+ ζ ( D u , u ) ] d s = 0 ,

where $\left(\tau ,\xi ,\eta ,\zeta \right)$ the unit vector of the outer normal to the surface S. The integral identity (Vorozhtsov  )

${∰}_{S}\left(\left[\tau A+\xi B+\eta C+\zeta D\right]u,u\right)\text{d}s=0$ ,

is called the energy integral for a symmetric system.

Problem 1. Consider the initial-boundary value problem for system (1)-(2) in the region $G=\left\{\left(t,x,y,z\right):0 with periodic boundary conditions

$u\left(t,0,y,z\right)=u\left(t,l,y,z\right)$ , ${B|}_{x=0}={B|}_{x=l}$ (4)

where $T,l$ are positive real numbers, and for any fixed $t,x$ ,

$\left(Сu,u\right)\underset{|y|,|z|\to +\infty }{\to }0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(Du,u\right)\underset{|y|,|z|\to +\infty }{\to }0$ (5)

with the initial data t = 0,

$u\left(0,x,y,z\right)={u}_{0}\left(x,y,z\right)\text{,0}\le x\le l\text{,}|y|<\infty \text{,}|z|<\infty .\text{}$ (6)

and ${u}_{0}\left(x,y,z\right)$ is a given function such that

${\int }_{0}^{l}{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }\left(A\left(0,x,y,z,{u}_{0}\left(x,y,z\right)\right){u}_{0}\left(x,y,z\right),{u}_{0}\left(x,y,z\right)\right)\text{d}z\text{d}y\text{d}x<+\infty$ .

Let us consider the solution of the symmetric t-hyperbolic system (1)-(2) in the domain G, and obtain the energy integrals

${∰}_{S}\left(\left[\tau A+\xi B+\eta C+\zeta D\right]u,u\right)\text{d}s=0.$

It is convenient to apply this identity not to the whole region G, but only ${t}_{1} , bounded by the planes $t={t}_{1},t={t}_{2}$ then we obtain

$\begin{array}{l}I\left({t}_{2}\right)-I\left({t}_{1}\right)+\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\left[-{\left(Bu,u\right)|}_{x=0}+{\left(Bu,u\right)|}_{x=l}\right]\text{d}t\text{d}y\text{d}z\\ +\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\underset{0}{\overset{l}{\int }}\underset{-\infty }{\overset{\infty }{\int }}{\left(Сu,u\right)|}_{-\infty }^{+\infty }\text{d}t\text{d}x\text{d}z+\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\underset{0}{\overset{l}{\int }}\underset{-\infty }{\overset{\infty }{\int }}{\left(Du,u\right)|}_{-\infty }^{+\infty }\text{d}t\text{d}x\text{d}y=0,\end{array}$

where $I\left(t\right)=\underset{t=const}{\iiint }\left(Au,u\right)\text{d}x\text{d}y\text{d}z$ .

From this equality, using the periodicity of the boundary conditions (4) and (5), we deduce the following equality

$I\left({t}_{2}\right)=I\left({t}_{1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\forall {t}_{1},{t}_{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le {t}_{1} (7)

Suppose that the matrix B has a special canonical form:

$B=diag\left({k}_{1},{k}_{2},\cdots ,{k}_{{n}^{+}},-{k}_{{n}^{+}+1},\cdots ,-{k}_{{n}^{+}+{n}^{-}}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{n}^{+}+{n}^{-}=n;\text{\hspace{0.17em}}\text{\hspace{0.17em}}{k}_{i}>0,\text{\hspace{0.17em}}i=1,\cdots ,n.$

Such a diagonal form of the matrix B is essentially used when specifying the boundary conditions for the following problems.

Problem 2. Let us consider the initial-boundary value problem for system (1)-(2) in the region G with boundary conditions for $x=0$

${u}^{I}=S{u}^{II},\text{}0 (8)

and for $x=l$

${u}^{II}=R{u}^{I},\text{}0 (9)

and with the initial data (6). It is assumed that the condition (5) is satisfied. Note that

$\begin{array}{l}u={\left({u}^{I},{u}^{II}\right)}^{\text{T}}\text{,}{u}^{I}={\left({u}_{1}\text{,}{u}_{2}\text{,}\cdots \text{,}{u}_{{n}^{+}}\right)}^{\text{T}}\text{,}{u}^{II}={\left({u}_{{n}^{+}+1}\text{,}{u}_{2}\text{,}\cdots \text{,}{u}_{n}\right)}^{\text{T}},\\ S=S\left(u,t,y,z\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}R=R\left(u,t,y,z\right),\end{array}$

are the rectangular matrices of dimension ${n}^{+}×{n}^{-}$ and ${n}^{-}×{n}^{+}$ , respectively.

