= Solution
At the rigid surface, the <no-slip boundary condition> and no penetration give
$$
\boxed{u=v=0\qquad(y=0).}
$$
At the horizontal <free surface>, the <kinematic boundary condition> and the <stress boundary condition> are
$$
\boxed{v=\dot h,
\qquad \sigma_{xy}=0,
\qquad \sigma_{yy}=-p_{
m air}
=-p_0+\frac12\rho_aE^2x^2
\qquad(y=h).}
$$
Here surface tension is absent and the inviscid air exerts no tangential traction.
The incompressible <Stokes equation> is
$$
-\nabla p+\mu\nabla^2\mathbf u=0,
\qquad \nabla\mathbin\cdot\mathbf u=0.
$$
Use the <Cartesian streamfunction> convention $u=\psi_y$, $v=-\psi_x$ and set
$$
\psi=xf(y).
$$
Then
$$
u=xf'(y),
\qquad v=-f(y),
$$
and the wall and zero-shear conditions become
$$
f(0)=f'(0)=0,
\qquad f''(h)=0.
$$
The two momentum equations read
$$
p_x=\mu x f'''(y),
\qquad p_y=-\mu f''(y).
$$
Equality of mixed derivatives gives $f^{(4)}=0$, so the boundary conditions imply
$$
f(y)=C\left(y^3-3hy^2\right).
$$
Integrating for the pressure and imposing the normal-stress condition at $y=h$ determines
$$
C=-\frac{\rho_aE^2}{6\mu}.
$$
The complete instantaneous velocity and pressure fields are therefore
$$
\boxed{u=\frac{\rho_aE^2}{2\mu}xy(2h-y),
\qquad
v=-\frac{\rho_aE^2}{6\mu}y^2(3h-y),}
$$
$$
\boxed{p=p_0-\frac{\rho_aE^2}{2}
\left(x^2+2hy-y^2+h^2\right).}
$$
Indeed, with the <viscous stress tensor> $\sigma=-pI+\mu(\nabla\mathbf u+\nabla\mathbf u^T)$, these expressions satisfy both traction conditions. Evaluating the vertical velocity at the material surface gives the <pressure-driven thinning of a uniform Stokes layer>:
$$
\dot h=v(h)=-\frac{\rho_aE^2}{3\mu}h^3.
$$
Thus
$$
\boxed{V=-\frac{dh}{dt}=\frac13\frac{\rho_a}{\mu}E^2h^3.}
$$
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