= Solution
Far from the inner edge of a steady <Keplerian accretion disk>, the inner angular-momentum flux is negligible compared with $\dot Mj$. Since $\Omega\propto R^{-3/2}$, the <viscous torque in an accretion disk> becomes $\mathcal G=3\pi\nu\Sigma j$. The steady balance $\mathcal G\simeq\dot Mj$ therefore gives $\nu\Sigma=\dot M/(3\pi)$. With the stated <surface density of a disk>,
$$
\boxed{\nu=\nu_0R^2,\qquad\nu_0\Sigma_0=\frac{\dot M}{3\pi}.}
$$
Steady <conservation of mass> gives $\dot M=-2\pi R\Sigma v_R$, and consequently
$$
\boxed{v_R=-\frac{\dot M}{2\pi R\Sigma}=-\frac{3\nu}{2R}=-\frac32\nu_0R.}
$$
Both relations use the far-inner-edge approximation, rather than extrapolating the stress-free boundary profile to the inner edge.
For the trace <passive scalar>, put $\sigma=\Sigma C$. Substitute this into the contaminant conservation equation and expand its advective flux:
$$
0=\partial_t(\Sigma C)+\nabla\cdot(\Sigma C u)-\nabla\cdot(k\Sigma\nabla C)
=C\{\partial_t\Sigma+\nabla\cdot(\Sigma u)\}+\Sigma\{\partial_tC+u\cdot\nabla C\}-\nabla\cdot(k\Sigma\nabla C).
$$
The braces multiplying $C$ vanish by background <conservation of mass>. Hence the <concentration> satisfies
$$
\boxed{\Sigma(C_t+u\cdot\nabla C)=\nabla\cdot(k\Sigma\nabla C).}
$$
The small contaminant fraction ensures it does not alter the ambient disk dynamics. The surface-density model already averages vertically; an axisymmetric background and ring-like injection give $C=C(R,t)$. More generally, rapid differential orbital motion winds azimuthal structure and facilitates mixing, but radial-only dependence remains a symmetry or mixing assumption, not a property of every initial injection.
For the axisymmetric <advection-diffusion equation>, cylindrical geometry gives
$$
\Sigma(C_t+v_RC_R)=\frac1R\partial_R(Rk\Sigma C_R).
$$
Here $k=\zeta\nu$ and $k\Sigma=\zeta\nu_0\Sigma_0$ are independent of radius. Thus
$$
C_t-\frac32\nu_0RC_R=\zeta\nu_0(R^2C_{RR}+RC_R).
$$
Let $Y=\ln(R/R_*)$, with any fixed reference length $R_*$. Since $C_Y=RC_R$ and $C_{YY}=R^2C_{RR}+RC_R$, this is the <tracer diffusion in a power-law accretion disk> equation
$$
\boxed{\frac1{\nu_0}C_t-\frac32C_Y=\zeta C_{YY}.}
$$
Assume $\zeta>0$, $\nu_0>0$, and write $v=3\nu_0/2$, $D=\zeta\nu_0$. The change $X=Y+vt$ turns this into the <heat equation> $C_t=DC_{XX}$. The radial <Dirac delta function> must be transformed with its Jacobian:
$$
\delta(R-R_0)=\frac1{R_0}\delta(Y-Y_0),\qquad Y_0=\ln(R_0/R_*).
$$
Convolution with the <Gaussian heat kernel> gives
$$
\boxed{C(R,t)=\frac{C_0}{R_0\sqrt{4\pi\zeta\nu_0t}}\exp\!\left[-\frac{\{\ln(R/R_0)+(3/2)\nu_0t\}^2}{4\zeta\nu_0t}\right].}
$$
This full-line logarithmic solution extends the far-edge power-law model to $0<R<\infty$. An actual finite inner boundary would require its own boundary condition and modified kernel.
The fraction must be computed with the contaminant mass measure, rather than with $C\,dR$. Indeed
$$
dM_c=2\pi R\sigma\,dR=2\pi\Sigma_0 C\,\frac{dR}R=2\pi\Sigma_0 C\,dY.
$$
The total contaminant mass is the conserved quantity $M_c=2\pi\Sigma_0 C_0/R_0$. Its normalized logarithmic radius is therefore a <normal distribution> with mean $Y_0-vt$ and <variance> $2Dt$. Integrating this tail gives, for $r>0$,
$$
\boxed{F(r,t)=\frac12\operatorname{erfc}\!\left(\frac{\ln r+(3/2)\nu_0t}{\sqrt{4\zeta\nu_0t}}\right).}
$$
The <complementary error function> accounts for the one-time mass fraction outside the specified radius.
For an outward target $r_1>1$, put $\ell=\ln r_1>0$. Since $\operatorname{erfc}$ is decreasing, maximize $F$ by minimizing $z(t)=(\ell+vt)/\sqrt{4Dt}$. Its derivative has the sign of $vt-\ell$, so the unique maximum occurs at $t_* =\ell/v=2\ln r_1/(3\nu_0)$. At this time $z(t_*)=\sqrt{v\ell/D}$. Thus the printed expression is correctly derived as the \b[maximum instantaneous exterior fraction]:
$$
\boxed{\max_{t>0}F(r_1,t)=\frac12\operatorname{erfc}\!\left(\sqrt{\frac{3\ln r_1}{2\zeta}}\right),\qquad r_1>1.}
$$
It is not, however, the probability of ever reaching the radius. A tracer can pass through the target and later return; its history is not counted by a one-time exterior fraction. The underlying logarithmic <diffusion process> is
$$
Y_t-Y_0=-vt+\sqrt{2D}\,B_t,
$$
where $B_t$ is standard <Brownian motion>. To compute its actual <hitting probability>, first put absorbing levels at $-A$ and $\ell$ relative to $Y_0$. Conditioning on the first small time step gives the backward equation $Dh''-vh'=0$ for the probability of reaching the upper level first, with $h(-A)=0$, $h(\ell)=1$. Solving gives
$$
h(y)=\frac{e^{vy/D}-e^{-vA/D}}{e^{v\ell/D}-e^{-vA/D}},\qquad
h(0)=\frac{1-e^{-vA/D}}{e^{v\ell/D}-e^{-vA/D}}.
$$
Letting the lower level tend to $-\infty$ exhausts all finite-time upward-hitting paths. Therefore the correct eventual probability in this idealized model is
$$
\boxed{P_{\mathrm{hit}}(r_1)=e^{-v\ell/D}=r_1^{-3/(2\zeta)},\qquad r_1>1.}
$$
For example, if $\ln r_1=2\zeta/3$, the maximum exterior fraction is $\tfrac12\operatorname{erfc}(1)\simeq0.07865$, whereas the actual <hitting probability> is $e^{-1}\simeq0.36788$. This is an explicit counterexample to the last inference in the question and an instance of <maximum occupation is not a hitting probability>.
For $r_1=1$, the target is the initial radius and the hitting probability is one, while $\sup_{t>0}F(1,t)=1/2$ with no attained maximum. For $0<r_1<1$, the inward drift ensures eventual hitting with probability one, and $\sup_{t>0}F(r_1,t)=1$ as $t\downarrow0$. The square-root expression involving $\ln r_1$ is therefore only meaningful for the outward-target case, and even there it describes occupancy rather than first passage.
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