= Solution
{c}
The <divergence> in <polar coordinates> and the given <streamfunction> convention yield
$$
\nabla\cdot\mathbf v
=\frac1r\partial_r(rv_r)+\frac1r\partial_\theta v_\theta
=\frac1r\left(\partial_r\partial_\theta\psi-\partial_\theta\partial_r\psi\right)=0.
$$
Thus the <incompressibility condition> holds wherever the mixed <partial derivatives> commute. The vertical <vorticity> is
$$
\zeta=\frac1r\left[\partial_r(rv_\theta)-\partial_\theta v_r\right]
=-\left(\partial_r^2+\frac1r\partial_r+\frac1{r^2}\partial_\theta^2\right)\psi
=-\nabla^2\psi.
$$
Neglecting both inertial terms in the <Navier-Stokes equation>, for constant <dynamic viscosity> and <mass density>, gives the <Stokes equation>:
$$
0=-\nabla p+\mu\nabla^2\mathbf v-\rho g\widehat{\mathbf z}.
$$
Its <curl> eliminates the <pressure gradient> and the uniform gravitational <body force>, so $\mu\nabla^2\zeta=0$. Hence the <biharmonic streamfunction> satisfies
$$
\boxed{\nabla^4\psi\approx0.}
$$
This conclusion requires the <quasi-steady Stokes approximation>. If local <acceleration> were retained, the corresponding planar <vorticity equation> would instead be $\partial_t\nabla^2\psi=\nu\nabla^4\psi$.
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