Problem 66 - Calculus for Biology and Medicine

Question

Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium. $$ A=\left[\begin{array}{rr} 3 & 6 \\ -1 & -4 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a saddle point and unstable.

Step 1: Identify Stability Criteria

To analyze stability, we need to find the eigenvalues of matrix \(A\) and see their real parts. The equilibrium point \((0, 0)\) will be stable if all eigenvalues have negative real parts. Conversely, if any eigenvalue has a positive real part, the equilibrium is unstable.

Step 2: Find Eigenvalues

We find eigenvalues by solving the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix. For matrix \(A = \begin{bmatrix} 3 & 6 \ -1 & -4 \end{bmatrix}\), construct \(A - \lambda I = \begin{bmatrix} 3-\lambda & 6 \ -1 & -4-\lambda \end{bmatrix}\).

Step 3: Compute Determinant

The determinant \(\det(A - \lambda I)\) is given by \[\det(\begin{bmatrix} 3-\lambda & 6 \ -1 & -4-\lambda \end{bmatrix}) = (3-\lambda)(-4-\lambda) - (-1)(6)\]. Simplify this to find the characteristic polynomial.

Step 4: Solve Characteristic Polynomial

Simplifying the determinant: \[ (3-\lambda)(-4-\lambda) + 6 = \lambda^2 + \lambda - 6.\]Solve for \(\lambda\) in the equation \(\lambda^2 + \lambda - 6 = 0\).

Step 5: Factor the Polynomial

Factor the polynomial: \(\lambda^2 + \lambda - 6 = (\lambda - 2)(\lambda + 3) = 0\). The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -3\).

Step 6: Evaluate Stability

The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -3\). Since \(\lambda_1 = 2\) is positive, the real part is positive, indicating that the equilibrium is unstable.

Step 7: Classify the Equilibrium

Since one eigenvalue is positive and the other is negative, the equilibrium point \((0, 0)\) is a saddle point.

Key Concepts

Eigenvalues

When exploring whether an equilibrium point in a system is stable or unstable, eigenvalues are crucial. These values are derived from the matrix associated with the system dynamics, such as matrix \(A\) provided in our problem. The eigenvalues help determine how the solutions will behave over time.In practical terms, eigenvalues indicate:

• The rate of expansion or contraction along a given direction in the system space.
• Changes in direction when passing through the equilibrium.

For matrix \(A\), the eigenvalues are calculated by finding solutions to the characteristic equation, revealing the nature of stability at the equilibrium. If all eigenvalues have negative real parts, the equilibrium is stable. Positive real parts signify instability, while complex eigenvalues can suggest oscillations.

Equilibrium Point

An equilibrium point is a state where the system doesn't change, meaning all derivatives of the system equal zero at that state. For the matrix \(A\), the equilibrium point under scrutiny is \((0,0)\).At this point:

• The system enters a state of rest where outward forces balance inward forces.
• The state remains constant over time if undisturbed.

Understanding an equilibrium point's stability helps predict how small disturbances will affect the system. This is essential for understanding the overall dynamics and behavior of the system in practical applications.

Characteristic Equation

The characteristic equation is vital when determining eigenvalues. It is derived from the transformation of a matrix \(A\) by subtracting \(\lambda I\), where \(\lambda\) is a scalar and \(I\) is the identity matrix. For our particular matrix \(A\), the characteristic equation is constructed from \( \det(A - \lambda I) = 0 \).Here's how to derive it:

• Subtract \(\lambda\) times the identity matrix from \(A\).
• Calculate the determinant of this new matrix.
• Solve the resulting polynomial equation for \(\lambda\).

Through this process, we obtained the characteristic polynomial \( \lambda^2 + \lambda - 6 \), which was then factored into \((\lambda - 2)(\lambda + 3) = 0\). Solving gives us the critical eigenvalues.

Saddle Point

A saddle point is a type of equilibrium point characterized by having both stable and unstable directions. This occurs when the matrix's eigenvalues have differing signs, indicating different types of stability in different directions.For the system associated with matrix \(A\):

• One eigenvalue is positive (\(\lambda_1 = 2\)).
• The other eigenvalue is negative (\(\lambda_2 = -3\)).

These mixed signs confirm that the equilibrium at \((0,0)\) is a saddle point. In this situation, trajectories approach the point in some directions (where the eigenvalue is negative) and diverge in others (where it is positive), creating a saddle-like shape in the phase space. Saddle points play significant roles in system behaviors, often being points of transition in dynamical systems.