Compute the linearization of an operator representing the Van der Pol oscillator

Let $G^{\mu}: C^{\infty}(\mathbb{R})\rightarrow C^{\infty}(\mathbb{R})$ be an operator defined as

$G^{\mu}[x]=x''+\mu(x^2-1)x'+x$

Where $C^{\infty}(\mathbb{R})$ is the space of infinitely-differentiable functions defined on $\mathbb{R}$. The case $G^{\mu}=0$ describes the Van der Pol oscillator.

Define the linearization of $G^{\mu}$ at $x$ to be the operator:

$L_x^{\mu}[v]= \lim_{h\to0} \frac{G^{\mu}(x+hv)-G^{\mu}(x)}{h} $

Question: compute $L_x^{\mu}[v]$ for any $x$, $v \in C^{\infty}(\mathbb{R})$ and $\mu \in \mathbb{R}$.

The operator $L_x^{\mu}[v]$ looks similar to the limit definition of the derivative, but I don't think I am simply meant to differentiate $G^{\mu}$. Finding $G^{\mu}(x+hv)-G^{\mu}(x)$ just gives a really long expression that doesn't look like it simplifies to anything useful.

Solution 1:

You can think that $x$ depends on a parameter $\lambda$ and write $x_\lambda$, with $x = x_0$, and $({\rm d}/{\rm d}\lambda)|_{\lambda=0} x_\lambda = v$. So compute $$L_x^\mu[v] = \frac{\rm d}{{\rm d}\lambda}\bigg|_{\lambda = 0}G^\mu[x_\lambda] = v''+2\mu xx'v+\mu(x^2-1)v'+v,$$with the product rule. The point is that $x \mapsto x'$ and $x\mapsto x''$ are linear operators in the variable $x$, so the total derivatives of those, at $x$, evaluated at $v$, are simply $v'$ and $v''$.