.30)

Thus the integral in (5.28) can be exactly obtained and

$\int_{\beta_{L}}^{\beta}\frac{1}{2c(0,\eta)}\frac{\partial
t}{\partial\alpha}(0,\eta)\cdot\Pi_{1}(0,\eta)\text{d}\eta=-\frac{\rho_{L}^{2}S_{1,L}^{\prime}}{8c_{L}^{3}}[c^{2}(0,\beta)-c_{L}^{2}]-\frac{1}{4}W_{x}(0)\left(\frac{1}{c(0,\beta)}-\frac{1}{c_{L}}\right),$
(5.31)

where (5.9) has been used. Substituting (5.29) and (5.31) into (5.28) yields
the expression of $\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta)$. The
second equality in (5.27) implies that

$2c(0,\beta)\cdot\frac{\partial\psi_{1}}{\partial
x}(0,\beta)=\Pi_{1}(0,\beta)-\left(\frac{\partial
t}{\partial\alpha}\right)^{-1}(0,\beta)\cdot\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta).$
(5.32)

Combining (5.32) with the first equality in (5.27), one gets

$\frac{\partial\psi_{1}}{\partial
t}(0,\beta)=-\left(\frac{u_{1}-c}{2c}\right)(0,\beta)\cdot\Pi_{1}(0,\beta)+\left(\frac{u_{1}+c}{2c}\right)(0,\beta)\cdot\left(\frac{\partial
t}{\partial\alpha}\right)^{-1}(0,\beta)\cdot\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta).$
(5.33)

Substituting the expression of
$\frac{\partial\psi_{1}}{\partial\alpha}(0,\beta)$ and (5.30) into (5.33)
gives (5.16). ∎

######  Lemma 5.2 (Resolution of the 1-rarefaction wave).

Assuming that the 1-rarefaction wave associated with $u_{1}-c$ moves to the
left, the limiting values $\frac{\mathcal{D}u_{1}}{\mathcal{D}t}(0,\beta)$ and
$\frac{\mathcal{D}p_{11}}{\mathcal{D}t}(0,\beta)$ satisfy

$\widetilde{a}_{1}(0,\beta)\cdot\frac{\mathcal{D}u_{1}}{\mathcal{D}t}(0,\beta)+\widetilde{b}_{1}(0,\beta)\cdot\frac{\mathcal{D}p_{11}}{\mathcal{D}t}(0,\beta)=\widetilde{d}_{1}(0,\beta),$
(5.34)

where

$\displaystyle\widetilde{a}_{1}(0,\beta)=1+\frac{u_{1}(0,\beta)}{c(0,\beta)},\quad\widetilde{b}_{1}(0,\beta)=\left(\frac{u_{1}}{3p_{11}}+\frac{1}{\rho
c}\right)(0,\beta),$
$\displaystyle\widetilde{d}_{1}(0,\beta)=\frac{\rho_{L}^{2}S_{1,L}^{\prime}\psi_{1,L}}{8c_{L}^{3}}[3c_{L}^{2}+c^{2}(0,\beta)]-\psi_{1,L}^{\prime}\psi_{1,L}-\frac{1}{2}\left(1+\frac{u_{1}(0,\beta)}{c(0,\beta)}\right)W_{x}(0).$

###### Proof.

Using (5.14) gets

$\frac{\partial u_{1}}{\partial t}+\frac{1}{\rho c}\frac{\partial
p_{11}}{\partial t}=\frac{\partial\psi_{1}}{\partial
t}-\frac{\rho^{2}}{2c}\frac{\partial S_{1}}{\partial t}.$ (5.35)

Substituting (2.8) into above equation obtains

$\left(1+\frac{u_{1}}{c}\right)\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\left(\frac{u_{1}}{3p_{11}}+\frac{1}{\rho
c}\right)\frac{\mathcal{D}p_{11}}{\mathcal{D}t}=\frac{\partial\psi_{1}}{\partial
t}-\frac{\rho^{2}}{2c}\frac{\partial S_{1}}{\partial
t}-\frac{u_{1}}{2c}W_{x}.$

Combining it with (5.15) and (5.16) gives (5.34). ∎

Taking $\beta=\beta_{\ast L}$ in (5.34) and using (2.8) obtains the first
equation for $\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}$ and
$\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$ as follows

$a_{1}\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+b_{1}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}=d_{1},$
(5.36)

where

$\begin{cases}a_{1}=\widetilde{a}_{1}(0,\beta_{\ast
L})=1+\frac{u_{1,\ast}}{c_{\ast L}},\\\ b_{1}=\widetilde{b}_{1}(0,\beta_{\ast
L})=\frac{u_{1,\ast}}{3p_{11,\ast}}+\frac{1}{\rho_{\ast L}c_{\ast L}},\\\
d_{1}=\widetilde{d}_{1}(0,\beta_{\ast
L})=\left[\frac{\rho_{L}^{2}S_{1,L}^{\prime}}{8c_{L}^{3}}(3c_{L}^{2}+c_{\ast
L}^{2})-\psi_{1,L}^{\prime}\right]\psi_{1,L}-\frac{1}{2}\left(1+\frac{u_{1,\ast}}{c_{\ast
L}}\right)W_{x}(0).\end{cases}$ (5.37)

######  Remark 5.3.

It is necessary to prove that both the limiting values of
$\frac{\mathcal{D}u_{1}}{\mathcal{D}t}$ and
$\frac{\mathcal{D}p_{11}}{\mathcal{D}t}$ do not change in the whole domain
between the 1-wave and the 6-wave, which are denoted by
$\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}$ and
$\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$, respectively.

To this end, we firstly prove that

$\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast
K}=\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast\ast
K},~{}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast
K}=\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast\ast K},~{}K=L,R.$
(5.38)

For $K=L$, because both $u_{1}$ and $p_{11}$ are the generalized Riemann
invariants for the 2-shear wave, one has

$\left(\frac{\mathcal{D}_{2}u_{1}}{\mathcal{D}t}\right)_{\ast
L}=\left(\frac{\mathcal{D}_{2}u_{1}}{\mathcal{D}t}\right)_{\ast\ast
L},~{}\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast
L}=\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast\ast L},$

where $\frac{\mathcal{D}_{2}}{\mathcal{D}t}:=\frac{\partial}{\partial
t}+\lambda_{2}\frac{\partial}{\partial x}$ and
$\lambda_{2}=u_{1,\ast}-\frac{c_{\ast L}}{\sqrt{3}}$. By (2.6) and (2.7), it
follows that

$\displaystyle\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast
L}-\frac{\lambda_{2}-u_{1,\ast}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast
L}=\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast\ast
L}-\frac{\lambda_{2}-u_{1,\ast}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast\ast
L},$ $\displaystyle\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast
L}-\rho_{\ast
L}(\lambda_{2}-u_{1,\ast})\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast
L}+\frac{1}{2}W_{x}(0)\right]=\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast\ast
L}-\rho_{\ast
L}(\lambda_{2}-u_{1,\ast})\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast\ast
L}+\frac{1}{2}W_{x}(0)\right].$

The above two equations can imply (5.38) for $K=L$, and the proof for $K=R$ is
similar. Besides, both $u_{1}$ and $p_{11}$ are also the generalized Riemann
invariants for the contact discontinuity, thus

$\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast\ast
R},~{}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast\ast R}.$ (5.39)

In virtue of (5.38) and (5.39), one finds that both the limiting values of
$\frac{\mathcal{D}u_{1}}{\mathcal{D}t}$ and
$\frac{\mathcal{D}p_{11}}{\mathcal{D}t}$ do not change in the whole domain
between the 1-wave and the 6-wave. Furthermore, combining (5.38) with (2.8),
one obtains

$\left(\frac{\partial u_{1}}{\partial t}\right)_{\ast K}=\left(\frac{\partial
u_{1}}{\partial t}\right)_{\ast\ast K},~{}\left(\frac{\partial
p_{11}}{\partial t}\right)_{\ast K}=\left(\frac{\partial p_{11}}{\partial
t}\right)_{\ast\ast K},~{}K=L,R,$

which means that both the limiting values of $\frac{\partial u_{1}}{\partial
t}$ and $\frac{\partial p_{11}}{\partial t}$ do not change across the 2-shear
wave and the 5-shear wave.

Up to now, we have established the first equation (5.36) that the limiting
values $\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}$ and
$\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$ satisfy. To
obtain their values, another equation is necessary by resolving the 6-shock
wave. For the shock wavs, one has

$\sigma=\frac{\rho u_{1}-\bar{\rho}\bar{u}_{1}}{\rho-\overline{\rho}},\quad
u_{1}=\overline{u}_{1}\pm\Phi(p_{11};\overline{p}_{11},\overline{\rho}),\quad\rho=H(p_{11};\overline{p}_{11},\overline{\rho}),$

where $(\rho,u_{1},p_{11})$ and
$(\overline{\rho},\overline{u}_{1},\overline{p}_{11})$ are the states ahead
and behind the shock wave with the speed $\sigma$, and in the second equality,
the ”$+$” is for 6-shock wave and ”$-$” is for 1-shock wave. According to the
discussion on the exact Riemann solver, one knows that

$\displaystyle\Phi(p_{11};\overline{p}_{11},\overline{\rho})=(p_{11}-\overline{p}_{11})\left[\frac{1}{\overline{\rho}(2p_{11}+\overline{p}_{11})}\right]^{\frac{1}{2}},$
$\displaystyle
H(p_{11};\overline{p}_{11},\overline{\rho})=\frac{2p_{11}+\overline{p}_{11}}{p_{11}+2\overline{p}_{11}}\overline{\rho}.$

Moreover, along the shock waves, there holds

$\frac{\mathcal{D}_{\sigma}\Gamma}{\mathcal{D}t}=0,$ (5.40)

where $\frac{\mathcal{D}_{\sigma}}{\mathcal{D}t}:=\frac{\partial}{\partial
t}+\sigma\frac{\partial}{\partial x}$ and
$\Gamma=u_{1}-[\overline{u}_{1}\pm\Phi(p_{11};\overline{p}_{11},\overline{\rho})]$
or $\Gamma=\rho-H(p_{11};\overline{p}_{11},\overline{\rho}).$ Taking
$\Gamma=u_{1}-[\overline{u}_{1}\pm\Phi(p_{11};\overline{p}_{11},\overline{\rho})]$
in (5.40) and utilizing (2.5a), (2.5b), (2.5d), (2.6), (2.7) and (2.8) yields

$\displaystyle\left[1\mp\rho(u_{1}-\sigma)\Phi_{1}\right]\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\left[\frac{u_{1}-\sigma}{3p_{11}}\mp\Phi_{1}\right]\frac{\mathcal{D}p_{11}}{\mathcal{D}t}$
$\displaystyle=$
$\displaystyle\pm\Phi_{3}(\sigma-\overline{u}_{1})\overline{\rho}^{\prime}+[\sigma-\overline{u}_{1}\mp
3\overline{p}_{11}\Phi_{2}\mp\overline{\rho}\Phi_{3}]\overline{u}_{1}^{\prime}+\left[-\frac{1}{\overline{\rho}}\pm(\sigma-\overline{u}_{1})\Phi_{2}\right]\overline{p}_{11}^{\prime}-\frac{1}{2}\left[\mp\rho(u_{1}-\sigma)\Phi_{1}+1\right]W_{x},$

where

$\begin{cases}\Phi_{1}=\frac{\partial\Phi}{\partial
p_{11}}=\frac{p_{11}+2\overline{p}_{11}}{2p_{11}+\overline{p}_{11}}\left[\frac{1}{\overline{\rho}(2p_{11}+\overline{p}_{11})}\right]^{\frac{1}{2}},\\\
\Phi_{2}=\frac{\partial\Phi}{\partial\overline{p}_{11}}=-\frac{5p_{11}+\overline{p}_{11}}{4p_{11}+2\overline{p}_{11}}\left[\frac{1}{\overline{\rho}(2p_{11}+\overline{p}_{11})}\right]^{\frac{1}{2}},\\\
\Phi_{3}=\frac{\partial\Phi}{\partial\overline{\rho}}=\frac{\overline{p}_{11}-p_{11}}{2\overline{\rho}}\left[\frac{1}{\overline{\rho}(2p_{11}+\overline{p}_{11})}\right]^{\frac{1}{2}}.\end{cases}$
(5.41)

Specifically, for the 6-shock wave, one has the following result.

