What I mean by that is, consider, say, the "Koch snowflake" curve. It is formed by repeatedly applying a substitution to the lines of a triangle to get the final curve in the limit.
What I am after is whether or not you can find a substitution and base figure, both of which can be described in "elementary" terms, such that the Julia sets, namely of the complex quadratic map $z \mapsto z^2 + c$, for at least some values of $c$, can be described in an analogous way. Is that possible for any value of $c$ for which the Julia set is both connected and fractal (i.e. has Hausdorff dimension > 1)? If so, what is an example, with attendant explicit description of substitution and base figure? If not, why is it not possible, and what is the proof?
For the dynamical system,
$$f_c(z_n)=z_{n+1}=z_{n}^2-c$$
The Julia set for a value $c$ is the collection of points $z_0$ such that the iterates stay bounded. The Mandelbrot set is the collection of points such that $z_0$ with $c=z_0$ such that the iterates stay bounded.
Denote the Julia set by $J_c$. Then assuming the attractor, fractal, for the dynamical system exists, the attractor for the system is $J_c$. By definition, this means,
$$f_c(J_c)=J_c$$
However, that also means,
$$J_c=f^{-1}_c(J_c)=\pm \sqrt{z+c}$$
Since the inverse is multivalued and we need all the values, we need to apply both, which is to say we take their union. This yields,
$$J_c=(\sqrt{J_c+c}) \cup (-\sqrt{J_c+c})=w_1(J_c) \cup w_2(J_c)$$
So these Julia sets can cast into the form of a nonlinear IFS with transformations given by,
$$\ w_1(z)=\sqrt{z+c}, \ w_2(z)=-\sqrt{z+c}; \ z \in C$$
With the Julia set $J_c$ as the attractor.