Derivation Rules
Going into Fourier space, we just have to replace derivatives via:
\[\frac{\partial}{\partial t}\longleftrightarrow-i\omega~\mathrm{and}~\frac{\partial}{\partial\mathbf{r}}\longleftrightarrow i\mathbf{k}.\]
How to show that \(\frac{\partial}{\partial t}\longrightarrow-i\omega\)
In order to show that \(\frac{\partial}{\partial t}\longrightarrow-i\omega\), we will start with the inverse Fourier transform for \(\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]\), obtaining:
\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt&_{P.I.}^{=}&\frac{1}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}|_{-\infty}^{\infty}-\frac{i\omega}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt\\&=&-\frac{i\omega}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt=-i\omega\mathbf{\overline{E}}\left(\mathbf{r},\omega\right) \end{eqnarray*}
To obtain a general rule for the \(n^{\textrm{th}}\)-derivative, we perform the second derivative:
\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial^{2}}{\partial t^{2}}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt&=&\frac{1}{2\pi}\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}|_{-\infty}^{\infty}-\frac{i\omega}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt\\&=&-\frac{i\omega}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}|_{-\infty}^{\infty}+\frac{\omega^{2}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt=\omega^{2}\mathbf{\overline{E}}\left(\mathbf{r},\omega\right) \end{eqnarray*}
Hence, we get for the \(n^{\textrm{th}}\)-derivative:
\[\frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial^{n}}{\partial t^{n}}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt=\frac{\left(-i\omega\right)^{n}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt=\left(-i\omega\right)^{n}\mathbf{\overline{E}}\left(\mathbf{r},\omega\right)\]
with \(n\in\mathbb{N}\)
How to show that \(\frac{\partial}{\partial\mathbf{r}}\longrightarrow i\mathbf{k}\)
To show that \(\frac{\partial}{\partial\mathbf{r}}\longrightarrow i\mathbf{k}\) we will use the same strategy as above. Starting with the Fourier transform for \(\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]\) we obtain:
\[\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}&_{P.I.}^{=}&\frac{1}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}|_{-\infty}^{\infty}+\frac{i\mathbf{k}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}\\&=&\frac{i\mathbf{k}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}=i\mathbf{k}\mathbf{E}\left(\mathbf{k},t\right) \end{eqnarray*}\]
Again, the second order derivative:
\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial^{2}}{\partial\mathbf{r}^{2}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}&_{P.I.}^{=}&\frac{1}{2\pi}\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}|_{-\infty}^{\infty}+\frac{i\mathbf{k}}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}\\&=&\frac{i\mathbf{k}}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}|_{-\infty}^{\infty}-\frac{\mathbf{k^{2}}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}=\mathbf{-k^{2}}\mathbf{E}\left(\mathbf{k},t\right) \end{eqnarray*}
We find analogously the general form:
\[\frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial^{n}}{\partial\mathbf{r}^{n}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}=\frac{\left(i\mathbf{k}\right)^{n}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}=\left(i\mathbf{k}\right)^{n}\mathbf{E}\left(\mathbf{k},t\right)\]
Hence, we can write in general:
\[\frac{\partial^{n}}{\partial t^{n}}\longleftrightarrow(-i\omega)^{n}~\mathrm{and}~\frac{\partial^{n}}{\partial\mathbf{r}^{n}}\longleftrightarrow(i\mathbf{k})^{n}.\]
Any linear partial differential equation transforms to an algebraic equation or set of equations in Fourier space, which can be solved conveniently. The general solution is time and space can be found by superposition and inverse Fourier transform.