Hongying Si, Weifeng Cui
School of Mathematics and Statistics, Shangqiu Normal University, Shangqiu, China.
DOI: 10.4236/jamp.2026.148157   PDF    HTML   XML   6 Downloads   46 Views   Citations

Abstract

In this paper, 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. Finally, the numerical results were given, and the stability and convergence of the method were verified.

Keywords

BBM Equation, The Mixed Finite Element, Full Discretization, Error Estimate

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1. Introduction

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.

2. Existence and Uniqueness of Generalized Solutions of the BBM Equation

Let $\Omega \subset R^2$ be a smooth, bounded region. In this article, we will consider the following nonlinear BBM equation:

$\{\begin{array}{ll}{u}_t-\Delta u_t=\nabla \cdot f\left(u\right),\hfill & \left(x,t\right)\in \Omega \times \left[0,T\right],\hfill \\ u\left(x,t\right)=0,\hfill & \left(x,t\right)\in \partial \Omega \times \left[0,T\right],\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & x\in \overline{\Omega }.\hfill \end{array}$ (1)

where $\Omega \subset R^2$ is a bounded domain with smooth boundary $\partial \Omega$ , $0<T<\infty$ and $f\left(u\right)=-\left(\frac{1}{2}{u}^2+u\right)$ . Let $H^m\left(\Omega \right)$ and $L^2\left(\Omega \right)=H^0\left(\Omega \right)$ denote the usual Sobolev space of real-valued functions defined on $\Omega$ . $\left(\cdot ,\cdot \right)$ denote the inner product in $L^2\left(\Omega \right)$ , The norm of this space is the usual Sobolev $H^m$ -norm and it will be denoted by ${\Vert ·\Vert }_m$ ,

${\Vert v\Vert }_m={\left[{\displaystyle \sum _{\left|\alpha \right|\le m}{\displaystyle {\int }_{\Omega }{\left|{\partial }^{\alpha }v\left(x\right)\right|}^2\text{d}x}}\right]}^{\frac{1}{2}},$

and Semi-norm

${\Vert v\Vert }_m={\left[{\displaystyle \sum _{\left|\alpha \right|=m}{\displaystyle {\int }_{\Omega }{\left|{\partial }^{\alpha }v\left(x\right)\right|}^2\text{d}x}}\right]}^{\frac{1}{2}}.$

Letting $\overrightarrow{p}=\nabla u_t$ , and using the Green’s formula, then we get the mixed variational form of problem (1): find $\left(u,\overrightarrow{p}\right):\left[0,T\right]\to H_0^1\left(\Omega \right)\times {\left(L^2\left(\Omega \right)\right)}^2$ , such that

$\{\begin{array}{ll}\left(u_t,v\right)+\left(\overrightarrow{p},\nabla v\right)=\left(\nabla \cdot f\left(u\right),v\right),\hfill & \forall v\in H_0^1\left(\Omega \right),\hfill \\ \left(\nabla u_t,\overrightarrow{q}\right)-\left(\overrightarrow{p},\overrightarrow{q}\right)=0,\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\hfill & \forall \overrightarrow{q}\in {\left(L^2\left(\Omega \right)\right)}^2,\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & x\in \overline{\Omega }.\hfill \end{array}$(2)

Theorem 2.1 Equation (2) has a unique solution $\left(u,\overrightarrow{p}\right)\in H_0^1\left(\Omega \right)\times {\left(L^2\left(\Omega \right)\right)}^2$ , and satisfy

${\Vert u\left(t\right)\Vert }_1={\Vert u\left(0\right)\Vert }_1,\forall t>0.$

Proof Since there be a unique solution $u$ in Equation (1), let $\left(u,\overrightarrow{p}\right)=\left(u,\nabla u_t\right)$ , be the solution of Equation (2), take $\overrightarrow{q}=\nabla u,v=u$ , then

$\left(u_t,u\right)+\left(\nabla u_t,\nabla u\right)=\left(\nabla \cdot f\left(u\right),u\right).$

Because

$\left(\nabla \cdot f\left(u\right),u\right)=\nabla \left(f\left(u\right),u\right)-\nabla \left(\phi \left(u\right)\right),$

where ${\phi }^{\prime }\left(u\right)=f\left(u\right)$ , according to the boundary value condition ${u|}_{\partial \Omega }=0$ and $\phi \left(0\right)=0$ , then

$\left(\nabla \cdot f\left(u\right),u\right)={\displaystyle {\int }_{\Omega }\nabla \cdot \left(f\left(u\right)u\right)\text{d}x}=0.$ (3)

we get

$\frac{\text{d}}{\text{d}t}{\Vert u\Vert }_0^2+\frac{\text{d}}{\text{d}t}{\Vert \nabla u\Vert }_0^2=0,$

then

$\frac{\text{d}}{\text{d}t}{\Vert u\Vert }_1^2=0.$ (4)

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

${\Vert u\left(t\right)\Vert }_1={\Vert u\left(0\right)\Vert }_1,\forall t>0.$ (5)

Now prove uniqueness, let $\left(u^{*},{\overrightarrow{p}}^{*}\right)$ be another solution of Equation (2), according to Equation (5), we get

${\Vert u^{*}\left(t\right)\Vert }_1={\Vert u^{*}\left(0\right)\Vert }_1,\forall t>0.$

