After working with some problems regarding flow, I came up to a similiar problem as the one presented here:


In solving the problem, we assume a laminar flow in steady state.

When using Navier-Stokes equations, to fully determine the velocity profile, I get a differential equation of order two. Meaning I need two boundary conditions in order to fully describe the velocity profile. First of all, if we assume the incline is stationary, no-slip forces us to set $v_x(0) = 0$ where $v_x$ is the velocity in the $x$-direction and a function of $y$. Now, since I've seen examples of solving this before, they also assumed that $v_x'(h)=0$, and it was somehow related to the shear stress.

First, I'd like to understand how certain partial derivatives of the velocity with combinations of different directions relate to shear stresses (i.e. derivative of $v_x$ with respect to $y$ or $v_y$ with respect to $x$).

Secondly, I want to understand why this boundary conditions holds. I assume that it has to do with friction and viscosity of the fluid. Since no fluid is above that region, we will have no shear force from any layer.


1 Answer 1


Recall that the viscosity stress tensor for an incompressible Newtonian fluid is (by definition): $$ \sigma=\eta(\nabla\otimes v+\nabla\otimes v^T) $$ You can derive this from a microscopic theory (transport phenomena) or phenomenologically (assume linearity, isotropy, and only dependence in first order spatial derivatives).

Btw, the resulting viscous shear forces at the boundary of a region $\Omega$ is: $$ F=\int \sigma \cdot d^2x $$ You obtain the Navier-Stokes by considering momentum balance in an infinitesimal domain or equivalently using Stokes’ formula.

In your case, using the equations give the shear stress of the free surface to be: $$ \sigma_{xy}=\eta(\partial_xv_y+\partial_yv_x) $$ Using the laminar hypothesis, this simplifies to: $$ \sigma_{xy}=\eta v_x’ $$ So the boundary condition of the free surface is equivalent to: $$ \sigma_{xy}(y=h)=0 $$

As you’ve written, this is justified physically. It essentially assumes that the air is inviscid so cannot apply a shear stress at the interface.

Hope this helps.

  • $\begingroup$ Thank you very much! This cleared it up very well for me, especially when you explained that the boundary conditions I asked about assumes inviscid air. I've reflected upon the shear stresses and their relation to the mixed partial derivatives and have come up with something I'd like to get feedback on: mixed partial derivatives of the velocity fields in laminar flow means that we study the change in velocity between fluid layers. A change in velocity occurs due to friction (Newtons 2nd law), where the friction comes from the viscous shear forces. Is this a good way of thinking? $\endgroup$
    – Tanamas
    May 5 at 17:44
  • 1
    $\begingroup$ You’re welcome. If I understand correctly, you are trying to build intuition for the link between shear stresses $\sigma$ and velocity $v$. There is no direct link with Newton’s second law though, unless you have a full microscopic theory. However, thinking in terms of laminar flow is instructive. The idea is that shear stresses opposes velocity gradients, so you’d expect in your case that $\sigma_{xy}$ has the same (strict) sign as $v_x’$ in order to have a negative feedback. I think that this is as far as you can get qualitatively. $\endgroup$
    – LPZ
    May 5 at 19:30
  • $\begingroup$ Yes, you understood me correctly. Thanks once again, this cleared it up alot for me! $\endgroup$
    – Tanamas
    May 6 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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