### 物理代写|理论力学代写theoretical mechanics代考|MATH3977

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学theoretical mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学theoretical mechanics代写方面经验极为丰富，各种代写理论力学theoretical mechanics相关的作业也就用不着说。

• 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

## 物理代写|理论力学代写theoretical mechanics代考|Infinite Periodic System. Plane Problem

The solution to the plane elasticity theory for the infinite periodic systems by the developed semi-analytical method is presented in $[16,17]$. Let us cite here only the properties of the kernel for respective integral equations and the discretization scheme.

As indicated above, it is necessary to consider the auxiliary integral equation, for which we should study the properties of its kernel $[16,17]$ :
\begin{aligned} &\frac{1}{2 a} \int_{-b}^{b} h(\eta) K(y-\eta) d \eta=1 ; K(y)=\sum_{n=1}^{\infty} L_{n} \cos \left(a_{n} y\right) ; L_{n}=\frac{R_{n}}{k_{2}^{2} q_{n}},|y|<b \ &q_{n}=\left[(\pi n / a)^{2}-k_{1}^{2}\right]^{1 / 2}, r_{n}=\left[(\pi n / a)^{2}-k_{2}^{2}\right]^{1 / 2}, \end{aligned}

$$R_{n}=\left[2 a_{n}^{2}-k_{2}^{2}\right]^{2}-4 r_{n} q_{n} a_{n}^{2}, \quad a_{n}=\pi n / a$$
Here $k_{1}, k_{2}$-wave numbers for the longitudinal and the transverse waves. Let us notice that $L_{n} \sim-2\left(1-c_{2}^{2} / c_{1}^{2}\right) a_{n}, n \rightarrow \infty$, where $c_{1}, c_{2}$-the speed of the longitudinal and the transverse wave, respectively. Then the expression for the kernel is transformed to the following form
\begin{aligned} K(y) &=-2\left(1-\frac{c_{2}^{2}}{c_{1}^{2}}\right) \sum_{n=1}^{\infty} a_{n} \cos \left(a_{n} y\right)+\sum_{n=1}^{\infty}\left[L_{n}+2\left(1-\frac{c_{2}^{2}}{c_{1}^{2}}\right) a_{n}\right] \cos \left(a_{n} y\right) \ &=-2\left(1-\frac{c_{2}^{2}}{c_{1}^{2}}\right) I(y)+K_{r}(y) \end{aligned}
Here the second sum is a certain regular function. The first one has both regular and singular parts: $I(y)=\left[I_{r}(y)+I_{s}(y)\right]$. After some transformations of the kernel (15) of the auxiliary integral Eq. (14) the regular and the singular parts become, respectively
$$I_{r}(y)=\frac{a}{\pi y^{2}}-\frac{\pi}{4 a \sin ^{2}(\pi y / 2 a)} ; I_{s}(y)=-\frac{a}{\pi y^{2}}$$

## 物理代写|理论力学代写theoretical mechanics代考|Finite Periodic System. Scalar Formulation

In order to solve the problem in the scalar case, we first consider the incidence of a plane wave upon a doubly-periodic system of rigid screens, which is finite in the both directions. In frames of the scalar acoustics, the wave equation for full acoustic pressure $p$ is reduced to the Helmholtz equation
$$\left(\Delta+k^{2}\right) \mathbf{p}=0$$
where $k$-the wave number of the acoustic wave, $\Delta$ denotes the two-dimensional Laplace operator, and the full wave pressure is a linear sum of the incident and the scattered field: $\mathbf{p}=\mathbf{p}^{i n c}+\mathbf{p}^{s c}$. To be more specific, let us restrict the consideration by the normal incidence of a plane wave, hence the incident wave field is $\mathbf{p}^{i n c}\left(y^{0}\right)=$ $\mathrm{e}^{i k y_{1}^{0}}$, where the two-dimensional point is $y^{0}=\left(y_{1}^{0}, y_{2}^{0}\right)$.

The boundary condition, in the case of acoustically hard boundary $\bar{L}$ has the form
$$\left.\frac{\partial \mathbf{p}}{\partial \mathbf{n}{y}}\right|{\tilde{L}}=0, \quad(y \in \tilde{L})$$

Here $\mathbf{n}{y}$ is the unit normal vector at the point $y$, and $\bar{L}=\sum{m=1}^{M} \tilde{l}_{m}$ represents itself the full set of boundary contours.