Substitute u into (2), then subtracting (2), we get

$\{\begin{array}{ll}\left(u_t-u_t^{*},v\right)+\left(\overrightarrow{p}-{\overrightarrow{p}}^{*},\nabla v\right)=\left(\nabla \cdot \left(f\left(u\right)-f\left(u^{*}\right)\right),v\right),\hfill & \forall v\in H_0^1\left(\Omega \right),\hfill \\ \left(\nabla \left(u_t-u_t^{*}\right),\overrightarrow{q}\right)-\left(\overrightarrow{p}-{\overrightarrow{p}}^{*},\overrightarrow{q}\right)=0,\text{}\hfill & \forall \overrightarrow{q}\in {\left(L^2\left(\Omega \right)\right)}^2.\text{}\text{}\text{}\text{}\text{}\text{}\text{}\hfill \end{array}$(6)

Taking $\overrightarrow{q}=\nabla \left(u-u^{*}\right),v=u_t-u_t^{*}$ in (6), according to (3), then

$\frac{\text{d}}{\text{d}t}{\Vert u-u^{*}\Vert }_0^2+\frac{\text{d}}{\text{d}t}{\Vert \nabla \left(u-u^{*}\right)\Vert }_0^2=0,$

Then

${\Vert u-u^{*}\Vert }_1^2={\Vert u\left(0\right)-u^{*}\left(0\right)\Vert }_1^2={\Vert u_0\left(x\right)-u_0\left(x\right)\Vert }_1^2=0,$

such that $u=u^{*}$ , let $\overrightarrow{q}=\overrightarrow{p}-{\overrightarrow{p}}^{*}$ , we have $\overrightarrow{p}={\overrightarrow{p}}^{*}$ .

This completes the proof.

3. Existence and Error Analysis of Semi-Discrete Mixed Finite Element Solutions

Let $T_{2h}$ is quasi-uniform rectangular partition of $\Omega$ , $T_{1h}$ is a sub-rectangle obtained by connecting the midpoints of the four sides of $T_{2h},T_1,T_2,T_3,T_4$ , as shown in Figure 1:

Figure 1. Rectangular partition.

$V_i\left(i=1,2\right)$ is a bi $i$ degree polynomial space defined on $T_{ih}$ , obviously, $V_i\subset H^1\left(\Omega \right)$ , Let $X_h=V_2\cap H_0^1\left(\Omega \right)$ , ${\overrightarrow{M}}_h=V_1$ , the finite element space is defined as follows

$X_h=\left\{v_h\in H_0^1\left(\Omega \right)\cap C^0\left(\overline{\Omega }\right):{v_h|}_K\in Q_2\left(K\right),K\in T_{2h}\right\},$

${\overrightarrow{M}}_h=\left\{\overrightarrow{q}=\left(q_1,q_2\right)\in {\left(L^2\left(\Omega \right)\right)}^2:{\overrightarrow{q}|}_{K^{\prime }}\in {\left(Q_1\left(K^{\prime }\right)\right)}^2,K^{\prime }\in T_{1h}\right\},$

where $Q_2\left(K\right)$ is a biquadratic polynomial space on a large element $K$ , $Q_1\left(K^{\prime }\right)$ is a bilinear polynomial space on a sub-element $K^{\text{'}}.$

Definition

${\Vert u\Vert }_K=\sqrt{{\displaystyle {\int }_K\left({\left(\frac{\partial u}{\partial x}\right)}^2+{\left(\frac{\partial u}{\partial y}\right)}^2\right)\text{d}x\text{d}y}},$

$u\left(a_i\right)=u_i.$

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] $\forall u\in C\left(\Omega \right)$ , then

${\Vert I_{2}u-I_{1}u\Vert }_a\le \sqrt{\frac{5}{6}}{\Vert I_{1}u\Vert }_a,$

where ${\Vert w\Vert }_{{}_a}^2=\left(\nabla w,\nabla w\right)={\left|w\right|}_1$ and $I_i:C^0\to V_i\left(i=1,2\right)$ is a be $i$ degree Lagrange interpolation operator defined on $T_{ih}$ .

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

$\underset{{\overrightarrow{q}}_h\in {\overrightarrow{M}}_h}{\sup }\frac{\left({\overrightarrow{q}}_h,\nabla v_h\right)}{{\Vert {\overrightarrow{q}}_h\Vert }_0}\ge \beta {\Vert v_h\Vert }_1,\forall v_h\in X_h.$ (7)

Proof From $v_h\in X_h$ , then $v_h\in H_1^0$ , and ${v_h|}_K\in Q_2\left(K\right)$ , according to Poincaré inequality, $\nabla v_h\in {\left(L^2\left(\Omega \right)\right)}^2$ and ${\nabla v_h|}_{K^{\prime }}\in {\left(Q_1\left(K^{\prime }\right)\right)}^2$ , we get

$\underset{{\overrightarrow{q}}_h\in {\overrightarrow{M}}_h}{\sup }\frac{\left({\overrightarrow{q}}_h,\nabla v_h\right)}{{\Vert {\overrightarrow{q}}_h\Vert }_0}\ge \frac{\left(\nabla v_h,\nabla v_h\right)}{{\Vert \nabla v_h\Vert }_0}={\left|v_h\right|}_1\ge \beta {\Vert v_h\Vert }_1,\forall v_h\in X_h.$

The proof is completed.

