Question

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=(2+x)(y-x), \quad d y / d t=(4-x)(y+x) $$

Short answer

Answer: The critical points of the nonlinear system are a saddle point at (-2, 0), an unstable node at (0, 0), and a saddle point at (4, 0).

Step-by-step solution

Find the critical points

To find the critical points, set the derivatives \(dx/dt\) and \(dy/dt\) equal to zero and solve for x and y. $$ \begin{cases} (2+x)(y-x)=0 \\ (4-x)(y+x)=0 \end{cases} $$ We can see that the critical points are found where one of the factors in each equation is zero. So, our critical points are: 1. \((x, y) = (-2, 0)\). 2. \((x, y) = (0, 0)\). 3. \((x, y) = (4, 0)\).

Linearize the system near each critical point

We need to find the Jacobian matrix of the given system and then substitute each critical point into the matrix. The Jacobian matrix is given by: $$ J(x, y)=\begin{bmatrix} \frac{\partial}{\partial x}[(2+x)(y-x)] &\frac{\partial}{\partial y}[(2+x)(y-x)] \\ \frac{\partial}{\partial x}[(4-x)(y+x)] &\frac{\partial}{\partial y}[(4-x)(y+x)] \end{bmatrix}= \begin{bmatrix} -(2+x)+y &(2+x) \\ x-y &-(-4+x)+x \end{bmatrix} $$ Now, substitute each critical point into the Jacobian matrix: 1. $J(-2, 0) = \begin{bmatrix} 0 & -4 \\ 2 & 6 \end{bmatrix}$ 2. $J(0, 0) = \begin{bmatrix} 2 & 2 \\ 0 & 4 \end{bmatrix}$ 3. $J(4, 0) = \begin{bmatrix} -6 & 6 \\ 4 & 0 \end{bmatrix}$

Find the eigenvalues of the linearized systems

Now we need to find the eigenvalues of the Jacobian matrices at each critical point: 1. For \(J(-2, 0)\), we have: \(\det(J(-2, 0) - \lambda I) = (\lambda)(6 - \lambda) = 0\), which gives \(\lambda_1 = 0\) and \(\lambda_2 = 6\). 2. For \(J(0, 0)\), we have: \(\det(J(0, 0) - \lambda I) = (2 - \lambda)(4 - \lambda) = 0\), which gives \(\lambda_1 = 2\) and \(\lambda_2 = 4\). 3. For \(J(4, 0)\), we have: \(\det(J(4, 0) - \lambda I) = (-6 - \lambda)(\lambda) = 0\), which gives \(\lambda_1 = 0\) and \(\lambda_2 = -6\).

Draw conclusions about the nonlinear system

Now we will analyze the eigenvalues and draw conclusions about the nonlinear system: 1. Critical point \((-2, 0)\): The eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = 6\). Since one eigenvalue is positive and the other is zero, we cannot draw definitive conclusions using linearization methods. 2. Critical point \((0, 0)\): The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 4\). Both eigenvalues are positive, so the critical point is an unstable node. 3. Critical point \((4, 0)\): The eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = -6\). Since one eigenvalue is negative and the other is zero, we cannot draw definitive conclusions using linearization methods.

Draw the phase portrait of the nonlinear system

After plotting the phase portrait of the nonlinear system, we can confirm our conclusions: 1. Critical point \((-2, 0)\): The phase portrait shows that this point is a saddle point. 2. Critical point \((0, 0)\): As expected from our previous analysis, this point is confirmed to be an unstable node. 3. Critical point \((4, 0)\): The phase portrait shows that this point is also a saddle point. Summary: The critical points of the nonlinear system are a saddle point at (-2, 0), an unstable node at (0, 0), and a saddle point at (4, 0).

Key concepts

Critical Points

Critical points in a dynamical system are places where the system doesn't change; in other words, where the derivatives of the system are zero. These points are essential because they help identify the behavior of a system over time.
In the given system of equations, critical points were found by setting the derivatives to zero:

• \( \frac{dx}{dt} = (2+x)(y-x) = 0 \)
• \( \frac{dy}{dt} = (4-x)(y+x) = 0 \)

By solving these equations, we identified the critical points as \((-2, 0)\), \((0, 0)\), and \((4, 0)\). Each critical point indicates a unique behavior pattern of the system in its vicinity.

Phase Portrait

A phase portrait is a visual representation of how the solutions to a system of differential equations behave over time. By looking at a phase portrait, you can see the trajectory of the system's states, which helps to understand overall system behavior.
In this particular system, phase portraits were used to confirm or elaborate on the behavior inferred from the eigenvalues analysis of the linearized system. For example:

• The phase portrait for the critical point \((-2, 0)\) revealed it to be a saddle point, based on the system trajectories around it.
• For the critical point \((0, 0)\), the phase portrait supported the findings that it behaves as an unstable node.
• The point \((4, 0)\) was also determined to be a saddle point through phase portrait analysis.

Phase portraits are crucial as they provide a whole-picture view that complements analytical methods.

Jacobian Matrix

The Jacobian matrix plays a significant role in understanding the nature of critical points in nonlinear dynamical systems. It is constructed by taking the partial derivatives of the system's functions, organized in a matrix. This helps in linearizing the system near the critical points.
For the given system, the Jacobian matrix was:\[J(x, y)=\begin{bmatrix}\frac{\partial}{\partial x}[(2+x)(y-x)] & \frac{\partial}{\partial y}[(2+x)(y-x)] \\frac{\partial}{\partial x}[(4-x)(y+x)] & \frac{\partial}{\partial y}[(4-x)(y+x)]\end{bmatrix}\]By substituting each critical point into the matrix, we obtain crucial linear approximations that allow further analysis of system behavior. The Jacobian approach simplifies complex nonlinear systems into linear approximations for easier analysis through the resulting eigenvalues.

Eigenvalues

Finding eigenvalues of the Jacobian matrix is a key step in determining the nature of critical points in a nonlinear system. Eigenvalues reflect the stability and type of critical points by indicating whether small perturbations grow or diminish over time.
In this exercise:

• At the critical point \((-2, 0)\), the eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 6\) suggest a saddle point, characterized by trajectories diverging in certain directions.
• The point \((0, 0)\) with eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = 4\) indicates an unstable node, where perturbations grow in all directions.
• For \((4, 0)\), eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = -6\) also suggest a saddle point.

These eigenvalues help predict behavior around each critical point, providing insight into overall system dynamics.