In order to develop the basic boundary integral equation, let us introduce a respective closed contour $l_{m}$ around the current screen. Obviously, for the given contours, in the case when the observation point $x$ is outside, the following standard integral representation is valid
$$p^{s c}\left(y^{0}\right)=\int_{L}\left(p(y) \frac{\partial \Phi}{\partial n_{y}}-\frac{\partial p(y)}{\partial n_{y}} \Phi\right) d L_{y}, \quad(y \in L)$$
where $\Phi=\Phi(r)$ is the Green’s function, which in the two-dimensional acoustic case is expressed through the Hankel function of the first kind $\Phi(r)=(i / 4) H_{0}^{(1)}(r), r=$ $\left|y-y^{0}\right|$

If each surrounding closed contour converges to the respective rigid screen located inside, then the second term in $(22)$ is cancelled, due to the boundary condition. The opposite sides of each obstacle are considered separately, being $l_{m}^{-}$и $l_{m}^{+}$, where the sign “plus” is related to the normal $\mathbf{n}{m}^{+}$, directed along the propagation of the incident wave, and the negative sign-oppositely. Then, the integral representation (22) can be reduced to the expression \begin{aligned} \mathbf{p}^{s c}\left(y^{0}\right) &=\sum{m=1}^{M}\left(\int_{\ell_{w}^{+}}\left(\mathbf{p}^{+}(y) \frac{\partial \Phi}{\partial \mathbf{n}{y}^{+}}\right) d \ell{y}^{+}+\int\left(\mathbf{p}^{-}(y) \frac{\partial \Phi}{\partial \mathbf{n}{y}^{-}}\right) d \ell{y}^{-}\right) \ &=\sum_{m=1}^{M} \int_{\ell_{m}^{+}} \mathbf{g}(y) \frac{\partial \Phi}{\partial \mathbf{n}{y}^{+}}\left(k\left|y-y^{0}\right|\right) d \ell{y}^{+}, \mathbf{g}(y)=\mathbf{p}^{+}(y)-\mathbf{p}^{-}(y), \quad\left(y \in l_{m}\right) \end{aligned}

## 物理代写|理论力学代写theoretical mechanics代考|Numerical Analysis

Let us perform a numerical analysis of the problems considered above, on example of the medium with the longitudinal wave speed $c_{1}=6000 \mathrm{~m} / \mathrm{s}$ (steel), and the ratio of the longitudinal and the transverse wave speeds is $c_{1} / c_{2}=1.87$.

To begin with, let us compare the moduli of reflection and transmission coefficients versus frequency parameter, between the three studied cases, for a single vertical array (see Figs. 2 and 3). With so doing, we assume that the longitudinal wave speed in the problem 2 is equal to the transverse wave speed of the problems 1 and 3. This condition shortens the one-mode frequency interval, whose limit from the right becomes $\pi / 1.87=1.680$, (see Figs. 2 and 3 ). In Figs. $4,5,6,7$ and 8 the comparative numerical analysis of the scalar problems 1 and 3 has been performed for the transverse incident wave. Let us notice that for all cases the filtration interval can be seen in the upper part of the one-mode frequency range. It is shown that lines 2 and 3 in Figs. 2 and 3, related to the scalar problems, are practically coinciding that takes place even for $N=10$ cracks in each vertical array. It should also be noted that line 1 related to the elastic problem, shows a significant domination of the filtration property, when compared with both infinite and finite scalar problems. Let us also notice that for two vertical arrays in the elastic problem a perfect filtration takes place for $a k \geq 0.7$, but for one vertical row this property is valid only for $a k \geq 1.5$; this also confirms the evident property that with the growth of the vertical rows the filtration becomes stronger.

Let us pass to the analysis of the grid size to the precision of the obtained results. It is stated that in the case of a single obstacle it is sufficient to take 10 grid nodes per each wavelength, to provide reliable results. With so doing, for the frequency $0.16 \mathrm{MHz}$ in this formulation the wavelength is $3.75 \mathrm{~cm}$, hence on the obstacle of the length $1.5 \mathrm{~cm}$ it is sufficient to take only 5 nodes. However, the complex geometry of the diffraction lattice requires greater number of nodes. It can be seen from Fig. 4 , which represents the results for the array of 10 vertical rows, each containing 10 obstacles, that with 10 nodes over each obstacle the calculations are correct only in the low-frequency case (for $k^{*} a<1$ ).

## 物理代写|理论力学代写theoretical mechanics代考|Infinite Periodic System. Plane Problem

12一个∫−bbH(这)ķ(是−这)d这=1;ķ(是)=∑n=1∞大号n因⁡(一个n是);大号n=Rnķ22qn,|是|<b qn=[(圆周率n/一个)2−ķ12]1/2,rn=[(圆周率n/一个)2−ķ22]1/2,

Rn=[2一个n2−ķ22]2−4rnqn一个n2,一个n=圆周率n/一个

ķ(是)=−2(1−C22C12)∑n=1∞一个n因⁡(一个n是)+∑n=1∞[大号n+2(1−C22C12)一个n]因⁡(一个n是) =−2(1−C22C12)我(是)+ķr(是)

## 物理代写|理论力学代写theoretical mechanics代考|Finite Periodic System. Scalar Formulation

(Δ+ķ2)p=0

∂p∂n是|大号~=0,(是∈大号~)

psC(是0)=∫大号(p(是)∂披∂n是−∂p(是)∂n是披)d大号是,(是∈大号)

psC(是0)=∑米=1米(∫ℓ在+(p+(是)∂披∂n是+)dℓ是++∫(p−(是)∂披∂n是−)dℓ是−) =∑米=1米∫ℓ米+G(是)∂披∂n是+(ķ|是−是0|)dℓ是+,G(是)=p+(是)−p−(是),(是∈l米)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。