For all $v\in H^1\left(\Omega \right),\overrightarrow{q}=\left(q_1,q_2\right)\in {\left(L^2\left(\Omega \right)\right)}^2$ , we define the interpolation operator ${\Pi }_h^2$ and ${\Pi }_h^1$ as follows:

${\Pi }_h^2:H^1\left(\Omega \right)\to X_h,{\Pi }_h^1:{\left(L^1\left(\Omega \right)\right)}^2\to {\overrightarrow{M}}_h.$

According to the projection property and duality principle, we get Lemma 3.3 and Lemma 3.4.

Lemma 3.3 [16] Existence of interpolation operator ${\Pi }_h^2:H^1\left(\Omega \right)\to X_h$ , such that

$\left(v-{\Pi }_h^2\left(v_h\right),v_h\right)=0,\forall v_h\in X_h,$

${\Vert {\Pi }_h^2\left(v\right)\Vert }_0\le {\Vert v\Vert }_0,\forall v\in H^1\left(\Omega \right),$

When $\forall v\in H^r\left(\Omega \right)$ , then

${\Vert v-{\Pi }_h^2\left(v\right)\Vert }_s\le C{h}^{r-s}{\left|v\right|}_r,s=0,1,s\le r\le m+1.$ (8)

Lemma 3.4 [17] Existence of an interpolation operator ${\Pi }_h^1:{\left(L^2\left(\Omega \right)\right)}^2\to {\overrightarrow{M}}_h$ , such that

$\left(p-{\Pi }_h^1\left(\overrightarrow{p}\right),{\overrightarrow{q}}_h\right)=0,\forall {\overrightarrow{q}}_h\in {\overrightarrow{M}}_h,$

${\Vert {\Pi }_h^1\left(\overrightarrow{p}\right)\Vert }_0\le {\Vert \overrightarrow{p}\Vert }_0,\forall \overrightarrow{p}\in {\left(L^2\left(\Omega \right)\right)}^2,$

When $\overrightarrow{p}\in {\overrightarrow{M}}_h$ , then

${\Vert {\Pi }_h^1\left(\overrightarrow{p}\right)\Vert }_s\le C{h}^{r-s}{\left|\overrightarrow{p}\right|}_r,s=0,1,s\le r\le m.$ (9)

Lemma 3.5 [18] (Gronwall Lemma) Suppose $g\left(t\right)$ be integrable on $\left[0,t_1\right]$ and positive everywhere, $c\ge 0$ is a constant. If $\psi \left(t\right)\in C^0\left(\left[0,t_1\right]\right)$ , and there is the following inequality

$0\le \psi \left(t\right)\le c+{\displaystyle {\int }_0^{t}g\left(s\right)\psi \left(s\right)\text{d}s},\forall t\in \left[0,t_1\right],$

then

$0\le \psi \left(t\right)\le c\cdot \exp \left({\displaystyle {\int }_0^{t}g\left(s\right)\text{d}s}\right),\forall t\in \left[0,t_1\right].$

Lemma 3.6 [19] Let $\left(a_n\right),\left(b_n\right),\left(c_n\right)$ be positive columns, $\left(c_n\right)$ monotonically increasing and satisfy

$a_n+b_n\le c_n+\lambda {\displaystyle \sum a_j},n\ge 1,\lambda \ge 0,$

and

$a_0+b_0\le c_0,$

then

$a_n+b_n\le c_n\cdot \exp \left(n\lambda \right),n\ge 0.$

Now, we consider the semi-discrete form of (2), find $\left(u_h,{\overrightarrow{p}}_h\right):\left[0,T\right]\to X_h\times M_h$ , such that

$\{\begin{array}{ll}\left(u_{ht},v_h\right)+\left({\overrightarrow{p}}_h,\nabla v_h\right)=\left(\nabla \cdot f\left(u_h\right),v_h\right)\hfill & \forall v_h\in X_h,\hfill \\ \left(\nabla u_{ht},{\overrightarrow{q}}_h\right)-\left({\overrightarrow{p}}_h,{\overrightarrow{q}}_h\right)=0,\text{}\hfill & \forall {\overrightarrow{q}}_h\in M_h,\text{}\text{}\text{}\text{}\text{}\text{}\hfill \\ u\left(x,0\right)={\Pi }_h^2{u}_0\left(x\right),\hfill & x\in \overline{\Omega }.\hfill \end{array}$(10)

Theorem 3.1 [7] Equation (10) has a unique solution

$\left(u_h,{\overrightarrow{p}}_h\right):\left[0,T\right]\to X_h\times M_h.$

Theorem 3.2 The solution $\left(u,\overrightarrow{p}\right),\left(u_h,{\overrightarrow{p}}_h\right)$ of (2) and (10), respectively, suppose $u,u_t\in H^3\left(\Omega \right)$ , $\overrightarrow{p}\in {\left(H^2\left(\Omega \right)\right)}^2$ , then there holds

${\Vert u-u_h\Vert }_1\le c{h}^2{\left[{\Vert u\Vert }_3^2+{\displaystyle {\int }_0^t\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }\right]}^{\frac{1}{2}},$ (11)

and

${\Vert \overrightarrow{p}-{\overrightarrow{p}}_h\Vert }_0\le c{h}^2{\left[\left({\Vert \overrightarrow{p}\Vert }_2^2+{\Vert u\Vert }_3^2\right)+{\displaystyle {\int }_0^t\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }\right]}^{\frac{1}{2}}.$(12)

