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. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {10} \\ {-1} & {-4}\end{array}\right) \mathbf{x} $$

Short answer

Question: Determine the eigenvalues for the given system and find the critical values of \(\alpha\) where the phase portrait changes its qualitative nature. Draw phase portraits for values of \(\alpha\) slightly below and above these critical values. Given system: $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {10} \\ {-1} & {-4}\end{array}\right) \mathbf{x} $$ Solution: Eigenvalues: $$ \lambda = \frac{(\alpha + 4) \pm \sqrt{(\alpha + 4)^{2}-4(4\alpha + 40)}}{2} $$ Critical values of \(\alpha\): \(\alpha_1=12\) and \(\alpha_2=-12\). Phase portraits: Draw phase portraits for different values of \(\alpha\) such as for \(\alpha\) slightly below -12, \(\alpha\) slightly above -12, \(\alpha\) slightly below 12 and \(\alpha\) slightly above 12. The nature of the phase portrait will change according to the discriminant of the eigenvalues' quadratic equation: - Real and distinct eigenvalues for \(\alpha < -12\) and \(\alpha > 12\) - Repeated real eigenvalue for \(\alpha = -12\) and \(\alpha = 12\) - Complex eigenvalues for \(-12 < \alpha < 12\).

Step-by-step solution

Find the eigenvalues

To find the eigenvalues, we need to solve the characteristic equation given by: $$ \text{det}(\mathbf{A}-\lambda \mathbf{I}) = 0, $$ where \(\text{A}\) is the given matrix and \(\mathbf{I}\) is the identity matrix. We have: $$ \text{det}\left(\begin{array}{cc}{\alpha-\lambda} & {10} \\ {-1} & {-4-\lambda}\end{array}\right)=0 $$ So we get: $$ (\alpha-\lambda)(-4-\lambda) + (-1) \cdot 10 = 0 $$

Calculate the characteristic equation

Now, we will expand the characteristic equation and solve for the eigenvalues: $$ -\lambda^2+(\alpha+4)\lambda-4\alpha-40 = 0 $$ This is a quadratic equation in \(\lambda\). We can simplify the equation as: $$ \lambda^{2} - (\alpha+4)\lambda + (4\alpha + 40) = 0 $$ Let's solve for \(\lambda\) using the quadratic formula: $$ \lambda=\frac{-b\pm \sqrt{b^{2}-4 a c}}{2 a} $$ Where \(a=1\), \(b=-(\alpha+4)\), and \(c=4\alpha + 40\). So, $$ \lambda = \frac{(\alpha + 4) \pm \sqrt{(\alpha + 4)^{2}-4(4\alpha + 40)}}{2} $$ Now we have the eigenvalues in terms of \(\alpha\).

Find the critical values of \(\alpha\)

The critical values of \(\alpha\) are found where the qualitative nature of the phase portrait changes. This occurs whenever the discriminant, which is the term inside the square root in the eigenvalue expression, is zero: $$ (\alpha+4)^{2}-4(4\alpha+40)=0 $$ Solve for \(\alpha\): $$ \alpha^{2}+8\alpha+16-16\alpha-160=0 $$ $$ \alpha^{2}-8\alpha-144=0 $$ Now solve the quadratic equation. Using the quadratic formula, we get: $$ \alpha_1=\frac{8+\sqrt{8^{2}+4 \cdot 144}}{2}=12 $$ $$ \alpha_2=\frac{8-\sqrt{8^{2}+4 \cdot 144}}{2}=-12 $$ So, the critical values of \(\alpha\) are \(\alpha_1=12\) and \(\alpha_2=-12\).

Draw phase portraits

Before drawing phase portraits, we need to examine the discriminant of the eigenvalues' quadratic equation. We found: $$ \text{discriminant} =(\alpha+4)^{2} - 4(4\alpha + 40) $$ We have three cases depending on the discriminant: - Case 1: Discriminant \(> 0\), we have real and distinct eigenvalues. - Case 2: Discriminant \(= 0\), we have a repeated real eigenvalue. - Case 3: Discriminant \(< 0\), we have complex eigenvalues. We already found the critical values \(\alpha_1=12\) and \(\alpha_2=-12\). So, the phase portraits will change between these values (below or above). Now, we will draw the phase portraits: - For \(\alpha < -12\), the discriminant is positive, resulting in real and distinct eigenvalues. - For \(\alpha = -12\), the discriminant is zero, resulting in a repeated real eigenvalue. - For \(-12 < \alpha < 12\), the discriminant is negative, resulting in complex eigenvalues. - For \(\alpha = 12\), the discriminant is zero, resulting in a repeated real eigenvalue. - For \(\alpha > 12\), the discriminant is positive, resulting in real and distinct eigenvalues. We can plot phase portraits for different values of \(\alpha\), such as for \(\alpha\) slightly below -12, \(\alpha\) slightly above -12, \(\alpha\) slightly below 12 and \(\alpha\) slightly above 12. In each case, the phase portraits will have a different qualitative nature, as described above.

Key concepts

Eigenvalues

In the context of linear algebra, eigenvalues are important because they show how a linear transformation affects vectors. They are derived from a matrix and tell us about the scaling factor by which the transformation changes the vector's direction. To find eigenvalues, we use the characteristic equation, which is derived from the given matrix of the system. Generally, for a matrix

• The eigenvalue equation is \( \text{det}(\mathbf{A}-\lambda \mathbf{I}) = 0 \), where \( \lambda \) represents the eigenvalues.
• Each eigenvalue corresponds to an eigenvector.
• Eigenvalues can be real or complex numbers.

Understanding eigenvalues helps us predict system behavior, especially in dynamic systems modeled by differential equations.

Phase Portrait

Phase portraits provide a visual representation of trajectories of a dynamical system in the phase plane. They are especially useful for understanding the behavior of systems of differential equations. Each point in the plot represents a state of the system. The phase portrait helps in studying stability, oscillations, and long-term behavior of the system. Here's a general guideline to interpret phase portraits:

• When eigenvalues are real and distinct, the trajectories show node-like behavior. It may be a sink or source depending on the sign of the eigenvalues.
• For complex eigenvalues, trajectories are spiral, signifying oscillations.
• If eigenvalues are real and repeated, the system shows a degenerate node.

Phase portraits change as parameters in the system change, marking different qualitative behaviors.

Characteristic Equation

The characteristic equation is crucial for finding eigenvalues. It is developed from the matrix of the system and the identity matrix. This equation takes the form of a polynomial equation, where the roots represent the eigenvalues:

• The general format is \( \text{det}(\mathbf{A}-\lambda \mathbf{I}) = 0 \), where \( \mathbf{A} \) is the matrix, \( \lambda \) is a placeholder for eigenvalues, and \( \mathbf{I} \) is the identity matrix.
• Determinants help in converting the matrix into a manageable polynomial form.
• Solving the characteristic equation often involves algebraic formulas like the quadratic formula.

The roots of this equation are the eigenvalues, which are essential in defining the nature of the system.

Critical Values

In analyzing systems through phase portraits and eigenvalues, critical values of parameters occur where the system exhibits a qualitative change in behavior. These are pivotal as they mark the thresholds where the stability and nature of the system move from one form to another:

• Critical values are derived from conditions where the properties of eigenvalues change (e.g., real to complex).
• Such values are determined by setting specific expressions equal to zero, indicating a shift.
• In the provided solution, critical values of \( \alpha \) are found where the discriminant of the characteristic equation becomes zero, signaling a transition in the eigenvalue nature.

Recognizing and calculating these values help in understanding how slight changes affect overall system behavior.