Author Topic: Q6 TUT 5101  (Read 2638 times)

Victor Ivrii

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Q6 TUT 5101
« on: November 17, 2018, 03:59:47 PM »
The coefficient matrix contains a parameter $\alpha$ . In each of these problems:

(a) Determine the eigenvalues in terms of $\alpha$.
(b)  Find the critical value or values of  $\alpha$  where the qualitative nature of the phase portrait for the system changes.
(c) Draw a phase portrait for a value of  $\alpha$ slightly below, and for another value slightly above, each critical value.
$$\mathbf{x}' =\begin{pmatrix}
2 &-5\\
\alpha & -2
\end{pmatrix}\mathbf{x}.$$

Jingze Wang

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Re: Q6 TUT 5101
« Reply #1 on: November 17, 2018, 04:00:25 PM »
First, try to find the eigenvalues with respect to the parameter


$A=\begin{bmatrix}
2&-5\\
\alpha&-2\\
\end{bmatrix}$


$det(A-rI)=(2-r)(-2-r)+5\alpha=0$


$r^2-4+5\alpha=0$


$r=\frac{\pm\sqrt{16-20\alpha}}{2}$


Notice that $-4+5\alpha$ determines the type of roots, so $\alpha=4/5$ is the critical value


Case 1


When $-4+5\alpha=0, \alpha=0$, there is a repeated eigenvalue 0 with one eigenvector



Case 2


When $-4+5\alpha>0, \alpha>4/5$, there are two distinct complex eigenvalues without real parts


Case 3


When $-4+5\alpha<0, \alpha<4/5$, there are two distinct real eigenvalues with different signs
« Last Edit: November 17, 2018, 04:08:35 PM by Jingze Wang »

Michael Poon

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Re: Q6 TUT 5101
« Reply #2 on: November 17, 2018, 04:07:45 PM »
a) Finding the eigenvalues:

Set the determinant = 0

\begin{align}
(2 - \lambda)(-2 - \lambda) - (-5)(\alpha) &= 0\\
\lambda^2 - 4 + 5\alpha &= 0\\
\lambda &= \pm \sqrt{4 - 5\alpha}
\end{align}

b)

Case 1: Eigenvalues real and opposite sign
when: $\alpha$ < $\frac{4}{5}$ (unstable saddle)

Case 2: Eigenvalues complex and opposite sign
when: $\alpha$ > $\frac{4}{5}$ (stable centre)

c) will post below:

Michael Poon

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Re: Q6 TUT 5101
« Reply #3 on: November 17, 2018, 04:08:43 PM »
Phase portrait attached
« Last Edit: November 25, 2018, 10:22:05 AM by Victor Ivrii »

Siran Wang

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Re: Q6 TUT 5101
« Reply #4 on: November 17, 2018, 04:23:23 PM »
(a)
 \begin{equation*}
  A-\lambda I=\begin{pmatrix}
  2-\lambda & -5\\
  \alpha & -2-\lambda
  \end{pmatrix}
  \end{equation*}
 (b)
  \begin{equation*}
  \det(A-\lambda I)=(2-\lambda)(-2-\lambda)+5\alpha=\lambda^2+5\alpha-4=0
  \end{equation*}
  \begin{equation*}
  \lambda=\frac{0\pm \sqrt{0^2-4(5\alpha-4)}}{2}=\frac{0\pm \sqrt{-4(5\alpha-4)}}{2}=\frac{0\pm 2\sqrt{-(5\alpha-4)}}{2}
  \end{equation*}
  \begin{equation*}
  \lambda_1=\sqrt{4-5\alpha}~~~~\lambda_2=-\sqrt{4-5\alpha}
  \end{equation*}
 
  \begin{equation*}\begin{split}
  b^2-4ac&\geq0\\
  4-5\alpha&\geq0\\
  -5\alpha&\geq-4\\
  \alpha&\leq\frac{4}{5}
  \end{split}
  \end{equation*}
  when $\alpha<\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are two different real numbers.
  when $\alpha>\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are solutions with complex numbers.
  when $\alpha=\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are repeated roots, which are 0.
 (c) in attachments

Victor Ivrii

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Re: Q6 TUT 5101
« Reply #5 on: November 25, 2018, 10:21:37 AM »
Jingze  is right
« Last Edit: November 25, 2018, 10:29:33 AM by Victor Ivrii »