notation in navier stokes equation

Question

I am reading navier stokes equation. And I got stuck at the very begining. It says when the vector field u is smooth and divergence-free, we have $$u\cdot \bigtriangledown u=div(u\otimes u)$$ And the navier stokes equation can be written as $$\partial_tu+div(u\otimes u)-v\bigtriangleup u=-\bigtriangledown p$$$$divu=0$$$$u|_{t=0}=u_0$$I don't quite understand the meaning of $\otimes$ here. So I also don't understand why we have $$u\cdot \bigtriangledown u=div(u\otimes u)$$ Thanks for any help!

Answer 1

The symbol $\otimes$ is the tensor product which you can read more about on Wikipedia. If you think of the vectors $u$ and $v$ as column vectors (3x1 matrices if we are talking about 3D space) then the tensor product $u\otimes v$ can be thought of as a square matrix (3x3 in the 3D case) whose entries are all the products of the individual entries of the vectors. So, in components: $$(u\otimes v)_{ij} = (u)_i(v)_j$$ The dot product between a vector and a tensor works like matrix multiplication, transposing the column vector to a row vector if it comes before the tensor, so that the dimensions work out. The result is a vector. (I will get back to how that looks for this case in a bit.)

The symbol $\nabla u$ is also a tensor, the gradient of a vector. Its components are the partial derivatives of the components of the vector, as so: $$(\nabla u)_{ij} = \frac {\partial (u)_j}{\partial x_i}$$ Viewing this tensor as a matrix, the divergence of the vector $u$ is equal to the trace of the tensor $\nabla u$. $$div(u) = \sum_{i=1}^3 (\nabla u)_{ii} = \sum_{i=1}^3 \frac {\partial (u)_i}{\partial x_i}$$

Finally, by analogy with this, the gradient of a tensor would be a higher-dimensional array of components (a 3x3x3 array) and the divergence of a tensor would be like the "trace" of this array - the result is a vector. For $div(u\otimes u)$, this ends up looking like this in components: $$(div(u\otimes u))_j = \sum_{i=1}^3 \frac {\partial (u\otimes u)_{ij}}{\partial x_i} = \sum_{i=1}^3 \frac {\partial ((u)_i(u)_j)}{\partial x_i}$$ If we find the components of $u\cdot \nabla u$ by doing the multiplication between the vector $u$ and the tensor $\nabla u$ as so: $$u\cdot \nabla u = u^T\nabla u$$ $$(u\cdot \nabla u)_j = \sum_{i=1}^3 {(u)_i(\nabla u)_{ij}} = \sum_{i=1}^3 {(u)_i\frac {\partial (u)_j}{\partial x_i}}$$ Then you should be able to see that $u\cdot \nabla u =div(u\otimes u)$ precisely when $div(u) = 0$.

Answer 2

If $\vec u$ is divergence free, then $\nabla \cdot \vec u=0$. Now, we have

$$\begin{align} \vec u\cdot \nabla \vec u&=\nabla \cdot(\vec u\otimes \vec u)\\\\ &=\left(\sum_{i=1}^3\hat x_i\frac{\partial }{\partial x_i}\right)\cdot \left(\sum_{j=1}^3\hat x_ju_j\sum_{j=1}^3\hat x_ku_k \right)\\\\ &=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\hat x_k(\hat x_i\cdot\hat x_j)\left(u_j\frac{\partial u_k}{\partial x_i}+u_k\frac{\partial u_j}{\partial x_i}\right)\\\\ &=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\hat x_k(\delta_{ij})\left(u_j\frac{\partial u_k}{\partial x_i}+u_k\frac{\partial u_j}{\partial x_i}\right)\\\\ &=\sum_{j=1}^3\sum_{k=1}^3\hat x_k\left(u_j\frac{\partial u_k}{\partial x_j}+u_k\frac{\partial u_j}{\partial x_j}\right)\\\\ &=\sum_{j=1}^3\sum_{k=1}^3\hat x_k\left(u_j\frac{\partial u_k}{\partial x_j}\right)+\sum_{j=1}^3\sum_{k=1}^3\hat x_k\left(u_k\frac{\partial u_j}{\partial x_j}\right)\\\\ &=\sum_{j=1}^3\left(u_j\frac{\partial }{\partial x_j}\right)\sum_{k=1}^3\hat x_ku_k+\sum_{j=1}^3\left(\frac{\partial u_j}{\partial x_j}\right)\sum_{k=1}^3\hat x_ku_k\\\\ &=\vec (u \cdot \nabla)\vec u+\vec u \nabla \cdot \vec u\\\\ &=\vec (u \cdot \nabla)\vec u \end{align}$$

as was to be shown!

Answer 3

It means a sort of "dyadic product" (it's really a tensor product, but that's not relevant): in index notation, $$ (u \times v)_{ij} = u_i v_j. $$ Therefore, what it means is that (in summation convention) $$ (\operatorname{div}(u \otimes u))_i = \partial_j (u_j u_i ) = (\partial_j u_j) u_i + u_j \partial_j u_i = ((\operatorname{div} u) u)_i + ((u \cdot \nabla) u)_i, $$ and the first of these is zero because we imposed $\operatorname{div}u=0$. The idea about $\otimes$ here, then, is that the product rule for derivatives looks like $$ \nabla \otimes (u \otimes v) = (\nabla \otimes u) \otimes v + u \otimes (\nabla \otimes v), $$ and then you take a trace to get the divergence.