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. $$ d x / d t=x-y^{2}, \quad d y / d t=y-x^{2} $$

Short answer

2. What is the Jacobian matrix? 3. What are the eigenvalues for each critical point? 4. What type of critical points are (0,0) and (1,1)? 5. How can we draw a phase portrait of the nonlinear system? Answers: 1. The critical points of the system are (0,0) and (1,1). 2. The Jacobian matrix is given by: $$ J(x,y) = \begin{pmatrix} 1 & -2y\\ -2x & 1 \end{pmatrix} $$ 3. The eigenvalues are λ₁ = 1 (double) for the critical point (0,0), and λ₂ = -1, λ₃ = 3 for the critical point (1,1). 4. The critical point (0,0) is an unstable improper node, and the critical point (1,1) is a saddle point, which is also unstable. 5. To draw a phase portrait of the nonlinear system, we can use computer software like MATLAB or Mathematica, or sketch the directions of the vector field while considering the behavior of trajectories around the critical points.

Step-by-step solution

Step 1. Determine all critical points:

To find the critical points, we need to set both dx/dt and dy/dt equal to zero and solve the system of equations. $$ \begin{cases} x - y^2 = 0\\ y - x^2 = 0 \end{cases} $$ Since both equations are equal to zero, we can set them equal to each another: $$ x^2 = y^2 $$ It implies that \(x = y\) or \(x = -y\). Substituting these conditions back into original equations, we obtain the critical points: \((0,0)\) and \((1,1)\).

Step 2. Find the corresponding linear system near each critical point:

To find the linear system near each critical point, we need to compute the Jacobian matrix of the original system: $$ J(x,y) = \begin{pmatrix} \frac{\partial (x-y^2)}{\partial x} & \frac{\partial (x-y^2)}{\partial y}\\ \frac{\partial (y-x^2)}{\partial x} & \frac{\partial (y-x^2)}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & -2y\\ -2x & 1 \end{pmatrix} $$ Evaluate the Jacobian matrix at the critical points \((0,0)\) and \((1,1)\): $$ J(0,0) = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}, \quad J(1,1) = \begin{pmatrix} 1 & -2\\ -2 & 1 \end{pmatrix} $$

Step 3. Find the eigenvalues of each linear system:

To find the eigenvalues of each linear system, we need to compute the determinant: $$ \det(J - \lambda I) = (1-\lambda)^2 - (-2y)(-2x) = \lambda^2 - 2\lambda + 1 - 4xy $$ For the critical points \((0,0)\) and \((1,1)\), we have: $$ \det(J(0,0) - \lambda I) = \lambda^2 - 2\lambda + 1\\ \det(J(1,1) - \lambda I) = \lambda^2 - 2\lambda - 3 $$ The eigenvalues are the roots of these characteristic equations: $$ \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2\\ \lambda^2 - 2\lambda - 3 = (\lambda+1)(\lambda-3) $$ Thus, eigenvalues are \(\lambda_1 = 1\) (double) for the critical point \((0,0)\) and \(\lambda_2 = -1, \lambda_3 = 3\) for the critical point \((1,1)\).

Step 4. Analyze the nonlinear system based on the eigenvalues:

Since eigenvalues for the critical point \((0,0)\) are positive and double, the point \((0,0)\) is an unstable improper node. For the critical point \((1,1)\), since one eigenvalue is positive and the other is negative, the point \((1,1)\) is a saddle point which is also unstable.

Step 5. Draw a phase portrait of the nonlinear system:

Drawing the phase portrait would confirm that \((0,0)\) is an unstable improper node and \((1,1)\) is a saddle point. In this case, the phase portrait can be drawn using computer software like MATLAB or Mathematica, or by sketching the directions of the vector field and considering the behavior of trajectories around the critical points. It would show that trajectories near \((0,0)\) move away from the point, and the system's state moves along its eigendirections at the saddle point \((1,1)\).

Key concepts

Critical Points: Finding Stability

In a system of differential equations, critical points are where the system's derivatives are zero.
This means there's no change at these points, indicating potential equilibrium.To identify critical points, set both derivatives equal to zero and solve the resulting algebraic system:

• Equation 1: \(x - y^2 = 0\)
• Equation 2: \(y - x^2 = 0\)

By substituting these into one another, we can determine the critical points.
In our case, solving this reveals points \((0,0)\) and \((1,1)\).
These points are essential as they help explore the behavior of the system around them in detail.

Linearization: Simplifying Complex Systems

Once we determine the critical points, the next step is linearization.
Linearization is a method that approximates a nonlinear system near a critical point.This involves creating a linear system that can easily be analyzed using the Jacobian matrix.
For our system, we determine the Jacobian matrix from the equations:\[J(x,y) = \begin{pmatrix}\frac{\partial (x-y^2)}{\partial x} & \frac{\partial (x-y^2)}{\partial y}\\frac{\partial (y-x^2)}{\partial x} & \frac{\partial (y-x^2)}{\partial y}\end{pmatrix}=\begin{pmatrix}1 & -2y\-2x & 1\end{pmatrix}\] Evaluate this matrix at the critical points to gain a simplified, linearized system you can easily analyze.
The Jacobian helps determine stability and predicts how small deviations near the critical points behave.

Eigenvalues: Unraveling System Characteristics

Eigenvalues provide insight into the system's stability properties around critical points.
They are calculated from the Jacobian matrix and show how trajectories behave as they approach equilibrium.To find eigenvalues, solve the characteristic equation \( \det(J - \lambda I) = 0 \).
Each solution, or eigenvalue, tells us about the growth or decay rate of the trajectories:For the critical point \((0,0)\):

• Eigenvalues are \(\lambda_1 = 1\) (double), indicating an unstable improper node.

For the critical point \((1,1)\):

• Eigenvalues are \(\lambda_2 = -1\) and \(\lambda_3 = 3\), indicating a saddle point, which is inherently unstable.

Analyzing eigenvalues is crucial as they guide us in predicting the dynamic behavior of systems near these points.

Phase Portrait: Visual Interpretation of Dynamics

A phase portrait is a graphical representation of trajectories of a dynamical system in the phase plane.
It helps visualizing the direction of the system’s state over time.The phase portrait provides an intuitive view of system dynamics of the equations \(d x / d t = x-y^{2}\) and \(d y / d t = y-x^{2}\):

• At \((0,0)\), trajectories move away due to the instability of the improper node.
• At \((1,1)\), the phase portrait illustrates a saddle point, with trajectories converging towards and diverging from this point.

Phase portraits are helpful for confirming analytical results from eigenvalues.
They provide an extended view where linearization might not offer conclusive insights about the nonlinear system.