Smooth round hills with an off-center peak I want smooth roundish hills, functions $f: \mathbb{D} \to \mathbb{R}$
on the unit disk, with


*

*a single peak at a given point $p$, e.g. $p = 0.5$

*tapering smoothly to $0$ on the boundary, $|z| = 1$.


Is there a "simple" formula $f( x, y ), x^2 + y^2 \le 1$, with these two properties ?
If $p = 0$, $hill(r) = 1 - 3\,r^2 + 2\,r^3$ is a nice symmetric hill. 
It would be nice if $f( x, y )$ reduced to this for $p = 0$, but not essential.
A possibly too-complicated idea is to first move $p$ to $0$,
i.e. find a smooth one-one function
$\qquad warp( z ): \mathbb{D} \to \mathbb{D} $ with
$\qquad warp( p ) = 0$
$\qquad warp( z ) = z $ for all $ |z| = 1$
and compose $f(z) = hill(\, |warp( z )| \, )$,
but so far I have no simple such $warp(z)$ .
 A: Do you require a function or just a process for creating such a hill?
You could construct lines from your peak to the edge then construct contours equi-distance along those constructed lines.
Something like this:

Other orientations can be obtained via rotation.
$$$$
Update: I just ran through the math for this and it probably doesn't fulfill your requirement for a "simple" function.
The line from $(0,p)$ through $(x,y)$ to the edge of the disk will intersect the disk at:
$$\left(\frac{xp(p-y)+x\sqrt{(p-y)^2+x^2(1-p^2)}}{x^2+(p-y)^2},\frac{x^3p+x(p-y)\sqrt{(p-y)^2+x^2(1-p^2)}}{x^2+(p-y)^2}\right)$$
The distance from $(0,p)$ to $(x,y)$ can then be divided by the distance from $(0,p)$ to the edge to get a normalized radius between $0$ and $1$ to then substitute into your previous function (or any other smooth one you line).
$$r=\frac{x^2+(p-y)^2}{p(p-y)+\sqrt{(p-y)^2+x^2(1-p^2)}}$$
Substituting this into you $f(r)=1-3r^2+2r^3$ gives:
$$f(x,y)=1-3\left(\frac{x^2+(p-y)^2}{p(p-y)+\sqrt{(p-y)^2+x^2(1-p^2)}}\right)^2+2\left(\frac{x^2+(p-y)^2}{p(p-y)+\sqrt{(p-y)^2+x^2(1-p^2)}}\right)^3$$
Some examples: 
Peak: $(0,\frac12)$

Peak: $(0,-\frac45)$

