MT 김신휘 Solving a Linear Ordinal Differential Equation -- Sage

MT Page 336 Problem 9.5 (선형 미분방정식 풀기)

Solve the system of linear differential equations $\mathbf{y}'=A\mathbf{y}$ with the initial condition $\mathbf{y}\left ( 0 \right )=\mathbf{y}_{0}$ , where

$A=\begin{bmatrix} 2 &1 &-1 \\ -3 &-1 &1 \\ 9 &3 &-4 \end{bmatrix}$ , $\mathbf{y}_{0}=\begin{bmatrix} -1\\ -1\\ 1 \end{bmatrix}$ .

(Citation : Jin Ho Kwak and Sungpyo Hong, 1997, Linear Algebra, 336p)

Sol)

The unique solution of a system of linear differential equations $\mathbf{y}'=A\mathbf{y}$ with initial condition $\mathbf{y}\left ( 0 \right )=\mathbf{y}_{0}$ is $\mathbf{y}\left ( t \right )=e^{tA}\mathbf{y}_{0}$ . ( Ref. Jin Ho Kwak and Sungpyo Hong, 1997, Linear Algebra, 230-238p )

Let $Q=\begin{bmatrix} \mathbf{u}_{1} & \mathbf{u}_{2} & \cdots & \mathbf{u}_{n} \end{bmatrix}$ .

$\mathbf{y}\left ( t \right )=e^{tA}\mathbf{y}_{0}=Qe^{tJ}\left ( Q^{-1}\mathbf{y}_{0} \right )=Q\begin{bmatrix} e^{tJ_{1}} & & &0 \\ &e^{tJ_{2}} & & \\ & & \ddots & \\ 0 & & &e^{tJ_{s}} \end{bmatrix}\left ( Q^{-1} \mathbf{y}_{0}\right )$ where $s$ : the number of jordan blocks.

Let $m_{1},m_{2},...,m_{s}$ be natural numbers such that $J_{i}$ is a $m_{i} \times m_{i}$ matrix.

=> $e^{tJ_{i}}=e^{\lambda _{i}t}\begin{bmatrix} 1 &t &t^{2}/2! &\cdots &t^{m_{i}-1}/(m_{i}-1)! \\ 0 &1 &t & \ddots & t^{m_{i}-2}/(m_{i}-2)!\\ & &1 &\ddots & \\ & & &\ddots &t \\ 0 & & & &1 \end{bmatrix}$ ( Ref. MT 김신휘 Computation of e^A using Jordan Canonical Form : Link )

=> $\mathbf{y}\left ( t \right )=\begin{bmatrix} \begin{bmatrix} \mathbf{u}_{n_{0}+1} & \mathbf{u}_{n_{0}+2} & \cdots & \mathbf{u}_{n_{1}} \end{bmatrix}e^{tJ_{1}} & \begin{bmatrix} \mathbf{u}_{n_{1}+1} & \mathbf{u}_{n_{1}+2} & \cdots & \mathbf{u}_{n_{2}} \end{bmatrix}e^{tJ_{2}} & \cdots & \begin{bmatrix} \mathbf{u}_{n_{s-1}+1} & \mathbf{u}_{n_{s-1}+2} & \cdots & \mathbf{u}_{n_{s}} \end{bmatrix}e^{tJ_{s}} \end{bmatrix}\begin{bmatrix} c_{1}\\ c_{2}\\ \vdots \\ c_{n} \end{bmatrix}$ where $Q^{-1}\mathbf{y}_{0}=\begin{bmatrix} c_{1}\\ c_{2}\\ \vdots \\ c_{n} \end{bmatrix}$ and $\forall i\in \left \{ 1,2,...,s \right \}, n_{i}:=\sum_{l=1}^{i}m_{l}, \; \; n_{0}:=0$

=> $\mathbf{y}\left ( t \right )=\begin{bmatrix} e^{\lambda_{1} t}\begin{bmatrix} \mathbf{u}_{n_{0}+1} & \mathbf{u}_{n_{0}+2} & \cdots & \mathbf{u}_{n_{1}} \end{bmatrix}\begin{bmatrix} 1 & \cdots & t^{m_{1}-1}/(m_{1}-1)!\\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \end{bmatrix} & \cdots & e^{\lambda_{s} t}\begin{bmatrix} \mathbf{u}_{n_{s-1}+1} & \mathbf{u}_{n_{s-1}+2} & \cdots & \mathbf{u}_{n_{s}} \end{bmatrix}\begin{bmatrix} 1 & \cdots & t^{m_{s}-1}/(m_{s}-1)!\\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \end{bmatrix} \end{bmatrix}\begin{bmatrix} c_{1}\\ c_{2}\\ \vdots \\ c_{n} \end{bmatrix}$

=> $\mathbf{y}\left ( t \right )=\begin{bmatrix} \sum_{j=1}^{s} (e^{\lambda_{j} t}\begin{bmatrix} \mathbf{u}_{n_{j-1}+1} & \mathbf{u}_{n_{j-1}+2} & \cdots & \mathbf{u}_{n_{j}} \end{bmatrix}\begin{bmatrix} 1 & \cdots & t^{m_{j}-1}/(m_{j}-1)!\\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \end{bmatrix}\begin{bmatrix} c_{n_{j-1}+1}\\ c_{n_{j-1}+2}\\ \vdots \\ c_{n_{j}} \end{bmatrix}) \end{bmatrix}$

=> $\mathbf{y}\left ( t \right )=\begin{bmatrix} \sum_{j=1}^{s} (e^{\lambda_{j} t}(\left ( \sum_{k=0}^{m_{j}-1} c_{k+n_{j-1}+1}\frac{t^{k}}{k!} \right )\mathbf{u}_{n_{j-1}+1}+\left ( \sum_{k=0}^{m_{j}-2} c_{k+n_{j-1}+2}\frac{t^{k}}{k!} \right )\mathbf{u}_{n_{j-1}+2}+\cdots +c_{n_{j}}\mathbf{u}_{n_{j}})) \end{bmatrix}$ .

Sage를 이용해서 $A=QJQ^{-1}$ 인 $A$ 의 Jordan form $J$ 와 transition matrix $Q$ 를 구하자.

이어서 $Q^{-1}\mathbf{y}_{0}=\begin{bmatrix} c_{1}\\ c_{2}\\ c_{3} \end{bmatrix}$ 를 구한다.

그러면 주어진 linear differential equation의 유일한 solution은

$\mathbf{y}\left ( t \right )=e^{-t}\left ( \left ( c_{1}+c_{2}t+\frac{c_3}{2}t^{2} \right )\mathbf{u}_{1}+\left ( c_{2}+c_{3}t \right )\mathbf{u}_{2}+c_{3}\mathbf{u}_{3} \right )$ 이다.