+ ζ ( D u , u ) ] d s = 0 ,

where ( τ , ξ , η , ζ ) the unit vector of the outer normal to the surface S. The integral identity (Vorozhtsov [1] )

S ( [ τ A + ξ B + η C + ζ D ] u , u ) 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 = { ( t , x , y , z ) : 0 < t T ; 0 < x < l , | y | < , | z | < } with periodic boundary conditions

u ( t , 0 , y , z ) = u ( t , l , y , z ) , B | x = 0 = B | x = l (4)

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

( С u , u ) | y | , | z | + 0 , ( D u , u ) | y | , | z | + 0 (5)

with the initial data t = 0,

u ( 0 , x , y , z ) = u 0 ( x , y , z ) ,0 x l , | y | < , | z | < . (6)

and u 0 ( x , y , z ) is a given function such that

0 l + + ( A ( 0 , x , y , z , u 0 ( x , y , z ) ) u 0 ( x , y , z ) , u 0 ( x , y , z ) ) d z d y d x < + .

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

S ( [ τ A + ξ B + η C + ζ D ] u , u ) d s = 0.

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

I ( t 2 ) I ( t 1 ) + t 1 t 2 [ ( B u , u ) | x = 0 + ( B u , u ) | x = l ] d t d y d z + t 1 t 2 0 l ( С u , u ) | + d t d x d z + t 1 t 2 0 l ( D u , u ) | + d t d x d y = 0 ,

where I ( t ) = t = c o n s t ( A u , u ) d x d y d z .

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

I ( t 2 ) = I ( t 1 ) , t 1 , t 2 , 0 t 1 < t < t 2 T . (7)

Suppose that the matrix B has a special canonical form:

B = d i a g ( k 1 , k 2 , , k n + , k n + + 1 , , k n + + n ) , n + + n = n ; k i > 0 , i = 1 , , 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 I I , 0 < t T , | y | < , | z | < (8)

and for x = l

u I I = R u I , 0 < t T , | y | < , | z | < (9)

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

u = ( u I , u I I ) T , u I = ( u 1 , u 2 , , u n + ) T , u I I = ( u n + + 1 , u 2 , , u n ) T , S = S ( u , t , y , z ) , R = R ( u , t , y , z ) ,

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 ( u 1 , u 2 , , u n ) satisfies the following inequality [26]

( B u , u ) | x = 0 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 ( u 1 , u 2 , , u n ) satisfies the following inequality [26]

( B u , u ) | x = l 0.

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

I ( t 2 ) I ( t 1 ) + t 1 t 2 [ ( B u , u ) | x = 0 + ( B u , u ) | x = l ] d t d y d z + t 1 t 2 0 l ( С u , u ) | + d t d x d z + t 1 t 2 0 l ( D u , u ) | + d t d x d y = 0 .

From this and using dissipative boundary conditions together with

( B u , u ) | x = 0 0 , ( B u , u ) | x = l 0 ,

we obtain the inequality

I ( t 2 ) I ( t 1 ) , t 1 , t 2 : 0 t 1 < t < t 2 T .

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 [27] , 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

t m = m Δ t , x i = i Δ x , y j = j Δ y , z k = k Δ z , m = 0 , , M , i = 0 , 1 , , N , | j | , | k | = 0 , 1 , , + , M Δ t = T , N Δ x = L ,

where Δ t , Δ x , Δ y , Δ 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 ( m Δ t , i Δ x , j Δ y , k Δ z ) = u i j k 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

[ U m + 1 + U ] ( u m + 1 u ) + r x U ( u i 1 / 2 L ) B + ( u i 1 / 2 L ) ( u u i 1 ) + r x U { [ B T + ( u i + 1 / 2 L ) u i + 1 / 2 L ] [ B T + ( u i 1 / 2 L ) u i 1 / 2 L ] } + r x U ( u i + 1 / 2 R ) B ( u i + 1 / 2 R ) ( u i + 1 u ) + r x U { [ B T ( u i + 1 / 2 R ) u i + 1 / 2 R ] [ B T ( u i 1 / 2 R ) u i 1 / 2 R ] } + r y U ( u j 1 / 2 L ) C + ( u j 1 / 2 L ) ( u u j 1 )

