物理代写|空气动力学代写Aerodynamics代考|Multi-dimensional Finite Element Schemes with Linear Elements

Finite element methods are very easy to define for triangular and tetrahedral meshes using piecewise linear trial solutions. Then we introduce basis functions $\phi_j$ that have the value unity at the $j$ th node and zero at all other nodes, as illustrated in Figure 3.8 for a triangular mesh. We can visualize the basis function of each node as a tent surrounding the node. Consider now Laplace’s equation in a triangulated domain $\mathcal{D}$ with Dirichlet boundary conditions on the boundary $\mathcal{B}$,
$$u_{x x}+u_{y y}=0 \text { in } \mathcal{D}$$
$u$ specified on $\mathcal{B}$.

The corresponding weak form is obtained by multiplying (3.45) by a test function $\psi(x, y)$ and integrating by parts to obtain
$$\int_{\mathcal{B}} \psi \nabla u_h \cdot \mathbf{n} d l-\int_{\mathcal{D}} \nabla u_h \cdot \nabla \psi d \mathcal{S}=0$$
where n is the unit normal to the boundary. The trial solution is
$$u_h=\sum_{j=1}^n u_j \phi_j(x, y)$$
Following the Galerkin method, we require $u_h$ to satisfy the weak form (3.46) for each test function $\psi=\phi_j$. With Dirichlet boundary conditions, we only need to solve for $u_j$ at the interior nodes, for which $\phi_j=0$ on $\mathcal{B}$ with the consequence that the boundary integral vanishes. Referring to Figure $3.8$, consider the equation for the node labeled zero. $\nabla \phi_0$ is nonzero only in the neighboring triangles, which are shaded in the figure. Moreover, both $\nabla u_h$ and $\nabla \phi_0$ are constant in each of these triangles. Thus, the residual equation for node 0 reduces to
$$r_0=\sum_{\text {neighbors }} \mathcal{S}k \nabla u{h k} \cdot \nabla \phi_{0 k}=0,$$
where $\mathcal{S}k$ is the area of the $k^{t h}$ triangle, and $\nabla u{h_k}$ and $\nabla \phi_{0_k}$ are the gradients of $u_h$ and $\phi_0$ in that triangle.

The gradient of any piecewise linear function $u_h$ on a triangular mesh can be evaluated very easily using Gauss’ theorem
$$\int_{\mathcal{D}} \frac{\partial u_h}{\partial x} d \mathcal{S}=\oint_{\mathcal{B}} u_h d y$$

物理代写|空气动力学代写Aerodynamics代考|Further Analysis of the Discrete Laplacian

If we consider the stencil illustrated in Figure 3.8, evaluation of formulas (3.47), (3.48), and (3.49) reduces the equation for node 0 to the form
$$r_0=\sum s_{k 0}\left(u_k-u_0\right),$$
where $s_{k 0}$ are the entries of the stiffness matrix between node $k$ and 0 . If we consider edge 20 , say, the total contribution of $u_2$, applying the formulas (3.47) and (3.48) to the triangles 012 and 023 with areas $\mathcal{S}1$ and $\mathcal{S}_2$ respectively, is $s{20}=\frac{1}{4 \mathcal{S}1}\left(\left(y_0-y_1\right)\left(y_2-y_1\right)+\left(x_0-x_1\right)\left(x_2-x_1\right)\right)$ Let $l{k 0}$ be the length of the edge $k 0$. Then, we find that
\begin{aligned} s_{20} &=\frac{l_{01} l_{21} \cos \theta_{012}}{2 l_{01} l_{21} \sin \theta_{012}}+\frac{l_{03} l_{23} \cos \theta_{032}}{2 l_{03} l_{23} \sin \theta_{032}} \ &=\frac{1}{2}\left(\cot \theta_{012}+\cot \theta_{032}\right), \end{aligned}
where $\theta_{012}$ and $\theta_{032}$ are the angles between edges $l_{01}$ and $l_{21}$ and $l_{03}$ and $l_{23}$, as illustrated in Figure 3.9.
In the case of the Poisson equation
$$u_{x x}+u_{y y}=f,$$

物理代写|空气动力学代写空气动力学代考|具有线性元素的多维有限元格式

$$u_{x x}+u_{y y}=0 \text { in } \mathcal{D}$$
$u$在$\mathcal{B}$上指定

$$\int_{\mathcal{B}} \psi \nabla u_h \cdot \mathbf{n} d l-\int_{\mathcal{D}} \nabla u_h \cdot \nabla \psi d \mathcal{S}=0$$
，其中n是边界的法线单位。试验解是
$$u_h=\sum_{j=1}^n u_j \phi_j(x, y)$$

$$r_0=\sum_{\text {neighbors }} \mathcal{S}k \nabla u{h k} \cdot \nabla \phi_{0 k}=0,$$
，其中$\mathcal{S}k$是$k^{t h}$三角形的面积，$\nabla u{h_k}$和$\nabla \phi_{0_k}$是该三角形中$u_h$和$\phi_0$的梯度 任何分段线性函数$u_h$在三角形网格上的梯度可以很容易地用高斯定理求出
$$\int_{\mathcal{D}} \frac{\partial u_h}{\partial x} d \mathcal{S}=\oint_{\mathcal{B}} u_h d y$$

物理代写|空气动力学代写空气动力学代考|离散拉普拉斯算子的进一步分析

$$r_0=\sum s_{k 0}\left(u_k-u_0\right),$$
，其中$s_{k 0}$是节点$k$和0之间的刚度矩阵项。如果我们考虑边20，假设$u_2$的总贡献，分别对面积为$\mathcal{S}1$和$\mathcal{S}2$的三角形012和023应用公式(3.47)和(3.48)，为$s{20}=\frac{1}{4 \mathcal{S}1}\left(\left(y_0-y_1\right)\left(y_2-y_1\right)+\left(x_0-x_1\right)\left(x_2-x_1\right)\right)$设$l{k 0}$为边$k 0$的长度。然后，我们发现
\begin{aligned} s{20} &=\frac{l_{01} l_{21} \cos \theta_{012}}{2 l_{01} l_{21} \sin \theta_{012}}+\frac{l_{03} l_{23} \cos \theta_{032}}{2 l_{03} l_{23} \sin \theta_{032}} \ &=\frac{1}{2}\left(\cot \theta_{012}+\cot \theta_{032}\right), \end{aligned}
，其中$\theta_{012}$和$\theta_{032}$是边$l_{01}$和$l_{21}$以及$l_{03}$和$l_{23}$之间的角度，如图3.9所示。在泊松方程
$$u_{x x}+u_{y y}=f,$$ 的情况下

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