Author Topic: TT2 # 4  (Read 6082 times)

Victor Ivrii

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TT2 # 4
« on: November 19, 2014, 08:47:30 PM »
Find the general solution and determine the type of behavior near the origin of the the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t= -6x + 5y\ , \\
&y'_t= -5x + 4y .
\end{aligned}\right.
\end{equation*}

Shuyang Wang

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Re: TT2 # 4
« Reply #1 on: November 19, 2014, 10:13:39 PM »
\begin{equation*} \textbf{x}'=\begin{pmatrix}\hphantom{-}-6 & 5\\\hphantom{-}-5 &4 \end{pmatrix}\textbf{x}\ . \end{equation*}

find eigenvalues

\begin{equation*} \det (A - rI) = \left|\begin{matrix}-6 - r &5\\-5&  4 - r\end{matrix}\right| =  r^2+ 2r + 1 = 0\implies r_1=r_2=-1\end{equation*}

then, find eigenvectors

\begin{equation*} \begin{pmatrix} -6 - r & \hphantom{-}5\\  \hphantom{-}-5 &4 -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}\end{equation*}

generalized eigenvector

\begin{equation*} \begin{pmatrix} -6 - r & \hphantom{-}5\\  \hphantom{-}-5 &4 -r\end{pmatrix}\mathbf{\xi}^2=\mathbf{\xi}^1 \end{equation*}
\begin{equation*} \mathbf{\xi}^2=\begin{pmatrix}0\\1/5\end{pmatrix}\end{equation*}

so

\begin{equation*}\mathbf{x}(t)= C_1e^{-t}\begin{pmatrix}1\\1\end{pmatrix}+ C_2e^{-t}\left(  t \begin{pmatrix}1\\1\end{pmatrix} + \begin{pmatrix}0\\1/5\end{pmatrix}\right)\end{equation*}

phase portrait(improper node, stable)

Chang Peng (Eddie) Liu

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Re: TT2 # 4
« Reply #2 on: November 20, 2014, 12:08:35 AM »
I just showed the steps for finding the second eigenvector..

After that, you would proceed to plug it into the standard general equation for a repeated eigenvalue question.

Chang Peng (Eddie) Liu

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Re: TT2 # 4
« Reply #3 on: November 20, 2014, 12:09:28 AM »
Graph of #4

Alex Hang

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Re: TT2 # 4
« Reply #4 on: November 20, 2014, 01:05:01 AM »
Eddie, shouldn't it be asymptotically stable?

Victor Ivrii

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Re: TT2 # 4
« Reply #5 on: November 20, 2014, 04:07:36 AM »
In the case of improper node one can also use clock-wise / counter-clock-wise criteria (sign of the top-right element in the matrix $A$).

Alex Hang

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Re: TT2 # 4
« Reply #6 on: November 20, 2014, 11:03:06 PM »
So professor, since the top right element is positive, does that mean it has clockwise orientation, so shouldn't it be asymptotically stable?
« Last Edit: November 20, 2014, 11:06:51 PM by Alex Hang »

Victor Ivrii

Completely correct solution was provided by Shuyang Wang. Orientation and stability are different questions: stability follows from the sign of the eigenvalue ($r<0$; $r>0$ means instability) and orientation from the sign of the top-right element (which in the framework of the focal point/center/improper node opposite to the sign of the bottom-left)