+ r y U { [ C T + ( u j + 1 / 2 L ) u j + 1 / 2 L ] [ C T + ( u j 1 / 2 L ) u j 1 / 2 L ] } + r y U ( u j + 1 / 2 R ) C ( u j + 1 / 2 R ) ( u j + 1 u ) + r y U { [ C T ( u j + 1 / 2 R ) u j + 1 / 2 R ] [ C T ( u j 1 / 2 R ) u j 1 / 2 R ] } + r z U ( u k 1 / 2 L ) D + ( u k 1 / 2 L ) ( u u k 1 ) + r z U { [ D T + ( u k + 1 / 2 L ) u k + 1 / 2 L ] [ D T + ( u k 1 / 2 L ) u k 1 / 2 L ] }

+ r z U ( u k + 1 / 2 R ) D ( u k + 1 / 2 R ) ( u k + 1 u ) + r z U { [ D T ( u k + 1 / 2 R ) u k + 1 / 2 R ] [ D T ( u k 1 / 2 R ) u k 1 / 2 R ] } = 0 ; m = 0 , 1 , , M 1 ; i = 0 , 1 , , N 1 ; | j | , | k | = { 0 , 1 , + } , (10)

and

u 0 j k m = u N j k m , u 1 j k m = u N 1 j k m m = 0 , 1 , , M , | j | , | k | = { 0 , 1 , , + } ( B T + ( u N 1 / 2 L ) u N 1 / 2 L , u N 1 ) + ( B T ( u N 1 / 2 R ) u N 1 / 2 R , u N ) = ( B T + ( u 1 / 2 L ) u 1 / 2 L , u 0 ) + ( B T ( u 1 / 2 R ) u 1 / 2 R , u 1 ) , ( C T + ( u j + 1 / 2 L ) u j + 1 / 2 L , u ) + ( C T ( u j + 1 / 2 R ) u j + 1 / 2 R , u j + 1 ) | j | + 0 , i , k , ( D T + ( u k + 1 / 2 L ) u k + 1 / 2 L , u ) + ( D T ( u k + 1 / 2 R ) u k + 1 / 2 R , u k + 1 ) | j | + 0 , i , k , (11)

u i j k 0 = u 0 ( i Δ x , j Δ y , k Δ z ) , i = 0 , 1 , , N , | j | , | k | = 0,1, , + , U = diag ( u 1 , u 2 , , u n ) , (12)

where

B ( u ) = B + ( u ) + B ( u ) , B + ( u ) 0 , B ( u ) 0 , C ( u ) = C + ( u ) + C ( u ) , C + ( u ) 0 , C ( u ) 0 , D ( u ) = D + ( u ) + D ( u ) , D + ( u ) 0 , D ( u ) 0 , u n , r x = Δ t / Δ x , r y = Δ t / Δ y , r z = Δ t / Δ z , B = B ( u i j k m , m Δ t , i Δ x , j Δ y , k Δ z ) ,

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 ( m Δ t , i Δ x , j Δ y , k Δ z )

u = u i j k m , u m + 1 = u i j k m + 1 , u i ± 1 = u i ± 1 j k m , u j ± 1 = u i j ± 1 k m , u k ± 1 = u i j k ± 1 m .

