Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 35 / 1

Use the half-plane-capacity parameterization in which $g_t(z)=z+2t/z+O(z^{-2})$ at infinity. The Chordal Loewner equation isIt holds up to the Loewner swallowing time of $z$. Although $\beta$ is nowhere differentiable, it is continuous: the equation for $g$ is an ordinary integral equation with a continuous time-dependent coefficient away from its pole.

For $x\in \mathbb{R}\setminus \{0\}$, the same real-valued equation has a unique solution until $g_t(x)-W_t$ first reaches zero. Its solution agrees with the boundary value of the conformal map from the unswallowed side. One can also obtain it by Schwarz reflection across an unswallowed real interval. Thusis the boundary-point swallowing time for a Loewner chain; it can be infinite. The collision definition agrees with membership in the closed hull. The initial point has $T(0)=0$. Swallowing a real point is not the same as visiting it: a curve can cut off an entire real interval in one step.

Before $T(x)$, the centered image satisfies the stochastic differential equationChanging the sign of the Brownian motion shows that $X/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa$. Here is a direct proof of the relevant hitting property, including finite lifetime.

Put $p=1-4/\kappa >0$. By the Itô formula, the drift of $X^p$ isThus $X^p$ is a local martingale and a scale function of a one-dimensional diffusion. Fix $R>x$. The nonnegative functionvanishes at both endpoints and satisfies $\mathcal{L}{u}_R=-1$, for $\mathcal{L}=(\kappa /2){\partial }_{vv}+(2/v){\partial }_v$. Stop first on $[ϵ,R]$, and also at a deterministic time. The Itô formula gives $\mathbb{E}(t\wedge {\tau }_{ϵ,R})\le u_R(x)$. Let $t\to \infty$ and then $ϵ\downarrow 0$. It follows that the first exit ${\tau }_{0,R}$ is finite almost surely, not merely that the path approaches a boundary at infinite time.

Optional stopping applied to $X^p$ on this bounded interval givesOn $\{{\tau }_0=\infty \}$ the process must therefore hit every $R>x$ before zero. Letting $R\to \infty$ makes that probability zero. ConsequentlyAt $\kappa =4$ the dimension is two; zero is not hit from a positive starting point. This endpoint must not be included in the finite-swallowing assertion.

For $\lambda >0$, define ${\tilde{g}}_t(z)={\lambda }^{-1/2}{g}_{\lambda t}(\sqrt{\lambda }z)$. Differentiating the Chordal Loewner equation givesBrownian scaling makes $\tilde{W}$ have the original driver law. The corresponding hull is ${\lambda }^{-1/2}{K}_{\lambda t}$, soThe time scaling applies to the whole coupled hull process. In particular, the joint law of swallowing times satisfies $(T(cx),T(cy))\stackrel{d}{=}{c}^2(T(x),T(y))$.

Thus $F(cx,cy)=F(x,y)$. A pair $0<x<y$ is determined up to dilation by $z=y/(y-x)>1$: dilating by $(y-x{)}^{-1}$ turns it into $(z-1,z)$. Define $f(z)=F(z-1,z)$. Then

Put $X_t=g_t(x)-W_t$, $Y_t=g_t(y)-W_t$ and $D_t=Y_t-X_t$. Before $T(x)$, order preservation gives $0<X_t<Y_t$ and $D_t>0$. The common noise cancels from their difference:Since $D$ has finite variation, applying the Itô formula to $Z=Y/D$ gives the SLE two-boundary-point ratio diffusionWith the strictly increasing clock $u(t)={\int }_0^t{D}_s^{-2}ds$, the Dambis-Dubins-Schwarz theorem yields a Brownian motion $B$ for which the time-changed process obeysThe sign of the new Brownian motion has absorbed the minus sign above.

The domain Markov property of a chordal Loewner chain says that, conditionally on the past before $T(x)$, the mapped future is a fresh SLE. Its boundary marked points are $X_t,Y_t$. HenceLocalize, for example, where $D_t\ge 1/n$, $Z_t\in [1+1/n,n]$ and $t\le n$. On each such stopped interval this is a bounded conditional-expectation martingale, and therefore a genuine martingale. No smoothness assumption on $f$ is needed for this argument. The exit calculation below identifies it explicitly.

Differentiate once less than might seem necessary: writing $v=h^{\prime }$ turns the equation intoUniqueness for this first-order linear equation also covers the solution $v\equiv 0$. Thus every solution has $h^{\prime }(z)=C[z(z-1){]}^{-4/\kappa }$. Since $4/\kappa <1$, the derivative is integrable at $1$. The specified boundary value fixes the additive constant, givingThe strict positivity forces $C>0$; $C=0$ gives the zero function. These are all the positive solutions. The function $s_{\kappa }$ is the increasing scale function of a one-dimensional diffusion for the ratio process.

As $v\to \infty$, $[v(v-1){]}^{-4/\kappa }\sim v^{-8/\kappa }$. ThereforeThus every positive solution is unbounded when κ ≥ 8, including the logarithmically divergent critical case. For $4<\kappa <8$, by contrast,For example the substitution $r=1-1/v$ identifies this normalizing constant as $B(1-4/\kappa ,8/\kappa -1)$, with both beta function parameters positive.

