Core idea

For smooth divergence-free velocity and constant ν\nu, the is equivalent to

tu+(uu+pIν(u+(u)T))=f.\partial_tu+\nabla\cdot \bigl(u\otimes u+pI-\nu(\nabla u+(\nabla u)^{\mathsf T})\bigr)=f.

The divergence of a matrix is taken row by row: (J)i=jjJij(\nabla\cdot J)_i=\sum_j\partial_jJ_{ij}.

Verification

The product rule yields

(uu)=(u)u+u(u),\nabla\cdot(u\otimes u)=(u\cdot\nabla)u+u(\nabla\cdot u),

while equality of mixed derivatives gives

(u+(u)T)=Δu+(u).\nabla\cdot(\nabla u+(\nabla u)^{\mathsf T}) =\Delta u+\nabla(\nabla\cdot u).

Both extra terms vanish under incompressibility. Also (pI)=p\nabla\cdot(pI)=\nabla p.

Simpler equivalent flux

For divergence-free velocity, uu+pIνuu\otimes u+pI-\nu\nabla u has the same divergence as the symmetric-stress flux. The flux tensors differ, but their local momentum equations coincide under this constraint.