We consider the following reconstruction

u i + 1 / 2 L = u + 1 2 Ψ ( R i ) ( u u i 1 ) , u i 1 / 2 R = u 1 2 Ψ ( 1 R i ) ( u i + 1 u ) , u j + 1 / 2 L = u + 1 2 Ψ ( R j ) ( u u j 1 ) , u j 1 / 2 R = u 1 2 Ψ ( 1 R j ) ( u j + 1 u ) , u k + 1 / 2 L = u + 1 2 Ψ ( R k ) ( u u k 1 ) , u k 1 / 2 R = u 1 2 Ψ ( 1 R k ) ( u k + 1 u ) ,

Ψ ( R ) = diag ( ψ ( R 1 ) , ψ ( R 2 ) , , ψ ( R n ) ) , Ψ ( 1 R ) = diag ( ψ ( 1 R 1 ) , ψ ( 1 R 2 ) , , ψ ( 1 R n ) ) , ( R q ) i = ( u q ) i + 1 ( u q ) i ( u q ) i ( u q ) i 1 , ( R q ) j = ( u q ) j + 1 ( u q ) j ( u q ) j ( u q ) j 1 , ( R q ) k = ( u q ) k + 1 ( u q ) k ( u q ) k ( u q ) k 1 ,

where ψ : R R is a continuous function which is called limiter.

Theorem: The solutions of the finite-difference scheme (10)-(12) is stable in energetic norm I m , where I m = Δ x Δ y Δ z i = 1 N 1 j = + k = + ( u i j k m , u i j k m ) .

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 = ( 1 , 1 , , 1 ) T

( [ U m + 1 + U ] ( u m + 1 u ) , E ) + r x ( U ( u i 1 / 2 L ) B + ( u i 1 / 2 L ) ( u u i 1 ) , E ) + r x ( U { [ B T + ( u i + 1 / 2 L ) u i + 1 / 2 L ] [ B T + ( u i 1 / 2 L ) u i 1 / 2 L ] } , E ) + r x ( U ( u i + 1 / 2 R ) B ( u i + 1 / 2 R ) ( u i + 1 u ) , E ) + r x ( U { [ B T ( u i + 1 / 2 R ) u i + 1 / 2 R ] [ B T ( u i 1 / 2 R ) u i 1 / 2 R ] } , E )

+ r y ( U ( u j 1 / 2 L ) C + ( u j 1 / 2 L ) ( u u j 1 ) , E ) + r y ( U { [ C T + ( u j + 1 / 2 L ) u j + 1 / 2 L ] [ C T + ( u j 1 / 2 L ) u j 1 / 2 L ] } , E ) + r y ( U ( u j + 1 / 2 R ) C ( u j + 1 / 2 R ) ( u j + 1 u ) , E ) + r y ( U { [ C T ( u j + 1 / 2 R ) u j + 1 / 2 R ] [ C T ( u j 1 / 2 R ) u j 1 / 2 R ] } , E ) + r z ( U ( u k 1 / 2 L ) D + ( u k 1 / 2 L ) ( u u k 1 ) , E )

+ r z ( U { [ D T + ( u k + 1 / 2 L ) u k + 1 / 2 L ] [ D T + ( u k 1 / 2 L ) u k 1 / 2 L ] } , E ) + r z ( U ( u k + 1 / 2 R ) D ( u k + 1 / 2 R ) ( u k + 1 u ) , E ) + r z ( U { [ D T ( u k + 1 / 2 R ) u k + 1 / 2 R ] [ D T ( u k 1 / 2 R ) u k 1 / 2 R ] } , E ) = 0 ; m = 0 , 1 , , M 1 ; i = 0 , 1 , , N 1 ; | j | , | k | = { 0 , 1 , , + } .

We transform each summand to give

1 ) ( [ U m + 1 + U ] ( u m + 1 u ) , E ) = ( [ u m + 1 u ] , [ u m + 1 + u ] ) = ( u m + 1 , u m + 1 ) ( u , u ) ;

2 ) ( U ( u i 1 / 2 L ) B + ( u i 1 / 2 L ) ( u u i 1 ) , E ) + ( U { [ B T + ( u i + 1 / 2 L ) u i + 1 / 2 L ] [ B T + ( u i 1 / 2 L ) u i 1 / 2 L ] } , E ) = ( B + ( u i 1 / 2 L ) ( u u i 1 ) , u i 1 / 2 L ) + ( { [ B T + ( u i + 1 / 2 L ) u i + 1 / 2 L ] [ B T + ( u i 1 / 2 L ) u i 1 / 2 L ] } , u )

