# Maxwell's equations and the wave equation

Let $E=(E^1, E^2, E^3)$ be the electric field and $B=(B^1, B^2,B^3)$ be the magnetic field satifying the Maxwell's equations: $\left\{ \begin{array}{ll} E_t= \text{curl} (B), \\ B_t=-\text{curl}(E),\\ \nabla \cdot B=\nabla \cdot E=0. \end{array} \right.$ Show that the components of the electric and magnetic fields satisfy the wave equation, i.e. $u_{tt}-\Delta u=0$ for $u=E^{i}$ or  $B^i$, $i=1,2,3.$

Calculate $\text{curl}(\text{curl } \textbf{E}) = \text{curl}(-\textbf{B}_t)$ $= \left( -\frac{\partial^2 B^3}{\partial y \partial t} + \frac{\partial^2 B^2}{\partial z \partial t}, \frac{\partial^2 B^3}{\partial x \partial t} - \frac{\partial^2 B^1}{\partial z \partial t}, -\frac{\partial^2 B^2}{\partial x \partial t} + \frac{\partial B^1}{\partial y \partial t} \right)$ $= -\frac{\partial}{\partial t} \text{curl } \textbf{B} -\frac{\partial}{\partial t} \textbf{E}_t = -\frac{\partial^2 \textbf{E}}{\partial t^2}.$ On the other hand $$\text{curl}(\text{curl } \textbf{E}) = \nabla (\text{div } \textbf{E})-\Delta \textbf{E} = -\Delta \textbf{E}.$$ Therefore, $$-\Delta \textbf{E}=-\frac{\partial^2 \textbf{E}}{\partial t^2} \Rightarrow \frac{\partial^2 \textbf{E}}{\partial t^2}-\Delta \textbf{E}=0,$$ which means $u=E^i$ solves $u_{tt}-\Delta u=0$ for $i=1,2,3$. Similarly, we calculate $\text{curl}(\text{curl } \textbf{B}) = \text{curl}(\textbf{E}_t)$ $= \left( \frac{\partial^2 E^3}{\partial y \partial t} - \frac{\partial^2 E^2}{\partial z \partial t}, -\frac{\partial^2 E^3}{\partial x \partial t} + \frac{\partial^2 E^1}{\partial z \partial t}, \frac{\partial^2 E^2}{\partial x \partial t} - \frac{\partial E^1}{\partial y \partial t} \right)$ $= \frac{\partial}{\partial t} \text{curl } \textbf{E} = \frac{\partial}{\partial t} \left(-\textbf{B}_t \right) = -\frac{\partial^2 \textbf{B}}{\partial t^2}$ Alternatively, $\text{curl}(\text{curl } \textbf{B}) = \nabla (\text{div } \textbf{B})-\Delta \textbf{B} = -\Delta \textbf{B}$. Therefore, $$-\nabla^2 \textbf{B}=-\frac{\partial^2 \textbf{B}}{\partial t^2} \Rightarrow \frac{\partial^2 \textbf{B}}{\partial t^2}-\Delta \textbf{B}=0,$$ which means $u=B^i$ solves $u_{tt}-\Delta u=0$ for $i=1,2,3$.