######  Lemma 5.4 (Resolution of the 6-shock wave).

Assume that the 6-shock wave associated with $u_{1}+c$ moves to the right. The
limiting values $\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}$
and $\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$ satisfy

$a_{2}\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+b_{2}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}=d_{2},$
(5.42)

where

$\begin{cases}a_{2}=1-\rho_{\ast
R}(u_{1,\ast}-\sigma_{R})\cdot\Phi_{1}^{R},\\\
b_{2}=\frac{u_{1,\ast}-\sigma_{R}}{3p_{11,\ast}}-\Phi_{1}^{R},\\\
d_{2}=L_{\rho}^{R}\cdot\rho_{R}^{\prime}+L_{u_{1}}^{R}\cdot
u_{1,R}^{\prime}+L_{p_{11}}^{R}\cdot
p_{11,R}^{\prime}-\frac{1}{2}W_{x}(0)\cdot L_{W}^{R},\end{cases}$ (5.43)

and

$\displaystyle
L_{\rho}^{R}=(\sigma_{R}-u_{1,R})\cdot\Phi_{3}^{R},~{}L_{u_{1}}^{R}=\sigma_{R}-u_{1,R}-3p_{11,R}\cdot\Phi_{2}^{R}-\rho_{R}\cdot\Phi_{3}^{R},$
$\displaystyle
L_{p_{11}}^{R}=-\frac{1}{\rho_{R}}+(\sigma_{R}-u_{1,R})\cdot\Phi_{2}^{R},~{}L_{W}^{R}=-\rho_{\ast
R}(u_{1,\ast}-\sigma_{R})\cdot\Phi_{1}^{R}+1$

with $\Phi_{i}^{R}:=\Phi_{i}(p_{11,\ast};p_{11,R},\rho_{R})$ ($i=1,2,3$).

######  Remark 5.5.

We now prove that

$\left(\frac{\partial\rho}{\partial t}\right)_{\ast
K}=\left(\frac{\partial\rho}{\partial t}\right)_{\ast\ast K},~{}K=L,R.$ (5.44)

(i) If $u_{1,\ast}=0$, then by (2.5a) and (2.6), one has
$\frac{\partial\rho}{\partial
t}=\frac{\rho}{3p_{11}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}$, and thus
(5.44) may be concluded from (5.38).

(ii) If $u_{1,\ast}\neq 0$, then since $\rho$ is a generalized Riemann
invariant for the 2-shear wave, one has

$u_{1,\ast}\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast
L}=u_{1,\ast}\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast\ast
L},$

thus by using (2.5a), (2.6) and (5.38), one can have (5.44) for $K=L$. The
proof for $K=R$ is similar, because $\rho$ is also a generalized Riemann
invariant for the 5-shear wave.

In virtue of (5.44), henceforth, we will not distinguish
$\left(\frac{\partial\rho}{\partial t}\right)_{\ast\ast K}$ from
$\left(\frac{\partial\rho}{\partial t}\right)_{\ast K}$ ($K=L,R$) any more.

######  Theorem 5.6 (Computing $(\frac{\partial u_{1}}{\partial t})_{\ast K}$,
$(\frac{\partial p_{11}}{\partial t})_{\ast K}$ and
$(\frac{\partial\rho}{\partial t})_{\ast K}$ ($K=L,R$)).

In the domain between the 1-rarefaction wave and the 6-shock wave, one has

$\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}=\frac{d_{1}b_{2}-d_{2}b_{1}}{a_{1}b_{2}-a_{2}b_{1}},\
\
\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}=\frac{d_{1}a_{2}-d_{2}a_{1}}{a_{2}b_{1}-a_{1}b_{2}},$
(5.45)

where $a_{1},b_{1},d_{1}$ are given in (5.37) and $a_{2},b_{2},d_{2}$ are
given in (5.43). Then by (2.8), one gets

$\begin{split}&\left(\frac{\partial u_{1}}{\partial t}\right)_{\ast
K}=\frac{u_{1,\ast}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}+\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast},\\\
&\left(\frac{\partial p_{11}}{\partial t}\right)_{\ast
K}=\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}+\rho_{\ast
K}u_{1,\ast}\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+\frac{1}{2}\rho_{\ast
K}u_{1,\ast}W_{x}(0)\end{split}$ (5.46)

with $K=L,R$. Moreover,

$\left(\frac{\partial\rho}{\partial t}\right)_{\ast L}=\frac{1}{c_{\ast
L}^{2}}\left[\left(\frac{\partial p_{11}}{\partial t}\right)_{\ast
L}+\rho_{L}^{2}S_{1,L}^{\prime}\rho_{\ast L}u_{1,\ast}\frac{c_{\ast
L}^{3}}{c_{L}^{3}}\right],$ (5.47)

and ${\rm\left(\frac{\partial\rho}{\partial t}\right)_{\ast R}}$ satisfies

$g_{\rho}^{R}\cdot\left(\frac{\partial\rho}{\partial t}\right)_{\ast
R}+g_{u_{1}}^{R}\cdot\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+g_{p_{11}}^{R}\cdot\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}=u_{1,\ast}\cdot
f_{R},$ (5.48)

where

$\displaystyle g_{\rho}^{R}=u_{1,\ast}-\sigma_{R},~{}g_{u_{1}}^{R}=\rho_{\ast
R}u_{1,\ast}(\sigma_{R}-u_{1,\ast})\cdot
H_{1}^{R},~{}g_{p_{11}}^{R}=\frac{\sigma_{R}}{c_{\ast R}^{2}}-u_{1,\ast}\cdot
H_{1}^{R},$ $\displaystyle f_{R}=(\sigma_{R}-u_{1,R})\cdot H_{2}^{R}\cdot
p_{11,R}^{\prime}+(\sigma_{R}-u_{1,R})\cdot
H_{3}^{R}\cdot\rho_{R}^{\prime}-\rho_{R}(H_{3}^{R}+c_{R}^{2}\cdot
H_{2}^{R})\cdot u_{1,R}^{\prime}-\frac{1}{2}\rho_{\ast
R}(\sigma_{R}-u_{1,\ast})W_{x}(0)\cdot H_{1}^{R}$

with $H_{i}^{R}=H_{i}(p_{11,\ast};p_{11,R},\rho_{R})$ ($i=1,2,3$) and

$H_{1}=\frac{\partial H}{\partial
p_{11}}=\frac{3\bar{\rho}\bar{p}_{11}}{(p_{11}+2\overline{p}_{11})^{2}},\quad
H_{2}=\frac{\partial
H}{\partial\overline{p}_{11}}=-\frac{3\overline{\rho}p_{11}}{(p_{11}+2\overline{p}_{11})^{2}},\quad
H_{3}=\frac{\partial
H}{\partial\overline{\rho}}=\frac{2p_{11}+\overline{p}_{11}}{p_{11}+2\overline{p}_{11}}.$
(5.49)

###### Proof.

In virtue of (5.36) and (5.42), the proofs of (5.45) and (5.46) are direct. At
the 1-rarefaction side, due to (5.12), one has

$\left(\frac{\partial p_{11}}{\partial t}\right)_{\ast L}=c_{\ast
L}^{2}\left(\frac{\partial\rho}{\partial t}\right)_{\ast L}+\rho_{\ast
L}^{3}\left(\frac{\partial S_{1}}{\partial t}\right)_{\ast L}.$

Combing it with the result obtained by taking $\beta=\beta_{\ast L}$ in (5.15)
implies (5.47).

At the shock wave side, by taking
$\Gamma=\rho-H(p_{11};\overline{p}_{11},\overline{\rho})$ in (5.40), one gets

$\frac{\partial\rho}{\partial t}+\sigma\frac{\partial\rho}{\partial
x}=H_{1}\cdot\left(\frac{\partial p_{11}}{\partial t}+\sigma\frac{\partial
p_{11}}{\partial
x}\right)+H_{2}\cdot\left(\frac{\partial\overline{p}_{11}}{\partial
t}+\sigma\frac{\partial\overline{p}_{11}}{\partial
x}\right)+H_{3}\cdot\left(\frac{\partial\overline{\rho}}{\partial
t}+\sigma\frac{\partial\overline{\rho}}{\partial x}\right).$ (5.50)

Then multiplying both sides of (5.50) by $u_{1}$, and utilizing (2.5a),
(2.5d), (2.6) and (2.7), one obtains

$\displaystyle(u_{1}-\sigma)\frac{\partial\rho}{\partial t}+\rho u_{1}(\sigma-
u_{1})H_{1}\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\left(\frac{\sigma}{c^{2}}-u_{1}H_{1}\right)\frac{\mathcal{D}p_{11}}{\mathcal{D}t}$
$\displaystyle=$ $\displaystyle
u_{1}\cdot\left[(\sigma-\overline{u}_{1})H_{2}\cdot\overline{p}_{11}^{\prime}+(\sigma-\overline{u}_{1})H_{3}\cdot\overline{\rho}^{\prime}-\overline{\rho}(H_{3}+\overline{c}^{2}H_{2})\cdot\overline{u}_{1}^{\prime}-\frac{1}{2}\rho(\sigma-
u_{1})W_{x}H_{1}\right],$ (5.51)

where $\overline{c}=\sqrt{\frac{3\overline{p}_{11}}{\overline{\rho}}}$.
Specifically, for the 6-shock wave, (5.51) implies (5.48). ∎

######  Remark 5.7.

For other possible cases, computing $(\frac{\partial u_{1}}{\partial t})_{\ast
K}$, $(\frac{\partial p_{11}}{\partial t})_{\ast K}$ and
$(\frac{\partial\rho}{\partial t})_{\ast K}$ ($K=L,R$) is presented in B.

####  5.1.2 Computing $(\frac{\partial u_{2}}{\partial t})_{\ast L}$,
$(\frac{\partial p_{12}}{\partial t})_{\ast L}$ and $(\frac{\partial
p_{22}}{\partial t})_{\ast L}$

This section gives the values of $(\frac{\partial u_{2}}{\partial t})_{\ast
L}$, $(\frac{\partial p_{12}}{\partial t})_{\ast L}$ and $(\frac{\partial
p_{22}}{\partial t})_{\ast L}$ by utilizing the remaining three generalized
Riemann invariants associated with the 1-rarefaction wave

$\psi_{2}:=u_{2}+\frac{\sqrt{3}p_{12}}{\sqrt{\rho
p_{11}}},\quad\psi_{3}:=\frac{p_{12}}{\rho^{3}},\quad
S_{2}:=\frac{\det(\mathbf{p})}{\rho^{4}}.$

From those, one has the following three total differentials

$\displaystyle\text{d}\psi_{2}=\text{d}u_{2}+\frac{\sqrt{3}}{\sqrt{\rho
p_{11}}}\text{d}p_{12}-\frac{\sqrt{3}p_{12}}{2\rho\sqrt{\rho
p_{11}}}\text{d}\rho-\frac{\sqrt{3}p_{12}}{2p_{11}\sqrt{\rho
p_{11}}}\text{d}p_{11},$ (5.52)
$\displaystyle\text{d}\psi_{3}=\frac{1}{\rho^{3}}\text{d}p_{12}-\frac{3p_{12}}{\rho^{4}}\text{d}\rho,$
(5.53)
$\displaystyle\text{d}S_{2}=\frac{p_{11}}{\rho^{4}}\text{d}p_{22}+\frac{p_{22}}{\rho^{4}}\text{d}p_{11}-\frac{2p_{12}}{\rho^{4}}\text{d}p_{12}-\frac{4\det(\mathbf{p})}{\rho^{5}}\text{d}\rho.$
(5.54)

Combing them with (2.5a) and (2.5c)-(2.5f) gives

$\frac{\partial\psi_{2}}{\partial t}+(u_{1}+c)\frac{\partial\psi_{2}}{\partial
x}=\Pi_{2},\quad\frac{\partial\psi_{3}}{\partial
t}+(u_{1}+c)\frac{\partial\psi_{3}}{\partial x}=\Pi_{3},\quad\frac{\partial
S_{2}}{\partial t}+u_{1}\frac{\partial S_{2}}{\partial x}=0,$

where

$\displaystyle\Pi_{2}:=\frac{2}{\rho}\frac{\partial p_{12}}{\partial
x}-\frac{3p_{12}}{2\rho^{2}}\frac{\partial\rho}{\partial
x}-\frac{3p_{12}}{2\rho p_{11}}\frac{\partial p_{11}}{\partial x},$ (5.55)
$\displaystyle\Pi_{3}:=\frac{c}{\rho^{3}}\frac{\partial p_{12}}{\partial
x}+\frac{p_{12}}{\rho^{3}}\frac{\partial u_{1}}{\partial
x}-\frac{p_{11}}{\rho^{3}}\frac{\partial u_{2}}{\partial
x}-\frac{3p_{12}c}{\rho^{4}}\frac{\partial\rho}{\partial x}.$ (5.56)

