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

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.