# Simple question on stoke's first problem

1. Oct 25, 2013

### joshmccraney

in stoke's first problem, we have parallel flow initially at rest and then, instantaneously (i know, unrealistic) one wall starts to move in the x-direction with no acceleration (yes, it magically becomes a velocity). the y-direction is perpendicular to the x direction and is directing toward the other wall. the z-direction is coming "out of the page".

okay, now time for the question.

conservation of mass for incompressible flows yields: $$\nabla \cdot \vec{U}=0 \Rightarrow \frac{dv}{dy}=0$$ since $\frac{du}{dx}=\frac{dw}{dz}$ are identically zero. my question is, why isn't $\frac{dv}{dy}$ identically zero? i know it equals zero from the above, but how does it exist at all? how is there velocity in the $y$ direction?

it makes sense to me that the other two are zero by inspection, but why not $y$? to me it seems $\frac{du}{dy}$ would not be zero but that all others would be. please help me understand how there exists y-direction velocity ($y$ )

2. Nov 8, 2013

### njoysci

Your intuition and majority of statements are accurate, just a little variable terminology confusion:
U = u i + v j + w k
For your example: 0<= u <= V where V = wall velocity; v = 0; w = 0
Therefore ∂v/∂x = ∂v/∂y = ∂v/∂z = 0 and ∂w/∂x = ∂w/∂y = ∂w/∂z = 0
∇ dot U = ∂u/∂x + ∂v/∂y + ∂w/∂z = 0 (conservation of mass for incompressible flow)
Since flow is constant in the x direction for a given y, then ∂u/∂x = 0
So indeed ∇ dot U = 0 based upon the above partials.
In summary, since v is the fluid velocity in the y direction, it is zero. Hence ∂v/∂y = 0.
The only fluid flow is in the x direction, and since u varies as a function of y and is not constant, then ∂u/∂y is not zero. The molecules of each horizontal layer collide and change the molecular velocity in the adjacent horizontal layer. For instance for a typical Newtonian fluid
∂u/∂y = - fluid shear stress / viscosity