Maxwell's equations, verification Equations f Maxwell:
$$
\begin{cases}
\nabla \cdot \bf E=0\\
\nabla \cdot \bf H = 0\\
\nabla \wedge \bf E =-μ_o \frac{\partial \bf H}{\partial t}\\
\nabla \wedge \bf H =ε_ο \frac{\partial \bf E}{\partial t}\\
\end{cases}
$$
I want to show that if $\bf u$ verifies the equations of waves:
$$
\begin{cases}
\frac{\partial^2 \bf E}{\partial t^2}\ =\ (C_0)^2Δ\bf E\\
\frac{\partial^2 \bf H}{\partial t^2}\ =\ (C_0)^2Δ\bf H\\
\end{cases}
$$
where $$(C_0)=\frac{1}{\sqrt{μ_οε_ο}}$$
then 
$$
\begin{cases}
\bf E=a\nabla\wedge(\nabla\wedge\bf u )\\
\bf H=aε_ο\nabla\wedge\left(\frac{\partial\bf u}{\partial t}\right)
\end{cases}
$$
and 
$$
\begin{cases}
\bf E= -βμ_ο\nabla\wedge\left(\frac{\partial \bf u}{\partial t}\right)\\
\bf H= β\nabla\wedge(\nabla\wedge \bf u)\\
\end{cases}
$$
are two solutions of the Maxwell equations, no matter the value of the constants $a, β$
 A: We recall the general identities from vector calculus
$\nabla \times (\nabla \times \mathbf v) = \nabla (\nabla \cdot \mathbf v) - \nabla^2 \mathbf v, \tag 1$
$\nabla \cdot (\nabla \times \mathbf v) = 0, \tag 2$
and we are given that $\mathbf u$ satisfies
$\mathbf u_{tt} = \dfrac{\partial^2 \mathbf u}{\partial t^2} = C_0^2 \nabla^2 \mathbf u. \tag 3$
We first use (1) and (2) to show that
$\mathbf E = \mathbf a  \nabla \times (\nabla \times \mathbf u), \tag 3$
$\mathbf H = \mathbf a \mathbf \varepsilon_0 \nabla \times \mathbf u_t \tag 4$
satisfy the Maxwell system.  It is evident via (2) that
$\nabla \cdot \mathbf E = 0 \tag 5$
and
$\nabla \cdot \mathbf H = 0; \tag 6$
also,
$\nabla \times \mathbf H = \mathbf a \varepsilon_0 \nabla \times (\nabla \times \mathbf u_t) = \mathbf a \varepsilon_0 (\nabla \times (\nabla \times \mathbf u))_t = \varepsilon_0 \mathbf E_t; \tag 7$
furthermore,
$\nabla \times \mathbf E = \mathbf a \nabla \times (\nabla \times (\nabla \times \mathbf u)) = \mathbf a \nabla(\nabla \cdot (\nabla \times \mathbf u))- \mathbf a \nabla^2(\nabla \times \mathbf u)$
$= -\mathbf a \nabla^2(\nabla \times \mathbf u) = -\mathbf a \nabla \times \nabla^2 \mathbf u = -\mathbf a \nabla \times C_0^{-2}u_{tt}= -\mathbf a \nabla \times \mu_0 \varepsilon_0 \mathbf u_{tt}$$
= -\mathbf a \mu_0 \varepsilon_0 \nabla \times \mathbf u_{tt} = -\mathbf a \mu_0 \varepsilon_0 (\nabla \times \mathbf u_t)_t = -\mu_0 \mathbf H_t.  \tag 8$
Next, we use (1) and (2) again to show that
$\mathbf E = -\beta \mu_0 \nabla \times u_t, \tag 9$
$\mathbf H = \beta \nabla \times (\nabla \times u), \tag{10}$
are aslo a Maxwellian pair of fields:  as before, (2) yields (5) and (6); from (9), 
$\nabla \times \mathbf E = -\beta \mu_0 \nabla \times (\nabla \times u_t) = -\beta \mu_0 (\nabla \times (\nabla \times u))_t = -\mu_0 \mathbf H_t; \tag{11}$
from (10),
$\nabla \times \mathbf H = \beta \nabla \times (\nabla \times (\nabla \times u)) = \beta \nabla(\nabla \cdot (\nabla \times \mathbf u)) - \beta \nabla^2(\nabla \times \mathbf u)$
$= -\beta \nabla \times \nabla^2 \mathbf u = -\beta \nabla \times C_0^{-2} u_{tt} = -\beta \mu_0 \varepsilon_0 \nabla \times u_{tt} = -\beta \mu_0 \varepsilon_0 (\nabla \times u_t)_t = \varepsilon_0 \mathbf E_t. \tag{12}$
A: There are two general equations that will help you here. One is the differential version of the rule that $W \cdot W \times V = 0$ (the cross-product of two vectors is perpendicular to both of them). The differential version is $$\nabla \cdot \nabla \times V = 0$$ The other is the "BAC-CAB" rule: $A\times (B \times C) = B(A\cdot C) - C(A \cdot B)$. The differential version is:
$$\nabla \times \nabla \times V = \nabla(\nabla \cdot V) - \nabla \cdot \nabla V = \nabla(\nabla \cdot V) - \Delta V$$
