Author Topic: Problem 3 (morning)  (Read 725 times)

Victor Ivrii

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Problem 3 (morning)
« on: November 19, 2019, 04:21:57 AM »
(a) Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix} -2 &1\\
-1 &0\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.

(b) Find the general solution
$$
\mathbf{x}'=\begin{pmatrix} -2 &1\\
-1 &0\end{pmatrix}\mathbf{x}+
\begin{pmatrix} 0  \\
\dfrac{e^{-t}} {t^2+1} \end{pmatrix}.
$$

Yiheng Bian

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Re: Problem 3 (morning)
« Reply #1 on: November 19, 2019, 04:28:19 AM »
(a):
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
-2-\lambda & 1  \\
-1 & -\lambda
\end{vmatrix}=2\lambda+{\lambda}^2+1=0
$$
So
$$
(\lambda+1)^2=0\\
\lambda_1=\lambda_2=-1
$$
Then:
$$
(A-{\lambda}I)x=0
$$
$$
\begin{pmatrix}
-1 & 1 \\
-1 & 1
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
RREF
$$
\begin{pmatrix}
-1 & 1 \\
0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
$$
\text{if let }x_2=t\\
x_1=x_2=t
$$
So
$$
t*{\begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad}
$$
Since we just have only one eigenvector:
$$
\begin{pmatrix}
-1 & 1 \\
-1 & 1
\end{pmatrix}
\quad= \begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad
$$
$$
\begin{pmatrix}
-1 & 1 \\
0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
1  \\
0
\end{pmatrix}
\quad
$$
$$
\text{So the other eigenvector is:}= \begin{pmatrix}
0  \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
y=c_1e^{-t}*{\begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad}+c_2e^{-t}[t*{\begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad} +{\begin{pmatrix}
0  \\
1
\end{pmatrix}
\quad}]
$$




(b):
$$
\phi = \begin{pmatrix}
e^{-t} & te^{-t} \\
e^{-t} & e^{-t}(t+1)
\end{pmatrix}
\quad
$$
$$
\phi * u' = g(t)
$$
$$
\begin{pmatrix}
e^{-t} & te^{-t} \\
e^{-t} & e^{-t}(t+1)
\end{pmatrix}
\quad *{\begin{pmatrix}
u_1'  \\
u_2'
\end{pmatrix}
\quad}=\begin{pmatrix}
0  \\
\frac{e^{-t}}{t^2 + 1 }
\end{pmatrix}
\quad
$$
Simplify we can get:
$$
u_1'=-\frac{t}{t^2 + 1}\\
u_2'=\frac{1}{t^2+1}
$$
Therefore:
$$
u_1=-0.5ln(t^2+1)+c_1\\
u_2=arctant+c_2
$$
Finally:
$$
x=\phi * u=(-0.5ln(t^2+1)+c_1)*{\begin{pmatrix}
e^{-t}  \\
e^{-t}
\end{pmatrix}
\quad} +(arctant+c_2)*\begin{pmatrix}
te^{-t}  \\
e^{-t}(t+1)
\end{pmatrix}
\quad
$$
« Last Edit: November 19, 2019, 06:09:53 PM by Yiheng Bian »

xuanzhong

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Re: Problem 3 (morning)
« Reply #2 on: November 19, 2019, 05:54:32 AM »
HERE'S THE SOLUTION FOR SKETCHING.

Ruojing Chen

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Re: Problem 3 (morning)
« Reply #3 on: November 19, 2019, 06:03:57 AM »
a)$$x'= \left( \begin{matrix}
-2 & 1 \\
-1 & 0 \\
\end{matrix} \right )x+\left( \begin{matrix}
0 \\
\frac{e^{-t}}{t^2+1} \\
\end{matrix} \right )$$

$$det(A-\lambda I)=0$$
$$(-2-\lambda)(-\lambda)+1=0$$
$$\lambda^2+2\lambda+1=0$$
$$\lambda_1=\lambda_2=-1$$