Proof Let

$u-u_h=\left(u-{\Pi }_h^{2}u\right)+\left({\Pi }_h^{2}u-u_h\right)=\eta +\xi ,$

$\overrightarrow{p}-{\overrightarrow{p}}_h=\left(\overrightarrow{p}-{\Pi }_h^1\overrightarrow{p}\right)+\left({\Pi }_h^1\overrightarrow{p}-{\overrightarrow{p}}_h\right)=\overrightarrow{\rho }+\overrightarrow{\theta }.$

For all $v_h\in X_h,{\overrightarrow{q}}_h\in M_h$ , the error equation is

$\{\begin{array}{l}\left({\xi }_t,v_h\right)+\left(\overrightarrow{\theta },\nabla v_h\right)=-\left({\eta }_t,v_h\right)-\left(\overrightarrow{\rho },\nabla v_h\right)+\left(\nabla \cdot \left(f\left(u\right)-f\left(u_h\right)\right),v_h\right),\\ \left(\nabla {\xi }_t,{\overrightarrow{q}}_h\right)-\left(\overrightarrow{\theta },{\overrightarrow{q}}_h\right)=\left(\overrightarrow{\rho },{\overrightarrow{q}}_h\right)-\left(\nabla {\eta }_t,{\overrightarrow{q}}_h\right).\end{array}$ (13)

Taking $v_h\in \xi ,{\overrightarrow{q}}_h\in \nabla \xi$ , and adding the two formulas of (13), we have

$\left({\xi }_t,\xi \right)+\left(\nabla {\xi }_t,\nabla \xi \right)=-\left({\eta }_t,\xi \right)-\left(\nabla {\eta }_t,\nabla \xi \right)+\left(\left(f\left(u\right)-f\left(u_h\right)\right),\nabla \xi \right)$

According to the Young inequality, the Cauchy-Schwarz inequality and the Lipschitz condition ${\Vert f\left(u\right)-f\left(u_h\right)\Vert }_0\le c{\Vert u-u_h\Vert }_0$ , we get

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\Vert \xi \Vert }_0^2+\frac{\text{d}}{\text{d}t}{\Vert \nabla \xi \Vert }_0^2\le c\left[{\Vert {\eta }_t\Vert }_0^2+{\Vert \nabla {\eta }_t\Vert }_0^2+\left({\Vert \xi \Vert }_0^2+{\Vert \eta \Vert }_0^2\right)\right]\\ +\epsilon {\Vert \xi \Vert }_0^2+2\epsilon {\Vert \nabla \xi \Vert }_0^2.\end{array}$ (14)

Since $\xi \left(0\right)=0$ , integrate (14) from 0 to $t$ , and then according to Lemma 3.5 and (8), we get

${\Vert \xi \Vert }_0^2+{\Vert \nabla \xi \Vert }_0^2\le c{h}^4{\displaystyle {\int }_0^t\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }.$ (15)

Using the triangle inequality and the interpolation theory, we get

${\Vert u-u_h\Vert }_1^2\le {\Vert \eta \Vert }_1^2+{\Vert \xi \Vert }_1^2\le c{h}^4\left[{\Vert u\Vert }_3^2+{\displaystyle {\int }_0^t\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }\right].$

The proof of (11) is completed. Taking $v_h={\xi }_t,{\overrightarrow{q}}_h=\nabla {\xi }_t$ , and adding the two formulas of (13), we have

$\left({\xi }_t,{\xi }_t\right)+\left(\nabla {\xi }_t,\nabla {\xi }_t\right)=-\left({\eta }_t,{\xi }_t\right)-\left(\nabla {\eta }_t,\nabla {\xi }_t\right)+\left(\left(f\left(u\right)-f\left(u_h\right)\right),\nabla {\xi }_t\right).$

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

${\Vert {\xi }_t\Vert }_0^2+{\Vert \nabla {\xi }_t\Vert }_0^2\le c{h}^4{\displaystyle {\int }_0^t\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }.$ (16)

Taking ${\overrightarrow{q}}_h=\overrightarrow{\theta }$ , then

${\Vert \overrightarrow{\theta }\Vert }_0^2=\left(\overrightarrow{\theta },\overrightarrow{\theta }\right)=\left(\nabla {\xi }_t,\overrightarrow{\theta }\right)+\left(\nabla {\eta }_t,\overrightarrow{\theta }\right)-\left(\overrightarrow{\rho },\overrightarrow{\theta }\right).$

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

${\Vert \overrightarrow{\theta }\Vert }_0^2\le c\left({\Vert \nabla {\xi }_t\Vert }_0^2+{\Vert \overrightarrow{\rho }\Vert }_0^2+{\Vert \nabla {\eta }_t\Vert }_0^2\right)+\epsilon {\Vert \overrightarrow{\theta }\Vert }_0^2.$

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

${\Vert \overrightarrow{\theta }\Vert }_0^2\le c{h}^4{\displaystyle {\int }_0^t\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }+c{h}^4\left({\Vert \overrightarrow{p}\Vert }_2^2+{\Vert u_t\Vert }_3^2\right).$

Thus, using trigonometric inequality and Lemma 3.4, we obtain

${\Vert \overrightarrow{p}-{\overrightarrow{p}}_h\Vert }_0^2\le {\Vert \overrightarrow{\rho }\Vert }_0^2+{\Vert \overrightarrow{\theta }\Vert }_0^2\le c{h}^4\left[\left({\Vert \overrightarrow{p}\Vert }_2^2+{\Vert u_t\Vert }_3^2\right)+{\displaystyle {\int }_0^t\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }\right].$

The proof of (12) is completed.

