Derivation of the Helmholtz equation

[redtree]

TL;DR Summary

Derivation from the Fourier transform

I am trying to understand the Helmholtz equation, where the Helmholtz equation can be considered as the time-independent form of the wave equation. It seems to me that the Helmholtz equation can be derived from the Fourier transform, such that it is part of a larger set of equations of varying order.
Given a differentiable function ##f(\vec{x})##, I note the differentiation property of the Fourier transform,

\begin{equation}

\begin{split}

\partial_{\vec{x}} f(\vec{x}) &= \partial_{\vec{x}} \int_{\mathbb{R}} e^{2 \pi i \vec{k} \cdot \vec{x}} \hat{f}(\vec{k}) d\vec{k}

\\

&= \int_{\mathbb{R}} \partial_{\vec{x}} e^{2 \pi i \vec{k} \cdot \vec{x}} \hat{f}(\vec{k}) d\vec{k}

\\

&= \int_{\mathbb{R}} 2 \pi i \vec{k} e^{2 \pi i \vec{k} \cdot \vec{x}} \hat{f}(\vec{k}) d\vec{k}

\\

&= \mathscr{F}^{-1} \left[2 \pi i \vec{k} \hat{f}(\vec{k}) \right]

\end{split}

\end{equation}
The converse is also true, such that

\begin{equation}

\begin{split}

2 \pi i \vec{k} \hat{f}(\vec{k}) &= \mathscr{F}\left[\partial_{\vec{x}} f(\vec{x}) \right]

\\

&= \int_{\mathbb{R}} e^{-2 \pi i \vec{k} \cdot \vec{x}} \bigg( \partial_{\vec{x}} f(\vec{x}) \bigg) d\vec{x}

\end{split}

\end{equation}
Regarding ##2 \pi i \vec{k} \hat{f}(\vec{k})##, I also note the following:

\begin{equation}

\begin{split}

2 \pi i \vec{k} \hat{f}(\vec{k}) &= 2 \pi i \vec{k} \mathscr{F}\left[f(\vec{x}) \right]

\\

&= 2 \pi i \vec{k} \int_{\mathbb{R}} e^{-2 \pi i \vec{k} \cdot \vec{x}} f(\vec{x}) d\vec{x}

\\

&= \int_{\mathbb{R}} e^{-2 \pi i \vec{k} \cdot \vec{x}} \bigg( 2 \pi i \vec{k} f(\vec{x}) \bigg) d\vec{x}

\end{split}

\end{equation}Combining these last two equations,

\begin{equation}

\begin{split}

\int_{\mathbb{R}} e^{-2 \pi i \vec{k} \cdot \vec{x}} \bigg( \partial_{\vec{x}} f(\vec{x}) \bigg) d\vec{x}&= \int_{\mathbb{R}} e^{-2 \pi i \vec{k} \cdot \vec{x}} \bigg( 2 \pi i \vec{k} f(\vec{x}) \bigg) d\vec{x}

\end{split}

\end{equation}

which simplifies to

\begin{equation}

\begin{split}

\vec{k} f(\vec{x}) &= -\frac{i}{(2 \pi) } \partial_{\vec{x}} f(\vec{x})

\end{split}

\end{equation}
This same holds true for any ##n##-th derivative, such that

\begin{equation}

\begin{split}

\big(\vec{k} \big)^n f(\vec{x}) &= \left(-\frac{i}{2 \pi} \right)^n \big(\partial_{\vec{x}} \big)^n f(\vec{x})

\end{split}

\end{equation}
Setting ##n=2## produces the Helmholtz equation

\begin{equation}

\begin{split}

\big( \vec{k} \big)^2 f(\vec{x}) &= -\frac{1}{(2 \pi)^2} \big( \partial_{\vec{x}} \big)^2 f(\vec{x})

\end{split}

\end{equation}

which would make the Helmholtz equation the member of order ##n=2## of a larger set of Helmholtz-type equations

 

[vanhees71]

That's a bit complicated. Just start from the wave equation for some field ##\Phi(t,\vec{x})##
$$\left (\frac{1}{c^2} \partial_t^2 - \Delta \right) \Phi(t,\vec{x})=0.$$
Here ##c## is the phase velocity of the waves.

