Linearization of a PDE I have been struggling with some linearization argument of the following paper: "M. Weinstein: Modulational stability of ground states of NLS". In order to give a bit of context to my question, let us consider the NLS equation $$ 2i\phi_t+\Delta\phi+\vert\phi\vert^{2\sigma}\phi=0, \quad 0<\sigma<\tfrac{2}{n-2}.$$ 
This equation has very interesting "localized" solutions of the form: $$\phi(t,x)=u(x)e^{it/2},$$
where $u(t,x)$ solves $\Delta u-u+\vert u\vert^{2\sigma}u=0$. Besides, the latter equation has an even more interesting real, positive and radial $H^1(\mathbb{R}^n)$ solution called "Ground state" and denoted by $R(x)$. 
Now let me try to explain my question. Consider the perturbed Initial Valued Problem (IVP): $$2i\phi_t^\varepsilon+\Delta \phi^\varepsilon+\vert \phi^\varepsilon\vert^{2\sigma}\phi^\varepsilon=\varepsilon F(\vert \phi^\varepsilon\vert)\phi^\varepsilon, \quad \phi^\varepsilon(t=0,x)=R(x)+\varepsilon S(x)$$
We will seek solutions of the previous equation of the form $$\phi^\varepsilon(t,x)=(R(x)+\varepsilon w_1+\varepsilon^2 w_2+...)e^{it/2}.$$
According to Weinstein if you reeplace this function into the perturbed equation and linearize you will get the following IVP for the linearized perturbation $w$: $$ 2iw_t+\Delta w-w+(\sigma+1)R^{2\sigma}w+\sigma R^{2\sigma}\overline{w}=F(R)R, \quad w(0,x)=0.$$
Now my problem is: I do not really understand how to obtain this linearization, can someone explain a little bit how to obtain it? Or recommend some references to learn about it.
Note: $H^1(\mathbb{R}^n)$ denotes the Sobolev space (also denoted by $W^{1,2}$).
Note2: The parameter $n$ denotes the dimension.
 A: The equation is of the form $L[\phi] + A(\phi) + \epsilon B(\phi) = 0$ where $L$ is the linear map
$$ L = 2i\partial_t + \Delta ,$$ and $A,B$ are
$$ A(\phi) = |\phi|^{2\sigma} \phi , \quad B(\phi) = - F(|\phi|)\phi.$$
Subbing in $\phi=Re^{it/2} + \epsilon we^{it/2} $ we obtain
$$ L[Re^{it/2}] + \epsilon L[we^{it/2}] + A(Re^{it/2} + \epsilon we^{it/2}) + \epsilon B(Re^{it/2} + \epsilon we^{it/2}) = 0.$$
To find the leading order behavior,  use the Taylor approximation 
$$ A(Re^{it/2} + \epsilon we^{it/2}) = A(Re^{it/2}) + \epsilon A'(Re^{it/2})[we^{it/2} ] + O(\epsilon^2), \\ B(Re^{it/2}+\epsilon we^{it/2}) = B(Re^{it/2}) + \epsilon  B'(Re^{it/2})[we^{it/2}] + O(\epsilon^2),$$
to find
$$ L[Re^{it/2}] + A(Re^{it/2}) + \epsilon \left[ L[we^{it/2}]+A'(Re^{it/2})[we^{it/2}] + B(Re^{it/2})\right] = O(\epsilon^2).$$
Note that the term involving $B'$ is $O(\epsilon^2)$. As noted in OP, $L[Re^{it/2}] + A(Re^{it/2}) =0$, so dividing through by $\epsilon$,
$$L[we^{it/2}]+A'(Re^{it/2})[we^{it/2}] + B(Re^{it/2}) = O(\epsilon).$$
Noting that
\begin{align}B(Re^{it/2})& = -F(R)Re^{it/2}, \\
 L[we^{it/2}] &= (2iw_t-w + \Delta w)e^{it/2},
\end{align}
and 
\begin{align}
A'(z)[h] &= \partial_z A(z) h +  \partial_{\bar z} A(z) \overline h, \phantom{\big)}\\
  \partial_z A(z) &=  \partial_z (z^{\sigma+1} \bar z^\sigma ) = (\sigma+1)|z|^{2\sigma},\\
\partial_{\bar z} A(z) &=  \partial_{\bar z} (z^{\sigma+1} \bar z^\sigma ) = \sigma|z|^{2\sigma-2}z^2,
\end{align}
which gives
$$
A'(Re^{it/2})[we^{it/2}] = (\sigma+1)R^{2\sigma}we^{it/2} + \sigma R^{2\sigma} e^{it} \bar we^{ - it/2} = \Big [(\sigma+1)R^{2\sigma}w + \sigma R^{2\sigma}\bar w \Big]e^{it/2},
$$
so we can divide through by $e^{it/2}$ to see
$$ 2iw_t-w + \Delta w + (\sigma+1)R^{2\sigma}w + \sigma R^{2\sigma}\bar w = F(R)R + O(\epsilon).$$
For a linearisation, you don't need to further expand $w = w_1 + \epsilon w_2 +\dots$, due to the $O(\epsilon)$ term.
