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In each of Problems 13 through 20 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. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {1} \\ {-1} & {\alpha}\end{array}\right) \mathbf{x} $$

Short Answer

Expert verified
Answer: The qualitative nature of the phase portrait for given matrix A(α) will always show a spiraling motion for any value of α.

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01

Find the Eigenvalues in terms of α

To find the eigenvalues, we will calculate the characteristic equation of the given matrix: $$ \det(A - \lambda I) = \det \begin{pmatrix} \alpha - \lambda & 1 \\ -1 & \alpha - \lambda \end{pmatrix} $$ Calculating the determinant: $$ (\alpha - \lambda)^2 - (-1)(1) = \lambda^2 - 2\alpha \lambda + \alpha^2 + 1 $$ This is a quadratic equation in terms of λ, so we will use the quadratic formula to solve for λ in terms of α: $$ \lambda_{1,2} = \frac{2\alpha \pm \sqrt{(-2\alpha)^2 - 4(\alpha^2 + 1)}}{2} $$ Simplifying the equation: $$ \lambda_{1,2} = \alpha \pm \sqrt{\alpha^2 - (\alpha^2 + 1)} $$
02

Determine Critical Values of α

To find the critical values where the qualitative nature of the phase portrait changes, we need to consider the real part of the eigenvalues and observe for what values of α the eigenvalues would have a different sign or when they become complex conjugate pairs. We examine the discriminant: $$ \alpha^2 - (\alpha^2 + 1) $$ The critical point happens when this discriminant changes its sign: $$ \Delta(\alpha) = \alpha^2 - (\alpha^2 + 1) = -1 $$ In this case, the discriminant is always negative, meaning that the eigenvalues will always be complex conjugate pairs. Hence, there are no critical values of α that would change the qualitative nature of the phase portrait.
03

Draw Phase Portraits for Values of α Slightly Above and Below the Critical Value

Since there is no critical value, the qualitative nature of the phase portrait will not change. As the eigenvalues are complex conjugate pairs, we know that the phase portrait will display a spiraling motion. By choosing any values for α, the spirals will always occur. For example, let's choose two random values for α, such as α = -1 and α = 1. For both of these cases, the phase portrait displays spiraling motion. In conclusion, by analyzing the eigenvalues in terms of α, we found that the qualitative nature of the phase portrait does not change, and it will always show a spiraling motion for any value of α.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Eigenvalues in Differential Equations
The task of finding eigenvalues is essential in understanding the behavior of differential equations, especially when parameters like \(\alpha\) are involved. To discover the eigenvalues of a matrix dependent on \(\alpha\), we first need to determine the characteristic equation. For our given matrix:
 \[ A = \begin{pmatrix} \alpha & 1 \ -1 & \alpha \end{pmatrix} \] 
we calculate the determinant of \(A - \lambda I\) where \(\lambda\) represents our eigenvalues. This involves determining:
  • \[ \text{det}(A - \lambda I) = (\alpha - \lambda)^2 - (-1)(1) \] which yields the quadratic equation
  • \[ \lambda^2 - 2\alpha \lambda + \alpha^2 + 1 = 0 \]
Solving this quadratic using the quadratic formula gives us the eigenvalues:
\[ \lambda_{1,2} = \alpha \pm \sqrt{\alpha^2 - (\alpha^2 + 1)} \]
These solutions are in terms of \(\alpha\) and help predict how changes in \(\alpha\) influence the system. Understanding these eigenvalues aids in drawing important conclusions about the overall behavior of the differential system.
Phase Portrait and Its Qualitative Nature
Phase portraits provide a geometric visualization of the trajectories of differential equations in a state-space diagram. For our system, the phase portrait is influenced by the nature of its eigenvalues.Given that the eigenvalues here are complex conjugates, as determined by the discriminant being negative \(\alpha^2 - (\alpha^2 + 1) = -1\), this means the matrix will always generate spirals. The system does not have a real part crossing zero, indicating no shifts in type or stability, and therefore the phase portrait remains consistent.This 'spiraling motion' commonly occurs in systems that characterize oscillatory behavior. In simpler terms, whichever value of \(\alpha\) you choose, the trajectories in our differential system's phase portrait revolve around a fixed point in a consistent loop-like pattern. This consistent nature could apply to situations like electrical circuits or mechanical oscillations where such repetitive behaviors are seen.
The Role of Parameter Analysis in Change Detection
Parameter analysis in differential equations involves determining how variations in a parameter, such as \(\alpha\), influence the solutions or the structure of these solutions. In our context, the parameter \(\alpha\) was closely examined for values that could lead to qualitative changes in the phase portrait.Typically, critical values occur when the discriminant of the characteristic equation reaches zero, causing the nature of the eigenvalues to shift from real to complex or vice versa. However, in our scenario, the discriminant calculation
  • \[ \Delta(\alpha) = \alpha^2 - (\alpha^2 + 1) \] results in a constant -1.
This tells us the eigenvalues remain as complex conjugate pairs irrespective of \(\alpha\), showcasing a constant spiraling motion without a shift in qualitative nature.Understanding these parameter-driven changes helps in many applications, as it clarifies stability shifts and behavior transitions that are pivotal when designing systems like control systems, where stability and predictability are key. Detailed analysis can ensure that operations continue seamlessly across a range of parameter values without unexpected transitions.

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Most popular questions from this chapter

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{cc}{1} & {i} \\ {-i} & {1}\end{array}\right) $$

Find the solution of the given initial value problem. Draw the trajectory of the solution in the \(x_{1} x_{2}-\) plane and also the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{3} & {9} \\ {-1} & {-3}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{2} \\ {4}\end{array}\right) $$

Find the solution of the given initial value problem. Draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) - space and also draw the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{-\frac{5}{2}} & {1} & {1} \\ {1} & {-\frac{5}{2}} & {1} \\ {1} & {1} & {-\frac{5}{2}}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{r}{2} \\ {3} \\\ {-1}\end{array}\right) $$

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{1} & {-1} \\ {5} & {-3}\end{array}\right) \mathbf{x} $$

Deal with the problem of solving \(\mathbf{A x}=\mathbf{b}\) when \(\operatorname{det} \mathbf{A}=0\) Suppose that det \(\mathbf{\Lambda}=0,\) and that \(\mathbf{x}=\mathbf{x}^{(0)}\) is a solution of \(\mathbf{A} \mathbf{x}=\mathbf{b} .\) Show that if \(\xi\) is a solution of \(\mathbf{A} \xi=\mathbf{0}\) and \(\alpha\) is any constant, then \(\mathbf{x}=\mathbf{x}^{(0)}+\alpha \xi\) is also a solution of \(\mathbf{A} \mathbf{x}=\mathbf{b} .\)

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