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In this paper, we investigate a numerical method for the generalized Novikov equation. We propose a conservative finite difference scheme and use Brouwer fixed point theorem to obtain the existence of the solution of the corresponding difference equation. We also prove the convergence and stability of the solution by using the discrete energy method. Moreover, we obtain the truncation error of the difference scheme which is .

The Camassa-Holm (CH) equation

m t + 2 u x m + u m x = 0 (1.1)

was derived by Camassa and Holm [

m t + 3 u x m + u m x = 0 (1.2)

It was regarded as a model for nonlinear shallow water dynamics and could be derived from the governing equation for water waves [

Notice that both the CH and DP equations had the quadratic nonlinearities and H 1 conservation laws. So it was of great interest to search for such type of equations with higher-order nonlinearities. To the best of our knowledge, two integrable equations with cubic nonlinearities had been proposed. One was a modified CH equation

m t + ( ( u 2 − u x 2 ) m ) x = 0 (1.3)

which was obtained by the tri-Hamiltonian duality approach from the bi-Hamiltonian structure of the mKdV equation [

m t + 3 u u x m + u 2 m x = 0 (1.4)

which was obtained by Novikov [

u t + u x x x + a 1 u 2 u x + a 2 u u x = 0

they deduced the following generalized CH equation with the cubic and quadratic nonlinearities [

m t + a 1 ( ( u 2 − u x 2 ) m ) x + a 2 ( 2 u x m + u m x ) = 0 (1.5)

This equation admitted Lax-pair and peaked solitons [

There are few studies on numerical solutions of the generalized Novikov equation. Therefore in this paper, we will construct a finite difference scheme for the initial boundary problem of equations as follows,

m t + a 1 ( 3 u u x m + u 2 m x ) + a 2 u x m + a 3 u m x = 0 , m = u − u x x , 0 < t ≤ T , 0 ≤ x ≤ L (1.6)

u ( 0 , t ) = u ( L , t ) , 0 < t ≤ T (1.7)

u ( 0 , x ) = u 0 ( x ) , 0 < x ≤ L (1.8)

where u ( x , t ) is a function of time t and a single spatial variable x , and a 1 , a 2 , a 3 are constants. It is clear that Equation (1.6) is reduced respectively to the Novikov Equation (1.4), the CH Equation (1.1) and the DP Equation (1.2), when a 1 = 1 , a 2 = 0 , a 3 = 0 ; a 1 = 0 , a 2 = 2 , a 3 = 1 ; a 1 = 0 , a 2 = 3 , a 3 = 1 . Clearly, Equation (1.6) is a linear combination of the Novikov equation, the CH equation, and the DP equation.

We shall give an energy conservative finite scheme for Equations (1.6)-(1.8) and obtain the module estimation. Moreover, we prove the convergence and stability of the finite scheme by use of discrete energy method.

For convenience, we denote Ω = { ( x , t ) | 0 ≤ x ≤ L , 0 < t ≤ T } in the following section. Let N , J be any positive integers and h = L J + 1 , τ = T N , x j = j h

for j = 0 , 1 , ⋯ , J + 1 . Denote t n = n τ , u j n = u ( x j , t n ) , u n = ( u 0 n , ⋯ , u J + 1 n ) for n = 0 , ⋯ , N and Z h 0 = { u = ( u 0 , ⋯ , u J + 1 ) | u 0 = u J + 1 = 0 } .

For simplicity, we introduce some notations as follows:

u x n = u j + 1 n − u j n h , u x ¯ n = u j n − u j − 1 n h , u x ^ n = u j + 1 n − u j − 1 n 2 h ,

u t n = u j n + 1 − u j n τ , u t ¯ n = u j n − u j n − 1 τ , u t ^ n = u j n + 1 − u j n − 1 2 τ ,

u ¯ = u j n + 1 + u j n − 1 2 , u n = u j n (2.1)

For u , v ∈ Z h 0 , we define a discrete inner product and the discrete L 2 -norm ‖ ⋅ ‖ h as

( u , v ) h = h ∑ j = 1 J u j v j , ‖ u ‖ h = ( h ∑ j = 1 J u j 2 ) 1 2 (2.2)

In order to obtain the module estimation, investigate the convergence and stability of the finite difference scheme, we need introduce two lemmas as follows,

Lemma 2.1. (Discrete Sobolev inequality [

‖ u n ‖ ∞ ≤ c 1 ‖ u n ‖ + c 2 ‖ u x n ‖ (2.3)

