# Deriving the Hemholtz equation from the wave equation

I am trying to derive the Hemholtz equation in electromagnetism. In the Lorentz gauge $$\partial_\mu A^\mu = 0$$ the inhomogeneous maxwell equations read: $$$$\label{maxwell inh} \square A^\mu = \mu_0 J^\mu \iff \bigg(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\bigg)A^\mu = \mu_0 J^\mu$$$$ Let us take a temporal Fourier transform: \begin{align} A_\mu(\textbf{r},t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \tilde{A}_\mu(\textbf{r},\omega) e^{-i\omega t}, \ J_\mu(\textbf{r},t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \tilde{J}_\mu(\textbf{r},\omega) e^{-i\omega t} \end{align} and substitute into the wave equation: \begin{align} &\int_{-\infty}^\infty\frac{d\omega}{2\pi} \bigg(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\bigg)\tilde{A}_\mu(\textbf{r},\omega) e^{-i\omega t} =\int_{-\infty}^\infty \frac{d\omega}{2\pi} \mu_0 \tilde{J}_\mu(\textbf{r},\omega) e^{-i\omega t}\\ &\implies \bigg(\nabla^2 + \frac{\omega^2}{c^2}\bigg)\tilde{A}_\mu = -\mu_0 \tilde{J}_\mu \end{align} The last equation is known as the Hemholtz equation. However, I'm not sure how to justify the last step since the integral over $$\mathbb{R}$$ of a non-zero quantity can still be zero i.e. an odd function.

• Note you are just taking Fourier transform, which is invertible for e.g., treating $A(x,-),\tilde{A}(x,-)$ as distributions. Commented Oct 16, 2021 at 11:42
• I don't see any problems. If both sides are zero the equation is satisfied trivially. Commented Oct 16, 2021 at 12:28