For deriving the week formulation of the stokes problem $$ \begin{align} -\frac{1}{Re} \Delta v + \nabla p &= f \text{ in }\Omega\\ \nabla \cdot v &= 0 \text{ in } \Omega\\ v &= 0 \text{ on } \partial\Omega \end{align} $$ one usually uses the sobolev room $H_0^1(\Omega)$ for the velocity and $$ L_0^2(\Omega) = \{p\in L^2(\Omega) |\ \int_\Omega p\ dx= 0\} $$ for the pressure.

The physical motivation behind $H_0^1(\Omega)$ is finite energy and dissipation. It also incorporates the no-slip boundary condition.

What is the physical motivation behind $L_0^2(\Omega)$?


1 Answer 1


I know this question is old, but I stumbled across it and since it's a quick answer decided to help in case you are still wondering.

I hope the $L^2(\Omega)$ part is hopefully not to bad from a practical standpoint. We do not have any derivatives of $p$ in the weak form of the Stokes equation. Thus, we do not need the pressure to even be weakly differentiable. Of course, one may ask whether it is appropriate to have a discontinuous physical quantity, but that starts to get outside the realm of math and more into the realm of philosophy.

However the normalization condition $\int_{\Omega} p \, d\mathbf{x}= 0$ has two possible explanations. First, look at the original (strong) formulation of the problem. There is no dependence on $p$ directly, only $\nabla p$. This means that for any particular solution $\hat{p}$, $\hat{p} + \lambda$ is also a solution for any real constant $\lambda$ (since $\nabla \lambda = 0$ for a constant). So, from a mathematical perspective, we impose that the mean of the pressure is zero, just so that we have a unique solution.

Ok, but why does any of this make physical sense? Well, remember that pressure is really an internal energy of sorts in this context--a driving force. And it turns out that what matters is not pressure, but the pressure drop or change in pressure. Imagine if this weren't the case. Let's say we have a pipe that is 100 ft long. We would expect, that in some sense each 10 ft section of pipe should act about the same, right? But the pressure is decreasing continuously down the length of pipe. So if the value of pressure mattered then we would have different behavior at 50 ft down the pipe than we have at 20 ft down the pipe. But rather the behavior is constant because pressure drop is constant.

Another way to think about this is this: The normalization condition tells us how to calibrate our pressure gauge. This is a common cause of confusion in engineering for example, where there is discussion of "absolute" pressure and "gauge" pressure. Absolute pressure gauges are calibrated to take into account things like atmospheric pressure. While gauge pressure is calibrated to read zero at atmospheric pressure. One can think of the normalized pressure as something similar to the gauge pressure.


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