An efficient computational approach for two-dimensional variant of nonlinear-dispersive model of shallow water wave

Abstract

A three-level linearized difference scheme for two-dimensional dispersive shallow water wave that is governed by the Rosenau-RLW equation is considered. It is proved that the proposed difference scheme is conservative, uniquely solvable and unconditionally convergent. The convergence order in maximum norm is $O(\tau ^2+h_1^2+h_2^2)$, where $\tau$ is the temporal grid size and $h_1, h_2$ are spatial grid sizes in the x- and y-directions, respectively. Some numerical examples are provided to demonstrate the efficiency and applicability of the method and to verify its rate of convergence. The numerical results are compared with exact solutions and other existing method. Comparison reveals that our method improves the accuracy of the space and time direction and shortens computation time largely.

Appendix

Lemma 1

For any grid functions $V^n\in {\mathcal {V}}_h,$ we have

$$\begin{aligned} ({\partial }_{{\bar{t}}}V^n,{\bar{V}}^n)& = \frac{1}{2} \partial _{{\bar{t}}}\Vert V^n\Vert ^2, \end{aligned}$$

(A.1)

$$\begin{aligned} (\Delta _h(\partial _{{\bar{t}}}V^n), {\bar{V}}^n)& = -\frac{1}{2} \partial _{{\bar{t}}}|V^n|_1^2, \end{aligned}$$

(A.2)

$$\begin{aligned} (\Delta _h^2(\partial _{{\bar{t}}}V^n), {\bar{V}}^n)& = \frac{1}{2} \partial _{{\bar{t}}}\Vert \Delta _hV^n\Vert ^2, \end{aligned}$$

(A.3)

$$\begin{aligned} ({\nabla }_h{\bar{V}}^n,{\bar{V}}^n)& = 0, \end{aligned}$$

(A.4)

$$\begin{aligned} (\phi (V^n,\bar{V^n}),{\bar{V}}^n)& = 0. \end{aligned}$$

(A.5)

Proof

For every $V^n\in {\mathcal {V}}_h,$ we have

$$\begin{aligned} ({\partial }_{{\bar{t}}}V^n,{\bar{V}}^n)& = h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\displaystyle \frac{V_{i,j}^{n+1}- V_{i,j}^{n-1}}{2\tau }\cdot \displaystyle \frac{V_{i,j}^{n+1}+ V_{i,j}^{n-1}}{2}\\& = \displaystyle \frac{h_1h_2}{2}\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2} \displaystyle \frac{(V_{i,j}^{n+1})^2- (V_{i,j}^{n-1})^2}{2\tau }\\& = \displaystyle \frac{1}{2} \partial _{{\bar{t}}}\Vert V^n\Vert ^2, \end{aligned}$$

and (A.1) follows. In view of difference properties and (2.5), we obtain for $V^n\in {\mathcal {V}}_h$

$$\begin{aligned} (\Delta _h({\partial }_{{\bar{t}}}V^n),{\bar{V}}^n)& = \displaystyle \frac{h_1h_2}{4\tau }\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}(\delta ^2_x+\delta ^2_y)({V_{i,j}^{n+1}- V_{i,j}^{n-1}})\\&\cdot ({V_{i,j}^{n+1}+ V_{i,j}^{n-1}})\\& = -\displaystyle \frac{h_1h_2}{4\tau }\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2} (\delta _x+\delta _y)({V_{i,j}^{n+1}- V_{i,j}^{n-1}})\\&\cdot (\delta _x+\delta _y)({V_{i,j}^{n+1}+ V_{i,j}^{n-1}})\\& = -\displaystyle \frac{1}{4\tau }(|V^{n+1}|_1^2-|V^{n-1}|_1^2) \\& = -\displaystyle \frac{1}{2} \partial _{{\bar{t}}}|V^n|_1^2. \end{aligned}$$

We get (A.2). Similarly, we have for $V^n\in {\mathcal {V}}_h$

$$\begin{aligned} (\Delta _h^2({\partial }_{{\bar{t}}}V^n),{\bar{V}}^n)& = (\Delta _h({\partial }_{{\bar{t}}}V^n),\Delta _h{\bar{V}}^n) \\& = \displaystyle \frac{h_1h_2}{4\tau }\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\Delta _h(V_{i,j}^{n+1}- V_{i,j}^{n-1})\cdot \Delta _h(V_{i,j}^{n+1}+ V_{i,j}^{n-1})\\& = \displaystyle \frac{1}{4\tau }(\Vert \Delta _hV^{n+1}\Vert ^2 -\Vert \Delta _hV^{n-1}\Vert ^2) \\& = \displaystyle \frac{1}{2} \partial _{{\bar{t}}}\Vert \Delta _h V^n\Vert ^2, \end{aligned}$$

and (A.3) follows. For any $V^n, W^n \in {\mathcal {V}}_h,$ we have

$$\begin{aligned} ({\nabla }_h{\bar{V}}^n,{\bar{W}}^n)=-({\bar{V}}^n, {\nabla }_h{\bar{W}}^n), \end{aligned}$$

in particular, if ${\bar{V}}^n={\bar{W}}^n$, then

$$\begin{aligned} ({\nabla }_h{\bar{V}}^n,{\bar{V}}^n)=-({\bar{V}}^n, {\nabla }_h{\bar{V}}^n). \end{aligned}$$