= ( [ u u i 1 ] , B T + ( u i 1 / 2 L ) u i 1 / 2 L ) + ( u , { [ B T + ( u i + 1 / 2 L ) u i + 1 / 2 L ] [ B T + ( u i 1 / 2 L ) u i 1 / 2 L ] } ) = ( u , B T + ( u i + 1 / 2 L ) u i + 1 / 2 L ) ( u i 1 , B T + ( u i 1 / 2 L ) u i 1 / 2 L ) = ( B T + ( u i + 1 / 2 L ) u i + 1 / 2 L , u ) ( B T + ( u i 1 / 2 L ) u i 1 / 2 L , u i 1 ) .

3 ) ( U ( u i + 1 / 2 R ) B ( u i + 1 / 2 R ) ( u i + 1 u ) , E ) + ( U { [ B T ( u i + 1 / 2 R ) u i + 1 / 2 R ] [ B T ( u i 1 / 2 R ) u i 1 / 2 R ] } , E ) = ( B ( u i + 1 / 2 R ) ( u i + 1 u ) , u i + 1 / 2 R ) + ( { [ B T ( u i + 1 / 2 R ) u i + 1 / 2 R ] [ B T ( u i 1 / 2 R ) u i 1 / 2 R ] } , u )

= ( [ u i + 1 u ] , B T ( u i + 1 / 2 R ) u i + 1 / 2 R ) + ( { [ B T ( u i + 1 / 2 R ) u i + 1 / 2 R ] [ B T ( u i 1 / 2 R ) u i 1 / 2 R ] } , u ) = ( u i + 1 , B T ( u i + 1 / 2 R ) u i + 1 / 2 R ) ( B T ( u i 1 / 2 R ) u i 1 / 2 R , u ) = ( B T ( u i + 1 / 2 R ) u i + 1 / 2 R , u i + 1 ) ( B T ( u i 1 / 2 R ) u i 1 / 2 R , u ) .

4 ) ( U ( u j 1 / 2 L ) C + ( u j 1 / 2 L ) ( u u j 1 ) , E ) + ( U { [ C T + ( u j + 1 / 2 L ) u j + 1 / 2 L ] [ C T + ( u j 1 / 2 L ) u j 1 / 2 L ] } , E ) = ( C T + ( u j + 1 / 2 L ) u j + 1 / 2 L , u ) ( C T + ( u j 1 / 2 L ) u j 1 / 2 L , u j 1 ) ;

5 ) ( U ( u j + 1 / 2 R ) C ( u j + 1 / 2 R ) ( u j + 1 u ) , E ) + ( U { [ C T ( u j + 1 / 2 R ) u j + 1 / 2 R ] [ C T ( u j 1 / 2 R ) u j 1 / 2 R ] } , E ) = ( C T ( u j + 1 / 2 R ) u j + 1 / 2 R , u j + 1 ) ( C T ( u j 1 / 2 R ) u j 1 / 2 R , u ) ;

6 ) ( U ( u k 1 / 2 L ) D + ( u k 1 / 2 L ) ( u u k 1 ) , E ) + ( U { [ D T + ( u k + 1 / 2 L ) u k + 1 / 2 L ] [ D T + ( u k 1 / 2 L ) u k 1 / 2 L ] } , E ) = ( D T + ( u k + 1 / 2 L ) u k + 1 / 2 L , u ) ( D T + ( u k 1 / 2 L ) u k 1 / 2 L , u k 1 ) ;

7 ) ( U ( u k + 1 / 2 R ) D ( u k + 1 / 2 R ) ( u k + 1 u ) , E ) + ( U { [ D T ( u k + 1 / 2 R ) u k + 1 / 2 R ] [ D T ( u k 1 / 2 R ) u k 1 / 2 R ] } , E ) = ( D T ( u k + 1 / 2 R ) u k + 1 / 2 R , u k + 1 ) ( D T ( u k 1 / 2 R ) u k 1 / 2 R , u ) ;

