### 数学代写|计算线性代数代写Computational Linear Algebra代考|MATHS 2104

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|A Two Point Boundary Value Problem

Consider the simple two point boundary value problem
$$-u^{\prime \prime}(x)=f(x), \quad x \in[0,1], \quad u(0)=0, u(1)=0$$
where $f$ is a given continuous function on $[0,1]$ and $u$ is an unknown function. This problem is also known as the one-dimensional (1D) Poisson problem. In principle it is easy to solve $(2.20)$ exactly. We just integrate $f$ twice and determine the two integration constants so that the homogeneous boundary conditions $u(0)=u(1)=$ 0 are satisfied. For example, if $f(x)=1$ then $u(x)=x(x-1) / 2$ is the solution.
Suppose $f$ cannot be integrated exactly. Problem $(2.20)$ can then be solved approximately using the finite difference method. We need a difference approximation to the second derivative. If $g$ is a function differentiable at $x$ then
$$g^{\prime}(x)=\lim _{h \rightarrow 0} \frac{g\left(x+\frac{h}{2}\right)-g\left(x-\frac{h}{2}\right)}{h}$$ and applying this to a function $u$ that is twice differentiable at $x$
\begin{aligned} u^{\prime \prime}(x) &=\lim {h \rightarrow 0} \frac{u^{\prime}\left(x+\frac{h}{2}\right)-u^{\prime}\left(x-\frac{h}{2}\right)}{h}=\lim {h \rightarrow 0} \frac{\frac{u(x+h)-u(x)}{h}-\frac{u(x)-u(x-h)}{h}}{h} \ &=\lim {h \rightarrow 0} \frac{u(x+h)-2 u(x)+u(x-h)}{h^{2}} \end{aligned} To define the points where this difference approximation is used we choose a positive integer $m$, let $h:=1 /(m+1)$ be the discretization parameter, and replace the interval $[0,1]$ by grid points $x{j}:=j h$ for $j=0,1, \ldots, m+1$. We then obtain approximations $v_{j}$ to the exact solution $u\left(x_{j}\right)$ for $j=1, \ldots, m$ by replacing the differential equation by the difference equation
$$\frac{-v_{j-1}+2 v_{j}-v_{j+1}}{h^{2}}=f(j h), \quad j=1, \ldots, m, \quad v_{0}=v_{m+1}=0$$
Moving the $h^{2}$ factor to the right hand side this can be written as an $m \times m$ linear system

The matrix $T$ is called the second derivative matrix and will occur frequently in this book. It is our second example of a tridiagonal matrix, $T=\operatorname{tridiag}\left(a_{i}, d_{i}, c_{i}\right) \in$ $\mathbb{R}^{m \times m}$, where in this case $a_{i}=c_{i}=-1$ and $d_{i}=2$ for all $i$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Diagonal Dominance

We want to show that $(2.21)$ has a unique solution. Note that $T$ is not strictly diagonally dominant. However, $T$ is weakly diagonally dominant in accordance with the following definition.

Definition $2.3$ (Diagonal Dominance) The matrix $A=\left[a_{i j}\right] \in \mathbb{C}^{n \times n}$ is weakly diagonally dominant if
$$\left|a_{i i}\right| \geq \sum_{j \neq i}\left|a_{i j}\right|, i=1, \ldots, n$$

We showed in Theorem $2.2$ that a strictly diagonally dominant matrix is nonsingular. This is in general not true in the weakly diagonally dominant case. Consider the 3 matrices
$$\boldsymbol{A}{1}=\left[\begin{array}{lll} 1 & 1 & 0 \ 1 & 2 & 1 \ 0 & 1 & 1 \end{array}\right], \quad \boldsymbol{A}{2}=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{array}\right], \quad \boldsymbol{A}{3}=\left[\begin{array}{rrr} 2 & -1 & 0 \ -1 & 2 & -1 \ 0 & -1 & 2 \end{array}\right]$$ They are all weakly diagonally dominant, but $A{1}$ and $A_{2}$ are singular, while $A_{3}$ is nonsingular. Indeed, for $A_{1}$ column two is the sum of columns one and three, $A_{2}$ has a zero row, and $\operatorname{det}\left(\boldsymbol{A}_{3}\right)=4 \neq 0$. It follows that for the nonsingularity and existence of an LU factorization of a weakly diagonally dominant matrix we need some additional conditions. Here are some sufficient conditions.

Theorem 2.4 (Weak Diagonal Dominance) Suppose $\boldsymbol{A}=\operatorname{tridiag}\left(a_{i}, d_{i}, c_{i}\right) \in$ $\mathbb{C}^{n \times n}$ is tridiagonal and weakly diagonally dominant. If in addition $\left|d_{1}\right|>\left|c_{1}\right|$ and $a_{i} \neq 0$ for $i=1, \ldots, n-2$, then $\boldsymbol{A}$ has a unique $L U$ factorization (2.15). If in addition $d_{n} \neq 0$, then $\boldsymbol{A}$ is nonsingular.

