Mapping a curve-sided quadrilateral to a rectangle

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I am currently investigating different ways of solving the Laplace equation $$\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial z^2} = 0 $$ numerically on the domain $\Omega$ shown as the shaded region in the figure below. As the figure shows, $\Omega$ is bounded by the vertical lines $x = 0$, $x = L$, the horizontal line $z = -h$ and the curve $(x, \eta(x))$ where $\eta(x)$ is a smooth, $L$-periodic function.

One strategy for solving the Laplace equation is to make a coordinate transformation $(x,z) \mapsto (r(x,z), s(x,z))$ and then solve the equation that the function $G(r(x,z), s(x,z)) = F(x,z)$ satisfies on the transformed domain. For this strategy to be practical, the coordinate transformation should map $\Omega$ to a rectangle.

Now, my questions (which is probably more of a request) is the following: Which coordinate tranformations from $\Omega$ to a rectangle exist? I am aware of the simple vertical stretching $(x,z) \mapsto (x, s(x,z))$ where $$ s(x,z) = \frac{2z + h - \eta(x)}{h + \eta(x)}$$ and in principal it works fine. In the mapped coordinates the function $G(x,s(x,z)) = F(x,z)$ solves the equation $$ \frac{\partial^2 G}{\partial x^2} + \bigg( \bigg[\frac{\partial s}{\partial x} \bigg]^2 + \bigg[ \frac{\partial s}{\partial z} \bigg]^2 \bigg) \frac{\partial^2 G}{\partial s^2} + 2 \frac{\partial s}{\partial x} \frac{\partial^2 G}{\partial x \partial s} + \frac{\partial^2 s}{\partial x^2} \frac{\partial G}{\partial s} = 0, \qquad 0 \leq x \leq L, \, \, \, -1 \leq s \leq 1$$ and while this certainly can be solved numerically, it would be nice to have a coordinate transformation which does not give rise to such a complicated equation. The ideal transformation would of course be a conformal one, since the Laplace equation is invariant under such a change of coordinates.

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