3
$\begingroup$

For droplet interactions in low-Reynolds number flow, solutions are available when the underlying flow can be written as linear compositions of strain and rotation, see Batchelor & Green (1972a).

When the background flow is assumed to be parabolic (i.e., pressure-driven), I have only seen boundary integral simulations, using various kinds of kernels, see e.g. Coulliette & Pozrikidis (1998). Also, there doesn't seem to be a "simple" physical interpretation of the resulting dynamics (simple in the sense that one can infer "easily" from one configuration to another).

Maybe this is very naive, but my question is what makes the parabolic flows so different from linear flows, mathematically?

To define the problem more specifically, one can consider the typical case where the equations are

$\partial_t {\bf u}=-\nabla p + \nu \Delta {\bf u} + {\bf f}, \quad \nabla \cdot {\bf u}= 0,$

where $\bf f$ is the forcing term giving the surface tension force, supported on the droplet interface. For simplicity, one can further assume the droplet is perfectly spherical, and the density and viscosity ratios are both unity.

Thanks in advance.

$\endgroup$
1
$\begingroup$

I will give a sketchy answer below to close this thread for now. Maybe experts in this field can give a better, more definite answer in the future.

The short answer is this: the linearity of the Stokes flow, combined with a linear velocity distribution in the far field, allows the manipulation of known solutions in simpler cases. This is only possible for linear flows; for nonlinear flows, even the simplest Poiseuille flows, no analytical solutions have been found.

The detailed arguments for the above statement is very technical. The interested reader can see e.g. Zinchenko (1983), a brilliant, but somehow unnoticed, soviet paper. Another practical reason for the lack of progress might be the vast development of numerical solvers in the last decades, which makes the solution more flexible. Finally, even with the analytical approach, numerical procedures have to be introduced at some point to evaluate some coefficients. It's not very intuitive anyway.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.