# 动态系统代写|DYNAMICAL SYSTEMS MATHS4074 University of Glasgow Assignment

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Stability analysis involves studying the behavior of solutions to differential equations or maps as time goes on. An equilibrium solution is a solution that doesn’t change over time, so it’s important to determine whether these solutions are stable (perturbations from the equilibrium decay over time) or unstable (perturbations grow over time). Depending on the system and the equilibrium solution, there are different methods for determining stability, including linearization, Lyapunov functions, and phase portraits.

Bifurcation theory is concerned with how the behavior of a dynamic system changes as parameters are varied. When a system undergoes a bifurcation, its behavior can change drastically, for example, an equilibrium solution may suddenly appear or disappear, or the system may exhibit periodic or chaotic behavior. Understanding the bifurcations that a system undergoes can provide insight into the underlying mechanisms that govern its behavior.

Overall, the topics covered in this course should provide you with a solid foundation for understanding the behavior of nonlinear systems and the methods used to analyze them.

Let $A \in \mathbb{C}^{n \times n}$, and $B \in \mathbb{C}^{m \times m}$. Show that $X(t)=e^{A t} X(0) e^{B t}$ is the solution to $\dot{X}=$ $A X+X B$

Recalling the definition of matrix exponential, $e^{A t}=\sum_{i=0}^{\infty} \frac{1}{i !}(A t)^i$, it is clear that, for any matrix $A, e^{A t}=I$ for $t=0$, and $d e^{A t} / d t=A e^{A t}=e^{A t} A$.
Hence,
\begin{aligned} \frac{d}{d t}\left(e^{A t} X(0) e^{B t}\right)=\left(\frac{d}{d t} e^{A t}\right) X(0) e^{B t}+e^{A t} & \left(\frac{d}{d t} X(0)\right) e^{B t}+e^{A t} X(0)\left(\frac{d}{d t} e^{B t}\right) \ & =A\left(e^{A t} X(0) e^{B t}\right)+0+\left(e^{A t} X(0) e^{B t}\right) B \end{aligned}
Furthermore, for $t=0$,
$$\left.\left(e^{A t} X(0) e^{B t}\right)\right|_{t=0}=X(0)$$
Hence we can conclude that the proposed function is in fact the solution to the initialvalue problem under consideration.

Use the projection theorem to solve the problem:
$$\min _{x \in \mathbb{R}^n}\left{x^T Q x: A x=b\right}$$
where $Q$ is a positive-definite $n \times n$ matrix, $A$ is a $m \times n$ real matrix, with rank $m<n$, and $b$ is a real $m$-dimensional vector. Is the solution unique?

(Note that $Q$ being positive-definite implies it is self-adjoint, i.e., Hermitian.) Let $x_0$ be such that $A x_0=b$, and consider the change of variables $z=x-x_0$. In the inner product space $\mathbb{R}^n$, with inner product $\langle u, v\rangle=u^T Q v$, it is desired to minimize $|x|^2=x^T Q x=\left|z+x_0\right|^2$, subject to the constraint that $z$ lies in the subspace $M:=\left{z \in \mathbb{R}^n: A z=0\right}$. Using the projection theorem, we know that an optimal solution $\hat{z}=\hat{x}-x_0$ must be such that $\left(\hat{z}+x_0=\hat{x}\right) \perp M$, i.e., $\langle\hat{x}, y\rangle=\hat{x}^T Q y=0$, for all $y \in M$. Summarizing, we know that
\begin{aligned} \hat{x}^T Q y & =0, \quad \forall A y=0 \ A x & =b . \end{aligned}
In order to satisfy the first equation for all $y$ such that $A y=0, \hat{x}$ must be of the form $\hat{x}=Q^{-1} A^T v$, for some $v \in \mathbb{R}^m$. The vector $v$ can be found using the constraint $A x=b$, i.e.,
$$A \hat{x}=A Q^{-1} A^T v=b,$$
and hence
$$v=\left(A Q^{-1} A^T\right)^{-1} b .$$
Concluding,
$$\hat{x}=Q^{-1} A^T\left(A Q^{-1} A^T\right)^{-1} b$$

Let $|A|<1$. Show that $\left|(I-A)^{-1}\right| \geq \frac{1}{1+|A|}$.

First of all, for any vector $x_0$, with $\left|x_0\right|=1$,
$$\left|(I-A) x_0\right| \geq\left|x_0\right|-\left|A x_0\right| \geq 1-|A|>0$$

which shows that the matrix $I-A$ is invertible, i.e., there is no vector $x_0$, with $\left|x_0\right|=1$, such that $(I-A) x_0=0$.
Furthermore, the following chain of inequalities holds:
\begin{aligned} 1=|I|=\left|(I-A)(I-A)^{-1}\right| & \leq|I-A| \cdot\left|(I-A)^{-1}\right| \ & \leq(|I|+|A|) \cdot\left|(I-A)^{-1}\right|=(1+|A|) \cdot\left|(I-A)^{-1}\right|, \end{aligned}
and the result follows. The definition of induced norm implies that $|I|=1$. The first inequality is due to the submultiplicative property of induced norms. The second inequality can be derived from the triangle inequality.