Proof The proof is similar to the proof of Theorem 2.2. The matrix $\boldsymbol{A}$ has an LU factorization if the $u_{k}$ ‘s in (2.16) are nonzero for $k=1, \ldots, n-1$. For this it is sufficient to show by induction that $\left|u_{k}\right|>\left|c_{k}\right|$ for $k=1, \ldots, n-1$. By assumption $\left|u_{1}\right|=\left|d_{1}\right|>\left|c_{1}\right|$. Suppose $\left|u_{k}\right|>\left|c_{k}\right|$ for some $1 \leq k \leq n-2$. Then $\left|c_{k}\right| /\left|u_{k}\right|<1$ and by (2.16) and since $a_{k} \neq 0$ $$\left|u_{k+1}\right|=\left|d_{k+1}-l_{k} c_{k}\right|=\left|d_{k+1}-\frac{a_{k} c_{k}}{u_{k}}\right| \geq\left|d_{k+1}\right|-\frac{\left|a_{k}\right|\left|c_{k}\right|}{\left|u_{k}\right|}>\left|d_{k+1}\right|-\left|a_{k}\right| .$$
This also holds for $k=n-1$ if $a_{n-1} \neq 0$. By (2.23) and weak diagonal dominance $\left|u_{k+1}\right|>\left|d_{k+1}\right|-\left|a_{k}\right| \geq\left|c_{k+1}\right|$ and it follows by induction that an LU factorization exists. It is unique since any LU factorization must satisfy (2.16). For the nonsingularity we need to show that $u_{n} \neq 0$. For then by Lemma $2.5$, both $\boldsymbol{L}$ and $\boldsymbol{U}$ are nonsingular, and this is equivalent to $\boldsymbol{A}=\boldsymbol{L} \boldsymbol{U}$ being nonsingular. If $a_{n-1} \neq 0$ then by (2.16) $\left|u_{n}\right|>\left|d_{n}\right|-\left|a_{n-1}\right| \geq 0$ by weak diagonal dominance, while if $a_{n-1}=0$ then again by (2.23) $\left|u_{n}\right| \geq\left|d_{n}\right|>0$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|The Buckling of a Beam

Consider a horizontal beam of length $L$ located between 0 and $L$ on the $x$-axis of the plane. We assume that the beam is fixed at $x=0$ and $x=L$ and that a force $F$ is applied at $(L, 0)$ in the direction towards the origin. This situation can be modeled by the boundary value problem
$$R y^{\prime \prime}(x)=-F y(x), \quad y(0)=y(L)=0,$$
where $y(x)$ is the vertical displacement of the beam at $x$, and $R$ is a constant defined by the rigidity of the beam. We can transform the problem to the unit interval $[0,1]$ by considering the function $u:[0,1] \rightarrow \mathbb{R}$ given by $u(t):=y(t L)$. Since $u^{\prime \prime}(t)=$ $L^{2} y^{\prime \prime}(t L)$, the problem $(2.24)$ then becomes
$$u^{\prime \prime}(t)=-K u(t), \quad u(0)=u(1)=0, \quad K:=\frac{F L^{2}}{R} .$$
Clearly $u=0$ is a solution, but we can have nonzero solutions corresponding to certain values of the $\mathrm{K}$ known as eigenvalues. The corresponding function $u$ is called an eigenfunction. If $F=0$ then $K=0$ and $u=0$ is the only solution, but if the force is increased it will reach a critical value where the beam will buckle and maybe break. This critical value corresponds to the smallest eigenvalue of (2.25). With $u(t)=\sin (\pi t)$ we find $u^{\prime \prime}(t)=-\pi^{2} u(t)$ and this $u$ is a solution if $K=\pi^{2}$. It can be shown that this is the smallest eigenvalue of (2.25) and solving for $F$ we find $F=\frac{\pi^{2} R}{L^{2}}$.

We can approximate this eigenvalue numerically. Choosing $m \in \mathbb{N}, h:=1 /(m+$
1) and using for the second derivative the approximation
$$u^{\prime \prime}(j h) \approx \frac{u((j+1) h)-2 u(j h)+u((j-1) h)}{h^{2}}, \quad j=1, \ldots, m,$$
(this is the same finite difference approximation as in Sect. 2.2) we obtain
$$\frac{-v_{j-1}+2 v_{j}-v_{j+1}}{h^{2}}=K v_{j}, \quad j=1, \ldots, m, h=\frac{1}{m+1}, \quad v_{0}=v_{m+1}=0$$

where $v_{j} \approx u(j h)$ for $j=0, \ldots, m+1$. If we define $\lambda:=h^{2} K$ then we obtain the equation
$$T v=\lambda v, \text { with } v=\left[v_{1}, \ldots, v_{m}\right]^{T}$$
and
The problem now is to determine the eigenvalues of $T$. Normally we would need a numerical method to determine the eigenvalues of a matrix, but for this simple problem the eigenvalues can be determined exactly. We show in the next subsection that the smallest eigenvalue of $(2.26)$ is given by $\lambda=4 \sin ^{2}(\pi h / 2)$. Since $\lambda=$ $h^{2} K=\frac{h^{2} F L^{2}}{R}$ we can solve for $F$ to obtain
$$F=\frac{4 \sin ^{2}(\pi h / 2) R}{h^{2} L^{2}}$$
For small $h$ this is a good approximation to the value $\frac{\pi^{2} R}{L^{2}}$ we computed above.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|A Two Point Boundary Value Problem

−在′′(X)=F(X),X∈[0,1],在(0)=0,在(1)=0

G′(X)=林H→0G(X+H2)−G(X−H2)H并将其应用于函数在在X

−在j−1+2在j−在j+1H2=F(jH),j=1,…,米,在0=在米+1=0

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Diagonal Dominance

|一个一世一世|≥∑j≠一世|一个一世j|,一世=1,…,n

|在ķ+1|=|dķ+1−lķCķ|=|dķ+1−一个ķCķ在ķ|≥|dķ+1|−|一个ķ||Cķ||在ķ|>|dķ+1|−|一个ķ|.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|The Buckling of a Beam

R是′′(X)=−F是(X),是(0)=是(大号)=0,

1) 并使用二阶导数的近似值

（这与第 2.2 节中的有限差分近似相同）我们得到

−在j−1+2在j−在j+1H2=ķ在j,j=1,…,米,H=1米+1,在0=在米+1=0

F=4罪2⁡(圆周率H/2)RH2大号2

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