Q4 problem 1

[Yuan Bian]

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

[Yuan Bian]

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

[Guang_Yang]

7.5 #5

[Tao Hu]

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]

no, two distinct real eigenvalues both >0, it's unstable node; two distinct real eigenvalues both <0, then it's stable node.

[Shuyang Wang]

\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):