Author Topic: TT2-P3  (Read 1500 times)

Victor Ivrii

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TT2-P3
« on: November 20, 2018, 05:46:11 AM »
(a) Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix}\hphantom{-}2 &\hphantom{-}1\\
-3 &-2\end{pmatrix}\mathbf{x}.$$

(b) Sketch corresponding trajectories. Describe the picture (stable/unstable, node, focus, center, saddle).

(c) Solve
$$
\mathbf{x}'=\begin{pmatrix}\hphantom{-}2 &\hphantom{-}1\\
-3 &-2\end{pmatrix}\mathbf{x} +
\begin{pmatrix}\hphantom{-} \frac{4}{e^t+e^{-t}} \\
-\frac{12}{e^t+e^{-t}}\end{pmatrix},\qquad
\mathbf{x}(0)=\begin{pmatrix} 0 \\
0\end{pmatrix}.
$$

Boyu Zheng

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Re: TT2-P3
« Reply #1 on: November 20, 2018, 07:39:51 AM »
here is my answer
« Last Edit: November 20, 2018, 01:01:42 PM by Boyu Zheng »

Zhanhao Ye

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Re: TT2-P3
« Reply #2 on: November 20, 2018, 09:51:51 AM »
The attachment is my solution.
« Last Edit: November 20, 2018, 11:20:35 AM by Ye Zhanhao »

Jingze Wang

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Re: TT2-P3
« Reply #3 on: November 20, 2018, 10:05:33 AM »
a)First, try to find the eigenvalues with respect to the parameter


$A=\begin{bmatrix}
2&1\\
-3&-2\\
\end{bmatrix}$


$det(A-rI)=(2-r)/(-2-r)+3=0$

$r^2-1=0$

$r=\pm1$

When r=-1, $3x_1+x_2=0$
We get $\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}$ is the corresponding eigenvector
When r=1,  $x_1+x_2=0$
We get $\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}$ is the corresponding eigenvector
Then the general solution is $$y=c_1\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}e^{-t}+c_2\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}e^{t}$$
« Last Edit: November 20, 2018, 11:14:01 AM by Jingze Wang »

Michael Poon

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Re: TT2-P3
« Reply #4 on: November 20, 2018, 12:12:59 PM »
Seems like no one has added a phase portrait yet.

I attached it below, it is a saddle, with eigenvalues real and opposite.

Jingze Wang

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Re: TT2-P3
« Reply #5 on: November 20, 2018, 02:35:45 PM »
here is my answer
Hi, my answer is same like yours except I got $C_1=-2ln2, C_2=0$
I am not confident about my answer, feel free to correct me :)

Tzu-Ching Yen

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Re: TT2-P3
« Reply #6 on: November 20, 2018, 02:47:25 PM »
I agree with Jingze's solution. It is obvious that Boyu's answer is not (0, 0) at $t =0$

Yulin WANG

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Re: TT2-P3
« Reply #7 on: November 20, 2018, 04:03:36 PM »
here is my answer
Hi, my answer is same like yours except I got $C_1=-2ln2, C_2=0$
I am not confident about my answer, feel free to correct me :)
My answer is the same as yours.

Zhihong Yin

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Re: TT2-P3
« Reply #8 on: November 20, 2018, 04:11:24 PM »
I am not sure.

Siran Wang

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Re: TT2-P3
« Reply #9 on: November 22, 2018, 02:24:55 PM »
a)
\begin{equation*}
  A-\lambda I=\begin{pmatrix}
  2-\lambda & 1\\
  -3 & -2-\lambda
  \end{pmatrix}
  \end{equation*}

  \begin{equation*}
  \det(A-\lambda I)=(2-\lambda)(-2-\lambda)+3=0
  \end{equation*}
  \begin{equation*}
  \end{equation*}
  \begin{equation*}
  \lambda_1=1~~~~\lambda_2=-1
  \end{equation*}

When $\lambda_1=1$
   \begin{equation*}
  x_1+x_2=0
  \end{equation*}
 The first eigenvector is $\begin{pmatrix}
1\\
-1
\end{pmatrix}$

When $\lambda_1=-1$
   \begin{equation*}
  3x_1+x_2=0
  \end{equation*}
 The second eigenvector is $\begin{pmatrix}
1\\
-3
\end{pmatrix}$
So, the general solution is $y=C_1e^t\begin{pmatrix}
1\\
-1
\end{pmatrix}+C_2e^{-t}\begin{pmatrix}
1\\
-3
\end{pmatrix}$

b) unstable, saddle point
graph in attachment

c)
\begin{equation*}
\phi=\begin{pmatrix}
e^t&e{-t}\\
-e^t&-3e^{-t}
\end{pmatrix}
\end{equation*}
\begin{equation*}
\begin{pmatrix}
e^t&e{-t}\\
-e^t&-3e^{-t}
\end{pmatrix}\begin{pmatrix}
U_1\\
U_2
\end{pmatrix}=\begin{pmatrix}
\frac{4}{e^t+e^{-t}}\\
\frac{-12}{e^t+e^{-t}}
\end{pmatrix}
 \end{equation*}
\begin{equation*}
e^tU_1+e^{-t}U_2=\frac{4}{e^t+e^{-t}}
 \end{equation*}
\begin{equation*}
-e^tU_1-3e^{-t}U_2= \frac{-12}{e^t+e^{-t}}
 \end{equation*}
\begin{equation*}
U_2= \frac{4}{1+e^{-2t}}-e^{2t}U_1
 \end{equation*}
\begin{equation*}
-e^tU_1-3e^{-t}(\frac{4}{1+e^{-2t}}-e^{2t}U_1)=\frac{-12}{e^t+e^{-t}}
 \end{equation*}
\begin{equation*}
U_1=0~~~U_2=\frac{4}{1+e^{-2t}}
 \end{equation*}
\begin{equation*}
V_1=\int U_1dt=\int0dt=0
 \end{equation*}
\begin{equation*}
V_2=\int U_2dt=\int\frac{4}{1+e^{-2t}}dt=2ln(1+e^{2t})
 \end{equation*}
\begin{equation*}\begin{pmatrix}
x_1\\
x_2
\end{pmatrix}=\begin{pmatrix}
C_1e^t+C_2e^{-t}+2e{-t}ln(1+e^{2t}\\
-C_1e^t-3C_2e^{-t}-6e{-t}ln(1+e^{2t}
\end{pmatrix}
 \end{equation*}
since $\begin{equation*}x(0)=\begin{pmatrix}
0\\
0
\end{pmatrix}
\end{equation*}$
\begin{equation*}
C_1+C_2+2ln2=0~~~-C_1-3C_2-6ln2=0
\end{equation*}\begin{equation*}
C_1=0~~~C_2=-2ln2
\end{equation*}

Victor Ivrii

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Re: TT2-P3
« Reply #10 on: November 25, 2018, 11:14:58 AM »
There is computer-generated picture