I want to solve the partial differential equation $\Delta^2 u = f$, with $u,f:\mathbb{R}^n \to \mathbb{R}$ for some $n\geq 3$, and assume that $f$ has compact support. I want to solve this using a Fourier transform, because this is also useful to solve the equation $\Delta u = f$. Taking the Fourier transforms twice with respect to $x_i$, for all $i=1,\ldots, n$, we get the equation $\hat{u}(\omega) = \frac{\hat{f}(\omega)}{|\omega|^4}$, where $\hat{u}$ and $\hat{f}$ are the Fourier transforms of respectively $u$ and $f$. By writing this as $\hat{u}(\omega) = \frac{\left(\frac{\hat{f}(\omega)}{|\omega|^2}\right)}{|\omega|^2}$, I am able to find an expression for $u$ as two nested convolution integrals. Is is also possible to express $u$ as as single convolution integral, or is this double integral inherent to the equation $\Delta^2u = f$?
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The solution you seem to have found is the one we get if we decompose the PDE as $\Delta u = g$ with $\Delta g = f$. This naturally leads to a double convolution: $u = (fh)h$ where $\hat{h} = -1/|\omega|^2$. However you can atleast formally write the solution as $u = f*H$ where $\hat{H} = 1/|\omega|^4$. As for finding the inverse transform see http://math.stackexchange.com/questions/48430/n-dimensional-inverse-fourier-transform-of-frac1-mathbf-omega?noredirect=1&lq=1 – Winther Mar 07 '17 at 02:51