Solve this PDE using Laplace transform :
$$ EI {\partial^4 y(x,t)\over\partial x^4}+\mu {\partial^2y(x,t)\over\partial t^2}= F(x,t) $$ $$F(x,t)= P\delta(x-u) / \int_{-\infty}^\infty\delta(x-x_0)f(x)dx = f(x_0)$$$ \delta $ is the dirac function , P is the amplitude of the applied load and $u=u(t)$ the position of the load
initial conditions :
For $x=0 , l$ :
$${\partial^3 y(x,t)\over\partial x^3}=k_ly(x,t)$$ ,
$${\partial^2 y(x,t)\over\partial x^2}=k_t{\partial y(x,t)\over\partial x}$$ ,
$$y(x,t)= {\partial y(x,t)\over\partial t}=0$$
$l$ is the beam length and $k_t$ and $k_l$ are constants
i already solved it using green functions and i wondered if there is a way to solve it using Laplace transform (because i did it when the equation = 0) ,
an attempt :
we take : $ y(x,t) = y(x).z(t) / z(t)= z_1sin(\omega t) + z_2sin(\omega t) $
$EIy^{(4)}(x) - \mu \omega^2y(x)= {F(x,t)\over z(t)} $
$EI\mathcal L(y^{(4)}(x)) - \mu \omega^2\mathcal L(y(x))= \mathcal L({F(x,t)\over z(t)}) $
$$\mathcal L({F(x,t)\over z(t)})=P{e^{-su}\over z(u)} : (1)$$
if $(1)$ is true then the whole problem is based on the calculation of : $$\mathcal L^{-1}({e^{-su}\over s^4-\lambda^4})$$
any hint ? is it even possible ?
the paper .
If I am not too rusty you can use partial fraction decomposition such as $$e^{-su}\bigg(\frac 1{s^4-\lambda^4}\bigg)=e^{-su}\bigg(-\frac 1{2\lambda^3} \frac \lambda{s^2+\lambda^2}+\frac 1{4\lambda^3}\frac 1{x-\lambda}-\frac 1{4\lambda^3}\frac 1{x+\lambda}\bigg)$$ $$\mathcal L^{-1}\Bigg(e^{-su}\bigg(-\frac 1{2\lambda^3} \frac \lambda{s^2+\lambda^2}+\frac 1{4\lambda^3}\frac 1{x-\lambda}-\frac 1{4\lambda^3}\frac 1{x+\lambda}\bigg)\Bigg)$$ Without the dirac function the inverse Laplace is $$\Rightarrow -\frac 1{2\lambda^3} \sin(\lambda t)+\frac 1{4\lambda^3}e^{\lambda t}-\frac 1{4\lambda^3}e^{-\lambda t}$$ and the final version $$\Rightarrow k_u(t)\bigg(-\frac 1{2\lambda^3} \sin(\lambda t-\lambda u)+\frac 1{4\lambda^3}e^{\lambda t-\lambda u}-\frac 1{4\lambda^3}e^{-\lambda t+\lambda u}\bigg)$$ where $$k_u(t\lt u)=0\text{ and }k_u(t\gt u)=1$$
Please check for typos