Index notation with Navier-Stokes equations

Question

This is an index-notation question rather then the NS one:

For incompressible flow and Newtonian fluid, the continuity equation is denoted with: $$\frac{\partial u_i}{\partial x_i} = 0, $$ which means ${\rm div} u = 0$. Which is fine.

But then in the momentum equation, the divergence in the convection is described via $$ \frac{\partial u_i}{\partial x_j}u_i, $$ Which means ${\rm div} uu$. Which is also fine. But why is $$ {\rm div}\neq\frac{\partial u_i}{\partial x_i}? $$

So I would be very grateful if somebody could explain this simply (meaning-using simple words and things). Or am I a lost cause?

Answer 1

The divergence is a vector operator. This simply means that it is a differential operator that acts only on vectors. In this particular case, $$ {\rm div}\equiv\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z} $$ Which, assuming an implicit summation, $$ {\rm div}\equiv\frac{\partial}{\partial x_i}\hat{x}_i\tag{1} $$ Since the velocity field is a vector, $$ \mathbf{u}:=(u,\,v,\,w)=u\hat{x}+v\hat{y}+w\hat{z}\tag{2} $$ then the divergence of this is the dot product of the velocity and the vector operator: \begin{align} {\rm div}\mathbf u&=\left(\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}\right)\cdot\left(u\hat{x}+v\hat{y}+w\hat{z}\right)\\ &=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}\\ &\equiv\frac{\partial u_i}{\partial x_i} \end{align} where $i$ is an index (1, 2, 3) that is implicitly summed over.

In the case of the Navier-Stokes equations, we have $$ \left(\frac{\partial}{\partial t}+\mathbf u\cdot{\rm div}\right)\mathbf u=\frac1\rho\nabla p+\nu\nabla^2\mathbf u + \frac1\rho\mathbf f $$ where the second term on the left is the one you are concerned with. First, you must take the dot-product of the velocity with the divergence operator (e.g., (2) dotted with (1)): $$ \mathbf u\cdot{\rm div}=u_j\frac{\partial}{\partial x_j} $$ which is a new operator. Then you can apply the vector $\mathbf u=u_i$ to this operator to get $$ \left(\mathbf u\cdot{\rm div}\right)\mathbf u=u_j\frac{\partial u_i}{\partial x_j} $$

Answer 2

This might be a really old question, but I'd like to answer so that someone that bumps into this might have another view on this. What is confusing you is the notation used, you are seeing is that for some reason $$\mathrm{div}\mathbf{u}\mathbf{u}=\frac{\partial u_i}{\partial x_j}u_i$$ Which is incorrect in Euclidean index notation. Now it depends on the left hand side of the equation, putting up a parenthesis we have $$\mathrm{div}(\mathbf{u})\mathbf{u}$$ In index notation it is simply $$u_j\frac{\partial u_i}{\partial x_i}$$ In Kyle Kanos' answer we also find some curious notation $$(\mathbf{u}\cdot\mathrm{div})\mathbf{u}=u_j\frac{\partial u_i}{\partial x_j}$$ This is NOT correct, since, simply expanding the operation on the left hand we get $$\mathbf{u}\mathrm{div}\mathbf{u}=u_j\frac{\partial u_i}{\partial x_i}$$ One way to understand it is simply passing to nabla notation, and remember how $\mathrm{div}[\cdot]=\nabla\cdot[\cdot]$. What has been written in the last equation is instead $$(\mathbf{u}\cdot\mathrm{grad})\mathbf{u}=(\mathbf{u}\cdot\nabla)\mathbf{u}=u_j\frac{\partial u_i}{\partial x_j}$$ In conclusion, to answer you properly $$\frac{\partial u_i}{\partial x_j}u_i=\nabla_j\mathbf{u}_i\cdot\mathbf{u}_i=J_{ij}u\ne\mathrm{div}\mathbf{u}\mathbf{u}$$ With $J_{ij}$ the Jacobian matrix of $u_i$. And, yes, indeed $$\mathrm{div}\ \mathbf{u}=\sum_{i=0}^{3}\frac{\partial u_i}{\partial x_i}=\partial_iu_i$$ A simple way to memorize this is using this notation conversion. Let $\mathbf{a}=a_i$ be some vector field which is derivable, then $$\begin{aligned} \mathrm{grad}\ \mathbf{a}&=\nabla\mathbf{a}=\partial_ja_i\\ \mathrm{div}\ \mathbf{a}&=\nabla\cdot\mathbf{a}=\partial_ia_i\\ \mathrm{curl}\ \mathbf{a}&=\nabla\wedge\mathbf{a}=\epsilon_{ijk}\partial_ja_k \end{aligned}$$

Answer 3

The Navier-Stokes equation includes a term of the form \begin{equation} (\vec{u}\cdot\nabla)\vec{u}\end{equation} which in index notation is written as \begin{equation} u_i \partial_i u_j \end{equation}

Now consider the quantity $\nabla \cdot (\vec{u}\vec{u})$. If we express this (rather less confusingly) in index notation and expand by the product rule we find \begin{equation} \partial_i(u_iu_j) = u_i\partial_i u_j + u_j\partial_iu_i\end{equation}

The first of these terms is the term which appears in the NS equation and the second is $\vec{u}(\nabla\cdot\vec{u})$. We already have, however, that $\nabla\cdot\vec{u} = 0$ so this term vanishes and means the two expressions you have for the term are equal in the incompressible case.

Answer 4

i think this reference about index notation (and vectors) should help (another one)

BTW $$\frac{\partial u_i}{\partial x_j}u_i$$

does not mean $${\rm div} uu$$

(as one can see in the reference) unless sth is a typo somewhere.

UPDATE:

Tha vector form of Navier-Stokes equations (general) is:

The term:

$$v \cdot \nabla v$$

in index notation is the inner (dot) product of the velocity field and the gradient operator applied to the velocity field. In index notation one would

use the kronecker delta tensor ($\delta_{ij} = 1$ if $i=j$, else $0$) to

formulate the term like this:

$$v \cdot \nabla v \implies \left(v_i \partial_i \right) v_k$$

So the full equation in index notation would be:

$$\rho \left( \partial_t v_k + \left(v_i \partial_i \right) v_k \right) = -\partial_k p + \partial_i T_i + f_k$$

NOTE: If one wants to be more correct (tensor-analysis kind of correct) the indexes in the summation should be on-top (contra-variant) in other words, the dual of the vector is used. Of course for usual (euclidean) vectors used (which are self-dual) there is no difference.

$$\rho \left( \partial_t v_k + \left(v_i \partial^i \right) v_k \right) = -\partial_k p + \partial_i T^i + f_k$$

Notation:

$$\partial_i = \frac{\partial}{\partial x_i}, \partial^i = \frac{\partial}{\partial x^i}, \partial_t = \frac{\partial}{\partial t}$$