Author Topic: Problem 4 (morning)  (Read 1026 times)

Victor Ivrii

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Problem 4 (morning)
« on: November 19, 2019, 04:23:46 AM »
Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
2 & -3\\
4 &-2\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.

Yiheng Bian

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Re: Problem 4 (morning)
« Reply #1 on: November 19, 2019, 04:28:39 AM »
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
2-\lambda & -3  \\
4 & -2-\lambda
\end{vmatrix}={\lambda}^2+8=0
$$
So
$$
\lambda_1=\sqrt{8}i\\
\lambda_2=-\sqrt{8}i
$$
$$
\text{when } \lambda=\sqrt{8}I\\
\begin{vmatrix}
2-\sqrt{8}i & -3  \\
4 & -2-\sqrt{8}i
\end{vmatrix} = \begin{vmatrix}
0  \\
0
\end{vmatrix}
$$
RREF:
$$
\begin{pmatrix}
2-\sqrt{8}i & -3 \\
0 & 0
\end{pmatrix}
\quad = \begin{pmatrix}
0 \\
0
\end{pmatrix}
\quad
$$
Let x_2=t
So we can get:
$$
{(2-\sqrt{8}i})x_1=3x_2=3t\\
x_1=\frac{3t}{2-\sqrt{8}i}
$$
So
$$
t*\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
e^{i\sqrt8t}\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad =
(cos(\sqrt8t)+isin(\sqrt8t))\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad=\begin{pmatrix}
\frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} +\frac{isin(\sqrt8t)+i\sqrt2cos(\sqrt8t)}{2}\\
cos(\sqrt8t )+ isin(\sqrt8t))
\end{pmatrix}
\quad
$$
So
$$
y=c_1\begin{pmatrix}
\frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} \\
cos(\sqrt8t)
\end{pmatrix}
\quad +c_2\begin{pmatrix}
\frac{sin(\sqrt8t)+\sqrt2cos(\sqrt8t)}{2}\\
sin(\sqrt8t))
\end{pmatrix}
\quad
$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln
« Last Edit: November 24, 2019, 10:58:29 AM by Victor Ivrii »

Lan Cheng

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Re: Problem 4 (morning)
« Reply #2 on: November 19, 2019, 05:59:51 AM »
a) First we should solve for eigenvalues:

let $det\begin{bmatrix}2-\lambda & -3\\
4 & -2-\lambda
\end{bmatrix}=0.\lambda_{1}=2\sqrt{2}i,\lambda_{2}=-2\sqrt{2}i.$

Second we need to solve for eigenvectors:

Let $(A-\lambda I)v=0.$ When $\lambda=2\sqrt{2}i,v=\begin{bmatrix}\sqrt{2}i+1\\
2
\end{bmatrix}.$

Therefore, $e^{(2\sqrt{2}i)t}\begin{bmatrix}\sqrt{2}i+1\\
2
\end{bmatrix}=(cos(2\sqrt{2}t)+isin(2\sqrt{2}t))\begin{bmatrix}\sqrt{2}i+1\\
2
\end{bmatrix}.$

$x(t)=C_{1}\begin{bmatrix}-\sqrt{2}sin(2\sqrt{2}t)+cos(2\sqrt{2}t)\\
2cos(2\sqrt{2}t)
\end{bmatrix}+C_{2}\begin{bmatrix}\sqrt{2}cos(2\sqrt{2}t)+sin(2\sqrt{2}t)\\
2sin(2\sqrt{2}t)
\end{bmatrix}.$

b) See photo below.

Ruojing Chen

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Re: Problem 4 (morning)
« Reply #3 on: November 19, 2019, 06:04:30 AM »
$$x'= \left (
\begin{matrix}
2 & -3 \\
4 & -2
\end{matrix}
\right ) x$$

$$det(A-\lambda I)=0$$
$$(2-\lambda)(-2-\lambda)-(-3)*4=0$$
$$-4+\lambda^2+12=0$$
$$\lambda^2=-8$$
$$\Rightarrow \lambda=\pm2\sqrt{2}i$$

$$when \lambda=2\sqrt{2}i$$
$$\left (
\begin{matrix}
2-2\sqrt{2}i & -3 \\
4 & -2-2\sqrt{2}i
\end{matrix}
\right ) \left (
\begin{matrix}
x_1 \\
x_2
\end{matrix}
\right ) =\left (
\begin{matrix}
0\\
0
\end{matrix}
\right ) $$
$$(2-2\sqrt{2}i)x_1=3x_2$$
$$\Rightarrow x=t
\left (
\begin{matrix}
3 \\
2-2\sqrt{2}i
\end{matrix}
\right )$$

$$e^{2\sqrt{2}it}=t
\left (
\begin{matrix}
3 \\
2-2\sqrt{2}i
\end{matrix}
\right )$$
$$=(cos2\sqrt{2}t+isin2\sqrt{2}t)\left (
\begin{matrix}
3 \\
2-2\sqrt{2}i
\end{matrix}
\right )$$

$$\therefore =c_1
\left(
\begin{matrix}
3cos2\sqrt{2} \\
2cos2\sqrt{2}+2\sqrt{2}sin2\sqrt{2}
\end{matrix}
\right )+ c_2\left(\begin{matrix}
3sin2\sqrt{2} \\
2sin2\sqrt{2}-2cos2\sqrt{2}
\end{matrix}
\right )$$
« Last Edit: November 19, 2019, 06:37:58 AM by Ruojing Chen »

Ruojing Chen

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Re: Problem 4 (morning)
« Reply #4 on: November 19, 2019, 06:17:30 AM »
a) sketch

Yue Sagawa

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Re: Problem 4 (morning)
« Reply #5 on: November 19, 2019, 06:18:55 AM »
I found the general real solution.

Yiran Wang

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Re: Problem 4 (morning)
« Reply #6 on: November 19, 2019, 12:58:03 PM »
This is the solution

Yiran Wang

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Re: Problem 4 (morning)
« Reply #7 on: November 19, 2019, 01:01:27 PM »
 :)

Victor Ivrii

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Re: Problem 4 (morning)
« Reply #8 on: November 24, 2019, 11:00:36 AM »
What everybody is missing:

we see that characteristic roots $k_{1,2}= \pm \sqrt{8}i$ are purely imaginary. So, it is  center  and with  counter-clock-wise  orientation  since the bottom-left element is positive.