Lemma 2.2. (Discrete Gronwall inequality [

ω ( k ) ≤ ρ ( k ) + c τ ∑ l = 0 k − 1 w ( l ) (2.4)

then

ω ( k ) ≤ ρ ( k ) c c τ k (2.5)

The Equation (1.6) can be rewritten in the following form:

u t − u x x t + a 1 ( 3 u u x ( u − u x x ) + u 2 ( u x − u x x x ) ) + a 2 u x ( u − u x x ) + a 3 u ( u x − u x x x ) = 0

which can be expressed as

u t − u x x t + 4 a 1 u 2 u x + ( a 2 + a 3 ) u u x = 3 a 1 u u x u x x + a 1 u 2 u x x x + a 2 u x u x x + a 3 u u x x x

Firstly, we construct an energy conservative finite scheme for the problem (1.6)-(1.8) as follows:

u t n − u t x x ¯ n + a 1 ( ( ( u n + 1 2 ) 3 ) x ¯ + ( u n + 1 2 ) 2 u x n + 1 2 ) + a 2 + a 3 3 ( ( ( u n + 1 2 ) 2 ) x ¯ + u n + 1 2 u x n + 1 2 ) = a 1 ( ( ( u n + 1 2 ) 2 u x x n + 1 2 ) x ¯ + u n + 1 2 u x n + 1 2 u x x n + 1 2 ) + a 3 ( ( u n + 1 2 u x x n + 1 2 ) x ¯ + u x n + 1 2 u x x n + 1 2 ) + ( a 2 − 2 a 3 ) ( ( ( u x n + 1 2 ) 2 ) x ¯ − u x n + 1 2 u x x ¯ n + 1 2 ) (3.1)

u j 0 = u 0 ( x j ) , 0 ≤ j ≤ J + 1 (3.2)

and

u 0 n = u J + 1 n = 0 , 0 ≤ n ≤ N (3.3)

Lemma 3.1. The difference scheme satisfies discrete conservative law as follows:

E n = ‖ u n ‖ 2 + ‖ u x n ‖ 2 = E n − 1 = ⋯ = E 0 (3.4)

Proof. Computing the inner product of the difference scheme with 2 u n + 1 2 , we have

( u t , 2 u n + 1 2 ) = h ∑ j = 1 J u t ⋅ ( 2 u n + 1 2 ) = h τ ∑ j = 1 J ( u j n + 1 − u j n ) ( u j n + 1 + u j n ) = 1 τ ( ‖ u n + 1 ‖ 2 − ‖ u n ‖ 2 )

( u t x x ¯ , 2 u n + 1 2 ) = h ∑ j = 1 J u t x x ¯ ⋅ ( 2 u n + 1 2 ) = − h τ ∑ j = 1 J ( u j n + 1 − u j n ) x ( u j n + 1 + u j n ) x = − 1 τ ( ‖ u x n + 1 ‖ 2 − ‖ u x n ‖ 2 )

( a 1 ( ( ( u n + 1 2 ) 3 ) x ¯ + ( u n + 1 2 ) 2 u x n + 1 2 ) , 2 u n + 1 2 ) = 2 a 1 h ∑ j = 1 J ( ( ( u j n + 1 2 ) 3 ) x ¯ ⋅ u j n + 1 2 + ( u j n + 1 2 ) 2 u x j n + 1 2 ⋅ u j n + 1 2 ) = 2 a 1 h ∑ j = 1 J ( ( ( u j n + 1 2 ) 3 ) x ¯ ⋅ u j n + 1 2 − ( ( u j n + 1 2 ) 3 ) x ¯ u j n + 1 2 ) = 0

( a 2 + a 3 3 ( ( ( u n + 1 2 ) 2 ) x ¯ + u n + 1 2 u x n + 1 2 ) , 2 u n + 1 2 ) = 2 ( a 2 + a 3 ) 3 h ∑ j = 1 J ( ( ( u j n + 1 2 ) 2 ) x ¯ ⋅ u j n + 1 2 + u j n + 1 2 u x j n + 1 2 ⋅ u j n + 1 2 ) = 2 ( a 2 + a 3 ) 3 h ∑ j = 1 J ( ( ( u j n + 1 2 ) 2 ) x ¯ ⋅ u j n + 1 2 − ( ( u j n + 1 2 ) 2 ) x ¯ u j n + 1 2 ) = 0

