Intuition about Stokes flow beween solid walls

[Hemmer]

Hi there,

I have a question about incompressible Stokes flow in a channel between solid walls (with no-slip boundary conditions at $y = 0, L_y$). It is my intuition that, if the flow direction is $x$ (periodic), and the direction normal to the walls is $y$, then there cannot be a net velocity in that direction, i.e. $\langle v_y \rangle = 0$, as this somehow implies fluid must be passing through the walls? Is this correct and if so how should I go about showing this? I guess it might require some integration of the incompressibility condition but I've not got anywhere yet. Full equations:

$$\eta\nabla^2 \textbf{v} - \nabla p = \textbf{f}, \qquad\nabla . \textbf{v}=0$$

Please let me know if you require any additional information, and any replies greatly appreciated!

 

[AndyW]

Hi there,

Thank you for your question. You are correct in your intuition that in incompressible Stokes flow in a channel between solid walls, there cannot be a net velocity in the normal direction to the walls. This is a result of the no-slip boundary condition at the walls, which means that the velocity of the fluid at the wall must be equal to the velocity of the wall itself. Therefore, there can be no net fluid passing through the walls.

To formally show this, we can use the incompressibility condition, which states that the divergence of the velocity field must be equal to zero. In this case, we can write this as:

$$\nabla \cdot \textbf{v} = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} = 0$$

Since the flow direction is in the $x$ direction and the walls are in the $y$ direction, we can assume that the velocity in the $x$ direction is non-zero while the velocity in the $y$ direction is zero. Therefore, we can rewrite the above equation as:

$$\frac{\partial v_x}{\partial x} = 0$$

This means that the velocity in the $x$ direction is constant and does not vary in the $x$ direction. However, since the flow is periodic, the velocity at one end of the channel must be equal to the velocity at the other end. Therefore, the only possible solution is that the velocity in the $x$ direction is zero everywhere. This also means that the average velocity in the $y$ direction must be zero, as you correctly stated.

I hope this helps to clarify your understanding. If you have any further questions, please don't hesitate to ask. Best of luck with your research!