Electromagnetic wave equation

Remarks

Derivation: from $\nabla \times E = -\partial_t B$ and $\nabla \times B = \frac{1}{c^2} \partial_t E$ one gets $\nabla \times (\nabla \times E) = -\frac{1}{c^2} \partial_t^2 E$, and $\nabla \times (\nabla \times E) = \nabla (\nabla \cdot E) - \Delta E = -\Delta E$ because $\nabla \cdot E = 0$. The same steps give the equation for $B$. With charges and currents one gets the inhomogeneous electromagnetic wave equations instead. The converse is false: a pair of solutions of the wave equation need not solve Maxwell's equations, which also require zero divergence and the coupling between $E$ and $B$. In a linear, homogeneous and isotropic medium with permittivity $\varepsilon$ and permeability $\mu$, $c$ is replaced by $1/\sqrt{\varepsilon \mu}$.

Example: $u(x, y, z, t) = (0, \cos(kx - \omega t), 0)$ with $k > 0$ and $\omega = ck$ is a solution, since $\frac{1}{c^2} \partial_t^2 u_y = -\frac{\omega^2}{c^2} u_y = -k^2 u_y = \Delta u_y$.

Non-example: the same field with $\omega = 2ck$ is not a solution, since then $\frac{1}{c^2} \partial_t^2 u_y - \Delta u_y = -3 k^2 u_y \neq 0$.