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  Why is a smooth weak solution strong for stationary linear Stokes problem with zero-traction boundary condition?

+ 3 like - 0 dislike

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

$-\nu \bigtriangleup v + \bigtriangledown q = f$ on $\Omega \subset \mathbb{R}^3$

$\bigtriangledown \cdot v = 0$ on $\Omega$

$S(v,q) = 0$ 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 $i=1,2,3$

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

$\int_{\partial \Omega} S(v,q) \cdot \phi = 0$

for all solenoidal $\phi \in H^1$. However, I can't quite figure out why this necessarily leads to $S(v,q)=0$.

This post imported from StackExchange MathOverflow at 2015-04-07 13:20 (UTC), posted by SE-user Navier_Stoked
asked Aug 31, 2010 in Resources and References by Navier_Stoked (0 points) [ no revision ]
retagged Jan 14, 2016
+1: Good question, great username.

This post imported from StackExchange MathOverflow at 2015-04-07 13:20 (UTC), posted by SE-user Cam McLeman
What do "smooth weak" and "smooth generalized" mean?

This post imported from StackExchange MathOverflow at 2015-04-07 13:20 (UTC), posted by SE-user Bob Terrell
What are the assumptions on $\Omega$ and $\partial \Omega$?

This post imported from StackExchange MathOverflow at 2015-04-07 13:20 (UTC), posted by SE-user Nilima Nigam

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