Similar to the derivation of (5.28), for $\psi_{2}$, one can also obtain

$\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta)=\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta_{L})+\int_{\beta_{L}}^{\beta}\frac{1}{2c(0,\eta)}\frac{\partial
t}{\partial\alpha}(0,\eta)\cdot\Pi_{2}(0,\eta)\text{d}\eta,$ (5.57)

where

$\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta_{L})=\frac{\partial
t}{\partial\alpha}(0,\beta_{L})\cdot(\Pi_{2,L}-2c_{L}\psi_{2,L}^{\prime}),$

with $\Pi_{2,L}$ and $\psi_{2,L}^{\prime}$ being given by (5.55) and (5.52),
respectively. Similar to (5.33), one has

$\frac{\partial\psi_{2}}{\partial
t}(0,\beta)=-\left(\frac{u_{1}-c}{2c}\right)(0,\beta)\cdot\Pi_{2}(0,\beta)+\left(\frac{u_{1}+c}{2c}\right)(0,\beta)\cdot\left(\frac{\partial
t}{\partial\alpha}\right)^{-1}(0,\beta)\cdot\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta).$
(5.58)

Note that if one wants to get the explicit expression of
$\frac{\partial\psi_{2}}{\partial t}(0,\beta)$ in a way similar to the
derivation of (5.16) in Lemma 5.1, one has to first obtain the explicit
expression of $\Pi_{2}(0,\beta)$ so as to exactly obtain the integral in
(5.57). Such task seems to be difficultly completed since it is hard to
explicitly represent the spatial derivatives of the primitive variables with
the characteristic coordinate $(\alpha,\beta)$. In practice, it may not be
necessary to exactly get the integral in (5.57) and a second-order numerical
approximation to the integral is enough. Specifically, applying the
trapezoidal rule in (5.57) obtains

$\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta)=\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta_{L})+\frac{\beta-\beta_{L}}{2}\left[\frac{1}{4c(0,\beta)}\Pi_{2}(0,\beta)+\frac{1}{4c_{L}}\Pi_{2,L}\right].$

Substituting it into (5.58) gets

$\frac{\partial\psi_{2}}{\partial
t}(0,\beta)=\left[\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta_{L})+\frac{\Pi_{2,L}}{8c_{L}}(\beta-\beta_{L})\right]\psi_{1,L}+\frac{1}{2c(0,\beta)}\left(\frac{\beta-\beta_{L}}{4}\psi_{1,L}-\beta\right)\cdot\Pi_{2}(0,\beta).$
(5.59)

Moreover, by using (5.12), (5.55) is rewritten into

$\Pi_{2}=\frac{2}{\rho}\frac{\partial p_{12}}{\partial x}-\frac{2p_{12}}{\rho
p_{11}}\frac{\partial p_{11}}{\partial
x}+\frac{\rho^{2}p_{12}}{2p_{11}}\frac{\partial S_{1}}{\partial x}.$

Substituting (2.7) and (2.10) into above expression yields

$\Pi_{2}=-2\frac{\mathcal{D}u_{2}}{\mathcal{D}t}+\frac{2p_{12}}{p_{11}}\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\frac{1}{2}W_{x}\right)+\frac{\rho^{2}p_{12}}{2p_{11}}\frac{\partial
S_{1}}{\partial x}.$ (5.60)

######  Lemma 5.8 (Resolution of the 1-rarefaction wave).

Assume that the 1-rarefaction wave associated with $u_{1}-c$ moves to the
left. Then $\frac{\mathcal{D}u_{2}}{\mathcal{D}t}(0,\beta)$ and
$\frac{\mathcal{D}p_{12}}{\mathcal{D}t}(0,\beta)$ satisfy

$\widetilde{a}_{3,L}(0,\beta)\cdot\frac{\mathcal{D}u_{2}}{\mathcal{D}t}(0,\beta)+\widetilde{b}_{3,L}(0,\beta)\cdot\frac{\mathcal{D}p_{12}}{\mathcal{D}t}(0,\beta)=\widetilde{d}_{3,L}(0,\beta),$
(5.61)

where

$\displaystyle\widetilde{a}_{3,L}(0,\beta)=1+\left(\frac{\sqrt{3}\rho
u_{1}}{\sqrt{\rho
p_{11}}}\right)(0,\beta)+\frac{1}{c(0,\beta)}\left(\frac{\beta-\beta_{L}}{4}\psi_{1,L}-\beta\right),$
$\displaystyle\widetilde{b}_{3,L}(0,\beta)=\left(\frac{u_{1}}{p_{11}}+\frac{\sqrt{3}}{\sqrt{\rho
p_{11}}}\right)(0,\beta),$
$\displaystyle\widetilde{d}_{3,L}(0,\beta)=\left[\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta_{L})+\frac{\Pi_{2,L}}{8c_{L}}(\beta-\beta_{L})\right]\psi_{1,L}+\frac{1}{2c(0,\beta)}\left(\frac{\beta-\beta_{L}}{4}\psi_{1,L}-\beta\right)\left\\{\left(\frac{2p_{12}}{p_{11}}\right)(0,\beta)\right.$
$\displaystyle\quad\quad\quad\left.\cdot\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)(0,\beta)+\frac{1}{2}W_{x}(0)\right]+\frac{S_{1,L}^{\prime}}{c_{L}}\left(\frac{c\rho^{2}p_{12}}{2p_{11}}\right)(0,\beta)\right\\}+\left(\frac{\sqrt{3}p_{12}}{2\rho\sqrt{\rho
p_{11}}}\frac{\partial\rho}{\partial t}\right)(0,\beta)$
$\displaystyle\quad\quad\quad+\left(\frac{\sqrt{3}p_{12}}{2p_{11}\sqrt{\rho
p_{11}}}\frac{\partial p_{11}}{\partial
t}\right)(0,\beta)+\left(\frac{2u_{1}p_{12}}{3p_{11}^{2}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)(0,\beta).$

###### Proof.

Due to (5.52), one has

$\frac{\partial u_{2}}{\partial t}+\frac{\sqrt{3}}{\sqrt{\rho
p_{11}}}\frac{\partial p_{12}}{\partial t}=\frac{\partial\psi_{2}}{\partial
t}+\frac{\sqrt{3}p_{12}}{2\rho\sqrt{\rho p_{11}}}\frac{\partial\rho}{\partial
t}+\frac{\sqrt{3}p_{12}}{2p_{11}\sqrt{\rho p_{11}}}\frac{\partial
p_{11}}{\partial t}.$

Substituting (2.11) into the above expression gives

$\left(1+\frac{\sqrt{3}\rho u_{1}}{\sqrt{\rho
p_{11}}}\right)\frac{\mathcal{D}u_{2}}{\mathcal{D}t}+\left(\frac{u_{1}}{p_{11}}+\frac{\sqrt{3}}{\sqrt{\rho
p_{11}}}\right)\frac{\mathcal{D}p_{12}}{\mathcal{D}t}-\frac{2u_{1}p_{12}}{3p_{11}^{2}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}=\frac{\partial\psi_{2}}{\partial
t}+\frac{\sqrt{3}p_{12}}{2\rho\sqrt{\rho p_{11}}}\frac{\partial\rho}{\partial
t}+\frac{\sqrt{3}p_{12}}{2p_{11}\sqrt{\rho p_{11}}}\frac{\partial
p_{11}}{\partial t}.$ (5.62)

Taking the limiting value at $(0,\beta)$ in (5.62), and combining it with
(5.59) and (5.60), one obtains (5.61). Note that in the expression of
$\widetilde{d}_{3,L}(0,\beta)$, (5.23) has been used. ∎

Taking $\beta=\beta_{\ast L}$ in (5.61), one obtains the first equation that
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast L}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast L}$ satisfy

$a_{3,L}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
L}+b_{3,L}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
L}=d_{3,L},$ (5.63)

with $a_{3,L}=\widetilde{a}_{3,L}(0,\beta_{\ast L})$,
$b_{3,L}=\widetilde{b}_{3,L}(0,\beta_{\ast L})$ and
$d_{3,L}=\widetilde{d}_{3,L}(0,\beta_{\ast L})$.

For $\psi_{3}$, similar to $\psi_{2}$, one can get the following results

$\displaystyle\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta)=\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta_{L})+\int_{\beta_{L}}^{\beta}\frac{1}{2c(0,\eta)}\frac{\partial
t}{\partial\alpha}(0,\eta)\cdot\Pi_{3}(0,\eta)\text{d}\eta,$ (5.64)
$\displaystyle\frac{\partial\psi_{3}}{\partial
t}(0,\beta)=-\left(\frac{u_{1}-c}{2c}\right)(0,\beta)\cdot\Pi_{3}(0,\beta)+\left(\frac{u_{1}+c}{2c}\right)(0,\beta)\cdot\left(\frac{\partial
t}{\partial\alpha}\right)^{-1}(0,\beta)\cdot\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta),$
(5.65)

where

$\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta_{L})=\frac{\partial
t}{\partial\alpha}(0,\beta_{L})\cdot(\Pi_{3,L}-2c_{L}\psi_{3,L}^{\prime}),$

with $\Pi_{3,L}$ and $\psi_{3,L}^{\prime}$ being given by (5.56) and (5.53),
respectively. Applying the trapezoidal rule in (5.64) obtains

$\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta)=\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta_{L})+\frac{\beta-\beta_{L}}{2}\left[\frac{1}{4c(0,\beta)}\Pi_{3}(0,\beta)+\frac{1}{4c_{L}}\Pi_{3,L}\right].$

Substituting it into (5.65) gets

$\frac{\partial\psi_{3}}{\partial
t}(0,\beta)=\left[\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta_{L})+\frac{\Pi_{3,L}}{8c_{L}}(\beta-\beta_{L})\right]\psi_{1,L}+\frac{1}{2c(0,\beta)}\left(\frac{\beta-\beta_{L}}{4}\psi_{1,L}-\beta\right)\cdot\Pi_{3}(0,\beta).$
(5.66)

Besides, by using (5.12), (2.10), (2.6), (2.9) and (2.7), one can rewrite
(5.56) into

$\Pi_{3}=\frac{1}{\rho^{3}}\left[-\rho
c\frac{\mathcal{D}u_{2}}{\mathcal{D}t}-\frac{p_{12}}{p_{11}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}+\frac{\mathcal{D}p_{12}}{\mathcal{D}t}+\frac{3p_{12}}{c}\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\frac{1}{2}W_{x}\right)\right]+\frac{3p_{12}}{\rho
c}\frac{\partial S_{1}}{\partial x}.$ (5.67)

######  Lemma 5.9 (Resolution of the 1-rarefaction wave).

Assume that the 1-rarefaction wave associated with $u_{1}-c$ moves to the
left. Then $\frac{\mathcal{D}u_{2}}{\mathcal{D}t}(0,\beta)$ and
$\frac{\mathcal{D}p_{12}}{\mathcal{D}t}(0,\beta)$ satisfy