The boundary conditions (8) given on the boundary $x=0$ are said to be dissipative at the points of this boundary if vector function $\left({u}_{1}\text{,}{u}_{2}\text{,}\cdots \text{,}{u}_{n}\right)$ satisfies the following inequality 

$-{\left(Bu,u\right)|}_{x=0}\ge 0.$

The boundary conditions (9) given on the boundary $x=l$ are said to be dissipative if at the points of this boundary the vector function $\left({u}_{1}\text{,}{u}_{2}\text{,}\cdots \text{,}{u}_{n}\right)$ satisfies the following inequality 

${\left(Bu,u\right)|}_{x=l}\ge 0.$

For the solution of the initial boundary value problems (1), (2), (6), (8), (9), we obtain the following equality

$\begin{array}{l}I\left({t}_{2}\right)-I\left({t}_{1}\right)+\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\left[-{\left(Bu,u\right)|}_{x=0}+{\left(Bu,u\right)|}_{x=l}\right]\text{d}t\text{d}y\text{d}z\\ +\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\underset{0}{\overset{l}{\int }}\underset{-\infty }{\overset{\infty }{\int }}{\left(Сu,u\right)|}_{-\infty }^{+\infty }\text{d}t\text{d}x\text{d}z+\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\underset{0}{\overset{l}{\int }}\underset{-\infty }{\overset{\infty }{\int }}{\left(Du,u\right)|}_{-\infty }^{+\infty }\text{d}t\text{d}x\text{d}y=0.\end{array}$

From this and using dissipative boundary conditions together with

$-{\left(Bu,u\right)|}_{x=0}\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(Bu,u\right)|}_{x=l}\ge 0,$

we obtain the inequality

$I\left({t}_{2}\right)\le I\left({t}_{1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\forall {t}_{1},{t}_{2}:0\le {t}_{1}

Remark: As an example, we consider a system of equations describing the three-dimensional motion of a gas, under the assumption that the gas is in viscid, not thermally conductive, and is in a state of local thermodynamic equilibrium. In  , it is shown that for the defined conditions the three-dimensional system of equations of gas dynamics could be represented in the form (1) (2).

3. Difference Scheme with Limiter

In this section, we describe a difference scheme, which can be used to approximately solve a dissipative boundary value problem. Then we obtain the estimates for the solutions of difference equations which is analog to estimates of energy integrals.

Let us consider the difference grid points

$\begin{array}{l}{t}_{m}=m\cdot \Delta t\text{},\text{}{x}_{i}=i\cdot \Delta x,{y}_{j}=j\cdot \Delta y,{z}_{k}=k\cdot \Delta z,\\ m=0,\cdots ,M,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=0,1,\cdots ,N,\text{\hspace{0.17em}}|j|,|k|=0,1,\cdots ,+\infty ,\\ M\cdot \Delta t=T,\text{\hspace{0.17em}}\text{\hspace{0.17em}}N\cdot \Delta x=L,\end{array}$

where $\Delta t,\Delta x,\Delta y,\Delta z$ are step size of the difference grid. For the value of the solution at the difference grid points, we introduce the following notation

$u\left(m\cdot \Delta t,i\cdot \Delta x,j\cdot \Delta y,k\cdot \Delta z\right)={u}_{ijk}^{m}=u.$

We assume that А is a unit matrix. The difference model for problem (1) (2) with initial-boundary value (6), (8), (9) is formulated as follows

$\begin{array}{l}\left[{U}^{m+1}+U\right]\left({u}^{m+1}-u\right)+{r}_{x}U\left({u}_{i-1/2}^{L}\right){B}^{+}\left({u}_{i-1/2}^{L}\right)\left(u-{u}_{i-1}\right)\\ \text{ }+{r}_{x}U\left\{\left[{B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L}\right]-\left[{B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L}\right]\right\}\\ \text{ }+{r}_{x}U\left({u}_{i+1/2}^{R}\right){B}^{-}\left({u}_{i+1/2}^{R}\right)\left({u}_{i+1}-u\right)\\ \text{ }+{r}_{x}U\left\{\left[{B}^{T}{}^{-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R}\right]-\left[{B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R}\right]\right\}\\ \text{ }+{r}_{y}U\left({u}_{j-1/2}^{L}\right){C}^{+}\left({u}_{j-1/2}^{L}\right)\left(u-{u}_{j-1}\right)\end{array}$

$\begin{array}{l}\text{ }+{r}_{y}U\left\{\left[{C}^{T}{}^{+}\left({u}_{j+1/2}^{L}\right){u}_{j+1/2}^{L}\right]-\left[{C}^{T}{}^{+}\left({u}_{j-1/2}^{L}\right){u}_{j-1/2}^{L}\right]\right\}\\ \text{ }+{r}_{y}U\left({u}_{j+1/2}^{R}\right){C}^{-}\left({u}_{j+1/2}^{R}\right)\left({u}_{j+1}-u\right)\\ \text{ }+{r}_{y}U\left\{\left[{C}^{T}{}^{-}\left({u}_{j+1/2}^{R}\right){u}_{j+1/2}^{R}\right]-\left[{C}^{T}{}^{-}\left({u}_{j-1/2}^{R}\right){u}_{j-1/2}^{R}\right]\right\}\\ \text{ }+{r}_{z}U\left({u}_{k-1/2}^{L}\right){D}^{+}\left({u}_{k-1/2}^{L}\right)\left(u-{u}_{k-1}\right)\\ \text{ }+{r}_{z}U\left\{\left[{D}^{T}{}^{+}\left({u}_{k+1/2}^{L}\right){u}_{k+1/2}^{L}\right]-\left[{D}^{T}{}^{+}\left({u}_{k-1/2}^{L}\right){u}_{k-1/2}^{L}\right]\right\}\end{array}$

