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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|Relation Between the Velocity and Deformation

The gradient of $\mathbf{x}$ with respect to $\mathbf{X}$ is called the deformation gradient. We shall denote it by $\mathbf{F}$ and write:
$$\mathbf{F}=\nabla_{\mathbf{X}} \mathbf{x}, \quad F_{\alpha}^{i}=\frac{\partial x^{i}}{\partial X^{\alpha}}=x_{, \alpha^{*}}^{i}$$
Note that the above definition means that
$$\mathbf{F}(\mathbf{X}, 0)=1$$
where 1 is the identity matrix. Next, due to the fact that $\operatorname{det} \mathbf{F}$ (det denotes the determinant) is the Jacobian of the mapping and hence the measure of the ratio of the volumes in the $\mathbf{X}$ and $\mathbf{x}$ spaces, we demand that $\operatorname{det} \mathbf{F}>0$. This cnsures that the mapping in Eq. (2.1.1) is not degenerate, i.e., the conservation of mass is assured, and that the inversion in Eq. (2.1.2) is locally possible, as already remarked.

Now, there is a very simple relation between the two tensors $\mathbf{F}$ and $\mathbf{L}$. This arises from the equality of the mixed partial derivatives:
$$\frac{\partial}{\partial X^{\alpha}} \frac{\partial M^{i}}{\partial t}=\frac{\partial}{\partial t} \frac{\partial M^{i}}{\partial X^{\alpha}} .$$
Equivalently,
$$\frac{\partial \hat{v}^{i}}{\partial X^{\alpha}}=\frac{\partial}{\partial t} F^{i} .$$
Expressing the velocity as a spatial field, we have
$$\frac{\partial \hat{v}^{i}}{\partial X^{\alpha}}=\frac{\partial v^{i}}{\partial x^{j}} \frac{\partial x^{j}}{\partial X^{\alpha}}=L^{i}{ }{j} F{\alpha}^{j} .$$
Using the convention of the superposed dot as the material derivative, Eqs. (2.2.4)(2.2.5) may be combined and rewritten as
$$\dot{\mathbf{F}}=\mathbf{I} \cdot \mathbf{F}$$

物理代写|流体力学代写Fluid Mechanics代考|Rigid Motion

From analytical mechanics, it is well known that a rigid body moves in such a way that it translates and rotates about an axis as it does so. In continuum mechanics, one says that a material particle experiences a rigid motion if
$$\mathbf{x}(\mathbf{X}, t)=\mathbf{Q}(t) \mathbf{X}+\mathbf{c}(t),$$
where $\mathbf{Q}(t)$ is a time-dependent orthogonal tensor signifying the rotation and $\mathbf{c}(t)$ is the translation vector. The velocity vector is given by
$$\hat{\mathbf{v}}(\mathbf{X}, t)=\dot{\mathbf{Q}}(t) \mathbf{X}+\dot{\mathbf{c}}(t),$$
which has the following spatial representation:
$$\mathbf{v}(\mathbf{x}, t)=\dot{\mathbf{Q}}(t) \mathbf{Q}^{T}(t)(\mathbf{x}-\mathbf{c}(t))+\dot{\mathbf{c}}(t)$$
from which it follows that the velocity gradient has the form $\mathbf{L}(t)=\dot{\mathbf{Q}}(t) \mathbf{Q}^{T}(t)$. Here, the superscript $T$ denotes the transpose. Since the orthogonality of $\mathbf{Q}(t)$ implies that $\mathbf{Q}(t) \mathbf{Q}^{T}(t)=\mathbf{1}$, where 1 is the identity tensor, one obtains the following:
$$\frac{d}{d t}\left(\mathbf{Q}(t) \mathbf{Q}^{T}(t)\right)=\mathbf{0}$$
for all $t$. Clearly,
$$\frac{d}{d t}\left(\mathbf{Q}(t) \mathbf{Q}^{T}(t)\right)=\left(\frac{d}{d t} \mathbf{Q}(t)\right) \mathbf{Q}^{T}(t)+\mathbf{Q}(t) \frac{d}{d t}\left(\mathbf{Q}^{T}(t)\right)=\mathbf{0}$$

力学代考

物理代写|流体力学代写Fluid Mechanics代考|Relation Between the Velocity and Deformation

$$\mathbf{F}=\nabla_{\mathbf{X} \mathbf{x}}, \quad F_{\alpha}^{i}=\frac{\partial x^{i}}{\partial X^{\alpha}}=x_{, \alpha^{*}}^{i}$$

$$\mathbf{F}(\mathbf{X}, 0)=1$$

$$\frac{\partial}{\partial X^{\alpha}} \frac{\partial M^{i}}{\partial t}=\frac{\partial}{\partial t} \frac{\partial M^{i}}{\partial X^{\alpha}} .$$

$$\frac{\partial \hat{v}^{i}}{\partial X^{\alpha}}=\frac{\partial}{\partial t} F^{i}$$

$$\frac{\partial \hat{v}^{i}}{\partial X^{\alpha}}=\frac{\partial v^{i}}{\partial x^{j}} \frac{\partial x^{j}}{\partial X^{\alpha}}=L^{i} j F \alpha^{j}$$

$$\dot{\mathbf{F}}=\mathbf{I} \cdot \mathbf{F}$$

物理代写|流体力学代写Fluid Mechanics代考|Rigid Motion

$$\mathbf{x}(\mathbf{X}, t)=\mathbf{Q}(t) \mathbf{X}+\mathbf{c}(t),$$

$$\hat{\mathbf{v}}(\mathbf{X}, t)=\dot{\mathbf{Q}}(t) \mathbf{X}+\dot{\mathbf{c}}(t),$$

$$\mathbf{v}(\mathbf{x}, t)=\dot{\mathbf{Q}}(t) \mathbf{Q}^{T}(t)(\mathbf{x}-\mathbf{c}(t))+\dot{\mathbf{c}}(t)$$

$$\frac{d}{d t}\left(\mathbf{Q}(t) \mathbf{Q}^{T}(t)\right)=\mathbf{0}$$

$$\frac{d}{d t}\left(\mathbf{Q}(t) \mathbf{Q}^{T}(t)\right)=\left(\frac{d}{d t} \mathbf{Q}(t)\right) \mathbf{Q}^{T}(t)+\mathbf{Q}(t) \frac{d}{d t}\left(\mathbf{Q}^{T}(t)\right)=\mathbf{0}$$

有限元方法代写

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MATLAB代写

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师