Question

Consider the autonomous system $$ \begin{aligned} &x^{\prime}=-x+x y+y \\ &y^{\prime}=x-x y-2 y \end{aligned} $$ This is the reduced system for the chemical reaction discussed in Exercise 19 of Section \(6.1\) with \(a(t)=x(t), c(t)=y(t), e_{0}=1\), and all rate constants set equal to 1 . (a) Show that this system has a single equilibrium point, \(\left(x_{e}, y_{e}\right)=(0,0)\). (b) Determine the linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\), and analyze its stability properties. (c) Show that the system is an almost linear system at equilibrium point \((0,0)\). (d) Use Theorem \(6.4\) to determine the equilibrium properties of the given nonlinear system at \((0,0)\).

Short answer

Answer: Yes, the given autonomous system is stable near its equilibrium point (0,0).

Step-by-step solution

Find the equilibrium point

To find the equilibrium point of the system, we need to set the given time derivatives equal to zero and solve for x and y: $$ \begin{aligned} &x' = -x + xy + y = 0\\ &y' = x - xy - 2y = 0 \end{aligned} $$ Now, we solve these equations simultaneously for x and y to find the equilibrium point: $$ x = xy + y \implies x(1 - y) = y $$ and $$ y = x - xy - 2y \implies y(1 + x) = x $$ From the first equation, we get two possibilities: 1. \(x = 0\) 2. \(1 - y = 0 \implies y = 1\) Substitute these possibilities into the second equation and solve for the corresponding values of y or x: 1. If \(x = 0\), then \(0 = 0\) (second equation is satisfied) 2. If \(y = 1\), then \(0 = x(2) \implies x = 0\) Thus, the equilibrium point is \((x_e, y_e) = (0, 0)\).

Linearize the system and analyze stability

Linearize the system by finding the Jacobian matrix A of the given autonomous system: $$ A = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} $$ Next, compute the partial derivatives and evaluate the Jacobian matrix at the equilibrium point \((0,0)\): $$ A = \begin{bmatrix} -y-1 & x+1\\ y+1 & -x-2 \end{bmatrix}_{(x=0,y=0)} = \begin{bmatrix} -1 & 1 \\ 1 & -2 \end{bmatrix} $$ Now, we find the eigenvalues of A to determine the stability properties of the linearized system: $$ \text{det}(\lambda I - A) = \begin{vmatrix} \lambda +1 & -1 \\ -1 & \lambda +2 \end{vmatrix} = (\lambda +1)(\lambda +2) -(-1)(-1) $$ Solving the characteristic equation, we find the eigenvalues: $$ \lambda^2 +3\lambda +1 = 0 $$ The eigenvalues are complex with negative real parts. Thus, the linearized system is stable.

Check if the system is almost linear at equilibrium point

We need to show that the sum of the first and second partial derivatives of the time derivatives with respect to x and y is continuous at the equilibrium point \((0,0)\) in order for the system to be almost linear at \((0,0)\). Compute the second partial derivatives: $$ \begin{aligned} \frac{\partial^2 x'}{\partial x^2} = 0, & \quad \frac{\partial^2 x'}{\partial x\partial y} = 1, & \quad \frac{\partial^2 x'}{\partial y^2} = 0 \\ \frac{\partial^2 y'}{\partial x^2} = 0, & \quad \frac{\partial^2 y'}{\partial x \partial y} = -1, & \quad \frac{\partial^2 y'}{\partial y^2} = 0 \end{aligned} $$ Since all of these second partial derivatives are continuous at \((0,0)\), the given system is an almost linear system at the equilibrium point \((0,0)\).

Use Theorem 6.4 to determine the stability properties of the nonlinear system

Theorem 6.4 states that if an almost linear system has a stable linearized system at equilibrium point, the nonlinear system will also be stable near that equilibrium point. Since we have already shown that the given system is an almost linear system and its linearized system is stable at the equilibrium point \((0,0)\), we can conclude that the given nonlinear system is also stable near the equilibrium point \((0,0)\).

Key concepts

Equilibrium Point

In the context of autonomous systems, an equilibrium point is a crucial concept. It is where the system remains at rest, meaning the change in both variables is zero. To find it, we need to set the derivatives equal to zero and solve for the variables involved.
For our system, this involves solving the equations:

• \(-x + xy + y = 0\)
• \(x - xy - 2y = 0\)

Because this system simplifies, we only have one equilibrium point: \((x_e, y_e) = (0,0)\). Understanding this point helps predict the system’s behavior around it.

Linearization

Linearization is used to approximate a non-linear system near an equilibrium point. It simplifies analysis and makes it possible to use linear algebra for understanding system behavior.
We achieve linearization by creating the Jacobian matrix, which captures the linear part of the system. For our problem, the Jacobian at the equilibrium point \((0,0)\) is:

• \(A = \begin{bmatrix}-1 & 1\ 1 & -2\end{bmatrix}\)

This matrix is a simplified version of the original system and helps in performing stability analysis.

Stability Analysis

Stability analysis examines how a system behaves when it is slightly disturbed from an equilibrium point. If the system returns to the equilibrium, it is considered stable.
For linear systems, this involves analyzing the eigenvalues of the Jacobian matrix.
In this problem, our linearized system has a Jacobian matrix eigenvalues that are complex with negative real parts.
This indicates that the equilibrium is stable, as any disturbances around \((0,0)\) will decay over time, eventually returning to the point.

Jacobians

The Jacobian matrix is critical in understanding system dynamics near equilibrium points. It is composed of the first partial derivatives of the system equations.
For our autonomous system, the Jacobian matrix aids in linearizing the system, making stability analysis feasible.
By calculating the derivatives:

• \(\frac{\partial x'}{\partial x} = -1,\ \frac{\partial x'}{\partial y} = 1\)
• \( \frac{\partial y'}{\partial x} = 1,\ \frac{\partial y'}{\partial y} = -2\)

This information constructs the Jacobian, which is pivotal in analyzing behavior around equilibrium.

Eigenvalues

Eigenvalues of the Jacobian matrix provide insights into the nature of the equilibrium point's stability. They determine whether perturbations grow or shrink over time.
In the given problem, by solving the characteristic equation

• \(\lambda^2 + 3\lambda + 1 = 0\)

We find eigenvalues with negative real parts, indicating stability. The presence of complex eigenvalues suggests oscillations that dampen over time. These evaluations confirm that small disturbances around the equilibrium point \((0,0)\) will die out, rendering the point stable.