Vector Calculus Operator $\vec{u} \cdot \nabla$

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!

• 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 $u$. – MathNewbie Jun 23 '15 at 5:32

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.