1

Going through a fluid mechanics book, I encountered this expression: $$\nabla \cdot (uu^T) = u \cdot \nabla u, $$ whereby $u$ is supposed to be the velocity vector field. I can not make sense of this. Considering $u$ as a vector, $uu^T$ will be a 3 by 3 matrix. For the left-hand side of the equation above I took the divergence by taking the divergence of the three column vectors of the matrix. For the right-hand side I do not know what the gradient of a vector field is defined and what the dot product there means. Can somebody explain all the terms and operations in the equation and prove it holds ? Thanks.

user996159
  • 876

1 Answers1

1

In terms of Cartesian coordinates and using the Einstein summation convention (sum over repeated indices):

$$\nabla \cdot(\mathbb{uu}^T) = \partial_i(u_iu_j) = (\partial_iu_i)u_j + u_i \partial_i u_j= (\nabla \cdot \mathbb{u})\mathbb{u} + \mathbb{u} \cdot \nabla \mathbb{u}$$

If the flow is incompressible, we have $\nabla \cdot \mathbb{u} = 0$ and we get

$$\nabla \cdot(\mathbb{uu}^T) =\mathbb{u} \cdot \nabla \mathbb{u}$$

RRL
  • 90,707
  • That is fine but I wanted to get this in terms of matrices. Can you show that ? In particular, I do not see how the "dot product" is used. Also, In the text I read, they did not assume the fluid is incompressible. – user996159 Mar 26 '24 at 18:34
  • @user996159: Are you familiar with the algebra of 2nd order tensors: dyadics? – RRL Mar 26 '24 at 18:36
  • The expression $\mathbb{u} \cdot \nabla \mathbb{u}$ is sometimes written as $(\mathbb{u} \cdot \nabla) \mathbb{u}$. Take the dot product of $u$ with the nabla operator like it is a vector. – RRL Mar 26 '24 at 18:39
  • No, I am not. I started by representing $uu^T$ as a matrix and then computed the divergence in the way I described above. I could not show this is equal to the right-hand side. The main obstacle was, I could not figure out how the vector $u$ and the matrix $\nabla u$ are multiplied via the dot product. – user996159 Mar 26 '24 at 18:40
  • So $\nabla \mathbb{u}$ is a tensor whose matrix with respect to standard Cartesian unit basis vectors is $[\partial_i u_j]$. Then in matrix form the expression $\mathbb{u} \cdot \nabla \mathbb{u}$ is the row vector with components $[u_1 u_2 u_3]$ times that matrix. – RRL Mar 26 '24 at 18:45
  • That's fine. How will you compute the left-hand side ? The way I did it, as described above, does not lead to the equality. – user996159 Mar 26 '24 at 18:53
  • It seems you are trying to draw some ad hoc analogy to matrices and vectors so I can't really be sure what to tell you. I made a guess in translating for you in the comment above for the right-hand side. There are strict definitions for divergence of a tensor or a dyadic product, etc. It would be advisable to learn it. This seems good. – RRL Mar 26 '24 at 19:15
  • Also the book applies a transpose to $u$. The Navier-Stokes equations are normally written with terms like $\nabla \cdot (\mathbb{u}\mathbb{u})$. I can't really unravel this alternate convention. – RRL Mar 26 '24 at 19:18
  • So, taking the divergence of each column in the matrix $uu^T$ is then wrong ? – user996159 Mar 26 '24 at 19:25
  • I used the definition of the divergence of a tensor field provided in that document, but it leads to the same result I got by computing div $(uu^T)$ as described in my post. So, I can not show the identity either way. – user996159 Mar 26 '24 at 19:44