Womersley flow

Example

Consider a Newtonian fluid of density $\rho$ and viscosity $\mu$ in a rigid straight pipe of circular cross-section with radius $R$, along the $z$-axis. Drive it by an oscillating pressure gradient $-\partial p / \partial z = \operatorname{Re}(A e^{i \omega t})$ with amplitude $A$ and angular frequency $\omega > 0$. Let $\alpha = R \sqrt{\omega \rho / \mu}$ be the Womersley number and $J_0$ the Bessel function of the first kind of order $0$. The Womersley flow is the axial velocity $$u(r, t) = \operatorname{Re} \left\{ \frac{A}{i \omega \rho} \left[ 1 - \frac{J_0(i^{3/2} \alpha \, r / R)}{J_0(i^{3/2} \alpha)} \right] e^{i \omega t} \right\},$$ where $r$ is the distance to the axis and $i^{3/2} = e^{3\pi i/4}$.

Check: for a fully developed axial flow the Navier–Stokes equations reduce to $\rho \, \partial_t u = -\partial_z p + \mu (u_{rr} + u_r / r)$. With $u = \operatorname{Re}(U(r) e^{i\omega t})$ this becomes $i \omega \rho U = A + \mu (U'' + U'/r)$. The constant $A/(i\omega\rho)$ is a particular solution. The function $U = J_0(\lambda r)$ satisfies $U'' + U'/r = -\lambda^2 U$, and the choice $\lambda = i^{3/2} \alpha / R$ gives $-\lambda^2 = i \alpha^2 / R^2 = i \omega \rho / \mu$, so it solves the homogeneous equation. The coefficient is chosen so that $U(R) = 0$, which is the no-slip condition.