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Write $\gamma =1-2a\ne 0$ and interpret the positive solution up to its first hit of zero, where the displayed drift is singular. For $s(x)=x^{\gamma }$,Thus $s$ is a scale function of a one-dimensional diffusion on every $[\epsilon ,\infty )$ with $\epsilon >0$. It is strictly decreasing if $a>1/2$; multiplying by $-1$ gives the conventionally increasing scale, without changing any hitting ratio.

Fix $0<\epsilon <x<R$. Exit from $(\epsilon ,R)$ is almost surely finite. Indeed the coefficients are locally Lipschitz there, so there is no lifetime obstruction in this compact interval. If the process stayed inside forever, the equation and $a/X>0$ would give $B_t\le R-x$ for every $t$, contradicting the almost sure unboundedness above of Brownian motion. Therefore part (a)'s bounded optional-stopping argument applies and givesIf $a>1/2$, then $\gamma <0$ and ${\epsilon }^{\gamma }\to \infty$, so ${\mathbb{P}}_x(T_{\epsilon }<T_R)\to 0$. Hitting zero before $R$ would entail hitting every $\epsilon$ before $R$; it therefore has probability zero. Every continuous path reaching zero in finite time is bounded up to that time, so it belongs to this event for some integer $R>x$. A countable union proves

If $0<a<1/2$, then $\gamma \in (0,1)$. Taking $\epsilon \downarrow 0$ in the probabilities alone does not yet exclude approaching zero only at infinite time. To establish finite-time access to zero for Bessel dimensions below two, defineIt is continuous, nonnegative, zero at $0,R$, and satisfies $\mathcal{L}{v}_R=-1$ on $(0,R)$, since $\mathcal{L}(y^{\gamma })=0$ and $\mathcal{L}(y^2)=1+2a$. Itô's formula stopped at ${\tau }_{\epsilon ,R}=T_{\epsilon }\wedge T_R$, followed first by expectation at $t\wedge {\tau }_{\epsilon ,R}$, givesLet $t\to \infty$ and then $\epsilon \downarrow 0$. The stopping times increase and have uniformly bounded expectation, so their limit ${\tau }_{0,R}$ is finite almost surely. On this finite interval $X\le R$, and the increasing drift integral is bounded by $R-x-{\inf }_{u\le {\tau }_{0,R}}{B}_u$. Hence it has a finite limit, and the stochastic equation makes $X$ extend continuously to ${\tau }_{0,R}$. If the upper endpoint was not reached, the lower stopped values tend to zero, so the terminal value is zero. Thus ${\tau }_{0,R}=T_0\wedge T_R$ genuinely denotes a finite hitting time.

The lower-exit events decrease to $\{T_0<T_R\}$ as $\epsilon \downarrow 0$, giving ${\mathbb{P}}_x(T_0<T_R)=1-(x/R{)}^{\gamma }$. Let $R\to \infty$ to obtainThere is no finite explosion to $+\infty$: on a finite horizon, any excursion above level $1$ has drift at most $a$, so its height is bounded by $\max \{x,1\}+2{\sup }_{u\le t}|B_u|+at$. The two alternatives therefore describe the zero-boundary lifetime, giving the Hitting-zero classification for a Bessel process of dimension $2a+1$.