# Instantaneity of Stokes Flow

Background Info:

A steady-state incompressible newtonian Stokes fluid occupying a domain $\Omega \in \mathbb{R}^2$ has flow velocity $\textbf{u}$ and pressure $p$ given by the following system of PDEs: \begin{align*} -\nu \Delta \textbf{u} + \frac{1}{\rho} \nabla p &= \textbf{f} \text{ in } \Omega, \\ \nabla \cdot \textbf{u} &= 0 \text{ in } \Omega. \label{eq:4} \end{align*} In order for this system to have a unique solution, values for $\textbf{u}$ (or its derivatives) must be prescribed at the boundaries, and some condition on $p$ must be enforced in order to overcome certain uncertainty (since $p$ appears through its gradient and is thus only determined up through an arbitrary constant). I am interested in solving this system of PDEs with (time-dependent) moving boundary conditions numerically (using a Finite Element Method). I am unsure if I should use the steady-state versus unsteady equations that describe the motion of Stokes fluid.

More relevant information directly pertaining to my question:

The wikipedia article on Stokes flow states that Stokes fluids are instantaneous, meaning there is no dependence on time in the system above (other than possibly through boundary conditions). However, the article also states that "occasionally one might consider the unsteady Stokes equations, in which the term $\rho {\frac{\partial \mathbf {u}}{\partial t}}$ is added to the left hand side of the momentum balance equation".

Specific questions:

In this case, wouldn't adding the time-dependent term $\rho \frac{\partial \textbf{u}}{\partial t}$ make the Stokes fluid non-instantaneous? Isn't the point of a "steady-state" fluid to be time-independent anyways? Why is it even mentioned that Stokes flow is instantaneous then? What exactly does adding the time-dependent term do if "given the boundary conditions of a Stokes flow, the flow can be found without knowledge of the flow at any other time." If I want to simulate Stokes flow in a box, where one of the boundaries is moving, do I need to include the $\rho \frac{\partial \textbf{u}}{\partial t}$ term?.

If the question is "Do I need to include the $\rho \frac{\partial \textbf{u}}{\partial t}$ term to simulate Stokes flow", the answer is no because (as you point out) that term is neglected in the definition of the assumption one makes for obtaining Stokes flow.