Do solutions to the forced Stokes equations for zero-Reynolds number flows always exist?

Question

I'm in the process of solving a forced Stokes flow problem in two dimensions, the governing equations of which are given by $$ \nabla p = \eta \Delta \mathbf{v} + \mathbf{f}, \quad \nabla\cdot \mathbf{v} = 0. $$ where $\mathbf{f} = f_x\mathbf{e}_x + f_y\mathbf{e}_y$ is known. I am attempting to solve the problem via a stream function: we let $$ \mathbf{v} = \frac{\partial\psi}{\partial y} \mathbf{e}_x - \frac{\partial\psi}{\partial x} \mathbf{e}_y. $$ By taking the curl of the Stokes equation it can be shown that $\psi$ then satisfies a bi-Poisson equation $$ \Delta^2 \psi = \frac1\eta \left[\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right]. $$ My question is are there any restrictions on the forcing term $\mathbf{f}$, other than differentiability, for the problem to be well-posed? In the case I'm working on (I haven't included the details as it is a messy boundary layer approach since $\mathbf{f}$ is a bunch of Bessel functions that must be approximated) I have successfully solved for the stream function but cannot integrate it up to find the pressure: backfilling into the original Stokes equation gives me an expression for $\nabla p$ that is not the gradient of a function. Can anyone suggest why this might be?

Answer 1

I don't have a yes/no answer, but it sounds like you have a vector function, $\eta\Delta v+f$, with curl zero but it is not a gradient. This can happen in a region that is not simply-connected. The classic example is $$ \frac{(-y,x)}{x^2+y^2}, \qquad \qquad (x,y)\ne(0,0)$$ which looks like the gradient of $\tan^{-1}(y/x)$ but isn't because the inverse tangent function is not continuous all the way around the origin. Maybe this kind of thing has happened if you are working with an external flow around an airfoil, for example. It doesn't mean there is no solution, but it might mean there is no stream function.