Inhomogeneous electromagnetic wave equation

Remarks

Every twice differentiable solution of Maxwell's equations in vacuum satisfies the equations for $E$ and $B$. For example, the identity $\nabla \times (\nabla \times E) = \nabla (\nabla \cdot E) - \Delta E$ with $\nabla \times E = -\partial_t B$ and $\nabla \cdot E = \rho / \varepsilon_0$ gives the equation for $E$. The converse needs care. A solution of the wave equations solves Maxwell's equations only together with the first-order equations, or with the gauge condition in the potential form.

Example: let $\rho = \rho_0$ be constant, $J = 0$, $\varphi(x, t) = -\rho_0 \abs{x}^2 / (6 \varepsilon_0)$ and $A = 0$. Then $\partial_t \varphi = 0$ and $\Delta \varphi = -\rho_0 / \varepsilon_0$, so $\square \varphi = \rho_0 / \varepsilon_0$; also $\square A = 0 = \mu_0 J$, and the Lorenz condition holds.

Non-example: let $\rho = 0$, $J = 0$, $\varphi(x, t) = t$ and $A = 0$. Both wave equations hold, since $\square \varphi = 0$ and $\square A = 0$. The Lorenz condition fails, since $\nabla \cdot A + \frac{1}{c^2} \partial_t \varphi = 1/c^2 \neq 0$, so this pair is not of the form required in the definition.

In the static case, $\partial_t = 0$, the equation for $\varphi$ is Poisson's equation $-\Delta \varphi = \rho / \varepsilon_0$.