How to find a function/mathematical model which will allows for the same two points to be intercepted while changing the convexity of the curve. I want to change the convexity with a single variable parameter. In the attached figure I have plotted the 2 extremes for the convexity of the curve, the blue line is one extreme with no convexity and the orange curve is close to a step (with maximum convexity) in green is an intermediate characteristic. https://imgur.com/a/XFmAGCR I have started by looking at a decaying exponential function like:
y = exp (-(x-h)) + k;
However I do not know how to change the parameters so that they will satisfy the 2 conditions that I mentioned above. Also I am not sure which part of mathematics this is, so if somebody could point me to some related literature I would be very grateful. I look forward to some creative answers.
If you only care about the part of the function between the two points, you can just use exponents; by setting A to (0,1) and B to (1,0) you can make the calculations easier and then map everything to where you want it with linear transformations. The zero-curvature case is just a line that goes through A and B, and the equation is $x+y-1=0$, and the corresponding function is $y=1-x$.
Now to again make things easier, let's assume the equation we want treats x and y the same way, we can imagine the equation could be derived from the simple, zero-curvature equation by applying some sort of transformation on x and y; we want this transformation to be driven by one parameter, and in one case to be the identity, but we also want it to keep zero and one the same in all cases.
Exponents are the kind of thing we're looking for, and the equation I got is $x^{1 \over a}+y^{1 \over a}-1=0$, and the associated function is $y=(1-x^{1\over a})^a$: if a is 1, we just get a line ($y=1-x$) and if it's 2 we get a quarter of a circle.
It works perfectly well as long as we stay in the $[0,1]$ range for x and we keep $a>0$ (by the way, if $0<a<1$ the shape becomes convex in the other direction).