Solving the differential equation of a beam under moving load using green functions i started working on this paper and i didnt understand one part of it , the  problem is : 

Solve this equation using green functions : 
$$ EI {\partial^4 y(x,t)\over\partial x^4}+\mu {\partial^2y(x,t)\over\partial t^2}= F(x,t) :(1)$$
  $$F(x,t)= P\delta(x-u)$$ 
   with $\int_{-\infty}^\infty\delta(x-x_0)f(x)dx = f(x_0)$

$\delta$ is the Dirac-delta function, P is the amplitude of the applied load, and $u(t)=v\, t$ the position of the load.

initial conditions : 
$${\partial^3 y(x,t)\over\partial x^3}=k_ly(x,t)$$ , 
$${\partial^2 y(x,t)\over\partial x^2}=k_t{\partial y(x,t)\over\partial x}$$ , 
For $x=0 , l$ :
$$y(x,t)= {\partial y(x,t)\over\partial t}=0$$
$l$ is the beam length and $k_t$ and $k_l$ are constants

and the paper said that : 

Using the dynamic Green function, the solution of
  Equation $(1)$ can be written as:
  $$y(x,t) = G(x,u)P : (2)$$
  where $G(x,u)$ is the solution of the differential equation:
  $${d^4y(x)\over dx^4} - \psi ^4 y(x) = \delta(x-u):(3) $$
  in witch : $\psi ^4 = {\omega ^2 \mu \over EI}$

My question is how did they simplify it to the form $(2)$ where $G$ is solution to $(3)$ ?
 A: The Fourier transform of $y\left(x,t\right)$ from the time to the frequency domain is given by $Y\left(x,\omega\right)=\int_{-\infty}^{\infty}y\left(x,t\right)e^{i\omega t}dt$ and satisfies the differential equation:
$$
EI\frac{\partial^{4}Y\left(x,\omega\right)}{\partial x^{4}}+\mu\left(i\omega\right)^{2}Y\left(x,\omega\right)=\int_{-\infty}^{\infty}P\delta\left(x-u\left(t\right)\right)e^{i\omega t}dt
$$
Now, use this property of the delta function:
$$
\delta\left(g\left(s\right)\right)=\sum_{i}\frac{\delta\left(s-s_{i}\right)}{\left|g'\left(s_{i}\right)\right|}
$$ 
where $s_{i}$ are the roots of the function, so that $g\left(s_{i}\right)=0$. 
$$
\int_{-\infty}^{\infty}P\delta\left(g\left(t\right)\right)e^{i\omega t}dt=\int_{-\infty}^{\infty}P\sum_{i}\frac{\delta\left(t-t_{i}\right)}{\left|g'\left(t_{i}\right)\right|}e^{i\omega t}dt=P\sum_{i}\frac{e^{i\omega t_{i}}}{\left|g'\left(t_{i}\right)\right|}
$$ 
Then we have:
$$
\frac{\partial^{4}Y\left(x,\omega\right)}{\partial x^{4}}-\frac{\mu\omega^{2}}{EI}Y\left(x,\omega\right)=\frac{P}{EI}\sum_{i}\frac{e^{i\omega t_{i}}}{\left|g'\left(t_{i}\right)\right|}
$$ 
The Green's function solution to this equation is:
$$
Y\left(x,\omega\right)=\int_{-\infty}^{\infty}\frac{P}{EI}\sum_{i}\frac{e^{i\omega t_{i}}}{\left|g'\left(t_{i}\right)\right|}G\left(x,x'\right)dx'
$$
  The Green's function satisfies the equation:
$$
\frac{\partial^{4}G\left(x,x'\right)}{\partial x^{4}}-\psi^{4}G\left(x,x'\right)=\delta\left(x-x'\right)
$$ 
Now we have to invert the Fourier transform.
$$
y\left(x,t\right) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{P}{EI}\sum_{i}\frac{e^{i\omega t_{i}}}{\left|g'\left(t_{i}\right)\right|}G\left(x,x'\right)e^{-i\omega t}dx'd\omega
 =$$    $$=\frac{P}{EI}\sum_{i}\left\{ \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{\left|g'\left(t_{i}\right)\right|}G\left(x,x'\right)e^{i\omega\left(t_{i}-t\right)}dx'd\omega\right\} 
 =$$    $$=\frac{P}{EI}\sum_{i}\frac{1}{\left|g'\left(t_{i}\right)\right|}\int_{-\infty}^{\infty}G\left(x,x'\right)\delta\left(t-t_{i}\right)dx'
$$ 
Where the definition: $\delta\left(t-t'\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega\left(t-t'\right)}d\omega$, has been used. So, this is for a general $g\left(t\right)$. Now, if $g\left(t\right)=x-vt$, then:
$$
y\left(x,t\right) = \frac{P}{EI}\frac{1}{v}\int_{-\infty}^{\infty}G\left(x,x'\right)\delta\left(t-\frac{x'}{v}\right)dx'
 = \frac{P}{EI}G\left(x,vt\right)
$$ 
Not sure why the paper does not have the $EI$ factor.