Taking all these transformations into account, we obtain

( u m + 1 , u m + 1 ) ( u , u ) + r x [ ( B T + ( u i + 1 / 2 L ) u i + 1 / 2 L , u ) ( B T + ( u i 1 / 2 L ) u i 1 / 2 L , u i 1 ) + ( B T ( u i + 1 / 2 R ) u i + 1 / 2 R , u i + 1 ) ( B T ( u i 1 / 2 R ) u i 1 / 2 R , u ) ] + r y [ ( C T + ( u j + 1 / 2 L ) u j + 1 / 2 L , u ) ( C T + ( u j 1 / 2 L ) u j 1 / 2 L , u j 1 ) + ( C T ( u j + 1 / 2 R ) u j + 1 / 2 R , u j + 1 ) ( C T ( u j 1 / 2 R ) u j 1 / 2 R , u ) ] + r z [ ( D T + ( u k + 1 / 2 L ) u k + 1 / 2 L , u ) ( D T + ( u k 1 / 2 L ) u k 1 / 2 L , u k 1 ) + ( D T ( u k + 1 / 2 R ) u k + 1 / 2 R , u k + 1 ) ( D T ( u k 1 / 2 R ) u k 1 / 2 R , u ) ] = 0. (13)

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

Δ x Δ y Δ z i = 1 N 1 j = + k = + ( u m + 1 , u m + 1 ) Δ x Δ y Δ z i = 1 N 1 j = + k = + ( u , u ) + Δ x Δ y Δ z i = 1 N 1 j = + k = + r x [ ( B T + ( u i + 1 / 2 L ) u i + 1 / 2 L , u ) ( B T + ( u i 1 / 2 L ) u i 1 / 2 L , u i 1 ) + ( B T ( u i + 1 / 2 R ) u i + 1 / 2 R , u i + 1 ) ( B T ( u i 1 / 2 R ) u i 1 / 2 R , u ) ]

+ Δ x Δ y Δ z i = 1 N 1 j = + k = + r y [ ( C T + ( u j + 1 / 2 L ) u j + 1 / 2 L , u ) ( C T + ( u j 1 / 2 L ) u j 1 / 2 L , u j 1 ) + ( C T ( u j + 1 / 2 R ) u j + 1 / 2 R , u j + 1 ) ( C T ( u j 1 / 2 R ) u j 1 / 2 R , u ) ] + Δ x Δ y Δ z i = 1 N 1 j = + k = + r z [ ( D T + ( u k + 1 / 2 L ) u k + 1 / 2 L , u ) ( D T + ( u k 1 / 2 L ) u k 1 / 2 L , u k 1 ) + ( D T ( u k + 1 / 2 R ) u k + 1 / 2 R , u k + 1 ) ( D T ( u k 1 / 2 R ) u k 1 / 2 R , u ) ] .

Then, using the following relations

1 ) Δ x Δ y Δ z i = 1 N 1 j = + k = + r x [ ( B T + ( u i + 1 / 2 L ) u i + 1 / 2 L , u ) ( B T + ( u i 1 / 2 L ) u i 1 / 2 L , u i 1 ) + ( B T ( u i + 1 / 2 R ) u i + 1 / 2 R , u i + 1 ) ( B T ( u i 1 / 2 R ) u i 1 / 2 R , u ) ] = Δ x Δ y Δ z j = + k = + r x [ ( B T + ( u N 1 / 2 L ) u N 1 / 2 L , u N 1 ) ( B T + ( u 1 / 2 L ) u 1 / 2 L , u 0 ) + ( B T ( u N 1 / 2 R ) u N 1 / 2 R , u N ) ( B T ( u 1 / 2 R ) u 1 / 2 R , u 1 ) ] = 0

2 ) Δ x Δ y Δ z i = 1 N 1 j = + k = + r y [ ( C T + ( u j + 1 / 2 L ) u j + 1 / 2 L , u ) ( C T + ( u j 1 / 2 L ) u j 1 / 2 L , u j 1 ) + ( C T ( u j + 1 / 2 R ) u j + 1 / 2 R , u j + 1 ) ( C T ( u j 1 / 2 R ) u j 1 / 2 R , u ) ] = 0 ;

