The BBM Equation was proposed by Benjamin, Bona, and Mahony when they studied the unidirectional propagation of nonlinear dispersive long waves in 1972, which is also called the regular long wave equation. The equation is an improvement of the KDV equation in describing the unidirectional propagation of small amplitude and long surface gravity waves in 1 + 1 dimensions. The BBM equation is widely used in linear optics and particle physics. Therefore, it is of great significance to study the numerical solution of the BBM equation. The study of this equation has attracted extensive attention from scholars at home and abroad. Khaled Omrani [1] presented the Galerkin finite element method for the BBM equation: the standard Galerkin method is used for space discretization, and the C-N scheme is used for time discretization. The convergence of the equation is proved. In paper [2], the existence and uniqueness of solutions of the BBM equation in one-dimensional case are discussed. In article [3], Feng Min fu uses the Crank-Nicolson difference method to discretize the equation. In [4]-[6], the finite difference method and the general method of the equation are studied. In paper [7], Tan Yanmei studied the fully discrete mixed finite element scheme on a triangular element, proved the existence and uniqueness of the finite element solution, and gave an error analysis.

The mixed finite element method was founded by Babuska and Brezzi in the 1970s [8][9]. In the early 1980s, the mixed finite element method was improved by Falk and Osborn [10][11]. So far, the mixed finite element method has been widely used to solve partial differential equations, such as bi-harmonic equations [12], parabolic equations [13], Stokes equations [14], and so on. However, the study of a fully discrete mixed finite method for the BBM equation on a rectangular element has not been reported.

On the basis of [8], the mixed finite element method is applied to two-dimensional nonlinear BBM equations on rectangular meshes. The mixed finite element schemes of semidiscretization and full discretization are established. The space is approximated by a continuous biquadratic finite element and a bilinear finite element, and the time is discretized by the backward Euler difference scheme. The existence and uniqueness of the finite element solution are proved, and the optimal error estimation is obtained. In the second part of this paper, the existence and uniqueness of a solution to the BBM equation are established. In the third part, the mixed finite element scheme for semi-discretization on a rectangular grid is established, the existence and uniqueness of the finite element solution are proved, and the error is analyzed. In the fourth part, the convergence of the backward Euler fully discrete mixed finite element scheme is discussed. In the fifth part, a numerical example was given to verify the correctness and effectiveness of this method.

Let Ω ⊂ R 2 be a smooth, bounded region. In this article, we will consider the following nonlinear BBM equation:

where Ω ⊂ R 2 is a bounded domain with smooth boundary ∂ Ω , 0 < T < ∞ and f ( u ) = − ( 1 2 u 2 + u ) . Let H m ( Ω ) and L 2 ( Ω ) = H 0 ( Ω ) denote the usual Sobolev space of real-valued functions defined on Ω . ( ⋅ , ⋅ ) denote the inner product in L 2 ( Ω ) , The norm of this space is the usual Sobolev H m -norm and it will be denoted by ‖ · ‖ m ,

and Semi-norm

Letting p → = ∇ u t , and using the Green’s formula, then we get the mixed variational form of problem (1): find ( u , p → ) : [ 0 , T ] → H 0 1 ( Ω ) × ( L 2 ( Ω ) ) 2 , such that

Theorem 2.1 Equation (2) has a unique solution ( u , p → ) ∈ H 0 1 ( Ω ) × ( L 2 ( Ω ) ) 2 , and satisfy

Proof Since there be a unique solution u in Equation (1), let ( u , p → ) = ( u , ∇ u t ) , be the solution of Equation (2), take q → = ∇ u , v = u , then

Because

where φ ′ ( u ) = f ( u ) , according to the boundary value condition u | ∂ Ω = 0 and φ ( 0 ) = 0 , then

we get

then

Integrating Equation (4) from 0 to t, we can get

Now prove uniqueness, let ( u * , p → * ) be another solution of Equation (2), according to Equation (5), we get

Substitute uinto (2), then subtracting (2), we get

Taking q → = ∇ ( u − u * ) , v = u t − u t * in (6), according to (3), then

Then

such that u = u * , let q → = p → − p → * , we have p → = p → * .

This completes the proof.

Let T 2 h is quasi-uniform rectangular partition of Ω , T 1 h is a sub-rectangle obtained by connecting the midpoints of the four sides of T 2 h , T 1 , T 2 , T 3 , T 4 , as shown in Figure 1:

Figure 1. Rectangular partition.

V i ( i = 1 , 2 ) is a bi i degree polynomial space defined on T i h , obviously, V i ⊂ H 1 ( Ω ) , Let X h = V 2 ∩ H 0 1 ( Ω ) , M → h = V 1 , the finite element space is defined as follows

where Q 2 ( K ) is a biquadratic polynomial space on a large element K , Q 1 ( K ′ ) is a bilinear polynomial space on a sub-element K ' .

