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# 数学代写|非线性动力系统代写Nonlinear Dynamics代考|CS-E5755 Linear differential equations

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## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|Linear differential equations

A homogeneous linear $N^{\text {th }}$ order $\mathrm{ODE}$,
$$\frac{d^N x}{d t^N}+c_{N-1} \frac{d^{N-1} x}{d t^{N-1}}+\ldots+c_1 \frac{d x}{d t}+c_0 x=0$$
may be written in matrix form, as
$$\frac{d}{d t}\left(\begin{array}{c} \varphi_1 \ \varphi_2 \ \vdots \ \varphi_N \end{array}\right)=\overbrace{\left(\begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0 \ 0 & 0 & 1 & \cdots & 0 \ \vdots & \vdots & \vdots & & \vdots \ -c_0 & -c_1 & -c_2 & \cdots & -c_{N-1} \end{array}\right)}^M\left(\begin{array}{c} \varphi_1 \ \varphi_2 \ \vdots \ \varphi_N \end{array}\right) .$$
Thus,
$$\dot{\varphi}=M \boldsymbol{\varphi},$$
and if the coefficients $c_k$ are time-independent, i.e. the ODE is autonomous, the solution is obtained by exponentiating the constant matrix $Q$ :
$$\boldsymbol{\varphi}(t)=\exp (M t) \boldsymbol{\varphi}(0)$$
the exponential of a matrix may be given meaning by its Taylor series expansion. If the ODE is not autonomous, then $M=M(t)$ is time-dependent, and the solution is given by the path-ordered exponential,
$$\varphi(t)=\mathcal{P} \exp \left{\int_0^t d t^{\prime} M\left(t^{\prime}\right)\right} \varphi(0),$$
As defined, the equation $\dot{\varphi}=\boldsymbol{V}(\boldsymbol{\varphi})$ is autonomous, since $g_t$ depends only on $t$ and on no other time variable. However, by extending the phase space from $\mathcal{M}$ to $\mathbb{R} \times \mathcal{M}$, which is of dimension $(N+1)$, one can describe arbitrary time-dependent ODEs.

## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|Lyapunov functions

For a general dynamical system $\dot{\varphi}=\boldsymbol{V}(\boldsymbol{\varphi})$, a Lyapunov function $L(\boldsymbol{\varphi})$ is a function which satisfies
$$\boldsymbol{\nabla} L(\boldsymbol{\varphi}) \cdot \boldsymbol{V}(\boldsymbol{\varphi}) \leq 0 .$$
There is no simple way to determine whether a Lyapunov function exists for a given dynamical system, or, if it does exist, what the Lyapunov function is. However, if a Lyapunov function can be found, then this severely limits the possible behavior of the system. This is because $L(\varphi(t))$ must be a monotonic function of time:
$$\frac{d}{d t} L(\boldsymbol{\varphi}(t))=\boldsymbol{\nabla} L \cdot \frac{d \boldsymbol{\varphi}}{d t}=\boldsymbol{\nabla} L(\boldsymbol{\varphi}) \cdot \boldsymbol{V}(\boldsymbol{\varphi}) \leq 0 .$$

Thus, the system evolves toward a local minimum of the Lyapunov function. In general this means that oscillations are impossible in systems for which a Lyapunov function exists. For example, the relaxational dynamics of the magnetization $M$ of a system are sometimes modeled by the equation
$$\frac{d M}{d t}=-\Gamma \frac{\partial F}{\partial M},$$
where $F(M, T)$ is the free energy of the system. In this model, assuming constant temperature $T, \dot{F}=F^{\prime}(M) \dot{M}=-\Gamma\left[F^{\prime}(M)\right]^2 \leq 0$. So the free energy $F(M)$ itself is a Lyapunov function, and it monotonically decreases during the evolution of the system. We shall meet up with this example again in the next chapter when we discuss imperfect bifurcations.

## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|Linear differential equations

$$\frac{d^N x}{d t^N}+c_{N-1} \frac{d^{N-1} x}{d t^{N-1}}+\ldots+c_1 \frac{d x}{d t}+c_0 x=0$$

$$\dot{\varphi}=M \varphi,$$

$$\varphi(t)=\exp (M t) \varphi(0)$$

\left 缺少或无法识别的分隔符

## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|Lyapunov functions

$$\boldsymbol{\nabla} L(\varphi) \cdot \boldsymbol{V}(\boldsymbol{\varphi}) \leq 0$$

$$\frac{d}{d t} L(\varphi(t))=\nabla L \cdot \frac{d \varphi}{d t}=\boldsymbol{\nabla} L(\varphi) \cdot \boldsymbol{V}(\varphi) \leq 0 .$$

$$\frac{d M}{d t}=-\Gamma \frac{\partial F}{\partial M},$$

## MATLAB代写

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