Mathematical Transformation in a paper I am currently reading this paper: https://ieeexplore.ieee.org/abstract/document/1707778
but I don't understand how they made the following mathematical transformation.
They have the following function:
\begin{equation*} w^{IV}(x)-N(w)w^{\prime \prime}(x)=F_{e}(w(x)) \tag{7} \end{equation*}
whith
\begin{equation*} N(w)=N_{1} \int_{0}^{1}(w^{\prime})^{2}dx+N_{0} \end{equation*}
Then they define:\begin{equation*} w(x)=\zeta\bar{w}(x). \tag{15} \end{equation*}
And derive:
\begin{equation*} (k_{0}+k_{1})\zeta+k_{2}\zeta^{3}=\mu f_{e}(\zeta) \tag{16} \end{equation*}
Don't bother about the right side of the equation, I am only interested to know how they derive the left side.
In the paper, they write:

The governing equation for ζ is derived by multiplying both sides of
  (7) by w¯, integrating over the beam span, and substituting into the
  resulting equation the approximate solution \begin{equation*} w(x)=\zeta\bar{w}(x). \tag{15} \end{equation*} After integrating by
  parts, the governing equation for ζ becomes

 A: The equation is
$$
w^{IV}-N w''=F_e(w).
$$
Multiplying both sides by $\bar{w}$
$$
\bar{w} w^{IV}-N \bar{w} w''=F_e(w) \bar{w},
$$
and integrating in $0\leq x \leq 1$,
$$
\int_0^1 \bar{w} w^{IV} dx - N \int_0^1 \bar{w} w'' dx= \int_0^1 F_e(w) \bar{w} dx.
$$
Now we substitute
$$
w=\zeta \bar{w},
$$
leading to
$$
\zeta \int_0^1 \bar{w} \bar{w}^{IV} dx - N \zeta \int_0^1 \bar{w} \bar{w}'' dx= \int_0^1 F_e(\zeta\bar{w}) \bar{w} dx,
$$
with $N$ now given by
$$
N = N_1 \zeta^2 \int_0^1 (\bar{w}')^2 dx + N_0.
$$
See that
$$\int_0^1 \bar{w} \bar{w}'' dx$$
can be integrated by parts using $u=\bar{w}$ and $dv=\bar{w}'' dx$, leading to
$$
\int_0^1 \bar{w} \bar{w}'' dx = \left.\bar{w} \bar{w}'\right|_0^1 - \int_0^1 (\bar{w}')^2 dx.
$$
If the boundary conditions $(2)$ or $(3)$ holds, the first term in the RHS is $0$, then
$$
\int_0^1 \bar{w} \bar{w}'' dx = - \int_0^1 (\bar{w}')^2 dx.
$$
The integral
$$\int_0^1 \bar{w} \bar{w}^{IV} dx $$
can be evaluated in the same way as
$$
\int_0^1 \bar{w} \bar{w}^{IV} dx = \left. \bar{w} \bar{w}^{III} \right|_0^1 - \int_0^1 \bar{w}' \bar{w}^{III} dx.
$$
Assuming, for the sake of simplicity, that boundary conditions $(3)$ holds, the first term in RHS vanishes and the integral in RHS can be evaluated again with integration by parts, leading to
$$
\int_0^1 \bar{w} \bar{w}^{IV} dx = -\left.\bar{w}'\bar{w}'' \right|_0^1 + \int_0^1 (\bar{w}'')^2 dx.
$$
The first term in RHS vanishes with any of the boundary conditions, leading finally to
$$
\int_0^1 \bar{w} \bar{w}^{IV} dx = \int_0^1 (\bar{w}'')^2 dx.
$$
The equation is now
$$
\zeta \int_0^1 (\bar{w}'')^2 dx + N_1 \zeta^3 \left(\int_0^1 (\bar{w}')^2 dx \right)^2  + N_0 \zeta \int_0^1 (\bar{w}')^2 dx= \int_0^1 F_e \bar{w} dx.
$$
Then, defining
$$
k_0 = \int_0^1 (\bar{w}'')^2 dx,
$$
$$
k_1 = N_0 \int_0^1 (\bar{w}')^2dx,
$$
$$
k_2 = N_1 \left(\int_0^1 (\bar{w}')^2 dx\right)^2,
$$
and
$$
f_e(\zeta) = \frac{1}{\mu} \int_0^1 F_e (\zeta \bar{w}) \bar{w} dx,
$$
the equation can be written as
$$
(k_0+k_1) \zeta + k_2 \zeta^3 = \mu f_e(\zeta),
$$
which is equation $(16)$ in the paper.
