### Author Topic: Q4 problem 1  (Read 3289 times)

#### Yuan Bian

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• Posts: 14
• Karma: 4
##### Q4 problem 1
« on: November 12, 2014, 08:56:10 PM »
7.5 p 407 # 5

Draw a full phase portrait and describe completely the type of the stationary point  indicating if it is stable or unstable and in the case of the center and focus indicate orientation (clockwise or counter-clockwise)
\begin{equation*}
\textbf{x}'=\begin{pmatrix}
-2 & \hphantom{-}1\\  \hphantom{-}1 &-2
\end{pmatrix}\textbf{x}\ .
\end{equation*}

sorry two picture can't see normally, there are two links of picture of q1 and q2
http://www.imagebam.com/image/f84041363999992
http://www.imagebam.com/image/64d11b364003377
« Last Edit: November 12, 2014, 09:43:55 PM by Victor Ivrii »

#### Yuan Bian

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• Posts: 14
• Karma: 4
##### Re: Q4
« Reply #1 on: November 12, 2014, 08:56:52 PM »
Q1: 7.5 #5
r2+4r+3=0
(r+3)(r+1)=0
r1=-3, r2=-1
b/c r1<r2<0, stable node
sorry I don't know how to upload picture...I tried, but I failed
now I share the link of picture in the first post

« Last Edit: November 12, 2014, 09:15:56 PM by Yuan Bian »

#### Guang_Yang

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• Posts: 2
• Karma: 1
##### Re: Q4 problem 1
« Reply #2 on: November 12, 2014, 09:53:00 PM »
7.5 #5

#### Tao Hu

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##### Re: Q4 problem 1
« Reply #3 on: November 13, 2014, 03:44:01 PM »
I think two real Eigenvalue with same sign should give us a stable node. The direction in this case would be towards origin, since both x1 and x2 approach 0 as t tends to infinity.

#### Yuan Bian

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• Posts: 14
• Karma: 4
##### Re: Q4 problem 1
« Reply #4 on: November 13, 2014, 08:55:11 PM »
no, two distinct real eigenvalues both >0, it's unstable node; two distinct real eigenvalues both <0, then it's stable node.
« Last Edit: November 13, 2014, 09:22:39 PM by Yuan Bian »

#### Shuyang Wang

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##### Re: Q4 problem 1
« Reply #5 on: November 13, 2014, 09:27:50 PM »
\begin{equation*} \textbf{x}'=\begin{pmatrix} -2 & \hphantom{-}1\\  \hphantom{-}1 &-2 \end{pmatrix}\textbf{x}\ . \end{equation*}
find eigenvalues
\begin{equation*} \det (A - rI) = \left|\begin{matrix}-2 - r &1\\1&  - 2 - r\end{matrix}\right| =  r^2+ 4r + 3 = 0\implies r_1=-3, r_2=-1\end{equation*}
then, find eigenvectors
\begin{equation*} \begin{pmatrix} -2 - r & \hphantom{-}1\\  \hphantom{-}1 &-2 -r\end{pmatrix}\begin{pmatrix}\mathbf{\xi}_1\\\mathbf{\xi}_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}
then
\begin{equation*}\mathbf{\xi}^1 =\begin{pmatrix}1\\-1\end{pmatrix} , \mathbf{\xi}^2 =\begin{pmatrix}1\\1\end{pmatrix}\end{equation*}
so
\begin{equation*}\mathbf{x}= C_1e^{-3t}\begin{pmatrix}1\\-1\end{pmatrix}+ C_2e^{-t}\begin{pmatrix}1\\1\end{pmatrix}\end{equation*}
plot(stable):
« Last Edit: November 13, 2014, 09:44:38 PM by Shuyang Wang »