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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

## 物理代写|流体力学代写Fluid Mechanics代考|Incompressible Potential Flows

As seen in Chap. 3, an incompressible flow satisfies the condition $D \rho / D t=0$ which, in conjunction with the continuity equation, leads to $\nabla \cdot \boldsymbol{V}=0$. Furthermore, we assume that the flow is irrotational with $\nabla \times V=0$ everywhere in the flow field. This assumption, which significantly simplifies the mathematical treatment of the flow field, allows introduction of a scalar function called the velocity potential $\Phi$, from which the velocity vector and its components are derived as the gradient of the potential $\Phi$ :
$$\boldsymbol{V}=\nabla \Phi, e_i V_i=e_i \frac{\partial \Phi}{\partial x_i} .$$
Expanding the index notation results in:
$$\boldsymbol{V}=e_i \frac{\partial \Phi}{\partial x_i}=e_1 \frac{\partial \Phi}{\partial x_1}+e_2 \frac{\partial \Phi}{\partial x_2}+e_3 \frac{\partial \Phi}{\partial x_3}$$
Inserting Eq. (6.1) into the continuity equation for incompressible flow $\nabla \cdot \boldsymbol{V}=0$, we arrive at:
$$\nabla \cdot \boldsymbol{V}=\nabla \cdot \nabla \Phi=\Delta \Phi=0$$
Equation (6.3) is the Laplace equation decomposed as:
$$\frac{\partial^2 \Phi}{\partial x_i \partial x_i}=0$$
The Laplace equation (6.4) is an elliptic, linear partial differential equation encountered in many branches of engineering and physics such as electromagnetism, heat conduction, and theory of elasticity. It can be solved using appropriate boundary conditions. The introduction of the velocity potential $\Phi$ in conjunction with the Bernoulli equation having a constant that has the same value everywhere in the flow field significantly reduces the solution efforts. The solution of the Laplace equation simultaneously satisfies the continuity condition $\nabla \cdot \boldsymbol{V}=0$ (no divergence) as well as the irrotationality condition $\nabla \times \boldsymbol{V}=0$. In addition, the solution has to satisfy the boundary conditions dictated by the solid surfaces that the potential flow is exposed to. As a simple example, we will consider a potential flow past a flat surface.

## 物理代写|流体力学代写Fluid Mechanics代考|Complex Potential for Plane Flows

Plane potential flows that satisfy the Laplace equation are treated most effectively using complex variables. These flows differ from other two-dimensional flows (with two independent variables) because two independent variables, $x$ and $y$, can be combined into one complex variable:
$$z=x+i y=r \cos \theta+i r \sin \theta=r(\cos \theta+i \sin \theta)=r e^{i \theta}$$
with $i=\sqrt{-1}$. The complex variable $\mathrm{z}$ and its conjugate complex $\bar{z}$ are shown in parts of the variable $z$.

Since every analytic function of the complex coordinate $z$ satisfies Laplace’s equation, the computation of both the direct and indirect problems becomes considerably easier. If we know the flow past a cylindrical body whose cross-sectional surface is simply connected (e.g. circular cylinder), then according to the Riemann mapping theorem, we can obtain the flow past any other cylinder using a conformal transformation. By this theorem, every simple connected region in the complex plane can be mapped into the inside of the unit circle. By doing this, in principle, we have solved the problem of flow past a body, and we only need to find a suitable mapping function.

The complex function $F(z)$ is called analytic (holomorphic), if it is complex differentiable at every point $z$, where the limit
$$\lim _{\Delta z \rightarrow 0} \frac{F(z+\Delta z)-F(z)}{\Delta z}=\frac{d F}{d z}$$
exists and is independent of the path from $z$ to $z+\Delta z$. If this requirement is not satisfied, the point is a singular point. Along a path parallel to the $x$ axis, the relation
$$\frac{d F}{d z}=\frac{\partial F}{\partial x}$$
holds and the same holds for the path parallel to the $y$ axis
$$\frac{d F}{d z}=\frac{\partial F}{\partial(i y)}$$
Since every complex function $F(z)$ is of the form
$$F(z)=\Phi(x, y)+i \psi(x, y) ; d F=d \Phi+i d \Psi$$
we then have
$$\frac{\partial F}{\partial x}=\frac{\partial \Phi}{\partial x}+i \frac{\partial \psi}{\partial x}=\frac{1}{i} \frac{\partial \Phi}{\partial y}+\frac{\partial \psi}{\partial y}=\frac{1}{i} \frac{\partial F}{\partial y}$$

# 力学代考

## 物理代写|流体力学代写流体力学代考|不可压缩势流

$$\boldsymbol{V}=\nabla \Phi, e_i V_i=e_i \frac{\partial \Phi}{\partial x_i} .$$

$$\boldsymbol{V}=e_i \frac{\partial \Phi}{\partial x_i}=e_1 \frac{\partial \Phi}{\partial x_1}+e_2 \frac{\partial \Phi}{\partial x_2}+e_3 \frac{\partial \Phi}{\partial x_3}$$

$$\nabla \cdot \boldsymbol{V}=\nabla \cdot \nabla \Phi=\Delta \Phi=0$$

$$\frac{\partial^2 \Phi}{\partial x_i \partial x_i}=0$$

## 物理代写|流体力学代写流体力学代考|平面流体的复数势能

. . . . . . . > . > .

$$z=x+i y=r \cos \theta+i r \sin \theta=r(\cos \theta+i \sin \theta)=r e^{i \theta}$$
with $i=\sqrt{-1}$。复变$\mathrm{z}$及其共轭复变$\bar{z}$显示在变量$z$的部分中

$$\lim _{\Delta z \rightarrow 0} \frac{F(z+\Delta z)-F(z)}{\Delta z}=\frac{d F}{d z}$$

$$\frac{d F}{d z}=\frac{\partial F}{\partial x}$$

$$\frac{d F}{d z}=\frac{\partial F}{\partial(i y)}$$

$$F(z)=\Phi(x, y)+i \psi(x, y) ; d F=d \Phi+i d \Psi$$
，我们就有
$$\frac{\partial F}{\partial x}=\frac{\partial \Phi}{\partial x}+i \frac{\partial \psi}{\partial x}=\frac{1}{i} \frac{\partial \Phi}{\partial y}+\frac{\partial \psi}{\partial y}=\frac{1}{i} \frac{\partial F}{\partial y}$$

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## MATLAB代写

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assignmentutor™您的专属作业导师
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