Question

In each exercise, locate all equilibrium points for the given autonomous system. Determine whether the equilibrium point or points are asymptotically stable, stable but not asymptotically stable, or unstable. $$ \begin{aligned} &y_{1}^{\prime}=2 y_{1}+y_{2}+y_{3} \\ &y_{2}^{\prime}=y_{1}+y_{2}+2 y_{3} \\ &y_{3}^{\prime}=y_{1}+2 y_{2}+y_{3} \end{aligned} $$

Short answer

Question: Locate the equilibrium points of the given autonomous system and determine their stability. System of equations: $$ \begin{aligned} &y_1' = 2 y_{1}+y_{2}+y_{3} \\ &y_2' = y_{1}+y_{2}+2 y_{3} \\ &y_3' = y_{1}+2 y_{2}+y_{3} \end{aligned} $$ Answer: The system has one equilibrium point at \((0,0,0)\), and it is unstable.

Step-by-step solution

Find the equilibrium points

To find the equilibrium points, set each equation to 0 and solve the resulting system of linear equations: $$ \begin{aligned} &2 y_{1}+y_{2}+y_{3} = 0 \\ &y_{1}+y_{2}+2 y_{3} = 0 \\ &y_{1}+2 y_{2}+y_{3} = 0 \end{aligned} $$ We can solve using Gaussian elimination, or any other method of solving linear equations. Here, we can easily solve for \(y_1\), and then find \(y_2\) and \(y_3\). From the second equation, we have \(y_1 = -y_2 - 2y_3\). Substituting this into the first equation, we get $$ -2y_2 - 3y_3 + y_2 + y_3 = 0 \implies -y_2 - 2y_3 = 0 $$ Finally, we can substitute our expressions for \(y_1\) and \(y_2\) into the third equation, resulting in: $$ -(y_2 + 2y_3) + 2y_2 + y_3 = 0 \implies y_2 - y_3 = 0 \implies y_2 = y_3 $$ From the last equation, we get that \(y_2 = y_3\). Combining with the result we had earlier for \(y_1\), we have the equilibrium point \((y_1, y_2, y_3) = (0,0,0)\).

Analyze the stability of the equilibrium point

To determine the stability of the equilibrium point, we need to find the Jacobian matrix and its eigenvalues. The Jacobian matrix for this system is given by $$ J = \begin{bmatrix} \frac{\partial y_1'}{\partial y_1} & \frac{\partial y_1'}{\partial y_2} & \frac{\partial y_1'}{\partial y_3} \\ \frac{\partial y_2'}{\partial y_1} & \frac{\partial y_2'}{\partial y_2} & \frac{\partial y_2'}{\partial y_3} \\ \frac{\partial y_3'}{\partial y_1} & \frac{\partial y_3'}{\partial y_2} & \frac{\partial y_3'}{\partial y_3} \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{bmatrix} $$ Now, we need to find the eigenvalues of the Jacobian matrix. The characteristic equation of a 3x3 matrix is given by $$ \det(J - \lambda I) = \begin{vmatrix} 2-\lambda & 1 & 1\\ 1 & 1-\lambda & 2\\ 1& 2 & 1-\lambda \end{vmatrix} = 0 $$ Computing the determinant, we get $$ (2-\lambda)((1-\lambda)^2 - 4) - 1(2(2-\lambda) - 4) = (\lambda^2 - 6\lambda + 8)(\lambda - 3) = (\lambda-2)(\lambda-4)(\lambda-3) = 0 $$ We have 3 eigenvalues: \(\lambda_1 = 2, \lambda_2 = 3, \lambda_3 = 4\). Since the real part of all eigenvalues is positive, the equilibrium point \((0,0,0)\) is unstable.

Key concepts

Equilibrium Points

Equilibrium points are specific values in a dynamical system where all derivatives are equal to zero. In an autonomous system, like the one given, these points can be seen as the resting spots where the system neither grows nor decays. To find them, you set each differential equation to zero, essentially creating a system of algebraic equations. Solving this system will give the equilibrium points. In our case, when we set:

• \(2 y_{1} + y_{2} + y_{3} = 0\)
• \(y_{1} + y_{2} + 2 y_{3} = 0\)
• \(y_{1} + 2 y_{2} + y_{3} = 0\)

We find the solution \((y_1, y_2, y_3) = (0, 0, 0)\). This point, \((0,0,0)\), represents a state where the system is perfectly balanced.

Stability Analysis

Stability analysis helps us understand how a system behaves near an equilibrium point when subjected to minor disturbances. If a system returns to the equilibrium point after a disturbance, it is asymptotically stable. If it stays near the point but doesn’t return, it is stable but not asymptotically stable. If the system moves away, it's unstable.
In practice, finding the eigenvalues of the Jacobian matrix will help us determine the stability. The sign of the real part of these eigenvalues is crucial.

• If all are negative, the system is stable and returns to equilibrium.
• If any are positive, the system is unstable and moves away from equilibrium.
• If zero, the stability is more complex to determine.

In this exercise, all eigenvalues were positive, indicating the equilibrium is unstable.

Jacobian Matrix

The Jacobian matrix is central to analyzing the stability of systems of differential equations. It's essentially a matrix of all first-order partial derivatives describing how each variable in a system affects the others. For our system, the Jacobian matrix is: \[ \begin{bmatrix} 2 & 1 & 1 \ 1 & 1 & 2 \ 1 & 2 & 1 \end{bmatrix} \] This matrix provides insight into how small changes in the variables \(y_1, y_2,\) and \(y_3\) affect the rate of change in the system. By analyzing the Jacobian at the equilibrium point, we can glean whether small perturbations will dissipate or escalate. This ultimately connects directly to the eigenvalue analysis.

Eigenvalues

Eigenvalues are vital for determining the stability of an equilibrium point in a dynamical system. They arise from solving the characteristic equation \(\det(J - \lambda I) = 0\), where \(J\) is the Jacobian matrix and \(\lambda\) are the eigenvalues. These values tell us about the nature of the trajectories near an equilibrium point. In simple terms:

• If all eigenvalues have negative real parts, perturbations diminish, making the point stable.
• If any eigenvalue has a positive real part, perturbations amplify, rendering the point unstable.
• Eigenvalues with zero real parts can result in more complex trajectory behaviors.

For the given system, with eigenvalues \(\lambda_1 = 2, \lambda_2 = 3, \lambda_3 = 4\), all are positive, confirming the instability of the equilibrium point \((0, 0, 0)\). This means disturbances will grow, pushing the system away from equilibrium.