Consider the forced Van der Pol oscillator equation:
$$\Large\bf\sf x''-\mu(1-x^2)x'+x\quad=\quad A \sin(\omega t)$$
with $\large \mu= 8.53$, $\large A = 1.2$ and $\large \omega = 2\pi/10$.

$\cancel{\text{1. Express the equation as a system of differential equations.}}$
2. Find the ﬁxed points of the system.
3. Linearize the system about the ﬁxed points.
4. Determine the linear stability of the ﬁxed points by ﬁnding the eigenvalues of the linearized system.
5. Solve the system using the 4th-order Runge-Kutta method using initial condition near and far the ﬁxed points.

Question

Consider the forced Van der Pol oscillator equation:
$$\Large\bf\sf x''-\mu(1-x^2)x'+x\quad=\quad A \sin(\omega t)$$
with $\large \mu= 8.53$, $\large A = 1.2$ and $\large \omega = 2\pi/10$.

$\cancel{\text{1. Express the equation as a system of differential equations.}}$
2. Find the ﬁxed points of the system.
3. Linearize the system about the ﬁxed points.
4. Determine the linear stability of the ﬁxed points by ﬁnding the eigenvalues of the linearized system.
5. Solve the system using the 4th-order Runge-Kutta method using initial condition near and far the ﬁxed points.

zepdrix · Answer

6. Compare the numerical results with the linear stability. How far from the ﬁxed points you need to get for the linear stability to be no longer valid.
7. Try other values of $\mu$.

zepdrix · Answer

Anyone good with MatLab? :(

zepdrix · Answer

I think I've got the first part figured out, thanks to Sith.$$\Large\bf\sf x_1=x$$$$\Large\bf\sf x_2=x'$$Which gives us the system,$$\Large\bf\sf x_1'=x_2$$$$\Large\bf\sf x_2'=A \sin(\omega t) + \mu (1-x_1^2)x_2 -x_1$$

zepdrix · Answer

For fixed points I set the left sides to zero I think?

zepdrix · Answer

$$\Large\bf\sf 0=x_2$$$$\Large\bf\sf 0=A \sin(\omega t) + \mu (1-x_1^2)x_2 -x_1$$

zepdrix · Answer

Plugging x_2 into the second equation..$$\Large\bf\sf x_1=A \sin(\omega t)$$Mmmm hopefully I'm on the right track here +_+