In order to see the energy dissipation property, it is more convenient to work with the following system representation

{ u t = v x , v t − v x x t = ∂ ∂ x ( δ G δ u ) − γ v , (5)

where δ G δ u is the Euler-Lagrange variational derivative [24] [25] defined by

δ G δ u = ∂ G ∂ u − ∂ ∂ x ( δ G δ u x ) ,   G = u 2 2 + u 3 3 .

Then, the improved Boussinesq equation with Stokes damping has the following mass conservation property.

Proposition 1. Let u be the solution of Equation (5). Then the total mass ∫ a b     u d x is independent of t. Namely,

d d t ∫ a b     u d x = 0. (6)

The conservation of mass can be easily proved with periodic boundary conditions.

Proposition 2. Let u be the solution of Equation (5). Then the total energy ∫ a b     G ˜ ( u , u t , v ) d x is independent of t. Namely,

d d t ∫ a b     G ˜ ( u , u t , v ) d x ≤ 0, (7)

where

G ˜ ( u , u t , v ) = v 2 2 + ( u t ) 2 2 + G ( u ) .

Proof. According to the chain rule, we have

d d t ∫ a b     G ˜ d x = ∫ a b ( v v t + u t u t t + δ G δ u u t ) d x

By the first equality of Equation (5), we can get

= ∫ a b ( v v t + v x v x t + δ G δ u v x ) d x

Using the integration by part formula, we have

= ∫ a b     v ( v t − v x x t − ∂ ∂ x ( δ G δ u ) ) d x

At last, according to the second equality of Equation (5), we can complete the proof

d d t ∫ a b     G ˜ d x = − γ ∫ a b     v 2 d x ≤ 0.

The region of interest [ a , b ] can be decomposed into a grid T h with nodes

a = x 0 < x 1 < x 2 < ⋯ < x N − 1 < x N = b . (8)

where x i − x i − 1 = h and h = ( b − a ) / N . Denote T h = { I i : I i = [ x i − 1 , x i ] , i = 1 , 2 , ⋯ , N } . Accordingly, we place a dual grid T h * with nodes

a = x 0 < x 1 / 2 < x 3 / 2 < ⋯ < x N − 3 / 2 < x N − 1 / 2 < x N = b , (9)

where x i − 1 / 2 = x i − h / 2 , i = 1 , 2 , ⋯ , N Denote I 0 ∗ = [ x 0 , x 1 / 2 ] , I i ∗ = [ x i − 1 / 2 , x i + 1 / 2 ] and I N ∗ = [ x N − 1 / 2 , x N ] .

Select the trial function space U h as the linear element space with respect to T h . The basis function of the node x i is

ϕ i ( x ) = { 1 − | x − x i | / h ,       x i − 1 ≤ x ≤ x i + 1 , 0 ,                                       elsewhere . (10)

Then numerical solution U for Equation (5) can be uniquely written as U = ∑ i = 1 N     U i ϕ i ( x ) , where U i = U ( x i , t ) . So in the element I i , we have

U = U i − 1 ( 1 − μ ) + U i μ , (11)

U ′ = ( U i − U i − 1 ) / h , (12)

where μ = ( x − x i − 1 ) / h .

The test function space V h corresponding to T h * is taken as the piecewise constant function space. The test function of the nodes x j is

ψ j ( x ) = { 1,     x j − 1 / 2 ≤ x ≤ x j + 1 / 2 , 0,           elsewhere , (13)

In addition, we take Δ t as the time step and the numerical solution can be denoted by U k n ≃ u ( k h , n Δ t ) .

Recall the definition G ˜ ( u , u t , v ) = v 2 2 + ( u t ) 2 2 + G ( u ) . According to the definition of Euler-Lagrange derivative, we have δ G ˜ δ u = δ G δ u , δ G ˜ δ u t = u t and δ G ˜ δ v = v . So the Equation (5) can be rewritten as

{ u t = ∂ ∂ x ( δ G ˜ δ v ) , ( 1 − ∂ x 2 ) v t = ∂ ∂ x ( δ G ˜ δ u ) − γ v , (14)

where ∂ x 2 = ∂ 2 ∂ x 2 .

Using any ψ ∈ V h and ψ ˜ ∈ V h to multiply both sides of Equation (14) and integrating on the interval [ a , b ] , we can obtain the following semi-discrete scheme

{ ( u t , ψ ) = ( ∂ ∂ x ( δ G ˜ δ v ) , ψ ) , ( ( 1 − ∂ x 2 ) v t , ψ ˜ ) = ( ∂ ∂ x ( δ G ˜ δ u ) − γ v , ψ ˜ ) , (15)

where ( ⋅ , ⋅ ) denotes the inner product of L 2 ( [ a , b ] ) .

