# 最优控制方案|MA30061/MA50061 Optimal Control solution代写

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By uniqueness for the initial value problem and the Intermediate Value Theorem, we have that $x(t)>0$ for all $t \in\left[0, t_{f}\right]$. It follows that $\sigma\left(t_{f}\right)>0$. If $r \geq 1$, then $\sigma \leq 0$, so that $\sigma$ is non-increasing, which combined with $\sigma\left(t_{f}\right)>0$ shows that $\sigma>0$ on $\left[0, t_{f}\right]$. We conclude that $u^{*} \equiv 0$ in that case. If $r \in(0,1)$, then $\dot{\sigma}>0$, so that $\sigma$ is strictly increasing. It follows that singular control is not possible and that there can be at most one switch.r.

This problem fits into the standard form of the optimal control problem with
$$\begin{gathered} n=2, m=1, \quad h(\tau, \xi, v)=\left[\begin{array}{c} \xi_{2} \ v \end{array}\right], \quad f(\tau, \xi, v)=\frac{E(\tau) I(\tau)}{2} v^{2}-q(\tau) \xi_{1}, \quad \phi(\tau, \xi)=0, \ t_{0}=a, \quad t_{f}=b, \quad x_{0}=\left[\begin{array}{l} 0 \ 0 \end{array}\right], \quad x\left(t_{f}\right)=\text { free, } \quad \mathscr{H}=\mathbb{R} . \end{gathered}$$
(2b) The Hamiltonian function is
$$\mathcal{H}(\tau, \xi, v, \rho)=\frac{E(\tau) I(\tau)}{2} v^{2}-q(\tau) \xi_{1}+\rho_{1} \xi_{2}+\rho_{2} v$$
We have
$$\frac{\partial \mathcal{H}}{\partial v}=E(\tau) I(\tau) v+\rho_{2}$$
From this we see that $v \mapsto \mathcal{H}(\tau, \xi, v, \rho)$ is minimized when
$$E(\tau) I(\tau) v+\rho_{2}=0$$
Therefore
$$u^{*}(t)=\frac{-1}{E(t) I(t)} p_{2}(t)$$
We also see that
$$\frac{\partial \mathcal{H}^{T}}{\partial \xi}=\left[\begin{array}{c} -q(\tau) \ \rho_{1} \end{array}\right], \quad \frac{\partial \phi^{T}}{\partial \xi}=0$$
The adjoint equation with transversality condition then is
$$\dot{p}=\left[\begin{array}{c} q \ -p_{1} \end{array}\right], \quad p\left(t_{f}\right)=0$$

## MA30061/MA50061 COURSE NOTES ：

Integrating twice gives
$$y(t)=\frac{t^{5}}{40}+A \frac{t^{4}}{12}+B \frac{t^{3}}{6}+C t+D$$
where $C, D \in \mathbb{R}$ are constants. The boundary conditions are
$$y(0)=0, \quad \dot{y}(0)=0, \quad \ddot{y}(1)=0, \quad\left(\frac{d}{d t} \frac{1}{t} \ddot{y}(t)\right)(1)=0 .$$
The first of these gives $D=0$ whereas the second gives $C=0$. From the third we obtain
$$\frac{1}{2}+A+B=0$$
whereas from the fourth we obtain
$$1+A=0$$
Hence we obtain $A=-1$ and $B=\frac{1}{2}$. Therefore
$$y(t)=\frac{t^{5}}{40}-\frac{t^{4}}{12}+\frac{t^{3}}{12}$$

# 非线性常微分方程的分析 | MA30062 Analysis of Nonlinear Ordinary Differential Equations代写

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There are various solutions that students might put forward. They should note the requirement for Diffiffiffie-Hellman key exchange and assumptions regarding the fact that attackers can not currently broadcast so that they are assuming that no man in the middle attack can take place. They should note that there is no way to avoid having to make this assumption and may even conclude that it increases risks to trust a system where this is a fundamental issue and that management should wait til summer.

