MIT18.06 Linear Algebra (23)

lec23

differential equation $\frac{\mathrm{d}u}{\mathrm{d}t}=Au$

e.g. Solve

$$\begin{array}{l}\frac{d u_{1}}{d t}=-u_{1}+2 u_{2} \\\frac{d u_{2}}{d t}=u_{1}-2 u_{2}\end{array}$$

$$\longrightarrow \frac{du}{dt} = Au$$

where

$$A=\left[\begin{array}{cc}-1 & 2 \\1 & -2\end{array}\right]$$

and

$$u(0)=\begin{bmatrix}u_1 \\ u_2\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$$

Find eigenvalues and eigenvectors of $A$ :

$$⁍$$

$$x_1 = \begin{bmatrix}2 \\ 1\end{bmatrix},x_2 = \begin{bmatrix}1 \\ -1\end{bmatrix}$$

Then the general solution is:

$$u(t)=c_1e^{\lambda_1t}x_1+c_2e^{\lambda_2t}x_2$$

= $c_1e^{\lambda_1t}x_1$ and $c_2e^{\lambda_2t}x_2$ are both solutions.

e.g. plug $u=e^{\lambda_1t}x_1$ in $\frac{\mathrm{d}u}{\mathrm{d}t}=Au$ ,

$$\lambda_1e^{\lambda_1t}x_1=Ae^{\lambda_1t}x_1 \\ \Rightarrow \lambda_1x_1=Ax_1$$

1. stability ( $u(t) \to 0$ ) : $\mathrm{Re} \lambda < 0$
2. steady state : $\lambda_1 = 0$ and other $\mathrm{Re} \lambda < 0$
3. blow up : $\forall \mathrm{Re}\lambda<0$

for $2\times 2$ matrices, how to get $\mathrm{Re}\lambda_1<0, \mathrm{Re}\lambda_2<0$ ?

$$\lambda_1+\lambda_2=\mathrm{tr}(A)<0\\ \lambda_1\lambda_2=\det(A)>0$$

Uncouple $\frac{\mathrm{d}u}{\mathrm{d}t}=Au$

Set

$$u=Sv$$

$$\frac{\mathrm{d}u}{\mathrm{d}t}=S\frac{\mathrm{d}v}{\mathrm{d}t}=Au=ASv$$

$$\Rightarrow \frac{\mathrm{d}v}{\mathrm{d}t}=S^{-1} A S v=\Lambda v$$

i.e.

$$\frac{\mathrm{d}v_1}{\mathrm{d}t}=\lambda_1 v_1\\\vdots$$

$v$’s are uncoupled.

So

$$⁍$$

$$\therefore u=Sv = Se^{\Lambda t}v(0)=Se^{\Lambda t}S^{-1}u(0) \\= e^{At}u(0)$$

upd: what $e^{\Lambda t}$ means? e^{\Lambda t}=\begin{bmatrix} e^{\lambda_1 t} & 0 & 0\ 0 & \ddots  & 0\ 0 & 0 & e^{\lambda_n t}\end{bmatrix}

Meaning of exponential of a matrix ( $Se^{\Lambda t}S^{-1} = e^{At}$ )

Taylor Series $e^{x}=\sum \frac{x^{n}}{n !}$ :

$$e^{A t}=I+A t+\frac{(A t)^{2}}{2}+\frac{(A t)^{3}}{6}+\cdots+\frac{(A t)^{n}}{n !}+\cdots$$

also, $\frac{1}{1-x}=\sum_{0}^{\infty} x^{n}$ : (for $|\lambda(At)|<1$ )

$$(I-A t)^{-1}=I+A t+(A t)^{2}+(A t)^{3} t+\cdots$$

Why $I+A t+\frac{(A t)^{2}}{2}+\frac{(A t)^{3}}{6}+\cdots+\frac{(A t)^{n}}{n !}+\cdots=Se^{\Lambda t}S^{-1}$ ?

$$\begin{aligned} e^{At} & = I+A t+\frac{(A t)^{2}}{2}+\frac{(A t)^{3}}{6}+\cdots+\frac{(A t)^{n}}{n !}+\cdots \\ & = SS^{-1}+S\Lambda S^{-1}t + \frac{S\Lambda^2 S^{-1}t^2}{2}+\cdots \\ & = Se^{\Lambda t}S^{-1} \end{aligned}$$

Because $\Lambda$ is diagonal, (use Taylor series)

$$e^{\Lambda t}=\begin{bmatrix} e^{\lambda_1 t} & 0 & 0\\ 0 & \ddots  & 0\\ 0 & 0 & e^{\lambda_n t}\end{bmatrix}$$

2-order differential equation

Solve

$$y''+by'+ky=0$$

let

$$u= \begin{bmatrix}y' \\ y\end{bmatrix}\\ u'= \begin{bmatrix}y'' \\ y'\end{bmatrix}=\begin{bmatrix}-b & -k \\1 & 0\end{bmatrix}\begin{bmatrix}y' \\ y\end{bmatrix}$$

that is in the form of $\frac{\mathrm{d}u}{\mathrm{d}t}=Au$