$\begin{array}{l}\text{ }+{r}_{z}U\left({u}_{k+1/2}^{R}\right){D}^{-}\left({u}_{k+1/2}^{R}\right)\left({u}_{k+1}-u\right)\\ \text{ }+{r}_{z}U\left\{\left[{D}^{T}{}^{-}\left({u}_{k+1/2}^{R}\right){u}_{k+1/2}^{R}\right]-\left[{D}^{T}{}^{-}\left({u}_{k-1/2}^{R}\right){u}_{k-1/2}^{R}\right]\right\}=0;\\ m=0,1,\cdots ,M-1;\text{}i=0,1,\cdots ,N-1;\text{}|j|,|k|\text{=}\left\{0,1,\cdots +\infty \right\}\text{,}\end{array}$ (10)

and

(11)

$\begin{array}{l}{u}_{ijk}^{0}={u}_{0}\left(i\cdot \Delta x,j\cdot \Delta y,k\cdot \Delta z\right),\text{}i=0,1,\cdots ,N,\\ |j|,|k|=\text{0,1,}\cdots ,\text{+}\infty ,U=\text{diag}\left({u}_{1},{u}_{2},\cdots ,{u}_{n}\right)\text{,}\end{array}$ (12)

where

$\begin{array}{l}B\left(u\right)={B}^{+}\left(u\right)+{B}^{-}\left(u\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}^{+}\left(u\right)\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}^{-}\left(u\right)\le 0,\\ C\left(u\right)={C}^{+}\left(u\right)+{C}^{-}\left(u\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}^{+}\left(u\right)\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}^{-}\left(u\right)\le 0,\\ D\left(u\right)={D}^{+}\left(u\right)+{D}^{-}\left(u\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}^{+}\left(u\right)\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}^{-}\left(u\right)\le 0,\\ \forall u\in {ℝ}^{n},\text{\hspace{0.17em}}{r}_{x}=\Delta t/\Delta x,\text{}{r}_{y}=\Delta t/\Delta y,\text{}{r}_{z}=\Delta t/\Delta z,\text{}\\ B=B\left({u}_{ijk}^{m},m\cdot \Delta t,i\cdot \Delta x,j\cdot \Delta y,k\cdot \Delta z\right),\end{array}$

with ${B}^{T+},{B}^{T-},{C}^{T+},{C}^{T-},{D}^{T+},{D}^{T-}$ are the corresponding transpose matrices. Here we have omitted the indices which is not having shifts relative to the points $\left(m\cdot \Delta t,i\cdot \Delta x,j\cdot \Delta y,k\cdot \Delta z\right)$

$u={u}_{ijk}^{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}^{m+1}={u}_{ijk}^{m+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{i±1}={u}_{i±1jk}^{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{j±1}={u}_{ij±1k}^{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{k±1}={u}_{ijk±1}^{m}.$

We consider the following reconstruction

$\begin{array}{l}{u}_{i+1/2}^{L}=u+\frac{1}{2}\Psi \left({R}_{i}\right)\left(u-{u}_{i-1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{i-1/2}^{R}=u-\frac{1}{2}\Psi \left(\frac{1}{{R}_{i}}\right)\left({u}_{i+1}-u\right),\\ {u}_{j+1/2}^{L}=u+\frac{1}{2}\Psi \left({R}_{j}\right)\left(u-{u}_{j-1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{j-1/2}^{R}=u-\frac{1}{2}\Psi \left(\frac{1}{{R}_{j}}\right)\left({u}_{j+1}-u\right),\\ {u}_{k+1/2}^{L}=u+\frac{1}{2}\Psi \left({R}_{k}\right)\left(u-{u}_{k-1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{k-1/2}^{R}=u-\frac{1}{2}\Psi \left(\frac{1}{{R}_{k}}\right)\left({u}_{k+1}-u\right),\end{array}$

$\begin{array}{l}\Psi \left(R\right)=\text{diag}\left(\psi \left({R}_{1}\right),\psi \left({R}_{2}\right),\cdots ,\psi \left({R}_{n}\right)\right),\\ \Psi \left(\frac{1}{R}\right)=\text{diag}\left(\psi \left(\frac{1}{{R}_{1}}\right),\psi \left(\frac{1}{{R}_{2}}\right),\cdots ,\psi \left(\frac{1}{{R}_{n}}\right)\right),\\ {\left({R}_{q}\right)}_{i}=\frac{{\left({u}_{q}\right)}_{i+1}-{\left({u}_{q}\right)}_{i}}{{\left({u}_{q}\right)}_{i}-{\left({u}_{q}\right)}_{i-1}},\text{\hspace{0.17em}}{\left({R}_{q}\right)}_{j}=\frac{{\left({u}_{q}\right)}_{j+1}-{\left({u}_{q}\right)}_{j}}{{\left({u}_{q}\right)}_{j}-{\left({u}_{q}\right)}_{j-1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left({R}_{q}\right)}_{k}=\frac{{\left({u}_{q}\right)}_{k+1}-{\left({u}_{q}\right)}_{k}}{{\left({u}_{q}\right)}_{k}-{\left({u}_{q}\right)}_{k-1}},\end{array}$

where $\psi :R\to R$ is a continuous function which is called limiter.

