Poynting relationship
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Problem:

Given
\begin{equation}\label{eqn:poynting:20}
\spacegrad \cross \BE
= -\BM_i – \PD{t}{\BB},
\end{equation}

and
\begin{equation}\label{eqn:poynting:40}
\spacegrad \cross \BH
= \BJ_i + \BJ_c + \PD{t}{\BD},
\end{equation}

expand the divergence of \( \BE \cross \BH \) to find the form of the Poynting theorem.

Solution:

First we need the chain rule for of this sort of divergence. Using primes to indicate the scope of the gradient operation

\begin{equation}\label{eqn:poynting:60}
\begin{aligned}
\spacegrad \cdot \lr{ \BE \cross \BH }
&=
\spacegrad’ \cdot \lr{ \BE’ \cross \BH }
–
\spacegrad’ \cdot \lr{ \BH’ \cross \BE } \\
&=
\BH \cdot \lr{ \spacegrad’ \cross \BE’ }
–
\BH \cdot \lr{ \spacegrad’ \cross \BH’ } \\
&=
\BH \cdot \lr{ \spacegrad \cross \BE }
–
\BE \cdot \lr{ \spacegrad \cross \BH }.
\end{aligned}
\end{equation}

In the second step, cyclic permutation of the triple product was used.
This checks against the inside front cover of Jackson [1]. Now we can plug in the Maxwell equation cross products.

\begin{equation}\label{eqn:poynting:80}
\begin{aligned}
\spacegrad \cdot \lr{ \BE \cross \BH }
&=
\BH \cdot \lr{ -\BM_i – \PD{t}{\BB} }
–
\BE \cdot \lr{ \BJ_i + \BJ_c + \PD{t}{\BD} } \\
&=
-\BH \cdot \BM_i
-\mu \BH \cdot \PD{t}{\BH}
–
\BE \cdot \BJ_i
–
\BE \cdot \BJ_c
–
\epsilon \BE \cdot \PD{t}{\BE},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:poynting:120}
\boxed{
0
=
\spacegrad \cdot \lr{ \BE \cross \BH }
+ \frac{\epsilon}{2} \PD{t}{} \Abs{ \BE }^2
+ \frac{\mu}{2} \PD{t}{} \Abs{ \BH }^2
+ \BH \cdot \BM_i
+ \BE \cdot \BJ_i
+ \sigma \Abs{\BE}^2.
}
\end{equation}

References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.