J=( 
∂x
∂ 
x
˙
 
​
 
∂x
∂ 
y
˙
​
 
​
 
​
  
∂y
∂ 
x
˙
 
​
 
∂y
∂ 
y
˙
​
 
​
 
​
 )=( 
ϵ 
1
​
 −2σ 
1
​
 X−α 
1
​
 Y
−α 
2
​
 Y
​
  
−α 
1
​
 X
ϵ 
2
​
 −2σ 
2
​
 Y−α 
2
​
 X
​
 ).
At the equilibrium point 
(
X
,
Y
)
(X,Y), the characteristic equation is found by solving 
det
⁡
(
J
−
λ
I
)
=
0
det(J−λI)=0, leading to the quadratic form:

To solve for the characteristic equation at equilibrium, let's break down the steps involved in finding the determinant and eventually arriving at the quadratic form.

1. **Jacobian Matrix \( J \):**
   The given Jacobian matrix \( J \) is:

   \[
   J = \begin{pmatrix} 
   \epsilon_1 - 2\sigma_1 X - \alpha_1 Y & -\alpha_2 Y \\
   -\alpha_1 X & \epsilon_2 - 2\sigma_2 Y - \alpha_2 X
   \end{pmatrix}
   \]

   where:
   - \( \epsilon_1, \epsilon_2 \) are constants related to the system,
   - \( \sigma_1, \sigma_2 \) are constants related to the stability at the equilibrium point,
   - \( \alpha_1, \alpha_2 \) are interaction terms,
   - \( X \) and \( Y \) are the variables for the equilibrium points.

2. **Characteristic Equation:**
   The characteristic equation is found by solving the determinant of \( J - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues of the system. Thus, we have:

   \[
   \text{det}(J - \lambda I) = 0
   \]

   This becomes:

   \[
   \text{det}\left( \begin{pmatrix} 
   \epsilon_1 - 2\sigma_1 X - \alpha_1 Y - \lambda & -\alpha_2 Y \\
   -\alpha_1 X & \epsilon_2 - 2\sigma_2 Y - \alpha_2 X - \lambda
   \end{pmatrix} \right) = 0
   \]

3. **Determinant Calculation:**
   The determinant of a \( 2 \times 2 \) matrix is given by:

   \[
   \text{det}(A) = ad - bc
   \]

   For our matrix:

   \[
   \begin{vmatrix} 
   \epsilon_1 - 2\sigma_1 X - \alpha_1 Y - \lambda & -\alpha_2 Y \\
   -\alpha_1 X & \epsilon_2 - 2\sigma_2 Y - \alpha_2 X - \lambda
   \end{vmatrix}
   \]

   The determinant will be:

   \[
   \left( \epsilon_1 - 2\sigma_1 X - \alpha_1 Y - \lambda \right) \left( \epsilon_2 - 2\sigma_2 Y - \alpha_2 X - \lambda \right) - (-\alpha_2 Y)(-\alpha_1 X)
   \]

4. **Simplifying the Determinant:**
   Expanding this determinant results in:

   \[
   \left( \epsilon_1 - 2\sigma_1 X - \alpha_1 Y - \lambda \right) \left( \epsilon_2 - 2\sigma_2 Y - \alpha_2 X - \lambda \right) - \alpha_2 \alpha_1 XY
   \]

5. **Quadratic Form:**
   The characteristic equation is a quadratic equation in \( \lambda \). After simplification, it takes the following form:

   \[
   \lambda^2 + a\lambda + b = 0
   \]

   where:
   - \( a \) is the trace of matrix \( J \),
   - \( b \) is the determinant of matrix \( J \).

In this case, \( a = -(\epsilon_1 + \epsilon_2) + 2\sigma_1 X + 2\sigma_2 Y + \alpha_1 Y + \alpha_2 X \), and \( b \) involves the product of the diagonal elements minus the interaction terms.

Would you like me to continue with the explicit solution for the eigenvalues \( \lambda \), or do you need help with a specific part of this?

### Related Questions:
1. How do interaction terms \( \alpha_1 \) and \( \alpha_2 \) affect the stability of the system?
2. What is the physical interpretation of the eigenvalues in this system?
3. How can we classify the equilibrium point based on the eigenvalues?
4. What are the roles of \( \epsilon_1 \), \( \epsilon_2 \), \( \sigma_1 \), and \( \sigma_2 \) in determining system behavior?
5. How can you apply this method to a higher-dimensional system?

### Tip:
Understanding how the trace and determinant of the Jacobian matrix relate to the stability of the equilibrium points is key in analyzing dynamical systems.

J=(∂x/∂x˙ ∂x/∂y˙ ∂y/∂x˙ ∂y/∂y˙)=(ϵ1−2σ1X−α1Y −α2Y −α1X ϵ2−2σ2Y−α2X). At the equilibrium point (X,Y), the characteristic equation is found by solving det(J−λI)=0, leading to the quadratic form.

Discover how to find the characteristic equation for a 2x2 Jacobian matrix at equilibrium. Learn the key steps of determining eigenvalues, calculating the determinant, and simplifying the quadratic form. This guide covers advanced concepts in dynamical systems, suitable for undergraduate students.

Math Problem Statement

−α 1 X ϵ 2 −2σ 2 Y−α 2 X ). At the equilibrium point ( X , Y ) (X,Y), the characteristic equation is found by solving det ⁡ ( J − λ I )

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