Author Topic: Problem 4 (main sitting)  (Read 1489 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2554
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Problem 4 (main sitting)
« on: November 19, 2019, 04:23:20 AM »
Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
3 & 3\\
-2 &-1\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.

Yiheng Bian

  • Full Member
  • ***
  • Posts: 29
  • Karma: 12
    • View Profile
Re: Problem 4 (main sitting)
« Reply #1 on: November 19, 2019, 04:30:28 AM »
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
3-\lambda & 3  \\
-2 & -1-\lambda
\end{vmatrix}={\lambda -1}^2+2=0
$$
WRONG So
$$
\lambda_1=1 +\sqrt{2}i\\
\lambda_2=1-\sqrt{2}i
$$
$$
\text{when } \lambda=1 +\sqrt{2}i\\
\begin{vmatrix}
2-\sqrt{2}i & 3  \\
-2 & -2-\sqrt{2}i
\end{vmatrix} = \begin{vmatrix}
0  \\
0
\end{vmatrix}
$$
RREF:
$$
\begin{pmatrix}
2-\sqrt{2}i & 3 \\
0 & 0
\end{pmatrix}
\quad = \begin{pmatrix}
0 \\
0
\end{pmatrix}
\quad
$$

Let x_2=t
So we can get:
$$
{(2-\sqrt{2}i})x_1=-3x_2=-3t\\
x_1=\frac{-3t}{2-\sqrt{2}i}
$$
So
$$
t*\begin{pmatrix}
-1-\frac{\sqrt{2}i}{2} \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
e^{1+i\sqrt2t}\begin{pmatrix}
-1-\frac{\sqrt{2}i}{2} \\
1
\end{pmatrix}
\quad =

e^t[\begin{pmatrix}
-cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\
cos(\sqrt{2}t)
\end{pmatrix}
\quad +
i\begin{pmatrix}
-sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\
sin(\sqrt{2}t)
\end{pmatrix}
\quad]
$$
So, general solution:
$$
y=c_1e^t\begin{pmatrix}
-cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\
cos(\sqrt{2}t)
\end{pmatrix}
\quad +
c_2e^t\begin{pmatrix}
-sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\
sin(\sqrt{2}t)
\end{pmatrix}
\quad
$$
« Last Edit: November 24, 2019, 10:40:38 AM by Victor Ivrii »

Lisa Dou

  • Newbie
  • *
  • Posts: 1
  • Karma: 0
    • View Profile
Re: Problem 4 (main sitting)
« Reply #2 on: November 19, 2019, 05:17:54 AM »
The Sketch
« Last Edit: November 19, 2019, 05:26:23 AM by Lisa Dou »

Aparna

  • Newbie
  • *
  • Posts: 2
  • Karma: 0
    • View Profile
Re: Problem 4 (main sitting)
« Reply #3 on: November 19, 2019, 08:47:30 AM »
Computer-generated sketch:

Sifan Shao

  • Newbie
  • *
  • Posts: 2
  • Karma: 0
    • View Profile
Re: Problem 4 (main sitting)
« Reply #4 on: November 19, 2019, 09:15:01 AM »
The Sketch
we can do root2i*R1+R2 to simplify the matrix

Ruodan Chen

  • Newbie
  • *
  • Posts: 4
  • Karma: 2
    • View Profile
Re: Problem 4 (main sitting)
« Reply #5 on: November 19, 2019, 10:41:19 AM »
4) $x'=\left(\begin{array}{cc}
3 & 3\\
-2 & -1
\end{array}\right)x$

a)

$det(A-\lambda I)=det(\begin{array}{cc}
3-\lambda & 3\\
-2 & -1-\lambda
\end{array})=(3-\lambda)(-1-\lambda)+6=0$

$\lambda^{2}-2\lambda+3=0$

$(\lambda-1)^{2}=-2$

\$\lambda=1\pm\sqrt{2}i$

\$\lambda=1+\sqrt{2}i$

$(\begin{array}{ccc}
2-\sqrt{2}i & 3 & 0\\
-2 & -2-\sqrt{2}i & 0
\end{array}) = (\begin{array}{ccc}
2-\sqrt{2}i & 3 & 0\\
0 & 0 & 0
\end{array}) $

