## 数学代写|数值分析代写numerical analysis代考|MATH3003

2023年3月30日

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## 数学代写|数值分析代写numerical analysis代考|Convergence Theory

The abstract Galerkin method for finding $v \in V$ where
$$a(u, v)=b(v) \text { for all } V,$$ is to pick a finite-dimensional space of approximations $V_h \subset V$ and find $v_h \in V_h$ where
(6.3.25) $a\left(u_h, v_h\right)=b\left(v_h\right) \quad$ for all $v_h \in V_h$.
The foundation of the convergence theory of finite-element methods is Céa’s inequality:

Theorem 6.14 (Céa’s inequality). If $a(u, v)$ is a continuous elliptic bilinear form on a Hilbert space $V$, then there is a constant $C$, depending only on $M$ and $\alpha$ for the elliptic form $a(\cdot, \cdot)$, where the true solution $u$ and the solution $u_h$ of $(6.3 .25)$ satisfy
$$\left|u-u_h\right|_V \leq C \min _{v \in V_h}|u-v|_V$$
That is, the error in the solution of the Galerkin method in the $V$ norm is within a constant factor of the approximation error in the $V$ norm by functions in $V_h$.

Proof Suppose that $a(u, u) \geq \alpha|u|_V^2$ for all $u \in V$ with $\alpha>0$ since $a(\cdot, \cdot)$ is elliptic. Suppose also that $|a(u, v)| \leq M|u|_V|v|_V$ since $a(\cdot, \cdot)$ is a continuous bilinear form. Then the Galerkin method (6.3.25) implies that $u_h \in V_h$ where
\begin{aligned} & a\left(u_h, v_h\right)=b\left(v_h\right) \quad \text { for all } v \in V_h \text {, while } \ & a(u, v)=b(v) \quad \text { for all } v \in V \text {. } \ & \end{aligned}
Then
\begin{aligned} \alpha\left|u_h-u\right|_V^2 & \leq a\left(u-u_h, u-u_h\right) \ & =a\left(u-u_h, u-v_h\right) \quad \text { for any } v_h \in V_h \text { since } \ a\left(u-u_h, u_h-v_h\right) & =a\left(u, u_h-v_h\right)-a\left(u_h, u_h-v_h\right) \ & =b\left(u_h-v_h\right)-b\left(u_h-v_h\right)=0 \quad \text { as } u_h-v_h \in V_h \subset V . \end{aligned}

## 数学代写|数值分析代写numerical analysis代考|Conditioning of the Linear Systems

The error bounds arising from Céa’s inequality (Theorem 6.14 ) give the impression that the only issue is the ability to approximate a solution $u \in V$ by functions $u_h \in V_h$. From this point of view, the more functions in $V_h$ the better. However, the linear systems can become extremely ill-conditioned if the basis for $V_h$ is close to being linearly dependent. For example, using the basis $\phi_j(x)=x^{j-1}$ for $j=1,2, \ldots, n$ on $\Omega=(0,1)$ will result in extremely ill-conditioned linear systems, with the condition number growing at least exponentially in $n$. Small errors in the formulation which can include floating point roundoff, or in the numerical solver, can result in potentially large errors in the solution.

Fortunately, well-shaped triangulations can avoid this exponential growth in the condition number under some very mild conditions. To see why, we start with the mass matrix, which we can use to identify near linear dependence of basis functions. For the Poisson problem using a basis $\left{\phi_1, \phi_2, \ldots, \phi_N\right}$, the mass matrix $M_h$ is given by
$$m_{j k}=\int_{\Omega} \phi_j(\boldsymbol{x}) \cdot \phi_k(\boldsymbol{x}) d \boldsymbol{x}$$
while the linear system to solve for $u_h=\sum_{j=1}^N u_j \phi_j$ has the matrix $A_h$ given by
$$a_{j k}=\int_{\Omega} \nabla \phi_j(\boldsymbol{x})^T \nabla \phi_k(\boldsymbol{x}) d \boldsymbol{x},$$
which is often called the stiffness matrix.
For a basis coming from an interpolation scheme on triangles, we look to the basis $\left{\widehat{\phi}1, \ldots, \widehat{\phi}{\widehat{N}}\right}$ for the reference triangle $\widehat{K}$. As these are linearly independent, the mass matrix $\widehat{M}$ for $\widehat{K}$ by itself,
$$\widehat{m}{r s}=\int{\widehat{K}} \widehat{\phi}_r(\widehat{\boldsymbol{x}}) \cdot \widehat{\phi}_s(\widehat{\boldsymbol{x}}) d \widehat{\boldsymbol{x}}$$ must be positive definite. The condition number $\kappa_2(\widehat{M})$ can be used as a measure of the quality of the basis on $\widehat{K}$ (smaller is better), but with a fixed basis on $\widehat{K}$, this is a known and finite quantity. For example, if $\widehat{K}$ is the triangle with vertices $(0,0),(1,0)$, and $(0,1)$, and the basis is linear $\widehat{\phi}_1(x, y)=x, \widehat{\phi}_2(x, y)=y$, and $\widehat{\phi}_3(x, y)=1-x-y$ then
$$\widehat{M}=\frac{1}{24}\left[\begin{array}{lll} 2 & 1 & 1 \ 1 & 2 & 1 \ 1 & 1 & 2 \end{array}\right] ; \quad \kappa_2(\widehat{M})=4$$

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|Convergence Theory

$a(u, v)=b(v)$ for all $V$

(6.3.25) $a\left(u_h, v_h\right)=b\left(v_h\right) \quad$ 对全部 $v_h \in V_h$.

$$\left|u-u_h\right|V \leq C \min {v \in V_h}|u-v|_V$$

$a\left(u_h, v_h\right)=b\left(v_h\right) \quad$ for all $v \in V_h$, while $\quad a(u$

$$\alpha\left|u_h-u\right|_V^2 \leq a\left(u-u_h, u-u_h\right) \quad=a\left(u-u_h\right.$$

## 数学代写|数值分析代写numerical analysis代考|Conditioning of the Linear Systems

$$m_{j k}=\int_{\Omega} \phi_j(\boldsymbol{x}) \cdot \phi_k(\boldsymbol{x}) d \boldsymbol{x}$$

$$a_{j k}=\int_{\Omega} \nabla \phi_j(\boldsymbol{x})^T \nabla \phi_k(\boldsymbol{x}) d \boldsymbol{x}$$

$\widehat{M}$ 为了 $\widehat{K}$ 通过它自己，
$$\widehat{m} r s=\int \widehat{K} \widehat{\phi}_r(\widehat{\boldsymbol{x}}) \cdot \widehat{\phi}_s(\widehat{\boldsymbol{x}}) d \widehat{\boldsymbol{x}}$$

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