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. $$ x^{\prime}=\left(\begin{array}{cc}{4} & {\alpha} \\ {8} & {-6}\end{array}\right) \mathbf{x} $$

Short answer

Answer: The phase portraits slightly below and above the critical values of α both exhibit a saddle point.

Step-by-step solution

Calculate the eigenvalues in terms of \(\alpha\)

Given the system, $$ x^{\prime}=\left(\begin{array}{cc}{4} & {\alpha} \\ {8} & {-6}\end{array}\right) \mathbf{x}, $$ we need to find the eigenvalues of the coefficient matrix. To find the eigenvalues, we need to solve the characteristic equation: $$ \text{det}(A - \lambda I) = 0, $$ where \(A\) is the coefficient matrix, \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix. In this case, we have: $$ \text{det}\left(\begin{array}{cc}{4-\lambda} & {\alpha} \\ {8} & {-6-\lambda}\end{array}\right) = (4-\lambda)(-6-\lambda) - 8\alpha = 0. $$ Expanding the equation, we have a quadratic equation in terms of \(\lambda\): $$ \lambda^2 - 2\lambda - 24 - 8\alpha = 0. $$ Now we can calculate the eigenvalues in terms of the parameter \(\alpha\). We'll use the quadratic formula, which gives: $$ \lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$ In our case, \(a=1\), \(b=-2\), and \(c=-24-8\alpha\). Applying the quadratic formula, we get: $$ \lambda_{1,2}(\alpha) = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-24 - 8\alpha)}}{2} = 1 \pm \sqrt{25 + 8\alpha}. $$ So, the eigenvalues of the matrix are: $$ \lambda_1(\alpha) = 1 + \sqrt{25 + 8\alpha}, $$ $$ \lambda_2(\alpha) = 1 - \sqrt{25 + 8\alpha}. $$

Determine critical values of \(\alpha\)

The qualitative nature of the phase portrait changes based on the signs and values of the real parts of the eigenvalues. We need to find the critical values of \(\alpha\) where these changes occur. Let's analyze the possible cases: Case a: Both eigenvalues are real and have the same sign (either positive or negative) Case b: Both eigenvalues are real and have opposite signs Case c: Both eigenvalues are complex with positive real parts Case d: Both eigenvalues are complex with negative real parts Case e: Both eigenvalues are complex with zero real parts We need to find the values of \(\alpha\) that will cause a transition between these cases. First, let's find the values of \(\alpha\) that will make either of the eigenvalues complex. This occurs when the discriminant of the quadratic equation is negative. In our case, the discriminant is \(25+8\alpha\). Thus, we need to find the \(\alpha\) for which: $$ 25+8\alpha < 0 \Rightarrow \alpha < -\frac{25}{8}. $$ Now let's consider the case when both eigenvalues are real. We know that the real parts of both eigenvalues are 1, so we don't need to analyze case c, d, or e further. Now we need to analyze cases a and b. If the eigenvalues have the same sign (either positive or negative), it means that their sum is either positive or negative: $$ \lambda_1 + \lambda_2 = 2. $$ Since 2 is positive, we know that both eigenvalues must be positive for a critical value. Therefore, we have case a: both eigenvalues are real and positive. The critical value for \(\alpha\) is when both eigenvalues are real and positive, which occurs when: $$ -\frac{25}{8} \leq \alpha. $$

Draw phase portraits for different values of \(\alpha\)

Next, we will draw phase portraits for values of \(\alpha\) slightly below and above the critical value that we found in Step 2. For this, we can choose \(\alpha_1 = -4\) (below critical value) and \(\alpha_2 = 0\) (above critical value). For \(\alpha_1 = -4\), the eigenvalues are: $$ \lambda_1(-4) = 1 + \sqrt{25 + 8(-4)} = 1 + 3 = 4, $$ $$ \lambda_2(-4) = 1 - \sqrt{25 + 8(-4)} = 1 - 3 = -2. $$ In this case, we have one positive and one negative eigenvalue, which implies a saddle point. For \(\alpha_2 = 0\), the eigenvalues are: $$ \lambda_1(0) = 1 + \sqrt{25 + 8(0)} = 1 + 5 = 6, $$ $$ \lambda_2(0) = 1 - \sqrt{25 + 8(0)} = 1 - 5 = -4. $$ In this case, we still have one positive and one negative eigenvalue, which again implies a saddle point. Therefore, the phase portraits slightly below and above the critical values of \(\alpha\) both exhibit a saddle point.

Key concepts

Phase Portrait

A phase portrait is a geometric representation of the trajectories that solutions to a system of differential equations can follow. Imagine plotting a graph where each point corresponds to the state of a system at a given time. The trajectory shows how the system evolves from one state to another.

• Phase portraits help visualize the behavior of systems described by differential equations.
• They provide insights into the stability and dynamics of the system.

Trajectories can be straight lines, indicating equilibrium or stable states, or they can be more complex, showing oscillations or decay. In our example, we look at phase portraits for values of \( \alpha \) below and above critical values to understand changes in system behavior.

Critical Values

In the analysis of differential equations, finding critical values is paramount. These are specific parameter values where the qualitative behavior of a system changes. They are tipping points or thresholds.

• Critical values can mark transitions between different types of system behavior.
• They are crucial for understanding bifurcations, where the structure of a system changes as parameters vary.

For example, we determined that a critical value is when eigenvalues transition from complex to real, impacting the system's dynamics.

Saddle Point

A saddle point is a type of equilibrium in a dynamical system where trajectories approach the point but then diverge. It resembles the shape of a horse saddle.

• Saddle points have both stable and unstable directions.
• The presence of a saddle point often indicates an unstable system.

In our exercise, for values of \( \alpha \) both slightly below and above the critical point, the system showed a saddle point, which shows instability since trajectories start close but eventually veer off.

Differential Equations

Differential equations involve functions and their derivatives, showcasing how a quantity changes over time or space. They are vital in modeling continuous processes in physics, engineering, and other sciences.

• Linear differential equations have a degree of one; they are easier and often solvable analytically.
• Some complex systems require numerical solutions for differential equations.

In our problem, a linear matrix differential equation describes how the state \( \mathbf{x} \) evolves, given the parameter \( \alpha \). Understanding how to manipulate and utilize these equations is crucial in mathematics.

Stability Analysis

Stability analysis in systems tells us whether solutions to a differential equation will converge to an equilibrium or diverge. It assesses if small deviations grow or shrink over time.

• Stable systems return to equilibrium after small disturbances.
• Unstable systems deviate away from equilibrium when perturbed.

In the exercise, analyzing eigenvalues gives insight into stability. Positive real parts indicate instability, often seen in saddle points, while negative parts suggest a stable focus or node. These insights are crucial for predicting long-term behavior.