= Solution
The <magnetic second-moment tensor> is $M_{ij}=B_iB_j$. This is the magnetic dyadic called the Maxwell tensor in the paper; the isotropic part of the usual <Maxwell stress tensor> is already contained in the <magnetohydrodynamic total pressure>. The <solenoidal magnetic-field constraint> gives
$$
\partial_jM_{ij}
=B_j\partial_jB_i+B_i\partial_jB_j
=B_j\partial_jB_i.
$$
Consequently the <magnetohydrodynamic momentum equation> in part <1/a> becomes $D_tu_i=-\partial_i p+\partial_jM_{ij}$, where $D_t=\partial_t+u_n\partial_n$ is the <material derivative>. Applying its product rule and the <ideal magnetohydrodynamic induction equation> gives
$$
\begin{aligned}
D_tM_{ij}
&=(D_tB_i)B_j+B_i(D_tB_j)\\
&=B_nB_j\partial_nu_i+B_iB_n\partial_nu_j
=\boxed{M_{nj}\partial_nu_i+M_{in}\partial_nu_j}.
\end{aligned}
$$
Thus the evolution uses only $\mathbf u,M,p$, together with $\partial_i u_i=0$. To determine $p$, take the <divergence> of the <magnetohydrodynamic momentum equation>. Since $\partial_t\partial_i u_i=0$ and $u_j\partial_j\partial_i u_i=0$,
$$
\boxed{\nabla^2p=\partial_i\partial_jM_{ij}
-(\partial_i u_j)(\partial_j u_i).}
$$
This is the <pressure Poisson equation for incompressible flow>, with <boundary conditions> supplied by the imposed normal <acceleration> and normal <stress>, or by periodicity or decay at infinity as appropriate. For example, the normal component of the <magnetohydrodynamic momentum equation> requires $\mathbf n\cdot\nabla p=\mathbf n\cdot(\nabla\cdot M-D_t\mathbf u)$ at a prescribed boundary. The <pressure> is determined up to a spatially constant function of <time>.
For equivalence to one resolved <magnetic field>, the initial $M$ must actually be $\mathbf B\mathbf B^T$ for a <divergence-free vector field>. In particular it is a symmetric <positive semidefinite matrix> of <matrix rank> at most one. The tensor equations also admit more general stresses; the averaged $M$ in part <1/d> need not have <matrix rank> one.
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