4. Backward Euler, Fully Discrete Mixed Finite Element Scheme

For any given positive integer $N$ , let $\Delta t=\frac{T}{N}$ , denote the size of the time discretization and $0=t_0<t_1<\cdots <t_N=T,\Delta t=\frac{T}{N},t_n=n\Delta t,n=0,1,\cdots ,N$ .

For smooth function ${\phi }^n=\phi \left(t_n\right)$ , let

${\partial }_t{\phi }^n=\frac{1}{\Delta t}\left({\phi }^n-{\phi }^{n-1}\right),{\partial }_t\nabla {\phi }^n=\frac{1}{\Delta t}\left(\nabla {\phi }^n-\nabla {\phi }^{n-1}\right).$

Now we consider the backward Euler full discretization scheme of (2), find $\left(u_h^n,{\overrightarrow{p}}_h^n\right):\left[0,T\right]\to X_h\times {\overrightarrow{M}}_h\left(n=1,2,\cdots ,N\right)$ , such that

$\{\begin{array}{ll}\left({\partial }_t{u}_h^n,v_h\right)+\left({\overrightarrow{p}}_h^n,\nabla v_h\right)=\left(\nabla \cdot f\left(u_h^n\right),v_h\right),\hfill & \forall v_h\in X_h,\hfill \\ \left({\partial }_t{u}_h^n,{\overrightarrow{q}}_h\right)-\left({\overrightarrow{p}}_h^n,{\overrightarrow{q}}_h\right)=0,\text{}\hfill & \forall {\overrightarrow{q}}_h\in M_h,\hfill \\ u\left(x,0\right)={\Pi }_h^2{u}_0\left(x\right),\hfill & x\in \overline{\Omega }.\hfill \end{array}$(17)

Theorem 4.1 The solution $u^n,{\overrightarrow{p}}^n,\left(u_h^n,{\overrightarrow{p}}_h^n\right)$ of (2) and (17), respectively, suppose $u,u_t,u_{tt}\in H^3\left(\Omega \right)$ , $\overrightarrow{p},{\overrightarrow{p}}_t\in {\left(H^2\left(\Omega \right)\right)}^2$ , then there holds

${\Vert u_n-u_h^n\Vert }_1^2\le C\left[h^4+{\left(\Delta t\right)}^2\right],$ (18)

and

${\Vert {\overrightarrow{p}}^n-{\overrightarrow{p}}_h^n\Vert }_0^2\le C\left[h^4+{\left(\Delta t\right)}^2\right].$ (19)

Proof Taking $t=t_n,v=v_h,\overrightarrow{q}={\overrightarrow{q}}_h$ in (2), then

$\{\begin{array}{ll}\left(u_t^n,v_h\right)+\left({\overrightarrow{p}}^n,\nabla v_h\right)=\left(\nabla \cdot f\left(u^n\right),v_h\right),\hfill & \forall v_h\in X_h,\hfill \\ \left(\nabla u_t^n,{\overrightarrow{q}}_h\right)-\left({\overrightarrow{p}}^n,{\overrightarrow{q}}_h\right)=0,\text{}\hfill & \forall {\overrightarrow{q}}_h\in M_h,\hfill \\ u\left(x,0\right)=u_0\left(x\right),\hfill & x\in \overline{\Omega }.\hfill \end{array}$(20)

Subtracting (17) from (20), we have

$\{\begin{array}{l}\left({\partial }_t{u}^n-{\partial }_t{u}_h^n,v_h\right)+\left({\overrightarrow{p}}^n-{\overrightarrow{p}}_h^n,\nabla v_h\right)=\left(\nabla \cdot \left(f\left(u^n\right)-f\left(u_h^n\right)\right),v_h\right)\\ +\left(R_1^n,v_h\right),\forall v_h\in X_h,\\ \left({\partial }_t\nabla u^n-{\partial }_t\nabla u_h^n,{\overrightarrow{q}}_h\right)-\left({\overrightarrow{p}}^n-{\overrightarrow{p}}_h^n,{\overrightarrow{q}}_h\right)=\left(R_2^n,{\overrightarrow{q}}_h\right),\text{}\text{}\forall {\overrightarrow{q}}_h\in M_h.\text{}\text{}\text{}\text{}\end{array}$ (21)

where $R_1^n={\partial }_t\left(u^n-u_t^n\right),R_2^n={\partial }_t\left(\nabla u^n-\nabla u_t^n\right)$ .

