# Why can the weak forms of the Stokes and continuity equations be combined into a single equation?

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?

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.

• 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?
– Hao
Commented Jan 22 at 15:27
• 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 Commented Jan 22 at 18:17
• @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. Commented Jan 22 at 19:55
• @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?
– Hao
Commented Jan 23 at 12:03
• 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 Commented Jan 23 at 16:17