A conservation of momentum explanation for why Euler flows are mean zero

Something that I often need to do in my convex integration calculations is justify why I can uniquely solve the inverse divergence. For example if we want to solve for a v such that

div v = f

where f is some given scalar valued function, if it’s possible to solve this (which it is for my fluid equations i.e. Euler 2-d, 3-d, and SQG) any solution v can have a constant vector added to it and still be a solution.

In order to ensure the uniqueness of a solution we like to impose a condition such as v must be a mean zero vector field (sometimes this act is referred to as fixing a gauge for the problem).

In my studies I was wanting of a physics explanation why any realistic fluid should satisfy a mean-zero condition and I found an answer! It’s due to the conservation of momentum and you can see in section 3 of the following notes, a calculation which justifies why a vector flow should be mean-zero (assuming a uniform density).

Roughly speaking, one checks that the time derivative of the momentum is 0, hence the momentum is constant. This means that the integral of the vector field is constant in time and up to subtracting this constant from vector field solution, we may wlog assume we have a vector field solution which is mean 0.

Conservation of momentum \implies \int v = 0.

Let the momentum =: \Pi = \int v. Then

0 = \partial_t \Pi^l = \partial_t \int v \implies \int v = C, a constant. By adjusting v by this constant i.e. v - \Pi, wlog v has \int v = 0

I just wanted to post this here so that I wouldn’t lose this reference as I go about my calculations.


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