TITLE: Apply Fourier transform to the equation $\nabla\cdot[\mathbf{F}\delta(\mathbf{r})]=\nabla^2p$
QUESTION [2 upvotes]: Consider an equation
$$\nabla\cdot[\mathbf{F}\delta(\mathbf{r})]=\nabla^2p,$$
in which $\mathbf{F}$ is a differentiable vector function, $\delta(\mathbf{r})$ is the Dirac delta function, $\nabla\cdot$ is a divergence operator, $\nabla^2$ is the Laplace operator, and $p$ is a differentiable scalar function.
I have difficult to apply Fourier transform to this equation in order to get $\mathrm{i}\mathbf{k}\cdot\mathbf{F}=k^2\hat{p}$, where $\hat{}$ denotes the transformed function.
What I have tried is as follows:
$$\mathcal{F} [\mathrm{LHS}]=\mathcal{F}[(\nabla\delta)\cdot \mathbf{F}+\delta\nabla\cdot\mathbf{F}]=\mathcal{F}[(\nabla\delta)\cdot \mathbf{F}]+\delta \mathcal{F}[\nabla\cdot\mathbf{F}],$$
$$\mathcal{F}[\mathrm{RHS}]=[(\mathrm{i}k_x)^2+(\mathrm{i}k_x)^2]\hat{p}=-(k_x^2+k_y^2)\hat{p}\equiv-k^2\hat{p}.$$
I don't know how to further evaluate the FT of the LHS. Thank you in advance.

Update (Aug.24,2020):
Applying the definition of FT: $\hat{f}(\mathbf{k})=\int_{-\infty}^{\infty}f(\mathbf{r})e^{-\mathrm{i}\mathbf{k}\cdot\mathbf{r}}\:\mathrm{d}\mathbf{r}$
on LHS and RHS, respectively:
$$\mathcal{F}\{\nabla\cdot [\mathbf{F}\delta(\mathbf{r})]\}=\mathrm{i}\mathbf{k}\cdot\mathcal{F}[\mathbf{F}\delta(\mathbf{r})]=\mathrm{i}\mathbf{k}\cdot\int_{-\infty}^{\infty}\mathbf{F}\delta(\mathbf{r})e^{-\mathrm{i}\mathbf{k}\cdot\mathbf{r}}\:\mathrm{d}\mathbf{r}=\mathrm{i}\mathbf{k}\cdot\left(\mathbf{F}e^{-\mathrm{i}\mathbf{k}\cdot\mathbf{r}} \right)\vert_{\mathbf{r}=\mathbf{0}}=\mathrm{i}\mathbf{k}\cdot\mathbf{F}(\mathbf{0}),$$
and
$$\mathcal{F}[\nabla^2p]=(\mathrm{i}\mathbf{k})\cdot(\mathrm{i}\mathbf{k})\int_{-\infty}^{\infty}p e^{-\mathrm{i}\mathbf{k}\cdot\mathbf{r}}\:\mathrm{d}\mathbf{r}=-k^2\hat{p}.$$
It follows that $\mathrm{i}\mathbf{k}\cdot\mathbf{F}(\mathbf{0})=-k^2\hat{p}$, which differs from the expected result by a negative sign.

REPLY [1 votes]: Note that we have
$$\begin{align}
\mathscr{F}\{\nabla \cdot (\vec F\delta)\}&=\mathscr{F}\{i\vec k\cdot \vec F\delta\}\\\\
&=i\vec k\cdot \vec F(0)
\end{align}$$
Alternatively, note that
$$\begin{align}
\mathscr{F}\{\nabla \cdot (\vec F\delta)\}&=\mathscr{F}\{\vec F\cdot \nabla (\delta)+\delta \nabla\cdot \vec F\}\\\\
&=-\left.\left(\nabla \cdot (e^{-i\vec k\cdot \vec r}\vec F(\vec r))\right)\right|_{\vec r=0}+\left.\left(\nabla \cdot (\vec F(\vec r))\right)\right|_{\vec r=0}\\\\
&=i\vec k\cdot \vec F(0)
\end{align}$$
as expected!
{"set_name": "stack_exchange", "score": 2, "question_id": 3800827}