= Solution
Farther than a few <edge> lengths from the root, replace the radial coordinate $r$ by a physical coordinate $x=ar$ and smooth the <master equation>. It has the same constant-coefficient <Fokker-Planck equation> as in (b). Under the conventional first-time-derivative truncation,
$$
D_0=\frac{a^2}{2\tau},\qquad
\boxed{C=\frac{a(2-z)}{z\tau}<0\quad\text{for }z>2.}
$$
The radial drift $v=-C=a(z-2)/(z\tau)$ is outward. For $z=2$ the tree is the infinite line, its radial distance is an unbiased reflected walk, and $C=0$.
The root cannot absorb <probability> or allow negative radial distance. Write the equation as a <continuity equation>, using the <Fokker-Planck probability current>
$$
\partial_tP=-\partial_xj,\qquad j=-D\partial_xP-CP.
$$
The appropriate continuum <boundary condition> is reflecting, or zero flux:
$$
\boxed{j(0,t)=0\quad\Longleftrightarrow\quad D\partial_xP(0,t)+CP(0,t)=0.}
$$
It is not generally $\partial_xP(0,t)=0$: that simpler condition is correct only when $C=0$. With the finite-bias fixed-step Gaussian convention from (b), use $D=D_{\rm fixed}=2a^2(z-1)/(z^2\tau)$ instead; the drift sign and zero-current <boundary condition> are unchanged. A continuum <boundary condition> represents the microscopic root region after coarse-graining, rather than reproducing each discrete root <probability> at individual step times.
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