How to define a surface which includes three coordinate axes?

Question

Imagine a plane with two crossing lines along $x$ and $y$ axes, and one line $y=x$. Now bend the $y=x$ line towards $z$ axis, dragging the part of that plane with it. When this line coincides with $z$ axis, you'll get a surface similar to this:

But this one was generated from $z=\text{gaussian}_{0.3a}(x)$, i.e. if continued it'd not have $z$ full axis as its subset.

I'd like the surface to be:

1. Differentiable (except maybe at origin)
2. Symmetric in the sense that for each axis $A$ there existed a rotation which placed another axis $B$ to original location of $A$, leaving the surface unchanged.
3. Appeared as a single bent & stretched plane

How could I find (analytically define) such a surface (or a family of surfaces)?

Answer 1

As requested by OP, an answer in stead of a comment. My answer is simply an intuitive approach to finding a single function with the desired properties, and does not at all hint at what the class of desired functions might look like.

As I know close to nothing about graphs or differentiable things, I restrict my attention to polynomials, things I am more familiar with. The symmetry requirement suggests considering a sum of a polynomial $f\in k[X,Y,Z]$ over its orbit under the action of $S_3$ or $A_3$, that is

$$F_S=\sum_{\sigma\in S_3}\sigma(f)\qquad\text{ or }\quad F_A=\sum_{\sigma\in A_3}\sigma(f).$$

The constructed polynomial is (almost) symmetric in $X$, $Y$ and $Z$, so for the graphs of $F_S$ and $F_A$ to contain the coordinate axes it suffices for $f$ to be zero on the $Z$-axis. Without loss of generality we may assume $f\in(XY)$. The simplest example would then be $f=XY$, yielding

$$F_A=XY+XZ+YZ=\tfrac{1}{2}\cdot F_S,$$

which is everywhere differentiable except at the origin. Unfortunately this does not meet the third requirement. To meet this requirement, it makes sense to consider a polynomial $f$ of odd degree.

The odd degree will make sure that the evaluation maps $k^2\ \longrightarrow\ k$ are surjective, for any of the three functions that $f$ induces on $k^2$ (corresponding to the three variables), resulting in a smoothly connected surface. Now the simplest example is $f=X^2Y$, yielding $$F_A=X^2Y+Y^2Z+Z^2X,$$ $$F_S=X^2Y+X^2Z+Y^2X+Y^2Z+Z^2X+Z^2Y,$$ which apparently do the job.

Answer 2

Consider the monkey saddle, equation $z = x^3 - 3 xy^2 $. Note that this graph has 6-fold rotational symmetry, and along the lines $y=0$, $y= \frac{\sqrt 3} 2 x$, and $y = \frac{ - \sqrt 3} 2 x$, increases like $z = r^3$. Hence, taking the cube root, $z = \sqrt[3]{x^3 - 3 xy^2}$ gives us a function which is linear along 3 equally spaced directions.

Compare this with the coordinate $x,y,z$-axes when viewed from the point $(1,1,1)$; this also looks linear along 3 equally spaced directions, but with slope $1/2$ in these directions. Hence it follows that the function
$ z = \dfrac 12 \cdot \sqrt[3]{x^3 - 3xy^2},$
rotated appropriately, is your desired function.