$\widetilde{a}_{4,L}(0,\beta)\cdot\frac{\mathcal{D}u_{2}}{\mathcal{D}t}(0,\beta)+\widetilde{b}_{4,L}(0,\beta)\cdot\frac{\mathcal{D}p_{12}}{\mathcal{D}t}(0,\beta)=\widetilde{d}_{4,L}(0,\beta),$
(5.68)

where

$\displaystyle\widetilde{a}_{4,L}(0,\beta)=\frac{1}{\rho^{2}(0,\beta)}\left[u_{1}(0,\beta)+\frac{1}{2}\left(\frac{\beta-\beta_{L}}{4}\psi_{1,L}-\beta\right)\right],$
$\displaystyle\widetilde{b}_{4,L}(0,\beta)=\frac{1}{\rho^{3}(0,\beta)}\left[1-\frac{1}{2c(0,\beta)}\left(\frac{\beta-\beta_{L}}{4}\psi_{1,L}-\beta\right)\right],$
$\displaystyle\widetilde{d}_{4,L}(0,\beta)=\left[\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta_{L})+\frac{\Pi_{3,L}}{8c_{L}}(\beta-\beta_{L})\right]\psi_{1,L}+\left(\frac{3p_{12}}{\rho^{4}}\frac{\partial\rho}{\partial
t}\right)(0,\beta)+\frac{1}{2c(0,\beta)}\left(\frac{\beta-\beta_{L}}{4}\psi_{1,L}-\beta\right)$
$\displaystyle\qquad\quad\cdot\left\\{\frac{1}{\rho^{3}(0,\beta)}\left[-\frac{p_{12}}{p_{11}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right.\left.+\frac{3p_{12}}{c}\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}+\frac{1}{2}W_{x}(0)\right)\right](0,\beta)+\frac{S_{1,L}^{\prime}}{c_{L}}\left(\frac{3p_{12}}{\rho}\right)(0,\beta)\right\\}.$

###### Proof.

By (5.53) and (2.11), one has

$\frac{\partial\psi_{3}}{\partial
t}=\frac{1}{\rho^{3}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}+\rho
u_{1}\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)-\frac{3p_{12}}{\rho^{4}}\frac{\partial\rho}{\partial
t}.$

Combining it with (5.67) and (5.66) gives (5.68). Similarly, in the expression
of $\widetilde{d}_{4,L}(0,\beta)$, (5.23) has been used. ∎

Taking $\beta=\beta_{\ast L}$ in (5.68), one obtains the second equation that
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast L}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast L}$ satisfy

$a_{4,L}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
L}+b_{4,L}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
L}=d_{4,L},$ (5.69)

with $a_{4,L}=\widetilde{a}_{4,L}(0,\beta_{\ast L})$,
$b_{4,L}=\widetilde{b}_{4,L}(0,\beta_{\ast L})$ and
$d_{4,L}=\widetilde{d}_{4,L}(0,\beta_{\ast L})$.

For $S_{2}$, with the similar derivation of (5.24), one can obtain

$\frac{\partial S_{2}}{\partial
t}(0,\beta)=-\frac{S_{2,L}^{\prime}}{c_{L}}(cu_{1})(0,\beta),$ (5.70)

with $S_{2,L}^{\prime}$ being given by (5.54). Moreover, by (5.54), one has

$\left(\frac{p_{11}}{\rho^{4}}\frac{\partial p_{22}}{\partial
t}\right)(0,\beta)=\frac{\partial S_{2}}{\partial
t}(0,\beta)-\left(\frac{p_{22}}{\rho^{4}}\frac{\partial p_{11}}{\partial
t}\right)(0,\beta)+\left(\frac{2p_{12}}{\rho^{4}}\frac{\partial
p_{12}}{\partial
t}\right)(0,\beta)+\left(\frac{4\det(\mathbf{p})}{\rho^{5}}\frac{\partial\rho}{\partial
t}\right)(0,\beta).$ (5.71)

######  Theorem 5.10 (Computing $(\frac{\partial u_{2}}{\partial t})_{\ast
L}$, $(\frac{\partial p_{12}}{\partial t})_{\ast L}$ and $(\frac{\partial
p_{22}}{\partial t})_{\ast L}$).

Assume that the 1-rarefaction wave associated with $u_{1}-c$ move to the left.
Then one has

$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
L}=\frac{d_{3,L}b_{4,L}-d_{4,L}b_{3,L}}{a_{3,L}b_{4,L}-a_{4,L}b_{3,L}},\ \
\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
L}=\frac{d_{3,L}a_{4,L}-d_{4,L}a_{3,L}}{b_{3,L}a_{4,L}-b_{4,L}a_{3,L}},$
(5.72)

and

$\begin{split}&\left(\frac{\partial u_{2}}{\partial t}\right)_{\ast
L}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
L}+\frac{u_{1,\ast}}{p_{11,\ast}}\left[\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
L}-\frac{2p_{12,\ast
L}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}\right],\\\
&\left(\frac{\partial p_{12}}{\partial t}\right)_{\ast
L}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast L}+\rho_{\ast
L}u_{1,\ast}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
L}.\end{split}$ (5.73)

The limiting value $\left(\frac{\partial p_{22}}{\partial t}\right)_{\ast L}$
can be obtained by

$\left(\frac{p_{11}}{\rho^{4}}\frac{\partial p_{22}}{\partial t}\right)_{\ast
L}=-\frac{S_{2,L}^{\prime}}{c_{L}}c_{\ast
L}u_{1,\ast}-\left(\frac{p_{22}}{\rho^{4}}\frac{\partial p_{11}}{\partial
t}\right)_{\ast L}+\left(\frac{2p_{12}}{\rho^{4}}\frac{\partial
p_{12}}{\partial t}\right)_{\ast
L}+\left(\frac{4\det(\mathbf{p})}{\rho^{5}}\frac{\partial\rho}{\partial
t}\right)_{\ast L}.$ (5.74)

###### Proof.

Combining (5.63) and (5.69) yields (5.72), and then can get (5.73) by (2.11).
Substituting (5.70) into (5.71) and taking $\beta=\beta_{\ast L}$ obtain
(5.74). ∎

######  Remark 5.11.

For the 1-shock wave, computing $(\frac{\partial u_{2}}{\partial t})_{\ast
L}$, $(\frac{\partial p_{12}}{\partial t})_{\ast L}$ and $(\frac{\partial
p_{22}}{\partial t})_{\ast L}$ is presented in C.

####  5.1.3 Computing $(\frac{\partial u_{2}}{\partial t})_{\ast R}$,
$(\frac{\partial p_{12}}{\partial t})_{\ast R}$ and $(\frac{\partial
p_{22}}{\partial t})_{\ast R}$

At the 6-shock wave side, the Rankine-Hugoniot jump conditions will be used to
derive the limiting values $(\frac{\partial u_{2}}{\partial t})_{\ast R}$,
$(\frac{\partial p_{12}}{\partial t})_{\ast R}$ and $(\frac{\partial
p_{22}}{\partial t})_{\ast R}$.

Across the shock wave with speed $\sigma$, the Rankine-Hugoniot jump condition
for $m_{2}$ implies

$\rho
u_{2}(u_{1}-\sigma)+p_{12}=\bar{\rho}\bar{u}_{2}(\overline{u}_{1}-\sigma)+\overline{p}_{12}.$

If denoting $\Gamma_{m_{2}}:=\rho u_{2}(u_{1}-\sigma)+p_{12}$, one has
$\Gamma_{m_{2}}=\Gamma_{\overline{m}_{2}}$ and

$\text{d}\Gamma_{m_{2}}=u_{2}(u_{1}-\sigma)\text{d}\rho+\rho
u_{2}\text{d}u_{1}+\rho(u_{1}-\sigma)\text{d}u_{2}+\text{d}p_{12}-\rho
u_{2}\text{d}\sigma.$

It follows that

$\frac{\mathcal{D}_{\sigma}\Gamma_{m_{2}}}{\mathcal{D}t}=u_{2}(u_{1}-\sigma)\frac{\mathcal{D}_{\sigma}\rho}{\mathcal{D}t}+\rho
u_{2}\frac{\mathcal{D}_{\sigma}u_{1}}{\mathcal{D}t}+\rho(u_{1}-\sigma)\frac{\mathcal{D}_{\sigma}u_{2}}{\mathcal{D}t}+\frac{\mathcal{D}_{\sigma}p_{12}}{\mathcal{D}t}-\rho
u_{2}\frac{\mathcal{D}_{\sigma}\sigma}{\mathcal{D}t},$

and then one can utilize the relation along the shock wave, i.e.,
$\frac{\mathcal{D}_{\sigma}\Gamma_{m_{2}}}{\mathcal{D}t}=\frac{\mathcal{D}_{\sigma}\Gamma_{\overline{m}_{2}}}{\mathcal{D}t}$,
to get the first equation that the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast K}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast K}$ ($K=L,R$)
satisfy.

Specifically, for the 6-shock wave, one has

$\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{m_{2,\ast
R}}}{\mathcal{D}t}=\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{m_{2,R}}}{\mathcal{D}t},$
(5.75)

where

$\displaystyle\sigma_{R}=u_{1,R}+\left(\frac{2p_{11,\ast}+p_{11,R}}{\rho_{R}}\right)^{\frac{1}{2}},$
(5.76) $\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{m_{2,\ast
R}}}{\mathcal{D}t}=u_{2,\ast
R}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast
R}+\rho_{\ast R}u_{2,\ast
R}\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast R}$
$\displaystyle\quad\quad+\rho_{\ast
R}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}u_{2}}{\mathcal{D}t}\right)_{\ast
R}+\left(\frac{\mathcal{D}_{\sigma_{R}}p_{12}}{\mathcal{D}t}\right)_{\ast
R}-\rho_{\ast R}u_{2,\ast
R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t},$ (5.77)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{m_{2,R}}}{\mathcal{D}t}=u_{2,R}(u_{1,R}-\sigma_{R})\frac{\mathcal{D}_{\sigma_{R}}\rho_{R}}{\mathcal{D}t}+\rho_{R}u_{2,R}\frac{\mathcal{D}_{\sigma_{R}}u_{1,R}}{\mathcal{D}t}$
$\displaystyle\quad\quad+\rho_{R}(u_{1,R}-\sigma_{R})\frac{\mathcal{D}_{\sigma_{R}}u_{2,R}}{\mathcal{D}t}+\frac{\mathcal{D}_{\sigma_{R}}p_{12,R}}{\mathcal{D}t}-\rho_{R}u_{2,R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}.$
(5.78)

By (2.6), one has

$\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast
R}=\left(\frac{\partial u_{1}}{\partial t}\right)_{\ast
R}-\frac{\sigma_{R}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}.$
(5.79)

By (2.9), one gets

$\left(\frac{\mathcal{D}_{\sigma_{R}}u_{2}}{\mathcal{D}t}\right)_{\ast
R}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}-\frac{\sigma_{R}-u_{1,\ast}}{p_{11,\ast}}\left[\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}-\frac{2p_{12,\ast
R}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}\right].$
(5.80)

By (2.10), one obtains

$\left(\frac{\mathcal{D}_{\sigma_{R}}p_{12}}{\mathcal{D}t}\right)_{\ast
R}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast R}-\rho_{\ast
R}(\sigma_{R}-u_{1,\ast})\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}.$ (5.81)

By (2.5) and (5.76), one has

$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\rho_{R}}{\mathcal{D}t}=(\sigma_{R}-u_{1,R})\rho_{R}^{\prime}-\rho_{R}u_{1,R}^{\prime},$
(5.82a)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}u_{1,R}}{\mathcal{D}t}=(\sigma_{R}-u_{1,R})u_{1,R}^{\prime}-\frac{1}{\rho_{R}}p_{11,R}^{\prime}-\frac{1}{2}W_{x}(0),$
(5.82b)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}u_{2,R}}{\mathcal{D}t}=(\sigma_{R}-u_{1,R})u_{2,R}^{\prime}-\frac{1}{\rho_{R}}p_{12,R}^{\prime},$
(5.82c)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}p_{11,R}}{\mathcal{D}t}=(\sigma_{R}-u_{1,R})p_{11,R}^{\prime}-3p_{11,R}u_{1,R}^{\prime},$
(5.82d)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}p_{12,R}}{\mathcal{D}t}=(\sigma_{R}-u_{1,R})p_{12,R}^{\prime}-2p_{12,R}u_{1,R}^{\prime}-p_{11,R}u_{2,R}^{\prime},$
(5.82e)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}=\frac{\mathcal{D}_{\sigma_{R}}u_{1,R}}{\mathcal{D}t}-\frac{\sqrt{2p_{11,\ast}+p_{11,R}}}{2\rho_{R}\sqrt{\rho_{R}}}\frac{\mathcal{D}_{\sigma_{R}}\rho_{R}}{\mathcal{D}t}+\frac{1}{2\sqrt{\rho_{R}(2p_{11,\ast}+p_{11,R})}}\frac{\mathcal{D}_{\sigma_{R}}p_{11,R}}{\mathcal{D}t}+\frac{1}{\sqrt{\rho_{R}(2p_{11,\ast}+p_{11,R})}}\frac{\mathcal{D}_{\sigma_{R}}p_{11,\ast}}{\mathcal{D}t}.$
(5.82f)