Theorem: The solutions of the finite-difference scheme (10)-(12) is stable in energetic norm $\sqrt{{I}_{m}}$ , where ${I}_{m}=\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}\left({u}_{ijk}^{m},{u}_{ijk}^{m}\right)$ .

Proof: Now we prove that the difference model (10)-(12) admits the presence of difference analog of the dissipative energy integral. This gives us the possibility to get the energetic estimation (the difference analog a priori estimation (7)), from which follows stability of the difference Scheme (10)-(12). We multiply the system (10) by the vector $E={\left(1,1,\cdots ,1\right)}^{\text{T}}$

$\begin{array}{l}\left(\left[{U}^{m+1}+U\right]\left({u}^{m+1}-u\right),E\right)\\ \text{ }+{r}_{x}\left(U\left({u}_{i-1/2}^{L}\right){B}^{+}\left({u}_{i-1/2}^{L}\right)\left(u-{u}_{i-1}\right),E\right)\\ \text{ }+{r}_{x}\left(U\left\{\left[{B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L}\right]-\left[{B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L}\right]\right\},E\right)\\ \text{ }+{r}_{x}\left(U\left({u}_{i+1/2}^{R}\right){B}^{-}\left({u}_{i+1/2}^{R}\right)\left({u}_{i+1}-u\right),E\right)\\ \text{ }+{r}_{x}\left(U\left\{\left[{B}^{T}{}^{-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R}\right]-\left[{B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R}\right]\right\},E\right)\end{array}$

$\begin{array}{l}\text{ }+{r}_{y}\left(U\left({u}_{j-1/2}^{L}\right){C}^{+}\left({u}_{j-1/2}^{L}\right)\left(u-{u}_{j-1}\right),E\right)\\ \text{ }+{r}_{y}\left(U\left\{\left[{C}^{T}{}^{+}\left({u}_{j+1/2}^{L}\right){u}_{j+1/2}^{L}\right]-\left[{C}^{T}{}^{+}\left({u}_{j-1/2}^{L}\right){u}_{j-1/2}^{L}\right]\right\},E\right)\\ \text{ }+{r}_{y}\left(U\left({u}_{j+1/2}^{R}\right){C}^{-}\left({u}_{j+1/2}^{R}\right)\left({u}_{j+1}-u\right),E\right)\\ \text{ }+{r}_{y}\left(U\left\{\left[{C}^{T}{}^{-}\left({u}_{j+1/2}^{R}\right){u}_{j+1/2}^{R}\right]-\left[{C}^{T}{}^{-}\left({u}_{j-1/2}^{R}\right){u}_{j-1/2}^{R}\right]\right\},E\right)\\ \text{ }+{r}_{z}\left(U\left({u}_{k-1/2}^{L}\right){D}^{+}\left({u}_{k-1/2}^{L}\right)\left(u-{u}_{k-1}\right),E\right)\end{array}$

$\begin{array}{l}\text{ }+{r}_{z}\left(U\left\{\left[{D}^{T}{}^{+}\left({u}_{k+1/2}^{L}\right){u}_{k+1/2}^{L}\right]-\left[{D}^{T}{}^{+}\left({u}_{k-1/2}^{L}\right){u}_{k-1/2}^{L}\right]\right\},E\right)\\ \text{ }+{r}_{z}\left(U\left({u}_{k+1/2}^{R}\right){D}^{-}\left({u}_{k+1/2}^{R}\right)\left({u}_{k+1}-u\right),E\right)\\ \text{ }+{r}_{z}\left(U\left\{\left[{D}^{T}{}^{-}\left({u}_{k+1/2}^{R}\right){u}_{k+1/2}^{R}\right]-\left[{D}^{T}{}^{-}\left({u}_{k-1/2}^{R}\right){u}_{k-1/2}^{R}\right]\right\},E\right)=0;\\ m=0,1,\cdots ,M-1;\text{}i=0,1,\cdots ,N-1;\text{}|j|,|k|=\left\{0,1,\cdots ,+\infty \right\}.\end{array}$

We transform each summand to give

$1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\left[{U}^{m+1}+U\right]\left({u}^{m+1}-u\right),E\right)=\left(\left[{u}^{m+1}-u\right],\left[{u}^{m+1}+u\right]\right)=\left({u}^{m+1},{u}^{m+1}\right)-\left(u,u\right);$

$\begin{array}{l}2\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(U\left({u}_{i-1/2}^{L}\right){B}^{+}\left({u}_{i-1/2}^{L}\right)\left(u-{u}_{i-1}\right),E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(U\left\{\left[{B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L}\right]-\left[{B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L}\right]\right\},E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({B}^{+}\left({u}_{i-1/2}^{L}\right)\left(u-{u}_{i-1}\right),{u}_{i-1/2}^{L}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\left(\left\{\left[{B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L}\right]-\left[{B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L}\right]\right\},u\right)\end{array}$

$\begin{array}{l}=\left(\left[u-{u}_{i-1}\right],{B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\left(u,\left\{\left[{B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L}\right]-\left[{B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L}\right]\right\}\right)\\ =\left(u,{B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L}\right)-\left({u}_{i-1},{B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L}\right)\\ =\left({B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L},u\right)-\left({B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L},{u}_{i-1}\right).\end{array}$

