### Author Topic: quiz6 lec0101  (Read 931 times)

#### yilin huang

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• Karma: 0 ##### quiz6 lec0101
« on: November 15, 2019, 05:56:37 PM »
Consider the equation: $\mathbf{x}' = \begin{bmatrix} 2 & -5 \\ \alpha & -2\\ \end{bmatrix}\mathbf{x}$.
Solution:
a). Assume the solution is $\mathbf{x} = \mathbf{\xi}e^{rt}$ to find the fundamental set of solutions.
Let $\mathbf{\xi} = \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}$, we obtain the following set of linear algebraic equations:
$$\begin{bmatrix} 2 - r & -5 \\ \alpha & -2 - r\\ \end{bmatrix}\mathbf{\xi} = \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix} = \mathbf{\xi} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$
In order to get the eigenvalue to let: $\begin{vmatrix} 2 - r & -5 \\ \alpha & -2 - r\\ \end{vmatrix} = 0$
\begin{align*}
(2-r)(-2-r) - (-5)\alpha & = 0\\
r^2 +5\alpha-4 & = 0\\
r = \pm \sqrt{4-5\alpha}
\end{align*}
So, the eigenvalues are $\pm \sqrt{4-5\alpha}$.

b) There are three cases for $\alpha$: i) $\alpha < \frac{4}{5}$, ii) $\alpha > \frac{4}{5}$, iii) $\alpha ＝ \frac{4}{5}$,
Case i): $\alpha < \frac{4}{5}$
Since $\alpha < \frac{4}{5}$, we have $4 - 5\alpha > 0$, such that $\pm \sqrt{4-5\alpha}$ are real.
Because eigenvalues are real and different, the equilibrium point $(0,0)$ is a saddle point and the stability of critical point is unstable.
Case ii): $\alpha > \frac{4}{5}$
Since $4 - 5\alpha < 0$, we have the eigenvalues has zero real part, such that the trajectories neither approach the origin nor become unbounded but repeatedly traverse a closed curve about the origin.
So the equilibrium point $(0,0)$ is a center and is the stable but not asymptotically stable.\\
In conclusion, the critical value of $\alpha$ where the qualitative nature of phase changes is $\alpha = \frac{4}{5}$

c)
When $\alpha = \frac{4}{5}$, we have the following graph:
« Last Edit: November 15, 2019, 08:08:29 PM by yilin huang »