For simplicity, we introduce the following notation:

$\begin{array}{l}{u}^n-u_h^n=\left(u^n-{\Pi }_h^2{u}^n\right)+\left({\Pi }_h^2{u}^n-u_h^n\right)={\eta }^n+{\xi }^n,\\ {\overrightarrow{p}}^n-{\overrightarrow{p}}_h^n=\left({\overrightarrow{p}}^n-{\Pi }_h^1{\overrightarrow{p}}^n\right)+\left({\Pi }_h^1{\overrightarrow{p}}^n-{\overrightarrow{p}}_h^n\right)={\overrightarrow{\rho }}^n+{\overrightarrow{\theta }}^n.\end{array}$

So (21) becomes

$\{\begin{array}{l}\left({\partial }_t{\xi }^n,v_h\right)+\left({\overrightarrow{\theta }}^n,\nabla v_h\right)=-\left({\partial }_t{\eta }^n,v_h\right)-\left({\overrightarrow{\rho }}^n,\nabla v_h\right)+\left(R_1^n,v_h\right)\\ +\left(\nabla \cdot \left(f\left(u^n\right)-f\left(u_h^n\right)\right),v_h\right),\forall v_h\in X_h,\\ \left({\partial }_t\nabla {\xi }^n,{\overrightarrow{q}}_h\right)-\left({\overrightarrow{\theta }}^n,{\overrightarrow{q}}_h\right)=-\left({\partial }_t\nabla {\eta }^n,{\overrightarrow{q}}_h\right)+\left({\overrightarrow{\rho }}^n,{\overrightarrow{q}}_h\right)\\ +\left(R_2^n,{\overrightarrow{q}}_h\right),\text{}\text{}\forall {\overrightarrow{q}}_h\in M_h.\text{}\text{}\text{}\text{}\end{array}$ (22)

Taking $v_h={\xi }^n,{\overrightarrow{q}}_h^n=\nabla {\xi }^n$ in (22), we get

$\begin{array}{l}\left({\partial }_t{\xi }^n,{\xi }^n\right)+\left({\partial }_t\nabla {\xi }^n,\nabla {\xi }^n\right)=-\left({\partial }_t{\eta }^n,{\xi }^n\right)-\left({\partial }_t\nabla {\eta }^n,\nabla {\xi }^n\right)\\ +\left(\nabla \cdot \left(f\left(u^n\right)-f\left(u_h^n\right)\right),{\xi }^n\right)\\ +\left(R_1^n,{\xi }^n\right)+\left(R_2^n,\nabla {\xi }^n\right).\end{array}$ (23)

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

$\left({\partial }_t{\xi }^n,{\xi }^n\right)=\frac{1}{2\Delta t}\left({\Vert {\xi }^n\Vert }_0^2-{\Vert {\xi }^{n-1}\Vert }_0^2+{\Vert {\xi }^n-{\xi }^{n-1}\Vert }_0^2\right),$

$\left({\partial }_t\nabla {\xi }^n,\nabla {\xi }^n\right)=\frac{1}{2\Delta t}\left({\Vert \nabla {\xi }^n\Vert }_0^2-{\Vert \nabla {\xi }^{n-1}\Vert }_0^2+{\Vert \nabla {\xi }^n-\nabla {\xi }^{n-1}\Vert }_0^2\right).$

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

$\left({\partial }_t{\eta }^n,{\xi }^n\right)+\left({\partial }_t\nabla {\eta }^n,\nabla {\xi }^n\right)\le C\left({\Vert {\partial }_t{\eta }^n\Vert }_0^2+{\Vert {\partial }_t\nabla {\eta }^n\Vert }_0^2\right)+\epsilon \left({\Vert {\xi }^n\Vert }_0^2+{\Vert \nabla {\xi }^n\Vert }_0^2\right),$ (24)

And

$\left(R_1^n,{\xi }^n\right)+\left(R_2^n,\nabla {\xi }^n\right)\le C\left({\Vert R_1^n\Vert }_0^2+{\Vert R_2^n\Vert }_0^2\right)+\epsilon \left({\Vert {\xi }^n\Vert }_0^2+{\Vert \nabla {\xi }^n\Vert }_0^2\right).$ (25)

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

$\begin{array}{c}\left(\nabla \cdot \left(f\left(u^n\right)-f\left(u_h^n\right)\right),{\xi }^n\right)\le C{\Vert u^n-u_h^n\Vert }_0^2+\epsilon {\Vert \nabla {\xi }^n\Vert }_0^2\\ \le C\left({\Vert {\xi }^n\Vert }_0^2+{\Vert {\eta }^n\Vert }_0^2\right)+\epsilon {\Vert \nabla {\xi }^n\Vert }_0^2.\end{array}$ (26)

According to the Taylor formula, we get

${\Vert R_1^n\Vert }_0^2+{\Vert R_2^n\Vert }_0^2\le C{\left(\Delta t\right)}^2\left({\Vert u_{tt}\left(t_{n+{\theta }_1}\right)\Vert }_0^2+{\Vert \nabla u_{tt}\left(t_{n+{\theta }_2}\right)\Vert }_0^2\right).$ (27)