Now we express the field as a Fourier integral with respect to time. Using the HEP physicists convention concerning the factors ##2 \pi## and the sign in the exponential this reads
$$\Phi(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} \omega \frac{1}{2 \pi} \tilde{\Phi}(\omega,\vec{x}) \exp(-\mathrm{i} \omega t) \; \Leftrightarrow \; \tilde{\Phi}(\omega,\vec{x}) = \int_{\mathbb{R}} \mathrm{d} t \tilde{\Phi}(\omega) \exp(+\mathrm{i} \omega t).$$
Now plug the first form of the Fourier transformation into the wave equation (just assuming you can integrate under the integral)
$$\left (\frac{1}{c^2} \partial_t^2 - \Delta \right) \Phi(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} \omega \frac{1}{2 \pi} (-k^2-\Delta) \tilde{\phi}(\omega,\vec{x})=0 \; \Rightarrow \; (k^2+\Delta) \tilde{\Phi}(\omega,\vec{x})=0.$$
Here ##k=\omega/c##, and that's the Helmholtz equation.

 

[redtree]

I rewrite the derivation you cite in slightly different notation as follows:

\begin{equation}

\begin{split}

\Phi(x_0,\vec{x}_3) &= \int_{\mathbb{R}} \hat{\phi}(k_0,\vec{x}_3) e^{- 2 \pi i k_0 x_0} dk_0

\end{split}

\end{equation}

where ##\vec{x}_4 = [x_0, x_1, x_2, x_3] = [x_0, \vec{x}_3]## and ##\vec{k}_4 = [k_0, k_1, k_2, k_3] = [k_0, \vec{k}_3]##
Given the wave equation:

\begin{equation}

\begin{split}

\bigg(\partial_{x_0}^2 - \partial_{\vec{x}_3}^2 \bigg) \Phi(x_0,\vec{x}_3) &= 0

\end{split}

\end{equation}

where ##\partial_{x_0} = \partial_t## and ##\partial_{\vec{x}_3}^2 = \Delta##
Combining the two equations

\begin{equation}

\begin{split}

\bigg(\partial_{x_0}^2 - \partial_{\vec{x}_3}^2 \bigg) \Phi(x_0,\vec{x}_3) &= \bigg(\partial_{x_0}^2 - \partial_{\vec{x}_3}^2 \bigg) \int_{\mathbb{R}} \hat{\phi}(k_0,\vec{x}_3) e^{- 2 \pi i k_0 x_0} dk_0

\\

&= \bigg(\left(2 \pi k_0\right)^2 - \partial_{\vec{x}_3}^2 \bigg) \hat{\Phi}(k_0,\vec{x}_3)

\\

&= 0

\end{split}

\end{equation}

which I summarize:

\begin{equation}

\begin{split}

\bigg(\left(2 \pi k_0\right)^2 - \partial_{\vec{x}_3}^2 \bigg) \hat{\Phi}(k_0,\vec{x}_3) &= 0

\end{split}

\end{equation}
Comparing this to the equation I derived (in similar notation)

\begin{equation}

\begin{split}

\bigg( \left(2 \pi \vec{k}_3\right)^2 - \partial_{\vec{x}_3}^2 \bigg) \Phi(\vec{x}_3) &= 0

\end{split}

\end{equation}
Thus, the equation I derived above is different from the Helmholtz equation. I get it. Thanks. Is there a name for the equation I derived. It seems correct. I don't see a mistake in its derivation. Am I missing something?

 

[vanhees71]

You forget the factor ##\mathrm{i}^2=-1## from the two time-derivatives! BTW it's very complicated to introduce the ##2 \pi## in the exponent and working with ##\nu## instead of ##\omega##. I've seen only one textbook (Weizel, Lehrbuch der Theoretischen Physik, an otherwise excellent textbook, but as far as I know not available in English), where this is done, and you suffer from a lot of factors ##2 \pi##.