Toronto Math Forum

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: