# Apply Fourier transform to the equation $\nabla\cdot[\mathbf{F}\delta(\mathbf{r})]=\nabla^2p$

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.

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!

• thank you for the answer. Could you give a little more details of the FT including $\delta$ function in any of your approaches? Please see my updated derivation using the definition of FT. Am I doing wrong somewhere?
– jsxs
Aug 24, 2020 at 2:02
• You're welcome. My pleasure. Our answers differ because of our different definitions of the Fourier Transform. I used the $e^{i\vec k\cdot \vec r}$ while you used $e^{-i\vec k \cdot \vec r}$. So just replace $i$ with $-i$ and our answers are the same. Aug 24, 2020 at 2:32
• could you review my update. I understand what you were talking about. However, there is still a difference in the sign. Thank you in advance.
– jsxs
Aug 24, 2020 at 5:00
• I did review you update and it is correct. Again the reason our answers differ by a factor of $-1$ is that we are using different conventions for defining the Fourier Transform. You're using the kernel $e^{-i \vec k \cdot \vec r}$ while I'm using the kernel $e^{+i \vec k \cdot \vec r}$. Aug 24, 2020 at 13:52
• I guess that I am doing wrong somewhere because my derivation is still different from the expected result by a sign even I stick to one convention. I have upvoted your answer and would accept one when the problem is solved.
– jsxs
Aug 24, 2020 at 14:06