$\begin{array}{l}3\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(U\left({u}_{i+1/2}^{R}\right){B}^{-}\left({u}_{i+1/2}^{R}\right)\left({u}_{i+1}-u\right),E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(U\left\{\left[{B}^{T}{}^{-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R}\right]-\left[{B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R}\right]\right\},E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({B}^{-}\left({u}_{i+1/2}^{R}\right)\left({u}_{i+1}-u\right),{u}_{i+1/2}^{R}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\left\{\left[{B}^{T}{}^{-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R}\right]-\left[{B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R}\right]\right\},u\right)\end{array}$

$\begin{array}{l}=\left(\left[{u}_{i+1}-u\right],{B}^{T-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\left(\left\{\left[{B}^{T}{}^{-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R}\right]-\left[{B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R}\right]\right\},u\right)\\ =\left({u}_{i+1},{B}^{T-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R}\right)-\left({B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R},u\right)\\ =\left({B}^{T-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R},{u}_{i+1}\right)-\left({B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R},u\right).\end{array}$

$\begin{array}{l}4\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(U\left({u}_{j-1/2}^{L}\right){C}^{+}\left({u}_{j-1/2}^{L}\right)\left(u-{u}_{j-1}\right),E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(U\left\{\left[{C}^{T}{}^{+}\left({u}_{j+1/2}^{L}\right){u}_{j+1/2}^{L}\right]-\left[{C}^{T}{}^{+}\left({u}_{j-1/2}^{L}\right){u}_{j-1/2}^{L}\right]\right\},E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({C}^{T}{}^{+}\left({u}_{j+1/2}^{L}\right){u}_{j+1/2}^{L},u\right)-\left({C}^{T}{}^{+}\left({u}_{j-1/2}^{L}\right){u}_{j-1/2}^{L},{u}_{j-1}\right);\end{array}$

$\begin{array}{l}5\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(U\left({u}_{j+1/2}^{R}\right){C}^{-}\left({u}_{j+1/2}^{R}\right)\left({u}_{j+1}-u\right),E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(U\left\{\left[{C}^{T}{}^{-}\left({u}_{j+1/2}^{R}\right){u}_{j+1/2}^{R}\right]-\left[{C}^{T}{}^{-}\left({u}_{j-1/2}^{R}\right){u}_{j-1/2}^{R}\right]\right\},E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({C}^{T}{}^{-}\left({u}_{j+1/2}^{R}\right){u}_{j+1/2}^{R},{u}_{j+1}\right)-\left({C}^{T}{}^{-}\left({u}_{j-1/2}^{R}\right){u}_{j-1/2}^{R},u\right);\end{array}$

$\begin{array}{l}6\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(U\left({u}_{k-1/2}^{L}\right){D}^{+}\left({u}_{k-1/2}^{L}\right)\left(u-{u}_{k-1}\right),E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(U\left\{\left[{D}^{T}{}^{+}\left({u}_{k+1/2}^{L}\right){u}_{k+1/2}^{L}\right]-\left[{D}^{T}{}^{+}\left({u}_{k-1/2}^{L}\right){u}_{k-1/2}^{L}\right]\right\},E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({D}^{T}{}^{+}\left({u}_{k+1/2}^{L}\right){u}_{k+1/2}^{L},u\right)-\left({D}^{T}{}^{+}\left({u}_{k-1/2}^{L}\right){u}_{k-1/2}^{L},{u}_{k-1}\right);\end{array}$

$\begin{array}{l}7\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(U\left({u}_{k+1/2}^{R}\right){D}^{-}\left({u}_{k+1/2}^{R}\right)\left({u}_{k+1}-u\right),E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(U\left\{\left[{D}^{T}{}^{-}\left({u}_{k+1/2}^{R}\right){u}_{k+1/2}^{R}\right]-\left[{D}^{T}{}^{-}\left({u}_{k-1/2}^{R}\right){u}_{k-1/2}^{R}\right]\right\},E\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({D}^{T}{}^{-}\left({u}_{k+1/2}^{R}\right){u}_{k+1/2}^{R},{u}_{k+1}\right)-\left({D}^{T}{}^{-}\left({u}_{k-1/2}^{R}\right){u}_{k-1/2}^{R},u\right);\end{array}$

