MAT244--2018F > Quiz-6

Q6 TUT 5101

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Victor Ivrii:
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

Jingze Wang:
First, try to find the eigenvalues with respect to the parameter





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

Michael Poon:
a) Finding the eigenvalues:

Set the determinant = 0

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


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:
Phase portrait attached

Siran Wang:
  A-\lambda I=\begin{pmatrix}
  2-\lambda & -5\\
  \alpha & -2-\lambda
  \det(A-\lambda I)=(2-\lambda)(-2-\lambda)+5\alpha=\lambda^2+5\alpha-4=0
  \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}
  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


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