 Solution

Since $\dot{u}(t)=a_t^2>0$, invert the image clock before exit and put ${\hat{\xi }}_u=W_{t(u)}$. Its quadratic variation is $6u$ and its initial value is zero, because $\Phi (0)=0$. By the Lévy characterization of Brownian motion, or the localized Dambis-Dubins-Schwarz theorem,for a standard Brownian motion up to the image exit time. If that lifetime is finite, the stopped Brownian motion can be extended beyond it; only the stopped law is used here.

In this clock the image mapping-out functions obey the ordinary Chordal Loewner equation with driver $\sqrt{6}{\hat{B}}_u$. Their generated trace therefore has the law of SLE with parameter six, stopped when it leaves $N^{\ast }$. Meanwhile the original image segment is exactly $\Phi (\gamma )$ up to exit from $N$. The clock change is increasing, so it disappears after passing to $[\gamma ]$. This proves the local stopped-curve equality required in part (a).

For completeness, the original law is scale invariant: the maps $r^{-1}{g}_{r^{2}t}(rz)$ have driver $r^{-1}{\xi }_{r^{2}t}$, whose law is again $\sqrt{6}{B}_t$ by Brownian scaling. Hence $(r^{-1}{\gamma }_{r^{2}t}{)}_{t\ge 0}$ has the original trace law. Together with the stopped conformal-change argument, this establishesThe conclusion concerns local unparameterized segments up to exit; the capacity parameter itself changes according to $du=h_t^{\prime }({\xi }_t{)}^{2}dt$.