Now we discretize the above semi-discrete scheme in time. To simplify the notation, we will use the difference operator δ n 〈 1 〉 u n = u n + 1 − u n − 1 2 Δ t . Additionally, we define the discrete version of Euler-Lagrange derivative δ G ˜ δ u = u 2 2 + u 3 3 , δ G ˜ δ u t = u t and δ G ˜ δ v = v by

δ G ˜ d δ ( u n + 1 , u n , u n − 1 ) = u n + u n ( u n + 1 + u n + u n − 1 ) 3 , (16)

δ G ˜ d δ ( δ n 〈 1 〉 u n ) = δ n 〈 1 〉 u n , (17)

δ G ˜ d δ ( v n ) = v n . (18)

At last, we can get the following energy dissipation finite volume element schemes

{ ( δ n 〈 1 〉 u n , ψ ) = ( ∂ ∂ x ( δ G ˜ d δ ( v n ) ) , ψ ) , ( ( 1 − ∂ x 2 ) δ n 〈 1 〉 v n , ψ ˜ ) = ( ∂ ∂ x ( δ G ˜ d δ ( u n + 1 , u n , u n − 1 ) ) − γ v n , ψ ˜ ) . (19)

Under the periodic boundary conditions, the Equation (19) has the following discrete mass conservation law and energy dissipation property.

Theorem 1. (Discrete mass conservation law). Under the discrete periodic boundary condition, then numerical solution by Equation (19) conserves mass

∫ a b     u n d x = c o n s t . (20)

Theorem 2. (Discrete energy dissipation property). Under the discrete periodic boundary condition, then numerical solution by Equation (19) conserves energy

1 Δ t ( ∫ a b     G ˜ d n + 1 d x − ∫ a b     G ˜ d n d x ) ≤ 0. (21)

Proof.

1 Δ t ∫ a b ( G ˜ d n + 1 − G ˜ d n ) d x = ( δ G ˜ d δ ( u n + 1 , u n , u n − 1 ) , δ n 〈 1 〉 u n ) + ( δ G ˜ d δ ( δ n 〈 1 〉 u n ) , δ n 〈 1 〉 u n + 1 − δ n 〈 1 〉 u n − 1 2 Δ t ) + ( δ G ˜ d δ ( v n ) , δ n 〈 1 〉 v n ) = ( δ G ˜ d δ ( u n + 1 , u n , u n − 1 ) , ∂ ∂ x ( δ G ˜ d δ ( v n ) ) ) + ( ∂ ∂ x ( δ G ˜ d δ ( v n ) ) , v x n + 1 − v x n − 1 2 Δ t ) + ( δ G ˜ d δ ( v n ) , δ n 〈 1 〉 v n ) = − ( ∂ ∂ x ( δ G ˜ d δ ( u n + 1 , u n , u n − 1 ) ) , δ G ˜ d δ ( v n ) ) + ( ( 1 − ∂ x 2 ) δ n 〈 1 〉 v n , δ G ˜ d δ ( v n ) ) = − ( γ v n , δ G ˜ d δ ( v n ) ) = − γ ∫ a b ( v n ) 2 d x ≤ 0

For comparison, we employ the standard finite volume element scheme constructed based on the

{ u t = v x , ( 1 − ∂ x 2 ) v t = ( u + u 2 ) x − γ v . (22)

Then we can obtain the following finite volume element schemes

{ ( δ n 〈 1 〉 u n , ψ ) = ( v x n , ψ ) , ( ( 1 − ∂ x 2 ) δ n 〈 1 〉 v n , ψ ˜ ) = ( ( u n + ( u n ) 2 ) x − γ v n , ψ ˜ ) . (23)

Theorem 3. (Discrete mass conservation law). Under the discrete periodic boundary condition, then numerical solution by Equation (23) conserves mass

∫ a b     u n d x = c o n s t   . (24)

The initial conditions are taken as

u ( x ,0 ) = α   sech 2 ( α 6 x − x 0 β ) , (25)

u t ( x , 0 ) = 2 α α 6   sech 2 ( α 6 x − x 0 β ) tanh ( α 6 x − x 0 β ) , (26)

where α is amplitude of the solitary wave which is initially centered at x = x 0 travelling with velocity β . In addition, we will take x 0 = 0 , α = 0.5 , γ = 0.04 and the region [ − 20,80 ] in the following.

As the exact solution for the considered examples is not available, we will compute maximum absolute error for the considered examples using the half mesh principle. The maximum absolute error is defined as follows,

E ( h , Δ t ) = max i | u h Δ t ( i , n ) − u h / 2 Δ t / 2 ( 2 i ,2 n ) | . (27)

The maximum absolute errors Equation (3) and solitary wave with the above parameters have been computed at t= 1. They are displayed in Table 1 for the new finite volume element schemes (NFVEM). Examination of the table shows that the error measures of the finite volume element scheme diminish approximately quadratically as the space step size and time step size are simultaneously halved. The error table illustrates that the new numerical schemes have second-order accuracy in space and time direction.

Furthermore, we compute the problem using standard finite volume element schemes (FVEM). The obtained maximum absolute errors are displayed in Table 2. From the table, we can find that the standard finite volume element schemes also have second-order accuracy in space and time direction. The maximum absolute errors computed by the two numerical schemes are consistent. In other words, the proposed new schemes do not reduce the accuracy of the finite volume method.

Now we consider the conservation properties of the new finite volume element schemes and standard finite volume element schemes. The initial conditions and parameters are the same as the above except γ .

Figure 1 shows the evolutions of discrete mass of the proposed new finite volume element schemes and standard finite volume element schemes with γ = 10 − 4 .