Linearize the equation $$x'' = -\alpha x-\rho x'+c \sin(t)$$

It is very easy when $c=0$ giving you a $$ x' = y $$$$ y' = -\alpha x -\rho y $$ giving you a very nice phase portrait.

However, if $c$ is non-zero, the linearization should be like $$ x' = y $$$$ y' = -\alpha x -\rho y +c\sin(t) $$ but this gives a very ugly phase portrait (the lines keep intersecting with themselves)

Is this still accurate or should I further use a Jacobian to linearize the matrix? (If so, can somebody provide a small hint on how to approach the result, take any initial state for initial condition)

  • $\begingroup$ This is a linear equation. What more do you want ? $\endgroup$ – Yves Daoust May 21 '18 at 22:20

First of all, the linearization for $c \neq 0$ is wrong. Indeed, the term $c\sin(t)$ has not been differentiated. However, it is difficult to do this since it does not depend on $x$ or $y$.

How can we deal with the term $\sin(t)$? The presence of a time-varying term implies that the order of the system is bigger than $2$.

Indeed, consider the following ODE:

$$z'' = -z.$$

For $z(0) = 0$ and $z'(0) = 1$, the solution is $\sin(t)$. Thus, we can rewrite the original system as follows:

$$\begin{cases} x'' = -\alpha x-\rho x'+c z\\ z'' = -z \end{cases}.$$

Therefore, you have a forth order system. Setting $y=x'$ and $w=z'$, it can be rewritten as:

$$\begin{cases} x' = y\\ y' = -\alpha x-\rho y+c z\\ z' = w\\ w' = -z \end{cases}.$$

Notice that the system is linear.

Finally, whichever are the initial conditions on $x$ and $y$, you must always use also $z(0) = 0$ and $z'(0) = w(0) = 1.$

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  • $\begingroup$ How do you ensure that the solutions to z are somehow dependent on the x since originally the x contained sin t? $\endgroup$ – mathnoob123 May 21 '18 at 21:57
  • $\begingroup$ @mathisbetter Wait! The solution of $x$ is dependent on $z$. $z$ is equal to $\sin(t)$ using the right initial conditions (i.e. $z(0) = 0$ and $z'(0) = 1$). Then, $x$ still depends on $\sin(t)$. $\endgroup$ – the_candyman May 21 '18 at 21:58
  • $\begingroup$ Ah perfect. Thanks. $\endgroup$ – mathnoob123 May 21 '18 at 22:00
  • $\begingroup$ Can you also show me how the phase plot looks like? I tried implementing the above equations but it still gives me an absurd plot. $\endgroup$ – mathnoob123 May 22 '18 at 8:30

The system $$x' = y$$

$$y' = -\alpha x -\rho y +c\sin(t)$$

is linear.

What you have is a non-autonomous, in-homogeneous system and that is the problem with the phase portrait.

When your system is non-autonomous, the phase portrait is better understood in three dimensions $(t,x,y)$ with the time dimension also present.

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