By (2.7), one obtains

$\frac{\mathcal{D}_{\sigma_{R}}p_{11,\ast}}{\mathcal{D}t}=\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}-\rho_{\ast
R}(\sigma_{R}-u_{1,\ast})\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+\frac{1}{2}W_{x}(0)\right].$
(5.83)

Substituting (5.82a), (5.82b), (5.82d) and (5.83) into (5.82f) can obtain the
value of $\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}$. Similarly,
substituting (5.82a), (5.82b), (5.82c), (5.82e) and (5.82f) into (5.78) may
get the value of
$\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{m_{2,R}}}{\mathcal{D}t}$. As for the
computation of
$\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast R}$, one
uses the relation across the shock wave, i.e.,
$\rho=H(p_{11};\overline{p}_{11},\overline{\rho})$, to obtain

$\frac{\mathcal{D}_{\sigma}\rho}{\mathcal{D}t}=H_{1}\cdot\frac{\mathcal{D}_{\sigma}p_{11}}{\mathcal{D}t}+H_{2}\cdot\frac{\mathcal{D}_{\sigma}\overline{p}_{11}}{\mathcal{D}t}+H_{3}\cdot\frac{\mathcal{D}_{\sigma}\overline{\rho}}{\mathcal{D}t}.$

Applying it to the 6-shock wave gets

$\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast
R}=H_{1}^{R}\cdot\frac{\mathcal{D}_{\sigma_{R}}p_{11,\ast}}{\mathcal{D}t}+H_{2}^{R}\cdot\frac{\mathcal{D}_{\sigma_{R}}p_{11,R}}{\mathcal{D}t}+H_{3}^{R}\cdot\frac{\mathcal{D}_{\sigma_{R}}\rho_{R}}{\mathcal{D}t}.$
(5.84)

######  Lemma 5.12 (Resolution of the 6-shock wave).

Assume that the 6-shock wave associated with $u_{1}+c$ moves to the right.
Then the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast R}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast R}$ satisfy

$a_{3,R}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}+b_{3,R}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}=d_{3,R},$ (5.85)

where

$\displaystyle a_{3,R}=2\rho_{\ast R}(u_{1,\ast}-\sigma_{R}),\quad
b_{3,R}=1+\frac{\rho_{\ast R}(u_{1,\ast}-\sigma_{R})^{2}}{p_{11,\ast}},$
$\displaystyle
d_{3,R}=\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{m_{2,R}}}{\mathcal{D}t}+\rho_{\ast
R}u_{2,\ast
R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}-u_{2,\ast
R}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast
R}$ $\displaystyle\quad-\rho_{\ast R}u_{2,\ast
R}\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast
R}+\frac{2\rho_{\ast R}p_{12,\ast
R}(u_{1,\ast}-\sigma_{R})^{2}}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}.$

Here the values of
$\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{m_{2,R}}}{\mathcal{D}t}$,
$\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}$,
$\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast R}$,
$\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast R}$ and
$\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$ are given by
(5.78), (5.82f), (5.84), (5.79) and (5.45), respectively.

###### Proof.

One can first substitute (5.80) and (5.81) into (5.77), and then combine
(5.77) with (5.75) to yield (5.85). ∎

Across the shock wave with speed $\sigma$, the Rankine-Hugoniot jump condition
for $E_{12}$ implies that $\Gamma_{E_{12}}=\Gamma_{\overline{E}_{12}}$ with

$\Gamma_{E_{12}}:=\frac{1}{2}(\rho
u_{1}u_{2}+p_{12})(u_{1}-\sigma)+\frac{1}{2}(p_{11}u_{2}+p_{12}u_{1}).$

It follows that

$\displaystyle\frac{\mathcal{D}_{\sigma}\Gamma_{E_{12}}}{\mathcal{D}t}=$
$\displaystyle\frac{1}{2}u_{1}u_{2}(u_{1}-\sigma)\frac{\mathcal{D}_{\sigma}\rho}{\mathcal{D}t}+\left(2E_{12}-\frac{1}{2}\rho
u_{2}\sigma\right)\frac{\mathcal{D}_{\sigma}u_{1}}{\mathcal{D}t}+\left(E_{11}-\frac{1}{2}\rho
u_{1}\sigma\right)\frac{\mathcal{D}_{\sigma}u_{2}}{\mathcal{D}t}$
$\displaystyle+\frac{1}{2}u_{2}\frac{\mathcal{D}_{\sigma}p_{11}}{\mathcal{D}t}+\left(u_{1}-\frac{1}{2}\sigma\right)\frac{\mathcal{D}_{\sigma}p_{12}}{\mathcal{D}t}-E_{12}\frac{\mathcal{D}_{\sigma}\sigma}{\mathcal{D}t}.$

Then one can use the relation along the shock wave, i.e.,
$\frac{\mathcal{D}_{\sigma}\Gamma_{E_{12}}}{\mathcal{D}t}=\frac{\mathcal{D}_{\sigma}\Gamma_{\overline{E}_{12}}}{\mathcal{D}t}$,
to build the second equation that the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast K}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast K}$ ($K=L,R$)
satisfy.

Specifically, for the 6-shock wave, one has

$\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{12,\ast
R}}}{\mathcal{D}t}=\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{12,R}}}{\mathcal{D}t},$
(5.86)

where

$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{12,\ast
R}}}{\mathcal{D}t}=\frac{1}{2}u_{1,\ast}u_{2,\ast
R}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast
R}+\left(2E_{12,\ast R}-\frac{1}{2}\rho_{\ast R}u_{2,\ast
R}\sigma_{R}\right)\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast
R}$ $\displaystyle\qquad+\left(E_{11,\ast R}-\frac{1}{2}\rho_{\ast
R}u_{1,\ast}\sigma_{R}\right)\left(\frac{\mathcal{D}_{\sigma_{R}}u_{2}}{\mathcal{D}t}\right)_{\ast
R}+\frac{1}{2}u_{2,\ast
R}\frac{\mathcal{D}_{\sigma_{R}}p_{11,\ast}}{\mathcal{D}t}$
$\displaystyle\qquad+\left(u_{1,\ast}-\frac{1}{2}\sigma_{R}\right)\left(\frac{\mathcal{D}_{\sigma_{R}}p_{12}}{\mathcal{D}t}\right)_{\ast
R}-E_{12,\ast R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t},$
(5.87)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{12,R}}}{\mathcal{D}t}=\frac{1}{2}u_{1,R}u_{2,R}(u_{1,R}-\sigma_{R})\frac{\mathcal{D}_{\sigma_{R}}\rho_{R}}{\mathcal{D}t}+\left(2E_{12,R}-\frac{1}{2}\rho_{R}u_{2,R}\sigma_{R}\right)\frac{\mathcal{D}_{\sigma_{R}}u_{1,R}}{\mathcal{D}t}$
$\displaystyle\quad+\left(E_{11,R}-\frac{1}{2}\rho_{R}u_{1,R}\sigma_{R}\right)\frac{\mathcal{D}_{\sigma_{R}}u_{2,R}}{\mathcal{D}t}+\frac{1}{2}u_{2,R}\frac{\mathcal{D}_{\sigma_{R}}p_{11,R}}{\mathcal{D}t}+\left(u_{1,R}-\frac{1}{2}\sigma_{R}\right)\frac{\mathcal{D}_{\sigma_{R}}p_{12,R}}{\mathcal{D}t}-E_{12,R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}.$
(5.88)

Substituting (5.82) into (5.88) can obtain the value of
$\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{12,R}}}{\mathcal{D}t}$.

######  Lemma 5.13 (Resolution of the 6-shock wave).

Assume that the 6-shock wave associated with $u_{1}+c$ moves to the right.
Then the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast R}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast R}$ satisfy

$a_{4,R}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}+b_{4,R}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}=d_{4,R},$ (5.89)

where

$\displaystyle a_{4,R}=E_{11,\ast R}+\rho_{\ast
R}(u_{1,\ast}-\sigma_{R})^{2}-\frac{1}{2}\rho_{\ast R}\sigma_{R}^{2},\quad
b_{4,R}=\left(E_{11,\ast R}-\frac{1}{2}\rho_{\ast
R}u_{1,\ast}\sigma_{R}\right)\frac{u_{1,\ast}-\sigma_{R}}{p_{11,\ast}}+u_{1,\ast}-\frac{1}{2}\sigma_{R},$
$\displaystyle
d_{4,R}=\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{12,R}}}{\mathcal{D}t}+E_{12,\ast
R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}-\frac{1}{2}u_{1,\ast}u_{2,\ast
R}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast
R}-\frac{1}{2}u_{2,\ast
R}\frac{\mathcal{D}_{\sigma_{R}}p_{11,\ast}}{\mathcal{D}t}$
$\displaystyle\qquad-\left(2E_{12,\ast R}-\frac{1}{2}\rho_{\ast R}u_{2,\ast
R}\sigma_{R}\right)\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast
R}+\left(E_{11,\ast R}-\frac{1}{2}\rho_{\ast
R}u_{1,\ast}\sigma_{R}\right)\frac{2p_{12,\ast
R}(u_{1,\ast}-\sigma_{R})}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}.$

Here the value of $d_{4,R}$ can be obtained by (5.88), (5.82f), (5.84),
(5.83), (5.79) and (5.45).

###### Proof.

The proof is easily completed by combining (5.80), (5.81), (5.87) and (5.86).
∎

Across the shock wave with speed $\sigma$, the Rankine-Hugoniot jump condition
for $E_{22}$ implies that $\Gamma_{E_{22}}=\Gamma_{\overline{E}_{22}}$ with

$\Gamma_{E_{22}}:=\frac{1}{2}(p_{22}+\rho
u_{2}^{2})(u_{1}-\sigma)+p_{12}u_{2}.$

It follows that

$\displaystyle\frac{\mathcal{D}_{\sigma}\Gamma_{E_{22}}}{\mathcal{D}t}=$
$\displaystyle\frac{1}{2}u_{2}^{2}(u_{1}-\sigma)\frac{\mathcal{D}_{\sigma}\rho}{\mathcal{D}t}+E_{22}\frac{\mathcal{D}_{\sigma}u_{1}}{\mathcal{D}t}+(2E_{12}-\rho
u_{2}\sigma)\frac{\mathcal{D}_{\sigma}u_{2}}{\mathcal{D}t}$
$\displaystyle+u_{2}\frac{\mathcal{D}_{\sigma}p_{12}}{\mathcal{D}t}+\frac{1}{2}(u_{1}-\sigma)\frac{\mathcal{D}_{\sigma}p_{22}}{\mathcal{D}t}-E_{22}\frac{\mathcal{D}_{\sigma}\sigma}{\mathcal{D}t}.$

Then one can utilize the relation along the shock wave, i.e.,
$\frac{\mathcal{D}_{\sigma}\Gamma_{E_{22}}}{\mathcal{D}t}=\frac{\mathcal{D}_{\sigma}\Gamma_{\overline{E}_{22}}}{\mathcal{D}t}$,
to compute the limiting value $\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast K}$, $K=L,R$.

Specifically, for the 6-shock wave, one has

$\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,\ast
R}}}{\mathcal{D}t}=\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,R}}}{\mathcal{D}t},$
(5.90)

where

$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,\ast
R}}}{\mathcal{D}t}=\frac{1}{2}u_{2,\ast
R}^{2}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast
R}+E_{22,\ast
R}\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast
R}+(2E_{12,\ast R}-\rho_{\ast R}u_{2,\ast
R}\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}u_{2}}{\mathcal{D}t}\right)_{\ast
R}$ $\displaystyle\qquad+u_{2,\ast
R}\left(\frac{\mathcal{D}_{\sigma_{R}}p_{12}}{\mathcal{D}t}\right)_{\ast
R}+\frac{1}{2}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}p_{22}}{\mathcal{D}t}\right)_{\ast
R}-E_{22,\ast R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t},$
(5.91)
$\displaystyle\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,R}}}{\mathcal{D}t}=\frac{1}{2}u_{2,R}^{2}(u_{1,R}-\sigma_{R})\frac{\mathcal{D}_{\sigma_{R}}\rho_{R}}{\mathcal{D}t}+E_{22,R}\frac{\mathcal{D}_{\sigma_{R}}u_{1,R}}{\mathcal{D}t}+(2E_{12,R}-\rho_{R}u_{2,R}\sigma_{R})\frac{\mathcal{D}_{\sigma_{R}}u_{2,R}}{\mathcal{D}t}$
$\displaystyle\qquad+u_{2,R}\frac{\mathcal{D}_{\sigma_{R}}p_{12,R}}{\mathcal{D}t}+\frac{1}{2}(u_{1,R}-\sigma_{R})\frac{\mathcal{D}_{\sigma_{R}}p_{22,R}}{\mathcal{D}t}-E_{22,R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}.$
(5.92)

By (2.5f), one has

$\frac{\mathcal{D}_{\sigma_{R}}p_{22,R}}{\mathcal{D}t}=(\sigma_{R}-u_{1,R})p_{22,R}^{\prime}-p_{22,R}u_{1,R}^{\prime}-2p_{12,R}u_{2,R}^{\prime}.$
(5.93)

Substituting (5.82a), (5.82b), (5.82c), (5.82e) and (5.93) into (5.92) may
obtain the value of
$\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,R}}}{\mathcal{D}t}$.