( a 1 ( ( ( u n + 1 2 ) 2 u x x n + 1 2 ) x ¯ + u n + 1 2 u x n + 1 2 u x x n + 1 2 ) , 2 u n + 1 2 ) = 2 a 1 h ∑ j = 1 J ( ( ( u j n + 1 2 ) 2 u x x j n + 1 2 ) x ¯ ⋅ u j n + 1 2 + u j n + 1 2 u x j n + 1 2 u x x j n + 1 2 ⋅ u j n + 1 2 ) = 2 a 1 h ∑ j = 1 J ( ( ( u j n + 1 2 ) 2 u x x j n + 1 2 ) x ¯ ⋅ u j n + 1 2 − ( ( u j n + 1 2 ) 2 u x x j n + 1 2 ) x ¯ u j n + 1 2 ) = 0

( a 3 ( ( u n + 1 2 u x x n + 1 2 ) x ¯ + u x n + 1 2 u x x n + 1 2 ) , 2 u n + 1 2 ) = 2 a 3 h ∑ j = 1 J ( ( u j n + 1 2 u x x j n + 1 2 ) x ¯ ⋅ u j n + 1 2 + u x j n + 1 2 u x x j n + 1 2 ⋅ u j n + 1 2 ) = 2 a 3 h ∑ j = 1 J ( ( u j n + 1 2 u x x j n + 1 2 ) x ¯ ⋅ u j n + 1 2 − u j n + 1 2 ( u j n + 1 2 u x x j n + 1 2 ) x ¯ ) = 0

( ( a 2 − 2 a 3 ) ( ( ( u x n + 1 2 ) 2 ) x ¯ − u x n + 1 2 u x x ¯ n + 1 2 ) , 2 u n + 1 2 ) = 2 ( a 2 − 2 a 3 ) h ∑ j = 1 J ( ( ( u x j n + 1 2 ) 2 ) x ¯ ⋅ u j n + 1 2 − u x j n + 1 2 u x x ¯ j n + 1 2 ⋅ u j n + 1 2 ) = 2 ( a 2 − 2 a 3 ) h ∑ j = 1 J ( − ( u x j n + 1 2 ) 2 ⋅ u x j n + 1 2 + u x j n + 1 2 u x j n + 1 2 ⋅ u x j n + 1 2 ) = 0

From the above formulas, we can simplify formula as

1 τ ( ‖ u n + 1 ‖ 2 − ‖ u n ‖ 2 ) + 1 τ ( ‖ u x n + 1 ‖ 2 − ‖ u x n ‖ 2 ) = 0 (3.5)

then

‖ u n + 1 ‖ 2 + ‖ u x n + 1 ‖ 2 = ‖ u n ‖ 2 + ‖ u x n ‖ 2 (3.6)

So the difference scheme satisfies the discrete conservative law as follows

E n = ‖ u n ‖ 2 + ‖ u x n ‖ 2 = E n − 1 = ⋯ = E 0 (3.7)

This completes the proof of Lemma 3.1.

Theorem 3.2. The solution of the difference scheme (3.1)-(3.3) satisfies:

‖ u n ‖ ≤ C , ‖ u x n ‖ ≤ C , ‖ u n ‖ ∞ ≤ C (3.8)

Proof. From Lemma 3.1, we know ‖ u n ‖ ≤ C and ‖ u x n ‖ ≤ C . Then by Lemma 2.1, we obtain ‖ u n ‖ ∞ ≤ C . This completes the proof of Theorem 3.2.

Theorem 3.3. There exists u n ∈ Z h 0 satisfying the difference scheme (3.1)-(3.3).

Proof. We shall use Mathematical Induction to prove Theorem 3.3.

When n = 1 , it is known that there exists u 1 satisfying the difference scheme from initial condition. Next, we need to prove the case of n > 1 .

Assume that there exists u n satisfies the difference scheme when n < N , then we need to prove that there exists u n + 1 satisfies the difference scheme.

Define a operator ω ( v ) in Z h 0 as follows:

ω ( v ) = 2 v − 2 u n − ( 2 v x x ¯ − 2 u x x ¯ n ) + a 1 τ ( ( v 3 ) x ¯ + v 2 v x ) + a 2 + a 3 3 τ ( ( v 2 ) x ¯ + v v x ) − a 1 τ ( ( v 2 v x x ) x ¯ + v v x v x x ) − a 3 τ ( ( v v x x ) x ¯ + v x v x x ) − ( a 2 − 2 a 3 ) τ ( ( ( v x ) 2 ) x ¯ − v x v x x ¯ ) (3.9)

It is clear that ω is continuous.

Computing the inner product of the operator ω ( v ) , we obtain

( 2 v , v ) = 2 ‖ v ‖ 2

( 2 v x x ¯ , v ) = − 2 ( v x , v x ) = − 2 ‖ v x ‖ 2

( ( v 3 ) x ¯ + v 2 v x , v ) = h ∑ j = 1 J ( ( v 3 ) x ¯ ⋅ v + v 2 v x ⋅ v ) = 0

( ( v 2 ) x ¯ + v v x , v ) = h ∑ j = 1 J ( ( v 2 ) x ¯ ⋅ v + v v x ⋅ v ) = 0

( ( v 2 v x x ) x ¯ + v v x v x x , v ) = h ∑ j = 1 J ( ( v 2 v x x ) x ¯ ⋅ v + v v x v x x ⋅ v ) = 0

( ( v v x x ) x ¯ + v x v x x , v ) = h ∑ j = 1 J ( ( v v x x ) x ¯ ⋅ v + v x v x x ⋅ v ) = 0

( ( ( v x ) 2 ) x ¯ − v x v x x ¯ , v ) = h ∑ j = 1 J ( − ( v x ) 2 ⋅ v x + v x v x ⋅ v x ) = 0

By using Cauchy-Schwartz inequality, we obtain

( 2 u n , v ) ≤ 2 ‖ u n ‖ ⋅ ‖ v ‖ (3.10)

( 2 u x x ¯ n , v ) = − 2 ( u x n , v x ) ≥ − 2 ‖ u x n ‖ ⋅ ‖ v x ‖ (3.11)

From above discussions, we get

( ω ( v ) , v ) ≥ 2 ‖ v ‖ 2 − 2 ‖ u n ‖ ⋅ ‖ v ‖ + 2 ‖ v x ‖ 2 − 2 ‖ u x n ‖ ⋅ ‖ v x ‖ ≥ ‖ v ‖ 2 − ‖ u n ‖ 2 − ‖ u x n ‖ 2 (3.12)

For ∀ v ∈ Z h 0 , then

‖ v ‖ 2 = ‖ u n ‖ 2 + ‖ u x n ‖ 2 + 1 (3.13)

So we get ( ω ( v ) , v ) > 0 . By Brouwer fixed point theorem, there exists v ∗ ∈ H such that ω ( v ∗ ) = 0 and ‖ v ∗ ‖ ≤ ∂ .

Let u n + 1 = 2 v ∗ − u n , it is easy to verify that u n + 1 satisfies the difference scheme for the problem.

So we complete the proof of Theorem 3.3.

In order to investigate the convergence and stability of the difference scheme, we need to obtain the truncation error of the scheme.

Theorem 4.1. If the solution u ( x , t ) of equation is sufficiently regular, then the truncation error of the difference scheme is R j n = O ( τ 2 + h 2 ) .

Proof. Firstly, we can use the Taylor expansion of u j n + 1 , u j n , u j + 1 n + 1 , u j + 1 n , u j − 1 n + 1 , u j − 1 n at the point ( x j , t n + 1 2 ) . Secondly, from the above Taylor expansions we reorganize difference equation at point ( x j , t n + 1 2 ) , then

( x j , t n + 1 2 ) u t n = u j n + 1 − u j n τ = 1 τ ( ( u j n + 1 2 + τ 2 ( ∂ u ∂ t ) j n + 1 2 + τ 2 8 ( ∂ 2 u ∂ t 2 ) j n + 1 2 + τ 3 48 ( ∂ 3 u ∂ t 3 ) j n + 1 2 + ⋯ ) − ( u j n + 1 2 − τ 2 ( ∂ u ∂ t ) j n + 1 2 + τ 2 8 ( ∂ 2 u ∂ t 2 ) j n + 1 2 + ⋯ ) ) = ( ∂ u ∂ t ) j n + 1 2 + τ 3 24 ( ∂ 3 u ∂ t 3 ) j n + 1 2 + ⋯

u t x x ¯ n = ( u x x ¯ ) j n + 1 − ( u x x ¯ ) j n τ = ( ( u x ¯ ) j + 1 n + 1 − ( u x ¯ ) j n + 1 ) − ( ( u x ¯ ) j + 1 n − ( u x ¯ ) j n ) τ h = 1 h 2 ( ( ( ∂ u ∂ t ) j + 1 n + 1 2 − 2 ( ∂ u ∂ t ) j n + 1 2 + ( ∂ u ∂ t ) j − 1 n + 1 2 ) + ( τ 2 24 ( ∂ 3 u ∂ t 3 ) j + 1 n + 1 2 − τ 2 12 ( ∂ 3 u ∂ t 3 ) j n + 1 2 + τ 2 24 ( ∂ 3 u ∂ t 3 ) j − 1 n + 1 2 ) + ⋯ ) = ( ∂ 3 u ∂ t ∂ x 2 ) j n + 1 2 + ⋯

