Author Topic: Problem 1, night sections  (Read 1119 times)

Victor Ivrii

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Problem 1, night sections
« on: November 20, 2013, 08:39:11 PM »
Find the general solution of the given system of equations and describe the behaviour of
the solution as $t\to \infty$:
\begin{equation*}
\mathbf{x}'=\begin{pmatrix}3 &-2\\2 &-2\end{pmatrix}\mathbf{x}.
\end{equation*}

Yangming Cai

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Re: Problem 1, night sections
« Reply #1 on: November 20, 2013, 09:23:47 PM »
 answer as follow

Victor Ivrii

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Re: Problem 1, night sections
« Reply #2 on: November 21, 2013, 05:12:03 AM »
One needs to distinguish cases of $t\to +\infty$ and $t\to -\infty$.

Yangming Cai

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Re: Problem 1, night sections
« Reply #3 on: November 21, 2013, 12:21:49 PM »
\begin{array}{l}\det (A - rI) = \left( {\begin{array}{*{20}{c}}{3 - r}&{ - 2}\\2&{ - 2 - r}\end{array}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {r^2} - 1 - 2 = 0\\r = 2,\;r =  - 1\\{\rm{when}}\;r = 2\\1{\xi _1} = 2{\xi _2}{\rm{and}}{\xi ^1} = \left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\\{\rm{when}}\;r =  - 1\\2{\xi _1} = 1{\xi _2}{\rm{and}}{\xi ^2} = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\\x = {c_1}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right){e^{ - t}}\end{array}
\begin{array}{l}{\rm{for }}{c_1} = 0,as\;t \to  + \infty ,x \to 0.as\;t \to  - \infty ,x \to  + \infty \\{\rm{for }}{c_1} \ne 0,as\;t \to  + \infty ,{\rm{the first term dominates, so }}x \to  + \infty .\\{\rm{               }}as\;t \to  - \infty ,{\rm{the second term dominates}},sox \to  + \infty \end{array}
« Last Edit: November 21, 2013, 12:53:00 PM by Yangming Cai »

Victor Ivrii

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Re: Problem 1, night sections
« Reply #4 on: November 21, 2013, 05:33:53 PM »
What a brain dead s.w wrote this code?

$$
\det (A - rI) = \left|\begin{matrix}3 - r &-2\\2&  - 2 - r\end{matrix}\right| =  r^2-r - 2 = 0\implies r_1=2, r_2=-1
$$
$r = 2\implies \xi_1=2\xi_2$ and we take $\mathbf{\xi}^1= \begin{pmatrix}2\\1\end{pmatrix}$,
$r=-1\implies 2\xi_1=\xi_2$  and we take $\mathbf{\xi}^1= \begin{pmatrix}1\\2\end{pmatrix}$.
Finally
$$
\mathbf{x}= C_1e^{2t}\mathbf{\xi}^1= \begin{pmatrix}2\\1\end{pmatrix}+ C_2e^{-t}\begin{pmatrix}1\\2\end{pmatrix}
$$

As $t\to +\infty$  $|\mathbf{x}(t)|\to \infty $ unless $C_1=0$; then $|\mathbf{x}(t)|\to 0$,
As $t\to -\infty$  $|\mathbf{x}(t)|\to \infty $ unless $C_2=0$; then $|\mathbf{x}(t)|\to 0$.

It is a saddle
« Last Edit: November 21, 2013, 05:37:07 PM by Victor Ivrii »