Question

The coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues 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 crititical value. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{\frac{5}{4}} & {\frac{2}{4}} \\\ {\alpha} & {\frac{5}{4}}\end{array}\right) \mathbf{x} $$

Short answer

Answer: The critical value of \(\alpha\) is \(\frac{9}{8}\).

Step-by-step solution

Find Eigenvalues

First, let's find the eigenvalues of the coefficient matrix: $$ \left(\begin{array}{ll}\frac{5}{4}-\lambda & \frac{2}{4} \\ \alpha & \frac{5}{4} - \lambda \end{array}\right) $$ Calculate the determinant and set it equal to zero: $$ \begin{aligned} \text{det}\left(\begin{array}{ll}\frac{5}{4}-\lambda & \frac{2}{4} \\ \alpha & \frac{5}{4} - \lambda \end{array}\right) &= \left(\frac{5}{4}-\lambda\right)\left(\frac{5}{4} - \lambda \right) - \left(\frac{2}{4}\right)\alpha \\ &= \frac{25}{16} - 2\frac{5}{4}\lambda + \lambda^2-\frac{1}{2}\alpha = 0 \end{aligned} $$ Now we have the characteristic equation in terms of \(\alpha\) and \(\lambda\): $$ \lambda^2 - \frac{5}{2}\lambda + \frac{25}{16} - \frac{1}{2}\alpha = 0 $$

Find Critical Values of \(\alpha\)

The critical values of \(\alpha\) are when we switch between having real and complex eigenvalues. This happens when the discriminant of the characteristic equation changes its sign. The discriminant, \(D\), is given by: $$ D = b^2 - 4ac $$ For our characteristic equation, the discriminant is: $$ D = \left(-\frac{5}{2}\right)^2 - 4\left(\frac{25}{16} - \frac{1}{2}\alpha\right) $$ Solve for when \(D = 0\) to find the critical values of \(\alpha\): $$ \frac{25}{4} - 4\left(\frac{25}{16} - \frac{1}{2}\alpha\right) = 0 $$ Rearranging, we get $$ \alpha = \frac{9}{8} $$

Analyze Phase Portraits for Different Values of \(\alpha\)

Now, we will analyze the phase portrait for values of \(\alpha\) slightly below and above its critical value. - When \(\alpha < \frac{9}{8}\), we have real eigenvalues since the discriminant \(D>0\). The system will have either an unstable or stable focus/node, depending on the signs of the eigenvalues. - When \(\alpha = \frac{9}{8}\), we have a double eigenvalue and the system is degenerate. - When \(\alpha > \frac{9}{8}\), we have complex eigenvalues since the discriminant \(D<0\). The system will have either an unstable or stable spiral, depending on the signs of the eigenvalues. Now, the student can draw phase portraits for each case, illustrating the different behavior of the system for different values of \(\alpha\).