( ( u n + 1 2 ) 3 ) x ¯ = ( ( u n + 1 2 ) 3 ) j − ( ( u n + 1 2 ) 3 ) j − 1 h = 3 ( u n + 1 2 ) 2 ( ∂ u ∂ x ) j n + 1 2 − h 2 ( 3 ( u n + 1 2 ) 2 ( ∂ 2 u ∂ x 2 ) j n + 1 2 + 6 u n + 1 2 ( ∂ u ∂ x ) j n + 1 2 + ⋯ )

( u n + 1 2 ) 2 u x n + 1 2 = ( u n + 1 2 ) 2 ⋅ u j + 1 n + 1 2 − u j n + 1 2 h = ( u n + 1 2 ) 2 ⋅ ( ( ∂ u ∂ x ) j n + 1 2 + h 2 ( ∂ 2 u ∂ x 2 ) j n + 1 2 + h 2 6 ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯ )

( ( u n + 1 2 ) 2 ) x ¯ = ( ( u n + 1 2 ) 2 ) j − ( ( u n + 1 2 ) 2 ) j − 1 h = 2 u n + 1 2 ( ∂ u ∂ x ) j n + 1 2 − h 2 ( 2 u n + 1 2 ( ∂ 2 u ∂ x 2 ) j n + 1 2 + 2 ( ∂ u ∂ x ) j n + 1 2 + ⋯ )

u n + 1 2 u x n + 1 2 = u n + 1 2 ⋅ u j + 1 n + 1 2 − u j n + 1 2 h = u n + 1 2 ⋅ ( ( ∂ u ∂ x ) j n + 1 2 + h 2 ( ∂ 2 u ∂ x 2 ) j n + 1 2 + h 2 6 ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯ )

u n + 1 2 u x n + 1 2 u x x n + 1 2 = u n + 1 2 ⋅ u j + 1 n + 1 2 − u j n + 1 2 h ⋅ u j + 2 n + 1 2 − 2 u j + 1 n + 1 2 + u j n + 1 2 h 2 = u n + 1 2 ⋅ ( ( ∂ u ∂ x ) j n + 1 2 + h 2 ( ∂ 2 u ∂ x 2 ) j n + 1 2 + ⋯ ) ⋅ ( ( ∂ 2 u ∂ x 2 ) j n + 1 2 + h ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯ )

( ( u n + 1 2 ) 2 u x x n + 1 2 ) x ¯ = ( u j n + 1 2 ) 2 u x x j n + 1 2 − ( u j − 1 n + 1 2 ) 2 u x x ( j − 1 ) n + 1 2 h = 2 u n + 1 2 ⋅ ( ( ∂ u ∂ x ) j n + 1 2 ⋅ ( ∂ 2 u ∂ x 2 ) j n + 1 2 + ( ∂ 3 u ∂ x 3 ) j n + 1 2 + h ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯ )

u x x + 1 2 u x x x + 1 2 = u j + 1 x + 1 2 − u j x + 1 2 h ⋅ u j + 2 x + 1 2 − 2 u j + 1 x + 1 2 + u j x + 1 2 h 2 = ( ( ∂ u ∂ x ) j n + 1 2 + h 2 ( ∂ 2 u ∂ x 2 ) j n + 1 2 + ⋯ ) ⋅ ( ( ∂ 2 u ∂ x 2 ) j n + 1 2 + h ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯ )

( u n + 1 2 u x x n + 1 2 ) x ¯ = u j n + 1 2 u x x j n + 1 2 − u j − 1 n + 1 2 u x x ( j − 1 ) n + 1 2 h = ( ∂ u ∂ x ) j n + 1 2 ⋅ ( ∂ 2 u ∂ x 2 ) j n + 1 2 + ( ∂ 3 u ∂ x 3 ) j n + 1 2 + h ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯

( ( u x n + 1 2 ) 2 ) x ¯ = u x j n + 1 2 u x j n + 1 2 − u x ( j − 1 ) n + 1 2 u x ( j − 1 ) n + 1 2 h = ( ∂ u ∂ x ) j n + 1 2 ⋅ ( ∂ 2 u ∂ x 2 ) j n + 1 2 + ( ∂ 3 u ∂ x 3 ) j n + 1 2 + h ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯

u x n + 1 2 u x x ¯ n + 1 2 = u j + 1 n + 1 2 − u j n + 1 2 h ⋅ u j + 1 n + 1 2 − 2 u j n + 1 2 + u j − 1 n + 1 2 h 2 = ( ( ∂ u ∂ x ) j n + 1 2 + h 2 ( ∂ 2 u ∂ x 2 ) j n + 1 2 + ⋯ ) ⋅ ( ( ∂ 2 u ∂ x 2 ) j n + 1 2 + h ( ∂ 3 u ∂ x 3 ) j n + 1 2 + ⋯ )

From the above expansions, we obtain the linear part of Equation (3.1) u t n − u t x x ¯ n at the point ( x j , t n + 1 2 ) satisfying

u t n − u t x x ¯ n = ( ∂ u ∂ t − ∂ 3 u ∂ t ∂ x 2 ) ( x j , t n + 1 2 ) + O ( τ 2 + h 2 )

On the other hand, the nonlinear parts at the point ( x j , t n + 1 2 ) satisfying

a 1 ( ( ( u n + 1 2 ) 3 ) x ¯ + ( u n + 1 2 ) 2 u x n + 1 2 ) = 4 a 1 u 2 ( ∂ u ∂ x ) j n + 1 2 + O ( τ 2 + h 2 )

a 2 + a 3 3 ( ( ( u n + 1 2 ) 2 ) x ¯ + u n + 1 2 u x n + 1 2 ) = ( a 2 + a 3 ) u ( ∂ u ∂ x ) j n + 1 2 + O ( τ 2 + h 2 )

a 1 ( ( ( u n + 1 2 ) 2 u x x n + 1 2 ) x ¯ + u n + 1 2 u x n + 1 2 u x x n + 1 2 ) = a 1 ( 3 u ( ∂ u ∂ x ⋅ ∂ 2 u ∂ x 2 ) j n + 1 2 + 2 u ( ∂ 3 u ∂ x 3 ) j n + 1 2 ) + O ( τ 2 + h 2 )

a 3 ( ( u n + 1 2 u x x n + 1 2 ) x ¯ + u x n + 1 2 u x x n + 1 2 ) = a 3 ( 2 ( ∂ u ∂ x ⋅ ∂ 2 u ∂ x 2 ) j n + 1 2 + ( ∂ 3 u ∂ x 3 ) j n + 1 2 ) + O ( τ 2 + h 2 )

( a 2 − 2 a 3 ) ( ( ( u x n + 1 2 ) 2 ) x ¯ − u x n + 1 2 u x x ¯ n + 1 2 ) = ( a 2 − 2 a 3 ) ( 2 ( ∂ u ∂ x ⋅ ∂ 2 u ∂ x 2 ) j n + 1 2 + ( ∂ 3 u ∂ x 3 ) j n + 1 2 ) + O ( τ 2 + h 2 )

Then we can easily obtain that truncation error of scheme (3.1) is R j n = O ( τ 2 + h 2 ) .

Theorem 4.2. The solution of the difference scheme (3.1)-(3.3) approaches to the solution of the original differential Equations (1.6)-(1.8) in L ∞ , and the corresponding truncation error is O ( τ 2 + h 2 ) .

Proof. Assume that u j n is the solution of the difference scheme (3.1)-(3.3), U j n is the solution of the original differential Equations (1.6)-(1.8). We can easily have e j n = U j n − u j n .

For the difference scheme (3.1),

R n = U t n − U t x x ¯ n + a 1 ( ( ( U n + 1 2 ) 3 ) x ¯ + ( U n + 1 2 ) 2 U x n + 1 2 ) + a 2 + a 3 3 ( ( ( U n + 1 2 ) 2 ) x ¯ + U n + 1 2 U x n + 1 2 ) − a 1 ( ( ( U n + 1 2 ) 2 U x x n + 1 2 ) x ¯ + U n + 1 2 U x n + 1 2 U x x n + 1 2 ) − a 3 ( ( U n + 1 2 U x x n + 1 2 ) x ¯ + U x n + 1 2 U x x n + 1 2 ) − ( a 2 − 2 a 3 ) ( ( U x n + 1 2 ) 2 ) x ¯ − U x n + 1 2 U x x ¯ n + 1 2 (4.1)

then (4,1) minus (3.1), yields

R n = e t n − e t x x ¯ n + I 1 + I 2 + I 3 + I 4 + I 5 (4.2)

where

I 1 = a 1 ( ( ( U n + 1 2 ) 3 ) x ¯ + ( U n + 1 2 ) 2 U x n + 1 2 − ( ( u n + 1 2 ) 3 ) x ¯ − ( u n + 1 2 ) 2 u x n + 1 2 )

