移动网格方法最新进展

最新进展

最近我们在移动网格方面的研究在以下几个方面,

不可压流体

对于不可压流体,由于速度场满足不可压条件,所以奇异性相对较小,要构造 控制函数比较困难。另外,将不可压的速度场从旧网格更新到新网格上,仍然 要满足不可压条件是比较困难的。

For imcompressible, the singularity is comparatively weak bacause the velocity is divergence free that the mechanism to construct control function is much more difficult than those numerical examples we mentioned before. And to update the velocity to the new mesh efficiently but still divergence free is still a problem.

我们研究了对不可压流构造控制函数的方案,并且作为我们提出来的一般的网格 间插值的公式,开发了一个新的旧网格到新网格插值的方案,使得不可压条件能 够保持。下面是两个算例:

We studied the stragety to construct monitor function for imcompressible flow and give a general frame to update the solution to the new mesh, while at the same time keep certain constraint of the solution. The following are two numerical examples:

Double Shear Flow

不可压Navier-Stokes方程

\[ \begin{array}{l} \vec{u}_{t}+(\vec{u}\cdot\nabla)\vec{u} +\nabla p =\frac{1}{Re} \Delta \vec{u}\\ \nabla \cdot \vec{u} = 0 \end{array} \]

我们使用原始变量进行求解,计算的区域是标准的正方形,我们取双周期边界条件

\[ \begin{array}{rcl} \vec{u}(-1,y;t) &=& \vec{u}(1,y;t) \\ \vec{u}(x,-1;t) &=& \vec{u}(x,1;t) \\ p(-1,y;t) &=& p(1,y;t) \\ p(-1,y;t) &=& p(1,y;t) \end{array} \]

初值为

\[ \begin{array}{rcl} u &=& \left\{\begin{array}{ll} \tanh(\rho(y - 0.25)) & \mathrm{for\ } y < 0.5;\\ \tanh(\rho(0.75 - y)) & \mathrm{for\ } y > 0.5. \end{array}\right.\\ v &=& \delta \sin(2\pi x) \end{array} \]

得到的一个计算结果的图形为:

double_shear_flow_mesh.gif

Mesh

double_shear_flow_velo.gif

Velocity

double_shear_flow_vort.gif

Contour of Vorticity

Boussinesq Flow

方程为

\[ \begin{array}{l} \rho_t + \vec{u}\cdot\nabla\rho = 0\\ \vec{u}_{t}+(\vec{u}\cdot\nabla)\vec{u} +\nabla p = \left(\begin{array}{c} 0 \\ \rho \end{array}\right) \\ \nabla \cdot \vec{u} = 0 \end{array} \]

双曲型守恒律

我们主要考虑的是Euler方程

\[ \begin{array}{l} \displaystyle\frac{\partial \rho }{\partial t}+\displaystyle\frac{\partial \rho u}{\partial x}+\displaystyle\frac{\partial \rho v}{\partial y} = 0 \\ \displaystyle\frac{\partial \rho u}{\partial t}+\displaystyle\frac{\partial \left( p+\rho u^{2}\right) }{\partial x}+\displaystyle\frac{\partial \rho uv}{\partial y} = 0 \\ \displaystyle\frac{\partial \rho v}{\partial t}+\displaystyle\frac{\partial \rho uv}{\partial x}+\displaystyle\frac{\partial \left( p+\rho v^{2}\right) }{\partial y} = 0 \\ \displaystyle\frac{\partial e}{\partial t}+\displaystyle\frac{\partial \left( u\left( p+e\right) \right) }{\partial x}+\displaystyle\frac{\partial \left( v\left( p+e\right) \right) }{\partial y} = 0 \end{array} \]

状态方程为

\[ e=\frac{p}{\gamma -1}+\frac{1}{2}\rho \left( u^{2} + v^{2} \right) \]

二维 Riemann 问题

初值为

\[ \left(\rho, u, v, p\right) = \left\{ \begin{array}{ll} (1.1, 0.0, 0.0, 1.1), & \rm{if\ } x>0.5,y>0.5, \\ (0.5065, 0.8939, 0.0, 0.35), & \rm{if\ } x<0.5, y>0.5, \\ (1.1, 0.8939, 0.8939, 1.1), & \rm{if\ } x<0.5, y<0.5, \\ (0.5065, 0.0, 0.8939, 0.35), & \rm{if\ } x>0.5, y<0.5. \end{array}\right. \]

euler_mesh.png
euler_density_contour.png

双马赫反射问题

The domain is a rectangle $ [0, 4]\times[0, 1] $. A right moving shock is initially positioned at $ (1/6, 0) $ and makes a $ 60 $ degree angle with the $ x $-axis. The inflow is with Mach number 10. The boundary condition at bottom is the exact post shock condition from $0$ to $ 1/6 $ and is reflective for the rest. At the top boundary, the flow values are set to describe the exact motion of the Mach 10 shock. On the left and right boundaries, the inflow and outflow boundary conditions are used respectively.

double_mach_mesh.gif
double_mach_density_contour.gif

球面上的网格移动

在球面上实现移动网格方法的算法是一件看起来自然但是却很不容易的事情,最近 这个方面的工作有比较大的进展,邸亚娜博士正在将得到的结果整理成为论文。
Generated on Fri Jun 29 16:17:00 2007 for Moving Mesh by  doxygen 1.4.7