### Author Topic: Q6 TUT 5102  (Read 4631 times)

#### Victor Ivrii

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##### Q6 TUT 5102
« on: November 17, 2018, 04:01:01 PM »
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

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##### Re: Q6 TUT 5102
« Reply #1 on: November 17, 2018, 04:35:52 PM »
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

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##### Re: Q6 TUT 5102
« Reply #2 on: November 17, 2018, 04:42:29 PM »
Phase portraits attached below:

Top: Eigenvalues real & same sign (+ve), stable

Middle: Eigenvalues real & opposite sign, saddle

Bottom: Eigenvalues complex & negative, unstable spiral
« Last Edit: November 17, 2018, 04:47:28 PM by Michael Poon »

#### Jiacheng Ge

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##### Re: Q6 TUT 5102
« Reply #3 on: November 18, 2018, 12:41:00 PM »
My solution is different.

#### Victor Ivrii

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##### Re: Q6 TUT 5102
« Reply #4 on: November 25, 2018, 09:49:09 AM »
When eigenvalues pass from real to complex conjugate, stability does not change. Jiacheng is right