Question

Consider the autonomous system $$ d x / d l=y, \quad d y / d t=x+2 x^{3} $$ (a) Show that the critical point \((0,0)\) is a saddle point. (b) Sketch the trajectories for the corresponding linear system by integrating the equation for \(d y / d x\). Show from the parametric form of the solution that the only trajectory on which \(x \rightarrow 0, y \rightarrow 0\) as \(t \rightarrow \infty\) is \(y=-x\). (c) Determine the trajectories for the nonlinear system by integrating the equation for \(d y / d x\). Sketch the trajectories for the nonlinear system that correspond to \(y=-x\) and \(y=x\) for the linear system.

Short answer

(b) What is the trajectory where \(x \rightarrow 0, y \rightarrow 0\) as \(t \rightarrow \infty\) for the linear system? (c) What are the trajectories corresponding to \(y=-x\) and \(y=x\) for the linear system? a) Yes, the critical point (0,0) is a saddle point for the given system because one eigenvalue is positive and the other is negative. b) The trajectory where \(x \rightarrow 0, y \rightarrow 0\) as \(t \rightarrow \infty\) for the linear system is \(y = -x\). c) Trajectories corresponding to \(y=-x\) and \(y=x\) for the linear system are heteroclinic orbits connecting the saddle point (0,0) to itself, with different behaviors for each trajectory.

Step-by-step solution

Find the linearization of the given system near the critical point (0, 0)

First, let's find the Jacobian matrix of the given system. The Jacobian matrix is given by $$ J(x, y)=\left[\begin{array}{cc} \frac{\partial f_{1}}{\partial x} & \frac{\partial f_{1}}{\partial y} \\ \frac{\partial f_{2}}{\partial x} & \frac{\partial f_{2}}{\partial y} \end{array}\right] $$ where \(f_1(x, y) = y\) and \(f_2(x, y) = x + 2x^3\). We can compute the partial derivatives and, therefore, the Jacobian matrix: $$ J(x, y)=\left[\begin{array}{cc} 0 & 1 \\ 1+6x^{2} & 0 \end{array}\right] $$ Now we need to evaluate this matrix at the critical point (0, 0): $$ J(0, 0)=\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] $$

Determine the eigenvalues of the linearized system

The eigenvalues of the linearized system will help us understand the behavior around the critical point. The characteristic equation of the matrix \(J(0,0)\) is: $$ \det(J- \lambda I)=\left|\begin{array}{cc} -\lambda & 1 \\ 1 & -\lambda \end{array}\right|= \lambda^{2} - 1 $$ The eigenvalues are the roots of this equation: $$ \lambda^{2} - 1=0 $$ Thus, \(\lambda = \pm 1\)

Determine the behavior of the critical point (0, 0)

We have found two eigenvalues \(\lambda = 1\) and \(\lambda = -1\). Since one eigenvalue is positive and the other is negative, the critical point (0, 0) is a saddle point. This answers the question in part (a).

Sketch the trajectories of the linear system and find the specific trajectory

The equation \(dy/dx\) can be determined from the given system as follows: $$ \frac{dy}{dx} = \frac{dy / dt}{dx / dt} = \frac{x + 2x^{3}}{y} $$ Now, using standard techniques for solving first-order ordinary differential equations, we can integrate this expression and find the parametric form of the solution. If \(y \ne 0\), we can divide both sides by \(y\) and obtain the following separable ODE: $$ \frac{dy}{y} = \frac{(1 + 2x^{2}) dx}{x} $$ Integrate both sides: $$ \int\frac{dy}{y} = \int\frac{(1 + 2x^{2}) dx}{x} $$ We get $$ \ln |y| = \ln{|x|} + x^{2} + C $$ or, equivalently, $$ y = Kx e^{x^{2}} $$ where \(K = e^C\) is an integration constant. As \(t \rightarrow \infty\), to ensure \(x \rightarrow 0\) and \(y \rightarrow 0\), we must have \(y = -x\). This is the trajectory we were asked to find in part (b).

