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I just want to double check on this operator and it's properties. It pops up in fluid mechanics often and I just want to be sure about my understanding:

1) $$(\vec u \cdot \nabla)\vec u$$

Is this equal to

$$ (u_x \frac{\partial u_x}{\partial x} + u_y \frac{\partial u_x}{\partial y}) \vec i + (u_x \frac{\partial u_y}{\partial x} + u_y \frac{\partial u_y}{\partial y}) \vec j$$

The part that gives me confusion is when the operator $(\vec u \cdot \nabla)$ acts on a vector. So the result would be a vector if I'm not mistaken?

2) How does the operator behave algebraically? For example, say I have a scalar A and a vector $\vec B$ in the following expression:

$$(\vec u \cdot \nabla)\vec u + \nabla[A(\vec u \cdot \nabla)(\nabla\cdot\vec B)]$$

Can these terms be combined and simplified to an expression which the operator $(\vec u \cdot \nabla)$ acts on? Basically, I want the operator to "factor out" and act on an equivalent expression in the form $(\vec u \cdot \nabla)(terms)$, if that makes sense. Thanks for your insight!

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  • $\begingroup$ I've seen the notation $\vec{u} \cdot \nabla$, though I'm not 100% sure on it. $u$ is a scalar field and that above expression is the divergence of $u$. $\endgroup$ – MathNewbie Jun 23 '15 at 5:32
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This operator is a scalar operator meaning it can act on a vector or a scalar. You have demonstrated this by having it act on both $\vec u$ (a vector) and $\nabla\cdot \vec B$, a scalar. But, because $\vec u\cdot \nabla$ is a differential operator, it does not commute directly. However, we can apply the product rule with the gradient to almost get the result you want: \begin{align*} (\vec u\cdot\nabla)\vec u+&\nabla\left[A(\vec u\cdot\nabla)(\nabla\cdot \vec B)\right]=(\vec u\cdot\nabla)\vec u+(\vec u\cdot\nabla)(\nabla\cdot\vec B)(\nabla A)+(\vec u\cdot\nabla)(\nabla^2\vec B)A\\ &\quad\quad\quad+\left[\nabla(\vec u\cdot\nabla)\right](\nabla\cdot\vec B)A\\ &=(\vec u\cdot\nabla)\left(\vec u+(\nabla\cdot\vec B)(\nabla A)+(\nabla^2\vec B)A\right)+\left[\nabla(\vec u\cdot\nabla)\right](\nabla\cdot\vec B)A\\ \end{align*} It is possible to distribute the gradient in the last term, but the expression gets exceedingly complicated, especially if we work in more than three dimensions.

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  • $\begingroup$ @ Alex When you say " it does not commute in the way you describe unless...", does is commute in another way (as opposed to my poor example)? I've edited my example...I know you can't distribute out the differential operator. $\endgroup$ – ThatsRightJack Jun 23 '15 at 7:29
  • $\begingroup$ @ThatsRightJack I have edited my answer to more directly (I think) address your question. Does this answer it for you? $\endgroup$ – Alex S Jun 23 '15 at 13:58
  • $\begingroup$ @ Alex Ahh yes...good old product rule! That atleast gives me hope. I will take a look at more today to see if that leads to somewhere. I'll leave this question open for the duration to see if any other suggestions arise, but I appreciate your help! $\endgroup$ – ThatsRightJack Jun 23 '15 at 20:05

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