Oseen equations

Remarks

Example: in the channel $-1 < y < 1$, with $U = (1, 0, 0)$ and $f = 0$, the field $u = (1 - y^2, 0, 0)$ with $p = -2 \nu x$ is a solution. Indeed $(U \cdot \nabla) u = \partial_x u = 0$, $\nu \Delta u = (-2\nu, 0, 0) = \nabla p$, and $\nabla \cdot u = 0$.

Non-example: $u = (x, 0, 0)$ fails the second equation for every $p$, since $\nabla \cdot u = 1$.

The equations come from the Navier–Stokes equations by writing the velocity as $U + u$ and dropping the term $(u \cdot \nabla) u$, quadratic in the disturbance. With $U = 0$ they reduce to the Stokes equations. The time-dependent version adds $\partial_t u$ on the left. The operator is not symmetric: the term $(U \cdot \nabla) u$ makes the fundamental solution decay faster upstream than downstream, which models a wake.