$$ \begin{align} (\bm{u} \cdot \nabla) \bm{u} = \nabla \left( \frac{|\bm{u}|^2}{2} \right) -\bm{u} \times (\nabla \times \bm{u}) \end{align} $$
を示したい．そのために，上式を整理して
$$ \begin{align} \bm{u} \times (\nabla \times \bm{u}) = \nabla \left( \frac{|\bm{u}|^2}{2} \right)-(\bm{u} \cdot \nabla) \bm{u} \end{align} $$
を示す．まず，$\bm{v} = \bm{u} \times (\nabla \times \bm{u})$とおくと
$$ \begin{align} v_i &= \varepsilon_{ijk} u_j (\nabla \times \bm{u})k\\ &= \varepsilon{ijk} u_j \varepsilon_{klm} \frac{\partial}{\partial x_l} u_m\\ &= \varepsilon_{kij} \varepsilon_{klm} u_j \frac{\partial u_m}{\partial x_l}\\ &= (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) u_j \frac{\partial u_m}{\partial x_l}\\ &= u_j \frac{\partial u_j}{\partial x_i} -u_j \frac{\partial u_i}{\partial x_j}\\ &= \frac{\partial}{\partial x_i} \left( \frac{u_j^2}{2} \right) -u_j \frac{\partial u_i}{\partial x_j}\\ \end{align} $$
と変形できる．これは(2)である．