3 ) Δ x Δ y Δ z i = 1 N 1 j = + k = + r z [ ( D T + ( u k + 1 / 2 L ) u k + 1 / 2 L , u ) ( D T + ( u k 1 / 2 L ) u k 1 / 2 L , u k 1 ) + ( D T ( u k + 1 / 2 R ) u k + 1 / 2 R , u k + 1 ) ( D T ( u k 1 / 2 R ) u k 1 / 2 R , u ) ] = 0 ;

we have

Δ x Δ y Δ z i = 1 N 1 j = + k = + ( u m + 1 , u m + 1 ) = Δ x Δ y Δ z i = 1 N 1 j = + k = + ( u , u ) , m = 0 , , M 1.

Hence

I m + 1 = I m , m = 0 , , M 1 , (14)

which means the stability of the difference scheme (10)-(12) in the energetic norm 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

f ( u ) = u 2 2 , f ( u ) = f + ( u ) + f ( u ) , d f + d u 0 , d f d u 0 , | u | | x | + 0 , u R , u i m + 1 = u i m Δ t Δ x [ f + ( u i + 1 / 2 L ) f + ( u i 1 / 2 L ) + f ( u i + 1 / 2 R ) f ( u i 1 / 2 R ) ]

Consider the following reconstruction:

u i + 1 / 2 L = u i m + 1 2 ψ ( R i m ) ( u i m u i 1 m ) , u i 1 / 2 R = u i m 1 2 ψ ( 1 R i m ) ( u i + 1 m u i m ) , R i m = u i + 1 m u i m u i m u i 1 m ,

where u i k 1 Δ x x i 1 / 2 x i + 1 / 2 u ( ξ , k Δ t ) d ξ .

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

u m + 1 u + 4 3 ( u m + 1 + u ) r × [ ( f + ( u i 1 / 2 L ) ( u i u i 1 ) ) + ( [ f + ( u i + 1 / 2 L ) f + ( u i 1 / 2 L ) ] u ) + ( f ( u i + 1 / 2 R ) ( u i + 1 u i ) ) + ( [ f ( u i + 1 / 2 R ) f ( u i 1 / 2 R ) ] u ) ] = 0 , r = Δ t / Δ x . (15)

We now prove that the difference scheme (15) admits the availability of difference analogue of the energy integral. Multiplying the system (15) by [ u m + 1 + u ] :

[ u m + 1 u ] [ u m + 1 + u ] + 4 3 r [ ( f + ( u i 1 / 2 L ) ( u u i 1 ) ) + ( [ f + ( u i + 1 / 2 L ) f + ( u i 1 / 2 L ) ] u ) + ( f ( u i + 1 / 2 R ) ( u i + 1 u ) ) + ( [ f ( u i + 1 / 2 R ) f ( u i 1 / 2 R ) ] u ) ] = 0.

Using formulas of difference differentiation, we obtain the following identity

[ ( f + ( u i 1 / 2 L ) ( u u i 1 ) ) + ( [ f + ( u i + 1 / 2 L ) f + ( u i 1 / 2 L ) ] u ) = f + ( u i + 1 / 2 L ) u f + ( u i 1 / 2 L ) u i 1 , ( f ( u i + 1 / 2 R ) ( u i + 1 u ) ) + ( [ f ( u i + 1 / 2 R ) f ( u i 1 / 2 R ) ] u ) = f ( u i + 1 / 2 R ) u i + 1 f ( u i 1 / 2 R ) u .

Taking into account all of these transformations, we have

( u m + 1 ) 2 ( u ) 2 + 4 3 r { f + ( u i + 1 / 2 L ) u f + ( u i 1 / 2 L ) u i 1 + f ( u i + 1 / 2 R ) u i + 1 f ( u i 1 / 2 R ) u } = 0. (16)

Multiply both sides of the (16) by Δ x and sum up over i from to + , and noting that the function u tends to zero at infinity and denoting the quantity Δ x i = ( u ) 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 , m = 1 , , M (17)

which means stability of the difference scheme in the norm I m .

