Suppose \begin{alignat*}{2} &\nabla\cdot E=0\quad\quad\quad&\nabla\cdot H=0 \\ &E=-\frac{\partial H}{\partial t}&H=-\frac{\partial E}{\partial t} \end{alignat*} I need to show that \begin{gather*} \nabla^2E=\frac{\partial^2E}{\partial t^2}\quad\quad\quad\nabla^2H=\frac{\partial^2H}{\partial t^2} \end{gather*} but I am getting problem even in the signs. Can you help me?
So far, I've noticed that \begin{align*} \nabla\times(\nabla\times H)&=(\nabla\cdot H)\nabla-\nabla^2H \\ &=-\nabla^2H\quad\quad\quad(\nabla\cdot H=0) \\ \nabla\times\left(-\frac{\partial E}{\partial t}\right)&=−\nabla^2H \end{align*} but then I don't know how to conclude $$\nabla\times\left(-\frac{\partial E}{\partial t}\right)=\frac{\partial^2E}{\partial t^2}$$ I am not understanding the term $\frac{\partial E}{\partial t}$.
Let's consider two vector fields $E, H \colon \mathbb{R}^3 \times \mathbb{R} \to \mathbb{R}^3$ with $E = E(x, t)$ and $H = H(x,t)$. Here $x$ is a spatial variable and $t$ denotes time. Let's assume $E, H \in C^2(\mathbb{R}^3 \times \mathbb{R}; \mathbb{R}^3)$ (this assumption can be weakend, but that's not the point here).
Assume $(E, H)$ solves Maxwell's equations, i.e., for all $(x, t) \in \mathbb{R}^3 \times \mathbb{R}$ there holds \begin{align} \nabla_x \cdot E(x, t) &= 0 & \nabla_x \cdot H(x, t) &= 0 \\ \frac{\partial E}{\partial t}(x, t) &= \nabla_x \times H(x,t) & \frac{\partial H}{\partial t} &= - \nabla_x \times E(x,t) \end{align} Here $\nabla_x := (\partial_{x_1}, \partial_{x_2}, \partial_{x_3})$.
We need to show that $E = (E_1, E_2, E_3)$ (and $H$) solve the wave equation. By convention, we set \begin{equation} \Delta E(x, t) = \nabla^2 E(x, t) := ( \Delta E_1(x, t), \Delta E_2(x, t), \Delta E_3(x, t))\, . \end{equation}
Applying the curl-operator on $\partial_t H = - \nabla_x \times E$, we find \begin{equation} \nabla_x \times \bigg( \frac{\partial H}{\partial t}\bigg) = - \nabla_x \times \Big( \nabla_x \times E\Big) = - \Big( - \Delta E + \nabla_x (\nabla_x \cdot E) \Big)\, . \quad (1) \end{equation} As $\nabla_x \cdot E = 0$, interchanging $\nabla_x$ and $\partial_t$ in $(1)$, we get \begin{equation} \Delta E = \frac{\partial}{\partial t}\big(\nabla_x \times H \big) = \frac{\partial}{ \partial t} \bigg( \frac{\partial E}{\partial t} \bigg) = \frac{\partial^2 E}{\partial t^2}\, .\quad (2) \end{equation} Equation $(2)$ is the wave equation, as expected.
The proof for $H$ is similar.