Taking all these transformations into account, we obtain

$\begin{array}{l}\left({u}^{m+1},{u}^{m+1}\right)-\left(u,u\right)+{r}_{x}\left[\left({B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L},u\right)-\left({B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L},{u}_{i-1}\right)\\ \text{ }+\left({B}^{T-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R},{u}_{i+1}\right)-\left({B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R},u\right)\right]+{r}_{y}\left[\left({C}^{T}{}^{+}\left({u}_{j+1/2}^{L}\right){u}_{j+1/2}^{L},u\right)\\ \text{ }-\left({C}^{T}{}^{+}\left({u}_{j-1/2}^{L}\right){u}_{j-1/2}^{L},{u}_{j-1}\right)+\left({C}^{T}{}^{-}\left({u}_{j+1/2}^{R}\right){u}_{j+1/2}^{R},{u}_{j+1}\right)-\left({C}^{T}{}^{-}\left({u}_{j-1/2}^{R}\right){u}_{j-1/2}^{R},u\right)\right]\\ \text{ }+{r}_{z}\left[\left({D}^{T}{}^{+}\left({u}_{k+1/2}^{L}\right){u}_{k+1/2}^{L},u\right)-\left({D}^{T}{}^{+}\left({u}_{k-1/2}^{L}\right){u}_{k-1/2}^{L},{u}_{k-1}\right)\\ \text{ }+\left({D}^{T}{}^{-}\left({u}_{k+1/2}^{R}\right){u}_{k+1/2}^{R},{u}_{k+1}\right)-\left({D}^{T}{}^{-}\left({u}_{k-1/2}^{R}\right){u}_{k-1/2}^{R},u\right)\right]=0.\end{array}$ (13)

We multiply both sides of the Equation (13) by $\Delta x\Delta y\Delta z$ and sum up over $i=1,\cdots ,N-1$ , over j from $-\infty$ to $+\infty$ and over k from $-\infty$ to $+\infty$

$\begin{array}{l}\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}\left({u}^{m+1},{u}^{m+1}\right)-\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}\left(u,u\right)\\ +\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}{r}_{x}\left[\left({B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L},u\right)-\left({B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L},{u}_{i-1}\right)\\ +\left({B}^{T-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R},{u}_{i+1}\right)-\left({B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R},u\right)\right]\end{array}$

$\begin{array}{l}+\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}{r}_{y}\left[\left({C}^{T}{}^{+}\left({u}_{j+1/2}^{L}\right){u}_{j+1/2}^{L},u\right)-\left({C}^{T}{}^{+}\left({u}_{j-1/2}^{L}\right){u}_{j-1/2}^{L},{u}_{j-1}\right)\\ +\left({C}^{T}{}^{-}\left({u}_{j+1/2}^{R}\right){u}_{j+1/2}^{R},{u}_{j+1}\right)-\left({C}^{T}{}^{-}\left({u}_{j-1/2}^{R}\right){u}_{j-1/2}^{R},u\right)\right]\\ +\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}{r}_{z}\left[\left({D}^{T}{}^{+}\left({u}_{k+1/2}^{L}\right){u}_{k+1/2}^{L},u\right)-\left({D}^{T}{}^{+}\left({u}_{k-1/2}^{L}\right){u}_{k-1/2}^{L},{u}_{k-1}\right)\\ +\left({D}^{T}{}^{-}\left({u}_{k+1/2}^{R}\right){u}_{k+1/2}^{R},{u}_{k+1}\right)-\left({D}^{T}{}^{-}\left({u}_{k-1/2}^{R}\right){u}_{k-1/2}^{R},u\right)\right].\end{array}$

Then, using the following relations

$\begin{array}{l}1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}{r}_{x}\left[\left({B}^{T}{}^{+}\left({u}_{i+1/2}^{L}\right){u}_{i+1/2}^{L},u\right)-\left({B}^{T}{}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1/2}^{L},{u}_{i-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({B}^{T-}\left({u}_{i+1/2}^{R}\right){u}_{i+1/2}^{R},{u}_{i+1}\right)-\left({B}^{T}{}^{-}\left({u}_{i-1/2}^{R}\right){u}_{i-1/2}^{R},u\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\Delta x\Delta y\Delta z\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}{r}_{x}\left[\left({B}^{T}{}^{+}\left({u}_{N-1/2}^{L}\right){u}_{N-1/2}^{L},{u}_{N-1}\right)-\left({B}^{T}{}^{+}\left({u}_{1/2}^{L}\right){u}_{1/2}^{L},{u}_{0}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({B}^{T-}\left({u}_{N-1/2}^{R}\right){u}_{N-1/2}^{R},{u}_{N}\right)-\left({B}^{T}{}^{-}\left({u}_{1/2}^{R}\right){u}_{1/2}^{R},{u}_{1}\right)\right]=0\end{array}$

$\begin{array}{l}2\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}{r}_{y}\left[\left({C}^{T}{}^{+}\left({u}_{j+1/2}^{L}\right){u}_{j+1/2}^{L},u\right)-\left({C}^{T}{}^{+}\left({u}_{j-1/2}^{L}\right){u}_{j-1/2}^{L},{u}_{j-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({C}^{T}{}^{-}\left({u}_{j+1/2}^{R}\right){u}_{j+1/2}^{R},{u}_{j+1}\right)-\left({C}^{T}{}^{-}\left({u}_{j-1/2}^{R}\right){u}_{j-1/2}^{R},u\right)\right]=0;\end{array}$

$\begin{array}{l}3\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}{r}_{z}\left[\left({D}^{T}{}^{+}\left({u}_{k+1/2}^{L}\right){u}_{k+1/2}^{L},u\right)-\left({D}^{T}{}^{+}\left({u}_{k-1/2}^{L}\right){u}_{k-1/2}^{L},{u}_{k-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({D}^{T}{}^{-}\left({u}_{k+1/2}^{R}\right){u}_{k+1/2}^{R},{u}_{k+1}\right)-\left({D}^{T}{}^{-}\left({u}_{k-1/2}^{R}\right){u}_{k-1/2}^{R},u\right)\right]=0;\end{array}$

we have

$\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}\left({u}^{m+1},{u}^{m+1}\right)=\Delta x\Delta y\Delta z\underset{i=1}{\overset{N-1}{\sum }}\underset{j=-\infty }{\overset{+\infty }{\sum }}\underset{k=-\infty }{\overset{+\infty }{\sum }}\left(u,u\right),\text{\hspace{0.17em}}m=0,\cdots ,M-1.$

Hence

${I}_{m+1}={I}_{m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m=0,\cdots ,M-1,$ (14)

which means the stability of the difference scheme (10)-(12) in the energetic norm $\sqrt{{I}_{m}}$ is established.

