= Solution
Assume $\kappa>0$ and $x\ne0$, since the initial driving point does not admit the displayed ordinary boundary flow. The sign of the driver in this part is negative, so
$$
d\bigl(g_t(x\sqrt\kappa)-\xi_t\bigr)
=\frac2{g_t(x\sqrt\kappa)-\xi_t}\,dt+\sqrt\kappa\,dW_t.
$$
After division by $\sqrt\kappa$, the <Boundary-point Bessel flow for SLE> is
$$
\boxed{dX_t=dW_t+\frac a{X_t}\,dt,\qquad
X_0=x,\qquad a=\frac2\kappa.}
$$
For $x>0$ this is a <Bessel process> of dimension $\delta=1+2a$. For $x<0$, $-X_t$ obeys the same equation driven by $-W_t$ until hitting zero. It suffices to treat a positive initial value.
The <infinitesimal generator> is $\mathcal Lf=\tfrac12f''+(a/u)f'$. An increasing <scale function of a one-dimensional diffusion> is
$$
s(u)=
\begin{cases}
u^{1-2a}/(1-2a),&a\ne1/2,\\
\log u,&a=1/2.
\end{cases}
$$
It satisfies $\mathcal Ls=0$. For $0<\varepsilon<x<b$, <optional stopping theorem> gives the <boundary hitting probability from a diffusion scale function>
$$
\mathbb P_x(\tau_\varepsilon<\tau_b)
=\frac{s(b)-s(x)}{s(b)-s(\varepsilon)}.
$$
When $a<1/2$, put $\beta=1-2a>0$. The inner boundary is reached in finite time: the nonnegative function
$$
v(u)=\frac{b^{2-\beta}u^\beta-u^2}{1+2a}
$$
vanishes at $0,b$ and satisfies $\mathcal Lv=-1$. Applying the <Itô formula> before exiting $(\varepsilon,b)$ gives $\mathbb E(\tau_\varepsilon\wedge\tau_b)\le v(x)$. As $\varepsilon\downarrow0$, these times increase to a finite limiting exit time almost surely. Thus the scale limit is an actual hitting event, not just asymptotic approach to zero, and
$$
\mathbb P_x(\tau_0<\tau_b)=1-(x/b)^\beta.
$$
Let $b\uparrow\infty$ to obtain $\mathbb P_x(\tau_0<\infty)=1$.
If $a>1/2$, then $s(\varepsilon)\to-\infty$, and the same formula makes the probability of hitting zero before any fixed $b$ equal to zero. A finite-time hit would occur before reaching some integer upper level, because the stopped path is continuous and bounded on a finite interval. Taking the countable union over those levels proves that no hit occurs. At $a=1/2$, $s(u)=\log u$ gives the same conclusion. Therefore
$$
\boxed{\tau_0<\infty\text{ a.s. if }a<\tfrac12;\qquad
\tau_0=\infty\text{ a.s. if }a\ge\tfrac12.}
$$
The source's $x=0$ case must be excluded: zero is then already the initial value, and the displayed singular flow is undefined.
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