Adaptive Mesh Method for Solving a Second-Order Hyperbolic Equation

Zhou Qin,1, Yang Yin2

 基金资助: 国家自然科学基金.  11671342湖南省教育厅科学研究项目基金.  18C1097湖南省自然科学基金.  2018JJ2374

 Fund supported: the NSFC.  11671342the Scientific Research Fund of Hunan Provincial Education Department.  18C1097the Natural Science Foundation of Hunan Province.  2018JJ2374

Abstract

In this paper, we study a class of second-order hyperbolic equations with small parameters. An adaptive moving mesh method for solving the equation with finite differencing scheme is proposed, and the moving mesh algorithm is given. The superiority of the method is verified by numerical experiments, and the result on uniform mesh is improved.

Keywords： Hyperbolic equation ; Difference scheme ; Adaptive moving mesh ; Mesh iteration

Zhou Qin, Yang Yin. Adaptive Mesh Method for Solving a Second-Order Hyperbolic Equation. Acta Mathematica Scientia[J], 2019, 39(4): 942-950 doi:

1 引言

$$$\left\{ \begin{array}{ll} \frac{\partial^2 u}{\partial t^2} = \varepsilon^2a(x, t)\frac{\partial^2 u}{\partial x^2}+f(x, t), \; &0\leq x \leq L, \; 0\leq t \leq T, \\ u(x, 0) = \varphi(x), \; \frac{\partial u}{\partial t}\mid_{t = 0} = \psi(x), \; &0\leq x \leq L, \\ u(0, t) = g_0(t), \; u(L, t) = g_1(t), \; &0\leq t \leq T, \end{array} \right.$$$

3 自适应网格的生成和解的更新

$x $$\xi 分别代表物理坐标系和计算坐标系,根据模型问题,有 x\in [0, L] .假定 \xi \in [0, 1] ,我们建立 x$$ \xi$之间的一一坐标变换: $x = x(\xi), \; \xi\in [0, 1]$,且$x(0) = 0, \; x(1) = L$.由等分布原理的标准形式

$$$(\omega x_\xi)_\xi = 0,$$$

$$$\omega(u(x, t)) = \sqrt{1+\rho(u_x(x, t))^2}$$$

$$$w(u^{[v]}_{j+\frac{1}{2}})(x^{[v]}_{j+1}-x^{[v+1]}_j)-w(u^{[v]}_{j-\frac{1}{2}}) (x^{[v+1]}_{j}-x^{[v]}_{j-1}) = 0.$$$

$$$x^{[v+1]}_{j} = \frac{w(u^{[v]}_{j+\frac{1}{2}})x^{[v]}_{j+1}+w(u^{[v]}_{j-\frac{1}{2}})x^{[v]}_{j-1}}{w(u^{[v]}_{j+\frac{1}{2}})+w(u^{[v]}_{j-\frac{1}{2}})}.$$$

$$$\tilde{u}_{j+\frac{1}{2}} = \frac{1}{\tilde{x}_{j+1}-\tilde{x}_j}\int _{\tilde{x}_j}^{\tilde{x}_{j+1}} u(\tilde{x}, t) {\rm d}\tilde{x}, \; \; u_{j+\frac{1}{2}} = \frac{1}{x_{j+1}-x_j}\int _{x_j}^{x_{j+1}} u(x, t) {\rm d}x.$$$

$\begin{eqnarray} \int _{\tilde{x}_j}^{\tilde{x}_{j+1}} u(\tilde{x}, t){\rm d}\tilde{x} & = & \int _{x_j}^{x_{j+1}} u(x-c(x), t)(1-c'(x)) {\rm d}x\\ &\approx&\int _{x_j}^{x_{j+1}} (u(x, t)-c(x)u_x(x, t))(1-c'(x)) {\rm d}x\\ &\approx&\int _{x_j}^{x_{j+1}} (u(x, t)-(cu)_x) {\rm d}x\\ & = &\int _{x_j}^{x_{j+1}} u(x, t){\rm d}x-((cu)_{j+1}-(cu)_{j}), \end{eqnarray}$

$$$(\tilde{x}_{j+1}-\tilde{x}_j)\tilde{u}_{j+\frac{1}{2}} = (x_{j+1}-x_j)u_{j+\frac{1}{2}}-((cu)_{j+1}-(cu)_{j}).$$$

$$$\tilde{u}_{j+\frac{1}{2}} = \frac{x_{j+1}-x_j}{\tilde{x}_{j+1}-\tilde{x}_j}{u}_{j+\frac{1}{2}} -\frac{1}{\tilde{x}_{j+1}-\tilde{x}_j}(({cu})_{j+1}-({cu})_{j})$$$

4 算法和数值实验

 $N\times M$ $t=0.5$ $r$ $t=1$ $r$ $t=1.5$ $r$ $t=2$ $r$ $50\times50$ 9.508E-05 2.797E-04 5.684E-04 9.174E-04 $100\times100$ 2.530E-05 1.91 8.225E-05 1.77 1.684E-04 1.76 2.673E-04 1.78 $200\times200$ 8.934E-06 1.50 2.225E-05 1.89 4.630E-05 1.86 7.321E-05 1.87

 $N\times M$ $t=0.5$ $r$ $t=1$ $r$ $t=1.5$ $r$ $t=2$ $r$ $50\times50$ 1.952E-04 7.678E-04 1.480E-03 2.196E-03 $100\times100$ 5.472E-05 1.83 2.433E-04 1.66 4.720E-04 1.65 6.970E-04 1.66 $200\times200$ 2.099E-05 1.38 6.838E-05 1.83 1.354E-04 1.80 1.999E-04 1.80

 $N\times M$ $t=0.5$ $t=1$ $t=1.5$ $t=2$ $50\times50$ 7.683E-04 2.571E-03 4.787E-03 6.949E-03 $100\times100$ 3.513E-04 1.113E-03 1.902E-03 2.499E-03 $200\times200$ 1.127E-04 3.276E-04 5.221E-04 6.652E-04

 $N\times M$ $t=0.5$ $t=1$ $t=1.5$ $t=2$ $50\times50$ 5.433E-03 1.813E-02 3.353E-02 4.808E-02 $100\times100$ 3.256E-03 1.001E-02 1.596E-02 1.834E-02 $200\times200$ 1.201E-03 2.842E-03 4.560E-03 5.417E-03

 $N\times M$ $t=0.5$ $r$ $t=1$ $r$ $t=1.5$ $r$ $t=2$ $r$ $50\times50$ 1.622E-04 2.400E-04 4.571E-04 6.861E-04 $100\times100$ 4.274E-05 1.92 5.581E-05 2.10 1.147E-04 1.99 1.750E-04 1.97 $200\times200$ 1.948E-05 1.13 1.698E-05 1.72 2.800E-05 2.03 4.703E-05 1.90

 $N\times M$ $t=0.5$ $r$ $t=1$ $r$ $t=1.5$ $r$ $t=2$ $r$ $50\times50$ 1.952E-04 7.678E-04 1.480E-03 2.196E-03 $100\times100$ 5.472E-05 1.83 2.433E-04 1.66 4.720E-04 1.65 6.970E-04 1.66 $200\times200$ 2.099E-05 1.38 6.838E-05 1.83 1.354E-04 1.80 1.999E-04 1.80

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