Phase plane gradient

If we have a state space model we can choose any two states and map out the state gradients with all other states fixed.

If we assume the general form of a non-linear model with two states of the form

\[ \begin{bmatrix}\dot{x}_1\\\dot{x}_2\end{bmatrix} =f\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) \]

If our phase space has axes for $x_1$ (horizontal) and $x_2$ (vertical) we can, for every point on the $x_1,x_2$ plane, compute the gradient of the trajectory.

This gradient is $\frac{d x_2}{d x_1}$ and we can expand it so

\[ \frac{d x_2}{d x_1}=\frac{d x_2}{d x_1}\frac{d t}{d t} = \frac{d x_2}{d t}\frac{d t}{d x_1} = \frac{d x_2}{d t}/\frac{d x_1}{d t} = \frac{\dot{x}_2}{\dot{x}_1} \]

van der Pol oscillators

First studied by the Dutch engineer/physicist Balthasar van der Pol, the equations emerge from an analysis of vacuum tube oscillators. Later used by Mary Cartwright and J.E. Littlewood to explore chaos.

The van der Pol equation (with zero input) expressed as a second order single equation is

\[ \frac{d^2y}{ dt^2}-\mu(1-y^2)\frac{dy}{dt}+y=0 \]

where $\mu$ is a constant value that controls behaviour.

Some observations are:

We can express the van der Pol equations in a state space form

\[ \begin{bmatrix}\dot{x_1}\\\dot{x_2}\\\end{bmatrix} = \begin{bmatrix}0 & 1\\-1 & \mu (1-x_1^2)\end{bmatrix} \begin{bmatrix}{x_1}\\{x_2}\\\end{bmatrix} \]

Although this looks like a linear system this is not the case. Why? (are all the values in the $A$ matrix constants?)

Figure 1: local gradients in the phase space of the van der Pol equation. Note change of behaviour at $x_1=\pm1$

Local gradients can be calculated at all points in the phase plane since

\[ \frac{dx_2}{dx_1}= \frac{dx_2}{dt}\frac{dt}{dx_1}= \frac{\dot{x}_2}{\dot{x}_1} \]

So for the van der Pol oscillator the local gradient for any point can be computed as

\[ \frac{\dot{x}_2}{\dot{x}_1} =\frac{{x}_2}{ - x_1 +\mu (1-x_1^2) x_2} \]

The above conditions are now