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. $$ d x / d t=1-x y, \quad d y / d t=x-y^{3} $$

Short answer

#tag_title# Short Answer To find the critical points, we set both dx/dt and dy/dt to 0 and solve the system of equations. There are two critical points: (1,1) and (1,-1). Then, we find the Jacobian matrix for each point, and calculate the eigenvalues. The eigenvalue results indicate that the critical point (1,1) is a stable spiral, and the critical point (1,-1) is a saddle. The phase portrait sketch would have trajectories spiraling inward towards point (1,1) and moving away or towards point (1,-1) based on their eigenvector direction.

Step-by-step solution

Find the critical points

To find the critical points, set dx/dt and dy/dt to 0, and solve the resulting system of equations for x and y: $$ 1-xy = 0 \\ x-y^3 =0 $$ We can rewrite the second equation as x = y^3. Substitute this into the first equation and solve for y: $$ 1-y(y^3) = 1-y^4 = 0 $$ We obtain y=1 or y= -1. For y=1, x=1 and for y=-1, x =1. So, there are two critical points: \((1,1)\) and \((1,-1)\).

Linearize the system

Find the Jacobian matrix of the given system and evaluate it at the critical points. The Jacobian matrix is: $$ J(x,y)=\begin{bmatrix} \frac{\partial (1-xy)}{\partial x} & \frac{\partial (1-xy)}{\partial y} \\ \frac{\partial (x-y^3)}{\partial x} & \frac{\partial (x-y^3)}{\partial y} \end{bmatrix}=\begin{bmatrix} -y & -x \\ 1 & -3y^2 \end{bmatrix} $$ Evaluate the Jacobian matrix at \((1,1)\) and \((1,-1)\): $$ J(1,1) = \begin{bmatrix} -1 & -1 \\ 1 & -3 \end{bmatrix}, \quad J(1,-1) = \begin{bmatrix} 1 & -1 \\ 1 & -3 \end{bmatrix} $$

Calculate the eigenvalues

To find the eigenvalues of each Jacobian matrix, we need to solve the characteristic equation for each matrix given by: $$ \det(J-\lambda I) = 0 $$ For J(1,1): $$ \det\begin{bmatrix} -1-\lambda & -1 \\ 1 & -3-\lambda \end{bmatrix}=(\lambda+1)(\lambda+3) - 1 = \lambda^2+4\lambda+2 = 0 $$ Solving this quadratic equation, we get complex eigenvalues \(\lambda_1 = -2+i\), \(\lambda_2 = -2-i\). Since the real part of both eigenvalues is negative, the critical point \((1,1)\) is a stable spiral. For J(1,-1): $$ \det\begin{bmatrix} 1-\lambda & -1 \\ 1 & -3-\lambda \end{bmatrix}=(\lambda-1)(\lambda+3) - 1 = \lambda^2+2\lambda-3 = 0 $$ Solving this quadratic equation, we get real eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = -3\). Since the eigenvalues have different signs, the critical point \((1,-1)\) is a saddle.

Phase portrait

To draw a phase portrait, we can use arrows to represent the direction of the vector field and the trajectories of the system. The critical point \((1,1)\) is a stable spiral, meaning that trajectories will spiral inward towards the point. The critical point \((1,-1)\) is a saddle, which means that trajectories will move away from the point along one eigenvector direction and move toward the point along the other eigenvector direction. Based on these observations, we would expect to see a phase portrait with trajectories spiraling in towards the point \((1,1)\), and trajectories moving away from or towards the point \((1,-1)\) depending on the direction of the eigenvector. A more accurate phase portrait can be drawn using a computer program or by plotting several trajectories, but this qualitative description should help confirm our conclusions.

Key concepts

Critical Points

In the study of nonlinear systems, critical points play a crucial role in understanding the system's dynamics. Critical points, also known as equilibrium points, are where the system does not change over time. To determine the critical points, we set the derivatives of the system to zero. This means finding where both \( dx/dt \) and \( dy/dt \) are equal to zero simultaneously.

For the given system:

• \( dx/dt = 1 - xy \)
• \( dy/dt = x - y^3 \)

We find the solutions to the equations:

• \( 1 - xy = 0 \)
• \( x - y^3 = 0 \)

By solving these equations, we find the critical points \((1, 1)\) and \((1, -1)\). These points are where the system's behavior changes, making them key to further analysis.

Jacobian Matrix

The Jacobian Matrix is a mathematical tool used to study the behavior of a nonlinear system near its critical points. It is composed of partial derivatives of the system's equations, representing how the system's variables impact each other around the critical points.

For our system, the Jacobian matrix is:\[J(x,y)=\begin{bmatrix}\frac{\partial (1-xy)}{\partial x} & \frac{\partial (1-xy)}{\partial y} \\frac{\partial (x-y^3)}{\partial x} & \frac{\partial (x-y^3)}{\partial y}\end{bmatrix}=\begin{bmatrix}-y & -x \1 & -3y^2\end{bmatrix}\]By evaluating the Jacobian matrix at the critical points \((1, 1)\) and \((1, -1)\), we can understand how small deviations will evolve.

This gives:

• For \((1, 1)\), the Jacobian is: \[J(1,1) = \begin{bmatrix}-1 & -1 \1 & -3\end{bmatrix}\]
• For \((1, -1)\), the Jacobian is: \[J(1,-1) = \begin{bmatrix}1 & -1 \1 & -3\end{bmatrix}\]

In essence, the Jacobian helps predict the local behavior of the system near the critical points.

Eigenvalues

Eigenvalues are critical for understanding the stability and type of equilibrium at critical points of a nonlinear system. They are derived from the Jacobian matrix and help predict how solutions behave over time.

To find the eigenvalues, solve the characteristic equation derived from the Jacobian matrix:\[\det(J-\lambda I) = 0\]For the critical point \((1,1)\), the eigenvalues are complex numbers:

• \(\lambda_1 = -2+i\)
• \(\lambda_2 = -2-i\)

The negative real parts indicate a stable spiral, where trajectories spiral inward.

For the critical point \((1,-1)\), the eigenvalues are real and have different signs:

• \(\lambda_1 = 1\)
• \(\lambda_2 = -3\)

This indicates a saddle point, where trajectories move away in one direction and towards the point in another.

Thus, understanding eigenvalues gives valuable insights into the system's stability and behavior at its critical points.

Phase Portrait

A phase portrait is a graphical representation that captures the trajectories of a dynamical system in the phase space, giving a visual overview of a system's behavior near its critical points. Its contours and arrows illustrate how solutions evolve over time.

For the given system:

• The critical point \((1,1)\) is depicted as a stable spiral in the phase portrait. This means that trajectories will tend to spiral inward towards this point over time.
• The critical point \((1,-1)\) appears as a saddle. Trajectories move towards the point along some paths and diverge away along others, creating a distinctive saddle shape in the portrait.

By examining a phase portrait, we can draw conclusions about the overall dynamics and stability of the nonlinear system in an intuitive and visual way, confirming or extending the linear analysis of the Jacobian matrix.