### Author Topic: Final review of 9.1  (Read 1679 times)

#### Meiyi Lu

• Jr. Member
• Posts: 5
• Karma: 1
##### Final review of 9.1
« on: November 27, 2018, 06:01:40 PM »
Determine the critical point $x=x_0$, and then classify its type and examine its stability by making the transformation $x=x^0 + u$.\\
\begin{equation*}
\frac{dx}{dt} = \begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}x +
\begin{bmatrix}
-1 \\ 5
\end{bmatrix}
\end{equation*}

#### Tianfangtong Zhang

• Full Member
• Posts: 16
• Karma: 15
##### Re: Final review of 9.1
« Reply #1 on: November 27, 2018, 06:09:55 PM »
\begin{equation*}
\begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}x^0 +
\begin{bmatrix}
-1 \\ 5
\end{bmatrix} = 0
\end{equation*}
\begin{equation*}
\begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}x^0 = -\begin{bmatrix}
-1 \\ 5
\end{bmatrix}
\end{equation*}
\begin{align*}
x_0 &=  \begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}^{-1}\begin{bmatrix}
1 \\ -5
\end{bmatrix}\\ &=
\begin{bmatrix}
-2 \\ 1
\end{bmatrix}
\end{align*}
Hence the critical point is $x_0 = -2$ and $y_0 = 1$

Let $x = x^0 + u$

Then $\frac{dx}{dt} = \frac{du}{dt}$ (as $\frac{dx^0}{dt} = 0$)

Then \begin{align*}
\frac{du}{dt} &= \begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}\begin{bmatrix}
-2 \\ 1
\end{bmatrix}  + \begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}u + \begin{bmatrix}
-1 \\ 5
\end{bmatrix} \\
&= -\begin{bmatrix}
-1 \\ 5
\end{bmatrix} +
\begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}u + \begin{bmatrix}
-1 \\ 5
\end{bmatrix}
\end{align*}
\begin{equation*}
\frac{du}{dt} = \begin{bmatrix}
-1 & -1 \\
2 & -1
\end{bmatrix}u
\end{equation*}

Let $u = ae^{rt}$
\begin{equation*}
\begin{bmatrix}
-1-r & -1 \\
2 & -1-r
\end{bmatrix}
\begin{bmatrix}
a_1 \\ a_2
\end{bmatrix} =
\begin{bmatrix}
0 \\ 0
\end{bmatrix}
\end{equation*}
\begin{equation*}
r^2 + 2r + 3 = 0
\end{equation*}
\begin{equation*}
r = -1 + \pm \sqrt{2}i
\end{equation*}

Thus the eigenvalue are $-1 + \pm \sqrt{2}i$, the critical point is asymptotically stable spiral point