I'm working through a section in a book I'm reading about delay differential equations (Semi-discretization for time delay systems, Springer), and the authors are discussing the following equation $$\left(\frac13ml^2-\frac14ml^2\cos^2\phi\right) \ddot{\phi} +\frac18mgl^2\dot{\phi}^2\sin(2\phi)- \frac12mgl\sin\phi\\ = -\frac12l\cos\phi (k_p\phi(t-\tau)+k_d\dot{\phi}(t-\tau)),$$where $m$, $l$, $g$, $\tau$, $k_d$ and $k_p$ are all constants.
They say that linearizing about $\phi = 0$ reduces the equation to $$\frac{1}{12}ml^2\ddot{\phi}-\frac12mgl\phi = -\frac12l(K_p\phi(t-\tau)+K_d\dot{\phi}(t-\tau)),$$
But I'm not really familiar with linearization of second order equations and I am having trouble finding useful resources regarding the matter. Can anyone show me how to linearize the first equation? Thanks!
Reducing the equation to values close to $0$ can be realized by setting $ϕ=\epsilon u$ for some small scale $ϵ\approx 0$ and then removing any terms that are of a higher order in $ϵ$. Insert and divide by $ϵ$, \begin{multline} ml^2\left(\frac13−\frac14\cos^2(ϵu)\right)\ddot u + \frac18mgl^2ϵ\dot u^2\sin(2ϵu)−\frac12mgl\frac{\sin(ϵu)}{ϵ} \\=−\frac12l\cos(ϵu)(k_pu(t−τ)+k_d\dot u(t−τ)). \end{multline} Now assuming that $u$ has moderate values, $|u|\ll ϵ^{-1}$, the Taylor series of the trigonometric functions provide $$\cos(ϵu)=1+O(ϵ^2), ~~ \frac{\sin(ϵu)}{ϵ}=u+O(ϵ^2), ~~ ϵ\sin(2ϵu)=O(ϵ^2).$$ So up to $ϵ^2$-small terms $$ ml^2\frac1{12}\ddot u −\frac12mglu=−\frac12l(k_pu(t−τ)+k_d\dot u(t−τ)). $$ or rescaled to the scale of $ϕ$, $$ ml^2\frac1{12}\ddot ϕ −\frac12mglϕ=−\frac12l(k_pϕ(t−τ)+k_d\dot ϕ(t−τ)). $$