I want to see the following identity that is used in the proof of the first version of Bernoulli's theorem without using index notation
$(\mathbf{u} \cdot \mathbf{\nabla})\mathbf{u} = (\nabla \times \mathbf{u}) \times \mathbf{u} + \nabla \left(\frac{\mathbf{|u|}^2}{2} \right)$
To do that I apply the identity $\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) – \mathbf{C}(\mathbf{A} \cdot \mathbf{B})$ to $\mathbf{u} \times (\nabla \times \mathbf{u})$, so
$(\nabla \times \mathbf{u}) \times \mathbf{u} = – \mathbf{u} \times (\nabla \times \mathbf{u}) = – [\nabla (\mathbf{u} \cdot \mathbf{u}) – \mathbf{u}(\mathbf{u} \cdot \nabla)] = (\mathbf{u} \cdot \nabla) \mathbf{u} – \nabla (\mathbf{|u|}^2)$
Then
$(\mathbf{u} \cdot \nabla)\mathbf{u} = (\nabla \times \mathbf{u}) \times \mathbf{u} + \nabla (\mathbf{|u|}^2)$
Clearly a factor 1/2 is missing in the last term. Where is the error in this reasoning?
Best Answer
You need to keep track of what's being differentiated. Let us write $$ (\dot\nabla\times\dot{\mathbf u})\times\mathbf u $$ so that it's clear that $\dot\nabla$ is differentiating only $\dot{\mathbf u}$. Now proceed as you did to get $$ (\dot\nabla\times\dot{\mathbf u})\times\mathbf u = -\mathbf u\times(\dot\nabla\times\dot{\mathbf u}) = \dot{\mathbf u}(\mathbf u\cdot\dot\nabla) - \dot\nabla(\mathbf u\cdot\dot{\mathbf u}). $$ The first term we can rearrange into standard differentiate-to-the-right notation: $$ \dot{\mathbf u}(\mathbf u\cdot\dot\nabla) = (\mathbf u\cdot\nabla)\mathbf u. $$ For the second term, we see $$ \nabla(|\mathbf u|^2) = \nabla(\mathbf u\cdot\mathbf u) = \dot\nabla(\dot{\mathbf u}\cdot\dot{\mathbf u}) = \dot\nabla(\dot{\mathbf u}\cdot\mathbf u) + \dot\nabla(\mathbf u\cdot\dot{\mathbf u}) = 2\dot\nabla(\mathbf u\cdot\dot{\mathbf u}). $$ Thus, all together $$ (\nabla\times\mathbf u)\times\mathbf u = \dot{\mathbf u}(\mathbf u\cdot\dot\nabla) - \dot\nabla(\mathbf u\cdot\dot{\mathbf u}) = (\mathbf u\cdot\nabla)\mathbf u - \frac12 \nabla(|\mathbf u|^2). $$