Triangular DGM

1. Basis functions

decomposing the domain \(\Omega\) into \(N_e\) conforming

non-overlapping triangular elements \(\Omega_e\).

\[\begin{equation}
\Omega = \bigcup_{e = 1}^{N_e} \Omega_e
\end{equation}\]

nonsingular mapping \(x = \Psi(\mathbf{\xi})\) which defines a transformation from the physical Cartesian coordinate system to the local reference coordinate system defined on the reference triangle.

local elementwise solution \(\mathbf{q}\) by an N th order polynomial in \(\mathbf{\xi}\) as

\[\begin{equation}
\mathbf{q}_N (\mathbf{\xi}) = \sum_{i = 1}^{M_N} \psi_i (\mathbf{\xi}) \mathbf{q}_N (\mathbf{\xi}_i)
\end{equation}\]

where \(\mathbf{\xi}_i\) represents \(M = \frac{1}{2} ( N + 1)( N + 2)\) interpolation points and \(\psi_i (\mathbf{\xi})\) are the associatedmultivariate Lagrange polynomials.

an explicit formula for the Lagrange basis —— reference to an easily constructed orthonormal PKD polynomial basis and the generalized Vandermonde matrix.

通过正交多项式和Vandermonde构造参考单元上Lagrange基函数。

2. Integration

2.1. Area integrals

\(\int_{\Omega_e} f(x) g(x) dx = \sum_{i = 1}^{M_C} \omega_i^e \left| J^e(\mathbf{\xi}_i) \right| f(\mathbf{\xi}_i) g(\mathbf{\xi}_i)\)

where \(M_C\) is a function of \(C\) which represents the order of the cubature approximation.

2.2. Boundary integrals

\(\int_{\Gamma_e} f(x) g(x) dx = \sum_{i = 0}^{Q} \omega_i^s \left| J^s(\mathbf{\xi}_i) \right| f(\mathbf{\xi}_i) g(\mathbf{\xi}_i)\)

where \(Q\) represents the order of the quadrature approximation. Using the Gauss quadrature, we

can use \(Q = N\) to achieve order \(2N\) accuracy.

3. Tangent and normal vectors of the element edges

4. Semi-discrete equations

5. Matrix form of the semi-discrete equations

5.1. Elimination of the mass matrix

将方程左乘质量矩阵的逆并除以雅克比系数,可得

\[\begin{equation}
\frac{\partial \mathbf{q}^e_i}{\partial t} + \left( \hat{D}_{ij}^{\xi} \xi_x^e + \hat{D}_{ij}^{\eta} \eta_x^e \right) \mathbf{f}_j^e + \left( \hat{D}_{ij}^{\xi} \xi_y^e + \hat{D}_{ij}^{\eta} \eta_y^e \right) \mathbf{g}_j^e - S_i^e = \frac{\left| J^s \right|}{\left| J^e \right|} \hat{M}_{ij}^s \left[ n_x^s \left( \mathbf{f}^e - \mathbf{f}^* \right)_j + n_y^s \left( \mathbf{g}^e - \mathbf{g}^* \right)_j \right]
\end{equation}\]

where the matrices are defined as

\[\begin{equation}
\begin{array}{lll}
\hat{D}_{ij}^{\xi} = M_{ik}^{-1} D_{kj}^{\xi}, & \hat{D}_{ij}^{\eta} = M_{ik}^{-1} D_{kj}^{\eta}, &
\hat{M}_{ij}^{s} = M_{ik}^{-1} M_{kj}^{\xi},
\end{array}
\end{equation}\]

where

\[\begin{equation}
\begin{array}{ll}
M_{ij} = \sum_{k = 1}^{M_C} \omega_k \psi_{ik} \phi_{jk}, & M_{ij}^s = \sum_{k = 1}^{M_Q} \omega_k \psi_{ik} \phi_{jk} \cr
D_{ij}^{\xi} = \sum_{k = 1}^{M_C} \omega_k \psi_{ik} \frac{\partial \phi_{jk}}{\partial \xi}, & D_{ij}^{\eta} = \sum_{k = 1}^{M_C} \omega_k \psi_{ik} \frac{\partial \phi_{jk}}{\partial \eta}
\end{array}
\end{equation}\]

\(M_C\) and \(M_Q\) denote the number of cubature (two dimensional) and quadrature (one dimensional) integration points required to achieve order 2N accuracy, and \(\psi_{ik}\) represents the function \(\psi\) at the \(i=1, \cdots,M_N\) interpolation points evaluated at the integration point k.