Find a number $\alpha>0$ such that
$$(T y)(t)=x_{0}+\int_{0}^{t} e^{s} y(s)+s d s$$Solution. following the lines of the proof of Proposition $2.7$ (first part) (unseen as an example)
For $t \in(-1,1)$ we observe
\begin{aligned} \left|(T y)(t)-x_{0}\right| & \leq\left|\int_{0}^{t} e^{s}\right| y(s)|+| s|d s| \ & \leq\left|\int_{0}^{t} e\left(\left|x_{0}\right|+1\right)+1 d s\right| \ &=\left(e\left(\left|x_{0}\right|+1\right)+1\right)|t| \end{aligned}
Setting $M:=e\left(\left|x_{0}\right|+1\right)+1$, we can set $\alpha_{0}:=\frac{1}{M}$ to obtain that $T$ maps $X_{\alpha_{0}}$ into itself. [2 points]

## MA30062 COURSE NOTES ：

Compute the flow of the ODE $x^{\prime}=-x^{3}$.Let $\phi$ be the flow for $x^{\prime}=-x^{3}$. Show that $\left|\phi^{t}(a)\right| \leq\left|\phi^{s}(a)\right|$ whenever $t \geq s$.Let $\phi$ be the flow for $x^{\prime}=-x^{3}$ and $a, b \in \mathbb{R}$. Use Gronwall’s lemma to show $\left|\phi^{t}(a)-\phi^{t}(b)\right| \leq|a-b| \exp \left(\left(|a|^{2}+|a||b|+|b|^{2}\right)|t|\right)$ for $t \leq 0$.
Hint: You may find the identity $x^{3}-y^{3}=\left(x^{2}+x y+y^{2}\right)(x-y)$ useful.Show that 0 is a globally attractive and Lyapunov stable equilibrium for $x^{\prime}=-x^{3}$ and that this cannot be identified using linearisation.

# 密码学| CM30173 Cryptography代写

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There are various solutions that students might put forward. They should note the requirement for Diffiffiffie-Hellman key exchange and assumptions regarding the fact that attackers can not currently broadcast so that they are assuming that no man in the middle attack can take place. They should note that there is no way to avoid having to make this assumption and may even conclude that it increases risks to trust a system where this is a fundamental issue and that management should wait til summer.

• $\mathcal{P}=\mathcal{C}=\mathbb{Z}_{2}$
• $\mathcal{K}=\mathbb{Z}{2}$ and we generate a keystream $k{1} k_{2} k_{3} \ldots$ with $k_{i} \in \mathcal{K}$.
• We encrypt plaintext $x=x_{1} x_{2} x_{3} \ldots$ with the keystream thus
$$e_{k_{i}}\left(x_{i}\right)=\left(x_{i}+k_{i}\right) \quad \bmod 2$$
(the exclusive-or of $x_{i}$ and $k_{i}$ this is also written $x_{i} \oplus k_{i}$ ) and decrypt
$$d_{k_{i}}\left(y_{i}\right)=\left(y_{i}+k_{i}\right) \quad \bmod 2$$

## CM30173 COURSE NOTES ：

We can use the cipher block chaining (CBC) mode for a block cipher with a fixed public $I V$ to create a MAC.
CBC mode a message $x=x_{1} x_{2} \ldots x_{n}$ split into blocks and calculates
\begin{aligned} y_{0} &=I V \ y_{i} &=e_{k}\left(y_{i-1} \oplus x_{i}\right) \quad i \geq 1 \end{aligned}
This idea is adapted to form CBC-MAC by carrying out the same series of calculations but only returning the output from the last loop:
Inputs: $x$, a block cipher such as DES
$$y_{0}=I V=00 \ldots 0$$
for $i$ from 1 to $n$ do
$$y_{i}=e_{k}^{\mathrm{DES}}\left(y_{i-1} \oplus x_{i}\right)$$
end do
return $y_{n}$

