Poiseuille flow

Example

Consider a Newtonian fluid of density $\rho > 0$ and dynamic viscosity $\mu > 0$ in the straight pipe $\set{(x, y, z) : x^2 + y^2 < R^2}$ of radius $R > 0$. Let $G > 0$ be a constant and let $r = \sqrt{x^2 + y^2}$. Poiseuille flow is the velocity field and pressure $$u = \frac{G}{4 \mu} (R^2 - r^2) \, e_z, \qquad p = p_0 - G z,$$ where $e_z = (0, 0, 1)$ and $p_0$ is a constant. It solves the steady Navier–Stokes equations $\rho (u \cdot \nabla) u = -\nabla p + \mu \Delta u$, $\nabla \cdot u = 0$.

Check: $u$ has only a $z$-component $w(r)$, which does not depend on $z$, so $\nabla \cdot u = \partial_z w = 0$ and $(u \cdot \nabla) u = w \, \partial_z u = 0$. In polar form $\Delta w = \frac{1}{r} \frac{d}{dr}\big( r w'(r) \big) = \frac{1}{r} \frac{d}{dr}\Big( -\frac{G r^2}{2 \mu} \Big) = -\frac{G}{\mu}$. So $\mu \Delta u = -G e_z = \nabla p$, and the momentum equation holds. On the wall $r = R$ the velocity is zero, so the no-slip condition holds.