4.10: Example - the Van der Pol oscillator
- Page ID
- 219167
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In the 1920, the Dutch physicist Balthasar van der Pol was employed by Philips to work on vacuum tubes. Vacuum tubes were electronic devices used to amplify signals in the days before transistors. Van der Pol found that under certain circumstances the vacuum tubes would spontaneously oscillate on their own. Uncontrolled self-oscillations are a very undesirable behavior, so he was interested in understanding why they happened. Van der Pol investigated the physical reasons for the oscillation and determined that the voltage swing across the vacuum tube could be modeled by the second order ODE
\[ \frac{d^2 y}{d t^2} - \epsilon(1 - y^2)\frac{d y}{d t} + y = 0 \label{eq:4.5}\]
where \(y\) is the voltage across the tube. The \(\epsilon\) term represents a non-linear damping since it involves the \(d y/d t\) term. When \(y < 1\) this term represents amplification: it imparts a positive value to \({d^2 y}/{d t^2}\) when \({d y}/{d t}\) is also positive, thereby the amplitude of the oscillation increases. When \(y > 1\) the opposite happens: the term adds a negative value to \({d^2 y}/{d t^2}\) for positive \({d y}/{d t}\) causing the amplitude of the oscillation to decrease. Therefore, the action of the \(\epsilon\) term is to create an oscillation whose amplitude remains bounded in time.
This second order ODE may be broken into two first order ODEs by defining
\[\begin{aligned} \nonumber u &= y \\ \nonumber v &= \frac{d y}{d t}\end{aligned}\]
Then, the ODE system is
\[\begin{aligned} \label{eq:VanDerPolOscSystem} \nonumber \frac{d u}{d t } &= v \\ \nonumber \frac{d v}{d t} &= \epsilon(1 - u^2) v - u\end{aligned}\]
Integrating this system using RK4 gives the plot shown in 4.16. Things to note about the results shown in 4.16:
- As predicted, after a quick start-up phase, the Van der Pol ODE produces a constant amplitude, periodic oscillation. We note that the oscillation looks nothing like the usual sin or cosine oscillations produced by linear oscillators. The strong nonlinearity introduced by the \(dy/dt\) term is the reason for behind this shape.
- Note that the \(h=0.2\) curve is displaced from the curves generated with smaller \(h\) values. This is a real effect. The Van der Pol equation (Eq. 4.5) is a so-called "stiff" ODE, which makes it difficult to solve numerically, at least using larger step sizes. This property will be detailed in section 5.3 below.
