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

+ 3 like - 0 dislike
150 views

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
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

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysi$\varnothing$sOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.