I Derivation of the Helmholtz equation

AI Thread Summary
The Helmholtz equation is derived as the time-independent form of the wave equation, which can be expressed using the Fourier transform. The discussion highlights the differentiation properties of the Fourier transform, leading to the conclusion that the Helmholtz equation can be represented as a member of a larger set of equations. Specifically, by setting the second derivative in the spatial domain, the Helmholtz equation emerges from the wave equation. The conversation also addresses the complexities introduced by different conventions, such as the use of factors of 2π and the representation of wave frequencies. Overall, the derivation process and the relationship between the wave equation and the Helmholtz equation are central to the discussion.
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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
 
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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.
 
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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?
 
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##.
 
vanhees71 said:
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.
vanhees71 do you mind going through all the relevant steps after plugging in the first form of the Fourier transformation into the wave equation? I don't understand how to reach the final form of the Helmholtz equation. Your help would be greatly appreciated.
 
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