Determine the trajectories for the nonlinear system

We have already found the equation for the nonlinear system in Step 4: $$ \frac{dy}{dx} = \frac{x + 2x^{3}}{y} $$ Now, use this equation to integrate the equation for \(dy/dx\): $$ \int\frac{dy}{y} = \int\frac{(1 + 2x^{2}) dx}{x} $$

Sketch the trajectories corresponding to \(y=-x\) and \(y=x\) for the linear system

The analysis for the solution of the nonlinear system shows that the only trajectory for which \(x \rightarrow 0\) and \(y \rightarrow 0\) as \(t \rightarrow \infty\) is \(y = -x\). Thus, the trajectory corresponding to \(y = -x\) for the linear system will be a heteroclinic orbit in the phase plane connecting the saddle point (0, 0) to itself. The trajectory corresponding to \(y = x\) for the linear system will be another heteroclinic orbit in the phase plane connecting the saddle point to itself but with a very different behavior from the \(y=-x\) trajectory. Finally, to sketch the trajectories, you can draw two heteroclinic orbits, one for \(y=-x\) and one for \(y=x\), both connecting the saddle point (0, 0) to itself, respecting the behaviors found in the analysis.

Key concepts

Critical points

Critical points are essential in analyzing autonomous systems because they give us information about the behavior of the system. In the given autonomous system, a critical point is where the derivatives are zero, meaning

• \( \frac{dx}{dl} = y = 0 \)
• \( \frac{dy}{dt} = x + 2x^3 = 0 \)

For these conditions to hold, both \( y \) and \( x \) must be zero. Thus, the point \((0,0)\) is identified as a critical point. When analyzing critical points, it is essential to determine whether they are stable, unstable, or saddle points, as each type has different implications for the trajectories around them. In this exercise, it is essential to demonstrate that \((0,0)\) is a saddle point, which means it has a characteristic with both attractive and repulsive attributes, leading to complex trajectory behavior around that point.

Jacobian matrix

The Jacobian matrix is a fundamental tool in analyzing more complex systems, as it provides the linear approximation of a nonlinear system near a critical point. For the formulas given:

• \( f_1(x, y) = y \)
• \( f_2(x, y) = x + 2x^3 \)

We calculate the Jacobian by finding the partial derivatives:

• \( \left[ \begin{array}{cc} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \ 1 + 6x^2 & 0 \end{array} \right] \)

By evaluating this matrix at the critical point \((0,0)\), we attain:

• \( J(0, 0) = \left[ \begin{array}{cc} 0 & 1 \ 1 & 0 \end{array} \right] \)

This result is crucial for linearizing the system and further understanding its behavior.

Eigenvalues

Eigenvalues play a pivotal role in understanding the nature of critical points. For a linearized system at a critical point, eigenvalues derived from the Jacobian matrix tell us whether the point is a source, sink, or saddle point. To find the eigenvalues of our system, we solve the characteristic equation: The characteristic equation is:

• \( \det(J - \lambda I) = \lambda^2 - 1 = 0 \)

Solving this gives eigenvalues:

• \( \lambda = \pm 1 \)

The presence of both a positive eigenvalue (\( \lambda = 1 \)) and a negative eigenvalue (\( \lambda = -1 \)) suggests that the critical point \((0,0)\) is a saddle point. In systems analysis, a saddle point implies that trajectories approach the point along one direction and repel away in another, indicating that it is inherently unstable, with directional stability only in limited pathways.

Linearization

Linearization simplifies the analysis of nonlinear differential systems by approximating them with linear ones near critical points. In this exercise, it involves taking the Jacobian matrix and evaluating it at the critical point. This helps formulate a linear system that approximates the nonlinear behavior around \((0,0)\). This approximation allows us to:

• Predict nearby trajectories' expectation.
• Utilize eigenanalysis to determine stability and the type of critical point.

The resulting linear system from the Jacobian matrix is simple:

• \(\frac{dy}{dx} = \frac{y}{x + 2x^3} \)

By integrating this linear expression, we can analyze trajectory behaviors near \((0,0)\). Simplifications such as linearization are indispensable when dealing with complex dynamics and predictions about system stability.

Trajectories

Trajectories depict the path that solutions will follow over time in the phase space of the system. For both the linear and nonlinear systems, these paths reveal the nature of the critical point and overall system behavior. In the current problem, we aim to sketch these trajectories to bring clarity by visualizing solution behaviors:

• The linear trajectory solution assesses how the system behaves near the saddle point (solution: \(y=-x\)).
• The nonlinear trajectories require integrating \(\frac{dy}{dx} = \frac{x + 2x^3}{y}\), leading to solutions like \(y = Kxe^{x^2}\).

Sketching helps clarify abstract concepts into tangible paths in the phase space. Through these depictions, it's observed that a specific trajectory where both variables approach zero over time is \(y = -x\). Practical sketches give insights into stability and potential paths systems may traverse, helping in the intuitive understanding of its evolution over time.