Author Topic: Final review of 9.1  (Read 336 times)

Meiyi Lu

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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

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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

Victor Ivrii

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Re: Final review of 9.1
« Reply #2 on: November 28, 2018, 04:34:33 AM »
You were asked to compare two solutions: with supplied initial conditions and slightly modified