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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 $(3)$ ?

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It seems that you have to do some Fourier transform on t, but the problem is that we don't know the Fourier transform of $\delta(x - u(t))$, because we don't know $u(t)$. $$EI {\partial^4 y(x,t)\over\partial x^4}+\mu {\partial^2y(x,t)\over\partial t^2}= P \delta(x -u(t))$$ By Fourier transform on t, you will obtain (knowing that $\omega$ has only possible discrete values due to the boudary conditions): $$EI {\partial^4 y(x,\omega)\over\partial x^4} -\mu \omega^2 y(x,\omega)= P* ~Fourier~ Transform [\delta(x - u(t)] (\omega)$$ –  Trimok Jun 4 '13 at 9:15
Why does this have closevotes? It is a legitimate technical question! –  Dilaton Jun 4 '13 at 14:37
we know that $v$ is a constant , so $u(t)=tv$ –  Lofaif Jun 4 '13 at 16:15

1 Answer 1

up vote 4 down vote accepted

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.

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+1 for the general formula. I think too there is a missing factor in the paper. –  Trimok Jun 5 '13 at 9:59
thank you for your answer sir , i have a small question .. why is $\psi ^4 = {\omega ^2 m \over EI }$ and not $ {\mu \omega ^2 \over EI} $ ?? –  Lofaif Jun 5 '13 at 19:33
i think it might be a typo. there are some others in that paper. also, i looked through ref 8 of the paper you posted and it has the correct $\psi$ definition. –  igphys Jun 6 '13 at 3:55

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