I 2 = a 2 + a 3 3 ( ( ( U n + 1 2 ) 2 ) x ¯ + U n + 1 2 U x n + 1 2 − ( ( u n + 1 2 ) 2 ) x ¯ − u n + 1 2 u x n + 1 2 )

I 3 = a 1 ( ( ( u n + 1 2 ) 2 u x x n + 1 2 ) x ¯ + u n + 1 2 u x n + 1 2 u x x n + 1 2 − ( ( U n + 1 2 ) 2 U x x n + 1 2 ) x ¯ − U n + 1 2 U x n + 1 2 U x x n + 1 2 )

I 4 = a 3 ( ( u n + 1 2 u x x n + 1 2 ) x ¯ + u x n + 1 2 u x x n + 1 2 − ( U n + 1 2 U x x n + 1 2 ) x ¯ − U x n + 1 2 U x x n + 1 2 )

I 5 = ( a 2 − 2 a 3 ) ( ( ( u x n + 1 2 ) 2 ) x ¯ − u x n + 1 2 u x x ¯ n + 1 2 − ( ( U x n + 1 2 ) 2 ) x ¯ + U x n + 1 2 U x x ¯ n + 1 2 )

Taking the inner product of (4.2) with 2 e n + 1 2 , we obtain

( R n , 2 e n + 1 2 ) = ( e t n , 2 e n + 1 2 ) − ( e t x x ¯ n , 2 e n + 1 2 ) + ( I 1 , 2 e n + 1 2 ) + ( I 2 , 2 e n + 1 2 ) + ( I 3 , 2 e n + 1 2 ) + ( I 4 , 2 e n + 1 2 ) + ( I 5 , 2 e n + 1 2 )

By computing ( e t n , 2 e n + 1 2 ) , ( e t x x ¯ n , 2 e n + 1 2 ) , ( I 1 , 2 e n + 1 2 ) , ( I 2 , 2 e n + 1 2 ) , ( I 3 , 2 e n + 1 2 ) , ( I 4 , 2 e n + 1 2 ) and ( I 5 , 2 e n + 1 2 ) respectively, and from Lemma 2.3 and Cauchy-Schwartz inequality, we have

( e t n , 2 e n + 1 2 ) = h ∑ j = 0 J e t j ⋅ 2 e j n + 1 2 = 1 τ ( ‖ e n + 1 ‖ 2 − ‖ e n ‖ 2 )

( e t x x ¯ n , 2 e n + 1 2 ) = h ∑ j = 0 J e t x x ¯ j ⋅ 2 e j n + 1 2 = 1 τ ( ‖ e x n + 1 ‖ 2 − ‖ e x n ‖ 2 )

( I 1 , 2 e n + 1 2 ) ≤ C ( ‖ e x n + 1 ‖ 2 + ‖ e x n ‖ 2 + ‖ e n + 1 ‖ 2 + ‖ e n ‖ 2 )

( I 2 , 2 e n + 1 2 ) ≤ C ( ‖ e x n + 1 ‖ 2 + ‖ e x n ‖ 2 + ‖ e n + 1 ‖ 2 + ‖ e n ‖ 2 )

( I 3 , 2 e n + 1 2 ) ≤ C ( ‖ e x n + 1 ‖ 2 + ‖ e x n ‖ 2 + ‖ e n + 1 ‖ 2 + ‖ e n ‖ 2 )

( I 4 , 2 e n + 1 2 ) ≤ C ( ‖ e x n + 1 ‖ 2 + ‖ e x n ‖ 2 + ‖ e n + 1 ‖ 2 + ‖ e n ‖ 2 )

( I 5 , 2 e n + 1 2 ) ≤ C ( ‖ e x n + 1 ‖ 2 + ‖ e x n ‖ 2 + ‖ e n + 1 ‖ 2 + ‖ e n ‖ 2 )

