Question

Use the information given about the nature of the equilibrium point at the origin to determine the value or range of permissible values for the unspecified entry in the coefficient matrix. Given \(\mathbf{y}^{\prime}=\left[\begin{array}{ll}-4 & \alpha \\ -2 & 2\end{array}\right] \mathbf{y}\), for what values of \(\alpha\) (if any) can the origin be an asymptotically stable spiral point?

Short answer

Based on the given system of differential equations, the equilibrium point at the origin will be an asymptotically stable spiral point as long as the unspecified entry \(\alpha\) in the coefficient matrix is not equal to \(-\frac{9}{8}\).

Step-by-step solution

Finding the eigenvalues of the matrix

First, we need to find the eigenvalues of the given coefficient matrix: $$\left[\begin{array}{ll}-4 & \alpha \\\ -2 & 2\end{array}\right]$$ To do this, we will compute the characteristic polynomial of the matrix by subtracting the eigenvalue (\(\lambda\)) along the diagonal and taking the determinant: $$\det\left(\begin{array}{cc}-4-\lambda & \alpha \\ -2 & 2-\lambda\end{array}\right)=(-4-\lambda)(2-\lambda)-(-2\alpha)=\lambda^{2} + 2 \lambda - 8 - 2\alpha$$

Compute the real and imaginary parts of eigenvalues

We will now compute the real and imaginary parts of the eigenvalues of the characteristic polynomial. Using the quadratic formula: $$\lambda = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with \(a=1\), \(b=2\), and \(c=-8-2\alpha\), we get: $$\lambda = \frac{-2\pm\sqrt{2^{2}-4(-8-2\alpha)}}{2}$$ This simplifies to: $$\lambda = -1\pm\sqrt{9+8\alpha}$$ The real part of the eigenvalues is \(-1\), and the imaginary part is \(\pm\sqrt{9+8\alpha}\).

Determine the conditions for an asymptotically stable spiral point

For the origin to be an asymptotically stable spiral point, the following conditions must be satisfied: 1. The real part of the eigenvalues should be negative: \(-1 < 0\) 2. The imaginary part of the eigenvalues should be non-zero: \(\sqrt{9+8\alpha}\neq 0\) Since the first condition is already satisfied, let's focus on the second condition. The imaginary part will be non-zero if the expression inside the square root is not equal to zero. So, we need to find the value of \(\alpha\) that satisfies: $$9+8\alpha\neq 0$$

Solve for the range of permissible values for \(\alpha\)

Solving the inequality, we get: $$8\alpha \neq -9$$ $$\alpha \neq -\frac{9}{8}$$

State the conclusion

The origin will be an asymptotically stable spiral point for the values of \(\alpha\) that satisfy: $$\alpha \neq -\frac{9}{8}$$

Key concepts

Eigenvalues of a Matrix

To understand the behavior of a system of linear differential equations, one often looks at the eigenvalues of the associated coefficient matrix. These are special numbers that give us valuable insights into the stability and dynamics of the system.

Eigenvalues are found by solving the characteristic equation of the matrix, which is derived from its characteristic polynomial. They allow us to predict if solutions grow, decay, oscillate, or combine some of these behaviors as time goes on. In particular, negative eigenvalues suggest that the system will return to equilibrium over time – a property known as stability. So, for a point to be an asymptotically stable spiral, we expect its eigenvalues to have negative real parts and non-zero imaginary parts, indicating a spiraling motion that decays toward the equilibrium.

Characteristic Polynomial

The characteristic polynomial of a matrix is a bridge linking linear algebra to differential equations. It is a polynomial equation that arises from the determinant of the matrix, obtained by subtracting an eigenvalue multiplied by the identity matrix from the original matrix. Here's the process simplified:

• Start with the coefficient matrix of the system.
• Subtract \(\lambda\) times the identity matrix from it, where \(\lambda\) is a scalar representing an eigenvalue.
• Determine the determinant of this new matrix.
• The resulting expression gives us the characteristic polynomial.

Then, we solve for \(\lambda\) to find the eigenvalues. The roots of this polynomial – which can be real or complex – provide deep insights into the system's behavior. With the right values, they tell us if the system is in for a stable, unstable, or undetermined future.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are essential in describing various phenomena in engineering, physics, biology, and economics, among other fields.

For example, in the context of linear differential equations, the role of a coefficient matrix is to dictate the evolution of the system over time. The way its solutions behave can be largely understood through eigenvalues and eigenvectors, leading to solutions that may exponentially grow, decay, or oscillate. In the exercise provided, by analyzing the coefficients through the eigenvalues, we can deduce the stability of the system and predict the possible long-term behavior around an equilibrium point, which in this case, we want to be an asymptotically stable spiral.