# SLE revision 2: Loewner’s Differential Equation

Last time I set up the geometric notions of probability that will be needed to proceed with the course material. Now we consider the deterministic differential equation due to Loewner (1923) which he used to make progress on the Bieberbach Conjecture, but which will also underpin the construction of SLE. This proof is adapted for this specific case from the slightly more general argument in Duren’s Univalent Functions (Section 3.3). Because in that setting the result concerns an infinite domain, readers should beware that though I am using identical notation, in about half the cases, my function are the inverse and my sets the complement of what they are in Duren.

To explain the construction, as with so many things, a picture speaks a thousand words. Unfortunately I have neither the software nor, right now, the time to produce the necessary diagrams, so the following will have to suffice. Consider a deterministic simple curve in the unit disc, $(\gamma(t): t\in[0,\infty))$. Removing initial segments of the curve gives the nested simply-connected regions: $U_t:=\mathbb{U}\backslash \gamma[0,t]$.

Then define as in the previous post, the unique conformal map $f_t: U_s\rightarrow \mathbb{U}$ such that $f_t(0)=0, f_t'(0)\in\mathbb{R}^+$,

and furthermore set $\xi_t$ to be the image of $\gamma(t)$ under this map. (Note that though the conformal map is not defined on the boundary, it must extend continuously). $f_t'(0)$ is increasing.

Very informally, this derivative records how much twisting is required at the origin to turn the slit domain into the open disk. Extending the path will demand further twisting. More rigorously, set: $g_t=f_t^{-1}:\mathbb{U}\rightarrow U_t$.

Then the $(g_t)$ are injective functions from the unit disc to itself which preserve the origin, so Schwarz’s lemma applies. They are clearly not rotations, so $|g_t'(0)|<1$.

By the inverse function theorem, $|f_t'(0)|>1$ (*). Now, given $t>s$, can decompose: $f_t=f_s\circ \tilde{f}$,

and $\tilde{f}$ has this useful Schwarz property (*) also. By applying the chain rule, noting that $f_s(0)=0$, we deduce that $|f_t'(0)|>|f_s'(0)|$.

This means we are free to demand that the curve has time parameter such that $|f_t'(0)|=e^t$.

A reminder of the statement of Schwarz’s Lemma: Given $f:\mathbb{U}\rightarrow\mathbb{U}, f(0)=0, |f(z)|\leq 1$,

then $|f(z)|\leq |z|\;\forall z\text{ and }|f'(0)|\leq 1$.

Furthermore, equality holds for either of these conditions (excluding z=0) only if f is a rotation.

Proof: This follows as an easy consequence of the maximum modulus principle when applied to the function $\frac{f(z)}{z}$. See Wikipedia for details.

Derivation of the Loewner Equation

The important consequence of this reparameterisation is that: $f_t(z)=e^sz+a_2z^2+\ldots$.

Now define the map $h_{s,t}:= f_t^{-1}\circ f_s=e^{s-t}z+\ldots: \mathbb{U}\rightarrow f_s(U_t)$ (1)

as shown above. $f_s$ collapses $\gamma[0,s]$ onto the boundary, but in the image there remains a path $f_s(\gamma(s,t]) = J_{s,t}$ say, which starts from $\xi_s$. Let $A_{s,t}$ denote the pre-image of this on the boundary of the disc: $A_{s,t}:= h_{s,t}^{-1}(J_{s,t})$.

Now $\phi(z):=\log\left(\frac{h_{s,t}(z)}{z}\right)$ on branch s.t. $\phi(0)=s-t$ by (1).

It is clear that $\Re \phi(z)=0\;\forall z\in\partial\mathbb{U}\backslash A_{s,t}$ and it is non-zero (indeed negative) on $A_{s,t}$.

Now $\phi$ is analytic, and so is defined on the whole unit disc by its behaviour on the boundary, through the Poisson kernel: $\phi(z)=\frac{1}{2\pi}\Re\left[\int_{-\pi}^\pi\frac{e^{i\theta}+z}{e^{-i\theta}-z}\phi(e^{i\theta})d\theta\right]$

Suppose the ends of the arc $A_{s,t}$ are $e^{ia},e^{ib}$. So we can replace the bounds on the Poisson integral with and b. In particular, $s-t=\phi(0)=\frac{1}{2\pi}\int_a^b \Re[\phi(e^{i\theta})]d\theta$ (2).

We now substitute $f_t(z)$ for z, remembering the definition (1). Then separate $\frac{e^{i\theta}+f_t(z)}{e^{i\theta}-f_t(z)}$ into real and imaginary parts, before applying the Mean Value Theorem to each, using (2) to obtain: $\log\frac{f_s(z)}{f_t(z)}=(s-t)\left[\Re\left(\frac{e^{i\sigma}+f_t(z)}{e^{i\sigma}-f_t(z)}\right)+\Im\left(\frac{e^{i\tau}+f_t(z)}{e^{i\tau}-f_t(z)}\right)\right]$

where $e^{i\sigma}, e^{i\tau}\in A_{s,t}$. Now we let $s\uparrow t$. We assume $\xi_s$ is continuous as a function of s (a non-trivial and non-easy assumption, but one I don’t want to explore further now). Then as $\xi_s\in A_{s,t}$, this means that by a simple ‘epsilon over two’ argument both $e^{i\sigma},e^{i\tau}\rightarrow \xi_t$.

The left hand side converges to the left-derivative, giving: $\frac{\partial}{\partial s}|_{s=t} \log f_s(z)=\frac{\xi(t)+f_t(z)}{\xi(t)-f_t(z)}$.

An entirely analogous calculation would give the right-derivative, and so we have Loewner’s differential equation, written in its conventional form as: $\frac{\partial}{\partial t}f_t(z)=-f_t(z)\frac{f_t(z)+\xi(t)}{f_t(z)-\xi(t)}$