Question

Locate the unique equilibrium point of the given nonhomogeneous system, and determine the stability properties of this equilibrium point. Is it asymptotically stable, stable but not asymptotically stable, or unstable? $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 3 & 2 \\ -4 & -3 \end{array}\right] \mathbf{y}+\left[\begin{array}{r} -2 \\ 2 \end{array}\right] $$

Short answer

Answer: The unique equilibrium point is $\mathbf{y}^{*}=\left[\begin{array}{r} 1 \\ 1/2 \end{array}\right]$ and it is unstable.

Step-by-step solution

Find the equilibrium point

To find the equilibrium point, we set the derivative of the system equal to 0: $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 3 & 2 \\ -4 & -3 \end{array}\right] \mathbf{y}+\left[\begin{array}{r} -2 \\ 2 \end{array}\right] = \mathbf{0} $$ Now, solve for \(\mathbf{y}\). Let $\mathbf{y}=\left[\begin{array}{r} y_1 \\ y_2 \end{array}\right]$, then we have: $$ \left[\begin{array}{rr} 3 & 2 \\ -4 & -3 \end{array}\right] \left[\begin{array}{r} y_1 \\ y_2 \end{array}\right] + \left[\begin{array}{r} -2 \\ 2 \end{array}\right] = \mathbf{0} $$ By writing the system of linear equations, we have: $$ \begin{cases} 3 y_1 + 2 y_2 = 2 \\ -4 y_1 - 3 y_2 = -2 \end{cases} $$ Solving this linear system, we find the equilibrium point: $$ \mathbf{y}^{*} = \left[\begin{array}{r} 1 \\ 1/2 \end{array}\right] $$

Determine the stability

To determine the stability of the equilibrium point, we need to analyze the eigenvalues of the matrix in the system: $$ \left[\begin{array}{rr} 3 & 2 \\ -4 & -3 \end{array}\right] $$ Find the eigenvalues by computing the determinant of \((A - \lambda I)\) and setting it to 0: $$ \text{det}\left(\left[\begin{array}{cc} 3-\lambda & 2 \\ -4 & -3-\lambda \end{array}\right]\right)= (3-\lambda)(-3-\lambda) - (2)(-4) = \lambda^2-9 $$ Set this determinant to 0 and solve for \(\lambda\): $$ \lambda^2-9=0 \Rightarrow \lambda_1 = 3, \lambda_2 = -3 $$ We have one positive eigenvalue (\(\lambda_1 = 3\)) and one negative eigenvalue (\(\lambda_2 = -3\)). Since we have both a positive and a negative eigenvalue, the equilibrium point is unstable. Therefore, the unique equilibrium point is $\mathbf{y}^{*}=\left[\begin{array}{r} 1 \\ 1/2 \end{array}\right]$ and it is unstable.

Key concepts

Stability Analysis

Stability analysis is a key concept in understanding the behavior of dynamical systems around their equilibrium points. When dealing with a system, like the one given, we primarily want to know if small perturbations or changes in conditions will cause the state of the system to return to equilibrium, or if they will lead to large deviations.
To assess this, we focus on determining whether an equilibrium point is:

• Asymptotically stable: Any deviation from the equilibrium shrinks over time, indicating the system naturally returns to equilibrium.
• Stable but not asymptotically stable: Deviations neither grow nor shrink, meaning the system can remain near equilibrium indefinitely.
• Unstable: Small changes in state could cause the system to move far away from equilibrium.

In the exercise given, we have a single equilibrium point found using the conditions for equilibrium. We then check its stability using eigenvalue analysis, which provides insights into how system states behave over time in the vicinity of this equilibrium.

Eigenvalues

Eigenvalues are fundamental in the stability analysis of linear systems. They originate from a matrix that represents the system, and each eigenvalue provides crucial information about system behavior near an equilibrium point.
In our exercise, the system matrix is A = \[\begin{array}{rr} 3 & 2 \ -4 & -3 \end{array}\]. To find eigenvalues, we solve the characteristic equation obtained from the determinant calculation \( \text{det}(A - \lambda I) = \lambda^2 - 9 \).
This results in:

• A positive eigenvalue \( \lambda_1 = 3 \) indicating a direction in which disturbances grow exponentially over time.
• A negative eigenvalue \( \lambda_2 = -3 \) suggesting a direction in which disturbances decay.

Since there is a mix of positive and negative eigenvalues, this suggests that some perturbations could grow, making the system unstable. The presence of any positive eigenvalue typically signifies an unstable system, even if other eigenvalues are negative.

Nonhomogeneous Systems

A nonhomogeneous system, as given in our exercise, includes terms that do not solely depend on the dependent variables, but also on constant or varying external inputs. The system is expressed as:
\[ \mathbf{y}^{\prime}=A\mathbf{y}+\mathbf{b} \]
where \( A \mathbf{y} \) represents the homogeneous part and \( \mathbf{b} \) is the nonhomogeneous part. The vector \( \mathbf{b} = \begin{array}{r} -2 \ 2 \end{array} \) introduces fixed alterations that affect the equilibrium point.

• In homogeneous systems, solutions tend to be relatively simpler and solutions deal primarily with the term \( A \mathbf{y} \).
• In contrast, nonhomogeneous systems like ours require adjusting for the constant vector \( \mathbf{b} \) when finding equilibrium points and analyzing stability.

These systems require different techniques to ascertain whether an equilibrium point is reached and to understand how they react over time. In our exercise, solving the equations where \( A\mathbf{y}+ \mathbf{b} = 0 \) determines the unique equilibrium point, which is then tested for stability through eigenvalue analysis.