Dear all,
I have been looking at this solution of the Navier Stokes equations in Fenics.
By multiplying Eq. (3.29) by \nu and integrating over \Omega, one obtains the term
\langle \nabla \sigma, \nu \rangle = \int _\Omega dx (\partial_i \sigma_{ij}) \nu_j. Dropping the n+1/2 superscript for simplicity and integrating by parts, this term can be rewritten as
\langle \nabla \sigma, \nu \rangle = \int_{\partial \Omega} dS \, \sigma_{ij} \hat{n}_j \nu_i - \langle \sigma , \epsilon(\nu) \rangle_\Omega = \int_{\partial \Omega} dS \, (\partial_i u_j + \partial_j u_i) \hat{n}_j \nu_i - \int_{\partial \Omega} dS \, p \hat{n}_i \nu_i - \langle \sigma , \epsilon(\nu) \rangle_\Omega, where dS is the surface element in \partial \Omega and \hat{n} its unit normal pointing outwards \Omega.
The term \int_{\partial \Omega} dS \, (\partial_i u_j + \partial_j u_i) \hat{n}_j \nu_i is supposed to be the last term in the left-hand side of Eq. (3.33), i.e., \langle \mu \nabla u \cdot n, \nu \rangle, which is written as
dot(mu*nabla_grad(U)*n, v)*ds
in the code. Here nabla_grad is non-symmetric, thus it appears that the second term in (\partial_i u_j + \partial_j u_i)n_j is missing in the webpage.
Is this because it is somehow assumed that the directional derivative of u along \hat{n} is zero on \partial \Omega?