Question

Consider the van der Pol system \(x^{\prime}=y, \quad y^{\prime}=-x+\mu\left(1-x^{2}\right) y\) where now we allow the parameter \(\mu\) to be any real number. a. Show that the origin is the only critical point. Determine its type, its stability property, and how these depend on \(\mu .\) b. Let \(\mu=-1 ;\) draw a phase portrait, and conclude that there is a periodic solution that surrounds the origin. Observe that this periodic solution is unstable. Compare your plot with Figure 9.7 .4. c. Draw a phase portrait for a few other negative values of \(\mu\). Describe how the shape of the periodic solution changes with \(\mu\). d. Consider small positive or negative values of \(\mu .\) By drawing phase portraits, determine how the periodic solution changes as \(\mu \rightarrow 0 .\) Compare the behavior of the van der Pol system as \(\mu\) increases through zero with the behavior of the system in Problem 16.

Short answer

Question: Show the origin as the only critical point of the van der Pol system and determine its type and stability, as well as its dependency on \(\mu\). Answer: The only critical point of the van der Pol system is at the origin (0,0). The type and stability of the critical point depend on the real parts of the eigenvalues of the Jacobian matrix: 1. If \(|\mu| < 2\), the critical point is a spiral. 2. If \(|\mu| = 2\), the critical point is a degenerate node. 3. If \(|\mu| > 2\), the critical point is a saddle.

Step-by-step solution

Find the critical points

To find the critical points, set both time-derivatives equal to zero and solve for \(x\) and \(y\): \(x'= y = 0\) \(y' = -x+\mu(1-x^2)y =0\) From the first equation, we have that \(y=0\). Substituting this into the second equation, we get \(-x=0\). The only critical point is at the origin \((0,0)\).

Linearize the system

Write the Jacobian matrix for the given system: \(J(x,y) = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1+\mu(1-3x^2)y & \mu(1-x^2) \end{pmatrix}\) Evaluate the Jacobian at the critical point (0,0): \(J(0,0) = \begin{pmatrix} 0 & 1 \\ -1 & \mu \end{pmatrix}\)

Determine type and stability properties of the critical point

Find the eigenvalues of the Jacobian matrix evaluated at the origin: \(\det(J-\lambda I) = \begin{vmatrix} -\lambda & 1\\ -1 & \mu-\lambda \end{vmatrix} = -\lambda(\mu-\lambda)+1 = \lambda^2-\mu\lambda+1\) Finding the eigenvalues, we have: \(\lambda = \frac{\mu \pm \sqrt{\mu^2-4}}{2}\) The stability properties of the origin depend on the real parts of the eigenvalues. We have three cases: 1. If \(|\mu| < 2\), the eigenvalues are complex, and the critical point is a spiral. 2. If \(|\mu| = 2\), the eigenvalues are real and equal, and the critical point is a degenerate node. 3. If \(|\mu| > 2\), the eigenvalues are real and opposite in sign, and the critical point is a saddle. b. Phase portrait for \(\mu=-1\)

Create the system for \(\mu=-1\)

Substitute \(\mu=-1\) in the van der Pol system: \(x'=y\) \(y'= -x -(1-x^2)y\)

Draw the phase portrait

Draw the phase portrait for the system (for example, using a software tool). You can observe that there is a periodic solution surrounding the origin and that it is unstable as trajectories move away from it. Compare the plot with Figure 9.7.4. c. Phase portraits for other negative values of \(\mu\)

Create the systems

Substitute different negative values of \(\mu\) into the van der Pol system, in order to observe the behavior of the periodic solution.

Draw phase portraits

Draw phase portraits for these different values of \(\mu\). Notice how the shape of the periodic solution changes as \(\mu\) becomes more negative. As \(\mu\) becomes more negative, the oscillations become more asymmetric and the amplitude increases. d. Change of the periodic solution as \(\mu \rightarrow 0\)

Create the systems

Consider small positive or negative values of \(\mu\) and substitute them into the van der Pol system.

Draw phase portraits

Draw phase portraits for these small values of \(\mu\). Observe how the periodic solution changes as \(\mu \rightarrow 0\). The periodic solution should approach a circle (phase portrait resembling that of a linear center) as \(\mu \rightarrow 0\). The behavior of the van der Pol system as \(\mu\) increases through zero is more complex than in Problem 16, where the central critical point became unstable for positive values of the parameter.

Key concepts

Critical Point Analysis

Understanding the behavior of dynamical systems often begins with identifying and analyzing critical points—the places where the system's state does not change over time. In the van der Pol system, we start by setting the time derivatives to zero. This yields the equations \(x'=y=0\) and \(y'=-x+\mu(1-x^2)y=0\). Through substitution and algebraic manipulation, we discover that the only critical point is at the origin \( (0,0) \). This point represents a state where the system is at rest, so to speak, and understanding its properties will give us insights into the overall dynamics of the system, especially its stability and long-term behavior.

Phase Portrait

A phase portrait is a graphical tool that represents all possible states of a dynamical system and the trajectories that the system follows through its state space. For the van der Pol system, when \(\mu=-1\), plotting the phase portrait illustrates that there is a closed trajectory, indicating a periodic solution, which surrounds the origin. This implies that the system exhibits oscillatory behavior. When drawing phase portraits for different negative values of \(\mu\), we observe a change in the geometry of the periodic solution, indicating that the system's behavior is sensitive to the parameter's value. In essence, a phase portrait charts the 'life cycle' of a system, mapping out its potential paths from one state to another.

Stability of Dynamical Systems

The concept of stability in dynamical systems refers to the tendency of a system to return to its equilibrium state after a small disturbance. A system can be stable, unstable, or exhibit mixed stability characteristics, depending on the nature of its critical points. In the case of the van der Pol system, the stability of the origin, which is the only critical point, varies with \(\mu\). For values of \(\mu\) within particular ranges, the critical point can be a spiral, a degenerate node, or a saddle—each scenario offering different stability behaviors. Understanding stability allows us to predict whether or not the system will settle into a steady state or continue to evolve over time.

Jacobian Matrix

The Jacobian matrix is a powerful tool in analyzing the local behavior of a dynamical system near its critical points. It's a matrix of first-order partial derivatives representing how a system's state variables change with respect to each other. For the van der Pol system, the Jacobian matrix, evaluated at the critical point \( (0,0) \), is \[ J(0,0) = \begin{pmatrix} 0 & 1 \ -1 & \mu \end{pmatrix} \]. This matrix is pivotal for determining not just the type of the critical point at the origin, but also its stability properties by examining the eigenvalues and the corresponding eigenvectors of this matrix.

Eigenvalues and Stability

Eigenvalues play a central role in deciphering the stability of a system's critical points. After calculating the determinant of the Jacobian matrix minus \(\lambda I\), the resulting characteristic equation leads us to the eigenvalues \(\lambda\). The stability properties of the origin in the van der Pol system depend on the real parts of these eigenvalues. When \(\mu\) is within certain bounds, the eigenvalues will be complex, implying oscillatory behavior; when \(\mu\) precisely matches these bounds, we have degenerate stability; and when \(\mu\) exceeds them, we're presented with instability due to a saddle point. Therefore, eigenvalues serve as indicators, revealing whether perturbations dampen out and return to equilibrium (stability) or amplify and lead to unbounded behavior (instability).