# Can the convolution theorem be applied in this case?

The kernel integration to extend the applicability range (or domain) of an equation $\mathcal{F}(\Psi,\mathbf{x},t)$ originally defined in $\Omega_f$ can be written as \begin{gather} \mathcal{F}_\epsilon(\Psi,\mathbf{x},t)=\int_{\Omega_f}\mathcal{F}(\Psi,\mathbf{x}-\mathbf{x_f},t)K_\epsilon(\mathbf{x_f})\mathrm{d}\mathbf{x_f} \end{gather}

where $\Psi$ are the unkowns of the equation, $K$ is the kernel function and $\epsilon$ denotes the equation extended to the full domain.

So, my question is how can this process be performed in spectral (Fourier) space? The convolution theorem allows the convolution in physical space to be computed as a multiplication in spectral space (does this apply here as well?), which is cheaper. I intend to do that but I am not entirely sure how to.

Imagine that $\mathcal{F}$ is just a constant value $\alpha$ everywhere in $\Omega_f$, then \begin{gather} \mathcal{F}_\epsilon(\Psi,\mathbf{x},t)=\int_{\Omega_f}\mathcal{F}(\Psi,\mathbf{x}-\mathbf{x_f},t)K_\epsilon(\mathbf{x_f})\mathrm{d}\mathbf{x_f}=\alpha \int_{\Omega_f}K_\epsilon(\mathbf{x_f})\mathrm{d}\mathbf{x_f} \end{gather} and this would result into a smooth decrease of $\alpha$ when approaching to the extended domain (and vanishing inside the extended domain). Now, if we apply the convolution theorem to this same case the following equation should be valid (?) \begin{gather} \hat{\mathcal{F}}_\epsilon(\Psi,\mathbf{k},t)=\hat{\mathcal{F}}(\Psi,\mathbf{k},t)\hat{K}^f_\epsilon(\mathbf{k}) \end{gather} where the hat denotes a Fourier transformed quantity. So, for a constant value of $\mathcal{F}$ in physical space, only the zero-th (mean) mode contains information in the Fourier space. Then, after doing the multiplication in Fourier space and an inverse transform to physical space, the result is just another constant value everywhere in the fluid domain. And this is obviously different from performing the convolution in physical space.

I cannot see what I am missing and any help would very greatly appreciated. Thanks.

• If $\mathbf 1$ denotes the constant function and $\int g=1$, then the convolution $\mathbf 1\star g$ is $\mathbf 1$. So it's not obviously different, in fact it's the same in that example. – Dap Sep 8 '17 at 5:51
• But if the convolution theorem is applied $\hat{\mathbf{1}}\hat{g}$ only gives a mean mode and the inverse transform does not provide the constant value, but another one everywhere in the domain. – b-fg Sep 8 '17 at 5:56
• you've given a perfectly good argument that $\delta_0 \hat g = \delta_0$, in other words $\hat g(0)=1$, for any kernel that behaves as you expect on constant functions. (Possibly with some extra factors of $2\pi$ depending on Fourier transform convention.) This doesn't seem at all contradictory. – Dap Sep 8 '17 at 7:02