# Toronto Math Forum

## MAT244-2014F => MAT244 Math--Tests => Quiz 4 => Topic started by: Yuan Bian on November 12, 2014, 08:56:10 PM

Title: Q4 problem 1
Post by: Yuan Bian 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*}

(http://<a href="http://www.imagebam.com/image/f84041363999992" target="_blank"><img src="http://thumbnails109.imagebam.com/36400/f84041363999992.jpg" alt="imagebam.com"></a>)
(http://<a href="http://www.imagebam.com/image/64d11b364003377" target="_blank"><img src="http://thumbnails111.imagebam.com/36401/64d11b364003377.jpg" alt="imagebam.com"></a>)
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
Title: Re: Q4
Post by: Yuan Bian 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

Title: Re: Q4 problem 1
Post by: Guang_Yang on November 12, 2014, 09:53:00 PM
7.5 #5
Title: Re: Q4 problem 1
Post by: Tao Hu 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.  ;)
Title: Re: Q4 problem 1
Post by: Yuan Bian 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.
Title: Re: Q4 problem 1
Post by: Shuyang Wang 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):