# 数学代写|数值分析代写numerical analysis代考|CIVL5458

## 数学代写|数值分析代写numerical analysis代考|Linear Iterative Methods

Definition $6.1$ (iterative method). Let $A \in \mathbb{C}^{n \times n}$ with $\operatorname{det}(\mathrm{A}) \neq 0$ and $f \in \mathbb{C}^n$. An iterative method to find an approximate solution to $A x=f$ is a process to generate a sequence of approximations $\left{\boldsymbol{x}k\right}{k=1}^{\infty}$ via an iteration of the form
$$x_k=\varphi\left(\mathrm{A}, f, x_{k-1}, \ldots, x_{k-r}\right),$$
given the starting values $\boldsymbol{x}0, \ldots, \boldsymbol{x}{r-1} \in \mathbb{C}^n$. Here,
$$\varphi(\cdot, \cdot \ldots \cdot \cdot): \mathbb{C}^{n \times n} \times \mathbb{C}^n \times \cdots \times \mathbb{C}^n \rightarrow \mathbb{C}^n$$
is called the iteration function. If $r=1$, we say that the process is a two-layer method; otherwise, we say it is a multilayer method.

Definition $6.2$ (linear iterative method). Let $\mathrm{A} \in \mathbb{C}^{n \times n}$ with $\operatorname{det}(\mathrm{A}) \neq 0$ and $\boldsymbol{f} \in \mathbb{C}^{\prime \prime}$. Set $\boldsymbol{x}=\mathrm{A}^{-1} \boldsymbol{f}$. I he two-layer iterative method
$$x_k=\varphi\left(\mathrm{A}, \boldsymbol{f}, \boldsymbol{x}{k-1}\right)$$ is said to be consistent if and only if $x-\varphi(\mathrm{A}, f, x)$, i.e., $x-\mathrm{A}^1 f$ is a fixed point of $\varphi(\mathbf{A}, \boldsymbol{f}, \cdot)$. The method is linear if and only if $$\varphi\left(\mathrm{A}, \alpha \boldsymbol{f}_1+\beta \boldsymbol{f}_2, \alpha \boldsymbol{x}_1+\beta \boldsymbol{x}_2\right)=\alpha \varphi\left(\mathrm{A}, \boldsymbol{f}_1, \boldsymbol{x}_1\right)+\beta \varphi\left(\mathrm{A}, \boldsymbol{f}_2, \boldsymbol{x}_2\right)$$ for all $\alpha, \beta \in \mathbb{C}$ and $\boldsymbol{f}_1, \boldsymbol{f}_2, \boldsymbol{x}_1, \boldsymbol{x}_2 \in \mathbb{C}^n$. Proposition $6.3$ (general form). Let $\mathrm{A} \in \mathbb{C}^{n \times n}$ with $\operatorname{det}(\mathrm{A}) \neq 0$ and $f \in \mathbb{C}^n$. Any two-layer, linear, and consistent method can be written in the form $$x{k+1}=x_k+\mathrm{Cr}\left(x_k\right)=x_k+\mathrm{C}\left(f-A x_k\right)$$
for some matrix $\mathrm{C} \in \mathbb{C}^{n \times n}$, where $r(z)=f-\mathrm{A} z$ is the residual vector.
Proof. A two-layer method is defined by an iteration function
$$\varphi(\cdot, \cdot, \cdot): \mathbb{C}^{n \times n} \times \mathbb{C}^n \times \mathbb{C}^n \rightarrow \mathbb{C}^n .$$
Given $\varphi$, define the operator
$$C=\varphi(\mathbf{A}, \mathbf{7}, \mathbf{0})$$

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

The theory in this section is based on properties of the spectrum $\sigma(\mathrm{T})$ of the error transfer matrix. In particular, with the tools developed in Section 4.1, we can provide conditions for a linear method to be convergent.

Theorem $6.8$ (convergence of linear methods). Suppose that $\mathrm{A}, \mathrm{B} \in \mathbb{C}^{n \times n}$ are invertible, $f, \boldsymbol{x}_0 \in \mathbb{C}^n$ are given, and $\boldsymbol{x}=\mathrm{A}^{-1} \boldsymbol{f}$.

