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. $$ d x / d t=-(x-y)(1-x-y), \quad d y / d t=x(2+y) $$

Short answer

Short Answer: To find the behavior of the given nonlinear system, we first found the critical points (0,0) and (0,-2). By linearizing the system using the Jacobian matrix and finding the eigenvalues, we concluded that (0,0) corresponds to a saddle point. The linear system analysis for the critical point (0,-2) was inconclusive, but by observing the nonlinear system's trajectories, we can determine that it exhibits a source-like behavior.

Step-by-step solution

Find the critical points of the system

To find the critical points, set both x and y derivatives equal to zero and solve for x and y: $$ -(x-y)(1-x-y) = 0 \\ x(2+y) = 0 $$ Solving these equations, we get two critical points: \((0,0)\) and \((0,-2)\).

Linearize the system near each critical point

To linearize the system, we need to find the Jacobian matrix of the system and evaluate it at the critical points. The Jacobian matrix for this system is: $$ J(x,y) = \begin{pmatrix} \frac{\partial}{\partial x}(-(x-y)(1-x-y)) & \frac{\partial}{\partial y}(-(x-y)(1-x-y)) \\ \frac{\partial}{\partial x}(x(2+y)) & \frac{\partial}{\partial y}(x(2+y)) \end{pmatrix} = \begin{pmatrix} -1+2x+2y & 1-2x \\ 2+y & x \end{pmatrix} $$ Now, we will evaluate the Jacobian matrix at the critical points: For the critical point \((0,0)\): $$ J(0,0) = \begin{pmatrix} -1 & 1 \\ 2 & 0 \end{pmatrix} $$ For the critical point \((0,-2)\): $$ J(0,-2) = \begin{pmatrix} -1 & 5 \\ 0 & 0 \end{pmatrix} $$

Find the eigenvalues and analyze the behavior of the nonlinear system

To find the eigenvalues of the linear systems, we need to compute the characteristic equation for each Jacobian matrix and solve for the eigenvalues. For the critical point \((0,0)\), the characteristic equation is: $$ \det(J(0,0) - \lambda I) = \det \begin{pmatrix} -1-\lambda & 1 \\ 2 & -\lambda \end{pmatrix} = \lambda^2 + \lambda - 2 = (\lambda + 2)(\lambda - 1) $$ The eigenvalues for this system are \(\lambda_1 = -2\) and \(\lambda_2 = 1\). Since one eigenvalue is positive and one is negative, the critical point \((0,0)\) corresponds to a saddle point. For the critical point \((0,-2)\), the characteristic equation is: $$ \det(J(0,-2) - \lambda I) = \det \begin{pmatrix} -1-\lambda & 5 \\ 0 & -\lambda \end{pmatrix} = \lambda^2 + \lambda = \lambda(\lambda + 1) $$ The eigenvalues for this system are \(\lambda_1 = 0\) and \(\lambda_2 = -1\). Since one eigenvalue is negative and the other is zero, the linear system does not provide definite information about the nonlinear system.

Draw the phase portrait of the nonlinear system

To draw the phase portrait, we need to observe the behavior of the trajectories of the nonlinear system and identify the type of the critical points. Without loss of generality, we will only sketch part of the phase diagram. - Near the critical point \((0,0)\), trajectories will exhibit a saddle point behavior as concluded from the eigenvalues analysis. - Near the critical point \((0,-2)\), trajectories behavior is unclear from linear analysis. However, by looking at the vector field and the nonlinear system equations, we can see that trajectories near \((0,-2)\) will move away from this point in the direction of the y-axis. Therefore, despite the linear system not providing definite information about the critical point \((0,-2)\), we can conclude that (based on the phase portrait) the behavior near \((0,-2)\) is like a source. In conclusion, the nonlinear system has a saddle point at \((0,0)\) and a source-like behavior near the point \((0,-2)\).

Key concepts

Critical Points

Critical points are essential features of a dynamical system, where the system's state doesn't change over time. They occur where the derivatives of the system are equal to zero. So, for the given system, setting both derivatives to zero, we have:

• \(-(x-y)(1-x-y) = 0\)
• \(x(2+y) = 0\)

Solving these equations gives us the critical points at \( (0, 0) \) and \( (0, -2) \). Understanding these points is crucial because they can tell us about the steady states of the system and help predict the long-term behavior of a dynamical system.

Jacobian Matrix

The Jacobian Matrix plays an important role in analyzing nonlinear dynamical systems. It shows how small changes in the variables may influence the system's evolution. For our system, the Jacobian is:\[J(x,y) = \begin{pmatrix} -1+2x+2y & 1-2x \ 2+y & x \end{pmatrix}\]Evaluating this matrix at each critical point helps us understand the local behavior around those points. At \( (0,0) \), the Jacobian becomes \[J(0,0) = \begin{pmatrix} -1 & 1 \ 2 & 0 \end{pmatrix}\],while at \( (0,-2) \), it changes to\[J(0,-2) = \begin{pmatrix} -1 & 5 \ 0 & 0 \end{pmatrix}\].The Jacobian matrix is your friend when you want to determine the local stability and type of equilibrium points.

Eigenvalues

Eigenvalues tell us about the stability of the critical points by examining the linear system derived from the nonlinear system. Computing these involves finding the characteristic equation of the Jacobian matrix:

• For \( J(0,0) = \begin{pmatrix} -1 & 1 \ 2 & 0 \end{pmatrix} \), we find \( \det(J(0,0) - \lambda I) = (\lambda + 2)(\lambda - 1) \)\, yielding eigenvalues \(-2\) and \(1\).
• For \( J(0,-2) = \begin{pmatrix} -1 & 5 \ 0 & 0 \end{pmatrix} \), the determinant \( \lambda(\lambda + 1) \) gives us eigenvalues \(0\) and \(-1\).

These eigenvalues tell us that there is a saddle point at \( (0,0) \) because of the opposite signs, indicating instability. At \( (0,-2),\) the presence of a zero eigenvalue means that the linear system alone cannot definitively describe the dynamics near this point, so we need to integrate other methods or tools.

Phase Portrait

Phase portraits are visual representations that help in understanding the behavior of dynamical systems graphically. By sketching the trajectories, critical points, and other behaviors, we can visualize how the system evolves over time:

• Near \( (0,0)\), based on the eigenvalues, we see saddle point behavior where trajectories head away along one direction and are attracted in another.
• The behavior near \( (0,-2) \) is not clear from linear analysis. However, by analyzing the vector fields or computing higher-order analysis, we conclude that the nonlinear behavior near this point resembles a source with trajectories moving away along the y-axis.

Phase portraits give us a "big picture" view of the system's dynamics, letting us visually assess the stability and nature of critical points. This can be especially helpful when linear approximations fall short in capturing the system's behavior.