Is there a smooth map from the square to the deltoid?

Question

Is there a $C^\infty$ map between a unit square in $\mathbb R^2$ and a deltoid like this one

The deltoid is obtained by varying the angles $\theta_1$, $\theta_2$ in the equations \begin{align} x_2 &= \frac13 \left( \cos\left(\theta_1\right) + \cos\left(\theta_2\right) + \cos\left(\theta_1-\theta_2\right) \right)\\ x_2 &= \frac13 \left( \sin\left(\theta_1\right) - \sin\left(\theta_2\right) - \sin\left(\theta_1-\theta_2\right)\right) \end{align}

Hence, the deltoid represents the range of the implicit map $$ d: \left(\theta_1, \theta_2 \right) \mapsto \left(x_1,x_2\right) $$ defined by the previous system of equations.
The question is to find a smooth map $f$ for which $$ f\left(\left[0,1\right]^2\right) = d\left( \left[0,\pi\right]^2\right). $$

I don't see anyway to obtain $f$.

Once I have $f$, I will use it to construct bivariate chebyshev polynomial on a square.
An analytic expression for $f$ will be good, but as long as the derivative can be obtained numerically, it's ok.

Answer 1

There is, the Riemann Mapping Theorem. The square goes to the disc using Schwarz-Christoffel. There is such a mapping between the disc and deltoid, but there may not be published versions available.