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I have a problem finding the Inverse Fourier transform of the

\begin{equation} \dfrac{1}{k^2+a^2}\dfrac{1}{k^2}e^{i\textbf{k}\cdot \bf{r}}(\textbf{k}\cdot \bf{v}) \bf{k} \end{equation} where $\bf{r}$ is the radius vector, $\bf{v}$ is the velocity,$\bf{k}$ is the variable in Fourier space and $a$ is a contant.

Any suggestion?

Jon Snow
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2 Answers2

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Let $F(k)=\frac{1}{k^2(k^2+a^2)}$. We can write $F(k)$ as

$$\begin{align} F(k)&=\frac{1}{k^2(k^2+a^2)}\\\\ &=\frac{1}{a^2}\left(\frac{1}{k^2}-\frac{1}{k^2+a^2}\right) \end{align}$$

Then, the Fourier Transform of $F$ can be written

$$\begin{align} \mathscr{F}\{F\}(\vec r)&=\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty F(k)\,e^{i\vec k \cdot \vec r}\,dk_x\,dk_y\,dk_z\\\\ &=\frac1{a^2}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\vec k \cdot \vec r}}{k^2}\,dk_x\,dk_y\,dk_z-\frac1{a^2}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\vec k \cdot \vec r}}{k^2+a^2}\,dk_x\,dk_y\,dk_z \end{align}$$

In THIS ANSWER, I evaluated the Fourier Transform of $\frac{1}{k^2+a^2}$ as $f(r;a)=2\pi^2\frac{e^{-|a|r}}{r}$.

so that

$$\begin{align} \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\left(\frac{(\vec v\cdot \vec k)\vec k}{k^2(k^2+a^2)}\right)\,e^{i\vec k \cdot \vec r}\,dk_x\,dk_y\,dk_z&=-(\vec v\cdot \nabla)\nabla\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\vec k \cdot \vec r}}{k^2(k^2+a^2)}\,dk_x\,dk_y\,dk_z\\\\ &=\frac{1}{a^2}(\vec v\cdot \nabla)\nabla(f(r;a)-f(r;0)) \end{align}$$

Community
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Mark Viola
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  • Dr. MV, once again thank you very much. I am impressed by the answer... – Jon Snow Mar 06 '17 at 20:21
  • You're welcome. My pleasure. -Mark – Mark Viola Mar 06 '17 at 20:22
  • What vote? I didn't remove anything. You helped me a lot and I voted for both your answers. – Jon Snow Mar 21 '17 at 19:47
  • @JonSnow Perhaps it was a system issue, but the accepted vote status appeared to have been removed and given to another answer. Just moments ago, I received notification that my answer was accepted. Curious indeed. -Mark – Mark Viola Mar 21 '17 at 19:49
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{\large\mathbb{R}^{3}}{1 \over k^{2} + a^{2}} \,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}} \pars{\mathbf{k}\cdot\mathbf{v}}\mathbf{k}\,\,\dd^{3}\mathbf{k} = -\ic\nabla\pars{\mathbf{v}\cdot\int_{\large\mathbb{R}^{3}} {1 \over k^{2} + a^{2}}\,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}} \mathbf{k}\,\,\dd^{3}\mathbf{k}} \\[5mm] = &\ -\nabla\pars{\mathbf{v}\cdot\nabla\int_{\large\mathbb{R}^{3}} {1 \over k^{2} + a^{2}}\,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}} \dd^{3}\mathbf{k}} \\[5mm] = &\ -\nabla\bracks{\mathbf{v}\cdot\nabla\int_{0}^{\infty} {1 \over k^{2} + a^{2}}\,{1 \over k^{2}}\, \pars{\int\expo{\ic\mathbf{k}\cdot\mathbf{r}} \,{\dd\Omega_{\mathbf{k}} \over 4\pi}}4\pi k^{2}\,\dd k} \\[5mm] = &\ -4\pi\nabla\bracks{\mathbf{v}\cdot\nabla\int_{0}^{\infty}{1 \over k^{2} + a^{2}} {\sin\pars{kr} \over kr}\,\dd k} = -4\pi\nabla\braces{\mathbf{v}\cdot\nabla\bracks{r\int_{0}^{\infty}{1 \over \xi^{2} + \pars{ar}^{2}} {\sin\pars{\xi} \over \xi}\,\dd\xi}} \\[5mm] = &\ -4\pi a\nabla_{\mathbf{R}}\braces{\mathbf{v}\cdot\nabla_{\mathbf{R}} \bracks{R\int_{0}^{\infty}{1 \over \xi^{2} + R^{2}} {\sin\pars{\xi} \over \xi}\,\dd\xi}}\,,\qquad\qquad \mbox{where}\ \mathbf{R} \equiv a\mathbf{r} \end{align}

With $\ds{\mrm{f}\pars{z} \equiv {1 - \expo{-z} \over z}}$,

\begin{align} &\int_{\large\mathbb{R}^{3}}{1 \over k^{2} + a^{2}} \,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}} \pars{\mathbf{k}\cdot\mathbf{v}}\mathbf{k}\,\,\dd^{3}\mathbf{k}= -2\pi^{2}a\,\nabla_{\mathbf{R}}\bracks{\mathbf{v}\cdot\nabla_{\mathbf{R}} \mrm{f}\pars{R}} = -2\pi^{2}a\,\nabla_{\mathbf{R}}\bracks{\mathbf{v}\cdot \mrm{f}'\pars{R}\,{\mathbf{R} \over R}} \\[5mm] = &\ -2\pi^{2}a\,\sum_{i}\hat{x}_{i}\,\partiald{}{x_{i}} \sum_{j}v_{j}\bracks{\mrm{f}'\pars{R}{x_{j} \over R}} = -2\pi^{2}a\,\sum_{ij}\hat{x}_{i}\,v_{j}\,\partiald{}{x_{i}} \bracks{x_{j}\,{\mrm{f}'\pars{R} \over R}} \\[5mm] = &\ -2\pi^{2}a\,\sum_{ij}\hat{x}_{i}\,v_{j} \braces{\delta_{ij}\,{\mrm{f}'\pars{R} \over R} + x_{j}\,{x_{i} \over R}\, \totald{}{R}\bracks{\mrm{f}'\pars{R} \over R}} \\[5mm] = &\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{% -2\pi a^{2}\,{\mrm{f}'\pars{R} \over R}\,\mathbf{v} - 2\pi a^{2}\braces{{1 \over R}\,\totald{}{R}\bracks{\mrm{f}'\pars{R} \over R}} \pars{\mathbf{v}\cdot\mathbf{R}} \mathbf{R}}} \end{align}

Felix Marin
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