In the Euler equations, there is a term of the form

$(\textbf{U}\cdot \nabla) \textbf{U}$

I've seen this expression replaced with the vector identity

$(\textbf{U}\cdot \nabla) \textbf{U} = \nabla(\frac{1}{2}\textbf{U}^2)+\nabla \times \textbf{U} \times \textbf{U}$

If I were to write out the terms for $(\textbf{U}\cdot \nabla) \textbf{U}$, say the $i^{th}$ component, that would give $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$. Apply a product rule to each term

$u\frac{\partial u}{\partial x} = \frac{\partial (uu)}{\partial x} - u\frac{\partial u}{\partial x}$

$v\frac{\partial u}{\partial y} = \frac{\partial (uv)}{\partial y} - u\frac{\partial v}{\partial y}$

$w\frac{\partial u}{\partial z} = \frac{\partial (uw)}{\partial z} - u\frac{\partial w}{\partial z}$

Collecting the terms gives

$\frac{\partial (uu)}{\partial x} + \frac{\partial (uv)}{\partial y}+\frac{\partial (uw)}{\partial z} -u(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}) $

Applying the same procedure to the $j^{th}$ and $k^{th}$ component gives

$\frac{\partial (vu)}{\partial x} + \frac{\partial (vv)}{\partial y}+\frac{\partial (vw)}{\partial z} -v(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}) $

$\frac{\partial (wu)}{\partial x} + \frac{\partial (wv)}{\partial y}+\frac{\partial (ww)}{\partial z} -w(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}) $

OK...so I have a full set of terms for the $i^{th}$, $j^{th}$, and $k^{th}$ components. I will collect them here

$\frac{\partial (uu)}{\partial x} + \frac{\partial (uv)}{\partial y}+\frac{\partial (uw)}{\partial z} -u(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}) $

$\frac{\partial (vu)}{\partial x} + \frac{\partial (vv)}{\partial y}+\frac{\partial (vw)}{\partial z} -v(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}) $

$\frac{\partial (wu)}{\partial x} + \frac{\partial (wv)}{\partial y}+\frac{\partial (ww)}{\partial z} -w(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}) $

Show how this set of equations is expressed in vector notation. With the aid of Einstein notation, show that the set of equations is equal to

$\textbf{e}_ju_i\frac{\partial u_i}{\partial x_j} + \textbf{e}_iu_j\frac{\partial u_i}{\partial x_j}-\textbf{e}_ju_i\frac{\partial u_i}{\partial x_j}$

(the identity proof is shown in @led23head's answer below).


closed as off-topic by Kyle Kanos, Jon Custer, sammy gerbil, Rococo, AccidentalFourierTransform Feb 16 '17 at 11:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, Jon Custer, Rococo, AccidentalFourierTransform
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Your calculation is correct. And the vector identity is also correct. However showing that one leads to another seems to require tedious index-juggling. $\endgroup$ – Deep Feb 13 '17 at 7:59
  • 1
    $\begingroup$ Write both the left- and the right-hand side of the identity in components, and show that the two sides are identical, after taking into account the continuity equation. That's really all that needs to be done here. Somewhat tedious plug-and-chug, I'll grant you, but it's certainly not rocket science. $\endgroup$ – Pirx Feb 13 '17 at 21:57
  • $\begingroup$ @Pirx I'm not asking about the identity, I know that's right. I think it might be how the index is interpreted. In the case $\nabla(\frac{1}{2}\textbf{U}^2)$, I think the index is $\nabla_i(\frac{1}{2}\textbf{U}_j^2)$, whereas when I apply the product rule, I think the indexing is $\nabla_j(\textbf{U}_i\textbf{U}_j)$, but again I'm not sure. In the end, yes they should be the same, but my intuition is telling me the indexing is different in each approach. I was hoping to confirm that was the case. I'll edit the question for further clarity. $\endgroup$ – ThatsRightJack Feb 13 '17 at 23:25
  • $\begingroup$ @ThatsRightJack Using your notation, $\nabla_i(\frac{1}{2}\mathbf{U}_j^2) = \mathbf{U}_j (\nabla_i \mathbf{U}_j$). This can be accomplished via either the product rule or the chain rule. Note that on both sides of this equation, there's an implied summation over $j$, but not $i$. $\endgroup$ – LedHead Feb 14 '17 at 8:08
  • $\begingroup$ @led23head I'm pretty sure $(\textbf{U}\cdot \nabla)\textbf{U}$ is conventionally written as $(\textbf{U}_j\cdot \nabla_j)\textbf{U}_i$ in index notation. In my edit, I asked how to write the two equalities in index notation. Can you help express those? $\endgroup$ – ThatsRightJack Feb 14 '17 at 21:26

I wrote this out using Einstein notation, except since it's an orthogonal space, I ignored upper indices. Anywhere you see a double index, there's an implied summation.