######  Theorem 5.14 (Computing $(\frac{\partial u_{2}}{\partial t})_{\ast
R}$, $(\frac{\partial p_{12}}{\partial t})_{\ast R}$ and $(\frac{\partial
p_{22}}{\partial t})_{\ast R}$).

Assume that the 6-shock wave associated with $u_{1}+c$ moves to the right.
Then the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast R}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast R}$ are

$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}=\frac{d_{3,R}b_{4,R}-d_{4,R}b_{3,R}}{a_{3,R}b_{4,R}-a_{4,R}b_{3,R}},\quad\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}=\frac{d_{3,R}a_{4,R}-d_{4,R}a_{3,R}}{b_{3,R}a_{4,R}-b_{4,R}a_{3,R}}.$
(5.94)

It follows that

$\begin{split}&\left(\frac{\partial u_{2}}{\partial t}\right)_{\ast
R}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}+\frac{u_{1,\ast}}{p_{11,\ast}}\left[\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}-\frac{2p_{12,\ast
R}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}\right],\\\
&\left(\frac{\partial p_{12}}{\partial t}\right)_{\ast
R}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast R}+\rho_{\ast
R}u_{1,\ast}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}.\end{split}$ (5.95)

Moreover, the limiting value $\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast R}$ is computed by

$g_{p_{22}}^{\ast R}\left(\frac{\partial p_{22}}{\partial t}\right)_{\ast
R}=f_{p_{22}}^{\ast R},$ (5.96)

where

$\displaystyle g_{p_{22}}^{\ast R}=\frac{1}{2}(u_{1,\ast}-\sigma_{R})^{2},$
$\displaystyle f_{p_{22}}^{\ast R}=u_{1,\ast}\cdot\widetilde{f}_{p_{22}}^{\ast
R}-\frac{(p_{11,\ast}p_{22,\ast R}-4p_{12,\ast
R}^{2})(u_{1,\ast}-\sigma_{R})\sigma_{R}}{6p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}-\frac{p_{12,\ast
R}\sigma_{R}(u_{1,\ast}-\sigma_{R})}{p_{11,\ast}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R},$

with

$\displaystyle\widetilde{f}_{p_{22}}^{\ast
R}=\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,R}}}{\mathcal{D}t}+E_{22,\ast
R}\frac{\mathcal{D}_{\sigma_{R}}\sigma_{R}}{\mathcal{D}t}-\frac{1}{2}u_{2,\ast
R}^{2}(u_{1,\ast}-\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast
R}$ $\displaystyle\qquad-E_{22,\ast
R}\left(\frac{\mathcal{D}_{\sigma_{R}}u_{1}}{\mathcal{D}t}\right)_{\ast
R}-(2E_{12,\ast R}-\rho_{\ast R}u_{2,\ast
R}\sigma_{R})\left(\frac{\mathcal{D}_{\sigma_{R}}u_{2}}{\mathcal{D}t}\right)_{\ast
R}-u_{2,\ast
R}\left(\frac{\mathcal{D}_{\sigma_{R}}p_{12}}{\mathcal{D}t}\right)_{\ast R}.$

###### Proof.

One can deduce (5.94) from (5.85) and (5.89), and then by (2.11), get (5.95).
To derive (5.96), one can first utilize (2.5f), (2.6) and (2.9) to obtain

$u_{1,\ast}\left(\frac{\mathcal{D}_{\sigma_{R}}p_{22}}{\mathcal{D}t}\right)_{\ast
R}=(u_{1,\ast}-\sigma_{R})\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast R}+\frac{(p_{11,\ast}p_{22,\ast R}-4p_{12,\ast
R}^{2})\sigma_{R}}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}+\frac{2p_{12,\ast
R}\sigma_{R}}{p_{11,\ast}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}.$ (5.97)

Multiplying both sides of (5.90) by $u_{1,\ast}$ yields

$u_{1,\ast}\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,\ast
R}}}{\mathcal{D}t}=u_{1,\ast}\frac{\mathcal{D}_{\sigma_{R}}\Gamma_{E_{22,R}}}{\mathcal{D}t}.$
(5.98)

Then substituting (5.91) and (5.92) into (5.98) and applying (5.97) gets
(5.96). ∎

######  Remark 5.15.

For the 6-rarefaction wave, computing $(\frac{\partial u_{2}}{\partial
t})_{\ast R}$, $(\frac{\partial p_{12}}{\partial t})_{\ast R}$ and
$(\frac{\partial p_{22}}{\partial t})_{\ast R}$ is in D.

####  5.1.4 Computing $\left(\frac{\partial u_{2}}{\partial
t}\right)_{\ast\ast K}$, $\left(\frac{\partial p_{12}}{\partial
t}\right)_{\ast\ast K}$ and $\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast\ast K}$ $(K=L,R)$

In this subsection, the generalized Riemann invariants associated with the 2-
and 5-shear waves are used to derive the limiting values $\left(\frac{\partial
u_{2}}{\partial t}\right)_{\ast\ast K}$, $\left(\frac{\partial
p_{12}}{\partial t}\right)_{\ast\ast K}$ and $\left(\frac{\partial
p_{22}}{\partial t}\right)_{\ast\ast K}$ $(K=L,R)$.

If denoting $\Theta_{2}:=u_{2}+\frac{p_{12}}{\sqrt{\rho p_{11}}}$ and
recalling that $\frac{\mathcal{D}_{2}}{\mathcal{D}t}:=\frac{\partial}{\partial
t}+\lambda_{2}\frac{\partial}{\partial x}$ with
$\lambda_{2}=u_{1,\ast}-\frac{c_{\ast L}}{\sqrt{3}}$, then

$\frac{\mathcal{D}_{2}\Theta_{2}}{\mathcal{D}t}=-\frac{p_{12}}{2\rho\sqrt{\rho
p_{11}}}\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}+\frac{\mathcal{D}_{2}u_{2}}{\mathcal{D}t}-\frac{p_{12}}{2p_{11}\sqrt{\rho
p_{11}}}\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}+\frac{1}{\sqrt{\rho
p_{11}}}\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}.$

Because $\Theta_{2}$ is a generalized Riemann invariant associated with the
2-shear wave, one has

$\frac{\mathcal{D}_{2}\Theta_{2,\ast\ast
L}}{\mathcal{D}t}=\frac{\mathcal{D}_{2}\Theta_{2,\ast L}}{\mathcal{D}t},$
(5.99)

where

$\displaystyle\frac{\mathcal{D}_{2}\Theta_{2,\ast\ast
L}}{\mathcal{D}t}=-\frac{p_{12,\ast\ast}}{2\rho_{\ast L}\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast\ast
L}+\left(\frac{\mathcal{D}_{2}u_{2}}{\mathcal{D}t}\right)_{\ast\ast
L}-\frac{p_{12,\ast\ast}}{2p_{11,\ast}\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast\ast
L}+\frac{1}{\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast\ast
L},$ (5.100) $\displaystyle\frac{\mathcal{D}_{2}\Theta_{2,\ast
L}}{\mathcal{D}t}=-\frac{p_{12,\ast L}}{2\rho_{\ast L}\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast
L}+\left(\frac{\mathcal{D}_{2}u_{2}}{\mathcal{D}t}\right)_{\ast
L}-\frac{p_{12,\ast L}}{2p_{11,\ast}\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast
L}+\frac{1}{\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast
L}.$ (5.101)

Because both $\rho$ and $p_{11}$ are the generalized Riemann invariants
associated with the 2-shear wave, one actually has
$\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast L}$ and
$\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast L}$. By
(2.7), one gets

$\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast
L}=\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}-\rho_{\ast
L}(\lambda_{2}-u_{1,\ast})\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+\frac{1}{2}W_{x}(0)\right].$

As for the value of
$\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast L}$, recalling
the equation (5.12), that is
$\text{d}p_{11}=c^{2}\text{d}\rho+\rho^{3}\text{d}S_{1}$, one has

$\left(\frac{\partial\rho}{\partial x}\right)_{\ast L}=\frac{1}{c_{\ast
L}^{2}}\left(\frac{\partial p_{11}}{\partial x}\right)_{\ast
L}-\frac{\rho_{\ast L}^{3}}{c_{\ast L}^{2}}\left(\frac{\partial
S_{1}}{\partial x}\right)_{\ast L}=-\frac{\rho_{\ast L}}{c_{\ast
L}^{2}}\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+\frac{1}{2}W_{x}(0)\right]-\frac{\rho_{\ast
L}^{3}S_{1,L}^{\prime}}{c_{\ast L}c_{L}},$

where (2.7) and (5.23) have been used in the second equality, and
$S_{1,L}^{\prime}$ is given in (5.17), then
$\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast
L}=\left(\frac{\partial\rho}{\partial t}\right)_{\ast
L}+\lambda_{2}\left(\frac{\partial\rho}{\partial x}\right)_{\ast L}$ can be
obtained.

######  Remark 5.16.

For the 1-shock wave, one can get the value of
$\left(\frac{\mathcal{D}_{\sigma_{L}}\rho}{\mathcal{D}t}\right)_{\ast L}$ in a
similar way to compute
$\left(\frac{\mathcal{D}_{\sigma_{R}}\rho}{\mathcal{D}t}\right)_{\ast R}$ in
Subsection 5.1.3, see (5.84). Thus

$\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast
L}=\left(1-\frac{\lambda_{2}}{\sigma_{L}}\right)\left(\frac{\partial\rho}{\partial
t}\right)_{\ast
L}+\frac{\lambda_{2}}{\sigma_{L}}\left(\frac{\mathcal{D}_{\sigma_{L}}\rho}{\mathcal{D}t}\right)_{\ast
L}.$

By (2.9), one has

$\displaystyle\left(\frac{\mathcal{D}_{2}u_{2}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}-\frac{\lambda_{2}-u_{1,\ast}}{p_{11,\ast}}\left[\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}-\frac{2p_{12,\ast\ast}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}\right],$
(5.102)
$\displaystyle\left(\frac{\mathcal{D}_{2}u_{2}}{\mathcal{D}t}\right)_{\ast
L}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
L}-\frac{\lambda_{2}-u_{1,\ast}}{p_{11,\ast}}\left[\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
L}-\frac{2p_{12,\ast
L}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}\right],$

where the notations
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}:=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast R}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}:=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast R}$ have been
used because both $u_{2}$ and $p_{12}$ are generalized Riemann invariants
associated with the contact discontinuity. By (2.10), one has

$\displaystyle\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast\ast
L}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}-\rho_{\ast
L}(\lambda_{2}-u_{1,\ast})\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast},$
(5.103)
$\displaystyle\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast
L}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast L}-\rho_{\ast
L}(\lambda_{2}-u_{1,\ast})\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
L}.$

Note that the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast L}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast L}$ have been
obtained in Subsection 5.1.2, see Theorem 5.10.