Definition

Let I 2 u be a biquadratic Lagrange interpolation polynomial, which is defined on the nodes of T , I 1 u be a piecewise bilinear Lagrange interpolation polynomial, which is defined on four sub-rectangles T 1 , T 2 , T 3 , T 4 , and there hold

Lemma 3.1 [12] ∀ u ∈ C ( Ω ) , then

where ‖ w ‖ a 2 = ( ∇ w , ∇ w ) = | w | 1 and I i : C 0 → V i ( i = 1 , 2 ) is a be i degree Lagrange interpolation operator defined on T i h .

Lemma 3.2 [15] The discrete BB condition holds, that is, there is a constant β > 0 independent of h , such that

Proof From v h ∈ X h , then v h ∈ H 1 0 , and v h | K ∈ Q 2 ( K ) , according to Poincaré inequality, ∇ v h ∈ ( L 2 ( Ω ) ) 2 and ∇ v h | K ′ ∈ ( Q 1 ( K ′ ) ) 2 , we get

The proof is completed.

For all v ∈ H 1 ( Ω ) , q → = ( q 1 , q 2 ) ∈ ( L 2 ( Ω ) ) 2 , we define the interpolation operator Π h 2 and Π h 1 as follows:

According to the projection property and duality principle, we get Lemma 3.3 and Lemma3.4.

Lemma 3.3 [16] Existence of interpolation operator Π h 2 : H 1 ( Ω ) → X h , such that

Lemma 3.4 [17] Existence of an interpolation operator Π h 1 : ( L 2 ( Ω ) ) 2 → M → h , such that

Lemma 3.5 [18] (Gronwall Lemma) Suppose g ( t ) be integrable on [ 0 , t 1 ] and positive everywhere, c ≥ 0 is a constant. If ψ ( t ) ∈ C 0 ( [ 0 , t 1 ] ) , and there is the following inequality

then

Lemma 3.6 [19] Let ( a n ) , ( b n ) , ( c n ) be positive columns, ( c n ) monotonically increasing and satisfy

and

then

Now, we consider the semi-discrete form of (2), find ( u h , p → h ) : [ 0 , T ] → X h × M h , such that

Theorem3.1 [7] Equation (10) has a unique solution

Theorem 3.2The solution ( u , p → ) , ( u h , p → h ) of (2) and (10), respectively, suppose u , u t ∈ H 3 ( Ω ) , p → ∈ ( H 2 ( Ω ) ) 2 , then there holds

and

Proof Let

For all v h ∈ X h , q → h ∈ M h , the error equation is

Taking v h ∈ ξ , q → h ∈ ∇ ξ , and adding the two formulas of (13), we have

According to the Young inequality, the Cauchy-Schwarz inequality and the Lipschitz condition ‖ f ( u ) − f ( u h ) ‖ 0 ≤ c ‖ u − u h ‖ 0 , we get

Since ξ ( 0 ) = 0 , integrate (14) from 0 to t , and then according to Lemma 3.5 and (8), we get

Using the triangle inequality and the interpolation theory, we get

The proof of (11) is completed. Taking v h = ξ t , q → h = ∇ ξ t , and adding the two formulas of (13), we have

Using the same method as (15), then there hold

Taking q → h = θ → , then

According to the Young inequality, the Cauchy-Schwarz inequality, we get

Then, according to Lemma 3.3, Lemma 3.4, and (16), we get

Thus, using trigonometric inequality and Lemma 3.4, we obtain

The proof of (12) is completed.

For any given positive integer N , let Δ t = T N , denote the size of the time discretization and 0 = t 0 < t 1 < ⋯ < t N = T , Δ t = T N , t n = n Δ t , n = 0 , 1 , ⋯ , N .

For smooth function φ n = φ ( t n ) , let

Now we consider the backward Euler full discretization scheme of (2), find ( u h n , p → h n ) : [ 0 , T ] → X h × M → h ( n = 1 , 2 , ⋯ , N ) , such that

Theorem 4.1 The solution u n , p → n , ( u h n , p → h n ) of (2) and (17), respectively, suppose u , u t , u t t ∈ H 3 ( Ω ) , p → , p → t ∈ ( H 2 ( Ω ) ) 2 , then there holds

and

Proof Taking t = t n , v = v h , q → = q → h in (2), then

Subtracting (17) from (20), we have

For simplicity, we introduce the following notation:

So (21) becomes

Taking v h = ξ n , q → h n = ∇ ξ n in (22), we get

From the left of the above formula, which can be rewritten as

Using the Cauchy-Schwarz inequality and the Young Inequality, we get

And

further, using the Cauchy-Schwartz inequality, the Young Inequality, and the Lipschitz condition, we get

According to the Taylor formula, we get

substitute (24), (25), (26) in (23), then

Multiply both ends of the above formula by 4 Δ t , taking ε = 1 6 Δ t , and add from 0 to N , we get

From Lemma 3.3 and Lemma 3.6, we have

According to the regularity of u, we have

Taking v h = ∂ t ξ n , q → h = ∂ t ∇ ξ n in (22), using the Lipschitz condition, we get

Multiply both ends of the above formula by 2, taking ε = 1 2 , then according to Lemma 3.3 and the regularity of u , we get

Then, using Lemma3.3 and the triangle inequality, the proof of (18) is completed. Letting q → h = θ → n the second formula of (22), we get