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: $$\begin{array}{rl}\text{det}\left(\begin{array}{ll}\frac{5}{4}-\lambda & \frac{2}{4}\\ \alpha & \frac{5}{4}-\lambda \end{array}\right)& =(\frac{5}{4}-\lambda )(\frac{5}{4}-\lambda )-\left(\frac{2}{4}\right)\alpha \\ & =\frac{25}{16}-2\frac{5}{4}\lambda +{\lambda }^2-\frac{1}{2}\alpha =0\end{array}$$ 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: $$D={(-\frac{5}{2})}^2-4(\frac{25}{16}-\frac{1}{2}\alpha )$$ Solve for when $D=0$ to find the critical values of $\alpha$: $$\frac{25}{4}-4(\frac{25}{16}-\frac{1}{2}\alpha )=0$$ 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 $\alpha$.

Key concepts

Eigenvalues

To better understand how differential equations behave, it's vital to grasp the concept of eigenvalues. Eigenvalues are a set of special scalars associated with a linear transformation represented by a matrix. In our problem, we derive them by first setting up the characteristic equation from the matrix $$\left(\begin{array}{ll}\frac{5}{4}-\lambda & \frac{2}{4}\\ alpha& \frac{5}{4}-\lambda \end{array}\right)$$.The characteristic equation is obtained by calculating the determinant set to zero:

• $$\text{det}\left(\begin{array}{ll}\frac{5}{4}-\lambda & \frac{2}{4}\\ alpha& \frac{5}{4}-\lambda \end{array}\right)=\frac{25}{16}-2\frac{5}{4}\lambda +{\lambda }^2-\frac{1}{2}\alpha =0$$

Finding the solutions for $\lambda$, termed the eigenvalues, reveals critical insights into the system's dynamic properties. We effectively determine how variables evolve over time. Changes in the eigenvalues, specifically switching between real and complex values, can indicate a shift in the system's nature.

Phase Portraits

Phase portraits are a crucial visualization tool in studying differential equations. They graphically represent trajectories of a dynamical system in the phase plane, providing deep insight into the stability and behavior of the system. For instance, a phase portrait may show lines spiraling outwards or inwards, indicating the presence of a spiral focus.For the given system:

• When $\alpha <\frac{9}{8}$, the eigenvalues are real and the portrait might display node-like behaviors, either stable or unstable depending on the sign of the eigenvalues.
• At the critical value $\alpha =\frac{9}{8}$, we witness a double eigenvalue, indicative of a degenerate or critically damped system.
• For $\alpha >\frac{9}{8}$, complex eigenvalues appear, leading to a portrait with a spiral, showing either a stable or unstable spiral motion.

By interpreting these phase portraits, one can grasp how initial conditions affect a system's evolution, showcasing behaviors such as convergence to equilibria, cyclical paths, or divergence.

Critical Values

Critical values of parameters within a differential equation indicate points where the qualitative nature of the system changes. In analyzing systems like ours, recognizing where these critical transitions occur is key to understanding system dynamics.In the exercise, critical values are determined by evaluating when the discriminant of the characteristic equation equals zero:

• The discriminant is $D={(-\frac{5}{2})}^2-4(\frac{25}{16}-\frac{1}{2}\alpha )$.
• Solving $D=0$ gives $\alpha =\frac{9}{8}$.

These points not only mark changes in eigenvalue types but also signal shifts in system behavior, from stable to unstable, spiral to node, and more. Recognizing these changes allows one to predict dynamic responses to parameter modifications, which is crucial for system control and optimization.