# 离散数学与汇编 Discrete Math with Comp MATH223101

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Just as in the treatment of the case $k=200$ above, we subtract the number of primes up to $10^{k-1}$ from the number of primes up to $10^{k}$. By the Prime Number Theorem, this number is about
$$\frac{10^{k}}{k \ln 10}-\frac{10^{k-1}}{(k-1) \ln 10}=\frac{(9 k-10) 10^{k-1}}{k(k-1) \ln 10} .$$
Since
$$\frac{9 k-10}{k-1}=9-\frac{1}{k-1}$$

is very close to 9 if $k$ is large, we get that the number of primes with $k$ digits is approximately
$$\frac{9 \cdot 10^{k-1}}{k \ln 10}$$
Comparing this with the total number of positive integers with $k$ digits, which we know is $10^{k}-10^{k-1}=9 \cdot 10^{k-1}$, we get
$$\frac{9 \cdot 10^{k-1}}{k \ln 10 \cdot 9 \cdot 10^{k-1}}=\frac{1}{(\ln 10) k} \approx \frac{1}{2.3 k} .$$

## MATH223101COURSE NOTES ：

Simplifying, we obtain
$$4 k^{2}-4 n k+n^{2}-n-2<0 .$$
Solving for $k$, we get that the left-hand side is nonpositive between the two roots:
$$\frac{n}{2}-\frac{1}{2} \sqrt{n+2} \leq k \leq \frac{n}{2}+\frac{1}{2} \sqrt{n+2}$$
So the first integer $k$ for which this is nonpositive is
$$k=\left\lceil\frac{n}{2}-\frac{1}{2} \sqrt{n+2}\right]$$

# 数学方法2|MA30059/MA40059/MA50059 Mathematical Methods 2代写

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In some sense, the unifying theme of the unit is partial differential equations (we shall see that
although variational principles such as those mentioned in Section 1.1.3 are not differential equations, they are intimately linked to them). You have met several PDEs over the last few years, but
we shall (hopefully) study PDEs in a different manner to how you may have done so up to this
point.

Fix $\mathbf{x} \in \mathbb{R}^{m}$.
(i) A distribution $T: C_{c}^{\infty}\left(\mathbb{R}^{m}\right) \rightarrow \mathbb{R}$ is called a fundamental solution of Laplace’s equation with respect to the point $\mathbf{x} \in \mathbb{R}^{m}$ if
$$\Delta T=\delta_{\mathrm{x}},$$
that is, equality as distributions. Here $\delta_{\mathbf{x}}$ is the shifted Dirac Delta distribution defined by $\delta_{\mathbf{x}}(\phi)=\phi(\mathbf{x})$ for all $\phi \in C_{c}^{\infty}\left(\mathbb{R}^{m}\right)$.
(ii) Let $\phi \in C_{c}^{\infty}\left(\mathbb{R}^{m}\right)$. As $\phi$ is compactly supported, there exists $\rho>0$ such that $|\mathbf{x}|<\rho$ and $\phi(\mathbf{y})=0$ for all $\mathbf{y} \in \mathbb{R}^{m}$ with $|\mathbf{y}| \geq \rho$. Choose $\Omega:=B(\mathbf{0}, \rho+1)$, so that
$\phi, \frac{\partial \phi}{\partial n}=0 \quad$ on $\partial \Omega$,
and $\mathbf{x} \in \Omega$. Since evidently $\phi \in C^{2}(\bar{\Omega})$, Green’s Integral Representation reduces to
$$\int_{\mathbb{R}^{m}} N_{\mathbf{x}}(\mathbf{y}) \Delta \phi(\mathbf{y}) \mathrm{d} \mathbf{y}=\phi(\mathbf{x}) .$$
As $\phi$ was arbitrary, in terms of distributions, this reads
$$\Delta T_{N_{\mathrm{x}}}=\delta_{\mathrm{x}}$$
where $T_{N_{\mathbf{x}}}$ is the distribution corresponding to $N_{\mathbf{x}}$, as required.

