= Solution
Writing $R=x^2+y^2$, direct differentiation gives
$$
\frac{\partial u}{\partial x}
=\frac{-2x(x^2-3y^2)}{R^3},
\qquad
\frac{\partial v}{\partial y}
=\frac{2x(x^2-3y^2)}{R^3},
$$
so $\nabla\mathbin{\cdot}\mathbf u=0$. Likewise,
$$
\frac{\partial v}{\partial x}
=\frac{-2y(3x^2-y^2)}{R^3}
=\frac{\partial u}{\partial y},
$$
so the two-dimensional <vorticity> $\partial_xv-\partial_yu$ vanishes. The flow is therefore <incompressible flow>[incompressible] and <irrotational flow>[irrotational] away from the origin.
Back to article page
