Question

Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{-1} & {-1} \\ {-\alpha} & {-1}\end{array}\right) \mathbf{x} $$ $$ \begin{array}{l}{\text { (a) Solve the system for } \alpha=0.5 \text { . What are the eigennalues of the coefficient mattix? }} \\ {\text { Classifith the equilitrium point a the natare the cigemalues of the coefficient matrix? Classify }} \\ {\text { the equilithessm for } \alpha \text { . What as the cigemalluce of the coefficient matrix Classify }} \\ {\text { the equilibrium poin at the oigin as to the styse. ematitue different types of behwior. }} \\\ {\text { (c) the parts (a) and (b) solutions of thesystem exhibit two quite different ypes of behwior. }}\end{array} $$ $$ \begin{array}{l}{\text { Find the eigenvalues of the coefficient matrix in terms of } \alpha \text { and determine the value of } \alpha} \\ {\text { between } 0.5 \text { and } 2 \text { where the transition from one type of behavior to the other occurs. This }} \\ {\text { critical value of } \alpha \text { is called a bifurcation point. }}\end{array} $$ $$ \begin{array}{l}{\text { Electric Circuits. Problems } 32 \text { and } 33 \text { are concerned with the clectric circuit described by the }} \\ {\text { system of differential equations in Problem } 20 \text { of Section } 7.1 \text { : }}\end{array} $$ $$ \frac{d}{d t}\left(\begin{array}{l}{l} \\\ {V}\end{array}\right)=\left(\begin{array}{cc}{-\frac{R_{1}}{L}} & {-\frac{1}{L}} \\ {\frac{1}{C}} & {-\frac{1}{C R_{2}}}\end{array}\right)\left(\begin{array}{l}{I} \\ {V}\end{array}\right) $$

Short answer

Based on the given step by step solution, the matrix of the given system is: $$ \begin{pmatrix} -1 & -1 \\ -\alpha & -1 \end{pmatrix} $$ For α = 0.5, the eigenvalues of this matrix are $$\lambda_{1}=-1.618$$ and $$\lambda_{2}=-0.382$$, which means that the equilibrium point at the origin is a stable node. The eigenvalues of the coefficient matrix, in terms of α, can be found from the quadratic equation: $$\lambda^2+2\lambda+(\alpha-1)=0$$. The bifurcation point, where the transition from one type of behavior to the other occurs, occurs at α = 2.

Step-by-step solution

Rewrite the System

We have the given system of linear differential equations: $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{-1} & {-1} \\ {-\alpha} & {-1}\end{array}\right) \mathbf{x} $$ Let's now rewrite it as follows: Let $$\mathbf{x}=\begin{pmatrix} x_1 \\ x_2\end{pmatrix}$$, and so the system becomes $$ \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -\alpha & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} $$

Solve the System for α = 0.5

For α = 0.5, the system becomes: $$ \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -0.5 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} $$ To find the eigenvalues of the coefficient matrix, we compute the determinant of the matrix times the identity matrix minus the eigenvalue variable, λ: $$ \text{det}\left(\begin{array}{cc} -1-\lambda & -1 \\ -0.5 & -1-\lambda \end{array}\right)=\left(-1-\lambda\right)\left(-1-\lambda\right)-\left(-0.5\right)(-1) $$ Solving for λ, we have: $$ \lambda^2 + 2\lambda -1 +0.5 = 0 $$ $$ \lambda^2+2\lambda-0.5=0 $$ The eigenvalues are $$\lambda_{1}=-1.618$$ and $$\lambda_{2}=-0.382$$.

Nature of the Equilibrium Point

To determine the nature of the equilibrium point at the origin, we can analyze the eigenvalues of the coefficient matrix: 1. If both eigenvalues are real and have the same sign, the equilibrium is a stable or unstable node. 2. If eigenvalues are complex conjugates, the equilibrium is either a stable or unstable spiral. 3. If eigenvalues are real with opposite signs, the equilibrium is a saddle point. In our case, both eigenvalues are real and negative. Hence, the equilibrium point at the origin is a stable node.

Eigenvalues in Terms of α

Now, let's determine the eigenvalues of the coefficient matrix in terms of α: $$ \text{det}\left(\begin{array}{cc} -1-\lambda & -1 \\ -\alpha & -1-\lambda \end{array}\right)=\left(-1-\lambda\right)\left(-1-\lambda\right)-\alpha\left(-1\right) $$ $$ \lambda^2 + 2\lambda -1+\alpha = 0 $$

Find the Bifurcation Point

To find the bifurcation point, we need to find the value of α between 0.5 and 2 where the transition from one type of behavior to the other occurs. We can do so by determining the discriminant (Δ) of our quadratic equation. The discriminant will indicate the nature of the eigenvalues: 1. If Δ > 0, both eigenvalues are real. 2. If Δ = 0, both eigenvalues are real and equal. 3. If Δ < 0, eigenvalues are complex conjugates. So, let's find the discriminant: $$ \Delta = b^2 - 4ac = (2)^2 - 4(1)(-1 + \alpha) = 4 + 4 - 4\alpha $$ Setting the discriminant equal to zero, we can find the critical value of α: $$ 4 + 4 - 4\alpha = 0 $$ Solving for α: $$ \alpha = \frac{8}{4}=2 $$ So, the bifurcation point occurs at α = 2. That is when the transition from one type of behavior to the other occurs.

Key concepts

Eigenvalues

When dealing with linear differential equations, the concept of **eigenvalues** becomes important. Eigenvalues provide insight into the dynamics of systems described by such equations.
Here's a simplified explanation:- They are derived from the system's coefficient matrix. By solving the characteristic equation, which involves setting the determinant of the matrix minus an identity matrix times a variable \( \lambda \) to zero, we can find these eigenvalues.- In our exercise, solving for \( \alpha=0.5 \), the eigenvalues obtained are \( \lambda_{1} = -1.618 \) and \( \lambda_{2} = -0.382 \).
- The signs of eigenvalues tell us about the system's stability; negative values indicate a tendency of the system to stabilize.

Equilibrium Point

An **equilibrium point** in differential equations represents the state where the system doesn't change, meaning it is steady over time.
To classify this equilibrium point, such as in our exercise:- We examine the nature of the eigenvalues to understand the equilibrium's stability.- When we have real and negative eigenvalues, like \( \lambda_{1} = -1.618 \) and \( \lambda_{2} = -0.382 \) for \( \alpha=0.5 \), the origin is classified as a stable node.
- At this point, the system remains balanced unless disturbed by an external factor.

Bifurcation Point

A **bifurcation point** marks a critical transition in the behavior of dynamical systems.
In essence:- It's the value of a parameter (normally \( \alpha \)) where the system undergoes a qualitative change in its dynamics.
- In our example, solving the characteristic equation takes us to the value \( \alpha = 2 \) where the discriminant \( \Delta = 0 \). This indicates that at \( \alpha = 2 \), the nature of eigenvalues changes, showing a bifurcation point.- Beyond this point, the system might switch from stable to unstable or vice versa.

Linear Stability Analysis

With **linear stability analysis**, we evaluate how small perturbations near equilibrium points evolve over time.
This analysis helps in:- Determining if such points are stable (the system returns to equilibrium) or unstable (the system diverges from equilibrium).- By linearizing the system near the equilibrium and examining the eigenvalues, stability can be confirmed.
- As seen in our example, for \( \alpha=0.5 \), the analysis showed negative eigenvalues. This is indicative of a stable linear system where disturbances diminish over time, allowing the system to return to its equilibrium state after perturbations.