## MA30059/MA40059/MA50059 COURSE NOTES ：

We need to find a $v$ satisfying $(1)-(3)$ above. Then we set $G(\mathbf{x}, \mathbf{y}):=N_{\mathbf{x}}(\mathbf{y})+v(\mathbf{x}, \mathbf{y})$ If $\mathbf{x}=0$, then (3) becomes
$$v(\mathbf{0}, \mathbf{y})=\frac{1}{4 \pi} \quad \forall y \in \partial \Omega_{0},$$
so we can just choose
$$v(\mathbf{0}, \mathbf{y})=\frac{1}{4 \pi} \quad \forall y \in \bar{\Omega}{0},$$ which satisfies Laplace’s equation in $\Omega$ (and is in $C^{2}(\bar{\Omega})$ ) since it is constant. We now consider the case $\mathbf{x} \neq 0$. In light of the key property of $\mathbf{r}(\mathbf{x})$, namely, $$|\mathbf{x}| \cdot|\mathbf{y}-\mathbf{r}(\mathbf{x})|=|\mathbf{y}-\mathbf{x}| \quad \forall \mathbf{x} \in \Omega{0} \backslash{\mathbf{0}}, \forall \mathbf{y} \in \partial \Omega_{0},$$
the requirement (3) becomes
$$v(\mathbf{x}, \mathbf{y})=\frac{1}{4 \pi} \frac{1}{|\mathbf{x}-\mathbf{y}|}=\frac{1}{4 \pi} \frac{1}{|\mathbf{x}| \cdot|\mathbf{y}-\mathbf{r}(\mathbf{x})|} \quad \forall \mathbf{y} \in \partial \Omega_{0}$$
Thus, we let $v$ equal the function on the right hand side of the above for all $\mathbf{y} \in \Omega_{0}$, which is well defined since $\mathbf{r}(\mathbf{x})-\mathbf{y} \neq \mathbf{0}$ for all $\mathbf{y} \in \bar{\Omega}{0}$ and $\mathbf{x} \in \Omega{0} \backslash{\mathbf{0}}$. Furthermore, since
$$v(\mathbf{x}, \mathbf{y})=-\frac{1}{|\mathbf{x}|} N_{\mathbf{r}(\mathbf{x})}(\mathbf{y}) \quad \forall \mathbf{y} \in \bar{\Omega}_{0},$$

# 环和多项式 Rings and Polynomials MATH202701

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Let us make another calculation of a greatest common divisor using the prime factorization approach, as a guide to formulating precisely what the approach is. We will compute $(8316,19800)$. The prime factorizations are
$$8316=2^{2} \times 3^{3} \times 7 \times 11$$
and
$$19800=2^{3} \times 3^{2} \times 5^{2} \times 11$$

Let us rewrite these factorizations as
$$8316=2^{2} \times 3^{3} \times 5^{0} \times 7^{1} \times 11^{1}$$
and
$$19800=2^{3} \times 3^{2} \times 5^{2} \times 7^{0} \times 11^{1} .$$

## MATH202701COURSE NOTES ：

These three statements can be summarized in equations. For example, the first statement can be written as
$$\mathcal{C}(0)+\mathcal{C}(1)=\mathcal{C}(1)$$
Write the two other statements as equations of the same type.
Perform analogous steps for multiplication: What is the product of an even integer and an odd integer? An even integer and an even integer? An odd integer and an odd integer? Summarize your answers in three equations of the form
$$\mathcal{C}(i) \times \mathcal{C}(j)=\mathcal{C}(k)$$
Observe that your sum and product rules turn the collection
$${\mathcal{C}(0), \mathcal{C}(1)}$$

# 差分方程建模 Modelling w/Differential Eqtns MATH140001

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Example. Consider $f(x)=\sqrt{|x|}$. Then $\left(x_{1}, x_{2}\right)=(0, \infty)$,
$$F(x)=2\left(\sqrt{x}-\sqrt{x_{0}}\right) \text {. }$$
and
$$\varphi(t)=\left(\sqrt{x_{0}}+\frac{t}{2}\right)^{2}, \quad-2 \sqrt{x_{0}}<t<\infty$$

Then $\phi \in C^{1}\left(\left(T_{-}, T_{+}\right)\right)$and
$$\lim {t \uparrow T{+}} \phi(t)=x_{2}, \quad \text { respectively } \quad \lim {t \downarrow T{-}} \phi(t)=x_{1} .$$
In particular, $\phi$ exists for all $t>0$ (resp. $t<0$ ) if and only if
$$T_{+}=\int_{x_{0}}^{x_{2}} \frac{d y}{f(y)}=+\infty$$

## MATH140001COURSE NOTES ：

$$s=\sigma(t), \quad y=\eta(t, x)$$
(which map the fibers $t=$ const to the fibers $s=$ const). Denoting the inverse transform by
$$t=\tau(s), \quad x=\xi(s, y),$$
a straightforward application of the chain rule shows that $\phi(t)$ satisfies
$$\dot{x}=f(t, x)$$
if and only if $\psi(s)=\eta(\tau(s), \phi(\tau(s)))$ satisfies
$$\dot{y}=\dot{\tau}\left(\frac{\partial \eta}{\partial t}(\tau, \xi)+\frac{\partial \eta}{\partial x}(\tau, \xi) f(\tau, \xi)\right)$$

# 线性代数入门 Introductory Linear Algebra MATH106001

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Let $F$ be the given hermitian matrix (3.4.14) and $w \in C$. Assume that
$d_{1}:=\operatorname{det} F(1, \ldots, n-1) \neq 0, \quad d_{2}:=F(2, \ldots, n) \neq 0, \quad d_{3}:=\operatorname{det} F(2, \ldots, n-1) \neq 0 .$
Then
$$\operatorname{det} F(w)=\frac{d_{1} d_{2}}{d_{3}}\left(1-\frac{\left|w-w_{0}\right|^{2} d_{3}^{2}}{d_{1} d_{2}}\right)$$

where
$$w_{0}=\frac{1}{w_{1 n}} \sum_{j=2}^{n-1} f_{n j} w_{j n}$$
and
$$[F(1, \ldots, n-1)]^{-1}=\left(w_{j k}\right)_{j, k=1}^{n-1} .$$

## MATH106001COURSE NOTES ：

It follows from these equations that
$$\rho_{n}=\frac{\operatorname{det} F(w)}{d_{2}} \text { and } \mu_{n}=\frac{\operatorname{det} F(w)}{d_{1}} .$$
Hence $\rho_{n}=\rho_{n-1}\left(1-\varphi_{n} \psi_{n}\right)$ and $\mu_{n}=\mu_{n-1}\left(1-\varphi_{n} \psi_{n}\right)$.
If we define (this is the same $w_{0}$ as in formula (3.4.15))
$$w_{0}=-\sum_{j=2}^{n-1} f_{n j} \nu_{j}=\overline{-\sum_{j=2}^{n-1} f_{1 j} \mu_{j}}$$
then $\delta_{n}=w-w_{0}, \Delta_{n}=\overline{w-w_{0}}$ and
$$\left|w-u_{0}\right|^{2}=\varphi_{n} \psi_{n} \rho_{n-1} \mu_{n-1} .$$

# 热能和热传递 Thermal power and heat transfer ME20015

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$$M e=M e_{(\lambda=1)} \cdot \lambda^{m}$$
In Anglo-Saxon countries, the following nomenclature is often used for the description of the fill characteristics curves:
$$\frac{K a V}{L}=f(L / G)=f(1 / \lambda)$$