1. The sequence $\left{\boldsymbol{x}k\right}{k=1}^{\infty}$ defined by the linear two-layer stationary iterative method (6.2) converges to $x$ for any starting point $x_0$ if and only if $\rho(T)<1$, where $T$ is the error transfer matrix $\mathrm{T}=\mathrm{I}_n-\mathrm{B}^{-1} \mathrm{~A}$.
2. A sufficient condition for the convergence of $\left{x_k\right}_{k=1}^{\infty}$ for any starting point $x_0$ is the condition that $|\mathrm{T}|<1$ for some induced matrix norm.
Proof. Before we begin the proof, observe that
$$\boldsymbol{e}k=T \boldsymbol{e}{k-1}=T^2 \boldsymbol{e}{k-2}=\cdots=T^k e_0 .$$ Also, observe that $\boldsymbol{x}_k \rightarrow \boldsymbol{x}=\mathrm{A}^{-1} \boldsymbol{f}$, as $k \rightarrow \infty$, if and only if $\boldsymbol{e}_k \rightarrow \mathbf{0}$, as $k \rightarrow \infty$. Suppose that $\boldsymbol{x}_k \rightarrow \boldsymbol{x}=\mathrm{A}^{-1} \boldsymbol{f}$, as $k \rightarrow \infty$, for any $\boldsymbol{x}_0$. Then $\boldsymbol{e}_k \rightarrow \mathbf{0}$, as $k \rightarrow \infty$, for any $\boldsymbol{e}_0$. Set $\boldsymbol{e}_0-\boldsymbol{w}$, where $(\lambda, \boldsymbol{w})$ is any eigenpair of $\mathrm{T}$, with $|\boldsymbol{w}|{\infty}-1$. Then
$$\boldsymbol{e}k=\lambda^k \boldsymbol{e}_0$$ and $$|\lambda|^k=|\lambda|^k|\boldsymbol{w}|{\infty}=\left|\boldsymbol{e}k\right|{\infty} \rightarrow 0 .$$
It follows that $|\lambda|<1$. Since $\lambda$ was arbitrary, $\rho(T)<1$.
If $\rho(T)<1$, appealing to Theorem $4.8$,
$$\lim {k \rightarrow \infty} e_k=\lim {k \rightarrow \infty} T^k e_0=\mathbf{0}$$
for any $\boldsymbol{e}_0$. Hence, $\boldsymbol{x}_k \rightarrow \boldsymbol{x}=\mathrm{A}^{-1} \boldsymbol{f}$, as $k \rightarrow \infty$, for any $\boldsymbol{x}_0$.

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|Linear Iterative Methods

$$x_k=\varphi\left(\mathrm{A}, f, x_{k-1}, \ldots, x_{k-r}\right),$$

$$\varphi(\cdot, \cdots \cdots): \mathbb{C}^{n \times n} \times \mathbb{C}^n \times \cdots \times \mathbb{C}^n \rightarrow \mathbb{C}^n$$

$$x_k=\varphi(\mathrm{A}, \boldsymbol{f}, \boldsymbol{x} k-1)$$

$$\varphi\left(\mathrm{A}, \alpha \boldsymbol{f}_1+\beta \boldsymbol{f}_2, \alpha \boldsymbol{x}_1+\beta \boldsymbol{x}_2\right)=\alpha \varphi\left(\mathrm{A}, \boldsymbol{f}_1, \boldsymbol{x}_1\right)+\beta \varphi\left(\mathrm{A}, \boldsymbol{f}_2, \boldsymbol{x}_2\right)$$

$$x k+1=x_k+\mathrm{Cr}\left(x_k\right)=x_k+\mathrm{C}\left(f-A x_k\right)$$

$$\varphi(\cdot, \cdot, \cdot): \mathbb{C}^{n \times n} \times \mathbb{C}^n \times \mathbb{C}^n \rightarrow \mathbb{C}^n .$$

$$C=\varphi(\mathbf{A}, \mathbf{7}, \mathbf{0})$$

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

1. 序列 $\$ \backslash$left{\boldsymbol{x}$k \mid$right${k=1} \wedge{$infty} definedbythelineartwo – layerstationaryiterativemethod(6.2) convergestoX foranystartingpointx_0ifandonlyi$\backslash \mathrm{rho}(\mathrm{T})<1$, where 吨 istheerrortransfermatrix$\backslash \operatorname{mathrm}{T}=\backslash \operatorname{mathrm{ 1} } \mathrm{n} \backslash \backslash \operatorname{mathrm}{B} \wedge{-1} \backslash \operatorname{mathrm}{\sim A} \$$。 2. 收敛的充分条件 \backslash left{{_k\right} {k \mathrm{k}=1}^{\wedge}{ {infty} } 对于任何起点 x_0 条件是 |\mathrm{T}|<1 对于一些诱导矩阵 范数。 证明。在我们开始证明之前，观察 \ \$$
$\backslash$ boldsymbol ${e} k=T \backslash$ boldsymbol ${e}{k-1}=T^{\wedge} 2 \backslash$ boldsymbol{e} ${k-2}=\backslash c$ dots $=T^{\wedge} k$ e_ 0 .
Also, observethat $\$ \boldsymbol{x}k \rightarrow \boldsymbol{x}=\mathrm{A}^{-1} \boldsymbol{f} \$$, as \ k \rightarrow \infty \$$, ifandonlyif $\$ \boldsymbol{e}_k \rightarrow \mathbf{0} \$$, as \ k \rightarrow \infty \backslash boldsymbol{e} k=\mid l a m b d a^{\wedge} k \backslash |boldsymbol{e}_O and \left.|| \backslash a m b d a\right|^{\wedge} k=\left.|| \backslash a m b d a\right|^{\wedge} k|\backslash b o l d s y m b o l{w}| \backslash infty }=\backslash left||lboldsymbol{e } k \mid right \mid \bigwedge infty } \rightarrow 0 。 Itfollowsthat \|\lambda|<1 \$$.Since $\$ \lambda \$$wasarbitrary \ \rho(T)<1 \$$. If $\$ \rho(T)<1 \$$, appealin \backslash \lim {k \backslash rightarrow \backslash infty } e k=\backslash lim {k \backslash rightarrow \backslash infty } T^{\wedge} k e_{-} 0=\backslash mathbf {0} \ \$$
对于任何 $\boldsymbol{e}_0$. 因此， $\boldsymbol{x}_k \rightarrow \boldsymbol{x}=\mathrm{A}^{-1} \boldsymbol{f}$ ， 作为 $k \rightarrow \infty$ ，对于任何 $\boldsymbol{x}_0$.

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