Jtdylktuy

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)=zetabar{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)=zetabar{w}(x). tag{15} end{equation*} After integrating by
parts, the governing equation for ζ becomes

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 $0leq x leq 1$,
$$
int_0^1 bar{w} w^{IV} dx - N int_0^1 bar{w} w'' dx= int_0^1 F_e(w) bar{w} dx.
$$
Now we substitute
$$
w=zeta bar{w},
$$
leading to
$$
zeta int_0^1 bar{w} bar{w}^{IV} dx - N zeta int_0^1 bar{w} bar{w}'' dx= int_0^1 F_e(zetabar{w}) bar{w} dx,
$$
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
$$
k_0 = int_0^1 (bar{w}'')^2 dx,
$$
$$
k_1 = N_0 int_0^1 (bar{w}')^2dx,
$$
$$
k_2 = N_1 left(int_0^1 (bar{w}')^2 dxright)^2,
$$
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.