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Far from the inner edge of a steady Keplerian accretion disk, the inner angular-momentum flux is negligible compared with $\dot{M}j$. Since $\Omega \propto R^{-3/2}$, the viscous torque in an accretion disk becomes $\mathcal{G}=3\pi \nu \Sigma j$. The steady balance $\mathcal{G}\simeq \dot{M}j$ therefore gives $\nu \Sigma =\dot{M}/(3\pi )$. With the stated surface density of a disk,Steady conservation of mass gives $\dot{M}=-2\pi R\Sigma v_R$, and consequentlyBoth 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:The braces multiplying $C$ vanish by background conservation of mass. Hence the concentration satisfiesThe 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 givesHere $k=\zeta \nu$ and $k\Sigma =\zeta {\nu }_0{\Sigma }_0$ are independent of radius. ThusLet $Y=\ln (R/R_{\ast })$, with any fixed reference length $R_{\ast }$. Since $C_Y=R{C}_R$ and $C_{YY}=R^2{C}_{RR}+R{C}_R$, this is the tracer diffusion in a power-law accretion disk equationAssume $\zeta >0$, ${\nu }_0>0$, and write $v=3{\nu }_0/2$, $D=\zeta {\nu }_0$. The change $X=Y+vt$ turns this into the heat equation $C_t=D{C}_{XX}$. The radial Dirac delta function must be transformed with its Jacobian:Convolution with the Gaussian heat kernel givesThis 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 $CdR$. IndeedThe 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 $r>0$,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 $\mathrm{erfc}$ 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 $t_{\ast }=\ell /v=2\ln r_1/(3{\nu }_0)$. At this time $z(t_{\ast })=\sqrt{v\ell /D}$. Thus the printed expression is correctly derived as the maximum instantaneous exterior fraction:

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 iswhere $B_t$ 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 $D{h}^{\prime \prime }-v{h}^{\prime }=0$ for the probability of reaching the upper level first, with $h(-A)=0$, $h(\ell )=1$. Solving givesLetting the lower level tend to $-\infty$ exhausts all finite-time upward-hitting paths. Therefore the correct eventual probability in this idealized model isFor example, if $\ln r_1=2\zeta /3$, the maximum exterior fraction is $\frac{1}{2}\mathrm{erfc}(1)\simeq 0.07865$, whereas the actual hitting probability is $e^{-1}\simeq 0.36788$. 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 $t\downarrow 0$. 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.