Here, $x_i$ are the coordinates, so $x_1=x$, $x_2=y$, $x_3=z$, and $\mathbf{e}_i$ are the corresponding unit vectors, $\mathbf{e}_1=\mathbf{\hat{x}}$, $\mathbf{e}_2=\mathbf{\hat{y}}$, $\mathbf{e}_3=\mathbf{\hat{z}}$. Using this notation, we write vectors and the $\nabla$ operator as

$\mathbf{U} = \mathbf{e}_i u_i $

$\nabla = \mathbf{e}_i \frac{\partial}{\partial x_i} $

Here are your quantities:

$(\mathbf{U} \cdot \nabla) \mathbf{U} = u_k (\mathbf{e}_k \cdot \mathbf{e}_j) \frac{\partial u_i}{\partial x_j} \mathbf{e}_i = u_k \delta_{jk} \frac{\partial u_i}{\partial x_j} \mathbf{e}_i = \mathbf{e}_i u_j \frac{\partial u_i}{\partial x_j}$

$\nabla( \frac{1}{2} \mathbf{U}^2 ) = \mathbf{e}_j \frac{\partial}{\partial x_j} \left( \frac{1}{2} u_i u_i \right) = \mathbf{e}_j u_i \frac{\partial u_i}{\partial x_j} $

$(\nabla \times \mathbf{U}) \times \mathbf{U} = \left( (\mathbf{e}_j \frac{\partial}{\partial x_j} )\times (u_k \mathbf{e}_k) \right) \times (u_m \mathbf{e}_m) = (\mathbf{e}_j\times \mathbf{e}_k \times \mathbf{e}_m) u_m \frac{\partial u_k}{\partial x_j} = \mathbf{e}_n \epsilon_{ijk}\epsilon_{imn} u_m \frac{\partial u_k}{\partial x_j} $

where $\epsilon_{ijk}$ is the Levi-Civita symbol. Using the following identity

$\epsilon_{ijk} \epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$

this simplifies to

$(\nabla \times \mathbf{U}) \times \mathbf{U} = \mathbf{e}_n \delta_{jm}\delta_{kn} u_m\frac{\partial u_k}{\partial x_j} - \mathbf{e}_n \delta_{jn}\delta_{km} u_m\frac{\partial u_k}{\partial x_j} = \mathbf{e}_k u_j\frac{\partial u_k}{\partial x_j} - \mathbf{e}_j u_k\frac{\partial u_k}{\partial x_j}$

Reindexing $k \rightarrow i$

$(\nabla \times \mathbf{U}) \times \mathbf{U} = \mathbf{e}_i u_j\frac{\partial u_i}{\partial x_j} - \mathbf{e}_j u_i\frac{\partial u_i}{\partial x_j}$

And now it's clear that

$(\mathbf{U} \cdot \nabla) \mathbf{U} = \nabla( \frac{1}{2} \mathbf{U}^2 ) + (\nabla \times \mathbf{U}) \times \mathbf{U}$

And your second identity is false. (OP has edited his original question to remove this second identity.)

Edit: After your last edit, I see that your very first step is wrong. Put in my notation, you're doing

$\mathbf{e}_i u_j \frac{\partial u_i}{\partial x_j} = \mathbf{e}_i\frac{\partial (u_i u_j)}{\partial x_j} - \mathbf{e}_i u_i \frac{\partial u_j}{\partial x_j}$

It's not invalid, it just doesn't take you in the correct direction. Note that none of the terms in the original section of my answer look like either of the terms on the right-hand side of this equation. For some reason you got stuck thinking that you should be able to go from there and make everything work out.

Rather, look at the single terms which represent $(\mathbf{U} \cdot \nabla)\mathbf{U}$ and $\nabla(\frac{1}{2}\mathbf{U}^2)$ and the two terms that represent $(\nabla \times \mathbf{U}) \times \mathbf{U}$, and see how they're related to each other. Really, most of the "magic" lies in manipulating $(\nabla \times \mathbf{U}) \times \mathbf{U}$ into the two terms given.

  • $\begingroup$ That is certainly helpful and more than I expected, so I want to thank you! When you say "your second identity is false", I'm assuming you're talking about #2 in my edit (I thought that might be wrong). Is there a "product rule" associated with $(\textbf{U}\cdot \nabla)\textbf{U}$? If so, what does it look like in Einstein notation? I'm going to add a second "edit" to build on what you've provided. $\endgroup$ – ThatsRightJack Feb 15 '17 at 1:50
  • $\begingroup$ @ThatsRightJack Pretty much any identity you want can be found here. The identity that I've worked with is essentially the product-rule identity $\nabla(\mathbf{A} \cdot \mathbf{B}) = (\mathbf{A} \cdot \nabla ) \mathbf{B} + (\mathbf{B} \cdot \nabla ) \mathbf{A} + \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})$. $\endgroup$ – LedHead Feb 15 '17 at 4:23
  • $\begingroup$ Again, sorry for the confusion. I added an edit 3. I'm trying to compare the two different approaches in presenting the Euler equations. In one case, you can use the identity (which you've been helpful in showing) and the other approach is to apply a product rule to each term in the set of equations. I believe both approaches give the same answer, which is what I'm trying to realize $\endgroup$ – ThatsRightJack Feb 15 '17 at 5:09
  • $\begingroup$ @ThatsRightJack The identity is derived from applying the product rule to $\nabla(\mathbf{A}\cdot\mathbf{B})$, so they're the same thing. Honestly, it's difficult to understand what you're trying to ask because (1) your question is overly verbose and meandering, (2) your math is imprecise, and (3) your notation is inconsistent (you use four separate notations for the same concepts). I think if you were more diligent about making your question precise, you would find the answer yourself. $\endgroup$ – LedHead Feb 15 '17 at 5:21
  • $\begingroup$ I couldn't agree more. I'm new to vector notation and know very little on Einstein notation. I went back and tried to clean up the question and remove all the clutter. Would you mind taking one more look at it. $\endgroup$ – ThatsRightJack Feb 15 '17 at 6:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.