######  Lemma 5.17.

Assume that the 2-shear wave associated with $u_{1}-\frac{c}{\sqrt{3}}$ moves
to the left. Then the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}$ satisfy

$a_{5,L}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}+b_{5,L}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}=d_{5,L},$
(5.104)

where

$\displaystyle a_{5,L}=2,\quad b_{5,L}=\frac{2}{\sqrt{\rho_{\ast
L}p_{11,\ast}}},$ $\displaystyle d_{5,L}=\frac{p_{12,\ast\ast}-p_{12,\ast
L}}{2\rho_{\ast L}\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast
L}+\left(\frac{\mathcal{D}_{2}u_{2}}{\mathcal{D}t}\right)_{\ast
L}+\frac{p_{12,\ast\ast}-p_{12,\ast L}}{2p_{11,\ast}\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast
L}$ $\displaystyle\qquad+\frac{1}{\sqrt{\rho_{\ast
L}p_{11,\ast}}}\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast
L}-\frac{2p_{12,\ast\ast}(\lambda_{2}-u_{1,\ast})}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}.$

###### Proof.

Substituting (5.100) and (5.101) into (5.99) and utilizing (5.102) and (5.103)
may obtain (5.104). ∎

If denoting $\Theta_{5}:=u_{2}-\frac{p_{12}}{\sqrt{\rho p_{11}}}$ and
$\frac{\mathcal{D}_{5}}{\mathcal{D}t}:=\frac{\partial}{\partial
t}+\lambda_{5}\frac{\partial}{\partial x}$ with
$\lambda_{5}=u_{1,\ast}+\frac{c_{\ast R}}{\sqrt{3}}$, then

$\text{d}\Theta_{5}=\frac{p_{12}}{2\rho\sqrt{\rho
p_{11}}}\text{d}\rho+\text{d}u_{2}+\frac{p_{12}}{2p_{11}\sqrt{\rho
p_{11}}}\text{d}p_{11}-\frac{1}{\sqrt{\rho p_{11}}}\text{d}p_{12}.$

Because $\Theta_{5}$ is a generalized Riemann invariant associated with the
5-shear wave, one has

$\frac{\mathcal{D}_{5}\Theta_{5,\ast\ast
R}}{\mathcal{D}t}=\frac{\mathcal{D}_{5}\Theta_{5,\ast R}}{\mathcal{D}t}.$

With similar derivation to build (5.104), one can obtain the following result.

######  Lemma 5.18.

Assume that the 5-shear wave associated with $u_{1}+\frac{c}{\sqrt{3}}$ moves
to the right. Then the limiting values
$\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}$ and
$\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}$ satisfy

$a_{5,R}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}+b_{5,R}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}=d_{5,R},$
(5.105)

where

$\displaystyle a_{5,R}=2,\quad b_{5,R}=-\frac{2}{\sqrt{\rho_{\ast
R}p_{11,\ast}}},$ $\displaystyle d_{5,R}=\frac{p_{12,\ast
R}-p_{12,\ast\ast}}{2\rho_{\ast R}\sqrt{\rho_{\ast
R}p_{11,\ast}}}\left(\frac{\mathcal{D}_{5}\rho}{\mathcal{D}t}\right)_{\ast
R}+\left(\frac{\mathcal{D}_{5}u_{2}}{\mathcal{D}t}\right)_{\ast
R}+\frac{p_{12,\ast R}-p_{12,\ast\ast}}{2p_{11,\ast}\sqrt{\rho_{\ast
R}p_{11,\ast}}}\left(\frac{\mathcal{D}_{5}p_{11}}{\mathcal{D}t}\right)_{\ast
R}$ $\displaystyle\qquad-\frac{1}{\sqrt{\rho_{\ast
R}p_{11,\ast}}}\left(\frac{\mathcal{D}_{5}p_{12}}{\mathcal{D}t}\right)_{\ast
R}-\frac{2p_{12,\ast\ast}(\lambda_{5}-u_{1,\ast})}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$

with the computation of
$\left(\frac{\mathcal{D}_{5}\rho}{\mathcal{D}t}\right)_{\ast R}$ being similar
to $\left(\frac{\mathcal{D}_{2}\rho}{\mathcal{D}t}\right)_{\ast L}$ for the
1-shock wave case, see Remark 5.16, and

$\displaystyle\left(\frac{\mathcal{D}_{5}p_{11}}{\mathcal{D}t}\right)_{\ast
R}=\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}-\rho_{\ast
R}(\lambda_{5}-u_{1,\ast})\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+\frac{1}{2}W_{x}(0)\right],$
$\displaystyle\left(\frac{\mathcal{D}_{5}u_{2}}{\mathcal{D}t}\right)_{\ast
R}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}-\frac{\lambda_{5}-u_{1,\ast}}{p_{11,\ast}}\left[\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}-\frac{2p_{12,\ast
R}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}\right],$
$\displaystyle\left(\frac{\mathcal{D}_{5}p_{12}}{\mathcal{D}t}\right)_{\ast
R}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast R}-\rho_{\ast
R}(\lambda_{5}-u_{1,\ast})\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast
R}.$

######  Remark 5.19.

For the 6-rarefaction wave case, similar to the 1-rarefaction wave case, one
has

$\left(\frac{\partial\rho}{\partial x}\right)_{\ast R}=-\frac{\rho_{\ast
R}}{c_{\ast
R}^{2}}\left[\left(\frac{\mathcal{D}u_{1}}{\mathcal{D}t}\right)_{\ast}+\frac{1}{2}W_{x}(0)\right]-\frac{\rho_{\ast
R}^{3}S_{1,R}^{\prime}}{c_{\ast R}c_{R}},$

where
$S_{1,R}^{\prime}=\frac{1}{\rho_{R}^{3}}(p_{11,R}^{\prime}-c_{R}^{2}\rho_{R}^{\prime})$.
Thus the value of $\left(\frac{\mathcal{D}_{5}\rho}{\mathcal{D}t}\right)_{\ast
R}=\left(\frac{\partial\rho}{\partial t}\right)_{\ast
R}+\lambda_{5}\left(\frac{\partial\rho}{\partial x}\right)_{\ast R}$ can be
obtained.

######  Theorem 5.20 (Computing $\left(\frac{\partial u_{2}}{\partial
t}\right)_{\ast\ast K}$ and $\left(\frac{\partial p_{12}}{\partial
t}\right)_{\ast\ast K}$ with $K=L,R$).

Assume that the 2-shear wave associated with $u_{1}-\frac{c}{\sqrt{3}}$ moves
to the left and the 5-shear wave associated with $u_{1}+\frac{c}{\sqrt{3}}$
moves to the right. Then one has

$\begin{split}&\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}=\frac{d_{5,L}b_{5,R}-d_{5,R}b_{5,L}}{a_{5,L}b_{5,R}-a_{5,R}b_{5,L}},\\\
&\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}=\frac{d_{5,L}a_{5,R}-d_{5,R}a_{5,L}}{a_{5,R}b_{5,L}-a_{5,L}b_{5,R}}.\end{split}$
(5.106)

It follows that

$\begin{split}&\left(\frac{\partial u_{2}}{\partial t}\right)_{\ast\ast
K}=\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}+\frac{u_{1,\ast}}{p_{11,\ast}}\left[\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}-\frac{2p_{12,\ast\ast}}{3p_{11,\ast}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}\right],\\\
&\left(\frac{\partial p_{12}}{\partial t}\right)_{\ast\ast
K}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}+\rho_{\ast
K}u_{1,\ast}\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast},\end{split}$
(5.107)

with $K=L,R$.

###### Proof.

(5.106) is the direct result from (5.104) and (5.105). Applying (2.11) may
obtain (5.107). ∎

Because $\det(\mathbf{p})$ is a generalized Riemann invariant for the 2-shear
wave, one has

$\frac{\mathcal{D}_{2}\det(\mathbf{p}_{\ast\ast
L})}{\mathcal{D}t}=\frac{\mathcal{D}_{2}\det(\mathbf{p}_{\ast
L})}{\mathcal{D}t}.$ (5.108)

Due to
$\text{d}(\det(\mathbf{p}))=p_{22}\text{d}p_{11}-2p_{12}\text{d}p_{12}+p_{11}\text{d}p_{22}$,
one has

$\begin{split}&\frac{\mathcal{D}_{2}\det(\mathbf{p}_{\ast\ast
L})}{\mathcal{D}t}=p_{22,\ast\ast
L}\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast\ast
L}-2p_{12,\ast\ast}\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast\ast
L}+p_{11,\ast}\left(\frac{\mathcal{D}_{2}p_{22}}{\mathcal{D}t}\right)_{\ast\ast
L},\\\ &\frac{\mathcal{D}_{2}\det(\mathbf{p}_{\ast
L})}{\mathcal{D}t}=p_{22,\ast
L}\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast L}-2p_{12,\ast
L}\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast
L}+p_{11,\ast}\left(\frac{\mathcal{D}_{2}p_{22}}{\mathcal{D}t}\right)_{\ast
L}.\end{split}$ (5.109)

Using (2.5f), (2.6) and (2.9) yields

$\begin{split}&u_{1,\ast}\left(\frac{\mathcal{D}_{2}p_{22}}{\mathcal{D}t}\right)_{\ast\ast
L}=(u_{1,\ast}-\lambda_{2})\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast\ast L}+\frac{\lambda_{2}(p_{11,\ast}p_{22,\ast\ast
L}-4p_{12,\ast\ast}^{2})}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}+\frac{2\lambda_{2}p_{12,\ast\ast}}{p_{11,\ast}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast},\\\
&u_{1,\ast}\left(\frac{\mathcal{D}_{2}p_{22}}{\mathcal{D}t}\right)_{\ast
L}=(u_{1,\ast}-\lambda_{2})\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast L}+\frac{\lambda_{2}(p_{11,\ast}p_{22,\ast L}-4p_{12,\ast
L}^{2})}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}+\frac{2\lambda_{2}p_{12,\ast
L}}{p_{11,\ast}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
L}.\end{split}$ (5.110)

Similarly, one has

$\displaystyle
u_{1,\ast}\left(\frac{\mathcal{D}_{5}p_{22}}{\mathcal{D}t}\right)_{\ast\ast
R}=(u_{1,\ast}-\lambda_{5})\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast\ast R}+\frac{\lambda_{5}(p_{11,\ast}p_{22,\ast\ast
R}-4p_{12,\ast\ast}^{2})}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}+\frac{2\lambda_{5}p_{12,\ast\ast}}{p_{11,\ast}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast},$
$\displaystyle
u_{1,\ast}\left(\frac{\mathcal{D}_{5}p_{22}}{\mathcal{D}t}\right)_{\ast
R}=(u_{1,\ast}-\lambda_{5})\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast R}+\frac{\lambda_{5}(p_{11,\ast}p_{22,\ast R}-4p_{12,\ast
R}^{2})}{3p_{11,\ast}^{2}}\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}+\frac{2\lambda_{5}p_{12,\ast
R}}{p_{11,\ast}}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast R}.$

######  Theorem 5.21 (Computing $\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast\ast K}$ $(K=L,R)$).

Assume that the 2-shear wave associated with $u_{1}-\frac{c}{\sqrt{3}}$ moves
to the left and the 5-shear wave associated with $u_{1}+\frac{c}{\sqrt{3}}$
moves to the right. Then $\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast\ast K}$ satisfies