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

$\begin{array}{l}\frac{1}{2\Delta t}\left[{\Vert {\xi }^n\Vert }_0^2+{\Vert \nabla {\xi }^n\Vert }_0^2-\left({\Vert {\xi }^{n-1}\Vert }_0^2+{\Vert \nabla {\xi }^{n-1}\Vert }_0^2\right)\right]\\ \le C\left({\Vert {\partial }_t{\eta }^n\Vert }_0^2+{\Vert {\partial }_t\nabla {\eta }^n\Vert }_0^2+{\Vert {\eta }^n\Vert }_0^2\right)\\ +C{\left(\Delta t\right)}^2\left({\Vert u_{tt}\left(t_{n+{\theta }_1}\right)\Vert }_0^2+{\Vert \nabla u_{tt}\left(t_{n+{\theta }_2}\right)\Vert }_0^2\right)+3\epsilon \left({\Vert {\xi }^n\Vert }_0^2+{\Vert \nabla {\xi }^n\Vert }_0^2\right).\end{array}$

Multiply both ends of the above formula by $4\Delta t$ , taking $\epsilon =\frac{1}{6\Delta t}$ , and add from 0 to $N$ , we get

$\begin{array}{l}{\Vert {\xi }^n\Vert }_0^2+{\Vert \nabla {\xi }^n\Vert }_0^2\le C\Delta t{\displaystyle \sum _{k=0}^N\left[{\Vert {\partial }_t{\eta }^k\Vert }_0^2+{\Vert {\partial }_t\nabla {\eta }^k\Vert }_0^2+{\Vert {\eta }^k\Vert }_0^2\right]}\\ +{\left(\Delta t\right)}^2\left({\Vert u_{tt}\left(t_{k+{\theta }_1}\right)\Vert }_0^2+{\Vert \nabla u_{tt}\left(t_{k+{\theta }_2}\right)\Vert }_0^2\right).\end{array}$

From Lemma 3.3 and Lemma 3.6, we have

${\Vert {\xi }^n\Vert }_0^2+{\Vert \nabla {\xi }^n\Vert }_0^2\le C\left[h^{2m}{\displaystyle {\int }_0^{t_n}\left({\Vert u\Vert }_3^2+{\Vert u_t\Vert }_3^2\right)\text{d}\tau }+{\left(\Delta t\right)}^2{\displaystyle {\int }_0^{t_n}{\Vert u_{tt}\Vert }_1^2\text{d}\tau }\right].$ (28)

According to the regularity of u, we have

${\Vert {\xi }^n\Vert }_1^2={\Vert {\xi }^n\Vert }_0^2+{\Vert \nabla {\xi }^n\Vert }_0^2\le C\left[h^4+{\left(\Delta t\right)}^2\right].$ (29)

Taking $v_h={\partial }_t{\xi }^n,{\overrightarrow{q}}_h={\partial }_t\nabla {\xi }^n$ in (22), using the Lipschitz condition, we get

$\begin{array}{l}{\Vert {\partial }_t{\xi }^n\Vert }_0^2+{\Vert {\partial }_t\nabla {\xi }^n\Vert }_0^2=-\left({\partial }_t{\eta }^n,{\partial }_t{\xi }^n\right)-\left({\partial }_t\nabla {\eta }^n,{\partial }_t\nabla {\xi }^n\right)\\ +\left(\nabla \cdot \left(f\left(u^n\right)-f\left(u_h^n\right)\right),{\partial }_t{\xi }^n\right)\\ +\left(R_1^n,{\partial }_t{\xi }^n\right)+\left(R_2^n,{\partial }_t\nabla {\xi }^n\right)\\ \le C\left({\Vert {\partial }_t{\eta }^n\Vert }_0^2+{\Vert {\partial }_t\nabla {\eta }^n\Vert }_0^2+{\Vert R_1^n\Vert }_0^2+{\Vert R_2^n\Vert }_0^2\right)\\ +\epsilon \left({\Vert {\partial }_t{\xi }^n\Vert }_0^2+{\Vert {\partial }_t\nabla {\xi }^n\Vert }_0^2\right).\end{array}$

Multiply both ends of the above formula by 2, taking $\epsilon =\frac{1}{2}$ , then according to Lemma 3.3 and the regularity of $u$ , we get

${\Vert {\partial }_t{\xi }^n\Vert }_0^2+{\Vert {\partial }_t\nabla {\xi }^n\Vert }_0^2\le C\left[h^4+{\left(\Delta t\right)}^2\right].$ (30)

Then, using Lemma 3.3 and the triangle inequality, the proof of (18) is completed. Letting ${\overrightarrow{q}}_h={\overrightarrow{\theta }}^n$ the second formula of (22), we get

$\begin{array}{c}{\Vert {\overrightarrow{\theta }}^n\Vert }_0^2=\left({\partial }_t\nabla {\xi }^n,{\overrightarrow{\theta }}^n\right)+\left({\partial }_t\nabla {\eta }^n,{\overrightarrow{\theta }}^n\right)+\left({\overrightarrow{\rho }}^n,{\overrightarrow{\theta }}^n\right)+\left(R_2^n,{\overrightarrow{\theta }}^n\right)\\ \le 4\epsilon {\Vert {\overrightarrow{\theta }}^n\Vert }_0^2+C\left({\Vert {\partial }_t\nabla {\xi }^n\Vert }_0^2+{\Vert {\partial }_t\nabla {\eta }^n\Vert }_0^2+{\Vert {\overrightarrow{\rho }}^n\Vert }_0^2+{\Vert R_2^n\Vert }_0^2\right).\end{array}$

Letting $\epsilon =\frac{1}{8}$ , from (29), Lemma 3.3, and Lemma 3.4, then there holds

${\Vert {\overrightarrow{\theta }}^n\Vert }_0^2\le C\left[h^4+{\left(\Delta t\right)}^2\right].$ (31)

then using the triangle inequality and the interpolation theorem, the proof of (19) is completed.