$$when \lambda =-1$$
$$(A-\lambda I)x=0$$
$$\left( \begin{matrix}
-1 & 1 \\
-1 & 1 \\
\end{matrix} \right )\left( \begin{matrix}
x_1 \\
x_2 \\
\end{matrix} \right )=\left( \begin{matrix}
0 \\
0 \\
\end{matrix} \right )$$
$$x_1=x_2 \Rightarrow x=t\left( \begin{matrix}
1 \\
1 \\
\end{matrix} \right )$$

$$\left( \begin{matrix}
-1 & 1 \\
-1 & 1 \\
\end{matrix} \right )\left( \begin{matrix}
x_1 \\
x_2 \\
\end{matrix} \right )=\left( \begin{matrix}
1 \\
1 \\
\end{matrix} \right )$$
$$x_1+x_2=1 \Rightarrow x=t\left( \begin{matrix}
0 \\
1 \\
\end{matrix} \right )$$

$$\therefore y=c_1e^{-t}\left( \begin{matrix}
1 \\
1 \\
\end{matrix} \right )+c_2e^{-t}(\left( \begin{matrix}
1 \\
1 \\
\end{matrix} \right )t+\left( \begin{matrix}
0 \\
1 \\
\end{matrix} \right ))$$

B)$$\phi u'=g(t)$$
$$\left( \begin{matrix}
e^{-t} & e^{-t}t \\
e^{-t} & e^{-t}t+e^{-t}\\
\end{matrix} \right ) \left( \begin{matrix}
u_1' \\
u_2' \\
\end{matrix} \right )=\left( \begin{matrix}
0 \\
\frac{e^{-t}}{t^2+1} \\
\end{matrix} \right )$$
$$
\left \{
             \begin{array}{lr}
             u_1'=-\frac{t}{t^2+1} &\\
             u_2'=\frac{1}{t^2+1}
             \end{array}
\right. $$
$$\Rightarrow \left \{
            \begin{array}{lr}
             u_1=-\frac{1}{2} \ln(t^2+1)&\\
             u_2=\arctan t
             \end{array}
\right. $$
$$x=\phi u$$
$$\therefore x=\left( \begin{matrix}
e^{-t} & e^{-t}t \\
e^{-t} & e^{-t}t+e^{-t}\\
\end{matrix} \right )\left( \begin{matrix}
 -\frac{1}{2}\ln(t^2+1)\\
\arctan t\\
\end{matrix} \right )$$

$$x=(-\frac{1}{2}\ln(t^2+1)+c_1)\left( \begin{matrix}
e^{-t} \\
e^{-t} \\
\end{matrix} \right )+(c_2+\arctan t)\left( \begin{matrix}
e^{-t}t \\
e^{-t}t+e^{-t} \\
\end{matrix} \right) $$

OK, but LaTeX sucks:
1) \det should be escaped as well
2) Text should not be  included in math formulae, or included  through  \tex{blah blah } to make it upright and properly spaced

3) Directions are opposite
« Last Edit: November 24, 2019, 09:51:05 AM by Victor Ivrii »

Yiran Wang

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Re: Problem 3 (morning)
« Reply #4 on: November 19, 2019, 12:17:53 PM »
Det(A-rI)=0
(-2-r)(-r)+1=2r+r^2+1=0
r1=r2=-1
(A-rI)=0
we have eigenvector (1 1)
Let (A-rI)=(1 1)
We have another eignvector (-1 0)
We have x=C1e^-t *(1 1)+ C2e^-t *[ t(1 1)+(-1 0)]
e^-t u1+(te^-t – e^-t)u2=0   e^-t u1+te^-t u2=e^-t/(t^2+1)
we have u1=-t/(t^2+1)
u2=1/(t^2+1)
U1=-0.5ln(t^2+1)+C1
U2=arctant+C2
X=U1(e^-t  e^-t)+ U2(te^-t  te^-t +e^-t)

Yiran Wang

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Re: Problem 3 (morning)
« Reply #5 on: November 19, 2019, 12:27:05 PM »
repeated root graph

Mingdi Xie

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Re: Problem 3 (morning)
« Reply #6 on: November 19, 2019, 02:47:01 PM »
 :)

Victor Ivrii

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Re: Problem 3 (morning)
« Reply #7 on: November 24, 2019, 09:55:08 AM »
What everybody is missing
In problem got lost "classify point $(0,0)$"

stable improper node; since the bottom left element is negative, it is clockwise