For any $C^2$ function $h$, the Itô formula and the ratio equation giveThe bracket is zero for the functions just found. On compact localized intervals the stochastic integrand is bounded, so the stochastic integral is a martingale. Removing localization proves that $h(Z_t)$, on $0\le t<T(x)$, is a local martingale. It is nonnegative because $Z_t>1$ and $C>0$.

In particular it is a nonnegative local martingale, hence a supermartingale after the usual stopping and extension by its endpoint limit. This is the justification for using its maximal inequality; it is not automatically an unstopped uniformly integrable martingale.

The time-changed diffusion has generator $\mathcal{A}=(\kappa /2){\partial }_{zz}+2(1/z+1/(z-1)){\partial }_z$. For $1<z<R$, its boundary hitting probability from a diffusion scale function isTo see this directly, stop the scale local martingale on $[1+ϵ,R]$ and pass to $ϵ\downarrow 0$. Exit from $(1,R)$ occurs in finite clock time: the scale density behaves as $(z-1{)}^{-4/\kappa }$ and the speed density as $(z-1{)}^{4/\kappa }$, so the usual scale-times-speed exit integral is finite at $1$. More explicitly, with $m(v)=2/(\kappa s_{\kappa }^{\prime }(v))$ the finite-interval diffusion exit Green kernel givesThe equation $\mathcal{A}u=-1$ follows by differentiation and the derivative jump at $v=z$, so stopping the corresponding Itô identity gives this exit bound. The integrand near $1$ is $O(v-1)$; elsewhere it is bounded. At large $z$ the drift is bounded, excluding explosion at infinity in finite clock time.

Here is why these exits describe the original swallowing event. In clock time,If $\hat{Z}$ hits $1$ at finite $u$, its integrated stochastic differential equation shows ${\int }_0^{{\tau }_1}({\hat{Z}}_v-1{)}^{-1}dv<\infty$: the Brownian term has a finite limit and both positive drift integrals must then have finite limits. Consequently $D({\tau }_1)>0$, $X$ hits zero and $Y$ stays positive. This gives $T(x)<T(y)$.

Conversely $T(x)$ is finite by part (b). If its clock ends at a finite value, the diffusion must hit $1$; an interior endpoint with positive $D$ would leave $X$ positive. If the clock runs forever without hitting $1$, the standard one-dimensional scale classification gives escape to infinity whenever $S_{\kappa }<\infty$. This cannot represent a strict swallowing event, which has $D\to Y_{T(x)}>0$ and $Z\to 1$. Hence the surviving event represents $T(x)=T(y)$. This reasoning also explains why a simultaneous collision must be treated separately rather than assigned the boundary value at $1$.

For $\kappa \ge 8$, the scale is unbounded. The maximal inequality for a nonnegative supermartingale gives $\mathbb{P}(\sup s_{\kappa }(\hat{Z})\ge s_{\kappa }(R))\le s_{\kappa }(z)/s_{\kappa }(R)$; letting $R\to \infty$ rules out escape. Finite-interval exit then forces a hit at $1$. For $4<\kappa <8$, taking $R\to \infty$ givesThe strict-event probability is also positive, since $0<s_{\kappa }(z)<S_{\kappa }$ for finite $z$.

For $\kappa \ge 8$, first intersect the probability-one strict-order events over rational $0<a<b$. The real boundary flow gives nondecreasing swallowing times on $(0,\infty )$. For any $0<x<y$, choose rational $x<a<b<y$; then $T(x)\le T(a)<T(b)\le T(y)$. This proves the simultaneous assertionThe range is positive points, as in the definition of $F$; reflection reverses the ordering on the negative axis. At the printed endpoint $\kappa =4$, each fixed positive point has $T=\infty$. Thus $\mathbb{P}(T(x)=T(y))=1$ if equality of extended times is allowed, but there is no finite simultaneous-swallowing assertion there. The preceding formulas assume $\kappa >4$.

Work in the intended $\kappa \ge 8$ regime. On one probability-one event, all positive points have finite swallowing times and the strict ordering from part (d) holds for all pairs. Finiteness for arbitrary points follows from finiteness for rational points and monotonicity. Reflection supplies the negative-axis version on the same event.

Fix $r>0$ and suppose the Loewner trace has not visited $r$ by $T(r)$. Its image up to this finite time is compact, so a small interval $I$ about $r$, together with a thin upper collar, is disjoint from the whole trace segment. Every point of that collar belongs to the same component of the complement at every earlier time. Thus all points in a smaller interval are disconnected from infinity at precisely the same time as $r$: before $T(r)$ none is swallowed, and at $T(r)$ all are. This contradicts strict ordering for two positive points in that interval.

Therefore $r$ is visited by time $T(r)$. It cannot be visited before its swallowing time, so the visit occurs at $T(r)$. The same deterministic argument works for every real $r<0$ by reflection, while $0$ is the starting point. HenceThe threshold matters. If part (e) is read as retaining only $\kappa >4$, it is false: for $4<\kappa <8$, part (d) gives positive probability of swallowing a nonempty interval at one time, although a continuous trace can visit at most one of its points at that instant. Interior points of that swallowed boundary interval cannot subsequently be reached in capacity time: a later visit, by continuity, would make the curve stay inside an already filled hull for a time interval, contradicting strictly increasing half-plane capacity. Thus the preceding $\kappa \ge 8$ conclusion is the necessary intended qualification.