5. Numerical Example

Positioning Consider the following BBM Equation:

$\begin{array}{l}{u}_t-u_{xxt}-u_{yyt}+u_x+u_y+u{u}_x+u{u}_y=0,\left(x,y,t\right)\in \Omega \times \left(0,T\right],\\ u\left(x,y,0\right)=u_0\left(x,y\right)=x^2{\left(1-x\right)}^2{y}^2{\left(1-y\right)}^2,\left(x,y\right)\in \partial \Omega .\end{array}$

where $\Omega =\left(0,1\right)\times \left(0,1\right)$ . The region $\Omega$ is divided into uniform rectangles along the X and Y directions, $Q_2$ element approximation is adopted for the velocity u space, and $Q_1$ element approximation is adopted for the P space.

Since the exact solution of problem (2) is unknown, we use the numerical solution on the fine grid as the exact solution to estimate the error. We take some different values of $\Delta t$ and $h$ to verify the format of this paper, as shown in the following table.

Table 1. $\Delta t=0.1,h=1/10,e=u^{*}-u_h,\eta =p^{*}-p_h$ .

$t$	${\Vert e\Vert }_{L^2\left(\Omega \right)}$	${\Vert e\Vert }_{H^1\left(\Omega \right)}$	${\Vert \eta \Vert }_{L^2\left(\Omega \right)}$
0.2	$6.31\times {10}^{-7}$	$3.71\times {10}^{-4}$	$1.31\times {10}^{-6}$
0.4	$6.80\times {10}^{-7}$	$4.75\times {10}^{-4}$	$8.20\times {10}^{-7}$
0.6	$7.19\times {10}^{-7}$	$4.91\times {10}^{-4}$	$6.27\times {10}^{-7}$
0.8	$7.20\times {10}^{-7}$	$5.01\times {10}^{-4}$	$4.83\times {10}^{-7}$
1.0	$7.01\times {10}^{-7}$	$5.44\times {10}^{-4}$	$3.04\times {10}^{-7}$

Table 2. $\Delta t=0.1,h=1/20,e=u^{*}-u_h,\eta =p^{*}-p_h$ .

$t$	${\Vert e\Vert }_{L^2\left(\Omega \right)}$	${\Vert e\Vert }_{H^1\left(\Omega \right)}$	${\Vert \eta \Vert }_{L^2\left(\Omega \right)}$
0.2	$4.00\times {10}^{-8}$	$8.71\times {10}^{-5}$	$9.28\times {10}^{-8}$
0.4	$5.00\times {10}^{-8}$	$1.35\times {10}^{-4}$	$5.49\times {10}^{-8}$
0.6	$6.00\times {10}^{-8}$	$1.52\times {10}^{-4}$	$4.23\times {10}^{-8}$
0.8	$6.00\times {10}^{-8}$	$1.73\times {10}^{-4}$	$3.58\times {10}^{-8}$
1.0	$5.00\times {10}^{-8}$	$1.86\times {10}^{-4}$	$2.67\times {10}^{-8}$

Table 3. $\Delta t=0.1,h=1/40,e=u^{*}-u_h,\eta =p^{*}-p_h$ .

$t$	${\Vert e\Vert }_{L^2\left(\Omega \right)}$	${\Vert e\Vert }_{H^1\left(\Omega \right)}$	${\Vert \eta \Vert }_{L^2\left(\Omega \right)}$
0.2	$2.00\times {10}^{-10}$	$2.31\times {10}^{-5}$	$9.28\times {10}^{-8}$
0.4	$1.00\times {10}^{-8}$	$3.23\times {10}^{-5}$	$5.49\times {10}^{-8}$
0.6	$1.00\times {10}^{-8}$	$3.78\times {10}^{-5}$	$4.23\times {10}^{-8}$
0.8	$1.00\times {10}^{-8}$	$4.24\times {10}^{-5}$	$3.58\times {10}^{-8}$
1.0	$2.00\times {10}^{-10}$	$4.75\times {10}^{-5}$	$2.70\times {10}^{-8}$

Table 4. $t=1,\Delta t=h^2,e=u^{*}-u_h,\eta =p^{*}-p_h$ .

$h$	${\Vert e\Vert }_{H^1\left(\Omega \right)}$	$rate$	${\Vert \eta \Vert }_{L^2\left(\Omega \right)}$	$rate$
1/4	$7.92\times {10}^{-4}$	-	$5.78\times {10}^{-7}$	-
1/8	$1.96\times {10}^{-4}$	2.11	$1.39\times {10}^{-7}$	2.01
1/16	$4.93\times {10}^{-5}$	1.97	$3.08\times {10}^{-8}$	2.05
1/32	$1.34\times {10}^{-5}$	1.87	$7.96\times {10}^{-9}$	1.96

The numerical results (Tables 1-4) show that the scheme is feasible, and further verify that the hybrid stabilization method presented in this paper is stable and convergent.

Funding

Key Scientific Research Project of Colleges and Universities in Henan Province; Project Name: Petrov-Galerkin Application Research of Finite Element Method in Nonlinear Equation; Project Number: 25A110013.