I‘ve heard people talk about the overdetermination of the Maxwell equations, which are of course:
$$ \nabla \cdot E =\frac{\rho}{\epsilon}\\ $$ $$ \nabla \cdot B=0 \\ $$ $$ \nabla \times E=-\partial_t B \\ $$ $$ \nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \partial_t E $$
I assume by overdetermined they mean that by applying the divergence operator to the last two and using $\nabla \cdot (\nabla \times F)=0$ (for twice differentiable vector fields F) it follows that:
$$ -\partial_t \nabla \cdot B = 0 $$ and $$ \mu_0 \nabla \cdot J + \mu_0 \epsilon_0 \partial_t \nabla \cdot E =0 $$ Now using the continuity equation $\nabla \cdot J = -\partial_t \rho$: $$ 0=-\mu_0 \partial_t \rho+\mu_0 \epsilon_0 \partial_t \nabla \cdot E =\mu_0 \partial_t(\epsilon_0 \nabla \cdot E- \rho) $$
This implies that $\nabla \cdot B$ and $\epsilon_0 \nabla \cdot E-\rho$ are time independent, so if the initial conditions fulfill $\nabla \cdot B=0$ and $\nabla \cdot E=\frac{\rho}{\epsilon_0}$ then these identities hold for all times, therefore the first two Maxwell equations are redundant.
I don’t understand why we were allowed to use the continuity equation $\nabla \cdot J=-\partial_t \rho$ here, doesn’t this add more information on how B and E are linked? Have I therefore cheated in my derivation above?