The Stagnation Point That Isn't Stable

physicsfluids/potential-flowdifficulty 4/5#potential-flow#stagnation#stability#vector-field

A steady, incompressible, irrotational 2-D flow near a stagnation point has velocity field
$\vec u = (a x,\,-a y)$ with $a>0$. A neutrally buoyant tracer is released near the origin.

(a) Show the field is both divergence-free and irrotational. (b) Write the tracer's
trajectory $y(x)$ from $\dot x = a x,\ \dot y = -a y$, and use it to decide whether the origin is a
stable equilibrium for the tracer. (c) A common claim is "a stagnation point is where the fluid
is at rest, so a tracer placed there stays there." Identify the false attractor in this claim and
state (1)'s verdict precisely.

-3 -2 -1 1 2 3 -2 -1.5 -1 -0.5 0.5 1 1.5 2 x y

The streamlines are the hyperbolae $xy = \text{const}$:

$$ y(x) = \frac{C}{x} $$(1)
Answer & solution

Answer

(a) $ \nabla\cdot\vec u = 0,\quad (\nabla\times\vec u)_z = 0. $

(b) $ xy = \text{const};\quad \text{the origin is a saddle — unstable along } x. $

(c)

"Zero velocity ⇒ stable" is the false attractor. Stability is set by the field's gradient
$\mathrm{diag}(a,-a)$, not by the velocity being zero: the origin is a saddle, so tracers are
swept away along $x$.

Solution

(a) $\nabla\cdot\vec u = a - a = 0$ and $(\nabla\times\vec u)_z = \partial_x(-ay) - \partial_y(ax) = 0$.

(b) $\dot x = ax \Rightarrow x = x_0 e^{at}$ (grows) while $\dot y = -ay \Rightarrow y = y_0 e^{-at}$
(decays), so $xy = x_0 y_0$ is constant — equation (1). The $x$-direction is repelling.

(c) The false attractor: "velocity zero ⇒ stable." The origin is a saddle; any tracer with
$x_0 \neq 0$ is swept away exponentially. Zero velocity at a point says nothing about the stability
of nearby trajectories — that is set by the field's gradient, here $\mathrm{diag}(a,-a)$.