4. Numerical Example

Example 1. As an example, we consider the Burger’s hyperbolic equation given by:

${u}_{t}+u{u}_{x}=0$ .

We introduce the following notations

$\begin{array}{l}f\left(u\right)=\frac{{u}^{2}}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(u\right)={f}^{+}\left(u\right)+{f}^{-}\left(u\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{d}{f}^{+}}{\text{d}u}\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{d}{f}^{-}}{\text{d}u}\le 0,\text{\hspace{0.17em}}|u|\underset{|x|\to +\infty }{\to }0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}u\in R,\\ {u}_{i}^{m+1}={u}_{i}^{m}-\frac{\Delta t}{\Delta x}\left[{f}^{+}\left({u}_{i+1/2}^{L}\right)-{f}^{+}\left({u}_{i-1/2}^{L}\right)+{f}^{-}\left({u}_{i+1/2}^{R}\right)-{f}^{-}\left({u}_{i-1/2}^{R}\right)\right]\end{array}$

Consider the following reconstruction:

$\begin{array}{l}{u}_{i+1/2}^{L}={u}_{i}^{m}+\frac{1}{2}\psi \left({R}_{i}^{m}\right)\left({u}_{i}^{m}-{u}_{i-1}^{m}\right),\\ {u}_{i-1/2}^{R}={u}_{i}^{m}-\frac{1}{2}\psi \left(\frac{1}{{R}_{i}^{m}}\right)\left({u}_{i+1}^{m}-{u}_{i}^{m}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{i}^{m}=\frac{{u}_{i+1}^{m}-{u}_{i}^{m}}{{u}_{i}^{m}-{u}_{i-1}^{m}},\end{array}$

where ${u}_{i}^{k}\approx \frac{1}{\Delta x}{\int }_{{x}_{i-1/2}}^{{x}_{i+1/2}}u\left(\xi ,k\Delta t\right)\text{d}\xi$ .

When $\psi =0$ corresponds to the scheme of the first order, $\psi =1$ is the one sided upwind scheme of second order. Consider the following modification of the obtained scheme

$\begin{array}{l}{u}^{m+1}-u+\frac{4}{3\left({u}^{m+1}+u\right)}r×\left[\left({f}^{+}\left({u}_{i-1/2}^{L}\right)\cdot \left({u}_{i}-{u}_{i-1}\right)\right)+\left(\left[{f}^{+}\left({u}_{i+1/2}^{L}\right)-{f}^{+}\left({u}_{i-1/2}^{L}\right)\right]\cdot u\right)\\ +\left({f}^{-}\left({u}_{i+1/2}^{R}\right)\cdot \left({u}_{i+1}-{u}_{i}\right)\right)+\left(\left[{f}^{-}\left({u}_{i+1/2}^{R}\right)-{f}^{-}\left({u}_{i-1/2}^{R}\right)\right]\cdot u\right)\right]=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\Delta t/\Delta x.\end{array}$ (15)

We now prove that the difference scheme (15) admits the availability of difference analogue of the energy integral. Multiplying the system (15) by $\left[{u}^{m+1}+u\right]$ :

$\begin{array}{l}\left[{u}^{m+1}-u\right]\cdot \left[{u}^{m+1}+u\right]+\frac{4}{3}r\cdot \left[\left({f}^{+}\left({u}_{i-1/2}^{L}\right)\cdot \left(u-{u}_{i-1}\right)\right)+\left(\left[{f}^{+}\left({u}_{i+1/2}^{L}\right)-{f}^{+}\left({u}_{i-1/2}^{L}\right)\right]\cdot u\right)\\ +\left({f}^{-}\left({u}_{i+1/2}^{R}\right)\cdot \left({u}_{i+1}-u\right)\right)+\left(\left[{f}^{-}\left({u}_{i+1/2}^{R}\right)-{f}^{-}\left({u}_{i-1/2}^{R}\right)\right]\cdot u\right)\right]=0.\end{array}$

Using formulas of difference differentiation, we obtain the following identity

$\begin{array}{l}\left[\left({f}^{+}\left({u}_{i-1/2}^{L}\right)\cdot \left(u-{u}_{i-1}\right)\right)+\left(\left[{f}^{+}\left({u}_{i+1/2}^{L}\right)-{f}^{+}\left({u}_{i-1/2}^{L}\right)\right]\cdot u\right)\\ ={f}^{+}\left({u}_{i+1/2}^{L}\right)u-{f}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1},\\ \left({f}^{-}\left({u}_{i+1/2}^{R}\right)\cdot \left({u}_{i+1}-u\right)\right)+\left(\left[{f}^{-}\left({u}_{i+1/2}^{R}\right)-{f}^{-}\left({u}_{i-1/2}^{R}\right)\right]\cdot u\right)\\ ={f}^{-}\left({u}_{i+1/2}^{R}\right){u}_{i+1}-{f}^{-}\left({u}_{i-1/2}^{R}\right)u.\end{array}$

