[Math] notation in navier stokes equation

partial differential equations

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!

Best Answer

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 $div(u) = 0$.

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