I'm trying to solve this problem from a textbook:
The transverse displalcement in an elastic bar is given by
$\dfrac{\partial^4w}{\partial x^4}+\dfrac{\partial^2w}{\partial t^2}=0 \tag{1}$
$-\infty < x < \infty$, $t > 0$
with initial conditions
$w(x,0)=f(x)$
$w_{t}(x,0)=0$
Using Fourier Transforms and convolution, show that the solution is given by $$ w(x,t) = \frac{1}{(4 \pi t)^{1/2}} \int_{-\infty}^{\infty} \cos\left(\frac{(x-u)^2}{4t} - \frac{\pi}{4}\right)f(u)\ du$$
Using this definition of the Fourier Transform: $$\mathcal{F}[f(t)](k) = \int_{-\infty}^{\infty} f(t) e^{-i k t} \, dt. $$
I'm guessing that the standard approach to this question is taking FT of both sides of $(1)$ but I'm not sure how to begin this. If the FT is defined as an integral of $dt$, how would I represent the FT of a fourth order partial derivative with respect to a different variable (in this case, $x$)?
Some tips with the convolution would also help greatly! Thanks.
EDIT: If my intuition is correct, equation $(1)$, after FT, should take the form
$\dfrac{\partial^4w}{\partial x^4} - k^2\bar W = 0$
but this doesn't seem easy to solve either.
IC: $w_{t}(0,x)=0$ changed to $w_{t}(x,0)=0$
Taking Fourier transforms (in the $x$ variable) we get $$ \frac{d^2\hat w}{dt^2}+k^4\,\hat w=0,\quad \hat w(0)=\hat f(k),\quad\frac{d\hat w}{dt}(0)=0. $$ The solution is $$ \hat w(k,t)=\cos(k^2t)\,\hat f(k). $$ Finally, $$ \mathcal F^{-1}\bigl(\cos(k^2)\bigr)(x)=\frac12\Bigl(\cos\frac{x^2}{4}+\sin\frac{x^2}{4}\Bigr)=\frac{\sqrt2}{2}\cos\Bigl(\frac{x^2}{4}-\frac{\pi}{4}\Bigr). $$