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