Question

Consider the nonhomogeneous linear system \(\mathbf{y}^{\prime}=A \mathbf{y}+\mathbf{g}_{0}\), where \(A\) is a real invertible \((2 \times 2)\) matrix and \(\mathbf{g}_{0}\) is a real \((2 \times 1)\) constant vector. (a) Determine the unique equilibrium point, \(\mathbf{y}_{e}\), of this system. (b) Show how Theorem \(6.3\) can be used to determine the stability properties of this equilibrium point. [Hint: Adopt the change of dependent variable \(\mathbf{z}(t)=\mathbf{y}(t)-\mathbf{y}_{e} .\) ]

Short answer

(a) To find the unique equilibrium point ye, you need to set dy/dt equal to 0 and solve for the vector y: 0 = A * ye + g0 => ye = -A^(-1) * g0 (b) To determine the stability properties of the equilibrium point ye, analyze the eigenvalues of matrix A following Theorem 6.3: Case 1: All eigenvalues have negative real parts → ye is asymptotically stable Case 2: Any eigenvalue has a positive real part → ye is unstable Case 3: All eigenvalues have non-positive real parts and no repeated eigenvalues with a zero real part → ye is stable

Step-by-step solution

Find the equilibrium point ye

To find the unique equilibrium point ye, set dy/dt equal to 0 and solve for the vector y: 0 = A * ye + g0 => A * ye = -g0 => ye = -A^(-1) * g0 Here, A^(-1) is the inverse of matrix A, which exists because A is given as invertible.

Change dependent variable to z(t)

Define a new variable z(t) as follows: z(t) = y(t) - ye Now, we want to find dz/dt. Using the chain rule, dz/dt = dy/dt - d(ye)/dt Since ye is a constant, d(ye)/dt = 0, so dz/dt = dy/dt Now substitute the given equation dy/dt = A * y + g0: dz/dt = A * y + g0 Express y in terms of z and ye: y = z + ye So, dz/dt = A * (z + ye) + g0

Homogenize the transformed system

Now we have the equation for dz/dt: dz/dt = A * (z + ye) + g0 Substitute the equilibrium point ye from Step 1: dz/dt = A * (z - A^(-1) * g0) + g0 Using the property of matrix multiplication, we can rewrite it as: dz/dt = A * z + (A * (-A^(-1) * g0) + g0) Since A * (-A^(-1) * g0) + g0 = 0, the transformed system equation is: dz/dt = A * z

Use Theorem 6.3 to determine stability properties

Theorem 6.3 states that the stability properties of a linear homogeneous system are determined by the real parts of the eigenvalues of the coefficient matrix. Now, the transformed system is a linear homogeneous system with matrix A. To determine the stability properties, we should analyze the eigenvalues of matrix A. Case 1: If all eigenvalues of A have negative real parts, the equilibrium point ye is asymptotically stable. Case 2: If any eigenvalue of A has a positive real part, the equilibrium point ye is unstable. Case 3: If all eigenvalues of A have non-positive real parts, and there are no repeated eigenvalues with a zero real part, then the equilibrium point ye is stable. The conclusion on the stability properties of ye will depend on the specific matrix A and its eigenvalues.

Key concepts

Equilibrium Point

In a nonhomogeneous linear system like \( \mathbf{y}^{\prime} = A \mathbf{y} + \mathbf{g}_{0} \), finding the equilibrium point is key to understanding the system's behavior. An equilibrium point, \( \mathbf{y}_{e} \), is a state where the system doesn't change over time, meaning \( \frac{d\mathbf{y}}{dt} = 0 \). To find this, we solve:

• Set \( A \mathbf{y}_{e} + \mathbf{g}_{0} = 0 \).
• This simplifies to \( A \mathbf{y}_{e} = -\mathbf{g}_{0} \).
• Given that \( A \) is invertible, multiply by the inverse of \( A \): \( \mathbf{y}_{e} = -A^{-1}\mathbf{g}_{0} \).

This solution gives us the unique state where no external changes force the system away from \( \mathbf{y}_{e} \). Understanding this helps us anticipate the natural state of the system without external influences. By knowing \( \mathbf{y}_{e} \), we can further examine its stability and behavior under different conditions.

Stability Analysis

Stability analysis is crucial in predicting how a system reacts to small disturbances. To analyze stability, we transform the system around the equilibrium point. This involves setting \( \mathbf{z}(t) = \mathbf{y}(t) - \mathbf{y}_{e} \), which leads to \( \frac{d\mathbf{z}}{dt} = A \mathbf{z} \), since \( \mathbf{y}_{e} \) is constant. Now, our focus shifts to this homogeneous system.### Steps to analyze stability:

• Determine \( \mathbf{z}(t) \): This change of variable assumes any deviation from \( \mathbf{y}_{e} \) is captured by \( \mathbf{z}(t) \).
• Acquire \( \frac{d\mathbf{z}}{dt} \): The equation \( \frac{d\mathbf{z}}{dt} = A \mathbf{z} \) is key.

Theorem 6.3 states that a system's stability is decided by the eigenvalues of the coefficient matrix \( A \). Different eigenvalue scenarios reflect different stability characteristics.
Analyzing these properties helps decide if the system will return to equilibrium, diverge, or remain change-free when slightly disturbed.

Eigenvalues

Eigenvalues play a pivotal role in understanding the dynamics of the system, specifically determining stability. In linear algebra, eigenvalues are scalars that influence how matrices act on vectors, crucial for systems described by \( \mathbf{z}^{\prime} = A \mathbf{z} \).### Key insights on eigenvalues:

• All eigenvalues of \( A \) with negative real parts indicate asymptotic stability, meaning solutions approach \( \mathbf{y}_{e} \) over time.
• Any eigenvalue with a positive real part signifies instability, implying solutions diverge away from \( \mathbf{y}_{e} \).
• If all eigenvalues have non-positive real parts, and no zeros are repeated, \( \mathbf{y}_{e} \) is stable but not necessarily attractive.

Real parts of eigenvalues guide us regarding how perturbations decay or grow without external energy being added to the system.
This enables predictions about how the system will behave in practical applications, emphasizing the importance of a detailed eigenvalue analysis.