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## 数学代写|常微分方程代写ordinary differential equation代考|The matrix exponential

We begin with the study of the autonomous linear first-order system
$$\dot{x}(t)=A x(t), \quad x(0)=x_0,$$
where $A$ is an $n$ by $n$ matrix. If we perform the Picard iteration we obtain
\begin{aligned} x_0(t) &=x_0 \ x_1(t) &=x_0+\int_0^t A x_0(s) d s=x_0+A x_0 \int_0^t d s=x_0+t A x_0 \ x_2(t) &=x_0+\int_0^t A x_1(s) d s=x_0+A x_0 \int_0^t d s+A^2 x_0 \int_0^t s d s \ &=x_0+t A x_0+\frac{t^2}{2} A^2 x_0 \end{aligned}
and hence by induction
$$x_m(t)=\sum_{j=0}^m \frac{t^j}{j !} A^j x_0$$
The limit as $m \rightarrow \infty$ is given by
$$x(t)=\lim {m \rightarrow \infty} x_m(t)=\sum{j=0}^{\infty} \frac{t^j}{j !} A^j x_0 .$$
In the one dimensional case $(n=1)$ this series is just the usual exponential and hence we will write
$$x(t)=\exp (t A) x_0,$$

where we define the matrix exponential by
$$\exp (A)=\sum_{j=0}^{\infty} \frac{1}{j !} A^j .$$
Hence, in order to understand our original problem, we have to understand the matrix exponential! The Picard iteration ensures convergence of $\exp (A) x_0$ for every vector $x_0$ and choosing the canonical basis vectors of $\mathbb{R}^n$ we see that all matrix elements converge. However, for later use we want to introduce a suitable norm for matrices and give a direct proof for convergence of the above series in this norm.

## 数学代写|常微分方程代写ordinary differential equation代考|Linear autonomous first-order systems

In the previous section we have seen that the solution of the autonomous linear first-order system (3.1) is given by
$$x(t)=\exp (t A) x_0 .$$
In particular, the map $\exp (t A)$ provides an isomorphism between all initial conditions $x_0$ and all solutions. Hence the set of all solutions is a vector space isomorphic to $\mathbb{R}^n$ (respectively $\mathbb{C}^n$ if we allow complex initial values).
In order to understand the dynamics of the system (3.1), we need to understand the properties of the function $\exp (t A)$. We will start with the case of two dimensions which covers all prototypical cases. Furthermore, we will assume $A$ as well as $x_0$ to be real-valued.

In this situation there are two eigenvalues, $\alpha_1$ and $\alpha_2$, which are either both real or otherwise complex conjugates of each other. We begin with the generic case where $A$ is diagonalizable and hence there are two linearly independent eigenvectors, $u_1$ and $u_2$, which form the columns of $U$. In particular,
$$U^{-1} A U=\left(\begin{array}{cc} \alpha_1 & 0 \ 0 & \alpha_2 \end{array}\right)$$
and the solution (3.23) is given by
$$x(t)=U \exp \left(t U^{-1} A U\right) U^{-1} x_0=U\left(\begin{array}{cc} \mathrm{e}^{\alpha_1 t} & 0 \ 0 & \mathrm{e}^{\alpha_2 t} \end{array}\right) U^{-1} x_0 .$$
Abbreviating $y_0=U^{-1} x_0=\left(y_{0,1}, y_{0,2}\right)$ we obtain
$$x(t)=y_{0,1} \mathrm{e}^{\alpha_1 t} u_1+y_{0,2} \mathrm{e}^{\alpha_2 t} u_2 .$$
In the case where both eigenvalues are real, all quantities in (3.26) are real. Otherwise we have $\alpha_2=\alpha_1^$ and we can assume $u_2=u_1^$ without loss of generality. Let us write $\alpha_1 \equiv \alpha=\lambda+\mathrm{i} \omega$ and $\alpha_2 \equiv \alpha^*=\lambda-\mathrm{i} \omega$.

# 常微分方程代写

## 数学代写|常微分方程代写ordinary differential equation代考|The matrix exponential

$$\dot{x}(t)=A x(t), \quad x(0)=x_0,$$

$$x_0(t)=x_0 x_1(t) \quad=x_0+\int_0^t A x_0(s) d s=x_0+A x_0 \int_0^t d s=x_0+t A x_0 x_2(t)=x_0+\int_0^t A x_1(s) d s=x_0+A x_0 \int_0^t d s+A^2 x_0 \int_0^t s d s \quad=x_0+$$

$$x_m(t)=\sum_{j=0}^m \frac{t^j}{j !} A^j x_0$$

$$x(t)=\lim m \rightarrow \infty x_m(t)=\sum j=0^{\infty} \frac{t^j}{j !} A^j x_0 .$$

$$x(t)=\exp (t A) x_0,$$

$$\exp (A)=\sum_{j=0}^{\infty} \frac{1}{j !} A^j .$$

## 数学代写|常微分方程代写ordinary differential equation代考|Linear autonomous first-order systems

$$x(t)=\exp (t A) x_0 .$$

$$x(t)=y_{0,1} \mathrm{e}^{\alpha_1 t} u_1+y_{0,2} \mathrm{e}^{\alpha_2 t} u_2 .$$

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