de Rham curve

Example

Let $d_0, d_1 \colon \C \to \C$ be two contraction mappings with fixed points $p_0$ and $p_1$, and assume $d_0(p_1) = d_1(p_0)$. The de Rham curve of $(d_0, d_1)$ is the continuous map $c \colon [0, 1] \to \C$ that solves the system of functional equations $$c\left(\frac{x}{2}\right) = d_0(c(x)), \qquad c\left(\frac{1 + x}{2}\right) = d_1(c(x)) \qquad \text{for all } x \in [0, 1].$$ Setting $x = 0$ in the first equation and $x = 1$ in the second gives $c(0) = p_0$ and $c(1) = p_1$. Both equations give a value at $1/2$: the first with $x = 1$ gives $d_0(p_1)$, the second with $x = 0$ gives $d_1(p_0)$. This is why the condition $d_0(p_1) = d_1(p_0)$ is needed.

A concrete case: let $a = \frac12 + \frac{\sqrt3}{6} i$, $d_0(z) = a \bar z$ and $d_1(z) = a + (1 - a) \bar z$. Both maps are contractions with ratio $\abs{a} = \abs{1 - a} = 1/\sqrt3 < 1$. The fixed point of $d_0$ is $p_0 = 0$, and the fixed point of $d_1$ is $p_1 = 1$, since $d_1(1) = a + 1 - a = 1$. The condition holds: $d_0(1) = a = d_1(0)$. The resulting curve from $0$ to $1$ is the Koch curve.

The simplest case is $d_0(z) = z/2$ and $d_1(z) = (z + 1)/2$, for which $c(x) = x$: indeed $c(x/2) = x/2 = d_0(x)$ and $c((1+x)/2) = (1 + x)/2 = d_1(x)$.