## 線性微分方程解的存在性與唯一性

$\displaystyle \frac{dx_i}{dt}=f_i(x_1,\ldots,x_n),~~i=1,\ldots,n$

$\displaystyle f_i(c_1,\ldots,c_n)=0,~~i=1,\ldots,n$

$\displaystyle x_i=c_i+y_i$

\displaystyle\begin{aligned} \frac{dy_i}{dt}&=\frac{dx_i}{dt}=f_i(c_1+y_1,\ldots,c_n+y_n)\\ &=f_i(c_1,\ldots,c_n)+\sum_{j=1}^n\frac{\partial f_i}{\partial x_j}y_j+\frac{1}{2}\sum_{j=1}^n\sum_{k=1}^n\frac{\partial^2 f_i}{\partial x_j\partial x_k}y_jy_k+\cdots\\ &=\sum_{j=1}^n\frac{\partial f_i}{\partial x_j}y_j+\frac{1}{2}\sum_{j=1}^n\sum_{k=1}^n\frac{\partial^2 f_i}{\partial x_j\partial x_k}y_jy_k+\cdots \end{aligned}

$x_i=c_i$$i=1,\ldots,n$，令 $\displaystyle a_{ij}=\frac{\partial f_i}{\partial x_j}$。如果忽略高階項，物理系統在均衡狀態 $(c_1,\ldots,c_n)$ 附近的行為可以用下列線性微分方程近似：

$\displaystyle \frac{dy_i}{dt}=\sum_{j=1}^na_{ij}y_j,~~i=1,2,\ldots,n$

$\displaystyle \begin{bmatrix} \displaystyle\frac{dy_1}{dt}\\[0.8em] \displaystyle\frac{dy_2}{dt}\\[0.3em] \vdots\\[0.3em] \displaystyle\frac{dy_n}{dt} \end{bmatrix}=\begin{bmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn} \end{bmatrix}\begin{bmatrix} y_1\\ y_2\\ \vdots\\ y_n \end{bmatrix}$

$\mathbf{y}=[y_i]$ 是一 $n$ 維向量且 $A=[a_{ij}]$ 是一 $n\times n$ 階矩陣。定義 $\displaystyle \left(\frac{d\mathbf{y}}{dt}\right)_i=\frac{dy_i}{dt}$，可得簡明的向量微分方程式

$\displaystyle \frac{d\mathbf{y}}{dt}=A\mathbf{y},~~\mathbf{y}(0)=\mathbf{c}'$

$\displaystyle \frac{d\mathbf{x}}{dt}=A(t)\mathbf{x},~~\mathbf{x}(0)=\mathbf{c}$

$\displaystyle \frac{dX}{dt}=A(t)X,~~X(0)=I$

$\displaystyle \frac{d\mathbf{x}}{dt}=\frac{dX\mathbf{c}}{dt}=\frac{dX}{dt}\mathbf{c}=A(t)X\mathbf{c}=A(t)\mathbf{x}$

$\displaystyle X=I+\int_0^tA(s)Xds$

\displaystyle\begin{aligned} X_0&=I\\ X_{k+1}&=I+\int_0^tA(s)X_kds,~~k=0,1,2,\ldots \end{aligned}

$\displaystyle X_{k+1}-X_k=\int_0^tA(s)(X_k-X_{k-1})ds,~~k=1,2,\ldots$

$\displaystyle m=\max_{0\le s\le t}\Vert A(s)\Vert$

$\displaystyle \Vert X_1-X_0\Vert=\left\Vert\int_0^tA(s)ds\right\Vert\le\int_0^t\Vert A(s)\Vert ds\le mt$

\displaystyle\begin{aligned} \Vert X_{k+1}-X_k\Vert&=\left\Vert\int_0^tA(s)(X_k-X_{k-1})ds\right\Vert\\ &\le\int_0^t\Vert A(s)\Vert\cdot\Vert X_k-X_{k-1}\Vert ds\\ &\le m\int_0^t\Vert X_k-X_{k-1}\Vert ds.\end{aligned}

$\displaystyle \Vert X_{k+1}-X_k\Vert\le\frac{m^{k+1}t^{k+1}}{(k+1)!}$

\displaystyle\begin{aligned} Y_0&=I\\ Y_{k+1}&=I+\int_0^tA(s)Y_kds,~~k=0,1,2,\ldots \end{aligned}

$\displaystyle X_{k+1}-Y_{k+1}=\int_0^tA(s)(X_k-Y_k)ds,~~k=0,1,2,\ldots$

$\displaystyle \frac{dx}{dt}=ax,~~x(0)=c$

$\displaystyle \frac{dX}{dt}=AX,~~X(0)=C$

$\displaystyle e^{At}=I+At+\frac{A^2t^2}{2!}+\frac{A^3t^3}{3!}+\cdots$

$\displaystyle \frac{\Vert A^nt^n\Vert}{n!}\le\frac{\Vert A^n\Vert\cdot\vert t\vert^n}{n!}\le\frac{\Vert A\Vert^n\cdot\vert t\vert^n}{n!}$

$\displaystyle e^{\Vert A\Vert\cdot\vert t\vert}=1+\Vert A\Vert\cdot\vert t\vert+\frac{\Vert A\Vert^2\cdot\vert t\vert^2}{2!}+\cdots$

\displaystyle\begin{aligned} \frac{de^{At}}{dt}&=\frac{d}{dt}\sum_{k=0}^\infty\frac{A^kt^k}{k!}=\sum_{k=0}^\infty \frac{A^k}{k!}\frac{dt^k}{dt}\\ &=A\sum_{k=1}^\infty \frac{A^{k-1}t^{k-1}}{(k-1)!}=Ae^{At},\end{aligned}

$\displaystyle \frac{dX}{dt}=A(t)X,~~X(0)=I$

$\displaystyle d\det X=\hbox{tr}\left((\hbox{adj}X)dX\right)$

\displaystyle\begin{aligned} \frac{\partial \det X}{\partial x_{ij}}&=\frac{\sum_{k=1}^nx_{ik}c_{ik}}{\partial x_{ij}}=\sum_{k=1}^n\delta_{jk}c_{ik}\\ &=c_{ij}=(\hbox{adj}X)_{ji},\end{aligned}

\displaystyle\begin{aligned} d\det X&=\sum_{i=1}^n\sum_{j=1}^n\frac{\partial \det X}{\partial x_{ij}}dx_{ij}\\ &=\sum_{i=1}^n\sum_{j=1}^n\left(\hbox{adj}X\right)_{ji} dx_{ij}\\ &=\hbox{tr}\left((\hbox{adj}X)dX\right) .\end{aligned}

\displaystyle\begin{aligned} \frac{d\det X}{dt}&=\hbox{tr}\left(\hbox{adj}X\frac{dX}{dt}\right)=\hbox{tr}\left((\hbox{adj}X)AX\right)\\ &=\hbox{tr}\left(X(\hbox{adj}X)A\right)=\hbox{tr}\left((\det X)IA\right)\\ &=(\det X)(\hbox{tr}A). \end{aligned}

$\displaystyle \det X(t)=e^{\int_0^t\textrm{tr}A(s)ds}\det X(0)$