Therefore,

( R n , 2 e n + 1 2 ) = ( e t n , 2 e n + 1 2 ) − ( e t x x ¯ n , 2 e n + 1 2 ) + ( I 1 , 2 e n + 1 2 ) + ( I 2 , 2 e n + 1 2 ) + ( I 3 , 2 e n + 1 2 ) + ( I 4 , 2 e n + 1 2 ) + ( I 5 , 2 e n + 1 2 ) = 1 τ ( ‖ e n + 1 ‖ 2 − ‖ e n ‖ 2 ) − 1 τ ( ‖ e x n + 1 ‖ 2 − ‖ e x n ‖ 2 ) + ( I 1 , 2 e n + 1 2 ) + ( I 2 , 2 e n + 1 2 ) + ( I 3 , 2 e n + 1 2 ) + ( I 4 , 2 e n + 1 2 ) + ( I 5 , 2 e n + 1 2 )

Taking Schwartz inequality, it follows that

( R n , 2 e n + 1 2 ) ≤ 2 ‖ R n ‖ ⋅ ‖ e n + 1 2 ‖ ≤ ‖ R n ‖ 2 + 2 ( ‖ e n + 1 ‖ 2 + ‖ e n ‖ 2 )

From the above discussion, we have

‖ e n + 1 ‖ 2 − ‖ e n ‖ 2 + ‖ e x n + 1 ‖ 2 − ‖ e x n ‖ 2 ≤ C τ ‖ R n ‖ 2 + C τ ( ‖ e n + 1 ‖ 2 + ‖ e n ‖ 2 + ‖ e x n + 1 ‖ 2 + ‖ e x n ‖ 2 ) (4.3)

Let φ n = ‖ e n ‖ 2 + ‖ e x n ‖ 2 , then (4.3) becomes to

φ n + 1 − φ n ≤ C τ ( φ n + 1 + φ n ) + C τ ‖ R n ‖ 2

Hence we have

φ n − φ n − 1 ≤ C τ ( φ n + φ n − 1 ) + C τ ‖ R n − 1 ‖ 2

φ n − 1 − φ n − 2 ≤ C τ ( φ n − 1 + φ n − 2 ) + C τ ‖ R n − 2 ‖ 2

⋮

φ 1 − φ 0 ≤ C τ ( φ 1 + φ 0 ) + C τ ‖ R 0 ‖ 2

From the above inequalities, one has

φ n ≤ φ 0 + C τ ∑ i = 0 n − 1 φ i + C τ ∑ i = 0 n − 1 ‖ R i ‖ 2

where k ∑ i = 0 n − 1 ‖ R i ‖ 2 ≤ n τ max 0 ≤ i ≤ n − 1 ‖ R i ‖ 2 ≤ T O ( τ 2 + h 2 ) 2 .

Since φ 0 = 0 , then φ n ≤ C τ ∑ i = 0 n − 1 φ i + C T O ( τ 2 + h 2 ) 2 . From discrete Gronwall inequality, φ n ≤ O ( τ 2 + h 2 ) 2 , that is

‖ e x n ‖ 2 + ‖ e n ‖ 2 ≤ O ( τ 2 + h 2 ) 2

Then we have

‖ e n ‖ ≤ O ( τ 2 + h 2 ) , ‖ e x n ‖ ≤ O ( τ 2 + h 2 )

From Theorem 3.2, we assert

‖ e x n ‖ ∞ ≤ O ( τ 2 + h 2 )

We complete the proof of Theorem 4.2.

Theorem 4.3. The solution of the difference scheme (3.1)-(3.3) is stable in L ∞ .

Proof. Assume that u j n is the solution of the difference scheme (3.1)-(3.3), U j n is the solution of original differential Equations (1.6)-(1.8). We can easily obtain that e j n = U j n − u j n .

Then from Theorem 4.2, the following inequality holds true

‖ e n ‖ 2 ≤ C ‖ U 0 − u 0 ‖ 2

Thus, we complete the proof of Theorem 4.3.

In this paper, we give a difference scheme for the generalized Novikov equation. In Section 2, we give some preparation knowledge. In Section 3, we propose a conservative finite difference scheme for the generalized Novikov equation and use Brouwer fixed point theorem to obtain the existence of the solution for the corresponding difference equation. In Section 4, we prove the convergence and stability of the solution by using the discrete energy method.

This research is supported by the National Nature Science Foundation of China (Grant Nos. 11501253, 11571141 and 11571140), and the Nature Science Foundation of Jiangsu Province (Grant No. BK20140525), and the Advanced talent of Jiangsu University (Grant Nos. 14JDG070, 15JDG079).

Chen, W.X., Zhu, Q.Q. and Yang, P. (2017) On the Conservative Finite Difference Scheme for the Generalized Novikov Equation. Journal of Applied Mathematics and Physics, 5, 1776-1790. https://doi.org/10.4236/jamp.2017.59150