(Solved): \[ \nabla \times((\mathbf{u} \cdot \nabla) \mathbf{u})=\mathbf{u} \cdot \nabla(\nabla \times \math ...

\[ \nabla \times((\mathbf{u} \cdot \nabla) \mathbf{u})=\mathbf{u} \cdot \nabla(\nabla \times \mathbf{u})+(\nabla \cdot \mathbf{u})(\nabla \times \mathbf{u})-(\nabla \times \mathbf{u}) \cdot \nabla \mathbf{u} \] (a) What is the order of the equation? (b) Prove the identity. Useful identities: \[ \begin{array}{l} \nabla \times(\nabla \phi)=0 \text { (Curl of a gradient is zero) } \\ \nabla \cdot(\nabla \times \vec{a})=0 \text { (Divergence of curl is zero) } \\ \vec{a} \times(\nabla \times \vec{a})=\nabla(\vec{a} \cdot \vec{a} / 2)-\vec{a} \cdot \nabla \vec{a} \\ \left.\left.\varepsilon_{i j k} \varepsilon_{k l m}=\delta_{i l} \delta_{j m}-\delta_{i m} \delta_{j l} \text { (Epsilon - Delta relation; FLOI }=\operatorname{first}_{(i l)}\right) \operatorname{last}\left({ }_{j m}\right) \text {-outer }\left({ }_{i m}\right) \operatorname{inner}\left({ }_{j l}\right)\right) \end{array} \]