Author Topic: Problem 4 (noon)  (Read 1492 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2540
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Problem 4 (noon)
« on: November 19, 2019, 04:24:35 AM »
Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
1 & 3\\
-2 &-3\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.

Yuying Chen

  • Jr. Member
  • **
  • Posts: 14
  • Karma: 8
    • View Profile
Re: Problem 4 (noon)
« Reply #1 on: November 19, 2019, 05:17:25 AM »
$\det(A-\lambda I)=0\\$
$\begin{vmatrix}
1-\lambda & 3 \\
-2 &  -3-\lambda\\
\end{vmatrix}=(1-\lambda)(-3-\lambda)+6=0\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad \therefore \lambda=-1\pm \sqrt 2 i\\$
$\text{when $\lambda =-1+ \sqrt 2 i$},\\$
$
\begin{pmatrix}
2-\sqrt 2 i & 3 \\
-2 & -2-\sqrt 2 i
\end{pmatrix}= \begin{pmatrix}
2+\sqrt 2 i \\
-2
\end{pmatrix}\\$
$e^{(-1+ \sqrt 2 i)t}\begin{pmatrix}
2+\sqrt 2 i \\
-2
\end{pmatrix}=e^{-t}\begin{pmatrix}
2+\sqrt 2 i \\
-2
\end{pmatrix}(\cos\sqrt2t+i\sin\sqrt 2t)\\$
$\qquad\qquad\qquad\qquad =e^{-t}\begin{pmatrix}
2\cos\sqrt 2t+2i\sin\sqrt2t+\sqrt2i\cos\sqrt2t-\sqrt2\sin\sqrt2t \\
-2\cos\sqrt2t-2i\sin\sqrt2t
\end{pmatrix}\\$
$\qquad\qquad\qquad\qquad =e^{-t}i\begin{pmatrix}
2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\
-2\sin\sqrt2t
\end{pmatrix}+e^{-t}\begin{pmatrix}
2\cos\sqrt 2t -\sqrt2\sin\sqrt2t\\
-2\cos\sqrt2t
\end{pmatrix}\\$
$\therefore x(t)=c_1e^{-t}\begin{pmatrix}
2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\
-2\sin\sqrt2t
\end{pmatrix}+c_2e^{-t}\begin{pmatrix}
2\cos\sqrt 2t -\sqrt2\sin\sqrt2t \\
-2\cos\sqrt2t
\end{pmatrix}\\$
« Last Edit: November 20, 2019, 06:17:05 PM by Yuying Chen »

Ziqian Qiu

  • Newbie
  • *
  • Posts: 3
  • Karma: 0
    • View Profile
Re: Problem 4 (noon)
« Reply #2 on: November 19, 2019, 05:24:11 AM »
this is my solution
« Last Edit: November 19, 2019, 05:40:29 AM by Ziqian Qiu »

Ziqian Qiu

  • Newbie
  • *
  • Posts: 3
  • Karma: 0
    • View Profile
Re: Problem 4 (noon)
« Reply #3 on: November 19, 2019, 05:30:32 AM »
nvm I just updated the previous post
« Last Edit: November 19, 2019, 05:32:55 AM by Ziqian Qiu »

xuanzhong

  • Jr. Member
  • **
  • Posts: 11
  • Karma: 1
    • View Profile
Re: Problem 4 (noon)
« Reply #4 on: November 19, 2019, 05:47:01 AM »
here's the solution including sketching.

Changhao Jiang

  • Jr. Member
  • **
  • Posts: 8
  • Karma: 0
    • View Profile
Re: Problem 4 (noon)
« Reply #5 on: November 19, 2019, 05:50:48 AM »
To find eigenvalues, $det(A-\lambda x)=(1-\lambda)(-3-\lambda)-(-2)(3)=0$,we can get $\lambda = -1-\sqrt{2}$ or $\lambda = -1+\sqrt{2}$
To find eigenvectors, when $\lambda=-1-\sqrt{2}$,
$\begin{pmatrix}
2+\sqrt{2} & 3 \\
-2 & -2+\sqrt{2}
\end{pmatrix}
~
\begin{pmatrix}
2 & 2-\sqrt{2} \\
0 & 0
\end{pmatrix}$
the eigenvector is $\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}$
so $e^{(-1-\sqrt{2})t}\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}
= e^{-t}(\cos\sqrt{2}t-i\sin\sqrt{2}t)\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}
= e^{-t} (\begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + i\begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix})$
therefore, the general solution is $x(t)=c_1 e^{-t} \begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix}$
« Last Edit: November 19, 2019, 05:53:28 AM by Changhao Jiang »

NANAC

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 2
    • View Profile
Re: Problem 4 (noon)
« Reply #6 on: November 19, 2019, 09:21:22 AM »
Please see the attachment for the answer

baixiaox

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 0
    • View Profile
Re: Problem 4 (noon)
« Reply #7 on: November 19, 2019, 05:46:00 PM »
answer for tt2 question4

Mingdi Xie

  • Jr. Member
  • **
  • Posts: 14
  • Karma: 0
    • View Profile
Re: Problem 4 (noon)
« Reply #8 on: November 20, 2019, 03:07:38 PM »
This is my solution. :)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2540
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 4 (noon)
« Reply #9 on: November 24, 2019, 11:04:24 AM »
What everybody is missing

we see that characteristic roots $k_{1,2}=-1\pm \sqrt{2}i$ are complex, with negative real part. So, it is  stable focus  and with  clock-wise  orientation  since the bottom-left element is negative.