$$\left\{ \begin{align} \text{div } \textbf{E} &=0, \\ \text{div } \textbf{H} & =0, \\ \text{curl } \textbf{E} & = \dfrac{-1}{c} \dfrac{\partial\textbf{H}}{\partial t}, \\\text{curl } \textbf{H} & =\dfrac{1}{c} \dfrac{\partial\textbf{E}}{\partial t}, \end{align} \right.$$ where $c$ is the speed of light.
Using the equations above, and the assumption that $\textbf{E}$ and $\textbf{H}$ are infinitely differentiable, show that the following identity holds true.
$$\nabla \text{x}(\nabla \text{x} \textbf{E})= \dfrac{-1}{c}\dfrac{\partial^2\textbf{E}}{\partial t^2}$$
$$\nabla \text{x} \textbf{E} =\dfrac{-1}{c}\dfrac{\partial\textbf{H}}{\partial t}$$
$$\nabla \text{x} \left(\dfrac{-1}{c}\dfrac{\partial\textbf{H}}{\partial t}\right) = \dfrac{-1}{c} \left(\nabla \text{x} \dfrac{\partial \textbf{H}}{\partial t}\right) $$
$$=\dfrac{-1}{c} \left( \dfrac{1}{c} \dfrac{\partial^2\textbf{E}}{\partial t^2} \right)$$
$$=\dfrac{-1}{c^2} \dfrac{\partial^2\textbf{E}}{\partial t^2}$$
Could someone please correct this if it's wrong? I don't know if taking the curl of a derivative equals the derivative of the curl...if not, I don't know how else to prove this.
Thank you.
Yes, the space and time derivatives commute so you can exchange curl and $\partial/\partial t$. This follows from Schwarz' theorem which tells you that if a function has continuous second partial derivatives at any point (which $\mathbf{E}$ and $\mathbf{H}$ do since they are infinitely differentiable) then you can exchange the order of partial differentiation.