= Stokes equation
{c}
{title2=$-\nabla p+\mu\nabla^2\mathbf u+\mathbf f=\mathbf 0$}
For a steady <incompressible flow> of a <Newtonian fluid> with constant <density> $\rho$ and constant <dynamic viscosity> $\mu>0$, neglecting <inertia> in the <Navier-Stokes equation> gives
$$
-\nabla p+\mu\nabla^2\mathbf u+\mathbf f=\mathbf 0,
\qquad
\nabla\cdot\mathbf u=0.
$$
The first equation is local <force> balance; the second is <mass conservation>. The term $\mu\nabla^2\mathbf u$ is the <divergence> of the viscous part of the <Newtonian fluid stress tensor>. The <body force> $\mathbf f$ has units of <force> per unit <volume>, $\mathrm{N\,m^{-3}}$, and equals $\rho\mathbf g$ for uniform <gravity>. Appropriate boundary data for <velocity> or <traction>, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the <no-slip boundary condition>.
With uniform <gravity>, write $p=p_h+p'$ where the <hydrostatic pressure> satisfies $\nabla p_h=\rho\mathbf g$. Then the disturbance equation is $-\nabla p'+\mu\nabla^2\mathbf u=\mathbf 0$. This removes <gravity> from the fluid disturbance equation, while the <hydrostatic pressure> still supplies <buoyancy> on an immersed body. Omitting it from the total <force> balance would give the wrong <Stokes settling velocity>.
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