I am struggling with the following proof. Let u be a vector field We need to use suffix notation to prove $\mathbf{u}\times (\nabla \times u)=\frac{1}{2}\nabla (\mathbf{u} . \mathbf{u}) - (\mathbf{u}.\nabla)\mathbf{u}$
I managed to find the second term of the RHS but am struggling to see where this $\frac{1}{2}$ of the first term of the LHS.
Thanks
On
Observe, \begin{align} \bf{u} \times (\nabla \times \bf{u}) &= (u_h \hat{e}_h) \times \left(\epsilon_{ijk} \frac{\partial u_j}{\partial x_i} \hat{e}_k \right)\\ &= \epsilon_{ijk} u_h \left(\frac{\partial u_j}{\partial x_i}\right) (\hat{e}_h \times \hat{e}_k)\\ &= \epsilon_{ijk} u_h \left(\frac{\partial u_j}{\partial x_i}\right)(\epsilon_{hkl} \hat{e}_l)\\ &= (\epsilon_{ijk} \epsilon_{lhk}) u_h \left(\frac{\partial u_j}{\partial x_i}\right) \hat{e}_l\\ &=(\delta_{il}\delta_{jh} - \delta_{ih}\delta_{jl}) u_h \left(\frac{\partial u_j}{\partial x_i}\right) \hat{e}_l\\ &= \left(u_j \left(\frac{\partial u_j}{\partial x_l}\right) - u_i \left(\frac{\partial u_l}{\partial x_i}\right)\right) \hat{e}_l \end{align} From which you need to verify that $\frac{1}{2} \nabla(\bf{u} \cdot \bf{u})= u_j \left(\frac{\partial u_j}{\partial x_l}\right) \hat{e}_l$ and $(\bf{u} \cdot \nabla) \bf{u} = u_i \left(\frac{\partial u_l}{\partial x_i}\right) \hat{e}_l$. Observe, \begin{align} \frac{1}{2} \nabla(\bf{u} \cdot \bf{u}) &= \frac{1}{2} \nabla (u_j u_j)\\ &= \frac{1}{2} \left( \frac{\partial (u_{jj}^2)}{\partial x_l} \hat{e}_l \right)\\ &= \frac{1}{2} \left( 2u_j \frac{\partial u_j}{\partial x_l} \hat{e}_l \right)\\ &= u_j \frac{\partial u_j}{\partial x_l} \hat{e}_l \end{align} and \begin{align} (\bf{u} \cdot \nabla) \bf{u} &= \left( u_i \frac{\partial}{\partial x_i} \right)(u_l \hat{e_l})\\ &= u_i \frac{\partial u_l}{\partial x_i} \hat{e}_l \end{align} Which completes the proof.
Hint: $\mathbf{u}.\mathbf{u}=\sum_{i=1}^{n}u_i^2$ and $\frac{\partial \mathbf{u}.\mathbf{u}}{\partial u_i}=2u_i$