Therefore,

$$\begin{aligned} ({\nabla }_h{\bar{V}}^n,{\bar{V}}^n)=0, \end{aligned}$$

we find (A.4). In view of difference properties and (2.5), we have

$$\begin{aligned} (V^n{\nabla }_h{\bar{V}}^n,{\bar{V}}^n)= ({\nabla }_h{\bar{V}}^n,V^n{\bar{V}}^n)= -({\bar{V}}^n, {\nabla }_h(V^n{\bar{V}}^n)). \end{aligned}$$

The above equality becomes

$$\begin{aligned} (\phi (V^n,\bar{V^n}),{\bar{V}}^n)=\displaystyle \frac{1}{3}(V^n{\nabla }_h{\bar{V}}^n,{\bar{V}}^n) +\displaystyle \frac{1}{3}({\nabla }_h(V^n{\bar{V}}^n),{\bar{V}}^n)=0. \end{aligned}$$

This completes the proof of the Lemma 1. $\square$

Lemma 2

For $V^n\in {\mathcal {V}}_h,$ we have

$$\begin{aligned} \Vert \nabla _h V^n\Vert\le & |V^n|_1, \end{aligned}$$

(A.6)

$$\begin{aligned} \Vert \nabla _h V^n\Vert ^2\le & \Vert V^n\Vert .\Vert \Delta _h V^n\Vert , \end{aligned}$$

(A.7)

$$\begin{aligned} \Vert V^n\Vert _{\infty }^2\le & C \Vert V^n\Vert \left( \Vert \Delta _h V^n\Vert +\Vert V^n\Vert \right) . \end{aligned}$$

(A.8)

Proof

For $V^n\in {\mathcal {V}}_h,$ we have

$$\begin{aligned} \Vert \nabla _h V^n\Vert ^2& = \Vert V_{{\hat{x}}}^n\Vert ^2+\Vert V_{{\hat{y}}}^n\Vert ^2\\& = h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\left( \displaystyle \frac{V_{i+1,j}^{n}- V_{i-1,j}^{n}}{2h_1}\right) ^2 \\&+ h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\left( \displaystyle \frac{V_{i,j+1}^{n}- V_{i,j-1}^{n}}{2h_2}\right) ^2 \\& = h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\left( \displaystyle \frac{V_{i+1,j}^{n}- V_{i,j}^{n}}{2h_1}+\displaystyle \frac{V_{i,j}^{n}- V_{i-1,j}^{n}}{2h_1} \right) ^2 \\&+ h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\left( \displaystyle \frac{V_{i,j+1}^{n}- V_{i,j}^{n}}{2h_2}+\displaystyle \frac{V_{i,j}^{n}- V_{i,j-1}^{n}}{2h_2} \right) ^2 \\\le & 2h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\left( \displaystyle \frac{V_{i+1,j}^{n}- V_{i,j}^{n}}{2h_1}\right) ^2\\&+2h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}\left( \displaystyle \frac{V_{i,j}^{n}- V_{i-1,j}^{n}}{2h_1} \right) ^2 \\&+ 2h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}(\displaystyle \frac{V_{i,j+1}^{n}- V_{i,j}^{n}}{2h_2})^2\\&+2h_1h_2\displaystyle \sum _{i=1}^{M_1}\displaystyle \sum _{j=1}^{M_2}(\displaystyle \frac{V_{i,j}^{n}- V_{i,j-1}^{n}}{2h_2} )^2. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert \nabla _h V^n\Vert ^2\le \displaystyle \frac{1}{2}\Vert V_x^n\Vert ^2+\displaystyle \frac{1}{2}\Vert V_{{\bar{x}}}^n\Vert ^2+\displaystyle \frac{1}{2}\Vert V_y^n\Vert ^2+\displaystyle \frac{1}{2}\Vert V_{{\bar{y}}}^n\Vert ^2. \end{aligned}$$

From the properties of differences and periodic boundary, we obtain

$$\begin{aligned} \Vert V_x^n\Vert ^2=\Vert V_{{\bar{x}}}^n\Vert ^2\hbox { and }\Vert V_y^n\Vert ^2=\Vert V_{{\bar{y}}}^n\Vert ^2. \end{aligned}$$

This yields that

$$\begin{aligned} \Vert \nabla _h V^n\Vert ^2\le \Vert V_x^n\Vert ^2+\Vert V_y^n\Vert ^2=|V^n|_1^2, \end{aligned}$$

and (A.6) follows. By the discrete Green formula, we have for $V^n\in {\mathcal {V}}_h$

$$\begin{aligned} |V^n|_1^2=-(V^n, \Delta _hV^n)\le \Vert V^n\Vert \cdot \Vert \Delta _hV^n\Vert . \end{aligned}$$

The claimed inequality (A.7) follows from (A.6) immediately.

We can see the proof of the inequality (A.8) in [35]. $\square$