数学建模|MTH2040 Mathematical modelling代写 monash代写

0

这是一份monash悉尼大学MTH2040的成功案例

数学建模|MTH2040 Mathematical modelling代写 monash代写


问题 1.

If $|\alpha|=\nu$, then $\lambda_{1}=\lambda_{2}=-\alpha$ and the corresponding generalized eigenspace is spanned by
$$
z^{1}=\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right], \quad z^{2}=\left[\begin{array}{l}
0 \
1
\end{array}\right]
$$
where $z^{1}$ is a real eigenvector and $z^{2}$ is a generalized eigenvector of second order. By (34) the corresponding (generalized) real eigenmotions are
$$
x^{1}(t)=e^{-\alpha t}\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right], \quad x^{2}(t)=e^{-\alpha t}\left[\begin{array}{l}
0 \
1
\end{array}\right]+t e^{-\alpha t}\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right]
$$


证明 .

If $|\alpha|<v$ (hence $z^{2}=\overline{z^{1}}$ ) then $\lambda_{1,2}=-\alpha \pm x \sqrt{\nu^{2}-\alpha^{2}}$ and the corresponding real modes are by $(44),(45)$ and $(50)$
$$
x^{1}(t)=e^{-\alpha t}\left(\cos \left(\sqrt{u^{2}-\alpha^{2}} t\right)\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right]-\sin \left(\sqrt{u^{2}-\alpha^{2}} t\right)[\sqrt{0}]\right)
$$






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MTH2040 COURSE NOTES :

is called the Fibonacci sequence and plays a role in various fields of mathematics. The eigenvalues of the matrix in (59) are $\lambda_{1,2}=(1 \pm \sqrt{5}) / 2$ and the corresponding real modes are
$$
z^{1}(t)=\left(\frac{1+\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c}
(1+\sqrt{5}) / 2 \
1
\end{array}\right], \quad z^{2}(t)=\left(\frac{1-\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c}
(1-\sqrt{5}) / 2 \
1
\end{array}\right]
$$
The solution of the initial value problem is of the form $x(t)=\alpha_{1} z^{1}(t)+\alpha_{2} z^{2}(t)$ for some $\left(\alpha_{1}, \alpha_{2}\right) \in \mathbb{R}^{2}$. Since $x(0)=[1,0]^{\top}$ we obtain $\alpha_{1}=1 / \sqrt{5}, \alpha_{2}=-1 / \sqrt{5}$ and hence the following analytic expression for the Fibonacci sequence
$$
\xi(t)=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{t+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{t+1}\right]
$$




















数学建模|MTH1003 Mathematical Modelling代写

0

这是一份exeter埃克塞特大学MTH1003作业代写的成功案

数学建模|MTH1003 Mathematical Modelling代写

$$
y^{+}=\frac{1}{K} \frac{\mu+\mu_{t}}{\mu} .
$$
For the k- $\varepsilon$ model of turbulence this distance is:
$$
y^{+}=\frac{\rho \sqrt[4]{C_{\mu}} \sqrt{k_{p}} \delta_{n p}}{\mu} .
$$
In the previous equation, $k_{P}$ is the kinetic energy of turbulence at the centre of the boundary cell, while $\sigma_{n s}$ denotes the normal distance from the centre of the boundary cell to the wall.
The viscous sub-layer thickness is defined as the larger root of the equation:
$$
y_{r}^{+}=\frac{1}{K} \ln \left(\varepsilon y_{r}^{+}\right)
$$

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MTH1003 COURSE NOTES :

$\int_{V} \operatorname{grad} \psi \mathrm{d} V=\int_{S} \psi \mathrm{ds} \Rightarrow \quad(\operatorname{grad} \psi){\mathrm{P}{0}} \approx \frac{1}{V_{\mathrm{P}{0}}} \sum{j=1}^{n_{f}} \psi_{j} \mathbf{s}{j}$ Here, $\psi{j}$ is the value of variable $\psi$ at the cell face centre.
The first term in the prototype equation is different to the others because it contains an integral with respect to time. If the equation is rearranged into the following form:
$$
\frac{d \Psi}{d t}=F(\phi)
$$
where
$$
\Psi=\int_{\mathrm{V}} \rho B_{\phi} \mathrm{d} V \approx\left(\rho B_{\phi} V\right){\mathrm{P}{0}} \text { and } \phi=\phi(\mathbf{r}, t)
$$



数学建模|Mathematical Modelling代写 MT2507

0

这是一份andrews圣安德鲁斯大学 MT2507作业代写的成功案例

数学建模|Mathematical Modelling代写 MT2507
问题 1.