Since the mass matrix is constant (i.e. not a function of x) then, using Equations above, we can move the mass matrix inside the summations which are the discrete representations of the continuous integrals. This then gives

\[\begin{equation}
\begin{array}{ll}
\hat{M}_{ij}^{s} = \sum_{k = 1}^{M_Q} \omega_k \hat{\psi}_{ik} \psi_{jk}, & \hat{D}_{ij}^{\xi} = \sum_{k = 1}^{M_C} \omega_k \hat{\psi}_{ik} \frac{\partial \psi_{jk}}{\partial \xi}, & \hat{D}_{ij}^{\eta} = \sum_{k = 1}^{M_C} \omega_k \hat{\psi}_{ik} \frac{\partial \psi_{jk}}{\partial \eta}
\end{array}
\end{equation}\]

where

\[\begin{equation}
\hat{\psi}_i = M_{ik}^{-1} \psi_k
\end{equation}\]

根据

\(D_{ij}^{\xi} = \sum_{k = 1}^{M_C} \omega_k \psi_{ik} \frac{\partial \psi_{jk}}{\partial \xi}\)

我们可以将 \(D_{ij}^{\xi}\) 写为如下矩阵相乘形式

\[\begin{equation}
D_{ij}^{\xi} = \begin{bmatrix}
\omega_1 \psi_{11}, \omega_2 \psi_{12}, \cdots, \omega_{M_C} \psi_{1{M_C}}
\end{bmatrix}
\begin{bmatrix}
\frac{\partial \psi_{11}}{\partial \xi} \cr \frac{\partial \psi_{12}}{\partial \xi} \cr
\cdots \cr
\frac{\partial \psi_{1{M_C}}}{\partial \xi}
\end{bmatrix}
\end{equation}\]

因此

\[D^{\xi} = \begin{bmatrix}
\omega_1 \psi_{11}, \omega_2 \psi_{12}, \cdots, \omega_{M_C} \psi_{1{M_C}} \cr
\omega_1 \psi_{21}, \omega_2 \psi_{22}, \cdots, \omega_{M_C} \psi_{2{M_C}} \cr
\cdots \cr
\omega_1 \psi_{{M_C}1}, \omega_2 \psi_{{M_C}2}, \cdots, \omega_{M_C} \psi_{{M_C}{M_C}} \cr
\end{bmatrix}
\begin{bmatrix}
\frac{\partial \psi_{11}}{\partial \xi}, & \frac{\partial \psi_{21}}{\partial \xi}, & \cdots & \frac{\partial \psi_{{M_C}1}}{\partial \xi} \cr \frac{\partial \psi_{12}}{\partial \xi}, & \frac{\partial \psi_{22}}{\partial \xi}, & \cdots & \frac{\partial \psi_{{M_C}2}}{\partial \xi} \cr
\cdots \cr
\frac{\partial \psi_{1{M_C}}}{\partial \xi}, & \frac{\partial \psi_{2{M_C}}}{\partial \xi}, & \cdots & \frac{\partial \psi_{{M_C}{M_C}}}{\partial \xi}
\end{bmatrix}\]

因此

\[\hat{D}^{\xi} = M^{-1} \begin{bmatrix}
\omega_1 \psi_{11}, \omega_2 \psi_{12}, \cdots, \omega_{M_C} \psi_{1{M_C}} \cr
\omega_1 \psi_{21}, \omega_2 \psi_{22}, \cdots, \omega_{M_C} \psi_{2{M_C}} \cr
\cdots \cr
\omega_1 \psi_{{M_C}1}, \omega_2 \psi_{{M_C}2}, \cdots, \omega_{M_C} \psi_{{M_C}{M_C}} \cr
\end{bmatrix}
\begin{bmatrix}
\frac{\partial \psi_{11}}{\partial \xi}, & \frac{\partial \psi_{21}}{\partial \xi}, & \cdots & \frac{\partial \psi_{{M_C}1}}{\partial \xi} \cr \frac{\partial \psi_{12}}{\partial \xi}, & \frac{\partial \psi_{22}}{\partial \xi}, & \cdots & \frac{\partial \psi_{{M_C}2}}{\partial \xi} \cr
\cdots \cr
\frac{\partial \psi_{1{M_C}}}{\partial \xi}, & \frac{\partial \psi_{2{M_C}}}{\partial \xi}, & \cdots & \frac{\partial \psi_{{M_C}{M_C}}}{\partial \xi}
\end{bmatrix}\]

Reference:

[1]: GIRALDO F X, WARBURTON T. A high-order triangular discontinuous Galerkin oceanic shallow water model[J]. International Journal for Numerical Methods in Fluids, 2008, 56: 899–925.

体积与边精确积分DGM方法的更多相关文章

  1. Window中C++进行精确计时的方法

    嗯,程序员一个永恒的追求就是性能吧? 为了衡量性能,自然需要计时. 奈何无论C标准库还是C++标准库,因为通用性的考虑,其time API精度都不高.基本都是毫秒级的. 所以如果要真正精确地衡量程序的 ...

  2. 减小delphi体积的方法

    1.关闭RTTI反射机制  自从Delphi2010中引入了新的RTTI反射机制后,编译出来的程序会变得很大,这是因为默认情况下 Delphi2010 给所有类都加上了反射机制.而我们的工程并不每每都 ...