The convective heat transfer coefficient is calculated based on the Lewisanalogy taking into account a correction for one-directional diffusion at the phase interface:
$$\frac{a}{\beta_{x}}=c_{p m t} \cdot L e^{(1-n)} \cdot \frac{\left(p_{\mathrm{S}, \mathrm{W}}^{\prime \prime}-p_{\mathrm{S}, \mathrm{A}}\right)}{\left(p-p_{\mathrm{S}, \mathrm{W}}^{\prime \prime}\right) \cdot \ln \left(p-p_{\mathrm{S}, \mathrm{A}} / p-p_{\mathrm{S}, \mathrm{W}}^{\prime \prime}\right)}$$
Here, $c_{\mathrm{pm}}$ is the specific heat capacity of wet air related to the dry air mass:
$$c_{p m}=c_{p, \mathrm{~A}}+c_{p, \mathrm{~S}} \cdot X_{\mathrm{A}}$$

## ME10305 COURSE NOTES ：

where the effective Nusselt number is
$$N u_{\mathrm{a}}=\frac{N u_{\mathrm{D}}}{\sqrt{a_{1} R e_{D}}}$$
the parameter $a_{1}$, the dimensionless strain rate, is $4.0$ for a cylinder in cross flow. The effective turbulence intensity, $\mathrm{Tu}{\lambda}$, is given by $$T u{\lambda}=\frac{T u_{\mathrm{a}} \sqrt{L_{\mathrm{a}}}}{\left(1+0.004 L_{\mathrm{a}}^{2}\right)^{5 / 12}}$$
the dimensionless length scale, $L_{\mathrm{a}}$, is given by
$$L_{\mathrm{a}}=\frac{\Lambda_{x}}{D} \sqrt{a_{1} R e_{D}}$$
where $\Lambda_{x}$ is the integral length scale based upon the streamwise velocity fluctuation component. The turbulence term is given as:
$$T u_{\mathrm{a}}=\frac{T u \sqrt{R e_{D}}}{\sqrt{a_{1}}}$$

# 数学 Mathematics 2 ME10305

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$f^{\prime}(x)$ and $g^{\prime}(x)$ exist near, but not necessarily at, $a$. Then, if
$$f^{\prime}(x) / g^{\prime}(x)$$
tends to a limit $\lambda$ as $x \rightarrow a$, the limit of $f(x) / g(x)$ exists and equals $\lambda$ also.
By Cauchy’s formula, if $a<z_{1}<x$,
$$\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^{\prime}\left(z_{1}\right)}{g^{\prime}\left(z_{1}\right)}$$

Hence, assuming the limit in question exists,
$$\lim {x \rightarrow a+0} \frac{f(x)}{g(x)}=\lim {z_{1} \rightarrow a+0} \frac{f^{\prime}\left(z_{1}\right)}{g^{\prime}\left(z_{1}\right)}=\lambda$$
Here it has been assumed that $x>a$ and so only the right-hand limits have been taken-the point $a$ being approached from above.
If $x<a$, that is if $x<z_{2}<a$, a similar argument yields
$$\lim {x \rightarrow a-0} \frac{f(x)}{g(x)}=\lim {z_{x} \rightarrow a-0} \frac{f^{\prime}\left(z_{2}\right)}{g^{\prime}\left(z_{2}\right)}=\lambda$$
where left-hand limits have been taken.

Complex numbers were introduced first by Cardan in his examination of the solutions of cubic equations. Hence, consider the cubic equation $x^{3}-1=0$. This may be written
$$(x-1)\left(x^{2}+x+1\right)=0 .$$
Hence, the roots of the original cubic equation are $x=1$ and the roots of the quadratic equation
$$x^{2}+x+1=0$$
$$x=1 / 2(-1 \pm V-3)$$