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Uniqueness in law means that, for a fixed initial value, every two weak stochastic solutions of the stochastic differential equation have the same distribution as entire solution paths, even if their filtered probability spaces and driving Brownian motions differ. For an equation on $(0,\infty )$, initially consider the stopped path and lifetime at the boundary.

We first justify that the boundary cannot intervene when $\gamma \ge 2$. For any solution of the displayed Bessel equation, let ${\tau }_{\epsilon ,L}$ be its first exit from $(\epsilon ,L)$, where $0<\epsilon <r<L$. Applying Itô to $R^2$ givesThus the annulus exit has finite expectation and is almost surely finite. The one-dimensional generator is $\frac{1}{2}{\partial }_{rr}+\frac{\gamma -1}{2r}{\partial }_r$. Its harmonic scale functions are $\log r$ for $\gamma =2$ and $r^{2-\gamma }$ for $\gamma >2$, as direct differentiation verifies. They are bounded on the stopped annulus, so optional sampling givesBoth tend to zero as $\epsilon \downarrow 0$. A continuous path hitting zero before ${\tau }_L$ would hit every inner level first, so that event has probability zero. Exhausting the upper levels proves that zero is not hit at any finite time. Nor can an upper explosion occur: localization of $R^2$ gives $\mathbb{P}({\tau }_L\le t)\le (r^2+\gamma t)/L^2\to 0$. These are the Hitting-zero classification for a Bessel process arguments needed for the dimension-at-least-two case.

Now suppose $\gamma$ is an integer and take a $\gamma$-dimensional Brownian motion $U$ from any deterministic vector $u$ with $|u|=r$. For $\rho =|U|>0$, differentiation of the Euclidean norm givesDefine the radial noise $\beta ={\sum }_i\int U_i/\rho d{W}_i$, filling the integrand by a fixed unit vector on the zero set, as in part (b). It has bracket $t$ and is a Brownian motion. the Itô formula, initially before a possible zero, givesThe boundary argument just given makes this a global positive weak stochastic solution. The assumed uniqueness in law therefore identifies every solution's path law with that of $|U|$. Orthogonal invariance of Brownian motion makes this radial law independent of the selected point $u$ on the sphere. ConsequentlyThis proves equality of process laws, not merely equality of one-time marginal distributions.

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 $2a+1$.

For $\kappa >0$, divide the Boundary-point Bessel flow for SLE by $\sqrt{\kappa }$ and replace $W$ by $-W$. The resulting Bessel process has dimensionsince its drift is $\frac{\delta -1}{2R_t}=2/(\kappa R_t)$. The Hitting-zero classification for a Bessel process suggests the threshold $\delta =2$, or $\kappa =4$. Here is a direct verification including accessibility in finite time.

The infinitesimal generator of $X$ is $\mathcal{L}=\frac{\kappa }{2}{\partial }_x^2+(2/x){\partial }_x$. An increasing scale function of a one-dimensional diffusion isIt satisfies $\mathcal{L}s=0$. For $0<\epsilon <b<R$, the boundary hitting probability from a diffusion scale function, or optional stopping theorem applied to $s(X)$, givesFor $0<\kappa <4$, $s(\epsilon )\to -\infty$, so this probability tends to zero as $\epsilon \downarrow 0$. A finite zero hit has a bounded path before the hit and hence precedes $T_R$ for some integer $R$; taking a countable union proves that zero is never hit.

For $\kappa >4$, write $a=1-4/\kappa >0$. Taking the inner boundary to zero gives $1-(b/R{)}^a$. This limit really concerns a finite zero hit: the Itô formula givesStopped on $[\epsilon ,R]$, it impliesuniformly in $\epsilon$. The increasing limit of these exit times is finite almost surely; the stopped squared process has a continuous extension, so its limiting lower endpoint is zero. ThusLet $R\to \infty$ to obtain ${\mathbb{P}}_b(T_0<\infty )=1$. When $\kappa =0$, the explicit solution is $X_t=\sqrt{b^2+4t}$, which stays positive.

Combining this with the SLE boundary swallowing criterion gives the critical parameter and the two regimes:

The boundary assertion is interpreted at positive finite times: a chordal SLE already starts on the domain boundary, and its marked target is another boundary point. The precise conclusion in the complex upper half-plane is $\gamma (0,\infty )\subseteq \mathbb{H}$ almost surely.

For $0<\kappa \le 4$ and a fixed real $x>0$, the centered Boundary-point Bessel flow for SLE $V_t=g_t(x)-U_t$ solvesThus $R_t=V_t/\sqrt{\kappa }$ is a Bessel process driven by $-B$ of dimensionThe Hitting-zero classification for a Bessel process says that, started positively, it never hits zero when $\delta \ge 2$. For $x<0$, reflect the equation and apply the same result. Therefore every fixed nonzero boundary point has infinite swallowing time. A countable intersection gives this simultaneously for all nonzero rational boundary points.

To extend this countable conclusion to every real $x>0$, choose rational $q$ with $0<q<x$. The difference of two centered boundary solutions satisfiesso it remains positive while both flows exist. On any finite time interval, $V_t(q)$ is bounded away from zero, hence so is $V_t(x)$. Moreover $V_t(x)=x-U_t+{\int }_0^{t}2/V_s(x)ds$ remains bounded there, so the differential equation continues for the whole interval. Its uniform separation from the Loewner driving function also allows each flow map to continue as a holomorphic function on a complex neighborhood of $x$. Thus $x$ is outside $K_t$ and cannot be visited by the Loewner trace. Reflection gives the negative real axis. This uses countably many Bessel processes followed by deterministic flow comparison, rather than an uncountable union of probability-zero events.