The coordinates of four boundary faces generated in two dimensions, with the help of adaptation methods, can be written in vector form as:
$$
\begin{array}{ll}
\mathbf{a}{l}(\eta)=\mathbf{r}\left(\xi{l}, \eta\right), & l=1,2 \
\mathbf{b}{l}(\xi)=\mathbf{r}\left(\xi, \eta{l}\right), & l=1,2
\end{array}
$$


证明 .

where the coordinates of the transformed computational coordinate system, $\xi$ and $\eta$, are:
$\xi=\frac{(i-1)}{(I-1)}$ and $\eta=\frac{(j-1)}{(J-1)}$
$i$ and $j$ denote point numbers in physical coordinates while $I$ and $J$ are the overall number of points on these coordinates. As defined by the transfinite mapping method, the coordinates of the interior points are given as:


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MT2507 COURSE NOTES :

The simplest method of obtaining blending functions is by Lagrange interpolation:
$$
\begin{array}{ll}
\alpha_{1}(\xi)=1-\xi & \alpha_{2}(\xi)=\xi \
\beta_{1}(\eta)=1-\eta & \beta_{2}(\eta)=\eta
\end{array}
$$
Applying (3.37) to equation (3.36) gives the inner mesh points as:
$$
\begin{aligned}
&x(\xi, \eta)=(1-\xi) X_{1}(\xi, \eta)+\xi X_{2}(\xi, \eta) \
&y(\xi, \eta)=(1-\eta) Y_{1}(\xi, \eta)+\eta Y_{2}(\xi, \eta)
\end{aligned}
$$





数学建模|Mathematical Modeling代写 MATH 456

0

这是一份umass麻省大学 MATH 456作业代写的成功案例

数学建模|Mathematical Modeling代写 MATH 456
问题 1.

As for $n=n_{t}(a, t)$, we see that
$$
\left(a^{\prime}-a\right) \wedge\left(a^{\prime \prime}-a\right)=e^{\prime} \wedge e^{\prime \prime}\left(=O\left(\lambda^{2}\right)\right)=n_{0} \delta \Gamma_{0}+O\left(\lambda^{3}\right),
$$
where $n_{0}=n_{0}(a, t)$ and $\delta \Gamma_{0}=\delta \Gamma_{0}(a, t)$ is the area of the parallelepiped constructed over $e^{\prime}$ and $e^{\prime \prime}$; hence
$$
n_{i} \frac{\partial \Phi_{i}}{\partial a_{\alpha}} \delta \Gamma_{t}=n_{0 \alpha} \operatorname{det} \mathbf{F} \delta \Gamma_{0}+O\left(\lambda^{3}\right),
$$

证明 .

and, in vector form,
$$
n_{0} \operatorname{det} \mathbf{F} \delta \Gamma_{0}=\mathbf{F}^{T} \cdot n_{t} \delta \Gamma_{t}+O\left(\lambda^{3}\right) .
$$
Since $n_{0}$ has norm 1 , this also gives
$$
(\operatorname{det} \mathbf{F})^{2}\left(\delta \Gamma_{0}\right)^{2}=n_{t}^{T} \cdot \mathbf{F} \cdot \mathbf{F}^{T} \cdot n_{t}\left(\delta \Gamma_{t}\right)^{2}+O\left(\lambda^{5}\right),
$$
or, since $B=\mathbf{F} \cdot \mathbf{F}^{T}$ (see Section 5.1),
$$
(\operatorname{det} \mathbf{F})^{2}\left(\delta \Gamma_{0}\right)^{2}=n_{t}^{T} \cdot B \cdot n_{t}\left(\delta \Gamma_{t}\right)^{2}+O\left(\lambda^{5}\right) .
$$


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MMATH 456 COURSE NOTES :

If we know that
$$
\frac{d}{d t} \int_{\Omega_{t}^{\prime}} C d x=\int_{\Omega_{t}^{\prime}} f d x, \quad \forall \Omega_{t}^{\prime} \subset \Omega_{t}
$$
then
$$
\frac{\partial C}{\partial t}+\operatorname{div}(C U)=f \quad \text { in } \Omega_{t}^{i}, \quad i=1,2,
$$
and
$$
\int_{\Sigma_{i}^{\prime}}[C V] \cdot N d \Gamma=0, \quad \forall \Sigma_{t}^{\prime} \subset \Sigma_{t} ;
$$
hence,
$$
[C V] \cdot N=0 \text { on } \Sigma_{t} .
$$
When $C$ is a scalar, setting $v=V \cdot N$, we rewrite the last relation as
$$
(C v){2}-(C v){1}=0 .
$$