Why is a smooth weak solution strong for stationary linear Stokes problem with zero-traction boundary condition? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T08:23:02Z http://mathoverflow.net/feeds/question/37306 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37306/why-is-a-smooth-weak-solution-strong-for-stationary-linear-stokes-problem-with-ze Why is a smooth weak solution strong for stationary linear Stokes problem with zero-traction boundary condition? Navier_Stoked 2010-08-31T19:57:39Z 2010-09-01T18:39:51Z <p>Can anyone provide me with a reference giving details on how smooth generalized solutions of the stationary linear Stokes problem can be shown to be classical solutions when a zero-traction boundary condition is present? That is, given a smooth generalized solution of</p> <p>$-\nu \bigtriangleup v + \bigtriangledown q = f$ on $\Omega \subset \mathbb{R}^3$ </p> <p>$\bigtriangledown \cdot v = 0$ on $\Omega$</p> <p>$S(v,q) = 0$ on $\partial \Omega$ where $S_i(v,q) =q n_i - \nu \sum_{j=1}^3 (\partial_i v_j + \partial_j v_i)n_j$ for $i=1,2,3$</p> <p>how can it be shown that the zero-traction boundary condition is met? It's not difficult to show that the first two equations are satisfied on $\Omega$ and using the relevant Green's formula one can obtain</p> <p>$\int_{\partial \Omega} S(v,q) \cdot \phi = 0$ </p> <p>for all solenoidal $\phi \in H^1$. However, I can't quite figure out why this necessarily leads to $S(v,q)=0$.</p>