Example 2. Consider the following Cauchy problem

u t + ( u 2 / 2 ) x = 0 , < x < , t > 0 , | u | | x | + 0 , u ( x , 0 ) = { 1 , x < 0 , 0 , x > 0. (18)

Rewrite the difference scheme (15) in the following form

u m + 1 = u + 2 3 ( u m + 1 + u ) r × [ ( f + ( u i 1 / 2 L ) ( u u i 1 ) ) + ( [ f + ( u i + 1 / 2 L ) f + ( u i 1 / 2 L ) ] u ) + ( f ( u i + 1 / 2 R ) ( u i + 1 u ) ) + ( [ f ( u i + 1 / 2 R ) f ( u i 1 / 2 R ) ] u ) ] . (19)

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

u ( s + 1 ) = u + 2 3 ( u ( s ) + u ) r × [ ( f + ( u i 1 / 2 L ) ( u i u i 1 ) ) + ( [ f + ( u i + 1 / 2 L ) f + ( u i 1 / 2 L ) ] u ) + ( f ( u i + 1 / 2 R ) ( u i + 1 u i ) ) + ( [ f ( u i + 1 / 2 R ) f ( u i 1 / 2 R ) ] u ) ] .

We have carried out a posteriori error analysis of the proposed scheme. Table 1 presents the exact and numerical solutions at points ( t m , x i ) 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.