$v=(\begin{array}{c}2+\sqrt{2}i\\-2\end{array})$

$e^{(1+\sqrt{2}i)t}(\begin{array}{c}
2+\sqrt{2}i\\
-2
\end{array})=e^{t}(cos\sqrt{2}t+isin\sqrt{2}t)(\begin{array}{c}2+\sqrt{2}i\\-2\end{array}) = e^{t}(\begin{array}{c}
2cos\sqrt{2}t+2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t-2isin\sqrt{2}t
\end{array})=e^{t}(\begin{array}{c}
2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t
\end{array})+ie^{t}(\begin{array}{c}
2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\
-2isin\sqrt{2}t
\end{array})$

$x(t)=c_{1}e{}^{t}(\begin{array}{c}
2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t
\end{array})+c_{2}e{}^{t}(\begin{array}{c}
2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\
-2isin\sqrt{2}t
\end{array})$
You should consider real solutions
« Last Edit: November 24, 2019, 10:44:11 AM by Victor Ivrii »

yueyangyu

  • Jr. Member
  • **
  • Posts: 9
  • Karma: 0
    • View Profile
Re: Problem 4 (main sitting)
« Reply #6 on: November 19, 2019, 12:52:21 PM »
4) $x'=\left(\begin{array}{cc}
3 & 3\\
-2 & -1
\end{array}\right)x$

a)

$det(A-\lambda I)=det(\begin{array}{cc}
3-\lambda & 3\\
-2 & -1-\lambda
\end{array})=(3-\lambda)(-1-\lambda)+6=0$

$\lambda^{2}-2\lambda+3=0$

$(\lambda-1)^{2}=-2$

$\lambda=1\pm\sqrt{2}i$

$\lambda=1+\sqrt{2}i$

$(\begin{array}{ccc}
2-\sqrt{2}i & 3 & 0\\
-2 & -2-\sqrt{2}i & 0
\end{array}) = (\begin{array}{ccc}
2-\sqrt{2}i & 3 & 0\\
0 & 0 & 0
\end{array}) $

$v=(\begin{array}{c}2+\sqrt{2}i\\-2\end{array})$

$e^{(1+\sqrt{2}i)t}(\begin{array}{c}
2+\sqrt{2}i\\
-2
\end{array})=e^{t}(cos\sqrt{2}t+isin\sqrt{2}t)(\begin{array}{c}2+\sqrt{2}i\\-2\end{array}) = e^{t}(\begin{array}{c}
2cos\sqrt{2}t+2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t-2isin\sqrt{2}t
\end{array})=e^{t}(\begin{array}{c}
2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t
\end{array})+ie^{t}(\begin{array}{c}
2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\
-2isin\sqrt{2}t
\end{array})$

$x(t)=c_{1}e{}^{t}(\begin{array}{c}
2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t
\end{array})+c_{2}e{}^{t}(\begin{array}{c}
2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\
-2isin\sqrt{2}t
\end{array})$

zhaorola

  • Newbie
  • *
  • Posts: 1
  • Karma: 0
    • View Profile
Re: Problem 4 (main sitting)
« Reply #7 on: November 19, 2019, 01:25:45 PM »
Computer-generated sketch:

This doesn't look right. It should instead be an unstable spiral. I've attached an accurate phase portrait.

nayan

  • Newbie
  • *
  • Posts: 2
  • Karma: 0
    • View Profile
Re: Problem 4 (main sitting)
« Reply #8 on: November 19, 2019, 01:53:29 PM »
Computer-generated sketch:

This doesn't look right. It should instead be an unstable spiral. I've attached an accurate phase portrait.

Yes this appears to be the correct phase portrait.

Mingdi Xie

  • Jr. Member
  • **
  • Posts: 14
  • Karma: 0
    • View Profile
Re: Problem 4 (main sitting)
« Reply #9 on: November 19, 2019, 09:58:57 PM »
Here is my solution

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2554
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 4 (main sitting)
« Reply #10 on: November 24, 2019, 10:45:37 AM »
What everybody is missing

it is  unstable focus  and with  clock-wise  orientation  since the bottom-left element is negative.

« Last Edit: November 24, 2019, 10:47:19 AM by Victor Ivrii »