1

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?

Franco Medina
  • 13

2 Answers2

2

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 $$ (\nabla\times\mathbf u)\times\mathbf u = \dot{\mathbf u}(\mathbf u\cdot\dot\nabla) - \dot\nabla(\mathbf u\cdot\dot{\mathbf u}) = (\mathbf u\cdot\nabla)\mathbf u - \frac12 \nabla(|\mathbf u|^2). $$

Nicholas Todoroff
  • 10,107
0

Hmmm. I remember using this formula you showed here though now I realised that one should not use it. Instead when solving this formula one should proof it using vector identity for gradients: $\nabla(\vec{a}\cdot\vec{b})=(\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+\vec{a}\times(\nabla\times\vec{b})+\vec{b}\times(\nabla\times\vec{a}),\quad\vec{a}=\vec{u},\quad\vec{b}=\vec{u}$ $\Rightarrow\nabla(\vec{u}\cdot\vec{u})=(\vec{u}\cdot\nabla)\vec{u}+(\vec{u}\cdot\nabla)\vec{u}+\vec{u}\times(\nabla\times\vec{u})+\vec{u}\times(\nabla\times\vec{u})$

Now simplifying this: $\nabla(\vec{u}\cdot\vec{u})=2(\vec{u}\cdot\nabla)\vec{u}+2\vec{u}\times(\nabla\times\vec{u})$

rearranging the values: $2\vec{u}\times(\nabla\times\vec{u})=\nabla(\vec{u}\cdot\vec{u})-2(\vec{u}\cdot\nabla)\vec{u}$

dividing both sides by two: $\vec{u}\times(\nabla\times\vec{u})=\frac{1}{2}\nabla(\vec{u}\cdot\vec{u})-(\vec{u}\cdot\nabla)\vec{u}\quad\Vert\vec{u}\cdot\vec{u}=|u|^{2}$

Thus we get: $\vec{u}\times(\nabla\times\vec{u})=\frac{1}{2}\nabla(u^{2})-(\vec{u}\cdot\nabla)\vec{u}$

Then using the cross product antisymmetry identity $\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}$

$\Rightarrow(\nabla\times\vec{u})\times\vec{u}=(\vec{u}\cdot\nabla)\vec{u}-\frac{1}{2}\nabla(u^{2})\quad\square$

Teemu Mäki-Maunus
  • 11