# Why are solutions to the Loewner equation analytic?

Consider the Loewner differential equation

$$\frac{\partial g_t(z)}{\partial t} = \frac{2}{g_t(z) - \sqrt{\kappa} B_t}$$

where $B_t$ is a 1-dimensional standard Brownian motion starting at the origin. The solutions $g_t(z)$ should give a family of conformal maps wherever they are defined. I don't see why a solution $g_t(z)$ has to be analytic in $z$. I'm guessing that something like Cauchy-Kowalevski has to be used somehow, but to apply such theorems, we require the function $2/(g - \sqrt{\kappa} B_t)$ to be analytic, which it is not.

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Where is the $=$ sign? –  doraemonpaul Feb 2 '13 at 16:08

The fact that $g_t(z)$ is analytic in $z$ is clear, since the right-hand side is analytic in $g_t(z)$.
Goluzin's book presents a self-contained treatment of the deterministic (radial) Loewner equation with piecewise continuous driving term $k(s)$. He actually works with the inverse map and proves its analyticity, along with existence and uniqueness, by following the proof of the Picard existence theorem. I think for the chordal process this would work too: let $g_{0}(z,t)=z$, and for $n\ge 1$ $$g_{n}(z,t) = z+\int_0^t \frac{2}{g_{n-1}(z,s)-k(s)}\,ds$$ Then each $g_n(\cdot, t)$ is holomorphic in $z$ on its domain, and $g_n(\cdot, t)\to g(\cdot, t)$ uniformly on compact subsets of the domain of $g(\cdot,t)$.
"The fact that $g_t(z)$ is analytic in $z$ is clear, since the right-hand side is analytic in $g_t(z)$." I don't understand this argument (sorry, I know next to nothing about PDEs). What theorem is this invoking? –  user61035 Feb 5 '13 at 4:39