= Solution
Use outward radial <mass flux> $\mathcal F=2\pi r\Sigma u_r$ and outward stress-carried angular-momentum flux $\mathcal G$. The PDF includes a second conservation equation omitted from the converted TeX:
$$
\partial_t(\Sigma h)+\frac1{2\pi r}\partial_r(\mathcal Fh+\mathcal G)=-T.
$$
Together with <mass conservation>, $\Sigma_t+(2\pi r)^{-1}\mathcal F_r=-S$, it fixes the signs of the wind terms. Subtract h times the mass equation, noting that the prescribed $h(r)$ is independent of time. One obtains
$$
\mathcal F\frac{dh}{dr}+\frac{\partial\mathcal G}{\partial r}=2\pi r(Sh-T).
$$
The <viscous torque in an accretion disk> in this outward-flux convention is $\mathcal G=-2\pi\bar\nu\Sigma r^3\Omega'$. Thus, where $h'\ne0$,
$$
\mathcal F=2\pi(h')^{-1}\left[\partial_r(\bar\nu\Sigma r^3\Omega')+r(Sh-T)\right].
$$
Putting this result back into mass conservation gives
$$
\boxed{\Sigma_t=-\frac1r\partial_r\left\{(h')^{-1}\left[\partial_r(\bar\nu\Sigma r^3\Omega')+r(Sh-T)\right]\right\}-S.}
$$
The derivation requires a fixed rotation law; it would acquire additional terms if h evolved independently in time.
For <Keplerian rotation>, $h=\sqrt{GMr}$, $h'=h/(2r)$, and $r^3\Omega'=-(3/2)h$. With $T=qSh$, the radial flux becomes
$$
\mathcal F=-6\pi\sqrt r\,\partial_r(\sqrt r\bar\nu\Sigma)-4\pi(q-1)r^2S.
$$
Consequently the <wind-loaded Keplerian disk evolution> equation is
$$
\boxed{\Sigma_t=\frac3r\partial_r\left[\sqrt r\,\partial_r(\sqrt r\bar\nu\Sigma)\right]+\frac1r\partial_r[2(q-1)r^2S]-S.}
$$
The final term removes mass locally. The extra flux term describes the disc's radial response when the outflow removes more or less than the local <specific angular momentum>. In particular, $q=1$ cancels this additional transport term without canceling the local mass sink.
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