Taking into account all of these transformations, we have

$\begin{array}{l}{\left({u}^{m+1}\right)}^{2}-{\left(u\right)}^{2}+\frac{4}{3}r\cdot \left\{{f}^{+}\left({u}_{i+1/2}^{L}\right)u-{f}^{+}\left({u}_{i-1/2}^{L}\right){u}_{i-1}\\ +{f}^{-}\left({u}_{i+1/2}^{R}\right){u}_{i+1}-{f}^{-}\left({u}_{i-1/2}^{R}\right)u\right\}=0.\end{array}$ (16)

Multiply both sides of the (16) by $\Delta x$ and sum up over i from $-\infty$ to $+\infty$ , and noting that the function u tends to zero at infinity and denoting the quantity $\Delta x\underset{i=-\infty }{\overset{\infty }{\sum }}{\left(u\right)}^{2}$ by ${I}_{m}$ we obtain the equality

${I}_{m+1}-{I}_{m}=0$ .

From this it is easily follows the energetic estimate

${I}_{m}={I}_{0}\text{,}m=1,\cdots ,M$ (17)

which means stability of the difference scheme in the norm $\sqrt{{I}_{m}}$ .

Example 2. Consider the following Cauchy problem

$\begin{array}{l}{u}_{t}+{\left({u}^{2}/2\right)}_{x}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\infty 0,\\ |u|\underset{|x|\to +\infty }{\to }0,\\ u\left(x,0\right)=\left\{\begin{array}{ll}1,\hfill & x<0,\hfill \\ 0,\hfill & x>0.\hfill \end{array}\end{array}$ (18)

Rewrite the difference scheme (15) in the following form

$\begin{array}{c}{u}^{m+1}=u+\frac{2}{3\left({u}^{m+1}+u\right)}r×\left[\left({f}^{+}\left({u}_{i-1/2}^{L}\right)\left(u-{u}_{i-1}\right)\right)+\left(\left[{f}^{+}\left({u}_{i+1/2}^{L}\right)-{f}^{+}\left({u}_{i-1/2}^{L}\right)\right]u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({f}^{-}\left({u}_{i+1/2}^{R}\right)\left({u}_{i+1}-u\right)\right)+\left(\left[{f}^{-}\left({u}_{i+1/2}^{R}\right)-{f}^{-}\left({u}_{i-1/2}^{R}\right)\right]u\right)\right].\end{array}$ (19)

In solving the difference scheme (19) we apply the iteration method with respect to the nonlinear coefficients

$\begin{array}{c}{u}^{\left(s+1\right)}=u+\frac{2}{3\left({u}^{\left(s\right)}+u\right)}r×\left[\left({f}^{+}\left({u}_{i-1/2}^{L}\right)\left({u}_{i}-{u}_{i-1}\right)\right)+\left(\left[{f}^{+}\left({u}_{i+1/2}^{L}\right)-{f}^{+}\left({u}_{i-1/2}^{L}\right)\right]u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({f}^{-}\left({u}_{i+1/2}^{R}\right)\left({u}_{i+1}-{u}_{i}\right)\right)+\left(\left[{f}^{-}\left({u}_{i+1/2}^{R}\right)-{f}^{-}\left({u}_{i-1/2}^{R}\right)\right]u\right)\right].\end{array}$

We have carried out a posteriori error analysis of the proposed scheme. Table 1 presents the exact and numerical solutions at points $\left({t}_{m},{x}_{i}\right)$ for the Cauchy problem (18).

Figures 1-4 exhibits the graphical results of the exact and numerical solutions. Comparisons between numerical and exact solutions are presented in Figures 1-4. It can be concluded that the scheme with limiter well modulates the jump. All results are obtained in Mathcad.

Table 1. Exact and numerical solution for problem (18).

Figure 1. Exact solution of the problem of (18).

Figure 2. Numerical solution by scheme (19).

Figure 3. Red circle and solid blue lines are the exact and numerical solutions respectively for t = 0.4.

Figure 4. Red circle and solid blue lines are the exact and numerical solutions respectively for t = 1.4.

5. Conclusion

The class of three-dimensional quasilinear hyperbolic systems is studied. The formulation of initial boundary value problem for this class of quasilinear hyperbolic systems in two variants is given. A priori estimate of the solution of initial boundary value problem is obtained by constructing an energy integral. Difference scheme with limiter is constructed and a priori estimate for its solution is obtained. Numerical results for the developed schemes show agreement with exact solution.

Acknowledgements

This work was supported by Research Management Center (RMC), Universiti Sains Islam Malaysia (USIM) under Research Grand PPP/USG-0216/FST/30/15316.

Fund

This research was supported in part under R. A. Welch Foundation Grant E-0608.

Conflicts of Interest

The authors declare no conflicts of interest.

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