$(\mathbf{v} \cdot \nabla) \mathbf{v}$ Notation in Euler equation (fluid dynamics)

Question

Main question: Can I interpret $(\mathbf{v} \cdot \nabla) \mathbf{v}$ as $\nabla v \cdot v$?

My question comes from the wikipedia page on euler equations. I captured the relevant parts below:

The part that confuses me is the notation

$${\partial\mathbf{u} \over \partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} = -\nabla w + \mathbf{g}$$

When I tried to derive the equation myself, using the chain rule, I got ${D\mathbf v(x,t) \over Dt}= \nabla v \cdot x'(t) + \frac{\partial}{\partial t} v(x,t) = \nabla v \cdot v + \frac{\partial}{\partial t} v(x,t) $.

Assuming I did not make mistakes, can I interpret $(\mathbf{v} \cdot \nabla) \mathbf{v}$ as $\nabla v \cdot v$?

Answer 1

To answer your main question, I would say “sort of.” The ambiguity when you write $\nabla v\cdot v$ is it doesn’t specify what part of $\nabla v$ the dot product is hitting. $\nabla v$ has two different “vector” parts (specifically it is a rank 2 tensor): the one from $\nabla$ and the one from $v$. It is important that the $v$ is being dotted with the $\nabla$, not the other $v$. That’s why it’s often written $(v\cdot\nabla)v$. The least ambiguous way to write it is in coordinates: $$(v\cdot\nabla v)_i=\sum_{i=1}^dv_j\partial_jv_i$$ if you are in $d$-dimensional space. The intuitive way to think about this operation is you are taking the directional derivative in the direction $v$ of each coordinate of $v_i$.

Answer 2

$(\mathbb{v} \cdot \nabla )$ is the operator

$ v_{x} \frac{\partial}{\partial x} + v_{y} \frac{\partial }{\partial y} + v_{z} \frac{\partial}{\partial z}$

When this operator is applied to the vector $\mathbb{v}$, you end up with

$(\mathbb{v} \cdot \nabla)\mathbb{v}=v_{x} \frac{\partial \mathbb{v}}{\partial x} + v_{y} \frac{\partial \mathbb{v} }{\partial y} + v_{z} \frac{\partial \mathbb{v}}{\partial z}$

Note that $\frac{\partial \mathbb{v}}{\partial x}$ is a vector whose components are the partial derivatives of $v_{x}$, $v_{y}$, and $v_{z}$ with respect to $x$.

Thus we end up with

$(\mathbb{v} \cdot \nabla)\mathbb{v}= \left[ \begin{array}{c} v_{x} \frac{\partial v_{x}}{\partial x}+ v_{y} \frac{\partial v_{x}}{\partial y} + v_{z} \frac{\partial v_{x}}{\partial z} \\ v_{x} \frac{\partial v_{y}}{\partial x}+ v_{y} \frac{\partial v_{y}}{\partial y} + v_{z} \frac{\partial v_{y}}{\partial z} \\ v_{x} \frac{\partial v_{z}}{\partial x}+ v_{y} \frac{\partial v_{z}}{\partial y} + v_{z} \frac{\partial v_{z}}{\partial z} \\ \end{array} \right]$

Answer 3

If $v=(v_{x},v_{y},v_{z}), u=(u_{x},u_{y},u_{z})$, the usual inner product of $v$ and $u$ is $v\cdot u = v_{x}u_{x}+v_{y}u_{y}+v_{z}u_{z}$. Now, the notation $v\cdot \nabla$ means that you're treating $\nabla$ as a vector (operator) $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$, so that: $$v\cdot \nabla = v_{x}\frac{\partial}{\partial x} + v_{y}\frac{\partial}{\partial y} + v_{z}\frac{\partial}{\partial z}$$