MAT244--2018F > Quiz-6
Q6 TUT 5102
(1/1)
Victor Ivrii:
The coefficient matrix contains a parameter $\alpha$.
(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.
$$\mathbf{x}' =\begin{pmatrix}
4 &\alpha\\
8 &-6
\end{pmatrix}\mathbf{x}.$$
Michael Poon:
a) Finding the eigenvalues:
Set the determinant = 0
\begin{align}
(4 - \lambda)(-6 - \lambda) - 8\alpha &= 0\\
\lambda^2 + 2\lambda - 24 - 8\alpha &= 0\\
\lambda &= -1 \pm \sqrt{25 + 8\alpha}
\end{align}
b)
Case 1: Eigenvalues real and same sign
when: $\alpha$ > $\frac{-25}{8}$ + 1
Case 2: Eigenvalues real and opposite sign
when: $\frac{-25}{8}$ < $\alpha$ < $\frac{-25}{8} + 1$
Case 3: Eigenvalues complex
when: $\alpha$ < $\frac{-25}{8}$
critical points: $\alpha$ = $\frac{-25}{8}$, $\frac{-25}{8}$ + 1
c) will be posted below:
Michael Poon:
Phase portraits attached below:
Top: Eigenvalues real & same sign (+ve), stable
Middle: Eigenvalues real & opposite sign, saddle
Bottom: Eigenvalues complex & negative, unstable spiral
Jiacheng Ge:
My solution is different.
Victor Ivrii:
When eigenvalues pass from real to complex conjugate, stability does not change. Jiacheng is right
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