$g_{p_{22}}^{\ast\ast K}\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast\ast K}=f_{p_{22}}^{\ast\ast K},\ K=L,R,$ (5.111)

where

$\displaystyle g_{p_{22}}^{\ast\ast
L}=p_{11,\ast}(u_{1,\ast}-\lambda_{2}),\quad g_{p_{22}}^{\ast\ast
R}=p_{11,\ast}(u_{1,\ast}-\lambda_{5}),$ $\displaystyle f_{p_{22}}^{\ast\ast
K}=u_{1,\ast}\cdot\widetilde{f}_{p_{22}}^{\ast\ast
K}+p_{11,\ast}\cdot\widehat{f}_{p_{22}}^{\ast\ast K},\quad K=L,R,$

and

$\displaystyle\widetilde{f}_{p_{22}}^{\ast\ast L}=(p_{22,\ast
L}-p_{22,\ast\ast
L})\left(\frac{\mathcal{D}_{2}p_{11}}{\mathcal{D}t}\right)_{\ast
L}-2p_{12,\ast L}\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast
L}+2p_{12,\ast\ast}\left(\frac{\mathcal{D}_{2}p_{12}}{\mathcal{D}t}\right)_{\ast\ast
L},$ $\displaystyle\widetilde{f}_{p_{22}}^{\ast\ast R}=(p_{22,\ast
R}-p_{22,\ast\ast
R})\left(\frac{\mathcal{D}_{5}p_{11}}{\mathcal{D}t}\right)_{\ast
R}-2p_{12,\ast R}\left(\frac{\mathcal{D}_{5}p_{12}}{\mathcal{D}t}\right)_{\ast
R}+2p_{12,\ast\ast}\left(\frac{\mathcal{D}_{5}p_{12}}{\mathcal{D}t}\right)_{\ast\ast
R},$ $\displaystyle\widehat{f}_{p_{22}}^{\ast\ast
L}=(u_{1,\ast}-\lambda_{2})\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast L}+\frac{\lambda_{2}}{3p_{11,\ast}^{2}}(p_{11,\ast}p_{22,\ast
L}-4p_{12,\ast L}^{2}-p_{11,\ast}p_{22,\ast\ast
L}+4p_{12,\ast\ast}^{2})\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$
$\displaystyle\qquad+\frac{2\lambda_{2}}{p_{11,\ast}}\left[p_{12,\ast
L}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
L}-p_{12,\ast\ast}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}\right],$
$\displaystyle\widehat{f}_{p_{22}}^{\ast\ast
R}=(u_{1,\ast}-\lambda_{5})\left(\frac{\partial p_{22}}{\partial
t}\right)_{\ast R}+\frac{\lambda_{5}}{3p_{11,\ast}^{2}}(p_{11,\ast}p_{22,\ast
R}-4p_{12,\ast R}^{2}-p_{11,\ast}p_{22,\ast\ast
R}+4p_{12,\ast\ast}^{2})\left(\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)_{\ast}$
$\displaystyle\qquad+\frac{2\lambda_{5}}{p_{11,\ast}}\left[p_{12,\ast
R}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast
R}-p_{12,\ast\ast}\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}\right],$

with

$\left(\frac{\mathcal{D}_{5}p_{12}}{\mathcal{D}t}\right)_{\ast\ast
R}=\left(\frac{\mathcal{D}p_{12}}{\mathcal{D}t}\right)_{\ast\ast}-\rho_{\ast
R}(\lambda_{5}-u_{1,\ast})\left(\frac{\mathcal{D}u_{2}}{\mathcal{D}t}\right)_{\ast\ast}.$

###### Proof.

Multiplying both sides of (5.108) by $u_{1,\ast}$ gets

$u_{1,\ast}\frac{\mathcal{D}_{2}\det(\mathbf{p}_{\ast\ast
L})}{\mathcal{D}t}=u_{1,\ast}\frac{\mathcal{D}_{2}\det(\mathbf{p}_{\ast
L})}{\mathcal{D}t}.$ (5.112)

Then substituting (5.109) into (5.112), and utilizing (5.110) yields (5.111)
for $K=L$. Because $\det(\mathbf{p})$ is also a generalized Riemann invariant
for the 5-shear wave, one can also complete the proof for $K=R$ with similar
derivation. ∎

###  5.2 Sonic case

When the $t$-axis is located inside a rarefaction fan, the sonic case happens,
and the results for the nonsonic case can not be applied directly. Without
loss of generality, consider the local wave configuration where $t$-axis is
within the 1-rarefaction wave. The local characteristic coordinates
$(\alpha,\beta)$ within the 1-rarefaction wave are also introduced similar to
those in Subsection 5.1.1.

Denote $\phi:=u_{1}-c$, then

$\text{d}\phi=\text{d}u_{1}+\frac{c}{2\rho}\text{d}\rho-\frac{3}{2\rho
c}\text{d}p_{11}=\text{d}u_{1}-\frac{1}{\rho
c}\text{d}p_{11}-\frac{\rho^{2}}{2c}\text{d}S_{1},$ (5.113)

where (5.12) has been used. By (2.5a), (2.5b) and (2.5d), one has

$\frac{\partial\phi}{\partial t}+(u_{1}-c)\frac{\partial\phi}{\partial
x}=\Pi_{1},$

where $\Pi_{1}=\frac{\rho^{2}}{2}\frac{\partial S_{1}}{\partial
x}-\frac{1}{2}W_{x}$. Denote $\mathbf{V}_{0}$ is the limiting value of
$\mathbf{V}$ when $t\rightarrow 0+$ along the $t$-axis. Then along this
characteristic curve which is tangential to $t$-axis, one has
$\beta_{0}=u_{1,0}-c_{0}=0$. It follows that

$\left(\frac{\partial\phi}{\partial
t}\right)_{0}=\left(\frac{\partial\phi}{\partial
t}\right)_{0}+(u_{1,0}-c_{0})\left(\frac{\partial\phi}{\partial
x}\right)_{0}=\left(\frac{\rho^{2}}{2}\frac{\partial S_{1}}{\partial
x}\right)_{0}-\frac{1}{2}W_{x}(0).$

Hence, by (5.113), one has

$\left(\frac{\partial u_{1}}{\partial
t}\right)_{0}-\frac{1}{\rho_{0}c_{0}}\left(\frac{\partial p_{11}}{\partial
t}\right)_{0}=-\frac{1}{2}W_{x}(0),$ (5.114)

where the equality $\left(\frac{\partial S_{1}}{\partial
t}\right)_{0}=-u_{1,0}\left(\frac{\partial S_{1}}{\partial
x}\right)_{0}=-c_{0}\left(\frac{\partial S_{1}}{\partial x}\right)_{0}$ has
been used. Taking the limit as $t\rightarrow 0$ in (5.35), and let
$\beta=\beta_{0}=0$ in (5.15) and (5.16) obtains

$\left(\frac{\partial u_{1}}{\partial
t}\right)_{0}+\frac{1}{\rho_{0}c_{0}}\left(\frac{\partial p_{11}}{\partial
t}\right)_{0}=\widetilde{d}_{L,0},$ (5.115)

where

$\widetilde{d}_{L,0}=\frac{\rho_{L}^{2}S_{1,L}^{\prime}}{4c_{L}^{3}}u_{1,0}(3c_{L}^{2}+u_{1,0}^{2})-2\psi_{1,L}^{\prime}u_{1,0}-\frac{1}{2}W_{x}(0),$

with the values of $S_{1,L}^{\prime}$ and $\psi_{1,L}^{\prime}$ being given in
(5.17).

######  Theorem 5.22 (Computing $\left(\frac{\partial u_{1}}{\partial
t}\right)_{0}$, $\left(\frac{\partial p_{11}}{\partial t}\right)_{0}$ and
$\left(\frac{\partial\rho}{\partial t}\right)_{0}$).

Assume that the $t$-axis is located inside the 1-rarefaction wave associated
with the $u_{1}-c$ characteristic family. Then

$\displaystyle\begin{split}&\left(\frac{\partial u_{1}}{\partial
t}\right)_{0}=\frac{1}{2}\left[\widetilde{d}_{L,0}-\frac{1}{2}W_{x}(0)\right],\\\
&\left(\frac{\partial p_{11}}{\partial
t}\right)_{0}=\frac{\rho_{0}c_{0}}{2}\left[\widetilde{d}_{L,0}+\frac{1}{2}W_{x}(0)\right],\end{split}$
(5.116) $\displaystyle~{}~{}\left(\frac{\partial\rho}{\partial
t}\right)_{0}=\frac{1}{c_{0}^{2}}\left[\left(\frac{\partial p_{11}}{\partial
t}\right)_{0}+\frac{S_{1,L}^{\prime}}{c_{L}}\rho_{0}^{3}u_{1,0}^{2}\right].$
(5.117)

###### Proof.

By (5.114) and (5.115), (5.116) may be obtained. One takes $\beta=\beta_{0}=0$
in (5.24) to get

$\left(\frac{\partial S_{1}}{\partial
t}\right)_{0}=-\frac{S_{1,L}^{\prime}}{c_{L}}u_{1,0}^{2}.$

Substituting it into

$\left(\frac{\partial\rho}{\partial
t}\right)_{0}=\frac{1}{c_{0}^{2}}\left[\left(\frac{\partial p_{11}}{\partial
t}\right)_{0}-\rho_{0}^{3}\left(\frac{\partial S_{1}}{\partial
t}\right)_{0}\right]$

which is derived from (5.12), one obtains (5.117). ∎

It is noted that for the sonic case, one has the relation
$p_{11,0}=\frac{1}{3}(\rho u_{1}^{2})_{0}$ which is derived from
$u_{1,0}-c_{0}=0$; and (2.11) induces that if $p_{11}\neq\rho u_{1}^{2}$, one
has

$\begin{split}&\frac{\mathcal{D}u_{2}}{\mathcal{D}t}=\frac{p_{11}}{p_{11}-\rho
u_{1}^{2}}\left(\frac{\partial u_{2}}{\partial
t}+\frac{2u_{1}p_{12}}{3p_{11}^{2}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}-\frac{u_{1}}{p_{11}}\frac{\partial
p_{12}}{\partial t}\right),\\\
&\frac{\mathcal{D}p_{12}}{\mathcal{D}t}=\frac{-p_{11}}{p_{11}-\rho
u_{1}^{2}}\left[\rho u_{1}\left(\frac{\partial u_{2}}{\partial
t}+\frac{2u_{1}p_{12}}{3p_{11}^{2}}\frac{\mathcal{D}p_{11}}{\mathcal{D}t}\right)-\frac{\partial
p_{12}}{\partial t}\right].\end{split}$ (5.118)

Hence, substituting (5.118) into (5.61) and (5.68), one can acquire the two
equations which $\frac{\partial u_{2}}{\partial t}(0,\beta)$ and
$\frac{\partial p_{12}}{\partial t}(0,\beta)$ satisfy inside the 1-rarefaction
wave, and taking $\beta=\beta_{0}=0$ yields the following system

$\begin{cases}\widetilde{a}_{3,L,0}\left(\frac{\partial u_{2}}{\partial
t}\right)_{0}+\widetilde{b}_{3,L,0}\left(\frac{\partial p_{12}}{\partial
t}\right)_{0}=\widetilde{d}_{3,L,0},\\\
\widetilde{a}_{4,L,0}\left(\frac{\partial u_{2}}{\partial
t}\right)_{0}+\widetilde{b}_{4,L,0}\left(\frac{\partial p_{12}}{\partial
t}\right)_{0}=\widetilde{d}_{4,L,0},\end{cases}$ (5.119)

where

$\displaystyle\widetilde{a}_{3,L,0}=1+\frac{\beta_{L}}{4},~{}\widetilde{a}_{4,L,0}=\frac{\beta_{L}c_{0}}{4\rho_{0}^{2}},~{}\widetilde{b}_{3,L,0}=\frac{u_{1,0}}{p_{11,0}}\left(1-\frac{\beta_{L}}{4}\right),~{}\widetilde{b}_{4,L,0}=\frac{1}{\rho_{0}^{3}}\left(1-\frac{\beta_{L}}{2}\right),$
$\displaystyle\widetilde{d}_{3,L,0}=\left[\frac{\partial\psi_{2}}{\partial\alpha}(0,\beta_{L})-\frac{\Pi_{2,L}\beta_{L}}{8c_{L}}\right]\psi_{1,L}-\frac{\beta_{L}}{4}\cdot\left\\{\left(\frac{2p_{12}}{p_{11}}\right)_{0}\cdot\left[\left(\frac{\partial
u_{1}}{\partial
t}\right)_{0}+\frac{1}{2}W_{x}(0)\right]+\frac{S_{1,L}^{\prime}}{c_{L}}\left(\frac{c\rho^{2}p_{12}}{2p_{11}}\right)_{0}\right\\}$
$\displaystyle\qquad+\left(\frac{\sqrt{3}p_{12}}{2\rho\sqrt{\rho
p_{11}}}\frac{\partial\rho}{\partial
t}\right)_{0}+\left(\frac{\sqrt{3}p_{12}}{2p_{11}\sqrt{\rho
p_{11}}}\frac{\partial p_{11}}{\partial t}\right)_{0},$
$\displaystyle\widetilde{d}_{4,L,0}=\left[\frac{\partial\psi_{3}}{\partial\alpha}(0,\beta_{L})-\frac{\Pi_{3,L}\beta_{L}}{8c_{L}}\right]\psi_{1,L}-\frac{\beta_{L}}{4}\cdot\left\\{\left(\frac{3p_{12}}{c\rho^{3}}\right)_{0}\left[\left(\frac{\partial
u_{1}}{\partial