  3. 浅析人脸检测之Haar分类器方法:Haar特征、积分图、 AdaBoost 、级联

    浅析人脸检测之Haar分类器方法 一.Haar分类器的前世今生 人脸检测属于计算机视觉的范畴,早期人们的主要研究方向是人脸识别,即根据人脸来识别人物的身份,后来在复杂背景下的人脸检测需求越来越大,人脸 ...

  4. 积分从入门到放弃<2>

    这部分重新从定积分学了 1,lnx 的导数就是x^(-1) = 1/x 那么求∫(1/x)dx = ln|x|+C  2,初值问题.就是求∫f(x)dx = F(x) + C 求C . 3,Houdi ...

  5. STM32 精确输出PWM脉冲数控制电机(转)

    STM32 精确输出PWM脉冲数控制电机 发脉冲两种目的1)速度控制2)位置控制 速度控制目的和模拟量一样,没有什么需要关注的地方发送脉冲方式为PWM,速率稳定而且资源占用少 stm32位置控制需要获 ...

  6. 浅谈人脸检测之Haar分类器方法

    我们要探讨的Haar分类器实际上是Boosting算法(提升算法)的一个应用,Haar分类器用到了Boosting算法中的AdaBoost算法,只是把AdaBoost算法训练出的强分类器进行了级联,并 ...

  7. JavaScript中判断对象类型方法大全1

    我们知道,JavaScript中检测对象类型的运算符有:typeof.instanceof,还有对象的constructor属性: 1) typeof 运算符 typeof 是一元运算符,返回结果是一 ...

  8. 浅析人脸检测之Haar分类器方法

    一.Haar分类器的前世今生 人脸检测属于计算机视觉的范畴,早期人们的主要研究方向是人脸识别,即根据人脸来识别人物的身份,后来在复杂背景下的人脸检测需求越来越大,人脸检测也逐渐作为一个单独的研究方向发 ...

  9. JavaScript中判断对象类型的种种方法

    我们知道,JavaScript中检测对象类型的运算符有:typeof.instanceof,还有对象的constructor属性: 1) typeof 运算符 typeof 是一元运算符,返回结果是一 ...

随机推荐

  1. 【UE4 C++】学习笔记汇总

    UE4 概念知识 基础概念--文件结构.类型.反射.编译.接口.垃圾回收.序列化[导图] GamePlay架构[导图] 类的继承层级关系[导图] 反射机制 垃圾回收机制/算法 序列化 Actor 的生 ...

  2. 264.丑数II

    题目 给你一个整数 n ,请你找出并返回第 n 个 丑数 . 丑数 就是只包含质因数 2.3 和/或 5 的正整数. 示例 1: 输入:n = 10 输出:12 解释:[1, 2, 3, 4, 5, ...

  3. PinPoint单节点部署及客户端配置方法

    在一次做项目中,需要涉及全链路压测,为了更好定位链路中某一节点可能会出现的问题,在繁忙之余,快速部署及应用了该链路工具,分享给大家~ 话不多说,开始部署~ 一.环境配置1.1 获取需要的依赖包进入ho ...

  4. pyqgis环境配置

    配置pyqgis开发环境时,很多网上教程写的非常繁琐,这里仅仅找了一个最简单的配置方法,使用pycharm的IDE,安装QGIS软件后,在pycharm的ProjectInterpreter里面填写Q ...

  5. shell脚本自学笔记

    一. 什么是Shell脚本 shell脚本并不能作为正式的编程语言,因为它是在linux的shell中运行的,所以称为shell脚本.事实上,shell脚本就是一些命令的集合. 假如完成某个需求需要一 ...

  6. 21.6.4 test

    \(NOI\) 模拟赛 太离谱了,碳基生物心态极限 \(T1\),字符串滚出OI,最后想了个区间dp,期望得分32pts,实际得分0pts,不知为啥挂了.正解是没学过的SAM. \(T2\),正解博弈 ...

  7. IM服务器:我的千万级在线聊天服务器集群

    一.服务器特点 01.傻瓜式部署,一键式启动: 02.单机支持10万以上在线用户聊天(8G内存,如果内存足够大,并发量可超过10万): 03.支持服务器集群,集群间高内聚.低耦合,可动态横向扩展IM服 ...

  8. 检查是否是BST 牛客网 程序员面试金典 C++ java Python

    检查是否是BST 牛客网 程序员面试金典  C++ java Python 题目描述 请实现一个函数,检查一棵二叉树是否为二叉查找树. 给定树的根结点指针TreeNode* root,请返回一个boo ...

  9. 2021CCPC华为云挑战赛 部分题题解

    CDN流量调度问题 题看了没多久就看出来是\(DP\)的题,然后就设了状态\(f[i][j]\)表示到前\(i\)个点时已经用了\(j\)个节点的最小总代价,结果发现转移时\(O(nm^2)\),但这 ...

  10. char* 和 char[] 的区别

    一.代码 有关下面代码,p和q的区别是什么: int main(int argc, char *argv[]) { char* p = "Hello World"; char q[ ...