How to define a surface $z = f(x,y)$ with flat region at centre and sigmoidally tapering towards the edges?

Question

How do we define a continuos function $f(x,y)$ within the bounded domain $x \in [a,b]$ and $y \in [c,d]$ so that $z=f(x,y)$ has a flat surface at the centre (flat means $f(x,y)= C$, $C$ being constant in that centred region),and the same surface gradually tapers toward the edges on all $4$ sides)?
A 1D equivalent that i can think of is the sigmoidal/logistic function. It would be elegant to define such a function, as simple as $\dfrac{1}{1+e^{-x}}$ to define the tapering towards the edges. However, also taking the centre flat region into account, how difficult is it to make the combination continuos and having a derivative at each instant.
P.S : Probably my question is analagous to this one https://stackoverflow.com/questions/5225061/exponential-decay-surrounding-bounding-box.
But my definition is clearer than what was posed before.The bounding box is the region which is the flat surface(constant response to all $(x,y)$ value pairs within the bounding box) and surrounding it has to taper away towards the edges.

Answer 1

The problem with the sigmoid function is that it never actually tapers off to a constant, so you'll have difficulty making the derivative continuous. What you can do, is use a function like: $$g(x) = \frac{6 x}{5}+\frac{3}{5}-\frac{8 x^3}{5}, \ x\in[-0.5,0.5]$$ And $g(x) = 0$ otherwise. This is just a normalized series approximation for the sigmoid function. $g(x)$ tapers off with derivative $0$ at the ends.

Now let's define a function $f(x,y)$ using this $g(x)$:

$$f(x,y) = 1, \ [x,y] \in [a,b]\times[c,d]$$ $$f(x,y) = g(x - (a -1/2)), \ y \in [c,d],\ x < a$$ $$f(x,y) = g((b +1/2) -x), \ y \in [c,d],\ x > b$$ With the same idea for the the other side. Now you have the problem of the "corners", e.g. $x > b, y > d$ within a radius of $r<1$. you can again use $g(x)$ for the corners: $$f(x,y) = g\left(\sqrt{(x - a)^2 + (y- c)^2}\right)$$