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Consider this Stokes equations, $$ \left\{ \begin{array}{r} - \mu \Delta \vec{u} + \nabla P = \vec{f} \\ \nabla \cdot \vec{u} = 0 \end{array} \right. $$

Weak form I is: $$ \left\{ \begin{array}{r} {\color{red}{\mu <\nabla \phi, \nabla \vec{u} > - <\nabla \cdot \phi,P>}} &=& {\color{red}{< \phi,\vec{f}>}} \\ {\color{blue}{<\psi,\nabla \cdot \vec{u}>}} &=& {\color{blue}{0}} \end{array} \right. $$

Weak form II is: $$ {\color{red}{ \mu <\nabla \phi, \nabla \vec{u} > - <\nabla \cdot \phi,P>}} {\color{blue}{ -<\psi,\nabla \cdot \vec{u}>}} = {\color{red}{< \phi,\vec{f}>}} $$

Why are the number of equations reduced, but Forms I and II are still equivalent?

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1 Answer 1

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To get from weak form I to II, just add the two equations of weak form I.

To see the other direction, note that weak form II has to hold for all test functions $\phi$ and $\psi$. That is, you can test with $\phi = 0$ to obtain the second part of weak form I. Similarly, testing with $\psi = 0$ yields the first part.

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  • $\begingroup$ Thank you for your reply. So "it holds for all $\phi$ and all $\psi$." is equivalent to "it holds for all $\phi$ when $\psi = 0$, and holds for all $\psi$ when $\phi = 0 $.", right? $\endgroup$
    – Hao
    Commented Jan 22 at 15:27
  • $\begingroup$ Another way to think about it is to notice that if $\langle \psi, \nabla\cdot u \rangle \neq 0$, the LHS of the combined weak form can be made arbitrarily large by varying just $\psi$ and the RHS remains fixed, so that term must be zero $\endgroup$
    – whpowell96
    Commented Jan 22 at 18:17
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    $\begingroup$ @Hao, that's not mathematically equivalent (you only get one direction): If it holds for all $(\phi, \psi)$, then it holds for all $(\phi, 0)$ and $(0, \psi)$ in particular. In other words: If a statement is true for all points in a 2D plane, then it is also true for every line in the plane (i.e., the coordinate axes), because these are just subsets. $\endgroup$
    – cos_theta
    Commented Jan 22 at 19:55
  • $\begingroup$ @cos_theta, I'm still a little confused. If it holds for the two base vectors and each (ϕ,ψ) is a linear combination of the two base vectors, the former should mathematically equivalent to the latter, shouldn't it? Perhaps, in this case, each point is not to be considered as a linear combination of base vectors? $\endgroup$
    – Hao
    Commented Jan 23 at 12:03
  • $\begingroup$ You are correct, but you have to show that it works for all linear combinations. Showing that it works on just the basis elements is not enough $\endgroup$
    – whpowell96
    Commented Jan 23 at 16:17

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