To exclude a return to the starting point, use the boundary extension of the inverse map for a continuous Loewner trace: $g_s^{-1}$ is continuous on the closed complex upper half-plane and maps $U_s$ to $\gamma (s)$. This follows from the continuous trace theorem and the Caratheodory boundary extension theorem; it does not assume simplicity. At each deterministic rational time $s$, the Conformal Markov property of SLE makes the mapped future a fresh SLE, so its trace lies in $\mathbb{H}\cup \{0\}$ by the preceding argument. Mapping back with $g_s^{-1}$ therefore showsIf a point $q=\gamma (r)$ were revisited at $t>r$, choose rational $s\in (r,t)$ with $\gamma (s)\ne q$. Such an $s$ exists because continuity and strictly increasing half-plane capacity forbid a constant trace on an interval. But $q\in K_s\cup \mathbb{R}$, contradicting the displayed inclusion. This excludes all self-contacts, including returns to zero, and proves the simple curve property needed when discussing later swallowing times.

For $\kappa =0$, the deterministic Loewner trace is $\gamma (t)=2i\sqrt{t}$, so the conclusion follows directly. Hence the boundary-intersection threshold for SLE givesMapping to another marked simply connected domain carries positive-time points into its interior. The literal statement including the starting point is false, since $\gamma (0)$ is prescribed to be on the domain boundary.

Let $U_t=(1,0,0)+(B_t^1,B_t^2,B_t^3)$, with three independent standard Brownian motions, and set $R_t=|U_t|$ and $M_t=R_t^{-1}$. The radial stochastic process $R$ is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions $u\mapsto |u{|}^{-1}$ is a harmonic function away from zero: direct differentiation gives $\Delta (|u{|}^{-1})=0$.

For the stopping times $T_n=\inf \{t:R_t\notin (1/n,n)\}$, $n\ge 2$, the Itô formula givesThe coefficients are bounded by $n^2$, so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus $M$ is a nonnegative continuous local martingale.

To see that $M$ is not a true martingale, integrate the three-dimensional multivariate normal density of $U_t$ in spherical coordinates. The angular integral yieldswhere $\Phi$ is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.

For $\kappa >0$ and $x>0$, before the boundary point is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $d{U}_t=\sqrt{\kappa }d{B}_t$ giveSince $-B$ is again a standard Brownian motion, this is the Boundary-point Bessel flow for SLE withFor $x<0$, it is $-X_t$ that is the nonnegative Bessel process, with the same dimension; the formula without this sign convention is a signed Bessel flow.

When $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X_t$ never reaches zero. The borderline $\delta =2$ uses the scale function of a one-dimensional diffusion $\log x$: for $0<r<x<R$,Thus the critical case also cannot hit zero in finite time, though its all-time infimum is zero.

Apply the non-swallowing assertion simultaneously to all nonzero rational boundary points. The order-preserving real Loewner flow then keeps every compact real interval away from the origin in the surviving boundary. A boundary contact away from the starting point would cut off a nonempty real interval, swallowing a rational point, so the trace avoids $\mathbb{R}\setminus \{0\}$.

The same argument can be applied to the future after every rational time, using the Conformal Markov property of SLE. If the trace revisited an earlier point, choose a rational time strictly between the two visits. In the domain with that initial segment removed, the later visit would be a contact with an old boundary point away from the current growing tip. Under the mapping-out map, such a contact cuts off a real interval and contradicts the preceding non-swallowing result. This is the usual crosscut argument converting boundary non-swallowing into absence of self-intersections. Taking the countable intersection over rational restart times provesFor $\kappa =0$, the equation is deterministic and gives the simple slit $(0,2i\sqrt{t}]$. The division by $\sqrt{\kappa }$ is unnecessary in that case.

For a real boundary point $x\ne 0$, setAfter a deterministic rescaling of time, the Boundary-point Bessel flow for SLE says that $X$ is a Bessel process of dimensionWhen $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X$ never reaches zero. Thus no nonzero real boundary point is swallowed. The standard Loewner trace criterion then implies that each new tip is attached only to the preceding tip and the trace never intersects its past, so it is simple. For $\kappa =0$, the equation is driven by zero and generates a vertical slit. Hence ${\mathrm{SLE}}_{\kappa }$ is simple for $0\le \kappa \le 4$.

For $x\in \mathbb{R}\setminus \{0\}$, the centered Loewner imageis the Boundary-point Bessel flow for SLE, a Bessel process of dimensionWhen $\kappa >4$, one has $d<2$, and the Hitting-zero classification for a Bessel process says that $Y$ hits zero almost surely. Thus every fixed nonzero boundary point is swallowed in finite time.

A hull generated by a continuous trace cannot swallow a real point while the trace remains strictly inside $\mathbb{H}$: before any contact with the real line, the trace is a crosscut-free interior curve and no boundary interval is disconnected from infinity. Consequently finite swallowing forces the trace to meet $\mathbb{R}$ away from its initial point. SLE therefore almost surely intersects the boundary for every $\kappa \in \Lambda$.

For the Bessel processthe scale function of a one-dimensional diffusion is $s(x)=x^{2-d}$ when $d\ne 2$. If $0<ϵ<x<R$, the boundary hitting probability from a diffusion scale function gives

If $d<2$, then $s(0)=0$ and $s(R)\to \infty$. Letting $ϵ\downarrow 0$ and then $R\to \infty$ shows that the process cannot escape to infinity before reaching zero. The exit time from each bounded interval is finite almost surely, so $T_0<\infty$ almost surely.

If $d>2$, then $s(ϵ)\to \infty$ in absolute value as $ϵ\downarrow 0$. Equivalently,Thus the process does not hit zero. The borderline case $d=2$ has scale function $\log x$ and also does not hit zero. This is the Hitting-zero classification for a Bessel process.