Across the two vertical faces of a control rectangle, the mass fluxes are

ρ u (y, t) δy

with opposite signs, since

u

is independent of

x

. There is no flux across the horizontal faces because the

y

-velocity is zero. The net mass flux is therefore zero, equivalently

\nabla \cdot u = \frac{\partial u}{\partial x} = 0.

ii

Solution

x

δ x \times δy

- \frac{\partial p}{\partial x} δ x δy + o (δ x δy) .

τ_{x y} = μ u_{y}

, so the net viscous force is

μ u_{yy} δ x δy + o (δ x δy) .

There is no convective acceleration. Dividing the momentum balance by the area and taking the limit gives

ρ u_{t} = - \frac{d p}{d x} + μ u_{yy} .

The

y

p_{y} = 0

, so the pressure depends only on

x

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