# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 03:59:47 PM

Title: Q6 TUT 5101
Post by: Victor Ivrii 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.
$$\mathbf{x}' =\begin{pmatrix} 2 &-5\\ \alpha & -2 \end{pmatrix}\mathbf{x}.$$
Title: Re: Q6 TUT 5101
Post by: Jingze Wang 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
Title: Re: Q6 TUT 5101
Post by: Michael Poon 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:
Title: Re: Q6 TUT 5101
Post by: Michael Poon on November 17, 2018, 04:08:43 PM
Phase portrait attached
Title: Re: Q6 TUT 5101
Post by: Siran Wang 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
Title: Re: Q6 TUT 5101
Post by: Victor Ivrii on November 25, 2018, 10:21:37 AM
Jingze  is right