How to combine properties of different functions into a new function?

Question

The logistic function has the differential equation:

dy / dt = ky(1 - (y / L)) Solving this differential equation by separation of variables and integration yields the Logistic Equation:

f(t) = L / [1 + b * e^(-k * t)]

in which b = ± e^(-C), where "C" is a constant of integration.

The maximum height of the "S" shaped curve is "L", "k" controls the steepness and "b" moves the curve horizontally. The "S" shaped curve models some biological phenomena well such as the spread of disease or the maximum carrying capacity of population. It rises to a maximum height "L" and stays there.

However when populations die off quickly the logistic curve does not model that very well because it stays at a constant maximum height "L" forever.

I have noticed that the Witch of Agnesi curve does a good job of illustrating decay from a maximum, which is governed by the function:

g(t) = (8 * a^3) / (t^2 + 4a^2), where "a" is the radius of the circle that forms this curve which depends upon the size of the circle which is variable.

if a = 0.5, and I combine these curves:

h(t) = f(t) * g(t) I get a curve preserves some of the steepness of the Logistic function and drops off like the Witch of Agnesi, but it is not as steep as the logistic function and its maximum is smaller than the logistic function, although it looks a "S" shaped dropping off from a maximum

Red Curve: Logistic Function: f(t)

Black Curve: Witch of Agnesi: g(t)

Blue Curve: h(t) = f(t) * g(t)

Desired curve: Logistic curve (red) dropping off in the "green" region like the Witch of Agnesi. (Sorry if the green curve is not drawn to scale. I meant to make it look like the decay of the black curve.) The logistic curve models the steepness "k" better than the Witch of Agnesi curve, but the logistic curve does not drop off.

What kind of mathematical skills are required to form such a function? Do there exist any functions like this?

Thanks.

Answer 1

What do you think of this function, more general than the logistic function : $$f(t)=\frac{L}{1+b \:e^{-k\,t}+c\: e^{h\,t}}$$ The new parameters allow to fit to what you are expecting. A few examples below :

Answer 2

This is not another answer but a comment too long to be edited in the comments section.

Since the proposed function is accepted, a straightforward calculus of regression can be proposed according to the method introduced in the paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales $$f(t)=\frac{L}{1+b e^{-kt}+c e^{hx}}$$

$$y(t)=\frac{1}{f(t)}=\frac{1}{L}+\frac{b}{L}e^{-kt}+\frac{c}{L}e^{ht}$$ $$y(x)=A+Be^{px}+Ce^{qx}$$ with $\quad x=t\quad;\quad A=\frac{1}{L}\quad;\quad B=\frac{b}{L}\quad;\quad C=\frac{c}{L}\quad;\quad p=-k\quad;\quad q=h$

Double exponential regression : See pages 71-72 of the paper referenced above.

Note that there is a constant parameter in addition of the two exponentials. As a consequence the 4X4 matrix becomes a 5X5 matrix as shown below.

Data : $(t_1\:,\:f_1)\:,\:(t_2\:,\:f_2)\:,\:...\:,\:(t_k\:,\:f_k)\:,\:...\:,\:(t_n\:,\:f_n)$

NUMERICAL EXAMPLE :

Remember : This method provides an approximative result without specified criteria of fitting. If a particular criteria is specified and imperative, then iterative calculus must be used, thanks to other methods and of course will leads to a better fitting depending on the criteria considered.