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.