References

[1] Vorozhtsov, E.V. (2016) Construction of Third-Order Schemes Using Lagrange-Burmann Expansions for the Numerical Integration of Inviscid Gas Equations. Computational Methods and Programming, 17, 21-43. (In Russian)
[2] Samarskii, A.A. and Popov, Y.P. (1980) Difference Schemes of Gas Dynamics Nauka, Moscow. (In Russian)
[3] Tolstykh, A.I. (1990) Compact Difference Schemes and Their Application to Aero-Hydrodynamic Problems. Nauka, Moscow. (In Russian)
[4] LeVeque, R.J. (1994) Numerical Methods for Conservation Laws. Birkh?user, Basel.
[5] Tolstykh, A.I. (1994) High Accuracy Non-Centered Compact Difference Schemes for Fluid Dynamics Applications. World Scientific Publishing, Singapore.
https://doi.org/10.1142/2269
[6] Pinchukov, V.I. and Shu, C.W. (2000) High Order Numerical Methods for the Problems of Aerodynamics. Doklady Akademii Nauk, Novosibirsk. (In Russian)
[7] Toro, E.F. (2000) Riemann Solvers and Numerical Methods for Fluid Dynamics: A Practical Introduction. Springer, Berlin.
[8] Volkov, K.N., Deryugin, Y.N., Emel’yanov, V.N., Kozelkov, A.S. and Teterina, I.V. (2014) Difference Schemes in Gas Dynamics on Unstructured Grids. Fizmatlit, Moscow. (In Russian)
[9] Boscheri, W., Balsara, D.S. and Dumbser, M. (2014) Lagrangian ADER-WENO Finite Volume Schemes on Unstructured Triangular Meshes Based on Genuinely Multidimensional HLL Riemann Solvers. Journal of Computational Physics, 267, 112-138.
https://doi.org/10.1016/j.jcp.2014.02.023
[10] Richtmyer, R. and Morton, K. (1972) Difference Methods for Initial Value Problems. Wiley, New York.
[11] Rozhdestvenskii, B.L. and Yanenko, N.N. (1983) Systems of Quasilinear Equations and Their Application to Gas Dynamics. Nauka, Moscow.
[12] Popov, I.V. and Fryazinov, I.V. (2014) Method of Adaptive Artificial Viscosity for Solving the Gas Dynamics Equations. Moscow. (In Russian).
[13] LeVeque, R.J. (2004) Finite-Volume Methods for Hyperbolic Problems. Cambridge University Press, Cambridge.
[14] Vorozhtsov, E.V. (2000) Exercises for the Theory of Difference Schemes. Novosibirsk State University Press, Novosibirsk. (In Russian)
[15] Lax, P.D. and Wendroff, B. (1960) Systems of Conservation Laws III. Communications on Pure and Applied Mathematics, 13, 217-237.
https://doi.org/10.1002/cpa.3160130205
[16] MacCormack, R. (2003) The Effect of Viscosity in Hypervelocity Impact Cratering. Journal of Spacecraft and Rockets, 40, 757-763.
[17] Rusanov, V.V. (1968) Difference Schemes of the Third-Order Accuracy for Continuous Computation of Discontinuous Solutions. Doklady Akademii Nauk SSS, 180, 1303-1305.
[18] Burstein, S.Z. and Mirin, A.A. (1970) Third Order Difference Methods for Hyperbolic Equations. Journal of Computational Physics, 5, 547-571.
https://doi.org/10.1016/0021-9991(70)90080-X
[19] Balakin, V.B. (1970) Methods of the Runge-Kutta Type for Gas Dynamics. USSR Computational Mathematics and Mathematical Physics, 10, 208-216.
https://doi.org/10.1016/0041-5553(70)90192-8
[20] Warming, R.F., Kutler, P. and Lomax, H. (1973) Second-and Third-Order Noncentered Difference Schemes for Nonlinear Hyperbolic Equations. AIAA Journal, 11, 189-196.
https://doi.org/10.2514/3.50449
[21] Harten, A. (1983) High Resolution Schemes for Hyperbolic Conservation Laws. Journal of Computational Physics, 49, 357-393.
https://doi.org/10.1016/0021-9991(83)90136-5
[22] Bona, C., Bona-Casas, C., and Terradas, J. (2009) Linear High-Resolution Schemes for Hyperbolic Conservation Laws: TVB Numerical Evidence. Journal of Computational Physics, 228, 2266-2281.
https://doi.org/10.1016/j.jcp.2008.12.010
[23] Liska, A. and Wendroff, B. (2003) Comparison of Several Difference Schemes on 1D and 2D Test Problems for the Euler Equations. SIAM Journal on Scientific Computing, 25, 995-1017.
https://doi.org/10.1137/S1064827502402120
[24] Roe, P.L. (1981) Approximate Riemann Solvers, Parameter Vectors and Difference Schemes. Journal of Computational Physics, 43, 357-372.
[25] Chakravarthy, S. and Osher, S. (1985) A New Class of High Resolution TVD Schemes for Hyperbolic Conservation Laws. 23rd Aerospace Sciences Meeting, Reno, 14-17 January 1985.
https://doi.org/10.2514/6.1985-363
[26] Godunov, S.K. (1979) Equations of Mathematical Physics. Nauka, Moscow, 372.
[27] Blokhin, A.M. and Aloev, R.D. (1993) Energy Integrals and Their Applications to the Study of the Stability of the Difference Schemes. Novosibirsk State University Press, Novosibirsk, 224.
[28] Aloev, R.D., Eshkuvatov, Z.K., Davlatov, S.O. and Nik Long, N.M.A. (2014) Sufficient Condition of Stability of Finite Element Method for Symmetric T-Hyperbolic Systems with Constant Coefficients. Computers and Mathematics with Applications, 68, 1194-1204.
https://doi.org/10.1016/j.camwa.2014.08.019
[29] Aloev, R.D., Blokhin, A.M. and Hudayberganov, M.U. (2014) One Class of Stable Difference Schemes for Hyperbolic System. American Journal of Numerical Analysis, 2, 85-89.
[30] Aloev, R.D., Davlatov, S.O., Eshkuvatov, Z.K. and Nik Long, N.M.A. (2016) Uniqueness Solution of the Finite Elements Scheme for Symmetric Hyperbolic Systems with Variable Coefficients. Malaysian Journal of Mathematical Sciences, 10, 49-60.

  
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